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Exercise :

Let $X$ be a reflexive Banach space and $Y$ a Banach space. Also, let $A \in \mathcal{L}(X,Y)$ and $D \subseteq X$ be a closed, convex and bounded space. Show that $A(D) \subseteq Y$ is closed.

Discussion :

I know that for $A(D)$ to be closed, theoritically one should show that every sequence in $A(D)$ converges in $A(D)$. Also, since $Y$ is Banach, if $A(D)$ is closed it should also be Banach, so that may be a way of showing that it is closed.

After doing some research, I came across some posts quoting the Banach-Alaoglu theorem, but this is something I haven't been taught, so I guess there should be a more elaborate way around.

Any hints or elaborations will be greatly appreciated.

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  • $\begingroup$ A proof not using Banach Alaoglu or weak compactness of balls in $X$ is likely to be quite involved. $\endgroup$ – Kavi Rama Murthy Mar 2 at 11:58
  • $\begingroup$ @KaviRamaMurthy We have talked about weak compactness of balls and also used them extensively, just have not mentioned the Banach Alaoglu Theorem. Any hints on how I could approach it ? $\endgroup$ – Rebellos Mar 2 at 11:59
  • $\begingroup$ You can deduce $D$ is weakly compact ($D$ is a weakly closed (since it is convex and norm closed) subset of a weakly compact set (take a ball containing $D$)). $A$ is weak-weak continuous. The continuous image of a compact set is compact, and compact sets in Hausdorff spaces are closed. $\endgroup$ – David Mitra Mar 2 at 17:33
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Let $\{d_n\} \subset D$ and $Ad_n \to y$. Since $D$ is bounded the sequence $\{d_n\}$ lies in a weakly compact set. Hence there is a subsnet $\{d_n'\}$ converging weakly to some point $x \in X$. But $D$ is weakly closed, so $x \in D$. Further any norm- norm bounded continuous linear map is weak - to weak continous. Hence it follows that $\{Ad_n'\} \to Ax$ weakly. Since $Ad_n \to y$ weakly it follows that $y=Ax \in A(D)$ so $A(D)$ is closed.

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    $\begingroup$ I have used the fact that a convex set is norm closed iff it is weakly closed. $\endgroup$ – Kavi Rama Murthy Mar 2 at 12:09
  • $\begingroup$ Thank you, sir. I forgot that the dimension of the spaces has to be finite. I removed my answer. $\endgroup$ – Sujit Bhattacharyya Mar 2 at 12:10
  • $\begingroup$ Thanks a lot for the input. I am currently reading through carefully to understand ever point. May I ask (probably foolishly), why does D bounded implies that $d_n$ lies in a weakly compact set ? $\endgroup$ – Rebellos Mar 2 at 12:18
  • $\begingroup$ In reflexive spaces closed balls are weakly compact. From your earlier comment I though you might know this fact already. $\endgroup$ – Kavi Rama Murthy Mar 2 at 12:20
  • $\begingroup$ @KaviRamaMurthy I do in fact, but I am a beginner in Reflexive spaces (just was introduced parallely to my Operator Theory classes). Thanks for clearing up! $\endgroup$ – Rebellos Mar 2 at 12:21

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