The Boundary Harnack Principle for Nonlocal Elliptic Operators in Non-divergence Form

Abstract

We prove a boundary Harnack inequality for nonlocal elliptic operators L in non-divergence form with bounded measurable coefficients. Namely, our main result establishes that if Lu₁ = Lu₂ = 0 in Ω ∩ B₁, u₁ = u₂ = 0 in B₁ ∖Ω, and u₁,u₂ ≥ 0 in ℝⁿ, then u₁ and u₂ are comparable in B_1/2. The result applies to arbitrary open sets Ω. When Ω is Lipschitz, we show that the quotient u₁/u₂ is Hölder continuous up to the boundary in B_1/2. These results will be used in forthcoming works on obstacle-type problems for nonlocal operators.

Appendix: Subsolution in Lipschitz Domains

We prove here a lower bound for positive solutions u in Lipschitz domains, namely u ≥ cd^2s−γ in Ω for some small γ > 0. This is stated in Lemma 4.5, which we prove below.

For this, we need to construct the following subsolution.

Lemma A.1

Lets ∈ (0, 1), ande ∈ S^n− 1. Givenη > 0, there is𝜖 > 0 dependingonly on n, s,ηandellipticity constants such that the following holds.

Define

$${\Phi}(x) := \left( e\cdot x- \eta |x| \left( 1- \frac{(e\cdot x)^{2}}{|x|^{2}} \right)\right)_{+}^{2s-\epsilon}$$

Then,

$$\left\{\begin{array}{ll} M^{-} {\Phi} \ge 0 \quad & \text{in }\mathcal C_{\eta} \\ {\Phi} = 0 \quad & \text{in }\mathbb{R}^{n} \setminus \mathcal C_{\eta} \end{array}\right. $$

where$\mathcal C_{\eta }$is the cone defined by

$$\mathcal C_{\eta}: = \left\{ x \in \mathbb{R}^{n}\ : e\cdot \frac{x}{|x|} > \eta \left( 1 - \left( e\cdot \frac{ x }{|x|} \right)^{2}\right) \right\}.$$

The constant 𝜖depends only on η, s, and ellipticity constants.

In particular Φ satisfies M⁻ Φ ≥ 0 in all of ℝⁿ.

Proof

By homogeneity it is enough to prove that, for 𝜖 small enough, we have M⁻ Φ ≥ 1 on points belonging to $e + \partial \mathcal { C}_{\eta }$, since all the positive dilations of this set with respect to the origin cover the interior of $\mathcal {\tilde C}_{\eta }$.

Let thus $P\in \partial \mathcal { C}_{\eta }$, that is,

$$e\cdot P- \eta \left( |P| - \frac{(e\cdot P)^{2}}{|P|} \right) = 0. $$

Consider

$$\begin{array}{ll} {\Phi}_{P}(x) &:= {\Phi}(P+e+x) \\ &= \left( e\cdot (P+e+x)- \eta \left( |P+e+x| - \frac{(e\cdot (P+e+x))^{2}}{|P+e+x|} \right)\right)_{+}^{2s-\epsilon} \\ &= \left( 1 +e\cdot x- \eta \left( |P+e+x| -|P|- \frac{(e\cdot (P+e+x))^{2}}{|P+e+x|} +\frac{(e\cdot P)^{2}}{|P|} \right)\right)_{+}^{2s-\epsilon} \\ &=\left( 1 +e\cdot x- \eta \psi_{P}(x) \right)_{+}^{s+\epsilon}, \end{array}$$

where we define

$$\psi_{P}(x) := |P+e+x| -|P|- \frac{(e\cdot (P+e+x))^{2}}{|P+e+x|} +\frac{(e\cdot P)^{2}}{|P|} .$$

Note that the functions ψ_P satisfy

$$|\nabla \psi_{P}(x)| \le C \quad \text{in } \mathbb{R}^{n} \setminus \{ -P-e\}, $$

and

$$ |D^{2} \psi_{P}(x)| \le C \quad \text{for }x \in B_{1/2}, $$

(A.1)

where C does not depend on P (recall that |e| = 1).

