Functional Analysis/Geometry of Banach spaces

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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-* topologies[edit]

Let \mathcal{X} be a normed space. Since X^* is a Banach space, there is a canonical injection \pi: \mathcal{X} \to \mathcal{X}^{**} given by:

(\pi x)f = f(x) for f \in \mathcal{X}^* and x \in \mathcal{X}.

One of the most important question in the study of normed spaces is when this \pi is surjective; if this is the case, \mathcal{X} is said to be "reflexive". For one thing, since \mathcal{X}^{**}, as the dual of a normed space, is a Banach space even when \mathcal{X} is not, a normed space that is reflexive is always a Banach space, since \pi becomes an (isometrical) isomorphism. (Since \pi(X) separates points in X^*, the weak-* topology is Hausdorff by Theorem 1.something.)

Before studying this problem, we introduce some topologies. The weak-* topology for \mathcal{X}^* is the weakest among topologies for which every element of \pi(\mathcal{X}) is continuous. In other words, the weak-* topology is precisely the topology that makes the dual of \mathcal{X}^* \pi(\mathcal{X}). (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 \mathcal{X} is the weakest of topologies for which every element of \mathcal{X}^* is continuous. (As before, the weak topology is Hausdorff.)

4 Theorem (Alaoglu) The unit ball of \mathfrak{B}^* is weak-* compact.
Proof: For every f, \operatorname{ran}f is an element of \mathbf{C}^\mathfrak{B}. With this identification, we have: \mathfrak{B}^* \subset \mathbf{C}^\mathfrak{B}. The inclusion in topology also holds; i.e., \mathfrak{B}^* is a topological subspace of \mathbf{C}^\mathfrak{B}. The unit ball of \mathfrak{B}^* is a subset of the set

E = \prod_{x \in \mathfrak{B}} \{ \lambda; \lambda \in \mathbf{C}, |\lambda| \le \|x\|_\mathfrak{B} \}.

Since E, 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) \square

4. Theorem Let \mathcal{X} be a TVS whose dual separates points in \mathcal{X}. Then the weak-* topology on \mathcal{X}^* is metrizable if and only if \mathcal{X} 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 E \subset \mathcal{X} is weakly closed.
Proof: Let x be in the weak closure of E. Suppose, if possible, that x \not\in E. By (the geometric form) of the Hanh-Banach theorem, we can then find f \in \mathcal{X}^* and real number c such that:

\operatorname{Re}f (x) < c < \operatorname{Re}f(y) for every y \in E.

Set V = \{ y; \operatorname{Re}f(y) < c \}. What we have now is: x \in V \subset E^c where V is weakly open (by definition). This is contradiction.\square

4. Corollary The closed unit ball of \mathcal{X} (resp. \mathcal{X}^*) is weakly closed (resp. weak-* closed).

4 Exercise Let B be the unit ball of \mathcal{X}. Prove \pi(B) is weak-* dense in the closed unit ball of \mathcal{X}^{**}. (Hint: similar to the proof of Lemma 4.something.)

4 Theorem A set E is weak-* sequentially closed if and only if the intersection of E 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 \mathcal{X} be a Banach space. The following are equivalent:

  • (i) \mathcal{X} is reflexive.
  • (ii) The closed unit ball of \mathcal{X} 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) \Rightarrow (ii) is immediate. For (iii) \Rightarrow (i), we shall prove: if \mathcal{X} 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) \Rightarrow (iii), it suffices to prove:

4 Lemma Let \mathcal{X} be a Banach space, x_j \in X a sequence and F be the weak closure of x_j. If F is weakly compact, then F is weakly sequentially compact.
Proof: By replacing X with the closure of the linear span of X, we may assume that \mathcal{X} admits a dense countable subset E. Then for u, v \in\mathcal{X}^*, u(x) = v(x) for every x \in E implies u = v by continuity. This is to say, a set of functions of the form u \mapsto u(x) with x \in E separates points in X, a fortiori, B, the closed unit ball of X^*. The weak-* topology for B is therefore metrizable by Theorem 1.something. Since a compact metric space is second countable; thus, separable, B admits a countable (weak-*) dense subset B'. It follows that B' separates points in X. In fact, for any x \in X with \|x\|=1, by the Hahn-Banach theorem, we can find f \in B such that f(x) = \|x\| = 1. By denseness, there is g \in B' that is near x in the sense: |g(x) - f(x)| < 2^{-1}, and we have:

|g(x)| \ge |f(x)| - |g(x) - f(x)| > 2^{-1}.

