# Does bounded and continuous implies Lipschitz?

If a function $f : \mathbb{R} \rightarrow \mathbb{R}$ is integrable, bounded and continuous, is it also Lipschitz continuous?

No. Let $$f(x)=\left\{ \begin{array}{ll} 0&, x\leq 0\\ \sqrt{x} &,x\in [0,1]\\ -x^2+2x &, x\in [1,2]\\ 0&, x\geq 2. \end{array}\right.$$ $$f$$ is continuous, bounded and integrable, but it is not Lipschitz, since $$f|_{[0,1]}$$ is not Lipschitz.
• It's not integrable. To make it integrable, put $f(x)=\max(0,2-x)$ for $x>1$. May 10, 2014 at 17:07