# Convergence of sequence in $C^{k, \alpha}$ composed with $C^\infty$ function

Is it true that if a sequence $u_n \to u$ in $C^{k, \alpha}$ norm, and if you have a function $f \in C^\infty$, then $f(u_n) \to f(u)$ in $C^{k, \alpha}$? I think so, since this is true for ordinary $C^k$ space so the "norm part" of the $C^{k, \alpha}$ norm converges, but 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 Hölder space, it'll work for parabolic Hölder space too. Parabolic Hölder space is defined as follows. The seminorm is defined $$[u]_{\alpha} = \sup_{(x,t), (y,s) \in Q} \frac{|u(x,t) - u(y,s)|}{(|x-y|^2 + |t-s|)^{\frac{\alpha}{2}}},$$ and norm on the parabolic Hölder space $C^{k, \alpha}$ is defined $$\lVert{u}\rVert_{\widetilde{C}^{k, \alpha}(\overline{Q})} = \sum_{i+2j \leq k} \lVert{\frac{\partial^{i+j}u}{\partial x^i \partial t^j}}\rVert_{C(\overline{Q})} + \sum_{i+2j = k} \bigg[\frac{\partial^{i+j}u}{\partial x^i \partial t^j}\bigg]_\alpha.$$

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You take the space $C^{k,\alpha}$, but where are the functions defined? And how do you define parabolic Hölder space? – Davide Giraudo Jul 19 '12 at 10:59
@DavideGiraudo Say the $u_n$ and $u$ are defined on a compact set. For $f$ assume anything. Parabolic Holder space, I'll write up the norm. – TagWoh Jul 19 '12 at 11:04
We can control these semi-norms use mean-value property, or fundamental theorem of analysis, using the fact that $f$ and its derivatives are bounded. – Davide Giraudo Jul 19 '12 at 11:13
@DavideGiraudo Please see my edit. The compact subset is a compact subset of $\mathbb{R}^2$. Thanks! – TagWoh Jul 20 '12 at 14:52
Thanks, now I will think on it. – Davide Giraudo Jul 20 '12 at 14:53