Is the subspace $$C^k([0,T]) \subset C^{k-n}([0,T])$$ compact? I think the answer is no. But since $C^k$ is compactly embedded in $C^{k-n}$, it seems like it should be yes in some way. Can I do anything here?

($C^k$ is the space of $k$ times continuously differentiable functions)

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    $\begingroup$ What does the notation mean? $\endgroup$ – Mariano Suárez-Álvarez Aug 20 '12 at 19:27
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    $\begingroup$ (Vector spaces are rarely compact!) $\endgroup$ – Mariano Suárez-Álvarez Aug 20 '12 at 19:27
  • $\begingroup$ @MarianoSuárez-Alvarez $C^k$ is the space of $k$ times continuously differentiable functions. Yeah it doesn't look good... $\endgroup$ – Court Aug 20 '12 at 19:36
  • $\begingroup$ You should have no problems finding a sequence which does not contain a convergent subsequence, then :-) $\endgroup$ – Mariano Suárez-Álvarez Aug 20 '12 at 19:37
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    $\begingroup$ Hint: Every compact in normed space is bounded. $\endgroup$ – Norbert Aug 20 '12 at 19:42

What is true, I think, is that the unit ball of $C^k([0,T])$ is compact in $C^{k-n}([0,T])$.

EDIT: that should be "relatively compact", not "compact".

  • $\begingroup$ It should be relatively compact by Arzelà-Ascoli, but I don't think it is closed (you have no control on the derivatives of order $k-n+1, \ldots, k$, do you?). $\endgroup$ – t.b. Aug 20 '12 at 20:21
  • $\begingroup$ Yes, true. It is not compact. $\endgroup$ – Robert Israel Aug 20 '12 at 21:01
  • $\begingroup$ This shows that the inclusion map $C^k \hookrightarrow C^{k-n}$ is compact, which is probably what the OP was thinking of. $\endgroup$ – Nate Eldredge Aug 20 '12 at 23:52

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