Functional Analysis/Geometry of Banach spaces

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
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 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
4 Theorem (Kakutani) ''Let $$\mathcal{X}$$ be a Banach space. The following are equivalent: 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, which shows how to do this. Finally, for (ii) $$\Rightarrow$$ (iii), it suffices to prove:
 * (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.)

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 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, )

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 Schauder basis $$e_j$$. $$\mathcal{X}$$ is reflexive if and only if $$e_j$$ satisfies:'' 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,
 * (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 $$f \in \mathcal{X}^*$$, $$\lim_{n \to \infty} \sup \{ |f(x)|; x = \sum_{j \ge n}^\infty a_j e_j, \|x\|=1 \} = 0$$.''
 * $$b_l(x) = \lim_{k \to \infty} b_l(x_{n_k}) = \lim_{k \to \infty} \sum_{j=1}^{n_k} a_j b_l(e_j) = a_l$$, 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 $$k \ge N$$.

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

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
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 &ne; {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 &cup; {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:'' 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
 * (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).
 * $$\langle y, v \rangle = \sum_{k=1}^\infty \langle x, u_k \rangle \langle u_k, v \rangle = \sum_{k=1}^\infty \langle x, \langle v, u_k \rangle u_k \rangle = \langle x, \sum_{k=1}^\infty \langle v, u_k \rangle u_k \rangle = \langle x, v \rangle$$

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:

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 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 M. Riesz extension theorem)