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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. □
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Last modified: November 6, 2024.
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