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Is the closure of the unit ball in $C^2[0,1]$ compact in $C^1[0,1]$?

I don't know how to represent the ball in $C^2[0,1]$ with norm $\|\cdot\|_\infty$ or some other norm, I think this problem should use Arzela-Ascoli Theorem.

Shine
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    Yes, Ascoli is the way to go. On $C^2([0,1])$, your norm is something equivalent to $$\lVert f\rVert_{C^2} = \max \left{\lVert f\rVert_\infty, \lVert f'\rVert_\infty, \lVert f''\rVert_\infty \right}.$$ – Daniel Fischer Jul 18 '14 at 10:29
  • Yes, in this way, we are done. – Shine Jul 18 '14 at 10:41

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For completeness, I'll modify the answer in Is there a reference for compact imbedding of Hölder space? to fit this situation.

Take a bounded family of functions $(u_n)$ in $C^2([0,1])$. This implies $u_n$, $u_n'$, and $u_n''$ are uniformly bounded. By the mean value theorem, $u_n$ and $u_n'$ are equicontinuous. By the Ascoli-Arzelà theorem, we can extract a subsequence $v_k = u_{n_k}$ such that $v_k$ and $v_k'$ converge uniformly. This subsequence is Cauchy in $C^1[0,1]$ and therefore converges.

Community
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