Fix some open bounded convex $U\subseteq\mathbb{C}$ and some $0<\alpha<1$. Let $F: U \to B$, where $B$ is some Banach space, define,
$$\lVert F \rVert_{\infty}=\sup_{z\in U} \lVert F(z)\rVert \qquad D_{\alpha} F=\sup_{0<|z-w|<1}\left\{\frac{\lVert F(z)-F(w)\rVert}{|z-w|^\alpha} \right\}$$
If $F$ is differentiable $r$ times, set
$$\lVert F\rVert_{r,\alpha}=\sum_{k=0}^r \lVert F^{(k)}\rVert_{\infty} + D_\alpha F^{(r)}$$
I need to prove that $C^{r,\alpha}(U)$ is a Banach space, where $C^{r,\alpha}(U)$ is the space of all $F:U\to B$ which are differetiable $r$-times in $U$ and for which $\lVert F\rVert_{r,\alpha}$ is finite.
I believe the proof is done by induction on $r$, so i already proved that $C^{0,\alpha}(U)$ is complete, but I'm having a little trouble finishing the proof.
Assuming that $C^{r,\alpha}(U)$ is a Banach space, so I want to get that $C^{r+1,\alpha}(U)$ is also a Banach space. What I thought was,
Let $\{F_n\}$ be a Cauchy sequence in $C^{r+1,\alpha}(U)$, then $\{F'_n\}$ is a Cauchy sequence in $C^{r,\alpha}(U)$, by hypothesis, I have $F'_n \to G$ for some $G \in C^{r,\alpha}(U)$, but now, I don't know how I could conclude that $F_n \to F$ in $C^{r+1,\alpha}(U)$, where i believe i would have $G'=F$.
I thought this might be the way to the proof, but I couldn't go much further than that. This seems to be a standard result, but I couldn't find it in the literature.... Any hints?