I recently came across [this unanswered MO question](https://mathoverflow.net/questions/242324/set-of-w-continuous-operators-closed-for-the-weak-topology-or-not) 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](https://math.stackexchange.com/questions/1832836/a-linear-map-sy-to-x-is-weak-continuous-if-and-only-if-s-t-for-so?noredirect=1&lq=1) 
$$
{}^\*(\mathcal B(X,Y))=\{S\in\mathcal B(Y^\*,X^\*)\,|\,S\text{ is weak${}^\*$-continuous}\}
$$
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](https://math.stackexchange.com/q/1818531) 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!



Is the set of weak*-continuous operators closed in the weak*-operator topology?
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 $Y^{\*\*}$.


