Download Functional Analysis Lecture Notes

Functional Analysis Lecture Notes
Mahmoud Filali and Juho Rosqvist
2007
Contents
1 Linear Spaces
Elementary Concepts . . . . . . .
Subspaces and Quotients . .
Hamel Basis and Dimension
Linear Operators . . . . . . . . .
Linear Functionals . . . . .
The Algebraic Dual . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Norms and Seminorms . . . . . . . . . . . . . . . . . . .
Subspaces and Quotients . . . . . . . . . . . . . . . . . .
Bounded Linear Operators . . . . . . . . . . . . . . . . .
Bounded Linear Functionals and the Banach Dual
A Brief Excursion into Baire Spaces . . . . . . . . . . .
Applications of the Baire Category Theorem . . . . . .
Uniform Boundedness Principle . . . . . . . . . . .
Open Mapping Theorem . . . . . . . . . . . . . . .
Finite-dimensional Normed Spaces . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
. . .
. . .
Dual
. . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
2 Normed Spaces
3 Topological Vector Spaces
Local Base of Balanced and Absorbing Sets . . . . . . .
Boundedness and Continuity of Linear Operators . . . .
Continuous Linear Functionals and the Continuous
Convexity . . . . . . . . . . . . . . . . . . . . . . . . . .
4 Hahn-Banach Theorem
Hahn-Banach Theorem . . . . . . . . . . . .
Annihilators . . . . . . . . . . . . . . . . . .
Hyperplanes . . . . . . . . . . . . . . . . . .
Geometric Forms of Hahn-Banach Theorem
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
5 Extreme Points and the Krein-Milman Theorem
6 Weak Topologies
General Weak Topologies . . . . . . . . .
Second Dual and the Canonical Mapping
The Weak Topology on Normed Spaces .
The Weak∗ Topology on Banach Duals . .
1
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
3
3
4
4
8
10
11
12
12
13
16
18
20
23
23
24
27
32
34
35
37
38
40
41
47
50
53
59
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
62
62
65
66
67
7 Reexive Spaces
Separability Considerations . . . . . . . . . . . .
Duals and Subspaces of Reexive Spaces . . . . .
Alaoglu's Theorem and Boundedness in the Dual
Goldstine's Theorem . . . . . . . . . . . . . . . .
Exercises
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
69
69
71
72
77
88
2
Chapter 1
Linear Spaces
Elementary Concepts
Denition. Let K be a eld and let 1 denote its unit element. An abelian group
(X, +) together with a scalar multiplication mapping (α, x) 7→ αx of K ×X onto
X , is called a linear space (or a vector space ) over K, if the following conditions
are satised:
(1) 1x = x for all x ∈ X ;
(2) α(x + y) = αx + αy for all α ∈ K and x, y ∈ X ;
(3) (α + β)x = αx + βx for all α, β ∈ K and x ∈ X ;
(4) α(βx) = (αβ)x for all α, β ∈ K and x ∈ X .
The elements of X are called vectors and the elements of K scalars .
¤
We shall denote by 0 both the additive neutral element of the eld K, and the
zero vector, that is, the neutral element of the abelian group X . Also, as seen
above, + is used to denote addition in the eld K as well as the group operation
in X . The meaning of the symbols should always be clear from the context.
1.1 Proposition. Let X be a linear space over K. Then
(1) 0x = 0 for all x ∈ X ;
(2) (−1)x = −x for all x ∈ X ;
(3) α0 = 0 for all α ∈ K;
(4) given α ∈ K and x, y ∈ X , if αx = αy and α 6= 0, then x = y ;
(5) given α, β ∈ K and x ∈ X , if αx = βx and x 6= 0, then α = β .
The previous proposition can be condensed into the following form which is
probably easier to memorize.
1.2 Corollary. Let X be a linear space over K. Given x ∈ X and α ∈ K,
(1) (−1)x = −x and
(2) αx = 0 if and only if α = 0 or x = 0.
3
Subspaces and Quotients
Denition. A subset Y of a linear space X is called a linear subspace of X
if Y is a linear space over the same eld as X with the addition and scalar
multiplication of X restricted to Y .
¤
1.3 Proposition. A non-empty subset Y of a linear space X over K is a linear
subspace of X if and only if the following conditions hold:
(1) y1 + y2 ∈ Y whenever y1 , y2 ∈ Y ;
(2) αy ∈ Y whenever α ∈ K and y ∈ Y .
The preceding two conditions can be combined:
1.4 Proposition. A non-empty subset Y of a linear space X over K is a linear
subspace of X if and only if y1 + αy2 ∈ Y whenever y1 , y2 ∈ Y and α ∈ K.
Since the intersection of any family of subspaces of a linear space X is a
subspace of X (Exercise 3), there exists, given a subset S of X , a minimal
subspace of X containing S (minimal of course in the sense that it is contained
in any subspace that contains S ).
Denition. Let S be a subset of a linear space X over K. The linear span of
S is the minimal subspace of X containing S , and it is denoted by Span(S). ¤
Next we look at one way of dening new linear spaces from a given linear
space X over K. The details of this construction are left as an exercise to the
reader. Let N be a subspace of X . A relation ∼ dened on X by
x ∼ y ⇐⇒ x − y ∈ N
is an equivalence, so it partitions X into disjoint equivalence classes. Since
y ∼ x if and only if y = x + n for some n ∈ N , the equivalence class of x is
x + N = {x + n : n ∈ N }. The equivalence classes are also called cosets of N
and the set of all cosets is denoted by X/N = {x + N : x ∈ X}. It is a simple
task to verify that dening addition on X/N by (x + N ) + (y + N ) = (x + y) + N
makes X/N an abelian group with N = 0 + N as the zero element. Proceeding
analogously, the denition of scalar multiplication by α(x + N ) = (αx) + N for
x + N ∈ X/N and α ∈ K makes X/N a linear space over K. The linear space
X/N is called the quotient space of X by N .
Hamel Basis and Dimension
Denition. A nite subset {x1 , x2 , . . . , xn } of a linear space X is linearly dependent when there exist scalars α1 , α2 , . . . , αn not all zero such that
α1 x1 + α2 x2 + · · · + αn xn = 0.
Otherwise, the subset is said to be linearly independent.
An innite subset of a linear space is said to be linearly independent when
all of its nite subsets are linearly independent. Otherwise, the set is linearly
dependent.
¤
4
1.5 Lemma. If S and T are two linearly independent subsets of a linear space
X and Span(S) ∩ Span(T ) = {0}, then S ∪ T is linearly independent.
Proof.
Let F be an arbitrary nite subset of S ∪ T , and suppose that
X
X
αs s +
βt t = 0
s∈S∩F
(1.1)
t∈T ∩F
for some set of scalars αs and βt . But then
X
X
Span(S) 3
αs s = −
βt t ∈ Span(T ),
s∈S∩F
which implies that
X
t∈T ∩F
X
αs s =
s∈S∩F
βt t = 0
t∈T ∩F
since Span(S)∩Span(T ) = {0}. The scalars αs and βt above are all zero because
S and T are linearly independent sets. We started from (1.1), so this proves
the linear independence of F .Thus S ∪ T is linearly independent since F was
arbitrary.
¥
1.6 Corollary. Let S be a linearly independent subset of a linear space X . If
x ∈ X \ Span(S), then S ∪ {x} is linearly independent.
Note that αx 6∈ Span(S) whenever α 6= 0, for otherwise x = α−1 (αx) ∈
Span(S). Thus Span({x}) ∩ Span(S) = {0}. The set {x} is linearly independent
because x 6= 0.
¥
Proof.
Lemma 1.5 and Corollary 1.6 shall be used without comment in the sequel.
Denition. We say that a subset S of a linear space X is a (Hamel ) basis for
X when S is linearly independent and Span(S) = X .
¤
1.7 Theorem. If L is a linearly independent subset of a linear space X , then
L is contained in some basis for X .
Proof. Let S be the set of all linearly independent subsets of X that contain L.
Then S 6= ∅ is partially ordered by set inclusion, i.e., the relation ⊆ is reexive,
transitive and antisymmetric on S .
Let C be a chain in S , i.e., C ⊆ S and if S, T S
∈ C then S ⊆ T or T ⊆ S . We
show that C has an upper bound, namely U := C . Clearly, L ⊆ U and every
S ∈ C satises S ⊆ U . To see that U is linearly independent, suppose that
α1 x1 + · · · + αn xn = 0
for some scalars α1 , . . . , αn , vectors x1 , . . . , xn ∈ U and n ∈ N. Then each xk
belongs to some Sk ∈ C , k = 1, . . . , n, and since C is a chain, one of these Sk 's, T
say, contains all the others. So x1 , . . . , xn ∈ T and T ∈ C is linearly independent,
hence α1 = · · · = αn = 0 and U ∈ S is an upper bound, as claimed.
Since every chain in S has an upper bound, there exists a maximal element
M ∈ S by Zorn's lemma, i.e., if S ∈ S and M ⊆ S then S = M . Now M must
be a basis for X : M is by denition linearly independent, and if there was some
x ∈ X with x 6∈ Span(M ), then M ∪ {x} would be linearly independent, which
is impossible because M is maximal.
¥
5
1.8 Corollary. Every nonzero linear space has a Hamel basis.
If x 6= 0, then {x} is linearly independent and can therefore be extended
to a basis for the whole space.
¥
Proof.
1.9 Corollary. Let S be any non-empty subset of a linear space X , and let L
be any linearly independent subset of S . Then there exists a maximal linearly
independent subset M of S that contains L.
Repeat the beginning of the proof of Theorem 1.7 with additional constraints on S : let S be the set of all linearly independent subsets of S that
contain L.
¥
Proof.
1.10 Theorem. Every basis of a given linear space has the same cardinality.
Proof. Let S and T be bases for a linear space X over K. We shall demonstrate
that there exists an injection Φ : S → T . This will be enough, since the roles
of S and T can be interchanged to produce an injection on T into S . Then
applying the Schröder-Bernstein theorem (Exercise 6) ensures that there exists
a bijection between S and T , which by denition means that the sets S and T
are of the same cardinality.
Consider the set F of functions dened as follows: ϕ ∈ F if and only if
(1) the domain Dϕ and range Rϕ of ϕ satisfy Dϕ ⊆ S and Rϕ ⊆ T ,
(2) ϕ : Dϕ → Rϕ is one-to-one, and
(3) the set Rϕ ∪ (S \ Dϕ ) is linearly independent.
We must check that F is non-empty. Pick any s ∈ S and choose t ∈ T so that
t 6∈ Span(S \ {s}). Such a t exists because T spans X , and T ⊆ Span(S \ {s})
would imply that X = Span(T ) ⊆ Span(S \ {s}) $ X , which is absurd. Now,
{t} ∪ (S \ {s}) is linearly independent due to the choice of t because S \ {s} is
linearly independent. Hence the function ϕ = {(s, t)} belongs to F . Now dene
a relation ¹ on F by
ϕ ¹ ψ ⇐⇒ ψ is an extension of ϕ,
that is, Dϕ ⊆ Dψ and ψ(x) = ϕ(x) for all x ∈ Dϕ . A moment's reection
assures one that ¹ is a partial order on F .
If F0 is a chain in F , let
[
Dϕ0 =
Dϕ
ϕ∈F0
and dene a function ϕ0 on Dϕ0 as follows. Given x ∈ Dϕ0 , there exists some
ϕ ∈ F0 such that x ∈ Dϕ and we set ϕ0 (x) = ϕ(x). This is well-dened: if
ψ ∈ F0 then one of the functions ϕ and ψ extends the other because F0 is a
chain, and hence they agree at every point x that belongs to the domains of
both functions. We prove that ϕ0 thus dened is an upper bound for F0 .
That (1) holds for ϕ0 is clear, because
³ [
´
[
[
Rϕ0 = ϕ0 (Dϕ0 ) = ϕ0
Dϕ =
ϕ0 (Dϕ ) =
Rϕ
ϕ∈F
ϕ∈F0
6
ϕ∈F0
and the chain F0 is a subset of F .
If x, y ∈ Dϕ0 and x 6= y , there exist ϕx , ϕy ∈ F0 such that x ∈ Dϕx and
y ∈ Dϕy . Since F0 is a chain, one of the functions ϕx and ϕy , call it ϕ, is
greater, i.e., ϕx ¹ ϕ and ϕy ¹ ϕ. Since ϕ is one-to-one, ϕ(x) 6= ϕ(y). Because
ϕ0 extends ϕ ∈ F0 , this implies that ϕ0 (x) 6= ϕ0 (y), and so ϕ0 satises (2).
As for (3), note that if A = {xk : k = 1, . . . , n} is an arbitrary nite subset
of Rϕ0 ∪ (S \ Dϕ0 ), then for each xk ∈ A, either there is some ϕk ∈ F0 such
that xk ∈ Rϕk , or xk ∈ S \ Dϕ for all ϕ ∈ F0 . Since A is nite, we may pick
ϕ ∈ F0 so that ϕ º ϕk for all k = 1, . . . , n (either take the greatest ϕk or if there
are none, any element of the chain). Then A ⊆ Rϕ ∪ (S \ Dϕ ), so A is linearly
independent as a subset of a linearly independent set. Thus Rϕ0 ∪ (S \ Dϕ0 )
is linearly independent and ϕ0 ∈ F . It is clear from the denition of ϕ0 that
ϕ ¹ ϕ0 for all ϕ ∈ F0 .
Having shown that each chain in F has an upper bound, we invoke Zorn's
lemma to ascertain the existence of a maximal element Φ ∈ F . It remains to
prove that DΦ = S .
Suppose the contrarythat S \ DΦ 6= ∅. Then it follows that T 6= RΦ ,
because Span(T ) = X and Span(RΦ ) ∩ (S \ DΦ ) = ∅ by property (3) of Φ. Let
t 0 ∈ T \ RΦ .
Then either t0 ∈ Span(RΦ ∪ (S \ DΦ )) or not.
In the former case
X
X
t0 =
αt t +
βs s,
t∈RΦ
(1.2)
s∈S\DΦ
where only nitely many scalars αt and βs are nonzero, and since t0 is linearly
independent of RΦ , at least one βs is not zero, say βs0 6= 0. Then the extension
Φ0 : DΦ ∪ {s0 } → RΦ ∪ {t0 }
of Φ dened by Φ0 (s0 ) = t0 obviously satises properties (1) and (2). Also
RΦ0 ∪ (S \ DΦ0 ) is linearly independent, because otherwise (1.2) would also
hold with βs0 = 0, which is impossible (see Exercise 5) since RΦ ∪ (S \ DΦ ) is
linearly independent. Thus Φ0 also satises (3) and so Φ0 ∈ F . But this is a
contradiction with the maximality of Φ because Φ ¹ Φ0 and Φ0 6= Φ.
Therefore the only possibility is the latter case:
t0 6∈ Span(RΦ ∪ (S \ DΦ )).
(1.3)
Pick any s0 ∈ S \ DΦ and dene again the extension Φ0 of Φ by Φ0 (s0 ) = t0 .
Then Φ0 clearly satises the properties (1) and (2). But (3) holds too, because
RΦ0 ∪ (S \ DΦ0 ) ⊆ {t0 } ∪ (RΦ ∪ (S \ DΦ )),
which is a linearly independent set by (3) and (1.3). Thus Φ0 ∈ F , which is
again impossible due to the maximality of Φ. Therefore we must have DΦ = S
and the proof is complete.
¥
The previous theorems 1.10 and 1.8 ensure that the following denition of dimension is well posed and that every linear space has a dimension. It
Remark.
7
is worth noting that both proofs depended on the axiom of choice, which was
used in the form of Zorn's lemma. In the case of nite-dimensional spaces (see
below for denition), the proofs for both theorems are very simple and do not
depend on the axiom of choice. The reader is advised to carry them out.
¤
Denition. The cardinal number of a basis for a nonzero linear space X (over
K) is called the Hamel dimension of X (over K) and denoted by dim X . The
dimension of the zero space {0} is dened to be 0. The space X is said to
be nite-dimensional if dim X is a natural number, and innite-dimensional
otherwise.
¤
When we talk about basis and dimension, we shall always mean Hamel basis and dimension, unless otherwise specied. Note that the scalar eld K of
X although often inferred from contextis not irrelevant when determining
dimension. For instance, the set of complex numbers C is a linear space over C
of dimension 1, but C is also a linear space over R of dimension 2 (think of a
few bases for each of these spaces).
Linear Operators
Denition. Let X and Y be linear spaces over the same eld K and T : X → Y
a function. We say that T is a linear operator (or a linear transformation ) when
(1) T (x1 + x2 ) = T (x1 ) + T (x2 ) for all x1 , x2 ∈ X , and
(2) T (αx) = αT (x) for all α ∈ K and x ∈ X .
The value of T at x ∈ X is often denoted plainly by T x. The range (or
image ) of T is denoted and dened by Range(T ) = T (X) = {T x : x ∈ X} and
the kernel of T is Ker(T ) = {x ∈ X : T x = 0}.
¤
1.11 Proposition. Let X and Y be linear spaces over K. Then T : X → Y is
a linear operator if and only if
T (x1 + αx2 ) = T x1 + αT x2
for all x1 , x2 ∈ X and α ∈ K.
Proof.
Exercise 10.
¥
1.12 Proposition. If T : X → Y is a linear operator, then Range(T ) is a
subspace of Y and Ker(T ) is a subspace of X .
Proof.
Exercise 11.
¥
1.13 Proposition. Let X and Y be linear spaces over K, and let L(X, Y )
be the set of all linear operators from X into Y equipped with the pointwise
operations, that is,
(S + T )(x) = Sx + T x
(αT )(x) = αT x
for all x ∈ X
for all x ∈ X
whenever S, T ∈ L(X, Y ) and α ∈ K. Then L(X, Y ) is a linear space over K.
8
Proof.
Exercise 12.
¥
The following result is often useful when constructing linear operators.
1.14 Lemma. Let X and Y be linear spaces over K, and let B be a basis for
X . If f : B → Y is any function and the function f˜ : X → Y is dened by
Ã
!
X
X
˜
f
αb b =
αb f (b)
b∈B∩F
b∈B∩F
whenever F is a nite subset of B and αb ∈ K for all b ∈ B , then f˜ is a linear
operator that extends the function f .
Proof.
Exercise 13.
¥
1.15 Proposition. Let T : X → Y be a linear operator. Then the inverse T −1
of T : X → Range(T ) exists if and only if Ker(T ) = {0}. When T −1 exists, it
is a linear operator.
1.16 Theorem. If T : X → Y is a linear operator, then
dim Ker(T ) + dim Range(T ) = dim X.
(1.4)
Proof. If Ker(T ) = {0}, then by Proposition 1.15, T is a linear bijection
between X and Range(T ), so they must have the same dimension: the image
of any basis for X under T is a basis for Range(T ). Since dim {0} = 0, (1.4)
holds.
Suppose then that Ker(T ) 6= {0} and let A be a basis for Ker(T ). Then by
Theorem 1.7, there exists a basis C for X such that A ⊆ C . Let
B = C \ A.
Then Span(A) ∩ Span(B) = {0} because C is linearly independent. We prove
that T (B) = {T b : b ∈ B} is a basis for Range(T ). First, T (B) spans Range(T )
because C = A ∪ B , T (C) obviously spans Range(T ), and T a = 0 for all
a ∈ A ⊆ Ker(T ). Second, if T b1 , . . . , T bn ∈ T (B) and
à n
!
n
X
X
0=
αk T bk = T
αk bk
k=1
k=1
P
for some scalars α1 , . . . , αn , then
αk bk ∈ Ker(T ) ∩ Span(B) = {0}. Since B
is linearly independent, α1 = · · · = αn = 0, so T (B) is linearly independent and
a basis for Range(T ).
A similar argument shows that T is injective on B and hence a bijection
between B and T (B), so |B| = |T (B)| = dim Range(T ). On the other hand,
|A| = dim Ker(T ) and |C| = dim X . This proves (1.4), because |A| + |B| = |C|
when C = A ∪ B and A ∩ B = ∅ (this is the denition of the sum of cardinal
numbers).
¥
1.17 Corollary. Let T : X → Y be a linear operator. Then
(1) dim Range(T ) ≤ dim X ;
(2) if T −1 : Range(T ) → X exists, then dim Range(T ) = dim X ;
9
(3) if dim X = dim Y < ∞, then T −1 exists iff Range(T ) = Y .
Proof.
Trivial from equation (1.4). See also Exercise 14.
¥
Denition. Two linear spaces X and Y are said to be isomorphic when there
exists a bijective linear operator T : X → Y . Then T is called a (linear )
isomorphism.
¤
Note that isomorphic linear spaces must have the same underlying eld.
1.18 Theorem. Linear spaces of the same nite dimension and over the same
eld K are all isomorphic.
Proof.
Exercise 15.
¥
Linear Functionals
Denition. Let X be a vector space over K. A linear operator T : X → K is
called a linear functional on X . The set of all linear functionals on X is called
the algebraic dual of X and denoted by X 0 . The value of x0 ∈ X 0 at the vector
x ∈ X is usually denoted by hx0 , xi.
¤
1.19 Proposition. Every linear functional on a linear space is either identically
zero or surjective.
Proof.
Exercise 16
¥
Denition. A subspace M of a linear space X is said to be maximal when the
only subspace of X that properly contains M is X itself.
¤
1.20 Theorem. If X is a linear space and x0 is a nonzero linear functional on
X , then Ker(x0 ) is a maximal subspace of X .
Since Ker(x0 ) is a subspace by Proposition 1.12, we only need to show
that it is maximal. Since x0 6= 0, there exists y ∈ X \ Ker(x0 ). Thus for every
x ∈ X,
hx0 , xi
x− 0
y ∈ Ker(x0 ),
hx , yi
Proof.
and therefore x ∈ Ker(x0 ) + Ky . Hence Span(Ker(x0 ) ∪ {y}) = X and Ker(x0 )
must be maximal.
¥
1.21 Lemma. Let x01 , . . . , x0n and x0 be linear functionals on a linear space
X . Then x0 is a linear combination of the functionals x01 , . . . , x0n if and only if
Ker(x01 ) ∩ · · · ∩ Ker(x0n ) ⊆ Ker(x0 ).
The suciency of the condition is clear. We prove the necessity by
induction on n. Let Pn be the proposition
that if x01 , . . . , x0k and x0 are linear
Tn
functionals on a linear space X and k=1 Ker(x0k ) ⊆ Ker(x0 ), then x0 is a linear
combination of x01 , . . . , x0n .
Let x0 and x01 satisfy the hypotheses of P1 . Note rst, that the desired
conclusion is trivial if either of the functionals x0 and x01 is zero. Hence we
can assume that they are both nonzero. Then by Theorem 1.20, Ker(x01 ) is a
Proof.
10
maximal subspace of X , and therefore Ker(x0 ) = Ker(x01 ). Pick z ∈ X \ Ker(x01 )
and dene a functional z 0 ∈ X 0 by
z 0 = x0 −
hx0 , zi 0
x .
hx01 , zi 1
Then Ker(x0 ) ⊆ Ker(z 0 ) and hz 0 , zi = 0, so the kernel of z 0 must be all of X due
to the maximality of Ker(x0 ). So z 0 is identically zero and therefore
x0 =
hx0 , zi 0
x ,
hx01 , zi 1
as required for P1 .
Let n ≥ 2, assume that Pk holds for all k = 1, . . . , n − 1, and suppose
0
that x01 , . . . , x0n and x0 satisfy the hypotheses of Pn . Let y10 , . . . , yn−1
and y 0
0
0
0
be the restrictions of the functionals x1 , . . . , xn−1 and x , respectively, to the
0
subspace Ker(x0n ) of X . Then Ker(y10 ) ∩ · · · ∩ Ker(yn−1
) ⊆ Ker(y 0 ), so y 0 =
0
0
α1 y1 + · · · + αn−1 yn−1 for some scalars α1 , . . . , αn−1 by Pn−1 . Let
z 0 = x0 −
n−1
X
αk x0k .
k=1
0
0
0
Then z is a linear functional that vanishes on Ker(x0n ), so Ker(x
Pn n ) ⊆ 0Ker(z ).
0
0
0
By P1 , we have z = αn xn for some scalar αn , that is, x = k=1 αk xk . Thus
Pn holds and the induction is complete.
¥
The Algebraic Dual
If we dene the sum x0 + y 0 of x0 ∈ X 0 and y 0 ∈ X 0 with
hx0 + y 0 , xi = hx0 , xi + hy 0 , xi
for all x ∈ X
(1.5)
and the scalar multiple αx0 of x0 ∈ X 0 by α ∈ K with
hαx0 , xi = αhx0 , xi
for all x ∈ X,
(1.6)
then X 0 becomes a linear space over K with these operations (Exercise 17).
Since X 0 is itself a linear space, we may consider its dual (X 0 )0 =: X 00 , which
we refer to as the second (algebraic ) dual of X .
Now, for every x ∈ X there corresponds an element ϕ(x) ∈ X 00 dened by
hϕ(x), x0 i = hx0 , xi
for all x0 ∈ X 0 .
(1.7)
The mapping ϕ(x) is sometimes referred to as the evaluation map at the point
x. It is easily checked (Exercise 18) that for every x ∈ X , ϕ(x) is linear, that
is, ϕ(x) ∈ X 00 . We have thus dened a mapping ϕ : X → Range(ϕ) ⊆ X 00
which is linear and injective (Exercise 18), so it has an inverse. Hence X and
Range(ϕ) are isomorphic, and we may identify X and Range(ϕ) and regard X
as a subspace of X 00 .
Denition. The mapping ϕ : X → X 00 with ϕ(x) dened by (1.7) is called the
canonical embedding (or natural embedding ) of X into X 00 . If ϕ is onto, we say
that X is algebraically reexive.
¤
11
Chapter 2
Normed Spaces
Norms and Seminorms
Denition. Let X be a real or complex linear space. A function p : X → R is
called a seminorm when it satises the following two conditions:
(1) p(x + y) ≤ p(x) + p(y) for all x, y ∈ X ;
(2) p(αx) = |α| p(x) for all α ∈ K and x ∈ X .
¤
2.1 Proposition. Let p be a seminorm on a linear space X . Then
(1) p(0) = 0,
(2) p(x) ≥ 0 for all x ∈ X , and
(3) |p(x) − p(y)| ≤ p(x ± y) ≤ p(x) + p(y) for all x, y ∈ X .
Moreover, if p(x) = 0 implies x = 0, then p is a norm.
Proof.
Exercise 20.
¥
A seminorm that vanishes only at the zero vector is called a norm. However, the following denition of a normthough clearly equivalent by Proposition 2.1is perhaps more familiar and common in the literature.
Denition. Let X be a linear space over K (K = R or K = C). A mapping
k·k : X → R is called a norm on X when the following conditions are satised
for all α ∈ K and x, y ∈ X :
(1) kxk ≥ 0,
(2) kxk = 0 if and only if x = 0,
(3) kαxk = |α| kxk ,
(4) kx + yk ≤ kxk + kyk .
We say that X endowed with a norm is a normed space. When X is complete
in the metric given by the norm, d(x, y) = kx − yk , we say that X is a Banach
space.
¤
12
When we speak about the norm topology of X , we mean the metric space topology induced by the metric d dened above (see Exercise 22). Unless otherwise
indicated, topological concepts are to be understood in the norm topology.
As in any metric space, if x ∈ X and r > 0, we denote the open ball of radius
r about x by B(x, r) = {y ∈ X : d(x, y) < r}, and the corresponding closed ball
by B̄(x, r) = {y ∈ X : d(x, y) ≤ r}. Note that although B̄(x, r) is closed, it is
not necessarily the closure of B(x, r) despite our notation. However, the closure
of B(x, r) in a normed space is B̄(x, r). See Exercise 25.
The seminorms are very useful in describing topologies on a vector space, but
we shall only concern ourselves with norms in this chapterafter the following
exception, which elaborates on our earlier denition of a norm.
2.2 Proposition. If p is a seminorm on a linear space X , then the set N =
{x ∈ X : p(x) = 0} is a linear subspace of X and X/N is a normed space with
the norm kx + N k = p(x).
Proof.
Exercise 21.
¥
2.3 Proposition. Let X be a normed space over K. Then the following map-
pings are continuous with respect to the norm topology on X and the product
topologies on X × X and K × X (K has the usual topology)
(1) a : X × X → X dened by a(x, y) = x + y ,
(2) s : K × X → X dened by s(α, x) = αx,
(3) n : X → R dened by n(x) = kxk .
Proof.
Exercise 23.
¥
The continuity of the vector space operations is a very important property of
normed spaces, and it is in fact the property we choose as a starting point when
we begin generalizing the notions of this chapter to topological vector spaces in
chapter 3.
