Let $X$ be a Banach space. By the weak operator topology on $B(X)$, we mean the locally convex topology implemented by the following semi-norms: 
$$B(X)\to[0,\infty) : T\to|\langle Tx,x^\*\rangle|$$
where $x\in X$ and $x^\*\in X^\*$.


Q. True or false: Assume that with respect to the weak operator topology, the unit ball of $B(X)$ is second countable. Then the Banach space $X^\*$ is separable.


I had two abortive attempts:


1st. *Yes it is true:* The wot-separability of $B(X)\_{\|.\|\leq1}$ guarantees existence two sequences $\{x\_n\}\subseteq X$ and $\{x^\*\_m\}\subseteq X^\*$ such that $N\_{n,m}$ forms a sub-basic nbhds (at 0) for the weak operator topology where 
$$N\_{n,m}=\{T\in B(X)\_{\|.\|\leq1}: |\langle Tx\_n,x^\*\_m\rangle|<1\}$$
Let us consider $A$=conv$\_r(\{x^\*\_m\})$, the rational convex hull of $\{x^\*\_m\}$, which is a countable set. If $A$ is not dense in $X^\*$ then one may find $\phi\in X^{\*\*}$ with $\phi(x^\*\_m)=0$ for all $m\geq1$. It implies that all bounded linear maps of the form of $f\otimes\phi\in B(X,X^{\*\*})$, given by $x\to f(x)\phi$, are contained in the intersection $\bigcap N\_{n,m}$. Note that there is no nonzero bounded linear map $T\in B(X)$ contained in the intersection $\bigcap N\_{n,m}$but likely there are some operators in $B(X,X^{\*\*})$ contained in the intersection $\bigcap N\_{n,m}$! 


2ed. To find a counterexample, let us put $X=\ell^1$. We have in general $B(X,Y^{\*})\simeq (X\hat{\otimes}Y)^\*$. Therefore $B(\ell^1)=(\ell^1\hat{\otimes}c\_0)^\*$. One may check easily that the inclusion $\iota: B(\ell^1)\to(\ell^1\hat{\otimes}c\_0)^\*$ is wot-weak star continues but not homeomorphism (to see this, it is enough to consider shifts $T\_n(e\_k)=e\_{n+k}$). Note that the unit ball of $(\ell^1\hat{\otimes}c\_0)^\*$ is $w^\*$-compact and metrizable. Since the inclusion is not hoemorphism, one may not conclude the unit ball of $B(\ell^1)$ is wot-compact and metrizable but probably it remains second countable at least!


Finally note if $X^\*$ is separable then the unit ball of $B(X)$ is wot-second countable metrizable space. To see this assertion, note that the projective tensor product $X\hat{\otimes}X^\*$ is also separable. Therefore the unit bal of dual space $(X\hat{\otimes}X^\*)^\*$ is weak-star compact and metrizable. Since the inclusion from ($B(X)\_{\|.\|\leq1}$,wot) into the unit ball of $(X\hat{\otimes}X^\*)^\*$ is a homeomorphism, one may conclude that the unit ball of $B(X)$ is wot-second countable metrizable space.



An equivalent condition for separability of $X^*$
I *think* this works. Suppose $B(X)$ with WOT is second countable, so there is a countable base to the topology $(U\_n)$. This means that if $(T\_i)$ is a net in $B(X)$, then $T\_i\rightarrow 0$ WOT if and only if, for each $n$ with $0\in U\_n$, there is $i\_n$ so that $i\geq i\_n \implies T\_i\in U\_n$.


Each WOT open set is of the form
$$ \mathcal U(T\_0,(x\_j),(x\_j^\*)) := \{ T\in B(X) : |\langle (T-T\_0)(x\_j), x\_j^\* \rangle|\leq 1 \ (j=1,\cdots,n)\} $$
where $T\_0\in B(X), (x\_j)\_{j=1}^n\subseteq X, (x\_j^\*)\_{j=1}^n\subseteq X^\*$.  

Let $Q\subseteq X^\*$ be the collection of all $x\_j^\*$ which occur for some $U\_n$. Thus $Q$ is countable, and so the rational linear combinations of $Q$ is also a countable set.


Towards a contradiction, suppose that $X^\*$ is not separable. Hence the closed linear span on $Q$ is not all of $X^\*$. Thus there is $f\_0\in X^\*$ and a (bounded) net $(x\_i)$ in $X$ such that $f\_0(x\_i)=1$ for each $i$ but $x^\*(x\_i)\rightarrow 0$ for each $x^\*\in Q$.


Choose some $f\_1\in X^\*$ and let $T\_i$ be the rank-one operator $x\mapsto f\_1(x) x\_i$. Let $U\_n$ have the form $\mathcal U(T\_0,(x\_j),(x\_j^\*))$. As each $x\_j^\*\in Q$ we have that
$$ \langle T\_i(x\_j), x\_j^\* \rangle = f\_1(x\_j) x\_j^\*(x\_i) \rightarrow 0 $$
and so as $0\in U\_n$, if $i$ is sufficiently large also $T\_i\in U\_n$.


This is a contradiction, as if $f\_1(x\_1)=1$ say, then $\langle T\_i(x\_1), f\_0 \rangle = f\_0(x\_i) = 1$ for all $i$, and so $T\_i\not\rightarrow 0$ WOT.


Thus $X^\*$ is separable.


