Let $K: X \to Y$ be a bounded linear operator, where $X$ and $Y$ are two Banach spaces. Further assume that the image $imK$ is a $\infty$-dim closed subspace of Y.

In my script they claim that in such a setting $K$ can never be compact because by the closed image theorem we have that $K(B)$ ($B$ closed unit ball in $X$) contains an open ball and hence has no compact closure because the dimension is $\infty$.

My questions:

I understand that the dimension of the closed unit ball $B$ characterizes the dimension of the underlying Banach space but I don't understand how this argument is used here. Also I don't quite get the thing with the open ball due to the closed image theorem and the thing that follows with the compact closure. How does one mash these things together the right way?


Ok I might have found another way to proof the above:

Define $K_1: X \to im(K)$ by $x \mapsto K(x)$. Then $K_1$ is surjective, hence $K_1(B_{open})$ ($B_{open}$ denotes the open unit ball in x) contains a small $\delta$-ball $B_{\delta}$ centered at the origin of $Y$ by the open mapping theorem. If we assume $cl(K_1(B_{open}))$ to be compact in $im(K)$ we get that $cl(B_{\delta})=\{y \in Y | \Vert y \Vert_Y \leq \delta\} \subset cl(K_1(B_{open}))$ and since Y is Banach it's also Hausdorff therefore it follows that $cl(B_{\delta})$ is compact and hence by scaling the unit ball in $Y$ is also compact which contradicts the $\infty$-dimensional property of $im(K)$.

Is this right?

  1. We will replace $Y$ be the image of $X$, so that we can assume the map is surjective (just for notational reasons).
  2. Hence, by the open mapping theorem, $K : X \to Y$ is open.
  3. Suppose that the image of the open unit ball $B_X$ in $X$ was relatively compact. Then the closure of $K(B_X)$ is compact, but also contains the open set $K(B_X)$. Since $K(B_X)$ is an open set containing the origin, there is some $\lambda \in \mathbb{R}$ so that $B_Y \subseteq \lambda K(B_X)$. This implies that $B_Y$ is relatively compact, since $\lambda K(B_X)$ remains relatively compact.
  4. Now apply the theorem that a Banach space in which the closed unit ball is compact is finite dimensional.
  • $\begingroup$ Whoops I think worked out the same proof, didn't see your answer while I was editting! So I may assume that my proof in the edit is correct? $\endgroup$ – noctusraid Jan 14 '16 at 10:24
  • $\begingroup$ @AlessioPellegrini I think so. One comment is that you don't need Hausdorffness for a closed subset of a compact to be compact. (Hausdorff implies that compacts are closed.) $\endgroup$ – Lorenzo Najt Jan 14 '16 at 10:26
  • $\begingroup$ Oh yeah right, I just need the limit to be in the subset again, so being closed in a compact set implies being compact. $\endgroup$ – noctusraid Jan 14 '16 at 10:31
  • $\begingroup$ @AlessioPellegrini Yes, that works. I like to think of it as: an open cover of a closed $C \subset X$ can be extended to an open cover of $X$ by adding $X \setminus C$ to the set of opens... $\endgroup$ – Lorenzo Najt Jan 14 '16 at 10:32
  • $\begingroup$ yeah that's way more general as you don't have any metrics involved! Thanks a lot. $\endgroup$ – noctusraid Jan 14 '16 at 10:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.