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Introduction to Functional Analysis

Vladimir V. Kisil
School of Mathematics, University of Leeds, Leeds LS2 9JT, UK
email: kisilv@maths.leeds.ac.uk
Web: http://v-v-kisil.scienceontheweb.net/

November 6, 2024

Abstract: This is lecture notes for several courses on Functional Analysis at School of Mathematics of University of Leeds. They are based on the notes of Dr. Matt Daws, Prof. Jonathan R. Partington, Dr. David Salinger, and Prof. Alex Strohmaier used in the previous years. Some sections are borrowed from the textbooks, which I used since being a student myself. However all misprints, omissions, and errors are only my responsibility. I am very grateful to Filipa Soares de Almeida, Eric Borgnet, Pasc Gavruta for pointing out some of them. Please let me know if you find more.

The notes are available also for download in PDF.

The suggested textbooks are [, , , ]. The other nice books with many interesting problems are [, ].

Exercises with stars are not a part of mandatory material but are nevertheless worth to hear about. And they are not necessarily difficult, try to solve them!

Contents

Notations and Assumptions

+, ℝ+ denotes non-negative integers and reals.
x,y,z,… denotes vectors.
λ,µ,ν,… denotes scalars.
z, ℑ z stand for real and imaginary parts of a complex number z.

Integrability conditions

In this course, the functions we consider will be real or complex valued functions defined on the real line which are locally Riemann integrable. This means that they are Riemann integrable on any finite closed interval [a,b]. (A complex valued function is Riemann integrable iff its real and imaginary parts are Riemann-integrable.) In practice, we shall be dealing mainly with bounded functions that have only a finite number of points of discontinuity in any finite interval. We can relax the boundedness condition to allow improper Riemann integrals, but we then require the integral of the absolute value of the function to converge.

We mention this right at the start to get it out of the way. There are many fascinating subtleties connected with Fourier analysis, but those connected with technical aspects of integration theory are beyond the scope of the course. It turns out that one needs a “better” integral than the Riemann integral: the Lebesgue integral, and I commend the module, Linear Analysis 1, which includes an introduction to that topic which is available to MM students (or you could look it up in Real and Complex Analysis by Walter Rudin). Once one has the Lebesgue integral, one can start thinking about the different classes of functions to which Fourier analysis applies: the modern theory (not available to Fourier himself) can even go beyond functions and deal with generalized functions (distributions) such as the Dirac delta function which may be familiar to some of you from quantum theory.

From now on, when we say “function”, we shall assume the conditions of the first paragraph, unless anything is stated to the contrary.

0 Motivating Example: Fourier Series

0.1  Fourier series: basic notions

Before proceed with an abstract theory we consider a motivating example: Fourier series.

0.1.1 2π-periodic functions

In this part of the course we deal with functions (as above) that are periodic.

We say a function f:ℝ→ℂ is periodic with period T>0 if f(x+T)= f(x) for all x∈ ℝ. For example, sinx, cosx, eix(=cos x+i sinx) are periodic with period 2π. For kR∖{0}, sinkx, coskx, and eikx are periodic with period 2π/|k|. Constant functions are periodic with period T, for any T>0. We shall specialize to periodic functions with period 2π: we call them 2π-periodic functions, for short. Note that cosnx, sinnx and einx are 2π-periodic for n∈ℤ. (Of course these are also 2π/|n|-periodic.)

Any half-open interval of length T is a fundamental domain of a periodic function f of period T. Once you know the values of f on the fundamental domain, you know them everywhere, because any point x in ℝ can be written uniquely as x=w+nT where n∈ ℤ and w is in the fundamental domain. Thus f(x) = f(w+(n−1)T +T)=⋯ =f(w+T) =f(w).

For 2π-periodic functions, we shall usually take the fundamental domain to be ]−π, π]. By abuse of language, we shall sometimes refer to [−π, π] as the fundamental domain. We then have to be aware that f(π)=f(−π).

0.1.2 Integrating the complex exponential function

We shall need to calculate ∫ab eikxd x, for k∈ℝ. Note first that when k=0, the integrand is the constant function 1, so the result is ba. For non-zero k, ∫ab eikxd x= ∫ab (coskx+isinkx) d x = (1/k)[ (sinkxicoskx)]ab = (1/ik)[(coskx+isinkx)]ab = (1/ik)[eikx]ab = (1/ik)(eikbeika). Note that this is exactly the result you would have got by treating i as a real constant and using the usual formula for integrating eax. Note also that the cases k=0 and k≠0 have to be treated separately: this is typical.

Definition 1 Let f:ℝ→ℂ be a -periodic function which is Riemann integrable on [−π, π]. For each n∈ℤ we define the Fourier coefficient f(n) by
    f(n) = 
1
π
−π
f(x) einxdx .
Remark 2
  1. f(n) is a complex number whose modulus is the amplitude and whose argument is the phase (of that component of the original function).
  2. If f and g are Riemann integrable on an interval, then so is their product, so the integral is well-defined.
  3. The constant before the integral is to divide by the length of the interval.
  4. We could replace the range of integration by any interval of length , without altering the result, since the integrand is -periodic.
  5. Note the minus sign in the exponent of the exponential. The reason for this will soon become clear.
Example 3
  1. f(x) = c then f(0) =c and f(n) =0 when n≠0.
  2. f(x) = eikx, where k is an integer. f(n) = δnk.
  3. f is periodic and f(x) = x on ]−π, π]. (Diagram) Then f(0) = 0 and, for n≠0,
          f(n) = 
    1
    π
    −π
    xeinxdx = 


    xeinx
     2π in



    π



    −π
    +
    1
    in
    1
    π
    −π
    einxdx = 
    (−1)ni
    n
    .
Proposition 4 (Linearity) If f and g are -periodic functions and c and d are complex constants, then, for all n∈ℤ,
    (cf + dg6) (n) = cf(n) + dĝ(n) .
Corollary 5 If p(x) is a trigonometric polynomial, p(x)= ∑kk cneinx, then p(n) = cn for |n|≤ k and =0, for |n|≥ k.
    p(x) = 
 
n∈ℤ
 p(n)einx .

This follows immediately from Ex. 2 and Prop.4.

Remark 6
  1. This corollary explains why the minus sign is natural in the definition of the Fourier coefficients.
  2. The first part of the course will be devoted to the question of how far this result can be extended to other -periodic functions, that is, for which functions, and for which interpretations of infinite sums is it true that
    f(x) = 
     
    n∈ℤ
     f(n)einx . (1)
Definition 7  ∑n∈ℤ f(n)einx is called the Fourier series of the -periodic function f.

For real-valued functions, the introduction of complex exponentials seems artificial: indeed they can be avoided as follows. We work with (1) in the case of a finite sum: then we can rearrange the sum as

  
f(0) + 
 
n>0
 (f(n) einx +f(−n)einx)
 =
f(0) + 
 
n>0
 [(f(n)+f(−n))cosnx +i(f(n)−f(−n))sin nx]
 =
a0
 2
 +
 
n>0
 (ancosnx +bnsinnx)

Here

  an=
(f(n)+f(−n)) =
1
π
−π
f(x)(einx+einx) dx
 =
1
π
π
−π
f(x)cosnxdx

for n>0 and

  bn =i((f(n)−f(−n))=
1
π
π
−π
f(x)sin nxdx

for n>0. a0 = 1/π∫−ππf(x) d x, the constant chosen for consistency.

The an and bn are also called Fourier coefficients: if it is necessary to distinguish them, we may call them Fourier cosine and sine coefficients, respectively.

We note that if f is real-valued, then the an and bn are real numbers and so ℜ f(n) = ℜ f(−n), ℑ f(−n) = −ℑf(n): thus f(−n) is the complex conjugate of f(n). Further, if f is an even function then all the sine coefficients are 0 and if f is an odd function, all the cosine coefficients are zero. We note further that the sine and cosine coefficients of the functions coskx and sinkx themselves have a particularly simple form: ak=1 in the first case and bk=1 in the second. All the rest are zero.

For example, we should expect the 2π-periodic function whose value on ]−π,π] is x to have just sine coefficients: indeed this is the case: an=0 and bn=i(f(n)−f(−n)) = (−1)n+12/n for n>0.

The above question can then be reformulated as “to what extent is f(x) represented by the Fourier series a0/2 + ∑n>0(ancosx + bnsinx)?” For instance how well does ∑(−1)n+1(2/n)sinnx represent the 2π-periodic sawtooth function f whose value on ]−π, π] is given by f(x) = x. The easy points are x=0, x=π, where the terms are identically zero. This gives the ‘wrong’ value for x=π, but, if we look at the periodic function near π, we see that it jumps from π to −π, so perhaps the mean of those values isn’t a bad value for the series to converge to. We could conclude that we had defined the function incorrectly to begin with and that its value at the points (2n+1)π should have been zero anyway. In fact one can show (ref. ) that the Fourier series converges at all other points to the given values of f, but I shan’t include the proof in this course. The convergence is not at all uniform (it can’t be, because the partial sums are continuous functions, but the limit is discontinuous.) In particular we get the expansion

  
π
2
 = 2(1−1/3+1/5−⋯)

which can also be deduced from the Taylor series for tan−1.

0.2 The vibrating string

In this subsection we shall discuss the formal solutions of the wave equation in a special case which Fourier dealt with in his work.

We discuss the wave equation

2y
 ∂ x2
 =
1
K2
2y
 ∂ t2
, (2)

subject to the boundary conditions

y(0, t) = y(π, t) = 0, (3)

for all t≥0, and the initial conditions

  y(x,0)=F(x),
  yt(x,0)=0.

This is a mathematical model of a string on a musical instrument (guitar, harp, violin) which is of length π and is plucked, i.e. held in the shape F(x) and released at time t=0. The constant K depends on the length, density and tension of the string. We shall derive the formal solution (that is, a solution which assumes existence and ignores questions of convergence or of domain of definition).

0.2.1 Separation of variables

We first look (as Fourier and others before him did) for solutions of the form y(x,t) = f(x)g(t). Feeding this into the wave equation (2) we get

  f′′(x) g(t) = 
1
K2
f(x) g′′(t)

and so, dividing by f(x)g(t), we have

f′′(x)
f(x)
 = 
1
K2
g′′(t)
g(t)
. (4)

The left-hand side is an expression in x alone, the right-hand side in t alone. The conclusion must be that they are both identically equal to the same constant C, say.

We have f′′(x) −Cf(x) =0 subject to the condition f(0) = f(π) =0. Working through the method of solving linear second order differential equations tells you that the only solutions occur when C = −n2 for some positive integer n and the corresponding solutions, up to constant multiples, are f(x) = sinnx.

Returning to equation (4) gives the equation g′′(t)+K2n2g(t) =0 which has the general solution g(t) = ancosKnt + bnsinKnt. Thus the solution we get through separation of variables, using the boundary conditions but ignoring the initial conditions, are

  yn(x,t) = sinnx(an cosKnt + bn sinKnt) ,

for n≥ 1.

0.2.2 Principle of Superposition

To get the general solution we just add together all the solutions we have got so far, thus

y(x,t) = 
n=1
sinnx(an cosKnt + bn sin Knt) (5)

ignoring questions of convergence. (We can do this for a finite sum without difficulty because we are dealing with a linear differential equation: the iffy bit is to extend to an infinite sum.)

We now apply the initial condition y(x,0) = F(x) (note F has F(0) =F(π) =0). This gives

  F(x) =  
n=1
ansinnx .

We apply the reflection trick: the right-hand side is a series of odd functions so if we extend F to a function G by reflection in the origin, giving

  G(x):=

      F(x),  if  0≤ x≤π;
      −F(−x),  if  −π<x<0.

we have

  G(x) = 
n=1
ansinnx ,

for −π≤ x ≤ π.

If we multiply through by sinrx and integrate term by term, we get

ar = 
1
 π
π
−π
G(x)sinrxdx

so, assuming that this operation is valid, we find that the an are precisely the sine coefficients of G. (Those of you who took Real Analysis 2 last year may remember that a sufficient condition for integrating term-by -term is that the series which is integrated is itself uniformly convergent.)

If we now assume, further, that the right-hand side of (5) is differentiable (term by term) we differentiate with respect to t, and set t=0, to get

0=yt(x,0) = 
n=1
bnKn sinnx. (6)

This equation is solved by the choice bn=0 for all n, so we have the following result

Proposition 8 (Formal) Assuming that the formal manipulations are valid, a solution of the differential equation (2) with the given boundary and initial conditions is
    y(x,t) = 
1
an sinnx cosKnt ,(2.11)
where the coefficients an are the Fourier sine coefficients
    an = 
1
 π
π
−π
G(x)sinnxdx
of the periodic function G, defined on ]−π, π] by reflecting the graph of F in the origin.
Remark 9 This leaves us with the questions
  1. For which F are the manipulations valid?
  2. Is this the only solution of the differential equation? (which I’m not going to try to answer.)
  3. Is bn=0 all n the only solution of (6)? This is a special case of the uniqueness problem for trigonometric series.

0.3 Historic: Joseph Fourier

Joseph Fourier, Civil Servant, Egyptologist, and mathematician, was born in 1768 in Auxerre, France, son of a tailor. Debarred by birth from a career in the artillery, he was preparing to become a Benedictine monk (in order to be a teacher) when the French Revolution violently altered the course of history and Fourier’s life. He became president of the local revolutionary committee, was arrested during the Terror, but released at the fall of Robespierre.

Fourier then became a pupil at the Ecole Normale (the teachers’ academy) in Paris, studying under such great French mathematicians as Laplace and Lagrange. He became a teacher at the Ecole Polytechnique (the military academy).

He was ordered to serve as a scientist under Napoleon in Egypt. In 1801, Fourier returned to France to become Prefect of the Grenoble region. Among his most notable achievements in that office were the draining of some 20 thousand acres of swamps and the building of a new road across the alps.

During that time he wrote an important survey of Egyptian history (“a masterpiece and a turning point in the subject”).

In 1804 Fourier started the study of the theory of heat conduction, in the course of which he systematically used the sine-and-cosine series which are named after him. At the end of 1807, he submitted a memoir on this work to the Academy of Science. The memoir proved controversial both in terms of his use of Fourier series and of his derivation of the heat equation and was not accepted at that stage. He was able to resubmit a revised version in 1811: this had several important new features, including the introduction of the Fourier transform. With this version of his memoir, he won the Academy’s prize in mathematics. In 1817, Fourier was finally elected to the Academy of Sciences and in 1822 his 1811 memoir was published as “Théorie de la Chaleur”.

For more details see Fourier Analysis by T.W. Körner, 475-480 and for even more, see the biography by J. Herivel Joseph Fourier: the man and the physicist.

What is Fourier analysis. The idea is to analyse functions (into sine and cosines or, equivalently, complex exponentials) to find the underlying frequencies, their strengths (and phases) and, where possible, to see if they can be recombined (synthesis) into the original function. The answers will depend on the original properties of the functions, which often come from physics (heat, electronic or sound waves). This course will give basically a mathematical treatment and so will be interested in mathematical classes of functions (continuity, differentiability properties).

1 Basics of Metric Spaces

1.1 Metric Spaces

1.1.1 Metric spaces: definition and examples

In Analysis and Calculus the definition of convergence was based on the notion of a distance between points, namely the standard distance between two real numbers is given by

d(x,y)=|xy|.

Similarly, the distance between two points in the plane, given by

d(x,y)=d((x1,x2),(y1,y2))=
(x1y1)2+(x2y2)2
.

A metric space formalises this notion. This will give us the flexibility to talk about distances on function spaces, for example, or introduce other notions of distance on spaces.

Definition 1 (Metric Space) A metric space (X,d) is a set X together with a function d: X × X → ℝ that satisfies the following properties
  1. d(x,y) ≥ 0; and d(x,y)=0  ⇐⇒ x=y (positive definite);
  2. d(x,y)=d(y,x) (symmetric);
  3. d(x,z) ≤ d(x,y)+d(y,z) (triangle inequality).
The function d is called the metric. The word distance will be used interchangeably with the same meaning.
Example 2
  1. X=ℝ. The standard metric is given by d1(x,y)=|xy|. There are many other metrics on , for example
    d(x,y)=|exey|;
    d(x,y)=


    |xy| if  |xy| ≤ 1, 
        1 if  |xy| ≥ 1.
  2. Let X be any set whatsoever, then we can define the discrete metric
    d(x,y) =


          1 if  x ≠ y, 
          0 if  x=y.
  3. X=ℝm. The standard metric is the Euclidean metric: if x=(x1,x2,…,xm) and y=(y1,y2,…,ym) then
      d2(x,y)=
    (x1y1)2+(x2y2)2+…+(xmym)2
    .
    This is linked to the inner-product (scalar product), x ·y=x1 y1+ x2 y2+…+xm ym, since it is just √(xy).(xy). We will study inner products more carefully later, so for the moment we won’t prove the (well-known) fact that it is indeed a metric.

    Other possible metrics include

    d(x,y)=max{|x1y1|,|x2y2|,…,|xmym|}.

    Another metric on ℝm comes from the generalisation of our first example:

      d1(x,y)=|x1y1|+|x2y2|+…+|xmym| .

    These metrics d1, d2, d are all translation-invariant (i.e., d(x+z,y+z)=d(x,y)), and positively homogeneous (i.e., d(kx,ky)=|k|d(x,y)), see Ex. 8 for further discussion.

  4. Take X=C[a,b]. Here are three metrics similar to above ones:
    d2(f,g)=
    b
    a
     | f(x)−g(x) |2   dx
    .

    Again, this is linked to the idea of an inner product, so we will delay proving that it is a metric.

    d1(f,g)=
    b
    a
     |f(x)−g(x)|   dx,

    the area between two graphs

    d(f,g)=max{ |f(x)−g(x)|: a ≤ x ≤ b},

    the maximum vertical separation between two graphs.

Example 3 On C[0,1] take f(x)=x and g(x)=x2 and calculate
d2(f,g)=



1
0
 (xx2)2   dx


1/2



 
=
1/30
,
d1(f,g)=
1
0
 |xx2|   dx = 1/6,  and
d(f,g)=
 
max
x ∈ [0,1]
|xx2|=1/4.
Remark 4 Any subset of a metric space is again a metric space its own right, by restricting the distance function to the subset.
Example 5
  1. The interval [a,b] with d(x,y)=|xy| is a subspace of .
  2. The unit circle {(x1,x2) ∈ ℝ2: x12+x22=1 } with d2(x,y)=√(x1y1)2+(x2y2)2 is a subspace of 2.
  3. The space of polynomials P is a metric space with any of the metrics inherited from C[a,b] above.
Definition 6 A normed space (V,||· ||) is a real vector space V with a map ||·||: V → ℝ (called norm) satisfying
  1. ||v|| ≥ 0, and (||v||=0 ⇔ v=0),
  2. ||λ v|| = |λ| || v|| ,
  3. ||v+w|| ≤ ||v||+ ||w||.
Exercise 7 Prove that V is a metric space with metric d(v,w):=||vw||.
Exercise 8
  1. Write norms ||·||1, ||·||2, ||·|| on m which produces metrics d1, d2, d from Ex. 2.3.
    Hint: see (11) and  (9) below.
  2. Show, that the following are norms on the vector space V=C[a,b]:
         
          || f ||1
    = 
    b
    a
     |f(x)|   dx,
             
    || f ||2
    = 
    b
    a
     |f(x)|2   dx,
             
    || f ||
    = 
     
    sup
    x∈[a,b]
     |f(x)|.
             
    Furthermore, these norms generate the respective metrics d1, d2 and d from Ex. 2(4) as indicated in the previous exercise.
Definition 9 An inner product space(V,⟨·, ·⟩) is a real vector space V with a map ⟨·, ·⟩: V × V → ℝ (called inner product) satisfying
  1. ⟨ λ v,w ⟩ = λ⟨ v,w,
  2. v1 +v2 ,w ⟩ = ⟨ v1,w ⟩+⟨ v2,w,
  3. v,w ⟩ = ⟨ w,v,
  4. v,v ⟩ ≥ 0, and (v,v ⟩ =0 ⇔ v=0).
Exercise 10
  1. Prove that the Cauchy–Schwarz inequality | ⟨ v,w ⟩ |2 ≤ ⟨ v,v ⟩ ⟨ w,w holds.
    Hint: start by considering the expression v + λ w ,v + λ w⟩ ≥ 0 and analyse the discriminant of the quadratic expression for λ.
  2. Then prove that V is a normed space with norm || v||:= ⟨ v,v1/2.
  3. Which of the above norms ||·||1, ||·||2, ||·|| from Ex. 8 can be obtained from an inner product as described in the previous item?

There is a natural name for a class of maps, which preserve metrics:

Definition 11 (Isometry) Let (X, dX) and (Y, dY) be two metric spaces. A map φ: XY is an isometry if
    dY(φ(x1), φ(x2)) =  dX(x1, x2)     for all  x1, x2 ∈ X.
A metric space (X,dX) is isometric to a metric space (Y,dY) if there is an isometry bijection between X and Y.

1.1.2 Open and closed sets

Definition 12 (Open and closed balls) Let (X,d) be a metric space, let xX and let r>0. The open ball centred at x, with radius r, is the set
Br(x)={y ∈ X: d(x,y)<r },
and the closed ball is the set
Br(x)
={y ∈ X: d(x,y) ≤ r }.

Trivial but useful observations are:

Note, that in ℝ with the usual metric the open ball is Br(x)=(xr,x+r), an open interval, and the closed ball is Br(x)=[xr,x+r], a closed interval.

For the d2 metric on ℝ2, the unit ball, B1(0), is disc centred at the origin, excluding the boundary. You may like to think about what you get for other metrics on ℝ2. What are balls in the discrete metric, Ex. 2.2?

Definition 13 (Open sets) A subset U of a metric space (X,d) is said to be open, if for each point xU there is an r>0 such that the open ball Br(x) is contained in U (“room to swing a cat").

Clearly X itself is an open set, that is the whole metric space is open in itself. Also the empty set ∅ is also considered to be open in a trivial way.

Remark 14 Note that the property “be open” of a set depends on the metric space. For example if we consider the set [0,1] it is open in the metric space [0,1] with the standard metric, but not open in the set with standard metric.
Proposition 15 Every “open ball" Br(x) is an open set.
Proof. For if yBr(x), choose δ=rd(x,y). We claim that Bδ(y) ⊂ Br(x).

If zBδ(y), i.e., d(z,y)<δ, then by the triangle inequality

d(z,x) ≤ d(z,y)+d(y,x) <  δ + d(x,y) = r.

So zBr(x). □

Definition 16 (Closed set) A subset F of (X,d) is said to be closed, if its complement XF is open.

Note that closed does not mean “not open". In a metric space the sets ∅ and X are both open and closed. In ℝ we have:

Remark 17 As it can be seen from the definitions the property of a subset F to be open or closed depends from the surrounding space X. For example:
Example 18 If we take the discrete metric,
d(x,y)=


        1 if  x ≠ y, 
        0 if  x=y,
then each point {x}=B1/2(x) so is an open set. Hence every set U is open, since for xU we have B1/2(x) ⊆ U. Hence, by taking complements, every set is also closed.
Theorem 19 In a metric space, every one-point set {x0} is closed.
Proof. We need to show that the set U={xX: xx0} is open, so take a point xU. Now d(x,x0)>0, and the ball Br(x) is contained in U for every 0<r< d(x,x0). □
Theorem 20 Let (Uα)α ∈ A be any collection of open subsets of a metric space (X,d) (not necessarily finite!). Then α ∈ A Uα is open. Let U and V be open subsets of a metric space (X,d). Then UV is open. Hence (by induction) any finite intersection of open subsets is open.
Proof. If x ∈ ∪α ∈ A Uα then there is an α with xUα. Now Uα is open, so Br(x) ⊂ Uα for some r>0. Then Br(x) ⊂ ∪α ∈ A Uα so the union is open.

If now U and V are open and xUV, then ∃ r>0 and s>0 such that Br(x) ⊂ U and B(x,s) ⊂ V, since U and V are open. Then B(x,t) ⊂ UV if t ≤ min(r,s). □

Remark 21 Here we used a common property, which is helpful to remember: the minimum of a finite set of positive numbers is always positive (bigger than 0). However, the infimum of an infinite set of positive numbers can be zero, e.g. inf{1/n: n∈ ℕ}=0. Therefore, a transition from a given infinite set to a suitable finite set will be a reacquiring theme in our course, cf. compact set later in the course.

Thereafter, the collection of open sets is preserved by arbitrary unions and finite intersections.

However, an arbitrary intersection of open sets is not always open; for example (−1/n,1/n) is open for each n=1,2,3,…, but ∩n=1(−1/n,1/n)= {0}, which is not an open set.

For closed sets we swap union and intersection.

Theorem 22 Let (Fα)α ∈ A be any collection of closed subsets of a metric space (X,d) (not necessarily finite!). Then α ∈ A Fα is closed. Let F and G be closed subsets of a metric space (X,d). Then FG is closed. Hence (by induction) any finite union of closed subsets is closed.
Proof.

To prove this we recall de Morgan’s laws. We use the notation Sc for the complement XS of a set SX.

x ∉
 
α
Aα
 ⇐⇒ x ∉Aα for all  α,  so  (Aα)c = Aαc.
x ∉
 
α
Aα
 ⇐⇒ x ∉Aα for some  α,  so  (Aα)c = Aαc.

Write Uα= Fαc =XFα which is open. So ∪α ∈ A Uα is open by Theorem 20. Now, by de Morgan’s laws, (∩α ∈ A Fα)c = ∪α ∈ A Fαc. This is just ∪α ∈ A Uα. Since the complement of ∩α ∈ A Fα is open, it is closed.

Similarly, the complement of FG is FcGc, which is the intersection of two open sets and hence open by Theorem 20. Hence FG is closed. □

Infinite unions of closed sets do not need to be closed. An example is

  
n=1
[
1
n
,∞)=(0,∞),

which is open but not closed in ℝ with standard metric.

Definition 23 (Closure of a set) The closure of S, written S, is the smallest closed set containing S, and is contained in all other closed sets containing S.

The above smallest closed set containing S does exist, because we can define

S
 = {F: F ⊃ S  and  F  closed },

the intersection of all closed sets containing S. There is at least one closed set containing S, namely X itself.

Example 24 In the metric space the closure of S=[0,1) is [0,1]. This is closed, and there is nothing smaller that is closed and contains S.
Exercise 25 Give an example of an open ball Br(x) and the respective closed ball Br(x) with the same centre and radius in a metric space X, such that Br(x) is not the closure of Br(x). Note the slight discontent on our notations, which shall not mislead us in future.
Definition 26 (Dense subset) A subset SX is dense in X if S=X.
Theorem 27 The set of rationals is dense in , with the usual metric.
Proof. Suppose that F is a closed subset of ℝ which contains ℚ: we claim that it F=ℝ.

For U=ℝ ∖ F is open and contains no points of ℚ. But an open set U (unless it is empty) must contain an interval Br(x) for some xU, and hence a rational number within it.

Our only conclusion is that U=∅ and F=ℝ, so that =ℝ. □

Definition 28 (Neighbourhood) We say that V is a neighbourhood (nbh) of x if there is an open set U such that xUV; this means that ∃ δ>0 s.t. Bδ(x) ⊆ V. Thus, a set is open precisely when it is a neighbourhood of each of its points.
Example 29 The half-open interval [0,1) is a neighbourhood of every point in it except for 0.
Theorem 30 For a subset S of a metric space X, we have xS iff VS ≠ ∅ for all nhds V of x (i.e., all neighbourhoods of x meet S).
Proof. If there is a neighbourhood of x that doesn’t meet S, then there is an open subset U with xU and US=∅.

But then XU is a closed set containing S and so SXU, and then xS because xU.

Conversely, if every neighbourhood of x does meet S, then xS, as otherwise XS is as open neighbourhood of x that doesn’t meet S. □

Definition 31 (Interior) The interior of S, intS, is the largest open set contained in S, and can be written as
intS = { U: U ⊂ S   and   U  open }.
the union of all open sets contained in S. There is at least open set within S, namely .

We see that S is open exactly when S=intS, otherwise intS is smaller.

Example 32
  1. In the metric space we have int[0,1)=(0,1); clearly this is open and there is no larger open set contained in [0,1).
  2. intℚ = ∅. For any non-empty open set must contain an interval Br(x) and then it contains an irrational number, so isn’t contained in .
Proposition 33 intS=X ∖ (XS).
Proof. By De Morgan’s laws,
intS={ U: U ⊂ S  and  U  open } 
 =X ∖ {Uc: U ⊂ S  and  U  open }  
 =X ∖ {F: F ⊃ (X ∖ S)  and  F  closed } 
 =
X ∖ (
X∖ S
).
This is because US if and only if Uc= (XU) ⊃ (XS). Also F=Uc is closed precisely when U is open. That is, there is a correspondence between open sets contained in S and closed sets containing its complement. □

1.1.3 Convergence and continuity

Let (xn) be a sequence in a metric space (X,d), i.e., x1,x2,…. (Sometimes we may start counting at x0.)

Definition 34 (Convergence) We say xnx (i.e., xn converges to x) if d(xn,x) → 0 as n → ∞.

In other words: xnx if for any ε>0 there exists N∈ℕ such that for all n>N we have d(x,xn) < ε.

This is the usual notion of convergence if we think of points in ℝd with the Euclidean metric.

Theorem 35 Let (xn) be a sequence in a metric space (X,d). Then the following are equivalent:
  1. xnx;
  2. for every open U with xU, there exists an N>0 such that (n>N) xnU;
  3. for every ε>0 there exists an N>0 such that (n>N) xnBε(x).
Proof. 12 If xnx and xU, then there is a ball Bε(x) ⊂ U, since U is open. But xnx so d(xn,x) < ε for n sufficiently large, i.e., xnU for n sufficiently large.

23 is obvious.

Finally, 31. If the 3 condition works for a given ε>0 and large n the inclusion xnBε(x) implies d(xn,x)<ε. □

Theorem 36 Let S be a subset of the metric space X. Then xS if and only if there is a sequence (xn) of points of S with xnx.
Proof. If xS, then for each n we have B1/n(x) ∩ S ≠ ∅ by Theorem 30. So choose xnB1/n(x) ∩ S. Clearly d(xn,x) → 0, i.e., xnx.

Conversely, if xS, then there is a neighbourhood U of x with US=∅. Now no sequence in S can get into U so it cannot converge to x. □

This can also be phrased as follows, characterising closed set in terms of sequences.

Corollary 37 (Closedness under taking limits) A subset YX of a metric space (X,d) is closed if and only if for every sequence (xn) in Y that is convergent in X its limit is also in Y.

Hence, the closure S is obtained from S by adding all possible limit points of sequences in S.

Example 38
  1. Take (ℝ2,d1), where d1(x,y)=|x1y1|+|x2y2|, where x=(x1,x2) and y=(y1,y2), and consider the sequence (1/n,2n+1/n+1). We guess its limit is (0,2). To see if this is right, look at
    d1





    1
    n
    ,
    2n+1
    n+1



    ,(0,2)


    =


    1
    n



    +


    2n+1
    n+1
    −2


    = 
    1
    n
     + 
    1
    n+1
    → 0
    as n → ∞. So the limit is (0,2).
  2. In C[0,1] let fn(t)=tn and f(t)=0 for 0 ≤ t ≤ 1. Does fnf, (a) in d1, and (b) in d?
    (a)
    d1(fn,f)=
    1
    0
    tn   dt = 
    1
    n+1
     → 0
    as n → ∞. So fnf in d1.
    (b)
    d(fn,f)=max{tn: 0 ≤ t ≤ 1}=1 ¬→0
    as n → ∞. So fn ¬→f in d.
    Note: Say
    gng pointwise on [a,b] as n → ∞ if gn(x) → g(x) for all x ∈ [a,b]. If we define g(x)= {
        0 for  0 ≤ x < 1, 
        1 for   x=1,
    then fng pointwise on [0,1]. But gC[0,1], as it is not continuous at 1.
  3. Take the discrete metric
    d0(x,y)=


            1 if  x ≠ y, 
            0 if  x=y.
    Then xnx  ⇐⇒ d0(xn,x) → 0. But since d0(xn,x)=0 or 1, this happens if and only if d0(xn,x)=0 for n sufficiently large. That is, there is an n0 such that xn=x for all nn0.

    All convergent sequences in this metric are eventually constant. So, for example d0(1/n,0) ¬→0.

A result on convergence in ℝm.

Proposition 39 Take 2 with any of the metrics d1, d2 and d. Then a sequence xn=(an,bn) converges to x=(a,b) if and only if ana and bnb.
Proof. A useful observation is that for any xn and x:
    d1(xn,x) ≥ d2(xn,x) ≥ d(xn,x).
If ana and bnb, then for any ε>0 there are Na and Nb such that for N> Na we have | ana |<ε/2 and for n>Nb | bnb |<ε/2. Thus for any n > N=max(Na, Nb):
    ε > 
ana
+
bnb
 =  d1(xn,x) ≥ d2(xn,x) ≥ d(xn,x) ,
which shows the convergence in all three metrics.

To show the opposite, WLOG assume towards a contradiction that an ¬→a, that is, there exits ε>0 such that for any N there exists n>N such that | ana |>ε. Then:

     
    d1(xn,x) ≥ d2(xn,x) ≥    d(xn,x)= max{
ana
,
bnb
}> 
ana
          

showing the divergence in all three norms.

A similar result holds for ℝm in general.

Now let’s look at continuous functions again.

Theorem 40 If fnf in (C[a,b],d), then fnf in (C[a,b],d1).

Informally speaking, d convergence is stronger than d1 convergence.

Proof. d(fn,f)=max{|fn(x)−f(x)|:   axb} → 0 as n → ∞, so, given ε>0 there is an N so that d(fn,f)<ε for nN. It follows that if nN then
d1(fn,f) = 
b
a
 |fn(x)−f(x)|   dx ≤ 
b
a
 ε   dx = ε(ba),
so d1(fn,f) → 0 as n → ∞. □
Remark 41 It is also true that if d(fn,f) → 0 then fnf point-wise on [a,b]. The converse is false, cf. 38(2).

Now we look at continuous functions between general metric spaces.

Definition 42 (Continuity) Let f: (X,dX) → (Y,dY) be a map between metric spaces. We say that f is continuous at xX if for each ε>0 there is a δε,x>0 such that dY(f(x′),f(x)) < ε for all x′∈ X whenever dX(x′,x) < δε,x.

Another way of saying the same is that for every ε>0 there exists a δ>0 such that

f(Bδ(x)) ⊂ Bε(f(x)).

The map f is continuous, if it is continuous at all points of X.

Theorem 43 (Sequential continuity) For f as above, f is continuous at a if and only if, whenever a sequence xna, then f(xn) → f(a).

In short, f is continuous at a if and only if f permutes with the limit:

f


 
lim
n→ ∞
xn


= 
 
lim
n→ ∞
   f
xn
(7)

for any sequence xna.

Proof. Same proof as in real analysis, more or less. If f is continuous at a and xna, then for each ε>0 we have a δ>0 such that dY(f(x),f(a)) < ε whenever dX(x,a) < δ.

Then there’s an n0 with d(xn,a)<δ for all nn0, and so d(f(xn),f(a))<ε for all nn0. Thus f(xn) → f(x).

Conversely, if f is not continuous at a, then there is an ε for which no δ will do, so we can find xn with d(xn,a)<1/n, but d(f(xn),f(a)) ≥ ε. Then xna but f(xn) ¬→f(a). □

But there is a nicer way to define continuity. For a mapping f: XY and a set UY, let f−1(U) be the set, called pre-image or inverse image

f−1(U)={ x ∈ X: f(x) ∈ U }.

This makes sense even if f−1 is not defined as a function.

Theorem 44 (Continuity and open sets) A function f: XY is continuous if and only if f−1(U) is open in X for every open subset UY. In short: the inverse image of an open set is open.
Proof. Suppose that f is continuous, that UY is open, and that x0f−1(U), so f(x0) ∈ U. Now there is a ball Bε(f(x0)) ⊂ U, since U is open, and then by continuity there is a δ>0 such that dY(f(x),f(x0)) < ε whenever dX(x,x0) < δ. This means that for d(x,x0)<δ, f(x) ∈ U and so xf−1(U). That is, f−1(U) is open.

Conversely, if the inverse image of an open set is open, and x0X, let ε>0 be given. We know that Bε(f(x0)) is open, so f−1(B(f(x0),ε)) is open, and contains x0. So it contains some Bδ(x0) with δ>0.

But now if d(x,x0)<δ, we have xBδ(x0) ⊂ f−1(Bε(f(x0))) so f(x) ∈ Bε(f(x0)) and we have d(f(x),f(x0))<ε. □

Remark 45 Note that for f continuous we do not expect f(U) to be open for all open subsets of X, for example f: ℝ → ℝ, f ≡ 0, then f(ℝ)={0}, not open.
Example 46 Let X=ℝ with the discrete metric, and Y any metric space. Then all functions f: XY are continuous! Indeed, in either way:
Exercise 47 Which functions from a metric space X to the discrete metric space are continuous? Which function from the discrete metric space to are continuous?
Proposition 48 Let X and Y be metric spaces.
  1. A function f : XY is continuous if and only if f−1(F) is closed whenever F is a closed subset of Y.
  2. If f: XY and g: YZ are continuous, then so is the composition gf: XZ defined by (gf)(x) = g(f(x)).
Proof.
  1. We can do this by complements, as if F is closed, then U=Fc is open, and f−1(F)=f−1(U)c (a point is mapped into F if and only if it isn’t mapped into U).

    Then f−1(F) is always closed when F is closed  ⇐⇒  f−1(U) is always open when U is open.

  2. Take UZ open; then (gf)−1(U) = f−1(g−1(U)); for these are the points which map under f into g−1(U) so that they map under gf into U.

    Now g−1(U) is open in Y, as g is continuous, and then f−1(g−1(U)) is open in X since f is continuous.

In many cases we may need a stronger notion.

Definition 49 (Uniform continuity) A function f: (X,dX) → (Y, dY) is called uniformly continuous if for each ε>0 there exists δε>0 such that whenever x,x′∈ X satisfy dX(x,x′)≤δε, we have that dY(f(x),f(x′))≤ε.

Note, that here the same δε shall work for all xX. Thus any uniformly continuous function is continuous at every point. On the other hand the function f(x)=1/x on (0,1) is continuous but not uniformly continuous.

1.2 Useful properties of metric spaces

Metric spaces may or may not have some useful properties which we are discussing in the following subsections: completeness and compactness.

1.2.1 Cauchy sequences and completeness

Recall that if (X,d) is a metric space, then a sequence (xn) of elements of X converges to xX if d(xn,x) → 0, i.e., if given ε>0 there exists N such that d(xn,x)< ε whenever nN. Thus, to show that a sequence is convergent from the definition we need to present its limit x which may not belong to the sequence (xn). It would be convenient to deduce convergence of (xn) just through its own properties without a reference to extraneous x. This is possible for complete metric spaces studied in this subsection.

Often we think of convergent sequences as ones where xn and xm are close together when n and m are large. This is almost, but not quite, the same thing in a general metric space.

Definition 50 (Cauchy Sequence) A sequence (xn) in a metric space (X,d) is a Cauchy sequence if for any ε>0 there is an N such that d(xn,xm)<ε for all n, mN.
Example 51 Take xn=1/n in with the usual metric. Now d(xn,xm)=|1/n−1/m|. Suppose that n and m are both at least as big as N; then d(xn,xm) ≤ 1/N. Hence if ε>0 and we take N>1/ε, we have d(xn,xm)≤ 1/N whenever n and m are both N.

In fact all convergent sequences are Cauchy sequences, by the following result.

Theorem 52 Suppose that (xn) is a convergent sequence in a metric space (X,d), i.e., there is a limit point x such that d(xn,x) → 0. Then (xn) is a Cauchy sequence.
Proof. Take ε>0. Then there is an N such that d(xn,x)<ε/2 whenever nN. Now suppose both nN and mN. Then
d(xn,xm) ≤ d(xn,x)+d(x,xm) = d(xn,x)+d(xm,x) < ε/2+ε/2=ε,
and we are done. □
Proposition 53 Every subsequence of a Cauchy sequence is a Cauchy sequence.
Proof. If (xn) is Cauchy and (xnk) is a subsequence, then given ε>0 there is an N such that d(xn,xm) < ε whenever n, mN. Now there is a K such that nkN whenever kK. So d(xnk,xnl)<ε whenever k, lK. □

Does every Cauchy sequence converge?

Example 54
  1. (X,d)=ℚ, as a subspace of with the usual metric. Take x0=2 and define xn+1=xn/2+1/xn. The sequence continues 3/2, 17/12, 577/408,… and indeed the sequence converges in as xnx where x=x/2+1/x, i.e., x2=2. But this isn’t in .

    Thus (xn) is Cauchy in , since it converges to 2 when we think of it as a sequence in . So it is Cauchy in , but doesn’t converge to a point of .

  2. Easier. Take (X,d)=(0,1). Then (1/n) is a Cauchy sequence in X (since it is Cauchy in , as seen above), and has no limit in X.
In each case there are “points missing from X”.
Definition 55 (Completeness) A metric space (X,d) is complete if every Cauchy sequence in X converges to a limit in X.
Theorem 56 The metric space is complete.
Remark 57 In parts of the literature is simply defined as the completion of . In this case one does not have to prove that is complete, but it is complete by construction. One then has to work a bit to show that it is also a field.

This is a result from the first year. Since its proof depends on the definition of ℝ we will not demonstrate it here.

Example 58
  1. Open intervals in are not complete; closed intervals are complete.
  2. What about C[a,b] with d1, d2 or d?

    Following our consideration in Ex. 38.2, define fn in C[0,2] by

      fn(x)=


    xn for  0 ≤ x ≤ 1, 
          1 for   1 ≤ x ≤ 2.

    [DIAGRAM]

    Then

    d1(fn,fm)=
    2
    0
     |fn(x)−fm(x)|   dx
     =
    1
    0
     |xnxm|   dx
     =
    1
    0
     (xmxn)   dx    if   n ≥ m
     =
    1
    m+1
    1
    n+1
     ≤ 
    1
    m+1
     → 0,

    and hence (fn) is Cauchy in (C[0,2],d1). Does the sequence converge?

    If there is an fC[0,2] with fnf as n → ∞, then 02 |fn(x)−f(x)|   dx → 0, so 01 and 12 both tend to zero. So fnf in (C[0,1],d1), which means that f(x)=0 on [0,1] (from an example we did earlier). Likewise, f=1 on [1,2], which doesn’t give a continuous limit.

  3. Similarly, (C[a,b],d1) is incomplete in general. Also it is incomplete in the d2 metric, as the same example shows (a similar calculation with squares of functions). We will see later that it is complete in the d metric.
Remark 59 Note that 2 is also complete with any of the metrics d1, d2 and d; since a Cauchy/ convergent sequence (vn)=(xn,yn) in 2 is just one in which both (xn) and (yn) are Cauchy/ convergent sequences in (cf. Prop. 39).

Similar arguments show that k is also complete for k=1,2,3,…, and (with the same proof as for Corollary) all closed subsets of k are complete.

If a metric space (X,d) is not complete one can always pass to its abstract completion in the following sense.

Proposition 60 (Abstract completion) Any metric space (X,d) is isometric to a dense subspace of a complete metric space, which is called its abstract completion if (X,d).
Proof.[Sketch of proof] We describe a metric space (X,d) in which X is isometric to a dense subset. Consider the space X′ of Cauchy sequences of X. We define an equivalence relation ∼ on X′ by
(xn) ∼ (yn) ⇔ d(xn,yn) → 0.
The set X is defined to be the set of equivalence classes [(xn)]. It has a well defined metric given by
d([(xn)],[(yn)]):= 
 
lim
n → ∞
d(xn,yn).
One checks easily that this is metric and is well defined (does not depend on the chosen representative xn of [(xn)]). Now there is an injective map XX defined by sending x to the constant sequence (x,x,x,…). This map is an isometry. We can therefore think of (X,d) as a subset of (X,d). This subset is dense because every Cauchy sequence can be approximated by a sequence of constant sequences. So the only difficult bit in this construction is to show that (X,d) is complete. We will sketch the construction of a limit here. It turns out that it verifies completeness on a dense set.
Lemma 61 Suppose that (X,d) is a metric space and let YX be a dense set with the property that every Cauchy sequence in Y has a limit in X. Then (X,d) is complete.
Proof. Let (xn) be a Cauchy sequence in X. Now replace xn with another sequence yn in Y such that d(xn,yn)<1/n. Then, by the triangle inequality, yn is again a Cauchy sequence and converges, by assumption, to some xX. Then also xn converges to x. □

Let us turn to the proof of completeness of X′. Suppose that (xn) is a Cauchy sequence in X. Then, in X′ this sequence has the form ((x1,x1,…),(x2,x2,…),(x3,x3,…),…). This sequence has a limit, namely, (xn) itself. □

Exercise 62 (Extension by continuity) Let (X,d) be a metric space and X1 be a dense subset of X. Let f: X1Y be a uniformly continuous function to a complete metric space (Y,d′). Show that there is a unique function f′: XY which satisfies two properties:
  1. restriction of f to X1 coincides with f, that is f′(x)=f(x) for all xX1;
  2. f is continuous on X.
Furthermore, it can be shown that f is uniformly continuous on X. We will call f the extension of f by continuity and will often keep the same letter f to denote f.

There are many important consequences of Ex. 62, in particular the following.

Corollary 63 All abstract completions of a metric space (X,d) are isometric, in other words, the abstract completions is unique up to isometry.

1.2.2 Compactness

Accordingly to a dictionary: compact—closely and firmly united or packed together. For a metric space a meaning of “closely and firmly united” can be defined in several different forms—through open coverings or convergent subsequences—and we will see that these interpretations are equivalent.

An open cover of a metric space (X,d) is a family of open sets (Uα)α ∈ I such that

 
α ∈ I
Uα=X.

A subcover of a cover is a subset I′ ⊂ I of the index set such that (Uα)α ∈ I is still a cover.

Definition 64 (Compactness) A metric space (X,d) is called compact if every open cover has a finite subcover.

Informally: a space is compact if any infinite open covering is excessive and can be reduced just to a finite one. An example of a compact set is [0,1] and example of non-compact—all reals or the open interval (0,1). An importance of this concept is clarified by Rem. 21.

Definition 65 (Sequential Compactness) A metric space (X,d) is called sequentially compact if every sequence (xn)n ∈ ℕ in X has a convergent subsequence.

The limit of a convergent sequence is called the accumulation point of {xn}. It is instructive to compare the definitions of:

     
    x is the limit of {xn}:    ∀ ε>0   ∃ N   ∀ n>N:   d(x, xn) < ε;         
x is an accumulation point of {xn}:     ∀ ε>0   ∀ N   ∃ n>N:   d(x, xn) < ε.          

Thereafter, x is not an accumulation point of {xn} if for some ε>0 and some N for all subsequent n>N we have d(x, xn)>ε.

Informally: a space is sequentially compact if there is no room to place infinite number of points sufficiently apart from each other to avoid their condensation to a limit. Taking the sequence xn=n shows that the set of all reals is not sequentially compact. On the other hand, we know from previous years that bounded closed set in ℝn every sequence has a convergent subsequence. Therefore, bounded closed sets in ℝn are sequentially compact.

Exercise 66 What are compact sets in a discrete metric space? What are sequentially compact sets in a discrete metric space?
Lemma 67 Let (X,d) be a sequentially compact metric space. Then for every ε >0 there exist finitely many points x1,…,xn such that {Bε(xi)∣ i=1,…,n} is a cover.
Proof. Suppose this were not the case. Then there would exist an ε>0 such that for any finite number of points x1,…,xn the collection of balls Bε(xi) does not cover, i.e.
  
n
i=1
Bε(xi) ≠ X.
Starting with n=1 and then inductively adding points that are in the complement of ∪i=1n Bε(xi) we end up with an infinite sequence of points xi such that d(xi,xk) ≥ ε. This sequence cannot have a Cauchy subsequence (required for convergence) in contradiction with the sequential compactness of X. □
Theorem 68 A metric space (X,d) is compact if and only if it is sequentially compact.
Proof. We show the two directions separately.

Compactness implies sequential compactness: Suppose that X is compact and let (xi)i ∈ ℕ be a sequence. We want to show that it has a convergent subsequence. Suppose (xi) did not have a convergent subsequence. Then no point x is an accumulation point. Therefore, for each xX there exists an ε(x)>0 such that only finitely many i ∈ ℕ for which xiBε(x). Since (Bε(x))xX is an open cover it has a finite subcover, that is a finite number of balls with a finite number of xi in each. This contradicts to the infinite number of elements in the sequence (xi).

Sequential compactness implies compactness: This implication is quite tricky. The proof is again by contradiction. Let us assume our space is sequentially compact and there exists a cover Uα that does not have a finite subcover. By the above lemma there are finitely many points x1,…,xN1 such that B1(xi) is a cover. Each of the balls B1(xi) is covered by Uα as well. Since our cover does not have a finite subcover one of the balls B1(xi) does not have a finite subcover. Denote the relevant point xi by z1.

Again there are finitely many points x1,…,xN2 such that B1/2(xi) is a cover of X. The collection of sets B1(z1) ∩ B1/2(xi), with i=1,…,N2 is also a covering of B1(z1). In the same way as before there is at least one of the xi (which we will again call z2), such that B1(z1) ∩ B1/2(z2) can not be covered by a finite subcover of Uα. Continuing like this we construct a sequence of points zi such that none of the sets

   B1(z1) ⋂ B
 
1
2
(z2) ⋂ … ⋂ B
 
1
N
(zN)

can be covered by a finite subcover of Uα.

By assumption the sequence (zi) has a convergent subsequence. Say z is a limit point of that subsequence. Since Uα is an open cover the point z is contained in one of the Uα and of course that means that an open ball Bε(z) around z is contained in Uα for some ε>0.

