Confused about which Hölder spaces are Banach 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
 A: 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 $C(\bar\Omega)\equiv C_b(\Omega)$.
A: Note that all $g\in C^{k,\alpha}(\Omega)$ and $\beta\in\mathbb{N}_0^k$ we have
$$
\Vert \partial^\beta g\Vert_{C_b(\Omega)}\leq\Vert g\Vert_{C^{k,\alpha}(\Omega)}\tag{1}
$$
$$
|\partial^\beta g|_{C^{0,\alpha}}\leq \Vert g\Vert_{C^{k,\alpha}(\Omega)}\tag{2}
$$
Let $\{f_n:n\in\mathbb{N}\}$ be a Cauchy sequence in $C^{k,\alpha}(\Omega)$, then for all $\varepsilon>0$, we have $N\in\mathbb{N}$ such that $m,n>N$ implies 
$$
\Vert f_n-f_m\Vert_{C^{k,\alpha}(\Omega)}<\varepsilon\tag{3}
$$ 
Fix multi-index $\beta\in\mathbb{N}_0^k$ such that $|\beta|\leq k$. From $(1)$ and $(3)$ it follows that for all $\varepsilon>0$, we have $N\in\mathbb{N}$ such that $m,n>N$ implies $\Vert \partial^\beta f_n-\partial^\beta f_m\Vert_{C_b(\Omega)}<\varepsilon$. This means that $\{\partial^\beta f_n: n\in\mathbb{N}\}$ is a Cauchy sequence in $C_b(\Omega)$. Since $C_b(\Omega)$ is complete, then $\{\partial^\beta f_n: n\in\mathbb{N}\}$ converges to some function in $C_b(\Omega)$. By $\varphi$ we denote limit of $\{ f_n:n\in\mathbb{N}\}$ in $C_b(\Omega)$.
Now we use the following standard result

Let $\{f_n:n\in\mathbb{N}\}\subset C(\Omega)$ be a family of differentiable functions, such that
1) the sequence $\{\partial_{x_i}f_n:n\in\mathbb{N}\}$ converges in $C_b(\Omega)$ to $g\in C_b(\Omega)$.
2) for some point $x\in\Omega$ the sequence $\{f_n(x_0):n\in\mathbb{N}\}$ is convergent
Then $\{f_n:n\in\mathbb{N}\}$ converges in $C_b(\Omega)$ to some differentiable function $f\in C_b(\Omega)$, and moreover $\partial_{x_i} f=g$.

Using this result and induction by multi-indices one can show that for all $\beta\in\mathbb{N}_0^k$ with $|\beta|\leq k$ the sequence $\{\partial^\beta f_n:n\in\mathbb{N}\}$ uniformly converges to $\partial^\beta \varphi$. This means that
$$
\lim\limits_{n\to\infty}\Vert f_n-\varphi\Vert_{C^k(\Omega)}=0\tag{4}
$$
From $(2)$, $(3)$ and deinition of $|\cdot|_{C^{0,\alpha}(\Omega)}$ it follows that for all $\beta\in\mathbb{N}_0^k$ with $|\beta|= k$ and all $x,y\in\Omega$ such that $x\neq y$ we have
$$
\frac{|\partial^\beta f_n(x) - \partial^\beta f_m(y)|}{|x-y|^\alpha}<\varepsilon
$$
Let's take $m\to\infty$ in this inequality, then we get
$$
\frac{|\partial^\beta f_n(x) - \partial^\beta \varphi(y)|}{|x-y|^\alpha}<\varepsilon
$$
Since $\beta\in\mathbb{N}_0^k$ with $|\beta|= k$ and $x,y\in\Omega$ such that $x\neq y$ are arbitrary we can say that 
$$
\max\limits_{|\beta|=k}|\partial^\beta f_n-\partial^\beta \varphi|_{C^{0,\alpha}(\Omega)}<\varepsilon
$$
Thus for all $\varepsilon>0$ we have $N\in\mathbb{N}$ such that $n>N$ implies
$$
\max\limits_{|\beta|=k}|\partial^\beta f_n-\partial^\beta \varphi|_{C^{0,\alpha}(\Omega)}<\varepsilon
$$
This means that
$$
\lim\limits_{n\to\infty}\max\limits_{|\beta|=k}|\partial^\beta f_n-\partial^\beta \varphi|_{C^{0,\alpha}(\Omega)}=0\tag{5}
$$
From $(4)$ and $(5)$ it follows that $\{f_n:n\in\mathbb{N}\}$ converges to $\varphi$ in $C^{k,\alpha}(\Omega)$. Since we showed that arbitrary Cauchy sequence in $C^{k,\alpha}(\Omega)$ is convergent, then $C^{k,\alpha}(\Omega)$ is complete.
Proof of the completeness of $C^{k,\alpha}(\overline{\Omega})$ is already discussed in comments.
