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Elements of Functional Analysis
A Series of Lecture Notes Compiled by Matthew R. Gamel
This is an eclectic series of notes compiled by Matthew Gamel. Much of the content
comes from the lecture notes of my Fall 2004-Spring 2005 real analysis courses taught by
Prof. Roger Smith at Texas A&M University and my lecture notes for my Fall
2006-Spring 2007 functional analysis course taught by Prof. George Androulakis at the
University of South Carolina. Some of the content comes from my solutions to
homework problems and from other sources as well.
This set of notes evolved out of my desire to organize an eclectic series of notes from
various courses. I hope that these notes will prove to be beneficial as a reference or a
supplement to a graduate level course in the subject.
This set of notes is outlined as follows (note that these notes are currently incomplete):
a. Introduction to Banach and Hilbert Spaces
b. Cornerstone Theorems of Functional Analysis
c. Weak and Weak∗ Topologies
d. Schauder Bases and Other Topics in Banach Space Theory
e. Function Spaces
I have also included some exercises for the interested reader at the end of each section.
If there are any mathematical or typographical errors, please send them to me at
[email protected].
1
1
Banach and Hilbert Spaces
Here we discuss the basic theory of Banach and Hilbert spaces. Unless otherwise
specified, all vector spaces are over C.
1.1
Normed Vector Spaces and Inner Product Spaces
Definition 1.1.0
A normed space is a vector space X equiped with a function || · || from X to R satisfying
the following properties:
(a) ||x|| ≥ 0 for all x ∈ X with equality iff x = 0.
(b) ||αx|| = |α|||x|| for all α ∈ C and x ∈ X
(c) ||x + y|| ≤ ||x|| + ||y|| for all x, y ∈ X
If X is a normed space, we will often write it’s norm as || · ||X to avoid confusion with
other spaces. We omit the subscript when the space in question is understood. In
addition, if we define ρ(x, y) := ||x − y|| then it is easy to see that (X, ρ) is a metric
space whose metric is induced by the norm.
If (xn ) is a sequence in X, we say that xn converges to x0 ∈ X, written xn → x0 , if we
have limn→∞ ||xn − x0 || = 0. Moreover, if the normed space X is equiped with two
norms || · ||1 and || · ||2 , we say that these norms are equivalent if there exists constants
C1 > 0 and C2 > 0 so that
C1 ||x||1 ≤ ||x||2 ≤ C2 ||x||1 for all x ∈ X
It is easy to see that a sequence (xn ) converges in || · ||1 if and only if it converges in
|| · ||2 provided that they are equivalent. Equivalent norms, therefore, generate
equivalent norm topologies.
When applying the theory of Banach and Hilbert spaces, the following inequality makes
life exceptionally easier. We shall make use of this later so we prove it here.
2
Proposition 1.1.1 (Young’s Inequality)
Let a, b ∈ R+ and suppose that p > 1 and q are such that
ab ≤
1
p
+
1
q
= 1. Then
ap b q
+
p
q
Proof. If either a or b is 0 (or both, possibly) then the result is obvious. Assume, then,
that they are nonzero. Set f (t) = p1 tp + 1q − t where 0 ≤ t < ∞ so that f 0 (t) = tp−1 − 1.
On (0, 1), f 0 (t) < 0 so f is decreasing from f (0) = 1q > 0 down to t = 1. On (1, ∞),
f 0 (t) > 0 so f is increasing and the point t = 1 is an absolute minimum. Hence f (t) ≥ 0
a
for all t ≥ 0. Setting t = bq−1
and simplifying yields the result (verify).
We now define the basic notion of an inner product space and then prove some other
useful results.
Definition 1.1.2
An inner-product space is a vector space X equiped with a function (·, ·) from X × X to
C satisfying the following properties:
(a) (x, x) ≥ 0 with equality iff x = 0 for all x ∈ X.
(b) (x, y) = (y, x) for all x, y ∈ X
(c) (αx + βy, z) = α(x, z) + β(y, z) for all α, β ∈ C and x, y, z ∈ X.
We say that x, y ∈ X are orthogonal, written x ⊥ y, provided that (x, y) = 0.
Notice that property (b) also implies that (x, x) must be a real number. Hence property
(a) makes sense. If our vector space is over R instead of C, then property (b) simply
reduces to (x, y) = (y, x). Also if we combine property (b) and property (c), we get
(x, αy + βz) = α(x, y) + β(x, z). This next result is essential.
Proposition 1.1.3 (Cauchy-Shwartz Inequality)
Let X be an inner-product space. Then, for all x, y ∈ X, we have
|(x, y)| ≤
p
p
(x, x) (y, y)
3
Proof. Let λ ∈ C and let x, y ∈ X. Then
0 ≤ (x + λy, x + λy) = (x, x) + 2Re[λ(x, y)] + |λ|2 (y, y)
If x = 0 or y = 0 (or both) then the inequality clearly holds. Now suppose that y is
nonzero and set
(y, x)
λ=−
(y, y)
Then, after some minor arithmetic, we get
0 ≤ (x, x) − 2
|(y, x)|2 |(y, x)|2
+
(y, y)
(y, y)
so the result clearly holds.
Just as we noted that a normed space is also a metric space, we here demonstrate that
an inner-product space is also a normed space.
Proposition 1.1.4
p
Let X be an inner-product space. Then ||x|| := (x, x) defines a norm on X.
Proof. The first few properties are obvious so we only verify that the triangle inequality
holds. We have
||x + y||2 =
≤
≤
=
(x + y, x + y) = (x, x) + 2Re[(x, y)] + (y, y)
||x||2 + 2|(x, y)| + ||y||2
||x||2 + 2||x||||y|| + ||y||2
(||x|| + ||y||)2
Definition 1.1.5
Let X be a normed space. We call X a Banach space if it is complete in norm; that is,
every Cauchy sequence in X converges to a point in X.
This next definition is really a special case of the previous definition but we elect to
state it here seperately due to it’s importance.
4
Definition 1.1.6
Let H be an inner-product space. We call H a Hilbert space if it is complete in the norm
induced by the inner-product.
Essentially, a Hilbert space can be thought of as a generalization of Euclidean space in
finite dimensions and consequently there are many nice geometric interpretations
associated with Hilbert spaces. We shall see this at a later time.
1.2
Operators, Linear Functionals, and the Hahn-Banach
Theorem
Again, we are assuming all normed vector spaces are over C unless otherwise specified.
We begin with a useful equivalent formulation of a Banach space.
Proposition 1.2.0
Let X be a normed space. The following are equivalent:
(a) X is a Banach space
(b) For each sequence xn ∈ X (∀n) with the property that
∞
X
||xk ||X < ∞
k=1
we have the partial sums sn =
Pn
k=1
xk converge in norm.
Proof. Suppose that X is a Banach space and suppose that xn is a sequence in X with
P
∞
k=1 ||xk ||X < ∞. Clearly
n
n
X
X
||sn − sm ||X = xk ≤
||xk ||X
m+1 m+1
X
so that the partial sums are Cauchy and, hence, must converge. Conversely, let xn be
Cauchy in norm. By dropping to a subsequence, we may without loss of generality
5
assume that ||xn − xn−1 ||X ≤
n ≥ 2. Then
n
X
1
2n
when n ≥ 2. Define y1 = x1 and yn = xn − xn−1 when
n
X
1
≤ ||x1 ||X + 1 < ∞
k
2
k=1
k=2
P∞
P
Thus, we get k=1 ||yk ||X < ∞. Clearly sn := n1 yk = xn so that by hypothesis,
sn = xn converges and X is a Banach space.
||yk ||X ≤ ||x1 ||X +
If X and Y are normed spaces and T : X → Y , we call T an operator. The following
equivalent conditions are useful and extensively used.
Proposition 1.2.1
Let X and Y be normed spaces and let T : X → Y be a linear operator. Then the
following are equivalent:
(a) T is continuous.
(b) T is continuous at 0.
(c) (Boundedness) There is a constant C > 0 so that ||T x||Y ≤ C||x||X for all x ∈ X.
Proof. It is easy to see that (a) and (b) are equivalent. It is also obvious that (c) implies
(a) and (b). We will now show that (b) implies (c). Assume (b) and suppose that no
such C > 0 exists. Then, by normalizing if necessary, we may find a sequence yn ∈ X
with ||yn ||X = 1 such that ||T yn ||Y ≥ n. Let xn = √1n yn so that ||xn ||X → 0 as n → ∞.
√
Then xn → 0 and T xn = √1n T yn so ||T xn ||Y ≥ √nn = n but T xn does not go to 0, a
contradiction.
Definition 1.2.2
Let X and Y be normed spaces. We denote by L(X, Y ) the set of all continuous linear
operators from X to Y . If Y = R (or C) then L(X, R) is called the dual of X and is
denoted by X ∗ = L(X, R). If φ ∈ X ∗ , we sometimes call φ a linear functional.
By the previous proposition, we shall employ the terms continuous and bounded
interchangably in the context of linear operators between normed spaces. If T : X → Y
is bounded, we define
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||T || := sup
x6=0
||T x||Y
= sup ||T x||Y
||x||X
||x||X =1
This is called the operator norm and if we equip the space of all continous linear
operators L(X, Y ) with this norm, we get a normed space and in some cases, a Banach
space.
Theorem 1.2.3
If X and Y are normed spaces, then so is L(X, Y ). Aditionally, if Y is a Banach space,
then so is L(X, Y ).
Proof. It is easy to check that L(X, Y ) is a normed vector space. Let Tn ∈ L(X, Y ) be
Cauchy. Then for each x ∈ X, ||(Tn − Tm )x||Y ≤ ||Tn − Tm || · ||x||X so that Tn x is
Cauchy in Y . Since Y is a Banach space, there is a y ∈ Y so that Tn x → y. Define
T x := y. We will now show that T is linear and that Tn → T in the operator norm.
First, if x1 , x2 ∈ X then Tn x1 → T x1 and Tn x2 → T x2 so that Tn x1 + Tn x2 → T x1 + T x2 .
But Tn x1 + Tn x2 = Tn (x1 + x2 ) → T (x1 + x2 ) so by uniqueness of limits,
T x1 + T x2 = T (x1 + x2 ). Similarly, T (λx) = λT x whenever x ∈ X so T is linear.
Let > 0. Then we may find an N0 so that ||Tn − Tm || < whenever n, m ≥ N0 . Then,
for each x ∈ X with ||x||X = 1, we get ||Tn x − Tm x||Y < when n, m ≥ N0 . By
continuity of the norm, we fix any n ≥ N0 and let m → ∞ to get ||Tn x − T x|| < . This
holds for each n ≥ N0 and x ∈ X with ||x||X = 1. Thus
||Tn − T || = sup ||Tn x − T x||Y < whenever n ≥ N0
||x||X =1
Finally, because ||T || < ||TN0 || + 1, we’re finished.
An immediate consequence of this is if Y = R (or C) then X ∗ is a Banach space. We
may also define the double dual of a normed space X given by
X ∗∗ := (X ∗ )∗ = L(X ∗ , F)
where F = R or C. In a likewise manner, X ∗∗ is a Banach space.
7
Definition 1.2.4 (Canonical Embeddings)
Let X be a real or complex normed space and for each x ∈ X, define a map
Jx (φ) := φ(x) for all φ ∈ X ∗ . We call Jx the canonical embedding of X into X ∗∗ .
The canonical embedding satisfies many nice properties; namely, it induces a continuous
injective isometry from X into X ∗∗ (that is, ||Jx ||X ∗∗ = ||x||X ). The details are left as an
exercise for the reader.
Definition 1.2.5 (Reflexivity)
A normed space X is called reflexive if the canonical embedding is surjective.
Not every space is reflexive as we shall later see and it turns out to be fairly tricky to
show that many spaces are reflexive. Before we move on, we introduce one more concept
relating to dual spaces.
Definition 1.2.6 (Adjoint Operators)
Let X and Y be normed spaces. If T : X → Y is linear then the adjoint of T , denoted
T ∗ is given by T ∗ φ := φ ◦ T for all φ ∈ Y ∗ .
It is obvious that the adjoint operator T ∗ : Y ∗ → X ∗ is linear and we leave it to the
reader to show that if T is bounded, then so is it’s adjoint with ||T || = ||T ∗ ||.
We will now prepare to prove our first major result. However, we need some definitions
before we proceed. At this point, we will now require that our normed space be over R
rather than C.
Definition 1.2.7 (Sublinear Functionals and Seminorms)
A sublinear functional on a real vector space X is a function p : X → R such that
(a) p(x + y) ≤ p(x) + p(y) for all x, y ∈ X
(b) If λ ≥ 0 then p(λx) = λp(x) for all x ∈ X.