Now we show that there exits an N ∈ ℕ such that B1/N(zN) is a subset of Uα (this will be the desired contradiction!). Indeed, choose N large enough so that d(zN,z) + 1/N<ε. Then xB1/N(zN) implies that d(x,z) ≤ d(zN,z) + d(x,zN) < d(zN,z) + 1/N<ε. This means in particular that

   B1(z1) ⋂ B
 
1
2
(z2) ⋂ … ⋂ B
 
1
N
(zN)

is a subset of Uα. Thus, there is a subcover of the set B1(z1) ∩ … ∩ B1/N(zN) consisting of one element Uα. This is a contradiction as we constructed the sequence of balls in such a way that these sets cannot be covered by a finite number of the Uα. □

Definition 69 (Boundedness) A subset AX of a metric space is called bounded if there exists x0X and C>0 such that for all xA we have d(x0,x) ≤ C.
Remark 70 One can easily see, using the triangle inequality, that the reference point x0 can be chosen as any point in X. This means if AX is bounded and x0X, then there exist a C>0 such that d(x0,x) ≤ C for any xA.
Theorem 71 Suppose that AX is a compact subset of a metric space. Then A is closed and bounded.
Proof. First we show A is bounded. Choose any x0X and note that the set Bn(x0) indexed by n ∈ ℕ is an open cover of A. Hence, there exists a finite sub-cover< Bn1(x0),…,BnN(x0). Hence, ABC(x0), where C= max{n1,…,cN}. Hence, A is bounded.

Next assume that (xk) is a sequence in A that converges in X. Since A is compact there exists a subsequence that converges in A. Hence, the limit of xk must also be in A. Therefore, A is closed. □

The converse of this statement is not correct in general. It is however famously correct in ℝm.

Theorem 72 (Heine–Borel) A subset K ⊂ ℝm is compact if and only if it is closed and bounded.
Proof. We just need to combine the above statements. We have already shown that compactness implies closedness and boundedness. If K is closed and bounded we know from Analysis that it is sequentially compact. Therefore it is compact. □

As an illustration of further nice properties of compact spaces we mention the following result:

Exercise 73
  1. Any continuous function on a compact set is bounded.
  2. Any continuous function f: KX from a compact space K to a metric space X is uniformly continuous.
Remark 74 Note that there are two different sorts of properties of metric spaces: Completeness and compactness are of the first sort, closedness is of the second, cf. Rem 17.

2 Basics of Linear Spaces

  A person is solely the concentration of an infinite set of interrelations with another and others, and to separate a person from these relations means to take away any real meaning of the life.

 Vl. Soloviev


A space around us could be described as a three dimensional Euclidean space. To single out a point of that space we need a fixed frame of references and three real numbers, which are coordinates of the point. Similarly to describe a pair of points from our space we could use six coordinates; for three points—nine, end so on. This makes it reasonable to consider Euclidean (linear) spaces of an arbitrary finite dimension, which are studied in the courses of linear algebra.

The basic properties of Euclidean spaces are determined by its linear and metric structures. The linear space (or vector space) structure allows to add and subtract vectors associated to points as well as to multiply vectors by real or complex numbers (scalars).

The metric space structure assign a distance—non-negative real number—to a pair of points or, equivalently, defines a length of a vector defined by that pair. A metric (or, more generally a topology) is essential for definition of the core analytical notions like limit or continuity. The importance of linear and metric (topological) structure in analysis sometime encoded in the formula:

Analysis  = Algebra  + Geometry .   (8)

On the other hand we could observe that many sets admit a sort of linear and metric structures which are linked each other. Just few among many other examples are:

It is a very mathematical way of thinking to declare such sets to be spaces and call their elements points.

But shall we lose all information on a particular element (e.g. a sequence {1/n}) if we represent it by a shapeless and size-less “point” without any inner configuration? Surprisingly not: all properties of an element could be now retrieved not from its inner configuration but from interactions with other elements through linear and metric structures. Such a “sociological” approach to all kind of mathematical objects was codified in the abstract category theory.

Another surprise is that starting from our three dimensional Euclidean space and walking far away by a road of abstraction to infinite dimensional Hilbert spaces we are arriving just to yet another picture of the surrounding space—that time on the language of quantum mechanics.

  The distance from Manchester to Liverpool is 35 miles—just about the mileage in the opposite direction!

A tourist guide to England


2.1 Banach spaces (basic definitions only)

The following definition generalises the notion of distance known from the everyday life.

Definition 1 A metric (or distance function) d on a set M is a function d: M× M →ℝ+ from the set of pairs to non-negative real numbers such that:
  1. d(x,y)≥0 for all x, yM, d(x,y)=0 implies x=y .
  2. d(x,y)=d(y,x) for all x and y in M.
  3. d(x,y)+d(y,z)≥ d(x,z) for all x, y, and z in M (triangle inequality).
Exercise 2 Let M be the set of UK’s cities are the following function are metrics on M:
  1. d(A,B) is the price of 2nd class railway ticket from A to B.
  2. d(A,B) is the off-peak driving time from A to B.

The following notion is a useful specialisation of metric adopted to the linear structure.

Definition 3 Let V be a (real or complex) vector space. A norm on V is a real-valued function, written ||x||, such that
  1. ||x||≥ 0 for all xV, and ||x||=0 implies x=0.
  2. ||λ x|| = | λ | ||x|| for all scalar λ and vector x.
  3. ||x+y||≤ ||x||+||y|| (triangle inequality).
A vector space with a norm is called a normed space.

The connection between norm and metric is as follows:

Proposition 4 If ||·|| is a norm on V, then it gives a metric on V by d(x,y)=||xy||.

  (a)    (b)    
Figure 1: Triangle inequality in metric (a) and normed (b) spaces.

Proof. This is a simple exercise to derive items 13 of Definition 1 from corresponding items of Definition 3. For example, see the Figure 1 to derive the triangle inequality. □

An important notions known from real analysis are limit and convergence. Particularly we usually wish to have enough limiting points for all “reasonable” sequences.

Definition 5 A sequence {xk} in a metric space (M,d) is a Cauchy sequence, if for every є>0, there exists an integer n such that k,l>n implies that d(xk,xl)<є.
Definition 6 (M,d) is a complete metric space if every Cauchy sequence in M converges to a limit in M.

For example, the set of integers ℤ and reals ℝ with the natural distance functions are complete spaces, but the set of rationals ℚ is not. The complete normed spaces deserve a special name.

Definition 7 A Banach space is a complete normed space.
Exercise* 8 A convenient way to define a norm in a Banach space is as follows. The unit ball U in a normed space B is the set of x such that ||x||≤ 1. Prove that:
  1. U is a convex set, i.e. x, yU and λ∈ [0,1] the point λ x +(1−λ)y is also in U.
  2. ||x||=inf{ λ∈ℝ+  ∣  λ−1xU}.
  3. U is closed if and only if the space is Banach.

(i)    (ii)   (iii)
Figure 2: Different unit balls defining norms in ℝ2 from Example 9.

Example 9 Here is some examples of normed spaces.
  1. l2n is either n or n with norm defined by
    ⎪⎪
    ⎪⎪
    (x1,…,xn)⎪⎪
    ⎪⎪
    2 = 

    x1
    2+
    x2
    2+ ⋯+ 
    xn
    2
    . (9)
  2. l1n is either n or n with norm defined by
    ⎪⎪
    ⎪⎪
    (x1,…,xn)⎪⎪
    ⎪⎪
    1 = 

    x1
    +
    x2
    + ⋯+ 
    xn
    . (10)
  3. ln is either n or n with norm defined by
    ⎪⎪
    ⎪⎪
    (x1,…,xn)⎪⎪
    ⎪⎪
     = max(

    x1
    ,
    x2
    , ⋯, 
    xn
    ). (11)
  4. Let X be a topological space, then Cb(X) is the space of continuous bounded functions f: X→ℂ with norm ||f||=supX | f(x) |.
  5. Let X be any set, then l(X) is the space of all bounded (not necessarily continuous) functions f: X→ℂ with norm ||f||=supX | f(x) |.
All these normed spaces are also complete and thus are Banach spaces. Some more examples of both complete and incomplete spaces shall appear later.

  —We need an extra space to accommodate this product!

A manager to a shop assistant


2.2 Hilbert spaces

Although metric and norm capture important geometric information about linear spaces they are not sensitive enough to represent such geometric characterisation as angles (particularly orthogonality). To this end we need a further refinements.

From courses of linear algebra known that the scalar product ⟨ x,y ⟩= x1 y1 + ⋯ + xn yn is important in a space ℝn and defines a norm ||x||2=⟨ x,x ⟩. Here is a suitable generalisation:

Definition 10 A scalar product (or inner product) on a real or complex vector space V is a mapping V× V → ℂ, written x,y, that satisfies:
  1. x,x ⟩ ≥ 0 and x,x ⟩ =0 implies x=0.
  2. x,y ⟩ = y,x in complex spaces and x,y ⟩ = ⟨ y,x in real ones for all x, yV.
  3. ⟨ λ x,y ⟩=λ ⟨ x,y, for all x, yV and scalar λ. (What is xy?).
  4. x+y,z ⟩=⟨ x,z ⟩ + ⟨ y,z, for all x, y, and zV. (What is x, y+z?).

Last two properties of the scalar product is oftenly encoded in the phrase: “it is linear in the first variable if we fix the second and anti-linear in the second if we fix the first”.

Definition 11 An inner product space V is a real or complex vector space with a scalar product on it.
Example 12 Here is some examples of inner product spaces which demonstrate that expression ||x||=√x,x defines a norm.
  1. The inner product for n was defined in the beginning of this section. The inner product for n is given by x,y ⟩=∑1n xj ȳj. The norm ||x||=√1n | xj |2 makes it l2n from Example 1.
  2. The extension for infinite vectors: let l2 be
    l2={ sequences  {xj}1 ∣ 
    1

    xj
    2 < ∞}. (12)
    Let us equip this set with operations of term-wise addition and multiplication by scalars, then l2 is closed under them. Indeed it follows from the triangle inequality and properties of absolutely convergent series. From the standard Cauchy–Bunyakovskii–Schwarz inequality follows that the series 1xjȳj absolutely converges and its sum defined to be x,y.
  3. Let Cb[a,b] be a space of continuous functions on the interval [a,b]∈ℝ. As we learn from Example 4 a normed space it is a normed space with the norm ||f||=sup[a,b]| f(x) |. We could also define an inner product:
    ⟨ f,g  ⟩=
    b
    a
    f(x)ḡ(x) dx  and  ⎪⎪
    ⎪⎪
    f⎪⎪
    ⎪⎪
    2=


    b
    a

    f(x) 
    2dx


    1/2



     
    . (13)

Now we state, probably, the most important inequality in analysis.

Theorem 13 (Cauchy–Schwarz–Bunyakovskii inequality) For vectors x and y in an inner product space V let us define ||x||=√x,x and ||y||=√y,y then we have

⟨ x,y  ⟩ 
≤ ⎪⎪
⎪⎪
x⎪⎪
⎪⎪
⎪⎪
⎪⎪
y⎪⎪
⎪⎪
,  (14)
with equality if and only if x and y are scalar multiple each other.
Proof. For simplicity we start from a real vector space. Let we have two vectors u and v and want to define an inner product on the two-dimensional vector space spanned by them. That is we need to know a value of ⟨ au+bv, cu+dv ⟩ for all possible scalars a, b, c, d.

By the linearity ⟨ au+bv, cu+dv ⟩ = acu,u ⟩ + (bc+ad)⟨ u,v ⟩ + dbv,v ⟩, thus everything is defined as soon as we know three inner products ⟨ u,u ⟩, ⟨ u,v ⟩ and ⟨ v,v ⟩. First of all we need to demand ⟨ u,u ⟩ ≥ 0 and ⟨ v,v ⟩ ≥ 0.

Furthermore, they shall be such that ⟨ au+bv, au+bv ⟩ ≥ 0 for all scalar a and b. If a=0, that is reduced to the previous case ⟨ v,v ⟩ ≥ 0. If a is non-zero we note ⟨ au+bv, au+bv ⟩ = a2u+(b/a)v, u+(b/a)v ⟩ and letting λ = b/a we reduce our consideration to the quadratic expression

      ⟨ u+λ v, u+λ v   ⟩ = λ 2⟨ v,v  ⟩+2λ ⟨ u,v  ⟩+⟨ u,u  ⟩.

The graph of this function of λ is an upward parabolabecause ⟨ v,v ⟩ ≥ 0. Thus, it will be non-negative for all λ if its lowest value is non-negative. From the theory of quadratic expressions, the latter is achieved at λ =−⟨ u,v ⟩/⟨ v,v ⟩ and is equal to

   
⟨ u,v  ⟩2
⟨ v,v  ⟩2
⟨ v,v  ⟩ − 2
⟨ u,v  ⟩
⟨ v,v  ⟩
 ⟨ u,v  ⟩+⟨ u,u  ⟩=−
⟨ u,v  ⟩2
⟨ v,v  ⟩
+⟨ u,u  ⟩

If −⟨ u,v2/⟨ v,v ⟩+⟨ u,u ⟩ ≥ 0 then ⟨ v,v ⟩⟨ u,u ⟩ ≥ ⟨ u,v2.

Therefore, the Cauchy-Schwarz inequality is necessary and sufficient condition for the non-negativity of the inner product defined by the three values ⟨ u,u ⟩, ⟨ u,v ⟩ and ⟨ v,v ⟩.

After the previous discussion it is easy to get the result for complex vector space as well. For any x, yV and any t∈ℝ we have:

    0< ⟨ x+ty,x+ty  ⟩= ⟨ x,x  ⟩+2t ℜ ⟨ y,x  ⟩+t2⟨ y,y  ⟩),

Thus, the discriminant of this quadratic expression in t is non-positive: (ℜ ⟨ y,x ⟩)2−||x||2||y||2≤ 0, that is | ℜ ⟨ x,y ⟩ |≤||x||||y||. Replacing y by eiαy for an arbitrary α∈[−π,π] we get | ℜ (eiαx,y ⟩) | ≤||x||||y||, this implies the desired inequality.

Corollary 14 Any inner product space is a normed space with norm ||x||=√x,x (hence also a metric space, Prop. 4).
Proof. Just to check items 13 from Definition 3. □

Again complete inner product spaces deserve a special name

Definition 15 A complete inner product space is Hilbert space.

The relations between spaces introduced so far are as follows:

Hilbert spacesBanach spacesComplete metric spaces
  
inner product spacesnormed spaces metric spaces.

How can we tell if a given norm comes from an inner product?


Figure 3: To the parallelogram identity.

Theorem 16 (Parallelogram identity) In an inner product space H we have for all x and yH (see Figure 3):
⎪⎪
⎪⎪
x+y⎪⎪
⎪⎪
2+⎪⎪
⎪⎪
xy⎪⎪
⎪⎪
2=2⎪⎪
⎪⎪
x⎪⎪
⎪⎪
2+2⎪⎪
⎪⎪
y⎪⎪
⎪⎪
2. (15)
Proof. Just by linearity of inner product:
    ⟨ x+y,x+y  ⟩+⟨ xy,xy  ⟩=2⟨ x,x  ⟩+2⟨ y,y  ⟩,
because the cross terms cancel out. □
Exercise 17 Show that (15) is also a sufficient condition for a norm to arise from an inner product. Namely, for a norm on a complex Banach space satisfying to (15) the formula
     
    ⟨ x,y  ⟩=
1
4

⎪⎪
⎪⎪
x+y⎪⎪
⎪⎪
2⎪⎪
⎪⎪
xy⎪⎪
⎪⎪
2+i⎪⎪
⎪⎪
x+iy⎪⎪
⎪⎪
2i⎪⎪
⎪⎪
xiy⎪⎪
⎪⎪
2
 
(16)
 =
1
4
3
0
ik⎪⎪
⎪⎪
x+iky⎪⎪
⎪⎪
2
 
defines an inner product. What is a suitable formula for a real Banach space?

  Divide and rule!

Old but still much used recipe


2.3 Subspaces

To study Hilbert spaces we may use the traditional mathematical technique of analysis and synthesis: we split the initial Hilbert spaces into smaller and probably simpler subsets, investigate them separately, and then reconstruct the entire picture from these parts.

As known from the linear algebra, a linear subspace is a subset of a linear space is its subset, which inherits the linear structure, i.e. possibility to add vectors and multiply them by scalars. In this course we need also that subspaces inherit topological structure (coming either from a norm or an inner product) as well.

Definition 18 By a subspace of a normed space (or inner product space) we mean a linear subspace with the same norm (inner product respectively). We write XY or XY.
Example 19
  1. Cb(X) ⊂ l(X) where X is a metric space.
  2. Any linear subspace of n or n with any norm given in Example 13.
  3. Let c00 be the space of finite sequences, i.e. all sequences (xn) such that exist N with xn=0 for n>N. This is a subspace of l2 since 1| xj |2 is a finite sum, so finite.

We also wish that the both inhered structures (linear and topological) should be in agreement, i.e. the subspace should be complete. Such inheritance is linked to the property be closed.

A subspace need not be closed—for example the sequence

x=(1, 1/2, 1/3, 1/4, …)∈ l2    because   1/k2 < ∞

and xn=(1, 1/2,…, 1/n, 0, 0,…)∈ c00 converges to x thus xc00l2.

Proposition 20
  1. Any closed subspace of a Banach/Hilbert space is complete, hence also a Banach/Hilbert space.
  2. Any complete subspace is closed.
  3. The closure of subspace is again a subspace.
Proof.
  1. This is true in any metric space X: any Cauchy sequence from Y has a limit xX belonging to Ȳ, but if Y is closed then xY.
  2. Let Y is complete and x∈ Ȳ, then there is sequence xnx in Y and it is a Cauchy sequence. Then completeness of Y implies xY.
  3. If x, y∈ Ȳ then there are xn and yn in Y such that xnx and yny. From the  triangle inequality:
          ⎪⎪
    ⎪⎪
    (xn+yn)−(x+y)⎪⎪
    ⎪⎪
    ≤ ⎪⎪
    ⎪⎪
    xnx⎪⎪
    ⎪⎪
    +⎪⎪
    ⎪⎪
    yny⎪⎪
    ⎪⎪
     → 0,
    so xn+ynx+y and x+y∈ Ȳ. Similarly x∈Ȳ implies λ x ∈Ȳ for any λ.

Hence c00 is an incomplete inner product space, with inner product ⟨ x,y ⟩=∑1xk ȳk (this is a finite sum!) as it is not closed in l2.


(a)     (b) 
Figure 4: Jump function on (b) as a L2 limit of continuous functions from (a).

Similarly C[0,1] with inner product norm ||f||=(∫01 | f(t) |2 dt)1/2 is incomplete—take the large space X of functions continuous on [0,1] except for a possible jump at 1/2 (i.e. left and right limits exists but may be unequal and f(1/2)=limt→1/2+ f(t). Then the sequence of functions defined on Figure 4(a) has the limit shown on Figure 4(b) since:

  ⎪⎪
⎪⎪
ffn⎪⎪
⎪⎪
=
1
2
+
1
n
1
2
1
n

ffn
2dt < 
2
n
 → 0. 

Obviously fC[0,1]C[0,1].

Exercise 21 Show alternatively that the sequence of function fn from Figure 4(a) is a Cauchy sequence in C[0,1] but has no continuous limit.

Similarly the space C[a,b] is incomplete for any a<b if equipped by the inner product and the corresponding norm:

     
⟨ f,g  ⟩ =
b
a
f(t)ḡ(t) dt 
(17)
⎪⎪
⎪⎪
f⎪⎪
⎪⎪
2
 =



b
a

f(t) 
2  dt


1/2



 
.  
(18)
Definition 22 Define a Hilbert space L2[a,b] to be the smallest complete inner product space containing space C[a,b] with the restriction of inner product given by (17).

It is practical to realise L2[a,b] as a certain space of “functions” with the inner product defined via an integral. There are several ways to do that and we mention just two:

  1. Elements of L2[a,b] are equivalent classes of Cauchy sequences f(n) of functions from C[a,b].
  2. Let integration be extended from the Riemann definition to the wider Lebesgue integration (see Section 13). Let L be a set of square integrable in Lebesgue sense functions on [a,b] with a finite norm (18). Then L2[a,b] is a quotient space of L with respect to the equivalence relation fg ⇔ ||fg||2=0 .
    Example 23 Let the Cantor function on [0,1] be defined as follows:
          f(t)=

              1,t∈ ℚ;
              0,t∈ ℝ∖ℚ.
    This function is not integrable in the Riemann sense but does have the Lebesgue integral. The later however is equal to 0 and as an L2-function the Cantor function equivalent to the function identically equal to 0.
  3. The third possibility is to map L2(ℝ) onto a space of “true” functions but with an additional structure. For example, in quantum mechanics it is useful to work with the Segal–Bargmann space of analytic functions on ℂ with the inner product [, , ]:
        ⟨ f1,f2  ⟩=
     


    f1(z) f2(z)e

    z
    2
     
    dz.
Theorem 24 The sequence space l2 is complete, hence a Hilbert space.
Proof. Take a Cauchy sequence x(n)l2, where x(n)=(x1(n), x2(n), x3(n), … ). Our proof will have three steps: identify the limit x; show it is in l2; show x(n)x.
  1. If x(n) is a Cauchy sequence in l2 then xk(n) is also a Cauchy sequence of numbers for any fixed k:
           
    xk(n)xk(m)
     ≤ 


    k=1

    xk(n)xk(m)
    2


    1/2



     
     = ⎪⎪
    ⎪⎪
    x(n)x(m)⎪⎪
    ⎪⎪
     → 0.
    Let xk be the limit of xk(n).
  2. For a given є>0 find n0 such that ||x(n)x(m)||<є for all n,m>n0. For any K and m:
           
    K
    k=1

    xk(n)xk(m)
    2 ≤  ⎪⎪
    ⎪⎪
    x(n)x(m)⎪⎪
    ⎪⎪
    22.
    Let m→ ∞ then ∑k=1K | xk(n)xk |2 ≤ є2.
    Let K→ ∞ then ∑k=1| xk(n)xk |2 ≤ є2. Thus x(n)xl2 and because l2 is a linear space then x = x(n)−(x(n)x) is also in l2.
  3. We saw above that for any є >0 there is n0 such that ||x(n)x||<є for all n>n0. Thus x(n)x.
Consequently l2 is complete. □

  All good things are covered by a thick layer of chocolate (well, if something is not yet–it certainly will)


2.4 Linear spans

As was explained into introduction 2, we describe “internal” properties of a vector through its relations to other vectors. For a detailed description we need sufficiently many external reference points.

Let A be a subset (finite or infinite) of a normed space V. We may wish to upgrade it to a linear subspace in order to make it subject to our theory.

Definition 25 The linear span of A, write Lin(A), is the intersection of all linear subspaces of V containing A, i.e. the smallest subspace containing A, equivalently the set of all finite linear combination of elements of A. The closed linear span of A write CLin(A) is the intersection of all closed linear subspaces of V containing A, i.e. the smallest closed subspace containing A.
Exercise* 26
  1. Show that if A is a subset of finite dimension space then Lin(A)=CLin(A).
  2. Show that for an infinite A spaces Lin(A) and CLin(A)could be different. (Hint: use Example 3.)
Proposition 27Lin(A)=CLin(A).
Proof. Clearly Lin(A) is a closed subspace containing A thus it should contain CLin(A). Also Lin(A)⊂ CLin(A) thus Lin(A)CLin(A)=CLin(A). Therefore Lin(A)= CLin(A). □

Consequently CLin(A) is the set of all limiting points of finite linear combination of elements of A.

Example 28 Let V=C[a,b] with the sup norm ||·||. Then:
Lin{1,x,x2,…}={all polynomials}
CLin{1,x,x2,…}=C[a,b] by the Weierstrass approximation theorem proved later.
Remark 29 Note, that the relation PCLin(Q) between two sets P and Q is transitive: if PCLin(Q) and QCLin(R) then PCLin(R). This observation is often used in the following way. To show that PCLin(R) we introduce some intermediate sets Q1, …, Qn such that PCLin(Q1), QjCLin(Qj+1) and QnCLin(R), see the proof of Weierstrass Approximation Thm. 17 or § 14.2 for an illustration.

The following simple result will be used later many times without comments.

Lemma 30 (about Inner Product Limit) Suppose H is an inner product space and sequences xn and yn have limits x and y correspondingly. Then xn,yn ⟩→⟨ x,y or equivalently:
    
 
lim
n→∞
⟨ xn,yn  ⟩=⟨ 
 
lim
n→∞
xn,
 
lim
n→∞
yn  ⟩.
Proof. Obviously by the Cauchy–Schwarz inequality:
    
⟨ xn,yn  ⟩−⟨ x,y  ⟩ 
=
    
⟨ xnx,yn  ⟩+⟨ x,yny  ⟩ 
 

⟨ xnx,yn  ⟩ 
+
⟨ x,yny  ⟩ 
 
⎪⎪
⎪⎪
xnx⎪⎪
⎪⎪
⎪⎪
⎪⎪
yn⎪⎪
⎪⎪
+⎪⎪
⎪⎪
x⎪⎪
⎪⎪
⎪⎪
⎪⎪
yny⎪⎪
⎪⎪
→ 0,
since ||xnx||→ 0, ||yny||→ 0, and ||yn|| is bounded. □

3 Orthogonality

  Pythagoras is forever!

 The catchphrase from TV commercial of Hilbert Spaces course


As was mentioned in the introduction the Hilbert spaces is an analog of our 3D Euclidean space and theory of Hilbert spaces similar to plane or space geometry. One of the primary result of Euclidean geometry which still survives in high school curriculum despite its continuous nasty de-geometrisation is Pythagoras’ theorem based on the notion of orthogonality1.

So far we was concerned only with distances between points. Now we would like to study angles between vectors and notably right angles. Pythagoras’ theorem states that if the angle C in a triangle is right then c2=a2+b2, see Figure 5 .


Figure 5: The Pythagoras’ theorem c2=a2+b2

It is a very mathematical way of thinking to turn this property of right angles into their definition, which will work even in infinite dimensional Hilbert spaces.

  Look for a triangle, or even for a right triangle

 A universal advice in solving problems from elementary geometry.


3.1 Orthogonal System in Hilbert Space

In inner product spaces it is even more convenient to give a definition of orthogonality not from Pythagoras’ theorem but from an equivalent property of inner product.

Definition 1 Two vectors x and y in an inner product space are orthogonal if x,y ⟩=0, written xy.

An orthogonal sequence (or orthogonal system) en (finite or infinite) is one in which enem whenever nm.

An orthonormal sequence (or orthonormal system) en is an orthogonal sequence with ||en||=1 for all n.

Exercise 2
  1. Show that if xx then x=0 and consequently xy for any yH.
  2. Show that if all vectors of an orthogonal system are non-zero then they are linearly independent.
Example 3 These are orthonormal sequences:
  1. Basis vectors (1,0,0), (0,1,0), (0,0,1) in 3 or 3.
  2. Vectors en=(0,…,0,1,0,…) (with the only 1 on the nth place) in l2. (Could you see a similarity with the previous example?)
  3. Functions en(t)=1/(√2π) eint , n∈ℤ in C[0,2π]:
    ⟨ en,em  ⟩= 
    0
    1
    einteimtdt = 

              1,n=m;
              0,n≠ m.
    (19)
Exercise 4 Let A be a subset of an inner product space V and xy for any yA. Prove that xz for all zCLin(A).
Theorem 5 (Pythagoras’) If xy then ||x+y||2=||x||2+||y||2. Also if e1, …, en is orthonormal then
  ⎪⎪
⎪⎪
⎪⎪
⎪⎪
n
1
akek⎪⎪
⎪⎪
⎪⎪
⎪⎪
2=⟨ 
n
1
akek,
n
1
ak ek  ⟩=
n
1

ak
2.
Proof. A one-line calculation. □

The following theorem provides an important property of Hilbert spaces which will be used many times. Recall, that a subset K of a linear space V is convex if for all x, yK and λ∈ [0,1] the point λ x +(1−λ)y is also in K. Particularly any subspace is convex and any unit ball as well (see Exercise 1).

Theorem 6 (about the Nearest Point) Let K be a non-empty convex closed subset of a Hilbert space H. For any point xH there is the unique point yK nearest to x.
Proof. Let d=infyK d(x,y), where d(x,y)—the distance coming from the norm ||x||=√x,x and let yn a sequence points in K such that limn→ ∞d(x,yn)=d. Then yn is a Cauchy sequence. Indeed from the parallelogram identity for the parallelogram generated by vectors xyn and xym we have:
  ⎪⎪
⎪⎪
ynym⎪⎪
⎪⎪
2=2⎪⎪
⎪⎪
xyn⎪⎪
⎪⎪
2+2⎪⎪
⎪⎪
xym⎪⎪
⎪⎪
2⎪⎪
⎪⎪
2xynym⎪⎪
⎪⎪
2.
Note that ||2xynym||2=4||xyn+ym/2||2≥ 4d2 since yn+ym/2∈ K by its convexity. For sufficiently large m and n we get ||xym||2d +є and ||xyn||2d +є, thus ||ynym||≤ 4(d2+є)−4d2=4є, i.e. yn is a Cauchy sequence.

Let y be the limit of yn, which exists by the completeness of H, then yK since K is closed. Then d(x,y)=limn→ ∞d(x,yn)=d. This show the existence of the nearest point. Let y′ be another point in K such that d(x,y′)=d, then the parallelogram identity implies:

  ⎪⎪
⎪⎪
yy⎪⎪
⎪⎪
2=2⎪⎪
⎪⎪
xy⎪⎪
⎪⎪
2+2⎪⎪
⎪⎪
xy⎪⎪
⎪⎪
2⎪⎪
⎪⎪
2xyy⎪⎪
⎪⎪
2≤ 4d2−4d2=0.

This shows the uniqueness of the nearest point. □

Exercise* 7 The essential rôle of the parallelogram identity in the above proof indicates that the theorem does not hold in a general Banach space.
  1. Show that in 2 with either norm ||·||1 or ||·|| form Example 9 the nearest point could be non-unique;
  2. Could you construct an example (in Banach space) when the nearest point does not exists?

  Liberte, Egalite, Fraternite!

 A longstanding ideal approximated in the real life by something completely different


3.2 Bessel’s inequality

For the case then a convex subset is a subspace we could characterise the nearest point in the term of orthogonality.

Theorem 8 (on Perpendicular) Let M be a subspace of a Hilbert space H and a point xH be fixed. Then zM is the nearest point to x if and only if xz is orthogonal to any vector in M.

  (i)   (ii)  
Figure 6: (i) A smaller distance for a non-perpendicular direction; and
(ii) Best approximation from a subspace

Proof. Let z is the nearest point to x existing by the previous Theorem. We claim that xz orthogonal to any vector in M, otherwise there exists yM such that ⟨ xz,y ⟩≠ 0. Then
    ⎪⎪
⎪⎪
xz−є y⎪⎪
⎪⎪
2
=
⎪⎪
⎪⎪
xz⎪⎪
⎪⎪
2−2є ℜ⟨ xz,y  ⟩+є2⎪⎪
⎪⎪
y⎪⎪
⎪⎪
2
 <
⎪⎪
⎪⎪
xz⎪⎪
⎪⎪
2,
if є is chosen to be small enough and such that є ℜ⟨ xz,y ⟩ is positive, see Figure 6(i). Therefore we get a contradiction with the statement that z is closest point to x.

On the other hand if xz is orthogonal to all vectors in H1 then particularly (xz)⊥ (zy) for all yH1, see Figure 6(ii). Since xy=(xz)+(zy) we got by the Pythagoras’ theorem:

    ⎪⎪
⎪⎪
xy⎪⎪
⎪⎪
2=⎪⎪
⎪⎪
xz⎪⎪
⎪⎪
2 + ⎪⎪
⎪⎪
zy⎪⎪
⎪⎪
2.

So ||xy||2≥ ||xz||2 and the are equal if and only if z=y. □

Exercise 9 The above proof does not work if xz,y is an imaginary number, what to do in this case?

Consider now a basic case of approximation: let xH be fixed and e1, …, en be orthonormal and denote H1=Lin{e1,…,en}. We could try to approximate x by a vector y1 e1+⋯ +λn enH1.

Corollary 10 The minimal value of ||xy|| for yH1 is achieved when y=∑1nx,eiei.
Proof. Let z=∑1nx,eiei, then ⟨ xz,ei ⟩=⟨ x,ei ⟩−⟨ z,ei ⟩=0. By the previous Theorem z is the nearest point to x. □

Figure 7: Best approximation by three trigonometric polynomials

Example 11
  1. In 3 find the best approximation to (1,0,0) from the plane V:{x1+x2+x3=0}. We take an orthonormal basis e1=(2−1/2, −2−1/2,0), e2=(6−1/2, 6−1/2, −2· 6−1/2) of V (Check this!). Then:
          z=⟨ x,e1  ⟩e1+⟨ x,e2  ⟩e2=


    1
    2
    ,−
    1
    2
    ,0


    +


    1
    6
    ,
    1
    6
    ,−
    1
    3



    =


    2
    3
    ,−
    1
    3
    ,−
    1
    3



    . 
  2. In C[0,2π] what is the best approximation to f(t)=t by functions a+beit+ceit? Let
        e0=
    1
    ,    e1=
    1
    eit,    e−1=
    1
    eit.  
    We find:
        ⟨ f,e0  ⟩=
    0
    t
    dt=




    t2
    2
    1










    0
    =
    2
    π3/2;
        ⟨ f,e1  ⟩=
    0
    teit
    dt=i
       (Check this!) 
            ⟨ f,e−1  ⟩=
    0
    teit
    dt=−i
       (Why we may not check this one?)
    Then the best approximation is (see Figure 7):
        f0(t)=⟨ f,e0  ⟩e0+⟨ f,e1  ⟩e1+⟨ f,e−1  ⟩e−1
     =
    2
    π3/2
    +ieitieit=π−2sint.
Corollary 12 (Bessel’s inequality) If (ei) is orthonormal then
    ⎪⎪
⎪⎪
x⎪⎪
⎪⎪
2≥ 
n
i=1

⟨ x,ei  ⟩ 
2.
Proof. Let z= ∑1nx,eiei then xzei for all i therefore by Exercise 4 xzz. Hence:
    ⎪⎪
⎪⎪
x⎪⎪
⎪⎪
2
=
⎪⎪
⎪⎪
z⎪⎪
⎪⎪
2+⎪⎪
⎪⎪
xz⎪⎪
⎪⎪
2
 
⎪⎪
⎪⎪
z⎪⎪
⎪⎪
2=
n
i=1

⟨ x,ei  ⟩ 
2.

  —Did you say “rice and fish for them”?

 A student question


3.3 The Riesz–Fischer theorem

When (ei) is orthonormal we call ⟨ x,en ⟩ the nth Fourier coefficient of x (with respect to (ei), naturally).

Theorem 13 (Riesz–Fisher) Let (en)1 be an orthonormal sequence in a Hilbert space H. Then 1λn en converges in H if and only if 1| λn |2 < ∞. In this case ||∑1λn en||2=∑1| λn |2.
Proof. Necessity: Let xk=∑1k λn en and x=limk→ ∞ xk. So ⟨ x,en ⟩=limk→ ∞xk,en ⟩=λn for all n. By the Bessel’s inequality for all k
    ⎪⎪
⎪⎪
x⎪⎪
⎪⎪
2≥ 
k
1

⟨ x,en  ⟩ 
2= 
k
1

λn
2, 
hence ∑1k | λn |2 converges and the sum is at most ||x||2.

Sufficiency: Consider ||xkxm||=||∑mk λn en||=(∑mk | λn |2)1/2 for k>m. Since ∑mk | λn |2 converges xk is a Cauchy sequence in H and thus has a limit x. By the Pythagoras’ theorem ||xk||2=∑1k | λn |2 thus for k→ ∞ ||x||2=∑1| λn |2 by the Lemma about inner product limit. □

Observation: the closed linear span of an orthonormal sequence in any Hilbert space looks like l2, i.e. l2 is a universal model for a Hilbert space.

By Bessel’s inequality and the Riesz–Fisher theorem we know that the series ∑1x,eiei converges for any xH. What is its limit?

Let y=x− ∑1x,eiei, then

⟨ y,ek  ⟩=⟨ x,ek  ⟩− 
1
⟨ x,ei  ⟩ ⟨ ei,ek  ⟩=⟨ x,ek  ⟩−⟨ x,ek  ⟩ =0   for all  k. (20)
Definition 14 An orthonormal sequence (ei) in a Hilbert space H is complete if the identities y,ek ⟩=0 for all k imply y=0.

A complete orthonormal sequence is also called orthonormal basis in H.

Theorem 15 (on Orthonormal Basis) Let ei be an orthonormal basis in a Hilber space H. Then for any xH we have
    x=
n=1
⟨ x,en  ⟩en    and    ⎪⎪
⎪⎪
x⎪⎪
⎪⎪
2= 
n=1

⟨ x,en  ⟩ 
2.
Proof. By the Riesz–Fisher theorem, equation (20) and definition of orthonormal basis. □

  There are constructive existence theorems in mathematics.

 An example of pure existence statement


3.4 Construction of Orthonormal Sequences

Natural questions are: Do orthonormal sequences always exist? Could we construct them?

Theorem 16 (Gram–Schmidt) Let (xi) be a sequence of linearly independent vectors in an inner product space V. Then there exists orthonormal sequence (ei) such that
  Lin{x1,x2,…,xn}=Lin{e1,e2,…,en},    for all  n.
Proof. We give an explicit algorithm working by induction. The base of induction: the first vector is e1=x1/||x1||. The step of induction: let e1, e2, …, en are already constructed as required. Let yn+1=xn+1−∑i=1nxn+1,eiei. Then by (20) yn+1ei for i=1,…,n. We may put en+1=yn+1/||yn+1|| because yn+1≠ 0 due to linear independence of xk’s. Also
    Lin{e1,e2,…,en+1}=    Lin{e1,e2,…,yn+1}
 =    Lin{e1,e2,…,xn+1}
 =    Lin{x1,x2,…,xn+1}.
So (ei) are orthonormal sequence. □
Example 17 Consider C[0,1] with the usual inner product (17) and apply orthogonalisation to the sequence 1, x, x2, …. Because ||1||=1 then e1(x)=1. The continuation could be presented by the table:
      e1(x)=1 
      y2(x)=x−⟨ x,1  ⟩1=x
1
2
,    ⎪⎪
⎪⎪
y2⎪⎪
⎪⎪
2=
1
0
(x
1
2
)2dx=
1
12
,    e2(x)=
12
(x
1
2
)
      y3(x)=x2−⟨ x2,1  ⟩1−⟨ x2,x
1
2
  ⟩(x
1
2
)· 12 ,   …,   e3=
y3
⎪⎪
⎪⎪
y3⎪⎪
⎪⎪
      …  …  …

  
Figure 8: Five first Legendre Pi and Chebyshev Ti polynomials

Example 18 Many famous sequences of orthogonal polynomials, e.g. Chebyshev, Legendre, Laguerre, Hermite, can be obtained by orthogonalisation of 1, x, x2, …with various inner products.
  1. Legendre polynomials in C[−1,1] with inner product
    ⟨ f,g  ⟩=
    1
    −1
      f(t)
    g(t)
    dt. (21)
  2. Chebyshev polynomials in C[−1,1] with inner product
    ⟨ f,g  ⟩=
    1
    −1
      f(t)
    g(t)
    dt
    1−t2
    (22)
  3. Laguerre polynomials in the space of polynomials P[0,∞) with inner product
          ⟨ f,g  ⟩=
    0
    f(t)
    g(t)
    etdt.
See Figure 8 for the five first Legendre and Chebyshev polynomials. Observe the difference caused by the different inner products (21) and (22). On the other hand note the similarity in oscillating behaviour with different “frequencies”.

Another natural question is: When is an orthonormal sequence complete?

Proposition 19 Let (en) be an orthonormal sequence in a Hilbert space H. The following are equivalent:
  1. (en) is an orthonormal basis.
  2. CLin((en))=H.
  3. ||x||2=∑1| ⟨ x,en ⟩ |2 for all xH.
Proof. Clearly 1 implies 2 because x=∑1x,enen in CLin((en)) and ||x||2=∑1x,enen by Theorem 15. The same theorem tells that 1 implies 3.

If (en) is not complete then there exists xH such that x≠ 0 and ⟨ x,ek ⟩=0 for all k, so 3 fails, consequently 3 implies 1.

Finally if ⟨ x,ek ⟩=0 for all k then ⟨ x,y ⟩=0 for all yLin((en)) and moreover for all yCLin((en)), by the Lemma on continuity of the inner product. But then xCLin((en)) and 2 also fails because ⟨ x,x ⟩=0 is not possible. Thus 2 implies 1. □

Corollary 20 A separable Hilbert space (i.e. one with a countable dense set) can be identified with either l2n or l2, in other words it has an orthonormal basis (en) (finite or infinite) such that
    x=
n=1
⟨ x,en  ⟩en    and    ⎪⎪
⎪⎪
x⎪⎪
⎪⎪
2= 
n=1

⟨ x,en  ⟩ 
2.
Proof. Take a countable dense set (xk), then H=CLin((xk)), delete all vectors which are a linear combinations of preceding vectors, make orthonormalisation by Gram–Schmidt the remaining set and apply the previous proposition. □

  Most pleasant compliments are usually orthogonal to our real qualities.

 An advise based on observations


3.5 Orthogonal complements

Orthogonality allow us split a Hilbert space into subspaces which will be “independent from each other” as much as possible.

Definition 21 Let M be a subspace of an inner product space V. The orthogonal complement, written M, of M is
M={x∈ V: ⟨ x,m  ⟩=0 ∀  m∈ M}.
Theorem 22 If M is a closed subspace of a Hilbert space H then M is a closed subspace too (hence a Hilbert space too).
Proof. Clearly M is a subspace of H because x, yM implies ax+byM:
    ⟨ ax+by,m  ⟩=    a⟨ x,m  ⟩+   b⟨ y,m  ⟩=0.
Also if all xnM and xnx then xM due to inner product limit Lemma. □
Theorem 23 Let M be a closed subspace of a Hilber space H. Then for any xH there exists the unique decomposition x=m+n with mM, nM and ||x||2=||m||2+||n||2. Thus H=MM and (M)=M.
Proof. For a given x there exists the unique closest point m in M by the Theorem on nearest point and by the Theorem on perpendicular (xm)⊥ y for all yM.

So x= m + (xm)= m+n with mM and nM. The identity ||x||2=||m||2+||n||2 is just Pythagoras’ theorem and MM={0} because null vector is the only vector orthogonal to itself.

Finally (M)=M. We have H=MM=(M)M, for any x∈(M) there is a decomposition x=m+n with mM and nM, but then n is orthogonal to itself and therefore is zero. □

4 Duality of Linear Spaces

  Everything has another side


Orthonormal basis allows to reduce any question on Hilbert space to a question on sequence of numbers. This is powerful but sometimes heavy technique. Sometime we need a smaller and faster tool to study questions which are represented by a single number, for example to demonstrate that two vectors are different it is enough to show that there is a unequal values of a single coordinate. In such cases linear functionals are just what we needed.

  –Is it functional?
–Yes, it works!


4.1 Dual space of a normed space

Definition 1 A linear functional on a vector space V is a linear mapping α: V→ ℂ (or α: V→ ℝ in the real case), i.e.
    α(ax+by)=aα(x)+bα(y),     for all   x,y∈ V  and   a,b∈ℂ.
Exercise 2 Show that α(0) is necessarily 0.

We will not consider any functionals but linear, thus below functional always means linear functional.

Example 3
  1. Let V=ℂn and ck, k=1,…,n be complex numbers. Then α((x1,…,xn))=c1x1+⋯+c2x2 is a linear functional.
  2. On C[0,1] a functional is given by α(f)=∫01 f(t) d t.
  3. On a Hilbert space H for any xH a functional αx is given by αx(y)=⟨ y,x.
Theorem 4 Let V be a normed space and α is a linear functional. The following are equivalent:
  1. α is continuous (at any point of V).
  2. α is continuous at point 0.
  3. sup{| α(x) |: ||x||≤ 1}< ∞, i.e. α is a bounded linear functional.
Proof. Implication 12 is trivial.

Show 23. By the definition of continuity: for any є>0 there exists δ>0 such that ||v||<δ implies | α(v)−α(0) |<є . Take є=1 then | α(δ x) |<1 for all x with norm less than 1 because ||δ x||< δ. But from linearity of α the inequality | α(δ x) |<1 implies | α(x) |<1/δ<∞ for all ||x||≤ 1.

31. Let mentioned supremum be M. For any x, yV such that xy vector (xy)/||xy|| has norm 1. Thus | α ((xy)/||xy||) |<M. By the linearity of α this implies that | α (x)−α(y) |<M||xy||. Thus α is continuous. □

Definition 5 The dual space X* of a normed space X is the set of continuous linear functionals on X. Define a norm on it by
⎪⎪
⎪⎪
α⎪⎪
⎪⎪
=
 
sup
⎪⎪
⎪⎪
x⎪⎪
⎪⎪
= 1

α(x) 
.  (23)
Exercise 6
  1. Show that the chain of inequalities:
          ⎪⎪
    ⎪⎪
    α⎪⎪
    ⎪⎪
     ≤ 
     
    sup
    ⎪⎪
    ⎪⎪
    x⎪⎪
    ⎪⎪
     ≤ 1

    α(x) 
    ≤ 
     
    sup
    x ≠ 0 

    α(x) 
    ⎪⎪
    ⎪⎪
    x⎪⎪
    ⎪⎪
     ≤ ⎪⎪
    ⎪⎪
    α ⎪⎪
    ⎪⎪
    . 
    Deduce that any of the mentioned supremums deliver the norm of α. Which of them you will prefer if you need to show boundedness of α? Which of them is better to use if boundedness of α is given?
  2. Show that | α(x) |≤ ||α||·||x|| for all xX, α ∈ X*.

The important observations is that linear functionals form a normed space as follows:

Exercise 7
  1. Show that X* is a linear space with natural (point-wise) operations.
  2. Show that (23) defines a norm on X*.

Furthermeore, X* is always complete, regardless of properties of X!

Theorem 8 X* is a Banach space with the defined norm (even if X was incomplete).
Proof. Due to Exercise 7 we only need to show that X* is complete. Let (αn) be a Cauchy sequence in X*, then for any xX scalars αn(x) form a Cauchy sequence, since | αm(x)−αn(x) |≤||αm−αn||·||x||. Thus the sequence has a limit and we define α by α(x)=limn→∞αn(x). Clearly α is a linear functional on X. We should show that it is bounded and αn→ α. Given є>0 there exists N such that ||αn−αm||<є for all n, mN. If ||x||≤ 1 then | αn(x)−αm(x) |≤ є, let m→∞ then | αn(x)−α(x) |≤ є, so
    
α(x) 
≤ 
αn(x) 
+є≤ ⎪⎪
⎪⎪
αn⎪⎪
⎪⎪
 + є,
i.e. ||α|| is finite and ||αn−α||≤ є, thus αn→α. □
Definition 9 The kernel of linear functional α, write kerα, is the set all vectors xX such that α(x)=0.
Exercise 10 Show that
  1. kerα is a subspace of X.
  2. If α≢0 then obviously kerα ≠ X. Furthermore, if X has at least two linearly independent vectors then kerα ≠ {0}, thus kerα is a proper subspace of X.
  3. If α is continuous then kerα is closed.

  Study one and get any other for free!

 Hilbert spaces sale


4.2 Self-duality of Hilbert space

Lemma 11 (Riesz–Fréchet) Let H be a Hilbert space and α a continuous linear functional on H, then there exists the unique yH such that α(x)=⟨ x,y for all xH. Also ||α||H*=||y||H.
Proof. Uniqueness: if ⟨ x,y ⟩=⟨ x,y′ ⟩ ⇔ ⟨ x,yy′ ⟩=0 for all xH then yy′ is self-orthogonal and thus is zero (Exercise 1).

Existence: we may assume that α≢0 (otherwise take y=0), then M=kerα is a closed proper subspace of H. Since H=MM, there exists a non-zero zM, by scaling we could get α(z)=1. Then for any xH:

    x=(x−α(x)z)+α(x)z,     with x−α(x)z∈ M, α(x)z∈ M.

Because ⟨ x,z ⟩=α(x)⟨ z,z ⟩=α(x)||z||2 for any xH we set y=z/||z||2.

Equality of the norms ||α||H*=||y||H follows from the Cauchy–Bunyakovskii–Schwarz inequality in the form α(x)≤ ||x||·||y|| and the identity α(y/||y||)=||y||. □

Example 12 On L2[0,1] let α(f)=⟨ f,t2 ⟩=∫01 f(t)t2d t. Then
    ⎪⎪
⎪⎪
α⎪⎪
⎪⎪
=⎪⎪
⎪⎪
t2⎪⎪
⎪⎪
=


1
0
(t2)2dt


1/2



 
 =
1
5
.

5 Fourier Analysis

  All bases are equal, but some are more equal then others.


As we saw already any separable Hilbert space posses an orthonormal basis (infinitely many of them indeed). Are they equally good? This depends from our purposes. For solution of differential equation which arose in mathematical physics (wave, heat, Laplace equations, etc.) there is a proffered choice. The fundamental formula: d/dx eax=aeax reduces the derivative to a multiplication by a. We could benefit from this observation if the orthonormal basis will be constructed out of exponents. This helps to solve differential equations as was demonstrated in Subsection 0.2.

  7.40pm Fourier series: Episode II

 Today’s TV listing


5.1 Fourier series

Now we wish to address questions stated in Remark 9. Let us consider the space L2[−π,π]. As we saw in Example 3 there is an orthonormal sequence en(t)=(2π)−1/2eint in L2[−π,π]. We will show that it is an orthonormal basis, i.e.

  f(t)∈ L2[−π,π]  ⇔   f(t)=
k=−∞
⟨ f,ek  ⟩ek(t),

with convergence in L2 norm. To do this we show that CLin{ek:k∈ℤ}=L2[−π,π].

Let CP[−π,π] denote the continuous functions f on [−π,π] such that f(π)=f(−π). We also define f outside of the interval [−π,π] by periodicity.

Lemma 1 The space CP[−π,π] is dense in L2[−π,π].

Figure 9: A modification of continuous function to periodic

Proof. Let fL2[−π,π]. Given є>0 there exists gC[−π,π] such that ||fg||<є/2. From continuity of g on a compact set follows that there is M such that | g(t) |<M for all t∈[−π,π].

We can now replace g by periodic g′, which coincides with g on [−π,π−δ] for an arbitrary δ>0 and has the same bounds: | g′(t) |<M, see Figure 9. Then

    ⎪⎪
⎪⎪
gg⎪⎪
⎪⎪
22=
π
π−δ

g(t)−g′(t) 
2dt ≤ (2M)2δ.

So if δ<є2/(4M)2 then ||gg′||<є/2 and ||fg′||<є. □

Now if we could show that CLin{ek: k ∈ ℤ} includes CP[−π,π] then it also includes L2[−π,π].

Notation 2 Let fCP[−π,π],write
fn=
n
k=−n
 ⟨ f,ek  ⟩ ek ,   for   n=0,1,2,… (24)
the partial sum of the Fourier series for f.

