New boundary Harnack inequalities with right hand side

We prove new boundary Harnack inequalities in Lipschitz domains for equations with a right hand side. Our main result applies to non-divergence form operators with bounded measurable coefficients and to divergence form operators with continuous coefficients, whereas the right hand side is in $L^q$ with $q>n$. Our approach is based on the scaling and comparison arguments of \cite{DS20}, and we show that all our assumptions are sharp. As a consequence of our results, we deduce the $\mathcal{C}^{1,\alpha}$ regularity of the free boundary in the fully nonlinear obstacle problem and the fully nonlinear thin obstacle problem.

1. Introduction 1.1.Background.The boundary Harnack inequality states that all positive harmonic functions with zero boundary condition are locally comparable as they approach the boundary, under appropriate assumptions on the domain.More precisely, if u and v are positive harmonic functions in Ω that vanish on ∂Ω, then with C depending on the dimension and u(p)/v(p) for a fixed interior point p.
Notice that such a result is most relevant in domains that are less regular than C 1,Dini , because otherwise the Hopf lemma combined with the C 1 (Ω) regularity of the solutions yields the same conclusion, see for example [LZ18].
The boundary Harnack inequality is known to be true for a broad class of domains and for solutions of more general elliptic equations.The classical case for harmonic functions was first proved by Kemper in Lipschitz domains in [Kem72].Operators in divergence form were first considered by Caffarelli, Fabes, Mortola and Salsa in [CFMS81] in Lipschitz domains, while the case of operators in non-divergence form was treated in [FGMS88] by Fabes, Garofalo, Marin-Malave and Salsa.Jerison and Kenig extended the same result to NTA domains in the case of divergence form operators in [JK82].On the other hand, the case of non-divergence operators in Hölder domains with α > 1/2 was treated with probabilistic techniques in [BB94] by Bass and Burdzy.Recently, De Silva and Savin found a simple and unified proof of all these previous results in [DS20].
Besides, Allen and Shahgholian recently proved the boundary Harnack for divergence form equations with right hand side in Lipschitz domains [AS19], under appropriate assumptions on the operator, the right hand side and the domain.In particular, in the case of the Laplacian, their result implies that if the L ∞ norm of the right hand side and the Lipschitz constant of the domain are small enough, then the boundary Harnack inequality still holds.This enables using the classical proof in [Caf98] due to Caffarelli (see also [PSU12,Section 6.2] or [FR20, Section 5.4]) of the regularity of the free boundary in the obstacle problem ∆u = χ {u>0} in the more general case ∆u = f χ {u>0} , with f Lipschitz; see [AS19, Section 1.4.2].
Here, we extend such boundary Harnack inequality to non-divergence equations with possibly unbounded right hand side in L q , with q > n. (This was only known in C 1,1 domains [Sir17,Sir20].)This allows us to use the classical proof of the free boundary regularity in the obstacle problem ∆u = f χ {u>0} to the case f ∈ W 1,q , and can also be applied to fully nonlinear free boundary problems of the form (1.1)Moreover, we also establish a boundary Harnack for equations with a right hand side in slit domains, and use it to establish the C 1,α regularity of the free boundary in the fully nonlinear thin obstacle problem, a question left open in [RS16].
1.2.Setting.In the following, L will denote either a non-divergence form elliptic operator with bounded measurable coefficients, Lu = Tr(A(x)D 2 u), with λI ≤ A(x) ≤ ΛI, (1.2) with 0 < λ ≤ Λ, or a divergence form elliptic operator with continuous coefficients, Lu = Div(A(x)∇u), with λI ≤ A(x) ≤ ΛI and A ∈ C 0 , (1.3) where A has modulus of continuity σ, and 0 < λ ≤ Λ.We will consider Lipschitz domains of the following form, where B ′ 1 is the unit ball of R n−1 .Definition 1.1.We say Ω is a Lipschitz domain with Lipschitz constant L if Ω is the epigraph of a Lipschitz function g : B ′ 1 → R, with g(0) = 0: 1.3.Main results.We present here our new boundary Harnack inequality.
We emphasize that the following result applies to both non-divergence and divergence form operators, and that the only regularity assumption on the coefficients is the continuity of A(x) in case of divergence-form operators.Throughout the paper, when we say L n -viscosity or weak solutions, we refer to L n -viscosity solutions in the case of non-divergence form operators (1.2), and to weak solutions in the case of divergence form operators (1.3).
Theorem 1.2.Let q > n and L as in (1.2) or (1.3).There exist small constants c 0 > 0 and L 0 > 0 such that the following holds.