Now for fixed $\tilde e \in \partial \mathcal { C}_{\eta }\cap \partial B_{1}$ let us compute

$$\lim_{t\uparrow +\infty} \psi_{t\tilde e}(x) = \lim_{t\uparrow +\infty} (|t\tilde e +e+x|-|t\tilde e|) - \lim_{t\uparrow +\infty} \left( \frac{(e\cdot (t\tilde e +e+x))^{2}}{|t\tilde e +e+x|} -\frac{(e\cdot t \tilde e)^{2}}{|t\tilde e |} \right). $$

On the one hand, we have

$$\lim_{t\uparrow +\infty} (|\tilde e t+e+x|-|\tilde e t|) = \tilde e\cdot(e+x). $$

On the other hand to compute for $f_{t}(y) := \frac {(e\cdot (t\tilde e + y))^{2}}{|t\tilde e +y|}$ we have

$$\partial_{y_{i}} f_{t}(y) = \frac{2(e\cdot (t\tilde e+ y)) e_{i}}{|t\tilde e +y|} - \frac{(e\cdot (t\tilde e + y))^{2}}{|t\tilde e +y|^{3}} (t\tilde e +y )_{i} $$

and hence

$$\lim_{t\uparrow +\infty} \partial_{y_{i}} f_{t}(y) = \left( 2(e\cdot \tilde e) e_{i}- (e\cdot\tilde e)^{2} \tilde e_{i}\right). $$

Therefore,

$$\lim_{t\uparrow +\infty} \left( \frac{(e\cdot (t\tilde e +e+x))^{2}}{|t\tilde e +e+x|}-\frac{(e\cdot t \tilde e)^{2}}{|t\tilde e |}\right) = \left( 2(e\cdot \tilde e)e - (e\cdot\tilde e)^{2}\tilde e\right) \cdot(e+x) . $$

We have thus found

$$\lim_{t\uparrow +\infty} \psi_{P}(x) = \left( \tilde e-2(e\cdot \tilde e)e + (e\cdot\tilde e)^{2}\tilde e\right)\cdot(e+x) $$

and

$$\lim_{t\uparrow +\infty} \left( 1+e\cdot x-\eta \psi_{P}(x)\right) = \left( e -\eta \tilde e + 2\eta(e\cdot \tilde e)e-\eta (e\cdot\tilde e)^{2}\tilde e\right)\cdot(e+x) $$

Note that for δ small enough (depending only on η), if we define

$$\mathcal C_{\tilde e}:= \left\{x\in \mathbb{R}^{n} \ :\ \frac{x+e}{|x+e|}\cdot \frac{e-(e\cdot\tilde e)\tilde e}{|e-(e\cdot\tilde e)\tilde e|} \ge (1-\delta)\right\} $$

satisfies

$$ \lim_{t\uparrow +\infty} \left( 1+e\cdot x-\eta \psi_{P}(x)\right) \ge c|x|\quad \text{for all }x\in \mathcal C_{\tilde e} $$

(A.2)

where c > 0. Indeed, the vector $e^{\prime } :=e-(e\cdot \tilde e)\tilde e$ is perpendicular to $\tilde e$ and has positive scalar product with e. Thus, we have

$$\left( e -\eta \tilde e + 2\eta(e\cdot \tilde e)e-\eta (e\cdot\tilde e)^{2}\tilde e\right)\cdot e^{\prime}>0 $$

Let us show now that for ε > 0 small enough the function Φ_P satisfies

$$ M^{-}{\Phi}_{P}(0) \ge 1. $$

(A.3)

We first prove (A.3) in the case |P| ≥ R with R large enough. Indeed let $P= t\tilde e$ for $t\uparrow +\infty$ and $\tilde e \in \partial \mathcal { C}_{\eta }\cap \partial B_{1}$. Let us denote

$$\delta^{2} u(x,y) = \frac{u(x+y)+u(x-y)}{2}-u(x). $$

Using (A.1), and Eq. A.2, and Φ_P ≥ 0 we obtain

$$\begin{array}{ll} \underset{t\to \infty}\lim M^{-}{\Phi}_{P} (0) &\ge {\int}_{\mathbb{R}^{n}} \left( (\delta^{2} u)_{+} \frac{\lambda}{|y|^{n + 2s}} - (\delta^{2} u)_{-} \frac{{\Lambda}}{|y|^{n + 2s}}\right) \,dy \\ &\ge {\int}_{\mathcal C_{\tilde e}} (c|y| -C)_{+}^{2s-\epsilon}\,\frac{dy}{|y|^{n + 2s}} - C {\int}_{\mathbb{R}^{n}} \min\{1,|y|^{2}\} \frac{dy}{|y|^{n + 2s}} \\ &\ge \frac{c}{\epsilon}-C. \end{array} $$

Thus Eq. A.3 follows for |P|≥ R with R large, provided that 𝜖 is taken small enough.

