Mean value property for harmonic functions

For every closed ball contained in the domain of a harmonic function, the value at its centre equals both its average over the ball and its average over the boundary sphere.

If $u\ge 0$ is a harmonic function on a neighbourhood of $\overline{B_{4r}(y)}$, thenFor $x,z\in B_r(y)$, use $B_r(x)\subset B_{3r}(z)\subset B_{4r}(y)$ and the mean value property for harmonic functions. Nonnegativity lets us enlarge the integral, and the ratio of volumes is $3^n$. This compares values without requiring derivative bounds.

For a nonnegative harmonic function on the unit disc, the displayed inequalities are sharp. Apply the Poisson integral on the unit disk on radius $R>|z|$, bound its positive kernel between $(R-|z|)/(R+|z|)$ and its reciprocal, and use the mean value property for harmonic functions. Letting $R\uparrow 1$ proves the assertion. Positive Poisson kernels with one boundary pole give equality in the appropriate radial directions.

For a positive harmonic function on the unit disc, choose a global harmonic conjugate and apply the Carathéodory derivative bound for the right half-plane to $u+iv$. This gives $|\nabla u(0)|\le 2u(0)$ by the Cauchy-Riemann equations. A disk automorphism sending zero to $z$ has derivative modulus $1-|z{|}^2$, so the chain rule gives the displayed bound. Composing a boundary-pole Poisson kernel with a disk automorphism attains equality, proving the coefficient is optimal at every point.

If a continuous function has the spherical mean value property at every point for some sequence of radii decreasing to zero, then it is harmonic. Comparison on each relatively compact ball with the harmonic function having the same boundary data proves the result by propagating any hypothetical interior extremum along one of the admissible spheres.