A new boundary Harnack principle (equations with right hand side)

We introduce a new boundary Harnack principle in Lipschitz domains for equations with right hand side. Our approach, which uses comparisons and blow-ups, will adapt to more general domains as well as other types of operators. We prove the principle for divergence form elliptic equations with lower order terms including zero order terms. The inclusion of a zero order term appears to be new even in the absence of a right hand side.

where C depends on space dimension and u(x 0 )/v(x 0 ) for a fixed x 0 in the domain. We are interested in extending this result to the case of equations with right hand side. Of course such a general result is doomed to fail, unless some further restrictions are imposed. This can be seen through a simple 2-dimensional example with in the cone {x 1 > x 2 > 0} with aperture π/4. Consequently, there cannot exist 2 C > 0 such that Cu ≥ v. Another simple (and discouraging) example is in the set {x 1 > 0}. Again, there cannot exist a constant C such that close to the boundary Cu ≥ v. There are two observations to make from the above examples: 1) In the first example the domain is too narrow, with a sharp corner at the boundary.
Much of this work was completed while the first author visited KTH Royal Institute of Technology. Shahgholian was supported by Swedish Research Council. 1 The boundary Harnack Principle holds in very general domains, such as NTA domains, and uniform domains. It also holds for solutions to a large class of elliptic equations. 2 One can actually prove the failure of boundary Harnack between harmonic and superharmonic functions, in the first quadrant. This is illustrated in Example 1.2 in [11], in terms of free boundary problems.
2) In the second example we try to show subharmonic functions can dominate harmonic functions, by multiplying with a constant; this in general fails.
To put things in perspective, let D = {x 1 > 0} ∩ {x 1 > x 2 }and consider the following function Now let u be the positive homogeneous harmonic function vanishing on the boundary of the cone D. Then for v = u − w we have that u(x) > v(x), since w > 0 in D (by definition), and v > 0 since w ≈ r 2 and u ≈ r a (a < 2), for r small. We also have ∆u = 0, ∆v = −1 in D, with zero boundary values. In particular we have a boundary Harnack principle u(x) > v(x) where a harmonic function dominates a superharmonic one. The difference between this example and the first example above is that the cone is wider. The question that naturally arises is whether such a behaviour can be structured through a general statement, and if so what are the conditions for such a boundary Harnack principle.
A further observation is that when the domain D is uniformly C 1,Dini then a boundary Harnack principle holds, between a positive harmonic and a superharmonic functions (with bounded r.h.s.), vanishing on the boundary (∂D) ∩ B 1 . This is an easy consequence of Hopf's boundary point lemma and C 1 -regularity of solutions. Indeed, let u be a non-negative harmonic function in D and v satisfy ∆v = −1 in D, both with zero Dirichlet data on (∂D) ∩ B 1 . Then by Hopf's boundary principle (applied to u) and that v is uniformly C 1 , we have where ν is the interior normal direction, on ∂D. Hence for k large, the function w k := ku − v satisfies Hence w k > 0 in D in a small neighbourhood of the ∂D ∩ B 1 and inside D; this is exactly the boundary Harnack we asked for.

Main Result.
Our main result is the boundary Harnack principle mentioned above between a superharmonic and a positive harmonic function, in Lipschitz domains. For instructional reasons and for the benefit of the non-expert reader, we first state and prove the theorem inside a cone; see Theorem 2.1. The proof in this case presents the main ideas for the general case. Next, we do the same for a Lipschitz domain; see Theorem 3.10. We then apply the ideas to divergence operators, see Theorem 4.7. Our result holds even when including zero-order terms. This result appears to be new even when considering the standard boundary Harnack principle without right hand side, i.e. for solutions rather than supersolutions. In proving our new boundary Harnack principle, we will often utilize the boundary Harnack principle (without right hand side) for two nonnegative solutions. To avoid confusion we will always refer to this as the "standard boundary Harnack principle". 1.3. Related results. As previously mentioned the boundary Harnack principle has been proven on very general domains as well as for various operators. Recently, De Silva and Savin in [5] generalized the result in a new direction by showing that improving the regularity of the boundary improves the regularity of the boundary Harnack inequality. Their result applies to elliptic operators in divergence form with appropriate smoothness assumptions on the coefficients. Roughly speaking, if u, v vanish on ∂Ω with Lu = 0 and Lv = f in Ω, and if u > 0 in Ω and ∂Ω ∈ C k,β , As a corollary of Theorem 3.10 we obtain a similar estimate on Lipschitz domains with small enough Lipschitz norm, see Corollary 3.12. The result in [5] illustrates the principle that improving the regularity of the boundary gives a boundary Harnack principle for solutions with a right hand side. Our result shows that Lipschitz regularity is a sufficient condition to obtain an estimate of the form (1.1) when β = 0, and the allowed behavior for f is determined by the Lipschitz constant.