2.4 Proposition. Let X be a normed space over K, and let y ∈ X and α ∈ K
be xed. Then the mappings x 7→ y + x and x 7→ αx are both homeomorphisms
of X onto itself.
Proof.
Exercise 24. This is also a special case of Theorem 3.1.
¥
Subspaces and Quotients
2.5 Theorem. Let X be a normed space and Y a linear subspace of X . Then
the following assertions hold.
(1) The closure Y of Y is also a linear subspace of X .
(2) Y is a normed space with the norm of X restricted to Y .
(3) If Y is complete, then Y is closed.
(4) If X is a Banach space, then Y is closed if and only if Y is complete.
13
Proof. (1) Let x, y ∈ Y and α ∈ K be arbitrary. Then there are sequences
∞
(xn )∞
n=1 and (yn )n=1 in Y converging to x and y , respectively, in X . Since Y is
a linear space, (xn + αyn )∞
n=1 is also a sequence in Y . Now,
k(xn + αyn ) − (x + αy)k ≤ kxn − xk + |α| kyn − yk −→ 0
as n −→ ∞, that is, xn + αyn −→ x + αy . Thus x + αy ∈ Y , and so Y is a
linear subspace of X by Proposition 1.4.
(2) This statement is obvious.
(3) Let (yn )∞
n=1 be a sequence in Y converging to some y in X . Since convergent
sequence in X is Cauchy, our sequence is also Cauchy in the subspace Y . Since
Y is complete, this sequence must converge in Y . Because limits in norm are
unique, y ∈ Y and thus Y is closed.
(4) We just proved that a complete subspace is closed. Assume now, that Y is
closed, and let (yn )∞
n=1 be a Cauchy sequence in Y . Then it is also a Cauchy
sequence in X and, X being a Banach space, has a limit y ∈ X . But Y is closed,
so y ∈ Y . Because we study convergence in norm, yn −→ y also in Y , and thus
Y is complete.
¥
2.6 Theorem. Let M be a closed subspace of a normed space X over K. Then
X/M is a normed space over K with the norm
|||x + M ||| = inf {kx + mk : m ∈ M } .
(2.1)
Further, if X is a Banach space, then so is X/M .
First we must check that X/M is a linear space over K. Remember
that the cosets x + M = {x + m : m ∈ M } form a partition of X (they are
pairwise disjoint and their union is X ), and operations on the quotient space
X/M = {x + M : x ∈ X} are dened in the following way:
Proof.
(x + M ) + (y + M ) = (x + y) + M
α(x + M ) = (αx) + M
(2.2)
(2.3)
when x, y ∈ X and α ∈ K. From the theory of groups, recall that X/M is an
abelian group with respect to addition (2.2) and M = 0 + M is its neutral (zero)
element. Also, if x1 + M = x2 + M , that is, x1 − x2 ∈ M , then α(x1 − x2 ) =
αx1 − αx2 ∈ M since M is a subspace, so (αx1 ) + M = (αx2 ) + M and the
second operation (2.3) is well-dened too. The distributive laws follow directly
from the corresponding laws in X . Finally, 1(x + M ) = (1x) + M = x + M for
all x + M ∈ X/M , as required.
Next we prove that (2.1) denes a norm on X/M . Since y ∈ x + M if and
only if y = x + m for some m ∈ M , we can write the norm of a coset in a slightly
dierent way:
|||x + M ||| = inf {kyk : y ∈ x + M } ,
(2.4)
which is independent of the representative x of the coset x + M , so ||| · ||| is
well-dened on X/M .
14
Obviously |||x+M ||| ≥ 0 for every coset x+M ∈ X/M . Next, it is easily seen
from (2.4) that if |||x + M ||| = 0, then for each n = 1, 2, . . . there is yn ∈ x + M
such that kyn k < 1/n. Since the norm on X is continuous, lim yn = 0. But
x + M is closed because M is closed and the mapping y 7→ x + y of X onto X
is a homeomorphism (a continuous bijection with a continuous inverse). Thus
0 = lim yn ∈ x + M , that is, x + M = M . On the other hand, |||M ||| = 0
because 0 ∈ M . Hence |||x + M ||| = 0 if and only if x + M = M .
If x + M ∈ X/M , then clearly
|||0(x + M )||| = |||M ||| = 0 = |0| |||x + M |||.
Let now α ∈ K \ {0}. Then M = αM , and so
|||α(x + M )||| = inf {kαx + mk : m ∈ M }
= inf {kαx + αmk : m ∈ M }
= |α| inf {kx + mk : m ∈ M }
= |α| |||x + M |||,
as required. For the triangle inequality, let x1 , x2 ∈ X and let x ∈ x1 + M be
arbitrary. Then
|||(x1 + M ) + (x2 + M )||| = |||(x + M ) + (x2 + M )|||
= inf {k(x + x2 ) + mk : m ∈ M }
≤ kxk + inf {kx2 + mk : m ∈ M }
= kxk + |||x2 + M |||,
whence we obtain the desired inequality
|||(x1 + M ) + (x2 + M )||| ≤ kx1 + M k + kx2 + M k
by taking inmum with x ∈ x1 + M and keeping (2.4) in mind. Thus X/M is
a normed space over K.
Suppose then that X is a Banach space and let (xn + M )∞
n=1 be a Cauchy
sequence in X/M . Note that if
|||(a + M ) − (b + M )||| = inf {ka − b + mk : m ∈ M } < ε,
then there is some b̃ ∈ b + M such that
°
°
°
°
°a − b̃° < ε.
(Since ka − b + m0 k < ε for some m0 ∈ M , we can choose b̃ = b − m0 ∈ b + M .)
We now form inductively a sequence (zk )∞
k=1 in X and a strictly increasing
sequence (nk )∞
of
positive
integers.
Since
(xn + M )∞
n=1 is Cauchy, we may
k=1
pick n1 ∈ N such that |||(xi + M ) − (xj + M )||| < 12 whenever i, j ≥ n1 . Choose
z1 = xn1 .
Suppose we have chosen positive integers n1 < n2 < · · · < nk with the
property that for every i = 1, 2, . . . , k ,
|||(xp + M ) − (xq + M )||| <
15
1
2i
whenever p, q ≥ ni ,
(2.5)
and a sequence of vectors z1 , z2 , . . . , zk ∈ X such that zi ∈ xni + M and
kzi − zi+1 k <
1
2i
for all i = 1, 2, . . . , k − 1.
(2.6)
Then, since (xn + M )∞
n=1 is a Cauchy sequence, we may pick nk+1 > nk such
that |||(xp + M ) − (xq + M )||| < 1/2k+1 whenever p, q ≥ nk+1 . Further, as we
reasoned earlier, there is some zk+1 ∈ xnk+1 + M such that kzk − zk+1 k < 1/2k ,
because zk ∈ xnk + M and
|||(zk + M ) − (xnk+1 + M )||| = |||(xnk + M ) − (xnk+1 + M )||| <
1
.
2k
Now, applying the triangle inequality and (2.6), we get for every p ∈ N the
following:
kzk − zk+p k ≤
p−1
X
i=k
∞
X
1
−→ 0 as k −→ ∞
kzi − zi+1 k ≤
2i
i=k
P∞
because the series k=1 1/2i is convergent. Thus (zk )∞
k=1 is a Cauchy sequence
in X and has a limit, say z ∈ X . Now
|||(xnk + M ) − (z + M )||| = |||(zk + M ) − (z + M )|||
= |||(zk − z) + M |||
≤ kzk − zk −→ 0 as k −→ ∞,
∞
so the Cauchy sequence (xn + M )∞
n=1 has a subsequence (xnk + M )k=1 which
converges to z + M and must therefore itself converge to z + M . This proves
the completeness of X/M .
¥
Bounded Linear Operators
Denition. A subset S of a normed space X is said to bounded if and only if
there exists M > 0 such that ksk ≤ M for all s ∈ S .
¤
Note that this is the usual denition of a bounded set in any metric space. However, the following denition diers from the usual one of a bounded function
on a metric space.
Denition. Let X and Y be normed spaces. A linear operator T : X → Y
is said to be bounded if there exists M > 0 such that kT xk ≤ M kxk for all
x ∈ X.
¤
In other words, a linear operator T : X → Y is bounded when it maps bounded
sets to bounded sets; note that the image of X under T is bounded if and only
if T is identically zero.
2.7 Theorem. Let X and Y be normed spaces and T : X → Y a linear operator. Then the following conditions are equivalent:
(1) T is bounded.
(2) T is continuous on X .
(3) T is continuous at some point x0 ∈ X .
16
Certainly a bounded operator is continuous at 0, so (1) implies (3).
Suppose that (3) holds and let ε > 0. Then there is δ > 0 such that
kT z − T x0 k < ε whenever kz − x0 k < δ . If x and y in X are such that kx − yk <
δ , then k(x − y + x0 ) − x0 k < δ and thus
Proof.
kT x − T yk = kT (x − y + x0 ) − T x0 k < ε.
So (3) implies (2).
Suppose then that T is continuous. If T is not bounded, there exists for
every n ∈ N a vector xn ∈ X such that
(2.7)
kT xn k > n kxn k .
Note that kxn k 6= 0 because T 0 = 0 6> 0. Consider now the vectors
yn :=
xn
.
n kxn k
Clearly yn −→ 0 as n −→ ∞. Since T is continuous, we know that T yn −→ 0
as n −→ ∞. But for each n ∈ N
kT yn k =
1
kT xn k > 1
n kxn k
when (2.7) holds. This is a contradiction, so T must be bounded and (2) implies
(1), which completes the proof.
¥
Denition. Let T : X → Y be a bounded linear operator. The norm of T is
dened by
kT k = inf {M : kT xk ≤ M kxk
for all x ∈ X} .
(2.8)
¤
This terminology is justied in Theorem 2.11.
2.8 Proposition. If T : X → Y is a bounded linear operator, then
kT xk ≤ kT k kxk
Proof.
for all x ∈ X.
Exercise 36.
¥
2.9 Proposition. If T : X → Y is a bounded linear operator, then
kT k = sup {kT xk : kxk ≤ 1, x ∈ X}
= sup {kT xk : kxk = 1, x ∈ X}
½
¾
kT xk
= sup
: x 6= 0, x ∈ X ,
kxk
where the last two expression are applicable only for spaces X 6= {0}.
Proof.
Exercise 36.
¥
2.10 Proposition. Let X and Y be normed spaces and T : X → Y a linear
operator. Then the inverse function T −1 : Range(T ) → X of T exists and is a
bounded linear operator if and only if there exists a constant m > 0 such that
m kxk ≤ kT xk for all x ∈ X .
17
Proof.
Exercise 37
¥
Denition. Let X and Y be normed spaces. A bijective linear operator T :
X → Y is a linear homeomorphism (or a topological isomorphism ) when there
exist constants m > 0 and M > 0 such that
m kxk ≤ kT xk ≤ M kxk
for all x ∈ X.
(2.9)
¤
In essence, a linear homeomorphism is a bijective continuous linear transformation that has a continuous inverse, i.e., a linear isomorphism that is also a
homeomorphism (see Proposition 2.10 and Theorem 2.7; also the more general
denition on page 32 and Theorem 3.16 may be of interest).
Denition. Let X and Y be normed spaces over K. A linear operator T :
X → Y is called a linear isometry when kT xk = kxk for all x ∈ X . If T is both
a linear isometry and a linear isomorphism, we call it an isometric isomorphism
and say that the spaces X and Y are isometrically isomorphic.
¤
Note that an isometry is always injective, so a surjective linear isometry is
an isometric isomorphism. Furthermore, a linear isometry is bounded and the
inverse of an isometry is an isometry, so a surjective linear isometry is always a
linear homeomorphism.
If the identity operator is a linear homeomorphism on a linear space X with
respect to two norms on X , these norms are said to be equivalent:
Denition. Norms k·k1 and k·k2 on a linear space X are equivalent when there
exist constants m > 0 and M > 0 such that
m kxk1 ≤ kxk2 ≤ M kxk1
for all x ∈ X.
(2.10)
¤
It is easy to verify (Exercise 41) that the relation of two norms on X being
equivalent really is an equivalence among all norms on X .
Bounded Linear Functionals and the Banach Dual
2.11 Theorem. Let L(X, Y ) be the set of all bounded linear operators from
a normed space X into a normed space Y . Then L(X, Y ) is a normed space
with operations dened pointwise and the norm T 7→ kT k given by 2.8. If Y is
a Banach space, then so is L(X, Y ).
Proof. Let S, T : X → Y be bounded linear operators and let α ∈ K, the
underlying eld of X and Y . Then
k(S + αT )(x)k = kSx + αT xk ≤ kSxk + |α| kT xk
≤ (kSk + |α| kT k) kxk
for all x ∈ X , so S + αT is bounded. Hence L(X, Y ) is a subspace of the linear
space L(X, Y ) of all linear operators from X into Y .
It is obvious that kT k ∈ [0, ∞[ whenever T ∈ L(X, Y ). Since S , T and α
just discussed were arbitrary, the above equation yields
kS + T k ≤ kSk + kT k
18
for all S, T ∈ L(X, Y ). Since we also have
kαT k = sup kαT xk = |α| sup kT xk = |α| kT k
kxk≤1
kxk≤1
whenever T ∈ L(X, Y ) and α ∈ K, the mapping T 7→ kT k is a seminorm on
L(X, Y ). Furthermore, if kT k = 0, then kT xk = 0 for all x ∈ X . This means
that T x = 0 for all x ∈ X , i.e., T = 0. Thus L(X, Y ) is a normed space by
Proposition 2.1.
Suppose then that Y is a Banach space, and let (Tn )∞
n=1 be a Cauchy sequence in L(X, Y ). Then (Tn x)∞
is
a
Cauchy
sequence
in
Y for every x ∈ X
n=1
because
kTn x − Tm xk = k(Tn − Tm )xk ≤ kTn − Tm k kxk ,
and Y being a Banach space, it has a limit in Y . Thus we may dene a function
T : X → Y by
T x = lim Tn x.
n→∞
Since the vector operations in Y are continuous,
T (x + αy) = lim Tn (x + αy) = lim Tn x + α lim Tn y = T x + αT y
n→∞
n→∞
n→∞
whenever x, y ∈ X and α ∈ K, so T is in fact a linear operator. Moreover, there
exists n1 such that kTn − Tn1 k < 1 for all n ≥ n1 because (Tn )∞
n=1 is Cauchy,
so
°
°
°
°
kT xk = ° lim Tn x° = lim kTn xk = lim kTn x − Tn1 x + Tn1 xk
n→∞
n→∞
n→∞
≤ lim kTn − Tn1 k kxk + kTn1 k kxk ≤ (1 + kTn1 k) kxk
n→∞
for all x ∈ X . Thus T is also bounded.
Finally, since (Tn ) is Cauchy, there exists for every ε > 0 an index nε such
that kTn − Tm k < ε whenever m, n ≥ nε . Hence for every x ∈ X with kxk ≤ 1
we have
°
°
°
°
k(Tn − T )xk = °Tn x − lim Tm x° = lim kTn x − Tm xk
n→∞
m→∞
≤ lim kTn − Tm k kxk ≤ lim kTn − Tm k ≤ ε
m→∞
m→∞
whenever n > nε , which implies that kTn − T k ≤ ε whenever n ≥ nε . Since
ε > 0 was arbitrary, Tn −→ T . This proves that L(X, Y ) is complete.
¥
The space L(X, K) is usually denoted by X ∗ , and it is a linear subspace of X 0 .
By Theorem 2.11, X ∗ is a Banach space since K = R and K = C are Banach
spaces (norm on K is the absolute value).
Denition. Let X be a normed space over K. The Banach space of all con-
tinuous linear functionals on X into K is denoted by X ∗ and called the Banach
dual or continuous dual or Banach conjugate or just dual or conjugate of X . ¤
Remember that a linear operator on a normed space is bounded if and only if
it is continuous.
19
A Brief Excursion into Baire Spaces
In order to prove the principle of uniform boundedness and the open mapping
theorem, we need the Baire category theorem. First, let us recall the relevant
denitions from topology.
Denition. Let X be a topological space and S ⊆ X . Then S is said to be
dense in X if its closure is X , that is, S = X , and S is said to be nowhere dense
if its closure has empty interior, that is, ˚
S = ∅.
¤
Denition. A subset S of a topological space X is said to be meagre (or of
rst category ), when S is a countable union of nowhere dense sets. Otherwise,
S is nonmeagre (or of second category ).
¤
Denition. A topological space is called a Baire space when every non-empty
open set is nonmeagre.
¤
Note that a topological space is by denition non-empty and open in itself, so
every Baire space is nonmeagre in itself.
Next we present a few characterizations of Baire spaces.
2.12 Theorem. Let X be a topological space. The following statements are
all equivalent.
(1) X is a Baire space.
(2) The complement of any meagre subset of X is dense.
(3) The intersection of any countable family of dense open sets of X is dense.
(4) The union of any countable family of nowhere dense subsets of X has an
empty interior.
(5) The union of the interiors of the sets of any countable closed cover of X
is dense.
Proof.
We shall demonstrate that (1) ⇒ (5) ⇒ (3) ⇒ (2) ⇒ (4) ⇒ (1).
(1) ⇒
(5): Let F = {Fn : n ∈ N} be a closed cover of X , i.e., each Fn is closed
S∞
and n=1 Fn = X . For each n ∈ N, let Gn = Fn \ F˚n . Then
◦
◦
z
{
{
z
◦
G˚n = Fn ∩ (X \ F˚n ) = Fn ∩ (X \ F˚n ) = Fn ∩ (X \ F˚n ) = ∅,
S∞
so each Gn is nowhere dense and hence M := n=1 Gn is meagre. Because
every subset of a meagre set is clearly meagre and the only meagre open set in
a Baire space is the empty set, the interior of M must be empty. Thus
◦
X = X \ M̊ = X \ M ,
that is, X \ M is dense. Since all supersets of a dense set are dense, it will be
enough to show that
∞
[
X \M ⊆
F˚n .
(2.11)
n=1
20
Indeed,
Ã
(X\M ) ∩
∞
[
X\
!
F˚n
Ã
=X\
M∪
=X\
!
F˚n
n=1
n=1
∞
[
∞
[
(Gn ∪ F˚n ) = X \
n=1
∞
[
Fn = ∅
n=1
because F is a cover of X , so (2.11) must be true.
(5) ⇒ (3): Suppose that (5) holds, and assume that there exists a countable
family {Gn : n ∈ N} of dense open subsets of X such that
∞
\
(2.12)
Gn 6= X.
n=1
Then
Ã
X=
Ã
=
∞
\
!
Gn
n=1
∞
\
Ã
∪
∪
n=1
!
Gn
Ã
⊆
n=1
!
Gn
X\
∞
\
∞
[
!
∞
\
Gn
Ã
∪
X\
n=1
∞
\
!
Gn
(2.13)
n=1
(X \ Gn ).
n=1
Put
F0 =
∞
\
Gn
and
Fn = X \ Gn
for n ≥ 1.
n=1
Then each Fn is closed, {Fn }∞
n=0 is a cover of X by (2.13), and for every n ≥ 1
F˚n = X \˚Gn = X \ Gn = ∅,
since Gn is dense. Thus by (5), the set
∞
[
F˚n = F˚0
n=0
must be dense in X . But this is a contradiction, since
F˚0 ⊆ F0 = F0 =
∞
\
Gn 6= X
n=1
by our assumption. Hence (2.12) must be false and (3) holds.
S∞
(3) ⇒ (2): Let M be a meagre subset of X , that is, M = n=1 Mn , where
˚ = ∅ for all n ∈ N. Then X = X \ M
˚ = X \ M , so each X \ M is open
M
n
n
n
n
and dense. Thus by (3) we have
X=
∞
\
(X \ Mn ) = X \
n=1
∞
[
Mn ⊆ X \
n=1
21
∞
[
n=1
Mn = X \ M ,
so X \ M is dense, as required for (2).
(2) ⇒ (4): Let {Nn : b ∈ N} be a countable family of nowhere dense subsets
˚ = ∅ for all n ∈ N the set S∞ N is meagre and by (2) its
of X . Since N
n
n=1 n
complement is dense. This means that
X=X\
so the interior of
S∞
∞
[
Nn = X \
n=1
∞˚
[
Nn ,
n=1
n=1 Nn must be empty and (4) holds.
(4) ⇒ (1): Let A be an open subset of X . If A is meagre, it is a countable
union of nowhere dense sets and therefore has an empty interior by (4). But A
is open, so A = Å = ∅. Thus the only open meagre set is the empty set, i.e., X
is a Baire space.
¥
The following theorem is known as Baire's theorem or Baire category theorem ; the latter name is due to Baire spaces being nonmeagreBaire's theorem
implies that every complete metric space is of second category.
2.13 Theorem (Baire). Every complete metric space is a Baire space.
Let (X, d) be a complete metric space. We shall prove that X satises
condition (3) of Theorem 2.12. To this end, let A = {An : n ∈ N} be any
countable family of dense open subsets of X . Our task is thus to show that
Proof.
B(x, r) ∩
∞
\
An 6= ∅
(2.14)
n=1
for each x ∈ X and every r > 0.
We begin by dening inductively a sequence (xn )∞
n=1 in X and a sequence
of
positive
real
numbers
that
satisfy
(rn )∞
n=1
B̄(xn+1 , rn+1 ) ⊆ An ∩ B(xn , rn )
1
rn <
n
(2.15)
(2.16)
for all n = 1, 2, . . . . Let x1 ∈ X and r1 > 0 be arbitrary. Since our goal is to
prove (2.14) for x = x1 and r = r1 , we may assume with no loss of generality
that 0 < r1 < 1. Suppose that points x1 , x2 , . . . , xk ∈ X and positive radii
r1 , r2 , . . . , rk have been chosen to satisfy (2.15) and (2.16). Now, Ak ∩ B(xk , rk )
is open and nonempty because Ak is an open dense set and B(xk , rk ) is a
1
nonempty open set. Hence there exists xk+1 ∈ X and a positive radius δ < k+1
δ
such that B(xk+1 , δ) ⊆ Ak ∩ B(xk , rk ). Take rk+1 = 2 . Then xk+1 and rk+1
clearly satisfy (2.15) and (2.16). This completes our denition of the sequences
(xn ) and (rn ).
From (2.15) one sees that (B(xn , rn ))∞
n=1 is a shrinking sequence of balls,
i.e., B(xm , rm ) ⊆ B(xn , rn ) whenever m ≥ n. Thus
d(xm , xn ) < rn <
22
1
n
for all m ≥ n. So (xn ) is a Cauchy sequence and, X being complete, it converges
in X . Let x∞ = lim xn . Now, if m > n ≥ 1, then xm ∈ B̄(xn+1 , rn+1 ), which is
a closed set, so x∞ ∈ B̄(xn+1 , rn+1 ). Hence by (2.15), x∞ ∈ An ∩ B(xn , rn ) ⊆
B(x1 , r1 ) ∩ An for all n = 1, 2, . . . , and consequently,
∞
\
x∞ ∈
∞
\
(B(x1 , r1 ) ∩ An ) = B(x1 , r1 ) ∩
n=1
An ,
n=1
as required for (2.14).
¥
We present one more result that will be useful later (Theorem 7.13).
2.14 Theorem. Any compact Hausdor space is a Baire space.
Proof. Let X be a compact Hausdor space. We use characterization (3) of
Theorem 2.12: suppose that G1 , G2 , . . . are dense open sets of X . Let U be
an arbitrary nonempty open set. Then U ∩ G1 6= ∅ because G1 is dense. Since
every compact Hausdor space is regular, there exists a nonempty open set V1
such that V1 ⊆ U ∩ G1 . Similarly, there is a nonempty open set V2 such that
V2 ⊆ V1 ∩ G2 . Continuing in this manner, we can pick for each n = 2, 3, . . . a
nonempty open set Vn such that Vn ⊆ Vn−1 ∩ Gn .
of nonempty
compact sets and
T∞Now V1 , V2 , . . . is a decreasing sequence
T∞
T∞
T∞ hence
V
must
be
nonempty.
But
V
⊆
(U
∩
G
)
=
U
∩
(
n
n
n
n=1
n=1
n=1
n=1 Gn )
T∞
T∞
so U ∩ ( n=1 Gn ) is nonempty. Because U was arbitrary, n=1 Gn must be
dense, and so X is a Baire space.
¥
The preceeding theorem can be generalized: every locally compact Hausdor
space is a Baire space. For a proof of this, see [8].
Applications of the Baire Category Theorem
In this section we prove the uniform boundedness principle and the open mapping theorem. These two are among the cornerstones of classical functional
analysis, and their proofs depend on the Baire category theorem we obtained in
the previous section.
Uniform Boundedness Principle
2.15 Uniform Boundedness Principle (Banach-Steinhaus). Let X and
Y be normed spaces and F a collection of bounded linear operators from X
into Y . If X is a Banach space and
sup kT xk < ∞
T ∈F
then
for every x ∈ X,
(2.17)
sup kT k < ∞.
T ∈F
For each n = 1, 2, . . . , let An = {x ∈ X : kT xk ≤ n for all T ∈ F}.
Then by hypothesis (2.17), we have
Proof.
X=
∞
[
n=1
23
An .
Since X is complete, it is a Baire space by Theorem 2.13, and therefore nonmeagre. Thus the sets An , n = 1, 2, . . . , cannot all be nowhere dense, say A˚m 6= ∅.
But Am is in fact closed: if (xk )∞
k=1 is a sequence in Am that converges to x ∈ X
and T ∈ F , then
kT xk = lim kT xk k ≤ m
k→∞
by the continuity of norm and T . This holds for every T ∈ F , so x ∈ Am , as
required.
Since A˚m = A˚m 6= ∅, the set Am contains some non-empty open ball, say
B(x0 , r). Put δ = r/2 and let T ∈ F be arbitrary. If x ∈ X is such that
kxk = 1, then δx + x0 ∈ B(x0 , r) ⊆ Am and so
kT xk =
Thus kT k ≤
4m
r
kT (δx)k
kT (δx + x0 )k + kT x0 k
2m
4m
≤
≤
=
.
δ
δ
δ
r
for all T ∈ F .
¥
2.16 Corollary. If (Tn )∞
n=1 is a pointwise convergent sequence of bounded lin-
ear operators from a Banach space X into a normed space Y , then the function
T : X → Y dened by
T (x) = lim Tn x
n→∞
is a bounded linear operator.
Proof.
First, note that the continuity of vector operations in Y implies that
T (x + αy) = lim Tn (x + αy) = lim Tn x + α lim Tn y = T (x) + αT (y)
n→∞
n→∞
n→∞
for all x, y ∈ X and every scalar α, so T is a linear operator. Furthermore,
convergent sequences (in Y ) are bounded, so sup {kTn xk : n ∈ N} < ∞ for
each x ∈ X . Since X is a Banach space, M := sup {kTn k : n ∈ N} < ∞ by
Uniform Boundedness Principle 2.15, and therefore
°
°
°
°
kT xk = ° lim Tn x° = lim kTn xk ≤ lim kTn k kxk ≤ M kxk
n→∞
n→∞
n→∞
for all x ∈ X , i.e., T is bounded.
¥
Open Mapping Theorem
Denition. Let X and Y be topological spaces. A mapping f : X → Y is said
to be open when f (U ) is open in Y whenever U is open in X .
¤
Note that if f is an open mapping from X into Y , then f is also open from X
onto its image f (X), but the converse is not true in general.
We engage our task of proving the open mapping theorem rst at the origin
of our space in the hopes of not cluttering the already technical proof with
nonessential details. The actual theorem then follows from the homogeneity of
the space under translations.
2.17 Lemma. Let X and Y be Banach spaces and T : X → Y a surjective
bounded linear operator. Then the image of the open unit ball of X under T
contains an open ball about 0 ∈ Y .
24
Let us introduce a shorthand notation for certain open balls about
0 ∈ X in this proof: we denote
©
ª
Bn = B(0, 2−n ) = x ∈ X : kxk < 2−n
Proof.
for n = 0, 1, 2, . . . , which makes B0 the open unit ball of X .
¡
¢
First we show that T (B1 ) = T B(0, 12 ) contains some open ball. Given any
x ∈ X , for suciently large k ∈ N , we have x ∈ kB1 = B(0, k2 ), so
X=
∞
[
kB1 .
k=1
Since T is linear and onto,
̰
!
∞
∞
[
[
[
Y =T
kB1 =
kT (B1 ) j
kT (B1 ).
k=1
k=1
k=1
Since Y is complete, it is a Baire space by Theorem 2.13, and hence nonmeagre.
Thus for some k ∈ N, the closed set kT (B1 ) is not nowhere dense, i.e., it has
a nonempty interior. By Proposition 2.4, the interior of k1 kT (B1 ) = T (B1 ) is
therefore also nonempty and it contains some open ball, say, B := B(y0 , r) ⊆
T (B1 ).