We want to show that ||ffn||2→ 0. To this end we define nth Fejér sum by the formula

Fn=
f0+f1+⋯+fn
n+1
, (25)

and show that

    ⎪⎪
⎪⎪
Fnf⎪⎪
⎪⎪
 → 0.

Then we conclude

  ⎪⎪
⎪⎪
Fnf⎪⎪
⎪⎪
2=


π
−π

Fn(t)−f
2


1/2



 
 ≤ (2π)1/2 ⎪⎪
⎪⎪
Fnf⎪⎪
⎪⎪
→ 0.

Since FnLin((en)) then fCLin((en)) and hence f=∑−∞f,ekek.

Remark 3 It is not always true that ||fnf||→ 0 even for fCP[−π,π].
Exercise 4 Find an example illustrating the above Remark.

The summation method used in (25) us useful not only in the context of Fourier series but for many other cases as well. In such a wider framework the method is known as .

  It took 19 years of his life to prove this theorem


5.2 Fejér’s theorem

Proposition 5 (Fejér, age 19) Let fCP[−π,π]. Then
     
    Fn(x)=
1
π
−π
f(t) Kn(xt) dt,     where   
(26)
    Kn(t)=
1
n+1
n
k=0
k
m=−k
eimt,  
(27)
is the Fejér kernel.
Proof. From notation (24):
    fk(x)=
k
m=−k
 ⟨ f,em  ⟩ em(x)
 =
k
m=−k
π
−π
f(t)
eimt
dt  
eimx
 =
1
π
−π
f(t)
k
m=−k
eim(xt)dt.
Then from (25):
    Fn(x)=
1
n+1
n
k=0
fk(x)
 =
1
n+1
1
n
k=0
π
−π
f(t)
k
m=−k
eim(xt)dt
 =
1
π
−π
f(t) 
1
n+1
n
k=0
k
m=−k
eim(xt)dt,
which finishes the proof. □
Lemma 6 The Fejér kernel is -periodic, Kn(0)=n+1 and can be expressed as:
Kn(t)=
1
n+1
sin2
(n+1)t
2
sin2
t
2
,    for  t∉2πℤ. (28)

   1   
  z−11z  
 z−2z−11zz2 
Table 1: Counting powers in rows and columns

Proof. Let z=eit, then:
    Kn(t)=
1
n+1
n
k=0
(zk+⋯+1+z+⋯+zk)
 =
1
n+1
n
j=−n
(n+1−
j
)
zj, 
by switch from counting in rows to counting in columns in Table 1. Let w=eit/2, i.e. z=w2, then
     
    Kn(t)=
1
n+1
(w−2n+2w−2n+2+⋯+(n+1)+nw2+⋯+w2n)
 
 =
1
n+1
(wn+wn+2+⋯+wn−2+wn)2 
(29)
 =
1
n+1



wn−1wn+1
w−1w



2



 
   Could you sum a geometric progression?
 
 =
1
n+1






2isin
(n+1)t
2
2isin
t
2






2






 
, 
 
if w≠ ± 1. For the value of Kn(0) we substitute w=1 into (29). □

Figure 10: A family of Fejér kernels with the parameter m running from 0 to 9 is on the left picture. For a comparison unregularised Fourier kernels are on the right picture.

The first eleven Fejér kernels are shown on Figure 10, we could observe that:

Lemma 7 Fejér’s kernel has the following properties:
  1. Kn(t)≥0 for all t∈ ℝ and n∈ℕ.
  2. −ππKn(t) d t=2π.
  3. For any δ∈ (0,π)
          
    −δ
    −π
    +
    π
    δ
    Kn(t) dt → 0    as    n→ ∞.
Proof. The first property immediately follows from the explicit formula (28). In contrast the second property is easier to deduce from expression with double sum (27):
    
π
−π
Kn(t) dt
=
π
−π
1
n+1
n
k=0
k
m=−k
eimtdt
 =
     
1
n+1
n
k=0
k
m=−k
π
−π
eimtdt
 =
     
1
n+1
n
k=0
 2π
 =2π,
since the formula (19).

Finally if | t |>δ then sin2(t/2)≥ sin2(δ/2)>0 by monotonicity of sinus on [0,π/2], so:

    0≤ Kn(t) ≤ 
1
(n+1) sin2(δ/2)

implying:

  0≤ 
 
δ≤
t
≤ π
Kn(t)  dt ≤ 
1(π−δ)
(n+1) sin2(δ/2)
→ 0   as  n→ 0.

Therefore the third property follows from the squeeze rule. □

Theorem 8 (Fejér Theorem) Let fCP[−π,π]. Then its Fejér sums Fn (25) converges in supremum norm to f on [−π,π] and hence in L2 norm as well.
Proof. Idea of the proof: if in the formula (26)
      Fn(x)=
1
π
−π
f(t) Kn(xt) dt, 
t is long way from x, Kn is small (see Lemma 7 and Figure 10), for t near x, Kn is big with total “weight” 2π, so the weighted average of f(t) is near f(x).

Here are details. Using property 2 and periodicity of f and Kn we could express trivially

      f(x)= f(x)
1
x
x−π
Kn(xt)  dt  =
1
x
x−π
f(x) Kn(xt) dt. 

Similarly we rewrite (26) as

      Fn(x)=
1
x
x−π
f(t) Kn(xt) dt, 

then

    
f(x)−Fn(x) 
=
    
1



x
x−π
 (f(x)−f(t)) Kn(xt) dt


 
1
x
x−π

f(x)−f(t) 
Kn(xt) dt.

Given є>0 split into three intervals: I1=[x−π,x−δ], I2=[x−δ,x+δ], I3=[x+δ,x+π], where δ is chosen such that | f(t)−f(x) |<є/2 for tI2, which is possible by continuity of f. So

     
1
 


I2

f(x)−f(t) 
Kn(xt) dt
є
2
1
 


I2
  Kn(xt) dt <
є
2
.

And

     
1
 


I1⋃ I3

f(x)−f(t) 
Kn(xt)  dt
2⎪⎪
⎪⎪
f⎪⎪
⎪⎪
1
 


I1⋃ I3
  Kn(xt)  dt
 =
⎪⎪
⎪⎪
f⎪⎪
⎪⎪
π
 
δ<
u
  Kn(u)  du
 <
є
2
, 

if n is sufficiently large due to property 3 of Kn. Hence | f(x)−Fn(x) |<є for a large n independent of x. □

Remark 9 The above properties 13 and their usage in the last proof can be generalised to the concept of approximation of the identity. See § 15.4 for a further example.

We almost finished the demonstration that en(t)=(2π)−1/2eint is an orthonormal basis of L2[−π,π]:

Corollary 10 (Fourier series) Let fL2[−π,π], with Fourier series
    
n=−∞
⟨ f,en  ⟩en=
n=−∞
cneint   where  cn=
⟨ f,en  ⟩
=
1
π
−π
f(t)eintdt.
Then the series −∞f,enen=∑−∞cneint converges in L2[−π,π] to f, i.e
    
 
lim
k→ ∞
⎪⎪
⎪⎪
⎪⎪
⎪⎪
f
k
n=−k
cneint⎪⎪
⎪⎪
⎪⎪
⎪⎪
2=0.
Proof. This follows from the previous Theorem, Lemma 1 about density of CP in L2, and Theorem 15 on orthonormal basis. □
Remark 11 There is a reason why we had used the Fejér kernel and the Cezàro summation Fn (25) instead of plain partial sums fn (24) of the Fourier series. It can be shown that point-wise convergence fnf does not hold for every continuous function f, cf. Cor. 41.

5.3 Parseval’s formula

The following result first appeared in the framework of L2[−π,π] and only later was understood to be a general property of inner product spaces.

Theorem 12 (Parseval’s formula) If f, gL2[−π,π] have Fourier series f=∑n=−∞cneint and g=∑n=−∞dneint, then
⟨ f,g  ⟩=
π
−π
f(t)
g(t)
dt=2π
−∞
cn
dn
. (30)

More generally if f and g are two vectors of a Hilbert space H with an orthonormal basis (en)−∞ then

    ⟨ f,g  ⟩=
k=−∞
cn
dn
,    where  cn=⟨ f,en  ⟩, dn=⟨ g,en  ⟩,

are the Fourier coefficients of f and g.

Proof. In fact we could just prove the second, more general, statement—the first one is its particular realisation. Let fn=∑k=−nn ckek and gn=∑k=−nn dkek will be partial sums of the corresponding Fourier series. Then from orthonormality of (en) and linearity of the inner product:
    ⟨ fn,gn  ⟩=⟨ 
n
k=−n
ckek,
n
k=−n
dkek  ⟩=
n
k=−n
ck
dk
.
This formula together with the facts that fkf and gkg (following from Corollary 10) and Lemma about continuity of the inner product implies the assertion. □
Corollary 13 A integrable function f belongs to L2[−π,π] if and only if its Fourier series is convergent and then ||f||2=2π∑−∞| ck |2.
Proof. The necessity, i.e. implication fL2 ⇒ ⟨ f,f ⟩=||f||2=2π∑| ck |2 , follows from the previous Theorem. The sufficiency follows by Riesz–Fisher Theorem. □
Remark 14 The actual rôle of the Parseval’s formula is shadowed by the orthonormality and is rarely recognised until we meet the wavelets or coherent states. Indeed the equality (30) should be read as follows:
Theorem 15 (Modified Parseval) The map W: Hl2 given by the formula [Wf](n)=⟨ f,en ⟩ is an isometry for any orthonormal basis (en).
We could find many other systems of vectors (ex), xX (very different from orthonormal bases) such that the map W: HL2(X) given by the simple universal formula
[Wf](x)=⟨ f,ex  ⟩ (31)
will be an isometry of Hilbert spaces. The map (31) is oftenly called wavelet transform and most famous is the Cauchy integral formula in complex analysis. The majority of wavelets transforms are linked with group representations, see our postgraduate course Wavelets in Applied and Pure Maths.

  Heat and noise but not a fire?

 Answer:


5.4 Some Application of Fourier Series

We are going to provide now few examples which demonstrate the importance of the Fourier series in many questions. The first two (Example 16 and Theorem 17) belong to pure mathematics and last two are of more applicable nature.

Example 16 Let f(t)=t on [−π,π]. Then
    ⟨ f,en  ⟩=
π
−π
teintdt=





        (−1)n
2π i
n
,
n≠ 0
        0,n=0
   (check!),
so f(t)∼ ∑−∞(−1)n (i/n) eint. By a direct integration:
    ⎪⎪
⎪⎪
f⎪⎪
⎪⎪
22=
π
−π
t2dt=
3
3
.
On the other hand by the previous Corollary:
  ⎪⎪
⎪⎪
f⎪⎪
⎪⎪
22=2π
 
n≠ 0



(−1)ni
n



2=4π
n=1
1
n2
.
Thus we get a beautiful formula
  
1
1
n2
=
π2
6
.

Here is another important result.

Theorem 17 (Weierstrass Approximation Theorem) For any function fC[a,b] and any є>0 there exists a polynomial p such that ||fp||.
Proof. Change variable: t=2π(xa+b/2)/(ba) this maps x∈[a,b] onto t∈[−π,π]. Let P denote the subspace of polynomials in C[−π,π]. Then eint$P_^$ for any n∈ℤ since Taylor series converges uniformly in [−π,π]. Consequently P contains the closed linear span in (supremum norm) of eint, any n∈ℤ, which is CP[−π,π] by the Fejér theorem. Thus $P_^$CP[−π,π] and we extend that to non-periodic function as follows (why we could not make use of Lemma 1 here, by the way?).

For any fC[−π,π] let λ=(f(π)−f(−π))/(2π) then f1(t)=f(t)−λ tCP[−π,π] and could be approximated by a polynomial p1(t) from the above discussion. Then f(t) is approximated by the polynomial p(t)=p1(t)+λ t. □

It is easy to see, that the rôle of exponents eint in the above prove is rather modest: they can be replaced by any functions which has a Taylor expansion. The real glory of the Fourier analysis is demonstrated in the two following examples.


Figure 11: The dynamics of a heat equation:
x—coordinate on the rod,
t—time,
T—temperature.

Example 18 The modern history of the Fourier analysis starts from the works of Fourier on the heat equation. As was mentioned in the introduction to this part, the exceptional role of Fourier coefficients for differential equations is explained by the simple formula x einx= ineinx. We shortly review a solution of the heat equation to illustrate this.

Let we have a rod of the length . The temperature at its point x∈[−π,π] and a moment t∈[0,∞) is described by a function u(t,x) on [0,∞)×[−π,π]. The mathematical equation describing a dynamics of the temperature distribution is:

∂ u(t,x)
∂ t
=
2 u(t,x)
∂ x2
  or, equivalently,  
t−∂x2
u(t,x)=0.  (32)

For any fixed moment t0 the function u(t0,x) depends only from x∈[−π,π] and according to Corollary 10 could be represented by its Fourier series:

    u(t0,x)=
n=−∞
⟨ u,en  ⟩en=
n=−∞
cn(t0)einx, 

where

    cn(t0)=
⟨ u,en  ⟩
=
1
π
−π
u(t0,x)einxdx, 

with Fourier coefficients cn(t0) depending from t0. We substitute that decomposition into the heat equation (32) to receive:

     
     
t−∂x2
u(t,x)
=
t−∂x2
n=−∞
cn(t)einx
         
 
=    
n=−∞

t−∂x2
cn(t)einx
         
 
=    
n=−∞
(cn(t)+n2cn(t))einx=0 . 
        (33)

Since function einx form a basis the last equation (33) holds if and only if

cn(t)+n2cn(t)=0   for all  n  and  t.  (34)

Equations from the system (34) have general solutions of the form:

cn(t)=cn(0)en2t    for all  t∈[0,∞), (35)

producing a general solution of the heat equation (32) in the form:

u(t,x)=
n=−∞
cn(0)en2teinx =
n=−∞
cn(0)en2t+inx, (36)

where constant cn(0) could be defined from boundary condition. For example, if it is known that the initial distribution of temperature was u(0,x)=g(x) for a function g(x)∈L2[−π,π] then cn(0) is the n-th Fourier coefficient of g(x).

The general solution (36) helps produce both the analytical study of the heat equation (32) and numerical simulation. For example, from (36) obviously follows that

The example of numerical simulation for the initial value problem with g(x)=2cos(2*u) + 1.5sin(u). It is clearly illustrate our above conclusions.


Figure 12: Two oscillation with unharmonious frequencies and the appearing dissonance. Click to listen the blue and green pure harmonics and red dissonance.


  
  
  
Figure 13: Graphics of G5 performed on different musical instruments (click on picture to hear the sound). Samples are taken from Sound Library.


Figure 14: Fourier series for G5 performed on different musical instruments (same order and colour as on the previous Figure)


(a)   (b)
(c)
Figure 15: Limits of the Fourier analysis: different frequencies separated in time

Example 19 Among the oldest periodic functions in human culture are acoustic waves of musical tones. The mathematical theory of musics (including rudiments of the Fourier analysis!) is as old as mathematics itself and was highly respected already in Pythagoras’ school more 2500 years ago.

The earliest observations are that

  1. The musical sounds are made of pure harmonics (see the blue and green graphs on the Figure 12), in our language cos and sin functions form a basis;
  2. Not every two pure harmonics are compatible, to be their frequencies should make a simple ratio. Otherwise the dissonance (red graph on Figure 12) appears.

The musical tone, say G5, performed on different instruments clearly has something in common and different, see Figure 13 for comparisons. The decomposition into the pure harmonics, i.e. finding Fourier coefficient for the signal, could provide the complete characterisation, see Figure 14.

The Fourier analysis tells that:

  1. All sound have the same base (i.e. the lowest) frequencies which corresponds to the G5 tone, i.e. 788 Gz.
  2. The higher frequencies, which are necessarily are multiples of 788 Gz to avoid dissonance, appears with different weights for different instruments.

The Fourier analysis is very useful in the signal processing and is indeed the fundamental tool. However it is not universal and has very serious limitations. Consider the simple case of the signals plotted on the Figure 15(a) and (b). They are both made out of same two pure harmonics:

  1. On the first signal the two harmonics (drawn in blue and green) follow one after another in time on Figure 15(a);
  2. They just blended in equal proportions over the whole interval on Figure 15(b).

This appear to be two very different signals. However the Fourier performed over the whole interval does not seems to be very different, see Figure 15(c). Both transforms (drawn in blue-green and pink) have two major pikes corresponding to the pure frequencies. It is not very easy to extract differences between signals from their Fourier transform (yet this should be possible according to our study).

Even a better picture could be obtained if we use windowed Fourier transform, namely use a sliding “window” of the constant width instead of the entire interval for the Fourier transform. Yet even better analysis could be obtained by means of wavelets already mentioned in Remark 14 in connection with Plancherel’s formula. Roughly, wavelets correspond to a sliding window of a variable size—narrow for high frequencies and wide for low.

6 Operators

  All the space’s a stage,
and all functionals and operators merely players!


All our previous considerations were only a preparation of the stage and now the main actors come forward to perform a play. The vectors spaces are not so interesting while we consider them in statics, what really make them exciting is the their transformations. The natural first steps is to consider transformations which respect both linear structure and the norm.

6.1 Linear operators

Definition 1 A linear operator T between two normed spaces X and Y is a mapping T:XY such that Tv + µ u)=λ T(v) + µ T(u). The kernel of linear operator kerT and image are defined by
    kerT ={x∈ X: Tx=0}    ImT={y∈ Y: y=Tx,  for some x∈ X}.
Exercise 2 Show that kernel of T is a linear subspace of X and image of T is a linear subspace of Y.

As usual we are interested also in connections with the second (topological) structure:

Definition 3 A norm of linear operator is defined:
⎪⎪
⎪⎪
T⎪⎪
⎪⎪
=sup{⎪⎪
⎪⎪
Tx⎪⎪
⎪⎪
Y: ⎪⎪
⎪⎪
x⎪⎪
⎪⎪
X≤ 1}. (37)

T is a bounded linear operator if ||T||=sup{||Tx||: ||x||}<∞.

Exercise 4 Show that ||Tx||≤ ||T||·||x|| for all xX.
Example 5 Consider the following examples and determine kernel and images of the mentioned operators.
  1. On a normed space X define the zero operator to a space Y by Z: x→ 0 for all xX. Its norm is 0.
  2. On a normed space X define the identity operator by IX: xx for all xX. Its norm is 1.
  3. On a normed space X any linear functional define a linear operator from X to , its norm as operator is the same as functional.
  4. The set of operators from n to m is given by n× m matrices which acts on vector by the matrix multiplication. All linear operators on finite-dimensional spaces are bounded.
  5. On l2, let S(x1,x2,…)=(0,x1,x2,…) be the right shift operator. Clearly ||Sx||=||x|| for all x, so ||S||=1.
  6. On L2[a,b], let w(t)∈ C[a,b] and define multiplication operator Mwf by (Mw f)(t)=w(t)f(t). Now:
          ⎪⎪
    ⎪⎪
    Mwf⎪⎪
    ⎪⎪
    2
    =
    b
    a

    w(t) 
    2
    f(t) 
    2dt
     
    K2
    b
    a
          
    f(t) 
    2dt,   where   K=⎪⎪
    ⎪⎪
    w⎪⎪
    ⎪⎪
    =
     
    sup
    [a,b]

    w(t) 
    ,
    so ||Mw||≤ K.
    Exercise 6 Show that for multiplication operator in fact there is the equality of norms ||Mw||2= ||w(t)||.
Theorem 7 Let T: XY be a linear operator. The following conditions are equivalent:
  1. T is continuous on X;
  2. T is continuous at the point 0.
  3. T is a bounded linear operator.
Proof. Proof essentially follows the proof of similar Theorem 4. □

6.2 Orthoprojections

Here we will use orthogonal complement, see § 3.5, to introduce a class of linear operators—orthogonal projections. Despite of (or rather due to) their extreme simplicity these operators are among most frequently used tools in the theory of Hilbert spaces.

Corollary 8 (of Thm. 23, about Orthoprojection) Let M be a closed linear subspace of a hilbert space H. There is a linear map PM from H onto M (the orthogonal projection or orthoprojection) such that
PM2=PM,    kerPM=M,    PM=IPM. (38)
Proof. Let us define PM(x)=m where x=m+n is the decomposition from the previous theorem. The linearity of this operator follows from the fact that both M and M are linear subspaces. Also PM(m)=m for all mM and the image of PM is M. Thus PM2=PM. Also if PM(x)=0 then xM, i.e. kerPM=M. Similarly PM(x)=n where x=m+n and PM+PM=I. □
Example 9 Let (en) be an orthonormal basis in a Hilber space and let S⊂ ℕ be fixed. Let M=CLin{en: nS} and M=CLin{en:n∈ ℕ∖ S}. Then
    
k=1
akek =     
 
k∈ S
akek +
 
kS
akek.
Remark 10 In fact there is a one-to-one correspondence between closed linear subspaces of a Hilber space H and orthogonal projections defined by identities (38).

6.3 B(H) as a Banach space (and even algebra)

Theorem 11 Let B(X,Y) be the space of bounded linear operators from X and Y with the norm defined above. If Y is complete, then B(X,Y) is a Banach space.
Proof. The proof repeat proof of the Theorem 8, which is a particular case of the present theorem for Y=ℂ, see Example 3. □
Theorem 12 Let TB(X,Y) and SB(Y,Z), where X, Y, and Z are normed spaces. Then STB(X,Z) and ||ST||≤||S||||T||.
Proof. Clearly (ST)x=S(Tx)∈ Z, and
    ⎪⎪
⎪⎪
STx⎪⎪
⎪⎪
≤ ⎪⎪
⎪⎪
S⎪⎪
⎪⎪
⎪⎪
⎪⎪
Tx⎪⎪
⎪⎪
⎪⎪
⎪⎪
S⎪⎪
⎪⎪
⎪⎪
⎪⎪
T⎪⎪
⎪⎪
⎪⎪
⎪⎪
x⎪⎪
⎪⎪
,
which implies norm estimation if ||x||≤1. □
Corollary 13 Let TB(X,X)=B(X), where X is a normed space. Then for any n≥ 1, TnB(X) and ||Tn||≤ ||T||n.
Proof. It is induction by n with the trivial base n=1 and the step following from the previous theorem. □
Remark 14 Some texts use notations L(X,Y) and L(X) instead of ours B(X,Y) and B(X).
Definition 15 Let TB(X,Y). We say T is an invertible operator if there exists SB(Y,X) such that
    ST= IX   and    TS=IY.
Such an S is called the inverse operator of T.
Exercise 16 Show that
  1. for an invertible operator T:XY we have ker T={0} and T=Y.
  2. the inverse operator is unique (if exists at all). (Assume existence of S and S, then consider operator STS.)
Example 17 We consider inverses to operators from Exercise 5.
  1. The zero operator is never invertible unless the pathological spaces X=Y={0}.
  2. The identity operator IX is the inverse of itself.
  3. A linear functional is not invertible unless it is non-zero and X is one dimensional.
  4. An operator n→ ℂm is invertible if and only if m=n and corresponding square matrix is non-singular, i.e. has non-zero determinant.
  5. The right shift S is not invertible on l2 (it is one-to-one but is not onto). But the left shift operator T(x1,x2,…)=(x2,x3,…) is its left inverse, i.e. TS=I but TSI since ST(1,0,0,…)=(0,0,…). T is not invertible either (it is onto but not one-to-one), however S is its right inverse.
  6. Operator of multiplication Mw is invertible if and only if w−1C[a,b] and inverse is Mw−1. For example M1+t is invertible L2[0,1] and Mt is not.

6.4 Adjoints

Theorem 18 Let H and K be Hilbert Spaces and TB(H,K). Then there exists operator T*B(K,H) such that
      ⟨ Th,k  ⟩K=⟨ h,T*k  ⟩H    for all  h∈ H, k∈ K.
Such T* is called the adjoint operator of T. Also T**=T and ||T*||=||T||.
Proof. For any fixed kK the expression h:→ ⟨ Th,kK defines a bounded linear functional on H. By the Riesz–Fréchet lemma there is a unique yH such that ⟨ Th,kK=⟨ h,yH for all hH. Define T* k =y then T* is linear:
     ⟨ h,T*1k12k2)  ⟩H=      ⟨ Th1k12k2  ⟩K
 =λ1⟨ Th,k1  ⟩K+λ2⟨ Th,k2  ⟩K
 =λ1⟨ h,T*k1  ⟩H+λ2⟨ h,T*k2  ⟩K
 =⟨ h1T*k12T*k2  ⟩H
So T*1k12k2)=λ1T*k12T*k2. T** is defined by ⟨ k,T**h ⟩=⟨ T*k,h ⟩ and the identity ⟨ T**h,k ⟩=⟨ h,T*k ⟩=⟨ Th,k ⟩ for all h and k shows T**=T. Also:
     ⎪⎪
⎪⎪
T*k⎪⎪
⎪⎪
2
=⟨ T*k,T*k  ⟩=⟨ k,TT*k  ⟩
 
⎪⎪
⎪⎪
k⎪⎪
⎪⎪
·⎪⎪
⎪⎪
TT*k⎪⎪
⎪⎪
⎪⎪
⎪⎪
k⎪⎪
⎪⎪
·⎪⎪
⎪⎪
T⎪⎪
⎪⎪
 ·⎪⎪
⎪⎪
T*k⎪⎪
⎪⎪
,
which implies ||T*k||≤||T||·||k||, consequently ||T*||≤||T||. The opposite inequality follows from the identity ||T||=||T**||. □
Exercise 19
  1. For operators T1 and T2 show that
          (T1T2)*=T2*T1*,   (T1+T2)*=T1*+T2*   (λ T)*=λT*.
  2. If A is an operator on a Hilbert space H then (kerA)= Im A*.

6.5 Hermitian, unitary and normal operators

Definition 20 An operator T: HH is a Hermitian operator or self-adjoint operator if T=T*, i.e. Tx,y ⟩=⟨ x,Ty for all x, yH.
Example 21
  1. On l2 the adjoint S* to the right shift operator S is given by the left shift S*=T, indeed:
          ⟨ Sx,y  ⟩=      ⟨ (0,x1,x2,…),(y1,y2,…)  ⟩
     =      x1ȳ2+x2y_3+⋯=⟨ (x1,x2,…),(y2,y3,…)  ⟩
     =⟨ x,Ty  ⟩.
    Thus S is not Hermitian.
  2. Let D be diagonal operator on l2 given by
          D(x1,x2,…)=(λ1x1, λ2x2, …).
    where k) is any bounded complex sequence. It is easy to check that ||D||=||(λn)||=supk| λk | and
          D* (x1,x2,…)=(λ1x1, λ2x2, …),
    thus D is Hermitian if and only if λk∈ℝ for all k.
  3. If T: ℂn→ ℂn is represented by multiplication of a column vector by a matrix A, then T* is multiplication by the matrix A*—transpose and conjugate to A.
Exercise 22 Show that for any bounded operator T operators Tr=1/2(T+ T*), Ti=1/2i(TT*), T*T and TT* are Hermitians. Note, that any operator is the linear combination of two hermitian operators: T=Tr+i Ti (cf. z= ℜ z + iz for z∈ℂ).

To appreciate the next Theorem the following exercise is useful:

Exercise 23 Let H be a Hilbert space. Show that
  1. For xH we have ||x||= sup { | ⟨ x,y ⟩ | for all yH such that ||y||=1}.
  2. For TB(H) we have
    ⎪⎪
    ⎪⎪
    T⎪⎪
    ⎪⎪
    = sup {  
    ⟨ Tx,y  ⟩ 
      for all  x,y∈ H  such that  ⎪⎪
    ⎪⎪
    x⎪⎪
    ⎪⎪
    =⎪⎪
    ⎪⎪
    y⎪⎪
    ⎪⎪
    =1}. (39)

The next theorem says, that for a Hermitian operator T the supremum in (39) may be taken over the “diagonal” x=y only.

Theorem 24 Let T be a Hermitian operator on a Hilbert space. Then
    ⎪⎪
⎪⎪
T⎪⎪
⎪⎪
 = 
 
sup
⎪⎪
⎪⎪
x⎪⎪
⎪⎪
 = 1

⟨ Tx,x  ⟩ 
.
Proof. If Tx=0 for all xH, both sides of the identity are 0. So we suppose that ∃ xH for which Tx≠ 0.

We see that | ⟨ Tx,x ⟩ |≤ ||Tx||||x|| ≤ ||T||||x2||, so sup||x|| =1 | ⟨ Tx,x ⟩ |≤ ||T||. To get the inequality the other way around, we first write s:=sup||x|| =1 | ⟨ Tx,x ⟩ |. Then for any xH, we have | ⟨ Tx,x ⟩ |≤ s||x2||.

We now consider

    ⟨ T(x+y),x+y  ⟩ =⟨ Tx,x  ⟩ +⟨ Tx,y  ⟩+⟨ Ty,x  ⟩ +⟨ Ty,y  ⟩ =  ⟨ Tx,x  ⟩ +2ℜ ⟨ Tx,y  ⟩ +⟨ Ty,y  ⟩

(because T being Hermitian gives ⟨ Ty,x ⟩=⟨ y,Tx ⟩ =Tx,y) and, similarly,

    ⟨ T(xy),xy  ⟩ = ⟨ Tx,x  ⟩ −2ℜ ⟨ Tx,y  ⟩ +⟨ Ty,y  ⟩.

Subtracting gives

     
    4ℜ ⟨ Tx,y  ⟩= ⟨ T(x+y),x+y  ⟩−⟨ T(xy),xy  ⟩         
 
s(⎪⎪
⎪⎪
x+y⎪⎪
⎪⎪
2 +⎪⎪
⎪⎪
xy⎪⎪
⎪⎪
2)
         
 
= 2s(⎪⎪
⎪⎪
x⎪⎪
⎪⎪
2 +⎪⎪
⎪⎪
y⎪⎪
⎪⎪
2),
         

by the parallelogram identity.

Now, for xH such that Tx≠ 0, we put y=||Tx||−1||x|| Tx. Then ||y|| =||x|| and when we substitute into the previous inequality, we get

    4⎪⎪
⎪⎪
Tx⎪⎪
⎪⎪
⎪⎪
⎪⎪
x⎪⎪
⎪⎪
=4ℜ⟨ Tx,y  ⟩  ≤ 4s⎪⎪
⎪⎪
x2⎪⎪
⎪⎪
,

So ||Tx||≤ s||x|| and it follows that ||T||≤ s, as required. □

Definition 25 We say that U:HH is a unitary operator on a Hilbert space H if U*=U−1, i.e. U*U=UU*=I.
Example 26
  1. If D:l2l2 is a diagonal operator such that D ekk ek, then D* ek=λk ek and D is unitary if and only if | λk |=1 for all k.
  2. The shift operator S satisfies S*S=I but SS*I thus S is not unitary.
Theorem 27 For an operator U on a complex Hilbert space H the following are equivalent:
  1. U is unitary;
  2. U is surjection and an isometry, i.e. ||Ux||=||x|| for all xH;
  3. U is a surjection and preserves the inner product, i.e. Ux,Uy ⟩=⟨ x,y for all x, yH.
Proof. 12. Clearly unitarity of operator implies its invertibility and hence surjectivity. Also
    ⎪⎪
⎪⎪
Ux⎪⎪
⎪⎪
2=⟨ Ux,Ux  ⟩=⟨ x,U*Ux  ⟩=⟨ x,x  ⟩=⎪⎪
⎪⎪
x⎪⎪
⎪⎪
2. 
23. Using the polarisation identity (cf. polarisation in equation (16)):
    4⟨ Tx,y  ⟩=    ⟨ T(x+y),x+y  ⟩+i⟨ T(x+iy),x+iy  ⟩
  −⟨ T(xy),xy  ⟩−i⟨ T(xiy),xiy  ⟩. 
 =
3
k=0
ik⟨ T(x+iky),x+iky  ⟩
Take T=U*U and T=I, then
    4⟨ U*Ux,y  ⟩=
3
k=0
ik⟨ U*U(x+iky),x+iky  ⟩
 =
3
k=0
ik⟨ U(x+iky),U(x+iky)  ⟩
 =
3
k=0
ik⟨ (x+iky),(x+iky)  ⟩
 =4⟨ x,y  ⟩.
31. Indeed ⟨ U*U x,y ⟩=⟨ x,y ⟩ implies ⟨ (U*UI)x,y ⟩=0 for all x,yH, then U*U=I. Since U is surjective, for any yH there is xH such that y=Ux. Then, using the already established fact U*U=I we get
    UU*y = UU*(Ux) =  U(U*U)x = Ux= y.
Thus we have UU*=I as well and U is unitary. □
Definition 28 A normal operator T is one for which T*T=TT*.
Example 29
  1. Any self-adjoint operator T is normal, since T*=T.
  2. Any unitary operator U is normal, since U*U=I=UU*.
  3. Any diagonal operator D is normal , since D ekk ek, D* ek=λk ek, and DD*ek=D*D ek=| λk |2 ek.
  4. The shift operator S is not normal.
  5. A finite matrix is normal (as an operator on l2n) if and only if it has an orthonormal basis in which it is diagonal.
Remark 30 Theorems 24 and 2 draw similarity between those types of operators and multiplications by complex numbers. Indeed Theorem 24 said that an operator which significantly change direction of vectors (“rotates”) cannot be Hermitian, just like a multiplication by a real number scales but do not rotate. On the other hand Theorem 2 says that unitary operator just rotate vectors but do not scale, as a multiplication by an unimodular complex number. We will see further such connections in Theorem 17.

7 Spectral Theory

  Beware of ghosts2 in this area!


As we saw operators could be added and multiplied each other, in some sense they behave like numbers, but are much more complicated. In this lecture we will associate to each operator a set of complex numbers which reflects certain (unfortunately not all) properties of this operator.

The analogy between operators and numbers become even more deeper since we could construct functions of operators (called functional calculus) in a way we build numeric functions. The most important functions of this sort is called resolvent (see Definition 5). The methods of analytical functions are very powerful in operator theory and students may wish to refresh their knowledge of complex analysis before this part.

7.1 The spectrum of an operator on a Hilbert space

An eigenvalue of operator TB(H) is a complex number λ such that there exists a nonzero xH, called eigenvector with property Txx, in other words x∈ker(T−λ I).

In finite dimensions T−λ I is invertible if and only if λ is not an eigenvalue. In infinite dimensions it is not the same: the right shift operator S is not invertible but 0 is not its eigenvalue because Sx=0 implies x=0 (check!).

Definition 1 The resolvent set ρ(T) of an operator T is the set
    ρ (T)={λ∈ℂ: T−λ I  is invertible}.
The spectrum of operator TB(H), denoted σ(T), is the complement of the resolvent set ρ(T):
    σ(T)={λ∈ℂ: T−λ I  is not invertible}.
Example 2 If H is finite dimensional the from previous discussion follows that σ(T) is the set of eigenvalues of T for any T.

Even this example demonstrates that spectrum does not provide a complete description for operator even in finite-dimensional case. For example, both operators in 2 given by matrices (

    00
00

) and (

    00
10

) have a single point spectrum {0}, however are rather different. The situation became even worst in the infinite dimensional spaces.

Theorem 3 The spectrum σ(T) of a bounded operator T is a nonempty compact (i.e. closed and bounded) subset of .

For the proof we will need several Lemmas.

Lemma 4 Let AB(H). If ||A||<1 then IA is invertible in B(H) and inverse is given by the Neumann series (C. Neumann, 1877):
(IA)−1=I+A+A2+A3+…=
k=0
Ak. (40)
Proof. Define the sequence of operators Bn=I+A+⋯+AN—the partial sums of the infinite series (40). It is a Cauchy sequence, indeed:
    ⎪⎪
⎪⎪
BnBm⎪⎪
⎪⎪
=
⎪⎪
⎪⎪
Am+1+Am+2+⋯+An⎪⎪
⎪⎪
      (if  n<m)
 
⎪⎪
⎪⎪
Am+1⎪⎪
⎪⎪
+⎪⎪
⎪⎪
Am+2⎪⎪
⎪⎪
+⋯+⎪⎪
⎪⎪
An⎪⎪
⎪⎪
 
⎪⎪
⎪⎪
A⎪⎪
⎪⎪
m+1+⎪⎪
⎪⎪
A⎪⎪
⎪⎪
m+2+⋯+⎪⎪
⎪⎪
A⎪⎪
⎪⎪
n
 
⎪⎪
⎪⎪
A⎪⎪
⎪⎪
m+1
1−⎪⎪
⎪⎪
A⎪⎪
⎪⎪
<є 
for a large m. By the completeness of B(H) there is a limit, say B, of the sequence Bn. It is a simple algebra to check that (IA)Bn=Bn(IA)=IAn+1, passing to the limit in the norm topology, where An+1→ 0 and BnB we get:
    (IA)B=B(IA)=I  ⇔  B=(IA)−1. 
Definition 5 The resolventof an operator T is the operator valued function defined on the resolvent set by the formula:
R(λ,T)=(T−λ I)−1.         (41)
Corollary 6
  1. If | λ |>||T|| then λ∈ ρ(T), hence the spectrum is bounded.
  2. The resolvent set ρ(T) is open, i.e for any λ ∈ ρ(T) then there exist є>0 such that all µ with | λ−µ |<є are also in ρ(T), i.e. the resolvent set is open and the spectrum is closed.
Both statements together imply that the spectrum is compact.
Proof.
  1. If | λ |>||T|| then ||λ−1T||<1 and the operator T−λ I=−λ(I−λ−1T) has the inverse
    R(λ,T)= (T−λ I)−1=−
    k=0
    λk−1Tk.       (42)
    by the previous Lemma.
  2. Indeed:
          T−µ I=T−λ I + (λ−µ)I
     =(T−λ I)(I+(λ−µ)(T−λ I)−1).
    The last line is an invertible operator because T−λ I is invertible by the assumption and I+(λ−µ)(T−λ I)−1 is invertible by the previous Lemma, since ||(λ−µ)(T−λ I)−1||<1 if є<||(T−λ I)−1||.
Exercise 7
  1. Prove the first resolvent identity:
    R(λ,T)−R(µ,T)=(λ−µ)R(λ,T)R(µ,T) (43)
  2. Use the identity (43) to show that (T−µ I)−1→ (T−λ I)−1 as µ→ λ.
  3. Use the identity (43) to show that for z∈ρ(t) the complex derivative d/dz R(z,T) of the resolvent R(z,T) is well defined, i.e. the resolvent is an analytic function operator valued function of z.
Lemma 8 The spectrum is non-empty.
Proof. Let us assume the opposite, σ(T)=∅ then the resolvent function R(λ,T) is well defined for all λ∈ℂ. As could be seen from the von Neumann series (42) ||R(λ,T)||→ 0 as λ→ ∞. Thus for any vectors x, yH the function f(λ)=⟨ R(λ,T)x,y) ⟩ is analytic (see Exercise 3) function tensing to zero at infinity. Then by the Liouville theorem from complex analysis R(λ,T)=0, which is impossible. Thus the spectrum is not empty. □
Proof.[Proof of Theorem 3] Spectrum is nonempty by Lemma 8 and compact by Corollary 6. □
Remark 9 Theorem 3 gives the maximal possible description of the spectrum, indeed any non-empty compact set could be a spectrum for some bounded operator, see Problem 23.

7.2 The spectral radius formula

The following definition is of interest.

Definition 10 The spectral radius of T is
    r(T)=sup{
λ 
: λ∈ σ(T)}.

From the Lemma 1 immediately follows that r(T)≤||T||. The more accurate estimation is given by the following theorem.

Theorem 11 For a bounded operator T we have
r(T)=
 
lim
n→∞
⎪⎪
⎪⎪
Tn⎪⎪
⎪⎪
1/n. (44)

We start from the following general lemma:

Lemma 12 Let a sequence (an) of positive real numbers satisfies inequalities: 0≤ am+nam+an for all m and n. Then there is a limit limn→∞(an/n) and its equal to infn(an/n).
Proof. The statements follows from the observation that for any n and m=nk+l with 0≤ ln we have amkan+la1 thus, for big m we got am/man/n +la1/man/n+є. □
Proof.[Proof of Theorem 11] The existence of the limit limn→∞||Tn||1/n in (44) follows from the previous Lemma since by the Lemma 12 log||Tn+m||≤ log||Tn||+log||Tm||. Now we are using some results from the complex analysis. The Laurent series for the resolvent R(λ,T) in the neighbourhood of infinity is given by the von Neumann series (42). The radius of its convergence (which is equal, obviously, to r(T)) by the Hadamard theorem is exactly limn→∞||Tn||1/n. □
Corollary 13 There exists λ∈σ(T) such that | λ |=r(T).
Proof. Indeed, as its known from the complex analysis the boundary of the convergence circle of a Laurent (or Taylor) series contain a singular point, the singular point of the resolvent is obviously belongs to the spectrum. □
Example 14 Let us consider the left shift operator S*, for any λ∈ℂ such that | λ | <1 the vector (1,λ,λ23,…) is in l2 and is an eigenvector of S* with eigenvalue λ, so the open unit disk | λ |<1 belongs to σ(S*). On the other hand spectrum of S* belongs to the closed unit disk | λ |≤ 1 since r(S*)≤ ||S*||=1. Because spectrum is closed it should coincide with the closed unit disk, since the open unit disk is dense in it. Particularly 1∈σ(S*), but it is easy to see that 1 is not an eigenvalue of S*.
Proposition 15 For any TB(H) the spectrum of the adjoint operator is σ(T*)={λ: λ∈ σ(T)}.
Proof. If (T−λ I)V=V(T−λ I)=I the by taking adjoints V*(T*λI)=(T*λI)V*=I. So λ ∈ ρ(T) implies λ∈ρ(T*), using the property T**=T we could invert the implication and get the statement of proposition. □
Example 16 In continuation of Example 14 using the previous Proposition we conclude that σ(S) is also the closed unit disk, but S does not have eigenvalues at all!

7.3 Spectrum of Special Operators

Theorem 17
  1. If U is a unitary operator then σ(U)⊆ {| z |=1}.
  2. If T is Hermitian then σ(T)⊆ ℝ.
Proof.
  1. If | λ |>1 then ||λ−1U||<1 and then λ IU=λ(I−λ−1U) is invertible, thus λ∉σ(U). If | λ |<1 then ||λ U*||<1 and then λ IU=UU*I) is invertible, thus λ∉σ(U). The remaining set is exactly {z:| z |=1}.
  2. Without lost of generality we could assume that ||T||<1, otherwise we could multiply T by a small real scalar. Let us consider the Cayley transform which maps real axis to the unit circle:
          U=(TiI)(T+iI)−1.
    Straightforward calculations show that U is unitary if T is Hermitian. Let us take λ∉ℝ and λ≠ −i (this case could be checked directly by Lemma 4). Then the Cayley transform µ=(λ−i)(λ+i)−1 of λ is not on the unit circle and thus the operator
          U−µ I=(TiI)(T+iI)−1−(λ−i)(λ+i)−1I= 2i(λ+i)−1(T−λ I)(T+iI)−1,
    is invertible, which implies invertibility of T−λ I. So λ∉ℝ.

The above reduction of a self-adjoint operator to a unitary one (it can be done on the opposite direction as well!) is an important tool which can be applied in other questions as well, e.g. in the following exercise.

Exercise 18
  1. Show that an operator U: f(t) ↦ eitf(t) on L2[0,2π] is unitary and has the entire unit circle {| z |=1} as its spectrum .
  2. Find a self-adjoint operator T with the entire real line as its spectrum.

8 Compactness

It is not easy to study linear operators “in general” and there are many questions about operators in Hilbert spaces raised many decades ago which are still unanswered. Therefore it is reasonable to single out classes of operators which have (relatively) simple properties. Such a class of operators more closed to finite dimensional ones will be studied here.

  These operators are so compact that we even can fit them in our course


8.1 Compact operators

Let us recall some topological definition and results.

Definition 1 A compact set in a metric space is defined by the property that any its covering by a family of open sets contains a subcovering by a finite subfamily.

In the finite dimensional vector spaces ℝn or ℂn there is the following equivalent definition of compactness (equivalence of 1 and 2 is known as Heine–Borel theorem):

Theorem 2 If a set E in n or n has any of the following properties then it has other two as well:
  1. E is bounded and closed;
  2. E is compact;
  3. Any infinite subset of E has a limiting point belonging to E.
Exercise* 3 Which equivalences from above are not true any more in the infinite dimensional spaces?
Definition 4 Let X and Y be normed spaces, TB(X,Y) is a finite rank operator if Im T is a finite dimensional subspace of Y. T is a compact operator if whenever (xi)1 is a bounded sequence in X then its image (T xi)1 has a convergent subsequence in Y.

The set of finite rank operators is denote by F(X,Y) and the set of compact operators—by K(X,Y)

Exercise 5 Show that both F(X,Y) and K(X,Y) are linear subspaces of B(X,Y).

We intend to show that F(X,Y)⊂K(X,Y).

Lemma 6 Let Z be a finite-dimensional normed space. Then there is a number N and a mapping S: l2NZ which is invertible and such that S and S−1 are bounded.
Proof. The proof is given by an explicit construction. Let N=dimZ and z1, z2, …, zN be a basis in Z. Let us define
    S: l2N → Z    by     S(a1,a2,…,aN)=
N
k=1
akzk,
then we have an estimation of norm:
    ⎪⎪
⎪⎪
Sa⎪⎪
⎪⎪
=
⎪⎪
⎪⎪
⎪⎪
⎪⎪
N
k=1
akzk⎪⎪
⎪⎪
⎪⎪
⎪⎪
 ≤   
N
k=1

ak
⎪⎪
⎪⎪
zk⎪⎪
⎪⎪
  
 



N
k=1
     
ak
2


1/2



 



N
k=1
    ⎪⎪
⎪⎪
zk⎪⎪
⎪⎪
2


1/2



 
.
So ||S||≤ (∑1N ||zk||2)1/2 and S is continuous.

Clearly S has the trivial kernel, particularly ||Sa||>0 if ||a||=1. By the Heine–Borel theorem the unit sphere in l2N is compact, consequently the continuous function a↦ ||∑1N ak zk|| attains its lower bound, which has to be positive. This means there exists δ>0 such that ||a||=1 implies ||Sa||>δ , or, equivalently if ||z||<δ then ||S−1 z||<1. The later means that ||S−1||≤ δ−1 and boundedness of S−1. □

Corollary 7 For any two metric spaces X and Y we have F(X,Y)⊂ K(X,Y).
Proof. Let TF(X,Y), if (xn)1 is a bounded sequence in X then ((Txn)1Z=Im T is also bounded. Let S: l2NZ be a map constructed in the above Lemma. The sequence (S−1T xn)1 is bounded in l2N and thus has a limiting point, say a0. Then Sa0 is a limiting point of (T xn)1. □

There is a simple condition which allows to determine which diagonal operators are compact (particularly the identity operator IX is not compact if dimX =∞):

Proposition 8 Let T is a diagonal operator and given by identities T enn en for all n in a basis en. T is compact if and only if λn→ 0.

Figure 16: Distance between scales of orthonormal vectors

Proof. If λn↛0 then there exists a subsequence λnk and δ>0 such that | λnk |>δ for all k. Now the sequence (enk) is bounded but its image T enknk enk has no convergent subsequence because for any kl:
    ⎪⎪
⎪⎪
λ nkenk−λ nlenl⎪⎪
⎪⎪
  =  (
λ nk
2 + 
λ nl
2)1/2≥ 
2
δ ,
i.e. T enk is not a Cauchy sequence, see Figure 16. For the converse, note that if λn→ 0 then we can define a finite rank operator Tm, m≥ 1—m-“truncation” of T by:
Tmen = 

        Tennen,1≤ n≤ m;
        0 ,n>m.
(45)
Then obviously
    (TTm) en = 

        0,1≤ n≤ m;
        λnen ,n>m,
and ||TTm||=supn>m| λn |→ 0 if m→ ∞. All Tm are finite rank operators (so are compact) and T is also compact as their limit—by the next Theorem. □
Theorem 9 Let Tm be a sequence of compact operators convergent to an operator T in the norm topology (i.e. ||TTm||→ 0) then T is compact itself. Equivalently K(X,Y) is a closed subspace of B(X,Y).

Figure 17: The є/3 argument to estimate | f(x)−f(y) |.


T1x1(1)T1x2(1)T1x3(1) T1xn(1)a1
T2x1(2)T2x2(2)T2x3(2) T2xn(2)a2
T3x1(3)T3x2(3)T3x3(3) T3xn(3)a3
  
Tnx1(n)Tnx2(n)Tnx3(n) Tnxn(n)an
 
       
       a
Table 2: The “diagonal argument”.

Proof. Take a bounded sequence (xn)1. From compactness
of T1⇒ ∃subsequence (xn(1))1 of (xn)1s.t.(T1xn(1))1 is convergent.
of T2⇒ ∃subsequence (xn(2))1 of (xn(1))1s.t.(T2xn(2))1 is convergent.
of T3⇒ ∃subsequence (xn(3))1 of (xn(2))1s.t.(T3xn(3))1 is convergent.

Could we find a subsequence which converges for all Tm simultaneously? The first guess “take the intersection of all above sequences (xn(k))1” does not work because the intersection could be empty. The way out is provided by the diagonal argument (see Table 2): a subsequence (Tm xk(k))1 is convergent for all m, because at latest after the term xm(m) it is a subsequence of (xk(m))1.

We are claiming that a subsequence (T xk(k))1 of (T xn)1 is convergent as well. We use here є/3 argument (see Figure 17): for a given є>0 choose p∈ℕ such that ||TTp||<є/3. Because (Tp xk(k))→ 0 it is a Cauchy sequence, thus there exists n0>p such that ||Tp xk(k)Tp xl(l)||< є/3 for all k, l>n0. Then:

    ⎪⎪
⎪⎪
Txk(k)Txl(l)⎪⎪
⎪⎪
=
⎪⎪
⎪⎪
(Txk(k)Tp xk(k))+(Tpxk(k)Tpxl(l))+(Tpxl(l)T xl(l))⎪⎪
⎪⎪
 
⎪⎪
⎪⎪
Txk(k)Tpxk(k)⎪⎪
⎪⎪
+ ⎪⎪
⎪⎪
Tpxk(k)Tpxl(l)⎪⎪
⎪⎪
+⎪⎪
⎪⎪
Tpxl(l)T xl(l)⎪⎪
⎪⎪
 є

Thus T is compact. □

8.2 Hilbert–Schmidt operators

Definition 10 Let T: HK be a bounded linear map between two Hilbert spaces. Then T is said to be Hilbert–Schmidt operator if there exists an orthonormal basis in H such that the series k=1||T ek||2 is convergent.
Example 11
  1. Let T: l2l2 be a diagonal operator defined by Ten=en/n, for all n≥ 1. Then ∑ ||Ten||2=∑n−22/6 (see Example 16) is finite.
  2. The identity operator IH is not a Hilbert–Schmidt operator, unless H is finite dimensional.

