Holder condition for $x^\beta$ Let $f(x)=x^\beta$ (for some fixed $0&lt;\beta&lt;1$) be defined on $(0,1)$. It's not hard to see that $f$ is $\beta$-Holder. 


  How can I prove that $x^\beta$ is not $\alpha$-Holder for $\beta&lt;\alpha&lt;1$?
If $\alpha&gt;\beta$, then
$$\sup_{0&lt;x&lt;y&lt;1}\frac{|y^\beta-x^\beta|}{|y-x|^\alpha}\ge \sup_{0&lt;x&lt;1}\frac{|(2x)^\beta-x^\beta|}{|2x-x|^\alpha}=(2^\beta-1)\sup_{0&lt;x&lt;1}x^{\beta-\alpha}=+\infty.$$
It's enough to use a counter example, say $x_1 = 0$, $x_2 = 0.5$ then we have that 

$$ |0^\beta -0.5^\beta| \leq |0.5|^\beta$$

but the LHS you will see equals to $0.5^\beta$. Then for any $\alpha &gt;\beta$ you'll see that $0.5^\beta &gt; 0.5^\alpha$, so it can't be $\alpha$ - Holder for $\beta &lt; \alpha &lt; 1$

Using the value of $0$ is ok because $x^\beta$ us also $\beta$ - continuous on $[0,1]$...

If you don't like that to see the argument still holds instead of $0$ use a small $\epsilon$ in it's place. Then if $f(x)$ is $\alpha$ - Holder then we would have 

$$ |\epsilon^\beta - 0.5^\beta| \leq |\epsilon - 0.5|^\alpha.$$

Take the limit as $\epsilon \rightarrow 0$ then you have a contradiction.