If all mixed partials of a $C^1$ function exist and are continuous, is the function $C^2$? [closed]

Question

For $n \geq 2$, let $f: \mathbb R^n \to \mathbb R$ be a $C^1$ function such that the mixed partial derivatives $\partial_i \partial_j f$ exist and are continuous for all $i \neq j$. Is it true that $f$ is $C^2$?

Answer 1

No. Let $f(x,y) = x|x|$. It is easy to check $df = |x| ~dx$ is continuous. The mixed partial $\partial^2_{xy} f = 0$ exists and is continuous. But the function is not $C^2$.