$(C^{0,\alpha}(\bar{\Omega}), \|\cdot\|_{C^{0,\alpha}(\bar{\Omega})})$
is a Banach space
Let $f_n$ be a cauchy sequence in $C^{0,\alpha}(\bar{\Omega})$. Therefore, for every
$\epsilon > 0$ there exits $M$ such that for all $n,m > M$, we have
$\|f_n-f_m\|_{C^{0,\alpha}(\bar{\Omega})} < \epsilon$. This gives that
$[f_n-f_m]_{C^{0,\alpha}(\bar{\Omega})} < \epsilon$ and $\|f_n -f_m\|_{C^0(\bar{\Omega})} < \epsilon$. From the latter we get that
$f_n$ converges (in the sup norm) to a bounded continuous function $f$ (on $\bar{\Omega}$). From the former we get that $[f_n-f]_{C^{0,\alpha}(\bar{\Omega})} < \epsilon$ by letting $m \to \infty$. Therefore $\|f_n-f\|_{C^{0,\alpha}(\bar{\Omega})}
\to 0$ as $n \to \infty$. It remains to show that $f \in {C^{0,\alpha}(\bar{\Omega})}$
but this follows easily by
\[[f]_{C^{0,\alpha}(\bar{\Omega})} = [f-f_n+f_n]_{C^{0,\alpha}(\bar{\Omega})} \leq [f-f_n]_{C^{0,\alpha}(\bar{\Omega})} + [f_n]_{C^{0,\alpha}(\bar{\Omega})} < \infty\]
Note that this result does not depend on $\alpha$.
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