On a real or complex vector space, a seminorm is a function p : X → [0, ∞) satisfying
(a) p(x + y) ≤ p(x) + p(y) for all x, y ∈ X
8
(b) p(λx) = |λ|p(x) for all x ∈ X and λ ∈ R (or C)
The only difference between a seminorm and a norm is that the condition p(x) = 0 if
and only if x = 0 is missing. We briefly digress to give an example.
Example (Minkowski Functionals)
Let K ⊂ R2 be convex and contain a neighborhood of the origin. Let
Kλ = {λx : x ∈ K} and define p(x) := inf{λ : x ∈ Kλ }. We shall prove that this
function, called the Minkowski Functional, is a sublinear functional.
Firstly, observe that p(cx) = cp(x) whenever c ≥ 0. Let > 0 and x, y ∈ K. Then there
is a λ and a µ so that p(x) ≤ λ < p(x) + and p(y) ≤ µ < p(y) + . Clearly, x ∈ Kλ and
y ∈ Kµ . Hence, there is a k1 , k2 ∈ K so that x = λk1 and y = µk2 . Then
µ
λ
k1 +
k2 = (λ + µ)k
x + y = λk1 + µk2 = (λ + µ)
λ+µ
λ+µ
for some k ∈ K by convexity. So, we have
p(x + y) ≤ λ + µ ≤ p(x) + p(y) + 2
Letting → 0+ yields the desired result.
♦
We now proceed to prove our first major result, the Hahn-Banach Theorem. This
enables us to extend linear functionals on a subspace of a vector space to the whole
space.
Theorem 1.2.8 (Hahn-Banach Theorem)
Let X be a real vector space with a sublinear functional p and Y ⊂ X a subspace.
Suppose that ψ : Y → R is a linear functional and that ψ(y) ≤ p(y) for all y ∈ Y . Then
ψ extends to a linear functional φ : X → R so that φ|Y = ψ and φ(x) ≤ p(x) for all
x ∈ X.
Proof. Our strategy will be to demonstrate the theorem in two basic steps. During the
first step, we will essentially attempt to extend ψ as far as we are able. Then, if our
extension is not defined on all of X, we extend it again to obtain a contradiction.
Denote by M the collection of all subspaces M ⊇ Y and linear functionals θ : M → R
so that θ(m) ≤ p(m) for m ∈ M and θ|Y = ψ. Order this set in the following way:
9
(M1 , θ1 ) ≺ (M2 , θ2 ) iff M1 ⊆ M2 and θ2 |M1 = θ1 .
Let U = {(Mα , θα ) : α ∈ A} be a totally ordered subset of M (i.e. any two elements in
U are comparable). Set
[
M :=
Mα
α∈A
It is now not difficult to see that M is a subspace of X containing Y . On M , define
θ(x) = θα (x) when x ∈ Mα . It is easy to see that this is well defined. For if x ∈ Mα and
x ∈ Mβ then without loss of generality, assume that Mα ⊂ Mβ . Since θβ |Mα = θα we get
θα (x) = θβ (x).
We also have θ(x) ≤ p(x) for all x ∈ M because θα (x) ≤ p(x) for all α ∈ A. Moreover, if
x1 , x2 ∈ M with x1 ∈ Mα1 and x2 ∈ Mα2 then without loss of generality, suppose that
Mα1 ⊆ Mα2 . Then
θ(x1 + x2 ) = θα2 (x1 + x2 ) = θα2 (x1 ) + θα2 (x2 )
So θ is a linear functional with (M, θ) ∈ M and (M, θ) (Mα , θα ) for all α. By Zorn’s
Lemma, there is a maximal element of M, call it (Z, ω).
We now have an extension so we now contend that Z = X. Suppose the contrary, that
Z is strictly contained in X but not equal to X. Fix an element x0 ∈ X with x0 ∈
/ Z.
We will extend ω : Z → R to φ : Z0 → R satisfying φ(x) ≤ p(x) for x ∈ Z0 and φ|Z = ω
where Z0 = {z + λx0 : z ∈ Z and λ ∈ R}. This will be a contradiction because (Z, ω)
will no longer be maximal.
The functional φ, if it exists, is determined by γ = φ(x0 ). We must have
φ(z + λx0 ) ≤ p(z + λx0 ) for all z ∈ Z and λ ∈ R
This implies that ω(z) + λγ ≤ p(z + λx0 ) for all z ∈ Z and λ ∈ R. This is true if λ = 0
so if we can choose a γ to make this true when λ > 0 and λ < 0 then we will have
constructed an extension.
If λ > 0 we have
ω
z λ
+γ ≤p
z
λ
+ x0
10
∀z ∈ Z, λ > 0
Put z1 =
z
λ
so we must have z1 ∈ Z. Letting z and λ vary gives all elements in Z. Then
γ ≤ p(z1 + x0 ) − ω(z1 ) ∀z1 ∈ Z
If λ < 0 set λ = −µ for µ > 0. We want ω(z) − µγ ≤ p(z − µx0 ) or
z
z
ω
−γ ≤p
− x0
µ
µ
Put z2 = µz to get γ ≥ w(z2 ) − p(z2 − x0 ) for all z2 ∈ Z. Combining this condition with
the one we obtained when λ > 0, we can make the choice if
ω(z2 ) − p(z2 − x0 ) ≤ p(z1 + x0 ) − ω(z1 ) ∀z1 , z2 ∈ Z
However, observe that
ω(z1 + z2 ) ≤ p(z1 + z2 ) = p((z1 + x0 ) + (z2 − x0 ))
≤ p(z1 + x0 ) + p(z2 − x0 )
which is precisely the condition we needed to satisfy. Thus, it is possible to choose such
a γ (there is at least one) in the given interval. Hence, φ(x) = ω(x) + λγ extends ω on Z
to all of Z0 contradicting the maximality of (Z, ω). It must therefore follow that Z = X.
We can modify the result for normed spaces as follows:
Corollary 1.2.9 (Hahn-Banach Theorem for Normed Spaces)
Let X be a normed real vector space, Y ⊂ X a subspace, and ψ ∈ Y ∗ . Then ψ extends
to a bounded linear functional φ ∈ X ∗ with ||φ|| = ||ψ||.
Proof. Exercise for the reader; use the Hahn-Banach Theorem above.
Notice that we do not require any completeness for the Hahn-Banach Theorem to apply;
we merely need a vector space or a normed space over R. With a minimal amount of
effort, we may prove a version of the Hahn-Banach Theorem for vector spaces over C.
We do this now:
11
Corollary 1.2.10 (Hahn-Banach Theorem for Complex Vector Spaces)
Let X be a vector space over C, Y ⊂ X be a complex subspace, and p a seminorm on
X. Let ψ : Y → C be a linear functional satisfying |ψ(y)| ≤ p(y) for all y ∈ Y . Then ψ
extends to φ : X → C such that |φ(x)| ≤ p(x) for all x ∈ X.
Proof. We can interpret X and Y to also be vector spaces over R, call them X̃ and Ỹ .
In this respect, let θ : Ỹ → R where θ(y) = Re(ψ(y)). It is not difficult to check and see
that θ is linear over Ỹ .
Additionally, we have |θ(y)| ≤ |ψ(y)| ≤ p(y) for all y ∈ Ỹ . By the real form of the
Hahn-Banach Theorem, θ extends to β : X̃ → R with β(x) ≤ p(x) for x ∈ X̃. Here β is
R-linear. Moreover, we have
θ(iy) = Re(ψ(iy)) = Re(iψ(y)) = −Im(ψ(y))
Thus ψ(y) = θ(y) − iθ(iy). Define φ(x) = β(x) − iβ(ix). Because β is R-linear, then so
is φ. We now need to check that it is C-linear so we must show that φ(ix) = iφ(x). We
have
φ(ix) = β(ix) − iβ(−x) = iβ(x) + β(ix) = iφ(x)
Finally, we must show that |φ(x)| ≤ p(x) for x ∈ X. Since φ(x) is a complex number, we
may write it as |φ(x)|eit0 for some 0 ≤ t0 ≤ 2π. Hence e−it0 φ(x) = |φ(x)| so that
φ(e−it0 x) = |φ(x)|. This means that φ(e−it0 x) is a real number so we must have
φ(e−it0 x) = β(e−it0 x). Since β(e−it0 x) ≤ p(e−it0 x) = |e−it0 |p(x) = p(x) so we have
|φ(x)| ≤ p(x) for all x ∈ X and this completes the proof.
Let us illustrate the use of the Hahn-Banach Theorem with an example.
Example 1.2.11
Let X be a normed space over R (or C) and Y ⊂ X a subspace. Set x0 ∈ X\Y . We will
use the Hahn-Banach Theorem to demonstrate the existence of a φ ∈ X ∗ so that
φ|Y = 0 and φ(x0 ) = d(x0 , Y )1
To see this, set p(x) = d(x, Y ). It is not difficult to see that this is a sublinear functional
on X. Consider the subspace S = {λx0 : λ ∈ R} of X and put ψ(λx0 ) = λp(x0 ). If
1
d(x0 , Y ) := inf y∈Y ||x0 − y||X
12
λ < 0 then we must have ψ(λx0 ) < 0 so that ψ(λx0 ) ≤ p(λx0 ). If λ ≥ 0 then
ψ(λx0 ) = λp(x0 ) = p(λx0 ) so p dominates ψ on S. Moreover, it is obvious that ψ(x) is
linear and continuous. By the Hahn-Banach Theorem, extend ψ to φ on X where
φ(x) ≤ p(x). Note that because
−φ(x) = φ(−x) ≤ p(−x) = inf || − x − y|| = inf || − x + z|| = p(x)
y∈Y
z∈Y
we get |φ(x)| ≤ p(x) so that φ vanishes on Y .
♦
The following result allows us to separate subspaces of a Banach space by a linear
functional.
Theorem 1.2.12 (Hahn-Banach Separation Theorem)
Let X be a normed space over C and A, B ⊆ X be convex, nonempty, disjoint sets with
A open. Then there is a φ ∈ X ∗ so that
Re φ(a) < Re φ(b) ∀a ∈ A, b ∈ B
Proof. Firstly, assume that X is over R. Fix an a0 ∈ A and b0 ∈ B and put
C := A − B + (b0 − a0 ) = {a − b + b0 − a0 : a ∈ A and b ∈ B}
Since A and B are convex, we see that C is also convex. Moreover, 0 ∈ C. Since A is
open, it follows that C is also open because
[
C=
(A − b + b0 − a0 )
b∈B
Let p(x) be the Minkowski functional on C. That is, define
p(x) := inf{λ ≥ 0 : x ∈ λC}
We now demonstrate that
C = {x ∈ X : p(x) < 1}
13
Choose x ∈ X with p(x) < 1. We may firstly assume that x 6= 0 so that p(x) > 0 since
we know that 0 ∈ C. Then there is a 0 < λ0 < 1 so that x ∈ λ0 C. Define y = λ10 x so
that y ∈ C. Since C is convex we have λ0 y + (1 − λ0 ) · 0 = x ∈ C. We now show reverse
inclusion. Fix an x ∈ C and define a map ηx : R → X by ηx (t) := tx. Clearly ηx is
continuous. Because ηx (1) = x and C is open, we can find an open ball about x
contained entirely in C. By continuity, we may find a δ > 0 so that ηx (t) ∈ C when
1
1 − δ ≤ t ≤ 1 + δ. In particular, ηx (1 + δ) ∈ C so that x ∈ 1+δ
C. This gives
p(x) ≤
1
1+δ
so that p(x) < 1 as desired.
Now, put Y := span(b0 − a0 ). On Y , define ψ : Y → R by ψ(y) = ψ(λ(b0 − a0 )) := λ.
Clearly, ψ is linear and we easily note that ψ(b0 − a0 ) = 1. Also, we have b0 − a0 ∈
/C
because 0 ∈
/ A − B as A ∩ B = . By what we did above, b0 − a0 ∈
/ C tells us that
p(b0 − a0 ) ≥ 1. Thus ψ(b0 − a0 ) = 1 ≤ p(b0 − a0 ) so that if λ ≥ 0, we have
ψ(λ(b0 − a0 )) = λψ(b0 − a0 ) ≤ λp(b0 − a0 ) = p(λ(b0 − a0 ))
If λ < 0 then ψ(λ(b0 − a0 )) < 0 but p(λ(b0 − a0 )) ≥ 0 so that ψ(y) ≤ p(y) for all y ∈ Y .
