# Math Help - Lipschitz continuity

1. ## [SOLVED] Lipschitz continuity

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

2. ## Re: Lipschitz continuity

Take for example the function $f(x) = \sqrt{|x|}$ on $(-1,1)$
According to this theorem, yes. We have that $\mathbb{R}$ is a Banach space and $A=(0,1)$ is a convex open subset of $\mathbb{R}$. Also, $\sqrt{|x|}=\sqrt{x}$ is continuous on the closure $\bar{A}=[0,1]$ of A and differentiable on A. Therefore, $\sqrt{x}$ is Lipschitz continuous on $\bar{A}$ iff $(\sqrt{x})'$ is bounded on A, which it is not.
However, it is easiest to show that $\sqrt{x}$ is not Lipschitz continuous on [0,1) by definition since $(\sqrt{x}-\sqrt{0})/(x-0)$ is not bounded on that interval.