# Composition of $C^{k, \alpha}$ function with $C^\infty$ function on a compact domain

(I asked this question on MSE but I did not receive an answer so I hope I can post here.)

Let $S$ be a compact set in $\mathbb{R}^2$ and let $C^{k, \alpha}(S)$ denote the usual Holder space with $k$ continuous derivatives and finite $k$-th order seminorms with exponent $\alpha$.

1) Is it true that if $f \in C^\infty(S)$ and $u \in C^{k, \alpha}(S)$, then $f(u) \in C^{k, \alpha}(S)$?

I don't know how to show that the seminorm part (which involves supremums over the composition divided by a distance involving the arguments) is finite.

2) Is it true that if a sequence $u_n \to u$ in $C^{k, \alpha}(S)$, and if $f \in C^\infty(S)$, then $f(u_n) \to f(u)$ in $C^{k, \alpha}(S)$?

I think so, since this is true for ordinary $C^k$ space so the "norm part" of the $C^{k, \alpha}$ norm converges, but again I am not sure how to show that the seminorm part of the $C^{k, \alpha}$ norm converges.

And I guess if this works for Holder space, it'll work for parabolic Holder space too. Parabolic Holder space is defined as follows. The space $\widetilde{C}^{k, \alpha}(S)$ has the seminorm $$u_\alpha = \sup_{(x,t), (y,s) \in S} \frac{|u(x,t) - u(y,s)|}{(|x-y|^2 + |t-s|)^{\frac{\alpha}{2}}},$$ and norm $$\lVert{u}\rVert_{\widetilde{C}^{k, \alpha}(\overline{S})} = \sum_{i+2j \leq k} \lVert{\frac{\partial^{i+j}u}{\partial x^i \partial t^j}}\rVert_{C(\overline{S})} + \sum_{i+2j = k} \bigg[\frac{\partial^{i+j}u}{\partial x^i \partial t^j}\bigg]_\alpha.$$

I'm grateful for any help. Thanks.

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From the mean value theorem, we have $$|f(u(x))-f(u(y))| \leq \max_{\xi\in U}|f'(\xi)| \cdot |u(x)-u(y)|,$$ where $U=[\min u, \max u]$. This argument applied to $f^{(k)}$ should give $f\circ u \in C^{k,\alpha}$.
But we should apply this to the derivatives of $f\circ u$ (with respect to $x_1$ and $x_2$), so it's more complicated. I guess Faà di Bruno formula will be useful here. – Davide Giraudo Jul 22 '12 at 15:10
@Davide: Sure. I guess one can avoid Faa di Bruno by induction on $k$. – timur Jul 22 '12 at 15:41