Let Ω be a Lipschitz domain as in Definition 1.1, with Lipschitz constant L < L 0 .Let u and v > 0 be solutions of in the L n -viscosity or the weak sense, with f L q (B 1 ) ≤ c 0 , g L q (B 1 ) ≤ c 0 . (1.4) Additionally, assume that v(e n /2) ≥ 1 and either u > 0 and u(e n /2) ≤ 1, or u L p (B 1 ) ≤ 1 for some p > 0.
The constants C, c 0 , L 0 and α > 0 depend only on the dimension, q, λ, Λ, as well as p and σ, when applicable.
Remark 1.3.All the hypotheses of the theorem are optimal in the following sense: • If the Lipschitz constant L 0 of the domain is not small, the theorem fails, even for q = ∞ and for L = ∆.• If q = n, the theorem fails for any c 0 > 0 and any L 0 > 0, even for L = ∆.
• The result fails in general for operators in divergence form with bounded measurable coefficients.We provide counterexamples to plausible extensions in this sense in Section 6.
When the two functions are positive, we recover the standard symmetric formulation of the boundary Harnack.
Corollary 1.4.Let q > n and L as in (1.2) or (1.3).There exist small constants c 0 > 0 and L 0 > 0 such that the following holds.Let Ω be a Lipschitz domain as in Definition 1.1, with Lipschitz constant L < L 0 .Let u, v be positive solutions of in the L n -viscosity or the weak sense, with f and g satisfying (1.4).Assume u, v are normalized in the sense that u(e n /2) = v(e n /2) = 1.Then, The positive constants C, c 0 , L 0 and α depend only on the dimension, q, λ, Λ, as well as σ, when applicable.(Lipschitz), or our new Theorem 1.2 if f ∈ W 1,q with q > n.
In the fully nonlinear setting (1.1), the derivatives of the solution satisfy a linear equation in non-divergence form, L(∂ ν u) = g in {u > 0}, with bounded measurable coefficients A(x), and then having our new boundary Harnack for non-divergence operators proves useful to deduce results on the regularity of the free boundary.
It is well known that the free boundary may exhibit singularities.Hence, we need to introduce the notion of a regular point.
Definition 1.5.Let x 0 be a free boundary point for the classical obstacle problem, i.e. x 0 ∈ ∂{u > 0} for a solution of (1.5).We say that x 0 is a regular free boundary point if there exists r k ↓ 0 such that for some γ > 0 and e ∈ S n−1 .
Our next application was already known by using perturbative arguments with slightly weaker assumptions [Bla01].We include this result to illustrate the arguments that we will use in the fully nonlinear problems in a more easily readable setting.
Corollary 1.6.Let u be a solution of (1.5) with f ≥ c 0 > 0 in W 1,q (B 1 ), with q > n, and assume the origin is a regular free boundary point in the sense of Definition 1.5.
The fully nonlinear obstacle problem can be presented in at least two different formulations.The following one was studied by Lee in [Lee98].
(1.6) Here, we impose the following conditions: • F is uniformly elliptic.
Then, under these hypotheses, v ∈ C 1,1 and the free boundary is C 1,α at regular points.For our purposes, we will say a free boundary point is regular in the sense of Definition 1.5, as in the classical obstacle problem.
More generally, one can study problems of the form This is a generalization of problem (1.6).Indeed, if we define This fully nonlinear obstacle problem (and more general ones without the sign condition on u) has been further studied by Lee, Shahgholian, Figalli, and more recently by Indrei and Minne in [LS01,FS14,IM16].They proved that if F is convex, f is Lipschitz and f ≥ τ 0 > 0, the free boundary ∂Ω is C 1 at regular points.
As a consequence of our new boundary Harnack inequality, we extend their result for (1.7) in two ways.We lower the Lipschitz regularity required for f to W 1,q with q > n, and we prove C 1,α regularity of the free boundary instead of C 1 .
Corollary 1.7.Let u be a solution of (1.7).Assume as well: (H1) F is uniformly elliptic and F (0, x) = 0 for all x ∈ Ω. (H2) F is convex and C 1 in the first variable, and W 1,q in the second variable for some q > n. (H3) f ∈ W 1,q for some q > n, and f ≥ τ 0 > 0.Then, if the origin is a regular free boundary point in the sense of Definition 1.5, the free boundary is a C 1,α graph in B r for some small r > 0 and α > 0.
1.5.Thin obstacle problems.The thin obstacle problem, also known as the Signorini problem, is a classical free boundary problem that admits several formulations, see [Fer20] for a nice introduction to the topic.One can write the problem as the following, given an obstacle ϕ defined on {x n = 0}: (1.8) The first results on regularity of the solution v were established in the seventies, in particular it was proved in [Caf79] that v ∈ C 1,α for a small α > 0. Free boundary regularity remained open for quite some time, until the first free boundary regularity result, [ACS08], establishing that the free boundary is C 1,α at regular points when ϕ ≡ 0. Further results have been obtained in [KPS15,DS16] among others, proving that the free boundary is real analytic at regular points provided that ϕ is analytic.
