Consider the following problem:
Let $\Omega \subset \mathbb R^N$ be a domain and $u \in C(\Omega)$. Show that $u$ is subharmonic in the Mean Value Inequality sense, that is,
$$
u(x) \leq \frac{1}{\omega_N r^N} \int_{B_r(x)} u (y) \ dy, \quad \forall x \in \Omega, \forall 0 < r < d(x, \partial \Omega).
$$
if and only if for every $x_0 \in \Omega$ and every $\phi \in C^2(\Omega)$ such that $u – \phi$ attains a local maximum at $x_0$ it holds that
$$
– \Delta \phi(x_0) \leq 0,
$$
that is, $u$ is a viscosity subsolution of the Laplace equation.
For the $\Rightarrow$ implication, I managed to show that $\phi$ is subharmonic in the sense of the (volumetric) Mean Value Inequality. Since it is $C^2$, it would suffice to show that it implies that $\phi$ is subharmonic in the classic sense. How to do this, if true?
For the $\Leftarrow$ implication I have no ideas, and any hints are welcome.
Thanks in advance.
EDIT
For the $\Rightarrow$ implication, it almost suffices to show that $\phi$ is subharmonic in the Mean Value Property sense. Indeed, suppose that $- \Delta \phi(x_0) > 0$. Then we would have the opposite inequality.
Best Answer
There are lots of ways to prove this, depending on what you know about PDEs and viscosity solutions. Here is one hint: Fix your ball $B_r(x)$ and solve Laplace's equation $\Delta v = 0$ on $B_r(x)$ with boundary conditions $v=u$ on $\partial B_r(x)$ (I assume you know how to construct the solution with Poisson's Integral Formula). Then $v$ is $C^2$, and the viscosity subsolution property for $u$ can be used to show that $u\leq v$ on $B_r(x)$ (this does not require the full comparison principle machinery of viscosity solutions, and is easy to prove by hand; I can add a comment on this if needed). Then since $v$ is harmonic and smooth, use the mean value property for $v$.