Compact operator and compact embeddings

Question

Here, all spaces are Banach spaces.

Definition: A map $S:X \to X$ is compact if for every bounded sequence $\{u_n\}$, there exists a subsequence $\{u_{n_k}\}$ such that $\{S(u_{n_k})\}$ converges in $X$.

Question: suppose $A$ is compactly embedded in $B$. Suppose a map $T:A \to A$ is continuous. Is there any chance that $T:A \to A$ is compact? ($T:B \to A$ is not definable or is ill-defined). If context is important: take $A=C^{2, \alpha}$ and $B=C^{0, \alpha}$, Hölder continuous functions. It is true that $A$ is compactly embedded in $B$ (the norms are different on $A$ and $B$ -- they are the standard norms on Wikipedia).

Thoughts: I don't think so in general. I can't see any way, unless there's some cool theorem I'm not aware of (and I'm not aware of a lot of things so maybe this is possible).

Motivation: want to show existence to a PDE problem.

What is your definition "compactly embedded". Are you assuming that $A\subset B$? — Norbert, Aug 20 '12 at 21:43
A somewhat related result: http://math.stackexchange.com/q/118300 — t.b., Aug 20 '12 at 21:49
@Norbert The definition is as on wiki page http://en.wikipedia.org/wiki/Compactly_embedded. Yeah $A \subset B$ here. — Court, Aug 20 '12 at 21:57
@tomasz Sorry I forgot to write that all these spaces are Banach spaces. — Court, Aug 21 '12 at 13:01
Maybe I am missing the point. Assume $T$ is the identity. Then $T$ is compact iff every bounded sequence in $A$ is relatively compact. Who cares of $B$? Are you sure that $T$ maps $A$ to itself, without any reference to $B$? — Siminore, Aug 21 '12 at 13:08
@tomasz Really?? There are a lot of Sobolev spaces that are (eg. $H^1$ is CE in $L^2$, etc..) — Court, Aug 21 '12 at 15:19
@Siminore In my context $T$ is a second order differential operator, and take A to be $C^2$ Holder functions and $B$ to be $C^0$ Holder functions. Then $A$ is CE in $B$. — Court, Aug 21 '12 at 15:21
@tomasz Ah, I see. I'm glad you said posted this as this point is something i miss sometimes (that subspace norms can be different in such embeddings). — Court, Aug 21 '12 at 15:22
@tomasz Done. I apologise for lack of detail. I thought an abstract setting might have been better. — Court, Aug 21 '12 at 15:49
What is the relation of $B$ to $T$? As stated, $B$ has nothing to do with $T:A\to A$ or $A$. — timur, Aug 21 '12 at 15:58
@timur There is no relation other than stated. I was hoping the compact embedding might give me something but I know it's very far-fetched. — Court, Aug 21 '12 at 16:01
No, there is nothing. However, given your motivation, you might be able to get what you want once you set things up correctly. — timur, Aug 21 '12 at 16:08
@Court: It's okay. An abstractly stated question is a good thing, but not to the point where it's hard to tell what you're talking about. And providing context (well, reasonable amounts of it) is always a good thing – even if someone gives you a satisfactory “abstract” answer, he might add some insight about applying it in your particular case. — tomasz, Aug 21 '12 at 16:34
@Court: That said, I think your question is a little too abstract. As you've written it, it seems to me that $T$ could very well be the zero operator, so obviously compact. — tomasz, Aug 21 '12 at 16:36
Well, take $T$ to be the identity map, which is not compact. — timur, Aug 21 '12 at 16:45

score 2 · Answer 1 · answered Aug 22 '12 at 13:22

If the embedding $e: A \to B$ is compact, and there is a continuous linear map $S:B \to A$ such that $T$ equals the composite $S \circ e$, then $T$ will be compact (since the composite of a compact operator and any other continuous linear map is always compact).

[This idea is used in elliptic PDE theory, to deduce compactness of the inverse of elliptic differential operators. The point (stated very roughly) is that applying the inverse of an elliptic operator should improve the differentiability class of a function.]

Without such a factorization, it's not clear how you would ever hope to link the behaviour of $T$ and the compactness of the embedding $e$.

Compact operator and compact embeddings

1 Answers1