A relation to compact operator is as follows.

Theorem 12 All Hilbert–Schmidt operators are compact. (The opposite inclusion is false, give a counterexample!)
Proof. Let TB(H,K) have a convergent series ∑ ||T en||2 in an orthonormal basis (en)1 of H. We again (see (45)) define the m-truncation of T by the formula
Tmen = 

        Ten,1≤ n≤ m;
0 ,n>m.
(46)
Then Tm(∑1ak ek)=∑1m ak ek and each Tm is a finite rank operator because its image is spanned by the finite set of vectors Te1, …, Ten. We claim that ||TTm||→ 0. Indeed by linearity and definition of Tm:
    (TTm)


n=1
anen


=
n=m+1
an (Ten).
Thus:
     
    ⎪⎪
⎪⎪
⎪⎪
⎪⎪
(TTm)


n=1
anen


⎪⎪
⎪⎪
⎪⎪
⎪⎪
=
⎪⎪
⎪⎪
⎪⎪
⎪⎪
n=m+1
an (Ten)⎪⎪
⎪⎪
⎪⎪
⎪⎪
 
(47)
 
  
n=m+1

an
⎪⎪
⎪⎪
(Ten)⎪⎪
⎪⎪
 
 



n=m+1

an
2


1/2



 



n=m+1
⎪⎪
⎪⎪
(Ten)⎪⎪
⎪⎪
2


1/2



 
 
 
⎪⎪
⎪⎪
⎪⎪
⎪⎪
n=1
anen⎪⎪
⎪⎪
⎪⎪
⎪⎪



n=m+1
⎪⎪
⎪⎪
(Ten)⎪⎪
⎪⎪
2


1/2



 
 
(48)
so ||TTm||→ 0 and by the previous Theorem T is compact as a limit of compact operators. □
Corollary 13 (from the above proof) For a Hilbert–Schmidt operator
    ⎪⎪
⎪⎪
T⎪⎪
⎪⎪



n=m+1
⎪⎪
⎪⎪
(Ten)⎪⎪
⎪⎪
2


1/2



 
 .  
Proof. Just consider difference of T and T0=0 in (47)–(48). □
Example 14 An integral operator T on L2[0,1] is defined by the formula:
(Tf)(x)=
1
0
K(x,y)f(y) dy,   f(y)∈L2[0,1], (49)
where the continuous on [0,1]×[0,1] function K is called the kernel of integral operator.
Theorem 15 Integral operator (49) is Hilbert–Schmidt.
Proof. Let (en)−∞ be an orthonormal basis of L2[0,1], e.g. (ei nt)n∈ℤ. Let us consider the kernel Kx(y)=K(x,y) as a function of the argument y depending from the parameter x. Then:
    (Ten)(x)=
1
0
K(x,y)en(y) dy=
1
0
Kx(y)en(y) dy= ⟨ Kxn  ⟩.
So ||T en||2= ∫01| ⟨ Kxn ⟩ |2d x. Consequently:
     
    
−∞
⎪⎪
⎪⎪
Ten⎪⎪
⎪⎪
2
 =
      
−∞
1
0

⟨ Kxn  ⟩ 
2dx
 
  =
  
1
0
1

⟨ Kxn  ⟩ 
2dx 
(50)
  =
  
1
0
⎪⎪
⎪⎪
Kx⎪⎪
⎪⎪
2dx
 
  =
  
1
0
1
0
  
K(x,y) 
2dxdy < ∞
 
Exercise 16 Justify the exchange of summation and integration in (50).
Remark 17 The definition 14 and Theorem 15 work also for any T: L2[a,b] → L2[c,d] with a continuous kernel K(x,y) on [c,d]×[a,b].
Definition 18 Define Hilbert–Schmidt norm of a Hilbert–Schmidt operator A by ||A||HS2=∑n=1||Aen||2 (it is independent of the choice of orthonormal basis (en)1, see Question 27).
Exercise* 19 Show that set of Hilbert–Schmidt operators with the above norm is a Hilbert space and find the an expression for the inner product.
Example 20 Let K(x,y)=xy, then
    (Tf)(x)=
1
0
 (xy)f(y) dy =x
1
0
f(y) dy −
1
0
yf(y) dy
is a rank 2 operator. Furthermore:
    ⎪⎪
⎪⎪
T⎪⎪
⎪⎪
HS2
=
1
0
1
0
(xy)2dxdy = 
1
0



(xy)3
3



1



x=0
dy
 =
1
0
(1−y)3
3
+
y3
3
dy= 


(1−y)4
12
+
y4
12



1



0
=
1
6
. 
On the other hand there is an orthonormal basis such that
    Tf=
1
12
⟨ f,e1  ⟩e1
1
12
⟨ f,e2  ⟩e2,
and ||T||=1/√12 and 12 ||Tek||2=1/6 and we get ||T||≤ ||T||HS in agreement with Corollary 13.

9 Compact normal operators

Recall from Section 6.5 that an operator T is normal if TT*=T*T; Hermitian (T*=T) and unitary (T*=T−1) operators are normal.

9.1 Spectrum of normal operators

Theorem 1 Let TB(H) be a normal operator then
  1. kerT =kerT*, so ker(T−λ I) =ker (T*λI) for all λ∈ℂ
  2. Eigenvectors corresponding to distinct eigenvalues are orthogonal.
  3. ||T||=r(T).
Proof.
  1. Obviously:
          x∈kerT⟨ Tx,Tx  ⟩=0 ⇔ ⟨ T*Tx,x  ⟩=0 
     ⟨ TT*x,x  ⟩=0 ⇔  ⟨ T*x,T*x  ⟩=0 
     x∈kerT*.
    The second part holds because normalities of T and T−λ I are equivalent.
  2. If Txx, Tyy then from the previous statement T* y =µy. If λ≠µ then the identity
          λ⟨ x,y  ⟩=⟨ Tx,y  ⟩ =⟨ x,T*y  ⟩=µ⟨ x,y  ⟩
    implies ⟨ x,y ⟩=0.
  3. Let S=T*T, then S is Hermitian (check!). Consequently, inequality
        ⎪⎪
    ⎪⎪
    Sx⎪⎪
    ⎪⎪
    2=⟨ Sx,Sx  ⟩=⟨ S2x,x  ⟩≤ ⎪⎪
    ⎪⎪
    S2⎪⎪
    ⎪⎪
    ⎪⎪
    ⎪⎪
    x⎪⎪
    ⎪⎪
    2
    implies ||S||2≤ ||S2||. But the opposite inequality follows from the Theorem 12, thus we have the equality ||S2||=||S||2 and more generally by induction: ||S2m||=||S||2m for all m.

    Now we claim ||S||=||T||2. From Theorem 12 and 18 we get ||S||=||T*T||≤ ||T||2. On the other hand if ||x||=1 then

        ⎪⎪
    ⎪⎪
    T*T⎪⎪
    ⎪⎪
    ≥ 
    ⟨ T*Tx,x  ⟩ 
    =⟨ Tx,Tx  ⟩=⎪⎪
    ⎪⎪
    Tx⎪⎪
    ⎪⎪
    2

    implies the opposite inequality ||S||≥||T||2. Only now we use normality of T to obtain (T2m)*T2m=(T*T)2m and get the equality

        ⎪⎪
    ⎪⎪
    T2m⎪⎪
    ⎪⎪
    2=⎪⎪
    ⎪⎪
    (T*T)2m⎪⎪
    ⎪⎪
    =⎪⎪
    ⎪⎪
    T*T⎪⎪
    ⎪⎪
    2m =⎪⎪
    ⎪⎪
    T⎪⎪
    ⎪⎪
    2m+1.

    Thus:

        r(T)=
     
    lim
    m→∞
    ⎪⎪
    ⎪⎪
    T2m⎪⎪
    ⎪⎪
    1/2m=
     
    lim
    m→∞
    ⎪⎪
    ⎪⎪
    T⎪⎪
    ⎪⎪
    2m+1/2m+1 = ⎪⎪
    ⎪⎪
    T⎪⎪
    ⎪⎪
    .

    by the spectral radius formula (44).

Example 2 It is easy to see that normality is important in 3, indeed the non-normal operator T given by the matrix (
      01
      00
) in has one-point spectrum {0}, consequently r(T)=0 but ||T||=1.
Lemma 3 Let T be a compact normal operator then
  1. The set of of eigenvalues of T is either finite or a countable sequence tending to zero.
  2. All the eigenspaces, i.e. ker(T−λ I), are finite-dimensional for all λ≠ 0.
Remark 4 This Lemma is true for any compact operator, but we will not use that in our course.
Proof.
  1. Let H0 be the closed linear span of eigenvectors of T. Then T restricted to H0 is a diagonal compact operator with the same set of eigenvalues λn as in H. Then λn→ 0 from Proposition 8 .
    Exercise 5 Use the proof of Proposition 8 to give a direct demonstration.
    Proof.[Solution] Or straightforwardly assume opposite: there exist an δ>0 and infinitely many eigenvalues λn such that | λn |>δ. By the previous Theorem there is an orthonormal sequence vn of corresponding eigenvectors T vnn vn. Now the sequence (vn) is bounded but its image T vnn en has no convergent subsequence because for any kl:
            ⎪⎪
    ⎪⎪
    λ kvk−λ lel⎪⎪
    ⎪⎪
      =  (
    λ k
    2 + 
    λl
    2)1/2≥ 
    2
    δ ,
    i.e. T enk is not a Cauchy sequence, see Figure 16. □
  2. Similarly if H0=ker(T−λ I) is infinite dimensional, then restriction of T on H0 is λ I—which is non-compact by Proposition 8. Alternatively consider the infinite orthonormal sequence (vn), Tvnvn as in Exercise 5.
Lemma 6 Let T be a compact normal operator. Then all non-zero points λ∈ σ(T) are eigenvalues and there exists an eigenvalue of modulus ||T||.
Proof. Assume without lost of generality that T≠ 0. Let λ∈σ(T), without lost of generality (multiplying by a scalar) λ=1.

We claim that if 1 is not an eigenvalue then there exist δ>0 such that

⎪⎪
⎪⎪
(IT)x⎪⎪
⎪⎪
≥ δ⎪⎪
⎪⎪
x⎪⎪
⎪⎪
. (51)

Otherwise there exists a sequence of vectors (xn) with unit norm such that (IT)xn→ 0. Then from the compactness of T for a subsequence (xnk) there is yH such that Txnky, then xny implying Ty=y and y≠ 0—i.e. y is eigenvector with eigenvalue 1.

Now we claim Im (IT) is closed, i.e. yIm(IT) implies yIm(IT). Indeed, if (IT)xny, then there is a subsequence (xnk) such that Txnkz implying xnky+z, then (IT)(z+y)=y by continuity of IT.

Finally IT is injective, i.e ker(IT)={0}, by (51). By the property 1, ker(IT*)={0} as well. But because always ker(IT*)=Im(IT) (by 2) we got surjectivity, i.e. Im(IT)={0}, of IT. Thus (IT)−1 exists and is bounded because (51) implies ||y||>δ ||(IT)−1y||. Thus 1∉σ(T).

The existence of eigenvalue λ such that | λ |=||T|| follows from combination of Lemma 13 and Theorem 3. □

9.2 Compact normal operators

Theorem 7 (The spectral theorem for compact normal operators) Let T be a compact normal operator on a Hilbert space H. Then there exists an orthonormal sequence (en) of eigenvectors of T and corresponding eigenvalues n) such that:
Tx=
 
n
 λn ⟨ x,en  ⟩ en,     for all x∈ H. (52)
If n) is an infinite sequence it tends to zero.

Conversely, if T is given by a formula (52) then it is compact and normal.

Proof. Suppose T≠ 0. Then by the previous Theorem there exists an eigenvalue λ1 such that | λ1 |=||T|| with corresponding eigenvector e1 of the unit norm. Let H1=Lin(e1). If xH1 then
⟨ Tx,e1  ⟩=⟨ x,T*e1  ⟩=⟨ x,λ1 e1  ⟩=λ1⟨ x,e1  ⟩=0, (53)
thus TxH1 and similarly T* xH1. Write T1=T|H1 which is again a normal compact operator with a norm does not exceeding ||T||. We could inductively repeat this procedure for T1 obtaining sequence of eigenvalues λ2, λ3, …with eigenvectors e2, e3, …. If Tn=0 for a finite n then theorem is already proved. Otherwise we have an infinite sequence λn→ 0. Let
    x=
n
1
 ⟨ x,ek  ⟩ek +yn  ⇒  ⎪⎪
⎪⎪
x⎪⎪
⎪⎪
2=
n
1

⟨ x,ek  ⟩ 
2 +⎪⎪
⎪⎪
yn⎪⎪
⎪⎪
2 ,    yn∈ Hn,
from Pythagoras’s theorem. Then ||yn||≤ ||x|| and ||T yn||≤ ||Tn||||yn||≤ | λn |||x||→ 0 by Lemma 3. Thus
    Tx =
 
lim
n→ ∞



n
1
 ⟨ x,en  ⟩ Ten + Tyn


= 
1
λn⟨ x,en  ⟩ en

Conversely, if T x = ∑1λnx,enen then

    ⟨ Tx,y  ⟩=
1
λn⟨ x,en  ⟩ ⟨ en,y  ⟩ =
1
⟨ x,en  ⟩ λn
⟨ y,en  ⟩
,

thus T* y = ∑1λny,enen. Then we got the normality of T: T*Tx=TT*x= ∑1| λn |2y,enen. Also T is compact because it is a uniform limit of the finite rank operators Tnx=∑1n λnx,enen. □

Corollary 8 Let T be a compact normal operator on a separable Hilbert space H, then there exists a orthonormal basis gk such that
    Tx=
1
λn⟨ x,gn  ⟩ gn,
and λn are eigenvalues of T including zeros.
Proof. Let (en) be the orthonormal sequence constructed in the proof of the previous Theorem. Then x is perpendicular to all en if and only if its in the kernel of T. Let (fn) be any orthonormal basis of kerT. Then the union of (en) and (fn) is the orthonormal basis (gn) we have looked for. □
Exercise 9 Finish all details in the above proof.
Corollary 10 (Singular value decomposition) If T is any compact operator on a separable Hilbert space then there exists orthonormal sequences (ek) and (fk) such that Tx=∑k µkx,ekfk where k) is a sequence of positive numbers such that µk→ 0 if it is an infinite sequence.
Proof. Operator T*T is compact and Hermitian (hence normal). From the previous Corollary there is an orthonormal basis (ek) such that T*T x= ∑n λnx,ekek for some positive λn=||T en||2. Let µn=||Ten|| and fn=Tenn. Then fn is an orthonormal sequence (check!) and
    Tx=
 
n
 ⟨ x,en  ⟩ Ten =
 
n
 ⟨ x,en  ⟩ µnfn.
Corollary 11 A bounded operator in a Hilber space is compact if and only if it is a uniform limit of the finite rank operators.
Proof. Sufficiency follows from 9.
Necessity: by the previous Corollary Tx =∑nx,en ⟩ µn fn thus T is a uniform limit of operators Tm x=∑n=1mx,en ⟩ µn fn which are of finite rank. □

10 Integral equations

In this lecture we will study the Fredholm equation defined as follows. Let the integral operator with a kernel K(x,y) defined on [a,b]×[a,b] be defined as before:

(Tφ)(x)=
b
a
K(x,y)φ(y) dy. (54)

The Fredholm equation of the first and second kinds correspondingly are:

Tφ=f     and     φ −λ Tφ=f, (55)

for a function f on [a,b]. A special case is given by Volterra equation by an operator integral operator (54) T with a kernel K(x,y)=0 for all y>x which could be written as:

(Tφ)(x)=
x
a
K(x,y)φ(y) dy. (56)

We will consider integral operators with kernels K such that ∫abab K(x,y) d xd y<∞, then by Theorem 15 T is a Hilbert–Schmidt operator and in particular bounded.

As a reason to study Fredholm operators we will mention that solutions of differential equations in mathematical physics (notably heat and wave equations) requires a decomposition of a function f as a linear combination of functions K(x,y) with “coefficients” φ. This is an continuous analog of a discrete decomposition into Fourier series.

Using ideas from the proof of Lemma 4 we define Neumann series for the resolvent:

(I−λ T)−1=I+λ T + λ2T2+⋯, (57)

which is valid for all λ<||T||−1.

Example 1 Solve the Volterra equation
    φ(x)−λ
x
0
y φ(y) dy=x2,     on  L2[0,1].
In this case I−λ T φ = f, with f(x)=x2 and:
    K(x,y)=

        y,0≤ y ≤ x;
        0,x< y ≤ 1.
Straightforward calculations shows:
    (Tf)(x)=
x
0
y· y2dy=
x4
4
,
    (T2f)(x)=
x
0
y
y4
4
dy=
x6
24
, …
and generally by induction:
    (Tnf)(x) = 
x
0
y
y2n
2n−1n!
dy=
x2n+2
2n(n+1)!
.
Hence:
    φ(x)=
0
λnTnf = 
0
λnx2n+2
2n(n+1)!
 =
2
λ
0
λn+1x2n+2
2n+1(n+1)!
 =
2
λ
(eλ x2/2−1)     for all  λ ∈ ℂ∖ {0},
because in this case r(T)=0. For the Fredholm equations this is not always the case, see Tutorial problem 29.

Among other integral operators there is an important subclass with separable kernel, namely a kernel which has a form:

K(x,y)=
n
j=1
gj(x)hj(y). (58)

In such a case:

    (Tφ)(x)=
b
a
n
j=1
gj(x)hj(y)φ(y) dy
 =
n
j=1
gj(x) 
b
a
hj(y)φ(y) dy,

i.e. the image of T is spanned by g1(x), …, gn(x) and is finite dimensional, consequently the solution of such equation reduces to linear algebra.

Example 2 Solve the Fredholm equation (actually find eigenvectors of T):
    φ(x)=
λ 
0
 cos(x+y)φ(y) dy
 =
λ 
0
 (cosxcosy − sinx siny)φ(y) dy.
Clearly φ (x) should be a linear combination φ(x)=Acos x+Bsinx with coefficients A and B satisfying to:
    A=
λ 
0
 cosy (Acosy+Bsiny) dy,
    B=
−λ 
0
 siny (Acosy+Bsiny) dy.
Basic calculus implies A=λπ A and B=−λπ B and the only nonzero solutions are:
    λ=π−1A ≠ 0B = 0
    λ=−π−1A = 0B ≠ 0

We develop some Hilbert–Schmidt theory for integral operators.

Theorem 3 Suppose that K(x,y) is a continuous function on [a,b]×[a,b] and K(x,y)=K(y,x) and operator T is defined by (54). Then
  1. T is a self-adjoint Hilbert–Schmidt operator.
  2. All eigenvalues of T are real and satisfy n λn2<∞.
  3. The eigenvectors vn of T can be chosen as an orthonormal basis of L2[a,b], are continuous for nonzero λn and
          Tφ=
    n=1
    λn ⟨ φ,vn  ⟩vn    where    φ=
    n=1
    ⟨ φ,vn  ⟩vn
Proof.
  1. The condition K(x,y)=K(y,x) implies the Hermitian property of T:
          ⟨ Tφ,ψ  ⟩=
    b
    a



    b
    a
    K(x,y)φ(y) dy


    ψ(x) dx
     =
    b
    a
    b
    a
    K(x,y)φ(y) ψ(x) dxdy
     =
    b
    a
     φ(y)


    b
    a
    K(y,x) ψ(x)
    dx


    dy
     =⟨ φ,Tψ  ⟩.
    The Hilbert–Schmidt property (and hence compactness) was proved in Theorem 15.
  2. Spectrum of T is real as for any Hermitian operator, see Theorem 2 and finiteness of ∑n λn2 follows from Hilbert–Schmidt property
  3. The existence of orthonormal basis consisting from eigenvectors (vn) of T was proved in Corollary 8. If λn≠ 0 then:
          vn(x1)−vn(x2)=λn−1((Tvn)(x1)−(Tvn)(x2))
     =
    1
    λn
    b
    a
     (K(x1,y)−K(x2,y))vn(y) dy
    and by Cauchy–Schwarz-Bunyakovskii inequality:
          
    vn(x1)−vn(x2) 
    ≤ 
    1

    λn
    ⎪⎪
    ⎪⎪
    vn⎪⎪
    ⎪⎪
    2
    b
    a

    K(x1,y)−K(x2,y) 
    dy
    which tense to 0 due to (uniform) continuity of K(x,y).
Theorem 4 Let T be as in the previous Theorem. Then if λ≠ 0 and λ−1∉σ(T), the unique solution φ of the Fredholm equation of the second kind φ−λ T φ=f is
φ=
1
⟨ f,vn  ⟩
1−λ λn
vn. (59)
Proof. Let φ=∑1an vn where an=⟨ φ,vn ⟩, then
    φ−λ Tφ=
1
an(1−λ λn) vn =f=
1
⟨ f,vn  ⟩vn
if and only if an=⟨ f,vn ⟩/(1−λ λn) for all n. Note 1−λ λn≠ 0 since λ−1∉σ(T).

Because λn→ 0 we got ∑1| an |2 by its comparison with ∑1| ⟨ f,vn ⟩ |2=||f||2, thus the solution exists and is unique by the Riesz–Fisher Theorem. □

See Exercise 30 for an example.

Theorem 5 (Fredholm alternative) Let TK(H) be compact normal and λ∈ℂ∖ {0}. Consider the equations:
     
      φ−λ Tφ=0(60)
      φ−λ Tφ=f (61)
then either
  1. the only solution to (60) is φ=0 and (61) has a unique solution for any fH; or
  2. there exists a nonzero solution to (60) and (61) can be solved if and only if f is orthogonal all solutions to (60).
Proof.
  1. If φ=0 is the only solution of (60), then λ−1 is not an eigenvalue of T and then by Lemma 6 is neither in spectrum of T. Thus I−λ T is invertible and the unique solution of (61) is given by φ=(I−λ T)−1 f.
  2. A nonzero solution to (60) means that λ−1∈σ(T). Let (vn) be an orthonormal basis of eigenvectors of T for eigenvalues (λn). By Lemma 2 only a finite number of λn is equal to λ−1, say they are λ1, …, λN, then
          (I−λ T)φ=
    n=1
    (1−λ λn)⟨ φ,vn  ⟩vn =
    n=N+1
    (1−λ λn)⟨ φ,vn  ⟩vn.
    If f=∑1f,vnvn then the identity (I−λ T)φ=f is only possible if ⟨ f,vn ⟩=0 for 1≤ nN. Conversely from that condition we could give a solution
        φ=
    n=N+1
    ⟨ f,vn  ⟩
    1−λ λn
    vn +φ0,     for any  φ0Lin(v1,…,vN),
    which is again in H because fH and λn→ 0.
Example 6 Let us consider
    (Tφ)(x)=
1
0
(2xyxy+1)φ(y) dy.
Because the kernel of T is real and symmetric T=T*, the kernel is also separable:
    (Tφ)(x)=x
1
0
(2y−1)φ(y) dy+
1
0
(−y+1)φ(y) dy,
and T of the rank 2 with image of T spanned by 1 and x. By direct calculations:
    
      T:1
1
2
      T:x
1
6
x + 
1
6
,
  or T is given by the matrix 







        
1
2
1
6
        0
1
6








According to linear algebra decomposition over eigenvectors is:
    λ1=
1
2
with vector


        1
0


,
    λ2=
1
6
with vector





        −
1
2
1





with normalisation v1(y)=1, v2(y)=√12(y−1/2) and we complete it to an orthonormal basis (vn) of L2[0,1]. Then

11 Banach and Normed Spaces

We will work with either the field of real numbers ℝ or the complex numbers ℂ. To avoid repetition, we use K to denote either ℝ or ℂ.

11.1 Normed spaces

Recall, see Defn. 3, a norm on a vector space V is a map ||·||:V→[0,∞) such that

  1. ||u||=0 only when u=0;
  2. ||λ u|| = | λ | ||u|| for λ∈K and uV;
  3. ||u+v|| ≤ ||u|| + ||v|| for u,vV.

Note, that the second and third conditions imply that linear operations—multiplication by a scalar and addition of vectors respectively—are continuous in the topology defined by the norm.

A norm induces a metric, see Defn. 1, on V by setting d(u,v)=||uv||. When V is complete, see Defn. 6, for this metric, we say that V is a Banach space.

Theorem 1 Every finite-dimensional normed vector space is a Banach space.

We will use the following simple inequality:

Lemma 2 (Young’s inequality) Let two real numbers 1<p,q<∞ are related through 1/p+1/q=1 then

ab
≤ 

a
p
p
 + 

b
q
q
, (62)
for any complex a and b.
Proof.[First proof: analytic] Obviously, it is enough to prove inequality for positive reals a=| a | and b=| b |. If p>1 then 0<1/p < 1. Consider the function φ(t)=tmmt for an 0<m<1. From its derivative φ(t)=m(tm−1−1) we find the only critical point t=1 on [0,∞), which is its maximum for m=1/p<1. Thus write the inequality φ(t)≤ φ(1) for t=ap/bq and m=1/p. After a transformation we get a· bq/p−1≤ 1/p(apbq−1) and multiplication by bq with rearrangements lead to the desired result. □
Proof.[Second proof: geometric] Consider the plane with coordinates (x,y) and take the curve y=xp−1 which is the same as x=yq−1. Comparing areas on the figure:
we see that S1+S2ab for any positive reals a and b. Elementary integration shows:
    S1=
a
0
xp−1dx=
ap
p
,    S2=
b
0
yq−1dy=
bq
q
.
This finishes the demonstration. □
Remark 3 You may notice, that the both proofs introduced some specific auxiliary functions related to xp/p. It is a fruitful generalisation to conduct the proofs for more functions and derive respective forms of Young’s inequality.
Proposition 4 (Hölder’s Inequality) For 1<p<∞, let q∈(1,∞) be such that 1/p + 1/q = 1. For n≥1 and u,v∈Kn, we have that
    
n
j=1

ujvj
 ≤ 


n
j=1

uj
p


1
p



 



n
j=1

vj
q


1
q



 
. 
Proof. For reasons become clear soon we use the notation ||u||p=( ∑j=1n | uj |p )1/p and ||v||q= ( ∑j=1n | vj |q )1/q and define for 1≤ in:
    ai=
ui
⎪⎪
⎪⎪
u⎪⎪
⎪⎪
p
   and       bi=
vi
⎪⎪
⎪⎪
v⎪⎪
⎪⎪
q
.
Summing up for 1≤ in all inequalities obtained from (62):
    
aibi
≤ 

ai
p
p
 + 

bi
q
q
,
we get the result. □

Using Hölder inequality we can derive the following one:

Proposition 5 (Minkowski’s Inequality) For 1<p<∞, and n≥ 1, let u,v∈Kn. Then
    


n
j=1

uj+vj
p


1/p



 
≤ 


n
j=1

uj
p


1/p



 
 + 


n
j=1

vj
p


1/p



 
.  
Proof. For p>1 we have:
n
1

uk+vk
p =  
n
1

uk

uk+vk
p−1  +  
n
1

vk

uk+vk
p−1. (63)
By Hölder inequality
    
n
1

uk

uk+vk
p−1 ≤  


n
1

uk
p


1
p



 



n
1

uk+vk
q(p−1)


1
q



 
.
Adding a similar inequality for the second term in the right hand side of (63) and division by (∑1n | uk+vk |q(p−1))1/q yields the result. □

Minkowski’s inequality shows that for 1≤ p<∞ (the case p=1 is easy) we can define a norm ||·||p on Kn by

   ⎪⎪
⎪⎪
u⎪⎪
⎪⎪
p = 


n
j=1

uj
p


1/p



 
   ( u =(u1,⋯,un)∈Kn ). 

See, Figure 2 for illustration of various norms of this type defined in ℝ2.

We can define an infinite analogue of this. Let 1≤ p<∞, let lp be the space of all scalar sequences (xn) with ∑n | xn |p < ∞. A careful use of Minkowski’s inequality shows that lp is a vector space. Then lp becomes a normed space for the ||·||p norm. Note also, that l2 is the Hilbert space introduced before in Example 2.

Recall that a Cauchy sequence, see Defn. 5, in a normed space is bounded: if (xn) is Cauchy then we can find N with ||xnxm||<1 for all n,mN. Then ||xn|| ≤ ||xnxN|| + ||xN|| < ||xN||+1 for nN, so in particular, ||xn|| ≤ max( ||x1||,||x2||,⋯,||xN−1||,||xN||+1).

Theorem 6 For 1≤ p<∞, the space lp is a Banach space.
Remark 7 Most completeness proofs (in particular, all completeness proof in this course) are similar to the next one, see also Thm. 24. The general scheme of those proofs has three steps:
  1. For a general Cauchy sequence we build “limit” in some point-wise sense.
  2. At this stage it is not clear either the constructed “limit” is at our space at all, that is shown on the second step.
  3. From the construction it does not follows that the “limit” is really the limit in the topology of our space, that is the third step of the proof.
Proof. We repeat the proof of Thm. 24 changing 2 to p. Let (x(n)) be a Cauchy-sequence in lp; we wish to show this converges to some vector in lp.

For each n, x(n)lp so is a sequence of scalars, say (xk(n))k=1. As (x(n)) is Cauchy, for each є>0 there exists Nє so that ||x(n)x(m)||p ≤ є for n,mNє.

For k fixed,

    
xk(n) − xk(m)  
 ≤


 
j

xj(n) − xj(m)  
p


1/p



 
= ⎪⎪
⎪⎪
x(n) − x(m)⎪⎪
⎪⎪
p ≤ є,  

when n,mNє. Thus the scalar sequence (xk(n))n=1 is Cauchy in K and hence converges, to xk say. Let x=(xk), so that x is a candidate for the limit of (x(n)).

Firstly, we check that xx(n)lp for some n. Indeed, for a given є>0 find n0 such that ||x(n)x(m)||<є for all n,m>n0. For any K and m:

    
K
k=1

xk(n)xk(m)
p ≤  ⎪⎪
⎪⎪
x(n)x(m)⎪⎪
⎪⎪
pp.

Let m→ ∞ then ∑k=1K | xk(n)xk |p ≤ єp.
Let K→ ∞ then ∑k=1| xk(n)xk |p ≤ єp. Thus x(n)xlp and because lp is a linear space then x = x(n)−(x(n)x) is also in lp.

Finally, we saw above that for any є >0 there is n0 such that ||x(n)x||<є for all n>n0. Thus x(n)x. □

For p=∞, there are two analogies to the lp spaces. First, we define l to be the vector space of all bounded scalar sequences, with the sup-norm (||·||-norm):

⎪⎪
⎪⎪
(xn)⎪⎪
⎪⎪
 = 
 
sup
n∈ℕ

xn
    ( (xn)∈ l ).   (64)

Second, we define c0 to be the space of all scalar sequences (xn) which converge to 0. We equip c0 with the sup norm (64). This is defined, as if xn→0, then (xn) is bounded. Hence c0 is a subspace of l, and we can check (exercise!) that c0 is closed.

Theorem 8 The spaces c0 and l are Banach spaces.
Proof. This is another variant of the previous proof of Thm. 6. We do the l case. Again, let (x(n)) be a Cauchy sequence in l, and for each n, let x(n)=(xk(n))k=1. For є>0 we can find N such that ||x(n)x(m)|| < є for n,mN. Thus, for any k, we see that | xk(n)xk(m) | < є when n,mN. So (xk(n))n=1 is Cauchy, and hence converges, say to xk∈K. Let x=(xk).

Let mN, so that for any k, we have that

    
xk − xk(m)  
 = 
 
lim
n→∞

xk(n) − xk(m)  
≤ є. 

As k was arbitrary, we see that supk | xkxk(m) | ≤ є. So, firstly, this shows that (xx(m))∈l, and so also x = (xx(m)) + x(m)l. Secondly, we have shown that ||xx(m)|| ≤ є when mN, so x(m)x in norm. □

Example 9 We can also consider a Banach space of functions Lp[a,b] with the norm
    ⎪⎪
⎪⎪
f⎪⎪
⎪⎪
p=




b


a

f(t) 
pdt




1/p





 
.
See the discussion after Defn. 22 for a realisation of such spaces.

11.2 Bounded linear operators

Recall what a linear map is, see Defn. 1. A linear map is often called an operator. A linear map T:EF between normed spaces is bounded if there exists M>0 such that ||T(x)|| ≤ M ||x|| for xE, see Defn. 3. We write B(E,F) for the set of operators from E to F. For the natural operations, B(E,F) is a vector space. We norm B(E,F) by setting

⎪⎪
⎪⎪
T⎪⎪
⎪⎪
 = sup



⎪⎪
⎪⎪
T(x)⎪⎪
⎪⎪
⎪⎪
⎪⎪
x⎪⎪
⎪⎪
 : x∈ E, x≠0 



.  (65)
Exercise 10 Show that
  1. The expression (65) is a norm in the sense of Defn. 3.
  2. We equivalently have
            ⎪⎪
    ⎪⎪
    T⎪⎪
    ⎪⎪
     = sup

    ⎪⎪
    ⎪⎪
    T(x)⎪⎪
    ⎪⎪
     : x∈ E, ⎪⎪
    ⎪⎪
    x⎪⎪
    ⎪⎪
    ≤1 

    = sup

    ⎪⎪
    ⎪⎪
    T(x)⎪⎪
    ⎪⎪
     : x∈ E, ⎪⎪
    ⎪⎪
    x⎪⎪
    ⎪⎪
    =1 

    . 
Proposition 11 For a linear map T:EF between normed spaces, the following are equivalent:
  1. T is continuous (for the metrics induced by the norms on E and F);
  2. T is continuous at 0;
  3. T is bounded.
Proof. Proof essentially follows the proof of similar Theorem 4. See also discussion about usefulness of this theorem there. □
Theorem 12 Let E be a normed space, and let F be a Banach space. Then B(E,F) is a Banach space.
Proof. In the essence, we follows the same three-step procedure as in Thms. 24, 6 and 8. Let (Tn) be a Cauchy sequence in B(E,F). For xE, check that (Tn(x)) is Cauchy in F, and hence converges to, say, T(x), as F is complete. Then check that T:EF is linear, bounded, and that ||TnT||→ 0. □

We write B(E) for B(E,E). For normed spaces E, F and G, and for TB(E,F) and SB(F,G), we have that ST=STB(E,G) with ||ST|| ≤ ||S|| ||T||.

For TB(E,F), if there exists SB(F,E) with ST=IE, the identity of E, and TS=IF, then T is said to be invertible, and write T=S−1. In this case, we say that E and F are isomorphic spaces, and that T is an isomorphism.

If ||T(x)||=||x|| for each xE, we say that T is an isometry. If additionally T is an isomorphism, then T is an isometric isomorphism, and we say that E and F are isometrically isomorphic.

11.3 Dual Spaces

Let E be a normed vector space, and let E* (also written E′) be B(E,K), the space of bounded linear maps from E to K, which we call functionals, or more correctly, bounded linear functionals, see Defn. 1. Notice that as K is complete, the above theorem shows that E* is always a Banach space.

Theorem 13 Let 1<p<∞, and again let q be such that 1/p+1/q=1. Then the map lq→(lp)*: u↦φu, is an isometric isomorphism, where φu is defined, for u=(uj)∈lq, by
    φu(x) = 
j=1
ujxj    
x=(xj)∈lp
. 
Proof. By Hölder’s inequality, we see that
    
φu(x) 
 ≤ 
j=1

uj

xj
≤ 


j=1

uj
q


1/q



 



j=1

xj
p


1/p



 
= ⎪⎪
⎪⎪
u⎪⎪
⎪⎪
q⎪⎪
⎪⎪
x⎪⎪
⎪⎪
p. 
So the sum converges, and hence φu is defined. Clearly φu is linear, and the above estimate also shows that ||φu|| ≤ ||u||q. The map u↦ φu is also clearly linear, and we’ve just shown that it is norm-decreasing.

Now let φ∈(lp)*. For each n, let en = (0,⋯,0,1,0,⋯) with the 1 in the nth position. Then, for x=(xn)∈lp,

    ⎪⎪
⎪⎪
⎪⎪
⎪⎪
x − 
n
k=1
xkek⎪⎪
⎪⎪
⎪⎪
⎪⎪
p = 


k=n+1

xk
p


1/p



 
 → 0, 

as n→∞. As φ is continuous, we see that

    φ(x) = 
 
lim
n→∞
n
k=1
 φ(xkek) = 
k=1
xk φ(ek). 

Let uk=φ(ek) for each k. If u=(uk)∈lq then we would have that φ=φu.

Let us fix N∈ℕ, and define

    xk = 




0, if  uk=0  or  k>N; 
      
uk

uk
q−2,
if  uk≠0 and  k≤ N. 

Then we see that

    
k=1

xk
p = 
N
k=1

uk
p(q−1) = 
N
k=1

uk
q, 

as p(q−1) = q. Then, by the previous paragraph,

    φ(x) = 
k=1
xkuk = 
N
k=1

uk
q. 

Hence

    ⎪⎪
⎪⎪
φ⎪⎪
⎪⎪
 ≥ 

φ(x) 
⎪⎪
⎪⎪
x⎪⎪
⎪⎪
p
= 


N
k=1

uk
q


1−1/p



 
= 


N
k=1

uk
q


1/q



 
. 

By letting N→∞, it follows that ulq with ||u||q ≤ ||φ||. So φ=φu and ||φ|| = ||φu|| ≤ ||u||q. Hence every element of (lp)* arises as φu for some u, and also ||φu|| = ||u||q. □

Loosely speaking, we say that lq = (lp)*, although we should always be careful to keep in mind the exact map which gives this.

Corollary 14 (Riesz–Frechet Self-duality Lemma 11)l2 is self-dual: l2=l2*.

Similarly, we can show that c0*=l1 and that (l1)*=l (the implementing isometric isomorphism is giving by the same summation formula).

11.4 Hahn–Banach Theorem

Mathematical induction is a well known method to prove statements depending from a natural number. The mathematical induction is based on the following property of natural numbers: any subset of ℕ has the least element. This observation can be generalised to the transfinite induction described as follows.

A poset is a set X with a relation ≼ such that aa for all aX, if ab and ba then a=b, and if ab and bc, then ac. We say that (X,≼) is total if for every a,bX, either ab or ba. For a subset SX, an element aX is an upper bound for S if sa for every sS. An element aX is maximal if whenever bX is such that ab, then also ba.

Then Zorn’s Lemma tells us that if X is a non-empty poset such that every total subset has an upper bound, then X has a maximal element. Really this is an axiom which we have to assume, in addition to the usual axioms of set-theory. Zorn’s Lemma is equivalent to the axiom of choice and Zermelo’s theorem.

Theorem 15 (Hahn–Banach Theorem) Let E be a normed vector space, and let FE be a subspace. Let φ∈ F*. Then there exists ψ∈ E* with ||ψ||≤||φ|| and ψ(x)=φ(x) for each xF.
Proof. We do the real case. An “extension” of φ is a bounded linear map φG:G→ℝ such that FGE, φG(x)=φ(x) for xF, and ||φG||≤||φ||. We introduce a partial order on the pairs (G, φG) of subspaces and functionals as follows: (G1, φG1)≼ (G2, φG2) if and only if G1G2 and φG1(x)=φG2(x) for all xG1. A Zorn’s Lemma argument shows that a maximal extension φG:G→ℝ exists. We shall show that if GE, then we can extend φG, a contradiction.

Let xG, so an extension φ1 of φ to the linear span of G and x must have the form

    φ1(x′+ax) = φ(x) + a α    (x′∈ G, a∈ℝ), 

for some α∈ℝ. Under this, φ1 is linear and extends φ, but we also need to ensure that ||φ1||≤||φ||. That is, we need


φ(x′) + aα 
 ≤ ⎪⎪
⎪⎪
φ⎪⎪
⎪⎪
⎪⎪
⎪⎪
x′+ax⎪⎪
⎪⎪
   (x′∈ G, a∈ℝ).  (66)

It is straightforward for a=0, otherwise to simplify proof put −a y=x′ in (66) and divide both sides of the identity by a. Thus we need to show that there exist such α that

    
α−φ(y) 
 ≤ ⎪⎪
⎪⎪
φ⎪⎪
⎪⎪
⎪⎪
⎪⎪
xy⎪⎪
⎪⎪
  for all   y∈ G, a∈ℝ, 

or

    φ(y)−⎪⎪
⎪⎪
φ⎪⎪
⎪⎪
⎪⎪
⎪⎪
xy⎪⎪
⎪⎪
≤ α ≤ φ(y)+⎪⎪
⎪⎪
φ⎪⎪
⎪⎪
⎪⎪
⎪⎪
xy⎪⎪
⎪⎪
.

For any y1 and y2 in G we have:

    φ(y1)−φ(y2)≤ ⎪⎪
⎪⎪
φ⎪⎪
⎪⎪
⎪⎪
⎪⎪
y1y2⎪⎪
⎪⎪
≤ ⎪⎪
⎪⎪
φ⎪⎪
⎪⎪
 (⎪⎪
⎪⎪
xy2⎪⎪
⎪⎪
+ ⎪⎪
⎪⎪
xy1⎪⎪
⎪⎪
).

Thus

    φ(y1)−⎪⎪
⎪⎪
φ⎪⎪
⎪⎪
⎪⎪
⎪⎪
xy1⎪⎪
⎪⎪
≤ φ(y2)+⎪⎪
⎪⎪
φ⎪⎪
⎪⎪
⎪⎪
⎪⎪
xy2⎪⎪
⎪⎪
.

As y1 and y2 were arbitrary,

    
 
sup
y∈ G
 (φ(y) − ⎪⎪
⎪⎪
φ⎪⎪
⎪⎪
⎪⎪
⎪⎪
y+x⎪⎪
⎪⎪
) ≤
 
inf
y∈ G
 (φ(y) + ⎪⎪
⎪⎪
φ⎪⎪
⎪⎪
⎪⎪
⎪⎪
y+x⎪⎪
⎪⎪
). 

Hence we can choose α between the inf and the sup.

The complex case follows by “complexification”. □

The Hahn-Banach theorem tells us that a functional from a subspace can be extended to the whole space without increasing the norm. In particular, extending a functional on a one-dimensional subspace yields the following.

Corollary 16 Let E be a normed vector space, and let xE. Then there exists φ∈ E* with ||φ||=1 and φ(x)=||x||.

Another useful result which can be proved by Hahn-Banach is the following.

Corollary 17 Let E be a normed vector space, and let F be a subspace of E. For xE, the following are equivalent:
  1. xF the closure of F;
  2. for each φ∈ E* with φ(y)=0 for each yF, we have that φ(x)=0.
Proof. 12 follows because we can find a sequence (yn) in F with ynx; then it’s immediate that φ(x)=0, because φ is continuous. Conversely, we show that if 1 doesn’t hold then 2 doesn’t hold (that is, the contrapositive to 21).

So, xF. Define ψ:{F,x}→K by

    ψ(y+tx) = t    (y∈ F, t∈K). 

This is well-defined, for y, y′∈ F if y+tx=y′+tx then either t=t′, or otherwise x = (tt′)−1(y′−y) ∈ F which is a contradiction. The map ψ is obviously linear, so we need to show that it is bounded. Towards a contradiction, suppose that ψ is not bounded, so we can find a sequence (yn+tnx) with ||yn+tnx||≤1 for each n, and yet | ψ(yn+tnx) |=| tn |→∞. Then || tn−1 yn + x || ≤ 1/| tn | → 0, so that the sequence (−tn−1yn), which is in F, converges to x. So x is in the closure of F, a contradiction. So ψ is bounded. By Hahn-Banach theorem, we can find some φ∈ E* extending ψ. For yF, we have φ(y)=ψ(y)=0, while φ(x)=ψ(x)=1, so 2 doesn’t hold, as required. □

We define E** = (E*)* to be the bidual of E, and define J:EE** as follows. For xE, J(x) should be in E**, that is, a map E*→K. We define this to be the map φ↦φ(x) for φ∈ E*. We write this as

  J(x)(φ) = φ(x)    (x∈ E, φ∈ E*). 

The Corollary 16 shows that J is an isometry; when J is surjective (that is, when J is an isomorphism), we say that E is reflexive. For example, lp is reflexive for 1<p<∞. On the other hand c0 is not reflexive.

11.5 C(X) Spaces

This section is not examinable. Standard facts about topology will be used in later sections of the course.

All our topological spaces are assumed Hausdorff. Let X be a compact space, and let CK(X) be the space of continuous functions from X to K, with pointwise operations, so that CK(X) is a vector space. We norm CK(X) by setting

⎪⎪
⎪⎪
f⎪⎪
⎪⎪
 = 
 
sup
x∈ X

f(x) 
    (f∈ CK(X)). 
Theorem 18 Let X be a compact space. Then CK(X) is a Banach space.

Let E be a vector space, and let ||·||(1) and ||·||(2) be norms on E. These norms are equivalent if there exists m>0 with

  m−1⎪⎪
⎪⎪
x⎪⎪
⎪⎪
(2) ≤ ⎪⎪
⎪⎪
x⎪⎪
⎪⎪
(1) ≤ m⎪⎪
⎪⎪
x⎪⎪
⎪⎪
(2)    (x∈ E). 
Theorem 19 Let E be a finite-dimensional vector space with basis {e1,…,en}, so we can identify E with Kn as vector spaces, and hence talk about the norm ||·||2 on E. If ||·|| is any norm on E, then ||·|| and ||·||2 are equivalent.
Corollary 20 Let E be a finite-dimensional normed space. Then a subset XE is compact if and only if it is closed and bounded.
Lemma 21 Let E be a normed vector space, and let F be a closed subspace of E with EF. For 0<θ<1, we can find x0E with ||x0||≤1 and ||x0y||>θ for yF.
Theorem 22 Let E be an infinite-dimensional normed vector space. Then the closed unit ball of E, the set {xE : ||x||≤ 1}, is not compact.
Proof. Use the above lemma to construct a sequence (xn) in the closed unit ball of E with, say, ||xnxm||≥1/2 for each nm. Then (xn) can have no convergent subsequence, and so the closed unit ball cannot be compact. □

12 Measure Theory

The presentation in this section is close to [, , ].

12.1 Basic Measure Theory

The following object will be the cornerstone of our construction.

Definition 1 Let X be a set. A σ-algebra R on X is a collection of subsets of X, written R⊆ 2X, such that
  1. XR;
  2. if A,BR, then ABR;
  3. if (An) is any sequence in R, then n AnR.

Note, that in the third condition we admit any countable unions. The usage of “σ” in the names of σ -algebra and σ-ring is a reference to this. If we replace the condition by

  1. if (An)1m is any finite family in R, then ∪n=1m AnR;

then we obtain definitions of an algebra.

For a σ-algebra R and A,BR, we have

A ⋂ B = X
X∖(A⋂ B)
= X ∖
(X∖ A)⋃(X∖ B) 
R.

Similarly, R is closed under taking (countably) infinite intersections.

If we drop the first condition from the definition of (σ-)algebra (but keep the above conclusion from it!) we got a (σ-)ring, that is a (σ-)ring is closed under (countable) unions, (countable) intersections and subtractions of sets.

Exercise 2
  1. Use the above comments to write in full the three missing definitions: of set algebra, set ring and set σ-ring.
  2. Show that the empty set belongs to any non-empty ring.

Sets Ak are pairwise disjoint if AnAm=∅ for nm. We denote the union of pairwise disjoint sets by ⊔, e.g. ABC.

It is easy to work with a vector space through its basis. For a ring of sets the following notion works as a helpful “basis”.

Definition 3 A semiring S of sets is a collection such that
  1. it is closed under intersection;
  2. for A, BS we have AB=C1⊔ … ⊔ CN with CkS.

Again, any non-empty semiring contain the empty set.

Example 4 The following are semirings but not rings:
  1. The collection of intervals [a,b) on the real line;
  2. The collection of all rectangles { ax < b, cy <d } on the plane.

As the intersection of a family of σ-algebras is again a σ-algebra, and the power set 2X is a σ-algebra, it follows that given any collection D⊆ 2X, there is a σ-algebra R such that DR, such that if S is any other σ-algebra, with DS, then RS. We call R the σ-algebra generated by D.

Exercise 5 Let S be a semiring. Show that
  1. The collection of all finite disjoint unions k=1n Ak, where AkS, is a ring. We call it the ring R(S) generated by the semiring S.
  2. Any ring containing S contains R(S) as well.
  3. The collection of all finite (not necessarily disjoint!) unions k=1n Ak, where AkS, coincides with R(S).

We introduce the symbols +∞, −∞, and treat these as being “extended real numbers”, so −∞ < t < ∞ for t∈ℝ. We define t+∞ = ∞, t∞ = ∞ if t>0 and so forth. We do not (and cannot, in a consistent manner) define ∞ − ∞ or 0·∞.

Definition 6 A measure is a map µ:R→[0,∞] defined on a (semi-)ring (or σ-algebra) R, such that if A=⊔n An for AR and a finite subset (An) of R, then µ (A) = ∑n µ(An). This property is called additivity of a measure.

The additivity property of a measure is rather demanding. For example, let us consider the decomposition [0,1)=[0,1/2) ⊔ [1/2,1) = [0,1/3) ⊔ [1/3,2/3) ⊔ [2/3,1), then additivity puts measures of those five intervals into equations:

   µ([0,
1
2
)) + µ( [
1
2
,1) ) = µ([0.1)) = µ([0,
1
3
)) + µ([
1
3
,
2
3
)) + µ([
2
3
,1)).

Similar equations appear from any other (out of infinitely many) decomposition of [0,1), thus measures of various intervals are highly interconnected and very far from being arbitrary.

Exercise 7 Show that the following two conditions are equivalent:
  1. µ(∅)=0.
  2. There is a set AR such that µ(A)<∞.
The first condition often (but not always) is included in the definition of a measure.

In analysis we are interested in infinities and limits, thus the following extension of additivity is very important.

Definition 8 In terms of the previous definition we say that µ is countably additive (or σ-additive) if for any countable infinite family (An) of pairwise disjoint sets from R such that A=⊔n AnR we have µ(A) = ∑n µ(An). If the sum diverges, then as it will be the sum of positive numbers, we can, without problem, define it to be +∞.

