ProofOfBanachAlaogluTheorem 2005-03-30 01:53:23 2006-09-19 08:16:47 Proof Banach-Alaoglu theorem 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here 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.