On the Boundary Harnack Principle in Holder domains

We investigate the Boundary Harnack Principle in H\"older domains of exponent $\alpha>0$ by the analytical method developed in our previous work"A short proof of Boundary Harnack Principle".


Introduction
In this paper we continue the study of the boundary Harnack principle for solutions to elliptic equations, based on the method developed in [DS].The classical boundary Harnack principle states that two positive harmonic functions that vanish on a portion of the boundary of a Lipschitz domain must be comparable up to a multiplicative constant, see for example [A, D, K, W].Further extensions to more general operators and more general domains were obtained in several subsequent works [Kr,CFMS,JK,BB1,BBB,KiS].
In particular, Bass and Burdzy [BB1,BB2] and Banuelos, Bass and Burdzy [BBB] provided sharp versions using probabilistic methods.They established the boundary Harnack principle for nondivergence elliptic operators in Hölder domains (or more general twisted Hölder domains) of exponent α > 1 2 , and for divergence operators in Hölder domains of arbitrary exponent α > 0. For the case of divergence operators, an analytical proof based on Green's function was given by Ferrari in [F].
To state precisely the boundary Harnack principle in Hölder domains, first we introduce some notation.Let g : B 1 → R be a C α Hölder function of n−1 variables with g(0) = 0, and α ∈ (0, 1).Denote by Γ ⊂ R n the graph of g, and by C r the cylinder above B r and at height r on top of Γ C r := {x ∈ B r , g(x ) < x n < g(x ) + r}.
We say that C 1 is a C α -Holder domain in a neighborhood of Γ.
The following version of the boundary Harnack principle is due to Banuelos, Bass and Burdzy, see [BBB].
Theorem 1.1.Let Lu = div(A(x)∇u) be a uniformly elliptic linear operator, and assume that u, v are two positive solutions to with C depending on n, α, g C α and the ellipticity constants of L.
The assumption that u Recently, in [DS] we found a direct analytical method of proof of the boundary Harnack principle based on an iteration scheme and Harnack inequality.In particular we established the corresponding results in Hölder domains of exponent α > 1 2 for general equations either in divergence or nondivergence form.
In the present paper we discuss further the case of Hölder domains of arbitrary exponent α > 0, and give a proof of Theorem 1.1 using the same ideas from [DS].We also consider some extensions of Theorem 1.1 to non-divergence equations whose coefficients remain constant in the vertical direction.
The paper is self-contained and is organized as follows.In Section 2 we give two lemmas concerning Harnack inequality outside domains of small capacity.In Section 3 we use these lemmas and employ the arguments from [DS] to prove Theorem 1.1.Finally in Section 4 we provide some extensions of Theorem 1.1 to more general divergence operators, and certain non-divergence or fully nonlinear operators.

Two lemmas
In this section we present two lemmas concerning solutions to divergence equations in domains whose complement in the unit cube Q 1 has small capacity.
Given a domain Ω and a compact set K ⊂ Ω, we say that two functions u, v ∈ H 1 (Ω) agree on K, and write u In particular, if L is a uniformly elliptic operator in divergence form then u solves (2.1) Lu = 0 in Ω \ K, and u = 0 on ∂Ω, u = 1 in K, means that Lu = 0 in the open set Ω \ K, and ), and η = 1 in a neighborhood of K. Notice that the solution u to (2.1) is a supersolution in Ω, i.e.Lu ≤ 0 in Ω.Let Q 1 denote the unit cube in R n centered at 0, and E ⊂ Q 1 a closed set.Set, The first lemma states that a solution to Lv = 0 in Q 1 \ E satisfies the Harnack inequality in measure if E has small capacity.Positive constants depending on the dimension n and the ellipticity constants λ, Λ are called universal.
for some δ small, universal.Then for some c 0 small.
The second lemma is standard and states that the weak Harnack inequality holds for a subsolution v ≥ 0 which vanishes on a set E of positive capacity.
Lemma 2.2.Assume that v ≥ 0 in Q 1 , and Proof of Lemma 2.1.Let ψ be the solution to 1).By hypothesis and weak Harnack inequality we find for some small c 0 .
Similarly as above we define φ to be the solution to We claim that if δ is chosen sufficiently small then, For this we let w be the solution to (2.3) when L = .The Dirichlet energies of φ and w are comparable since ˆ(∇(φ − w)) T A∇φ dx = 0, hence By Poincaré inequality we find They satisfy the same equation in , and in a neighborhood of K by the continuity of v we have 2v ≥ 1 ≥ ψ − φ.

