In the previous chapter we studied a Banach space having a special geometric property: that is, a Hilbert space. This chapter continues this line of the study. The main topics of the chapter are (i) the notion of reflexibility of Banach spaces (ii) weak-* compactness, (iii) the study of a basis in Banach spaces and (iv) complemented (and uncomplemented) subspaces of Banach spaces. It turns out those are a geometric property,
Let
be a normed space. Since
is a Banach space, there is a canonical injection
given by:
for
and
.
One of the most important question in the study of normed spaces is when this
is surjective; if this is the case,
is said to be "reflexive". For one thing, since
, as the dual of a normed space, is a Banach space even when
is not, a normed space that is reflexive is always a Banach space, since
becomes an (isometrical) isomorphism. (Since
separates points in
, the weak-* topology is Hausdorff by Theorem 1.something.)
Before studying this problem, we introduce some topologies. The weak-* topology for
is the weakest among topologies for which every element of
is continuous. In other words, the weak-* topology is precisely the topology that makes the dual of
. (Recall that it becomes easier for a function to be continuous when there are more open sets in the domain of the function.)
The weak topology for
is the weakest of topologies for which every element of
is continuous. (As before, the weak topology is Hausdorff.)
4 Theorem (Alaoglu) The unit ball of
is weak-* compact.
Proof: For every
,
is an element of
. With this identification, we have:
. The inclusion in topology also holds; i.e.,
is a topological subspace of
. The unit ball of
is a subset of the set
.
Since
, a product of disks, is weak-* compact by w:Tychonoff's theorem (see Chapter 1), it suffices to show that the closed unit ball is weak-* closed. This is easy once we have the notion of nets, which will be introduced in the next chapter. For the sake of completeness, we give a direct argument here. (TODO)
4. Theorem Let
be a TVS whose dual separates points in
. Then the weak-* topology on
is metrizable if and only if
has a at most countable Hamel basis.
Obviously, all weakly closed sets and weak-* closed sets are closed (in their respective spaces.) The converse in general does not hold. On the other hand,
4. Lemma Every closed convex subset
is weakly closed.
Proof: Let
be in the weak closure of
. Suppose, if possible, that
. By (the geometric form) of the Hanh-Banach theorem, we can then find
and real number
such that:
for every
.
Set
. What we have now is:
where
is weakly open (by definition). This is contradiction.
4. Corollary The closed unit ball of
(resp.
) is weakly closed (resp. weak-* closed).
4 Exercise Let
be the unit ball of
. Prove
is weak-* dense in the closed unit ball of
. (Hint: similar to the proof of Lemma 4.something.)
4 Theorem A set
is weak-* sequentially closed if and only if the intersection of
and the (closed?) ball of arbitrary radius is weak-* sequentially closed.
Proof: (TODO: write a proof using PUB.)
4 Theorem (Kakutani) Let
be a Banach space. The following are equivalent:
- (i)
is reflexive.
- (ii) The closed unit ball of
is weakly compact.
- (iii) Every bounded set admits a weakly convergent subsequence. (thus, the unit ball in (ii) is actually weakly sequentially compact.)
Proof: (i)
(ii) is immediate. For (iii)
(i), we shall prove: if
is not reflexive, then we can find a normalized sequence that falsifies (iii). For that, see [1], which shows how to do this. Finally, for (ii)
(iii), it suffices to prove:
4 Lemma Let
be a Banach space,
a sequence and
be the weak closure of
. If
is weakly compact, then
is weakly sequentially compact.
Proof: By replacing
with the closure of the linear span of
, we may assume that
admits a dense countable subset
. Then for
,
for every
implies
by continuity. This is to say, a set of functions of the form
with
separates points in
, a fortiori,
, the closed unit ball of
. The weak-* topology for
is therefore metrizable by Theorem 1.something. Since a compact metric space is second countable; thus, separable,
admits a countable (weak-*) dense subset
. It follows that
separates points in
. In fact, for any
with
, by the Hahn-Banach theorem, we can find
such that
. By denseness, there is
that is near
in the sense:
, and we have:
.
Again by theorem 1.something,
is now metrizable.
Remark: Lemma 4.something is a special case of w:Eberlein–Šmulian theorem, which states that every subset of a Banach space is weakly compact if and only if it is weakly sequentially compact. (See [2], [3])
In particular, since every Hilbert space is reflexive, either (ii) or (iii) in the theorem always holds for all Hilbert spaces. But for (iii) we could have used alternatively:
4 Exercise Give a direct proof that (iii) of the theorem holds for a separable Hilbert space. (Hint: use an orthonormal basis to directly construct a subsequence.)
4 Corollary A Banach space
is reflexive if and only if
is reflexive.'
4 Theorem Let
be a Banach space with a w:Schauder basis
.
is reflexive if and only if
satisfies:
- (i)
converges in
.
- (ii) For any
,
.
Proof:
(
): Set
. By reflexivity,
then admits a weakly convergent subsequence
with limit
. By hypothesis, for any
, we can write:
with
. Thus,
, and so
.
This proves (i). For (ii), set
.
Then (ii) means that
for any
. Since
is a weakly closed subset of the closed unit ball of
, which is weakly compact by reflexivity,
is weakly compact. Hence, there is a sequence
such that:
for any
. It follows:

