Functional Analysis/Geometry of Banach spaces
←Chapter 3: Hilbert spaces | Functional Analysis Chapter 4: Geometry of Banach spaces |
Chapter 5: Operational calculus and representation theory→ |
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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-* topologies[edit]
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 spaces[edit]
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 spaces[edit]
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)
References[edit]
- SEPARABLE BANACH SPACE THEORY NEEDS STRONG SET EXISTENCE AXIOMS
- Functional Analysis and Infinite-dimensional Geometry