# Is the set of weak*-continuous operators closed in the weak*-operator topology?

I recently came across this unanswered MO question an answer to which I would also be interested in. However the formulation of said question is somewhat imprecise and lacking detail in my opinion so I figured I would post this question in my own words (if this is against this site's etiquette please let me know!).

Given normed spaces $$X,Y$$ (as usual over a complete field $$\mathbb F$$) one can consider the map $${}^*:\mathcal B(X,Y)\to\mathcal B(Y^*,X^*)$$ which maps any bounded operator $$T$$ to its adjoint operator $$T^*$$ (defined via $$T^*(y)=y\circ T$$ for all $$y\in Y^*$$). The map $${}^*$$ is known to be a linear isometry but in general it is not surjective. In fact one can show that $${}^*(\mathcal B(X,Y))=\{S\in\mathcal B(Y^*,X^*)\,|\,S\text{ is weak{}^*-continuous}\}$$ so $$\mathcal B(X,Y)\simeq \{S\in\mathcal B(Y^*,X^*)\,|\,S\text{ is weak{}^*-continuous}\}$$ by means of $${}^*$$. Here weak$${}^*$$-continuity refers to continuity of above $$S$$ as a map $$S:(Y^*,\sigma(Y^*,Y))\to (X^*,\sigma(X^*,X))$$ (i.e. continuity when equipping domain and codomain with the respective weak$${}^*$$-topology).

To ask about this set being closed we quickly have to recall some available topologies on $$\mathcal B(Y^*,X^*)$$: aside from the usual operator norm, strong operator and weak operator topology on this space one can equip it with the weak$${}^*$$-operator topology $$\tau_w^*$$ which is locally convex topology induced by the seminorms $$\{T\mapsto |(Ty)(x)|\}_{x\in X,y\in Y^*}$$. Equivalently $$\tau_w^*$$ is the coarsest topology on $$\mathcal B(Y^*,X^*)$$ such that all maps $$\{T\mapsto (Ty)(x)\}_{x\in X, y\in Y^*}$$ are continuous and a neighborhood basis of $$\tau_w^*$$ at $$T\in\mathcal B(Y^*,X^*)$$ is given by $$\{N^*(T,A,B,\varepsilon)\,|\,A\subset X\text{ and }B\subset Y^*\text{ both finite, }\varepsilon>0\}\quad\text{ where}\\ N^*(T,A,B,\varepsilon):= \{S\in\mathcal B(Y^*,X^*)\,|\,|(Ty)(x)-(Sy)(x)|<\varepsilon\text{ for all }x\in A,y\in B\}\,.$$ The idea behind this construction is to obtain a topology $$\tau_w^*$$ which is weaker than the weak operator topology (on $$\mathcal B(Y^*,X^*)$$) which is indeed the case; as expected these topologies coincide if $$X$$ is reflexive.

Now for some applications a desirable result would be the following: if a net $$(T_i)_{i\in I}$$ in $$\mathcal B(Y^*,X^*)$$ of weak$${}^*$$-continuous operators converges to $$T\in\mathcal B(Y^*,X^*)$$ with respsect to $$\tau_w^*$$ then $$T$$ is weak$${}^*$$-continuous itself.

In other words: is $${}^*(\mathcal B(X,Y))=\{T^*\,|\,T\in\mathcal B(X,Y)\}$$ closed in $$(\mathcal B(Y^*,X^*),\tau_w^*)$$?

This was also asked on math.SE in 2016 but the only answer given there is flawed because there is no reason for weak$${}^*$$-convergent nets to be bounded. In fact from my own attempts that seems to be the only thing which prevents a direct proof (e.g., showing that $$\mathcal B(Y^*,X^*)\setminus{}^*(\mathcal B(X,Y))$$ is open in $$\tau_w^*$$ using the neighborhood basis).

If this were true this would---as an immediate consequence---imply that the weak$${}^*$$-operator topology "mirrors" the weak operator topology (on $$\mathcal B(X,Y)$$) in the following sense:

Consider a subset $$A\subset {}^*(\mathcal B(X,Y))$$ with pre-dual set $$A_0\subset\mathcal B(X,Y)$$, i.e. $$(A_0)^*=A$$. Then $$A$$ is closed in the weak$${}^*$$-operator topology if and only if $$A_0$$ is closed in the weak operator topology.

An easy counterexample can be found as follows: Let $$X = \mathbb{F}$$ and let $$Y$$ be a non-reflexive Banach space. Then $$\mathcal{B}(Y^*,X^*)$$ is simply the bi-dual $$Y^{**}$$, and $${}^*(\mathcal{B}(X,Y))$$ is precisely the image $$j(Y)$$ of $$Y$$ in $$Y^{**}$$ under the evaluation map $$j: Y \to Y^{**}$$.
The topology $$\tau^*_w$$ on $$\mathcal{B}(Y^*,X^*) = Y^{**}$$ is simply the weak$${}^*$$-topology on $$Y^{**}$$, so $$j(Y)$$ is $$\tau^*_w$$-dense in $$Y^{**}$$, but not equal to $$Y^{**}$$ (since $$Y$$ is non reflexive). Hence, $$j(Y)$$ is not $$\tau^*_w$$-closed in $$Y^{**}$$.