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!