Next we prove that B − y0 = B(0, r) ⊆ T (B0 ). Since B − y0 ⊆ T (B1 ) − y0 ,
it will be enough to show that
T (B1 ) − y0 ⊆ T (B0 ).
(2.18)
Let y ∈ T (B1 ) − y0 . Then y + y0 ∈ T (B1 ). Since y0 ∈ T (B1 ) also, there exist
∞
sequences (vn )∞
n=1 and (wn )n=1 in B1 such that
lim T vn = y0
n→∞
and
lim T wn = y + y0 .
n→∞
The ball B1 has center 0 and radius 12 , so kwn − vn k ≤ kwn k +kzn k < 12 + 21 = 1
for all n ∈ N. Thus (wn − vn )∞
n=1 is a sequence in B0 , and the continuity of T
implies that
y = lim T wn − T vn = lim T (wn − vn ) ∈ T (B0 ).
n→∞
n→∞
This proves (2.18), because y ∈ T (B1 ) − y0 was arbitrary.
Having shown that
B(0, r) = B − y0 ⊆ T (B0 ),
(2.19)
we point out that 2−n T (B0 ) = T (2−n B0 ) = T (Bn ) for every n ∈ N, because T
is linear. If we denote Vn = B(0, 2−n r) ⊆ Y , we obtain from (2.19) that
Vn = 2−n B(0, r) ⊆ 2−n T (B0 ) ⊆ T (Bn )
(2.20)
for all n ∈ N.
Finally, we show that
³ r´
⊆ T (B0 ).
V1 = B 0,
2
25
(2.21)
Let y ∈ V1 be arbitrary. Then y ∈ T (B1 ) by (2.20), so there exists a u1 ∈ T (B1 )
such that ky − u1 k < 2−2 r. This means, that for some x1 ∈ B1 , we have
u1 = T x1 , and so
r
ky − T x1 k < 2 .
2
Suppose we have chosen points x1 , x2 , . . . , xk ∈ X such that
°
°
°
°
n
X
°
°
r
°y −
°
T
x
j ° < n+1
°
2
°
°
j=1
(2.22)
Pk
and xn ∈ Bn for all n = 1, 2, . . . , k . Now, if yk+1 := y − j=1 T xj , then
yk+1 ∈ Vk+1 by (2.22) and hence yk+1 ∈ T (Bk+1 ) by (2.20). Thus, like before,
there is a uk+1 ∈ T (Bk+1 ), i.e., xk+1 ∈ Bk+1 with T xk+1 = uk+1 such that
kyk+1 − uk+1 k < 2−(k+2) r. In other words, there is xk+1 ∈ Bk+1 such that
condition (2.22) is satised for n = k + 1. Thus there exists a sequence (xj )∞
j=1
in X satisfying
Pn(2.22) for all n ∈ N .
Let zn = j=1 xj for all n ∈ N. Since xj ∈ Bj , we have kxj k < 2−j for all
j ∈ N. The sequence (zn )∞
n=1 in X is Cauchy, because for every p ∈ N
kzn − zn+p k ≤
n+p
X
kxj k ≤
j=n
∞
X
1
1
= n−1 −→ 0
j
2
2
j=n
as n −→ ∞. Since X is complete, (zn ) converges, say zn −→ z ∈ X . But z ∈ B0
because
°
°
° n
°
∞
X
°X °
1 1
1
° ≤ kx1 k +
kzk = lim °
< + =1
x
j
°
°
j
n→∞
2
2 2
° j=1 °
j=2
and B0 has radius 1. Further, T zn −→ y due to (2.22) and the linearity of T .
Since T is continuous, T zn −→ T z , so we must have y = T z ∈ T (B0 ). This
conrms (2.21).
¥
The following theorem is also known as Banach-Schauder theorem. Due to
the important Corollary 2.19, it is also called the bounded inverse theorem.
2.18 Open Mapping Theorem. A surjective and bounded linear operator T
from a Banach space X onto a Banach space Y is an open mapping.
Let U be an arbitrary non-empty open set in X . We prove that every
point of T (U ) is an interior point, that is, T (U ) is open.
Let y ∈ T (U ) be arbitrary, and pick x ∈ U such that T x = y . Since U is
open, U − x is also open by Proposition 2.4, and therefore contains some open
ball about 0 ∈ X , say B(0, r) ⊆ U − x. Take ρ = 1/r. Then rρ = 1, and so
B(0, 1) ⊆ ρ(U − x).
By Lemma 2.17, there is some open ball about 0 ∈ Y , say B , contained in
the image under T of the unit ball of X . Then
¡
¢
¡
¢
¡
¢
rB ⊆ rT B(0, 1) ⊆ rT ρ(U − x) = rρ T (U ) − T x = T (U ) − y
Proof.
26
since T is linear. But B , and hence rB , is an open ball about 0 ∈ Y , so y + rB
is an open ball about y and
¡
¢
y + rB ⊆ y + T (U ) − y = T (U ).
Thus y is an interior point of T (U ) as claimed.
¥
2.19 Corollary (Bounded Inverse). The inverse T −1 of a bijective bounded
linear operator T from a Banach space X onto a Banach space Y is a bounded
linear operator.
Proposition 1.15 assures that T −1 is linear. The preimage of any open
set U ⊆ X under T −1 is T (U ) which, according to the open mapping theorem, is
open. Thus T −1 : Y → X is continuous and hence bounded by Theorem 2.7. ¥
Proof.
2.20 Corollary. A bijective bounded linear operator between Banach spaces
is a topological isomorphism.
−1
Let operator T : X → Y satisfy the description above.
Then T
° −1
°
is also a bounded linear operator by Corollary 2.19 and kxk = °T (T x)° ≤
° −1 °
°
°
°T ° kT xk for every x ∈ X . We choose m = °T −1 °−1 and M = kT k and
obtain m kxk ≤ kT xk ≤ M kxk .
¥
Proof.
2.21 Corollary. Every nonzero bounded linear functional on a Banach space
is open.
Every nonzero linear functional is surjective and R and C are Banach
spaces, so the open mapping theorem applies.
¥
Proof.
Finite-dimensional Normed Spaces
Let K = R or K = C. Whenever n = 1, 2, 3, . . . , we denote by `1 (n) the linear
space Kn over K equipped with the norm k·k1 that is dened by
k(α1 , . . . , αn )k1 =
n
X
|αk |
k=1
for (α1 , . . . , αn ) ∈ Kn .
We shall rst study the space `1 (n) because the norm k·k1 is technically
easy to handle, and then we prove that this is all we have to do, as far as
nite-dimensional normed spaces are concerned.
2.22 Proposition. Given n ∈ N, `1 (n) is a Banach space.
Proof.
Exercise 44.
¥
2.23 Proposition. Given n ∈ N, the closed unit ball of `1 (n) is compact.
Proof.
Exercise 45.
¥
2.24 Corollary. Given n ∈ N, all closed bounded subsets of `1 (n) are compact.
27
Let C be a closed and bounded subset of `1 (n). Then there is a constant
1
x is a
M > 0 such that kxk ≤ M for all x ∈ C . Since the mapping x 7→ M
homeomorphism of X onto itself by Proposition 2.4, the image of C under this
mapping is a closed subset of the closed unit ball B̄(0, 1) of X . This image is
compact because it is a closed subset of the compact set B̄(0, 1). Thus C , being
homeomorphic to a compact set, must itself be compact.
¥
Proof.
2.25 Theorem. Any two nite-dimensional normed spaces of the same nite
dimension and over the same eld are topologically isomorphic.
Let X be a normed space over K with dim X =: n > 0 (the case n = 0
is trivial). Since topological isomorphy is an equivalence among normed spaces,
it is enough to show that X is topologically isomorphic to the linear space `1 (n),
that is, Kn equipped with the norm
Proof.
k(α1 , . . . , αn )k1 =
n
X
|αk | .
k=1
Fix a basis B := {x1 , . . . , xn } for X and dene a mapping T : X → `1 (n) by
à n
!
X
T
αk xk = (α1 , . . . , αn ),
(2.23)
k=1
whenever α1 , . . . , αn ∈ K. This denition is well posed, because each x ∈ X has
a unique representation as a linear combination of vectors of B . For the same
reason, T is a bijection. Furthermore, T is easily seen to be linear, so we have
established that X and K are linearly isomorphic.
Denote the norm on X by k·k and let x =: α1 x1 + · · · + αn xn ∈ X be
arbitrary. Then
kxk ≤
n
X
|αk | kxk k ≤ M
k=1
n
X
|αk | = M kT xk1 ,
(2.24)
k=1
where M = max{kx1 k , . . . , kxn k} is a positive constant.
On the other hand, to prove that m kT xk1 ≤ kxk for some m > 0, it is
enough to consider the cases where kT xk1 = 1, for if T x = 0 there is nothing
to prove, and if T x 6= 0 we consider the vector y = kT 1xk x. Then kT yk1 = 1,
1
and m kT yk1 ≤ kyk implies m kT xk1 ≤ kxk . So we assume with no loss of
generality, that kT xk1 = 1 and dene a function f : `1 (n) → R by
°
°
f (α) = °T −1 α° .
(2.25)
Then f is continuous on `1 (n), because for any α, β ∈ Kn we have
¯°
° °
°¯ °
°
|f (α) − f (β)| = ¯°T 1 α° − °T 1 β ° ¯ ≤ °T −1 α − T −1 β °
°
°
= °T −1 (α − β)° ≤ M kα − βk
1
by triangle inequality and (2.24). The surface S(0, 1) of the unit ball in `1 (n) is
compact by Corollary 2.24, so f attains its minimum m, say, on S(0, 1). That
means there is some ᾰ ∈ S(0, 1) such that
m = f (ᾰ) = min {f (α) : α ∈ S(0, 1)} .
28
(2.26)
Therefore, whenever kT xk1 = 1, we have
°
°
m kT xk1 = m ≤ f (T x) = °T −1 T x° = kxk .
(2.27)
°
°
It remains to check that m > 0. If m = 0, then f (ᾰ) = °T −1 ᾰ° = 0, which
implies that ᾰ = 0 because T −1 is linear by Proposition 1.15 and injective. But
this is a contradiction, because ᾰ ∈ S(0, 1). Thus, combining (2.24) and (2.27),
we have found positive constants m and M such that
m kT xk1 ≤ kxk ≤ M kT xk1
(2.28)
for every x ∈ X .
¥
2.26 Corollary. Finite-dimensional normed spaces are all Banach spaces.
Let X be a normed space of dimension n ∈ N (case n = 0 is again
n
trivial), and let (xn )∞
n=1 be a Cauchy sequence in X . Let T : X → K be a
∞
n
topological isomorphism. Then (T xn )n=1 is a Cauchy sequence in K because
T is bounded. Since Kn is complete, there is y ∈ Kn such that limn T xn = y .
Let x = T −1 y ∈ X , and observe that
³
´
¡
¢
lim xn = lim T −1 T xn = T −1 lim T xn = T −1 y = x.
Proof.
n
n
n
This proves that X is complete
¥
2.27 Corollary. Finite-dimensional subspaces of normed spaces are closed.
Proof.
Apply Corollary 2.26 and Theorem 2.5, statement (3).
¥
2.28 Corollary. All norms on a given nite-dimensional linear space are equivalent.
Proof.
Exercise 41.
¥
2.29 Corollary. Let X and Y be normed spaces. If X is nite-dimensional,
then every linear operator T : X → Y is continuous.
Proof.
Exercise 42.
¥
We cannot hope for the previous corollary to hold without requiring dim X < ∞,
for there are always discontinuous linear operators on an innite-dimensional
normed space when the codomain in nontrivial (Exercise 43).
2.30 Corollary. Any bounded closed subset of a nite-dimensional normed
space is compact.
Let T be a topological isomorphism on a normed n-dimensional (n > 0;
case n = 0 remains trivial) space X onto `1 (n) and let C be a closed bounded
subset of X . Then T (C) is closed and bounded subset of `1 (n), and hence compact by Corollary 2.24. Since T −1 is continuous, C = T −1 (T (C)) is compact.¥
Proof.
Next we prove the converse of Corollary 2.30. First, we need a result stating,
in geometric terms, that given a proper closed subspace Y of a normed space,
there is a point on the surface of the unit ball whose distance to Y is as close
to 1 as we wish. However, we cannot have this distance equal to 1 in general,
even for Banach spaces (Exercise 46).
29
2.31 Lemma (Riesz's Lemma). Let Y be a proper closed subspace of a
normed space X . Whenever 0 < r < 1, there is an x0 in X such that
kx0 k = 1
and
ky − x0 k ≥ r
for all y ∈ Y .
Proof.
Fix any x1 ∈ X \ Y and let
(2.29)
δ = d(x1 , Y ) = inf {ky − x1 k : y ∈ Y } .
Then δ > 0 because Y is closed and x1 6∈ Y . Also, if 0 < r < 1, then
so there is y1 ∈ Y such that
λ := kx1 − y1 k <
δ
r
δ
.
r
> δ and
(2.30)
Again, λ > 0, so we may choose
x0 = λ−1 (x1 − y1 )
(2.31)
to obtain kx0 k = 1. Now, whenever y ∈ Y , we have λy + y1 ∈ Y . Thus, for
every y ∈ Y ,
ky − x0 k = λ−1 k(λy + y1 ) − x1 k ≥ λ−1 δ
by the choice of δ . This completes the proof, since λ−1 >
r
δ
by (2.30).
¥
2.32 Theorem. The closed unit ball of a normed space X is (norm) compact
if and only if dim X < ∞.
The suciency of the condition was proved in Corollary 2.30, because
the closed unit ball
Proof.
B := B̄(0, 1) = {x ∈ X : kxk ≤ 1}
(2.32)
is obviously closed and bounded.
To prove the necessity, suppose that dim X = ∞. Start with any nonzero
x1 ∈ B , and let X1 be the subspace of X spanned by x1 . That is, X1 =
{αx1 : α ∈ K}, where K is the scalar eld of X . Then X1 is a proper closed
subspace of dimension 1 that is closed by Corollary 2.27. According to Riesz's
lemma (2.31), we can pick x2 ∈ X such that kx2 k = 1 and kx2 − xk ≥ 21 for
every x ∈ X1 . In particular,
kx2 − x1 k ≥
1
.
2
(2.33)
Next, consider the subspace X2 generated by x1 and x2 , that is, X2 =
{αx1 + βx2 : α, β ∈ K}. Again, we apply Riesz's lemma to the 2-dimensional
closed subspace X2 and pick x3 ∈ X such that kx3 k = 1 and kx3 − xk ≥ 12 for
all x ∈ X2 . In particular,
kx3 − xm k ≥
1
2
30
for m = 1, 2.
Proceeding inductively, we obtain a sequence (xn )∞
n=1 in B with the following
properties: kxn k = 1 for every n, and
kxn − xm k ≥
1
2
whenever n 6= m.
(2.34)
This sequence has no Cauchy subsequences and therefore cannot have a convergent subsequence. This shows that the closed unit ball B is not compact.
¥
31
Chapter 3
Topological Vector Spaces
In this chapter we investigate linear spaces endowed with a more general topology than the metric topology given by a norm on the space. Reader should
pay attention to the fact that the concepts in this chapter that have already
been dened in the context of normed spaces are true generalizations, i.e., the
denitions are equivalent when talking about a norm topology.
Recall that a topological eld is a eld K endowed with a topology such
that the addition and multiplication as mappings from K × K onto K and the
inversion map α 7→ α−1 from K \{0} onto itself are all continuous. The additive
inversion map x 7→ −x is also continuous because −x = (−1)x holds for all x in
a eld.
To rephrase: a eld K with a given topology τ is a topological eld when
(K, +) and (K \ {0}, · ) are both topological groups with respect to τ .
Denition. A linear space X over a topological eld K is a topological linear
space (or a topological vector space ) when X is a equipped with a topology that
makes the vector addition mapping X × X → X and the scalar multiplication
mapping K × X → X continuous.
¤
Because −x = −1x, the mapping x 7→ −x is also continuous and hence a
topological linear space is a topological group with respect to the vector addition.
Denition. Let X and Y be topological linear spaces over K. A function
T : X → Y is called a linear homeomorphism (or a topological isomorphism )
between X and Y when T is both a linear isomorphism and a homeomorphism.
In other words, a linear homeomorphism is a bijective and continuous open
linear transformation. When there exists a linear homeomorphism between X
and Y , these two spaces are said to be topologically (and linearly ) isomorphic.
This shall be denoted by X ∼
¤
=Y.
We shall see in a while that this natural denition is consistent with our earlier
terminology for normed spaces (Theorem 3.16).
3.1 Theorem. Let X be a topological vector space over K. Given a xed
vector a ∈ X and a xed scalar α 6= 0 in K, the mappings
ta : x 7→ a + x
and
32
sα : x 7→ αx
of X onto itself are both homeomorphisms. Furthermore, sα is even a linear
homeomorphism.
Since α 6= 0, both mappings ta and sα are clearly bijections from X
onto X . The linearity of sα follows directly from the distributive laws of vector
spaces.
Let x ∈ X be arbitrary and let U be an open neighborhood of a + x ∈ X .
Since the mapping (v, w) 7→ v + w of X × X onto X is continuous, there is a
basic open neighborhood, say V × W , of (a, x) ∈ X × X that is mapped into
U , that is, V + W ⊆ U . But then a ∈ V and W is an open neighborhood of
x ∈ X , so ta (W ) = a + W ⊆ V + W ⊆ U . Thus ta is continuous.
Let again x ∈ X be arbitrary and let U be an open neighborhood of αx ∈ X .
Like before, the mapping (λ, v) 7→ λv of K × X onto X is continuous, so there
exist open neighborhoods Λ ⊆ K and V ⊆ X of α and x, respectively, such that
ΛV ⊆ U . Hence sα (V ) = αV ⊆ ΛV ⊆ U , as required for the continuity of sα .
To nish the proof, it is enough to check that the inverse functions of ta
and sα are continuous. But this is already done because (ta )−1 = t−a and
(sα )−1 = sα−1 .
¥
Proof.
Due to these homogeneity properties, the topology in a topological linear space
can be completely characterized with neighborhoods of the origin. These can
be translated at will: U is an open neighborhood of 0 ∈ X if and only if x + U
is an open neighborhood of x ∈ X .
The following lemmas are familiar from the theory of topological groups.
3.2 Lemma. If U is an open set in a topological vector space X , then S + U
is an open set whenever S ⊆ X .
Proof.
By Theorem 3.1, x + U is open for every x ∈ X . Hence
[
S+U =
s+U
s∈S
is open as a union of open sets.
¥
3.3 Lemma. If U is a neighborhood of 0 in a topological vector space, then
there exists an open neighborhood V of 0 such that V + V ⊆ U .
Since the vector addition is continuous and 0 + 0 = 0, there exist basic
open neighborhoods V1 and V2 of 0 such that V1 + V2 ⊆ U . Take V = V1 ∩ V2 .¥
Proof.
3.4 Lemma. Let C and D be subsets of a topological vector space X . If C is
compact and D is closed, then C + D is closed.
Proof. Suppose that x ∈ C + D . Then there exist nets (cλ )λ and (cλ )λ in C
and D, respectively, such that cλ + dλ −→ x. Since C is compact, the net (cλ )λ
has a subnet, say (cλµ )µ , that converges to some c ∈ C . Since (cλµ + dλµ )µ is a
subnet of a net converging to x, we must have cλµ + dλµ −→ x. Let us denote
the vector x − c by d. Then we get
µ
¶
d = x − c = lim cλµ + dλµ − c = lim(cλµ + dλµ − c)
µ
µ
= lim(cλµ − c) + lim dλµ = lim dλµ ∈ D
µ
µ
µ
33
because the
is continuous and D is closed. Thus the vector
¢
¡ vector addition
x = limµ cλµ + dλµ = c + d belongs to C + D, which proves that C + D is
closed.
¥
3.5 Theorem. If Y is a linear subspace of a topological vector space X , then Y
and its closure Y are both topological vector spaces with the subspace topology
inherited from X .
Since the restriction of any continuous function is continuous in the
subspace topology, it is clear that Y is a topological vector space. Therefore it
is enough to show that Y is a linear subspace of X .
Given x, y ∈ Y , there exist nets (xλ )λ and (yλ )λ in Y converging to x and
y , respectively. Since Y is a subspace, (xλ + αyλ )λ∈Λ is a net in Y for any xed
scalar α. Furthermore, xλ + αyλ −→ x + αy because the vector operations are
continuous. Thus x+αy ∈ Y and so Y is a linear subspace by Proposition 1.4.¥
Proof.
Local Base of Balanced and Absorbing Sets
Denition. A subset A of a real or complex linear space X is absorbing when
for every x ∈ X there exists a real number rx > 0 such that x ∈ αA whenever
α is a scalar that satises |α| ≥ rx .
¤
Equivalently, A is absorbing when for every x ∈ X , there exists a real number
rx > 0 such that βx ∈ A whenever 0 < |β| ≤ rx . It follows immediately from
the denition that the origin belongs to every absorbing set.
Denition. A non-empty subset B of a real or complex linear space X is
balanced when αB ⊆ B for every scalar α that satises |α| ≤ 1. The balanced
hull of a subset S of X is the intersection of all balanced sets that contain S .¤
It is again evident that every balanced set contains the origin.
3.6 Proposition. A subset B of a real or complex linear space X is balanced
if and only if γB ⊆ δB for all scalars γ and δ that satisfy |γ| ≤ |δ|.
Proof.
Exercise 50.
¥
3.7 Proposition. Let X and Y be real or complex linear spaces, T : X → Y
a linear operator, α a scalar and B a balanced subset of X . Then
(1) αB = |α| B and αB is a balanced set;
(2) T (B) is a balanced set in Y .
Proof.
Exercise 50.
¥
Balanced absorbing sets behave nicelyalmost like balls in normed spaces.
Fortunately, they are abundant in topological linear spaces:
3.8 Theorem. A real or complex topological vector space X has a base at 0
consisting of balanced absorbing sets.
34
We must show that every neighborhood of the origin contains a balanced absorbing open neighborhood of the origin. Let K be the scalar eld and
let U be an arbitrary neighborhood of 0 ∈ X .
We begin by showing that there is a balanced neighborhood B of 0 contained
in U . Since the mapping s : K × X → X , s(α, x) = αx is continuous and
s(0, 0) = 0, there exists a basic open neighborhood of (0, 0) in K × X that s
maps into U , that is, there exists some δ > 0 and an open neighborhood V of 0
in X such that s(Dδ × V ) = Dδ V ⊆ U , where Dδ = {λ ∈ K : |λ| < δ}. Let
[
B = Dδ V =
λV.
Proof.
λ∈K
|λ|<δ
Since 0 ∈ B and each λV is open by Theorem 3.1, B is an open neighborhood
of 0 in X .
If α ∈ K and |α| ≤ 1, then for every λ ∈ Dδ we have |αλ| = |α| |λ| < δ
so αDδ ⊆ Dδ . Thus αB = αDδ V ⊆ Dδ V = B whenever |α| ≤ 1 and so B is
balanced.
Next, we prove that B is in fact absorbing. Let x ∈ X . Since the mapping
s is continuous on K × X , the mapping sx dened on K by sx (α) = s(α, x) =
αx is also continuous. As sx (0) = 0, there is a basic neighborhood Dr =
{λ ∈ K : |λ| < r}, of 0 ∈ K, where r > 0, such that sx (Dr ) ⊆ B . In other
words, λx ∈ B whenever |λ| < r. Hence B is absorbing, as claimed.
¥
3.9 Corollary. Every neighborhood of the origin in a real or complex topological vector space is absorbing.
Proof.
Every neighborhood of 0 contains an absorbing neighborhood.
¥
Boundedness and Continuity of Linear Operators
Denition. A subset S of a topological linear space is said to be bounded when
for every neighborhood U of 0, there exists a scalar α such that S ⊆ αU .
¤
The word neighborhood in the previous denition and in Proposition 3.10 could be replaced with basic neighborhood or subbasic neighborhood without altering the concept of boundedness (see Exercise 59).
¤
Remark.
3.10 Proposition. A subset S of a real or complex topological vector space is
bounded if and only if for every neighborhood U of 0, there exists a real number
r > 0 such that S ⊆ rU .
The necessity is clear, so we shall prove the suciency: suppose that
S is bounded, and let U be any neighborhood of 0. Applying Theorem 3.8, pick
a balanced neighborhood V of 0 such that V ⊆ U . By the boundedness of S ,
there exists a scalar α such that S ⊆ αV . Let r = |α|. Then by Proposition 3.7,
S ⊆ αV = rV ⊆ rU , as required.
¥
Proof.
Of course, to maintain a consistent language, we should check that this concept
of boundedness is the same we dened earlier within normed spaces:
3.11 Proposition. A subset S of a normed space X is bounded if and only if
there exists M > 0 such that ksk ≤ M for all s ∈ S .
35
Suppose that M > 0 and ksk ≤ M for all s ∈ S , that is, S ⊆ B(0, M ).
For any neighborhood U of 0, there exists r > 0 such that B(0, r) ⊆ U . Let
α> M
r . Then αr > M and S ⊆ B(0, M ) ⊆ B(0, αr) = αB(0, r) ⊆ αU , so S is
bounded.
On the other hand, if S ⊆ X is bounded, then there is a scalar α such that
S ⊆ αB(0, 1). Choose M = |α|. If s ∈ S , there is x ∈ B(0, 1) such that s = αx.
Thus ksk = |α| kxk ≤ |α| ≤ M , as required.
¥
Proof.
This demonstrates that for a normed space X , the concept of a bounded set is
(as it should be) the same whether we look at X as a topological linear space
or as a metric space.
3.12 Lemma. If B1 , . . . , Bn are bounded subsets of a topological linear space
X and α1 , . . . , αn are scalars, then
α1 B1 + · · · + αn Bn
is a bounded subset of X .
Proof.
Exercise 60
¥
3.13 Corollary. Let B be a subset of a topological vector space X and let
x ∈ X . Then B is bounded if and only if x + B is bounded.
If X =
6 {0}, then {0} cannot be a neighborhood of 0 (see Exercise 48),
so it is clear that {x} and {−x} are bounded. Otherwise, the statement is
trivial.
¥
Proof.
3.14 Theorem. Let X and Y be topological linear spaces. A linear operator
T : X → Y is continuous if and only if T is continuous at one point x0 ∈ X .
Proof. A continuous operator T is by denition continuous at every point.
Conversely, suppose that T is continuous at x0 ∈ X , and let x ∈ X be arbitrary.
Theorem 3.1 will be used without notice in the following. If V is a neighborhood
of T x in Y , then T x0 − T x + V is a neighborhood of T x0 in Y . Since T
is continuous at x0 , there exists a neighborhood U of x0 such that T (U ) ⊆
T x0 − T x + V . Now x − x0 + U is a neighborhood of x in X and T (x − x0 + U ) =
T x−T x0 +T (U ) ⊆ T x−T x0 +T x0 −T x+V = V , as required for the continuity
of T at x. Hence T is continuous on X .
¥
Denition. Let X and Y be topological vector spaces. A linear operator T :
X → Y is said to be bounded when the image of every bounded subset of X
under T is a bounded subset of Y . Otherwise, T is unbounded.
¤
Again, the rst thing to check is the consistency with the normed space terminology:
3.15 Theorem. Let X and Y be normed spaces. A linear operator T : X → Y
is bounded if and only if there exists M > 0 such that kT xk ≤ M kxk for all
x ∈ X.
Suppose that T is bounded. The open ball B(0, 2) of X is bounded by
Proposition 3.11, so T (B(0, 2)) is bounded in Y , i.e., there exists a real number
Proof.
36
M > 0 such that kT xk ≤ M for all x ∈ B(0, 2). Let x ∈ X \ {0} be arbitrary.
−1
Then kxk x ∈ B(0, 2) and thus
°
°
°
°
−1
kT xk = kxk °T (kxk x)° ≤ M kxk ,
as required.
Conversely, suppose that there is M > 0 such that kT xk ≤ M kxk for all
x ∈ X . If B is any bounded set in X , there is R > 0 such that kxk ≤ R for
all x ∈ B . Thus kT xk ≤ M kxk ≤ M R for all x ∈ B , so T (B) is bounded and
hence T is bounded too.
¥
While we are at it, we verify that our concepts of linear homeomorphism (otherwise known as topological isomorphism) agree with our denition for normed
spaces:
3.16 Theorem. Let X and Y be normed spaces. A bijective linear operator
T : X → Y is a linear homeomorphism if and only if there exist constants m > 0
and M > 0 such that
m kxk ≤ kT xk ≤ M kxk
for all x ∈ X.