Again by theorem 1.something, F is now metrizable.\square

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 \mathcal{X} is reflexive if and only if \mathcal{X}^* is reflexive.'

4 Theorem Let \mathcal{X} be a Banach space with a w:Schauder basis e_j. \mathcal{X} is reflexive if and only if e_j satisfies:

  • (i) \sup_n \left\| \sum_{j=1}^n a_j e_j \right\| < \infty \Rightarrow \sum_{j=1}^\infty a_j e_j converges in \mathcal{X}.
  • (ii) For any f \in \mathcal{X}^*, \lim_{n \to \infty} \sup \{ |f(x)|; x = \sum_{j \ge n}^\infty a_j e_j, \|x\|=1 \} = 0.

Proof: (\Rightarrow): Set x_n = \sum_{j=1}^n a_j e_j. By reflexivity, x_n then admits a weakly convergent subsequence x_{n_k} with limit x. By hypothesis, for any x \in \mathcal{X}, we can write: x = \sum_{j=1}^\infty b_j(x) e_j with b_j \in \mathcal{X}^*. Thus,

b_l(x) = \lim_{k \to \infty} b_l(x_{n_k}) = \lim_{k \to \infty} \sum_{j=1}^{n_k} a_j b_l(e_j) = a_l, and so x = \sum_{j=1}^\infty a_j e_j.

This proves (i). For (ii), set

E_n = \{ x; x \in \mathcal{X}, \|x\| = 1, b_1(x) = ... b_{n-1}(x) = 0 \}.

Then (ii) means that \sup_{E_n} |f| \to 0 for any f \in \mathcal{X}^*. Since E_n is a weakly closed subset of the closed unit ball of \mathcal{X}^*, which is weakly compact by reflexivity, E_n is weakly compact. Hence, there is a sequence x_n such that: \sup_{E_n} |f| = |f(x_n)| for any f \in \mathcal{X}^*. It follows:

\lim_{n \to \infty} |f(x_n)| = |f(\lim_{n \to \infty} x_n)| = |f(\sum_{j=1}^\infty b_j(\lim_{n \to \infty} x_n) e_j)| = 0

since \lim_{n \to \infty} b_j(x_n) = 0. (TODO: but does \lim_{n \to \infty} x_n exist?) This proves (ii).
(\Leftarrow): Let x_n be a bounded sequence. For each j, the set \{ b_j(x_n) ; n \ge 1 \} is bounded; thus, admits a convergent sequence. By Cantor's diagonal argument, we can therefore find a subsequence x_{n_k} of x_n such that b_j(x_{n_k}) converges for every j. Set a_j = \lim_{n \to \infty} b_j(x_{n_k}). Let K = 2 \sup_n \|x_n\| and s_n = \sup \{ |f(y)|; y = \sum_{j=m+1}^\infty c_j e_j, \|y\| \le K \}. By (ii), \lim_{n \to \infty} s_n = 0. Now,

|f(\sum_{j=1}^m b_j(x_{n_k}) e_j)| \le |f(\sum_{j=1}^\infty b_j(x_{n_k})e_j)| + |f(\sum_{j=m+1}^\infty b_j(x_{n_k}) e_j)| \le \|f\| \sup_n \|x_n\| + s_m for f \in \mathcal{X}^*.

Since s_m is bounded, \sup_m |f(\sum_{j=1}^m a_j e_j)| < \infty for every f and so \sup_m \| \sum_{j=1}^m a_j e_j \| < \infty. By (i), \sum_{j=1}^m a_j e_j therefore exists. Let \epsilon > 0 be given. Then there exists m such that s_m < \epsilon / 2. Also, there exists N such that:

\sum_{j=1}^m (a_j - b_j(x_{n_k})) f(e_j) < \epsilon / 2 for every k \ge N.


|f(x_{n_k}) - f(\sum_{j=1}^\infty a_j e_j)| \le |\sum_{j=1}^m (a_j - b_j(x_{n_k})) f(e_j) | + |f(\sum_{j=m+1}^\infty (a_j - b_j(x_{n_k})) e_j| < \epsilon.

