Show that $f(x)=\sqrt{x}:[0,1]\rightarrow \mathbb{R}$ is not a Lipschitz function.

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?


$|\sqrt{x}-\sqrt{y}|\leq C |x-y|\Rightarrow \big|\frac{\sqrt{x}-\sqrt{y}}{x-y}\big| \leq C \Rightarrow 1 \leq C |\sqrt{x}+\sqrt{y}|$

Can you now find given $C>0$ real numbers $x,y$ which violates this law...

It should not be so difficult i guess....

Tags:

Real Analysis

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