Another related result is found in [12] where it is shown a superharmonic function is comparable to the first eigenfunction for a domain in R 2 with finitely many corners and with an interior cone condition.
1.4. Applications. We present two applications of our boundary Harnack principle: to the Hele-Shaw flow and to the obstacle problem.
1.4.1. Hele-Shaw Flow. The Hele-Shaw flow may be formulated as follows: For t > 0 we define u t (x) as the solution of where Ω 0 is the initial domain (filled with liquid) and Ω t = {u t > 0}, and z ∈ Ω 0 is the liquid-injection point. The Dirac mass at the point z means we inject more fluid at that point. This formulation is obtained after a Baiocchi-transformation, which is easily found in the classical literature for free boundary problems. Now suppose we do the Hele-Shaw experiment on a "table" having corners of various angels. More exactly suppose we consider where D ⊃ Ω 0 is a given domain. The zero boundary data on ∂D means that the liquid, when reaching the edge of the table, will fall to the ground. The question is whether the liquid will reach all points of the boundary of D in finite time t ≤ t 0 < ∞. By a barrier argument (see [11]) one can show that in two dimensions corners with angle smaller than or equal to π/2 will not get wet; an analogous result for higher dimensions is a consequence of Theorem 2.3 in this paper. Now the question is what can happen when the angle of the corner is larger than π/2; will the liquid reach such a corner? The answer to this question is yes, see [11].
We now consider D with Lipschitz boundary with Lipschitz constant L < 1. As a consequence of Theorem 3.10 we now show that every point of the boundary will get wet. We need to show for large values of t that u t > 0 in D ∩ B r (z 1 ), for any where h t is the harmonic function in B r (z 1 ) ∩ D with boundary values h t = u t , and k (t-independent) satisfies with zero boundary values on ∂(B r (z 1 ) ∩ D). It suffices then to show that for large values of t we have h t ≥ k in B r (z 1 ) ∩ D. By Theorem 3.10, for fixed t there exists some large constant C t such that C t h t ≥ k. By the standard boundary Harnack principle , A related question to the Hele-Shaw flow reaching corners, is whether for D ⊂ R n , with 0 ∈ ∂D, and ∂B 1 ∩ D = ∅, a solution to will be non-negative in D ∩ B 1 for m large enough. The answer to this question is already given in the discussion for Hele-Shaw experiment above.

Obstacle Problem.
A further application can be made to the regularity theory of the free boundary for the obstacle problem, that is formulated as a solution to with h ≥ c 0 > 0 Lipschitz, and a prescribed Dirichlet data on ∂B 1 . In proving regularity of the free boundary for the obstacle problem one can show that if a free boundary point z ∈ ∂{v > 0} ∩ B 1/2 is not a cusp point, then for some r > 0 and direction e, v e > 0 in the set {v > 0} ∩ B r (z), and that the free boundary is Lipschitz in B r (z). One can actually show that Lipschitz norm can be taken as small as one wishes, by taking the neighbourhood of z smaller. The proof for (non-uniform) Lipschitz regularity is actually much simpler than proving uniform regularity. We refer to [10] for background and details as well as other related original references.
The boundary Harnack principle in this paper allows us to deduce C 1,α -regularity of the free boundary for the obstacle problem, in an elementary way. 3 Indeed, we may consider the function H(x) = v e1 − Cv, which satisfies 4 We will now apply Corollary 3.12 to the functions H = v e1 − Cv and v e for any direction e orthogonal to e 1 . Both H and v e vanish on the free boundary ∂{v > 0}. By taking a neighborhood of z ∈ ∂{v > 0} ∩ B 1/2 small enough, the Lipschitz norm of ∂{v > 0} ∩ B 1/2 can be made arbitrarily small. Since ∆H ≤ 0 and |∆H| is bounded, we may apply Corollary 3.12 with γ = 0 to conclude that is C α inside {v > 0} (close to the free boundary point z).