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 T \in B(\mathfrak H). Then T(\overline{B}(0, 1)) is closed.
Proof: Since \overline{B}(0, 1) is weakly compact and T(\overline{B}(0, 1)) is convex, it suffices to show T is weakly continuous. But if x_n \to 0 weakly, then (Tx_n|y) =  (x_n|T^*y) \to 0 for any y. This shows that T is weakly continuous on \overline{B}(0, 1) (since bounded sets are weakly metrizable) and thus on \mathfrak H.\square

Since T is compact, it suffices to show that T(\overline{B}(0, 1)) is closed. But since T(\overline{B}(0, 1)) is weakly closed and convex, it is closed.

3 Lemma If T \in B(\mathfrak H) is self-adjoint and compact, then either \| T \| or -\| T \| is an eigenvalue of T.
Proof: First we prove that \|T\|^2 is an eigenvalue of T^2. Since T is compact, by the above lemma, there is a x_0 in the unit ball such that \|T\| = \|Tx_0\|. Since \langle T^2 x_0, x_0 \rangle = \|T\|^2,

\|T^2 x - \|T\|^2 x\|^2 \le \|T\|^2 - 2\|T\|^2 + \|T\|^2

Thus, T^2 x_0 = \|T\|^2 x_0. Since (T^2 - \|T\|^2I)x_0 = (T + \|T\|I)(T - \|T\|I)x_0, we see that (T - \|T\|I)x_0 is either zero or an eigenvector of T with respect to -\|T\|. \square

3 Theorem If T is normal; that is, T^*T = TT^*, 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.\square

3 Corollary (polar decomposition) Every compact operator K can be written as:

K = R|K|

where R is a partial isometry and |K| is the square root of K^*K

For T \in \mathcal{L}(\mathfrak{H}), let \sigma(T) be the set of all complex numbers \lambda such that T - \lambda I is not invertible. (Here, I is the identity operator on \mathfrak{H}.)

3 Corollary Let T \in B(\mathfrak H) be a compact normal operator. Then

\|T\| = \max_{\|x\|=1} \|(Tx | x)\| = \sup \{ |\lambda| | \lambda \in \sigma(T) \}

3 Theorem Let T be a densely defined operator on \mathfrak{H}. Then T is positive (i.e., \langle Tx, x \rangle \ge 0 for every x \in \operatorname{dom}T) if and only if T = T^* and \sigma(T) \subset [0, \infty).
Partial proof: (\Rightarrow) We have:

\langle Tx, x \rangle = \overline{\langle T^*x, x \rangle} for every x \in \operatorname{dom}T

But, by hypothesis, the right-hand side is real. That T = T^* follows from Lemma 5.something. The proof of the theorem will be completed by the spectrum decomposition theorem in Chapter 5.\square

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 u_k is an orthonormal sequence in a Hilbert space \mathfrak{H}, then

\sum_{k=1}^\infty |\langle x, u_k \rangle|^2 \le \|x\|^2 for any x \in \mathfrak{H}.

Proof: If \langle x, y \rangle = 0, then \|x + y \|^2 = \|x\|^2 + \|y\|^2. Thus,

\| x - \sum_{k=1}^n \langle x, u_k \rangle \|^2 = \|x\|^2 - 2 \operatorname{Re} \sum_{k=1}^n |\langle x, u_k \rangle |^2 + \sum_{k=1}^n | \langle x, u_k \rangle |^2 = \|x\|^2 - \sum_{k=1}^n | \langle x, u_k \rangle |^2.

Letting n \to \infty completes the proof. \square.

3 Theorem (Parseval) Let u_k be a orthonormal sequence in a Hilbert space \mathfrak{H}. Then the following are equivalent:

  • (i) \operatorname{span} \{ u_1, u_2, ... \} is dense in \mathfrak{H}.
  • (ii) For each x \in \mathfrak{H}, x = \sum_{k=1}^\infty \langle x, u_k \rangle u_k.
  • (iii) For each x, y \in \mathfrak{H}, \langle x, y \rangle = \sum_{k=1}^\infty \langle x, u_k \rangle \overline{\langle y, u_k \rangle}.
  • (iv) \|x\|^2 = \sum_{k=1}^\infty | \langle x, u_k \rangle |^2 (the Parseval equality).

