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.
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.