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}),


\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.