Alaoglu's Theorem

Monday, February 6, 2006:

Theorem 1.9.1   Alaoglu's Theorem ($2^{nd}$ most important)
Let $X$ be a normed space. Then $\{ f \in X^*: \Vert f\Vert \leq 1 \}$ is $wk^*$-compact.

Proof. We use Tychanoff's Theorem.
For $x \in B_1(0) \subset X$, let

$$D_x := \{ \alpha \in \mathbb{F}: \vert\alpha\vert \leq 1 \},$$

a compact set. Let

$$D = \prod_{x \in B_1(0)} D_x := \{ g : B_1(0) \rightarrow D_x \}$$

$D$, with the product topology, is compact by Tychanoff. Define $\tau_x: D \rightarrow D_x$ by $f \mapsto f(x)$. That is, $\tau(f)$ is $f\vert _{B_1(0)}$.

Claim 1: $\tau$ is injective
If $f,g \in B_1^*$. and $\tau(f) = \tau(g)$, then $f$ and $g$ agree on $\overline{B_1(0)}$ and, since $f$ and $g$ are linear, $f = g$.

Claim 2: $\tau$ is a ($wk^*$,product topology) homeomorphism onto its range
Suppose $f_\lambda \rightarrow f$ in the $wk^*$ topology in $X^*$. This occurs iff for all $x \in X$, $f_\lambda(x) \rightarrow f(x)$ iff $\forall x \in \overline{B_1(0)}$, $f_\lambda(x) \rightarrow f(x)$, iff $\forall x \in \overline{B_1(0)}$ $\pi_x(\tau(f_\lambda)) \rightarrow \pi_x(\tau(f))$, iff $\tau(f_\lambda) \rightarrow \tau(f)$ in the product topology.

Claim 3: $\tau(B_1^*)$ is (product topology) closed
Suppose $\tau(f_\lambda)$ is a net in $\tau(B_1^*)$ converging to $\phi \in D$. For $x,y \in \overline{B_1(0)}$ with $x+y \in \overline{B_1(0)}$,

$$\phi(x+y) = \lim_\lambda \tau(f_\lambda)(x+y) = \lim_\lambda \tau(f_\lambda)(x) + \lim_\lambda \tau(f_\lambda(y)) = \phi(x) + \phi(y)$$

Similarly,

$$\phi(\alpha x ) = \alpha \phi(x)$$

for $\alpha \in \mathbb{F}$, $x \in \overline{B_1(0)}$ such that $\alpha x \in \overline{B_1(0)}$. Extend $\phi$ to $f:X \rightarrow \mathbb{F}$ by

$$f(x) = \Vert x\Vert \phi\left( \frac{x}{\Vert x\Vert} \right)$$

Claim 4: $f \in X^*$
That $f$ is linear is an exercise
To see that $f$ is bounded, observe that

$$f(\overline{B_1(0)}) \subset \{ \alpha \in F : \vert\alpha\vert \leq 1 \}$$

Finally, $\tau(f) = g$

As a closed subset of a compact set, $\tau(B_1^*)$ is compact and, as $\tau$ is a homeomorphism, $B_1^*$ is compact.

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Corollary 1.9.2   Every Banach space is isometrically isomorphic to a closed subspace of $C(X)$ for some compact Hausdorff space $X$.

Proof. Let $B$ be a Banach space. Let $X$ be $\{ f \in B^* : \Vert f\Vert \leq 1 \}$ with the $wk^*$-topology. Define $\Delta : B \rightarrow C(X)$ by

$$(\Delta(b))(f) = f(b)$$

Fact: $\Delta$ is an isometric isomorphism.
To see isometric, observe, for $b \in B$, $\exists f \in B^*$, $\Vert f\Vert \leq 1$ such that $f(b) = \Vert b\Vert$ (Corollary of the Hahn Banach theorem).

Since $f \in X$, so

$$\Vert\Delta(b)\Vert _\infty = \sup_{g \in X} \vert\Delta(b)(g)\vert = \sup_g \in X \vert g(b)\vert = \Vert b \Vert$$

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Theorem 1.9.3   Let $S$ be a $wk^*$-closed subset of $X^*$, $X$ a Banach space. Then $S$ is $wk^*$-compact if and only if $S$ is norm-bounded

Proof.

$\rightarrow$

If $S$ is norm-bouned, then there is $r > 0$ such that $S \subset B_R(0)$ a $wk^*$-compact set, by Alaoglu's Theorem.

