Exercise :

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

Attempt :

In a previous exercise, I have shown that if $A \in \mathcal{L}(X,Y)$, then $A(D) $ is closed. Now, $A$ is a compact operator.

Since $D$ is bounded, I know that a compact operator transfers bounded spaces to relatively compact spaces (aka compact closure spaces). We have t show that $A(D)$ is not only relatively compact, but compact in general.

Since I haven't handled a lot of problems regarding reflexiveness and convexity in my functional analysis and operator theory courses, I fail to get an intuition inspired by these two properties. On the other hand, in order for $A(D)$ to be compact, it needs to be closed and bounded. Since $A$ is a compact operator though, wouldn't boundedness and closedness of $D$ be transfered to $Y$ in the image $A(D)$ ?

Any hints or elaborations will be greatly appreciated.


You have shown that $A(D)$ is a closed subset. The fact thT $A(D)$ is relatively compact is equivalent to saying that the adherence $\overline{A(D)}$ of $A(D)$ is compact. $\overline{A(D)}=A(D)$ since $A(D)$ is closed, we deduce that $A(D)$ is compact.

  • $\begingroup$ Well only seeking the mathematic rigorousness, it is $\mathcal{L}_c(X,Y) \subset \mathcal{L}(X,Y)$ which I guess makes it safe to say that if $A(D)$ is closed unded a bounded operator, it will also be closed under a compact, correct ? $\endgroup$ – Rebellos Mar 5 at 16:14
  • $\begingroup$ A compact operator is bounded. $\endgroup$ – Tsemo Aristide Mar 5 at 16:16
  • $\begingroup$ Yeah, that's all. Thanks, it was rather straightforward. Generally, in the whole exercise, the essence of $X$ being reflexive and convex wasn't as complex as I though in the end ! $\endgroup$ – Rebellos Mar 5 at 16:17

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