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If we suppose that a function is differentiable at every point on its domain (but place no more restrictions than that), does it follow that the function is Lipschitz on its domain? I think that it does not, and I suspect that it is because we do not require that the function is continuously differentiable. But I'm not sure how to prove my thought - specifically, I can't come up with a counterexample.

If I am right that it is not Lipschitz, can we say if it will be locally Lipschitz at every point?

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    $\begingroup$ $f(x)=\sqrt x$ on $(0,1)$? $\endgroup$ – David Mitra Dec 3 '20 at 6:59
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If $f\colon\Bbb R\longrightarrow\Bbb R$ is the function defined by $f(x)=x^2$, then $f$ is differentiable and even a $C^\infty$ function, but it is not Lipschitz continuous.

On the other hand, if you define$$\begin{array}{rccc}f\colon&\Bbb R&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x^2\sin\left(\frac1{x^2}\right)&\text{ if }x\ne0\\0&\text{ if }x=0,\end{cases}\end{array}$$then $f$ is not Lipschitz continuous on any neighborhood of $0$. This has to do with the fact that $f'$ is unbounded (which is stronger than being discontinuous).

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  • $\begingroup$ So it is the boundedness of the function (or its derivative), at least in a neighbourhood, that allows us to conclude Lipschitz continuity? $\endgroup$ – F McA Dec 3 '20 at 7:11
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    $\begingroup$ Every continuous function is locally bounded. But if the derivative of a function is unbounded, then $f$ may not be locally Lipschitz continuous. $\endgroup$ – José Carlos Santos Dec 3 '20 at 7:12

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