On the other hand
which by (2.2),(2.4)yields the desired conclusion.
Proof of Lemma 2.2.Assume that v L ∞ = 1.Then, by the maximum principle we have 1 − v ≥ φ, with φ as in (2.3) above.It suffices to show that φ ≥ c(δ) on ∂Q 7/8 which by the maximum principle implies the desired conclusion φ(0) ≥ c(δ) small.Since all the values of φ are comparable near ∂Q 7/8 by the Harnack inequality, we need to show that φ ≥ c (δ) at some point on ∂Q 7/8 .Assume by contradiction that |φ| ≤ µ is very close to 0 on ∂Q 7/8 .The Caccioppoli inequality (we think that φ is extended to 0 outside Q 1 ) implies This together with (2.5) implies that the Dirichlet energy of φ in Q 1 is bounded above by Cµ 2 , and we reach a contradiction.

Proof of Theorem 1.1
This section is devoted to the proof of Theorem 1.1.We recall that Γ denotes the graph of a C α function g, with α ∈ (0, 1), and C r denotes the cylinders of size r on top of Γ The main idea of the proof is to show through an iterative procedure that a solution w which vanishes on Γ and is mostly positive in C r , becomes positive near the origin.We denote by the points in the cylinder C r at height greater than r β on top of Γ, for some β > 1.
We divide the proof in three steps.
Step 1.We show that, there exist C 0 , β > 1 depending on n, α, g C α , and the ellipticity constants of L, such that if w is a solution to Lw = 0 in C r (possibly changing sign) which vanishes on Γ, w ≥ −a on C r 2 , for some small a = a(r) > 0, as long as r ≤ r 0 universal.
The conclusion can be iterated and we obtain that if the hypotheses are satisfied in C r0 then w > 0 on the line segment {te n , 0 < t < r 0 }.
Since g is Hölder continuous, we can apply interior Harnack inequality to w + 1 in a chain of balls and need to connect a point in A r/2 with a point in A r .We conclude that Next we take a point on Γ := {x n = g(x )}, say 0 for simplicity, and consider the cubes of size r β/α centered on the e n axis, i.e Q r β/α (te n ) (see Figure 2).
When t > Cr β the cube is in the interior of the domain and when t < −Cr β the cube is in the complement.There are Cr γ stacked cubes which connect the domain with its complement.The graph property of the domain implies that the capacity of the complement E = {x n ≤ g(x )} in Q r β/α (te n ) is decreasing with t.By continuity we can find a cube centered at t 0 e n such that, after a rescaling of factor r −β/α , cap 3/4 (E) = δ in that cube, with δ as in Lemma 2.1.For all cubes centered at te n with t ≥ t 0 we can apply Lemma 2.1 repeatedly for w + 1 and obtain an inequality in measure as in (3.3), ).Thus {w − = 0} has positive density in all cubes centered at te n with t ≥ t 0 .
Now we notice that we can apply weak Harnack inequality for w − in all cubes, in the top cubes with t ≥ t 0 because of the density property, and in the bottom ones with t ≤ t 0 because of Lemma 2.2.Hence w − decreased by a fixed factor on the e n axis passing through a point on Γ, with respect to its maximum over all cubes of size r β/α centered on that axis.As we move each time a r β/α distance inside the domain from the sides of C r , sup w − decreases geometrically hence, for c 0 small universal.We choose a(r) := e −c0r 1−β/α , and in view of (3.3), our claim is satisfied for all r small.
Step 2. [Carleson estimate] We show that, with C 2 universal.We apply an iterative argument similar to the one in Step 1.Since u(e n /2) = 1, the interior Harnack inequality gives that with C 1 universal.With the same notation as Step 1, we wish to prove that if r is smaller than a universal r 0 and for some y ∈ C 1/2 , then we can find Since |z − y| ≤ Cr α , we see that for r small enough, we can build a convergent sequence of points On the other hand the extension of u by 0 below Γ is a subsolution in a neighborhood of Γ. Therefore u is bounded above, and we reach a contradiction.
To show the existence of the point z we let By (3.4) we know that By Lemma 2.1 this estimate can be extended in measure for the cubes of size r β/α with t ≥ t 0 since the capacity of the complement is bounded above.More precisely, as in Step 1, in each cube of size r β/α we have that either {w = 0} has positive density (for the cubes with t ≥ t 0 ), or positive capacity (for the cubes with t ≤ t 0 ).Moreover, if our claim is not satisfied then we apply Weak Harnack inequality for w repeatedly as in Step 1 above.As we move inside the domain from the sides of C r (y , g(y )) we obtain In particular and we reach a contradiction.
Step 3. We prove the theorem using the Steps 1 and 2 above.After multiplication by a constant we may assume that u = v = 1 at 1 2 e n .It suffices to show that for a large constant C 3 > 0 universal, 0 }, provided that C 3 is chosen sufficiently large.Here f , r 0 and β are as in Step 1.
We conclude by Step 1 that w ≥ 0 on the line {te n , 0 < t < 3/4}.We can repeat the argument at all points on Γ ∩ C 1/2 , and the theorem is proved.