Consider now the fully nonlinear thin obstacle problem.
Milakis and Silvestre proved in [MS08] that solutions u are C 1,α in the symmetric case (even functions with respect to x n ).More recently, Fernández-Real extended the result to the non-symmetric case in [Fer16].The first result on free boundary regularity is due to the first author and Serra [RS16], where they proved the C 1 regularity of the free boundary near regular points.Here, we will prove for the first time that the free boundary is actually C 1,α .
To do this, we need to adapt Theorem 1.2 to the case of slit domains.We present here a simplified version, see Section 4 for a more general result.
Theorem 1.8.Let q > n and let L be as in (1.2).There exists small c 0 > 0 such that the following holds. Let Let u and v > 0 be L n -viscosity solutions of with f and g satisfying (1.4).Assume in addition that v(e n /2) ≥ 1, v(−e n /2) ≥ 1, and either u > 0 in B 1 \ K and max{u(e n /2), u(−e n /2)} ≤ 1, or u L p (B 1 ) ≤ 1 for some p > 0.Then, The positive constants C, c 0 , and α depend only on the dimension, q, λ, Λ, as well as p, when applicable.
Using this new boundary Harnack, we can prove the following.
Corollary 1.9.Assume that 0 is a regular free boundary point for (1.9) in the sense of [RS16], with F ∈ C 1 and ϕ ∈ W 3,q for some q > n.Then, there exists ρ > 0 such that the free boundary is a This is new, even when ϕ ∈ C ∞ .The higher regularity of the free boundary remains a challenging open question.
1.6.Plan of the paper.The paper is organized as follows.
In Section 2, we recall some classical results and tools, such as the ABP estimate and the weak Harnack inequality.Then, in Section 3 we prove our new boundary Harnack inequality for elliptic equations with right hand side, Theorem 1.2, by scaling and barrier arguments.Section 4 is devoted to adapting the result to slit domains.In Section 5, we prove the C 1,α regularity of the free boundary in the fully nonlinear obstacle problem, Corollary 1.7, and in the fully nonlinear thin obstacle problem, Corollary 1.9.Finally, in Section 6, we present two counterexamples that show the sharpness of our new boundary Harnack and in Section 7 we introduce a Hopf lemma for equations with right hand side.

Preliminaries
In this section we recall some classical tools and results that will be used throughout the paper.We will denote the Pucci extremal operators, see [CC95] or [FR20] for their properties.
2.1.L n -viscosity and weak solutions.In this work we are considering linear elliptic equations of the form Lu = f , with f ∈ L q , with q ≥ n.The most appropriate notion of solutions for a divergence form equation are the well-known weak solutions.
For the non-divergence form case, one could consider strong (W 2,n loc , solving the PDE in the a.e.sense) solutions, but all the arguments of the proof are equally viable for L n -viscosity solutions, which are more general.We present the minimal definition for the linear case.
and L in non-divergence form.We say u is a L n -viscosity subsolution (resp.supersolution) if, for all ϕ ∈ W 2,n loc (Ω) such that u − ϕ has a local maximum (resp.minimum) at x 0 , ess lim inf We will say equivalently that u is a solution of Lu ≤ (≥)f .When u is both a subsolution and a supersolution, we say u is a solution and write Lu = f .L n -viscosity solutions coincide with strong, viscosity or even classical solutions when they have the required regularity, and satisfy the maximum and comparison principles, but are more flexible, for example, allowing to compute limits under some reasonable hypotheses, and are thus preferred in some contexts.
Throughout this paper, the Dirichlet boundary conditions must be understood in the pointwise sense when we are dealing with L n -viscosity solutions, and in the trace sense when we are dealing with weak solutions.
2.2.Interior estimates.The Alexandroff-Bakelmann-Pucci estimate is one of the main tools in regularity theory for non-divergence form elliptic equations.We refer to [CC95, Theorem 3.2] and [CCKS96, Proposition 3.3] for the full details and a proof.
Theorem 2.2 (ABP Estimate).Assume that Ω ⊂ R n is a bounded domain.Let L be a non-divergence form operator as in (1.2) and let u ∈ C(Ω) satisfy Lu ≥ f in the L n -viscosity sense, with f ∈ L n (Ω).Assume that u is bounded on ∂Ω.
Then, sup with C only depending on the dimension, λ and Λ.
In the case of divergence form equations, the global boundedness of weak solutions is known in more generality.For our purposes, it is sufficient to consider the case p = n.