(3.1)
Suppose that T is a linear homeomorphism. Then T and T −1 are both
continuous, and by theorems 2.7 and 3.15,
there °exist constants r > 0 and
°
M > 0 such that kT xk ≤ M kxk and °T −1 (T x)° ≤ r kT xk for all x ∈ X .
Hence we obtain (3.1) by choosing m = r−1 .
Conversely,
Then T is bounded, and so is T −1
° −1 °assume
° that−1(3.1)
° holds.
1 °
1
°
°
°
because T y ≤ m T (T y) = m kyk for all y ∈ Y . Thus T is open and
continuous.
¥
Proof.
3.17 Theorem. Let X and Y be topological vector spaces. If a linear operator
T : X → Y is continuous, then T is bounded.
Proof. Let B be a bounded set in X and let V be an arbitrary neighborhood of
the origin in Y . Since T 0 = 0 and T is continuous, there exists a neighborhood
U of the origin in X such that T (U ) ⊆ V . Since B is bounded in X , there is a
scalar α such that B ⊆ αU . Therefore T (B) ⊆ T (αU ) = αT (U ) ⊆ αV because
T is linear. Hence T (B) is bounded in Y , as required.
¥
Although the converse of Theorem 3.17 holds if X and Y are normed
spaces, it is not true in general.
¤
Remark.
Continuous Linear Functionals and the Continuous Dual
3.18 Theorem. Let X be a real or complex topological vector space, and let
x0 be a linear functional on X . Then the following conditions are equivalent:
(1) x0 is continuous;
(2) x0 is continuous at some point x0 ∈ X ;
(3) x0 is bounded on some neighborhood U0 of the origin;
(4) Ker(x0 ) is closed.
37
Let K be the underlying eld of X . As a rst observation we note
that all the above statements are clearly true for the zero functional, so we shall
assume in the following that x0 6= 0.
If x0 is continuous, then clearly Ker(x0 ) is closed as a preimage of the closed
set {0} in K, so (1) =⇒ (4).
Suppose then that Ker(x0 ) is closed. Since x0 was assumed to be nonzero,
there exists y ∈ X\Ker(x0 ) and hence, by Theorem 3.8, a balanced neighborhood
U0 of 0 such that y + U0 ⊆ X \ Ker(x0 ). By Proposition 3.7, the image of U0
under x0 is balanced in K. It is easy to see that such a set is either bounded or
the whole of K. The latter is out of the question: for all u ∈ U0 , hx0 , yi 6= hx0 , ui
because y − u 6∈ Ker(x0 ). Thus we have shown that the image of U0 is bounded,
and so (4) =⇒ (3).
Suppose that x0 is bounded on some neighborhood U0 of the origin 0 ∈ X ,
that is, there exists a constant M > 0 such that |hx0 , ui| ≤ M for all u ∈ U0 .
ε
Then, given ε > 0 and a basic neighborhood B(0, ε) of 0 in K, the set M
U0 is a
0
neighborhood of the origin whose image under x is contained in B(0, ε). Hence
x0 is continuous at 0 ∈ X and (3) =⇒ (2).
Finally, the implication (2) =⇒ (1) is a special case of Theorem 3.14. ¥
Proof.
Recall that for a normed space X , we dened the continuous dual X ∗ as the
(normed) linear space of all bounded linear functionals on X . Since functional
on a normed space is continuous if and only if it is bounded, the following
terminology is consistent.
Denition. Let X be a topological vector space. The subspace of the algebraic
dual X 0 consisting of all the continuous linear functionals is called the continuous
dual of X and denoted by X ∗ .
¤
Convexity
Denition. A subset K of a real or complex linear space X is said to be convex
if
αx + (1 − α)y ∈ K
whenever x, y ∈ K and 0 ≤ α ≤ 1.
¤
In other words, convex set is a set which contains the line segments between all
its points.
Denition. The convex hull of a subset S of a linear space X is the intersection
of all convex subsets of X that contain S . The convex hull of S is denoted by
co S .
¤
Denition. The closed convex hull of a subset S of a topological vector space
is the intersection of all closed convex sets that contain S . The closed convex
hull of S is denoted by co S .
¤
Note that a linear space is always convex, and that the intersection of any family
of convex sets is a convex set (Exercise 51). Hence every subset of a (topological)
linear space has a (closed) convex hull and it is a (closed) convex set.
38
3.19 Proposition. Let S be a subset of a topological vector space X . Then
the closed convex hull of S is the closure of the convex hull of S , that is,
co S = co S.
It is clear from the denitions that co S ⊆ co S . Since co S is closed, it
follows that co S ⊆ co S . On the other hand, the closure of the convex set co S
is closed and convex (see Exercise 52) and contains S , so co S ⊆ co S .
¥
Proof.
3.20 Lemma. If C1 , . . . , Cn are convex subsets of a real or complex linear space
X and α1 , . . . , αn are scalars, then
α1 C1 + · · · + αn Cn
is a convex set.
We prove rst, that αC is convex when C is convex and α is any scalar.
Given x, y ∈ αC , there are v, w ∈ C such that x = αv and y = αw. Since C is
convex,
¡
¢
λx + (1 − λ)y = α λv + (1 − λ)w ∈ αC
Proof.
whenever 0 ≤ λ ≤ 1, so αC is also convex, as claimed.
Next we show that given convex sets C and D, the set C + D is convex.
Suppose that x, y ∈ C + D, that is, x = c1 + d1 and y = c2 + d2 with c1 , c2 ∈ C
and d1 , d2 ∈ D. Then
¡
¢ ¡
¢
λx + (1 − λ)y = λc1 + (1 − λ)c2 + λd1 + (1 − λ)d2 ∈ C + D
whenever 0 ≤ λ ≤ 1, as required.
Armed with these two results, completing the proof is only a simple application of mathematical induction and is left to the reader.
¥
Denition. A locally convex (topological linear ) space is a real or complex
topological vector space that has a base at 0 consisting of convex sets.
¤
3.21 Proposition. The unit ball B(0, 1) = {x ∈ X : kxk < 1} of any normed
space X is convex.
Proof.
Exercise 56.
¥
3.22 Theorem. All normed spaces are locally convex T4 topological vector
spaces.
To see that any normed space is a T4 topological vector space, apply
Proposition 2.3 and remember that any metric space is T4 . The local convexity
follows from Proposition 3.21 and Lemma 3.20.
¥
Proof.
39
Chapter 4
Hahn-Banach Theorem
Let X be a normed space. Recall that the algebraic dual X 0 of X was dened
as the set of all linear functionals x0 on X with pointwise operations. Also, the
value of x0 ∈ X 0 at point x was denoted by hx0 , xi.
We have also dened X ∗ , the Banach dual of X , to be the linear subspace of
0
X which consists of all bounded (or, equivalently, continuoussee Theorem 2.7)
linear functionals x0 , i. e., for some constant M , |hx0 , xi| ≤ M kxk whenever
x ∈ X.
4.1 Theorem. Let X be a normed space. Then X ∗ is a Banach space with the
norm
Proof.
kx0 k = sup {|hx0 , xi| : x ∈ X, kxk = 1} .
See Theorem 2.11 and Proposition 2.9.
¥
4.2 Theorem. If X is a nite-dimensional normed space, then its algebraic
and continuous duals coincide (as linear spaces) and have the same dimension
as X .
Let n = dim X > 0 (case n = 0 is trivial: {0}0 = {0}) and x a
basis
Pn {x1 , . . . , xn } for X . Then every x ∈ X is represented uniquely as x =
k=1 αk (x)xk , with scalars α1 (x), . . . , αn (x) depending on x of course. Dene
a mapping k·k1 : X → R by
Proof.
kxk1 =
n
X
|αk (x)| .
(4.1)
k=1
Then k·k1 is a norm on X and therefore equivalent to the original norm k·k by
Corollary 2.28, so there is M > 0 such that
kxk1 =
n
X
|αk (x)| ≤ M kxk
(4.2)
k=1
whenever x ∈ X .
Now, for each k = 1, . . . , n, let x0k : X → K be dened by
hx0k , xi = αk (x).
40
(4.3)
Then x0k is clearly a linear functional, i.e., x0k ∈ X 0 . We also see from (4.2) that
|hx0k , xi| = |αk (x)| ≤ kxk1 ≤ M kxk
for all x ∈ X . Thus x0k is bounded, i.e., x0 ∈ X ∗ .
Furthermore,
set B 0 := {x01 , . . . , x0n } is linearly independent. To see this,
Pthe
n
observe that if k=1 βk x0k = 0 for some β1 , . . . , βn ∈ K, that is,
0=h
n
X
βk x0k , xi =
k=1
n
X
βk hx0k , xi
for all x ∈ X,
k=1
then in particular for each l = 1, . . . , n we have, due to our denition of αk 's,
0=
n
X
βk hx0k , xl i =
k=1
n
X
βk αk (xl ) = βl ,
k=1
as required.
If x0 ∈ X 0 , then by our denition (4.3) of the functionals x0k ,
0
0
hx , xi = hx ,
n
X
αk (x)xk i =
k=1
n
X
0
αk (x)hx , xk i =
k=1
n
X
hx0 , xk ihx0k , xi,
k=1
showing that Span(B 0 ) = X 0 . Thus B 0 is a basis for X 0 and dim X 0 = n =
dim X . Moreover, since each x0k is bounded, it is easy to see that X 0 ⊆ X ∗ ,
namely, for all x ∈ X
|hx0 , xi| ≤
n
X
|hx0 , xk i| |hx0k , xi| ≤ L
k=1
n
X
|hx0k , xi| ≤ L kxk1 ≤ LM kxk ,
k=1
0
0
where M and L = max{|hx , x1 i| , . . . , |hx , xn i|} are independent of x. Hence
X 0 = X ∗ as sets, which completes the proof.
¥
4.3 Corollary. If X is a nite-dimensional normed space, then X and X ∗ are
topologically isomorphic.
Proof.
Glance at theorems 4.1, 4.2 and 2.25.
¥
Hahn-Banach Theorem
Denition. A real-valued function f on a linear space X is said to be subadditive when
f (x + y) ≤ f (x) + f (y) for all x, y ∈ X.
¤
Denition. Let X be a real or complex linear space. We say that a functional
f : X → K is positively homogeneous when
f (αx) = αf (x) for all α ≥ 0 in R ∩ K and all x ∈ X,
and absolutely homogeneous when
f (αx) = |α| f (x) for all α ∈ K and all x ∈ X.
41
¤
Denition. Let X be a real or complex linear space. A subadditive and positively homogeneous functional on X is called a sublinear functional.
¤
Note that a seminorm is a subadditive and absolutely homogeneous functional. Also, every norm is a seminorm, and every seminorm is a sublinear
functional.
Recall that given sets A, B and C with A ⊆ B and functions f : A → C and
F : B → C , we say that F is an extension of f to B , or equivalently, f is the
restriction of F to A (denoted f = F |A ), when F (x) = f (x) for all x ∈ A.
4.4 Lemma. Let M be a proper subspace of a real linear space X , let x0 ∈
X \ M , and denote by N the subspace spanned by M ∪ {x0 }. If p is a sublinear
functional on X and f is a linear functional on M such that
f (x) ≤ p(x) for all x ∈ M ,
then there exists a linear extension F of f to N that satises
F (x) ≤ p(x) for all x ∈ N .
Proof.
Since f is linear and dominated by p, we have for every y1 , y2 ∈ M
f (y1 ) − f (y2 ) = f (y1 − y2 ) ≤ p(y1 − y2 ) ≤ p(y1 + x0 ) + p(−y2 − x0 ).
Regrouping terms, we obtain
−p(−y2 − x0 ) − f (y2 ) ≤ p(y1 + x0 ) − f (y1 )
(4.4)
for all y1 , y2 ∈ M . Denote
a = sup {−p(−y2 − x0 ) − f (y2 ) : y2 ∈ M }
b = inf {p(y1 + x0 ) − f (y1 ) : y1 ∈ M } .
By virtue of (4.4), the real numbers a and b exist and satisfy a ≤ b. Let c0 be
a real number such that a ≤ c0 ≤ b. Then for any y ∈ M
−p(−y − x0 ) − f (y) ≤ c0 ≤ p(y + x0 ) − f (y).
(4.5)
Let us now turn our attention to extending f . Since x0 6∈ M , there exist for
every x ∈ N = Span(M ∪ {x0 }) a unique y ∈ M and a unique α ∈ R such that
x = y + αx0 .
(4.6)
Indeed, the existence of such a representation is clear, and if y1 + α1 x0 = x =
y2 +α2 x0 are two such representations, then (α1 −α2 )x0 = y2 −y1 ∈ M , since M
is a subspace. Thus (α1 − α2 )x0 = 0, and since x0 6= 0, we must have α1 = α2 ,
and consequently y1 = y2 , as required.
Having assured the existence and uniqueness of the representation (4.6) for
all x ∈ N , we may use it to dene a functional F on N : set
F (y + αx0 ) = f (y) + αc0
for all y ∈ M and all α ∈ R.
42
(4.7)
The linearity of F is rather evident because f is linear on the subspace M : if
x1 = y1 + α1 x0 and x2 = y2 + α2 x0 with y1 , y2 ∈ M and α1 , α2 ∈ R, and β ∈ R,
then
F (x1 + βx2 ) = F ((y1 + βy2 ) + (α1 + βα2 )x0 )
= f (y1 + βy2 ) + (α1 + βα2 )c0
= (f (y1 ) + α1 c0 ) + β (f (y2 ) + α2 c0 )
= F (x1 ) + βF (x2 ).
Also, F is obviously an extension of f , as x ∈ M implies α = 0 in the representation (4.6).
It remains to verify that p dominates F on N . Let x = y + αx0 ∈ N be
arbitrary with y ∈ M and α ∈ R. There are three possibilities: α is either zero,
negative or positive.
If α = 0, we are done since F |M = f .
If α < 0, then αy ∈ M and thus
³ y
´
³y´
−p − − x0 − f
≤ c0
α
α
by the left half of the inequality (4.5). Multiplying through by α < 0 yields
³ y
´
³y´
αc0 ≤ −αp − − x0 − αf
= p(y + αx0 ) − f (y)
α
α
because p is positively homogeneous and f is linear. Hence
F (x) = f (y) + αc0 ≤ p(y + αx0 ) = p(x),
(4.8)
as required.
Similarly, if α > 0, then the right half of(4.5) yields αc0 ≤ p(y + αx0 ) − f (y),
whence the required inequality (4.8) again follows.
¥
The astute reader will notice that if X has a nite basisand sometimes
even if X is innite-dimensional, depending on M one can apply Lemma 4.4
consecutively nitely many times to obtain a linear extension F of f to the whole
space X . Hovewer, the spaces of interest are often innite-dimensional where
this process cannot generally be guaranteed to terminate. To always ensure the
existence of such an extension, we must once again invoke the axiom of choice.
4.5 Hahn-Banach Theorem. Let M be a subspace of a real vector space X .
If p is a sublinear functional on X and f is a linear functional on M such that
f (x) ≤ p(x) for all x ∈ M ,
then there exists a linear extension F of f to X such that
F (x) ≤ p(x) for all x ∈ X.
Proof. If ϕ is a functional on some subset of X , we denote its domain by Dϕ in
this proof. Consider the set F of all linear extensions ϕ of f that are dominated
by p, that is,
ϕ(x) ≤ p(x) for all x ∈ Dϕ .
43
Certainly f ∈ F , so this set is nonempty. Note also, that the domain of each
ϕ ∈ F is a linear subspace of X . We impose a partial order ¹ on F in the
familiar way: given linear functionals ϕ and ψ in F ,
ϕ ¹ ψ ⇐⇒ ψ extends ϕ,
(4.9)
that is, if and only if Dϕ ⊆ Dψ and ψ|Dϕ = ϕ.
Suppose C is a chain in F , and let
[
Dϕ .
Du =
ϕ∈C
Then every x ∈ Du belongs to Dϕ for some ϕ ∈ C . We dene a functional u on
Du by
u(x) = ϕ(x) if x ∈ Dϕ and ϕ ∈ C.
(4.10)
This is well-dened (independent of the choice of ϕ) because C is a chain. By
denition, u extends every ϕ ∈ C , and u(x) ≤ p(x) for all x ∈ Du .
To see that u is linear, let α ∈ R and x, y ∈ Du be arbitrary. Then x ∈ Dϕ
and y ∈ Dψ for some ϕ, ψ ∈ C . Since C is a chain, ϕ extends ψ or vice versa;
suppose the former. Then x, y ∈ Dϕ , which is a subspace, so x+αy ∈ Dϕ ⊆ Du .
This shows that Du is a linear subspace of X . Further,
u(x + αy) = ϕ(x + αy) = ϕ(x) + αϕ(y) = u(x) + αu(y),
so u is linear. Hence u ∈ F is an upper bound for C .
Since every chain in F has an upper bound, Zorn's lemma is applicable:
there exists a maximal element F ∈ F . To complete the proof for the real case,
we only need to show that the domain DF of F is X .
Suppose, otherwise, that there exists a vector x0 ∈ X \ DF . Since F is a
linear functional, its domain is a linear subspace of X that is proper by our
assumption. The conditions of Lemma 4.4 are thus fullled, so there exists a
linear extension Φ of F to N = Span(DF ∪ {x0 }) such that Φ(x) ≤ p(x) for all
x ∈ N . Thus Φ ∈ F , which contradicts the maximality of F (because F ¹ Φ
and F 6= Φ). Therefore we must have DF = X , as claimed.
¥
4.6 Hahn-Banach Theorem (Generalized). Let X be a real or complex
linear space and M a subspace of X . If p is a seminorm on X and f is a linear
functional on M such that
|f (x)| ≤ p(x) for all x ∈ M ,
then there exists a linear extension F of f to X that satises
|F (x)| ≤ p(x) for all x ∈ X.
Assume rst that X is a real vector space. The seminorm p is in
particular a sublinear functional, and |f (x)| ≤ p(x) implies f (x) ≤ p(x), so the
conditions of the Hahn-Banach theorem 4.5 are fullled. Thus there exists a
linear functional F on X such that F |M = f and F (x) ≤ p(x) for all x ∈ X .
But p is absolutely homogeneous, so we also have
Proof.
−F (x) = F (−x) ≤ p(−x) = |−1| p(x) = p(x)
44
and hence |F (x)| ≤ p(x) for all x ∈ X . Thus F is the required extension.
Assume then that X is a complex vector space. Write
f (x) = r(x) + iq(x) for x ∈ M ,
where r and q are real-valued (the real and imaginary parts of f ). It is clear
that we can regard X also as a linear space over R, which makes M a proper
real linear subspace. Given x, y ∈ M and α ∈ R, we have
(r(x) + αr(y)) + i (q(x) + αq(y)) = f (x) + αf (y) = f (x + αy)
= r(x + αy) + iq(x + αy).
Equating the real and imaginary parts reveals that r and q are R-linear on M .
Furthermore,
i (r(x) + iq(x)) = if (x) = f (ix) = r(ix) + iq(ix)
for all x ∈ M . The real parts on both sides must be equal, that is, q(x) = −r(ix)
for all x ∈ M . Thus we have the representation
f (x) = r(x) − ir(ix) for x ∈ M .
(4.11)
Since the real part of any complex number cannot exceed the number itself
in absolute value, r satises |r(x)| ≤ |f (x)| ≤ p(x) for all x ∈ M . Hence we
can apply the already proved part of this theorem for real vector spaces: there
exists some R-linear functional R on X such that R|M = r and |R(x)| ≤ p(x)
for all x ∈ M . Dene a functional F on X by
F (x) = R(x) − iR(ix).
(4.12)
Then F is clearly R-linear, but we assert more: F is the desired extension of
f . Since R extends r, it is clear from (4.11) that F extends f . Thus our task is
twofold: we need to show that F is C-linear and dominated by p.
For the linearity, we already know that F (x + y) = F (x) + F (y) for all
x, y ∈ X because R has this property. If α = a + ib is any complex number with
real and imaginary parts a and b, respectively, then by the R-linearity of R we
have for any x ∈ X
F (αx) = R(ax + ibx) − iR(iax − bx)
= aR(x) + bR(ix) − iaR(ix) + ibR(x)
= (a + ib)R(x) − (a + ib)iR(ix)
= αF (x).
Thus F is C-linear.
If F (x) = 0, then |F (x)| ≤ p(x) because a seminorm is nonnegative (see
Exercise 20). Suppose then that F (x) 6= 0. Using the polar representation for
complex numbers, we can write F (x) = |F (x)| eiθ , or equivalently,
|F (x)| = F (x)e−iθ = F (e−iθ x)
which is a real number. Hence F (e−iθ x) must be equal to its real part, so
¯
¯
|F (x)| = R(e−iθ x) ≤ p(e−iθ x) = ¯e−iθ ¯ p(x) = p(x).
Thus |F (x)| ≤ p(x) for all x ∈ X , which completes the proof.
45
¥
4.7 Hahn-Banach Theorem (Normed Spaces). If m0 is a bounded linear
functional on a subspace M of a normed space X , then there exists a bounded
linear extension x0 of m0 to X such that
kx0 k = km0 k .
Dene a seminorm p on X by p(x) = km0 k kxk . Then m0 satises
|hm , xi| ≤ p(x) for all x ∈ M , so we may apply the Hahn-Banach Theorem 4.6:
there exists a linear functional x0 on X which extends m0 and satises |hx0 , xi| ≤
p(x) for all x ∈ X . But because of the way we dened p, this implies that x0 is
bounded and kx0 k ≤ km0 k .
On the other hand, for every ε > 0 there exists some mε ∈ M such that
kmε k ≤ 1 and km0 k − ε < |hm0 , mε i| = |hx0 , mε i|. Hence kx0 k ≥ km0 k .
¥
Proof.
0
4.8 Corollary. Let Z be a proper subspace of a normed space X . If x0 ∈ X
and δ = d(x0 , Z) = inf {kx0 − zk : z ∈ Z} > 0, then there exists a bounded
linear functional x0 on X such that
x 0 |Z = 0
and
kx0 k = 1
and hx0 , x0 i = δ.
Now x0 6∈ Z because δ > 0. Denote M = Span(Z ∪ {x0 }). Every
m ∈ M can be uniquely represented as m = z + αx0 with z ∈ Z and α ∈ K,
the scalar eld of X (cf. the proof of Lemma 4.4). Let m0 be the functional on
M dened by
m0 : z + αx0 7→ αδ for z ∈ Z and α ∈ K.
Proof.
Then m0 is clearly well-dened and linear. If m = z + αx0 ∈ M with z ∈ Z and
a nonzero α ∈ K, then
°
x°
°
°
|hm0 , mi| = |αδ| = |α| δ = |α| inf kx0 − xk = |α| inf °x0 + °
x∈Z
x∈Z
α
°
z°
°
°
≤ |α| °x0 + ° = kz + αx0 k = kmk
α
since Z is a subspace. Also hm0 , zi = 0 for all z ∈ Z , so m is bounded and
km0 k ≤ 1.
On the other hand, for every z ∈ Z ,
δ = hm0 , x0 − zi ≤ |hm0 , x0 − zi| ≤ km0 k kx0 − zk ,
and therefore
δ ≤ km0 k inf {kx0 − zk : z ∈ Z} = km0 k δ.
Since δ > 0, this implies km0 k ≥ 1.
We have constructed a functional m0 ∈ M ∗ such that m0 |Z = 0, km0 k = 1
and hm0 , x0 i = δ . By Hahn-Banach Theorem 4.7, there is an extension x0 ∈ X ∗
of m0 with the same properties.
¥
4.9 Corollary. Let Z be a proper closed subspace of a normed space X . If
x0 6∈ M , then there exists a bounded linear functional x0 on X such that
x0 |Z = 0 and
kx0 k = 1 and hx0 , x0 i 6= 0.
46
Proof.
Since Z is closed, δ = d(x0 , Z) > 0 in Corollary 4.8.
¥
4.10 Corollary. If x0 is a nonzero vector of a normed linear space X , then
there exists x0 ∈ X ∗ such that
kx0 k = 1 and hx0 , x0 i = kx0 k .
Proof.
Apply Corollary 4.8 with Z = {0}.
¥
The following consequence is important. It tells us that there are enough
bounded linear functionals on a normed space to separate pointsa property
which ensures us that the weak topologies we dene later in chapter 6 are strong
enough (Hausdor). The property is in fact necessary for this to happen (see
Theorem 6.5). Locally convex Hausdor spaces also share this property, and
the proof goes again via Hahn-Banach theorem (see Corollary 4.29).
4.11 Corollary. Let X be a normed space and let x, y ∈ X . If x 6= y , then
there exists x0 ∈ X ∗ such that
hx0 , xi =
6 hx0 , yi.
Proof.
Apply Corollary 4.10 with x0 = x − y .
¥
Next we describe the norm of an element of a normed space with bounded
linear functionals.
4.12 Corollary. If x is an element of a normed space X , then
kxk = sup {|hx0 , xi| : x0 ∈ X ∗ , kx0 k ≤ 1}
= sup {|hx0 , xi| : x0 ∈ X ∗ , kx0 k = 1}
½ 0
¾
|hx , xi|
0
∗
0
= sup
:
x
∈
X
,
x
=
6
0
.
kx0 k
Proof. The statement is obvious for x = 0, so we assume x 6= 0. If x0 ∈ X ∗ ,
then |hx0 , xi| ≤ kx0 k kxk . Thus the suprema in the assertion cannot exceed kxk .
On the other hand, since x 6= 0, there is by Corollary 4.10 a bounded linear
functional x0 ∈ X ∗ such that kx0 k = 1 and hx0 , xi = kxk , which completes the
proof.
¥
4.13 Corollary. If x is an element of a normed space X such that hx0 , xi = 0
for all x0 ∈ X ∗ satisfying kx0 k = 1, then x = 0.
Proof.
Clear from Corollary 4.12 because x = 0 if and only if kxk = 0.
¥
Annihilators
We now investigate certain subspaces of the dual X ∗ , namely the sets of functionals that vanish on a given subspace M of X , and related quotient spaces.
47
Denition. Let S be a non-empty subset of a normed space X . The annihilator
of S is denoted by S ⊥ and dened by
S ⊥ = {x0 ∈ X ∗ : hx0 , si = 0
for all s ∈ S} .
(4.13)
¤
4.14 Proposition. If S is a subset of a normed space X , then its annihilator
S ⊥ is a closed subspace of X ∗ .
Proof.
Exercise 62.
¥
4.15 Theorem. If M is a subspace of a normed space X , then
X ∗ /M ⊥ ∼
= M∗
isometrically.
Since M ⊥ is closed, X ∗ /M ⊥ is a normed space by Theorem 2.6. For
any given coset x0 + M ⊥ ∈ X ∗ /M ⊥ , we dene m0 : M → K by m0 = x0 |M , that
is,
hm0 , mi = hx0 , mi for all m ∈ M.
(4.14)
Proof.
This is well-dened because y 0 + M ⊥ = x0 + M ⊥ if and only if y 0 − x0 ∈ M ⊥ ,
that is, 0 = hy 0 − x0 , mi = hy 0 , mi − hx0 , mi for all m ∈ M , or equivalently,
y 0 |M = x0 |M .
The linearity of m0 follows directly from that of x0 . Also, if y 0 ∈ x0 + M ⊥ ,
we have |hm0 , mi| = |hy 0 , mi| ≤ ky 0 k kmk for every m ∈ M , which shows that
m0 ∈ M ∗ . Further, taking inmum yields the upper bound
©
ª
|hm0 , mi| ≤ inf ky 0 k : y 0 ∈ x0 + M ⊥ kmk = |||x0 + M ⊥ ||| kmk ,
from which we deduce that
km0 k ≤ |||x0 + M ⊥ |||.
(4.15)
On the other hand, Hahn-Banach Theorem 4.7 gives us an extension y 0 ∈ X ∗
of m0 such that ky 0 k = km0 k . In other words, y 0 ∈ x0 + M ⊥ and ky 0 k = km0 k .
Thus |||x0 + M ⊥ ||| ≤ km0 k . Combining this with (4.15) gives
km0 k = |||x0 + M ⊥ |||.
(4.16)
Now we dene T : X ∗ /M ⊥ → M ∗ by
¡
¢
T x0 + M ⊥ = x0 |M ,
(4.17)
where x0 |M = m0 was dened by (4.14). Having just proved in (4.16) that T is
an isometry, it only remains to check that T is linear and bijective. Since linear
isometry is always injective, it suces to prove that T is linear and surjective.
