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We know that when $I$ is a bounded interval and $1<p\leq \infty$ that the injection $W^{1,p}\subset C(\overline{I})$ is compact.

The proof of this fact uses the Arzela-Ascoli theorem on the unit ball $\mathcal{H}$ of $W^{1,p}(I)$. Is it true that the embedding of $W^{2,p}$ onto $C^1(\overline{I})$ is also compact?

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Suppose that $u\in W^{2,p}$ then $u'\in W^{1,p}$ and thus $u'\in C(\overline{I})$ and therefore $u\in C^1(\overline{I})$.

Now let $\mathcal{H}$ be the unit ball in $W^{2,p}$. Then $\mathcal{H}$ is equicontinuous since for all $u\in \mathcal{H}$, $$|u(x)-u(y)|=\left|\int_{y}^xu'(t)dt\right|\leq \|u'\|_p|x-y|^{1/p'}\leq |x-y|^{1/p'}.$$ Hence by Ascoli-Arzela, $\mathcal{H}$ has compact closure in $C(\overline{I})$. Since $\mathcal{H}\subset C^1(\overline{I})\subset C(\overline{I})$ then $\mathcal{H}$ has compact closure in $C^1(I).$ Is there anything wrong with the argument?

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