By Hahn-Banach, ψ extends to a linear map φ : R → X with φ|Y = ψ and φ(x) ≤ p(x)
for all x ∈ X. By what we have above, φ(x) ≤ 1 for all x ∈ C. Thus |φ(x)| ≤ 1 for each
x ∈ C ∩ (−C) so that φ is bounded in an open neighborhood of 0. Thus |φ(x)| ≤ M for
all x in a neighborhood of the unit ball of X. Thus φ ∈ X ∗ .
Now, for each a ∈ A and b ∈ B, we get φ(a − b + b0 − a0 ) < 1 so that
φ(a − b) + ψ(b0 − a0 ) < 1 or φ(a − b) < 0 so that φ(a) < φ(b).
Now, assume that X is over C. We may interpret X as a vector space over R so that
there is an R-linear functional φ ∈ X ∗ satisfying φ(a) < φ(b) for all a ∈ A and b ∈ B.
Define ψ(x) := φ(x) − iφ(ix). Then ψ(x) is our desired functional (we leave it to the
reader to verify).
14
1.3
Finite Dimensional Normed Spaces
In this section, we present some basic theory pertaining to finite dimensional normed
spaces. In the study of Banach space theory, finite dimensional spaces are rather
uninteresting because they satisfy so many nice properties as we shall see. It is when we
begin to study infinite dimensional spaces that entirely new and interesting phenomenon
occur.
Theorem 1.3.0
Let X be a finite dimensional normed linear space over R (or C). Then every closed,
bounded set in X is compact.
Proof. Let X be a finite dimensional normed space. Let dim(X) = n and select a basis
for X, say {x1 , x2 , ..., xn }. Define a map L : Rn → X by
L(ξ) =
n
X
ξk xk
k=1
where ξ = (ξ1 , ..., ξn ) ∈ Rn . We shall work with || · ||∞ in Rn . We know that L is
continuous because
n
n
n
X
X
X
||xk ||X
|ξk − ζk |||xk ||X ≤ ||ξ − ζ||∞
||Lξ − Lζ||X = (ξk − ζk )xk ≤
k=1
X
k=1
k=1
If Y ⊂ X is closed and bounded then A := L−1 (Y ) is also closed by continuity of L.
Because L is bijective (check), we see that Y = L(A). If we can show that A ⊂ Rn is
compact, then Y will also be compact because L is continuous.
Consider the boundary of the `∞ unit ball in Rn , ∂B1∞ (0). Since f (ξ) = ||Lξ||X is
continuous and ∂B1∞ (0) is compact, we know that f attains a minimum, m. Let ξ0 be
such that ||ξ0 ||∞ = 1 and f (ξ0 ) = ||Lξ0 ||X = m. Then
n
X
ξk0 xk m = k=1
X
Each xi is linearly independent and ξ0 6= 0 so we must have m > 0. Since Y is bounded,
there is a M > 0 with ||y||X ≤ M for all y ∈ Y . If ξ ∈ Rn then ξ/||ξ||∞ is a unit vector
in `∞ so we must have
15
ξ
L
≥ m
||ξ||∞ X
so that ||Lξ||X ≥ m||ξ||∞ (clearly this also holds if ξ = 0). For each ξ ∈ A we have
Lξ ∈ Y so m||ξ||∞ ≤ ||Lξ|| ≤ M or ||ξ||∞ ≤ M/m. Thus, A is bounded and we are
done. Technically, we proved this result for a real normed space but the proof is
identical for a complex normed space.
It turns out that finite dimensional spaces may be completely characterized by this
compactness property. In other words, the converse is also true: if X is a normed space
and every closed bounded set in X is compact, then X is finite dimensional. To prove
this, we first present a lemma.
Lemma 1.3.1 (Riesz’s Lemma)
Let X be a normed space and suppose that K is a closed, proper, nontrivial subspace of
X. If 0 < µ < 1 then there exists a point x ∈ X so that ||x|| = 1 and d(x, K) > µ.
Proof. Since K is proper, we may find an x0 ∈ K c ∩ X so that d(x0 , K) > 0. Then there
exists a k ∈ K so that ||x0 − k|| < µ−1 d(x0 , K). Define x := (x0 − k)/||x0 − k|| so that
||x|| = 1 and
d(x, K) =
d(x0 , K)
d(x0 − k, K)
=
>µ
||x0 − k||
||x0 − k||
which is what we wished to show.
We are now ready to prove the converse of Theorem 1.3.0.
Theorem 1.3.2
Suppose that X is a normed space and that every closed, bounded subset of X is
compact. Then X is finite dimensional.
Proof. It suffices to prove that if the closed unit ball is compact, then X is finite
dimensional. Suppose that X is not finite dimensional. Choose an x1 with ||x1 || = 1 and
let U1 := span(x1 ). Then U1 is a closed, proper subspace of X. By Riesz’s Lemma,
choose an x2 so that ||x2 || = 1 and d(x2 , U1 ) > 1/2. Then ||x2 − x1 || > 1/2. Let
16
U2 := span(x1 , x2 ). Again, by Riesz’s Lemma, choose an x3 so that d(x3 , U2 ) > 1/2 so
that ||x3 − x1 || > 1/2 and ||x3 − x2 || > 1/2. Continue this process inductively to generate
a sequence (xn ) so that ||xn − xj || > 1/2 whenever j < n and ||xn || = 1. But then (xn ) is
a sequence in the closed unit ball that has no convergent subsequence, a contradiction.
We may therefore reformulate a stronger version of Theorem 1.3.0 as follows:
Let X be a normed space over R or C. Then X is finite dimensional if and only if every
closed bounded subset of X is compact.
We now deduce some interesting corollaries to this statement.
Corollary 1.3.3
Let X be a finite dimensional normed space over R (or C). Then all norms on X are
equivalent.
Proof. Let || · || be any norm on X. Fix a basis B := {x1 , x2 , ..., xn } and without loss of
generality, assume that
P ||xj || = 1 for each j. Then any x ∈ X has the unique
representation x = nk=1 ck xk where each ck ∈ R (or C). Define the norm
||x||B :=
n
X
|ck |
k=1
By uniqueness of basis representations, this is well defined.
Consider S := {x ∈ X : ||x||B = 1} and define f (x) := ||x|| on S. By the previous
theorem S is closed and bounded and, hence, compact in the topology induced by
|| · ||B .We also note that for each x, y ∈ X, we have
|f (x) − f (y)| ≤ ||x − y|| ≤ ||x − y||B
Thus, f (x) is continuous in the topology induced by || · ||B . Because S is compact, there
exists positive constants m and M such that
m ≤ ||x|| ≤ M ∀x ∈ S
Then for any x ∈ X, we have x/||x||B ∈ S so that
17
m||x||B ≤ ||x|| ≤ M ||x||B
Thus, any two norms are equivalent to || · ||B and, hence, each other.
Norm equivalence tells us that there is one and only one norm topology on X. We shall
later see that something stronger is true: there is precisely one linear Hausdorff topology
on a finite dimensional space. Of course, this linear T2 topology must be the norm
topology.
The following is an easy corollary.
Corollary 1.3.4
Let X be a finite dimensional normed space over R (or C). Then X is a Banach space in
any norm that one may impose on X.
Proof. Exercise for the reader.
We now make a few comments about the dual of a finite dimensional normed space.
Let X be a finite dimensional space where dim(X) = n. Choose a basis for X, say
B = {x1 , x2 , ..., xn }. By Hahn-Banach, there exists a linear functional φ1 ∈ X ∗ so that
φ1 (xj ) = δ1j . Continue this process to choose finitely many functionals φi ∈ X ∗
(1 ≤ i ≤ n) so that φi (xj ) = δij and put
B ∗ := {φ1 , ..., φn }
It is clear that the functionals in B ∗ are linearly independent. Moreover, if x =
then φj (x) = cj for 1 ≤ j ≤ n. Then if ψ ∈ X ∗ , we see that
ψ(x) =
n
X
ck ψ(xk ) =
k=1
n
X
Pn
1
ck xk ,
φk (x)ψ(xk )
k=1
That is, B ∗ forms a basis for the dual space X ∗ and dim(X) = dim(X ∗ ). We refer to B ∗
as the dual basis for X ∗ . Things in general are not always so nice and we will see this
later. We conclude this section with a simple observation.
18
Proposition 1.3.5
Let X be a real or complex finite dimensional normed space. Then X is reflexive.
Proof. It is necessary to demonstrate that the canonical embedding
is surjective. Let
Pn
∗∗
ψ ∈ X . Then by the discussion above, we may write ψ = k=1 ck ψk where
ψk (φj ) = δkjP
and the ψj ’s and φj ’s form the dual bases
for X ∗∗ and X ∗ respectively.
P
Define x := nk=1 ck xk . Then if φ ∈ X ∗ , write φ = nk=1 bk φk so that
ψ(φ) =
n
X
ck ψk (φ) =
k=1
n X
n
X
ck bj ψk (φj ) =
k=1 j=1
n
X
ck bk = φ(x) = Jx (φ)
k=1
This is what we wished to show.
In general, it is not a trivial task to prove that a space is reflexive despite the simplicity
of the previous argument. We will later revisit finite dimensional spaces when we discuss
weak topologies.
To conclude, we here present a pithy summary of this section: all finite dimensional
normed spaces are norm-equivalent, reflexive Banach spaces that are topologically the
same as Rn . If life were only this simple in practice.
1.4
Inner-Product and Hilbert Spaces
Recall that a Hilbert space is a special type of Banach space; namely, an inner-product
space induces a natural norm ||x|| = (x, x)1/2 so a Hilbert space is a Banach space in
this particular norm. Note that all Hilbert and inner-product spaces will be taken over
C unless we explicitly specify that they are over R.
Let’s begin with a basic definition.
Definition 1.4.0
Let H be an inner product space. We say that a pair of vectors xα , xβ for α, β ∈ A and
α 6= β are orthogonal if (xα , xβ ) = 0. A set of vectors {xα : α ∈ A} is called orthonormal
if each vector is orthogonal and ||xα ||H = 1 for all α ∈ A.
If two vectors x, y ∈ H are orthogonal, we will often write x ⊥ y. We now present some
basic results.
19
Proposition 1.4.1 (The Pythagorean Theorem)
Let H be an inner product space and suppose that {x1 , x2 , ..., xn } ⊂ H are orthogonal.
Then
2
n
n
X
X
xk =
||xk ||2H
k=1
k=1
H
Proof. We have
n
n
X 2
X
X
X
X
xk ,
xk ) =
xk = (
(xh , xk ) =
(xk , xk )
k=1
h,k
H
k=1
Proposition 1.4.2
Let H be an inner product space and suppose that {x1 , x2 , ..., xn } ⊂ H are orthonormal.
Then
2
n
n
X
X
|ak |2 for all ak ∈ C
ak x k =
(a) k=1
(b)
n
X
H
k=1
|(x, xk )|2 ≤ ||x||2H (Bessel’s Inequality)
k=1
Proof. For (a) we have
2
n
X
ak x k k=1
=
n
X
ak x k ,
k=1
H
=
n X
n
X
n
X
!
ak xk
k=1
ak ah (xk , xh )
k=1 h=1
=
n
X
|ak |2 (xk , xk )
k=1
=
n
X
k=1
20
|ak |2
P
P
To obtain (b), set y = n1 (x, xk )xk so that by (a) we have ||y||2H = n1 |(x, xk )|2 . Set
z = y − x so that for each j, we have
(z, xj ) = (y, xj ) − (x, xj ) = (x, xj ) · (xj , xj ) − (x, xj ) = 0
Since z ⊥ xk for each k, we must also have z ⊥ a for each a ∈ span(x1 , x2 , ..., xn ). Hence
z ⊥ y. By the Pythagorean Theorem, ||y − z||2 = ||y||2 + ||z||2 . Hence
||x||2 = ||y||2 + ||z||2 ≥
n
X
|(x, xk )|2
k=1
If we are given a finite collection of linearly independent vectors in an innder product
space, we may always construct an orthonormal set of vectors that has the same span as
our original set of vectors. This is called the Gramm-Schmidt orthonormalization
process and we discuss this now.
