Let $X, Y$ be Banach spaces. Moreover, let $\mathcal{L}(X,Y)$ be the space of bounded linear operators equipped with the standard operator norm topology, and $\mathcal{L}_{\mathrm{s}}(X,Y)$ the same space but with the strong operator topology.
When do Borel $\sigma$-algebras generated by the operator norm topology and the strong operator topology coincide? In other words $$\mathfrak{B}(\mathcal{L}(X,Y)) = \mathfrak{B}(\mathcal{L}_{\mathrm{s}}(X,Y)).$$
Some partial results:
$\mathfrak{B}(\mathcal{L}(X,Y)) \supset \mathfrak{B}(\mathcal{L}_{\mathrm{s}}(X,Y)) $ is trivial;
Under additional assumptions; that $X,Y$ are separable; it can be found along the lines of Lemma A.2 in A semi-invertible operator Oseledets theorem; C. González-Tokman & A. Quas that $\{T\in \mathcal{L}(X,Y) : \|T\|\le r \}\in\mathfrak{B}(\mathcal{L}_{\mathrm{s}}(X,Y))$ for all $r$. Therefore, if in addition $\mathcal{L}(X,Y)$ is also separable (with respect to operator norm topology) then $\mathfrak{B}(\mathcal{L}(X,Y)) = \mathfrak{B}(\mathcal{L}_{\mathrm{s}}(X,Y))$ should hold. Since, any open set will be the countable union of balls;
Another approach is to look at a special case when $Y=\mathbb{R}$. Then $\mathcal{L}_{\mathrm{s}}(X,\mathbb{R})$ will become $X^*$ with the weak-* topology. Since, any Hausdorff topology weaker than the norm topology (such as the weak-* topology) is equal to the norm topology on a compact subset; $\mathfrak{B}(\mathcal{L}(X,\mathbb{R})){\restriction}_{A} = \mathfrak{B}(\mathcal{L}_{\mathrm{s}}(X,\mathbb{R})){\restriction}_{A}$ for any compact $A$ in $\mathcal{L}(X,\mathbb{R})$.
I'm rather interested in some more general results concerning only assumption on $X$ and $Y$, not $\mathcal{L}(X,Y)$. References welcome, thanks in advance.