Proof: Let \mathcal{M} = \operatorname{span} \{ u_1, u_2, ... \}. If v \in \mathcal{M}, then it has the form: v = \sum_{k=1}^\infty \alpha_k u_k for some scalars \alpha_k. Since \langle v, u_j \rangle = \sum_{k=1}^\infty a_j \langle u_k, u_j \rangle = a_j we can also write: v = \sum_{k=1}^\infty \langle v, u_k \rangle u_k. Let y =\sum_{k=1}^\infty \langle x, u_k \rangle u_k. Bessel's inequality and that \mathfrak{H} is complete ensure that y exists. Since

\langle y, v \rangle = \sum_{k=1}^\infty \langle x, u_k \rangle \langle u_k, v \rangle = \sum_{k=1}^\infty \langle x, \langle v, u_k \rangle u_k \rangle = \langle x, \sum_{k=1}^\infty \langle v, u_k \rangle u_k \rangle = \langle x, v \rangle

for all v \in \mathcal{M}, we have x - y \in \mathcal{M}^\bot = \{0\}, proving (i) \Rightarrow (ii). Now (ii) \Rightarrow (iii) follows since

|\langle x, y \rangle - \sum_{k=1}^n \langle x, u_k \rangle \overline {\langle y, u_k \rangle}| = | \langle x, y - \sum_{k=1}^n \langle y, u_k \rangle u_k \rangle | \to 0 as n \to \infty

To get (iii) \Rightarrow (iv), take x = y. To show (iv) \Rightarrow (i), suppose that (i) is false. Then there exists a z \in (\operatorname{span \{ u_1, u_2, ... \}})^{\bot} with z \ne 0. Then

\sum_{k=1}^\infty | \langle z, u_k \rangle |^2 = 0 < \|z\|^2.

Thus, (iv) is false.\square

3 Theorem Let x_k be an orthogonal sequence in a Hilbert space (\mathfrak{H}, \|\cdot\| = \langle \cdot, \cdot \rangle^{1/2}). Then the series \sum_{k=1}^\infty x_k converges if and only if the series \sum_{k=1}^\infty \langle x_k, y \rangle converges for every y \in \mathfrak{H}.
Proof: Since

\sum_{k=1}^\infty | \langle x_k, y \rangle | \le \|y\| \sum_{k=1}^\infty \| x_k \| and \sum_{k=1}^\infty \| x_k \| = \left\| \sum_{k=1}^\infty x_k \right\|

by orthogonality, we obtain the direct part. For the converse, let E = \left\{ \sum_{k=1}^n x_k ; n \ge 1 \right\}. Since

\sup_E |\langle \cdot, y \rangle| = \sup_n |\sum_{k=1}^n \langle x_k, y \rangle| < \infty for each y

by hypothesis, E is bounded by Theorem 3.something. Hence, \sum_{k=1}^\infty \|x_k\| < \infty and \sum_{k=1}^n x_k 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 \mathfrak{H} 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 \mathcal{X} is reflexive if and only if every element of \mathcal{X} attains its maximum on the closed unit ball of \mathcal{X}.

4 Corollary (Krein-Smulian) Let \mathcal{X} be a Banach space and K \subset \mathcal{X} a weakly compact subset of \mathcal{X}. then \overline{co}(K) is weakly compact.
Proof: [4]

A Banach space is said to be uniformly convex if

\|x_n\| \le 1, \|y_n\| \le 1 and \|x_n + y_n\| \to 0 \Rightarrow \|x_n - y_n\| \to 2

Clearly, Hilbert spaces are uniformly convex. The point of this notion is the next result.

4 Theorem Every uniformly convex space \mathfrak{B} is reflexive.
Proof: Suppose, if possible, that \mathfrak{B} is uniformly convex but is not reflexive. \square

4 Theorem Every finite dimensional Banach space is reflexive.
Proof: (TODO)

4 Theorem Let \mathfrak{B}_1, \mathfrak{B}_2 be Banach spaces. If  \mathfrak{B}_1 has a w:Schauder basis, then the space of finite-rank operators on \mathfrak{B}_1 is (operator-norm) dense in the space of compact operators on \mathfrak{B}_1.

5 Theorem L^p spaces with 1 < p < \infty are uniformly convex (thus, reflexive).
Proof: (TODO)

5 Theorem (M. Riesz extension theorem) (see w:M. Riesz extension theorem)