The Gramm-Schmidt Orthonormalization Process
Let φ1 , φ2 , ..., φn be linearly independent vectors in an inner product space V . Then
there exists n orthonormal vectors ψ1 , ψ2 , ..., ψn so that
span(φ1 , ..., φn ) = span(ψ1 , ..., ψn )
Proof. We prove by induction. If n = 1 then set ψ1 = ||φφ11||V . Assume this is true for
some n and let φ1 , ..., φn , φn+1 be linearly independent. Apply the inductive hypothesis
to get span(φ1 , ..., φn ) = span(ψ1 , ..., ψn ) where ψ1 , ..., ψn are orthonormal. Set
sn (f ) =
n
X
(f, ψk )ψk
k=1
Clearly, φn+1 ∈
/ span(ψ1 , ..., ψn ). Set ϕ = φn+1 − sn (φn+1 ) 6= 0. Then for 1 ≤ j ≤ n we
have
(ϕ, ψj ) = (φn+1 , ψj ) −
n
X
(φn+1 , ψk )(ψk , ψj ) = (φn+1 , ψj ) − (φn+1 , ψj ) = 0
k=1
21
So ϕ is orthogonal to each ψj . Now let ψn+1 =
ϕ
||ϕ||V
and we’re finished.
We will now focus our attention on ways of representing elements in a Hilbert space. We
begin with a simple convergence result.
Proposition 1.4.3
Let H be a Hilbert space and (xk )∞
k=1
Pnbe an orthonormal sequence in H and let αk ∈ C
for
P∞all k.2 Then the sequence yn := 1 αk xk converges in || · ||H if and only if
1 |αk | < ∞.
Proof. Exercise for the reader.
If yn :=
∞
X
n
X
αk xk converges to a y ∈ H, we shall henceforth denote it’s limit y as
1
αk xk . Thus, convergence will always be refering to norm convergence unless it is
k=1
otherwise clear from the context.
Definition 1.4.4
An orthonormal basis for a Hilbert space H is an orthonormal set X = {xα }α∈A such
that the linear span, span(X), is dense in H.
We will now demonstrate the existance of such a basis for any Hilbert space but we first
make an observation that will prove to be useful.
Remark: Let’s consider an arbitrary collection of orthonormal vectors {xα }α∈A in H. If
we fix an x ∈ H and let Sn = {xα : |(x, xα )| ≥ n1 } then it turns out that Sn must be
finite. If not, let (xm ) be a sequence of distinct points in Sn . Then
1
≤ |(x, xm )|2 for all m
2
n
By Bessel’s Inequality, it follows that
m
m
X
m X 1
=
≤
|(x, xm )|2 ≤ ||x||2H
2
n2
n
k=1
k=1
22
for
S∞all m, a contradiction. Hence each Sn must be finite. But then, we must have
n=1 Sn is countable so (x, xα ) = 0 for all but a countable number of α’s in A.
♦
Theorem 1.4.5
Let H be a Hilbert space. Then the following hold:
(a) H has an orthonormal basis
(b) If K is a closed subspace of H then there is an orthonormal basis for H containing
an orthonormal basis for K.
Proof. Let U := {X : X ⊂ H is orthonormal}. Order U by set inclusion and select a
totally ordered subset U0 = {Xα : α ∈ A and Xα is an orthonormal set}. Let us put
[
Xα
X0 :=
α∈A
Observe firstly that X0 is an orthonormal set and a maximal element of U0 . Hence by
Zorn’s Lemma, there is a maximal orthonormal set in H. Denote this orthonormal set as
X := {xα : α ∈ A}. We will now demonstrate that X forms an orthonormal basis for H.
Suppose not. Select an x0 ∈ H\span(X). Let S := {α : (x0 , xα ) = 0} and
T := {α : (x0 , xα ) 6= 0}. Then T is countable by the previous remark so let {xk }∞
k=1 be
n
X
|(x0 , xk )|2 ≤ ||x0 ||2H so by
an enumeration of T . By Bessel’s Inequality, we have
Proposition 1.4.3, there is a y0 ∈ H with y0 =
∞
X
k=1
(x0 , xk )xk . Set y := x0 − y0 and put
k=1
sn :=
n
X
(x0 , xk )xk
k=1
For each α ∈ S, we have
(y, xα ) = (x0 , xα ) − lim
n
X
n→∞
23
k=1
(x0 , xk )(xk , xα ) = 0
Similarly, for each fixed j we have
(y, xj ) = (x0 , xj ) − lim
n→∞
n
X
!
(x0 , xk )xk , xj
= (x0 , xj ) − (x0 , xj ) = 0
k=1
So y ⊥ xα for all α ∈ A. If y = 0, then limn→∞ ||x0 − sn ||H = 0 so x0 ∈ span(X), a
contradiction. If y 6= 0, then X ∪ { ||y||y H } is a larger orthonormal set, contradicting the
maximality of X. Hence, span(X) = H.
The proof of part (b) is left as an exercise for the reader.
We will now present a list of equivalent conditions that are extremely useful, especially
in Fourier analysis.
Proposition 1.4.6
Let {xk }∞
k=1 be a countable orthonormal set of vectors in a Hilbert space H. Then the
following are equivalent:
(a) {xk }∞
k=1 forms an orthonormal basis for H.
(b) (Completeness) If x ∈ H and (x, xk ) = 0 for all k then x = 0.
P
(c) Every x ∈ H may be uniquely written as x = ∞
k=1 (x, xk )xk
P
2
(d) (Parseval’s Identity) ||x||2H = ∞
k=1 |(x, xk )| for all x ∈ H.
Proof. (a) =⇒ (b). Fix x ∈ H and suppose that x ⊥ xk for all k. Then clearly we get
x ⊥ s for all s ∈ span(X). By assumption, there is a sequence (sn ) in the linear span so
that sn → x in norm. Hence 0 = (sn , x) → (x, x) = ||x||2H so x = 0.
Pn
2
2
(b) =⇒ (c). Fix
an
x
∈
H.
By
Bessel’s
Inequality,
we
see
that
k=1 |(x, xk )| ≤ ||x||H
P∞
2
for all n. Thus
P∞ k=1 |(x, xk )| < ∞ so by Proposition 1.4.3, we see that there is a y ∈ H
so that y = k=1 (x, xk )xk . P
It is easy to see that (x − y) ⊥ xk for all k so by hypothesis,
x = y. Additionally, if x = ∞
k=1 ak xk for ak ∈ C for all k, then taking the inner product
of both sides gives ak = (x, xk ).
(c) =⇒ (d). It easily follows from the definition of convergence that
24
2
n
X
(x, xk )xk → ||x||2H as n → ∞
k=1
H
Now, apply Proposition 1.4.2(a) to obtain (d).
(d) =⇒ (a). Fix x ∈ H. Then for all n we see that
n
n
X
X
|(x, xk )|2 → 0 as n → ∞
(x, xk )xk = ||x||2H −
x −
k=1
k=1
H
This concludes the proof.
Note that in the above proof, we could have replaced our countable orthonormal basis
with an arbitrary one {xα }α∈A because for each x, we know that (x, xα ) = 0 for all but
countably many α ∈ A.
Proposition 1.4.7
Let H be a Hilbert space and suppose that H has a countable orthonormal basis. Then
any other orthonormal basis must be countable.
Proof. Suppose that we have a countable orthonormal basis for a Hilbert space H.
Enumerate our orthonormal basis by the sequence (xn ) in H. Suppose that
B = {xα : α ∈ A} is another orthonormal basis for H. For each n ∈ N, define
An := {α ∈ A : (xn , xα ) 6= 0}
S
Then each An must be countable so that ∞
1 An is countable. We now claim that
A=
∞
[
An
n=1
S
∞
Suppose, by contradiction, that ∞
1 An is strictly contained in A. Let β ∈ A\ ∪1 An .
Then (xn , xβ ) = 0 for all n. Since all of the xn ’s forms an orthonormal basis, we must
have xβ = 0 and the result then follows.
25
If we have a closed subspace of a Hilbert space, we can decompose the space into a
direct sum of the closed space with the set of all elements orthogonal to it. If M ⊂ H
where H is an inner product space, we set
M ⊥ := {x ∈ H : (x, y) = 0 for all y ∈ M }
By continuity of the inner product and linearity, it follows that M ⊥ is a closed subspace
of H. We now recall a definition from linear algebra.
If V is a vector space and W1 and W2 are subpaces, then we write V = W1 ⊕ W2 if
W1 ∩ W2 = {0} and v = w1 + w2 for w1 ∈ W1 and w2 ∈ W2 . We say that V is the direct
sum of W1 and W2 .
Theorem 1.4.8 (Orthogonal Decomposition of a Hilbert Space)
Let M ⊂ H be a (topologically) closed subspace of a Hilbert space H. Then
H = M ⊕ M ⊥ and if x = y + z where y ∈ M and z ∈ M ⊥ then z and y are unique and
are at a minimal distance to x.
Proof. We first observe that for any x, y ∈ H that
||x + y||2H + ||x − y||2H = 2||x||2H + 2||y||2H
This is called the parallelogram law and the proof is not difficult (see the exercises). Set
x ∈ H and let
δ := inf ||x − y||H
y∈M
We then know that there is a sequence yn ∈ M (∀n) so that ||x − yn ||H → δ. By the
parallelogram law, we have
||yn − ym ||2H + ||yn + ym − 2x||2H = 2 ||yn − x||2H + ||ym − x||2H
We then manipulate the above equation so that
1
||yn − ym ||2H = 2 ||yn − x||2H + ||ym − x||2H − 4|| (yn + ym ) − x||2H
2
26
It is important to note that because M is a subspace, we have (1/2)(yn + ym ) ∈ M .
Hence, we get
||ym − yn ||2H ≤ 2 ||yn − x||2H + ||ym − x||2H − 4δ 2
The above inequality tells us that (yn ) is Cauchy in || · ||H so it must converge to some
y ∈ M since M is closed. Hence ||x − y||H = δ.
Put z := x − y. Fix v ∈ M . We can always make (z, v) real by multiplying v by an
appropriate constant so without loss of generality, we assume this is real. Consider the
function
g(t) := ||tv + z||2H = t2 ||v||2H + 2t(z, v) + ||z||2H
for real valued t. g(t) attains it’s minimum value of δ 2 when t = 0 since
tv + z = x − (y − tv) but y − tv is in M for any t ∈ R. Hence g 0 (0) = 2(z, v) = 0 so that
z ∈ M ⊥.
If v ∈ M ∩ M ⊥ then (v, m) = 0 for all m ∈ M . Since v ∈ M , we get ||v||2H = (v, v) = 0
so that v = 0. Hence M ∩ M ⊥ = {0}. To get uniqueness, let x = y0 + z0 where y0 ∈ M
and z0 ∈ M ⊥ . Then y − y0 = z − z0 are elements in M ∩ M ⊥ so y = y0 and z = z0 .
Finally, we establish that z and y are the elements closest to x. Let y0 ∈ M . Then
||x − y0 ||2H = ||(x − y) + (y − y0 )||2H = ||(x − y)||2H + ||y − y0 ||2H ≥ ||x − y||2H
since x − y = z ∈ M ⊥ and y − y0 ∈ M (apply the Pythagorean Theorem). Notice that
we have equality iff y = y0 . Exactly the same argument can be used to show that z is
the element of M ⊥ closest to x.
The next theorem is an application of this orthogonal decomposition and allows us to
characterize all linear, continuous functionals on a Hilbert space H.
Theorem 1.4.9 (Riesz Representation Theorem)
Let H be a Hilbert space and let φ ∈ H ∗ . Then there exists a unique x0 ∈ H so that
φ(x) = (x, x0 ) with ||φ|| = ||x0 ||H .
27
Proof. Let M := ker(φ). Then M is a closed subspace of H and by the decomposition
theorem above, we get H = M ⊕ M ⊥ . If M ⊥ is trivial, then H = M so that φ = 0. In
this case, let x0 = 0. Assume, then, that M ⊥ is nontrivial. Choose z ∈ M ⊥ with
||z||H = 1 and define u := φ(x)z − φ(z)x for x ∈ H. Clearly u ∈ M . Then
0 = (u, z) = φ(x)||z||2H − φ(z)(x, z) = φ(x) − (x, φ(z)z)
so setting x0 = φ(z)z gives φ(x) = (x, x0 ). Moreover, ||x0 ||H = |φ(z)| ≤ ||φ|| and
|φ(x)| ≤ ||x||H · ||x0 ||H giving ||φ|| ≤ ||x0 ||H yielding equality. Finally, if (x, x1 ) = (x, x2 )
for all x ∈ H, we get (x, x1 − x2 ) = 0 for all x so that x1 = x2 giving uniqueness.
We may now completely characterize the dual of any Hilbert space. We have:
Corollary 1.4.10
Let H be a Hilbert space. Then H ∼
= H ∗ (we employ the symbol ∼
= to mean
2
isometrically isomorphic ).
Proof. Clearly each x0 ∈ H gives an element of the dual, namely (x, x0 ). The Riesz
Representation Theorem tells us that we may identify elements of the dual with
elements of H and that the identification is actually surjective. Moreover, the fact that
||φ|| = ||x0 ||H for φ ∈ H ∗ tells us that this relationship is an isometry.
