# Harnack inequality implies Holder continuity

We will show a basic argument why Harnack inequality implies Holder continuity. Assume ${u \in H^1(B_2)}$ is a weak solution to a uniform elliptic equation

$\displaystyle -\partial_i(a^{ij} \partial_j u) = 0,$

in ${B_2}$. Assume we have Harnack inequality (Indeed we do have due to Moser), i.e. for any positive weak solution ${u}$ we will have

$\displaystyle \sup_{B_1} u \le C \inf_{B_1} u,$

where ${C}$ is a constant independent of ${u}$.

First one will know ${u \in L^\infty(B_1)}$, see this note.

If ${u}$ is a weak solution, so are ${\sup_{B_2} u - u}$ and ${u - \inf_{B_2} u}$. And they are strictly positive in ${B_1}$ due to the strong maximum principle, otherwise ${u}$ will be a constant. By Harnack inequality, we will have

$\displaystyle \sup_{B_2} u - \inf_{B_1} u \le C(\sup_{B_2} u - \sup_{B_1} u),$

$\displaystyle \sup_{B_1} u - \inf_{B_2} u \le C(\inf_{B_1} u - \inf_{B_2} u).$

Adding these two inequalities up we will have

$\displaystyle \text{osc}_{B_2} u + osc_{B_1} u \le C(osc_{B_2} u - osc_{B_1}),$

hence

$\displaystyle osc_{B_1} u \le \frac{C-1}{C+1} osc_{B_2} u.$

Denote ${\theta = \frac{C-1}{C+1} < 1}$, and choose ${\alpha > 0}$ such that ${\theta = 2^{- \alpha}}$, by interation we know

$\displaystyle osc_{B_{2^{-k}}} u \le \theta^k osc_{B_1} u \le 2 \theta^k \|u\|_{L^\infty(B_1)}.$

Then for any ${0 < r < 1}$, choose ${k \in {\mathbb N}}$ such that ${2^{-k-1} < r \le 2^{-k}}$, we will have

$\displaystyle osc_{B_r} u \le osc_{B_{2^{-k}}} u \le 2^{-\alpha k + 1} \|u\|_{L^\infty(B_1)} = 2^{1+\alpha} \cdot 2^{- \alpha ( k +1)} \|u\|_{L^\infty(B_1)}< 2^{1+\alpha} \|u\|_{L^\infty(B_1)} r^\alpha.$

So Holder continuity follows.

Theorem Suppose ${v \in W^{1,2}(B_R)}$ is a subsolution of ${-\partial_i (a^{ij}(x) \partial_j u) =0}$, ${\lambda I \le (a^{ij}(x)) \le \Lambda I}$. Then for all ${p>0}$, ${0<\theta <1}$,
$\displaystyle \sup_{B_{\theta R}} v \le C(n, \lambda, \Lambda, p) (1- \theta)^{- \frac{n}{p}} \left( \frac{1}{|B_R|} \int_{B_R} (v^+)^p \right)^{\frac{1}{p}}.$