Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 12 / 5 / ii

A local form of the Harnack inequality for harmonic functions is: if $u\ge 0$ is harmonic on a neighbourhood of $\overline{B_{4r}(y)}$, thenFor $x,z\in B_r(y)$, the triangle inequality gives $B_r(x)\subset B_{3r}(z)\subset B_{4r}(y)$. The ball mean value property for harmonic functions and nonnegativity therefore implyTaking the supremum over $x$ and infimum over $z$ proves the inequality, including the case where the infimum is zero. No strict positivity assumption was used.

On a connected domain $\Omega$, the equivalent compact-set statement is that for each nonempty compact $K⋐\Omega$ there exists $C(K,\Omega ,n)$ withHere is the passage from balls to this statement. Cover $K$ by finitely many small balls $B_{r_j}(y_j)$ whose quadrupled closed balls lie in $\Omega$. Join their centres to a fixed centre by paths in $\Omega$. Each of the finitely many paths has compact image and positive distance from the boundary, so subdividing it gives a finite chain of overlapping small balls whose quadrupled closures also lie in $\Omega$. The local inequality compares any two points in each ball; using a point in each successive overlap propagates comparisons along the chain. The number of comparisons is bounded uniformly over this finite collection of chains and covering balls. Multiplying the factors $3^n$ gives one finite constant for all pairs $x,z\in K$, proving the compact-set inequality. In particular a nonnegative harmonic function vanishing at one interior point vanishes throughout a connected domain.