# $A(D) \subseteq Y$ is closed, if $X$ is reflexive, $Y$ is Banach and $D \subseteq X$ is closed, convex and bounded.

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.

• A proof not using Banach Alaoglu or weak compactness of balls in $X$ is likely to be quite involved. – Kavi Rama Murthy Mar 2 at 11:58
• @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 ? – Rebellos Mar 2 at 11:59
• 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. – David Mitra Mar 2 at 17:33

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.
• 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 ? – Rebellos Mar 2 at 12:18