# Does $f\colon \Omega \to \mathbb R$ differentiable imply $f$ locally Lipschitz?

Let $f\colon \Omega \subseteq \mathbb R^n \to \mathbb R$ be a differentiable function. Is it true that $f$ is locally Lipschitz, i.e. Lipschitz on every compact $K \subset \Omega$?

If $f$ were continuously differentiable, the answer would clearly be affirmative, by the mean value theorem and Weierstrass. What if we ask only for $f$ to be differentiable? I think I've found a counterexample with $n=1$: $$f \colon \mathbb R \ni x \mapsto \begin{cases}x^2\sin\left(\frac{1}{x^2}\right) & x\ne 0 \\ 0 & x=0 \end{cases}$$ The function $f$ is differentiable for every $x \in \mathbb R$, but I think that $$f'(x) = \begin{cases} 2x\sin\left(\frac{1}{x^2}\right) - \frac{2}{x}\cos\left(\frac{1}{x^2}\right) & x\ne 0 \\ 0 & x=0 \end{cases}$$ is not bounded in every neighbourhood of $0$. What do you think? Is my counterexample correct? Thanks.

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## 1 Answer

Yes, your counterexample is correct. While the Lipschitz property does not imply differentiability, it provides a uniform bound for the derivative if the latter exists. So, a function with unbounded derivative cannot be Lipschitz, and a function whose derivative is not bounded locally cannot be locally Lipschitz.

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Sorry for the silly previous comment, +1. – 1015 Feb 24 '13 at 15:12
Thank you very much. – Romeo Feb 24 '13 at 18:12