Remark: It is somewhat customary to write H = H ∗ instead of H ∼
= H ∗ . Throughout
∗
∼
these notes, I shall employ the notation H = H .
♦
Corollary 1.4.11 (Reflexivity of Hilbert Space)
Every Hilbert space is reflexive.
Proof. Define S : H ∗ → H by S(φ) = x0 where x0 is the unique element in H so that
φ(x) = (x, x0 )H . It is obvious that H ∗ is a Banach space but we also contend that it is a
Hilbert space. Define (φ, ϕ) := (Sϕ, Sφ)H . It is clear that (·, ·) is an inner product on
H ∗ . This inner product induces the operator norm because
(φ, φ) = (Sφ, Sφ)H = ||Sφ||2H = ||x0 ||2H = ||φ||2
2
Two spaces are X and Y are isometrically isomorphic if there exists a bounded, linear, bijective
isometry T : X → Y .
28
Let ψ ∈ H ∗∗ . H ∗ is a Hilbert space as we have just seen so by the Riesz Representation
Theorem, there is a φ0 ∈ H ∗ so that ψ(φ) = (φ, φ0 ) = (Sφ0 , Sφ)H . But φ(x) = (x, x0 )H
where x0 = Sφ. Set z0 := Sφ0 . Then
ψ(φ) = (Sφ0 , Sφ)H = (z0 , x0 )H = φ(z0 ) = Jz0 (φ)
Thus the canonical embedding is surjective and we are finished.
We conclude this section by proving a slight generalization of the Riesz Representation
Theorem for real Hilbert spaces that has a wide variety of uses, particularly in the study
of partial differential equations3 .
Definition 1.4.12
Let X be a real or complex normed space. A bilinear form is a map B : X × X → R
that is linear in both variables. A bilinear form is continuous (or bounded) if there is a
constant M ≥ 0 so that
|B(u, v)| ≤ M ||u||X ||v||X
A bilinear form is called coercive on a subspace Y ⊆ X if there is a C > 0 so that
B(v, v) ≥ C||v||2X
Finally, a bilinear form is called symmetric if B(x, y) = B(y, x) for all x, y ∈ X.
At this point, we will attempt to demonstrate that if φ ∈ H ∗ and B is a bounded,
coercive bilinear form, then there exists a unique u0 ∈ H so that φ(v) = B(v, u0 ) for all
v ∈ H. It is not difficult to prove that a unique solution to this problem exists when B
is symmetric.
This is because B(u, v) forms an inner product in this case (verify) in which H is
complete (again, verify). The main difficulty in solving this problem occurs when our
bilinear form fails to be symmetric. This, however, is of little consequence in the long
run due to the following result.
3
As someone with an interest in PDEs, it seems almost immoral to leave out the Lax-Milgram theorem
whilst discussing Hilbert spaces.
29
Theorem 1.4.13 (Lax-Milgram)
Let H be a real Hilbert space and B be a continuous, coercive, bilinear form on H. Fix
a φ ∈ H ∗ . Then there is a unique u0 ∈ H so that B(u0 , v) = φ(v) for all v ∈ H.
Proof. If u ∈ H, define a functional Tu (v) = B(u, v). Then for each u, Tu ∈ H ∗ (verify).
Then by the Riesz Representation Theorem, there is a u0 ∈ H so that
Tu (v) = B(u, v) = (u0 , v) for all v ∈ H. For each u ∈ H, define an operator T : H → H
by T (u) := u0 . Then by definition, Tu (v) = (T u, v) = B(u, v) for all v ∈ H. We will now
demonstrate that T ∈ L(H). Firstly, observe that when c1 , c2 ∈ R and u1 , u2 ∈ H, we get
(T (c1 u1 + c2 u2 ), v) = B(c1 u1 + c2 u2 , v) = c1 B(u1 , v) + c2 B(u2 , v)
= c1 (T u1 , v) + c2 (T u2 , v) = (c1 T u1 + c2 T u2 , v)
This holds for all v ∈ H to give linearity. By definition, we get
||T u||2H = B(u, T u) ≤ M ||u||H ||T u||H
for some M ≥ 0 so that ||T u||H ≤ M ||u||H and T is bounded. Furthermore, by
coercivity we get
C||u||2H ≤ B(u, u) = (T u, u) ≤ ||T u||H ||u||H
so that C||u||H ≤ ||T u||H (here, C is the coercivity constant). This obviously tells us
that T is injective. Moreover, if (yn ) is a sequence in R(T ) (the range of T ) where
yn → y0 in norm, then there is a sequence (xn ) in H so that yn = T (xn ). Since yn
converges, it is Cauchy. Then for any > 0 there is an N0 so that
C||xn − xm ||H ≤ ||T (xn ) − T (xm )||H < when n, m ≥ N0 . That is, (xn ) is Cauchy in H so xn → x0 for some x0 ∈ H. Then
yn → T (x0 ) so that y0 = T (x0 ) and R(T ) is a closed subspace of H. By the orthogonal
decomposition theorem, H = R(T ) ⊕ R(T )⊥ . We claim that R(T ) = H. To see this, let
u0 ∈ R(T )⊥ . Then C||u0 ||2H ≤ B(u0 , u0 ) = (T u0 , u0 ) = 0 so that R(T )⊥ = {0}.
By the Riesz Representation Theorem, there is a v0 ∈ H so that φ(v) = (v0 , v) for all
v ∈ H. Since R(T ) = H, we can find a u0 ∈ H so that T (u0 ) = v0 . Then
30
B(u0 , v) = (T u0 , v) = (v0 , v) = φ(v)
It remains to show that this u0 is unique. Suppose that there are two solutions u1 and
u2 . Then B(u1 , v) = B(u2 , v) for all v ∈ H. Then B(u1 − u2 , v) = 0 for all v ∈ H. In
particular,
C||u1 − u2 ||2H ≤ B(u1 − u2 , u1 − u2 ) = 0
so that u1 = u2 . This completes the proof.
We remark that the solution to the variational problem satisfies
||u0 ||H ≤
1
||φ||
C
We leave this trivial estimate as an exercise for the interested reader.
1.5
Exercises
All normed linear spaces are over C unless otherwise specified.
1. Let (X, ρ) be a metric space. Prove that (X, σ) is a metric space where
σ(x, y) :=
ρ(x, y)
1 + ρ(x, y)
In addition, show that (X, ρ) is complete if and only if (X, σ) is complete.
2. Exhibit a concrete example of a metric space whose metric does not induce a norm.
3. On a normed space X, prove that the norm || · ||X is a continuous function.
31
In exercises 4-6, you will consider the space of k-times continuously differentiable
functions f : [a, b] → R. We shall denote this space as C k [a, b] where k ∈ Z+ . Here,
C 0 [a, b] = C[a, b] simply denotes the space of continuous real valued functions on [a, b].
4. Consider C[a, b] and let
||f ||C[a,b] := max |f (x)|
a≤x≤b
(a) Prove that || · ||C[a,b] defines a norm and that C[a, b] equipped with this norm
forms a separable Banach space.
(b) Show that this norm does not induce an inner-product so that C[a, b] cannot
and the
be a Hilbert space. Suggestion: Consider f (x) ≡ 1 and g(x) = x−a
b−a
parallelogram law for inner-product spaces (see Exercise 10).
5. Consider C 1 [a, b]. Show that
||f ||C 1 [a,b] := ||f ||C[a,b] + ||f 0 ||C[a,b]
defines a norm on C 1 [a, b] and that C 1 [a, b] is a Banach space if equipped with this
norm. Here || · ||C[a,b] is the norm given in the previous exercise. How might we
generalize this for C k [a, b]?
6. Consider C[a, b] and define
Z
||f || :=
b
|f (x)|dx
a
Prove that this defines a norm on C[a, b] but that C[a, b] with this norm does not
form a Banach space.
32
7. Let `∞ be the space of all bounded sequences and put
||sn ||`∞ := sup |sn |
n∈N
Prove that this is a non-separable Banach space. Hint: Consider the space of all
sequences with either a 0 or 1 as an entry.
8. Prove from scratch the Hahn-Banach theorem for finite dimensional real normed
spaces without using Zorn’s Lemma.
9. Let H be an inner product space and let x1 , x2 , ..., xn ∈ H be nonzero orthonormal
vectors. Prove that they are linearly independent.
10. Let H be an inner-product space and let x, y ∈ H. Prove that
||x + y||2H + ||x − y||2H = 2 · ||x||2H + 2 · ||y||2H
This is called the parallelogram law for inner-product spaces.
11. P
Let `2 be the space of all complex sequences sn with the property that
∞
2
2
n=1 |sn | < ∞. Show that ` is a vector space and an inner product space with
inner product
(zn , wn )`2 :=
∞
X
zn wn
n=1
In addition, show that `2 is a separable Hilbert space.
12. In this exercise, you will demonstrate that every finite dimensional normed space is
complete regardless of the norm imposed on the space. There are several different
methods of acomplishing this task and you will explore two such ways here.
(a) Prove this by using norm equivalence.
33
(b) Prove this by induction on dimension. To do this, you may find it helpful to
utilize the example presented after the proof of the Hahn-Banach Theorem.
13. Prove Proposition 1.4.3.
14. Prove that if H is a Hilbert space and K ⊂ H is a closed subspace, then H has an
orthonormal basis consisting of an orthonormal basis for K.
15. Prove the Hahn-Banach Theorem for normed spaces.
16. Let X be a normed space and let x0 ∈ X with x0 6= 0. Then there exists a φ ∈ X ∗
such that ||φ|| = 1 and φ(x0 ) = ||x0 ||X .
17. Let X and Y be normed spaces and T ∈ L(X, Y ). Prove that T ∗ ∈ L(Y ∗ , X ∗ )
with ||T || = ||T ∗ ||.
18. Prove that a Hilbert space H is separable if and only if H has a countable
orthonormal basis.
19. Let X be a real or complex normed space. Prove that the canonical embedding of
X into X ∗∗ is a continuous, injective isometry.
20. Prove the conclusion of the Lax-Milgram Theorem when we make the stronger
assumption that B(x, y) is symmetric.
34
2
The Cornerstone Theorems
In functional analysis, there are a handful of results that are foundational to the subject.
We have already seen one such result in the previous section: the Hahn-Banach
Theorem. In this section, we prove the remaining results.
We first elaborate on notation. If X is a normed space, let
BX (x0 , r) := {x ∈ X : ||x − x0 ||X < r}
be an open ball centered at x0 with radius r. If the underlying space X is understood,
we will write B(x0 , r). If x0 = 0 and r = 1, we will simply write B or BX .
2.1
The Baire Category Theorem
The first result that we will demonstrate will be needed later and it is not difficult to
prove.
Theorem 2.1.0 (Baire Category Theorem)
Let (X, ρ) be a complete metric space and let Vn be open, dense sets in X. Then
∞
\
Vn
n=1
is dense in X.
Proof. Consider an arbitrary open ball B(x0 , r) := {x ∈ X : ρ(x, x0 ) < r} in X. We will
show that
∞
\
!
Vn
∩ B(x0 , r) 6= n=1
Now V1 is open and dense so there is a ball B(x1 , 2r1 ) ⊂ V1 ∩ B(x0 , r) with r1 < 1. In a
similar manner, V2 is open and dense so there is a ball B(x2 , 2r2 ) ⊂ V2 ∩ B(x1 , r1 ) with
r2 < 21 . In general, Vj is open and dense so there is a ball B(xj , 2rj ) ⊂ Vj ∩ B(xj−1 , rj−1 )
with rj < 1j . If n > m, then ρ(xn , xm ) < rm < m1 and if n ≤ m then ρ(xn , xm ) < rn < n1
so that (xn ) is Cauchy. Thus, there is a y ∈ X so that xn → y.
35
Note that y ∈ B(xn , rn ) for all n because B(xn , rn ) contains all but the first n − 1 terms
of the sequence. But then
!
n
\
y ∈ B(xn , rn ) ⊂ B(xn , 2rn ) ⊂
Vk ∩ B(x0 , r) ∀n
k=1
We may interpret this in the following way: a set is called nowhere dense if int A = .
The Baire Category Theorem then tells us that a complete metric
S∞space cannot be
written as a countable
union
of
nowhere
dense
sets.
For
if
X
=
1 Ak where int Ak = T∞
c
c
for all k, then = 1 (Ak ) . But each (Ak ) is open and dense which is a contradiction.
