# Proof of the Alaoglu Theorem

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.

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

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 $$$$\varphi_{x}:B_{X^{\ast}} \longrightarrow‎‎ D_{x} , \varphi x⁢(f)=f⁢(x) , \varphi:B_{X^{\ast}}\longrightarrow D\\$$$$ by $$$$\varphi=\prod_{x\in X⁢} \varphi{x},$$$$ so that $$$$\varphi⁢\left(f \right)=\left(f⁢(x)\right) x\in X.\\$$$$ Obviously, $$\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$$ is continuous and so is its inverse $$$$\varphi^{-1}:\varphi\left(B_{X^{*}}\right) \longrightarrow B_{X^{*}}\\$$$$ It remains to show that $$\varphi \left(B_{X*}\right)$$ is closed. If $$$$\varphi \left(f_{α}\right)$$$$ is a net in $$\varphi \left(B_{X*}\right)$$, converging to a point $$d=(d_{x})$$, $$x\in X\in D$$, we can define a function $$$$f: X \longrightarrow C , f⁢(x)=d_{x}.\\$$$$ As $$\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.

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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.