Harnack inequality on a ball

Fix $d=\mathrm{\mid }x-a\mathrm{\mid }$ and $d<s<R$. The closed ball of radius $s-d$ centered at $x$ lies in $B_R(a)$, and its open ball lies in $B_s(a)$. Nonnegativity and the ball mean property yield $u(x)\le \mathrm{\mid }{B}_{s-d}{\mathrm{\mid }}^{-1}{\int }_{B_s(a)}u=(s\mathrm{/}(s-d){)}^{n}u(a)$.

Let $s\uparrow R$ to obtain the first inequality. It is also valid for $x=a$, with equality. No integral at radius $R$ is needed.

For $\mathrm{\mid }x-a\mathrm{\mid }\le r<R$, divide the segment from $a$ to $x$ into $m$ equal steps. Each has length at most $(R-r)\mathrm{/}4$. Every segment point centers a ball of radius $\rho =(R-r)\mathrm{/}2$ inside $B_R(a)$. Apply the first inequality from each endpoint of a step to the other, using these radius-$\rho$ balls: either value is at most $2^n$ times the other. Multiplying along the $m$ steps proves both comparisons with $C$. This uses no division by a value of $u$, so also covers $u(a)=0$.

An additional consequence is Gantumur’s growth estimate. If an entire harmonic $u$ satisfies $u(x)\ge -p(\mathrm{\mid }x\mathrm{\mid })$ for a nonnegative nondecreasing function $p$, fix $r=\mathrm{\mid }x\mathrm{\mid }>0$ and apply the first inequality to $u+p(2r)\ge 0$ on $B_{2r}(0)$. It gives $u(x)\le 2^n(u(0)+p(2r))-p(2r)\le 2^n(u(0)+p(2r))$. At $x=0$ the sharper expression after subtracting $p(0)$ gives the same conclusion, since $u(0)+p(0)\ge 0$.