# Bounded partial derivatives implies uniform continuity

As is well-known, by mean value theorem, if $f$ is continuously differentiable on a convex domain $U$, and $Df$ is bounded, then $f$ is Lipschitz continuous (and thus uniformly continuous). However, if $U$ is not convex, could we find a counterexample that this statement is not right.

That is, I want to construct an domain $U$ and a continuously differentiable function $f:U\to\Bbb R$ such that $f$ has bounded partial derivative, but $f$ is not uniformly continuous.

## 1 Answer

Let $f(x) = \begin{cases} 0 , & x <1 \\ 1, & x > 1\end{cases}$ on $U= \mathbb{R} \setminus\{1\}$.

Note that convexity is not the issue per se. If $U$ is a connected domain, then it is straightforward by taking a polygonal path and using the mean value theorem to show that $f$ is Lipschitz (assuming that $Df$ is bounded, of course).

The real issue is that when the connected components of $U$ are not uniformly bounded 'away' from each other, then the above sort of behaviour is possible.