Next fix a level surface 5 v = l, and denote the free boundary as a Lipschitz graph as l → 0. Hence G e is C α .
1.5. Future directions. It seems plausible that the results presented in this paper can be generalized to other operators, as well as more complicated domains. Here we have chosen to treat the problem in Lipschitz domains only. In the final section we consider second order elliptic equations of divergence form. The coefficients are variable and assumed to be only bounded and measurable.
Key elements of our approach is the standard boundary Harnack principle, barrier arguments, as well as scaling and blow-up invariance. Since our approach is indirect and uses scalings, the core idea is to look at nonnegative solutions on global domains. The technical difficulties that seem to arise for generalizitions of our result concern the invariance of the domains in scaling.
The methods presented should also work to prove a boundary Harnack principle for the positivity set of a solution to the thin obstacle problem as long as it is assumed a priori that the free boundary is Lipschitz. Then we may argue in a similar way as above for the thin obstacle problem, with equations having Lipschitz right hand side; see [1] in combination with our result.

Boundary Harnack in Cones
We let C be any open cone in R n , with vertex at the origin such that C ∩ S n−1 is connected. If u is any harmonic function on C with u = 0 on ∂C, then in spherical coordinates where f k are the eigenfunctions to the Laplace-Beltrami on C ∩∂B 1 . If u is harmonic on C and nonnegative, then u is unique up to multiplicative constant to r α1 f 1 (θ).
Theorem 2.1 is a boundary Harnack principle, but with a right hand side. Clearly, a harmonic solution will control a subsolution. The significance of Theorem 2.1 is that a harmonic solution can control a supersolution, and that the allowed behavior for the right hand side depends on the opening of the cone or more explicitly on α 1 . When the opening of the cone is large (so that α 1 is small), then negative values for γ are allowed, and the right hand side can have singular behavior near the boundary. When the opening of the cone is small (so that α 1 is large), then γ must be positive and large, so that the right hand side must decay as it approaches the boundary.
In order to prove Theorem 2.1, we will need the following convergence result.
then there exists a subsequence v k → v uniformly on C ∩ B R−ǫ with v inheriting the above properties.
Proof. Since C ∩ ∂B R is a Lipschitz submanifold, C is compactly contained in a slightly larger cone C ′ ⊃ C, and the unique nonnegative harmonic solution with zero Dirichlet data on ∂C ′ is given as Using v as a barrier, the convergence result will follow by standard techniques. 7 The assumption that ∂C ∩ S n−1 is an (n − 2)-dimensional manifold of class C 1,α is not necessary. As we will see in Section 3, ∂C ∩ S n−1 may be a Lipshitz manifold provided the Lipschitz constant is small enough depending on γ, that appears in (2.2) An alternate proof of the above lemma, for a more general domain and more general operator, is given in the proof of Lemma 4.5.
We now give a proof of the main theorem in this Section.
Proof of Theorem 2.1. Fix x 0 ∈ C ∩ B 1/2 , and consider the nonnegative homogeneous solution u = r α1 f 1 (θ). By the standard boundary Harnack principle, any solution u as given in the statement of Theorem 2.1 will be comparable, and consequently bounded from below up to a multiplicative constant, by r α1 f (θ). Thus, we consider u = r α1 f (θ). Furthermore, any function v as given in the statement of Theorem 2.1 will be bounded from above by a constant multiple of the superharmonic function defined by Thus, it suffices to prove the theorem for v as defined above and with u = r α1 f (θ). In the following we use r as a scaling parameter which may coincide with r as the polar axis for homogeneous functions. We first show that there exists some constant We note that h is continuous away from the origin since v is continuous. For 0 < r < 1, we consider the rescaled functions defined on B 1/r . We point out that w r (x 0 ) = 0 and sup B1∩C |w r | = 1. We also define Since lim sup t→0 h(t) = +∞, it follows that By the definition of w r we have We use the triangle inequality to obtain (2.4) To bound the first term in the above inequality, we have by definition that To bound the second term in the inequality, we note that Also, since there exists C depending on x 0 such that if |x| = 1 then u(rx) ≤ Cu(rx 0 ). If we utilize the homogeneity of u, we conclude that We use this and the homogeneity of u to obtain We will now bound the Laplacian of w r k . From (2.3) we may apply Lemma A.1 to the sequence {a k } to conclude that there exists a subsequence k l such that for any j ∈ N, By choosing r k l and combining the above inequality with (2.4), (2.5), and (2.6) we conclude sup From the choice of a k l in Lemma A.1 and the fact that a k = +∞ it follows that eventually a k l ≥ l −2 k , so that if r k l = 2 −k l , then We now use the assumption 2 − α 1 + γ > 0. Since ∆u = 0 and ∆v = −|x| γ we have that for a sequence r k l → 0 there holds as long as 2 − α 1 + γ > 0. By Lemma 2.2 there exists a subsequence w k → w uniformly in B R ∩ C for any R > 0. Furthermore w will satisfy By property (ii) we have that w is not identically zero. By property (iii) we have that w changes sign, so that by (2.1) we have sup BR |w| ≥ cR α2 for R ≥ 1. Since |x| α1 ln(|x| + 1) < |x| α2 , for large x, this contradicts property (iv). Thus, we have shown that for any x 0 ∈ C there exists a constant C depending on x 0 such that v(rx 0 ) ≤ Cu(rx 0 ) for any 0 < r ≤ 1/2.