Theorem 2.3 ([GT98, Theorem 8.16]).Assume that Ω ⊂ R n is a bounded domain.Let L be a divergence form operator as in (1.3) and let u ∈ C(Ω) be a weak solution of Lu ≥ f , with f ∈ L p (Ω), p > n/2.Assume that u is bounded on ∂Ω.
Then, sup with C only depending on the dimension, |Ω|, p, λ and Λ.
We will need the two estimates that are classically combined to obtain the Krylov-Safonov Harnack inequality.The first one is the following weak Harnack inequality, valid for L n -viscosity solutions of non-divergence form equations.We refer to [Tru80, Theorem 2] and [KS09, Theorem 4.5].
Theorem 2.4 (Weak Harnack inequality).Let L be a non-divergence form operator as in (1.2).Let u satisfy Lu ≤ 0 in Ω in the L n -viscosity sense and let B R (y) ⊂ Ω.Then, for all σ < 1, where p and C are positive and depend only on the dimension, σ and Λ/λ.Now, combining this theorem with the ABP estimate, applied to the function 1−u, we obtain the following result.This is also valid for divergence form equations, and sometimes known as De Giorgi oscillation lemma in that setting.The case with f = 0 is found in [CS05, Theorem 11.2], and we can extend it easily to the general case using Theorem 2.3.
The second estimate is the upper bound in Harnack inequality, also valid for L n -viscosity solutions of non-divergence form equations [Tru80, Theorem 1], [KS12] and weak solutions of divergence form equations [DG57,LZ17].In the divergence form case, we can add the right hand side using Theorem 2.3.Theorem 2.6 (L ∞ bound for subsolutions).Let p > 0 and let L be as in (1.2) or (1.3).Let Lu ≥ f in B 1 , in the L n -viscosity or the weak sense.Then, where C p > 0 depends only on the dimension, p, λ and Λ.
3. Proof of Theorem 1.2 3.1.Nondegeneracy.To study solutions of Lu = f in a Lipschitz domain it is useful to know their behaviour in a cone.In this first part of the proof we show that, much like solutions of elliptic equations with zero Dirichlet boundary conditions separate linearly from the boundary of the domain in domains with the interior ball condition (Hopf lemma), the solutions of elliptic equations with zero Dirichlet boundary conditions separate as a power of the distance at corners, and the exponent approaches 1 as the corners become wider.
Let u be any solution of where C η is the cone defined as Then, u(te n ) ≥ t β , ∀t ∈ (0, 1).
Proof.Assume without loss of generality that β ∈ (1, 2), because if the inequality holds for β > 1, it holds also for all β ′ > β.We will use the comparison principle with a subsolution that has the desired behaviour.Let ε ∈ (0, 1/20) to be chosen later.Notice that √ 1 + ε − √ ε > 4/5.Define the subsolution ϕ as: We can readily check that ϕ(x) = 0 for x ∈ ∂C η .It is also clear that ϕ(x) ≤ 1 in {x n = 1} ∩ C η , and that ϕ > 0 in C η .Now, we need some estimates on f ε and its derivatives.For t ∈ [0, 1), In the last inequality we used that Then, we will make ε small and then η small in such a way that Lϕ ≥ c(β).To make the computations easier, we will use the Pucci operator M − , and we will denote t = η|x ′ |/x n .On the one hand, we can check that Now, we compute the discriminant of the second order polynomial that we found: Hence, the second order polyonmial is always positive and attains a minimum Now we need to compute the crossed derivatives.We begin with And finally, taking i = j in {1, . . ., n − 1}, Define H(x) = D 2 ϕ(x), and also H 0 (x) to be the matrix with ∂ 2 ϕ/∂x 2 n at the lower right corner and zeros in all other entries.On the one hand, by the definition of M − : M − (H 0 ) ≥ λx β−2 n c β .Moreover, using that H − H 0 is bounded by the sum of the coefficients, where Since ε > 0 is fixed, we choose η small enough such that F (η) ≥ λc β /2.To end the proof, where we use that x n ≤ 1 and β − 2 < 0. By the comparison principle, we conclude that u(te Remark 3.2.The constant L 0 in Theorem 1.2 is limited, in fact, by the value of η from this lemma, because the domain must contain wide enough cones, so the Lipschitz constant of the boundary must be small enough. To prove the nondegeneracy property for solutions of divergence form equations, we proceed by approximation.The continuity assumption on the coefficients in (1.3) is necessary, see Proposition 6.3.
The following lemma is a natural approximation property of divergence form equations.
