Let $\Omega\subset \mathbb{R}^n$ be an open subset. Let $u\colon \Omega\to [-\infty,+\infty)$ be an upper semi-continuous function.


If $u$ is subharmonic then for any point $x\in \Omega$ and any $C^2$-smooth function $\phi$ defined near $x$ and such that $u\leq \phi$ and $u(x)= \phi(x)$ one has
$$\Delta\phi(x)\geq 0,$$
where $\Delta$ is the Laplacian.


**Is the converse true? I.e. does the above property characterize subharmonic functions in the class of upper semi-continuous functions?
If such a characterization is true in the class of continuous functions, that would also be of interest to me.**



A possible characterization of subharmonic functions
By "$u$ is subharmonic" do you mean it is so in the comparison sense, namely: given every closed ball $B\subseteq \Omega$, and every harmonic $\phi$ on $B$ with $\phi|\_{\partial B} \geq u|\_{\partial B}$, then $u|\_B \leq \phi|\_B$? If so, it is known that this definition is equivalent to viscosity subharmonicity (the second description you gave) for USC functions.


Sketch of proof by contrapositive:


Suppose there exists a closed ball $B\subseteq \Omega$ and a harmonic function $\phi:B\to \Omega$ with $\phi|\_{\partial B} \geq u|\_{\partial B}$, such that there is some $y\in \mathring{B}$ such that $\phi(y) < u(y)$ (in other words, suppose that $u$ is not comparison subharmonic).


We can first make $\phi$ strictly superharmonic:
Denote by $c$ the center of the ball $B$, and $r$ its radius. Consider the function
$$ \phi\_\epsilon(z) = \phi(z) + \epsilon (r^2 - |z-c|^2) $$
For sufficiently small $\epsilon$ we have $\phi\_\epsilon(y) < u(y)$ still. Therefore since $u$ is upper semi continuous $\sup (u-\phi\_\epsilon)$ is strictly positive and hence attained in the interior of $B$; call this point $\mathring{y}$ and this maximum value of $\sup (u-\phi\_\epsilon) = \eta$.


Now examine the function
$$ \psi(z) = \phi\_\epsilon(z) + \eta + \frac\epsilon2 |z - \mathring{y}|^2 $$


1. By definition $\psi(\mathring{y}) = u(\mathring{y})$.
2. $\Delta \psi(z) = -\epsilon \cdot n < 0$
3. $\psi(z) - u(z) = \frac{\epsilon}2|z - \mathring{y}|^2 + \underbrace{\eta + \phi\_\epsilon(z) - u(z)}\_{\geq 0} > 0$ away from $\mathring{y}$.


This shows that $u$ is not viscosity subharmonic.