The former is easy:
¡
¢
hT (x0 + M ⊥ ) + α(y 0 + M ⊥ ) , mi = hx0 + αy 0 , mi
= hx0 , mi + αhy 0 , mi = hT (x0 + M ⊥ ), mi + αhT (y 0 + M ⊥ ), mi
48
for all m ∈ M , which means that
¡
¢
T (x0 + M ⊥ ) + α(y 0 + M ⊥ ) = T (x0 + M ⊥ ) + αT (y 0 + M ⊥ ).
For the latter, let m0 ∈ M ∗ be arbitrary. By Hahn-Banach Theorem 4.7, there
exists an extension x0 ∈ X ∗ of m0 , which obviously satises T (x0 + M ⊥ ) = m0 .¥
4.16 Theorem. If M is a closed subspace of a normed space X , then
(X/M )∗ ∼
= M⊥
isometrically.
Proof. Since M is closed, (X/M ) is a normed space by Theorem 2.6, and the
assertion is plausible. For brevity, we shall denote the elements of (X/M )∗ with
ϕ and ψ instead of the usual (x + M )0 in this proof. Dene T : (X/M )∗ → M ⊥
by
hT ψ, xi = hψ, x + M i for all x ∈ X.
(4.18)
First we check that T is really a mapping into M ⊥ . Let ψ ∈ (X/M )∗ be
arbitrary. If α ∈ K and x, y ∈ X , then
hT ψ, x + αyi = hψ, (x + αy) + M i = hψ, (x + M ) + α(y + M )i
= hψ, x + M i + αhψ, y + M i = hT ψ, xi + αhT ψ, yi.
Thus T ψ is linear. For the boundedness of T ψ , observe that for all x ∈ X ,
|hT ψ, xi| = |hψ, x + M i| ≤ kψk |||x + M ||| ≤ kψk kxk
(4.19)
by the denition of the quotient norm, so T ψ ∈ X ∗ . Lastly, if x ∈ M , then
x + M = M is the zero element of X/M and thus
hT ψ, xi = hψ, M i = 0
because ψ is linear. Hence T ψ ∈ M ⊥ , as claimed.
The mapping T is linear: if α ∈ K and ϕ, ψ ∈ (X/M )∗ , then
hT (ϕ + αψ), xi = hϕ + αψ, x + M i = hϕ, x + M i + αhψ, x + M i
= hT ϕ, xi + αhT ψ, xi
for all x ∈ X , meaning that T (ϕ + αψ) = T ϕ + αT ψ .
Taking supremum for x ∈ X with kxk = 1 in equation (4.19) yields
kT ψk ≤ kψk
for any ψ ∈ (X/M )∗ ,
(4.20)
that is, T is bounded and kT k ≤ 1. Hence, to demonstrate that T is an isometry,
it suces to show that
kψk ≤ kT ψk
for every ψ ∈ (X/M )∗ .
(4.21)
Let r > 1, and for every coset x + M with |||x + M ||| = 1, let x̃ be an element
of x + M satisfying kx̃k < r. Such an element exists, because |||x + M ||| =
49
inf {kyk : y ∈ x + M } < r. Then x̃ + M = x + M and thus
kψk = sup {|hψ, x + M i| : x ∈ X, |||x + M ||| = 1}
= sup {|hψ, x̃ + M i| : x ∈ X, |||x + M ||| = 1}
= sup {|hT ψ, x̃i| : x ∈ X, |||x + M ||| = 1}
≤ sup {kT ψk kx̃k : x ∈ X, |||x + M ||| = 1}
≤ r kT ψk .
Since this holds for all r > 1 we can let r −→ 1 to obtain (4.21).
Since a linear isometry is injective, it remains only to check that T is surjective. Let x0 ∈ M ⊥ . Then dene ψ : X/M → K by
hψ, x + M i = hx0 , xi
for all x ∈ X.
(4.22)
This is well dened: if x + M = y + M , then x − y ∈ M and
0 = hx0 , x − yi = hx0 , xi − hx0 , yi
since x0 ∈ M ⊥ , so in our denition (4.22), hx0 , xi = hx0 , yi independently of the
choice of the representatives x and y of the coset x + M .
The linearity of ψ follows easily from that of x0 . Furthermore, given a coset
x + M,
|hψ, x + M i| = |hψ, (x + m) + M i| = |hx0 , x + mi| ≤ kx0 k kx + mk
for every m ∈ M , which implies that
|hψ, x + M i| ≤ kx0 k inf kx + mk = kx0 k |||x + M |||,
m∈M
showing that ψ is bounded and indeed in (X/M )∗ . Since we chose ψ in (4.22)
so that T ψ = x0 , and x0 was arbitrary, we have proven that T is onto, and the
proof is complete.
¥
Note that the proof of Theorem 4.15 relied on the Hahn-Banach
theorem, while the proof of Theorem 4.16 did not; we had no need to extend
functionals when dening the isomorphism in the latter proof.
¤
Remark.
Hyperplanes
Our aim for the rest of this chapter is to formulate the Hahn-Banach theorem
in geometric terms; hyperplanes and convex sets to be precise. We'll also derive
forms that are sometimes called Hahn-Banach separation theorems. Since we
are already acquainted with convex sets, it is time to introduce hyperplanes.
Recall that a subspace M of a linear space X is maximal when X is the only
subspace that properly contains M .
Denition. Let M be a maximal linear subspace of a linear space X . When
x0 ∈ X , we call H = x0 + M a hyperplane in X . Two hyperplanes are said to
be parallel when they are translates of the same maximal subspace.
¤
Note that a hyperplane H is a (maximal) subspace if and only if 0 ∈ H .
50
4.17 Theorem. A subspace M of a linear space X is maximal if and only if
M = Ker(x0 ) for some nonzero linear functional x0 ∈ X 0 .
Let K be the scalar eld of X , as usual. Suppose M is a maximal subspace of X . Since M is a proper subspace, there exists y ∈ X \M . Furthermore,
every x ∈ X can be uniquely written in the form x = m + αy with m ∈ M and
α ∈ K because M is maximal. Dene a functional x0 on X by
Proof.
hx0 , m + αyi = α
for all m ∈ M and α ∈ K.
Then x0 is evidently linear, nonzero, and Ker(x0 ) = M as required.
Conversely, Ker(x0 ) is a maximal subspace of X for all nonzero x0 ∈ X 0 by
Theorem 1.20, so we are done.
¥
4.18 Corollary. A subset H of a linear space X is a hyperplane if and only if
H = x0 + Ker(x0 )
for some x0 ∈ X and some nonzero x0 ∈ X 0 .
Proof.
Trivial after Theorem 4.17.
¥
4.19 Corollary. A subset H of a linear space X is a hyperplane if and only if
H = {x ∈ X : hx0 , xi = α}
(4.23)
for some scalar α and some nonzero x0 ∈ X 0 . Moreover, if H is a hyperplane, it
contains the origin if and only if α = 0.
Let α be a scalar, let x0 be a nonzero linear functional on X and
suppose that H is the set dened in (4.23) via x0 . Since x0 6= 0, there is x1 ∈ X
such that hx0 , x1 i = β 6= 0. Let x0 = (α/β)x1 . Then hx0 , x0 i = α, and hence
Proof.
x0 + Ker(x0 ) = {x0 + z : z ∈ X, hx0 , zi = 0}
= {x : x − x0 ∈ X, hx0 , x − x0 i = 0}
= {x ∈ X : hx0 , xi = hx0 , x0 i}
= {x ∈ X : hx0 , xi = α}
= H.
Thus H is a translate of Ker(x0 ). Conversely, the preceeding calculation shows
that the translate of the kernel of any nonzero linear functional x0 by any vector
x0 ∈ X can be written in the form (4.23) by choosing α = hx0 , x0 i. The rst
statement now follows from Corollary 4.18.
Knowing this, the second statement is easy to prove: if H is a hyperplane,
that is, of the form (4.23), then 0 ∈ H if and only if α = hx0 , 0i = 0.
¥
4.20 Proposition. Let X be a linear space over K. Let H = x0 + M be a
hyperplane in X , where x0 ∈ X and M is a maximal subspace of X . If
H = {x ∈ X : hx0 , xi = α} = {x ∈ X : hy 0 , xi = β} ,
where x0 , y 0 ∈ X 0 and α, β ∈ K, then there exists a nonzero c ∈ K such that
y 0 = cx0
and β = cα.
Furthermore, M = Ker(x0 ) = Ker(y 0 ), α = hx0 , x0 i and β = hy 0 , x0 i,
51
Proof.
First, for every m ∈ M ,
hx0 , mi = hx0 , x0 + mi − hx0 , x0 + 0i = α − α = 0
because H = x0 + M . On the other hand, x0 cannot be the zero functional,
because oterwise we would have H = ∅ or H = X (depending on α), and
neither of these is a hyperplane. Fix any z ∈ X \ M . Because M is maximal,
Span(M ∪ {z}) = X , and hence every x ∈ X can be uniquely represented as
x = m x + λx z
with mx ∈ M and λx ∈ K.
(4.24)
Consequently, hx0 , xi = 0 + λx hx0 , zi and thus hx0 , zi 6= 0 lest x0 be the zero
functional. Hence Ker(x0 ) = M . Similarly, Ker(y 0 ) = M .
The equalities hx0 , x0 i = α and hy 0 , x0 i = β are clear because x0 ∈ H .
Finally, keeping the representation (4.24) in mind, we have for every x ∈
X \ M the following:
hy 0 , xi
hy 0 , mx i + hy 0 , λx zi
0 + λx hy 0 , zi
hy 0 , zi
= 0
=
= 0
=: c
0
0
0
hx , xi
hx , mx i + hx , λx zi
0 + λx hx , zi
hx , zi
which is a constant. Since x0 and y 0 vanish on M , we have y 0 = cx0 . It then
follows that β = hy 0 , x0 i = hcx0 , x0 i = cα, and the proof is complete.
¥
4.21 Corollary. Let X be a real linear space, and let
H = {x ∈ X : hx0 , xi = α} = {x ∈ X : hy 0 , xi = β}
be a hyperplane in X with x0 , y 0 ∈ X 0 and α, β ∈ R. If we denote
H + = {x ∈ X : hx0 , xi > α}
H − = {x ∈ X : hx0 , xi < α}
L+ = {x ∈ X : hy 0 , xi > α}
L− = {x ∈ X : hy 0 , xi < α} ,
then either
or
H + = L+
and H − = L−
H + = L−
and H − = L+ .
By Proposition 4.20, there is a real number c 6=
and β = cα. This will suce: suppose rst that c < 0.
hx0 , xi > α ⇐⇒ chx0 , xi < cα ⇐⇒ hy 0 , xi < β ⇐⇒ x
By reversing the inequality signs, we obtain H − = L+ .
c > 0 are equally easy and left to the reader.
Proof.
0 such that y 0 = cx0
Then x ∈ H + ⇐⇒
∈ L− , so H + = L− .
The calculations for
¥
This assures that the following denition is unambiguous.
Denition. Let X be a real linear space and let H = {x ∈ X : hx0 , xi = α} be
a hyperplane in X with x0 ∈ X 0 and α ∈ R, The sets
{x ∈ X : hx0 , xi < α}
and
{x ∈ X : hx0 , xi > α}
are called the half-spaces corresponding to H .
52
¤
This terminology is due to the obvious fact that X is the disjoint union of H
and these two half-spaces.
4.22 Theorem. Let K be a convex subset of a real linear space X and let H
be a hyperplane in X . Then K ∩ H = ∅ if and only if K is contained in one of
the half-spaces corresponding to H .
Let x0 ∈ X 0 and α ∈ R be such that H = {x ∈ X : hx0 , xi = α}.
If K is contained in one of the half-spaces, then it is clear that K ∩ H = ∅,
as hx0 , xi = α cannot happen simultaneously with hx0 , xi < α or hx0 , xi > α.
Suppose then that K is not contained in one of the half spaces. Since X is
the disjoint union of H and the half-spaces, there are two possibilities: either
K ∩ H 6= ∅, a case in which we are done, or there exist x, y ∈ K such that
Proof.
hx0 , xi < α
and hx0 , yi > α.
Dene a function ϕ : [0, 1] → R by
ϕ(λ) = hx0 , λx + (1 − λ)yi = λhx0 , xi + (1 − λ)hx0 , yi.
Then ϕ is clearly continuous, ϕ(1) = hx0 , xi < α and ϕ(0) = hx0 , yi > α.
Thus there exists β ∈]0, 1[ such that ϕ(β) = α by the well-known intermediate
value theorem. But then z = βx + (1 − β)y ∈ K because K is convex, and
hx0 , zi = ϕ(β) = α, so z ∈ H . Hence K ∩ H 6= ∅.
¥
Note that while the concept of a hyperplane is sensible in a complex linear
space X , a complex hyperplane cannot be used to split X in two half-spaces
{x ∈ X : hx0 , xi < α} and {x ∈ X : hx0 , xi > α} where x0 ∈ X ∗ \ {0}. This is
due to C not being an ordered eld and all nonzero linear functionals being
surjective (and hence not real-valued). However, if X is a linear space over C,
it can also be regarded as a linear space over R, so our considerations are not
necessarily constrained to real linear spaces only. Also, the real part of x0 could
be used (see Exercise 64).
Denition. Let X be a linear space over R or C. By a real linear functional
on X we mean a real-valued function on X that is a linear functional on the
linear space X over R. By a real hyperplane in X we mean a hyperplane in the
linear space X over R.
¤
Note that a real linear functional f on a linear space X over C cannot be linear
(with respect to C), that is f 6∈ X 0 , and we shall refrain from using the prime
notation for f in this case to avoid confusion.
Geometric Forms of Hahn-Banach Theorem
To make the transition from our analytic form of the Hahn-Banach theorem to
the desired geometric formulations, we employ a specic family of seminorms:
Denition. Let A be an absorbing subset of a linear space X over R or C.
The Minkowski functional (or the gauge functional ) of A is the functional
pA : X → [0, ∞[ dened by
pA (x) = inf {r : r > 0 and x ∈ rA} .
53
¤
4.23 Lemma. Let K be a convex balanced absorbing subset of a real or complex linear space X . Then the Minkowski functional pK of K is a seminorm on
X.
Let x, y ∈ X . If α and β are positive real numbers such that pK (x) < α
and pK (y) < β , then x ∈ αK and y ∈ βK by Proposition 3.7 because K is
balanced, so there are u, v ∈ K such that x = αu and y = βv . Since K is
convex,
1
α
β
(x + y) =
u+
v ∈ K,
α+β
α+β
α+β
Proof.
that is, x + y ∈ (α + β)K . Hence pK (x + y) ≤ α + β , and because this holds for
all α > pK (x) and β > pK (y), we must have pK (x + y) ≤ pK (x) + pK (y). Thus
pK is sublinear.
Because K is balanced, αK = |α| K for all scalars α. Hence for all x ∈ X
and α 6= 0, one has
pK (αx) = inf {r : r > 0, αx ∈ rK}
n
o
−1
= |α| inf |α| r : r > 0, x ∈ α−1 rK
n
o
−1
−1
= |α| inf |α| r : r > 0, x ∈ |α| rK
= |α| inf {s : s > 0, x ∈ sK}
= |α| pK (x).
As the equality pK (αx) = |α| pK (x) is trivial for α = 0, we are done.
¥
4.24 Lemma. Let K be a convex absorbing subset of a real linear space X and
let pK be the Minkowski functional of K . If L is a hyperplane in X such that
K ∩ L = ∅, then there exists x0 ∈ X 0 such that
(1) L = {x ∈ X : hx0 , xi = 1} and
(2) −pK (x) ≤ hx0 , xi ≤ pK (x) for all x ∈ X .
Proof. First note that since K is absorbing, 0 ∈ K and thus 0 6∈ L as K∩L = ∅.
By Corollary 4.19, there exists a nonzero linear functional y 0 ∈ X 0 and α ∈ R
such that L = {x ∈ X : hy 0 , xi = α}. Furthermore, α 6= 0, because 0 6∈ L. Thus
x0 = α−1 y 0 is a linear functional on X such that L = {x ∈ X : hx0 , xi = 1}. By
Theorem 4.22, K lies completely inside one of the half-spaces corresponding to
L. Since 0 ∈ K and hx0 , 0i = 0 < 1, we must have
K ⊆ {x ∈ X : hx0 , xi < 1} .
Let x ∈ X . If β > 0 is such that x ∈ βK , then β −1 x ∈ K and we have
hx , xi = βhx0 , β −1 xi < β . Thus hx0 , xi ≤ inf {β > 0 : x ∈ βK} = pK (x). In
turn, this implies that −hx0 , xi = hx0 , −xi ≤ pK (−x) = pK (x) because pK is a
seminorm. Hence −pK (x) ≤ hx0 , xi ≤ pK (x) for all x ∈ X , as required.
¥
0
Now we are ready to prove the geometric counterpart of Hahn-Banach Theorem 4.6.
54
4.25 Hahn-Banach Theorem (Geometric). Let K be a convex absorbing
subset of a real or complex linear space X and let pK be the Minkowski functional of K . If L0 = x0 + Y0 , where x0 ∈ X and Y0 is a linear subspace of X
such that K ∩ L0 = ∅, then there exists a real linear functional F on X such
that L = {x ∈ X : F (x) = 1} is a real hyperplane in X ,
(1) F (x) ≤ pK (x) for all x ∈ X ,
(2) L0 ⊆ L,
(3) K ⊆ {x ∈ X : F (x) ≤ 1}.
Again, 0 belongs to the absorbing set K so x0 6∈ Y0 . Let Y be the
subspace of the linear space X over R that is spanned by Y0 ∪ {x0 }. Then Y0 is
a maximal subspace of the real linear space Y , and so L0 is a hyperplane in Y .
Now consider K ∩ Y as a subset of the real linear space Y . As a subspace of
the real linear space X , Y is convex and thus K ∩ Y is convex as an intersection
of convex sets. Also, K is clearly absorbing in the real linear space X , so K ∩ Y
must be absorbing in Y . Furthermore, the Minkowski functional of K ∩ Y in Y
and the Minkowski functional of K in X satisfy
Proof.
pK∩Y (y) = inf {β > 0 : y ∈ β(K ∩ Y )}
= inf {β > 0 : y ∈ (βK) ∩ Y }
= inf {β > 0 : y ∈ βK}
= pK (y)
for all y ∈ Y . Since (K ∩ Y ) ∩ L0 = ∅, we may apply Lemma 4.24 for the
real linear space Y : there exists a real linear functional f on Y such that
L0 = {y ∈ Y : f (x) = 1} and f (y) ≤ pK∩Y (y) = pK (y) for all y ∈ Y .
Since pK is a seminorm on X , there exists, by the Hahn-Banach theorem
4.5, a real linear extension F of f to X such that F (x) ≤ pK (x) for all x ∈
X . If we set L = {x ∈ X : F (x) = 1}, it is clear that L0 ⊆ L. Also, since
L0 is a hyperplane in Y , f and consequently its extension F must be nozero
functionals, and hence L is a real hyperplane in X by Corollary 4.19. Finally,
F (x) ≤ pK (x) ≤ 1 for all x ∈ K , and thus K ⊆ {x ∈ X : F (x) ≤ 1}.
¥
Next we turn our attention to topological linear spaces. Remember, that
given a real or complex topological vector space X , a linear functional x0 on X
is continuous if and only if Ker(x0 ) is closed (see Theorem 3.18).
4.26 Proposition. Let X be a real or complex topological linear space, let
x0 ∈ X 0 be nonzero and α ∈ K. The hyperplane H = {x ∈ X : hx0 , xi = α} is
closed if and only if x0 is continuous.
Since H is a translate of Ker(x0 ) by Proposition 4.20 and translations
are homeomorphisms in X , the statement follows from Theorem 3.18.
¥
Proof.
4.27 Hahn-Banach Theorem (Topological Vector Spaces, 1st). Let K
be a convex absorbing subset of a real or complex topological vector space X
and suppose that K has a nonempty interior. If L0 = x0 + Y0 , where x0 ∈ X
and Y0 is a linear subspace of X such that K ∩ L0 = ∅, then there exists a closed
real hyperplane L = {x ∈ X : F (x) = 1}, where F is a continuous real linear
functional on X , such that
55
(1) L0 ⊆ L
(2) K̊ ⊆ {x ∈ X : F (x) < 1}.
(3) K ⊆ {x ∈ X : F (x) ≤ 1}.
By the previous geometric Hahn-Banach theorem 4.25, there exists
a real linear functional F on X such that L = {x ∈ X : F (x) = 1} is a real
hyperplane, L0 ⊆ L, F (x) ≤ pK (x) for all x ∈ X , and
Proof.
K ⊆ {x ∈ X : F (x) ≤ 1} .
We regard X as a topological linear space over R. If we can prove that L
is closed, the rest will be easy. Suppose, on the contrary, that L is not closed.
Since F is linear and nonzero, the range of F is R. Pick x1 , x2 ∈ X so that
F (x1 ) = 1 and F (x2 ) = 2. Denote the kernel of F by M , let L1 = L and
L2 = x2 + M . Applying Proposition 4.20, we have
L1 = x1 + M = {x ∈ X : F (x) = 1}
L2 = x2 + M = {x ∈ X : F (x) = 2} .
Since the translations are homeomorphisms of X , we know that L = L1 =
x1 + M . Thus L = L if and only if M = M . Since M is a subspace of X
that contains the maximal subspace M , either M = M or M = X . Because we
assumed that L is not closed, neither is M , so M must be dense in X . Thus
L2 = x2 + M is also dense, and so K̊ ∩ L2 6= ∅ because K has a nonempty
interior. But this is absurd, because F (x) ≤ 1 for all x ∈ K and F (x) = 2 for
all x ∈ L2 . Hence L must be closed.
Now, F is continuous by Proposition 4.26 and hence, as a pre-image of
a closed set, {x ∈ X : F (x) ≤ 1} = F −1 (] − ∞, 1]) is a closed set in X that
contains K . Thus K ⊆ {x ∈ X : F (x) ≤ 1}, as required.
Since K̊ ⊆ K ⊆ {x ∈ X : F (x) ≤ 1}, it only remains to show that an interior
point x of K cannot satisfy F (x) = 1. Suppose the contrary: x ∈ K̊ , F (x) = 1
and let U be a basic neighborhood of 0 such that x + U ⊆ K̊ . By Theorem 3.8,
we may assume that U is absorbing. Then there exists ε > 0 such that εx ∈ U .
Hence x + εx ∈ x + U ⊆ K , and so F (x + εx) ≤ 1. But F is real linear, so we
must have F (x + εx) = (1 + ε)F (x) = 1 + ε > 1, a contradiction. Thus F (x) < 1
for all x ∈ K̊ , and the proof is complete.
¥
4.28 Hahn-Banach Theorem (Topological Vector Spaces, 2nd). Let A
and B be convex nonempty subsets of a real or complex topological vector space
X . If A is open and A ∩ B = ∅, then there exists a continuous linear functional
x0 ∈ X ∗ and a real number β such that
Rehx0 , ai < β ≤ Rehx0 , bi
for all a ∈ A and b ∈ B .
Let K be the underlying scalar eld.
Let a0 ∈ A and b0 ∈ B be xed. Put x0 = b0 − a0 and K = x0 + A − B .
Then x0 6∈ K , for otherwise we would have 0 ∈ A S
− B which is impossible
because A ∩ B = ∅. Furthermore, K = x0 + A − B = b∈B (x0 + A + b) is open
Proof.
56
because each of the sets x0 + A + b is open by Theorem 3.1. It is clear that
0 = x0 + a0 − b0 ∈ K , so K is absorbing as a neighborhood of 0 by Corollary 3.9.
Let Y0 = {0} be the trivial linear subspace of X and put L0 = x0 + Y0 . Now
K is a convex absorbing set with nonempty interior and K∩L0 = ∅, which means
that we can apply the geometric Hahn-Banach theorem 4.27 for the topological
vector space X . Thus there exists a continuous real linear functional F on X
such that
{x0 } = L0 ⊆ {x ∈ X : F (x) = 1}
K = K̊ ⊆ {x ∈ X : F (x) < 1} .
Due to the choice of K this means that 1 > F (x0 +a−b) = F (x0 )+F (a)−F (b) =
1 + F (a) − F (b), which yields F (a) < F (b) for all a ∈ A and b ∈ B . By choosing
β = inf {F (b) : b ∈ B} we obtain
F (a) ≤ β ≤ F (b) for all a ∈ A and b ∈ B,
so it only remains to check that F (a) cannot be equal to β for any a ∈ A.
Suppose, on the contrary, that F (a1 ) = β for some a1 ∈ A and consider the
mapping g : R → X dened by g(t) = a1 +tx0 . Then g is clearly continuous and
g(0) = a1 ∈ A. Since A is open, there exists an ε > 0 such that g(]−2ε, 2ε[) ⊆ A;
in particular, g(ε) = a1 + εx0 ∈ A. But this leads to contradiction: β + ε =
F (a1 ) + εF (x0 ) = F (a1 + εx0 ) ≤ β . Hence we must have
F (a) < β ≤ F (b) for all a ∈ A and b ∈ B.
If K = R, we are done after choosing x0 = F . If K = C, we choose (see
Exercise 64) the unique linear functional x0 ∈ X 0 whose real part is F , namely
hx0 , xi = F (x) − iF (ix)
to obtain the desired inequality. The continuity of x0 follows from that of F . ¥
The previous formulation of the Hahn-Banach theorem makes it easy to prove
that the continuous dual of a locally convex Hausdor space is suciently large
to separate points in the space (compare with Corollary 4.11 to the Hahn-Banach
theorem for normed spaces).
4.29 Corollary. Let X be a real or complex locally convex Hausdor space,
and let x, y ∈ X . If x 6= y , then there exists x0 ∈ X ∗ such that
hx0 , xi =
6 hx0 , yi.
Since X is Hausdor, there exists an open neighborhood A of x such
that y 6∈ A. We can in fact assume that A is convex because X is locally
convex. Let B = {y}. Then B is clearly convex and A ∩ B = ∅ so there
exists, by Hahn-Banach Theorem 4.28, a continuous linear functional x0 such
that Rehx0 , xi < Rehx0 , yi and hence hx0 , xi 6= hx0 , yi, as required.
¥
Proof.
The reader might wonder why X was assumed to be Hausdor in the previous
theorem, but the proof used only the T0 separation property of X . The reason is that replacing Hausdor by T0 does not really relax the conditions
57
because every topological
completely regular, which
T3 21 equivalent conditions
regularity of a topological
group (and hence every topological vector space) is
makes the separation properties T0 , T1 , T2 , T3 and
for X . The somewhat tedious proof of the complete
group can be found for instance in [6].
4.30 Hahn-Banach Theorem (Locally Convex Spaces). Let C and D be
convex nonempty subsets of a real or complex locally convex space X . If C
is compact, D is closed and C ∩ D = ∅, then there exists a continuous linear
functional x0 ∈ X ∗ and real numbers γ and δ such that
Rehx0 , ci ≤ γ < δ ≤ Rehx0 , di
for all c ∈ C and all d ∈ D.
Let B = D −C . Since C and D are convex, so is D −C by Lemma 3.20.
According to Lemma 3.4, B is closed because C is compact and D is closed.
Furthermore, 0 6∈ B because C ∩ D = ∅ so X \ B is an open neighborhood of
0. Since X is locally convex, there exists a convex open neighborhood A of 0
contained in X \B . Now A∩B = ∅ so we may apply Hahn-Banach Theorem 4.28
and obtain a real number β and a continuous linear functional x0 ∈ X ∗ such
that
Rehx0 , ai < β ≤ Rehx0 , bi
Proof.
for all a ∈ A and b ∈ B .
The number β must be positive because 0 ∈ A and hx0 , 0i = 0. Thus
0 < β ≤ Rehx0 , d − ci = Rehx0 , di − Rehx0 , ci
and consequently
Rehx0 , ci + β ≤ Rehx0 , di
(4.25)
0
for all c ∈ C and d ∈ D. Let γ = sup {Rehx , ci : c ∈ C} and choose δ = γ + β .
Then it follows from (4.25) that
Rehx0 , ci ≤ γ < γ + β ≤ Rehx0 , di
for all c ∈ C and d ∈ D, as required.
¥
58
Chapter 5
Extreme Points and the
Krein-Milman Theorem
Denition. Let C be a convex subset of a linear space X . A point p ∈ C is
called an extreme point of C when p cannot be represented as
p = λx + (1 − λ)y
with x, y ∈ C , x 6= y and 0 < λ < 1. The set of all extreme points of C is
denoted by ext C .
¤
Denition. Let C be a convex subset of a linear space X . A convex nonempty
subset S of C is called a supporting set in C when for all x, y ∈ C , the existence
of a λ ∈]0, 1[ such that λx + (1 − λ)y ∈ S implies that x ∈ S and y ∈ S .
¤
In other words, a supporting set S in a convex set C is a convex subset of C
with the following property: if S contains a point of a line segment lying in C
that is not an endpoint, then S contains the entire line segment. The following
propositions are easy to prove.
