# Functional Analysis/Geometry of Banach spaces

 Functional Analysis Chapter 4: Geometry of Banach spaces
 (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

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

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 $k \ge N$.

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 ≠ {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)