We now show that the assumption that 2 − α 1 + γ > 0 is essential. We first consider the easier case when 2 − α 1 + γ < 0, and show that Theorem 2.1 cannot possibly hold. For clarity of exposition we restrict the analysis to the case when γ = 0, so that the right hand side is constant. Theorem 2.3. Let u and v be as in the statement of Theorem 2.1, and assume ∆v = −1 with 2 − α 1 < 0. Then for any C > 0, there exists ρ > 0 such that Proof. We note that by the standard boundary Harnack inequality, it is enough to consider u = r α1 f 1 (θ). We normalize f 1 so that sup f 1 = 1. Fix C > 0, and let z ∈ B r ∩ C. Define h(x) := |x − z| 2 /(2n). We note that h is constant on ∂B r (z). If there exists y ∈ ∂B r (z) with h(y) ≤ Cu(y) − v(y), then For small enough r, the above inequality cannot hold since 2 < α 1 . Then for small enough r, we have h ≥ Cu − v on ∂B r (z), and hence also on ∂(B r (z) ∩ C). By the Since C was arbitrary, for any δ > 0, we may choose r 0 such that if r < r 0 and x ∈ B r ∩ C, We can show that Theorem 2.1 is sharp by considering the critical case when 2 − α 1 = 0. When dimension n = 2 this result was shown in [11].
Theorem 2.4. Let C be a cone in R n with α 1 = 2. Then the boundary Harnack principle with right hand side does not hold.
Proof. Let v be as in the statement of the Theorem, and let u = r α1 f 1 (θ). Suppose that there exists C > 0 such that We now use a Weiss-type monotonicity formula for superharmonic functions as in [9]. The function is monotonically increasing in r and is constant if and only if v is homogeneous. We now consider the rescaled functions v r (x) := v(rx)/r 2 . From (2.9) and the fact that u is homogeneous of degree 2 we have that for any fixed x ∈ C ∩ B 1 , there exists C x such that C −1 x r 2 ≤ v r (x) ≤ C x r 2 for any 0 < r < 1. (2.10) From Lemma 2.2 we obtain that for a sequence r k → 0, v r k → v 0 uniformly in C ∩ B R for any R > 0. Furthermore, we will show that v 0 satisfies (i) v 0 is homogeneous of degree 2, Property (i) follows from the Weiss-type monotonicity formula in the following manner. One may easily check that W (ρr, v) = W (ρ, v r ). Since W (r, v) is monotone in r it follows that Since W (ρ, v 0 ) is constant, then v 0 is homogeneous of degree 2, see [9]. Properties (ii)-(iv) follow easily from the definition of v r and the uniform convergence. Finally, property (v) follows from (2.10). Then v 0 = r 2 g where the spherical piece g satisfies 2ng + ∆ θ g = −1. We now utilize the Fredholm Alternative for existence for weak solutions, see Chapter 6 in [6]. Since f 1 (in (2.1)) is a nontrivial solution, the solution g exists if and only if for all h such that 2nh + ∆ θ h = 0 (since the operator 2n + ∆ θ is self-adjoint). We recall that 2nf 1 + ∆ θ f 1 = 0. Then a necessary condition for g to exist is that Since f 1 > 0 in C ∩∂B 1 , the above equality cannot be true. Consequently, a solution g cannot exist, and we have a contradiction.

