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?
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.
