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I have a question concerning the definition of the space of Hölder continuous functions $C^{k,\alpha}(\Omega)$, where $\Omega\subset\mathbb{R}^n$ is an open set and $k\in\mathbb{N}$ and $\alpha\in (0,1]$.

Often the following definition is used: $f\in C^{k,\alpha}(\Omega)$ if and only if $f\in C^{k}(\Omega)$ such that all its derivatives up to order $k$ are bounded functions and the $k-th$ order derivatives are Hölder continuous.

I am wondering why only $D^{\beta}f:U\rightarrow \mathbb{R}$ with $\vert \beta \vert=k$ has to be Hölder continuous? Does it maybe follow that $f$ and its derivatives up to order $k-1$ have to be Hölder continuous as well? Otherwise the space of Hölder continuous functions would contain functions which are not Hölder continuous, which sounds very awkward to me.

Best wishes

el_tenedor
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Thorsten
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    Interesting question. Until now, I have only been dealing with Hölder spaces of bounded continuous functions with bounded derivatives on convex domains $\Omega$. For them, only Hölder continuity of the highest order derivative is relevant. All derivatives of lower order are not only Hölder but Lipschitz as you can easily prove by using the mean value inequality. On non-convex domains it is not evident why the Hölder continuity of the derivatives should be inherited by lower order derivatives. – el_tenedor Feb 11 '17 at 22:06
  • At least in this post, there is an example of a function which has a Hölder continuous derivative is but not Hölder itself: http://math.stackexchange.com/questions/734361 A main point in the constructed example was to use a non-convex domain: a slit annulus. – el_tenedor Feb 11 '17 at 22:08

1 Answers1

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It is possible to further weaken the assumptions given on $\Omega$ as being "only" a bounded domain with $C^1$ boundary (even Lipschitz should still work).

It suffices to prove that if $u \in C^{1,\alpha}_b(\Omega)$, i.e. $u$, and $\partial_i u$ are bounded and $\partial_i u$ are $\alpha$-Hölder, then $u$ is $\alpha$-Hölder as well.

Since $\Omega$ is a bounded $C^1$-domain, there exist finitely many bounded diffeomorphisms $\Phi_i \colon \mathbb R^n \to \mathbb R^n$ such that $$\Phi_i(\Omega \cap U_i) \subset B_i^+, \quad \Phi(\partial\Omega \cap U_i) = B_i^+ \cap (\mathbb R^{n-1} \times \{0\})$$ Where $(U_i)_{i = 1,\dots,N}$ is an open covering of $\partial\Omega$ and $$ B_i^+ := \{ x \in \mathbb R^n \colon |x| < r_i, x_n \geq 0\}. $$ Correct me if something seems odd here, as this should be standard for $C^1$-submanifolds.

Now, take $x,y \in \Omega$ and let $\delta > 0$ be the Lebesgue number of the covering $(U_i)$. Now we distinguish the following cases

  1. If $|x - y| < \delta$ and $x,y \in \overline{B_\delta(x)} \subset \Omega$: In this case, the mean value theorem is applicable and yields $$ |u(x) - u(y)| \leq \|\nabla u\|_\infty |x - y|. $$

  2. If $|x - y| < \delta$ and $x,y \in \Omega \cap U_i$: Consider the convex hull $$ W := \{(1 - t) \Phi_i(x) + t \Phi(y) \colon t \in [0,1]\} \subset B_i^+. $$ Then \begin{align*} |u(x) - u(y)| &= |u(\Phi_i^{-1}\circ \Phi_i(x)) - u(\Phi_i^{-1}\Phi_i(y))| \\ &= \Big|\int_0^1 \frac{d}{dt} \Big((u\circ\Phi_i^{-1})\big((1 - t)\Phi_i(x) + t \Phi_i(y)\big) \Big) d t \Big| \\ &\leq \| \nabla(u \circ \Phi_i^{-1})\|_\infty \cdot |\Phi_i(x) - \Phi_i(y)| \\ &\leq \| \nabla u\|_{\infty} \cdot \| D\Phi_i^{-1}\|_\infty \cdot \|D\Phi_i\|_{\infty} |x - y|, \end{align*} where we applied the mean value theorem to $\Phi_i$.

  3. If $|x - y| \geq \delta$, then $$ |u(x) - u(y)| \leq 2 \|u\|_\infty \cdot 1 \leq 2 \|u\|_{\infty} \frac{|x - y|}{\delta}. $$

Hence, $$ \sup_{x, y \in \Omega, x \neq y} \frac{|u(x) - u(y)|}{|x - y|} \leq C (\|u\|_\infty + \|\nabla u\|_\infty) $$ for some constant $C > 0$ and thus $u$ is Lipschitz, i.e. in particular $\alpha$-Hölder.

el_tenedor
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