PlanetMath: proof of Banach-Alaoglu theorem

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proof of Banach-Alaoglu theorem

(Proof)

For any $x\in X$ , let $D_x=\{z\in\mathbb{C}: \vert z\vert\leq \Vert x\Vert\}$ and $D=\Pi_{x\in X} D_x$ . Since is a compact subset of $\mathbb{C}$ , is compact in product topology by Tychonoff theorem.

We prove the theorem by finding a homeomorphism that maps the closed unit ball $B_{X^*}$ of onto a closed subset of . Define $\Phi_x:B_{X^*}\to D_x$ by $\Phi_x(f)=f(x)$ and $\Phi:B_{X^*}\to D$ by $\Phi=\Pi_{x\in X}\Phi_x$ , so that $\Phi(f)=(f(x))_{x\in X}$ . Obviously, $\Phi$ is one-to-one, and a net $(f_\alpha)$ in $B_{X^*}$ converges to in weak-* topology of iff $\Phi(f_\alpha)$ converges to $\Phi(f)$ in product topology, therefore $\Phi$ is continuous and so is its inverse $\Phi^{-1}:\Phi(B_{X^*})\to B_{X^*}$ .

It remains to show that $\Phi(B_{X^*})$ is closed. If $(\Phi(f_\alpha))$ is a net in $\Phi(B_{X^*})$ , converging to a point $d=(d_x)_{x\in X}\in D$ , we can define a function $f:X\to \mathbb{C}$ by . As $\lim_\alpha \Phi(f_\alpha(x))=d_x$ for all $x\in X$ by definition of weak-* convergence, one can easily see that is a linear functional in $B_{X^*}$ and that $\Phi(f)=d$ . This shows that is actually in $\Phi(B_{X^*})$ and finishes the proof.