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.


1 Answer 1


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.

  • $\begingroup$ It is probably worth pointing out that what I really proved here was that if $X$ with the weak topology is 2nd countable, then $X^*$ is separable. $\endgroup$ Dec 6, 2017 at 10:50

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