Note, that this property may be stated as a sort of continuity of an additive measure, cf. (7):

       µ


 
lim
n→∞
n
k=1
Ak


= 
 
lim
n→∞
 µ


n
k=1
Ak


.
Example 9
  1. Fix a point a∈ℝ and define a measure µ by the condition µ(A)=1 if aA and µ(A)=0 otherwise.
  2. For the ring obtained in Exercise 5 from semiring S in Example 1 define µ([a,b))=ba on S. This is a measure, and we will show its σ-additivity.
  3. For ring obtained in Exercise 5 from the semiring in Example 2, define µ(V)=(ba)(dc) for the rectangle V={ ax < b, cy <d } S. It will be again a σ-additive measure.
  4. Let X=ℕ and R=2, we define µ(A)=0 if A is a finite subset of X=ℕ and µ(A)=+∞ otherwise. Let An={n}, then µ(An)=0 and µ(⊔n An)=µ(ℕ)=+∞≠ ∑n µ(An)=0. Thus, this measure is not σ-additive.

We will see further examples of measures which are not σ-additive in Section 12.4.

Definition 10 A measure µ is finite if µ(A)<∞ for all AR.

A measure µ is σ-finite if X is a union of countable number of sets Xk, such that for any AR and any k∈ ℕ the intersection AXk is in R and µ(AXk)<∞.

Exercise 11 Modify the example 1 to obtain
  1. a measure which is not finite, but is σ-finite. (Hint: let the measure count the number of integer points in a set).
  2. a measure which is not σ-finite. (Hint: assign µ(A)=+∞ if aA.)
Proposition 12 Let µ be a σ-additive measure on a σ-algebra R. Then:
  1. If A,BR with AB, then µ(A)≤µ(B) [we call this property “monotonicity of a measure”];
  2. If A,BR with AB and µ(B)<∞, then µ(BA) = µ(B) − µ(A);
  3. If (An) is a sequence in R, with A1A2A3 ⊆⋯. Then
          
     
    lim
    n→∞
     µ(An) = µ
    nAn
    . 
  4. If (An) is a sequence in R, with A1A2A3 ⊇⋯. If µ(Am)<∞ for some m, then
     
    lim
    n→∞
     µ(An) = µ
    nAn
    .  (67)
Proof. The two first properties are easy to see. For the third statement, define A=∪n An, B1=A1 and Bn=AnAn−1, n>1. Then An=⊔k=1n Bn and A=⊔k=1Bn. Using the σ-additivity of measures µ(A)=∑k=1µ(Bk) and µ(An)=∑k=1n µ(Bk). From the theorem in real analysis that any monotonic sequence of real numbers converges (recall that we admit +∞ as limits’ value) we have µ(A)=∑k=1µ(Bk)=limn→ ∞k=1n µ(Bk) = limn→ ∞ µ(An). The last statement can be shown similarly. □
Exercise 13 Let a measure µ on be defined by µ(A)=0 for finite A and µ(A) = ∞ for infinite A. Check that µ is additive but not σ-additive. Therefore give an example that µ does not satisfies 3.

12.2 Extension of Measures

From now on we consider only finite measures, an extension to σ-finite measures will be done later.

Proposition 14 Any measure µ′ on a semiring S is uniquely extended to a measure µ on the generated ring R(S), see Ex. 5. If the initial measure was σ-additive, then the extension is σ-additive as well.
Proof. If an extension exists it shall satisfy µ(A)=∑k=1n µ′(Ak), where AkS. We need to show for this definition two elements:
  1. Consistency, i.e. independence of the value from a presentation of AR(S) as A=⊔k=1n Ak, where AkS. For two different presentation A=⊔j=1n Aj and A=⊔k=1m Bk define Cjk=AjBk, which will be pair-wise disjoint. By the additivity of µ′ we have µ′(Aj)=∑kµ′(Cjk) and µ′(Bk)=∑jµ′(Cjk). Then
          
     
    j
     µ′(Aj)=
     
    j
     
    k
     µ′(Cjk) =
     
    k
     
    j
     µ′(Cjk)=
     
    k
     µ′(Bk).
  2. Additivity. For A=⊔k=1n Ak, where AkR(S) we can present Ak=⊔j=1n(k) Cjk, CjkS. Thus A=⊔k=1nj=1n(k) Cjk and:
          µ(A)=
    n
    k=1
    n(k)
    j=1
    µ′(Cjk)= 
    n
    k=1
      µ(Ak).
Finally, show the σ-additivity. For a set A=⊔k=1Ak, where A and AkR(S), find presentations A=⊔j=1n Bj, BjS and Ak=⊔l=1m(k) Blk, BlkS. Define Cjlk=BjBlkS, then Bj=⊔k=1l=1m(k) Cjlk and Ak= ⊔j=1nl=1m(k) Cjlk. Then, from σ-additivity of µ′:
     
    µ(A)
=
n
j=1
 µ′(Bj)= 
n
j=1
k=1
m(k)
l=1
 µ′(Cjlk)= 
k=1
n
j=1
m(k)
l=1
 µ′(Cjlk) = 
k=1
µ(Ak),
         
where we changed the summation order in series with non-negative terms. □

In a similar way we can extend a measure from a semiring to corresponding σ-ring, however it can be done even for a larger family. The procedure recall the famous story on Baron Munchausen saves himself from being drowned in a swamp by pulling on his own hair. Indeed, initially we knew measure for elements of semiring S or their finite disjoint unions from R(S). For an arbitrary set A we may assign a measure from an element of R(S) which “approximates” A. But how to measure such approximation? Well, to this end we use the measure on R(S) again (pulling on his own hair)!

Coming back to exact definitions, we introduce the following notion.

Definition 15 Let S be a semi-ring of subsets in X, and µ be a measure defined on S. An outer measure µ* on X is a map µ*:2X→[0,∞] defined by:
    µ*(A)=inf



 
k
 µ(Ak),  such that A⊆ ⋃kAk,   Ak∈ S



.
Proposition 16 An outer measure has the following properties:
  1. µ*(∅)=0;
  2. if AB then µ*(A)≤µ*(B), this is called monotonicity of the outer measure;
  3. if (An) is any sequence in 2X, then µ*(∪n An) ≤ ∑n µ*(An).

The final condition says that an outer measure is countably sub-additive. Note, that an outer measure may be not a measure in the sense of Defn. 6 due to a luck of additivity.

Example 17 The Lebesgue outer measure on is defined out of the measure from Example 2, that is, for A⊆ℝ, as
    µ*(A) = inf



j=1
(bjaj) : A⊆ ⋃j=1[aj,bj) 



. 
We make this definition, as intuitively, the “length”, or measure, of the interval [a,b) is (ba).

For example, for outer Lebesgue measure we have µ*(A)=0 for any countable set, which follows, as clearly µ*({x})=0 for any x∈ℝ.

Lemma 18 Let a<b. Then µ*([a,b])=ba.
Proof. For є>0, as [a,b] ⊆ [a,b+є), we have that µ*([a,b])≤ (ba)+є. As є>0, was arbitrary, µ*([a,b]) ≤ ba.

To show the opposite inequality we observe that [a,b)⊂[a,b] and µ*[a,b) =ba (because [a,b) is in the semi-ring) so µ*[a,b]≥ ba by 2. □

Our next aim is to construct measures from outer measures. We use the notation AB=(AB)∖ (AB) for symmetric difference of sets.

Definition 19 Given an outer measure µ* defined by a measure µ on a semiring S, we define AX to be Lebesgue measurable if for any ε >0 there is a finite union B of elements in S (in other words: BR(S) by Lem. 3), such that µ*(AB)<ε .

Figure 18: Approximating area by refined simple sets arrangements.

See Fig. 18 for an illustration of the concept of measurable sets. Obviously all elements of S and R(S) are measurable.

Exercise 20
  1. Define a function of pairs of Lebesgue measurable sets A and B as the outer measure of the symmetric difference of A and B:
    d(A,B)=µ*(A▵ B).  (68)
    Show that d is a metric on the collection of equivalence classes with respect to the equivalence relation: AB if d(A,B)=0. Hint: to show the triangle inequality use the inclusion:
        A▵ B ⊆ (A▵ C) ⋃ (C▵ B)
  2. Let a sequence n)→ 0 be monotonically decreasing. For a Lebesgue measurable A there exists a sequence (An)⊂ R(S) such that d(A,An)< εn for each n. Show that (An) is a Cauchy sequence for the distance d (68).

An alternative definition of a measurable set is due to Carathéodory.

Definition 21 Given an outer measure µ*, we define EX to be Carathéodory measurable if
    µ*(A) = µ*(A⋂ E) + µ*(A∖ E),
for any AX.

As µ* is sub-additive, this is equivalent to

  µ*(A) ≥ µ*(A⋂ E) + µ*(A∖ E)    (A⊆ X), 

as the other inequality is automatic.

Exercise* 22
  1. Show that for a Lebesgue measurable set A and any ε>0 there exist two elements B1 and B2 of the ring R(S) such that B1AB2 and µ(B2B1) < ε, cf. areas shadowed in darker and lighter colours on Fig. 18.
    Hint: For a set BR(S) such that µ*(AB)<ε/2 from Defn. 19 shall exists CR(S) such that CAB and µ(C) < µ*(AB)+ε/2. Put B1 = BC and B2 = BC.
  2. Let µ(X)<∞ show that A is Lebesgue measurable if and only if µ(X) = µ*(A)+µ*(XA).
  3. Show that measurability by Lebesgue and Carathéodory are equivalent.

Suppose now that the ring R(S) is an algebra (i.e., contains the maximal element X). Then, the outer measure of any set is finite, and the following theorem holds:

Theorem 23 (Lebesgue) Let µ* be an outer measure on X defined by a semiring S, and let L be the collection of all Lebesgue measurable sets for µ*. Then L is a σ-algebra, and if µ′ is the restriction of µ* to L, then µ′ is a measure. Furthermore, µ′ is σ-additive on L if µ is σ-additive on S.
Proof.[Sketch of proof] Clearly, R(S)⊂ L. Now we show that µ*(A)=µ(A) for a set AR(S). If A⊂ ∪k Ak for AkS, then µ(A)≤ ∑k µ(Ak), taking the infimum we get µ(A)≤µ*(A). For the opposite inequality, any AR(S) has a disjoint representation A=⊔k Ak, AkS, thus µ*(A)≤ ∑k µ(Ak)=µ(A).

Now we will show that R(S) with the distance d (68) is an incomplete metric space, with the measure µ being uniformly continuous functions. Measurable sets make the completion of R(S) (cf. Ex. 2) with µ being continuation of µ* to the completion by continuity, cf. Ex. 62.

Then, by the definition, Lebesgue measurable sets make the closure of R(S) with respect to this distance.

We can check that measurable sets form an algebra. To this end we need to make estimations, say, of µ*((A1A2)▵ (B1B2)) in terms of µ*(AiBi). A demonstration for any finite number of sets is performed through mathematical inductions. The above two-sets case provide both: the base and the step of the induction.

Now, we show that L is σ-algebra. Let AkL and A=∪k Ak. Then for any ε>0 there exists BkR(S), such that µ*(AkBk)<ε/2k. Define B=∪k Bk. Then

    
kAk
▵     
kBk
⊂    ⋃k
Ak ▵ Bk
implies  µ*(A▵ B)<ε.

We cannot stop at this point since B=∪k Bk may be not in R(S). Thus, define B1=B1 and Bk=Bk∖ ∪i=1k−1 Bi, so Bk are pair-wise disjoint. Then B=⊔k Bk and BkR(S). From the convergence of the series there is N such that ∑k=Nµ(Bk)<ε . Let B′=∪k=1N Bk, which is in R(S). Then µ*(BB′)≤ ε and, thus, µ*(AB′)≤ 2ε.

To check that µ* is measure on L we use the following

Lemma 24  | µ*(A)−µ*(B) |≤ µ*(AB), that is µ* is uniformly continuous in the metric d(A,B) (68).
Proof.[Proof of the Lemma] Use inclusions AB∪(AB) and BA∪(AB). □

To show additivity take A1,2L , A=A1A2, B1,2R(S) and µ*(AiBi)<ε. Then µ*(A▵(B1B2))<2ε and | µ*(A) − µ*(B1B2) |<2ε. Thus µ*(B1B2)=µ(B1B2)=µ (B1) +µ (B2)−µ (B1B2), but µ (B1B2)=d(B1B2,∅)=d(B1B2,A1A2)<2ε. Therefore

    
µ*(B1⋃ B2)−µ (B1) −µ (B2) 
<2ε.

Combining everything together we get (this is a sort of ε/3-argument):

     
    

µ*(A)−µ*(A1)−µ*(A2) 
  
          
=

µ*(A)−µ*(B1⋃ B2) +µ*(B1⋃ B2) −(µ (B1) +µ (B2)) 
         
 
   +µ (B1) +µ (B2)−µ*(A1)−µ*(A2)
         

µ*(A)−µ*(B1⋃ B2) 
+
µ*(B1⋃ B2)−(µ (B1) +µ (B2)) 
         
 
  +
µ (B1) +µ (B2)−µ*(A1)−µ*(A2) 
         
6ε.          

Thus µ* is additive on L.

Check the countable additivity for A=⊔k Ak. The inequality µ*(A)≤ ∑kµ*(Ak) follows from countable sub-additivity. The opposite inequality is the limiting case of the finite inequality µ*(A)≥ µ*(⊔k=1N Ak)=∑k=1Nµ*(Ak) following from monotonicity and additivity of µ*. □

Corollary 25 Let E⊆ℝ be open or closed. Then E is Lebesgue measurable.
Proof. As σ-algebras are closed under taking complements, we need only show that open sets are Lebesgue measurable. For the latter we will use a common trick, using the density and the countability of the rationals.

Intervals (a,b) are Lebesgue measurable because they are countable unions of measurable half-open intervals from the semiring, e.g.:

    (0,1) =  
k=1
  


1
k+1
, 
1
k



.

Now let U⊆ℝ be open. For each xU, there exists ax<bx with x∈(ax,bx)⊆ U. By making ax slightly larger, and bx slightly smaller, we can ensure that ax,bx∈ℚ. Thus U = ∪x (ax, bx). Each interval is measurable, and there are at most a countable number of them (endpoints make a countable set) thus U is the countable (or finite) union of Lebesgue measurable sets, and hence U is Lebesgue measurable itself. □

We perform now an extension of finite measure to σ-finite one. Let µ be a σ-additive and σ-finite measure defined on a semiring in X=⊔k Xk, such that the restriction of µ to every Xk is finite. Consider the Lebesgue extension µk of µ defined within Xk. A set AX is measurable if every intersection AXk is µk measurable. For a such measurable set A we define its measure by the identity:

  µ(A)=
 
k
 µk(A⋂ Xk).

We call a measure µ defined on L complete if whenever EX is such that there exists FL with µ(F)=0 and EF, we have that EL. Measures constructed from outer measures by the above theorem are always complete. On the example sheet, we saw how to form a complete measure from a given measure. We call sets like E null sets: complete measures are useful, because it is helpful to be able to say that null sets are in our σ-algebra. Null sets can be quite complicated. For the Lebesgue measure, all countable subsets of ℝ are null, but then so is the Cantor set, which is uncountable.

Definition 26 If we have a property P(x) which is true except possibly xA and µ(A)=0, we say P(x) is almost everywhere or a.e..

12.3 Complex-Valued Measures and Charges

We start from the following observation.

Exercise 27 Let µ1 and µ2 be measures on a same σ-algebra. Define µ12 and λµ1, λ>0 by 12)(A)=µ1(A)+µ2(A) and (λµ1)(A)=λ(µ1(A)). Then µ12 and λµ1 are measures on the same σ-algebra as well.

In view of this, it will be helpful to extend the notion of a measure to obtain a linear space.

Definition 28 Let X be a set, and R be a σ-ring. A real- (complex-) valued function ν on R is called a charge (or signed measure) if it is countably additive as follows: for any AkR the identity A=⊔k Ak implies the series k ν(Ak) is absolute convergent and has the sum ν(A).

In the following “charge” means “real charge”.

Example 29 Any linear combination of σ-additive measures on with real (complex) coefficients is real (complex) charge.

The opposite statement is also true:

Theorem 30 Any real (complex) charge ν has a representation ν=µ1−µ2 (ν=µ1−µ2+iµ3iµ4), where µk are σ-additive measures.

To prove the theorem we need the following definition.

Definition 31 The variation of a charge on a set A is | ν |(A)=sup ∑k| ν(Ak) | for all disjoint splitting A=⊔k Ak.
Example 32 If ν=µ1−µ2, then | ν |(A)≤ µ1(A)+µ2(A). The inequality becomes an identity for disjunctive measures on A (that is there is a partition A=A1A2 such that µ2(A1)=µ1(A2)=0).

The relation of variation to charge is as follows:

Theorem 33 For any charge ν the function | ν | is a σ-additive measure.

Finally to prove the Thm. 30 we use the following

Proposition 34 For any charge ν the function | ν |−ν is a σ-additive measure as well.

From the Thm. 30 we can deduce

Corollary 35 The collection of all charges on a σ-algebra R is a linear space which is complete with respect to the distance:
    d12)=
 
sup
AR

ν1(A)−ν2(A) 
.

The following result is also important:

Theorem 36 (Hahn Decomposition) Let ν be a charge. There exist A,BL, called a Hahn decomposition of (X,ν), with AB=∅, AB= X and such that for any EL,
    ν (A⋂ E) ≥ 0,   ν(B⋂ E)≤ 0. 
This need not be unique.
Proof.[Sketch of proof] We only sketch this. We say that AL is positive if
    ν(E⋂ A)≥0    (EL), 
and similiarly define what it means for a measurable set to be negative. Suppose that ν never takes the value −∞ (the other case follows by considering the charge −ν).

Let β = infν(B0) where we take the infimum over all negative sets B0. If β=−∞ then for each n, we can find a negative Bn with ν(Bn)≤ −n. But then B=∪n Bn would be negative with ν(B)≤ −n for any n, so that ν(B)=−∞ a contradiction.

So β>−∞ and so for each n we can find a negative Bn ν(Bn) < β+1/n. Then we can show that B = ∪n Bn is negative, and argue that ν(B) ≤ β. As B is negative, actually ν(B) = β.

There then follows a very tedious argument, by contradiction, to show that A=XB is a positive set. Then (A,B) is the required decomposition. □

12.4 Constructing Measures, Products

Consider the semiring S of intervals [a,b). There is a simple description of all measures on it. For a measure µ define

Fµ(t)=



      µ([0,t))if  t>0,
      0if  t=0,
      −µ([t,0))if  t<0,
    
(69)

Fµ is monotonic and any monotonic function F defines a measure µ on S by the by µ([a,b))=F(b)−F(a). The correspondence is one-to-one with the additional assumption F(0)=0.

Theorem 37 The above measure µ is σ-additive on S if and only if F is continuous from the left: F(t−0)=F(t) for all t∈ℝ.
Proof. Necessity: F(t)−F(t−0)=limε→ 0µ([t−ε,t))=µ(limε→ 0[t−ε,t))=µ(∅)=0, by the continuity of a σ-additive measure, see 4.

For sufficiency assume [a,b)=⊔k [ak,bk). The inequality µ([a,b))≥ ∑k µ([ak,bk)) follows from additivity and monotonicity. For the opposite inequality take δk s.t. F(b)−F(b−δ)<ε and F(ak)−F(ak−δk)<ε/2k (use left continuity of F). Then the interval [a,b−δ] is covered by (ak−δk,bk), due to compactness of [a,b−δ] there is a finite subcovering. Thus µ([a,b−δ ))≤∑j=1N µ([akj−δkj,bkj)) and µ([a,b))≤∑j=1N µ([akj,bkj))+2ε . □

Exercise 38
  1. Give an example of function discontinued from the left at 1 and show that the resulting measure is additive but not σ-additive.
  2. Check that, if a function F is continuous at point a then µ({a})=0.
Example 39
  1. Take F(t)=t, then the corresponding measure is the Lebesgue measure on .
  2. Take F(t) be the integer part of t, then µ counts the number of integer within the set.
  3. Define the Cantor function as follows α(x)=1/2 on (1/3,2/3); α(x)=1/4 on (1/9,2/9); α(x)=3/4 on (7/9,8/9), and so for. This function is monotonic and can be continued to [0,1] by continuity, it is know as Cantor ladder. The resulting measure has the following properties:
    • The measure of the entire interval is 1.
    • Measure of every point is zero.
    • The measure of the Cantor set is 1, while its Lebesgue measure is 0.

Another possibility to build measures is their product. In particular, it allows to expand various measures defined through (69) on the real line to ℝn.

Definition 40 Let X and Y be spaces, and let S and T be semirings on X and Y respectively. Then S× T is the semiring consisting of { A× B : AS, BT } (“generalised rectangles”). Let µ and ν be measures on S and T respectively. Define the product measure µ×ν on S× T by the rule (µ× ν)(A× B)=µ(A) ν(B).
Example 41 The measure from Example 3 on the semiring of half-open rectangles is the product of two copies of pre-Lebesgue measures from Example 2 on the semiring of half-open intervals.

13 Integration

We now come to the main use of measure theory: to define a general theory of integration.

13.1 Measurable functions

From now on, by a measure space we shall mean a triple (X,L,µ), where X is a set, L is a σ-algebra on X, and µ is a σ-additive measure defined on L. We say that the members of L are measurable, or L-measurable, if necessary to avoid confusion.

Definition 1 A function f:X→ℝ is measurable if
    Ec(f)={x∈ X: f(x)<c}    that is    Ec(f)=f−1((−∞,c))
is in L (that is Ec(f) is a measurable set) for any c∈ℝ.

A complex-valued function is measurable if its real and imaginary parts are measurable.

Lemma 2 The following are equivalent:
  1. A function f is measurable;
  2. For any a<b the set f−1((a,b)) is measurable;
  3. For any open set U⊂ ℝ the set f−1(U) is measurable.
Proof. To show 1 ⇒  2 we note that
    f−1((a,b))  = Eb(f)∖ 


 
n
Ea+1/n(f) 


.
For 2 ⇒  3 use that any open set U⊂ ℝ is a union of countable set of intervals (a,b), cf. proof of Cor. 25.

The final implication 3 ⇒  1 directly follows from openness of (−∞,a). □

Corollary 3 Let f: X → ℝ be measurable and g: ℝ → ℝ be continuous, then the composition g(f(x)) is measurable.
Proof. The preimage of the open set (−∞,c) under a continuous g is an open set, say U. The preimage of U under f is measurable by Lem. 3. Thus, the preimage of (−∞,c) under the composition gf is measurable, thereafter gf is a measurable function. □
Theorem 4 Let f,g:X→ℝ be measurable. Then af (a∈ℝ), f+g, fg, max(f,g) and min(f,g) are all measurable. That is measurable functions form an algebra and this algebra is closed under convergence a.e.
Proof. Use Cor. 3 to show measurability of λ f, | f | and f2. The measurability of a sum f1 + f2 follows from the relation
    Ec(f1+f2)=⋃r∈ℚ (Er(f1)⋂ Ecr(f2)).
Next use the following identities:
    f1f2=
(f1+f2)2−(f1f2)2
4
,
    max(f1,f2)=
(f1+f2)+
f1f2
2
.

If (fn) is a non-increasing sequence of measurable functions converging to f. Than Ec(f)=∪n Ec(fn).

Moreover any limit can be replaced by two monotonic limits:

 
lim
n→ ∞
fn(x)=
 
lim
n→ ∞
 
lim
k→ ∞
 max (fn(x), fn+1(x),…,fn+k(x)). (70)

Finally if f1 is measurable and f2=f1 almost everywhere, then f2 is measurable as well. □

We can define several types of convergence for measurable functions.

Definition 5 We say that sequence (fn) of functions converges
  1. uniformly to f (notated fnf) if
          
     
    sup
    x∈ X

    fn(x)−f(x) 
     → 0;
  2. almost everywhere to f (notated fna.e.f) if
          fn(x)→ f(x)    for all  x∈ X∖ A,  µ(A)=0;
  3. in measure µ to f (notated fnµf) if for all ε>0
    µ({x∈ X: 
    fn(x)−f(x) 
    >ε }) → 0. (71)

Clearly uniform convergence implies both convergences a.e and in measure.

Theorem 6 On finite measures convergence a.e. implies convergence in measure.
Proof. Define An(ε)={xX: | fn(x)−f(x) |≥ ε}. Let Bn(ε)=∪kn Ak(ε). Clearly Bn(ε)⊃ Bn+1(ε), let B(ε)=∩1Bn(ε). If xB(ε) then fn(x)↛f(x). Thus µ(B(ε))=0, but µ(B(ε))=limn→ ∞µ(Bn(ε)), cf. (67). Since An(ε)⊂ Bn(ε) we see that µ(An(ε))→ 0 as required for (71) □

Note, that the construction of sets Bn(ε) is just another implementation of the “two monotonic limits” trick (70) for sets.

Exercise 7 Present examples of sequences (fn) and functions f such that:
  1. fnµf but not fna.e.f.
  2. fna.e.f but not fnf.

However we can slightly “fix” either the set or the sequence to “upgrade” the convergence as shown in the following two theorems.

Theorem 8 (Egorov) If fna.e.f on a finite measure set X then for any σ>0 there is EσX with µ(Eσ)<σ and fnf on XEσ.
Proof. We use An(ε) and Bn(ε) from the proof of Thm. 6. Observe that | f(x)−fk(x) |< ε uniformly for all xXBn(ε) and k>n. For every ε>0 we seen that µ(Bn(ε))→ 0, thus for each k there is N(k) such that µ(BN(k)(1/k))<σ/2k. Put Eσ=∪k BN(k)(1/k). □
Theorem 9 If fnµf then there is a subsequence (nk) such that fnka.e.f for k→ ∞.
Proof. In the notations of two previous proofs: for every natural k take nk such that µ(Ank(1/k))< 1/2k, which is possible since µ(An(ε))→ 0. Define Cm=∪k=mAnk(1/k) and C=∩ Cm. Then, µ(Cm)=1/2m−1 and, thus, µ(C)=0 by (67). If xC then there is such N that xAnk(1/k) for all k>N. That means that | fnk(x)−f(x) |<1/k for all such k, i.e fnk(x)→ f(x). Thus, we have the point-wise convergence everywhere except the zero-measure set C. □

It is worth to note, that we can use the last two theorem subsequently and upgrade the convergence in measure to the uniform convergence of a subsequence on a subset.

Exercise 10 For your counter examples from Exercise 7, find
  1. a subsequence fnk of the sequence from 1 which converges to f a.e.;
  2. a subset such that sequence from 2 converges uniformly.
Exercise 11 Read about Luzin’s C-property.

13.2 Lebesgue Integral

First we define a sort of “basis” for the space of integral functions.

Definition 12 For AX, we define χA to be the indicator function of A, by
    χA(x) = 


1: x∈ A, 
0: xA. 

Then, if χA is measurable, then χA−1( (1/2,3/2) ) = AL; conversely, if AL, then XAL, and we see that for any U⊆ℝ open, χA−1(U) is either ∅, A, XA, or X, all of which are in L. So χA is measurable if and only if AL.

Definition 13 A measurable function f:X→ℝ is simple if it attains only a countable number of values.
Lemma 14 A function f:X→ℝ is simple if and only if
f = 
k=1
tk χAk(72)
for some (tk)k=1⊆ℝ and AkL. That is, simple functions are linear combinations of indicator functions of measurable sets.

Moreover in the above representation the sets Ak can be pair-wise disjoint and all tk≠ 0 pair-wise different. In this case the representation is unique.

Notice that it is now obvious that

Corollary 15 The collection of simple functions forms a vector space: this wasn’t clear from the original definition.
Definition 16 A simple function in the form (72) with disjoint Ak is called summable if the following series converges:
k=1

tk
 µ(Ak)   if f has the above unique representation   f = 
k=1
tk χAk . (73)

It is another combinatorial exercise to show that this definition is independent of the way we write f.

Definition 17 We define the integral of a simple function f=∑k tk χAk (72) over a measurable set A by setting
    
 


A
f  d µ = 
k=1
tk µ(AkA).

Clearly the series converges for any simple summable function f. Moreover

Lemma 18 The value of integral of a simple summable function is independent from its representation by the sum of indicators (72). In particular, we can evaluate the integral taking the canonical representation over pair-wise disjoint sets having pair-wise different values.
Proof. This is another slightly tedious combinatorial exercise. You need to prove that the integral of a simple function is well-defined, in the sense that it is independent of the way we choose to write the simple function. □
Exercise 19 Let f be the function on [0,1] which take the value 1 in all rational points and 0—everywhere else. Find the value of the Lebesgue integral [0,1] f,dµ with respect to the Lebesgue measure on [0,1]. Show that the Riemann upper- and lower sums for f converges to different values, so f is not Riemann-integrable.
Remark 20 The previous exercise shows that the Lebesgue integral does not have those problems of the Riemann integral related to discontinuities. Indeed, most of function which are not Riemann-integrable are integrable in the sense of Lebesgue. The only reason, why a measurable function is not integrable by Lebesgue is divergence of the series (73). Therefore, we prefer to speak that the function is summable rather than integrable. However, those terms are used interchangeably in the mathematical literature.

We will denote by S(X) the collection of all simple summable functions on X.

Proposition 21 Let f, g:X→ ℝ be in S(X) (that is simple summable), let a, b∈ ℝ and A is a measurable set. Then:
  1. A af+bgd µ = aA fd µ + bA gd µ, that is S(X) is a linear space;
  2. The correspondence f→ ∫A fd µ is a linear functional on S(X);
  3. The correspondence A → ∫A fd µ is a charge;
  4. If fg then X fd µ ≤ ∫X gd µ, that is integral is monotonic;
  5. The function
    d1(f,g)=
     


    X

    f(x)−g(x) 
    d µ(x) (74)
    has all properties of a metric (distance) on S(X) probably except separation, but see the next item.
  6. For f≥ 0 we have X fd µ=0 if and only if µ( { xX : f(x)≠0 } ) = 0. Therefore for the function d1 (74):
          d1(f,g)=0    if and only if    f
    a.e.
    =
     
    g.
  7. The integral is uniformly continuous with respect the above metric d1 (74):
          




     


    A
    f(x) d µ(x)−
     


    A
    g(x) d µ(x) 




     ≤ d1(f,g).
Proof. The proof is almost obvious, for example the Property 1 easily follows from Lem. 18.

We will outline 3 only. Let f is an indicator function of a set B, then A→ ∫A fd µ=µ(AB) is a σ-additive measure (and thus—a charge). By the Cor. 35 the same is true for finite linear combinations of indicator functions and their limits in the sense of distance d1. □

We can identify functions which has the same values a.e. Then S(X) becomes a metric space with the distance d1 (74). The space may be incomplete and we may wish to look for its completion. However, if we will simply try to assign a limiting point to every Cauchy sequence in S(X), then the resulting space becomes so huge that it will be impossible to realise it as a space of functions on X.

Exercise 22 Use ideas of Ex. 1 to present a sequence of simple functions which has the Cauchy property in metric d1 (74) but does not have point-wise limits anywhere.

To reduce the number of Cauchy sequences in S(X) eligible to have a limit, we shall ask an additional condition. A convenient reduction to functions on X appears if we ask both the convergence in d1 metric and the point-wise convergence on X a.e.

Definition 23 A function f is summable by a measure µ if there is a sequence (fn)⊂S(X) such that
  1. the sequence (fn) is a Cauchy sequence in S(X);
  2. fna.e. f.

Clearly, if a function is summable, then any equivalent function is summable as well. Set of equivalent classes of summable functions will be denoted by L1(X).

Lemma 24 If the measure µ is finite then any bounded measurable function is summable.
Proof. Define Ekn(f)={xX: k/nf(x)< (k+1)/n} and fn=∑k k/n χEkn (note that the sum is finite due to boundedness of f).

Since | fn(x)−f(x) |<1/n we have uniform convergence (thus convergence a.e.) and (fn) is the Cauchy sequence: d1(fn,fm)=∫X| fnfm | d µ≤ (1/n+1/m)µ(X). □

Remark 25 This Lemma can be extended to the space of essentially bounded functions L(X), that is functions which are bounded a.e. In other words, L(X)⊂L1(X) for finite measures.

Another simple result, which is useful on many occasions is as follows.

Lemma 26 If the measure µ is finite and fnf then d1(fn,f)→ 0.
Corollary 27 For a convergent sequence fna.e. f, which admits the uniform bound | fn(x) |<M for all n and x, we have d1(fn,f)→ 0.
Proof. For any ε>0, by the Egorov’s theorem 8 we can find E, such that
  1. µ(E)< ε/2M; and
  2. from the uniform convergence on XE there exists N such that for any n>N we have | f(x)−fn(x) |<ε /2µ(X).
Combining this we found that for n>N, d1(fn,f)< M ε/2M + µ(X) ε /2µ(X) < ε . □
Exercise 28 Convergence in the metric d1 and a.e. do not imply each other:
  1. Give an example of fna.e. f such that d1(fn ,f)↛0.
  2. Give an example of the sequence (fn) and function f in L1(X) such that d1(fn ,f)→ 0 but fn does not converge to f a.e.

To build integral we need the following

Lemma 29 Let (fn) and (gn) be two Cauchy sequences in S(X) with the same limit a.e., then d1(fn,gn)→ 0.
Proof. Let φn=fngn, then this is a Cauchy sequence with zero limit a.e. Assume the opposite to the statement: there exist δ>0 and sequence (nk) such that ∫x| φnk | d µ>δ. Rescaling-renumbering we can obtain ∫x| φn | d µ>1.

Take quickly convergent subsequence using the Cauchy property:

    d1nknk+1)≤ 1/2k+2.

Renumbering agian assume d1kk+1)≤ 1/2k+2.

Since φ1 is a simple, take the canonical presentation φ1=∑k tk χAk, then ∑k | tk | µ(Ak)=∫X | φ1 | d µ≥ 1. Thus, there exists N, such that ∑k=1N | tk | µ(Ak)≥ 3/4. Put A=⊔k=1N Ak and C=max1≤ kN| tk |=maxxA| φ1(x) |.

By the Egorov’s Theorem 8 there is EA such that µ(E)<1/(4C) and φn⇉ 0 on B=AE. Then

    
 


B

φ1
d µ= 
 


A

φ1
d µ−
 


E

φ1
d µ≥
3
4
1
4C
· C=
1
2
.

By the triangle inequality for d1:

    




 


B

φn
d µ−
 


B

φn+1
d µ 




 ≤ d1nn+1)≤ 
1
2n+2

we get

    
 


B

φn
d µ≥ 
 


B

φ1
d µ−
n−1
k=1





 


B

φn
d µ−
 


B

φn+1
d µ 




1
2
n−1
1
1
2k+2
>
1
4
.

But this contradicts to the fact ∫B | φn | d µ → 0, which follows from the uniform convergence φn⇉ 0 on B. □

It follows from the Lemma that we can use any Cauchy sequence of simple functions for the extension of integral.

Corollary 30 The functional IA(f)=∫A f(x) d µ(x), defined on any AL on the space of simple functions S(X) can be extended by continuity to the functional on L1(X,µ).
Definition 31 For an arbitrary summable fL1(X), we define the Lebesgue integral
    
 


A
f  d µ =
 
lim
n→ ∞
 


A
fn  d µ,
where the Cauchy sequence fn of summable simple functions converges to f a.e.
Theorem 32
  1. L1(X) is a linear space.
  2. For any measurable set AX the correspondence f↦ ∫A fd µ is a linear functional on L1(X).
  3. For any fL1(X) the value ν(A)=∫A fd µ is a charge.
  4. d1(f,g)=∫A | fg |  d µ is a distance on L1(X).
Proof. The proof is follows from Prop. 21 and continuity of extension. □

Summing up: we build L1(X) as a completion of S(X) with respect to the distance d1 such that elements of L1(X) are associated with (equivalence classes of) measurable functions on X.

13.3 Properties of the Lebesgue Integral

The space L1 was defined from dual convergence—in d1 metric and point-wise a.e. Can we get the continuity of the integral from the convergence almost everywhere alone? No, in general. However, we will state now some results on continuity of the integral under convergence a.e. with some additional assumptions. Finally, we show that L1(X) is closed in d1 metric.

Theorem 33 (Lebesgue on dominated convergence) Let (fn) be a sequence of µ-summable functions on X, and there is φ∈L1(X) such that | fn(x) |≤ φ(x) for all xX, n∈ℕ.

If fna.e. f, then fL1(X) and for any measurable A:

    
 
lim
n→∞
 


A
fn  d µ    =    
 


A
f  d µ.

Proof. For any measurable A the expression ν(A)=∫A φ  d µ defines a finite measure on X due to non-negativeness of φ and Thm. 32.
Lemma 34 (Change of variable) If g is measurable and bounded then fg is µ-summable and for any µ-measurable set A we have
 


A
f  d µ= 
 


A
g  d ν. (75)
Proof.[Proof of the Lemma] Let M be the set of all g such that the Lemma is true. M includes any indicator functions gB of a measurable B:
      
 


A
f  d µ=
 


A
 φχB  d µ = 
 


A⋂ B
 φ  d µ =ν(A⋂ B)=
 


A
 χBd ν=
 


A
gd ν.      
Thus M contains also finite linear combinations of indicators. For any n∈ℕ and a bounded g two functions g(x)=1/n[ng(x)] and g+(x)=g+1/n are finite linear combinations of indicators and are in M. Since g(x)≤ g(x)≤ g+(x) we have
      
 


A
gd ν= 
 


A
 φ gd µ≤ 
 


A
φ gd µ≤ 
 


A
 φ g+d µ=
 


A
g+d ν. 
By squeeze rule for n→ ∞ we have the middle term tenses to ∫Agd ν, that is gM.

Note, that formula (75) is a change of variable in the Lebesgue integral of the type: ∫f(sinx) cosxd x = ∫f(sinx)  d (sinx). □

For the proof of the theorem define:
    gn(x)=


        fn(x)/φ(x),if  φ(x)≠ 0,
        0,if  φ(x)= 0,
    g(x)=


        f(x)/φ(x),if  φ(x)≠ 0,
        0,if  φ(x)= 0.
  
Then gn is bounded by 1 and gna.e. g. To show the theorem it will be enough to show limn→ ∞A gnd ν=∫A gd ν. For the uniformly bounded functions on the finite measure set this can be derived from the Egorov’s Thm. 8, see an example of this in the proof of Lemma 29. □

Note, that in the above proof summability of φ was used to obtain the finiteness of the measure ν, which is required for Egorov’s Thm. 8.

Exercise 35 Give an example of fna.e. f such that X fnd µ ≠ ∫X fd µ. For such an example, try to find a function φ such that | fn | ≤ φ for all n and check either φ is summable.
Exercise 36 (Chebyshev’s inequality) Show that: if f is non-negative and summable, then
µ{x∈ X: f(x)>c} < 
1
c
 


X
fd µ. (76)
Theorem 37 (B. Levi’s, on monotone convergence) Let (fn) be monotonically increasing sequence of µ-summable functions on X. Define f(x)=limn→∞ fn(x) (allowing the value +∞).
  1. If all integrals X fnd µ are bounded by the same value C<∞ then f is summable and X fd µ=limn→∞X fnd µ.
  2. If limn→∞X fnd µ=+∞ then function f is not summable.
Proof. Replacing fn by fnf1 and f by ff1 we can assume fn≥ 0 and f≥ 0. Let E be the set where f is infinite, then
    E=⋂NnENn,    where   ENn={x∈ X: fn(x)≥ N}.
By Chebyshev’s inequality (76) we have
    Nµ(ENn) <  
 


ENn
fnd µ ≤ 
 


X
fnd µ≤ C,
then µ(ENn)≤ C/N . Thus µ(E)=limN→∞limn→∞ µ(ENn)=0.

Thus f is finite a.e.

Lemma 38 Let f be a measurable non-negative function attaining only finite values. f is summable if and only if sup∫A fd µ<∞, where the supremum is taken over all finite-measure set A such that f is bounded on A.
Proof.[Proof of the Lemma] Necessity: if f is summable then for any set AX we have ∫A fd µ≤ ∫X fd µ<∞, thus the supremum is finite.

Sufficiency: let sup∫A fd µ=M<∞, define B={xX: f(x)=0} and Ak={xX: 2kf(x)<2k+1, k∈ℤ}, by (76) we have µ(Ak)<M/2k and X=B⊔(⊔k=0Ak). Define

      g(x)=


          2k,if  x∈ Ak,
          0,if  x∈ B,
      fn(x)=


          f(x),if  x∈ ⊔nnAn,
          0,otherwise.
    

Then g(x)≤ f(x) < 2g(x). Function g is a simple function, its summability follows from the estimate ∫nn Ak gd µ≤∫nn Ak fd µ≤ M which is valid for any n, taking n→ ∞ we get summability of g. Furthermore, fna.e. f and fn(x)≤ f(x) <2g(x), so we use the Lebesgue Thm. 33 on dominated convergence to obtain the conclusion. □

Let A be a finite measure set such that f is bounded on A, then

    
 


A
fd µ
Cor. 27
=
 
 
lim
n→∞
 


A
fnd µ≤
 
lim
n→∞
 


X
fnd µ≤ C.

This show summability of f by the previous Lemma. The rest of statement and (contrapositive to) the second part follows from the Lebesgue Thm. 33 on dominated convergence. □

Now we can extend this result dropping the monotonicity assumption.

Lemma 39 (Fatou) If a sequence (fn) of µ-summable non-negative functions is such that: then f is µ-summable and X fd µ≤ C.
Proof.Let us replace the limit fnf by two monotonic limits. Define:
    gkn(x)=min(fn(x),…,fn+k(x)),
    gn(x)=
 
lim
k→ ∞
gkn(x).
Then gn is a non-decreasing sequence of functions and limn→ ∞ gn(x)=f(x) a.e. Since gnfn, from monotonicity of integral we get ∫X gnd µ≤ C for all n. Then Levi’s Thm. 37 implies that f is summable and ∫X fd µ≤ C. □
Remark 40 Note that the price for dropping monotonicity from Thm. 37 to Lem. 39 is that the limit X fnd µ → ∫X fd µ may not hold any more.
Exercise 41 Give an example such that under the Fatou’s lemma condition we get limn→∞X fnd µ ≠ ∫X fd µ.

Now we can show that L1(X) is complete:

Theorem 42 L1(X) is a Banach space.
Proof. It is clear that the distance function d1 indeed define a norm ||f||1=d1(f,0). We only need to demonstrate the completeness. We again utilise the three-step procedure from Rem. 7.

Take a Cauchy sequence (fn) and building a subsequence if necessary, assume that its quickly convergent that is d1(fn,fn+1)≤ 1/2k. Put

    φ1=f1  and     φn=fnfn−1  for   n>1. Then  fn=
n
k=1
 φk .

The sequence ψn(x)=∑1n | φk(x) | is monotonic, integrals ∫X ψnd µ are bounded by the same constant ||f1||1+1. Thus, by the B. Levi’s Thm. 37 and its proof, ψn→ ψ for a summable essentially bounded function ψ. Therefore, the series ∑φk(x) converges as well to a value f(x) of a function f. But, this means that fna.e. f (the first step is completed).

We also notice | fn(x) |≤| ψ(x) |. Thus by the Lebesgue Thm. 33 on dominated convergence fL1(X) (the second step is completed).

Furthermore,

0≤
 
lim
n→ ∞
 


X

fnf
d µ≤
 
lim
n→ ∞
k=n
⎪⎪
⎪⎪
φk⎪⎪
⎪⎪
=0.

That is, fnf in the norm of L1(X). (That completes the third step and the whole proof). □

The next important property of the Lebesgue integral is its absolute continuity.

Theorem 43 (Absolute continuity of Lebesgue integral) Let fL1(X). Then for any ε>0 there is a δ>0 such that | ∫A fd µ |<ε if µ(A)<δ.
Proof. If f is essentially bounded by M, then it is enough to set δ=ε/M. In general let:
    An=
{x∈ X: n
f(x) 
< n+1},
    Bn=0nAk,
    Cn=X∖ Bn.
Then ∫X| f | d µ=∑0Ak| f | d µ, thus there is an N such that ∑NAk| f | d µ=∫CN| f | d µ<ε/2. Now put δ =ε/2N+2, then for any AX with µ(A)<δ:
    




 


A
fd µ 




 


A

f
d µ=
 


A⋂ BN

f
d µ+
 


A⋂ CN

f
d µ < 
ε
2
+
ε
2
=ε.

13.4 Integration on Product Measures

It is well-known geometrical interpretation of an integral in calculus as the “area under the graph”. If we advance from “area” to a “measure” then the Lebesgue integral can be treated as theory of measures of very special shapes created by graphs of functions. This shapes belong to the product spaces of the function domain and its range. We introduced product measures in Defn. 40, now we will study them in same details using the Lebesgue integral. We start from the following

Theorem 44 Let X and Y be spaces, and let S and T be semirings on X and Y respectively and µ and ν be measures on S and T respectively. If µ and ν are σ-additive, then the product measure ν× µ from Defn. 40 is σ-additive as well.
Proof. For any C=A× BS× T let us define fC(x)=χA(x)ν(B). Then
    (µ×ν)(C)=µ(A)ν(B)=
 


X
fCd µ.
If the same set C has a representation C=⊔k Ck for CkS× T, then σ-additivity of ν implies fC=∑k fCk. By the Lebesgue theorem 33 on dominated convergence:
    
 


X
fCd µ=
 
k
 


X
fCkd µ.
Thus
    (µ×ν)(C)=
 
k
(µ×ν)(Ck).

The above correspondence CfC can be extended to the ring R(S× T) generated by S× T by the formula:

  fC=
 
k
fCk,    for  C=⊔kCk∈ R(S× T).

We have the uniform continuity of this correspondence:

  ⎪⎪
⎪⎪
fC1fC2⎪⎪
⎪⎪
1≤ (µ×ν)(C1▵ C2)=d1(C1,C2)

because from the representation C1=A1B and C2=A2B, where B=C1C2 one can see that fC1fC2=fA1fA2, fC1C2=fA1+fA2 together with | fA1fA2 |≤ fA1+fA2 for non-negative functions.

Thus the map CfC can be extended to the map of σ-algebra L(X× Y) of µ×ν-measurable set to L1(X) by the formula flimn Cn=limn fCn.

Exercise 45 Describe topologies where two limits from the last formula are taken.

The following lemma provides the geometric interpretation of the function fC as the size of the slice of the set C along x=const.

Lemma 46 Let CL(X× Y). For almost every xX the set Cx={yY: (x,y)∈ C} is ν-measurable and ν(Cx)=fC(x).
Proof. For sets from the ring R(S× T) it is true by the definition. If C(n) is a monotonic sequence of sets, then ν(limn Cx(n))=limn ν(Cx(n)) by σ-additivity of measures. Thus the property ν(Cx)=fx(C) is preserved by monotonic limits. The following result of the separate interest:
Lemma 47 Any measurable set can be received (up to a set of zero measure) from elementary sets by two monotonic limits.
Proof.[Proof of Lem. 47] Let C be a measurable set, put CnR(S× T) to approximate C up to 2n in µ×ν. Let C′=∩n=1k =1Cn+k, then
      (µ× ν)
C∖ ⋃k=1Cn+k
=0  and   (µ× ν)
k=1Cn+k∖ C
=21−n.
Then (µ×ν)(C′▵ C)≤ 21−n for any n∈ℕ. □
Coming back to Lem. 46 we notice that (in the above notations) fC=fC almost everywhere. Then:
    fC(x)
a.e
=
 
fC(x)=ν(Cx)=ν(Cx).

The following theorem generalizes the meaning of the integral as “area under the graph”.

Theorem 48 Let µ and ν are σ-finite measures and C be a µ×ν measurable set X× Y. We define Cx={yY: (x,y)∈ C}. Then for µ-almost every xX the set Cx is ν-measurable, function fC(x)=ν(Cx) is µ-measurable and
(µ×ν)(C)=
 


X
fCd µ, (77)
where both parts may have the value +∞.
Proof. If C has a finite measure, then the statement is reduced to Lem. 46 and a passage to limit in (77).

If C has an infinite measure, then there exists a sequence of CnC, such that ∪n Cn=C and (µ×ν)(Cn)→ ∞. Then fC(x)=limn fCn (x) and

    
 


X
fCnd µ=(µ×ν)(Cn)→ +∞.

Thus fC is measurable and non-summable. □

This theorem justify the well-known technique to calculation of areas (volumes) as integrals of length (areas) of the sections.

Remark 49
  1. The role of spaces X and Y in Theorem 48 is symmetric, thus we can swap them in the conclusion.
  2. The Theorem 48 can be extended to any finite number of measure spaces. For the case of three spaces (X,µ), (Y,ν), (Z,λ) we have:
    (µ×ν×λ )(C)=
     


    X× Y
     λ(Cxy) d (µ×ν)(x,y)=
     


    Z
     (µ×ν)(Cz) d λ(z), (78)
    where
        Cxy={z∈ Z: (x,y,z)∈ C},
        Cz={(x,y)∈ X× Y: (x,y,z) ∈ C}.
Theorem 50 (Fubini) Let f(x,y) be a summable function on the product of spaces (X,µ) and (Y,ν). Then:
  1. For µ-almost every xX the function f(x,y) is summable on Y and fY(x)=∫Y f(x,y) d ν(y) is a µ-summable on X.
  2. For ν-almost every yY the function f(x,y) is summable on X and fX(y)=∫X f(x,y) d µ(x) is a ν-summable on Y.
  3. There are the identities:
         
           
     


    X× Y
    f(x,y) d (µ×ν)(x,y)
    =
     


    X





     


    Y
    f(x,y) d ν(y)




    dµ(x)
    (79)
     =
     


    Y





     


    X
    f(x,y) d µ(x)




    dν(y).  
     
  4. For a non-negative functions the existence of any repeated integral in (79) implies summability of f on X× Y.
Proof. From the decomposition f=f+f we can reduce our consideration to non-negative functions. Let us consider the product of three spaces (X,µ), (Y,ν), (ℝ,λ), with λ=dz being the Lebesgue measure on ℝ. Define
    C={(x,y,z)∈ X× Y× ℝ: 0≤ z≤ f(x,y)}.
Using the relation (78) we get:
    Cxy={z∈ ℝ: 0≤ z≤ f(x,y)},    λ(Cxy)=f(x,y)
    Cx=
{(y,z)∈ Y× ℝ: 0≤ z≤ f(x,y)},    (ν× λ)(Cx)=
 


Y
f(x,y) d ν(y).
the theorem follows from those relations. □
Exercise 51

13.5 Absolute Continuity of Measures

Here, we consider another topic in the measure theory which benefits from the integration theory.