Definition 2.1.1
Let (X, T ) be a topological space. A set in X is said to be of first category if it can be
written as a countable union of nowhere dense sets. If a set is not first category, it is
said to be of second category.
Thus, the Baire Category Theorem says the following: Every complete metric space is of
second category.
2.2
The Open Mapping Theorem
Most of the cornerstone theorems are simple applications of the Baire Category
Theorem. Yet, these theorems are so important and so widely used that we devote an
entire section to them. We begin with the open mapping theorem.
If f : X → Y where X and Y are topological spaces, we say that f is an open (closed)
mapping if f (U ) is open (closed) in Y whenever U is open (closed) in X. From basic
complex function theory, we know that every nonconstant holomorphic function is an
open mapping. We prove a similar result for Banach Spaces but first, we need a lemma.
Lemma 2.2.0
Let T ∈ L(X, Y ) where X is a Banach space and Y is a normed space. Assume that for
some > 0 we get BY ⊆ T (BX ). Then T (BX ) ⊆ 2T (BX ).
36
Proof. Let y ∈ T (BX ). We will find an x ∈ 2BX so that T x = y. There exists an
x1 ∈ BX so that ||T x1 − y||Y < /2. Then y − T x1 ∈ 2 BY ⊆ 12 T (BX ) by assumption.
Likewise, there is an x2 ∈ 21 BX so that
||y − T x1 − T x2 ||Y <
22
so that y − T x1 − T x2 ∈
1
BY ⊆ 2 T (BX )
2
2
2
1
Continue this construction inductively to obtain a sequence (xn ) with ||xn ||X < n−1 for
2
∞
n
X
X
all n. Thus
||xk ||X < ∞ so that sn :=
xk → x ∈ X in norm.
k=1
Note also that ||x||X ≤
k=1
P
||xk ||X ≤ 2. Hence
y − T
!
xk < n → 0
2
k=1
n
X
X
as n → ∞. Thus y = T x.
Theorem 2.2.1 (Open Mapping Theorem)
Let X and Y be Banach spaces. If T : X → Y is a bounded, linear, surjective operator,
then T is open.
Proof. Observe that
Y = T (X) =
∞
[
T (kBX ) =
k=1
∞
[
T (kBX )
k=1
By the Baire Category Theorem, there exists an n so that int T (nBX ) 6= . Thus, there
is an open W ⊂ T (nBX ). Choose y0 ∈ W and η > 0 so that y0 + y ∈ W whenever
||y||Y < η. Then, there are sequences (aj ) and (bj ) in nBX so that T aj → y0 and
T bj → y0 + y as j → ∞. Setting cj := bj − aj gives ||cj ||X < 2n with T cj → y. Thus, for
all y ∈ ηBY , there is a sequence (cj ) in 2nBX so that T cj → y. In other words,
ηBY ⊆ T (2nBX ). Then, we may choose an appropriate > 0 so that BY ⊆ T (BX ).
37
Let U be an arbitrary open set in X. For x ∈ U , we have T x ∈ T (U ). Choose a δ > 0 so
that x + δBX ⊆ U and T x + δT (BX ) = T (x + δBX ) ⊂ T (U ). Then by the Lemma
above, we have
Tx +
δ
δ
BY ⊆ T x + T (BX ) ⊆ T x + δT (BX ) ⊆ T (U )
2
2
This is what we wished to show.
We obtain the following useful result as an easy corollary.
Corollary 2.2.2 (Inverse Mapping Theorem)
Let X and Y be Banach spaces where T : X → Y is a bounded, linear bijection. Then
T −1 ∈ L(Y, X).
Proof. We know an inverse exists and it is not terribly difficult to check to see that T −1
is linear. Continuity of T −1 is equivalent to openness of T so we are finished.
2.3
Uniformly Bounded Operators and the Closed Graph
Theorem
In this section, we prove the remaining two cornerstone theorems: the uniform
boundedness principle and the closed graph theorem. We begin with the uniform
boundedness principle.
Theorem 2.3.0 (Uniform Boundedness Principle/Banach-Steinhaus Theorem)
Let X be a Banach space and Y be a normed space. Let {Tα } be a collection of
bounded, linear operators from X to Y . Then
sup ||Tα x||Y < ∞ ∀x ∈ X implies sup ||Tα || < ∞
α
α
Proof. Write
38
X=
∞
[
Ωn where Ωn := {x ∈ X : sup ||Tα x||Y ≤ n}
α
n=1
=
\
{x ∈ X : ||Tα x||Y ≤ n}
α
Clearly each Ωn is closed. By the Baire Category Theorem, there is an m so that
int Ωm 6= . Then there is an x0 ∈ X and an > 0 so that B(x0 , ) ⊆ Ωm . For all y ∈ X
with ||y||X = 1, we get x0 + (/2)y ∈ B(x0 , ) so that
||Tα y||Y − ||Tα x||Y ≤ ||Tα (x0 + (/2)y)||Y ≤ m ∀α
2
But this implies that
2
||Tα || ≤
m + sup ||Tβ x0 ||Y
< ∞ ∀α
β
We now move on to the closed graph theorem. Firstly, note that if X and Y are normed
spaces, then we may make the product space X × Y a normed space via
||(x, y)||p :=

 (||x||pX + ||y||pY )1/p 1 ≤ p < ∞

max (||x||X , ||y||Y ) p = ∞
It does not matter which norm we choose because they each induce the same topology
(verify). Our last cornerstone theorem is useful when attempting to show that an
operator is bounded. First, we need a definition.
Definition 2.3.1
Let T : X → Y be an operator. Define the graph of T to be
G(T ) := {(x, T x) : x ∈ X} ⊆ X × Y
39
Theorem 2.3.2 (Closed Graph Theorem)
Let X and Y be Banach spaces and suppose that T : X → Y is a linear operator. If the
graph of T , G(T ), is a closed subset of X × Y , then T is bounded.
Proof. Let πX : X × Y → X be the projection map πX (x, y) = x onto X. Similarly, let
πY be the projection map from X × Y onto Y . Now
||πX (x, y)||X = ||x||X ≤ ||(x, y)||2
so that πX is bounded and linear. Similarly for πY . Consider pX := πX |G where
G = G(T ) is the graph. Since G is closed, it is a Banach space. Clearly, pX is bijective
and since it is continuous, the inverse mapping theorem tells us that p−1
X is a bounded,
−1
−1
linear operator. But pX (x) = (x, T x). Hence T = πY ◦ pX so that T is continuous.
2.4
Exercises
1. Let (X, T ) be a topological space. Show that a set A ⊂ X is nowhere dense if and
only if (A)c is dense in X.
2. Suppose that T : X → Y is linear and that φ ◦ T ∈ X ∗ for all φ ∈ Y ∗ . Prove that
T is bounded.
3. Suppose that X and Y are Banach spaces and that (Tn ) is a sequence in L(X, Y )
so that (Tn x) converges in Y for all x ∈ X. If T x := lim Tn x, then prove that T is
a bounded, linear operator.
4. Let X and Y be normed spaces. Show that each ||(x, y)||p are equivalent norms on
X × Y for 1 ≤ p ≤ ∞. Additionally, if X and Y are Banach spaces, show that
X × Y is a Banach space in ||(x, y)||p for 1 ≤ p ≤ ∞.
5. Let X be a normed space that is complete in norms || · ||1 and || · ||2 with
||x||1 ≤ ||x||2 for all x ∈ X. Show that || · ||1 and || · ||2 are equivalent.
6. Let X and Y be Banach spaces with X ∼
= Y . Prove that X ∗ ∼
= Y ∗ (Suggestion:
Consider the adjoint of our isometry from X to Y ).
40
3
Weak and Weak* Topologies
In the last few sections, we have primarily concerned ourselves with the norm topology
on a normed space X. However, there are many times when it may be more convinient
to consider other topologies on X aside from the norm topology. In this section, we shall
introduce the weak and weak* topologies on a normed vector space. For our purposes,
weak topologies can tell us quite a few things about reflexivity. It is assumed that the
reader is familiar with the basics of general point-set topology.
3.1
Basic Ideas
We first introduce some notation. All vector spaces we consider will either be over R or
C. We shall, therefore, employ the notation F to denote either R or C. If it is necessary
to make the field explicit, we shall do so.
Definition 3.1.0
Let X be a vector space over F. A topological vector space (TVS or linear topology) is a
Hausdorff topology on X so that addition and scalar multiplication are continuous on
X × X and F × X respectively.
Some authors require that a TVS be T1 rather than T2 but most linear topologies
encountered in practive are T2 .
Definition 3.1.1
A set V ⊂ X is called balanced (or circled) if for any x ∈ V and λ ∈ F with |λ| ≤ 1, we
get λx ∈ V .
The following is easy but is quite useful:
Proposition 3.1.2
Let X be a topological vector space. Then every neighborhood V of 0 contains a
balanced neighborhood of 0.
Proof. Let V be a neighborhood of 0. By continuity of (α, x) 7→ αx at (0, 0) we are
guaranteed the existance of a neighborhood W of 0 and some λ0 > 0 so that |λ| ≤ λ0
and x ∈ W implies λx ∈ V . Set
41
U :=
[
λW
|λ|≤λ0
It is clear that U is a neighborhood of 0 and U ⊂ V . Also, if α ∈ R (or C) with |α| ≤ 1
and y ∈ U , then y = λ1 y1 where y1 ∈ W and |λ1 | ≤ λ0 . Then, αy = αλ1 y1 ∈ U as
|αλ1 | ≤ λ0 .
There is much more we can say about linear topologies but we withold these things until
or unless we need them. The concept of a weak topology is actually fairly general so we
here present the general definition.
Definition 3.1.3 (Weak Topologies)
Let X be a set and (Xα , Tα )α∈A a collection of topological spaces. For each α ∈ A, let
fα : X → Xα and set F := {fα : α ∈ A}. The weak topology on X induced by F is the
weakest topology on X making each f ∈ F continuous.
We will now present a fairly standard illustration of the weak topology in a general
setting.
Example (Product Topologies)
Let (Xα , Tα ) be a collection
of topological spaces. Normally, we define the product
Q
topology Tw on X
Q:= α Xα in the following way: let B be a base for Tw consisting of
sets of the form α Uα where each Uα is open in Xα and Uα = Xα for all but finitely
many α ∈ A. Note that each
Y
α
Uα =
n
\
πα−1k (Uαk )
k=1
where each πα : X → Xα is a projection map and Uα = Xα except when α = αj for some
1 ≤ j ≤ n. Thus, the weak topology is the topology whose subbase consists of sets of the
form πα−1 (Uα ).
It is clear that in this topology, each πα is continuous. However, this topology is the
weakest topology that makes the projection maps continuous. That is, if T is another
topology that makes the projection maps continuous, then clearly Tw ⊂ T .
42
It is important to realize that this topology is fundamentally Q
different than the box
topology. The box topology has as a base all sets of the form α Uα where each Uα is
open (i.e. we omit the condition that all but finitely many must be the factor space).
The box topology does, however, coincide with the product topology when the product
space has only finitely many factors.
♦
In general, it is easily verified that the weak topology is a topology whose subbase is
given by C := {fα−1 (Uα ) : Uα ∈ Tα , α ∈ A}.
3.2
Weak Topologies on Normed Spaces
Before we begin our discussion of weak and weak* topologies, it is necessary to recall a
few things. For each x ∈ X, let Jx = Jx be the canonical embedding of X into X ∗∗
(recall that (Jx)(φ) := φ(x) for all φ ∈ X ∗ ). Then we may identify X with J(X) ⊂ X ∗∗
so that X ,→ J(X) ⊂ X ∗∗ .
Definition 3.2.0
If X is a normed space, let σ(X, X ∗ ) be the weak topology on X induced by X ∗ . We
denote by σ(X ∗ , X) the weak* topology, the weakest topology so that the set of all
canonical embeddings {Jx : x ∈ X} are continuous on X ∗ .
Let > 0 and F ⊂ X ∗ be finite. Define
V,F (x0 ) := {x ∈ X : |φ(x) − φ(x0 )| < , φ ∈ F }
We then leave it to the reader to verify that an open neighborhood base about x0 ∈ X is
given by
Bx0 := {V,F (x0 ) : > 0, F ⊂ X ∗ is finite}
Likewise, if
∗
V,F
(φ0 ) := {φ ∈ X ∗ : |φ(x) − φ0 (x)| < , x ∈ F }
where F ⊂ X is finite and > 0 then an open neighborhood base about φ0 ∈ X ∗ is given
by
43
∗
Bφ∗ 0 := {V,F
(φ0 ) : > 0, F ⊂ X is finite}
We now deduce an elementary but important result concerning convergence in the weak
and weak* topology.