since
. (TODO: but does
exist?) This proves (ii).
(
): Let
be a bounded sequence. For each
, the set
is bounded; thus, admits a convergent sequence. By Cantor's diagonal argument, we can therefore find a subsequence
of
such that
converges for every
. Set
. Let
and
. By (ii),
. Now,
for
.
Since
is bounded,
for every
and so
. By (i),
therefore exists. Let
be given. Then there exists
such that
. Also, there exists
such that:
for every
.
Hence,
.
4 Exercise Prove that every infinite-dimensional Banach space contains a closed subspace with a Schauder basis. (Hint: construct a basis by induction.)
3 Lemma Let
. Then
is closed.
Proof: Since
is weakly compact and
is convex, it suffices to show
is weakly continuous. But if
weakly, then
for any y. This shows that T is weakly continuous on
(since bounded sets are weakly metrizable) and thus on
.
Since T is compact, it suffices to show that
is closed. But since
is weakly closed and convex, it is closed.
3 Lemma If
is self-adjoint and compact, then either
or
is an eigenvalue of T.
Proof: First we prove that
is an eigenvalue of
. Since
is compact, by the above lemma, there is a
in the unit ball such that
. Since
,

Thus,
. Since
, we see that
is either zero or an eigenvector of
with respect to
.
3 Theorem If T is normal; that is,
, then there exists an orthonormal basis consisting of eigenvectors of T.
Proof: Since we may assume that T is self-adjoint, the theorem follows from the preceding lemma by transfinite induction. By Zorn's lemma, select U to be a maximal subset of H with the following three properties: all elements of U are eigenvectors of T, they have norm one, and any two distinct elements of U are orthogonal. Let F be the orthogonal complement of the linear span of U. If F ≠ {0}, it is a non-trivial invariant subspace of T, and by the initial claim there must exist a norm one eigenvector y of T in F. But then U ∪ {y} contradicts the maximality of U. It follows that F = {0}, hence span(U) is dense in H. This shows that U is an orthonormal basis of H consisting of eigenvectors of T.
3 Corollary (polar decomposition) Every compact operator K can be written as:

where R is a partial isometry and
is the square root of
For
, let
be the set of all complex numbers
such that
is not invertible. (Here, I is the identity operator on
.)
3 Corollary Let
be a compact normal operator. Then

3 Theorem Let
be a densely defined operator on
. Then
is positive (i.e.,
for every
) if and only if
and
.
Partial proof:
We have:
for every 
But, by hypothesis, the right-hand side is real. That
follows from Lemma 5.something. The proof of the theorem will be completed by the spectrum decomposition theorem in Chapter 5.
More materials on compact operators, especially on their spectral properties, can be found in a chapter in the appendix where we study Fredholm operators.
3 Lemma (Bessel's inequality) If
is an orthonormal sequence in a Hilbert space
, then
for any
.
Proof: If
, then
. Thus,
.
Letting
completes the proof.
.
3 Theorem (Parseval) Let
be a orthonormal sequence in a Hilbert space
. Then the following are equivalent:
- (i)
is dense in
.
- (ii) For each
,
.
- (iii) For each
,
.
- (iv)
(the Parseval equality).
Proof: Let
. If
, then it has the form:
for some scalars
. Since
we can also write:
. Let
. Bessel's inequality and that
is complete ensure that
exists.
Since

for all
, we have
, proving (i)
(ii). Now (ii)
(iii) follows since
as 
To get (iii)
(iv), take
. To show (iv)
(i), suppose that (i) is false. Then there exists a
with
. Then
.
Thus, (iv) is false.
3 Theorem Let
be an orthogonal sequence in a Hilbert space
. Then the series
converges if and only if the series
converges for every
.
Proof: Since
and 
by orthogonality, we obtain the direct part. For the converse, let
. Since
for each 
by hypothesis,
is bounded by Theorem 3.something. Hence,
and
converges by completeness.
The theorem is meant to give an example. An analogous issue in the Banach space will be discussed in the next chapter.
4 Theorem A Hilbert space
is separable if and only if it has an (countable) orthonormal basis.
It is plain that a Banach space is separable if it has a Schauder basis. Unfortunately, the converse is false.
4 Theorem (James) A Banach space
is reflexive if and only if every element of
attains its maximum on the closed unit ball of
.
4 Corollary (Krein-Smulian) Let
be a Banach space and
a weakly compact subset of
. then
is weakly compact.
Proof: [4]
A Banach space is said to be uniformly convex if
and 
Clearly, Hilbert spaces are uniformly convex. The point of this notion is the next result.
4 Theorem Every uniformly convex space
is reflexive.
Proof: Suppose, if possible, that
is uniformly convex but is not reflexive.
4 Theorem Every finite dimensional Banach space is reflexive.
Proof: (TODO)
4 Theorem Let
be Banach spaces. If
has a w:Schauder basis, then the space of finite-rank operators on
is (operator-norm) dense in the space of compact operators on
.
5 Theorem
spaces with
are uniformly convex (thus, reflexive).
Proof: (TODO)
5 Theorem (M. Riesz extension theorem) (see w:M. Riesz extension theorem)