Consider the following setting:
Let $X, Y$ be vector spaces over the field $\mathbb{K} \in \{\mathbb{R}, \mathbb{C}\}$. Furthermore, let $\tau_1, \tau_2$ be locally convex topologies on $X$ and $\sigma_1, \sigma_2$ locally convex topologies on $Y$ such that their topological duals coincide, i.e. $${X}^{*_{\tau_1}} = X^{*_{\tau_2}}\quad\text{and}\quad {Y}^{*_{\sigma_1}} = Y^{*_{\sigma_2}}. \tag{$\ast$}$$
Under which additional assumptions do we have that their linear continuous operators coincide, i.e. $$ \mathfrak{L}_{\tau_1, \sigma_1}(X,Y) = \mathfrak{L}_{\tau_2, \sigma_2}(X,Y) ? $$
I have a proof for this property once I assume there exista a closed graph theorem for the pairs $(\tau_1, \sigma_1)$ and $(\tau_2, \sigma_2)$. See below for a proof.
I suppose that the assumptions can be weakened further if we only assume one pair, e.g. $(\tau_1, \sigma_1)$ to have a closed graph theorem available and the second pair as weak topologies relative to $\tau_1$ and $\sigma_1$. In this case, I'm sure that one can adapt the proof as in this post or that other one.
What do you think?
My Idea: I think that we can build on the fact that $(\ast)$ gives us that the closure of convex sets coincides, i.e. if $C \subseteq X$ and $K \subset Y$ are convex sets, we have $$ \overline{C}^{\tau_1} = \overline{C}^{\tau_2} \quad\text{and}\quad \overline{K}^{\sigma_1} = \overline{K}^{\sigma_2}. $$
Furthermore, we need some closed graph theorem for locally convex spaces. This should give us the first restriction to Fréchet spaces (can we say more? I was hoping to get a result which would solve this problem, but afaik strong and weak operator topologies are not Fréchet). Then, I would argue as follows:
If $T \in \mathfrak{L}_{\tau_1, \sigma_1}(X,Y)$, then its graph is closed w.r.t. the product topology $\tau_1 \times \sigma_1$ on $X\times Y$. This is once again a locally convex space and $$ (X \times Y)^{*_{\tau_1 \times \sigma_1}} = X^{*_{\tau_1}} \times Y^{*_{\sigma_1}} = X^{*_{\tau_2}} \times Y^{*_{\sigma_2}} = (X \times Y)^{*_{\tau_2 \times \sigma_2}}. $$ Furthermore, since the graph of a linear operator is a convex set, we have $$ \Gamma(T) = \overline{\Gamma(T)}^{*_{\tau_1 \times \sigma_1}} = \overline{\Gamma(T)}^{*_{\tau_2 \times \sigma_2}}. $$ Once again, applying the closed graph theorem but in the other direction gives $T \in \mathfrak{L}_{\tau_2, \sigma_2}(X,Y)$.
