Let $\Omega\subseteq\mathbb{R}^n$ open and convex. It is elementary that if $u\in C^2(\Omega)$ then
$$u \text{ is convex}\iff D^2u\geq0 \ \text{ in } \Omega$$
I was looking for a similar characterization for $u\in C(\Omega)$. If I remember correctly it can be formalized using distribution theory, but I was wondering if it can also be done somehow using viscosity theory. I don't know exactly how to formalize it and prove it(if it's true, of course).


Can you give me some hint or reference?



Viscosity characterization of convex functions
You can always consider the second (distribution) derivative of a continuous function and require that it is non-negative, which means that for all $T\in \mathbb R^n$ and all $\phi\in \mathscr D(\Omega;\mathbb R\_+)$
\begin{multline}
\sum\_{1\le j,k\le n}\langle \frac{\partial^2u}{\partial x\_j\partial x\_k}(x)T\_k T\_j, \phi(x)\rangle\_{\mathscr D'(\Omega), \mathscr D(\Omega)}
=
\sum\_{1\le j,k\le n}T\_k T\_j\langle u(x),\frac{\partial^2\phi} {\partial x\_j\partial x\_k}(x)\rangle\_{\mathscr D'(\Omega), \mathscr D(\Omega)}
\\
=\sum\_{1\le j,k\le n}T\_k T\_j\int u(x)\frac{\partial^2\phi} {\partial x\_j\partial x\_k}(x)dx\ge 0.
\end{multline}
Note also that non-negative distributions are actually Radon measures, so that, for all $T\in \mathbb R^n$,
$$
\sum\_{1\le j,k\le n}\frac{\partial^2u}{\partial x\_j\partial x\_k}(x)T\_k T\_j
\quad\text{ is a non-negative Radon measure.}
$$
Now, if $\rho$ is a non-negative compactly supported smooth function with integral 1 and $
\rho\_\epsilon(x)=\rho(x/\epsilon)\epsilon^{-n},
$
you may require that
$$
u\ast \langle (D^2\rho\_\epsilon) T\cdot T\rangle\ge 0,
$$
which is a pointwise condition when $u$ is continuous.