Lemma 3.3.Let Ω be a bounded Lipschitz domain and K ⊂ Ω a compact subset.Let L 1 , L 2 be divergence form operators, and let u 1 , u 2 ∈ H 1 (Ω) be the solutions of the following Dirichlet problems with g ∈ H 1 (Ω) and Then, , where C > 0 and τ ∈ (0, 1) depend only on K, Ω, g and the ellipticity constants.
and thus Hence, using that the H 1 norm of u 2 can be bounded by a constant depending on the domain, the ellipticity constants and the boundary data, This, combined with the Poincaré inequality, yields On the other hand, let δ = d(K, ∂Ω) and define the enlarged compact set . By the De Giorgi-Nash-Moser theorem, we have , where α and C 3 depend only on the domain, the dimension and the ellipticity constants, thus v C 0,α (K ′ ) ≤ 2C 3 .
Let p ∈ K such that |v| reaches its maximum, and assume without loss of generality that v(p) > 0.Then, for all This presents us with two cases.When L ∞ (Ω) }, and the result follows.
As a consequence, we can prove the analogue of Lemma 3.1 for divergence form equations.
Lemma 3.4.Let L be in divergence form with continuous coefficients, with modulus of continuity σ as in (1.3).Let β ′ > 1.There exists sufficiently small η ′ > 0 such that the following holds.
Let u be a solution of The constants t σ and η ′ are positive and depend only on the dimension, β ′ , σ, λ and Λ.
Proof.We will assume without loss of generality that β ′ ∈ (1, 2) and that Lu = 0 in Let η > 0 be the one provided by Lemma 3.1 with exponent β.Let η ′ < η/8 and k 0 ∈ Z + , to be chosen later.We will prove by induction that u(2 −k e n ) ≥ c2 −kγ for all integer k ≥ k 0 and some c > 0. Notice that this implies that u(te n ) ≥ c ′ t γ for some smaller c ′ > 0 by a direct application of interior Harnack.To end the proof, notice that if t is small enough, since We proceed now with the induction.First, we define the following auxiliary functions.
We claim that there exists c > 0 such that, for all integer For the first k 0 , first observe that u ≥ 0 everywhere by the maximum principle.Then, apply the interior Harnack inequality to the cylinder and let v and v 0 the solutions of the following Dirichlet problems and Observe that v = v 0 = 0 on the lateral boundary of the cone C 2 −k ,η ′ .Then, it is clear that u ≥ v from the boundary conditions.Furthermore, by a rescaling of Lemma 3.3, Hence, we can apply a rescaled Lemma 3.1 to the normalized 2 kγ v 0 , because L 0 has constant coefficients and is also a non-divergence form operator to obtain Then, passing the information on where for the last inequality we first choose a small η ′ such that with δ > 0, and then take k large and use that A is continuous.Hence, if η ′ is small enough and k 0 is large enough in the first place, the inductive step holds.Now we are ready to prove Theorem 1.2.We divide the proof into three parts: an upper bound, a lower bound, and the proof of the C 0,α regularity of the quotient.
3.2.Upper bound.We follow the arguments of [DS20]; see also [Saf10].The first lemma is a geometric fact that will make subsequent computations easier.Lemma 3.5.Let Ω be a Lipschitz domain as in Definition 1.1, with Lipschitz constant L < 1/16.Let where δ ∈ (0, 1/3).Then A is star-shaped with respect to the point e n /2.
).We distinguish the upper and the lower boundaries of A as: The first step is proving that ∂ l A is a Lipschitz graph with the same or lower Lipschitz constant.For this, consider the set which contains the points above ∂ l A. For every vertical line l passing through (x ′ , 0), with x ′ ∈ B ′ 1−δ , the set l ∩ Ω δ is not empty, so we can define h : Then, for a given x ′ ∈ B ′ 1−δ , (x ′ , y) ∈ Ω δ for all y > h(x ′ ).Indeed, for every point In any case, we have proven that Moreover, this shows that ∂ l A is a subset of the graph of h.Now we want to see that h is Lipschitz.Notice that we can also define h with the complement set, This can be seen as the superior envolvent of a union of spheres of radius δ centered at every point of ∂Ω, hence Since this is a supremum of equi-Lipschitz functions, h is also Lipschitz with the same or lower constant, L ′ ≤ L < 1/16.From g(0) = 0 we can also derive h(0) ≥ δ, and h(0) ≤ δ √ L 2 + 1 < 1.02δ.Now we will see that A is star-shaped with center at e n /2, constructing a segment from e n /2 to every point in A that lies entirely inside A. Let p = e n /2 be a point in A, and let q = (q ′ , q n ) be the intersection of the line through p and e n /2 and ∂A, that lies on the side of p and is furthest from e n /2.We will see later that there is only one intersection at each side, but considering the furthest is enough for now.
