On bounded sets of $L^2$ the weak topology is metriziable

Question

I just read the weak topology is metriziable on bounded sets of $L^2$. Does that mean if $h_n\in B$ and $h_n\to h$ weakly in $L^2$ then $h_n\to h$ w.r.t. $d$? Does this metric have an explicit formula?

Answer 1

This works if your measure space is such that $L^2$ is separable. We have the following fact.

Fact: if $\{y_n\}$ is a dense subset of $B$ and $\{h_n\}\subset B$, where $B$ is the unit ball, then $h_n\to 0$ weakly if and only if $\langle h_n,y_k\rangle\xrightarrow[n]{}0$.

Using the fact, we can define the metric $$ d(x,y)=\sum_k 2^{-n}|\langle x-y,y_k\rangle|. $$

It's crucial for everything to work that we work on a bounded set.

Also, a bit of reflexion on the proof should make it clear that what one is really showing is that if a Banach space has separable dual, then the weak topology is metrizable on bounded sets.

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Proof of the "Fact". Suppose that $\langle h_n,y_k\rangle\to0$ for all $k$. Fix $x$ and $\varepsilon>0$. Then there exists $m$ such that $\|x-y_m\|<\varepsilon$. We have $|\langle h_n,y_k\rangle|\leq \|h_n\|\,\|y_k\|\leq1$ for all $n,k$. Thus \begin{align} |\langle h_n,x\rangle| &\leq|\langle h_n,x-y_m\rangle|+|\langle h_n,y_m\rangle|\\[0.3cm] &\leq\|h_n\|\,\|x-y_m\|+|\langle h_n,y_m\rangle|\\[0.3cm] &\leq\varepsilon+|\langle h_n,y_m\rangle|\\[0.3cm] \end{align} From this we get $$ \limsup_n|\langle h_n,x\rangle|\leq\varepsilon. $$ And as $\varepsilon$ was arbitrary, the limit is zero.