Boundary Harnack in Lipschitz domains
In this section we consider Lipschitz domains Ω L,R where We will assume g(0) = 0, and will write Ω L when R = 1 and Ω L,∞ if R = ∞. Also, we define u ∈ S(Ω L,R ) if ∆u(x) = 0 in Ω L,R , u(x) = 0 on Ω c L,R ∩ B R , and for γ ∈ R we define u ∈ S(Ω L,R , d γ ) if It will be necessary to use the solutions on right circular cones as barriers. Consequently, we define If u ∈ S(C L,∞ ) and u ≥ 0, then as noted in Section 2, we have u = r α1 f 1 (θ) (and unique up to multiplicative constant), and we will denote r α1 f 1 by u L . We note that the definition of C −L still makes sense when −L < 0; although, the cone C −L is not convex. In this situation we write C −L and similarly u −L for the nonnegative solution with zero boundary data on C −L . In order to prove a boundary Harnack principle with a right hand side for Lipschitz domains we will adapt the proof from Section 2 in the following manner: • We will employ compactness methods and thus need a convergence result provided by Lemma 3.2. • We will need to bound the behavior of a nonnegative harmonic function at the boundary from above and below which is given in Lemma 3.3. • We will need a Liouville type result which is given in Lemma 3.6.
• We will then adapt the proof of Theorem 2.1 (again using compactness techniques) to obtain the proof of Theorem 3.10. We give later a proof of a more general version of this lemma; see Lemma 4.4 in Section 4.

Lemma 3.2.
Let Ω L k ,R k be a sequence of domains with L k ≤ M , R k ≥ 1, and 0 ∈ ∂Ω L k . Let u k ∈ S(Ω L k ,R k , d γ ), and let α be the degree of homogeneity for u M , and assume 2 − α + γ > 0. We further assume either sup Br u k ≤ Cr β for r ≤ 1 and some constants C, β > 0.
Then there exists a subsequence with a limiting domain Ω L0,R0 and a limiting func- for all r < R 0 .
We give later a proof of the more general Lemma 3.2 in Section 4 that implies Lemma 3.2.  in From the comparison principle we have that u ≤ v on B 1 . By considering −v we obtain a similar bound from below to conclude the proof.  Proof. Consider w = u + v. Then w ≥ u. Let c ≥ 1 be the largest constant such that cu ≤ w. Then there exists a sequence {x k } ∈ Ω L,∞ such that We now invoke the standard boundary Harnack principle for Lipschitz domains on the nonnegative harmonic functions w − cu and w. There exists C 1 > 0 such that with r k := max{2|x k |, k}. From (3.1) the right hand side of (3.2) goes to zero. Then w ≡ cu, and the result follows. then v = cu for some c ∈ R.
Remark 3.8. The significance of Lemma 3.7 is that we do not require v ≥ 0.
Since the constants are independent of r ≥ 1, we conclude that v + ≤ C 1 u in Ω L,∞ . The same argument holds for v − so that |v| ≤ C 1 u in Ω L,∞ .
Let w = C 1 u − v ≥ 0. Then from Lemma 3.6 we have that w = cu for some constant c, so that v = (C 1 − c)u.
We will need an improvement over the previous lemma. then v = cu for some c ∈ R.
In Section 4 we give a proof of a more general result in Theorem 4.2.
Proof. It will suffice to assume ∆u = 0 in Ω L,2 and show that v ≤ Cu in B 1/2 . We initially prove the theorem for a fixed x 0 ∈ C M ∩ ∂B 1/2 . Suppose by way of contradiction that the theorem is not true. Then there exists u k ∈ S(Ω L k ,2 ) and v k ∈ S(Ω L k ,2 , d γ ) with ∆v k ≤ 0 and v k (x 0 ) = u k (x 0 ) = 1 and points x k ∈ Ω L k ,2 ∩ B 1/2 such that Harnack chains work on Lipschitz domains; therefore, from the interior Harnack inequality, by multiplying v k and u k by constants (uniformly bounded above and below), we may assume that v k (x ′ k , e n ) = u k (x ′ k , e n ) = 1.
