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As the title says, the question is how to prove

Let $X,Y$ be Banach spaces. A linear map $S:Y^*\to X^*$ is weak$^*$ continuous if and only if $S=T^*$ for some $T\in B(X,Y)$

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  • $\begingroup$ It is not entirely clear to me what the purpose of this question is. Do you want someone to verify your proof or are you looking for other proofs? $\endgroup$ – user342207 Jun 23 '16 at 16:40
  • $\begingroup$ It was about this question, that was closed while I was typing my answer (by users with no reputation in functional analysis) and remained closed for several days. By now, the question has been closed again, as a duplicate. $\endgroup$ – Martin Argerami Jun 23 '16 at 17:56
  • $\begingroup$ I am sorry to hear about that. I have the feeling something similar happened with this question: math.stackexchange.com/questions/1830609/…, which I answered. $\endgroup$ – user342207 Jun 23 '16 at 18:00
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    $\begingroup$ There is a strong tendency by some (many?) users to close questions that should no effort or context. This criterion is probably fine for calculus, say, because of the amount of homework questions posed. But it makes no sense for topics like functional analysis or c$^*$-algebras that belong to graduate courses. Regarding your answer, the downvote is probably due to the fact that linking to an image is frown-upon. The spirit of SE is to type the solutions so they are stored by SE (and it makes sense, because many links don't live for long, so eventually that question will have no answer). $\endgroup$ – Martin Argerami Jun 23 '16 at 18:16
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If $S=T^*$, and $f_j\to f$ weak$^*$ in $Y^*$, then for any $x\in X$ $$ Sf_j(x)=f_j(Tx)\to f(Tx)=T^*f(x)=Sf(x). $$ So $S$ is weak$^*$-continuous.

Conversely, assume that $S$ is weak$^*$-continuous. We know what our $T$ should satisfy if it exists: $Sf(x)=f(Tx)$. So let us use this to define $T$.

For any $x\in X$, consider the functional on $Y^*$ given by $\alpha_x:f\longmapsto Sf(x)$. By hypothesis this is weak$^*$-continuous; the weak$^*$-continuous functionals are bounded, so $\alpha_x\in Y^{**}$. But, also, a weak$^*$-continuous functional on the dual is in the predual. So there exists $y\in Y$ with $\alpha_x(f)=f(y)$ for all $f\in Y^*$. Define $Tx=y$. Note that $y$ is unique because $Y^*$ separates points in $Y$. From there we deduce that $T$ is linear. Finally, we have that $S $ is bounded; indeed, if $f_j\to f $ in $Y^* $ and $Sf_j\to g $ in $X^*$ (both in norm), then $f_j\to f $ in the weak$^*$ topology, so $Sf_j\to Sf $ weak$^*$; thus $Sf=g $ and $S $ is bounded by the Closed Graph Theorem. Then \begin{align} \|Tx\|&=\sup\{|f(Tx)|:\ f\in Y^*,\ \|f\|=1\}\\ \ \\ &=\sup\{|Sf(x)|:\ f\in Y^*,\ \|f\|=1\}\leq\|S\|\,\|x\|. \end{align}So $T$ is bounded with $\|T\|\leq\|S\|$.

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    $\begingroup$ you "should" have waited a little bit before posting your answer to your question so you can see others solutions and compare it to your own to see whose is the right answer, etc... $\endgroup$ – DeepSea Jun 20 '16 at 5:05
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    $\begingroup$ Then maybe the site shouldn't offer the "Answer your own question" button when one is posting the question. $\endgroup$ – Martin Argerami Jun 20 '16 at 5:13
  • $\begingroup$ Very nice answer ...But I am not able to understand the last line of your proof...Are you using $\|Sf(x)\|\leq \|Sf\|\|x\|\leq \|S\|\|f\||x\|$? Then you have to show that $S$ is norm continuous. $\endgroup$ – Black-horse Nov 28 '16 at 6:20
  • $\begingroup$ Indeed. I have edited the answer. $\endgroup$ – Martin Argerami Nov 28 '16 at 8:09

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