TITLE: Is the Space $C^\alpha$ separable?
QUESTION [5 upvotes]: Let $C^\alpha$be the space of continuous functions f(x) on [0,1]. Such that $ sup \frac{|f(x_1)-f(x_2|}{|x_1-x_2|^\alpha}$  ( $ 0\le x_1 \le x_2 \le 1$. Introduce in this space the norm $||f||_C^\alpha = |f(0)|$ + $ sup \frac{|f(x_1)-f(x_2|}{|x_1-x_2|^\alpha}$  ( $ 0\le x_1 \le x_2 \le 1$.
I need to know is it separable or not and why ?
I think it's not separable but don't know why ?

REPLY [11 votes]: For $a\in(0,1)$, define $u_{a}(t)=0$ for $t\in[0,a]$ and $u_{a}(t)=(t-a)^{\alpha}$ for $t\in(a,1]$, then for $0<a<b<1$, we have 
\begin{align*}
\|u_{a}-u_{b}\|&=\sup_{0\leq x_{1}\ne x_{2}\leq 1}\dfrac{|(u_{a}-u_{b})(x_{1})-(u_{a}-u_{b})(x_{2})|}{|x_{1}-x_{2}|^{\alpha}}\\
&\geq\dfrac{|(u_{a}-u_{b})(b)-(u_{a}-u_{b})(a)|}{|b-a|^{\alpha}}\\
&=\dfrac{(b-a)^{\alpha}}{|b-a|^{\alpha}}\\
&=1,
\end{align*}
but the set of all functions $u_{a}$, $a\in(0,1)$ is uncountable, thus the space is not separable.