Harnack constant from the poisson kernel ratio

The ball mean property and continuous mean-value theorem give smoothness. Fix $\mathrm{\mid }x\mathrm{\mid }<s<R$. The smooth trace on $\mathrm{\partial }{B}_s$ and harmonic replacement represent $u$ by the kernel there; at the center the same formula gives ${\int }_{\mathrm{\partial }{B}_s}u={\omega }_{n-1}{s}^{n-1}u(0)$.

For $\mathrm{\mid }y\mathrm{\mid }=s$, the inequalities $s-\mathrm{\mid }x\mathrm{\mid }\le \mathrm{\mid }x-y\mathrm{\mid }\le s+\mathrm{\mid }x\mathrm{\mid }$ bound the positive kernel above and below. Integrating against the nonnegative trace gives $\frac{1-q}{(1+q{)}^{n-1}}u(0)\le u(x)\le \frac{1+q}{(1-q{)}^{n-1}}u(0)$, where $q=\mathrm{\mid }x\mathrm{\mid }\mathrm{/}s<1$.

Let $s\uparrow R$ so $q\to t<1$. This proves the two bounds. If $u(0)=0$, ball Harnack already gives $u\equiv 0$ on the ball, and both displayed inequalities are equalities. At $x=0$, both factors equal one.