# Characterization of convex functions

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?

• You can prove both by mollification, since convolution with a positive function preserves convexity Mar 2, 2017 at 14:20