Separability of unit ball in strong operator topology

Question

Let $X$ be a separable Banach space. Let $B_1(X)$ be the set of all bounded linear operators $X \to X$ with operator norm $\leq 1$. Does $B_1(X)$ have to be separable in the strong operator topology?

Answer 1

It turns out the answer is yes. Let $(x_k)_{k \geq 1}$ be a countable dense subset of $X$ and consider the set $A = \{ (Tx_k)_{k \geq 1} : T \in B_1(X) \} \subseteq X^{\mathbb{N}}$. Since $X$ is separable and metrizable, $X^{\mathbb{N}}$ is also separable and metrizable, and therefore $A$ is also separable. Let $(T_n)_{n \geq 1}$ be an enumeration of the operators corresponding to a countable dense subset of $A$.

The set $\{T_n\}_{n \geq 1}$ is dense in $B_1(X)$ for the strong operator topology. To see this, fix $T \in B_1(X)$ and consider the sequence $(Tx_k)_{k \geq 1} \in A$. By construction there is a subsequence $(T_{n_j})_{j \geq 1}$ such that $(T_{n_j} x_k)_{k \geq 1} \to (Tx_k)_{k \geq 1}$ in $A$ as $j \to \infty$. In particular this implies that $T_{n_j} x_k \to T x_k$ as $j \to \infty$ for each fixed $k$. Finally because $\{x_k\}_{k \geq 1}$ is dense in $X$ this implies that $T_{n_j} \to T$ in the strong operator topology.

Answer 2

Let me add that it is not separable in the uniform operator topology. The example is in Section 1.2 in 1.

Stochastic Equations in Infinite Dimensions , DA PRATO, ZABCZYK