Because of the interior Harnack inequality, in order for (3.4) to occur, it is necessary that dist(x k , ∂Ω L k ) → 0. We let y k = (x ′ k , e n ), and similar to the proof of Theorem 2.1 we define For any N > 0 with N ∈ N, there existsÑ ∈ N, such that if m l ≥Ñ , then there is a k = k(m l ) such that a k,m l satisfies If we let r k = 2 −m l , then for those chosen k we have that Also we have that with the last inequality following from the bounds in Lemma 3.3. Then combining estimates (3.6) and (3.7) we obtain We now use Lemma 3.2 as r k → 0 to obtain limiting functions and domains. By choosing a subsequence we first consider the limit function We also obtain a limiting global Lipschitz domain Ω on which u is the unique (up to multiplicative constant) nonnegative harmonic function that vanishes on the boundary. A further subsequence guarantees y k → y 0 ∈ Ω. As in the proof of Theorem 2.1, as r k → 0 we have that |∆w r k ,k | → 0. Then by picking a further subsequence, as r k → 0 we obtain a limiting global Lipschitz domain function w with the following properties (i) ∆w = 0 in Ω and w = 0 on Ω c , From property (iv) Lemma 3.9 guarantees w(x) = cu(x), but then both properties (ii) and (iii) cannot hold since u(y 0 ) > 0. An interior Harnack inequality with a Harnack chain will also give the result for x 0 ∈ B 1/2 ∩ Ω L and the constant C depending on dist(x 0 , ∂Ω L ).
Proof. The argument for how the boundary Harnack principle implies Hölder regularity is now standard (see [1]). We outline how to adapt to the case when w solves ∆w = −1 in Ω L , It is now a standard argument (see [1]) that (3.10) implies that there exists β depending on M and M − L such that w u C 0,β (Ω L,1/2 ) ≤ C w(e n /2) u(e n /2) .
Since |v| ≤ w we obtain a Hölder growth bound for v/w at ∂Ω L,1/2 . The interior Hölder estimates for both v and u combined with the fact that |v|/u is bounded give interior Hölder estimates for v/u. The interior Hölder estimates combined with the boundary Hölder estimates for v/u are combined in a standard way to conclude (3.9).

Second-Order Elliptic Operators
The techniques employed in Section 3 can be applied to other elliptic operators, and in this section we extend the results of Section 3 to second order linear elliptic operators in divergence form on Lipschitz domains. Specifically, we consider operators of the form Lu = (a ij u i ) j + b i u i + cu We assume the following ellipticity conditions for some Λ > 0 and for all nonzero ξ ∈ R n . Furthermore, a ij (x) is a real n × n matrix. For the lower order terms we assume |c(x)|, |b i (x)| ≤ Λ − 1 and that c(x) ≤ 0.
We will continue with the notation of Section 3; however, we now write u ∈ S L (Ω L,R ) if To apply the Hölder continuity estimates for elliptic operators we will require that γ > −2/n; see Theorem 8.27 in [7]. Since the boundary is Lipschitz, this will ensure the correct integrability assumptions for the right hand side. We will assume these bounds throughout the section whenever referencing S L (Ω L,R , d γ ).
From the forthcoming Lemma 4.1, it will follow that if u ≥ 0 and u ∈ S L (C L , ∞), then u is unique up to multiplicative constant and we again denote u by u L ; however, u L will not necessarily be homogeneous. We recall that C L is defined although not convex when L < 0. To emphasize when −L < 0, we again write C −L and u −L when u ∈ S L (C −L ). We will follow the same outline as in Section 3.
In Section 3 we utilized the standard boundary Harnack principle. Since the standard boundary Harnack principle is unavailable when considering the zeroorder term c(x), we prove the next two Lemmas under the situation b i , c ≡ 0.
then there exists c ∈ R such that v ≡ cu.
Proof. When b i , c ≡ 0, there is a standard Boundary Harnack principle for divergence form equations [3]; therefore, the proof of Lemma 3.6 holds in this situation, and so the proof of Lemma 3.7 also holds as well.
Theorem 4.2. Assume L has no lower order terms; i.e, Lw = ∂ j (a ij u i ). Let v, u ∈ S L (Ω L,∞ ) with u ≥ 0, and assume b i , c ≡ 0. If there exists C > 0 such that then v(x) = cu(x) for some c ∈ R.