$\leftarrow$

If $S$ is $wk^*$-compact, so is the image of $S$ under any $wk^*$-continuous map. In particular, for all $x \in X$, the map $\hat x : S \rightarrow \mathbb{F}$, $f \mapsto f(x)$ is compact in $\mathbb{F}$ and hence, $\forall x \in X$, $\{ f(x) : f \in S \}$ is bounded. By the Principle of Uniform Boundedness (III.14.4), $S$ is norm-bounded.

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Let $\hat {}:X \rightarrow X^{**}$ be the canonical inclusion.

Proposition 1.9.4   The image of $B_1(0)$ under $\hat{}$ is $\sigma(X^{**},X^*)$-dense in $\{ \phi \in X^{**} : \Vert\phi\Vert \leq 1 \} =: B_1^{**}$

Proof. Let $B$ be the $\sigma(X^{**},X^*)$-closure of the unit ball of $X$ under $\hat{}$. Suppose $F \in B_1^{**}\setminus B$.
Claim: There exists $f \in X^*$, $\alpha \in \mathbb{R}$, $\epsilon > 0$ such that $\forall G \in B$, Re $G(f) \leq \alpha < \alpha + \epsilon \leq$    Re $F(f)$.
To see this, use the Hahn-Banach theorem (compact set version) to $X^{**}$ with the $\sigma(X^{**},X^*)$-topology. A continuous functional on this space is an element of $X^*$.
WLOG, $\alpha = 1$.
Then, $\vert F(f)\vert \geq 1 + \epsilon > 1$ but $\Vert F\Vert \leq 1$, so $\Vert f\Vert \leq 1$ by the claim.

    Re $$f(x) \leq 1, \forall x \in B_1(0) \leq X, \Vert f\Vert \leq 1.$$

$\qedsymbol$

Wednesday, February 8, 2006:

This above proposition is sometimes called Goldstine's Theorem.

Theorem 1.9.5   Let $X$ be a Banach space. The following are equivalent:

1. $X$ is reflexive.
2. $X^*$ is reflexive.
3. $\sigma(X^*,X) = \sigma(X^*,X^{**})$.
4. $B_1(0) \subset X$ is weakly compact.
5. Every subspace of $X$ is reflexive.
6. Every quotient of $X$ is reflexive.

Proof.

1. [ $1 \rightarrow 3$] Trivial
   
2. [ $4 \rightarrow 1$] $\sigma(X^{**},X^*)\vert _X = \sigma(X,X^*)$. By 4, $B_1(0)$ is $\sigma(X^{**},X^*)$-closed in the ball $X^{**}$. By Goldstine, $B_1(0)$ is $\sigma(X^{**},X^*)$-dense in the ball $X^{**}$. Thus, $B_1(0) =$ball $X^{**}$.
   Thus, by linearity, the canonical inclusion is onto; i.e., 1 holds.
   
3. [ $3 \rightarrow 2$] By Alaoglu, ball $X^*$ is $\sigma(X^*,X)$-compact, so by 3, it is $\sigma(X^*,X^{**})$-compact. In other words, 4 holds for $X^*$. By $4 \rightarrow 1$, $X^*$ is reflexive.
   
4. [ $2 \rightarrow 1$] $\overline{B_1(0)}$ is norm closed in $X^{**}$ so $\overline{B_1(0)}$ is $\sigma(X^{**},X^{***})$ closed (as norm closed implies weakly closed for convex sets). However, $X^{***} = X^*$, so $\overline{B_1(0)}$ is $\sigma(X^{**},X^*)$ closed in ball $X^{**}$. By Goldstine, it is $\sigma(X^{**},X^{*})$-dense, and so $\overline{B_1(0)} = B_1^{**}(0)$ and so $X$ is reflexive.
5. [ $1 \rightarrow 4$] By Alaoglu (for $X^*$), ball $X^{**}$ is $\sigma(X^{**},X^*)$-compact. But $X^{**} = X$, so $\overline{B_1(0)} =$ ball $X^{**}$. Hence, $\overline{B_1(0)}$ is $\sigma(X,X^*)$-compact.
   
6. [ $5 \rightarrow 1$] $X \leq X$, so $X$ is reflexive.
7. [ $6 \rightarrow 1$] Also trivial.
8. [ $1 \rightarrow 5$] Let $M \leq X$. ball $M$ = $M \cap \overline{B_1(0)}$. $M$ is a weakly closed subspace and $\overline{B_1(0)}$ is wealky compact (by 1). Thus, ball $M$ is weakly closed in $X$. But $\sigma(X,X^*)\vert _M = \sigma(M,M^*)$ (by Hahn-Banach).
   So, ball $M$ is $\sigma(M,M^*)$-compact in $M$ and so $M$ is reflexive.
9. [ $1 \rightarrow 6$] Homework.

