Functional Analysis/Geometry of Banach spaces

Functional Analysis
Chapter 4: Geometry of Banach spaces
0% developed  as of May 27, 2008 (May 27, 2008) This is not even a draft.

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,

Weak and weak-* topologiesEdit

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.)

Reflexive Banach spacesEdit

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.)

Compact operators on Hilbert spacesEdit

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)

ReferencesEdit