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