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(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?

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  • $\begingroup$ Does this question answer your question? math.stackexchange.com/questions/1246979/… $\endgroup$
    – PhoemueX
    May 20, 2021 at 10:34
  • $\begingroup$ 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) $\endgroup$
    – Muselive
    May 20, 2021 at 10:40
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    $\begingroup$ 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. $\endgroup$ May 20, 2021 at 14:38
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    $\begingroup$ Does this answer your question? A linear map $S:Y^*\to X^*$ is weak$^*$ continuous if and only if $S=T^*$ for som $T\in B(X,Y)$ $\endgroup$ May 21, 2021 at 0:24
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    $\begingroup$ Noted, thanks for the comment (retracted close vote) $\endgroup$ May 21, 2021 at 0:33

1 Answer 1

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

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  • $\begingroup$ 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? $\endgroup$
    – Muselive
    May 20, 2021 at 22:33
  • $\begingroup$ Top of the page 4.5.4 b $\endgroup$
    – Muselive
    May 20, 2021 at 22:34
  • $\begingroup$ 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. $\endgroup$ May 20, 2021 at 22:49
  • $\begingroup$ 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'' $\endgroup$
    – Muselive
    May 20, 2021 at 22:49
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    $\begingroup$ 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. $\endgroup$ May 20, 2021 at 22:50

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