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

Question

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.

Thanks in advance for any answer or comment!

Answer 1

The answer is "no", in general.

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 $Y^{**}$.