# Convex functions with non-singular hessian measure are continuously differentiable?

It is known that every convex function $f: \Omega\to \mathbb{R}$, $\Omega$ convex subset of $\mathbb{R}^n$, has a weak derivative of bounded variation $Df\in BV_{loc}(\mathbb{R}^n)$ (e.g. Evans and Gariepy 1992), and therefore a distributional Hessian $\mu_{ij}:=[D_i D_j f]$ which is a non-negative definite Radon measure which can be decomposed as $\mu_{ij}=\mu_{ij}^a+\mu_{ij}^s$ where $\mu^s$ is the singular part and $\mu^a$ is the absolutely continuous part with respect to the Lebesgue measure $\mathcal{L}^n$.

My question is: suppose that $\mu_{ij}^s$ vanishes, does $Df$ admit a continuous representative, and so is $f$ $C^1$?

Here are some comments. For $n=1$ the question is trivial, because $\mu^s=0$ is equivalent to the absolute continuity of $D f$.

For $n>1$ the condition that $Df$ has weak derivatives in $L^1$ does not imply that $Df$ is continuous. Perhaps by adding the convexity of $f$ one can conclude that $f$ is $C^1$?

I also know that the singular part decomposes into Cantor and Jump parts $\mu^s=\mu^c+\mu^j$ (e.g.\ Ambrosio, Fusco, Pallara, 2000), however it is seems to me that as it is defined the Jump set can be empty for discontinuous $Df$. Under convexity of $f$ the condition $\mu^j=0$ is sufficient for the continuous differentiability of $f$?

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I now realize that the answer is probably negative. The function $f(x,y)=\sqrt{x^2+y^2}$ seems to be a counterexample. Although $Df$ is indeed discontinuous at the origin the Jump set is empty because, contrary to its misleading name, the Jump set is made discontinuities of special type where just two special values of $Df$ are involved. The cantor part also vanishes. So the fact that $\mu^s=0$ does not improve the differentiability properties of $f$? – Ettore Minguzzi May 11 '14 at 2:03