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.
Best Answer
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 $$ \psi(z) = \phi_\epsilon(z) + \eta + \frac\epsilon2 |z - \mathring{y}|^2 $$
This shows that $u$ is not viscosity subharmonic.