Definition 52 Let X be a set with σ-algebra R and σ-finite measure µ and finite charge ν on R. The charge ν is absolutely continuous with respect to µ if µ(A)=0 for AR implies ν(A)=0. Two charges ν1 and ν2 are equivalent if two conditions | ν1 |(A)=0 and | ν2 |(A)=0 are equivalent.

The above definition seems to be not justifying “absolute continuity” name, but this will become clear from the following important theorem.

Theorem 53 (Radon–Nikodym) Any charge ν which absolutely continuous with respect to a measure µ has the form
    ν(A)=
 


A
fd µ,
where f is a function from L1. The function fL1 is uniquely defined by the charge ν.
Proof.[Sketch of the proof] First we will assume that ν is a measure. Let D be the collection of measurable functions g:X→[0,∞) such that
    
 


E
g  d µ ≤ ν(E)    (EL). 
Let α = supgDX gd µ ≤ ν(X) < ∞. So we can find a sequence (gn) in D with ∫X gnd µ → α.

We define f0(x) = supn gn(x). We can show that f0=∞ only on a set of µ-measure zero, so if we adjust f0 on this set, we get a measurable function f:X→[0,∞). There is now a long argument to show that f is as required.

If ν is a charge, we can find f by applying the previous operation to the measures ν+ and ν (as it is easy to verify that ν+⋘µ).

We show that f is essentially unique. If g is another function inducing ν, then

    
 


E
fg  d µ = ν(E) − ν(E) = 0    (EL). 

Let E = {xX : f(x)−g(x)≥ 0}, so as fg is measurable, EL. Then ∫E fgd µ =0 and fg≥0 on E, so by our result from integration theory, we have that fg=0 almost everywhere on E. Similarly, if F = {xX : f(x)−g(x)≤ 0}, then FL and fg=0 almost everywhere on F. As EF=X, we conclude that f=g almost everywhere. □

Corollary 54 Let µ be a measure on X, ν be a finite charge, which is absolutely continuous with respect to µ. For any ε>0 there exists δ>0 such that µ(A)<δ implies | ν |(A)<ε .
Proof. By the Radon–Nikodym theorem there is a function fL1(X,µ) such that ν(A)=∫A fd µ. Then | ν |(A)=∫A | f | d µ ad we get the statement from Theorem 43 on absolute continuity of the Lebesgue integral. □

14 Functional Spaces

In this section we describe various Banach spaces of functions on sets with measure.

14.1 Integrable Functions

Let (X,L,µ) be a measure space. For 1≤ p<∞, we define Lp(µ) to be the space of measurable functions f:X→K such that

  
 


X

f
p  d µ < ∞. 

We define ||·||p : Lp(µ)→[0,∞) by

⎪⎪
⎪⎪
f⎪⎪
⎪⎪
p = 




 


X

f
p  d µ 




1/p





 
    (f∈ Lp(µ)). 

Notice that if f=0 almost everywhere, then | f |p=0 almost everywhere, and so ||f||p=0. However, there can be non-zero functions such that f=0 almost everywhere. So ||·||p is not a norm on Lp(µ).

Exercise 1 Find a measure space (X,µ) such that lp=Lp(µ), that is the space of sequences lp is a particular case of function spaces considered in this section. It also explains why the following proofs are referencing to Section 11 so often.
Lemma 2 (Integral Hölder inequality) Let 1<p<∞, let q∈(1,∞) be such that 1/p + 1/q=1. For fLp(µ) and gLq(µ), we have that fg is summable, and
 


X

fg
  d µ ≤ ⎪⎪
⎪⎪
f⎪⎪
⎪⎪
p⎪⎪
⎪⎪
g⎪⎪
⎪⎪
q.  (80)
Proof. Recall that we know from Lem. 2 that
    
ab
 ≤ 

a
p
p
 + 

b
q
q
    (a,b∈K).  
Now we follow the steps in proof of Prop. 4. Define measurable functions a,b:X→K by setting
    a(x) = 
f(x)
⎪⎪
⎪⎪
f⎪⎪
⎪⎪
p
,   b(x) = 
g(x)
⎪⎪
⎪⎪
g⎪⎪
⎪⎪
q
    (x∈ X). 
So we have that
    
a(x) b(x) 
 ≤ 

f(x) 
p
p⎪⎪
⎪⎪
f⎪⎪
⎪⎪
pp
 +

g(x) 
q
q⎪⎪
⎪⎪
g⎪⎪
⎪⎪
qq
    (x∈ X). 
By integrating, we see that
    
 


X

ab
  d µ ≤ 
1
p⎪⎪
⎪⎪
f⎪⎪
⎪⎪
pp
 


X

f
p  d µ + 
1
q⎪⎪
⎪⎪
g⎪⎪
⎪⎪
qq
 


X

g
q  d µ = 
1
p
 + 
1
q
 = 1. 
Hence, by the definition of a and b,
    
 


X

fg
 ≤ ⎪⎪
⎪⎪
f⎪⎪
⎪⎪
p⎪⎪
⎪⎪
g⎪⎪
⎪⎪
q, 
as required. □
Lemma 3 Let f,gLp(µ) and let a∈K. Then:
  1. ||af||p = | a | ||f||p;
  2. || f+g ||p ≤ ||f||p + ||g||p.
In particular, Lp is a vector space.
Proof. Part 1 is easy. For 2, we need a version of Minkowski’s Inequality, which will follow from the previous lemma. We essentially repeat the proof of Prop. 5.

Notice that the p=1 case is easy, so suppose that 1<p<∞. We have that

     
 


X

f+g
p  d µ
= 
 


X

f+g
p−1
f+g
  d µ 
         
 
≤ 
 


X

f+g
p−1

f
 + 
g

d µ 
         
 
= 
 


X

f+g
p−1
f
  d µ + 
 


X

f+g
p−1
g
  d µ.
         

Applying the lemma, this is

    ≤ ⎪⎪
⎪⎪
f⎪⎪
⎪⎪
p




 


X

f+g
q(p−1)  d µ 




1/q





 
+ ⎪⎪
⎪⎪
g⎪⎪
⎪⎪
p




 


X

f+g
q(p−1)  d µ 




1/q





 
. 

As q(p−1)=p, we see that

    ⎪⎪
⎪⎪
f+g⎪⎪
⎪⎪
pp ≤ 
⎪⎪
⎪⎪
f⎪⎪
⎪⎪
p + ⎪⎪
⎪⎪
g⎪⎪
⎪⎪
p
⎪⎪
⎪⎪
f+g⎪⎪
⎪⎪
pp/q. 

As pp/q = 1, we conclude that

    ⎪⎪
⎪⎪
f+g⎪⎪
⎪⎪
p ≤ ⎪⎪
⎪⎪
f⎪⎪
⎪⎪
p + ⎪⎪
⎪⎪
g⎪⎪
⎪⎪
p, 

as required.

In particular, if f,gLp(µ) then af+gLp(µ), showing that Lp(µ) is a vector space. □

We define an equivalence relation ∼ on the space of measurable functions by setting fg if and only if f=g almost everywhere. We can check that ∼ is an equivalence relation (the slightly non-trivial part is that ∼ is transitive).

Proposition 4 For 1≤ p<∞, the collection of equivalence classes Lp(µ) / ∼ is a vector space, and ||·||p is a well-defined norm on Lp(µ) / ∼.
Proof. We need to show that addition, and scalar multiplication, are well-defined on Lp(µ)/∼. Let a∈K and f1,f2,g1,g2Lp(µ) with f1f2 and g1g2. Then it’s easy to see that af1+g1af2+g2; but this is all that’s required!

If fg then | f |p = | g |p almost everywhere, and so ||f||p = ||g||p. So ||·||p is well-defined on equivalence classes. In particular, if f∼ 0 then ||f||p=0. Conversely, if ||f||p=0 then ∫X | f |pd µ=0, so as | f |p is a positive function, we must have that | f |p=0 almost everywhere. Hence f=0 almost everywhere, so f∼ 0. That is,

    

f∈ Lp(µ) : f∼ 0 

= 

f∈ Lp(µ) : ⎪⎪
⎪⎪
f⎪⎪
⎪⎪
p=0 

. 

It follows from the above lemma that this is a subspace of Lp(µ).

The above lemma now immediately shows that ||·||p is a norm on Lp(µ)/∼. □

Definition 5 We write Lp(µ) for the normed space (Lp(µ)/∼ , ||·||p).

We will abuse notation and continue to write members of Lp(µ) as functions. Really they are equivalence classes, and so care must be taken when dealing with Lp(µ). For example, if fLp(µ), it does not make sense to talk about the value of f at a point.

Theorem 6 Let (fn) be a Cauchy sequence in Lp(µ). There exists fLp(µ) with ||fnf||p→ 0. In fact, we can find a subsequence (nk) such that fnkf pointwise, almost everywhere.
Proof. Consider first the case of a finite measure space X. We again follow the three steps scheme from Rem. 7. Let fn be a Cauchy sequence in Lp(µ). From the Hölder inequality (80) we see that ||fnfm||1≤ ||fnfm||p (µ(X))1/q. Thus, fn is also a Cauchy sequence in L1(µ). Thus by the Theorem 42 there is the limit function fL1(µ). Moreover, from the proof of that theorem we know that there is a subsequence fnk of fn convergent to f almost everywhere. Thus in the Cauchy sequence inequality
    
 


X

fnk −fnm  
pd µ <ε
we can pass to the limit m→ ∞ by the Fatou Lemma 39 and conclude:
    
 


X

fnk −f  
pd µ <ε.
So, fnk converges to f in Lp(µ), then fn converges to f in Lp(µ) as well.

For a σ-finite measure µ we represent X=⊔k Xk with µ(Xk)<+∞ for all k. The restriction (fn(k)) of a Cauchy sequence (fn)⊂Lp(X,µ) to every Xk is a Cauchy sequence in Lp(Xk,µ). By the previous paragraph there is the limit f(k)Lp(Xk,µ). Define a function fLp(X,µ) by the identities f(x)=f(k) if xXk. By the additivity of integral, the Cauchy condition on (fn) can be written as:

    
 


X

fnfm
pd µ=
k=1
 


Xk

fn(k)fm(k)
pd µ<ε. 

It implies for any M:

    
M
k=1
 


Xk

fn(k)fm(k)
pd µ<ε. 

In the last inequality we can pass to the limit m→ ∞:

    
M
k=1
 


Xk

fn(k)f(k)
pd µ<ε. 

Since the last inequality is independent of M we conclude:

    
 


X

fnf
pd µ=
k=1
 


Xk

fn(k)f(k)
pd µ<ε. 

Thus we conclude that fnf in Lp(X,µ). □

Corollary 7Lp(µ) is a Banach space.
Example 8 If p=2 then Lp(µ)=L2(µ) can be equipped with the inner product:
⟨ f,g  ⟩ =
 


X
fḡ d µ. (81)
The previous Corollary implies that L2(µ) is a Hilbert space, see a preliminary discussion in Defn. 22.
Proposition 9 Let (X,L,µ) be a measure space, and let 1≤ p<∞. We can define a map Φ:Lq(µ) → Lp(µ)* by setting Φ(f)=F, for fLq(µ), 1/p+1/q=1, where
    F:Lp(µ)→K,   g ↦ 
 


X
fg  d µ    (gLp(µ)).  
Proof. This proof very similar to proof of Thm. 13. For fLq(µ) and gLp(µ), it follows by the Hölder’s Inequality (80), that fg is summable, and
    




 


X
fg  d µ  




 ≤ 
 


X

fg
  d µ ≤ ⎪⎪
⎪⎪
f⎪⎪
⎪⎪
q⎪⎪
⎪⎪
g⎪⎪
⎪⎪
p. 

Let f1,f2Lq(µ) and g1,g2Lp(µ) with f1f2 and g1g2. Then f1g1 = f2g1 almost everywhere and f2g1 = f2g2 almost everywhere, so f1g1 = f2g2 almost everywhere, and hence

    
 


X
f1g1  d µ =  
 


X
f2g2  d µ. 

So Φ is well-defined.

Clearly Φ is linear, and we have shown that ||Φ(f)|| ≤ ||f||q.

Let fLq(µ) and define g:X→K by

    g(x) = 




f(x) 

f(x) 
q−2
: f(x)≠0, 
      0: f(x)=0. 

Then | g(x) | = | f(x) |q−1 for all xX, and so

    
 


X

g
p  d µ = 
 


X

f
p(q−1)  d µ = 
 


X

f
q  d µ, 

so ||g||p = ||f||qq/p, and so, in particular, gLp(µ). Let F=Φ(f), so that

    F(g) = 
 


X
fg  d µ = 
 


X

f
q  d µ = ⎪⎪
⎪⎪
f⎪⎪
⎪⎪
qq. 

Thus ||F|| ≥ ||f||qq / ||g||p = ||f||q. So we conclude that ||F|| = ||f||q, showing that Φ is an isometry. □

Proposition 10 Let (X,L,µ) be a finite measure space, let 1≤ p<∞, and let FLp(µ)*. Then there exists fLq(µ), 1/p+1/q=1 such that
    F(g) = 
 


X
fg  d µ    (gLp(µ)). 
Proof.[Sketch of the proof] As µ(X)<∞, for EL, we have that ||χE||p = µ(E)1/p < ∞. So χELp(µ), and hence we can define
    ν(E) = FE)    (EL). 
We proceed to show that ν is a signed (or complex) measure. Then we can apply the Radon-Nikodym Theorem 53 to find a function f:X→K such that
    FE) = ν(E) = 
 


E
f  d µ    (EL). 

There is then a long argument to show that fLq(µ), which we skip here. Finally, we need to show that

    
 


X
fg  d µ = F(g) 

for all gLp(µ), and not just for gE. That follows for simple functions with a finite set of values by linearity of the Lebesgue integral and F. Then, it can be extended by continuity to the entire space Lp(µ) in view in the following Prop. 14. □

Proposition 11 For 1<p<∞, we have that Lp(µ)* = Lq(µ) isometrically, under the identification of the above results.
Remark 12
  1. For p=q=2 we obtain a special case of the Riesz–Frechét theorem 11 about self-duality of the Hilbert space L2(µ).
  2. Note that L* is not isomorphic to L1, except finite-dimensional situation. Moreover if µ is not a point measure L1 is not a dual to any Banach space.
Exercise 13 Let µ be a measure on the real line.
  1. Show that the space L(ℝ,µ) is either finite-dimensional or non-separable.
  2. Show that for pq neither Lp(ℝ,µ) nor Lq(ℝ,µ) contains the other space.

14.2 Dense Subspaces in Lp

We note that fLp(X) if and only if | f |p is summable, thus we can use all results from Section 13 to investigate Lp(X).

Proposition 14 Let (X,L,µ) be a finite measure space, and let 1≤ p<∞. Then the collection of simple bounded functions attained only a finite number of values is dense in Lp(µ).
Proof.Let fLp(µ), and suppose for now that f≥0. For each n∈ℕ, let
    fn = min(n, 
1
n
 ⌊ nf ⌋). 
Then each fn is simple, fnf, and | fnf |p→0 pointwise. For each n, we have that
    0 ≤ fn ≤ f  0 ≤ ffn ≤ f, 
so that | ffn |p ≤ | f |p for all n. As ∫| f |pd µ<∞, we can apply the Dominated Convergence Theorem to see that
    
 
lim
n
 


X

fnf
p  d µ = 0, 
that is, ||fnf||p → 0.

The general case follows by taking positive and negative parts, and if K=ℂ, by taking real and imaginary parts first. □

Corollary 15 Let µ be the Lebesgue measure on the real line. The collection of simple bounded functions with compact supports attained only a finite number of values is dense in Lp(ℝ,µ).
Proof. Let fLp(ℝ,µ), since ∫ | f |pd µ = ∑k=−∞[k,k+1) | f |pd µ there exists N such that ∑k=−∞N + ∑Nk=∞[k,k+1) | f |pd µ < ε . By the previous Proposition, the restriction of f to [−N,N] can be ε-approximated by a simple bounded function f1 with support in [−N,N] attained only a finite number of values. Therefore f1 will be also (2ε)-approximation to f as well. □
Definition 16 A function f:ℝ→ ℂ is called step function if it a linear combination of a finite number of indicator functions of half-open disjoint intervals: f=∑k ck χ[ak,bk).

The regularity of the Lebesgue measure allows to make a stronger version of Prop. 14.

Lemma 17 The space of step functions is dense in Lp(ℝ).
Proof. By Prop. 14, for a given fLp(ℝ) and ε>0 there exists a simple function f0=∑k=1n ck χAk such that ||ff0||p<ε/2. Let M=||f0|| < ∞. By measurability of the set Ak there is Ck=⊔jmk [ajk,bjk) a disjoint finite union of half-open intervals such that µ(CkAk)<ε/2n3 M. Since Ak and Aj are disjoint for kj we also obtain by the triangle inequality: µ(CjAk)<ε/2n3 M and µ(CjCk)<2ε/2n3 M. We define a step function
  f1=
n
k=1
ck χCk=
n
k=1
mk
j
ck χ[ajk,bjk).
Clearly
   f1(x)=ck    for all  x ∈ Ak∖ ((Ck▵ Ak) ⋃(⋃j≠ kCj)).
Thus:
   µ({x ∈ ℝ  ∣  f0(x)≠ f1(x)}) ≤ n· n· 
ε
2n3M
 = 
ε
2 nM
.
Then ||f0f1||pnM· ε/2 n M=ε/2 because ||f1||< nM. Thus ||ff1||p<ε. □
Corollary 18 The collection of continuous function belonging to Lp(ℝ) is dense in Lp(ℝ).
Proof. In view of Rem. 29 and the previous Lemma it is enough to show that the characteristic function of an interval [a,b] can be approximated by a continuous function in Lp(ℝ). The idea of such approximation is illustrated by Fig. 4 and we skip the technical details. □

We will establish denseness of the subspace of smooth function in § 15.4.

Exercise 19 Show that every fL1(ℝ) is continuous on average, that is for any ε>0 there is δ>0 such that for all t such that | t |<δ we have:
 



f(x)−f(x+t) 
dx < ε . (82)

Here is an alternative demonstration of a similar result, it essentially encapsulate all the above separate statements. Let ([0,1],L,µ) be the restriction of Lebesgue measure to [0,1]. We often write Lp([0,1]) instead of Lp(µ).

Proposition 20 For 1≤ p<∞, we have that CK([0,1]) is dense in Lp([0,1]).
Proof. As [0,1] is a finite measure space, and each member of CK([0,1]) is bounded, it is easy to see that each fCK([0,1]) is such that ||f||p<∞. So it makes sense to regard CK([0,1]) as a subspace of Lp(µ). If CK([0,1]) is not dense in Lp(µ), then we can find a non-zero FLp([0,1])* with F(f)=0 for each fCK([0,1]). This was a corollary of the Hahn-Banach theorem 15.

So there exists a non-zero gLq([0,1]) with

    
 


[0,1]
fg  d µ = 0    (f∈ CK([0,1])). 

Let a<b in [0,1]. By approximating χ(a,b) by a continuous function, we can show that ∫(a,b) gd µ = ∫ g χ(a,b)d µ = 0.

Suppose for now that K=ℝ. Let A = { x∈[0,1] : g(x)≥0 } ∈ L. By the definition of the Lebesgue (outer) measure, for є>0, there exist sequences (an) and (bn) with A ⊆ ∪n (an,bn), and ∑n (bnan) ≤ µ(A) + є.

For each N, consider ∪n=1N (an,bn). If some (ai,bi) overlaps (aj,bj), then we could just consider the larger interval (min(ai,aj), max(bi,bj)). Formally by an induction argument, we see that we can write ∪n=1N (an,bn) as a finite union of some disjoint open intervals, which we abusing notations still denote by (an,bn). By linearity, it hence follows that for N∈ℕ, if we set BN = ⊔n=1N (an,bn), then

    g χBN  d µ = g χ(a1,b1)⊔⋯⊔(aN,bN)  d µ = 0. 

Let B=∪n (an,bn), so AB and µ(B) ≤ ∑n (bnan) ≤ µ(A)+є. We then have that

    
g χBN  d µ − g χB  d µ  
 = 
g χB∖ (a1,b1)⊔⋯⊔(aN,bN)  d µ  
. 

We now apply Hölder’s inequality to get

     
    
χB∖ (a1,b1)⋃⋯⋃(aN,bN)  d µ 
1/p ⎪⎪
⎪⎪
g⎪⎪
⎪⎪
q
= µ(B∖ (a1,b1)⊔⋯⊔(aN,bN))1/p⎪⎪
⎪⎪
g⎪⎪
⎪⎪
q
         
 
≤ 


n=N+1
(bnan) 


1/p



 
⎪⎪
⎪⎪
g⎪⎪
⎪⎪
q. 
         

We can make this arbitrarily small by making N large. Hence we conclude that

    g χB  d µ=0. 

Then we apply Hölder’s inequality again to see that

    
gχA  d µ  
 = 
gχA  d µ − gχB  d µ  
  = 
g χB∖ A  d µ  
 ≤ ⎪⎪
⎪⎪
g⎪⎪
⎪⎪
q µ(B∖ A)1/p ≤ ⎪⎪
⎪⎪
g⎪⎪
⎪⎪
q є1/p. 

As є>0 was arbitrary, we see that ∫A gd µ=0. As g is positive on A, we conclude that g=0 almost everywhere on A.

A similar argument applied to the set {x∈[0,1] : g(x)≤0} allows us to conclude that g=0 almost everywhere. If K=ℂ, then take real and imaginary parts. □

14.3 Continuous functions

Let K be a compact (always assumed Hausdorff) topological space.

Definition 21 The Borel σ-algebra, B(K), on K, is the σ-algebra generated by the open sets in K (recall what this means from Section 11.5). A member of B(K) is a Borel set.

Notice that if f:K→K is a continuous function, then clearly f is B(K)-measurable (the inverse image of an open set will be open, and hence certainly Borel). So if µ:B(K)→K is a finite real or complex charge (for K=ℝ or K=ℂ respectively), then f will be µ-summable (as f is bounded) and so we can define

  φµ:CK(K) → K,   φµ(f) = 
 


K
f  d µ    (f∈ CK(K)). 

Clearly φµ is linear. Suppose for now that µ is positive, so that

  
φµ(f) 
 ≤ 
 


K

f
  d µ ≤ ⎪⎪
⎪⎪
f⎪⎪
⎪⎪
 µ(K)    (f∈ CK(K)). 

So φµCK(K)* with ||φµ||≤ µ(K).

The aim of this section is to show that all of CK(K)* arises in this way. First we need to define a class of measures which are in a good agreement with the topological structure.

Definition 22 A measure µ:B(K)→[0,∞) is regular if for each AB(K), we have
     
    µ(A)
= sup

µ(E) : E⊆ A  and E is compact 

         
 
= inf

µ(U) : A⊆ U  and U is open 

.
         
A charge ν=ν+−ν is regular if ν+ and ν are regular measures. A complex measure is regular if its real and imaginary parts are regular.

Note the similarity between this notion and definition of outer measure.

Example 23
  1. Many common measures on the real line, e.g. the Lebesgue measure, point measures, etc., are regular.
  2. An example of the measure µ on [0,1] which is not regular:
          µ(∅)=0,   µ(
    {
    1
    2
    }
    )=1,    µ(A)=+∞,
    for any other subset A⊂[0,1].
  3. Another example of a σ-additive measure µ on [0,1] which is not regular:
          µ(A)=

              0, if A is at most countable;
              +∞otherwise.

The following subspace of the space of all simple functions is helpful.

As we are working only with compact spaces, for us, “compact” is the same as “closed”. Regular measures somehow interact “well” with the underlying topology on K.

We let M(K) and M(K) be the collection of all finite, regular real or complex charges (that is, signed or complex measures) on B(K).

Exercise 24 Check that, M(K) and M(K) are real or complex, respectively, vector spaces for the obvious definition of addition and scalar multiplication.

Recall, Defn. 31, that for µ∈ MK(K) we define the variation of µ

  ⎪⎪
⎪⎪
µ⎪⎪
⎪⎪
 = sup



n=1

µ(An) 




, 

where the supremum is taken over all sequences (An) of pairwise disjoint members of B(K), with ⊔n An=K. Such (An) are called partitions.

Proposition 25 The variation ||·|| is a norm on MK(K).
Proof. If µ=0 then clearly ||µ||=0. If ||µ||=0, then for AB(K), let A1=A, A2=KA and A3=A4=⋯=∅. Then (An) is a partition, and so
    0 = 
n=1

µ(An) 
 = 
µ(A) 
 + 
µ(K∖ A) 
. 
Hence µ(A)=0, and so as A was arbitrary, we have that µ=0.

Clearly ||aµ|| = | a |||µ|| for a∈K and µ∈ MK(K).

For µ,λ∈ MK(K) and a partition (An), we have that

    
 
n

(µ+λ)(An) 
 = 
 
n

µ(An)+λ(An) 
≤ 
 
n

µ(An) 
+
 
n

λ(An) 
 ≤ ⎪⎪
⎪⎪
µ⎪⎪
⎪⎪
 + ⎪⎪
⎪⎪
λ⎪⎪
⎪⎪
. 

As (An) was arbitrary, we see that ||µ+λ|| ≤ ||µ|| + ||λ||. □

To get a handle on the “regular” condition, we need to know a little more about CK(K).

Theorem 26 (Urysohn’s Lemma) Let K be a compact space, and let E,F be closed subsets of K with EF=∅. There exists f:K→[0,1] continuous with f(x)=1 for xE and f(x)=0 for xF (written f(E)={1} and f(F)={0}).
Proof. See a book on (point set) topology. □
Lemma 27 Let µ:B(K)→[0,∞) be a regular measure. Then for UK open, we have
    µ(U) = sup





 


K
f  d µ  : f∈ C(K), 0≤ f≤χU





. 
Proof. If 0≤ f≤χU, then
    0 = 
 


K
 0  d µ ≤ 
 


K
f  d µ ≤ 
 


K
 χU  d µ = µ(U). 
Conversely, let F=KU, a closed set. Let EU be closed. By Urysohn Lemma 26, there exists f:K→[0,1] continuous with f(E)={1} and f(F)={0}. So χEf ≤ χU, and hence
    µ(E) ≤ 
 


K
f  d µ ≤ µ(U). 
As µ is regular,
    µ(U) = sup

µ(E) : E⊆ U closed 

≤ sup





 


K
f  d µ : 0≤ f≤χU





≤ µ(U). 
Hence we have equality throughout. □

The next result tells that the variation coincides with the norm on real charges viewed as linear functionals on C(K).

Lemma 28 Let µ∈ M(K). Then
    ⎪⎪
⎪⎪
µ⎪⎪
⎪⎪
 = ⎪⎪
⎪⎪
φµ⎪⎪
⎪⎪
 := sup










 


K
f  d µ 




 : f∈ C(K), ⎪⎪
⎪⎪
f⎪⎪
⎪⎪
≤ 1 





. 
Proof. Let (A,B) be a Hahn decomposition (Thm. 36) for µ. For fC(K) with ||f||≤ 1, we have that
     





 


K
f  d µ 




≤ 




 


A
f  d µ 




+




 


B
f  d µ 




= 




 


A
f  d µ+




+




 


B
f  d µ




         
 
≤ 
 


A

f
  d µ+ + 
 


B

f
  d µ ≤ ⎪⎪
⎪⎪
f⎪⎪
⎪⎪

µ(A) − µ(B)
≤ ⎪⎪
⎪⎪
f⎪⎪
⎪⎪
⎪⎪
⎪⎪
µ⎪⎪
⎪⎪
, 
         
using the fact that µ(B)≤0 and that (A,B) is a partition of K.

Conversely, as µ is regular, for є>0, there exist closed sets E and F with EA, FB, and with µ+(E)> µ+(A)−є and µ(F)>µ(B)−є. By Urysohn Lemma 26, there exists f:K→[0,1] continuous with f(E)={1} and f(F)={0}. Let g=2f−1, so g is continuous, g takes values in [−1,1], and g(E)={1}, g(F)={−1}. Then

     
    
 


K
g  d µ
= 
 


E
 1  d µ + 
 


F
 −1  d µ + 
 


K∖ (E⋃ F)
g  d µ 
         
 
= µ(E) − µ(F) + 
 


A∖ E
g  d µ + 
 


B∖ F
g  d µ 
         
           

As EA, we have µ(E) = µ+(E), and as FB, we have −µ(F)=µ(F). So

     
 


K
g  d µ
> µ+(A)−є + µ(B) − є + 
 


A∖ E
g  d µ + 
 


B∖ F
g  d µ 
         
 
≥ 
µ(A) 
 + 
µ(B) 
 − 2є − 
µ(A∖ E) 
 − 
µ(B∖ F) 
         
 
≥ 
µ(A) 
 + 
µ(B) 
 − 4є. 
         

As є>0 was arbitrary, we see that ||φµ|| ≥ | µ(A) |+| µ(B) |=||µ||. □

Thus, we know that M(K) is isometrically embedded in C(K)*.

14.4 Riesz Representation Theorem

To facilitate an approach to the key point of this Subsection we will require some more definitions.

Definition 29 A functional F on C(K) is positive if for any non-negative function f≥ 0 we have F(f)≥0.
Lemma 30 Any positive linear functional F on C(X) is continuous and ||F||=F(1), where 1 is the function identically equal to 1 on X.
Proof. For any function f such that ||f||≤ 1 the function 1−f is non negative thus: F(1)−F(f)=F(1−f)>0, Thus F(1)>F(f), that is F is bounded and its norm is F(1). □

So for a positive functional you know the exact place where to spot its norm, while a linear functional can attain its norm in an generic point (if any) of the unit ball in C(X). It is also remarkable that any bounded linear functional can be represented by a pair of positive ones.

Lemma 31 Let λ be a continuous linear functional on C(X). Then there are positive functionals λ+ and λ on C(X), such that λ=λ+−λ.
Proof. First, for fC(K) with f≥0, we define
     
    λ+(f)
= sup

λ(g) : g∈ C(K), 0≤ g≤ f

≥0, 
         
λ(f)
= λ+(f) − λ(f) = sup

λ(g)−λ(f): g∈ C(K), 0≤ g≤ f

         
 
= sup

λ(h): h∈ C(K), 0≤ h+f≤ f

         
 
= sup

λ(h): h∈ C(K), −f ≤ h ≤ 0 

≥ 0. 
         
In a sense, this is similar to the Hahn decomposition (Thm. 36).

We can check that

    λ+(tf) = tλ+(f),   λ(tf) = tλ(f)    (t≥0, f≥0). 

For f1,f2≥ 0, we have that

     
λ+(f1+f2)
= sup

λ(g): 0≤ g ≤ f1+f2

         
 
= sup

λ(g1+g2): 0≤ g1+g2 ≤ f1+f2

         
 
≥ sup

λ(g1) + λ(g2): 0≤ g1f1, 0 ≤ g2 ≤ f2

         
 = λ+(f1) + λ+(f2).           

Conversely, if 0≤ gf1+f2, then set g1 = min(g,f1), so 0≤ g1f1. Let g2 = gg1 so g1g implies that 0≤ g2. For xK, if g1(x)=g(x) then g2(x) = 0 ≤ f2(x); if g1(x)=f1(x) then f1(x)≤ g(x) and so g2(x) = g(x)−f1(x) ≤ f2(x). So 0 ≤ g2f2, and g = g1 + g2. So in the above displayed equation, we really have equality throughout, and so λ+(f1+f2) = λ+(f1) + λ+(f2). As λ is additive, it is now immediate that λ(f1+f2) = λ(f1) + λ(f2)

For fC(K) we put f+(x)=max(f(x),0) and f(x)=−min(f(x),0). Then f±≥ 0 and f=f+f. We define:

    λ+(f) = λ+(f+) − λ+(f),   λ(f) = λ(f+) − λ(f). 

As when we were dealing with integration, we can check that λ+ and λ become linear functionals; by the previous Lemma they are bounded. □

Finally, we need a technical definition.

Definition 32 For fC(K), we define the support of f, written supp(f), to be the closure of the set {xK : f(x)≠0}.
Theorem 33 (Riesz Representation) Let K be a compact (Hausdorff) space, and let λ∈ CK(K)*. There exists a unique µ∈ MK(K) such that
    λ(f) = 
 


K
f  d µ    ( f∈ CK(K) ). 
Furthermore, ||λ|| = ||µ||.
Proof. Let us show uniqueness. If µ12MK(K) both induce λ then µ = µ1−µ2 induces the zero functional on CK(K). So for fC(K),
     
0
= ℜ 
 


K
f  d µ = 
 


K
f  d µr
         
 
= ℑ
 


K
f  d µ = 
 


K
f  d µi. 
         
So µr and µi both induce the zero functional on C(K). By Lemma 28, this means that ||µr|| = ||µi||=0, showing that µ = µr + iµi = 0, as required.

Existence is harder, and we shall only sketch it here. Firstly, we shall suppose that K=ℝ and that λ is positive.

Motivated by the above Lemmas 27 and 28, for UK open, we define

    µ*(U) = sup

λ(f):  f∈ C(K),       0≤ f≤χU,  supp(f)⊆ U

. 

For AK general, we define

    µ*(A) = inf

µ*(U):  U⊆ K  is open,       A⊆ U

. 

We then proceed to show that

If λ is not positive, then by Lemma 31 represent it as λ=λ+−λ for positive λ±. As λ+ and λ are positive functionals, we can find µ+ and µ positive measures in M(K) such that

    λ+(f) = 
 


K
f  d µ+,   λ(f) = 
 


K
f  d µ    (f∈ C(K)). 

Then if µ = µ+ − µ, we see that

    λ(f) = λ+(f) − λ(f) = 
 


K
f  d µ    (f∈ C(K)). 

Finally, if K=ℂ, then we use the same “complexification” trick from the proof of the Hahn-Banach Theorem 15. Namely, let λ∈ C(K)*, and define λr, λiC(K)* by

    λr(f) = ℜ  λ(f),   λi(f) = ℑ λ(f)    ( f∈ C(K) ). 

These are both clearly ℝ-linear. Notice also that | λr(f) | = | ℜ λ(f) | ≤ | λ(f) | ≤ ||λ|| ||f||, so λr is bounded; similarly λi.

By the real version of the Riesz Representation Theorem, there exist charges µr and µi such that

    ℜ λ(f) = λr(f) = 
 


K
f  d µr,   ℑλ(f) = λi(f) = 
 


K
f  d µi    (f∈ C(K) ). 

Then let µ=µr+iµi, so for fC(K),

     
 


K
f  d µ
= 
 


K
f  d µr + i
 


K
f  d µi
         
 
= 
 


K
 ℜ (f)  d µr + i
 


K
 ℑ(f)  d µr  + i
 


K
 ℜ (f)  d µi − 
 


K
 ℑ(f)  d µi
         
 = λr(ℜ (f)) + iλr(ℑ(f)) + iλi(ℜ (f)) − λi(ℑ(f))          
 = ℜ  λ(ℜ (f)) + iℜ λ(ℑ(f)) +iℑλ(ℜ (f)) − ℑλ(ℑ(f))          
 = λ(ℜ (f) + iℑ(f)) = λ(f),          

as required. □

Notice that we have not currently proved that ||µ|| = ||λ|| in the case K=ℂ. See a textbook for this.

15 Fourier Transform

In this section we will briefly present a theory of Fourier transform focusing on commutative group approach. We mainly follow footsteps of []*Ch. IV.

15.1 Convolutions on Commutative Groups

Let G be a commutative group, we will use + sign to denote group operation, respectively the inverse elements of gG will be denoted −g. We assume that G has a Hausdorff topology such that operations (g1,g2)↦ g1+g2 and g↦ −g are continuous maps. We also assume that the topology is locally compact, that is the group neutral element has a neighbourhood with a compact closure.

Example 1 Our main examples will be as follows:
  1. G=ℤ the group of integers with operation of addition and the discrete topology (each point is an open set).
  2. G=ℝ the group of real numbers with addition and the topology defined by open intervals.
  3. G=T the group of Euclidean rotations the unit circle in 2 with the natural topology. Another realisations of the same group:
    • Unimodular complex numbers under multiplication.
    • Factor group ℝ/ℤ, that is addition of real numbers modulo 1.
    There is a homomorphism between two realisations given by z=ei t, t∈[0,1), | z |=1.

We assume that G has a regular Borel measure which is invariant in the following sense.

Definition 2 Let µ be a measure on a commutative group G, µ is called invariant (or Haar measure) if for any measurable X and any gG the sets g+X and X are also measurable and µ(X)=µ(g+X)=µ(−X).

Such an invariant measure exists if and only if the group is locally compact, in this case the measure is uniquely defined up to the constant factor.

Exercise 3 Check that in the above three cases invariant measures are:
Definition 4 A convolution of two functions on a commutative group G with an invariant measure µ is defined by:
(f1*f2)(x)=
 


G
f1(xy) f2(y) d µ(y)=
 


G
f1(y) f2(xy) d µ(y). (83)
Theorem 5 If f1, f2L1(G,µ), then the integrals in (83) exist for almost every xG, the function f1*f2 is in L1(G,µ) and ||f1*f2||≤ ||f1||·||f2||.
Proof. If f1, f2L1(G,µ) then by Fubini’s Thm. 50 the function φ(x,y)=f1(x)*f2(y) is in L1(G× G, µ× µ) and ||φ||=||f1||· ||f2||.

Let us define a map τ: G× GG× G such that τ(x,y)=(x+y,y). It is measurable (send Borel sets to Borel sets) and preserves the measure µ×µ. Indeed, for an elementary set C=A× BG× G we have:

    (µ×µ)(τ(C))=
 


G× G
χτ(C)(x,y) d µ(x) d µ(y)
 =
 


G× G
 χC(xy,y) d µ(x) d µ(y)
 =
 


G





 


G
 χC(xy,y) d µ(x)




dµ(y)
 =
 


B
 µ(A+y) d µ(y)=µ(A)× µ(B)=(µ×µ)(C).

We used invariance of µ and Fubini’s Thm. 50. Therefore we have an isometric isomorphism of L1(G× G,µ× µ) into itself by the formula:

    Tφ(x,y)=φ(τ(x,y))=φ(xy,y).

If we apply this isomorphism to the above function φ(x,y)=f1(x)*f2(y) we shall obtain the statement. □

Definition 6 Denote by S(k) the map S(k): fk*f which we will call convolution operator with the kernel k.
Corollary 7 If kL1(G) then the convolution S(k) is a bounded linear operator on L1(G).
Theorem 8 Convolution is a commutative, associative and distributive operation. In particular S(f1)S(f2)=S(f2)S(f1)=S(f1*f2).
Proof. Direct calculation using change of variables. □

It follows from Thm. 5 that convolution is a closed operation on L1(G) and has nice properties due to Thm. 8. We fix this in the following definition.

Definition 9L1(G) equipped with the operation of convolution is called convolution algebra L1(G).

The following operators of special interest.

Definition 10 An operator of shift T(a) acts on functions by T(a): f(x)↦ f(x+a).
Lemma 11 An operator of shift is an isometry of Lp(G), 1≤ p≤∞.
Theorem 12 Operators of shifts and convolutions commute:
    T(a)(f1*f2)=T(a)f1*f2=f1*T(a)f2,
or
    T(a)S(f)=S(f)T(a)=S(T(a)f).
Proof. Just another calculation with a change of variables. □
Remark 13 Note that operator of shifts T(a) provide a representation of the group G by linear isometric operators in Lp(G), 1≤ p≤ ∞. A map fS(f) is a representation of the convolution algebra

There is a useful relation between support of functions and their convolutions.

Lemma 14 For any f1, f2L1(G) we have:
    supp(f1*f2)⊂supp(f1)+supp(f2).
Proof. If xsupp(f1)+supp(f2) then for any ysupp(f2) we have xysupp(f1). Thus for such x convolution is the integral of the identical zero. □
Exercise 15 Suppose that the function f1 is compactly supported and k times continuously differentiate in , and that the function f2 belongs to L1(ℝ). Prove that the convolution f1* f2 has continuous derivatives up to order k.
[
Hint: Express the derivative d/d x as the limit of operators (T(h)−I)/h when h→ 0 and use Thm. 12.]

15.2 Characters of Commutative Groups

Our purpose is to map the commutative algebra of convolutions to a commutative algebra of functions with point-wise multiplication. To this end we first represent elements of the group as operators of multiplication.

Definition 16 A character χ: G→ T is a continuous homomorphism of an abelian topological group G to the group T of unimodular complex numbers under multiplications:
    χ(x+y)=χ(x)χ(y).

Note, that a character is an eigenfunction for a shift operator T(a) with the eigenvalue χ(a). Furthermore, if a function f on G is an eigenfunction for all shift operators T(a), aG then the collection of respective eigenvalues λ(a) is a homomorphism of G to ℂ and f(a)=α λ(a) for some α∈ℂ. Moreover, if T(a) act by isometries on the space containing f(a) then λ(a) is a homomorphism to T.

Lemma 17 The product of two characters of a group is again a character of the group. If χ is a character of G then χ−1=χ is a character as well.
Proof. Let χ1 and χ2 be characters of G. Then:
    χ1(gh2(gh)=χ1(g1(h2(g2(h)
 =    (χ1(g2(g))(χ1(h2(h))∈T.
Definition 18 The dual group Ĝ is collection of all characters of G with operation of multiplication.

The dual group becomes a topological group with the uniform convergence on compacts: for any compact subset KG and any ε>0 there is N∈ℕ such that | χn(x)−χ(x) |<ε for all xK and n>N.

Exercise 19 Check that
  1. The sequence fn(x)=xn does not converge uniformly on compacts if considered on [0,1]. However it does converges uniformly on compacts if considered on (0,1).
  2. If X is a compact set then the topology of uniform convergence on compacts and the topology uniform convergence on X coincide.
Example 20 If G=ℤ then any character χ is defined by its values χ(1) since
χ(n)=[χ(1)]n. (84)
Since χ(1) can be any number on T we see that $ℤ^_$ is parametrised by T.
Theorem 21 The group $ℤ^_$ is isomorphic to T.
Proof. The correspondence from the above example is a group homomorphism. Indeed if χz is the character with χz(1)=z, then χz1χz2z1 z2. Since ℤ is discrete, every compact consists of a finite number of points, thus uniform convergence on compacts means point-wise convergence. The equation (84) shows that χzn→ χz if and only if χzn(1)→ χz(1), that is znz. □
Theorem 22 The group $T^_$ is isomorphic to .
Proof. For every n∈ℤ define a character of T by the identity
χn(z)=zn,   z∈T. (85)
We will show that these are the only characters in Cor. 26. The isomorphism property is easy to establish. The topological isomorphism follows from discreteness of $T^_$. Indeed due to compactness of T for nm:
    
 
max
z∈T

χn(z)−χm(z) 
2=
 
max
z∈T

1−ℜ zmn
2=22=4.
Thus, any convergent sequence (nk) have to be constant for sufficiently large k, that corresponds to a discrete topology on ℤ. □

The two last Theorem are an illustration to the following general statement.

Principle 23 (Pontryagin’s duality) For any locally compact commutative topological group G the natural map G→ Ĝ, such that it maps gG to a character fg on Ĝ by the formula:
fg(χ)=χ(g),    χ∈Ĝ, (86)
is an isomorphism of topological groups.
Remark 24
  1. The principle is not true for commutative group which are not locally compact.
  2. Note the similarity with an embedding of a vector space into the second dual.

In particular, the Pontryagin’s duality tells that the collection of all characters contains enough information to rebuild the initial group.

Theorem 25 The group $ℝ^_$ is isomorphic to .
Proof. For λ∈ℝ define a character χλ∈$ℝ^_$ by the identity
χλ(x)=e2π i λ x,    x∈ℝ. (87)
Moreover any smooth character of the group G=(ℝ, +) has the form (87). Indeed, let χ be a smooth character of ℝ. Put c=χ′(t)|t=0∈ ℂ. Then χ′(t)=cχ(t) and χ(t)=ect. We also get ciℝ and any such c defines a character. Then the multiplication of characters is: χ1(t2(t)=ec1tec2t=e(c2+c1)t. So we have a group isomorphism.

For a generic character we can apply first the smoothing technique and reduce to the above case.

Let us show topological homeomorphism. If λn→ λ then χλn→ χλ uniformly on any compact in ℝ from the explicit formula of the character. Reverse, let χλn→ χλ uniformly on any interval. Then χλn−λ(x)→ 1 uniformly on any compact, in particular, on [0,1]. But

      
 
sup
[0,1]

χλn−λ(x)− 1 
=
      
 
sup
[0,1]

sinπ (λn−λ)x
 =




          1,
if  
λn−λ 
≥ 1/2,
          sinπ 
λn−λ 
,
if  
λn−λ 
≤ 1/2.

Thus λn→ λ. □

Corollary 26 Any character of the group T has the form (85).
Proof. Let χ∈$T^_$, consider χ1(t)=χ(ei t) which is a character of ℝ. Thus χ1(t)=ei λ t for some λ∈ℝ. Since χ1(1)=1 then λ=n∈ℤ. Thus χ1(t)=ei n t, that is χ(z)=zn for z=ei t. □
Remark 27 Although $ℝ^_$ is isomorphic to there is no a canonical form for this isomorphism (unlike for ℝ→ $ℝ^_$). Our choice is convenient for the Poisson formula below, however some other popular definitions are λ→ ei λ x or λ→ ei λ x.

We can unify the previous three Theorem into the following statement.

Theorem 28 Let G=ℝn× ℤk× Tl be the direct product of groups. Then the dual group is Ĝ=ℝn× Tk× ℤl.

15.3 Fourier Transform on Commutative Groups

Definition 29 Let G be a locally compact commutative group with an invariant measure µ. For any fL1(G) define the Fourier transform f by
f(χ)=
 


G
f(x) χ(x) d µ(x),   χ∈Ĝ. (88)

That is the Fourier transform f is a function on the dual group Ĝ.

Example 30
  1. If G=ℤ, then fL1(Z) is a two-sided summable sequence (cn)n∈ℤ. Its Fourier transform is the function f(z)=∑n=−∞cn zn on T. Sometimes f(z) is called generating function of the sequence (cn).
  2. If G=T, then the Fourier transform of fL1(T) is its Fourier coefficients, see Section 5.1.
  3. If G=ℝ, the Fourier transform is also the function on given by the Fourier integral:
    f(λ)=
     


    f(x)  e−2πiλ xdx. (89)

The important properties of the Fourier transform are captured in the following statement.

Theorem 31 Let G be a locally compact commutative group with an invariant measure µ. The Fourier transform maps functions from L1(G) to continuous bounded functions on Ĝ. Moreover, a convolution is transformed to point-wise multiplication:
(f1*f2)^ (χ)=f1(χ)·f2(χ), (90)
a shift operator T(a), aG is transformed in multiplication by the character fa∈Ĝ:
(T(a)f)^ (χ)=fa(χ)·f(χ),   fa(χ)=χ(a) (91)
and multiplication by a character χ∈Ĝ is transformed to the shift T−1):
(χ· f)^ (χ1)=T−1)f(χ1)=f(χ−1χ1). (92)
Proof. Let fL1(G). For any ε>0 there is a compact KG such that ∫GK | f | d µ<ε. If χn→ χ in Ĝ, then we have the uniform convergence of χn→ χ on K, so there is n(ε) such that for k>n(ε) we have | χk(x)−χ(x) | <ε for all xK. Then
    
f(χn)−f(χ) 
 


K

f(x) 

χn(x)− χ(x) 
d µ(x)+
 


G∖ K

f(x) 

χn(x)− χ(x) 
d µ(x)
 
ε⎪⎪
⎪⎪
f⎪⎪
⎪⎪
+2ε.
Thus f is continuous. Its boundedness follows from the integral estimations. Algebraic maps (90)–(92) can be obtained by changes of variables under integration. For example, using Fubini’s Thm. 50 and invariance of the measure:
    (f1*f2)^ (χ )=
 


G
 


G
f1(s) f2(ts) dsχ(t) dt
 =
 


G
 


G
f1(s) χ(s)f2(ts) χ(ts) dsdt
 =f1(χ)f2(χ).

15.4 The Schwartz space of smooth rapidly decreasing functions

We say that a function f is rapidly decreasing if limx→ ±∞ | xkf(x) |=0 for any k∈ℕ.

Definition 32 The Schwartz space denoted by S or space of rapidly decreasing functions on Rn is the space of infinitely differentiable functions such that:
S = 



f∈ C∞ (ℝ):  
 
sup
x∈ ℝ

xα f(β) (x) 
<∞   ∀ α ,β ∈ ℕ



. (93)
Example 33 An example of a rapidly decreasing function is the Gaussian e−π x2.

It is worth to notice that SLp(ℝ) for any 1<p<∞. Moreover, S is dense in Lp(ℝ), for p=1 this can be shown in the following steps (other values of p can be done similarly but require some more care). First we will show that S is an ideal of the convolution algebra L1(ℝ).

Exercise 34 For any gS and fL1(ℝ) with compact support their convolution f*g belongs to S. [Hint: smoothness follows from Ex. 15.]

Define the family of functions gt(x) for t>0 in S by scaling the Gaussian:

  gt(x) = 
1
t
e−π (x/t)2.
Exercise 35 Show that gt(x) satisfies the following properties, cf. Lem 7:
  1. gt(x)>0 for all x∈ℝ and t>0.
  2. gt(x) d x=1 for all t>0. [Hint: use the table integral e−π x2d x=1.]
  3. For any ε>0 and any δ>0 there exists T>0 such that for all positive t< T we have:
          0< 
    −δ
    −∞
    +
    δ
    gt(x) dx < ε.

It is easy to see, that the above properties 13 are not unique to the Gaussian and a wide class have them. Such a family a family of functions is known as approximation of the identity [] due to the next property (94).

Exercise 36
  1. Let f be a continuous function with compact support, then
     
    lim
    t→ 0
    ⎪⎪
    ⎪⎪
    fgt*f⎪⎪
    ⎪⎪
    1=0 . (94)
    [Hint: use the proof of Thm. 8.]
  2. The Schwartz space S is dense in L1(ℝ). [Hint: use Prop. 20, Ex. 34 and (94).]

15.5 Fourier Integral

We recall the formula (89):

Definition 37 We define the Fourier integral of a function fL1(ℝ) by
f(λ)=
 


f(x) e−2π i λ xdx. (95)

We already know that f is a bounded continuous function on ℝ, a further property is:

Lemma 38 If a sequence of functions (fn)⊂L1(ℝ) converges in the metric L1(ℝ), then the sequence (fn) converges uniformly on the real line.
Proof. This follows from the estimation:
    
fn(λ)−fm(λ) 
 



fn(x)−fm(x) 
dx.
Lemma 39 The Fourier integral f of fL1(ℝ) has zero limits at −∞ and +∞.
Proof. Take f the indicator function of [a,b]. Then f(λ )=1/−2π i λ (e−2π i ae−2π i b), λ≠ 0. Thus limλ→ ±∞ f(λ)=0. By continuity from the previous Lemma this can be extended to the closure of step functions, which is the space L1(ℝ) by Lem. 17. □
Lemma 40 If f is absolutely continuous on every interval and f′∈L1(ℝ), then
    (f′)^=2πi λ f.
More generally:
(f(k))^=(2πi λ)kf. (96)
Proof. A direct demonstration is based on integration by parts, which is possible because assumption in the Lemma.

