# Embeddings between Hölder spaces $C^{0,\beta} \hookrightarrow C^{0, \alpha} .$

Let $\Omega \subset \mathbb R^n$ be an open subset and let $0 < \alpha < \beta \leq 1.$ We consider the space of Hölder continuous functions $C^{0, \alpha}$ which is a Banach space endowed with the norm

$$\| f\|_{C^{0, \alpha}} := \| f \|_{\infty} + \sup_{ x,y \in \Omega \\ x \neq y} \frac{ |f(x) -f(y)|}{|x-y|^\alpha}.$$

My questions has to do with the embedding $C^{0,\beta} \hookrightarrow C^{0, \alpha} .$

If $\Omega$ is bounded, then I can prove the estimate $\| f\|_{C^{0, \alpha}} \leq \text{diam}(\Omega)^{\beta -\alpha} \| f\|_{C^{0, \beta}} ,$ which in turn implies that the embedding is bounded, i.e. continuous.

Question: How can I show that the embedding is still continuous in the case where $\Omega$ is unbounded ?

Any help would be really appreciated.

• @GiuseppeNegro those are holder embedding and the domain does not matter very often see my answer – Guy Fsone Jan 23 '18 at 13:48
• @DanielFischer is your function at least bounded. ? Obviously no .As I said this embedding does care about the boundedness of the domain. the classical proof below – Guy Fsone Jan 23 '18 at 13:52
• For unbounded $\Omega$, you don't have $C^{0,\beta}(\Omega) \subset C^{0,\alpha}(\Omega)$ for $\alpha < \beta$. You then must look at the subspace of bounded Hölder-continuous functions, $C^{0,\beta}(\Omega) \cap L^{\infty}(\Omega)$. – Daniel Fischer Jan 23 '18 at 13:53
• @DanielFischer by defintion the Holder spaces are already bounded otherwise it is senseless to put use the sup-norm in the definition. check below – Guy Fsone Jan 23 '18 at 13:55
• @GuyFsone I only looked at the name of the spaces, and $C^{0,\beta}(\Omega)$ doesn't say the functions are bounded. (For bounded $\Omega$, they automatically are.) For the subspace of bounded functions, yes, you have the embedding. – Daniel Fischer Jan 23 '18 at 13:55

Note that Embeddings between Holder space do not care about the boundedness of the domain

Patently we have $$\sup_{ x,y \in \Omega \\ x \neq y} \frac{ |f(x) -f(y)|}{|x-y|^\alpha}\le \sup_{ x,y \in \Omega \\ |x -y|\le1} \frac{ |f(x) -f(y)|}{|x-y|^\alpha}+\sup_{ x,y \in \Omega \\ |x -y|\ge1} \frac{ |f(x) -f(y)|}{|x-y|^\alpha}$$

But since $|x-y|^\alpha\ge 1$ for $|x-y|\ge 1$. we obtain $$\sup_{ x,y \in \Omega \\ |x -y|\ge1} \frac{ |f(x) -f(y)|}{|x-y|^\alpha} \le \sup_{ x,y \in \Omega \\ |x -y|\ge1} |f(x) -f(y)| \le 2\|f\|_\infty$$

whereas if $|x-y|\le 1$ and $0 < \alpha < \beta \leq 1.$ then

$$|x-y|^{\beta-\alpha}\le1\implies |x-y|^{\beta}\le|x-y|^{\alpha}$$

and hence, $$\sup_{ x,y \in \Omega \\ |x -y|\le1} \frac{ |f(x) -f(y)|}{|x-y|^\alpha}\le \sup_{ x,y \in \Omega \\ |x -y|\le1} \frac{ |f(x) -f(y)|}{|x-y|^\beta}\le \sup_{ x,y \in \Omega \\x \neq y} \frac{ |f(x) -f(y)|}{|x-y|^\beta}$$

It plainly follows that $$\color{red}{\sup_{ x,y \in \Omega \\ x \neq y} \frac{ |f(x) -f(y)|}{|x-y|^\alpha}\le \sup_{ x,y \in \Omega \\ x \neq y} \frac{ |f(x) -f(y)|}{|x-y|^\beta}+2\|f\|_\infty\le 2\|f\|_{C^{0,\beta}}}$$

• Dear Guy Fsone: Thank you very much for your time and the nice solution! – passenger Jan 23 '18 at 14:39
• @passenger passenger you are welcome – Guy Fsone Jan 23 '18 at 14:46