Proving that a Hölder space is a Banach space

We have to show the following: given a Cauchy sequence $(u_n)_{n\in \mathbb N}$ in $C^{k,\gamma}(\overline U),$ there is a $u \in C^{k,\gamma}(\overline U)$ such that $\lim_{n\rightarrow\infty}u_n = u$ in $C^{k,\gamma}(\overline U).$

Note that the Hölder norm is the "sum" of the $C^k$ norm (i.e. sup-norm up to the $k$-th derivatives) and the Hölder condition with parameter $\gamma \in (0,1)$ for the $k$-th derivatives. From the $C^k$ part of the Hölder norm and the completeness of $C^k(\overline U),$ we get a $u \in C^k(\overline U)$ with $\lim_{n\rightarrow\infty}u_n = u$ in $C^k(\overline U).$ Now let's look at the Hölder condition. Let's fix a multi-index $\alpha$ with $|\alpha| = k$ and write $$ v := D^\alpha u\qquad and \qquad v_n := D^\alpha u_n $$ for short. Then we have for fixed $x\neq y$ $$ \begin{align} \frac{|v(x) - v(y)|}{|x-y|^\gamma} & = \frac{|v(x) - v_m(x) + v_m(x) - v_m(y) + v_m(y) - v(y)|}{|x-y|^\gamma} \\ & \leq \frac{|v(x) - v_m(x)|}{|x-y|^\gamma} + \frac{|v_m(x) - v_m(y)|}{|x-y|^\gamma} + \frac{|v_m(y) - v(y)|}{|x-y|^\gamma} \end{align} $$ for arbitrary $m \in \mathbb N.$ Here, the first and third summands on the r.h.s. can be made arbitrarily small by choosing $m$ large enough, since $v_m \rightarrow v$ uniformly, and the second summand on the r.h.s. is bounded independent of $m$ and $x\neq y$ since the sequence $u_n$ is Cauchy, hence bounded, in the Hölder norm. From all this we get a constant $M_\alpha$ such that $$ \frac{|v(x) - v(y)|}{|x-y|^\gamma} \leq M_\alpha \qquad independent\ of\ x\neq y. $$ This shows that we have $$ u\in C^{k,\gamma}(\overline U). $$ So far so good. Finally, we have to show that $u_n \rightarrow u$ in $C^{k,\gamma}(\overline U).$ From the construction of $u,$ we already have $u_n \rightarrow u$ in $C^k(\overline U),$ so we only have to show convergence in the Hölder seminorms of the $k$-th derivatives. We use the notation $\alpha,v,v_n$ from above. We have $$ \begin{align} \frac{|(v-v_m)(x) - (v-v_m)(y)|}{|x-y|^\gamma} & = \frac{|v(x) - v_m(x) - v(y) + v_m(y)|}{|x-y|^\gamma} \\ & = \frac{|(\lim_{k\rightarrow\infty}v_k(x)) - v_m(x) - (\lim_{k\rightarrow\infty}v_k(y)) + v_m(y)|}{|x-y|^\gamma} \\ & = \lim_{k\rightarrow\infty}\frac{|v_k(x) - v_m(x) - v_k(y) + v_m(y)|}{|x-y|^\gamma} \\ & = \lim_{k\rightarrow\infty}\frac{|(v_k - v_m)(x) - (v_k - v_m)(y)|}{|x-y|^\gamma}. \end{align} $$ Here, the r.h.s. can be made arbitrarily small independent of $x\neq y$ by choosing $m$ large enough, since the sequence $u_n$ is Cauchy in the Hölder norm. So we find $v_n \rightarrow v$ in the Hölder seminorm. Since there are only finitely many multi-indices $\alpha$ (or equivalently, $k$-th derivatives) to consider, we can conclude $$ u_n \rightarrow u\qquad in\ C^{k,\gamma}(\overline U), $$ as desired.

Since your comment to my first answer is tricky to read, let me restate it. The follow-up question is: how does $$ \frac{|v(x)-v(y)|}{|x-y|^\gamma} \le M_\alpha < \infty \qquad independent\ of\ x\neq y \qquad\qquad\qquad(*)$$ for every multi-index $\alpha$ of order $k$ imply $$ u \in C^{k,\gamma}(\bar{U})? $$ The answer is as follows: in order to show $u \in C^{k,\gamma}(\bar{U}),$ we have to show that all the ingredients of the Hölder norm of $u$ can be calculated, and the value of the Hölder norm thus obtained is finite.

By construction of $u,$ we have $$ \sum_{|\beta|\leq k} \|D^\beta u\|_{C(\overline U)} < \infty. $$ Moreover, (*) above implies that for a fixed multi-index $\alpha$ of order $k,$ we have $$ [D^\alpha u]_{C^{,\gamma}(\overline U)} \leq M_\alpha < \infty. $$ From this we get $$ \sum_{|\alpha| = k}[D^\alpha u]_{C^{,\gamma}(\overline U)} \leq \sum_{|\alpha| = k} M_\alpha < \infty. $$ So we have all in all $$ \|u\|_{C^{k,\gamma}(\bar{U})} = \sum_{|\beta|\leq k} \|D^\beta u\|_{C(\overline U)} + \sum_{|\alpha| = k}[D^\alpha u]_{C^{,\gamma}(\overline U)} < \infty $$ and thus $$ u \in C^{k,\gamma}(\bar{U}), $$ as desired.