Proposition 3.2.1
A net (xλ ) in X converges to x0 in σ(X, X ∗ ) if and only if φ(xλ ) → φ(x0 ) for all φ ∈ X ∗ .
Likewise, a net (φλ ) in X ∗ converges to φ0 in the weak* topology if and only if
φλ (x) → φ0 (x) for all x ∈ X.
Proof. Suppose that xλ → x0 weakly. Then for all > 0, V,φ (x0 ) is an open
neighborhood of x0 (note that in this case F = {φ} so we omit the set brackets). Then,
there is a λ0 so that xλ ∈ V,φ (x0 ) when λ λ0 . That is, |φ(xλ ) − φ(x0 )| < when
λ λ0 giving the result. Conversely, let V be an open neighborhood of x0 . Then there
is an > 0 and a finite F ⊂ X ∗ so that V,F ⊂ V . But φ(xλ ) → φ(x0 ) for all φ ∈ X ∗ by
assumption. Then, there exists λj so that |φj (xλ ) − φj (x0 )| < when λ λj and
F = {φ1 , ..., φn }. Since each λj is a member of a directed set, we may take the maximal
element of all such λj (1 ≤ j ≤ n), call it λ0 . Then xλ ∈ V,F (x0 ) ⊂ V for all λ λ0 .
That is, xλ → x0 and we’re finished. The proof for weak* convergence is similar.
If (xn ) is a sequence in a normed space X, we say that xn converges strongly to x0 if it is
norm convergent, i.e. ||xn − x0 ||X → 0 as n → ∞. It is easily verified that strong
sequential convergence implies weak sequential convergence. The following example
illustrates that the converse is untrue.
Example (Weak Convergence)
If ψ ∈ L2 (0, 2π)∗ then by the Riesz Representation Theorem, there is a g ∈ L2 (0, 2π) so
that
Z 2π
ψ(f ) =
f (x)g(x)dx
0
Consider the sequence fn (x) := cos(nx). By the Riemann-Lebesgue Lemma,
Z 2π
ψ(cos(nx)) =
cos(nx)g(x)dx → 0 as n → ∞
0
44
Z
so fn → 0 weakly. However, one may easily verify that
2π
cos2 (nx)dx = π for all n so
0
that fn does not converge to 0 in L2 .
♦
Weak convergence, as the name seems to imply, is hence generally weaker than strong
convergence. Even so, we may nevertheless say some nice things about weakly
convergent sequences.
Proposition 3.2.2
Let (xn ) be a sequence in a normed space X that converges weakly to x0 . Then:
(a) (xn ) is bounded in norm.
(b) ||x0 ||X ≤ lim inf ||xn ||X
n→∞
Proof. For (a), we without loss of generality assume that x0 = 0. Apply the canonical
embedding from X to X ∗∗ to each xn . Then Jxn (φ) = φ(xn ) where φ ∈ X ∗ so that
|Jxn (φ)| ≤ M where M = M (φ). By the uniform boundedness principle, ||Jxn || ≤ C for
all n. But ||Jxn || = ||xn ||X giving the result. For (b), use the Hahn-Banach Theorem to
choose a φ ∈ X ∗ so that φ(x0 ) = ||x0 ||X and ||φ|| = 1. Then |φ(xn )| ≤ ||xn ||X for all n.
Taking the limit infimum of both sides gives the desired result.
Remark on Notation: If a sequence (xn ) in X converges weakly to x0 , we will write
w
xn → x0 . Likewise, if a sequence (φn ) in X ∗ converges to φ0 in the weak* topology, we
w∗
will write φn → φ0 .
♦
We will now turn our attention to some of the more topological properties of weak and
weak* topologies.
Proposition 3.2.3
If X is a normed space, then σ(X, X ∗ ) and σ(X ∗ , X) are Hausdorff, topological vector
spaces.
Proof. We show that addition is continuous and continuity of scalar multiplication will
follow similarly. Let (xλ , yλ ) be a net that converges to (x0 , y0 ) in X × X. Then by
considering the projection maps, xλ → x0 and yλ → y0 in X. Then
45
φ(xλ + yλ ) = φ(xλ ) + φ(yλ ) → φ(x0 ) + φ(y0 ) = φ(x0 + y0 )
That is, xλ + yλ → x0 + y0 weakly so addition is continuous.
To see that it is T2 , let x, y ∈ X with x 6= y. By the Hahn-Banach Theorem, there is a
φ ∈ X ∗ so that φ(x − y) > 0 or φ(x) 6= φ(y). As the real line with the standard topology
is T2 , we may find disjoint open intervals I1 and I2 in R containing x and y respectively.
Then φ−1 (I1 ) and φ−1 (I2 ) are open and do the trick. The proof for σ(X ∗ , X) is
analagous and left to the reader.
We will now prepare to prove our first major result in this section. Our goal is to
demonstrate that the weak topology of a normed space is metrizable if and only if X is
finite dimensional. The next theorem demonstrates one direction of this.
Theorem 3.2.4 (Uniquness of T2 Topologies)
Every finite dimensional normed space admits a unique T2 linear topology.
Proof. Let T||·|| be the norm topology on X. Because the norm topology is T2 and linear,
it suffices to show that any other T2 linear topology T must be the norm topology. Fix a
basis B = {x1 , ..., xn } and without loss of generality (by norm equivalence), assume that
||x|| =
n
X
|cj | where x =
j=1
n
X
cj xj
j=1
P
(j)
(j)
Let (xλ ) be a net in X with xλ → 0. Then xλ = nj=1 cλ xj so that cλ → 0 in R (or C).
Consider the identity map I : (X, T||·|| ) → (X, T ). Then
Ixλ = xλ =
n
X
(j)
in T
cλ xj → =
j=1
n
X
0xj = 0
j=1
because T is a linear topology. Hence I is continuous and T ⊂ T||·|| . Now, consider the
unit ball about the origin in X, B = B(0, 1). Let S := {x ∈ X : ||x|| = 1} be the unit
sphere in X. Now X is finite dimensional so S is T||·|| -compact. But by the inclusion
T ⊂ T||·|| , we see that S is also T -compact. Because X is T2 , we see that S is T -closed4 .
4
Recall that a compact subset of a T2 space is closed.
46
Now 0 ∈
/ S = S (where the closure is the T -closure) so we may find a T -open set V with
0 ∈ V so that V ∩ S = . We may, without loss of generality, assume that V is balanced
(if not, choose a balanced subset). We claim that V ⊂ B(0, 1). If not, there is an x ∈ V
with ||x|| ≥ 1. But then x/||x|| ∈ V ∩ S, a contradiction. Thus, B(0, 1) is a T
neighborhood of 0.
If U is a T||·|| -open set and x0 ∈ U , then there is an r > 0 so that B(x0 , r) ⊂ U . But
rV + x0 ⊂ B(x0 , r) ⊂ U where rV + x0 is T -open. Hence U is T -open so that T||·|| ⊂ T
yielding equality.
We are almost ready to prove our metrizability theorem but first, we need a lemma.
Lemma 3.2.5
Let φ1 , φ2 , ...., φn ∈ X ∗ where X is a normed space. Then φ ∈ span(φ1 , ..., φn ) if and only
if
n
\
ker φj ⊂ ker φ
j=1
T
Proof. The direction =⇒ is trivial. Assume, then, that nj=1 ker φj ⊂ ker φ. Define
Ψ : X → Rn by Ψ(x) := (φ1 (x), ..., φn (x)). On the range Ψ(X), set ϕ : Ψ(X) → R by
ϕ(φ1 (x), ..., φn (x)) := φ(x). If yj = (φ1 (xj ), ..., φn (xj )) for j = 1, 2 so that y1 = y2 then
y1 − y2 = 0 so that x1 − x2 ∈ ker φj for all j. By hypothesis, φ(x1 − x2 ) = 0 or
φ(x1 ) = φ(x2 ). Thus ϕ is well defined.
Clearly Ψ(X) ⊂ Rn is a subspace. Then ϕ extends to a ϕ
e on all of Rn (in a linear
algebra sense by tweaking basis values). But
there is a λ = (λ1 , ..., λn ) ∈ Rn so that
Pthen
n
n
ϕ(y)
e
= (λ, y) for all y ∈ R (here, (x, y) = 1 xk yk is the inner product on Rn ). This
holds, in particular, when y ∈ Ψ(X).
We are now ready to demonstrate our theorem.
Theorem 3.2.6 (Metrizability of the Weak Topology)
Let X be a normed space. Then the weak topology on X is metrizable if and only if X
is finite dimensional.
47
Proof. Suppose that dim(X) < ∞. Because σ(X, X ∗ ) is a T2 linear topology, the
uniqueness theorem for T2 linear topologies tells us that σ(X, X ∗ ) = T||·|| . Hence, the
weak topology is metrizable.
Now, assume that the weak topology is metrizable and that X is not finite dimensional.
Then the weak topology is first countable. Let B0 := {Bi : i ∈ N} be a countable
neighborhood base at 0. Then there is an 1 > 0 and finite F1 ⊂ X ∗ so that
V1 ,F1 (0) ⊂ B1 . We may assume that 1 = 1 by replacing each φ ∈ F1 by (1/1 )φ. Let
V1 := V1 ,F1 (0). Likewise, there is an 2 > 0 and finite F2 ⊂ X ∗ so that V2 ,F2 (0) ⊂ B2 .
Again, we may assume that 2 = 1. Put V2 := V2 ,F1 ∪F2 (0) ⊂ V2 ,F2 (0). Continue this
process inductively to obtain a countable collection of open neighborhoods {Vn : n ∈ N}.
By construction, the collection of all such Vn forms a countable neighborhood base at 0
with Vn+1 ⊂ Vn . Enumerate the union of all such Vk by a sequence (φk ) by first
enumerating the elements in F1 , F2 , ... and so forth.
n
\
We first claim that
ker φj is nontrivial for all n. For if it were trivial for some m, then
j=1
m
\
ker φj = {0} ⊂ ker φ for any φ ∈ X ∗
j=1
By our lemma, φ ∈ span(φ1 , ..., φm ) for all φ ∈ X ∗ . Thus, X ∗ is finite dimensional so
that X ∗∗ is finite dimensional. But X may be embedded as a subspace of X ∗∗ implying
n
\
that X is finite dimensional, a contradiction. Thus,
ker φj 6= {0} for all n.
j=1
For each n, choose a nonzero
xn ∈
sn
\
ker φj
j=1
where sn denotes the position in the sequince (φk ) containg all elements up to Fn .
Assume also, without loss of generality, that ||xn ||X = n for all n. Then xn ∈ Vn for all
n. Let V be any weakly open neighborhood of 0. Then there is an m so that Vm ⊂ V .
But this implies that xn ∈ V for n ≥ m so that xn → 0 weakly. But ||xn ||X → ∞ as
n → ∞, a contradiction because (xn ) must be bounded in norm.
48
Note that this theorem did not require any condition of completeness (this property is
somewhat superfluous anyway since all finite dimensional normed spaces are Banach
spaces). Our goal in the later sections will be to show that in a general Banach space X,
weak compactness behaves like sequential compactness. This is a remarkable result
because the weak topology is almost never metrizable in practice as our theorem here
shows.
We now prove a theorem concerning metrizability in the weak* topology.
Theorem 3.2.7
Let X be a separable normed space and K ⊂ X ∗ be weak* compact. Then K is
metrizable in the weak* topology.
Proof. Let D := {xn : n ∈ N} be a countable dense subset of X and define the following:
ρ(φ, ϕ) :=
∞
X
2−k min(1, |φ(xk ) − ϕ(xk )|) for all φ, ϕ ∈ X ∗
k=1
The reader may verify that ρ is indeed a metric (one uses the fact that D is norm dense
in X to show this). Let Tρ be the topology generated by ρ. Firstly, we demonstrate that
Tρ ⊂ σ(X ∗ , X). To demonstrate this, we will show that every Tρ closed set is also weak*
closed.
w∗
Let A be closed in Tρ and let (φλ ) be a net in A with φλ → φ0 . It is necessary to show
that φ0 ∈ A. For any > 0 and j ∈ N, consider the open neighborhood of φ0 :
∗
Vj,
:= {φ ∈ X ∗ : |φ(xn ) − φ0 (xn )| < , 1 ≤ n ≤ j}
∗
Let > 0 and fix j ∈ N. Then there is a λ0 ∈ Λ so that φλ ∈ Vj,
when λ λ0 . Thus
ρ(φλ , φ0 ) < j
X
k=1
−k
2
+
∞
X
−k
2
k=j+1
= ( − 1)
j
X
2−k + 1 < k=1
whenever λ λ0 . But this, in turn, implies that φλ → φ0 in ρ. But A is closed in Tρ so
that φ0 ∈ A.
