how to prove $f(x) = x^s$ Lipschitz continuous 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)?
 A: As $0 < s \leq 1$, so we have $$f^\prime(x) = sx^{s-1} > 0 \ \mbox{ for all } x \in (0, +\infty),$$ and therefore we can conclude that (since the derivative of $f$ is positive on $(0, +\infty)$) $f$ is a strictly increasing function on $[0, +\infty)$. 
Now let $a \in \mathbb{R}$ such that $a > 0$. If $a \leq x < y$, then $f(x) < f(y)$ and by the Mean Value Theorem for Derivatives we can find a real number $c \in (x, y)$ such that  $$ \left\vert f(x) - f(y) \right\vert = f(y) - f(x) = f^\prime(c) (y-x).$$ 
But as $0 < a \leq x < c < y$, so we also have $$f^\prime(c) = sc^{s-1} \leq c^{s-1} < a^{s-1},$$ and so 
$$\left\vert f(x) - f(y) \right\vert \leq a^{s-1} (y-x) = a^{s-1} \left\vert x-y \right\vert.$$ Hence $f$ is Lipschizian on $[a, +\infty)$. 
On the other hand, if $0 < x < y$, then we have $f(x) < f(y)$, but as $$\lim_{c \to 0+} f^\prime(c) = \lim_{c \to 0+} \frac{sc^s}{c} = +\infty,$$ so we cannot find any real number $M$ which can serve as an upper bound for $f^\prime$. So $f$ is not Lipschizian on $(0, +\infty)$. 
A: 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
$$(x^s+y^s)(y^{1-s}-x^{1-s})\leq 0.$$
A: A. $f(x)=x^s$ is Lipschitz continuous in $[a,\infty)$.
By Mean Value Theorem, for every $x,y\in [a,\infty)$,
$$
f(x)-f(y)=f'(\xi)(x-y)=\frac{s}{xi^{1-s}}(x-y)
$$
where $\xi\in (x,y)\subset [a,\infty)$, and hence
$$
|\,f(x)-f(y)|=\frac{s}{\xi^{1-s}}|x-y|\le \frac{s}{a^{1-s}}|x-y|.
$$
Thus, Lipschitz continuity with $L=s/a^{1-s}$.
B. $\,f(x)=x^s$ is not Lipschitz continuous is $[0,\infty)$.
Suppose there exists an $L>0$, such that
$$
|\,f(x)-f(y)|\le L|x-y|,\quad\text{for all $x,y\in[0,\infty)$.}
$$
In particular, for $y=0$, we would have
$$
x^s=|\,f(x)-f(0)|\le L|x-0|=Lx,
$$
and hence, we would have that
$$
\frac{1}{L}\le x^{1-s},
$$ 
for all $x>0$. Contradiction, for example for $\,x=(2L)^{-\frac{1}{1-s}}.$
