# Why is every continuously differentiable function with a uniform bounded derivative lipschitz continuous

I only know how to prove this for functions on a convex set by using the mean value theorem, but is this also true for this general case when nothing is said about the domain of the function besides the fact that it is a subset of $\mathbb{R}^n$?

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sorry spelling mistake – user66906 Nov 17 '13 at 16:35
Even for connected $U \subset \mathbb{R}^n$ it need not be true. Consider $(r\cos\varphi,r\sin\varphi) \mapsto \varphi$ on $\{(x,y) : x^2+y^2 > 1, y\neq 0 \lor x > 0\}$ with $-\pi < \varphi < \pi$. – Daniel Fischer Nov 17 '13 at 16:44
One can prove this for quasiconvex domains (which are more general than convex): see here. And I think this is as far as one can go. If the domain is not quasiconvex, then the intrinsic distance function has bounded derivative but is not Lipschitz. – user103402 Nov 17 '13 at 16:46

The function $f(x) = \operatorname{sgn} x$ defined on the domain $D = \mathbb{R} \setminus \{0\}$ is differentiable at every point of $D$ and the derivative is continuous and bounded (it's identically zero), but it is not Lipschitz.
If the domain is not connected the assertion is incorrect. Define a function on the union of intervals $[2n, 2n+1]$ to be equal to $n^2$ on the corresponding interval $[2n, 2n+1]$. This is continuously differentiable but not Lipschitz.