The reverse inclusion will follow given the validity of the following claim: if T1 and T2
are two topologies on K with T1 ⊂ T2 , K is T2 -compact, and T1 is Hausdorff, then
49
T2 ⊂ T1 . Let A be T2 closed and let xλ → x0 in T1 where xλ ∈ A for all λ. Since A ⊂ K
and K is T2 compact, it follows that A is T2 compact as well. Thus, there is a subnet
(xλµ ) with xλµ → x1 ∈ A in T2 . But net convergence in T2 implies net convergence in T1
so that xλµ → x1 in T1 . But T1 is Hausdorff so limits are unique and x0 = x1 ∈ A and
we’re finished.
This last theorem will play an important role when we later discuss the Banach-Mazure
Theorem. We now turn to the issue of compactness and will revisit this idea as we
further develop the theory of Banach spaces. We will especially need many of these
ideas as we explore the idea of reflexivity.
3.3
Compactness and Reflexivity
Our next theorem in some ways justifies the usefulness of the weak* topology. Before we
present this result, we first recall a major result from topology.
Theorem (Tychonoff ’s Theorem)
A nonempty product space is compact (in the product topology) if and only if each
factor space is compact.
We are now ready to prove the Alaoglu Theorem:
Theorem 3.3.0 (Banach-Alaoglu Theorem)
Let X be a normed space. Then the closed unit ball B ∗ := {φ ∈ X ∗ : ||φ|| ≤ 1} is
compact in the weak* topology.
Q
Proof. For each x ∈ X, set Ax := {z ∈ C : |z| ≤ ||x||X } and put A := x∈X Ax . By the
Tychonoff Theorem, A is compact in the product topology. We may interpret the
elements of A as the set of all φ : X → C so that |φ(x)| ≤ ||x||X for all x ∈ X. Now B ∗
contains all such elements that are linear. Moreover, we know that the product topology
on A is the weakest topology so that the projection maps πx : A → Ax are continuous.
But each πx (φ) = φ(x) = Jx (φ) is really the canonical embedding from X into X ∗∗ so
the topology on A is really the weak* topology.
Because B ∗ ⊂ A and A is compact in the weak* topology, it suffices to show that B ∗ is
(relatively) weak* closed. Let (φλ ) be a net in B ∗ with φλ → φ0 . Then ||φλ || ≤ 1 for all
50
λ. By assumption, we know that φλ (x) → φ0 (x) for all x ∈ X. Then |φλ (x)| ≤ ||x|| for
all λ and all x ∈ X so that |φ0 (x)| ≤ ||x|| for all x ∈ X. Hence ||φ0 || ≤ 1 so φ0 ∈ B ∗ and
we’re finished.
Note that the Alaoglu Theorem is true for any closed ball, not just the unit closed ball.
We could use the uniform boundeness principle to demonstrate the following but we
treat it as a corollary to the Alaoglu Theorem.
Corollary 3.3.1
w∗
Let (φn ) be a sequence in X ∗ so that φn → φ0 . Then ||φ0 || ≤ lim inf ||φn ||.
n→∞
Proof. Set M := lim inf n→∞ ||φn ||. Fix > 0. Then there exists a subsequence (φnj ) of
(φn ) so that ||φnj || ≤ M + for all j. Because the norm-ball of radius M + is weak*
compact and because the weak* topology is T2 , we see that the norm-ball of radius
M + is weak* closed. Thus ||φ0 || ≤ M + for all > 0. Letting → 0+ yields the
desired result.
If (X, T ) is a topological space and K ⊂ X recall that K is called sequentially compact
if every sequence has a convergent subsequence. If X is metrizable, then sequential
compactness is equivalent to compactness but in a general topological space,
compactness is typically stronger property. This next theorem asserts that any norm
bounded subset of a Hilbert space is weakly sequentially compact. Let us prove this fact
now.
Theorem 3.3.2
Every norm bounded sequence in a Hilbert space H contains a weakly convergent
subsequence.
Proof. Suppose that ||xn ||H ≤ M for all n. Denote by Y the closure of the linear span of
each term in the sequence. Then the sequence of real numbers (x1 , xn ) has a convergent
subsequence, call it s1n := (x1 , x1n ). Now, (x2 , x1n ) has a convergent subsequence, call it
s2n := (x2 , x2n ). Continue this process to obtain, for each j, a convergent sequence
m
sjn := (xj , xjn ). Consider the diagonal sequence (xm
m ) in H. For each j, (xj , xm ) converges
because whenever m > j, it is a subsequence of the convergent sequence (xj , xjm ). It is
clear that (x, xm
m ) converges as m → ∞ whenever x is a finite sum of the form
51
x=
N
X
ck xk
k=1
But finite sums of this form are dense in Y so that if y ∈ Y , there is a sequence of finite
sums (yn ) so that ||yn − y||H → 0 as n → ∞. Also, (yn , xm
m ) converges for all n by our
remark above. But
m
m
(y, xm
m ) = (yn , xm ) + (y − yn , xm )
where |(y − yn , xm
m )| ≤ M ||y − yn ||H . But then
m
m
lim sup(y, xm
m ) = lim (yn , xm ) + lim sup(y − yn , xm )
m→∞
m→∞
m→∞
We could also take the limit infimum instead so that
m
m
lim sup(y, xm
m ) = lim inf (y, xm ) = lim sup lim (yn , xm )
m→∞
m→∞
n→∞
m→∞
Thus, (y, xm
m ) converges for all y ∈ Y .
By virtue of H being a Hilbert space, we have H = Y ⊕ Y ⊥ . Thus if x ∈ H, then
x = y + yb where y ∈ Y and yb ∈ Y ⊥ . Then
m
m
(x, xm
y , xm
m ) = (y, xm ) + (b
m ) = (y, xm )
m
m
as (b
y , xm
m ) = 0 for all m. In other words, (x, xm ) converges for all x ∈ H. Put wm := xm
and define
ϕ0 (x) := lim (wm , x) for all x ∈ H
m→∞
It is clear that ϕ0 ∈ H ∗ . By the Riesz Representation Theorem, there exists an x0 ∈ H
so that ϕ0 (x) = (x, x0 ) for all x ∈ H. If φ ∈ H ∗ , there is an x1 ∈ H so that
φ(x) = (x, x1 ) and
φ(wm ) = (wm , x1 ) → ϕ0 (x1 ) = (x1 , x0 ) = φ(x0 )
52
w
That is, wm → x0 which is what we wished to show.
In the later sections, we will prove the Eberlein-S̆mulian theorem which that tells us
that in a Banach space, weak compactness is equivalent to weak sequential compactness.
Thus, given the validity of Eberlein-S̆mulian, the theorem we just proved tells us that
the unit ball in a Hilbert space is weakly compact. It turns out that this very condition
is equivalent to a space being reflexive and we shall explore this now.
To save time, the notation B := {x ∈ X : ||x||X ≤ 1} and B ∗∗ := {ψ ∈ X ∗∗ : ||ψ|| ≤ 1}
will be employed. As usual, J = Jx will denote the canonical embedding of X into X ∗∗ .
Theorem 3.3.3 (Goldstein)
J(B) is weak* dense in B ∗∗ .
Proof. Note that if x ∈ B, then ||x||X ≤ 1 so that ||Jx || = ||x||X ≤ 1 and Jx ∈ B ∗∗ . That
is, J(B) ⊂ B ∗∗ . By the Alaoglu Theorem, B ∗∗ is weak* compact (here, we take X ∗ as
our normed space). Because the weak* topology is T2 , we know that B ∗∗ is weak* closed
and hence J(B) ⊂ B ∗∗ where the closure is the weak* closure. We will now demonstrate
the reverse inclusion.
Let ψ0 ∈ B ∗∗ . It is only necessary to show that for any > 0 and φ1 , ..., φN ∈ X ∗ with
||φk || = 1 that the open neighborhood
V := {ψ ∈ X ∗∗ : |ψ(φk ) − ψ0 (φk )| < , 1 ≤ k ≤ N }
meets J(B). It is enough to show the existance of an x ∈ X so that ||x||X < 1 + with
(Jx − ψ0 )(φk ) = 0 for each 1 ≤ k ≤ N . Given the validity of this claim for the moment,
we would then see that (1 + )−1 x belongs to B and for each 1 ≤ k ≤ N we would have
1
1
1 + Jx (φk ) − ψ0 (φk ) = 1 + Jx (φk ) − Jx (φk ) ≤ 1 + ||Jx || < We will now demonstrate the claim. Let S := span(φ1 , ..., φN ) so that dim(S) < ∞. It is
then easy to see that J : X → S ∗ is surjective. Let Ψ := ψ0 |S . Then Ψ ∈ S ∗ so there is
an x ∈ X so that Jx = Ψ. It is clear that (Jx − ψ0 )(φk ) = 0 for each k. Because
||x||X = ||Jx || = ||Ψ|| ≤ 1, it is clear that ||x||X < 1 + for all > 0.
53
We are now ready to prove our useful characterisation of reflexivity to which we earlier
alluded.
Corollary 3.3.4 (Characterization of Reflexive Spaces)
The closed unit ball of a normed space X is weakly compact if and only if X is reflexive.
Proof. Note that the canonical embedding J : X → X ∗∗ is continuous in the weak
topology of X. Thus if B is weakly compact, then J(B) is weak* compact and, hence,
weak* closed. By Goldstein’s Theorem, J(B) = J(B) = B ∗∗ . This easily implies that
the canonical embedding is surjective. On the other hand, suppose that X is reflexive.
Then (X, weak topology) and (X ∗∗ , weak* topology) are homeomorphic via J. But B ∗∗
is weak* compact by Alaoglu’s Theorem and the unit ball in X is consequently weakly
compact.
We will now conclude this section with an important example due to Robert C. James
in 1950. James constructed an example of a Banach space that is not reflexive. The
space he constructed is given by
n
o
J := (xn ) : lim xn = 0 and ||(xn )||J < ∞
n→∞
where
||(xn )||J := sup
n
X
!1/2
(xp2k−1 − xp2k )2 + (xp2k+1 )2
k=1
and the supremum is taken over all n and all increasing sequences of positive integers
p1 , p2 , ..., p2n+1 . In 1951, he utilized a similar non-reflexive space but constructed an
isomorphism to its double dual. Thus even if X ∼
= X ∗∗ it does not necessarily follow
that X is reflexive.
3.4
Exercises
1. Let X be a finite dimensional normed space. Prove that a sequence (xn ) in X
converges weakly to x0 if and only if it converges strongly to x0 .
54
2. This exercise helps further familiarize you with the weak and weak* topologies.
(a) Convince yourself that Bx0 and Bφ∗ 0 form neighborhood bases about x0 and φ0
respectively in the weak and weak* topology.
(b) Prove that the weak* topology is the topology of pointwise convergence; that
is, show that a net (φλ ) in X ∗ converges to φ0 in the weak* topology if and
only if φλ (x) → φ0 (x) for all x ∈ X.
3. Let X be a normed space. Prove that the norm-closed ball B = {x ∈ X : ||x|| ≤ 1}
is weakly closed. Additionally, show that if E ⊂ X is norm bounded, then it’s
weak closure is also norm bounded.
Exercises 4 and 5 are for those who are familiar with duality of Lp and C(K). We will
explore these topics later.
4. Let 1 < p < ∞. Show that a sequence (fn ) in Lp (0, 1) converges weakly to f if and
only if:
(a) ||fn ||Lp ≤ M for some M > 0.
Z b
Z b
f (t)dt for all (a, b) ⊂ (0, 1).
fn (t)dt =
(b) lim
n→∞
a
a
5. Let K ⊂ R be compact. Prove that a sequence (fn ) in C(K) converges weakly to
f if and only if:
(a) |f (x)| ≤ M for all x ∈ K
(b) fn (x) → f (x) pointwise for each x ∈ K
55
4
Bibliography
This needs to be organized, cleaned up, and formatted.
1. Gerald Folland, Real Analysis
2. Albiac and Kalton, Topics in Banach Space Theory
3. Charalambos D. Aliprantis and Kim C. Border, Infinite Dimensional Analysis: A
Hitchhiker’s Guide
4. James, R. C., Bases and Reflexivity of Banach Spaces, Ann. Math., 52, 518-527
(1950).
5. James, R. C., A Non-Reflexive Banach Space Isometric with Its Second Conjugate
Space, Proc. Nat. Acad. Sci., 37, 174-177 (1951).
56