# Can a function be differentiable everywhere on its domain, but not Lipschitz on its domain?

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?

• $f(x)=\sqrt x$ on $(0,1)$? – David Mitra Dec 3 '20 at 6:59

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).
• Every continuous function is locally bounded. But if the derivative of a function is unbounded, then $f$ may not be locally Lipschitz continuous. – José Carlos Santos Dec 3 '20 at 7:12