If q lies in ∂ l A, q n = h(q ′ ).If q lies in ∂ u A, the point is above ∂ l A and q n > h(q ′ ).In any case, we have always q n ≥ h(q ′ ).It is clear that the segment (e n /2)q, that can be parametrised by We will prove that it lies entirely above ∂ l A (except maybe in the point q), so it has not other intersections with ∂A besides q.We distinguish two cases: If q n ≥ 7/16, for any point tq ′ inside the segment joining 0 and q ′ in B ′ 1−δ (this means t ∈ (0, 1)), using that h is Lipschitz, Moreover, the height of the segment (e n /2)q above the point tq ′ is (1 − t)/2 + tq n ≥ 0.5 + (q n − 0.5)t ≥ 0.5 + (h(q ′ ) − 0.5)t ≥ 0.5 − t/16, and 0.5 − t/16 > 0.34 + t/16 because t/8 < 1/8 < 0.5 − 0.34 = 0.16.Combining the two inequalities, h(tq ′ ) < (1 − t)/2 + tq n as required.
On the other hand, if q n < 7/16, h(q ′ ) < 7/16 as well.Since h is Lipschitz, The height of the segment (e n /2)q above the point tq ′ is and 1/2 − (1/2 − q n )t > (q n + 1/16) − (1/16)t for t ∈ (0, 1) by a simple calculation.Hence, in any case the segment joining e n /2 and q crosses ∂A at q for the first time, implying A is star-shaped.Now, we derive an interior Harnack inequality for domains with the shape we want to consider.Lemma 3.6.Let Ω be a Lipschitz domain as in Definition 1.1, with Lipschitz constant L < 1/16.Let δ ∈ (0, 1/3).Let L be as in (1.2) or (1.3).Let u be a positive solution, in the L n -viscosity or the weak sense, of with C depending on the dimension, δ, λ and Λ, but not on the particular shape of Ω.
Take x 0 = x, . . ., x m = y a uniform partition on the segment xy.It is clear that |x i+1 − x i | < δ/4.Then, consider the balls B δ (x i ).We apply the interior Harnack inequality to obtain that sup In particular, u( ), and iterating this, u(y Now take x, z ∈ A, and apply the inequalities between u(x) and u(y) to u(y) and u(z).We can put them together finally to get Finally, notice that C, m and C ′′ do not depend on the shape of Ω.The next step is the following lemma, that shows that the condition u > 0 and u(e n /2) ≤ 1 implies u L p (B 1 ) ≤ c p in Theorem 1.2.Lemma 3.7.Let Ω a Lipschitz domain as in Definition 1.1, with Lipschitz constant L < 1/16.Let L be as in (1.2) or (1.3).Let u be a positive solution, in the L n -viscosity or the weak sense, of such that u(e n /2) ≤ 1, with f ∈ L n (B 1 ).Then, there exist p, C p > 0 such that with p and C p only depending on the dimension, λ, Λ and f L n (B 1 ) .
Proof.We will prove that there exist a sequence {a k } and some positive c and b This means roughly that sup u grows at most like d −K for some big K, and then u p will be integrable if p > 0 is small enough.First, by Lemma 3.6 applied with δ = 1/8, together with the fact that u ≤ 1 in at least a point of A 3 , sup Now, we will show that a k+1 ≤ c 1 a k + c 2 , with c i > 0. This easily implies by induction that a k ≤ cb k for some b, c > 0.
Take x ∈ A k+1 .We will prove that there exists a close y ∈ A k such that d(x, y) < 2 −k+3 .In fact, let z = (z ′ , z n ) be the intersection of ∂A k with the segment (e n /2)x.Proving d(x, z) < 2 −k+3 will suffice, because there are points in A k arbitrarily close to z.

and applying the interior Harnack inequality we get
). Iterating this inequality, u(y) ≤ c 1 u(x) + c 2 , for some constants c 1 , c 2 only depending on the dimension and f L n (B 1 ) .Now we know that sup u ≤ cb k in A k .Let p = log b √ 2, and compute the L p norm of u: where we have used that |A j+1 \ A j | ≤ 2 −j V (n).We will prove it now.
On the one hand, where |∂B 1 | is the (n − 1)-dimensional measure of the boundary of the ball B 1 .On the other hand, the second set is a subset of the thickening of ∂Ω in the e n direction, with height 2 −j √ L 2 + 1 at each side: The measure of this second set is 2 The previous Lemma 3.7 implies that u ∈ L p (B 1 ).Then, we can use Theorem 2.6 to obtain the following L ∞ bound on u: Proposition 3.8.Let Ω be a Lipschitz domain as in Definition 1.1, L as in (1.2) or (1.3) and r ∈ (0, 1).Let u be a L n -viscosity or weak solution of Then, for all p > 0, Proof.Denote v = u + , extending v by zero in B 1 \ Ω, and extend f by zero in B 1 \ Ω.Then, it is easy to check that Lv ≥ f in B 1 .Now use Theorem 2.6 and a covering argument to get sup The conclusion trivially follows.
