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Homeproof of Banach-Alaoglu theorem

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# 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.

## Mathematics Subject Classification

46B10*no label found*

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