Show that $f(x)=\sqrt{x}:[0,1]\rightarrow \mathbb{R}$ is not a Lipschitz function. A function $f:D\rightarrow \mathbb{R}$ is said to be a Lipschitz function provided that there is a nonnegative number $C$ such that
$|f(u)-f(v)|\le C|u-v|$ for all $u,v\in D$

We want to show there are there exist $u,v\in [0,1]$ such that $|\sqrt{u}-\sqrt{v}|\le C|u-v|$ is false, but I can't find anything that works. Any suggestions?
 A: Suppose $f$ is lipschitz. Then, there exists $C\ge 0$ such that for all $u,v\in [0,1]$ it holds that $|\sqrt u - \sqrt v|\leq C|u-v|$.
This inequality needs to hold for all $u,v$, in particular for $u\neq v$, thus, under this hypothesis it holds that $C\ge \left|\dfrac{\sqrt u-\sqrt v}{u-v}\right|=\left|\dfrac{\sqrt u-\sqrt v}{(\sqrt u-\sqrt v)(\sqrt u+\sqrt v)}\right|=\left|\dfrac{1}{\sqrt u+\sqrt v}\right|=\dfrac{1}{\sqrt u+\sqrt v}$.
Can you conclude?
A: $|\sqrt{x}-\sqrt{y}|\leq C |x-y|\Rightarrow \big|\frac{\sqrt{x}-\sqrt{y}}{x-y}\big| \leq C \Rightarrow 1 \leq C |\sqrt{x}+\sqrt{y}|$
Can you now find given $C>0$ real numbers $x,y$ which violates this law... 
It should not be so difficult i guess....
A: Suppose there exists $C\geq 0$ such that for all $x,y\in [0,1]$, $x\neq y$, $$\frac{|f(x)-f(y)|}{|x-y|}\leq C$$
By Mean Value Theorem, this means that $$|f'(\xi)|\leq C$$, where $\xi$ lies between $x$ and $y$.
However, $f'(x)=\frac{1}{2}x^{-1/2}$ is unbounded on $[0,1]$, so we can always choose $\xi$ sufficiently close to 0 such that the above inequality is violated. Contradiction.
