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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?

TheMobiusLoops
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3 Answers3

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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?

Git Gud
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  • I guess I'm just not seeing how this works out. No matter what $u$ and $v$ I pick, I'm always finding a $C$ that holds the inequality. – TheMobiusLoops Nov 06 '13 at 09:44
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    @TheMobiusLoops You're mixing the order of the quantifiers. A function $f$ is lipschitz if, and only if, $$\exists C\ge 0\forall x,y\in D_f(|f(x)-f(y)|\leq C|x-y|).$$ You need to pick a $C$ that words for ALL $x,y$. You can't first pick $x,y$ and then pick $C$. – Git Gud Nov 06 '13 at 09:50
  • OH! Thank you! That makes things work. – TheMobiusLoops Nov 06 '13 at 09:53
  • @TheMobiusLoops No problem. – Git Gud Nov 06 '13 at 09:54
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$|\sqrt{x}-\sqrt{y}|\leq C |x-y|\Rightarrow \big|\frac{\sqrt{x}-\sqrt{y}}{x-y}\big| \leq C \Rightarrow 1 \leq C |\sqrt{x}+\sqrt{y}|$

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

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

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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.

yoyostein
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