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

(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?

• Does this question answer your question? math.stackexchange.com/questions/1246979/… May 20, 2021 at 10:34
• Hm, ''Hence, ∃!y∈Y such that B(f)(x)=f(y)∀f∈Y∗ (This is a short lemma, that perhaps has been proved before in the textbook)?'' <- This may not be true in general normed spaces? (Its certainly true that if $Y$ is reflexive) May 20, 2021 at 10:40
• I think it works in any normed linear space. What does not, however, is the next step using the closed graph theorem, which needs completeness. May 20, 2021 at 14:38
• Does this answer your question? A linear map $S:Y^*\to X^*$ is weak$^*$ continuous if and only if $S=T^*$ for som $T\in B(X,Y)$ May 21, 2021 at 0:24
• Noted, thanks for the comment (retracted close vote) May 21, 2021 at 0:33

(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.

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 $$T\not\in \mathcal L(X,Y)$$.

• people.math.ethz.ch/~salamon/PREPRINTS/funcana-ams.pdf- The question appears here on page 190 (202 in the pdf). Could I be misunderstanding/misreading? May 20, 2021 at 22:33
• Top of the page 4.5.4 b May 20, 2021 at 22:34
• No, you are not misunderstanding. Either they are wrong, or my example is wrong. Note that in Conway's book the exercise requires both $X$ and $Y$ to be Banach. May 20, 2021 at 22:49
• Also I'm finding it difficult to find where exactly you need completeness of $Y$ in your proof (math.stackexchange.com/questions/1832836/…). A comment suggests it is in the invoking of closed graph theorm, but I thought dual spaces where always Banach spaces? (en.wikipedia.org/wiki/… under ''Dual spaces'' May 20, 2021 at 22:49
• One uses that $Y$ is Banach to say that any weak$^*$-continuous functional on $Y^{**}$ is given by evaluation at a point in $Y$. This is not true if $Y$ is not complete. May 20, 2021 at 22:50