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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 ?

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1 Answer 1

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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.

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