# Does separability of the strong operator topology imply separability of the underlying space?

Let $X$ be a Banach space and $B(X)$ be the space of bounded operators on $X$.

Suppose that the strong operator topology on $B(X)$ is separable and that the cardinal number of $B(X)$ is continuum.

Question. Can we conclude the norm topology on $X$ is separable?

Remark. It was proved by Tomek Kania that the assumption $|B(X)|=\mathfrak{c}$ is not solely enough to get the separability assertion.

Let $y$ be a norm-one vector in $X$. Consider the evaluation map $T\mapsto Ty$, which is a (linear) surjection from $B(X)$ onto $X$. This map is SOT-norm continuous. Indeed, suppose that $T_n\to T$ in SOT. In particular, $\|T_ny - Ty\|\to 0$ as $n\to \infty$.
A continuous image of a separable space is separable, hence $X$ is separable.
• So it looks like $|B(X)|=\frak c$ is superfluous. – Fan Zheng Jun 9 '18 at 4:17