5.1 Proposition. Let S be a supporting set in a convex subset C of a real or
complex linear space and suppose that R ⊆ S . Then R is a supporting set in S
if and only if R is a supporting set in C .
5.2 Proposition. Let S be a supporting set in a convex subset C of a real or
complex linear space. Then a point s ∈ S is an extreme point of S if and only
if s is an extreme point of C . In particular, a singleton {p} is a supporting set
in C if and only if p is an extreme point of S .
For example, consider a solid rectangle R in the plane. The extreme points
of R are precisely its four corners. Any side of R is a supporting set in A, as
is any corner of R, but the circumference of R, for instance, is not supporting
because it is not convex.
For another example, consider the closed unit ball B of the Euclidean threespace. The only supporting sets in B are the singletons of points on the unit
sphere S (the surface of B ) and B itself. The set of extreme points of B is S .
59
In both examples above, note that the set under consideration is the closed
convex hull of its extreme points. This is not a coincidence but a special case
of the Krein-Milman theorem, which we shall now prove.
5.3 Lemma. Let C be a non-empty compact convex subset of a real or complex
Hausdor topological vector space X , and let f be a continuous real linear
functional on X . Then the set
Cmax := {x ∈ C : f (x) = θ} ,
where θ = max {f (c) : c ∈ C}, is a closed supporting set in C .
As a continuous function, f attains its maximum θ on the compact set
C . The constant functions are always continuous, so the set Cmax is closed in
C by Lemma 7.7. Since X is Hausdor, the compact set C is closed, and hence
the set Cmax is also closed in X .
Because f is a real linear functional, it is easy to see that Cmax is convex:
Proof.
f (λx + (1 − λ)y) = λf (x) + (1 − λ)f (y) = λθ + (1 − λ)θ = θ
whenever x, y ∈ Cmax and 0 ≤ λ ≤ 1.
Finally, suppose that z = λx + (1 − λ)y ∈ Cmax for some x, y ∈ C and
0 < λ < 1. If f (x) < θ, then
f (z) = λf (x) + (1 − λ)f (y) < λθ + (1 − λ)θ = θ
because 0 < λ < 1. This is a contradiction, since z ∈ Cmax . Similarly, we cannot
have f (y) < θ, so the only remaining possibility must be true: f (x) = f (y) = θ.
Hence x, y ∈ Cmax and Cmax is a supporting set in C .
¥
5.4 Theorem (Krein-Milman). If C is a compact convex subset of a real or
complex locally convex Hausdor space X , then C is the closed convex hull of
its extreme points.
The case C = ∅ is trivial, so we assume that C is nonempty. Note that
it is implicit in our claim that ext C 6= ∅. We begin by proving this fact, because
we need it later on when applying the Hahn-Banach theorem.
Since X is Hausdor, C is closed and hence a closed supporting set in itself.
Thus the family S of all closed supporting sets in C is a non-empty set that
is partially ordered by set inclusion. Suppose that S is a chain in S. Since
S is a chain of nonempty
T subsets of C , it clearly has the nite intersection
property. But then L = S 6= ∅, because C is compact. It is obvious that L
is a supporting set in C and hence a lower bound for the chain S in S. Thus
every chain in S has a lower bound, and there exists a minimal element M in
S by Zorn's lemma. Note that M is a compact convex set.
Suppose that there exist two distinct points x, y ∈ M . Regarding X for the
moment as a real topological vector space, we know by Corollary 4.29 that there
exists a continuous real linear functional f on X such that f (x) 6= f (y). The
set
½
¾
Mmax := z ∈ M : f (z) = max f (m)
Proof.
m∈M
60
is a closed supporting set in M by Lemma 5.3 and hence also in C by Proposition 5.1. Since f is not constant on M , we must have Mmax ( M , which
contradicts the minimality of M . Hence there are no distinct points in M . As
a singleton supporting set, M consists of an extreme point of C .
It is clear that co(ext C) ⊆ C because ext C ⊆ C and C is a closed convex
set. Note that co(ext C) is compact as a closed subset of C . Furthermore, we
just proved that it is nonempty.
Suppose, contrary to our claim, that co(ext C) 6= C . Then there exists a
point x0 ∈ C \ co(ext C). Obviously the singleton {x0 } is a convex set, which is
also closed because X is Hausdor. By Hahn-Banach Theorem 4.30 there exist
real numbers γ and δ and a continuous linear functional x0 ∈ X ∗ such that
Rehx0 , yi ≤ γ < δ ≤ Rehx0 , x0 i
for all y ∈ co(ext C). The mapping Re x0 is a continuous real linear functional
on X and therefore
½
¾
Cmax := x ∈ C : hRe x0 , xi = maxhRe x0 , ci
c∈C
is a closed supporting set in C by Lemma 5.3. So Cmax is a non-empty compact
convex set and thus it contains an extreme point, p say, by the rst part of this
proof. But this is a contradiction since
Rehx0 , yi < Rehx0 , x0 i ≤ Rehx0 , pi
in particular for all y ∈ ext C , and p ∈ ext C by Proposition 5.2. Hence we must
have C = co(ext C) and the proof is complete.
¥
The next corollary was actually presented in the previous proof, but we state it
as a reminder.
5.5 Corollary. Every non-empty compact convex subset C of a real or complex
locally convex Hausdor space X contains an extreme point.
Proof.
This is evident because C = co(ext C) is nonempty and co ∅ = ∅.
61
¥
Chapter 6
Weak Topologies
General Weak Topologies
Denition. Let X be a nonempty set and let τ1 and τ2 be topologies on X . If
τ1 ⊆ τ2 , then τ1 is said to be weaker (or coarser ) than τ2 , and τ2 is said to be
stronger (or ner) than τ1 .
¤
Note that given two dierent topologies on X 6= ∅, neither is necessarily weaker
or stronger. Consider for instance the set of real numbers with the topologies
τ1 = {∅, {1}, R} and τ2 = {∅, {2}, R}. However, τ1 ∩ τ2 is the trivial topology
on R and it is weaker than both τ1 and τ2 . This is not a coincidence:
6.1 Proposition.
If X is a nonempty set and T is a family of topologies on
T
X , then
Proof.
T is a topology on X .
This follows easily from the denition of topology.
¥
Knowing this, we can speak of the weakest topology having a certain property
that is preserved under intersections (see Exercise 71).
6.2 Lemma. Let τ1 and τ2 be topologies on a non-empty set X . Then τ1
τ
1
is stronger than τ2 if and only if for every net (xλ )λ in X , xλ −→
x implies
τ2
xλ −→ x.
Proof. Suppose that τ2 ⊆ τ1 , and let (xλ )λ be a net in X converging to x
with respect to the topology τ1 . Since every τ2 -neighborhood of x is also a
τ2
τ1 -neighborhood, it is clear that xλ −→
x.
τ1
τ2
Conversely, suppose that xλ −→ x implies xλ −→
x whenever (xλ )λ is a net
τ1
in X . Let U ∈ τ2 . If (xλ )λ is a net in X \ U such that xλ −→
x, then x ∈ X \ U
τ2
because X \ U is a τ2 -closed set and xλ −→ x. Thus X \ U is a τ1 -closed set,
that is, U ∈ τ1 . Hence τ2 ⊆ τ1 , as required.
¥
Denition. Let X be a non-empty set, Y a topological space and F a non-
empty set of functions from X into Y . The weak topology on X induced by F
is the weakest topology on X in which every function f ∈ F is continuous. ¤
The following lemma will give us a choice between a few useful subbases for
weak topologies.
62
6.3 Lemma. Let Y be a topological space, X a non-empty set and F a nonempty collection of functions from X to Y . If B is the topology on Y , or a base
or a subbase for it, then the set
©
ª
β := f −1 (B) : f ∈ F and B ∈ B
is a subbase for the weak topology on X induced by F .
Let τ be the weak topology on X induced by F . Denote by hβi the
topology on X that has β as a subbase, that is, hβi is the topology on X for
which the set of all nite intersections of elements of β is a base. Since every
topology is a base for itself, and every base for a topology is a subbase for
the same topology, it is sucient to prove the statement under the standing
assumption that B is a subbase for the topology on Y .
Since each f ∈ F is continuous with respect to τ , and each subbasic set in
Y is open, it is clear that β ⊆ τ , and therefore hβi ⊆ τ . Since τ is the weakest
topology on X in which every f ∈ F is continuous, it is enough to show that
each f ∈ F is continuous with respect to hβi.
Let f ∈ F be arbitrary. If SU is open in Y , there exists a collection U of
basic open sets such that U = U. Moreover, for every T
V ∈ U, there exists a
nite subcollectionTBV of the subbase B such that V = BV . Now for every
V ∈ U, f −1 (V ) = B∈BV f −1 (B) ∈ hβi as a nite intersection of sets from β ,
S
and thus also f −1 (U ) = V ∈U f −1 (V ) ∈ hβi. This proves that f is continuous
with respect to hβi and nishes the proof.
¥
Proof.
6.4 Theorem. Let Y be a topological space, let X be a non-empty set endowed
with the weak topology induced by a collection F of functions from X into Y ,
and let (xλ )λ∈Λ be a net in X . Then xλ −→ x if and only if f (xλ ) −→ f (x) in
Y for all f ∈ F .
Suppose that xλ −→ x, and let f ∈ F be arbitrary. If V is any
neighborhood of f (x) in Y , then f −1 (V ) is a neighborhood of x because f
is by denition continuous. Since (xλ )λ is eventually in f −1 (V ), (f (xλ ))λ is
eventually in V , and thus f (xλ ) −→ f (x).
Conversely, suppose that f (xλ ) −→
© −1
ª f (x) for every f ∈ F . Remember that
f (V ) : f ∈ F and V is open in Y is a subbase for the weak topology induced by F . Therefore, if U is a basic neighborhood of x, there are functions
f1 , . . . , fn ∈ F and open sets V1 , . . . , Vn in Y such that
Proof.
U = f1−1 (V1 ) ∩ · · · ∩ fn−1 (Vn ).
If k ∈ {1, . . . , n}, then Vk is a neighborhood of fk (x) and fk (xλ ) −→ fk (x),
so there exists λk ∈ Λ such that fk (xλ ) ∈ Vk whenever λ ≥ λk . Since Λ is
a directed set, there exists λ0 ∈ Λ such that λ0 ≥ λk for every k = 1, . . . , n.
Hence fk (xλ ) ∈ Vk for all k = 1, . . . , n provided λ ≥ λ0 , so (xλ )λ is eventually
in U . This proves that xλ −→ x.
¥
After these general considerations we study certain weak topologies on linear
spaces. We begin with two (still fairly general) results, rst of which will give
us a solid foundation to work on later.
63
Denition. Let X and Y be sets. A family F of functions from X into Y is
said to separate points when for every pair of distinct points x1 and x2 in X ,
there exists a function f ∈ F such that f (x1 ) 6= f (x2 ).
¤
6.5 Theorem. Let X be a real or complex vector space, and let F be any
family of linear functionals on X . Then X is a locally convex space with the
weak topology induced by F . Furthermore, this topology is Hausdor if and
only if F separates points.
Let x and y be any pair of vectors in X , and let ((xλ , yλ ))λ be a net
in X × X converging to (x, y). Then (xλ )λ and (yλ )λ are nets in X converging
to x and y , respectively. Given any x0 ∈ F , we have
Proof.
hx0 , xλ + yλ i = hx0 , xλ i + hx0 , yλ i −→ hx0 , xi + hx0 , yi = hx0 , x + yi
because x0 is linear and by denition continuous in the weak topology induced by
F . This shows that the vector addition (x, y) 7→ x + y is a continuous mapping.
Similarly, if α is a scalar, x ∈ X and ((αλ , xλ ))λ∈Λ is a net in K × X converging
to (α, x), where K is the scalar eld of X , then
hx0 , αλ xλ i = αλ hx0 , xλ i −→ αhx0 , xi = hx0 , αxi
whenever x0 ∈ F . Thus the scalar multiplication mapping is also continuous,
and so X is a topological vector space.
To see that our topology is locally convex, consider the family of sets
V (x, x0 , ε) := {y ∈ X : |hx0 , yi − hx0 , xi| < ε} ,
(6.1)
where x ∈ X , x0 ∈ F and ε > 0. This family is a subbase for the weak topology
on X induced by F . Note that z + V (x, x0 , ε) = V (z + x, x0 , ε) whenever z ∈ X .
Since any intersection of convex sets is convex and each basic open set is an
intersection of (nitely many) sets of the form (6.1), it suces to show that
V (0, x0 , ε) is a convex set whenever x0 ∈ F and ε > 0. Indeed, if x, y ∈ V (0, x0 , ε)
and 0 ≤ λ ≤ 1, then
|hx0 , λx + (1 − λ)yi| ≤ λ |hx0 , xi| + (1 − λ) |hx0 , yi| < ε,
as required.
Suppose that F separates points and let x, y ∈ X with x 6= y . Then there
exists x0 ∈ F such that hx0 , xi 6= hx0 , yi, so ε := |hx0 , xi − hx0 , yi| > 0, and
therefore V (x, x0 , ε/2) and V (y, x0 , ε/2) are disjoint subbasic neighborhoods of
x and y as the preimages of disjoint open sets under the continuous mapping
x0 . Thus the weak topology on X induced by F is Hausdor when F separates
points.
Suppose then that F does not separate points, that is, there exist distinct
points x and y in X such that hx0 , xi = hx0 , yi for all x0 ∈ F . If V (z, x0 , ε)
is any subbasic neighborhood of x with z ∈ X , x0 ∈ F and ε > 0, then
|hx0 , yi − hx0 , zi| = |hx0 , xi − hx0 , zi| < ε, and so y ∈ V (z, x0 , ε). Hence y belongs
to any intersection of subbasic neighborhoods of x, in particular to each basic
neighborhood of x. Thus X cannot be Hausdor, and the proof is complete. ¥
6.6 Proposition. Let X be a real or complex vector space. If F is any innitedimensional linear subspace of the algebraic dual of X , then every non-empty
open set in the weak topology on X induced by F is unbounded.
64
By Corollary 3.13, it is enough to show that an arbitrary neighborhood
U of 0 in X is unbounded. Let f1 , . . . , fn ∈ F and let V1 , . . . , Vn be neighborhoods of 0 in K, the underlying eld of X , such that f1−1 (V1 )∩· · ·∩fn−1 (Vn ) ⊆ U .
If S = Ker(f1 ) ∩ · · · ∩ Ker(fn ), then clearly S ⊆ U , so it is enough to show that
S is unbounded. Since dim F = ∞, there exists f ∈ F that is not a linear
combination of f1 , . . . , fn . Then, by Lemma 1.21, S 6⊆ Ker(f ), so there exists
some x ∈ S \ Ker(f ). Since S is a subspace, nx ∈ S for every n ∈ N, so f (S) is
unbounded in K because |f (nx)| = n |f (x)| −→ ∞ as n → ∞. Hence S cannot
be bounded because f is bounded as a continuous functional by Theorem 3.17.¥
Proof.
Second Dual and the Canonical Mapping
Before bringing forth the main subjects of this chapter, the weak and weak∗
topologies, we should revisit the the concepts of canonical embeddings and duals
of normed spaces.
Let X be a normed space. Then X ∗ is a Banach space (Theorem 4.1) and
we may consider its Banach dual (X ∗ )∗ , which we denote simply by X ∗∗ . The
Banach space X ∗∗ is called the second conjugate, the second dual or the bidual
of X .
The next denition is well-posed byTheorem 6.7.
Denition. Given an element x of a normed space X , let xb be the functional
on X ∗ dened by
hb
x, x0 i = hx0 , xi for x0 ∈ X ∗ .
(6.2)
The mapping
J0 : X → X ∗∗
x 7→ x
b
is called the canonical mapping or canonical embedding of X into X ∗∗ . We
b.
denote the image of X under J0 by X
¤
6.7 Theorem. The canonical mapping J0 is a linear isometry from a normed
space X into its second dual X ∗∗ .
Proof. As suggested by our notation, x
b is linear for every x ∈ X : given
x0 , y 0 ∈ X ∗ and a scalar α,
hb
x, x0 + αy 0 i = hx0 + αy 0 , xi = hx0 , xi + αhy 0 , xi = hb
x, x0 i + αhb
x, y 0 i.
Furthermore, equation (6.2) yields
kb
xk = kxk
(6.3)
by the corollary 4.12, so x
b is also bounded, and in fact an isometry. Thus
x
b ∈ X ∗∗ and J0 is really a mapping into X ∗∗ .
Finally, J0 is itself linear because given x, y ∈ X and a scalar α, we have
hJ0 (x + αy), x0 i = hx0 , x + αyi
= hx0 , xi + αhx0 , yi
= hJ0 x, x0 i + αhJ0 y, x0 i
= hJ0 x + αJ0 y, x0 i
for all x0 ∈ X ∗ , that is, J0 (x + αy) = J0 x + αJ0 y .
65
¥
The Weak Topology on Normed Spaces
In the context of normed spaces, the weak topology always refers to a particular
weak topology, unless explicitly stated otherwise.
Denition. The weak topology on a normed space X is the weak topology
w
induced by the set of bounded linear functionals on X . We write xλ −→ x
when a net (xλ )λ converges to x in this topology, and say that (xλ )λ converges
weakly to x. Moreover, any topological concept used together with the word
weak or weakly means that the concept is to be understood with respect to the
weak topology.
¤
We continue to denote the convergence in norm by a plain arrow, that is, we
write xλ −→ x when kxλ − xk −→ 0. The norm topology is sometimes called
the strong topology in contrast to the weak topology, as it is, indeed, stronger:
6.8 Proposition. The weak topology on a normed space X is weaker than the
norm topology.
Since each x0 ∈ X ∗ is norm continuous, this is clear from the denition
of the weak topology on X .
¥
Proof.
6.9 Proposition. Every normed space is a locally convex Hausdor space with
respect to the weak topology.
Since the bounded linear functionals separate points by Corollary 4.10,
this result follows directly from Theorem 6.5.
¥
Proof.
6.10 Proposition. Let (xλ )λ be a net and x a vector in a normed space X .
w
Then xλ −→ x if and only if hx0 , xλ i −→ hx0 , xi for every x0 ∈ X ∗ .
Proof.
This is a special case of Theorem 6.4.
¥
6.11 Proposition. The norm and weak topologies on a normed space X are
the same if and only if X is nite-dimensional.
Proof. Suppose that X is nite-dimensional. Since there is only one topology
on {0}, we can assume that dim X = n ≥ 1. Let {x1 , . . . , xn } be a basis
0
0
for X and
Pn let {x1 , . . . , xn } be the corresponding coordinate functionals, that
is, hx0k , j=1 αj xj i = αk whenever α1 , . . . , αn are scalars. These are clearly
linear and thus in fact x01 , . . . , x0n ∈ X ∗ because all linear functionals on a nitedimensional normed space are continuous by Corollary 2.29. Thus, if (xλ )λ is a
w
net in X such that xλ −→ x, then hx0k , xλ i −→ hx0k , xi for all k = 1, . . . , n, and
consequently
n
n
X
X
0
xλ =
hxk , xλ ixk −→
hx0k , xixk = x
k=1
k=1
because the vector operations are continuous in norm. Hence the weak topology
is stronger than the norm topology by Lemma 6.2, and so the topologies are the
same by Proposition 6.8.
Conversely, suppose that X is innite-dimensional. Then, according to
Proposition 6.6, the weakly open sets are all weakly unbounded and therefore
cannot be bounded in the stronger norm topology. Since the open unit ball of
X is norm bounded, it cannot be weakly open. It is, of course, norm open, so
the norm topology must be dierent from the weak topology.
¥
66
6.12 Corollary. Let (xλ )λ∈Λ be a net and x a vector in a normed space X .
w
(1) If xλ −→ x, then xλ −→ x.
w
(2) If dim X < ∞, then xλ −→ x if and only if xλ −→ x.
Proof.
With Lemma 6.2 in mind, this is clear.
¥
The followingperhaps surprisingresult is a consequence of the uniform
boundedness principle.
6.13 Theorem. A subset B of a normed space X is weakly bounded if and
only if B is norm bounded.
Since the norm topology is stronger than the weak topology, it is clear
that norm bounded sets are weakly bounded.
Conversely, suppose that B is weakly bounded and non-empty, and consider
the canonical mapping J0 : X → X ∗∗ . Remember that J0 is an isometry Now
J0 (B) is a non-empty collection of linear functionals on the Banach space X ∗
indexed by B . Furthermore, for each x0 ∈ X ∗ ,
Proof.
sup {|hJ0 x, x0 i| : x ∈ B} = sup {|hx0 , xi| : x ∈ B} < ∞
because, with respect to the weak topology, x0 is continuous and hence bounded
by Theorem 3.17. Thus the Uniform Boundedness Principle 2.15 yields that
sup {kJ0 xk : x ∈ B} < ∞. Since J0 is an isometry by Theorem 6.7, we ultimately obtain sup {kxk : x ∈ B} < ∞, as required for the norm boundedness
of B .
¥
The Weak∗ Topology on Banach Duals
Denition. Let X ∗ be the Banach dual of a normed space X . The weak∗
b,
topology (read weak star topology) on X ∗ is the weak topology induced by X
the canonical image of X in X ∗∗ . We write
w∗
x0λ −→ x0
to indicate the convergence of a net (x0λ )λ to x0 in this topology, and say that
(x0λ )λ is weak∗ convergent to x0 .
¤
6.14 Proposition. The Banach dual X ∗ of a normed space X is a locally
convex Hausdor space with respect to the weak∗ topology.
If x0 , y 0 ∈ X ∗ are distinct points, there is by denition some x ∈ X
b separates points of
such that hx0 , xi 6= hy 0 , xi, that is, hb
x, x0 i 6= hb
x, y 0 i. Thus X
∗
X and we can apply Theorem 6.5 to nish the proof.
¥
Proof.
6.15 Proposition. Let (x0λ )λ be a net in the Banach dual X ∗ of a normed
w∗
space X . Then x0λ −→ x0 if and only if hx0λ , xi −→ hx0 , xi for every x ∈ X .
Proof.
Apply Theorem 6.4 and remember that hb
x, x0λ i = hx0λ , xi.
67
¥
6.16 Proposition. The weak∗ topology on the Banach dual X ∗ of a normed
space X is weaker than the weak topology on X ∗ .
Proof.
b ⊆ X ∗∗ .
This is immediate from the denitions, as X
¥
It is equally evident that weak and weak∗ topologies on X ∗ coincide when
b = X ∗∗ . The converse is also true, but proving it is much more challenging;
X
we will conclude this matter eventually in Theorem 7.19.
Since weak topology is always weaker than the norm topology, we can summarize the relations between norm, weak and weak∗ convergence:
6.17 Proposition. Let X ∗ be the Banach dual of a normed space X and (x0λ )λ
a net in X ∗ .
w
(1) If x0λ −→ x0 , then x0λ −→ x0 .
w
w∗
(2) If x0λ −→ x0 , then x0λ −→ x0 .
Proof.
(1) was done in Corollary 6.12, and (2) in Proposition 6.16.
68
¥
Chapter 7
Reexive Spaces
Let X be a normed space. Consider the canonical mapping J0 of X into its
second dual X ∗∗ , which maps each x ∈ X to the bounded linear functional x
b
on X ∗ that is dened by
hb
x, x0 i = hx0 , xi for all x0 ∈ X ∗ .
We have already seen in Theorem 6.7, that J0 is an isometry, i.e., kb
xk = kxk
b for J0 (X).
for all x ∈ X . Expanding our notation J0 x = x
b, we also write X
b , we can identify X
Since J0 is an isometric isomorphism between X and X
∗∗
b
b
with X as normed spaces. This subspace X of X may or may not be proper,
and this question is actually the core of the present chapter.
Denition. A normed space X is said to be reexive when Xb = X ∗∗ , that is,
when the canonical mapping is onto.
¤
Let us begin with a simple observation:
7.1 Proposition. Reexive spaces are all Banach spaces.
Proof. Let X be a reexive space. By Theorem 6.7, X is isometrically isob = X ∗∗ = (X ∗ )∗ , which is a Banach space by Theorem 4.1, and
morphic to X
hence X itself must be a Banach space.
¥
The converse of Proposition 7.1 is false. For example, `1 is a Banach
space, but `∗1 is topologically isomorphic to the nonseparable space `∞ . By
Corollary 7.3, which we prove in the next section, the separable space `1 cannot
be reexive because `∗1 is not separable.
¤
Remark.
It may also happen that X is isometrically isomorphic to X ∗∗ but X is not
reexive. See [3] for an example.
Separability Considerations
7.2 Theorem. Let X be a normed space. If X ∗ is separable, then so is X .
69
Proof. The trivial space {0} is separable; suppose that X is nontrivial. Let
{x01 , x02 , . . . } be a countable dense subset of X ∗ . By Proposition 2.9, kx0 k =
sup {|hx0 , xi| : kxk = 1}, so there exists, for every n ∈ N, xn ∈ X such that
kxn k = 1
and
|hx0n , xn i| ≥
kx0n k
.
2
(7.1)
Let M be the closed subspace of X generated by these vectors, that is,
M = Span({x1 , x2 , . . . }) ⊆ X.
Then M is separable. Indeed, if we take
( k
)
X
N=
αn xn : k ∈ N, αn ∈ Q + iQ for n = 1, . . . , k ,
n=1
then N = M , and N is countable (this requires the axiom of choice) because it
can be written as a countable union of countable sets. Thus it suces to show
that M = X .
We argue by contradiction: suppose that there exists x0 ∈ X \ M . Then,
by Corollary 4.9, there exists x0 ∈ X ∗ such that kx0 k = 1, hx0 , x0 i 6= 0 and
hx0 , xi = 0 for all x ∈ M . In particular,
hx0 , xn i = 0
for n = 1, 2, . . . .
(7.2)
Since {x01 , x02 , . . . } is dense in X ∗ , there is an n ∈ N such that kx0n − x0 k < 14 .
Thus by (7.1) and (7.2),
1
1 0
kxn k ≤ |hx0n , xn i| = |hx0n , xn i − hx0 , xn i| ≤ kx0n − x0 k kxn k <
2
4
since kxn k = 1. But this implies that kx0n k ≤
1
2
and hence
1 = kx0 k ≤ kx0 − x0n k + kx0n k <
which is absurd.
1 1
3
+ = ,
4 2
4
¥
The converse of Theorem 7.2 is false : the dual of a separable space
need not be separable. For an example, consider `1 which we know to be separable (Exercise 72). The Banach dual `∗1 is isometrically isomorphic to `∞
(Exercise 67) which is not separable.
¤
Remark.
However, we can sometimes use the theorem to determine that a given space is
nonreexive:
7.3 Corollary. A separable normed space X with a nonseparable dual X ∗
cannot be reexive.
Proof. Let X be a separable normed space. If X is reexive, then it is isometrically isomorphic to X ∗∗ by Theorem 6.7. Thus X ∗∗ is also separable.
But X ∗∗ is the dual of X ∗ , so Theorem 7.2 asserts that X ∗ is separablea
contradiction.
¥
70
Duals and Subspaces of Reexive Spaces
7.4 Theorem. If X is a reexive space, then so is X ∗ .
Let J0 : X → X ∗∗ and J1 : X ∗ → X ∗∗∗ be our canonical mappings.
Our aim is thus to prove that J1 (X ∗ ) = X ∗∗∗ .
Let x000 ∈ X ∗∗∗ be arbitrary. Consider a functional x0 dened on X by
Proof.
x0 = x000 ◦ J0 .
Then x0 is linear and bounded as a composition of bounded linear mappings, so
x0 ∈ X ∗ .
Let x00 ∈ X ∗∗ be arbitrary and let x ∈ X be the preimage of x00 under J0 ,
that is, J0 x = x00 . Such an x exists, because X is reexive. Now,
hJ1 x0 , x00 i = hx00 , x0 i
= hJ0 x, x000 ◦ J0 i
= hx000 ◦ J0 , xi
= hx000 , J0 xi
= hx000 , x00 i,
and since x00 ∈ X ∗∗ was arbitrary, this implies that J1 x0 = x000 . Thus J1 is onto,
and the proof is complete.
¥
7.5 Theorem. A Banach space X is reexive if and only if X ∗ is reexive.
If X is reexive, then X ∗ is reexive by Theorem 7.4.
Assume that X ∗ is reexive. We argue by contradiction: suppose that X is
b is a proper linear subspace of X ∗∗ . Since X is
not reexive. Then J0 (X) = X
b is complete and hence closed by Theorem 2.5.
complete and J0 is an isometry, X
Thus there exists, by Corollary 4.9, a functional x000 ∈ X ∗∗∗ such that kx000 k = 1
and hx000 , J0 xi = 0 for all x ∈ X . Since X ∗ is reexive, there exists x0 ∈ X ∗
such that J1 x0 = x000 . Accordingly,
Proof.
