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

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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.

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    $\begingroup$ 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

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