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.


Behavior of harmonic functions comparing with a Green’s function

Theorem Let {E \subset \subset B_1}, {u \in C^0 (\bar{B_1} \setminus E) \cap C^2 (B_1 \setminus E)} satisfies

\displaystyle - \Delta u =0, \quad \text{in } B_1 \setminus E, \text{ with } u \ge 0 \text{ on } \partial B_1,

If there exists {G \in C^2(\bar{B_1} \setminus E)} such that

\displaystyle - \Delta G =0, \quad \text{in } B_1 \setminus E \text{ and } \lim_{dist(x,E) \rightarrow 0} G(x) = +\infty.

If {\frac{u^-}{G(x)} \rightarrow 0} as {dist(x,E) \rightarrow 0}, then {u \ge 0} in {(B_1 \setminus E)}.

Proof: Without loss of generality assume {G \ge 0} on {\partial B_1}, or we can add a large constant on {G} to make it positive on boundary. {\forall \varepsilon >0}, there exists {0 < \delta(\epsilon) < \epsilon} such that

\displaystyle \frac{u^-}{G(x)} \le \varepsilon \text{ whenever } 0 < dist(x,E) \le \delta.

Since {\frac{-u}{G} \le \frac{\max(0, -u)}{G} = \frac{u^-}{G} \le \varepsilon}, we have

\displaystyle u(x) \ge - \varepsilon G(x) \text{ for } 0 < dist(x,E) \le \delta.

Also {u \ge - \varepsilon G(x)} on {\partial B_1} since {G} is non-negative on boundary. Then maximum principle tells { u \ge - \varepsilon G} in {B_1 \setminus E_\delta}, where {E_\delta = \{x\in B_1: dist(x,E) \le \delta \}}. Then we can first take {\delta \rightarrow 0^+} and then take {\varepsilon \rightarrow 0^+} to get {u \ge 0} in {B_1 \setminus E}. \Box

Remark The condition {\frac{u^-}{G(x)} \rightarrow 0} is essential, otherwise one can take {-G} plus a large constant as a counterexample.