We now concentrate in the case |P| < R. In this case we use that, taking δ > 0 small enough (depending on η) and defining the cone

$$\mathcal C_{e}:= \left\{x\in \mathbb{R}^{n} \ :\ \frac{x}{|x|}\cdot e \ge (1-\delta)\right\} $$

we have

$$e\cdot (P+e+x)- \eta \left( |P+e+x| - \frac{(e\cdot (P+e+x))^{2}}{|P+e+x|} \ge c |x|\right) $$

for $x\in \mathcal C_{e}$ with |x| ≥ L with L large enough (depending on R).

Thus, reasoning similarly as above but now integrating in $\mathcal C_{e} \cap \{ |x|>L\}$ instead of on $\mathcal C_{\tilde e}$ we prove (A.3) also in the case P ≥ R, provided that 𝜖 is small enough. Therefore the lemma is proved. □

Finally, we give the:

Proof of Lemma 4.5

Note that we only need to prove the conclusion of the Lemma for r > 0 small enough, since the conclusion for non-small r follows from the interior Harnack inequality.

Recall that ${\Omega }\subset \mathbb {R}^{n}$ is assumed to Lipschitz domain, with 0 ∈ ∂Ω. Then, for some e ∈ S^n− 1, η > 0 (typically large), and r₀ > 0 depending on (the Lipschitz regularity of) Ω we have

$$\tilde{\mathcal C}_{\eta} \cap B_{2r_{0}}\subset{\Omega} $$

where $\tilde {\mathcal C}_{\eta }$ is the cone of Lemma A.1, which is very sharp for η large.

Let Φ and 𝜖 > 0 be the subsolution and the constant in Lemma A.1. We now take

$$\tilde {\Phi} = \left( {\Phi} -(|x|/r_{0})^{2} \right) \chi_{2r_{0}}. $$

By Lemma (A.1) we have

$$M^{-}\tilde {\Phi} \ge - C \quad \text{in }B_{r_{0}} $$

while clearly $\tilde {\Phi }\le 0$ outside $B_{r_{0}}$.

Now we take observe that, for c₁ > 0 small enough we have

$$M^{-}(c_{1}\tilde{\Phi} + \chi_{D_{1}}) \ge -c_{1} C + c \ge c/2>0 $$

in $B_{r_{0}}$ — not that $B_{r_{0}}\cap D_{1} = \varnothing$ since r₀ is small.

Then, taking δ ∈ (0, c/2) we have

$$M^{-}(u-c_{1}\tilde{\Phi} + \chi_{D_{1}}) \le 0 \quad \text{ in }B_{r_{0}} $$

while

$$u-c_{1}\tilde{\Phi} + \chi_{D_{1}} \ge 0-c_{1}\tilde{\Phi} + 0 \ge 0\quad\text{in}(\mathbb{R}^{n}\setminus B_{r_{0}})\setminus {D_{1}} $$

and

$$u-c_{1}\tilde{\Phi} + \chi_{D_{1}} = (u-1)-c_{1}\tilde{\Phi} \ge 0-c_{1}\tilde{\Phi} \ge 0\quad \text{in } (\mathbb{R}^{n}\setminus B_{r_{0}})\cap {D_{1}}. $$

Then, by the maximum principle we obtain

$$u-c_{1}\tilde{\Phi} = u-c_{1}\tilde{\Phi} + \chi_{D_{1}} \ge 0 \quad \text{in } B_{r_{0}} $$

and hence

$$u(x)\ge c_{1}{\Phi}(x) - C |x|^{2} \quad\text{for }x\in B_{r_{0}} $$

which clearly implies the Lemma (taking γ = 𝜖). □