It may be also interesting to mention that the operation of differentiation D can be expressed through the shift operatot Ta:

D=
 
lim
Δ t → 0
TΔ t− I
Δ t
. (97)

By the formula (91), the Fourier integral transforms 1/Δ t(TΔ tI) into 1/Δ tλt)− 1). Providing we can justify that the Fourier integral commutes with the limit, the last operation is multiplication by χ′λ(0)=2πi λ. □

Corollary 41 If f(k)L1(ℝ) then
    
f 
=

(f(k))^ 

2π λ  
k
→ 0    as  λ → ∞,
that is f decrease at infinity faster than | λ |k.
Lemma 42 Let f(x) and xf(x) are both in L1(ℝ), then f is differentiable and
    f′=(−2 π ixf)^.
More generally
f(k)=((−2π ix)kf)^. (98)
Proof. There are several strategies to prove this results, all having their own merits:
  1. The most straightforward uses the differentiation under the integration sign.
  2. We can use the intertwining property (92) of the Fourier integral and the connection of derivative with shifts (97).
  3. Using the inverse Fourier integral (see below), we regard this Lemma as the dual to the Lemma 40.
Corollary 43 The Fourier transform of a smooth rapidly decreasing function is a smooth rapidly decreasing function.
Corollary 44 The Fourier integral of the Gaussian e−π x2 is e−π λ 2.
Proof.[] Note that the Gaussian g(x)=e−π x2 is a unique (up to a factor) solution of the equation g′+2π x g=0. Then, by Lemmas 40 and 42, its Fourier transform shall satisfy to the equation 2πi λ ĝ+ iĝ′=0. Thus, ĝ=c· e−π λ 2 with a constant factor c, its value 1 can be found from the classical integral ∫ e−π x2d x=1 which represents ĝ(0). □

The relation (96) and (98) allows to reduce many partial differential equations to algebraic one, see § 0.2 and 5.4. To convert solutions of algebraic equations into required differential equations we need the inverse of the Fourier transform.

Definition 45 We define the inverse Fourier transform on L1(ℝ):
f(λ)=
 


f(x) e2π i λ xdx.  (99)

We can notice the formal correspondence f(λ)=f(−λ)=f(λ), which is a manifestation of the group duality $ℝ^_$=ℝ for the real line. This immediately generates analogous results from Lem. 38 to Cor. 44 for the inverse Fourier transform.

Theorem 46 The Fourier integral and the inverse Fourier transform are inverse maps. That is, if g=f then f.
Proof.[Sketch of a proof] The exact meaning of the statement depends from the spaces which we consider as the domain and the range. Various variants and their proofs can be found in the literature. For example, in [, § IV.2.3], it is proven for the Schwartz space S of smooth rapidly decreasing functions.

The outline of the proof is as follows. Using the intertwining relations (96) and (98), we conclude the composition of Fourier integral and the inverse Fourier transform commutes both with operator of multiplication by x and differentiation. Then we need a result, that any operator commuting with multiplication by x is an operator of multiplication by a function f. For this function, the commutation with differentiation implies f′=0, that is f=const. The value of this constant can be evaluated by a Fourier transform on a single function, say the Gaussian e−π x2 from Cor. 44. □

The above Theorem states that the Fourier integral is an invertible map. For the Hilbert space L2(ℝ) we can show a stronger property—its unitarity.

Theorem 47 (Plancherel identity) The Fourier transform extends uniquely to a unitary map L2(ℝ)→ L2(ℝ):
 



f
2dx=
 



f 
2d λ. (100)
Proof. The proof will be done in three steps: first we establish the identity for smooth rapidly decreasing functions, then for L2 functions with compact support and finally for any L2 function.
  1. Take f1 and f2S be smooth rapidly decreasing functions and g1 and g2 be their Fourier transform. Then (using Fubini’s Thm. 50):
          
     


    f1(t) f2(t) dt
    =
          
     


       
     


    g1(λ )  e2π i λ td λ   f2(t) dt
     =
     


       g1(λ )
     


       e2π i λ t   f2(t) dtd λ
     =
     


       g1(λ )  ḡ2(λ ) d λ
    Put f1=f2=f (and therefore g1=g2=f) we get the identity ∫| f |2d x=∫| f |2d λ.

    The same identity (100) can be obtained from the property (f1f2)^=f1*f2, cf. (90), or explicitly:

          
     


    f1(x) f2(x) e−2π iλ xdx=
     


    f1(t) f2(λ−t) dt.

    Now, substitute λ=0 and f2=f1 (with its corollary f2(t)=f1(−t)) and obtain (100).

  2. Next let fL2(ℝ) with a support in (−a,a) then fL1(ℝ) as well, thus the Fourier transform is well-defined. Let fnS be a sequence with support on (−a,a) which converges to f in L2 and thus in L1. The Fourier transform gn converges to g uniformly and is a Cauchy sequence in L2 due to the above identity. Thus gng in L2 and we can extend the Plancherel identity by continuity to L2 functions with compact support.
  3. The final bit is done for a general fL2 the sequence
          fn(x)=



              f(x),
    if  
    x
    <n,
              0, otherwise;
    of truncations to the interval (−n,n). For fn the Plancherel identity is established above, and fnf in L2(ℝ). We also build their Fourier images gn and see that this is a Cauchy sequence in L2(ℝ), so gng.
If fL1L2 then the above g coincides with the ordinary Fourier transform on L1. □

We note that Plancherel identity and the Parseval’s identity (30) are cousins—they both states that the Fourier transform L2(G)→ L2(Ĝ) is an isometry for G=ℝ and G=T respectively. They may be combined to state the unitarity of the Fourier transform on L2(G) for the group G=ℝn× ℤk× Tl cf. Thm. 28.

Proofs of the following statements are not examinable Thms. 23, 36, 53, 33, 46, Props. 14, 20.

16 Advances of Metric Spaces

16.1 The Stone–Weierstrass Theorem

Density in metric spaces is an important concept since it allows to approximate any element by an element from the dense set. Furthermore, we can extend a uniformly continuous function from a dense subset by continuity, see Ex. 62. Thus, it is convenient to have some supply of manageable dense subsets of common metric spaces.

A famous case of density is the Theorem of Stone–Weierstrass, which in the original and most known form says that any continuous function on a compact interval can be uniformly approximated by a sequence of polynomials. Polynomials have many nice properties which make this dense subset particularly useful: easy computation, derivation and integration, etc. Yet, we will prove here a more general version of the Stone–Weierstrass theorem that applies to general compact metric spaces.

Theorem 1 (Stone–Weierstrass) Suppose that X is a compact metric space and let C(X,ℝ) be the Banach space of real valued continuous functions on X with norm || · ||. Suppose that AC(X,ℝ) is a unital subalgebra of C(X,ℝ), i.e. Suppose furthermore that A separates points, i.e. for any two x,yX with xy there exists a function fA such that f(x) ≠ f(y). Then, A is dense in C(X,ℝ).

This is an interestingly sounding theorem: it states that a subset of C(X,ℝ) which is closed under algebraic operations and separates points automatically has topological property—it is dense. Its consequences are striking. Before we prove this theorem let’s look at some of them.

Corollary 2[Weierstrass approximation theorem] The space of polynomials ℝ[x] is dense in C([a,b],ℝ) for any compact interval [a,b] in the in the || · || norm.

In other words, any continuous function can be approximated with arbitrary accuracy by a polynomial.

Corollary 3 The space of polynomials ℝ[x1,…,xn] is dense in C(K,ℝ) for any compact subset K of n in the || · || norm.

This is the higher dimensional version of the above theorem and states that a continuous functions of n-variables can be approximated by polynomials in n variables.

Corollary 4 Let C(S1,ℝ) be the space of continuous functions on the unit circle, or, equivalently, the space of 2 π-periodic real valued functions on . Then the finite linear span of the set
  
 
m ∈ ℕ
 {1, sin(mx), cos(mx)}
is dense in C(S1,ℝ).

The Stone–Weierstrass theorem is actually a consequence of the following theorem by Stone. This is a good illustration to inventor’s paradox stated by Polya [].

Here is some notation first for two functions f and g:

f ∧ g =  min{f,g},
f ∨ g = max{f,g}

Note that of f and g are continuous, then so are fg and fg (demonstrate this!).

Theorem 5 (Stone’s Theorem) Let X be a compact metric space and suppose that there is a subset A of C(X,ℝ) such that Then, A is dense in C(X,ℝ) in the topology induced by the norm || · || (the uniform topology).
Proof. We need to prove that any function g can be approximated by elements in A. For each two points x,y choose a function fx,yA such that fx,y(x)=g(x) and fx,y(y)=g(y). Such a function exists by our hypothesis for every pair of points. Now, for an ε>0 the sets
  Ox,y={z ∈ X ∣  fx,y(z) < g(z) + ε}
are open and form a cover of X even if we fix x. This is because Ox,y contains both x and y together with some neighbourhoods of these points. Now find a finite subcover for each fixed x. That is there are finitely many points y1, …, yn such that Ox,yi is an open cover. Now define the function
  fx= fx,y1 ∧ … ∧ fx,yn.
By hypothesis fx is in A for any xX and it has the property that
  fx(z) < g(z) + ε
but now for all z. Moreover, fx(x) = g(x). Again, the sets
  Ox={z ∈ X ∣  fx(z) > g(z) − ε}
make an open cover and therefore there is a finite subcover. This means there are finitely many points x1,…,xk such that Oxi is an open cover of X. Now the function
  f = fx1 ∨ … ∨ fxk
is in A and satisfies
  g(x)−ε < f(x) < g(x) + ε
for all x, or in other words || fg || < ε. □

It may be not obvious why conditions of Stone’s theorem 5 are more general than in Thm. 1. This will be seen from the following proof. We employ what Polya called leading particular case [, § 4.4]—we will show that the particular algebra of polynomials on [0,1] approximate the particular function √x and then reduce the general situation to it.

Proof.[Sketch of proof of the Stone Weierstrass theorem 1.] First we observe that if B is the closure in C(X,ℝ) of A from Thm. 1, then B will also be a unital point separating subalgebra of C(X,ℝ) (exercise!).

Step 1: If f is non-negative and in B, so is √f. To see this note that it is enough to show this for 0≤ f < 1 because in case f ≠0 we can compute √f by

  
f
 =
2
|| f||
1
2 || f ||
f
.

Now the Taylor series ∑k=0an xk for √1−x

1−x
=  
k=0
ak  xk = 1−
1
2
x
1
8
x2
1
16
x3
5
128
x4
7
256
x5−…

converges absolutely and uniformly on any interval [0,1−δ). Therefore, the series

  
k=0
ak (1−f−δ)k

converges in the Banach space B because all the partial sums are actually in B as B is a subalgebra. The limit of this sequence is, of course, √f. If we let δ go to zero, we can see that also √fB. This works because

  | 
f(x) + δ
 − 
f(x)
 | = δ ( 
f(x) + δ
 + 
f(x)
)−1 ≤ 
δ
  

so that the approximation is uniform.

Step 2: Since | f | = √f2 we have that fB implies | f | ∈ B. Since moreover,

   f ∧ g = 
f + g
2
 − 
| f −g |
2
,
   f ∨ g = 
f + g
2
 + 
| f −g |
2

we conclude from this that B is closed under the operations ∧ and ∨.

Step 3: Assume that xy are points in X and assume that a,b are real numbers. Then, by assumption there is an element f in B such that

   f(x) ≠ f(y).

Since B is a subspace that contains the constant functions, the function

     
    g(z)
= a + (ba)
f(x)−f(z)
f(x)−f(y)
         
 
= 
bf(x)−af(y)
f(x)−f(y)
 −
ba
f(x)−f(y)
f(z)
         

is also in B and it satisfies g(x)=a and g(y)=b.

Final Step: As we can see all the conditions of Stone’s theorem are satisfied and therefore B is dense in C(X,ℝ). Since B is closed in C(X,ℝ) this means that B=C(X,ℝ). Thus, A is dense in C(X,ℝ). □

We can extend the result from real scalars to complex one through identities ℜ z = 1/2(z +z) and ℑ z = 1/2(zz).

Corollary 6[Stone–Weierstrass (complex version)] Suppose that X is a compact metric space and let C(X,ℂ) be the complex Banach space of complex valued continuous functions on X with norm || · ||. Suppose that AC(X,ℂ) is a unital *-subalgebra of C(X,ℂ), i.e. Suppose furthermore that A separates points, i.e. for any two x,yX with xy there exists a function fA such that f(x) ≠ f(y). Then, A is dense in C(X,ℂ).

Using this complex version of the theorem (or simply the Euler identity ei φ = cosφ +i sinφ) we obtain the complex version of Cor. 4:

Corollary 7 The linear span of the set { ei m ϕm ∈ ℤ} is dense in C(S1,ℂ).

Note that we need both positive and negative values of m in ei m ϕ, the set { ei m ϕm ∈ ℕ0} is not dense in C(S1,ℂ).

The Stone–Weierstrass also says something about the separability of certain Banach spaces. Remember what it means for a topological space to be separable.

Definition 8 (Separable Metric Space) A metric space X is called separable if there exists a countable dense subset of X.

In other words, a separable metric space consists of accumulation points of a single sequence.

Suppose K ⊂ ℝn is compact. Then the space of polynomials with real coefficients and n-variables is dense in the space of continuous functions C(K). Of course every polynomial with real coefficients may be approximated by one with rational coefficients. Thus the set of rational polynomials ℚ[x1,…,xn] is dense in C(K). However, the space of rational polynomials is a countable set. In this way one obtains

Corollary 9 Let K be a compact subset of n then the Banach space C(K) is separable.

The following statement shows that continuous functions make only a tiny fraction of all bounded functions:

Exercise 10 Let X be an infinite set, show that the space B(X) of bounded functions on X is not separable. (Hint: present a set of disjoint balls of radius 1/2 parametrised by all real numbers.)

16.2 Contraction mappings and fixed point theorems

16.2.1 The Banach fixed point theorem

An important tool in numerical Analysis, but also in constructions of solutions of differential equations are fixed point approximations. In order to understand this, suppose that (X,d) is a metric space and f: XX a self-map. Then a point xX is called fixed point of f if f(x) = x. For example the function cos defines a self-map on the interval [0,1], and by starting with x1=0 and inductively computing xn+1 = cosxn one converges to the value roughly 0.739085 which is a fixed point of cos, i.e. solves the equation cos(x) = x. Under certain conditions one can show that such sequences always converge to a fixed point. This is the statement of the Banach fixed point theorem (contraction mapping principle).

Definition 11 (Contraction Mapping) Let (X,d) be a metric space. Then a map f: XX is called contraction if there exists a constant C<1 such that
  d(f(x),f(y)) ≤ Cd(x,y).

Note that any contraction is (uniformly) continuous.

Theorem 12 (Banach Fixed Point Theorem) Suppose that f: XX is a contraction on a complete metric space (X,d). Then f has a unique fixed point y. Moreover, for any xX the sequence (xn) defined recursively by
  xn+1 = f(xn),   x1 = x,
converges to y.
Proof. Let us start with uniqueness. If x,y are both fixed points in X, then since f is a contraction:
  d(x,y) ≤ Cd(x,y)
for some constant C<1. Hence, d(x,y)=0 and therefore x=y.

To prove the remaining claims we start with any x in X and we will show that the sequence xn defined by x1=x and xn+1=f(xn) converges. Since f is continuous the limit of (xn) must be a fixed point. Since (X,d) is complete we only need to show that (xn) is Cauchy. To see this note that

  d(xn+1,xn) ≤ Cd(xn,xn−1)

and therefore inductively,

  d(xn+1,xn) ≤ Cn−1d(x2,x1).

By the triangle inequality we have for any n,m>0

  d(xN+m,xN) ≤ (CN−1 + CN + … CN+m−2) d(x2,x1) ≤ CN−1
1
1−C
d(x2,x1) .

Since C<1 this can be made arbitrarily small by choosing N large enough. □

Corollary 13 Suppose that (X,d) is a complete metric space and f: XX a map such that fn is a contraction for some n ∈ ℕ. Then f has a unique fixed point.
Proof. Since fn is a contraction it has a unique fixed point xX, i.e.
   
 
 
f ∘ f … ∘ f


n−times
(x) =x.
Now note that
fn(f(x)) =  fn ∘ f (x) =   fn+1(x) = f∘ fn(x) = f(fn(x))=f(x)
and therefore f(x) is also a fixed point of fn. By uniqueness we must have f(x)=x. □

The question arises how to show that a given map f is a contraction. In subsets of ℝm there is a simple criterion. Recall that an open set U ⊂ ℝ is called convex if for any two points x,yU the line { t x + (1−t) yt ∈ [0,1] } is contained in U.

Theorem 14 (Mean Value Inequality) Suppose that U ⊂ ℝm is an open set with convex closure U and let f: U → ℝm be a C1-function. Let d f be the total derivative (or Jacobian) understood as a function on U with values in m × m-matrices. Suppose that || df(x) || ≤ M for all xU. Then f: U → ℝm satisfies
 || f(x) −f(y) || ≤ M || x − y ||
for all x,yU.
Proof. Given x,yU let γ(t) = t x + (1−t )y. Then d/dt γ(t) = xy.
  f(x) − f(y) = 
1
0
d
dt
f(γ(t)) dt = 
1
0
 (df) · 
d γ
dt
(t) dt.
Using the triangle inequality (this can be used for Riemann integrals too because these are limits of finite sums), one gets
  || f(x) −f (y) || ≤ 
1
0
 ||  (df) · 
d γ
dt
(t) || dt ≤ M
1
0
 || x − y || dt = M || xy||.
By continuity this inequality extends to U. □
Example 15 Consider the map f: ℝ2B1(0)B1(0),   (x,y) ↦ (x2/4+y/3+1/3,y2/4−x/2). Then
  df = 







x
2
1
3
1
2
y
2








.
The operator norm || df|| can be estimated by the Hilbert–Schmidt norm. Recall || A ||HS = (tr(A* A))1/2, so we get
  || df|| ≤ || df||HS = (
1
4
(x2+y2) + 
1
4
 + 
1
9
)1/2 <1.
Therefore f is a contraction. We can find the fixed point by starting, for example, with the point (0,0) and iterating. We get iterations:
(0,0), (0.333333, 0.), (0.361111, −0.166667), 
 (0.310378, −0.173611), (0.299547, −0.147654), 
 (0.306547, −0.144323), (0.308719, −0.148066),
 (0.307805, −0.148878), (0.307393, −0.148361),
 (0.307502, −0.148194), (0.307575, −0.148261), 
 (0.307564, −0.148292), (0.307551, −0.148284), 
(0.307552, −0.148279), (0.307554, −0.148279).
  
Example 16 Put a map of the country of your current presence on the floor, there’s a point on the map that is touching the actual point it refers to!

16.2.2 Applications of fixed point theory: The Picard-Lindelöf Theorem

Let f: K → ℝ be a function on a compact rectangle of the form K=[T1,T2] × [L1,L2] in ℝ2. Consider the initial value problem (IVP)

dy
dt
 = f(t,y),   y(t0) = y0, (101)

where y: [T1,T2] → ℝ, ty(t) is a function. The function f and the initial value y0 ∈ [L1,L2], and t0 ∈ [T1,T2] are given and we are looking for a function y satisfying the above equations.


  
  
Figure 19: Vector fields and their integral curves from Ex. 1720.

Example 17 Let f(t,x)=x and y0 =1, t0=0. Then the initial value problem is
  
dy
dt
 = y,   y(0) =1.
We know from other courses that there is a unique solution y(t) = et, see Fig. 19 top-left.
Example 18 Let f(t,x)=x2 and y0 =1, t0=0. Then the initial value problem is
  
dy
dt
 = y2,   y(0) =1.
We know from other courses that there is a unique solution y(t) = 1/1−t which exists only on the interval (−∞,1), see Fig. 19 top-right.
Example 19 Let f(t,x)=x2t and y0 =1, t0=0. Then the initial value problem is
  
dy
dt
 = y2t,   y(0) =1.
One can show that there exists a solution for small |t|, however this solution cannot be expressed in terms of elementary functions, see Fig. 19 bottom-left.
Example 20 Let f(t,x)=x2/3 and y0 =0, t0=0. Then the initial value problem is
  
dy
dt
 = y
2
3
 
,   y(0) =0.
It has at least two solutions, namely y=0 and y=t3/27, see Fig. 19 bottom-right.

Hence, there are two fundamental questions here: existence and uniqueness of solutions. The following theorem is one of the basic results in the theorem of ordinary differential equation and establishes existence and uniqueness under rather general assumptions.

Theorem 21 (Picard–Lindelöf theorem) Suppose that f: [T1,T2] × [y0C,y0+C] → ℝ is a continuous function such that for some M>0 we have
 |f(t,y1) − f(t,y2)| ≤ M | y1y2|   (Lipschitz condition)
for all t ∈ [T1,T2], y1,y2 ∈ [y0C,y0+C]. Then, for any t0 ∈ [T1,T2] the initial value problem
dy
dt
(t) = f(t,y(t)),   y(t0) = y0,
has a unique solution y in C1[a,b], where [a,b] is the interval [t0R, t0+R] ∩ [T1,T2], where
R= ||f||−1C.
(The solution exists for all times t such that |tt0| ≤ R).
Remark 22 Note, that the Lipschitz condition implies uniform continuity and is significantly stronger requirement.
Proof. Using the fundamental theorem of calculus we can write the IVP as a fixed point equation F(y) = y for a map defined by
  F(y) (t) = y0 + 
t
t0
f(s,y(s)) ds.
This is a map that will send a continuous function yC[T1,T2] to a continuous function F(y) ∈ C[T1,T2]. As a metric space we take
X = C([a,b], [y0 −C, y0 +C])
that is, the set of continuous functions on [a,b] taking values in the interval [y0C, y0 +C]. This is a closed (why?) subset of the Banach space C[a,b] and is therefore a complete metric space.

First we show that F: XX, i.e. F maps X to itself. Indeed,

 | F(y)(t) − y0 | = 


t
t0
f(s,y(s)) ds


≤ R || f ||≤ C.

Next we show that FN is a contraction for N large enough and thus establish the existence of a unique fixed point. It is the place to use the Lipschitz condition. Observe that for two functions y, y′ ∈ X we have

    | F(y)(t) − F(y′)(t) |
=  


t
t0
f(s,y(s)) − f(s,y′(s)) ds


 
≤  
t
t0
 | f(s,y(s)) − f(s,y′(s)) | ds ≤ |tt0| M || y − y′ ||.
(102)

We did not assume that (tt0) MRM <1, so F will in general not be a contraction. There are several ways to resolve this situations. For example, we can argue in either of the following two manners:

  1. We use both the result and the method from (102) to compute distances for higher powers of F, starting from the squares:
         
      | F2(y)(t) − F2(y′)(t) |
    ≤ 
    t
    t0
     | f(s,F(y)(s)) − f(s,F(y′)(s)) |  ds
             
     
    ≤ 
    t
    t0
     |st0| · M · || F(y) − F(y′)||  ds
             
     
    ≤ 
    t
    t0
     |st0| · M2 · || y − y′ ||  ds
             
     
     = 
    |tt0|2
    2
    M2  || y − y′ ||,
             
    and iterating this gives for any natural N:
     || FN(y) − FN(y′) ||≤  
    |tt0|N
    N !
    MN  || y − y′ ||.
    Since the factorial will overgrow the respective power, for N large enough, FN is a contraction and we deduce the existence of a unique solution from Cor. 13. This solution is in C1 since it can be written as the integral of a continuous function.
  2. The inequality (102) shows existence and uniqueness of solution only in the space of functions C([t0r,t0+r], [y0C,y0+C]) where r < M−1 and therefore |tt0| M< 1 in (102). Now suppose we have two solutions y and y′. They coincide at t0. Application of (102) to other initial points where the solutions coincide shows that the set E={ x ∈ [a,b] ∣ y(x) = y′(x)} is open. It is also the pre-image of the closed set {0} under the continuous map yy′. So we have that E is a closed and open subset of [a,b] that is non-empty. It must therefore be [a,b]. Hence, we get y = y′, establishing uniqueness in the whole C[a,b].

Note that this not only gives uniqueness and existence, but also gives a constructive method to compute the solution by iterating the map F starting for example with the constant function y(t)=y0. The iteration

yn+1(t) = y0 + 
t
t0
f(s,yn(s)) ds

is called Picard iteration. It will converge to the solution uniformly. See Fig. 20 for an illustration of few first iterations for the exponent functions.


Figure 20: Few initial Picard iterations for the differential equation y′=y: constant f0, linear f1, quadratic f2, etc.

Remark 23 The proof also gives a bound on the solution, namely if the assumptions are satisfied one gets | y(t) − y0 | ≤ C for t ∈ [a,b].
Remark 24 The proof works in the same way if y takes values in m and therefore f : ℝ × ℝm ⊃ [T1,T2] × BC(0) → ℝm. In fact, the target space may even be a Banach space (the derivative for Banach space-valued functions appropriately defined). Higher order differential equations may be written as systems of first order equations and hence the theorem applies to these as well. For example y″(t) + y(t) =0, y(0) = 1, y′(0) = 0 can be written as
  
d
dt


y
w


= 

w
y


,   

y
w


(0) = 

1 
0 


.
So here the function f is f(t,(x1,x2)) = (x2, −x1).
Example 25 Consider the IVP
  
dy
dt
 = y2t +1,   y(0) = 1.
Hence, f(t,x) = x2t +1. If we take f to be defined on the square [−T,T] × [1−C,1+C] then we obtain ||f||= (1+C)2 T +1 (the value at the top-right corner). In this case the solution will exist up to time
  min



T, 
C
(1+C)2T +1




.
If we choose, for example C=2 and T=1/2 we get that a unique solution exists up to time | t | ≤ 4/11. This solution will then satisfy | y(t) −1 | ≤ 2 for | t | ≤ 4/11.

In fact one can show that the solution can be expressed in a complicated way in terms of the Airy-Bi-function and it blows up at t=1.

16.2.3 Applications of fixed point theory: Inverse and Implicit Function Theorems

It is an easy exercise in Analysis to show that if a function fC1[a,b] has nowhere vanishing derivative, then f is invertible on its image. To be more precise, f−1: Im(f) → [a,b] exists and has derivative (f′(x))−1 at the point y=f(x). In higher dimensions a statement like this can not be correct as the following counterexample shows. Let 0<a<b and define

f: [a,b] × ℝ → ℝ2,
 (r,θ) ↦ (r cosθ, r sinθ).

This maps has invertible derivative

f′(r,θ) = 

cosθr sinθ 
sinθr cosθ. 


,   detf′(r,θ) = r2 >0.

at any point, the map is however not injective, see Fig. 21 for a cartoon illustration of the difference between one- and two-dimensional cases. However, for any point we can restrict domain and co-domain, so that the restriction of the function is invertible. In such a case we say that f is locally invertible. This concept will be explained in more detail below.


Figure 21: Flat and spiral staircases: can we return to the same value going just in one way?

Definition 26 (Local Invertibility) Suppose U1, U2 ⊂ ℝm are open subsets of m. Then a map f: U1U2 is called locally invertible at xU1 if there exists an open neighbourhood U of x such that f |U : Uf(U) is invertible. The function f is said to be locally invertible it it is locally invertible at x for any xU1.

Often, say for differential equations, we need a map which preserves differentiability of functions in both directions.

Definition 27 (Diffeomorphism) Suppose U1, U2 ⊂ ℝm are open subsets of m. Then a map f: U1U2 is called Ck-diffeomorphism if fCk(U1,U2) and if there exists a gCk(U2,U1) such that
   f ∘ g = 1U2,   g ∘ f = 1U1,
where 1U1 and 1U2 are the identity maps on U1 and U2 respectively.

There is also a local version of the above definition.

Definition 28 (Local Diffeomorphism) Suppose U1, U2 ⊂ ℝm are open subsets of m. Then a map f: U1U2 is called a local-Ck- diffeomorphism at xU1 if there exists an open neighbourhood U of x such that f |U: Uf(U) is a Ck-diffeomorphism. It is called a local-Ck- diffeomorphism if it is a local diffeomorphism at any point xU1.

Not every invertible Ck-map is a diffeomorphism. An example is the function f(x) = x3 whose inverse g(x) = x1/3 fails to be differentiable.

Theorem 29 (Inverse Function Theorem) Let U ⊂ ℝm be an open subset and suppose that fCk(U,ℝm) such that f′(x) is invertible at every point xU. Then f is a local Ck-diffeomorphism.

Before we can prove this theorem we need a Lemma, which basically says that under the assumptions of the inverse function theorem an inverse function must be in C1. That is, differentiability is the leading particular case [, § 4.4] for the general case of k-differentiable functions.

Lemma 30 Suppose that fC1(U1,U2) is bijective with continuous inverse. Assume that the derivative of f is invertible at any point, then f is a C1-diffeomorphism, and g′(f(x)) = (f′(x))−1.
Proof. Denote the inverse of f by g: U2U1. The continuity of f and g imply that xnx0 if and only if f(xn) → f(x0). We will show that g is differentiable at the point y0 = f(x0). If y=f(x) is very close to y0 (so that the line interval between x and x0 is contained in U1) then, by the MVT there exists a ξ on this line such that yy0 = f(x) − f(x0) = f′(ξ) · (xx0). Therefore, g(y)−g(y0) = (f′(ξ))−1 · (yy0). If y tends to y0, then ξ will tend to x0, and therefore, by continuity of f′ the value of (f′(ξ))−1 will tend to (f′(x0))−1. Thus, the partial derivatives of g exist and are continuous, so gC1. Note that we have used here that matrix inversion is continuous. □

Now we can proceed with the general situation.

Proof.[Proof of the Inverse Function Theorem 29] Let x0U and let y0=f(x0). We need to show that there exists an open neighborhood U1 of f(x0) such that f: f−1(U1) → U1 is a Ck-diffeomorphism. As a first step we construct a continuous inverse. Since f′(x0)=A is an invertible m × m-matrix we can change coordinates x = A−1 y + x0, so that we can assume without loss of generality that f′(x0)= 1 and x0=0. Replacing f by fy0 we also assume w.l.o.g. that y0=0. Since f′(x) is continuous there exists an ε>0 such that || f′(x) − 1 || ≤ 1/2 for all xBε(0). This ε>0 can also be chosen such that Bε(0) ⊂ U. Thus, || xf(x) || ≤ 1/2|| x|| for all xBε(0) by MVT, and for each yBε/2(0) the map
     x ↦ x + y − f(x)
is a contraction on Bε(0). Indeed, by MVT again:
      ||x + y − f(x) − (x′ + y − f(x′))||= ||x − f(x) − (x′ − f(x′))|| 
 = ||(f′(ξ) − 1) (xx′)||
 
≤ 
1
2
 ||xx′||,
(103)
where ||·|| is the norm of vectors in ℝm. Consider the complete metric space X=C(Bε/2(0),Bε(0)) and define the map
     F: X → X,   u ↦ F(u),   F(u)(y) = u(y) + y − f(u(y)). 
By the above this map is well defined and it also is a contraction
     
      || F(u)(y) −F(v)(y) ||
= || u(y) − f(u(y)) − 
v(y) −f(v(y)) 
|| 
         
 
≤ 
1
2
 || u(y) − v(y) ||  
 [by (103)]       
 
≤  
1
2
 || u − v ||. 
         
Hence, there exists a unique fixed point g. This fixed point yields a continuous inverse g of f|U defined on U =Bε/2(0) ∩ f−1(Bε/2(0)). By the previous Lemma this implies that g is differentiable. Now simply note that g′ = (f′)−1g. Since matrix inversion is smooth and f′ is in Ck−1 this implies that for mk−1 we get the conclusion (gCm) (gCm+1). Hence, g is in Ck. □

The implicit function theorem is actually a rather simple consequence of the inverse function theorem. It gives a nice criterion for local solvability of equations in many variables.

Theorem 31 (Implicit Function Theorem) Let U1 ⊂ ℝn × ℝm and U2 ⊂ ℝm be open subsets and let
   F: U1 → U2,   (x1, …, xn, y1,…,ym) ↦ F(x1, …, xn, y1,…,ym)
be a Ck-map. Suppose that F(x0,y0)=0 for some point (x0,y0) ∈ U1 and that the m × m-matrix y F(x0,y0) is invertible. Then there exists an neighborhood U of (x0,y0) ∈ ℝn × ℝm, an open neighborhood V of x0 in n, and a Ck-function f: V → ℝm such that
   {  (x,y) ∈ U ∣ F(x,y) =0 } =   {  (x , f(x)) ∈ U ∣ x ∈ V }.
The function f has derivative
   f′(x0)=−(∂yF(x0,y0))−1 ∂x  F(x0,y0)
at x0.
Proof. This is proved by reducing it to the inverse function theorem. Just design the map
  G :  U1 →  ℝn × ℝm,   (x,y) ↦ (x, F(x,y))
and then note that
  G′(x0,y0) = 

10 
xF(x0,y0)yF(x0,y0) 


is invertible with inverse
 (G′(x0,y0))−1 = 

10 
−(∂yF(x0,y0))−1 ∂x  F(x0,y0)(∂yF(x0,y0))−1


.
By the inverse function theorem there exists a local inverse G−1: U3U4, where U3 is an open neighborhood of 0 and U4 an open neighborhood of (x0,y0). Now define f by (x,f(x)) = G−1(x,0). □
Example 32 Consider the system of equations
  x12 + x22 + y12 + y22 = 2,
  x1 + x23  + y1 + y23 =2.
We would like to know if this system implicitly determines functions y1(x1,x2) and y2(x1,x2) near the point (0,0,1,1), which solves the equation. For this one simply applies the implicit function theorem to
  F(x1,x2,y1,y2) = ( x12 + x22 + y12 + y22 − 2,  x1 + x23  + y1 + y23 − 2).
The derivatives are
  ∂xF = 

2 x12 x2
13 x22


,   ∂yF = 

2 y12 y2
13 y22


The values of these derivatives at the point (0,0,1,1) are
  ∂xF(0,0,1,1) = 

00 
10 


,   ∂yF(0,0,1,1) = 

22 
13  


The latter matrix is invertible and one computes
  −(∂yF(x0,y0))−1 ∂x  F(x0,y0)(0,0,1,1) = 

1/20 
−1/20 


.
We conclude that there is an implicitly defined function (y1,y2)= f(x1,x2) whose derivative at (0,0) is given by
  

1/20 
−1/20 


.
The geometric meaning is that near the point (0,0,1,1) the system defines a two-dimensional manifold that is locally given by the graph of a function. Its tangent plane is spanned by the vectors (1/2,0,1,0) and (−1/2,0,0,1).
Example 33 Consider the system of equations
  x2 + y2 + z2 = 1,
  x + yz + z3 =1.
This is the intersection of a sphere (drawn in light green on Figure 22) with some cubic surface defined by the second equation (drawn in light blue). The point (0,0,1) solves the equation and is pictured as a little orange dot. By the implicit function theorem the intersection is a smooth curve (drawn in red) near this point which can be parametrised by x coordinate. Indeed, we can express y and z along the curve as functions of x because the resulting matrix
(y,z)F(0,1)=

2y2z
zy+3z2
  



 


y=0,z=1
 =  

02 
13 
  

is invertible.

Figure 22: Example of the implicit theorem: the intersection (red) of the unit sphere (green) and a cubic surface (blue).

Exercise 34 Fig. 22 suggests that the intersection curve can be alternatively parametrised by the coordinates y and cannot by z (why?). Check these claims by verifying conditions of Thm. 31.

16.3 The Baire Category Theorem and Applications

We are going to see another example of an abstract result which has several non-trivial consequences for real analysis.

16.3.1 The Baire’s Categories

Let us first prove the following result and then discuss its meaning and name.

Theorem 35 (Baire’s category theorem) Let (X,d) be a complete metric space and Un a sequence of open dense sets. Then the intersection S=∩n Un is dense.
Proof. The proof is rather straightforward. We need to show that any ball Bε(x0) contains an element of S. Let us therefore fix x0 and ε>0. Since U1 is dense the intersection of Bε(x0) with U1 is non-trivial. Thus there exists a point x1Bε(x0) ∩ U1. Now choose ε1 < ε/2 so that Bε1(x1)Bε(x) ∩ U1 (note the closure of the ball). Since U2 is dense, the intersection Bε1(x1) ∩ U2Bε(x0) ∩ U1U2 is non-empty. Choose a point x2 and ε2 < ε1 /2 such that Bε2(x2)Bε1(x1) ∩ U2Bε(x0) ∩ U1U2. Continue inductively, to obtain a sequence xn such that
Bεn(xn)
 ⊂ Bεn−1(xn−1) ⋂ Un  ⊂  Bε(x0) ⋂ U1 ⋂ U2 ⋂ … ⋂ Un,
and εn < 2n ε. In particular, for any n>N we have
  xn ∈ B2Nε(xN),
which implies that xn is a Cauchy sequence. Hence xn has a limit x, by completeness of (X,d). Consequently, x is contained in the closed ball BεN(xN) for any N, and therefore it is contained in Bε(x0) ∩ (∩n Un), as claimed. □

Completeness is essential here. For example, the conclusion does not hold for the metric space ℚ: take bijection ψ: ℕ → ℚ, and consider the open dense sets

  Un = { ψ(1), ψ(1), …, ψ(n)}c = {ψ(n+1), ψ(n+2),… }.

The intersection ∩n Un is empty.

The following historic terminology, due to Baire, is in use.

Definition 36 (Baire’s categories) A subset Y of a metric space X is called
  1. nowhere dense if the interior of Y is empty;
  2. of first category if there is a sequence (Yk) of nowhere dense sets with Y = ∪k Yk;
  3. of second category if it is not of first category.

Example of nowhere dense sets are ℤ ⊂ ℝ, the circle in ℝ2, or the set { 1/nn ∈ ℕ } ⊂ ℝ. Note that the complement of a nowhere dense set is a dense open set.

Corollary 37 In a complete metric space the complement of a set of the first category is dense.
Proof. Follows from relations for complements
  Yc = (⋃kYk)c = ⋂kYkc ⊃  ⋂k
Yk
c
and the fact that Ykc is dense. □

The following corollary is also called Baire’s category theorem in some sources:

Corollary 38 A complete metric space is of second category in itself, or plainly speaking it is never the union of a countable number of nowhere dense sets.

The theorem is often used to show abstract existence results. Here is an example.

Theorem 39 There exists a function fC[0,1] that is nowhere differentiable.
Proof. For each n ∈ ℕ define
  Un = 



f ∈ C[0,1]  s.t.  sup






f(x+h)−f(x)
h



  over  0 < |h| ≤ 
1
n




> n, ∀ x ∈ [0,1]



.
We will show that the Un are open and dense. By the Category theorem their intersection is also dense.

Un is open: Let fUn. For each x ∈ [0,1] choose δx>0 such that

 sup






f(x+h)−f(x)
h



  over  0 < |h| ≤ 
1
n




> n + δx,

hence there is a hx < 1/n with




f(x+hx)−f(x)
hx



 > n + δx.

By continuity of f there is an open neighborhood Ix of x such that




f(y+hx)−f(y)
hx



> n + δx.

for all yIx. These Ix form an open cover. We choose a finite subcover (Ixk)k=1,…,N. Let δ= min{δx1, …, δxN} > 0 . Then, for yIxk:




f(y+hxk)−f(y)
hxk



 > n + δ.

Now let gBε(f), where ε>0 is chosen so that ε < 1/2 δ hxk for all k. Then by an ε/3-style argument:




g(y+hxk) − g(y)
hxk



 ≥  


f(y+hxk) − f(y)
hxk



 − 2 
|| fg||
hxk
 > n + δ − 2 ε hxk−1 >n,

and therefore gUn. We conclude that Un is open.

Un is dense: For each ε>0 and fC[0,1] choose a polynomial p such that || fp || < ε/2 and a sequence of continuous function gmC[0,1] such that || g ||< ε/2 and such that for all x ∈ [0,1]:

 sup



gm(x+h)−gm(x)
h
  over  0 < |h| ≤ 
1
n




> m

by using a “zigzag” function. Then, for large enough m we have p+gmUn. □

The above proof actually shows much more, namely that the set of nowhere differentiable functions is dense in C[0,1]. It is also useful to compare it with the construction of the continuous nowhere differentiable Weierstrass function and identify some common elements.

16.3.2 Banach–Steinhaus Uniform Boundedness Principle

Another consequence of the Baire Category theorem is the Banach–Steinhaus uniform boundedness principle. Recall that, if X and Y are normed spaces, T: XY is called a bounded operator if it is a bounded linear map.

Theorem 40 (Banach–Steinhaus Uniform Boundedness Principle) Let X be a Banach space and Y a normed space, and let (Tα)α ∈ I be a family of bounded operators Tα: XY. Suppose that
  ∀ x ∈ X: 
 
sup
α
 || Tαx || < ∞.
Then we have supα || Tα|| < ∞, i.e. the family Tα is bounded in the set B(X,Y) of bounded operators from X to Y.
Proof. Define Xn = {xX ∣ supα || Tαx || ≤ n }. By assumption X = ∪n Xn. Note that all the Xn are closed. By the Baire category theorem at least one of these sets must have non-empty interior, since otherwise the Banach space X would be a countable union of nowhere dense sets. Hence, there exists N ∈ ℕ, yXN, and ε>0 such that Bε(y)XN. Now XN is symmetric under reflections x↦ −x and convex. So we get the same statement for −y. Hence, xBε(0) implies
x = 
1
2

(x + y) + (xy) 
∈ 
1
2

XN + XN
⊂ XN. (104)
This means that || x || ≤ ε implies || Tαx || ≤ N, and therefore || Tα|| ≤ ε−1 N for all α ∈ I. □

Recall that the Fourier series of a C1-function on a circle (identified with 2 π-periodic functions) converges uniformly to the function. We will now show that a statement like that can not hold for continuous functions.

Corollary 41 There exist continuous periodic functions whose Fourier series do not converge point-wise.
Proof. We will show that there exists a continuous function whose Fourier series does not converge at x=0. Suppose by contradiction such functions would not exist, so we would have point-wise convergence of the Fourier series
  
1
2
a0 + 
m =1
am cos(mx) + bm sin(mx)
for every fC(S1) = Cper(ℝ). Here we identify continuous functions on the unit circle with continuous 2 π-periodic functions Cper(ℝ). Hence we have a map
  Tn : C(S1) → ℝ, f ↦   
1
2
a0  + 
n
m =1
am
by mapping the function f to the n-th partial sum of its Fourier series at x=0. This is a family of bounded operators Tn: C(S1) → ℝ and by assumption we have for every f that
  
 
sup
n
 | Tn(f) | < ∞.
By Banach–Steinhaus theorem we have supn || Tn || = supn, || f ||=1 | Tn(f) | < ∞. Now one computes the norm of the map
  Tn : C(S1) → ℝ,    f ↦ 
1
π
π
−π
f(x) 


1
2
+
n
k=1
 cos(kx) 


dx =  
1
2 π
  
π
−π
f(x) Dn(x) dx
where
  Dn(x) = 
sin


(n+
1
2
) x


sin


x
2



is the Dirichlet kernel , cf. Lem. 6. This norm equals 1/2 π ∫−ππ| Dn(x) | dx = 1/2 π ∫02 π | Dn(x) | dx (Exercise) which goes to ∞ as n → ∞. Indeed, using sin(x/2) ≤ x/2 and substituting we get
     
   
2 π
0
 | Dn(x) | dx
≥ 
0
|sin((n+
1
2
) x)|
x/2
dx  
  [since  sins ≤ s]       
 
= 
(2n+1)π
0
|sin(t)|
t
dt  
 
 [change of variables    t=(n+
1
2
) x]
       
 
≥ 
2n
k=0
(k+1) π
k π
|sint|
t
dt  
  [split integral into intervals]       
 
 ≥ 


2n
k=0
π
0
sint
(k+1)
dt  


  
  [since  t ≤ k+1  for  t∈ (k,k+1) ]       
 
 = 2 
2n
k=0
1
k+1
  
  [evaluating the integral],        
which is the harmonic series divergent as n → ∞. This gives a contradiction. □

Another corollary of the Banach–Steinhaus principle is an important continuity statement. Recall that of X and Y are normed spaces them so is the Cartesian product X × Y equipped with the norm || (x,y) || = ( || x ||X2 + || y ||Y2 )1/2. It is easy to see that a sequence (xn,yn) converges to (x,y) in this norm if and only if xnx and yny.

Theorem 42 Suppose that X, Y are Banach spaces and suppose that B: X × Y → ℝ is a bilinear form on X × Y that is separately continuous, i.e. B(·, y) is continuous on X for every yY and B(x,·) is continuous on Y for every xX. Then B is continuous.
Proof. Suppose that (xn,yn) is a sequence that converges to (x,y). First note that
  B(xnx,yny)= B(xn,yn) − B(xn,y) − B(x,yn) + B(x,y),
where B(xn,y) → B(x,y) as well as B(x,yn) → B(x,y). So it is sufficient to show that B(xnx,yny) → 0 or, equivalently, B(xn,yn) → 0 for any xn→ 0 and yn→ 0. Now. the linear mappings Tn(x)= B(x,yn): X → ℝ are bounded, by assumption. Since ||yn||→ 0 the sequence Tn(x)→ 0 and is bounded for every xX. Then, by the Banach–Steinhaus theorem there exists a constant C such that ||Tn|| ≤ C for all n. That is |Tn(x)| = B(x, yn) ≤ C ||x|| for all n and xX. Therefore, |B(xn,yn)| ≤ C ||xn|| → 0. □
Remark 43 Recall that already on 2 separate continuity does not imply joint continuity for any function. The standard example from Analysis is the function
  f(x,y) = 





xy
x2 + y2
(x,y) ≠ (0,0) 
0(x,y) = 0,
which is continuous in x or y separately but is not jointly continuous.

16.3.3 The open mapping theorem

Recall that for a continuous map the pre-image of any open set is open. This does of course not mean that the image of any open set is open (for example, sin: ℝ → ℝ has image [−1,1], which is not open). A map f: XY between metric space is called open if the image of every open set is open. If a map is invertible then it is open if and only if its inverse is continuous. We start with a simple observation for linear maps. We will denote open balls in normed spaces X and Y by BrX(x) and BsY(y) respectively, or simply BrX and BsY if they are centred at the origin.

Lemma 44 Let X and Y be normed spaces. Then a linear map T: XY is open if and only if there exists ε>0 such that BεY(0) ⊂ T (B1X(0)), i.e. the image of the unit ball contains a zero’s neighbourhood.
Proof. If the map T is open it clearly has this property. Suppose conversely, that BεY(0) ⊂ T (B1X(0)) for some ε>0. Then, by scaling, Bε δY(0) ⊂ T (BδX(0)) for any δ>0. Suppose that U is open. Suppose that yf(U), that is there exists xU such that y=f(x). Then there exists δ>0 with x + BδX(0) ⊂ U and therefore
    TU ⊃ TBδX(x) = { Tx } +  TBδX(0) ⊃  { y } +  Bδ εY(0) =Bδ εY(y) .
Theorem 45 (Open Mapping Theorem) Let T : XY be a continuous surjective linear operator between Banach spaces. Then T is open.
Proof. Since T is surjective we have Y = ∪n T BnX. Therefore trivially, Y = ∪n T BnX. By the Baire category theorem one of the T BnX must have an interior point. Rescaling implies that T B1X has an interior point y0. Since T B1X is symmetric under reflection y→ −y, the point −y0 must also be an interior point. Therefore, by convexity of T B1X there exists a δ>0 with BδYT B1X, cf. (104). By linearity this means Bδ 2nYT B2nX for any natural n.

We will show that T B1XT B2X, with the implication from above that BδYT B2X, which will complete the proof by the previous Lemma. So, let yTB1X be arbitrary. Then, there exists x1B1X such that yT x1Bδ/2YTB1/2X. Repeating this, there exists x2B1/2X such that yT x1T x2Bδ/4Y.

Continuing inductively, we obtain a sequence (xn) with the property that || xn || < 2n+1 and

y − 
n
k=1
Txn ∈ Bδ 21−nY. (105)

By completeness of X, the absolute convergent series ∑1n xn converges to an element xX of norm || x|| < 2. By linearity an continuity of T we get from (105) that y = T x. Thus yTB2. □

If the map T is also injective (and, therefore, bijective with the inverse T−1) we can quickly conclude continuity of T−1.

Corollary 46 Suppose that T: XY is a bijective bounded linear map between Banach spaces. Then T has a bounded inverse T−1.

It is not rare that we may have two different norms ||·|| and ||·||* on the same Banach space X. We say that ||·|| and ||·||* are equivalent if there are constants c>0 and C>0 such that:

c ||x|| ≤ ||x||* ≤ C ||x||     for all  x ∈ X. (106)
Exercise 47
  1. Check that (106) defines an equivalence relations on the set of all norms on X.
  2. If a sequence is Cauchy/convergent/bounded in a norm then it is also Cauchy/convergent/bounded in any equivalent norm.

The Cor. 46 implies that if the identity map (X,||·||)→ (X,||·||*) is bounded then both norms are equivalent.

Corollary 48 Let (X,||·||) be a Banach space and ||·||* be a norm on X in which X is complete. If ||·|| ≤ C ||·||* for some C>0 the norms are equivalent.

16.3.4 The closed graph theorem

Suppose that X, Y are Banach spaces and suppose that DX is a linear subspace (not necessarily closed). Now suppose that T : DY is a linear operator. Then the graph gr(T) is defined as the subset {(x,Tx) ∣ xD} ⊂ X × Y. This is a linear subspace in the Banach space X × Y, which can be equipped with the norm ||(x,y)||2 = ||x||X2 + || y||Y2. One often uses the equivalent norm ||(x,y)|| = ||x||X + || y||Y but the first choice makes sure that the product X × Y is also a Hilbert space if X and Y are Hilbert spaces. We will refer to T as an operator from X to Y with domain D.

Definition 49 The operator T is called closed if and only if its graph is a closed subset of X × Y.

