Here is my effort to show this fact and I will use ball $X^*$ to denote the closed unit ball of $X^*$.

To show ball $X^*$ is weak star metrizable, we only have to show there is a metric $d$ on ball $X^*$ such that the topology induced by $d$ is the weak-star topology on ball $X^*$. Since $X$ is separable, ball $X$ is also separable. Thus there exists a countable dense subset $\{x_n\}$ in ball $X$. Now define the metric $d$ on ball $X^*$ by

$$d(x^*,y^*)=\sum_{n=1}^{\infty}\frac{|\langle x_n,x^*-y^*\rangle|}{2^n},$$

where $x^*,y^*\in\text{ball $X^*$}$. Let $T$ be the topology induced by $d$ and $wk^*$ be the weak-star topology on $\text{ball $X^*$}$. Then we need to show $T=wk^*$. And I try to use net convergence to show topology equivalence.

Let $x^*\in\text{ball $X^*$}$ and let $x_i^*$ be a net in $\text{ball $X^*$}$ such that $x_i^*\overset{wk^*}{\longrightarrow} x^*$. Then $\langle x_n,x_i^*\rangle\rightarrow\langle x_n,x^*\rangle$ for all $n$. Now Let $x^*\in\text{ball $X^*$}$ and let $x_i^*$ be a net in $\text{ball $X^*$}$ such that $x_i^*\rightarrow x^*$ in $(X,T)$. Then for each $\epsilon>0$, there exists $i_\epsilon\in\mathbb{N}$ such that $d(x_i^*,x^*)<\epsilon$ for all $i\geqslant i_\epsilon$; that is, for each $\epsilon>0$,

$$d(x_i^*,x^*)=\sum_{n=1}^{\infty}\frac{|\langle x_n,x_i^*-x^*\rangle|}{2^n}<\epsilon.$$

Here I want to show $x_i^*\overset{wk^*}{\longrightarrow} x^*$ if and only if $x_i^*\rightarrow x^*$ in $(X,T)$. But I forget some knowledge in Calculus. Can somebody help me to show this please? Thank you so much!!

  • $\begingroup$ maybe the fact that your ball in $X^*$ is weak* compact (by Banach-Alaoglu) helps. $\endgroup$ – noctusraid Sep 10 '17 at 15:05

Let $I$ denote the identity map from $(X^*,\text{weak*})$ onto $(X^*,d)$. Since $I$ is a bijection from a compact space (Banach-Alaoglu) into a Hausdorff space, we only need to show that $I$ is continuous.

To this end, fix a net $(x^*_\alpha)$ in $B_{X^*}$ such that $x^*_\alpha\to x^*$ in the weak* topology on $B_{X^*}$ for some $x^*\in B_{X^*}$. We want to show that $x^*_\alpha=I(x^*_\alpha)\to I(x^*)=x^*$ in the metric $d$; that is, we want: $$ \sum_{n=1}^\infty \frac{|\langle x_n,x_\alpha^*-x^*\rangle|}{2^n} \longrightarrow 0. $$ To this end, fix $\varepsilon>0$. Since each $x_n$ is in the unit ball of $X$ and $x^*$ and each $x^*_\alpha$ are in the unit ball of $X^*$, we deduce $$ \frac{|\langle x_n,x_\alpha^*-x^*\rangle|}{2^n} \le \frac{|x_\alpha^*(x_n)|+|x^*(x_n)|}{2^n} \le \frac{2}{2^n} = \frac{1}{2^{n-1}}. $$ Since $\sum_{n=1}^\infty2^{1-n}$ converges, there exists $N\in\mathbb{N}$ such that $\sum_{n=N+1}^\infty 2^{1-n}<\varepsilon/2$. Now, using the fact that $x_\alpha^*\to x^*$ in the weak* topology on $B_{X^*}$, for each $n\in\{1,\ldots,N\}$ there exists $\alpha_n$ in the directed set such that $$ |\langle x_n,x_\alpha^*-x^*\rangle| < \frac{2^n\varepsilon}{2N} $$ whenever $\alpha\ge\alpha_0$. Taking $\alpha_0$ in the direct set such that $\alpha_0\ge\alpha_1,\ldots,\alpha_N$, we obtain \begin{align*} \sum_{n=1}^\infty\frac{|\langle x_n,x_\alpha^*-x^*\rangle|}{2^n} &= \sum_{n=1}^N\frac{|\langle x_n,x_\alpha^*-x^*\rangle|}{2^n} +\sum_{n=N+1}^\infty\frac{|\langle x_n,x_\alpha^*-x^*\rangle|}{2^n} \\ &\le \sum_{n=1}^N \frac{\varepsilon}{2N} + \varepsilon/2 \\ &= \varepsilon \end{align*} whenever $\alpha\ge\alpha_0$. This completes the proof.