Remark 4.3. The proof given below for Theorem 4.2 depends on the function g being slowly varying at ∞. Thus, the same proof will actually show a stronger result: If for every ǫ > 0 there exists C ǫ such that |v(x)| ≤ C ǫ |x| ǫ u(x) for |x| ≥ 1, Proof. Suppose by way of contradiction that v is not a constant multiple of u. Then by Lemma 4.1 we have If h(R) = sup BR |v|/u, let g be the concave envelope of h. By assumption g(R) ≤ C ln(R + 1) for R ≥ 1. The function g satisfies, lim R→∞ g(CR) g(R) = 1 for any C > 0. There also exist infinitely many R k such that h(R k ) = g(R k ). We define From ( and that u k converges uniformly to w 1 on compact sets. From (4.2) we also have that lim with uniform convergence on compact sets and with |w 2 | ≤ Cw 1 . There also exists a limiting operator L 0 and limiting Lipschitz domainΩ such that w 1 , w 2 ∈ S L0 (Ω L,∞ ). Then from Lemma 4.1 we conclude that w 2 = cw 1 for some c ∈ R.
Without loss of generality we assume c = 1. Then for any ǫ > 0, there exists N ǫ ∈ N such that if k ≥ N ǫ , then By the standard boundary Harnack principle We fix x 1 ∈ Ω L,∞ . We also have Then for large enough k we have |v k (R −1 k x 1 )| ≤ u k (R −1 k x 1 )/2. Finally, we conclude then that The C in the above estimate only depends on the ellipticity constants of a ij and the Lipschitz constant for the domain. Consequently, we may choose ǫ small enough so that 2Cǫ < 1, which implies u(x 1 ) = 0 which is a contradiction since u > 0 in Ω L,∞ .
For the remainder of the section we no longer assume that the lower order terms are zero. Proof. Since u ≥ 0 and Lu ≤ 0 and u ∈ S L (Ω L,R , d γ ), this is an application of the interior weak Harnack inequality as well as uniform Hölder continuity up to the boundary, see [7].
We now state the analogue of Lemma 3.2.

Lemma 4.5.
Let Ω L k ,R k be a sequence of domains with L k ≤ M , R k ≥ 1, and 0 ∈ ∂Ω L k . Let u k ∈ S L (Ω L k ,R k , d γ ), and assume γ > −2/n. Further assume either (1) u k ≥ 0 and sup
Then there exists a subsequence with a limiting domain Ω L0,R0 and a limiting function u 0 ∈ S(Ω L0,R0 , d γ ) such that sup Br |u k − u 0 | → 0, for all r < R 0 .
Proof. This is an application of uniform Hölder continuity up to the boundary, see [7].

sup
Br (x) u ≤ c 2 u(e n /2)r β , for any r ≤ 1/4. (2) is just uniform Hölder continuity up to the boundary. Assume by way of contradiction that (1) is not true. Then there exists a sequence satisfying (1) u k ∈ S L k (Ω L k , d γ )
We rescale and let w k = u k (r k x) sup Br k u k .
From Lemma 4.5 we have that w k → w 0 uniformly and there is a limiting Lipschitz domain Ω L0 and limiting elliptic operator L 0 such that u ∈ S L0 (Ω L0 ). Now w 0 ≥ 0, and from the definition of w k we conclude that w 0 is not identically zero. However, w 0 (e n /4) = 0 which contradicts the strong maximum principle.
With the previous result, the proof of Theorem 4.7 proceeds exactly as in the case of Theorem 3.10.
Proof. The proof proceeds exactly as the proof of Theorem 3.10. We only highlight how the lower order terms vanish in the blow-up regime. The rescaled functions where for a function f we have Thus, in the blow-up regime the lower order terms disappear. As in the proof of Theorems 2.1 and 3.10 we may bound the denominator in (4.4) from below by r α k [ln(1/r k )] 2 for the constructed sequence of r k → 0. Using that the numerator is bounded from above by r 2−γ k , we have that in the blow-up regime the right hand side and lower order terms vanish. We then apply Lemma 4.1 and Theorem 4.2 as in the proof of Theorem 3.10.
Since g(k) ≥ a k and the series a k diverges, then the series g(k) diverges. By the ratio test it follows that This proves when j = 1. By induction we assume it holds true for j. Then Then by (A.2) as well as the induction hypothesis we conclude that lim k→∞ j+1 i=1 g(k − i) g(k) ≤ 1 + j.
Finally, we use that a k l = g(k l ) and a k l −i ≤ g(k l − i) to conclude (A.1).