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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. □
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Last modified: November 6, 2024.
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