Let $f$ be a continuous functions from $[0,1]$ to $\mathbb{R}$. Then, $f$ is not necessarily lipschitz.

Is the above statement true?

I thought since $f$ is continuous on a compact metric space, $f$ is bounded. Thus $f'$ is also bounded, which implies that $f$ is lipschitz.

Is my argument correct? If not, where did I get wrong? Help me.

  • 1
    $\begingroup$ You're assuming that $f$ is differentiable $\endgroup$ – DGRasines Jun 9 '14 at 23:13
  • $\begingroup$ In addition, derivatives need not be bounded. $\endgroup$ – David Mitra Jun 9 '14 at 23:39

A continuous function is not necessarily differentiable. In any case, your argument would work if the function was continuously differentiable. For a counter example, take $\sqrt x$, for instance.


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