Converse of the mean value property

Remarks

Proof idea: let $\eta_\varepsilon$ be a smooth radial mollifier supported in $B(0, \varepsilon)$ with integral $1$. Polar coordinates and the mean value property on each sphere give $u * \eta_\varepsilon = u$ at the points at distance more than $\varepsilon$ from the complement of $U$. So $u$ is smooth there. For a smooth function, the average over $\partial B(x, r)$ minus $u(x)$ equals $\frac{r^2}{2n} \Delta u(x) + o(r^2)$; since it is $0$, the Laplacian vanishes. Together with the mean value property of harmonic functions, this shows that, among continuous functions, harmonicity is the same as the mean value property on spheres. The same holds with averages over balls in place of spheres. For the version with ball averages, local integrability of $u$ is enough, because a function that equals its ball averages is automatically continuous. Example: in $\R^2$, $u(x, y) = x y$ is continuous and its average over each circle about $(a, b)$ is $a b$, so $u$ is harmonic, as one checks directly from $\Delta(xy) = 0$.

Going deeper

For a student

The averaging definition needs no derivatives, which makes it the natural starting point for harmonic functions on general spaces and for subharmonic functions. It is also the key step in proving that a locally uniform limit of harmonic functions is harmonic: the mean value property passes to uniform limits, and continuity is preserved. A stronger result, Koebe's theorem, only asks for the mean value property on all small enough spheres about each point.