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Definition 1.9.6   A sequence $(x_n)$ in a Banach space $X$ is weakly Cauchy if for all $f \in X^*$, $(f(x_n))$ is Cauchy in $\mathbb{F}$. We say $X$ is weakly sequentially complete if every weakly Cauchy sequence converges.

Example: A Banach space that is not weakly sequentiall complete. Use $C([0,1])$ with the sup-norm.
Let

$$f_n(t) = \begin{cases}1 - nt, & t \in [0,\frac1n] \ 0 \end{cases}$$

For $\mu \in M[0,1] = C[0,1]^*$,

$$\int f_n d\mu \rightarrow \mu(\{0\})$$

by MCT. So, $(f_n)$ is weakly Cauchy. But $f_n$ does not have a limit point in $C([0,1])$. Why?
Suppose $f_n \rightarrow f$ weakly. Then,

$$\int f_n d\mu \rightarrow \int f d\mu.$$

So if $\mu$ is a point-mass measure at $x$ in $[0,1]$ then $f_n(x) \rightarrow f(x)$. But the pointwise limit of $f_n$ is $\chi_{\{ 0 \}} \notin C[0,1]$.

Friday, February 10, 2006:

Remark: Why is ball $X$ closed in $X^{**}$? ball $X$ is really $\hat X \cap$ ball $X^{**}$. Since $\hat X$ is isometric, it is bounded below and so has closed range.

Corollary 1.9.7 (of Alaoglu)  
If $X$ is a reflexive Banach space, then $X$ is weakly sequentially complete.

Proof. Let $(x_n)$ be a weakly Cauchy sequence. Let $U$ be a neighborhood of 0. WLOG, $U$ is basic, i.e. $\exists f_1, \ldots, f_n \in X^*, \epsilon _1, \ldots, \epsilon _n > 0$ such that

$$U := \{ x \in X : \vert f_i(x)\vert < \epsilon _i, i = 1, \ldots, n \}$$

Notice each $\{f_i(x_n) : n \in \mathbb{N}\}$ is bounded (it's a Cauchy sequence in $\mathbb{F}$).
So there is $\epsilon > 0$ such that

$$\epsilon \{ x_n : n \in \mathbb{N}\} \subset U \}$$

Thus, $\{ x_n : n \in \mathbb{N}\}$ is weakly bounded. By PUB, $\exists M$ such that $\Vert x_n\Vert \leq M$. By reflexivity and Alaoglu, $B_M(0)$ in $X$ is weakly compact. Thus, $(x_n)$ has a cluster point $x \in \overline{B_m(0)}$ For each $f \in X^*$, $$\lim_{n \rightarrow \infty} f(x_n)$$ exists and since $f(x)$ is a cluster point for the $f(x_n)$, $$\lim_{n \rightarrow \infty} f(x_n) = f(x)$$. That is, $x_n \rightarrow x$ weakly. $\qedsymbol$

Definition 1.9.8   Let $M \leq X$, $X$ a normed space. We say $M$ is proximal if for all $x \in X$, there exists $m \in M$ such that $\Vert x - m \Vert =$   dist$(x,M)$.

Theorem 1.9.9   Every subspace of a reflexive Banach space is proximal.

Proof. Let $M \leq X$, $x \in X/M$, $d =$   dist$(x,M)$. Them map $y \mapsto \Vert x-y\Vert$ is weakly lower semicontinuous (HW). As $M \cap \overline{B_{2d}(x)}$ is weakly compact, and a lower semicontinuous function on a compact set attains its minimum, there is a $m_0 \in M$ such that

$$\Vert x - m_0 \Vert = \inf \{ \Vert x - m \Vert : m + M \cap \overline{B_{2d}(x)} \}$$

$$= \inf \{ \Vert x - m \Vert : m \in M \} =: dist(x,M)$$

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Remark:
For any Banach space $X$ and any finite dim $M \leq X$, $M$ is proximal.
Question - What if $\dim(X/M)$ is finite? No (last semester's homework).

Proposition 1.9.10   For $X$ a Banach space, $f \in X^*$, $\ker f$ is proximal iff $\exists x \in X, \Vert x\Vert = 1$ such $f(x) = \Vert f\Vert$.

Proof. Variation on Assign 5, Q5 from last semester - read text. $\qedsymbol$