proving that $f(x) = x^s$ is holder continuous with holder exponent s I want to show that $$f: \mathbb{R}_{\geq0} \to \mathbb{R}_{\geq0}$$ $$x \mapsto x^s$$ is Holder continuous with Holder exponent $s \in \mathbb{R}$, where $0<s \leq 1$. So what I want to show is that $\exists \hspace{2 mm} C \in \mathbb{R}_{\geq0}$ sucht that for all $ x,y \in \mathbb{R}_{\geq0},  $
$$|x^s -y^s| \leq C|x-y|^s$$
and therefore, assuming, wlog $\hspace{1mm} x>y$
$$(x^s -y^s) \leq C(x-y)^s.$$
I thought about Bernoulli's inequality but couldn't make that work.I thought about the binomial theorem, but didn't know how to handle the fact that $s \in \mathbb{R}$.
 A: Assume $x>y$.
$$x^s-y^s \leq C(x-y)^s$$
$$\Leftrightarrow x^s \leq C(x-y)^s+y^s$$
Claim: this holds for all $(x,y)$ when $C=1$. Proof: Because $0<s\leq 1$, for all $a,b \geq 0$
$$(\frac{a}{a+b})^s+(\frac{b}{a+b})^s \geq 1$$
$$\Leftrightarrow a^s+b^s \geq (a+b)^s$$
Now set $a=x-y$ and $b=y$.
A: $$|x^s - y^s||x^{1-s} + x^{1-2s}y^s + ... + y^{1-2s}| = |x - y|$$
$\forall{x,y},\,  \exists{m\gt0}\;\text{such that} \;m\le|x^{1-s} + x^{1-2s}y^s + ... + y^{1-2s}|$
If $|x - y|\lt 1$, $|x-y|\le |x-y|^s$.
Hence,
$$|x^s-y^s|\le\frac{1}{m}|x-y|\le\frac{1}{m}|x-y|^s$$
If, however, $|x-y|\gt1$, then $|x-y|\ge |x-y|^s$.
By the Archimede's theorem (or the Archimedean property of $\mathbb{R}$), 
$\exists{k}\;\text{such that}\;k|x-y|^s\ge|x-y|$.
Then, 
$$|x^s-y^s|\le\frac{1}{m}|x-y|\le\frac{k}{m}|x-y|^s$$
Notice that $k\ge1$, so $\frac{k}{m}|x-y|^s\ge\frac{1}{m}|x-y|^s$
Therefore, $$\forall{x,y \in \mathbb{R}_{\geq0}},\;|x^s-y^s|\le\frac{k}{m}|x-y|^s$$.
If we let $C=\frac{k}{m}$, then this becomes 
$$|x^s-y^s|\le C|x-y|^s$$
Hence proven.
