# Compact operator and compact embeddings

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 $A\subset B$? –  Norbert Aug 20 '12 at 21:43
A somewhat related result: math.stackexchange.com/q/118300 –  t.b. Aug 20 '12 at 21:49
@Norbert The definition is as on wiki page en.wikipedia.org/wiki/Compactly_embedded. Yeah $A \subset B$ here. –  Court Aug 20 '12 at 21:57
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
Well, take $T$ to be the identity map, which is not compact. –  timur Aug 21 '12 at 16:45

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).
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 $e$.