# $A(D) \subseteq Y$ is compact if $A \in \mathcal{L}_c(X,Y)$, $X$ reflexive, $Y$ Banach and $D$ closed, convex and bounded.

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.
• 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 ? – Rebellos Mar 5 at 16:14
• 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 ! – Rebellos Mar 5 at 16:17