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 $s \in \mathbb{R}$.
Best Answer
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 $b=y$.