# If locally convex topologies exhibit the same dual spaces, do they exhibit the same continuous linear operators?

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 $T \in \mathfrak{L}_{\tau_2, \sigma_2}(X,Y)$.

• To see that you need some additional hypotheses, consider $Y = X$, $\sigma_1 = \sigma_2 = \tau_1 = \tau(X,X^{\ast})$ and $\tau_2 = \sigma(X, X^{\ast})$. Then the identity is $\tau_1$-$\sigma_1$ continuous, but not $\tau_2$-$\sigma_2$ continuous. – Daniel Fischer Feb 9 '17 at 16:46
• @DanielFischer: What do you mean by $\tau(X,X^*)$ and $\sigma(X,X^*)$? – el_tenedor Feb 9 '17 at 16:56
• $\tau(E,F)$ is the Mackey topology of the dual system $E,F$, that is, the strongest locally convex topology on $E$ such that $F$ is the (topological) dual of $E$, and $\sigma(E,F)$ is the weak topology of the dual system. Consider a Banach space, then the Mackey topology is the norm topology. – Daniel Fischer Feb 9 '17 at 17:01
• Alright. I suspected that my claim wouldn't hold without any additional hypotheses. Have read through my "proof"? I found a crucial hypothesis to be that some kind of closed graph theorem holds. That's why I restricted the rest of the proof to Fréchet spaces. Question is: Is my proof correct, and does my claim hold under weaker assumptions (maybe with a different proof)? – el_tenedor Feb 9 '17 at 17:07
• Your argument shows that the families of closed operators (i.e., with closed graph) are the same. If you have a closed graph theorem for $\tau_k$-$\sigma_k$ for both $k = 1$ and $k = 2$, then you have the same continuous operators. But closed graph theorems are rare. – Daniel Fischer Feb 9 '17 at 17:13