Harnack's inequality

Remarks

For $\rho$ with $r < \rho < R$, the Poisson formula on the disc of radius $\rho$ gives $u(z)$ as an average of $u$ on the circle $\abs{w - a} = \rho$ against the Poisson kernel $(\rho^2 - r^2)/\abs{w - z}^2$. Since $\rho - r \le \abs{w - z} \le \rho + r$, this kernel lies between $(\rho - r)/(\rho + r)$ and $(\rho + r)/(\rho - r)$, and the average of $u$ over the circle is $u(a)$. Letting $\rho \to R$ gives the inequality. The hypothesis $u \ge 0$ is necessary: $u(z) = \operatorname{Re} z$ on the unit disc has $u(0) = 0$ but takes nonzero values. In $\R^n$ with $n \ge 2$, the same argument gives $\dfrac{R^{n-2}(R - r)}{(R + r)^{n-1}}\, u(a) \le u(x) \le \dfrac{R^{n-2}(R + r)}{(R - r)^{n-1}}\, u(a)$ for $\abs{x - a} = r < R$.

Going deeper

For a student

A chain of discs gives the general form on a connected open set $\Omega$ with a compact subset $K$. There is a constant $C$ such that $\sup_K u \le C \inf_K u$ for every nonnegative harmonic $u$ on $\Omega$. This is the key estimate behind Harnack's principle on increasing sequences of harmonic functions. It is used in Perron's method for the Dirichlet problem and in the study of positive harmonic functions, such as the Herglotz representation theorem. Moser extended Harnack's inequality to solutions of elliptic equations in divergence form with measurable coefficients.