0 = hx000 , J0 xi = hJ1 x0 , J0 xi = hJ0 x, x0 i = hx0 , xi
for all x ∈ X , meaning that x0 is the zero functional. But then J1 x0 = 0, which
is a contradiction because kJ1 x0 k = kx000 k = 1.
¥
7.6 Theorem. A closed subspace of a reexive space is reexive.
Let Y be a closed subspace of a reexive space X , and let J0 : X → X ∗∗
and J : Y → Y ∗∗ be the canonical mappings. We assume that Y is a proper
subspace, as the case Y = X is clear. Let y 00 ∈ Y ∗∗ be arbitrary. We must
prove that y 00 = Jy for some y ∈ Y .
Note that for every x0 ∈ X ∗ , the restriction x0 |Y of x0 to Y is clearly linear
and bounded on Y , that is, x0 |Y ∈ Y ∗ . Thus we can dene a functional x00 on
X ∗ by
hx00 , x0 i = hy 00 , x0 |Y i for x0 ∈ X ∗ .
(7.3)
Proof.
71
Remember that the operations in X ∗ are dened pointwise, so for all x01 , x02 ∈ X ∗
and every scalar α we have (x01 + αx02 )|Y = x01 |Y + αx02 |Y . Thus x00 is linear on
X ∗ because y 00 is linear on Y ∗ . Furthermore,
|hx00 , x0 i| = |hy 00 , x0 |Y i| ≤ ky 00 k kx0 |Y k ≤ ky 00 k kx0 k
because the norm of an operator cannot decrease in extensions. Hence x00 is
bounded and thus x00 ∈ X ∗∗ .
Since X is reexive, there exists a y ∈ X such that x00 = J0 y . If x0 ∈ X ∗ is
such that x0 |Y = 0, then
hx0 , yi = hJ0 y, x0 i = hx00 , x0 i = hy 00 , x0 |Y i = 0.
This means that we must have y ∈ Y , for otherwise, since Y is closed, there
would exist by Corollary 4.9 a functional x0 ∈ X ∗ such that x0 |Y = 0 and
hx0 , yi =
6 0, contrary to what we just saw.
Now, let y 0 ∈ Y ∗ be arbitrary. Pick any extension x0 ∈ X ∗ of y 0 (the
existence of such an x0 is guaranteed by Hahn-Banach Theorem 4.7) to see that
hJy, y 0 i = hy 0 , yi = hx0 , yi = hJ0 y, x0 i = hx00 , x0 i = hy 00 , x0 |Y i = hy 00 , y 0 i. (7.4)
Since (7.4) holds for all y 0 ∈ Y ∗ , we conclude that Jy = y 00 .
¥
Alaoglu's Theorem and Boundedness in the Dual
Next we prove the Alaoglu's theorem that is also known as the Banach-Alaoglu
theorem. We begin with a few reminders from general topology.
7.7 Lemma. If f and g are continuous functions from a topolgical space X
into a Hausdor topological space Y , then the set
{x ∈ X : f (x) = g(x)}
is closed in X .
Proof. Denote by S the subset of X on which f and g agree. If s ∈ S , there
exists a net (sλ )λ∈Λ in S converging to s. Since f and g are continuous, the
net (f (sλ ))λ∈Λ = (g(sλ ))λ∈Λ converges to f (s) and g(s) in Y . But limits in the
Hausdor space Y are unique, so f (s) = g(s) and s ∈ S . Hence S ⊆ S and S is
closed.
¥
Tychono's Theorem. Let {Xλ : λ ∈ Λ} be a family of topological spaces.
Then the product space
Y
Xλ
λ∈Λ
is compact if and only if Xλ is compact for every λ ∈ Λ.
The proof of Tychono's theorem is outside the scope of this text. It can be
found e.g. in [8].
Q Remember that given λ0 ∈ Λ, the λ0 th projection is the mapping πλ0 :
λ∈Λ Xλ → Xλ0 dened by πλ0 (f ) = f (λ0 ). The product topology is the weak
topology induced by the set of all projections πλ , λ ∈ Λ.
72
7.8 Theorem (Alaoglu). The norm closed unit ball of the Banach dual of a
normed space is weak∗ compact.
Proof.
Let X be a normed space over K. Write (with some abuse of notation)
B 0 = {x0 ∈ X ∗ : kx0 k ≤ 1}
for the norm closed unit ball of X ∗ . Our strategy is to embed B 0 as a closed
subspace into some compact space.
For any x ∈ X let
Dx = {α ∈ K : |α| ≤ kxk} ,
a closed disk if K = C, or a closed interval if K = R. Then, Dx is a compact
subspace of K whenever x ∈ X , and hence the product space
Y
D=
Dx
x∈X
is compact by Tychono 's theorem.
Note that, if x0 ∈ B 0 , then |hx0 , xi| ≤ kx0 k kxk ≤ kxk because kx0 k ≤ 1, so
0
hx , xi ∈ Dx for all x ∈ X . Hence we may dene a mapping
Φ : B0 → D
by Φ(x0 ) = x0 .
(7.5)
When x ∈ X we denote, as usual, the xth projection map by πx , that is,
πx : D → Dx , πx (g) = g(x). The following observation is useful:
(πx ◦ Φ)(x0 ) = hx0 , xi = hb
x, x0 i for all x ∈ X and x0 ∈ B 0 ,
(7.6)
πx ◦ Φ = x
b|B 0
(7.7)
that is,
whenever x ∈ X .
We claim that Φ is in fact a homeomorphism between B 0 (having the weak∗
subspace topology) onto Φ(B 0 ) (having the subspace topology from D). Obviously Φ is injective, and it is continuous because πx ◦ Φ is continuous for every
x ∈ X by the denition of the weak∗ topology and (7.7). To show that Φ is
open, we check that Φ−1 : Φ(B 0 ) → B 0 is continuous. Indeed, if (Φ(x0λ ))λ is a
net converging to Φ(x0 ), then (πx ◦ Φ)(x0λ ) −→ (πx ◦ Φ)(x0 ) for all x ∈ X because
the projections πx are continuous. But this means that hb
x, x0λ i −→ hb
x, x0 i for
0 w∗
0
−1
all x ∈ X , that is, by Proposition 6.15, xλ −→ x . Hence Φ is continuous,
and Φ is a homeomorphism between B 0 and Φ(B 0 ), as claimed.
Therefore, to prove that B 0 is weak∗ compact, it suces to show that
Φ(B 0 ) ⊆ I is closed, because a closed subspace of a compact space is compact.
We shall prove that

 

\
\


Φ(B 0 ) = 
A(x, y) ∩ 
x ∈ XS(α, x)
(7.8)
α
∈K
x∈X
y∈X
with
A(x, y) = {g ∈ D : πx+y (g) = πx (g) + πy (g)}
S(α, x) = {g ∈ D : παx (g) = απx (g)} .
73
This will be enough, because the projection mappings πx are continuous on
D, the sums and multiples of K-valued functions are continuous, and K is a
Hausdor space, so the sets A(x, y) and S(α, z) are closed in D by Lemma 7.7
for all x, y ∈ X and all α ∈ K.
We can write the sets A(x, y) and S(α, x) in a more familiar way:
A(x, y) = {g ∈ D : g(x + y) = g(x) + g(y)}
S(α, z) = {g ∈ D : g(αz) = αg(z)} .
This makes proving (7.8) rather simple: obviously Φ(B 0 ) is contained in the
0
0
intersection on the right hand side of (7.8) because Φ(x
S ) = x is linear on X
for every x0 ∈ B 0 . On the other hand, if g : X → x∈X Dx is any function
that belongs to the said intersection, then g must be linear on X and its norm
0
cannot
S exceed 1 because g(x) ∈ Dx for all x ∈ X . Hence g ∈ B (remember
that x∈X Dx ⊆ K) and g = Φ(g) ∈ Φ(B 0 ).
¥
7.9 Corollary. Any norm closed ball in the Banach dual of a normed space is
weak∗ compact.
Let X be a normed space. Regard X ∗ as a topological linear space with
the weak∗ topology. Then scaling and translation are both homeomorphisms
of X ∗ by Theorem 3.1. Thus, if C 0 = {x0 ∈ X ∗ : kx0 − x00 k ≤ r} is any norm
closed ball with center x00 ∈ X ∗ and radius r > 0 and B 0 is the norm closed unit
ball of X ∗ , then ϕ : B 0 → C 0 dened by ϕ(x0 ) = x00 + rx0 is a homeomorphism
as a composition of two homeomorphisms. Since B 0 is compact by Theorem 7.8,
so is C 0 .
¥
Proof.
7.10 Corollary. Let X be a normed space over K. Then there is a compact
space K such that X is isometrically isomorphic to a subspace of the normed
space C(K) of all continuous K-valued functions on K . If X is a Banach space,
then this subspace is closed.
Let K be the norm closed unit ball of X ∗ with subspace topology
inherited from the the weak∗ topology on X ∗ . Then K is compact by Theorem 7.8. The canonical embedding J0 : X → X ∗∗ is a linear isomorphism by
Theorem 6.7, and J0 x = x
b is by denition continuous on X ∗ with respect to the
∗
weak topology for every x ∈ X . The restriction x
b|K is therefore continuous on
K for every x ∈ X . The mapping J dened on X by Jx = x
b|K is thus a linear
isomorphism between X and J(X) ⊆ C(K). By Proposition 1.12, J(X) is a
subspace of C(K). Moreover, for every x ∈ X ,
Proof.
kJxk = sup {|(Jx)(x0 )| : x0 ∈ K}
= sup {|hb
x, x0 i| : x0 ∈ K}
= sup {|hx0 , xi| : x0 ∈ X ∗ , kx0 k ≤ 1} = kxk
by Corollary 4.12, so J is an isometric isomorphism, as required.
If X is a Banach space, then J(X) is complete and hence closed by Theorem 2.5.
¥
The following interesting consequence of the Alaoglu and Krein-Milman theorems can sometimes be used to show that a given normed space cannot be the
dual of any other normed space (see Exercise 79).
74
7.11 Corollary. The norm closed unit ball of the Banach dual X ∗ of any
normed space X is the weak∗ closed convex hull of its extreme points and hence
contains an extreme point.
The weak∗ topology on X ∗ is locally convex and Hausdor by Proposition 6.14, and the norm closed unit ball of X ∗ is convex by Proposition 3.21
and weak∗ compact by Alaoglu's theorem 7.8, so the statement follows from the
Krein-Milman theorem 5.4.
¥
Proof.
Remember that a subset F of a topological space is said to be relatively
compact (or conditionally compact ) when its closure F is compact.
7.12 Theorem. Let F be a subset of the Banach dual X ∗ of a normed space
X , and consider the following statements.
(1) F is norm bounded.
(2) F is weakly bounded.
(3) F is relatively weak∗ compact.
(4) F is weak∗ bounded.
Then either of the statements (1) and (2) implies all the other statements, and
(4) follows from any of the other statements. When X is a Banach space all
four statements are equivalent.
We already know by Theorem 6.13 that (1) ⇐⇒ (2), so the rst claim
will be proved by showing that (1) =⇒ (3) =⇒ (4). After that, the second
claim follows by proving that if X is complete, then (4) =⇒ (1).
For (1) =⇒ (3), suppose that F is norm bounded, that is, F ⊆ C 0 for some
norm closed ball C 0 in X ∗ by Proposition 3.11. Now regard X ∗ as a topological
linear space with the weak∗ topology. By Corollary 7.9, C 0 is compact and hence
also closed, because X ∗ is Hausdor. Thus the closure F ⊆ C 0 is compact as
a closed subset of a compact set, meaning that F is relatively compact, as
required.
For (3) =⇒ (4), consider again X ∗ with the weak∗ topology. Suppose that
F is relatively compact, that is, F is compact, and let K be the scalar eld,
as usual. For each x ∈ X , the mapping ϕx = x
b is continuous on X ∗ by the
∗
denition of the weak topology. Since the continuous image of a compact set
is compact, ϕ(F) is a compact subset of K. Thus ϕ(F) ⊆ ϕ(F) is a bounded
set, that is,
{hx0 , xi : x0 ∈ F} ⊆ K
Proof.
is bounded for every x ∈ X . Let U be any subbasic neighborhood of 0 ∈ X ∗ .
Then there exist x ∈ X and ε > 0 such that
{x0 ∈ X ∗ : |hb
x, x0 i| < ε} ⊆ U.
Since there is Mx > 0 such that |hx0 , xi| < Mx for all x0 ∈ F ,
¯
¯
¯
ε
ε
ε 0 ¯¯
0
¯hb
¯ x, 2Mx x i¯ = 2Mx |hx , xi| ≤ 2 < ε
75
x
for all x0 ∈ F , which means that α−1 F ⊆ U with α = 2M
ε . Hence F ⊆ αU ,
and F is bounded, as required.
Finally, suppose that X is a Banach space and F is weak∗ bounded. Let
x ∈ X be arbitrary. Since
V = {x0 ∈ X ∗ : |hb
x, x0 i| < 1}
is a weak∗ neighborhood of 0 ∈ X ∗ , there exists a nonzero scalar α such that
F ⊆ αV , or equivalently, α−1 F ⊆ V . This means that
¯
¯
¯
¯
|hx0 , xi| = |α| ¯α−1 hx0 , xi¯ = |α| ¯hb
x, α−1 x0 i¯ < |α|
for all x0 ∈ F , so
sup {|hx0 , xi| : x0 ∈ F } < ∞.
(7.9)
But x was arbitrary, so (7.9) holds for all x ∈ X . Since X is a Banach space,
the principle of uniform boundedness 2.15 applies, and we know that
sup {kx0 k : x0 ∈ F } < ∞,
or in other words, that F is norm bounded. Thus (4) =⇒ (1) when X is a
Banach space.
¥
The assumption of the completeness of X in Theorem 7.12 cannot
be discarded if we want the implication (4) =⇒ (1) to hold, that is, all the
four statements to be equivalent. Proving this is Exercise 78.
¤
Remark.
Furthermore, if F is convex, then (3) also implies all the other statements, as
we shall see:
7.13 Theorem. Any weak∗ compact convex subset C of the Banach dual X ∗
of a normed space X is norm bounded.
Since X ∗ is a metric space, it can be exhausted with norm closed balls
centered at 0, say Bn0 = B 0 (0, n) = {x0 ∈ X ∗ : kx0 k ≤ 1}, where n = 1, 2, . . . .
Thus we can write
∞
∞
[
[
0
C=C∩
B (0, n) =
(C ∩ Bn0 ) ,
(7.10)
Proof.
n=1
n=1
∗
which is a countable union of weak closed subsets of X ∗ . To see this, remember
that C is weak∗ compact and weak∗ topology is Hausdor, so C is weak∗ closed.
Also, each Bn0 is weak∗ compact by Corollary 7.9 and therefore weak∗ closed.
Now regard C with the weak∗ subspace topology. Since C is a compact
Hausdor space, it is a Baire space by Theorem 2.14. Because every Baire
space is nonmeagre, the sets C ∩ Bn0 cannot all be nowhere dense, say
◦
◦
∅ 6= C ∩
Bn0
z
{
= C ∩ Bn0 .
◦
z
{
Fix x ∈ C ∩ Bn0 and let U be a balanced neighborhood of 0 ∈ X ∗ such that
0
◦
{
z
(x + U ) ∩ C ⊆ C ∩ Bn0 ⊆ Bn0 .
0
76
As a weak∗ compact set, C is weak∗ bounded by Theorem 7.12 and hence
C − C is weak∗ bounded by Lemma 3.12. For this reason, there exists a nonzero
scalar β such that
β(C − C) ⊆ U.
Fix now a scalar α so that 0 < α < min{1, |β|}. Then we have
α(C − C) ⊆
α
α
β(C − C) ⊆ U ⊆ U
β
β
because |α/β| ≤ 1 and U is balanced, so
(1 − α)x0 + αC = x0 − αx0 + αC ⊆ x0 + α(C − C) ⊆ x0 + U.
Since C is convex,
(1 − α)x0 + αC ⊆ (x0 + U ) ∩ C ⊆ Bn0 .
Thus
C⊆
¯
µ ¯
¶
¯α − 1¯ 0
α−1 0
¯ kx k + n
x + α−1 Bn0 ⊆ B 0 0, ¯¯
α
α ¯
α
and so C is norm bounded by Proposition 3.11.
¥
Goldstine's Theorem
We derive rst a technical result from the geometric Hahn-Banach theorem.
7.14 Theorem (Helly). Let X be a normed space over K, let d > 0 be a real
number, c1 , . . . , cn ∈ K and x01 , . . . , x0n ∈ X ∗ . Then the following two statements
are equivalent:
(1) For every ε > 0 there exists xε ∈ X such that kxε k ≤ d + ε and
hx0k , xε i = ck
for all k = 1, . . . , n.
(2) For any choice of scalars α1 , . . . , αn ∈ K,
¯
¯
° n
°
n
¯X
¯
°X
°
¯
¯
°
0°
αk ck ¯ ≤ d °
αk xk ° .
¯
¯
¯
°
°
k=1
Proof.
Then
k=1
Suppose that statement (1) holds, and let α1 , . . . , αn ∈ K be arbitrary.
¯
¯ ¯
¯ ¯
¯
n
n
n
¯X
¯ ¯X
¯ ¯X
¯
¯
¯ ¯
¯
¯
¯
αk ck ¯ = ¯
αk hx0k , xε i¯ = ¯h
αk x0k , xε i¯
¯
¯
¯ ¯
¯ ¯
¯
k=1
k=1
k=1
°
°
°
°
n
n
°X
°
°X
°
°
°
0°
0°
≤°
αk xk ° kxε k ≤ (d + ε) °
αk xk °
°
°
°
°
k=1
k=1
for every ε > 0, and statement (2) follows.
77
Suppose then that statement (2) holds. Assume, for the moment, that
x01 , . . . , x0n are linearly independent. Consider the mapping T : X → Kn dened
by
T x = (hx01 , xi, . . . , hx0n , xi).
Then T is clearly linear. Furthermore, T must be onto. Otherwise, the linear
subspace T (X) of Kn is contained in some maximal subspace M of Kn . By
Theorem 4.17 there exists a nonzero linear functional a0 ∈ (Kn )0 such that
M = Ker(a0 ). Let {e1 , . . . , en } be the natural basis for Kn , that is, ek is the
vector whose co-ordinates are all zero except for the k th coordinate, which is
one. Let ak = ha0 , ek i ∈ K for k = 1, . . . , n. Then
ha0 , (γ1 , . . . , γn )i =
n
X
a k γk
k=1
for all (γ1 , . . . , γn ) ∈ Kn . Since a0 is nonzero, the scalars a1 , . . . , an cannot all
be equal to zero. Now, as T (X) ⊆ Ker(a0 ), we have
h
n
X
ak x0k , xi =
n
X
ak hx0k , xi = ha0 , T xi = 0
k=1
k=1
for all x ∈ X and hence
n
X
ak x0k = 0,
k=1
which contradicts the linear independence of x01 , . . . , x0n . Hence T must be onto.
Given ε > 0, dene
Kε = {x ∈ X : kxk < d + ε} .
Then Kε is clearly a convex balanced neighborhood of 0 and consequently absorbing by Corollary 3.9. Since T is linear, T (Kε ) is also convex and balanced
(see exercises 53 and 50). Furthermore, T is surjective so T (Kε ) is absorbing in
Kn (see Exercise 49).
Henceforth we regard Kn as a normed space with the norm
k(γ1 , . . . , γn )k∞ = sup {|γk | : k = 1, . . . , n} ,
and prove that T (Kε ) has a non-empty interior. Since T is onto, we can choose
for every vector ek in the natural basis for Kn introduced earlier, a vector
xk ∈ X such that T xk = ek . Let
δ=
d+ε
.
n sup {kxk k : k = 1, . . . , n}
Now if b = (b1 , . . . , bn ) ∈ B∞ (0, δ) = {a ∈ Kn : kak∞ < δ}, then |bk | < δ for
all k = 1, . . . , n and so
°
°
n
n
n
°X
° X
X
°
°
bk xk ° ≤
|bk | kxk k <
δ kxk k ≤ d + ε
°
°
°
k=1
k=1
k=1
78
due to the choice of δ . Thus
n
X
bk xk ∈ Kε
k=1
and consequently
b=
n
X
k=1
Ã
bk ek = T
n
X
!
bk xk
∈ T (Kε ).
k=1
Hence B∞ (0, δ) ⊆ T (Kε ) and 0 is an interior point of T (Kε ).
Put c = (c1 , . . . , cn ). Now if (1) does not hold, i.e., for some ε > 0 there is
no xε ∈ Kε such that hx0k , xε i = ck for all k = 1, . . . , n, then there is no xε ∈ Kε
such that T x = c, i.e., c 6∈ T (Kε ). Let L0 = c + {0}. Then T (Kε ) ∩ L0 = ∅
and T (Kε ) is convex balanced absorbing set with nonempty interior. By the
Hahn-Banach theorem 4.27, there exists a nonzero real linear functional R on
Kn such that
L0 ⊆ L = {a ∈ Kn : R(a) = 1}
T (Kε ) ⊆ {a ∈ Kn : R(a) ≤ 1} .
(7.11)
(7.12)
Note that if a ∈ T (Kε ), then −a = −1a is also in T (Kε ) because it is a balanced
set. Thus −R(a) = R(−a) ≤ 1 as R is real linear, and so |R(a)| ≤ 1 for every
a ∈ T (Kε ).
If K = R, we let αk = R(ek ) for k = 1, . . . P
, n, where {e1 , . . . , en } is still the
n
natural basis for Kn . Then R(γ1 , . . . , γn ) = k=1 αk γk for all (γ1 , . . . , γn ) ∈
Kn . Hence it follows from the the denition of T , (7.12) and (7.11), that for all
x ∈ Kε ,
¯
¯
¯
¯
n
n
¯X
¯
¯X
¯
¯
¯
¯
¯
0
αk hxk , xi¯ = |R(T x)| ≤ 1 = R(c) = |R(c)| = ¯
αk ck ¯
¯
¯
¯
¯
¯
k=1
k=1
because c ∈ L0 .
If K = C, the situation is a bit more complicated: we regard Kn as a real
linear space, and choose a basis {e1 , . . . , e2n } for Kn , where e1 , . . . , en are like
before and en+k = iek for k = 1, . . . , n. Set βk = R(ek ) ∈ R for k = 1, . . . , 2n.
Let x ∈ Kε be arbitrary. Denote, for k = 1, . . . , n, the real and imaginary
parts of hx0k , xi by rk and rn+k , respectively, so that hx0k , xi = rk + irn+k . As
before, c = (c1 , . . . , cn ), so we set similarly ck = qk + iqn+k with qk , qn+k ∈ R
for k = 1, . . . , n. Let F be the unique linear functional on the complex linear
space Kn whose real part is R, that is,
F (x) = R(x) − iR(ix) for all x ∈ X
(see Exercise 64 and the proof of Hahn-Banach Theorem 4.6 for an almost
complete solution). Put, analogously to the real case, αk = F (ek ) ∈ C for
79
k = 1, . . . , n. Then, like in the real case,
¯
¯
¯ 2n
¯
n
2n
¯X
¯
¯X
¯
X
¯
¯
¯
¯
0
αk hxk , xi¯ = |F (T x)| = |R(T x) − iR(iT x)| = ¯
βk rk − i
iβk rk ¯
¯
¯
¯
¯
¯
k=1
k=1
k=1
¯ 2n
¯
¯ X
¯
¯
¯
= ¯2
βk rk ¯ = 2 |R(T x)| ≤ 2 · 1 = 2R(c) = 2 |R(c)|
¯
¯
k=1
¯ 2n
¯
2n
¯X
¯
X
¯
¯
¯
¯
= ¯R(c) − i2 R(c)¯ = ¯
βk qk − i
iβk qk ¯
¯
¯
k=1
¯ n k=1 ¯
¯X
¯
¯
¯
= |R(c) − iR(ic)| = |F (c)| = ¯
αk ck ¯
¯
¯
k=1
because x ∈ Kε and c ∈ L0 .
So, in both cases, K = R and K = C, we have found scalars α1 , . . . , αn ∈ K
such that
¯
¯ ¯
¯
n
n
¯X
¯ ¯X
¯
¯
¯ ¯
¯
0
αk hxk , xi¯ ≤ ¯
αk ck ¯ for all x ∈ Kε .
(7.13)
¯
¯
¯ ¯
¯
k=1
k=1
Remember that R and, consequently, F are nonzero, so the scalars α1 , . . . , αn
cannot all be zero. Since x01 , . . . , x0n were assumed to be linearly independent,
P
n
0
k=1 αk xk is a nonzero functional on X . Thus
°
°
°
°
¯
¯
n
n
n
°X
°
°X
°
¯X
¯
°
°
¯
¯
0°
0°
0
0 < d°
αk xk ° < (d + ε) °
αk xk ° = sup ¯ hαk xk , (d + ε)xi¯
°
°
°
° kxk≤1 ¯
¯
k=1
k=1
k=1
¯
¯
¯
¯
n
n
¯X
¯ ¯X
¯
¯
¯ ¯
¯
= sup ¯
αk hx0k , xi¯ ≤ ¯
αk ck ¯
¯
¯
¯
¯
x∈Kε
k=1
k=1
by (7.13), contradicting statement (2). Hence our assumption must be false,
i.e., statement (1) holds.
Finally, assume that statement (2) holds and x01 , . . . , x0n are linearly dependent. Note that the case x01 = · · · = x0n = 0 is trivial because it forces
c1 = · · · = cn = 0 and thus the choice xε = 0 will do for any ε > 0. So we can
assume that the functionals x0k are not all zero. Let {x01 , . . . , x0m } be some maximal linearly independent subset of {x01 , . . . , x0n } (the ordering is insignicant).
It follows from our assumptions that 0 < m < n.
Since we already proved the theorem in case the functionals are linearly
independent, we know that for every ε > 0 there exists xε ∈ X such that
kxε k ≤ d + ε and hx0k , xε i = ck for all k = 1, . . . , m.
If m < p ≤ n, then x0p is linearly dependent on the functionals x1 , . . . , xm ,
that is, there exist scalars γ1 , . . . , γm ∈ K such that
x0p
=
m
X
γk x0k .
k=1
Therefore
hx0p , xε i =
m
X
γk hx0k , xε i =
k=1
m
X
k=1
80
γk ck .
Thus choosing αk = γk for k = 1, . . . , m and αp = −1 and ak for every other k ,
and applying statement (2) yields
¯ °
¯
°
m
m
¯
¯ °
°
X
¯
¯ ° 0 X
0°
γk ck ¯ ≤ °−xp +
γk xk ° = k0k = 0,
¯−cp +
¯
¯ °
°
k=1
k=1
so
cp =
m
X
γk ck = hx0p , xε i,
k=1
as required.
¥
7.15 Corollary. Let X be a normed space, x00 ∈ X ∗∗ and x01 , . . . , x0n ∈ X ∗ .
Then for every ε > 0 there exists xε ∈ X such that kxε k ≤ kx00 k + ε and
hx0k , xε i = hx00 , x0k i for every k = 1, . . . , n.
If x00 = 0, the statement is trivial. Assume that x00 6= 0. Choose
ck = hx , x0k i for every k = 1, . . . , n and let d = kx00 k . Then d > 0 and
¯
¯ ¯
¯
°
°
°
°
n
n
n
n
¯X
¯ ¯X
¯
°X
°
°X
°
¯
¯ ¯
°
00
0 ¯
00 °
0°
0°
αk ck ¯ = ¯
αk hx , xk i¯ ≤ kx k °
αk xk ° = d °
αk xk °
¯
¯
¯ ¯
¯
°
°
°
°
Proof.
00
k=1
k=1
k=1
k=1
for any choice of scalars α1 , . . . , αn . Now apply Helly's theorem 7.14.
¥
c1 of the norm closed unit
7.16 Theorem (Goldstine). The canonical image B
ball B1 of a normed space X is weak∗ dense in the norm closed unit ball B100 of
the second dual X ∗∗ .
To clarify our somewhat abusive notation: B1 = {x ∈ X : kxk ≤ 1} is
the norm closed unit ball of X , and B100 = {x00 ∈ X ∗∗ : kx00 k ≤ 1} is the norm
closed unit ball of X ∗∗ . Let J0 : X → X ∗∗ and J1 : X ∗ → X ∗∗∗ once again be
c1 = J0 (B1 )
the canonical mappings. Our task is to show that the closure of B
∗
∗∗
00
in the weak topology of X is B1 .
