# Compact subspace

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)

• What does the notation mean? – Mariano Suárez-Álvarez Aug 20 '12 at 19:27
• (Vector spaces are rarely compact!) – Mariano Suárez-Álvarez Aug 20 '12 at 19:27
• @MarianoSuárez-Alvarez $C^k$ is the space of $k$ times continuously differentiable functions. Yeah it doesn't look good... – Court Aug 20 '12 at 19:36
• You should have no problems finding a sequence which does not contain a convergent subsequence, then :-) – Mariano Suárez-Álvarez Aug 20 '12 at 19:37
• Hint: Every compact in normed space is bounded. – 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 $C^{k-n}([0,T])$.
• 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?). – t.b. Aug 20 '12 at 20:21
• This shows that the inclusion map $C^k \hookrightarrow C^{k-n}$ is compact, which is probably what the OP was thinking of. – Nate Eldredge Aug 20 '12 at 23:52