(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.


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 $f\circ u \in C^{k,\alpha}$.

Something similar can be done also for the second question.

  • 1
    $\begingroup$ 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. $\endgroup$ – Davide Giraudo Jul 22 '12 at 15:10
  • 1
    $\begingroup$ @Davide: Sure. I guess one can avoid Faa di Bruno by induction on $k$. $\endgroup$ – timur Jul 22 '12 at 15:41
  • 1
    $\begingroup$ Thanks for the answer @timur. @Davide I certainly appreciated your answer on MSE and I hope you don't think I was being rude, it's just that I did not feel comfortable with chain rule formula for the case I was looking at. $\endgroup$ – user25266 Jul 22 '12 at 21:29
  • 1
    $\begingroup$ @timur. I thought to that too, but I'm not sure it's a straightforward step. Can you detail it? $\endgroup$ – Davide Giraudo Aug 2 '12 at 22:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.