Some extensions
In this section we state a few variants of the Theorem 1.1.First we remark that the proof applies to operators involving lower order terms.
Theorem 4.1.The statement of Theorem 1.1 holds for general uniformly elliptic operators with the constant C depending also on q, b L q , d L q/2 .
Indeed, we only need to check that the statements of Section 2 continue to hold in the small cubes of size r β/α .After a dilation this corresponds to proving Lemmas 2.1 and 2.2 for operators L as above with b L q , d L q/2 sufficiently small.The proofs are identical since the presence of such lower order terms does not affect the energy estimates.
A counterexample of Bass and Burdzy in [BB2] shows that Theorem 1.1 does not hold in general for nondivergence equations when α < 1 2 .Here we remark that Theorem 1.1 remains valid with α > 0 for nondivergence linear operators which are translation invariant in the vertical direction.
Theorem 4.2.The statement of Theorem 1.1 holds for linear nondivergence uniformly elliptic operators of the form In this theorem we assume for simplicity that the coefficient matrix A depends continuously on its argument.Since u, v might not be continuous at all points on Γ, the hypothesis that u, v vanish on the boundary is understood in the sense that their extensions with 0 below Γ are bounded subsolutions for L, see [DS].
In this case we provide the corresponding lemmas of Section 2 by defining the capacity (with respect to L) as where φ solves Then Lemma 2.2 follows directly from the definition of the capacity, with c(δ) = δ.For Lemma 2.1 we see that (2.4) is satisfied since by the Weak Harnack inequality the set {φ > c 0 } must have small measure in Q 1 if δ is sufficiently small.The rest of the proof is the same.
The arguments of Section 3 can be repeated in the same way.The invariance of the operator L with respect to the vertical direction and the graph property of the boundary imply that the capacity of the complement E in the cubes Q r β/α (te n ) is monotone in t.We can apply again Lemma 2.1 for the top cubes with t ≥ t 0 and Lemma 2.2 for the bottom cubes with t ≤ t 0 , and carry on as before.
We also discuss the case of fully nonlinear operators (4.1) with F uniformly elliptic with constants λ, Λ, and homogenous of degree 1.
We can prove the lemmas of Section 2 for the operator F by using as capacity the definition above with Lφ = F (−D 2 φ).Then Lemma 2.2 follows again directly from the definition.For the proof of Lemma 2.1 we choose the function ψ to satisfy M − λ/n,Λ (ψ) = 0.Here as usual, M − denotes the extremal Pucci operator, with A a symmetric matrix whose eigenvalues belong to [λ, Λ].Then ψ − φ is a subsolution ψ) ≥ 0, and the rest of proof remains as before.However in the proof of the main Theorem 1.1 only Step 2, the Carleson estimate, can be carried out in this setting, since for Step 1 we need the lemmas of Section 2 to hold not only for solutions of the operator F but for the difference of two such solutions as well.
Finally we mention that in R 2 Theorem 1.1 holds under very mild assumptions on the domain and the operator.Here we state a version for L ∞ graphs and linear operators.
Theorem 4.4.Assume Γ ⊂ R 2 is the closure of the graph of a function g with g L ∞ ≤ 1/4.Then the statement of Theorem 1.1 holds for uniformly elliptic linear operators L in divergence or nondivergence form with constant C depending only on the ellipticity constants of L.
We only sketch Steps 1 and 2 of Section 3 in this setting which can be adapted to more general situations.They are based on topological considerations and do not require an iterative argument.

" ¥÷iIm
To prove this, we assume by contradiction that there is a connected component U of {w < 0} which intersects C 1/2 .This connected component must exit C 1 and since u( 1 2 e 2 ) 1, U must stay close to Γ. Thus we can find a nonintersecting polygonal line , say included in which connects the two lateral sides x 1 = 1 2 and x 1 = 3 4 (see Figure 3).The line splits the cylinder into two disjoint sets, and we define w to be equal to w on the set "above" and w = min{w, 0} on the set "below" .Then w is a supersolution of L in D. Since w is sufficiently large in a ball above , and w ≥ −1 in D we find that w ≥ 0 on the segment We reached a contradiction at the point where intersects this segment.
This follows similarly as in Step 1.If {u > C} has a connected component that intersects C 1/2 , then we can find a polygonal line as above where u is large.Thus min{u, C} extended by C below is a supersolution for L in D, and the maximum principle implies that u is large at the point (5/8, 1/2).Therefore by Harnack inequality u(0, 1/2) is large as well, and we reach a contradiction.