If $\Omega$ is an open set in $\mathbb{R}^n$, is the Hölder space $C^{k, \alpha}(\Omega)$ Banach? Or is it only that $C^{k, \alpha}(\overline{\Omega})$ is Banach, like with ordinary continuous functions? If not, why is that???
Norms are
$$|f|_{C^{0,\alpha}} = \sup_{x,y \in \Omega ,\ x \neq y} \frac{| f(x) – f(y) |}{|x-y|^\alpha},$$
$$|f|_{C^{k, \alpha}} = \|f\|_{C^k}+\max_{| \beta | = k} | D^\beta f |_{C^{0,\alpha}}$$
where
$$|f|_{C^k} = \max_{| \beta | \leq k} \, \sup_{x\in\Omega} |D^\beta f (x)|$$
Thanks for any help
Best Answer
Just so that you will be aware, if both notations $C^{k, \alpha}(\Omega)$ and $C^{k, \alpha}(\bar{\Omega})$ are used in the same book, there is a chance that they are actually different. The latter is the Banach space you have in mind. The former may mean a locally Hölder space, that is $$ C^{k, \alpha}(\Omega) = \{u\in C(\Omega):\textrm{for any compact set }K\subset\Omega, u\in C^{k, \alpha}(K)\}. $$ This is a Fréchet space, and the situation is of course analogous to the difference between $C(\Omega)$ and $C(\bar\Omega)\equiv C_b(\Omega)$.