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Let $\Omega$ be a convex open subset of $\mathbb R^n$ and let $f:\Omega\rightarrow \mathbb R$ be a convex function. Since $f$ is continuous, it can be considered as a distribution on $\Omega$ and then I claim that $$ \text{Hess}(f)=f''\text{ is a non-negative symmetric matrix of Radon measures, i.e.,} $$ $$\text{for every $\phi\in C^\infty_c(\Omega; \mathbb R_+), T\in \mathbb R^n$,} \qquad \sum_{1\le j,k\le n}\langle \frac{\partial^2 f}{\partial x_j\partial x_k}, \phi\rangle_{\mathscr D'(\Omega),\mathscr D(\Omega)} T_jT_k\ge 0. \tag 1$$ Note that this inequality implies that each ${\partial^2 f}/{\partial x_j\partial x_k}$ is a Radon measure (i.e. a distribution with order 0).

Another claim and my question: let $\Omega$ be as above and let $f\in \mathscr D'(\Omega)$ such that (1) holds true. Then I claim that $f$ is a convex function and my question is why?

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    $\begingroup$ You can prove both by mollification, since convolution with a positive function preserves convexity $\endgroup$ – Deane Yang Mar 2 '17 at 14:20
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This has been shown by Dudley, see https://www.jstor.org/stable/24490947.

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