Lipschitz continuity

• Dec 14th 2011, 12:53 AM
[SOLVED] Lipschitz continuity
I want to find out if a function is lipschitz continuous.
Take for example the function $\displaystyle f(x) = \sqrt{|x|}$ on $\displaystyle (-1,1)$
I know it's not lipschitz continuous since the graph's slope goes to +/-infinity when x approaches 0, but is this argument sufficient?
• Dec 14th 2011, 05:46 AM
emakarov
Re: Lipschitz continuity
Quote:

Originally Posted by madflame991
Take for example the function $\displaystyle f(x) = \sqrt{|x|}$ on $\displaystyle (-1,1)$
I know it's not lipschitz continuous since the graph's slope goes to +/-infinity when x approaches 0, but is this argument sufficient?

According to this theorem, yes. We have that $\displaystyle \mathbb{R}$ is a Banach space and $\displaystyle A=(0,1)$ is a convex open subset of $\displaystyle \mathbb{R}$. Also, $\displaystyle \sqrt{|x|}=\sqrt{x}$ is continuous on the closure $\displaystyle \bar{A}=[0,1]$ of A and differentiable on A. Therefore, $\displaystyle \sqrt{x}$ is Lipschitz continuous on $\displaystyle \bar{A}$ iff $\displaystyle (\sqrt{x})'$ is bounded on A, which it is not.

However, it is easiest to show that $\displaystyle \sqrt{x}$ is not Lipschitz continuous on [0,1) by definition since $\displaystyle (\sqrt{x}-\sqrt{0})/(x-0)$ is not bounded on that interval.