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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.

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    What is your definition "compactly embedded". Are you assuming that $A\subset B$?2012-08-20
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    A somewhat related result: http://math.stackexchange.com/q/1183002012-08-20
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    @Norbert The definition is as on wiki page http://en.wikipedia.org/wiki/Compactly_embedded. Yeah $A \subset B$ here.2012-08-20
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    @t.b. Thanks will check out.2012-08-20
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    @tomasz Sorry I forgot to write that all these spaces are Banach spaces.2012-08-21
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    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 $B$?2012-08-21
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    @tomasz Really?? There are a lot of Sobolev spaces that are (eg. $H^1$ is CE in $L^2$, etc..)2012-08-21
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    @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 $B$.2012-08-21
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    @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).2012-08-21
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    @tomasz Done. I apologise for lack of detail. I thought an abstract setting might have been better.2012-08-21
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    What is the relation of $B$ to $T$? As stated, $B$ has nothing to do with $T:A\to A$ or $A$.2012-08-21
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    @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.2012-08-21
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    No, there is nothing. However, given your motivation, you might be able to get what you want once you set things up correctly.2012-08-21
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    @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.2012-08-21
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    @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.2012-08-21
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    Well, take $T$ to be the identity map, which is not compact.2012-08-21

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