It is easy to see that T is closed if an only if xnx and T xny imply that T xnT x. Note the difference with continuity of T!!!

If T is an operator T : DY then its graph is a subset of X × Y. If we close this subset the resulting set may fail to be the graph of an operator. If the closure is the graph as well, we say that T is closable and its closure is the operator whose graph is obtained by closing the graph of T.

Differential operators are often closed but not bounded. Let L2[a,b] be the Hilbert space obtained by abstract completion of (C[a,b],|| ·||2), cf. Prop. 60. Then D=C1[a,b] is a dense subspace in L2[a,b] and the operator d/dx: C1[a,b] → L2[a,b] is of the above type. This operator is not closed, however it is closable and its closure therefore defines a closed operator with dense domain. We have already seen that this operator is unbounded and therefore it cannot be continuous.

Of course, the map D → (x,Tx) is a bijection from D to gr(T). We can use the norm on gr(T) to define a norm on D, which is then

 || x ||D = 
|| x ||X2 + || Tx ||Y2
1
2
 
.

Obviously, T is closed if and only of D with norm ||·||D is a Banach space. We are now ready to state the closed graph theorem. It is easy to check that T continuously maps (D, || · ||D) to Y.

Theorem 50 (Closed Graph Theorem) Suppose that X and Y are Banach spaces and suppose that T: XY is closed. Then T is bounded.
Proof. Since in this case we have D=X with have two norms ||·||X and || · ||D on X that are both complete. Clearly,
  ||·||X ≤ ||·||D,
and by Cor. 48 the norms are therefore equivalent. Hence,
  || Tx ||Y ≤ || x ||D ≤ C || x ||X
for some constant C>0. □

16.4 Semi-norms and locally convex topological vector spaces

Definition 51 (Semi-Norm) Let X be a vector space, then a map p: X → ℝ is called semi-norm if
  1. p(x) ≥ 0 for all xX,
  2. px) = |λ| p(x), for all λ ∈ ℝ, xX,
  3. p(x+y) ≤ p(x) + p(y), for all x,yX.

An example of a semi-norm on C1[0,1] is p(f):=|| f ′ ||. If (pα)α is a family of semi-norms with the property that

 ( ∀ α ∈ I, pα(x) =0 )  x=0

then we say X with that family is a locally convex topological vector space. There is a topology (that is, a description of all open sets) on such a vector space, by declaring a subset UX to be open if and only if for every point xU and any index α ∈ I there exists ε>0 such that { ypα(yx) < ε } ⊂ U. The notion of convergence one gets is xnx if and only of pα(xnx) → 0 for all α. The topology of point-wise convergence on the space of functions S → ℝ is for example of this type, with the family of semi-norms given by (px)s xS, px(f) = | f(x) |.

Another example is the vector space C(ℝm) with the topology of uniform convergence of all derivatives on compact sets. Here the family of semi-norms pα,K is indexed by all multi-indices α ∈ ℕ0m and all compact subsets K ⊂ ℝ and is given by

pα, K(f) = 
 
sup
x ∈ K
 | ∂αf(x) |.

If the family of semi-norms is countable then this topology is actually coming from a metric (so the space is a metric space)

d(x,y) = 
k=1
1
2k
pk(xy)
1+pk(xy)
.

Such a metric space is called Frechet space. Note that C(ℝm) is a Frechet space because the family of semi-norms above can be replaced by a countable one by taking a countable exhaustion of ℝm by compact subsets.

A Tutorial Problems

These are tutorial problems intended for self-assessment of the course understanding.

A.1 Tutorial problems I

All spaces are complex, unless otherwise specified.

 1 Show that ||f||=|f(0)|+sup|f′(t)| defines a norm on C1[0,1], which is the space of (real) functions on [0,1] with continuous derivative.
 2 Show that the formula ⟨ (xn),(yn)⟩ =∑n=1xnyn/n2 defines an inner product on l, the space of bounded (complex) sequences. What norm does it produce?
 3 Use the Cauchy–Schwarz inequality for a suitable inner product to prove that for all fC[0,1] the inequality



1
0
f(x)xdx


≤ C


1
0
 |f(x)|2  dx


1/2



 
holds for some constant C>0 (independent of f) and find the smallest possible C that holds for all functions f (hint: consider the cases of equality).
 4 We define the following norm on l, the space of bounded complex sequences:
||(xn)||= 
 
sup
n ≥ 1
 |xn|.
Show that this norm makes l into a Banach space (i.e., a complete normed space).
 5 Fix a vector (w1,…,wn) whose components are strictly positive real numbers, and define an inner product on n by
⟨ x,y ⟩ = 
n
k=1
wkxk
y
k.
Show that this makes n into a Hilbert space (i.e., a complete inner-product space).

A.2 Tutorial problems II

 6 Show that the supremum norm on C[0,1] isn’t given by an inner product, by finding a counterexample to the parallelogram law.
 7 In l2 let e1=(1,0,0,…), e2=(0,1,0,0,…), e3=(0,0,1,0,0,…), and so on. Show that Lin (e1,e2,…)=c00, and that CLin (e1,e2,…)=l2. What is CLin (e2,e3,…)?
 8 Let C[−1,1] have the standard L2 inner product, defined by
⟨ f, g⟩ = 
1
−1
f(t) 
g(t)
  dt.
Show that the functions 1, t and t2−1/3 form an orthogonal (not orthonormal!) basis for the subspace P2 of polynomials of degree at most 2 and hence calculate the best L2-approximation of the function t4 by polynomials in P2.
 9 Define an inner product on C[0,1] by
⟨ f,g⟩=
1
0
t
  f(t)  
g(t)
  dt.
Use the Gram–Schmidt process to find the first 2 terms of an orthonormal sequence formed by orthonormalising the sequence 1, t, t2, ….
 10 Consider the plane P in 4 (usual inner product) spanned by the vectors (1,1,0,0) and (1,0,0,−1). Find orthonormal bases for P and P, and verify directly that (P)=P.

A.3 Tutorial Problems III

 11 Let a and b be arbitrary real numbers with a < b. By using the fact that the functions 1/√einx, n ∈ ℤ, are orthonormal in L2[0,2π], together with the change of variable x=2π(ta)/(ba), find an orthonormal basis in L2[a,b] of the form en(t)=α ei n λ t, n ∈ ℤ, for suitable real constants α and λ.
 12 For which real values of α is
  
n=1
nαeint
the Fourier series of a function in L2[−π,π]?
 13 Calculate the Fourier series of f(t)=et on [−π,π] and use Parseval’s identity to deduce that
n=−∞
1
n2+1
 = 
π 
tanhπ
.
 14 Using the fact that (en) is a complete orthonormal system in L2[−π,π], where en(t)=exp(int)/√, show that e0,s1,c1,s2,c2,… is a complete orthonormal system, where sn(t)=sinnt/√π and cn(t)= cosnt/√π. Show that every L2[−π,π] function f has a Fourier series
a0+
n=1
ancosnt + bn sinnt,
converging in the L2 sense, and give a formula for the coefficients.
 15 Let C(T) be the space of continuous (complex) functions on the circle  T={ z ∈ ℂ: |z|=1 } with the supremum norm. Show that, for any polynomial f(z) in C(T)
 


|z|=1
f(z)  dz=0.
Deduce that the function f(z)=z is not the uniform limit of polynomials on the circle (i.e., Weierstrass’s approximation theorem doesn’t hold in this form).

A.4 Tutorial Problems IV

 16 Define a linear functional on C[0,1] (continuous functions on [0,1]) by α(f)=f(1/2). Show that α is bounded if we give C[0,1] the supremum norm. Show that α is not bounded if we use the L2 norm, because we can find a sequence (fn) of continuous functions on [0,1] such that ||fn||2 ≤ 1, but fn(1/2) → ∞.
 17 The Hardy space H2 is the Hilbert space of all power series f(z)=∑n=0an zn, such that n=0|an|2 < ∞, where the inner product is given by



n=0
anzn, 
n=0
bnzn


= 
n=0
an
bn
.
Show that the sequence 1, z, z2, z3, … is an orthonormal basis for H2.

Fix w with |w|<1 and define a linear functional on H2 by α(f)=f(w). Write down a formula for the function g(z) ∈ H2 such that α(f)=⟨ f, g. What is ||α||?

 18 The Volterra operator V: L2[0,1] → L2[0,1] is defined by
(Vf)(x)=
x
0
f(t)  dt.
Use the Cauchy–Schwarz inequality to show that |(Vf)(x)| ≤ √x||f||2 (hint: write (Vf)(x)=⟨ f, Jx where Jx is a function that you can write down explicitly).

Deduce that ||Vf||22 ≤ 1/ 2||f||22, and hence ||V|| ≤ 1/√2.

 19 Find the adjoints of the following operators:
  1. A:l2l2, defined by A(x1,x2,…)=(0,x1 / 1, x2/ 2, x3/ 3, …);

    and, on a general Hilbert space H:

  2. The rank-one operator R, defined by Rx=⟨ x,yz, where y and z are fixed elements of H;
  3. The projection operator PM, defined by PM(m+n)=m, where mM and nM, and H=MM as usual.
 20 Let UB(H) be a unitary operator. Show that (Uen) is an orthonormal basis of H whenever (en) is.

Let l2(ℤ) denote the Hilbert space of two-sided sequences (an)n=−∞ with

||(an)||2=
n=−∞
|an|2 < ∞.

Show that the bilateral right shift, V:l2(ℤ) → l2(ℤ) defined by V((an))=(bn), where bn=an−1 for all n∈ ℤ, is unitary, whereas the usual right shift S on l2=l2(ℕ) is not unitary.

A.5 Tutorial Problems V

 21 Let fC[−π,π] and let Mf be the multiplication operator on L2(−π,π), given by (Mfg)(t)=f(t) g(t), for gL2(−π,π). Find a function f′ ∈ C[−π,π] such that Mf*=Mf.

Show that Mf is always a normal operator. When is it Hermitian? When is it unitary?

 22 Let T be any operator such that Tn=0 for some integer n (such operators are called nilpotent). Show that IT is invertible (hint: consider I+T+T2+…+Tn−1). Deduce that IT is invertible for any λ ≠ 0.

What is σ(T)? What is r(T)?

 23 Let n) be a fixed bounded sequence of complex numbers, and define an operator on l2 by T((xn))=((yn)), where ynnxn for each n. Recall that T is a bounded operator and ||T||=||(λn)||. Let Λ={λ12,…}. Prove the following:
  1. Each λk is an eigenvalue of T, and hence is in σ(T).
  2. If λ ∉Λ, then the inverse of T−λ I exists (and is bounded).

Deduce that σ(T)=Λ. Note, that then any non-empty compact set could be a spectrum of some bounden operator.

 24 Let S be an isomorphism between Hilbert spaces H and K, that is, S: HK is a linear bijection such that S and S−1 are bounded operators. Suppose that TB(H). Show that T and STS−1 have the same spectrum and the same eigenvalues (if any).
 25 Define an operator U: l2(ℤ) → L2(−π,π) by U((an))=∑n=−∞an eint/√. Show that U is a bijection and an isometry, i.e., that ||Ux||=||x|| for all xl2(ℤ).

Let V be the bilateral right shift on l2(ℤ), the unitary operator defined on Question 20. Let fL2(−π,π). Show that (UVU−1f)(t)=eitf(t), and hence, using Question 24, show that σ(V)=T, the unit circle, but that V has no eigenvalues.

A.6 Tutorial Problems VI

 26 Show that K(X) is a closed linear subspace of B(X), and that AT and TA are compact whenever TK(X) and AB(X). (This means that K(X) is a closed ideal of B(X).)
 27 Let A be a Hilbert–Schmidt operator, and let (en)n≥ 1 and (fm)m≥ 1 be orthonormal bases of A. By writing each Aen as Aen=∑m=1Aen, fmfm, show that
n=1
||Aen||2=
m=1
||A*fm||2.
Deduce that the quantity ||A||HS2=∑n=1||Aen||2 is independent of the choice of orthonormal basis, and that ||A||HS=||A*||HS. (||A||HS is called the Hilbert–Schmidt norm of A.)
 28
  1. Let TK(H) be a compact operator. Using Question 26, show that T*T and TT* are compact Hermitian operators.
  2. Let (en)n≥ 1 and (fn)n ≥ 1 be orthonormal bases of a Hilbert space H, let n)n ≥ 1 be any bounded complex sequence, and let TB(H) be an operator defined by
    Tx=
    n=1
    αn ⟨ x, en ⟩ fn.
    Prove that T is Hilbert–Schmidt precisely when n) ∈ l2. Show that T is a compact operator if and only if αn → 0, and in this case write down spectral decompositions for the compact Hermitian operators T*T and TT*.
 29 Solve the Fredholm integral equation φ−λ Tφ=f, where f(x)=x and
(Tφ)(x)=
1
0
xy2 φ(y)  dy    (φ ∈ L2(0,1)),
for small values of λ by means of the Neumann series.

For what values of λ does the series converge? Write down a solution which is valid for all λ apart from one exception. What is the exception?

 30 Suppose that h is a -periodic L2(−π,π) function with Fourier series n=−∞an eint. Show that each of the functions φk(y)=eiky, k ∈ ℤ, is an eigenvector of the integral operator T on L2(−π,π) defined by
(Tφ)(x)=
π
−π
h(xy) φ(y)  dy,
and calculate the corresponding eigenvalues.

Now let h(t)=−log(2(1−cost)). Assuming, without proof, that h(t) has the Fourier series n ∈ ℤ, n ≠ 0 eint/|n|, use the Hilbert–Schmidt method to solve the Fredholm equation φ−λ Tφ=f, where f(t) has Fourier series n=−∞cn eint and 1/λ ∉σ(T).

A.7 Tutorial Problems VII

 31 Use the Gram–Schmidt algorithm to find an orthonormal basis for the subspace X of L2(−1,1) spanned by the functions t, t2 and t4.

Hence find the best L2(−1,1) approximation of the constant function f(t)=1 by functions from X.

 32 For n=1,2,… let φn denote the linear functional on l2 defined by
φn(x)=x1+x2+…+xn,
where x=(x1,x2,…) ∈ l2. Use the Riesz–Fréchet theorem to calculate ||φn||.
 33 Let T be a bounded linear operator on a Hilbert space, and suppose that T=A+iB, where A and B are self-adjoint operators. Express T* in terms of A and B, and hence solve for A and B in terms of T and T*.

Deduce that every operator T can be written T=A+iB, where A and B are self-adjoint, in a unique way.

Show that T is normal if and only if AB=BA.

 34 Let Pn be the subspace of L2(−π,π) consisting of all polynomials of degree at most n, and let Tn be the subspace consisting of all trigonometric polynomials of the form f(t)=∑k=−nn ak eikt. Calculate the spectrum of the differentiation operator D, defined by (Df)(t)=f′(t), when
  1. D is regarded as an operator on Pn, and
  2. D is regarded as an operator on Tn.
Note that both Pn and Tn are finite-dimensional Hilbert spaces.

Show that Tn has an orthonormal basis of eigenvectors of D, whereas Pn does not.

 35 Use the Neumann series to solve the Volterra integral equation φ−λ Tφ=f in L2[0,1], where λ∈ ℂ, f(t)= 1 for all t, and (Tφ)(x)=∫0x t2φ(t)  d t. (You should be able to sum the infinite series.)

B Solutions of Tutorial Problems

0=0Solutions of the tutorial problems will be distributed due in time on the paper.

1<0

B.1 Solution of Tuitorial Problem I

 1 Clearly the norm is non-negative. If ||f||=0, then f is constantly zero (since then f(0)=0 and f′(t)=0 everywhere). Also
||λ f ||=|λ f(0)|+ sup|λ f′(t)| 
 =|λ| |f(0)|+|λ|sup|  f′(t)| =|λ| ||f||,
and
||f+g||=|f(0)+g(0)| +sup|f′(t)+g′(t)| 
 |f(0)|+|g(0)|+sup|f′(t)|+sup|g′(t)| 
 =||f||+||g||.
 2 Clearly the sum is absolutely convergent (since ∑1/n2 < ∞), and checking the conditions:

y,x ⟩ = x,y,   ⟨λ x, y⟩ = λ ⟨ x,y,   x+y, z ⟩=⟨ x,z⟩ + ⟨ y,z,   and

x,x ⟩ > 0 except when x=0, when x,x⟩ =0 ,

is fairly straightforward algebra.

Also ||(xn)|| = ⟨(xn),(xn)⟩1/2 = (∑n=1|xn|2/n2)1/2.

 3 With the usual inner product f,g⟩=∫f ḡ d x, we have that |⟨ f,g⟩ | ≤ ||f|| ||g||, where g is the function g(x)=x. Now ||g||=1/√3 and this is the smallest possible constant C, since we do have g,g⟩ =||g|| ||g||.
 4 I’ll omit the proof that this is a norm (but if it gives any trouble, ask me). The proof of completeness is a bit like the l2 proof, only simpler. Suppose that (x(n)) is a Cauchy sequence in l. Then, for each coordinate k, |xk(n)xk(m)| ≤ ||x(n)x(m)||, and so (xk(n)) is a Cauchy sequence of complex numbers, converging to xk, say. Also |xk(n)xk(m)| < є for n and m greater than or equal to Nє, say. Letting m→∞ we get that |xk(n)xk| ≤ є for nNє, so (x(n)x) ∈ l and hence xl. Also ||x(n)x|| ≤ є for nNє, and so x(n)x.
 5 Clearly y, x⟩=∑k=1n wk yk xk = x,y, and the properties x+y, z⟩=⟨ x,z⟩ + ⟨ y, z and ⟨ λ x, y⟩ =λ ⟨ x, y are also straightforward to check. The norm produced is
||x||=⟨ x,x1/2=


n
k=1
wk |xk|2


1/2



 
,
which is strictly positive unless each xk is zero. For the completeness, note that a Cauchy sequence (x(m)) has the property that, for each є>0 there is a number Mє with ||x(m)x(p)||<є for m, pMє. That is,
n
k=1
wk |xk(m)xk(p)|22. (107)
Thus wk|xk(m)xk(p)|2 < є2, which is enough to show that in the kth coordinate we have a Cauchy sequence of complex numbers.

Define a vector x ∈ ℂn by xk=limm → ∞ xk(m) for each k. Now we have x(m)x, since

n
k=1
wk |xk(m)xk|2≤ є2,

for mMє, as we see on letting p → ∞ in (107).

2<0

B.2 Solutions of Tutorial Problems II

 6Lin(e1, e2, …) = c00 because a sequence is a finite linear combination of the ei if and only if it has finitely many nonzero terms. Taking the closure we get all of l2 since anything in l2 is the limit of a sequence in c00. This is so, because
||(x1, x2, …) − (x1, x2, …, xN, 0, 0, …)||2 = 
n=N+1
|xn|2,
which tends to zero as N → ∞. Finally CLin(e2,e3,…) is the same except that we only get sequences whose first term is zero, i.e. we get {x=(xn) ∈ l2: x1 = 0}.
 7 Calculate ⟨ 1,t, ⟨ 1,t2−1/3⟩ and t,t2−1/3⟩: they are all zero. Clearly the set is a basis for P2. Normalise the functions, to get e1(t)=1/√2, e2(t)=t3/2 and e3(t)=√45/8(t2−1/3), an orthonormal sequence. It now follows that, writing f(t)=t4, the best approximation is
g(t)=⟨f,e1⟩ e1+⟨ f,e2⟩ e2+⟨ f,e3⟩ e3
 =(2/5)(1/2)+(0)t+(45/8)(16/105)(t2−1/3) = (−3/35)+(6/7)t2.
As a cross-check, note that fg is indeed orthogonal to g.
 8  ⟨ 1,1⟩ =2/3, so e1(t)=√3/2. Now form f(t)=t−⟨ t,e1e1=t−⟨ t,1⟩ (3/2)=t−3/5. Since f,f⟩ =8/175 we take e2(t)=√175/8(t−3/5).
 9 The Gram–Schmidt process gives e1=(1,1,0,0)/√2, then y2=(1,0,0,−1)−(1,1,0,0)/2=(1/2,−1/2,0,−1), and then e2=y2/||y2||=(1,−1,0,−2)/√6.

The plane P consists of all vectors orthogonal to (1,1,0,0) and (1,0,0,−1), and hence is

{(x,y,z,w)∈ ℂ4: x+y=0, xw=0},

with general solution (a,−a,b,a), and basis (1,−1,0,1) and (0,0,1,0), which are already orthogonal.

We can thus take e3=(1,−1,0,1)/√3 and e4=(0,0,1,0) as a basis for P.

Finally, (P) consists of all vectors orthogonal to (1,−1,0,1) and (0,0,1,0), namely

{(x,y,z,w)∈ ℂ4: xy+w=0, z=0},

to which the general solution is (−a+b,b,0,a), with basis (−1,0,0,1) and (1,1,0,0). We are clearly back at P.

3<0

B.3 Solutions of Tutorial Problems III

 10 The given change of variables
x=
2π(ta) 
ba
     or     t=a+
(ba)x
 2π
takes t∈ [a,b] to x ∈ [0,2π]. We know that
1 


0
einxeimxdx=


1 if   n=m, 
0 if   n ≠ m,
so we obtain
1 
 2π
b


a
ein λ (ta) eim λ (ta)
2 π
ba
  dt=


1 if  n=m, 
0 if   n ≠ m,
where λ = 2π/(ba). Hence the functions en(t)=1 / √ba einλ t form an orthonormal set in L2[a,b]. They are even an orthonormal basis, since the same coordinate change shows that their closed linear span contains all fC[a,b] such that f(a)=f(b).
 11 By the Riesz–Fischer theorem, cnen converges to an L2 function iff ∑|cn|2 < ∞. Here cn=nα so we require n < ∞, i.e. α<−1/2.
 12 We calculate
⟨ f,en⟩ =
π
−π
eteintdt/
= (−1)n
eπe−π
 (1−in)
.
Also, by Parseval’s identity
n=−∞
|⟨ f,en⟩ |2 = ||f||22 = 
π
−π
e2tdt  = (e − e−2π)/2.
That is,
n=−∞
(eπe−π)2
 (1+n2)2π
 = 
(ee−2π)
2
.
Thus
n=−∞
1
 1+n2
= π 
e − e−2π
 (eπe−π)2
,
which gives the result.
 13 To check that the sequence is orthonormal it is probably easiest to write
sn=
2
(enen)/2i    and    cn=
2
(en+en)/2,
and use the orthonormality of (en) to calculate sn, sm, sn, cm and cn, cm.

Since exp(± int)=cos(ntisin(nt) and also cos(nt)=(exp(int)−exp(−int))/2 and sin(nt)=(exp(int)−exp(−int))/2i, the linear spans of (en) and {e0,s1,c1,s2,c2,…} are the same. Hence their closed linear spans are the same, so {e0,s1,c1,s2,c2,…} also forms an o.n.b.

As we have an orthonormal basis we have an expansion

f=⟨ f,e0⟩ e0 +
1
⟨ f,cn⟩ cn + 
1
⟨ f,sn⟩ sn

converging in L2. Hence

a0=(1/2π)
π
−π
f(t)  dt,  an = (1/π) 
π
−π
f(t) cosnt  dt and  bn = (1/π) 
π
−π
f(t) sinnt  dt.
 14  Complex analysis method: Clearly there is a polynomial g such that f(z)=g′(z). By the fundamental theorem of the calculus, g′ = 0 because the contour is closed. Or you could use Cauchy’s theorem.

Direct method:



0
f(eiθ) ieiθd θ = 
n
k=0
ak


0
   iei(k+1)θd θ=0,

where f(z)=∑k=0n akzk.

Now

 


|z|=1
z  dz =
 


|z|=1
 (1/z)  dz = 


0
eiθieiθd θ= 2π i.

(Again, there are several ways of doing this, as above.) But if fnf uniformly then fn → ∫f, which is impossible if fn are polynomials and f(z)=z.

4<0

B.4 Solutions of Tutorial Problems IV

 15 Since |α(f)|=|f(1/2)|≤ sup[0,1]|f(x)|=||f||, we have that ||α|| ≤ 1 (actually it equals 1) when we use the supremum norm.

Suppose now we define fn (starting at n=2, say) to be zero except on

[1/2−1/n, 1/2+1/n],

piecewise linear on [1/2−1/n, 1/2] and [1/2, 1/2+1/n], with fn(1/2)=An, where An is a positive number that we’ll choose in a minute. (The graph is going to be a thin steep triangle.) Now ||fn||22 ≤ (2/n) × An2, since fn is zero except on a set of length 2/n and always at most An (we could work it out exactly, but why bother?) So if we choose An=√n/2 we get ||fn||2 ≤ 1, and α(fn)=An so α(fn) → ∞, which means that α is unbounded in the L2 norm.

 16 Clearly we get orthonormality—just compute zk, zl. The fact that it is an orthonormal basis (i.e., complete) follows since the only function orthogonal to every zk has all its coefficients zero, so is the 0 function.

Now α(∑anzn)=∑an wn, and this is an bn only if we take bn=wn for each n. Hence

g(z)=
n=0
w
nzn = 1 /(1−
w
z).

Finally ||α||=||g||, so compute the H2 norm of g to get




n=0
|
w
|2n


1/2



 
=


1
 1−|w|2



1/2



 
.
 17 The function Jx must be χ[0,x], where
χ[0,x](t)=


1 on   [0,x],
0elsewhere.
It has L2 norm equal to the square root of 0x 12d t, i.e., x. Hence
|(Vf)(x)| ≤ ||χ[0,x]||2   ||f||2 =
x
  ||f||2.

Now integrate

1


0
 |(Vf)(x)|2  dx ≤ 
1


0
x ||f||22   dx = 
1 
 2
 ||f||22,  as required.
 18 The strategy here is to solve the equation Ax, y⟩ = ⟨ x, A* y, etc.

(i) Ax, y ⟩=x1y2/1+x2y3/2+…. This must be x, A* y, and so

A*y=(y2/1, y3/2, y4/3, …).

(ii) Rx, u ⟩ = ⟨ x,y ⟩  ⟨ z, u⟩ = ⟨ x, R* u, where R* u = z, u y = ⟨ u, zy.

(iii) Let’s take two vectors m+n and m′+n, with m, m′∈ M and n, n′∈ M. Then

⟨ PM(m+n), m′+n′ ⟩ = ⟨ m, m′+n′⟩= ⟨ m,m′⟩.

This is the inner product m+n,m′⟩ so PM*(m′+n′)=m, which means that PM*=PM again.

 19 Clearly
⟨ Uen, Uem⟩=


1 if  m=n, 
0 if  m ≠ n,
since U preserves the inner product (see notes). Also, if x,Uen ⟩=0 for all n, then U*x, en ⟩=0 for all n, so U*x=0, because (en) is an o.n.b., and so x=0. Hence (Uen) is an o.n.b.

To show that the bilateral right shift V is unitary, you can check any of the equivalent definitions. It is perhaps easiest just to observe that V is clearly a surjection and that ||Vx||=||x|| for all x. Alternatively, you can check that V* is the bilateral left shift, i.e., V*=V−1, by using the identity




n=−∞
xnen , V*
m=−∞
ymem


=



V
n=−∞
xnen ,
m=−∞
ymem


 =



V
n=−∞
xnen+1 , 
m=−∞
ymem


 =
n=−∞
xn
yn+1
,

which tells you that

V*
m=−∞
ymem =
m=−∞
ym+1em.

We saw in the lectures that S is not unitary, since SS*I.

5<0

B.5 Solutions of Tutorial Problems V

 20 Use the definition of adjoint:
⟨ Mfg, h ⟩ = ⟨ g, Mf*h ⟩
for g, hL2(−π,π), which means that
⟨ g, Mf*h ⟩ =
π
−π
f(t) g(t) 
h(t)
  dt.
This is the inner product between g and the function taking values f(t)h(t), so that Mf* g = Mfg, where f′(t)=f(t).

Clearly Mf Mf* g = Mf* Mf g, and is the function whose value at t is f(t) f(t) g(t).

Now Mf is Hermitian if and only if Mf=Mf*, or f=f. So f must be real-valued.

Also Mf is unitary if and only if Mf*=(Mf)−1, which means that f(t)f(t)=1 for all t, i.e. |f(t)|=1 for all t.

 21 Calculate: (I+T+…+Tn−1)(IT)=(IT)(I+T+…+Tn−1)=ITn=I. So we have an inverse for IT.

Of course T is also nilpotent, so IT is invertible, and so (multiplying by λ, which is nonzero), we have λ IT invertible, and λ ∉σ(T).

The spectrum is nonempty, so can only be {0}; indeed it’s obvious that T is not invertible when Tn=0. Hence r(T)=0 as well.

 22 (i) Since Tekk ek, where (en) is the usual orthonormal basis of l2, we see that λk is an eigenvalue, with eigenvector ek. Eigenvalues are always in the spectrum.

(ii) T−λ I takes (xn) to ((λn−λ)xn), and so its inverse must take (yn) to (yn/(λn−λ)). This is a bounded operator, since the sequence n−λ)−1 is bounded when λ ∉Λ.

Now σ(T) is a closed set. It contains Λ, so contains Λ. Indeed σ(T)=Λ, as it contains no points outside Λ, by (ii). Thus any nonempty compact set is the spectrum of some operator!

 23 Note that STS−1−λ I = S(T−λ I)S−1, and so STS−1−λ I is invertible if and only if T−λ I is invertible—indeed in that case its inverse would be S(T−λ I)−1S−1. Hence σ(T)=σ(STS−1).

Also, if Tuu, then STS−1 (Su)=STuSu, and Su ≠ 0 if u ≠ 0. Thus any eigenvector u of T corresponds to an eigenvector Su of STS−1, and vice-versa.

 24 The fact that U is a bijection and an isometry follows from the fact that the functions en(t)=eint/√, n ∈ ℤ form an orthonormal basis of L2(−π,π) (see notes), so that a function f is in L2(−π,π) if and only if f(t)=∑−∞an en, where (an) ∈ l2, and also ||f||2=||(an)||2 (Parseval).

Now UVU−1f= UVU−1n=−∞an en = ∑n=−∞an en+1, because V is the shift.

But if f(t)=∑n=−∞an en(t), then the function n=−∞an en+1(t) is just f(t)eit, since en(t)eit=en+1(t) for all t.

Now we work with the operator T=UVU−1 = Me on L2(−π,π), where e(t)=eit. Using Question 21, we see that this is a unitary operator and so σ(T) ⊆ T, but we can argue more directly.

The operator (T−λ I) is multiplication by eit−λ and its inverse, if it exists, is multiplication by hλ(t)=1/(eit−λ). For λ ∉T, hλC[−π,π] and so T−λ I has a bounded inverse. However, if λ ∈ T, then multiplication by hλ does not give a bounded operator (indeed, Mhλ e0=hλ/√, which is not even in L2). Hence σ(V)=σ(T)=T.

Also T has no eigenvalues, as, no matter which λ ∈ ℂ we choose, there will be no nonzero function f such that f(t)eitf(t) for all t. Hence V has no eigenvalues either, by Question 24.

6<0

B.6 Solutions of Tutorial Problems VI

 25 If T1 and T2 are compact, and (xn) is bounded, then we can find a subsequence (xn(k)) of (xn) such that (T1xn(k)) converges, and a further subsequence (xn(k(l))) such that both (T1xn(k(l))) and (T2xn(k(l))) converge. Then ((a1T1+a2T2)xn(k(l))) converges for any a1, a2 ∈ ℂ, so a1T1+a2T2 is compact. Since the norm limit of compact operators is compact, they form a closed subspace.

Given (xn) bounded, we can find a subsequence (xn(k)) such that (Txn(k)) converges, and hence so does (ATxn(k)), since A is continuous; hence AT is compact. Also (Axn) is bounded so there is a subsequence of (TAxn) that converges, and TA is compact.

 26  ∑n=1||Aen||2 = ∑n=1m=1|⟨ Aen, fm ⟩|2, since (fm) is an o.n.b. This equals
n=1
m=1
|⟨ en, A*fm ⟩|2,
or, summing over n first, m=1||A*fm||2, since (en) is also an o.n.b. Since the right hand side of the displayed formula doesn’t mention (en) it clearly makes no difference if we replace (en) by a different o.n.b. It is also clear that ||A||HS=||A*||HS as the LHS is just ||A||HS2 and the RHS is ||A*||HS2.
 27 (a) T*Tx,y⟩=⟨ Tx, Ty⟩=⟨ x,T*T y, so T*T is Hermitian.

Also TT*x,y⟩=⟨ T*x,T*y⟩ = ⟨ x,T**T*y⟩=⟨ x,TT*y, since T=T**, and hence TT* is Hermitian. Both are compact, since the product of a compact operator and a bounded operator is always compact (by Question 1).

(b) The point is that Tennfn, and so ∑||Ten||2 = ∑|αn|2< ∞ if and only if n) ∈ l2.

If αn → 0, then T is the limit of finite rank operators Tmx=∑n=1m αnx,enfn (cf. what we proved in the course about diagonal operators), and if αn ¬→0, then, for some δ>0, ||Ten(k)||=|αn(k)| ≥ δ, and (Ten(k)) has no convergent subsequence—again, see how we did this for diagonal operators.

⟨ Ten, fm ⟩ = ⟨ en, T*fm ⟩=


αn if  n=m, 
0 otherwise.

Hence T* maps fm to αmem, so T*x=∑m=1αmx,fmem.

This gives

T*Tx=
n=1
αn ⟨ x,enT*fn=
n=1
n|2 ⟨ x,en⟩ en,

and

TT*x=
m=1
αm
⟨ x,fmTem=
m=1
m|2⟨ x,fm⟩ fm.
 28 The Neumann series is
(I−λ T)−1=1+λ T + λ T2 + …,
valid for sufficiently small λ (e.g. |λ|  ||T|| < 1).

Taking f(x)=x, we find that (Tf)(x)=∫01 xy3dy=x/4, and in general (Tnf)(x)=x/4n.

The solution we obtain is φ=(1−λ T)−1f, which gives

φ(x)=x+λ x/4 + λ2x2/16 + …,

which converges to

φ(x)=x/(1−λ/4)=4x/(4−λ),

at least for |λ| < 4. It is easily seen that this solution is valid for all λ ≠ 4.

 29 Calculate
(Tφk)(x)=
π
−π
h(xy) eikydy.
Make the change of variables t=xy to get
(Tφk)(x)=
x


x−π
h(t)eik(xt)  dt = 2π akeikx,
using orthogonality and periodicity properties, so that φk is an eigenvector with eigenvalue ak.

Now T is a Hilbert–Schmidt operator with an orthonormal basis of eigenvectors, namely (ek)=(φk)/√. We can now work with either the (ek) or the k). If φ has Fourier series n=−∞dn eint, then φ−λ Tφ has Fourier series

n=−∞
cneint=
n=−∞
dn(1−λ λn) eint,

where

λn=


0 if  n=0,
2π /|n|if n ≠ 0,

so the solution is

φ(t)=
n=−∞
cn
 1−λ λn
eint.

B.7 Solutions of Tutorial Problems VII

 30 Take e1=t/||t||, and ||t||2=∫−11 t2d t = 2/3, so e1(t)=√3/2t.

Let w2(t)=t2−⟨ t2,e1e1=t2, and normalize to get e2(t)=√5/2t2.

Let w3(t)=t4−⟨ t4,e1e1 − ⟨ t4,e2e2 = t4 − 0 − (2/7)(5/2)t2 = t4− 5t2/7. Now

||w3||2=
1


−1
 (t8−10t6/7+25t4/49)  dt= (2/9)−(20/49)+(10/49)= 8/441,

so we take e3(t)=(21/√8)(t4−5t2/7).

The best approximation in X to f is g=∑k=13f,ekek, giving

g(t)=0 + (2/3)(5/2)t2 + (−8/105)(441/8)(t4−5t2/7),

which reduces to g(t)=14t2/3−21t4/5.

As a check, note that fg is orthogonal to each of the functions t, t2 and t4.

 31 We see that φn(x)=⟨ x, un, where un=(1,1,…,1,0,0,…), with n nonzero terms. Now ||φn||=||un||=√n.
 32 To get the adjoint calculate
⟨ (A+iB)x,y⟩ = ⟨ Ax,y⟩ + i⟨ Bx,y⟩ = ⟨ x,Ay⟩ + i⟨ x,By⟩ = ⟨ x, (AiB)y ⟩,
so T*=AiB.

Now A=(T+T*)/2 and B=(TT*)/(2i) (very like the formulae for real and imaginary parts of a complex number).

Since these formulae do define self-adjoint operators A and B, it is clear that every operator T has a unique decomposition as T=A+iB.

Note

T*T=(AiB)(A+iB)=A2iBA+iAB+B2

and

TT*=(A+iB)(AiB)=A2+iBAiAB+B2,

so that T*TTT*=2i(ABBA), and T is normal if and only if AB=BA.

 33 All we need to do is look for eigenvalues, as the spaces are finite-dimensional.

(i) Dff is impossible unless λ=0, since the degree of Df is lower than the degree of f. So σ(D)={0}. Indeed Dn+1=0, so D is nilpotent, which also implies that its spectrum is {0}, see earlier examples sheets. The only eigenvectors in Pn are constant functions, so we do not get a basis of eigenvectors.

(ii) D(eikt)=ikeikt, so σ(D)={0,± i, ± 2i, …, ± ni}. Now D has (2n+1) distinct eigenvalues, and Tn has an orthonormal basis of eigenvectors, namely (eikt/√)k=−nn.

 34 We get φ=(I−λ T)−1f=fTf + λ2 T2 f + ….

Now Tf(x)=∫0x t2d t=x3/3, (T2f)(x)=∫0x (t5/3)  d t=x6/18, ….

In general (Tn f)(x)=x3n/(3n n!). Summing the series we find that

φ(x)=exp(λ x3/3),

and the series converges for all λ ∈ ℂ.

C Course in the Nutshell

C.1 Some useful results and formulae (1)

 1 A norm on a vector space, ||x||, satisfies ||x||≥ 0, ||x||=0 if and only if x=0, ||λ x||=|λ|  ||x||, and ||x+y|| ≤ ||x|| + ||y|| (triangle inequality). A norm defines a metric and a complete normed space is called a Banach space.
 2 An inner-product space is a vector space (usually complex) with a scalar product on it, x,y⟩ ∈ ℂ such that x,y⟩=y,x, ⟨ λ x,y⟩=λ⟨ x,y, x+y,z⟩ =⟨ x,z⟩ +⟨ y,z, x,x⟩ ≥ 0 and x,x⟩ =0 if and only if x=0. This defines a norm by ||x||2=⟨ x,x. A complete inner-product space is called a Hilbert space. A Hilbert space is automatically a Banach space.
 3 The Cauchy–Schwarz inequality. |⟨ x,y⟩ | ≤ ||x|| ||y|| with equality if and only if x and y are linearly dependent.
 4 Some examples of Hilbert spaces. (i) Euclidean n. (ii) l2, sequences (ak) with ||(ak)||22=∑|ak|2 < ∞. In both cases ⟨ (ak),(bk)⟩=∑akbk. (iii) L2[a,b], functions on [a,b] with ||f||22=∫ab |f(t)|2dt < ∞. Here f,g ⟩=∫ab f(t) g(t)d t. (iv) Any closed subspace of a Hilbert space.
 5 Other examples of Banach spaces. (i) Cb(X), continuous bounded functions on a topological space X. (ii) l(X), all bounded functions on a set X. The supremum norms on Cb(X) and l(X) make them into Banach spaces. (iii) Any closed subspace of a Banach space.
 6 On incomplete spaces. The inner-product (L2) norm on C[0,1] is incomplete. c00 (sequences eventually zero), with the l2 norm, is another incomplete i.p.s.
 7 The parallelogram identity. ||x+y||2 + ||xy||2 = 2||x||2 + 2||y||2 in an inner-product space. Not in general normed spaces.
 8 On subspaces. Complete =⇒ closed. The closure of a linear subspace is still a linear subspace. Lin (A) is the smallest subspace containing A and CLin (A) is its closure, the smallest closed subspace containing A.
 9From now on we work in inner-product spaces.
 10 The orthogonality. xy if x,y⟩ =0. An orthogonal sequence has en,em⟩ =0 for nm. If all the vectors have norm 1 it is an orthonormal sequence (o.n.s.), e.g. en=(0,…,0,1,0,0,…) ∈ l2 and en(t)=(1/√) eint in L2(−π,π).
 11 Pythagoras’s theorem: if xy then ||x+y||2=||x||2+||y||2.
 12 The best approximation to x by a linear combination k=1nλkek is k=1nx,ekek if the ek are orthonormal. Note that x,ek is the Fourier coefficient of x w.r.t. ek.
 13  Bessel’s inequality. ||x||2 ≥ ∑k=1n |⟨ x,ek⟩ |2 if e1,…,en is an o.n.s.
 14  Riesz–Fischer theorem. For an o.n.s. (en) in a Hilbert space, ∑λn en converges if and only if ∑|λn|2 < ∞; then ||∑λn en ||2 = ∑|λn|2.
 15 A complete o.n.s. or orthonormal basis (o.n.b.) is an o.n.s. ( en) such that if y,en⟩ =0 for all n then y=0. In that case every vector is of the form ∑λn en as in the R-F theorem. Equivalently: the closed linear span of the (en) is the whole space.
 16  Gram–Schmidt orthonormalization process. Start with x1, x2, … linearly independent. Construct e1, e2, … an o.n.s. by inductively setting yn+1=xn+1−∑k=1nxn+1,ekek and then normalizing en+1=yn+1/||yn+1||.
 17 On orthogonal complements. M is the set of all vectors orthogonal to everything in M. If M is a closed linear subspace of a Hilbert space H then H=MM. There is also a linear map, PM the projection from H onto M with kernel M.
 18  Fourier series. Work in L2(−π,π) with o.n.s. en(t)=(1/√)eint. Let CP(−π,π) be the continuous periodic functions, which are dense in L2. For fCP(−π,π) write fm=∑n=−mmf,enen, m ≥ 0. We wish to show that ||fmf||2 → 0, i.e., that (en) is an o.n.b.
 19 The Fejér kernel. For fCP(−π,π) write Fm=(f0+…+fm)/(m+1). Then Fm(x)=(1/2π) ∫−ππf(t) Km(xt)  d t where Km(t)=(1/(m+1)) ∑k=0mn=−kk eint is the Fejér kernel. Also Km(t)=(1/(m+1)) [sin2 (m+1)t/2] / [sin2 t/2].
 20  Fejér’s theorem. If fCP(−π,π) then its Fejér sums tend uniformly to f on [−π,π] and hence in L2 norm also. Hence CLin ((en)) ⊇ CP(−π,π) so must be all of L2(−π,π). Thus (en) is an o.n.b.
 21Corollary. If fL2(−π,π) then f(t)=∑cn eint with convergence in L2, where cn=(1/2π) ∫−ππf(t)eintd t.
 22  Parseval’s formula. If f, gL2(−π,π) have Fourier series cn eint and dn eint then (1/2π)⟨ f,g⟩ = ∑cn dn.
 23  Weierstrass approximation theorem. The polynomials are dense in C[a,b] for any a<b (in the supremum norm).

C.2 Some useful results and formulae (2)

 24 On dual spaces. A linear functional on a vector space X is a linear mapping α:X → ℂ (or to in the real case), i.e., α(ax+by)=aα(x)+bα(y). When X is a normed space, α is continuous if and only if it is bounded, i.e., sup{|α(x)|: ||x|| ≤ 1} < ∞. Then we define ||α|| to be this sup, and it is a norm on the space X* of bounded linear functionals, making X* into a Banach space.
 25  Riesz–Fréchet theorem. If α:H → ℂ is a bounded linear functional on a Hilbert space H, then there is a unique yH such that α(x)=⟨ x,y for all xH; also ||α||=||y||.
 26 On linear operator. These are linear mappings T: XY, between normed spaces. Defining ||T||=sup{||T(x)||: ||x|| ≤ 1}, finite, makes the bounded (i.e., continuous) operators into a normed space, B(X,Y). When Y is complete, so is B(X,Y). We get ||Tx|| ≤ ||T||   ||x||, and, when we can compose operators, ||ST|| ≤ ||S||   ||T||. Write B(X) for B(X,X), and for TB(X), ||Tn|| ≤ ||T||n. Inverse S=T−1 when ST=TS=I.
 27 On adjoints. TB(H,K) determines T*B(K,H) such that Th, kK = ⟨ h, T*kH for all hH, kK. Also ||T*||=||T|| and T**=T.
 28 On unitary operator. Those UB(H) for which UU*=U*U=I. Equivalently, U is surjective and an isometry (and hence preserves the inner product).

Hermitian operator or self-adjoint operator. Those TB(H) such that T=T*.

On normal operator. Those TB(H) such that TT*=T*T (so including Hermitian and unitary operators).

 29 On spectrum. σ(T)={λ ∈ ℂ: (T−λ I) is not invertible in B(X)}. Includes all eigenvalues λ where Txx for some x ≠ 0, and often other things as well. On spectral radius: r(T)=sup{|λ|: λ∈ σ(T)}. Properties: σ(T) is closed, bounded and nonempty. Proof: based on the fact that (IA) is invertible for ||A|| < 1. This implies that r(T) ≤ ||T||.
 30 The spectral radius formula. r(T)=infn ≥ 1 ||Tn||1/n = limn → ∞ ||Tn||1/n.

Note that σ(Tn)={λn: λ ∈ σ(T)} and σ(T*)={λ: λ ∈ σ(T)}. The spectrum of a unitary operator is contained in {|z|=1}, and the spectrum of a self-adjoint operator is real (proof by Cayley transform: U=(TiI)(T+iI)−1 is unitary).

 31 On finite rank operator. TF(X,Y) if ImT is finite-dimensional.

On compact operator. TK(X,Y) if: whenever (xn) is bounded, then (Txn) has a convergent subsequence. Now F(X,Y) ⊆ K(X,Y) since bounded sequences in a finite-dimensional space have convergent subsequences (because when Z is f.d., Z is isomorphic to l2n, i.e., S:l2nZ with S, S−1 bounded). Also limits of compact operators are compact, which shows that a diagonal operator Tx=∑λnx,enen is compact iff λn → 0.

 32  Hilbert–Schmidt operators. T is H–S when ∑ ||Ten||2 < ∞ for some o.n.b. (en). All such operators are compact—write them as a limit of finite rank operators Tk with Tkn=1anen=∑n=1k an (Ten). This class includes integral operators T: L2(a,b)→ L2(a,b) of the form
(Tf)(x)=
b
a
K(x,y) f(y) dy,
where K is continuous on [a,b] × [a,b].
 33 On spectral properties of normal operators. If T is normal, then (i) kerT=kerT*, so Txx =⇒ T*x=λx; (ii) eigenvectors corresponding to distinct eigenvalues are orthogonal; (iii) ||T||=r(T).

If TB(H) is compact normal, then its set of eigenvalues is either finite or a sequence tending to zero. The eigenspaces are finite-dimensional, except possibly for λ=0. All nonzero points of the spectrum are eigenvalues.

 34 On spectral theorem for compact normal operators. There is an orthonormal sequence (ek) of eigenvectors of T, and eigenvalues k), such that Tx=∑k λkx,ekek. If k) is an infinite sequence, then it tends to 0. All operators of the above form are compact and normal.

Corollary. In the spectral theorem we can have the same formula with an orthonormal basis, adding in vectors from kerT.

 35 On general compact operators. We can write Tx=∑µkx, ekfk, where (ek) and (fk) are orthonormal sequences and k) is either a finite sequence or an infinite sequence tending to 0. Hence TB(H) is compact if and only if it is the norm limit of a sequence of finite-rank operators.
 36 On integral equations. Fredholm equations on L2(a,b) are Tφ=f or φ−λ Tφ=f, where (Tφ)(x)=∫ab K(x,y)φ(y) d y. Volterra equations similar, except that T is now defined by (Tφ)(x)=∫ax K(x,y)φ(y) d y.
 37  Neumann series. (I−λ T)−1=1+λ T2 T2 + …, for ||λ T ||<1.

On separable kernel. K(x,y)=∑j=1n gj(x)hj(y). The image of T (and hence its eigenvectors for λ≠ 0) lies in the space spanned by g1,…,gn.

 38  Hilbert–Schmidt theory. Suppose that KC([a,b]× [a,b]) and K(y,x)=K(x,y). Then (in the Fredholm case) T is a self-adjoint Hilbert–Schmidt operator and eigenvectors corresponding to nonzero eigenvalues are continuous functions. If λ≠ 0 and 1/λ ∉σ(T), the the solution of φ−λ Tφ=f is
φ=
k=1
⟨ f,vk ⟩
1−λλk
vk.
 39  Fredholm alternative. Let T be compact and normal and λ≠ 0. Consider the equations (i) φ−λ Tφ=0 and (ii) φ−λ Tφ=f. Then EITHER (A) The only solution of (i) is φ=0 and (ii) has a unique solution for all f OR (B) (i) has nonzero solutions φ and (ii) can be solved if and only if f is orthogonal to every solution of (i).

D Supplementary Sections

D.1 Reminder from Complex Analysis

The analytic function theory is the most powerful tool in the operator theory. Here we briefly recall few facts of complex analysis used in this course. Use any decent textbook on complex variables for a concise exposition. The only difference with our version that we consider function f(z) of a complex variable z taking value in an arbitrary normed space V over the field ℂ. By the direct inspection we could check that all standard proofs of the listed results work as well in this more general case.

Definition 1 A function f(z) of a complex variable z taking value in a normed vector space V is called differentiable at a point z0 if the following limit (called derivative of f(z) at z0) exists:
f′(z0)=
 
lim
Δ z→ 0
f(z0z)−f(z0)
Δ z
. (108)
Definition 2 A function f(z) is called holomorphic (or analytic) in an open set Ω⊂ℂ it is differentiable at any point of Ω.
Theorem 3 (Laurent Series) Let a function f(z) be analytical in the annulus r<z<R for some real r<R, then it could be uniquely represented by the Laurent series:
f(z)=
k=−∞
ckzk,     for some ck∈ V. (109)
Theorem 4 (Cauchy–Hadamard) The radii r and R, (r′<R) of convergence of the Laurent series (109) are given by
r′= 
 
liminf
n→ ∞
⎪⎪
⎪⎪
cn⎪⎪
⎪⎪
1/n   and   
1
R
=
 
limsup
n→ ∞
⎪⎪
⎪⎪
cn⎪⎪
⎪⎪
1/n. (110)

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1
Some more “strange” types of orthogonality can be seen in the paper Elliptic, Parabolic and Hyperbolic Analytic Function Theory–1: Geometry of Invariants.
2
Spectrum—ghost, spirit(lat.).
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