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


1 Answer 1


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.

  • 2
    $\begingroup$ It's not integrable. To make it integrable, put $f(x)=\max(0,2-x)$ for $x>1$. $\endgroup$
    – tomasz
    May 10, 2014 at 17:07
  • $\begingroup$ @tomasz, thanks and fixed! $\endgroup$
    – user73454
    May 11, 2014 at 2:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.