Let x00 ∈ B100 be arbitrary and let
Proof.
U = U (x00 , ε; x01 , . . . , x0n ) = {y 00 ∈ X ∗∗ : |hJ1 x0k , y 00 − x00 i| < ε for k = 1, . . . , n}
= {y 00 ∈ X ∗∗ : |hy 00 − x00 , x0k i| < ε for k = 1, . . . , n}
be a basic weak∗ neighborhood of x00 with ε > 0 and x01 , . . . , x0n ∈ X ∗ . We wish
c1 6= ∅, so we can assume without loss of generality that some
to show that U ∩ B
of the functionals x0k is nonzero. Let
δ=
ε
1 + ε + max kx0k k
and y 00 = (1 − δ)x00 .
(7.14)
1≤k≤n
Then 0 < δ < 1 so ky 00 k = |1 − δ| kx00 k ≤ |1 − δ| < 1. Thus 12 (1 − ky 00 k) > 0,
and by the previous corollary 7.15 to the Helly's theorem, there exists x ∈ X
such that kxk ≤ ky 00 k + 12 (1 − ky 00 k) < 1 and
hx0k , xi = hy 00 , x0k i for all k = 1, . . . , n.
81
(7.15)
Therefore we have
|hb
x − x00 , x0k i| = |hb
x, x0k i − hx00 , x0k i|
= |hx0k , xi − hx00 , x0k i| = |hy 00 , x0k i − hx00 , x0k i|
= |(1 − δ)hx00 , x0k i − hx00 , x0k i| = δ |hx00 , x0k i|
≤ δ kx00 k kx0k k ≤ δ kx0k k < ε
for all k = 1, . . . , n by (7.15) and the choice of δ . Thus x
b ∈ U and x ∈ B1 so
c1 6= ∅, as required.
U ∩B
¥
Looking back at the proof, we see that the canonical image of the
open unit ball of X is also weak∗ dense in the closed unit ball of X ∗∗ because
during the proof we could choose x ∈ X to satisfy kxk < 1.
¤
Remark.
7.17 Corollary. The canonical image Xb of a normed space X in the second
dual X ∗∗ is weak∗ dense.
00
b ). Then x00
Let x00 ∈ X ∗∗ be arbitrary but nonzero (clearly 0 ∈ X
kx k
is in the norm closed unit ball of X ∗∗ , and by Goldstine's theorem 7.16 there
00
w∗
exists a net (xλ )λ in the norm closed unit ball of X such that x
cλ −→ kxx00 k . Now
00 k x ) is a net in X
\
b , and
(kx
λ λ
Proof.
00 k x i = kx00 k hJ x0 , x
\
hJ1 x0 , kx
cλ i −→ kx00 k hJ1 x0 ,
λ
1
x00
i = hJ1 x0 , x00 i
kx00 k
w∗
00 k x −→
\
for every x0 ∈ X ∗ , meaning that kx
x00 .
λ
¥
7.18 Theorem (Kakutani). A normed space X is reexive if and only if the
norm closed unit ball of X is weakly compact.
Consider the canonical embedding J0 : X → X ∗∗ . We already know
b , but it is also a homeomorphism with
that J0 is a bijection between X and X
b.
respect to the weak topology on X and the weak∗ (subspace) topology on X
To see this, let (xλ )λ be a net in X and let x ∈ X . Then
Proof.
w
xλ −→ x ⇐⇒ hx0 , xλ i −→ hx0 , xi for all x0 ∈ X ∗
⇐⇒ hc
xλ , x0 i −→ hb
x, x0 i for all x0 ∈ X ∗
⇐⇒ hJ1 x0 , J0 xλ i −→ hJ1 x0 , J0 xi for all x0 ∈ X ∗
w∗
⇐⇒ J0 xλ −→ J0 x,
b is a
which implies that J0 and its inverse are both continuous so J0 : X → X
homeomorphism.
Let B1 = {x ∈ X : kxk ≤ 1} and B100 = {x00 ∈ X ∗∗ : kx00 k ≤ 1} be the norm
closed unit balls of X and X ∗∗ like before. Applying the Goldstine's theorem
7.16 gives
(7.16)
J0 (B1 ) ⊆ B100 ⊆ J0 (B1 ),
where the closure is taken in the weak∗ topology of X ∗∗ and the rst inclusion
is due to J0 being an isometry.
82
Suppose now that the norm closed unit ball B1 of X is weakly compact. Since
J0 is a homeomorphism as discussed above, J0 (B1 ) must be weak∗ compact in
J0 (X) and thus also in X ∗∗ . Since the weak∗ topology is Hausdor, J0 (B1 ) is
weak∗ closed. Hence the weak∗ closure J0 (B1 ) = J0 (B1 ) and so it follows from
(7.16) that
J0 (B1 ) = B100 .
−1
This will suce: if x00 ∈ X ∗∗ is nonzero, then kx00 k x00 ∈ B100 so there is y ∈ B1
−1
such that yb = J0 y = kx00 k x00 . Now x = kx00 k y ∈ X and x
b = kx00 k J0 y = x00 ,
∗∗
b = X , that is, X is reexive.
so X
Conversely, suppose that X is reexive. Then again J0 (B1 ) = B100 because
J0 : X → X ∗∗ is a bijective isometry. By Alaoglu's theorem 7.8, B100 is weak∗
compact. Since J0−1 is a homeomorphism from X ∗∗ with the weak∗ topology
onto X with the weak topology, B1 = J0−1 (B100 ) is weakly compact.
¥
7.19 Theorem. A normed space X is reexive if and only if it is a Banach
space and the weak topology on X ∗ coincides with the weak∗ topology on X ∗ .
b = X ∗∗ , so it is clear from the denition of
If X is reexive, then X
weak topologies that the weak and weak∗ topologies on X ∗ are the same: the
weak topology is the weak topology induced by X ∗∗ and the weak∗ topology is
b.
the weak topology induced by X
Conversely, suppose that X is a Banach space and that the two topologies in
question coincide on X ∗ . By Alaoglu's theorem 7.8, the norm closed unit ball in
X ∗ is weak∗ compact and thus weakly compact by our assumption. Kakutani's
theorem 7.18 then implies that X ∗ is reexive, so the Banach space X must also
be reexive by Theorem 7.5.
¥
Proof.
83
Bibliography
[1] A. L. Brown and A. Page. Elements of functional analysis. Van Nostrand
Reinhold Company, London-New York-Toronto, Ont., 1970. The New University Mathematics Series.
[2] Robert C. James. Linear functionals as dierentials of a norm. Math. Mag.,
24:237244, 1951.
[3] Robert C. James. A non-reexive Banach space isometric with its second
conjugate space. Proc. Nat. Acad. Sci. U. S. A., 37:174177, 1951.
[4] J. L. Kelley and Isaac Namioka. Linear topological spaces. With the collaboration of W. F. Donoghue, Jr., Kenneth R. Lucas, B. J. Pettis, Ebbe Thue
Poulsen, G. Baley Price, Wendy Robertson, W. R. Scott, Kennan T. Smith.
The University Series in Higher Mathematics. D. Van Nostrand Co., Inc.,
Princeton, N.J., 1963.
[5] Robert E. Megginson. An introduction to Banach space theory, volume 183
of Graduate Texts in Mathematics. Springer-Verlag, New York, 1998.
[6] L. S. Pontryagin. Topological groups. Translated from the second Russian
edition by Arlen Brown. Gordon and Breach Science Publishers, Inc., New
York, 1966.
[7] Angus Ellis Taylor and David C. Lay. Introduction to functional analysis.
John Wiley & Sons, New York-Chichester-Brisbane, second edition, 1980.
[8] Stephen Willard. General topology. Dover Publications Inc., Mineola, NY,
2004. Reprint of the 1970 original [Addison-Wesley, Reading, MA].
84
Index
absolutely homogeneous, 41
absorbing set, 34
Alaoglu's theorem, 73
algebraic dual, 10
annihilator, 48
convex
hull, 38
set, 38
closed hull, 38
dense, 20
nowhere , 20
dimension
of a linear space, 8
dual, 19
algebraic , 10
Banach , 19
continuous , 19, 38
Baire
category theorem, 22
space, 20
balanced
hull, 34
set, 34
Banach
conjugate, 19
dual, 19
space, 12
Banach-Alaoglu theorem, 73
Banach-Schauder theorem, 26
Banach-Steinhaus theorem, 23
basis, 5
bounded
linear operator, 16, 36
set, 16, 35
embedding
canonical , 65
equivalent
norms, 18
extreme point, 59
eld
topological , 32
functional, see linear functional
gauge , 53
Minkowski , 53
real linear , 53
sublinear , 42
canonical
embedding, 11, 65
mapping, 65
category
Baire theorem, 22
rst , 20
second , 20
closed convex hull, 38
compact
conditionally , 75
relatively , 75
conjugate, 19
Banach , 19
continuous
dual, 19
linear functional, 19
linear operator, 16
gauge functional, 53
Goldstine's theorem, 81
Hölder's inequality, 90
half space, 52
Hamel
basis, 5
dimension, 8
Helly's theorem, 77
homeomorphism
linear , 18, 32
homogeneous
absolutely , 41
positively , 41
85
hull
open
mapping, 24
mapping theorem, 26
balanced , 34
closed convex , 38
convex , 38
hyperplane, 50
real , 53
positively homogeneous, 41
principle of uniform boundedness, 23
quotient
norm, 13
space, 4, 13
inequality
Hölder's , 90
Minkowski's , 91
Young's , 90
isometry
linear , 18
isomorphism
isometric , 18
linear , 10
topological , 18, 32
range
of a linear operator, 8
real
hyperplane, 53
linear functional, 53
reexive space, 69
relatively compact, 75
Riesz's lemma, 30
Kakutani's theorem, 82
kernel
of a linear operator, 8
Krein-Milman theorem, 60
scalar, 3
seminorm, 12
separating family of functions, 63
set
absorbing , 34
balanced , 34
bounded , 35
convex , 38
supporting , 59
space
linear, 3
vector, 3
Baire , 20
Banach , 12
half , 52
locally convex , 39
normed , 12
topological linear , 32
topological vector , 32
span, 4
subadditive, 41
sublinear functional, 42
subspace
linear , 4
maximal , 10
supporting set, 59
linear
functional, 10
homeomorphism, 18, 32
independence, 4
operator, 8
space, 3
span, 4
linear operator
bounded , 16, 36
continuous , 16
unbounded , 36
locally convex space, 39
mapping
canonical , 65
maximal subspace, 10
meagre, 20
Minkowski
functional, 53
inequality, 91
nonmeagre, 20
norm, 12
equivalent , 18
of a linear operator, 17
semi, 12
normed space, 12
nowhere dense, 20
theorem
Alaoglu's , 73
Baire's , 22
Banach-Alaoglu , 73
86
Banach-Schauder , 26
Banach-Steinhaus , 23
Goldstine's , 81
Helly's , 77
Kakutani's , 82
Krein-Milman , 60
open mapping , 26
Tychono's , 72
topological
eld, 32
isomorphism, 18, 32
linear space, 32
vector space, 32
topology
coarser than, 62
weak , 62, 66
weak∗ , 67
weaker than, 62
Tychono's theorem, 72
unbounded
linear operator, 36
uniform boundedness principle, 23
vector, 3
space, 3
weak
convergence, 66
topology, 62, 66
weak∗
convergence, 67
topology, 67
Young's inequality, 90
87
Exercises
Exercises for Chapter 1
1 Exercise
Prove the basic properties of a linear space listed in Proposition 1.1.
2 Exercise
Prove the subspace criterions Proposition 1.3 and Proposition 1.4.
3 Exercise
Let X be a linear space over K and S ⊆ X . Show that
(1) the intersection of any family of subspaces of X is a subspace of X ;
(2) the linear span of S ⊆ S
X is the set of all linear combinations of vectors of
∞
S , that is, Span(S) = N =0 SN , where
(
SN =
N
X
)
αn sn : α1 , . . . , αN ∈ K and s1 , . . . , sN ∈ S
.
n=1
If S is nite, then ∞ in the union can be replaced with |S| or dim Span(S).
Prove that the relation used in dening the quotient of a linear space
X by a subspace N on page 4 is an equivalence on X , and that the operations
on X/N are well-dened and make X/N a linear space over K.
4 Exercise
Let B be a linearly independent subset of a linear space X over K.
Show that if x ∈ X can be represented as a linear combination of vectors of B ,
then this representation is unique. That is, if
X
X
αb b =
βb b,
x=
5 Exercise
b∈B
b∈B
where only nitely many αb , βb ∈ K are nonzero (so the sum can always be
interpreted as nite and makes sense), then αb = βb for all b ∈ B . In particular,
this implies that if B is a basis for X , then every x ∈ X has a unique representation as a (nite) linear combination of vectors of B . The scalars of this
representation are called the coordinates of x in the basis B .
Prove the Schröder-Bernstein theorem : if X and Y are non-empty
sets and there exist injections f : X → Y and g : Y → X , then there exists a
bijection between X and Y . In terms of cardinalities, this means that |X| ≤ |Y |
and |Y | ≤ |X| together imply |X| = |Y |.
6 Exercise
88
Let Y be a non-empty set and let X be any set. We denote by Y X
the set of all functions f : X → Y . Prove the following statements. If K is a
eld, then KX is a linear space over K with operations dened pointwise:
7 Exercise
(f + g)(x) = f (x) + g(x)
(αf )(x) = αf (x)
(7.17)
(7.18)
for all x ∈ X ; f, g ∈ KX and α ∈ K.
Let K be R or C, let 1 ≤ p <P
∞ and let `p be the set of all K-valued
∞
p
sequences (xn )∞
n=1 for which the series
n=1 |xn | converges. Prove that `p is
a linear space with pointwise operations dened by (xn ) + (yn ) = (xn + yn ) and
α(xn ) = (αxn ) for all (xn ), (yn ) ∈ `p and all α ∈ K.
8 Exercise
Let K be R or C. Show that the set of all bounded K-valued sequences is a linear space with the pointwise operations dened in exercise 8.
This space is denoted by `∞ .
9 Exercise
Prove Proposition 1.11given linear spaces X and Y over K, a function f : X → Y is linear if and only if f (x1 + αx2 ) = f (x1 ) + αf (x2 ) for all
x1 , x2 ∈ X and α ∈ K.
10 Exercise
Prove Proposition 1.12the range and kernel of a linear operator
are subspaces of its codomain and domain, respectively.
11 Exercise
Prove Proposition 1.13the set of linear operators from a linear
space to another is itself a linear space with pointwise operations.
12 Exercise
Prove Lemma 1.14any mapping of a basis of a linear space X into
a linear space Y can be extended to a linear operator.
13 Exercise
Prove Corollary 1.17 without using Theorem 1.16. Show that the
last statement is not true if dim X = dim Y is not nite.
14 Exercise
Prove Theorem 1.18linear spaces over the same eld and having
the same nite dimension are isomorphic.
15 Exercise
Prove Proposition 1.19every linear functional on a linear space is
either surjective or identically zero.
16 Exercise
Show that the algebraic dual X 0 of a linear space X over K endowed
with the operations dened pointwise by equations (1.5) and (1.6) is a linear
space over K.
17 Exercise
Show that the canonical embedding ϕ : X → X 00 , discussed and
dened on page 11 by hϕ(x), x0 i = hx0 , xi, really is an embedding:
18 Exercise
(1) the mappings ϕ(x) are linear on X 0 for each x ∈ X ;
(2) the mapping ϕ is linear;
(3) ϕ is injective.
89
Prove the following special case of the Riesz representation theorem: If K is R or C, X = Kn and x0 ∈ X 0 , then there exists a vector
x = (x1 , . . . , xn ) ∈ X such that hx0 , yi is the inner product hx | yi of x and
y for all y = (y1 , . . . , yn ) ∈ X , that is,
19 Exercise
hx0 , yi = hx | yi =
n
X
x̄k yk
for all y ∈ X,
k=1
where bar denotes complex conjugation.
Exercises for Chapter 2
Show that a seminorm p on a linear space X is nonnegative and
p(0) = 0; hence p is a norm on X if p(x) = 0 implies x = 0. Prove also the
complete triangle inequality (statement (3) of Proposition 2.1) for the seminorm
p.
20 Exercise
Let p be a seminorm on a linear space X over K. Prove that N :=
{x ∈ X : p(x) = 0} is a linear subspace of X , and X/N is a normed space over
K with the norm kx + N k = p(x).
21 Exercise
Show that a normed space X is a metric space with the metric dened
by d(x, y) = kx − yk for x, y ∈ X .
22 Exercise
Prove Proposition 2.3the vector space operations on a normed
space, as well as the norm itself, are continuous.
23 Exercise
Prove Proposition 2.4translations and scaling by xed elements
are homeomorphisms of a normed space onto itself.
24 Exercise
Let (X, d) be a metric space, x ∈ X and r > 0. Show that the
open ball B(x, r) = {y ∈ X : d(x, y) < r} is an open set and the closed ball
B̄(x, r) = {y ∈ X : d(x, y) ≤ r} is a closed set. Is it true that the closure of
the open ball B(x, r) is the corresponding closed ball B̄(x, r)? What if X is a
normed space?
25 Exercise
26 Exercise
Prove that `∞ dened in Exercise 9 is a Banach space with the norm
k(xn )k∞ = sup {|xn | : n ∈ N} .
Let 1 < p < ∞ and let q be the conjugate
= 1. Prove that if a, b ≥ 0, then
27 Exercise (Young's inequality)
exponent of p, that is,
1
p
+
1
q
ab ≤
bq
ap
+ .
p
q
Let 1 < p < ∞ and let q
be the conjugate exponent of p, that is, p1 + 1q = 1. Prove that if (xn )∞
n=1 and
∞
(yn )n=1 are sequences of complex numbers, then
̰
! p1 Ã ∞
! q1
∞
X
X
X
p
q
|xn yn | ≤
|xn |
|yn |
.
28 Exercise (Hölder's inequality for sequences)
n=1
n=1
n=1
90
Let 1 < p < ∞ and let
∞
(xn )∞
and
(y
)
be
sequences
of
complex
numbers.
Prove that
n n=1
n=1
29 Exercise (Minkowski inequality for sequences)
Ã
∞
X
! p1
|xn + yn |
p
Ã
≤
n=1
∞
X
! p1
|xn |
p
Ã
+
n=1
∞
X
! p1
|yn |
p
n=1
whenever the series on the right-hand side converge.
Prove that whenever 1 ≤ p < ∞, the space `p dened in Exercise 8
is a Banach space with the norm
30 Exercise
Ã
k(xn )kp =
∞
X
! p1
p
|xn |
.
n=1
Prove that if 1 ≤ p < q ≤ ∞, then `p ( `q . (Warning: the analogous
statement Lp ⊆ Lq is not true!)
31 Exercise
32 Exercise
Let c0 = {(xn )∞
n=1 ∈ `∞ : limn→∞ xn = 0} and prove that
[
`p ( c0 .
1≤p<∞
Let X be a real normed space over K = R. Prove the following
statements.
33 Exercise
(1) If x, y ∈ X are linearly dependent, then kx + yk = kxk ± kyk .
(2) Statement (1) need not hold if X is a complex normed space.
(3) If X is a real inner product space, the converse of statement (1) is also true.
That is, x, y ∈ X are linearly dependent if and only if kx + yk = kxk±kyk .
(4) The norm k(xn )k∞ = supn |xn | on the real normed space `∞ cannot arise
from an inner product.
Let X be any set and denote by B(X) the set of all bounded Kvalued functions on X , where K is either R or C. Show that B(X) is a Banach
space with pointwise operations and the norm dened by
34 Exercise
kf k := sup {|f (x)| : x ∈ X} .
(7.19)
Let C(X) denote the set of all continuous complex-valued functions
on a topological space X . Prove that if X is a compact space, then C(X)
with pointwise operations and the uniform norm dened by equation (7.19) is
a Banach space.
35 Exercise
Prove Proposition 2.9 and Proposition 2.8if X and Y are normed
spaces and T : X → Y is a bounded linear operator, then kT xk ≤ kT k kxk
for all x ∈ X , and kT k = sup {kT xk : kxk = 1} = sup {kT xk : kxk ≤ 1} =
sup {kT xk / kxk : x 6= 0}.
36 Exercise
91
Let X and Y be normed spaces and T : X → Y a linear operator.
Show that T −1 : Range(T ) → X exists and is a bounded linear operator if and
only if there exists m > 0 such that m kxk ≤ kT xk for all x ∈ X .
37 Exercise
38 Exercise
Prove that a meagre subset of a Baire space has empty interior.
Prove that if A is a meagre subset of a Baire space X , then X \ A is
nonmeagre.
39 Exercise
40 Exercise
Prove that an open subspace of a Baire space is a Baire space.
Prove Corollary 2.28any two norms on a nite-dimensional vector
space are equivalent. Prove also, that being equivalent really is an equivalence
relation in the set of all norms on a given linear space.
41 Exercise
Let X and Y be normed spaces and A : X → Y a linear operator.
Show that if dim X < ∞, then A is continuous. (Automatic continuity)
42 Exercise
Let X be an innite-dimensional normed space and let Y be any
nonzero normed space over the same eld as X . Prove that there exists a
discontinuous linear operator D : X → Y .
43 Exercise
44 Exercise
Prove Proposition 2.22`1 (n) is a Banach space for n = 1, 2, 3, . . . .
45 Exercise
Prove Proposition 2.23the closed unit ball of `1 (n) is compact.
Let C([0, 1]) be as dened in exercise 35. (X is the closed interval [0, 1] with the subspace topology inherited from R.) Prove the following
assertions:
46 Exercise
(1) E := {f ∈ C([0, 1]) : f (0) = 0} is a Banach space.
n
o
R1
(2) F := f ∈ E : 0 f (x) dx = 0 is a proper closed subspace of E .
(3) The Riesz's lemma (2.31) is not true for r = 1.
Give another proof of Theorem 2.32 by showing that if the unit
sphere S(0, 1) = {x ∈ X : kxk = 1} of a normed space X is compact, then
dim X < ∞. Or prove it by some other means!
47 Exercise
Exercises for Chapter 3
Show that any vector space is a topological vector space with the
trivial topology, but no other topological vector space than the zero space {0}
can have the discrete topology.
48 Exercise
Let X and Y be real or complex linear spaces and T : X → Y a
linear operator. Prove that if A is an absorbing subset of X , then T (A) is an
absorbing subset of Range(T ). Give an example of a situation where T (A) is
not absorbing in Y
49 Exercise
Let B be a subset of a real or complex linear space X . Prove the
following statements.
50 Exercise
92
(1) B is balanced if and only if αB ⊆ γB whenever |α| ≤ |γ| and α, γ ∈ K.
(2) If B is balanced, then αB is balanced and for all α ∈ K.
(3) If B is balanced, then αB = |α| B for all α ∈ K.
(4) If B is balanced, Y is another linear space over K and T : X → Y is a
linear operator, then T (B) is balanced in Y .
Show that the intersection of any family of convex subsets of a linear
space X is convex.
51 Exercise
52 Exercise
is convex.
Show that the closure of a convex subset of a topological vector space
If T : X → Y is a linear operator and K ⊆ X is convex, then T (C)
is a convex subset of Y .
53 Exercise
Let S be a nonempty subset of a linear space X . Show that the the
convex hull of S is convex. Demonstrate that the convex hull of S is generally
not the set T = {αx + (1 − α)y : x, y ∈ S, 0 ≤ α ≤ 1}. Find a necessary and
sucient condition for T to be the convex hull of S . Prove that co αS = α co S
whenever α is a scalar.
54 Exercise
Prove that if S ⊆ X , then
( n
)
X
λk xk : n ∈ N, λk ∈ [0, 1], xk ∈ S for k = 1, . . . , n .
co S =
55 Exercise
k=1
56 Exercise
Prove Proposition 3.21the unit ball of a normed space is convex.
Let S be a nonempty subset of a linear space X . Show that the
balanced hull of S is balanced and equal to the set {αx : α ∈ K, |α| ≤ 1, x ∈ S}.
Show also that the balanced hull of a balanced set B is B .
57 Exercise
Prove that if K is a convex subset of a linear space and α, β ≥ 0 are
scalars, then αS + βS = (α + β)S . Show that this equality does not hold in
general if we allow K to be nonconvex or one of the scalars to be negative.
58 Exercise
Let S be a subset of a topological linear space X . Then the following
statements are equivalent:
59 Exercise
(1) S is bounded.
(2) For every basic neighborhood U of 0, there exists a scalar α such that
S ⊆ αU .
(3) For every subbasic neighborhood U of 0, there exists a scalar α such that
S ⊆ αU .
60 Exercise
Prove Lemma 3.12a nite sum of bounded sets is bounded.
93
Exercises for Chapter 4
Prove this purely algebraic version of the Hahn-Banach theorem concerning general linear operators: If X and Y are linear spaces, M is a proper
subspace of X , and f : M → Y is a linear operator, then there exists a linear
operator F : X → Y such that F |M = f .
61 Exercise
Prove Proposition 4.14the annihilator S ⊥ of a subset S of a normed
space X is a closed subspace of X ∗ .
62 Exercise
Let X be a complex linear space and ϕ a real linear functional on
X . Show that ϕ 6∈ X 0 .
Regard C as a complex linear space. Show that {ir : r ∈ R} is a real hyperplane in C but it is not a hyperplane.
63 Exercise
64 Exercise
Let X be a complex linear space. Prove the following statements:
(1) If f ∈ X 0 , then the real part Re f and the imaginary part Im f of f are
real linear functionals on X .
(2) If r is a real linear functional on X , then there exists a unique (complex)
linear functional f ∈ X 0 such that Re f = r; actually, f (x) = r(x) −ir(ix).
(3) If r is a real linear functional on X , then there exists a unique (complex)
linear functional f ∈ X 0 such that Im f = r (how is f represented in terms
of r?).
Additionally, in the last two cases, if X is a normed space and r is bounded,
then so is f and kf k = krk .
Show that any hyperplane is convex. Show that the two half-spaces
corresponding to a given real hyperplane are convex.
65 Exercise
Let c0 be the set of all such K-valued sequences (K = R or K = C)
that converge to 0. Show that c0 equipped with pointwise operations and the
norm
k(xn )k∞ = sup {|xn | : n ∈ N}
66 Exercise
is a Banach space and nd its dual c∗0 . Here, nding the dual of a normed
space X means nding a more familiar Banach space Y that is (preferably
isometrically) isomorphic to X ∗ .
67 Exercise
Find `∗1 .
68 Exercise
Find `∗p when 1 < p < ∞.
Exercises for Chapter 5
Let X and Y be real or complex linear spaces, T : X → Y a linear
operator and S ⊆ X . Prove that if x0 is an extreme point of S , then T x0 is an
extreme point of T (S). Thus linear isomorphisms preserve extreme points.
69 Exercise
Let X be a real locally convex space and C a compact convex subset
of X . Then every continuous linear functional assumes its maximum on C at
an extreme point.
70 Exercise
94
Exercises for Chapter 6
Prove Proposition 6.1, and show that the denition of the weak topology on X 6= ∅ induced by a given set of functions from X to a given topological
space Y is well posedthat is, weak topology exists and is unique.
71 Exercise
Exercises for Chapter 7
Prove that the Banach space `p is separable whenever 1 ≤ p < ∞.
Prove also, that the Banach space `∞ is not separable.
72 Exercise
73 Exercise
Prove that `∗∞ is not isomorphic to `1 .
Let X and Y be normed spaces. Prove that if X and Y are (isometrically) isomorphic, then X ∗ and Y ∗ are also (isometrically) isomorphic.
74 Exercise
75 Exercise
Show that `p is reexive when 1 < p < ∞.
76 Exercise
Show that c0 , `1 and `∞ are not reexive.
77 Exercise
Show that c00 is not reexive.
(Hypothesis of Theorem 7.12) Let c00 be the linear space of all realvalued sequences x = (xn )∞
n=0 with only nitely many nonzero terms, and equip
it with the `1 norm
∞
X
kxk1 =
|xn | .
78 Exercise
n=0
(αn )∞
α=0
be a real-valued sequence with α0 = 0, αn > 0 for n = 1, 2, . . .
Let
and limn αn = +∞. For each n = 1, 2, . . . , dene a functional fn : c00 → R by
fn (x) = xn , and let f0 = 0 (the zero functional on c00 ). Prove that
(1) the normed space (C00 , k·k1 ) is not complete,
(2) F = {αn fn : n = 0, 1, 2, . . .} is a subset of c∗00 that is not norm bounded,
(3) F is weak∗ bounded, and
(4) F is weak∗ compact.
Can F be convex?
Show that c0 cannot be (isomorphic to) the Banach dual of any
normed space.
79 Exercise
95