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Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be a function that is twice-differentiable almost everywhere. That is, $f$ is differentiable almost everywhere, and its $\nabla f$ is also differentiable almost everywhere. Furthermore $\nabla^2 f = 0$ where defined, and so the derivatives of $\nabla f$ agrees as you approach the non-differential measure zero set from any direction.

My end goal is to show that $f$ is convex. The approach I'm taking is showing that the Hessian is PSD. If $f$ were twice differentiable everywhere, this would be sufficient. However $f$ is not even differentiable everywhere, it is differentiable almost everywhere, and where its Hessian is defined $\nabla^2 f = 0$. Does the conclusion that $f$ is convex still hold?

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    $\begingroup$ is the Heaviside step function a counterexample? $\endgroup$ – LinAlg Dec 16 '18 at 3:02
  • $\begingroup$ Isn't $f:x\mapsto -|x|$ a concave non-convex function $\mathbb R\to\mathbb R$ that satisfies your conditions? $\endgroup$ – kimchi lover Dec 16 '18 at 3:41

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