3.3.Lower bound.The next step is to construct an iteration to see that solutions of Lu = f that are sufficiently positive away from the boundary, and not very negative near it, are actually positive everywhere.As we have a right hand side f ∈ L q , we need to be careful with the scaling, so we cannot use directly interior Harnack estimates to prove positivity, and we will need the nondegeneracy estimates in Lemmas 3.1 and 3.4.
Let Ω be a Lipschitz domain as in Definition 1.1 with constant L < L * .Let f be such that f L q (B 1 ) ≤ c 0 .Let u be a L n -viscosity or weak solution of for some sufficiently small ρ, ε, δ, c 0 ∈ (0, 1), with ρ > 2δ, only depending on the dimension, κ, q, λ, Λ, as well as σ, when applicable.
Let h > 0 to be chosen later.For x 0 = (x ′ 0 , x 0n ) in {x ∈ B ρ : d(x, ∂Ω) > ρδ}, define the cone Here, we distinguish the upper and the lateral boundaries, respectively, and the upper half cone 0 ) + h/2}.Now, take L * = min{η/2, 1/16}.Hence, the slope of ∂Ω will be at most half of the slope of C, so the cone separates from the boundary.By some geometric computations, we find Furthermore, making h = 4δ 4 + η 2 , we will have ∂ u C ⊂ {x ∈ B 1 : d(x, ∂Ω) > δ}, and also C + ⊂ {x ∈ B 1 : d(x, ∂Ω) > δ}.Note that this forces δ to be small, but we will choose it at the end, so this is not a problem.
Define ũ(x) = u(x ′ 0 , g(x ′ 0 ) + hx) + ε.Let ũ = v + w, where By the ABP estimate, Theorem 2.2, in the non-divergence form case, or by Theorem 2.3 in the divergence form case, w L ∞ (Cη ) ≤ C ′ f L n (Cη) ≤ C ′ c 0 .On the other hand, v ≥ 0 on ∂C η , and v ≥ 1 on ∂ u C η and C + η (defining the upper boundary and the upper half anologously).Hence, we can apply Lemma 3.1 or a rescaled Lemma 3.4 to v to conclude that v(te n ) ≥ t β , possibly only for small t < t σ .
In order to do it, we need f + /ε to be small enough, to have f + /ε L n (B 1 ) ≤ δ(c L ) in the notation of the theorem.
We will iterate this reasoning with the functions To end the proof we only need to choose j such that (1−γ) j < ρ κ , and then make δ small until 1 − 3jδ ≥ ρ.Finally, notice that we need f + /((1 − γ) i ε) L n (B 1 ) ≤ δ(c L ) for i = 1, . . ., j to be able to apply successively Theorem 2.5.This is possible choosing c 0 accordingly once we know j.Now, we iterate the lemma to obtain the desired result.
Let Ω be a Lipschitz domain as in Definition 1.1 with constant L < L * .Let u be a solution of (3.1) with f such that f L q (B 1 ) ≤ c 0 .Then, Moreover, for all t ∈ (0, 1), The constants L * , ε, δ and c 0 depend only on the dimension, κ, q, λ, Λ, as well as σ, when applicable.
Proof.We will iterate the previous Lemma 3.9.Assume without loss of generality that κ < 2 − n/q.Let u 0 = u, f 0 = f , and define the scalings: Define Ω j to be the rescaled domains of the u j .Observe that the Lipschitz constant of the domains is the same or smaller, and that Lu j = f j .Now we will see that the right hand side is bounded as we need.Indeed, since ρ 2−κ < ρ n/q , and then f j L q (B 1 ) ≤ c 0 for all j.
Let Ω be a Lipschitz domain as in Definition 1.1 with constant L < L * .Let f such that f L q (B 1 ) ≤ c 0 .Let v be a positive L n -viscosity or weak solution of Lv = f , with v(e n /2) ≥ 1.Then, for all for some sufficiently small c 0 , c 1 > 0, only depending on the dimension, κ, q, λ, Λ, as well as σ, when applicable.Remark 3.12.We can also write x n − g(x ′ ) instead of d(x, ∂Ω), since the two quantities are comparable.
