# proof of Banach-Alaoglu theorem

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

We prove the theorem  by finding a homeomorphism that maps the closed unit ball $B_{X^{*}}$ of $X^{*}$ onto a closed subset of $D$. 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 $f$ in weak-* topology   of $X^{*}$ 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 $f(x)=d_{x}$. As $\lim_{\alpha}\Phi(f_{\alpha}(x))=d_{x}$ 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 $\Phi(f)=d$. This shows that $d$ is actually in $\Phi(B_{X^{*}})$ and finishes the proof.

Title proof of Banach-Alaoglu theorem ProofOfBanachAlaogluTheorem 2013-03-22 15:10:03 2013-03-22 15:10:03 Mathprof (13753) Mathprof (13753) 12 Mathprof (13753) Proof msc 46B10