A few comments on your post:

You write "Since $X$ is separable, ball $X$ is separable". This is a true statement, but only because $X$ is a metric space. In general topological spaces, subsets of separable spaces aren't necessarily separable.

It's important to show that $d$ is a metric. It is here that the density of $\{x_n : n\in\mathbb{N}\}$ in the unit ball of $X$ is actually needed.

You wrote "Then for each $\varepsilon>0$, there exists $i_\varepsilon\in\mathbb{N}$ such that $d(x_i^*,x^*)<\varepsilon$ for all $i\ge i_\varepsilon$." The problem is that $i_\varepsilon$ cannot be assumed to be a natural number, since your net may be indexed by some directed set other than $\mathbb{N}$.

  • $\begingroup$ Thank you so much for your help! Can you show me why $X$ is separable, ball $X$ is separable? And I sorry I don't see the density of $\{x_n\}$ is using in the proof of show $d$ is a metric. Can you prove it if possible!! $\endgroup$ – Answer Lee Sep 10 '17 at 15:43
  • $\begingroup$ @AnswerLee For metric spaces, separability is equivalent to second countability. This, and the fact that subsets of second countable spaces are second countable, shows that subsets of a separable metric space are separable. (We say that separability is hereditary for metric spaces, but not in general.) $\endgroup$ – John Griffin Sep 10 '17 at 15:46
  • $\begingroup$ @AnswerLee The density of $\{x_n\}$ is used in showing that $d(x^*,y^*)=0$ implies $x^*=y^*$. $\endgroup$ – John Griffin Sep 10 '17 at 15:47
  • $\begingroup$ Awesome!! Actually can I ask you some questions about my homework by sending you an email? Because your answer is so great and I can totally understand! Appreciate!! $\endgroup$ – Answer Lee Sep 10 '17 at 15:55
  • $\begingroup$ @AnswerLee I wouldn't feel comfortable communicating via email. Whenever I'm not busy with something else, I usually have MSE open and am answering questions, so you can just post here. One tip for getting better responses would be to accept answers you are satisfied with. I've noticed that you haven't accepted any answers in all of the questions you've asked here. This may prevent some people from putting a lot of effort into answering your questions. $\endgroup$ – John Griffin Sep 10 '17 at 16:13

You can do this directly: after checking that $d$ is a metric (it is), let $\tau$ be the topology induced by it. Then,

$(1). \ \sigma (X^*,X)\subseteq\tau :$

Let $S$ be the closed unit ball in $X^*,\ $ fix $f_0\in S$ and take $\left \{ f\in S:d(f_o,f) <\epsilon \right \}\subseteq \tau.$ There is an integer $k$ such that $\frac{1}{2^{k}}<\frac{\epsilon }{2}.$ Then, $\ U=\left \{ f\in S:|(f-f_0,x_j)|< \frac{\epsilon }{2};\ 0\le j\le k \right \}$ is weak* open in $S,$ contains $f_0$ and satisfies:

$\forall f\in U,\ d(f,f_0)=\sum_{n=0}^{k}\frac{(f-f_0,x_n)}{2^{n}}+\sum_{n> k}\frac{(f-f_0,x_n)}{2^{n}}<\frac{\epsilon }{2}+\frac{1}{2^{k}}<\epsilon.$

$(2).\ \tau \subseteq\sigma (X^*,X) :$

Take $U=\left \{ f\in S:|(f-f_0,y_j)|<\epsilon;\ 0\le j\le k \right \}$, a weak* neighborhood of $f_0.$ Let $M=\max \left \{ \|y_j\|:0\le j\le k \right \}.$ There are integers $n_j$ such that for any $\delta>0,\ \|x_{n_j}-y_j/M\|<\delta.$ And there is an $r>0,$ such that $M2^{n_j}r<\frac{\epsilon }{2}.$

Then, if $d(f,f_0)<r,$ we have of course, $\sum_{n=0}^{\infty }\frac{|(f-f_0,x_n)|}{2^{n}}<r$ so that $|(f-f_0,x_{n_j})|\le 2^{n_j}r$ and therefore,

$|(f-f_0,y_j)|=M|(f-f_0,y_j/M)|\le M|(f-f_0,x_{n_j})|+M|(f-f_0,x_{n_j}-y_j/M)|\le M(2^{n_j}r+2\delta)\le \epsilon /2+2M\delta.$

Thus, $f\in U$ as soon as $\delta$ is small enough.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.