# The Alaoglu's Theorem

I am trying to prove the Alaoglu's Theorem. But my professor told me there is something wrong with my proof. Can anyone help me? Thank you!

The Alaoglu's Theorem: Let $X$ be a normed space. Then ball $X^*$ is weak-star compact.

Proof:Let $D_x=\{\alpha\in\mathbb{F}:|\alpha|\leqslant1\}$ for each $x\in$ ball $X$. Let $$D=\prod\{D_x:x\in\textrm{ball } X\}.$$ Since $D_x$ is compact, by Tychonoff's Theorem, $D$ is compact. For each $x\in$ ball $X$, define $\tau:\textrm{ball } X^*\rightarrow D$ by $\tau(x^*)(x)=\langle x,x^*\rangle$. That is, $\tau(x^*)$ is the element of the product space $D$ whose $x$ coordinate is $\langle x,x^*\rangle$. Let $x^*\in \textrm{ball } X^*$ and let $(x_i^*)$ be a net in ball $X^*$ with $x_i^*\rightarrow x^*$. Then for each $x\in$ ball $X$, $$\tau(x_i^*)(x)=\langle x,x_i^*\rangle\rightarrow\langle x,x^*\rangle=\tau(x^*)(x).$$ That is, each coordinate of $\{\tau(x_i^*)\}$ converges to $\tau(x^*)$. Hence, $\tau(x_i^*)\rightarrow\tau(x^*)$ in $D$. Therefore, $\tau:\textrm{ball } X^*\rightarrow D$ is continuous.

Let $\tau(x_1^*)=\tau(x_2^*)$ for $x_1^*,x_2^*\in \textrm{ball } X^*$. Then $\langle x,x_1^*\rangle=\langle x,x_2^*\rangle$ for each $x\in$ ball $X$, and thus $x_1^*=x_2^*$. Hence, $\tau$ is injective. It implies that $\tau^{-1}$ exists. Let $\tau(x^*)\in D$ and let $\tau(x_i^*)$ be a net in $D$ with $\tau(x_i^*)\rightarrow\tau(x^*)$. Then for all $x\in \textrm{ball } X$, $\tau(x_i^*)(x)\rightarrow\tau(x^*)(x)$. Thus $\langle x,x_i^*\rangle\rightarrow\langle x,x^*\rangle$ for all $x\in \textrm{ball } X$. Hence, $x_i^*\rightarrow x^*$ in ball $X^*$ for all $x\in \textrm{ball } X$. Therefore, $\tau^{-1}:D\rightarrow\textrm{ball } X^*$ is continuous. Since the map from $\textrm{ball } X^*$ to $\tau(\textrm{ball } X^*)$ is bijective, $\tau$ is a homeomorphism from $\textrm{ball } X^*$ onto $\tau(\textrm{ball } X^*)$.

Let $f\in D$ and let $(x_i^*)$ be a net in ball $X^*$ such that $\tau(x_i^*)\rightarrow f$ in $D$. Then $f(x)=\lim\langle x,x_i^*\rangle$ exists for every $x\in$ ball $X$. Now we extend $f$ to $F:X\rightarrow\mathbb{F}$. Let $\alpha>0$ be such that $\|\alpha x\|\leqslant1$ for all $x\in X$. Then define $F(x)=\alpha^{-1}f(\alpha x)$. If also $\beta>0$ such that $\|\beta x\| \leqslant1$ for all $x\in X$, then $$\alpha^{-1}f(\alpha x) = \alpha^{-1} \lim \langle\alpha x, x_i^*\rangle=\beta^{-1}\lim\langle\beta x, x_i^*\rangle = \beta^{-1}f(\beta x).$$ Hence, $F(x)$ is well defined. For any $x_1,x_2\in X$ and $a\in\mathbb{F}$, we have \begin{align*} F(ax_1+x_2)&=\alpha^{-1}f(\alpha (ax_1+x_2))\\&=\alpha^{-1}\lim\langle\alpha ax_1+\alpha x_2,x_i^*\rangle\\&=a\alpha^{-1}\lim\langle\alpha x_1,x_i^*\rangle+\alpha^{-1}\lim\langle\alpha x_2,x_i^*\rangle\\&=a\alpha^{-1}f(\alpha x_1)+\alpha^{-1}f(\alpha x_2)\\&=aF(x_1)+F(x_2). \end{align*} Hence, $F$ is linear. Since $f\in D$, $f(x)\in D_x$ for all $x\in$ ball $X$. Thus $|f(x)|\leqslant1$ for all $x\in \textrm{ball } X$. Now let $\alpha=\frac{1}{\|x\|}$ for $x\in X-\{0\}$. Then $F(x)=\|x\|f(\frac{x}{\|x\|})$. Since $\frac{x}{\|x\|}\in \textrm{ball } X$, $|f(\frac{x}{\|x\|})|\leqslant1$. Thus $\|F(x)\|\leqslant\|x\|$ for $x\in X-\{0\}$. If $x=0$, $F(x)=0$. Hence, $\|F(x)\|\leqslant\|x\|$ for all $x\in X$. It implies that $F:X\rightarrow\mathbb{F}$ is a linear bounded functional and $\|F\|\leqslant1$; that is, $F\in\textrm{ball } X^*$. For any $x\in \textrm{ball } X$, we have $\tau(F)(x)=\langle x,F\rangle=f(x)$. It follows that $f\in \tau(\textrm{ball } X^*)$. Since $\tau(x_i^*)\rightarrow f$, $\tau(\textrm{ball } X^*)$ is closed in $D$. Thus $\tau(\textrm{ball } X^*)$ is compact. Therefore, ball $X^*$ is weak-star compact.

• Do you have more specific information about what part is supposed to be wrong? Here is a similar question: math.stackexchange.com/questions/1397655/… – s.harp Jul 4 '17 at 21:34
• Something wrong with the proof of \tau and \tau^{-1}. And my professor said the proof of letting \alpha=1/||x|| is redundant. I don't know why I thought it is true. – Answer Lee Jul 5 '17 at 3:13
• The part with $\alpha=1/\|x\|$ is redundant because if you have an element $f$ of $X^*$ so that it sends every element of the unit ball of $X$ into $D$ then $\|f\|≤1$ automatically. When you looked at $\tau^{-1}$ you wrote that it is a function $D\to B_{X^*}$. This is wrong, it is a function $\tau(B_X)\to B_{X^*}$. – s.harp Jul 5 '17 at 10:54
• Thanks for your reply!! But I am sorry I am not quite get it. Why $\tau^{-1}$ is from $\tau(B_X)$ to $B_{X^*}$? And my professor also said the proof of $F$ is linear and the last line of showing $\tau$ continuous are unclear at all. – Answer Lee Jul 5 '17 at 14:40
• A stupid example to illustrate what I mean: look at the inclusion $i:\Bbb N\to \Bbb Z$. This is injective, but the inverse is not defined for $-1$, only for the positive integers, ie $i(\Bbb N)$. For the same reason the inverse map here is only defined on the image of the map. For continuity of $\tau$ you verify continuity along nets. You use that $x_i^*\to x^*$ iff $x_i^*(x)\to x^*(x)$ for all $x$. Since $(\pi_x\circ \tau)(x_i^*)=x_i^*(x)$ this implies that $\pi_x\circ\tau$ is continuous for any $x$, thus $\tau$ is continuous. I don't see a problem here. – s.harp Jul 5 '17 at 15:30