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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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    What does the notation mean?2012-08-20
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    (Vector spaces are rarely compact!)2012-08-20
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    @MarianoSuárez-Alvarez $C^k$ is the space of $k$ times continuously differentiable functions. Yeah it doesn't look good...2012-08-20
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    You should have no problems finding a sequence which does not contain a convergent subsequence, then :-)2012-08-20
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    **Hint:** Every compact in normed space is bounded.2012-08-20
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    Probably it would be helpful for someone who reads this question also to know what do you mean with _compact_ . I think the definition is the following: https://en.wikipedia.org/wiki/Compact_space2016-01-11

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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".

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    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?).2012-08-20
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    Yes, true. It is not compact.2012-08-20
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    This shows that the inclusion map $C^k \hookrightarrow C^{k-n}$ is compact, which is probably what the OP was thinking of.2012-08-20