# Differentiability + Lipschitz = C^1?

It is well know that:

Let $E \subset \mathbb{R}^N$, let $f \colon E \to \mathbb{R}$, let $x_0 \in E^o$. Assume that there exists $r>0$ such that $B(x_0,r) \subset E$ and the partial derivatives $\frac{\partial f}{\partial x_i}$, $i=1 \dots N$, exist for every $x \in B(x_0,r)$ and are continuous at $x_0$. Then f is differentiable at $x_0$.

It is also known that the converse is not true, meaning that differentiability in one point does not imply that the partial derivatives are continuous at that point (here is a counterexample: http://web.mit.edu/watko/Public/024/notes/n03.pdf)

I was wondering if it is possible to proof a partial converse of the theorem, more precisely, my question is: is one of the two following statements true?

$1)$ Lipschitz continuity and differentiability in $B(x_0,r)$ imply continuity of the partial derivatives at $x_0$.

$2)$ Lipschitz continuity and differentiability in one point imply continuity of the partial derivatives in the same point.

Proofs and counterexamples are welcome!

• @user98130 It is indeed. – Daniel Fischer Oct 6 '13 at 9:48

Lipschitz continuity and differentiability do not imply the continuity of the partial derivatives.

We have the one-dimensional counterexample

$$f(x) = \begin{cases}x^2 \sin \tfrac1x &, x \neq 0\\ \quad 0 &, x = 0 \end{cases}$$

which is differentiable and locally Lipschitz-continuous everywhere. The derivative

$$f'(x) = \begin{cases}2x\sin \tfrac1x - \cos \tfrac1x &, x \neq 0\\ \qquad 0 &, x = 0\end{cases}$$

is not continuous in $0$. It is bounded on every bounded interval, whence $f$ is locally Lipschitz-continuous.

Higher-dimensional examples can be trivially obtained from this.