Proof of the Alaoglu Theorem

Question

I was reading through the proof of the Alaoglu theorem which states

Let $X$ be a normed space Then the unit ball in $X^*=B^*$ is compact with respect to the $weak^*$ topology.

The proof goes as follows.

First define $D_x = \{z \in \mathbb C : |z|\leq ||x|| \}$ Then we construct $$\tau : B^* \rightarrow \Pi_x D_x$$ $$ f \mapsto (f(x))_x$$

The range is compact by Tychonoff theorem and the map defined above is a continuous injection. After this they identity $B^*$ with a subspace of $ \Pi_x D_x$ by saying the inverse map from the image to $B^*$ is continuous. I was having difficulty seeing this fact since they loosely stated it. Also it should be noted that one must use the fact that it is $weak^*$ topology at this stage since the result is false for strong topology.

So here's how i attempted to show it. I'll show the map is a closed map. Let $Z\subset B^*$ is $weak^*$ closed . Then i look at its image. Suppose i get a convergent net $ \tau ( f_i) \rightarrow \alpha$ with $f_i \in Z$ Then we have $$ f_i(x) \rightarrow \alpha_x\ \forall x\in X$$

and hence by UBP $\exists f\in B^* $ such that $f(x)=\alpha_x$.

This shows that $f_i(x) \rightarrow f(x) \ \forall x$ and hence $f_i \rightarrow f$ in $weak^*$ topology. Thus $f \in Z$ and hence the image of $Z$ is closed. Thus we have $f$ is a homeomorphism onto its image and $f(B^*)$ being closed is compact. So $B^*$ is $weak^*$ compact.

Kindly point out if there are any gaps in the above argument.

Answer 1

For any $x\in X$, let \begin{equation} D_{x}=\lbrace z\in ℂ:\vert z\vert‎\leqslant‎\Vert x\Vert \rbrace\\ \end{equation}

and $D=\prod_ {x\in X} ⁢D_{x}$. Since $D_{x}$ is a compact subset of ℂ, D is compact in product topology by Tychonoff theorem.

We prove the theorem by finding a homeomorphism that maps the closed unit ball $ B_{{X^\ast}} $ of $X^{*}$ onto a closed subset of D. Define \begin{equation} \varphi_{x}:B_{X^{\ast}} \longrightarrow‎‎ D_{x} , \varphi x⁢(f)=f⁢(x) , \varphi:B_{X^{\ast}}\longrightarrow D\\ \end{equation} by \begin{equation} \varphi=\prod_{x\in X⁢} \varphi{x}, \end{equation} so that \begin{equation} \varphi⁢\left(f \right)=\left(f⁢(x)\right) x\in X.\\ \end{equation} Obviously, $\varphi $ is one to one, and a net $\left(f_{α}\right)\in B_{X^{*}}$ converges to f in weak$^{*}$ topology of $X^{*}$ if $\varphi \left(f_{α}\right)$ converges to $\varphi⁢(f)$ in product topology, therefore $ \varphi $ is continuous and so is its inverse \begin{equation} \varphi^{-1}:\varphi\left(B_{X^{*}}\right) \longrightarrow B_{X^{*}}\\ \end{equation} It remains to show that $\varphi \left(B_{X*}\right)$ is closed. If \begin{equation} \varphi \left(f_{α}\right) \end{equation} is a net in $\varphi \left(B_{X*}\right)$, converging to a point $ d=(d_{x}) $, $ x\in X\in D$, we can define a function \begin{equation} f: X \longrightarrow C , f⁢(x)=d_{x}.\\ \end{equation} As $\lim_{α}\varphi⁢\left(f{α⁢(x)}\right)=d_{x}$ for all $ x\in X$ by definition of weak$^{*}$ convergence, one can easily see that f is a linear functional in $ B_{X*}$ and that $\varphi(f)=d$. This shows that d is actually in $\varphi(B_{X^{*}})$ and finishes the proof.

Answer 2

From your proof it is not entirely clear why $f$ exists and is an element of $B^*$.

Indeed, define $f : X \to \mathbb{C}$ by $f(x) = \alpha_x$. We have that $f$ is linear:

$$f(\lambda x + \mu y) = \lim_{i} f_i(\lambda x + \mu y) = \lambda \lim_i f_i(x) + \mu \lim_i f_i(y) = \lambda \alpha_x + \mu\alpha_y = \lambda f(x) + \mu f(y)$$

Also for every $x \in X$ we have $f(x) = \alpha_x \in D_x$ so $\|f(x)\| \le \|x\|$ so $f$ is bounded and $\|f\| \le 1$.

Therefore $f \in B^*$. By construction for every $x \in X$ we have $f_i(x) \to \alpha_x = f(x)$ so $f_i \to f$ in the weak$^*$ topology. Since $Z$ weak$^*$ closed, it follows $f \in Z$ and then $(\alpha_x)_{x \in X} = \tau(f) \in \tau(Z)$.

We conclude that $\tau(Z)$ is closed.