A bounded linear operator from $X ^*\rightarrow Y^*$ is weak*-weak* continuous iff It is the adjoint of some Bounded linear operator $X\rightarrow Y$

Question

(this question is not a duplicate of this one since the latter only addresses the situation in the case of Banach spaces)

Let $X,Y$ be normed vector spaces and $B:Y^*\rightarrow X^*$ a linear operator. We want to show that $B$ is $weak*-weak*$ continuous iff $B=A^*$ for some $A \in \mathcal{L}(X,Y)$.

My intial idea was to set $A=\iota^{-1}_Y\circ B^{*}\circ\iota_X$ where $\iota:X \rightarrow X^{**}$ is the canonical embedding $x \mapsto ev_x$, the evaluation map of $x$ ie $\iota(x)f=f(x)$. I think this will work except how can I know $\iota^{-1}$ is defined (that is, how can I guarantee $B^*(\iota(x)) \in \iota (Y))?$ If this where Banach space I would be done, but I don't know what to do in this setting. Am I even on the right track?

EDIT: I found A linear map $S:Y^*\to X^*$ is weak$^*$ continuous if and only if $S=T^*$ for some $T\in B(X,Y)$ but I'm not clear on the setting. It looks to me (admitly naively) they are assuming reflexivity of $Y$ (and, or that $Y$ Banach, which I don't have?

Answer 1

(Edited 2024-02-03)

The result is true even if $Y$ is not complete. See this answer for a proof. The proof uses the Closed Graph Theorem, but it does so on the duals, which are Banach even if $X,Y$ aren't.

The counterexample in my original answer was wrong. The subtlety, noted in the comments by jantomico, is that while we have results like the dual of a normed space and the dual of its closure is the same, the weak$^*$-topology does depend on whether one takes the closure or not. In other words, in the context of the wrong example below, $\sigma(X^*,X)$ and $\sigma(X^*,\overline X)$ are not the same topology when $X$ is not complete.

I'm leaving the wrong example below, because it is mentioned in many comments, so that the conversation makes sense.

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Warning: The example below is WRONG. The map $S$ constructed in the example is $\sigma(Y^*,\overline Y)-\sigma(X^*,\overline X)$ continuous, but it is not $\sigma(Y^*,Y)-\sigma(X^*,X)$ continuous, which is what was asked.

For instance take $X=Y\subset\ell^1$ be $$ X=Y=\{x\in\ell^1:\ \exists n_0:\ n\geq n_0\implies x(n)=0\}. $$ Because $X$ and $Y$ are dense in $\ell^1$, we have $X^*=Y^*=\ell^\infty$.

Define $S:Y^*\to X^*$, that is $S:\ell^\infty\to\ell^\infty$ by $$ Sw=\big(\sum_n\frac{w(n)}{n^2},0,0,\ldots\big). $$ If $w_j\to0$ weak$^*$, this means that $\sum_nw_j(n)y(n)\to0$ for all $y\in Y$. In particular $\sum_n\frac{w_j(n)}{n^2}\to0$, and it follows $S$ is weak$^*$-weak$^*$ continuous.

If we had $S=T^*$, with $T\in \mathcal L(X,Y)$ this would mean that, for each $w\in\ell^\infty$ and $x\in X$, $$ (Sw)x=w(Tx). $$ This translates to $$ \sum_n\frac{w(n)x(1)}{n^2}=\sum_nw(n)\,(Tx)(n). $$ As this should work for all $w\in\ell^\infty$, it follows that we need $$ Tx=\bigg(\frac{x(1)}{n^2}\bigg)_n. $$ But then $Tx\not\in Y$ for any nonzero $x$, and so $T\not\in \mathcal L(X,Y)$.