I want to show that for $0<s \leq 1$ $$f: \mathbb{R}_{\geq0} \to \mathbb{R}_{\geq0}$$ $$x \mapsto x^s$$ is Lipschitz continuous on$[a,\infty)$ for any $a \in\mathbb{R}_{>0}$, and that it is NOT on $[0,\infty)$.
So what I want to show is that $\exists \hspace{2 mm} C \in \mathbb{R}_{\geq0}$ sucht that for all $ x,y \in [a,\infty) $
$$|x^s -y^s| \leq C|x-y|$$
and therefore, assuming, wlog $\hspace{1mm} x>y$
$$(x^s -y^s) \leq C(x-y),$$
And that $\nexists $ such $C$ on $[0,\infty).$
Once again my (lack of ) mastery with inequalities is not yet sufficient for this type of problem. Can someone lead the way (without recourse to concave/convex functions)?
Best Answer
A purely algebraic proof for $s= \tfrac12$ is immediate: suppose wlog that $x>y$, then $$\sqrt{x}-\sqrt{y}=(\sqrt{x}-\sqrt{y})\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}\leq \frac{x-y}{2\sqrt{a}}$$ because $x,y\geq a$.
This idea can be adapted to $0<s\leq 1$ as well: \begin{align} x^s-y^s&=(x^s-y^s)\frac{x^{1-s}+y^{1-s}}{x^{1-s}+y^{1-s}}\\ &=\frac{x-y+x^sy^{1-s}-y^sx^{1-s}}{x^{1-s}+y^{1-s}}\\ &\leq \frac{2(x-y)}{x^{1-s}+y^{1-s}}\leq\frac{x-y}{a^{1-s}} \end{align} where we used that $$ x^sy^{1-s}-y^sx^{1-s}\leq x-y$$ which follows from $$(x^s+y^s)(y^{1-s}-x^{1-s})\leq 0.$$