Boundary Harnack inequality
Let e = (1, 0, \dots, 0). Consider a function u satisfying
\begin{cases} \La u = 0 & \inn \Omega \cap B _2 \\ u \ge 0 & \inn \Omega \cap B _2 \\ u = 0 & \onn \partial \Omega \cap B _2 \end{cases}Theorem. If u, v satisfies above then
- \max _{\Omega \cap B _1} u \le C u (e)
- \sup _{\Omega \cap B _1} \frac uv \le C \frac{u(e)}{v(e)}
- \left[\frac uv\right] _{C ^\alpha (\Omega \cap B _1)} \le C \frac{u (e)}{v (e)}
Recall that:
- Interior Harnack inequality: if x can be connected to e by a chain of N balls, then u(x) \le C ^N u (e).
- Therefore, if the boundary is Lipschitz (or more generally called nontangentially accessible domains) then c d (x) ^q u (e) \le u (x) \le C d (x) ^{-q} u (e). Here d (x) = d (x, \partial \Omega)
- Zero extension is subharmonic, for which weak Harnack inequality holds: if \La u \ge 0 in B _1, u \le 1 in B _1, \abs{\set{u \le 0} \cap B _1} \ge \mu > 0 then \max _{B _\frac12} u \le 1- \theta. \theta can becomes large (close to 1) if \mu is close to \abs{B _1}.
- Weak Harnack inequality means: if \La u \ge 0 in B _1, u (x) \ge 1 at some x \in B _\frac12, \abs{\set{u \le 0} \cap B _1} \ge \mu > 0, then \max _{B _\frac12} u \ge 1 + \theta.
Proof of 1.
If it fails, then for arbitrarily large A we can find x _1 \in \Omega \cap B _1 with u (x _1) > A.
- x _1 must be \delta-close to \partial \Omega, otherwise we can use regular Harnack
- by weak Harnack, there exists x _2 with d(x _1, x _2) \le 2 d (x _1) such that u (x _2) \ge (1 + \theta) u (x _1). Reason: take r = d (x _1). Then B _{2 r} (x _1) has a lot of zeros.
- Then d (x _2) \le 3 d (x _1). We can find x _3 such that d (x _2, x _3) \le 2 d (x _2), and u (x _3) \ge u (x _2) (1 + \theta) ^2.
Then
C d (x _j) ^{-q} u (e) \ge u (x _j) \ge u (x _1) (1 + \theta) ^j \ge A (1 + \theta) ^j.So d (x _j) ^q \le \pthf{C u (e)}A (1 + \theta) ^{-j} \to 0. Then u (x _j) \to 0, contradicting to u (x _j) \ge A.
Proof of 2.
First we show the lemma.
Lemma. There exists \delta, \varepsilon > 0. If
\begin{cases} \La u = 0 & \inn \Omega \cap B _1 \\ u \ge 1 & \inn \Omega _\delta \cap B _1 \\ u \ge -\varepsilon & \inn \Omega \cap B _1 \\ u = 0 & \onn \partial \Omega \cap B _1 \end{cases}then u \ge 0 in \Omega \cap B _{\frac12}.
From Lemma to 2: WLOG u (e) = v (e) and u, v \le 1 in B _2. Apply to u - \varepsilon _0 v. Regular Harnack implies u - \varepsilon _0 v \ge c.
Lelemmamma. There exists \delta, a > 0. If
\begin{cases} \La u = 0 & \inn \Omega \cap B _1 \\ u \ge 1 & \inn \Omega _\delta \cap B _1 \\ u \ge -\varepsilon & \inn \Omega \cap B _1 \\ u = 0 & \onn \partial \Omega \cap B _1 \end{cases}then u \ge a in \Omega _{\frac\delta2} \cap B _{\frac12}, and u \ge -a \varepsilon in \Omega \cap B _\frac12.
Lemma is an extreme version of Lelemmamma with a = 0.
From Lelemmamma to Lemma: Iterate it to get a cone of positivity, which is the union \Omega _{\delta / 2 ^k} \cap B _{2 ^{-k}}.
Proof of Lelemmamma. Let v = u + \varepsilon _0. v \ge 0 and v \ge 1 in \Omega _\delta \cap B _\frac12. So using Harnack once we can get v \ge 2 a in \Omega _{\delta/2} \cap B _\frac12. Apply weak Harnack inequality to u _- = \max (0, -u). Recall the \theta can be large: \max _{B _\frac12} u _- \le C \nor{u _-} _{L ^1 (B _1)}. Since u _- is only nonzero in \Omega \setminus \Omega _\delta, it is a thin strip with very little measure.
Proof of 3.
Harnack inequality implies oscillation decay, which implies Hölder.
By part 2, \frac1A \le \frac uv \le A in B _1. So
\frac1A u \le v \le A u.Now A u - v and A v - u are both harmonic and nonnegative in B _1 and vanish on \partial \Omega, so by boundary Harnack
\frac1C \le \frac{A u - v}{A v - u} \le C \qquad \text{ in } B _{\frac12} \cap \Omega.So
\frac{C + A}{C + \frac1A} \cdot \frac1A u \le \frac{A + C}{A C + 1} u \le v \le \frac{A C + 1}{A + C} u = \frac{C + \frac1A}{C + A} \cdot A u.We have oscillation decay!