A function $f$ is $\alpha$-H$\ddot{\text{o}}$lder continuous if $|f(x) - f(y)| \leq C |x - y|^\alpha $ for some $C \geq 0$ and $\forall$ $x, y \in \mathbb{R}$ such that $|x - y| < 1$.
Define the norm: $||f||_{C^\alpha} = ||f||_\infty + \sup \limits_{\substack{x, y \in \mathbb{R} \\ |x - y| < 1}} \frac{|f(x) - f(y)|}{|x - y|^\alpha}$
Let $C^\alpha(\mathbb{R})$ = the set of $\alpha$-H$\ddot{\text{o}}$lder continuous functions = $\{ f : ||f||_{C^\alpha} < \infty \}$.
Show that $\left ( C^\alpha(\mathbb{R}) , ||.||_{C^\alpha} \right ) $ is a Banach space for any $0 < \alpha \leq 1$.
I already showed $C^\alpha$ is a normed vector space. Need to show completeness as follows:
Consider a cauchy sequence $\left ( f_n \in C^\alpha(X) \right )_{n \in \mathbb{N}}$. For every $\epsilon > 0$, there exists $N(\epsilon)$ such that for all $n, m > N$, we have $||f_n - f_m||_{C^\alpha} < \epsilon$ $\implies $ $$||f_n - f_m||_\infty + \sup_{\substack{x, y \in X \\ |x - y| < 1}} \frac{|f_n(x) - f_m(x) - f_n(y) + f_m(y)|}{|x - y|^\alpha} < \epsilon$$
Need some help/pointers to proceed further. I also know that $\alpha$-H$\ddot{\text{o}}$lder continuous functions are uniformly continuous for $0 < \alpha \leq 1$
You must at least have $\|f_{n}-f_{m}\|_{\infty}<\epsilon$, so for any $x$, \begin{align*} |f_{n}(x)-f_{m}(x)|<\epsilon,~~~~n,m>N, \end{align*} so the sequence of scalars $(f_{n}(x))$ is Cauchy and hence $f(x)=\lim_{n\rightarrow\infty}f_{n}(x)$ exists, and \begin{align*} |f_{n}(x)-f(x)|\leq\epsilon,~~~~n>N. \end{align*} Also for any $x,y$ such that $d(x,y)<1$, \begin{align*} \dfrac{|f_{n}(x)-f_{m}(x)-f_{n}(y)+f_{m}(y)|}{|x-y|^{\alpha}}<\epsilon,~~~~n,m>N, \end{align*} taking $m\rightarrow\infty$ we have \begin{align*} \dfrac{|f_{n}(x)-f(x)-f_{n}(y)+f(y)|}{|x-y|^{\alpha}}\leq\epsilon,~~~~n>N. \end{align*} The rest is easy.