Proof.Assume, after dividing by a constant if necessary, that v(e n /2) = 1.Let v ′ (x) = v(2x) and Ω ′ the corresponding scaled domain.By a version of Lemma 3.6, we have that if for any δ > 0, some small c 2 > 0 that depends only on δ, the dimension, q and the ellipticity constants.Now, apply the previous Proposition 3.10 to v ′ /c 2 in the balls B 1 (x 0 ) for any x 0 ∈ B 1 (we may need to ask that c 0 is smaller to do so).Hence, v ′ (x 0 + te n )/c 2 ≥ t κ , and this implies v(x 0 + (t/2)e n ) ≥ c 2 (t/2) κ .To end the proof, notice that d(x 0 + (t/2)e n , ∂Ω) ∈ [t/2, t √ L 2 + 1/2], so we can absorb the factor needed to change t for d(x, ∂Ω) in the constant c 1 .
3.4.Proof of the main result.Now we have all that we need to prove Theorem 1.2.Observe that Corollary 1.4 is a direct consequence.We divide the proof in two parts: in the first one we prove the inequality, and in the second we deduce the C 0,α regularity of u/v.

Define now
with ε > 0 to be determined later.We will show that w > 0 in B 1/2 , and therefore, taking By construction, w ≥ v/m ≥ 1 in A, and w ≥ −ε in B 3/4 .To apply Proposition 3.10 (rescaled to the ball B 3/4 ), we need to estimate Lw: Let w(x) = w(3x/4).Then, w ≥ 1 in Ω ∩ {x ∈ B 1 : d(x, ∂Ω) ≥ δ} and w ≥ −ε in Ω ∩ B 1 .Choosing sufficiently small ε, c 0 > 0 to apply Proposition 3.10, we get w > 0 in B 2/3 , thus w > 0 in B 1/2 .Now, for the boundary C 0,α regularity of the quotient u/v, we will first prove the regularity for the boundary points, and then we will extend it to the whole closed domain Ω ∩ B 1/2 , where u/v is extended by continuity on ∂Ω.These arguments are the standard ones found in the literature, but we have to be careful with some calculations to take into account the right hand side of the equations.Additionally, let c * 0 be the value for c 0 found in the first part of the proof.We will adjust the final value of c 0 in terms of this c * 0 .By a covering argument, the inequality u ≤ C ′ v is valid in Ω ∩ B 3/4 with an appropriate constant C ′ .Since either u > 0 or we can interchange u by −u and the hypotheses still hold, we have u ≥ −C ′ v as well.Let First, we will show by induction that there exist sequences {a j }, {b j } such that, for every integer j ≥ 2, For j = 2 we take a j = −C ′ , b j = C ′ , with the constant from the covering argument.Now, to perform the inductive step, we define two new functions: These functions are positive solutions of Lw i = f i in Ω ∩ B 2 −j (x 0 ), vanish continuously at ∂Ω, and w 1 + w 2 = v.Therefore, for one of them (the biggest in the point), 2w i (x 0 + e n /2 j+1 ) ≥ v(x 0 + e n /2 j+1 ).To apply the boundary Harnack, we define the following rescaled functions, with c 1 > 0 from Corollary 3.11 in order to have ṽ(e n /2) ≥ 1.Let ṽ Now we must check fi L q (B 1 ) ≤ c * 0 .Indeed, choosing c 0 appropriately, The same works for f 2 .
Applying a rescaled version of the boundary Harnack inequality to the functions 2w i , v, we get that Either ãj+1 > a j or bj+1 < b j .We cannot choose them yet as a j+1 , b j+1 , because we need to ensure 1 − θ ≥ 2 1−κ .This is done by choosing After this, a j ≤ u/v ≤ b j in Ω ∩ B 2 −j (x 0 ), and then sup (3.2) We can extend u/v by continuity at x 0 as the limit of the a j (or the b j ), and for any point Then, for every point x 0 on the boundary we have for some uniform constant C > 0, for any p ∈ B 1/2 ∩ Ω.Now, for the interior points, let Case 2. If the points are far compared with the distance to the boundary, in the sense that d ≥ δ i /4 for at least one of them, let y be a point in the boundary such that d(x i , y) < 8d for both of them (for example, in the case δ 1 ≤ 4d, let y be the closest point in the boundary to x 1 , so that d(x 2 , y) Case 3. When the points are close, i.e. d < 1/16 and d < min(δ 1 , δ 2 )/4, suppose without loss of generality 0 < δ 1 ≤ δ 2 .Let Now, we introduce an auxiliary function w = u − µv, with µ to be determined later.
, and v L ∞ (B r/2 (x 2 )) ≥ c 1 r κ by Corollary 3.11, then the right hand side of the previous inequality is bounded by some constant C 2 that only depends on n, q, λ, Λ.Hence, ), so that w(x 2 ) = 0. Let k 0 be the maximum positive integer such that δ 2 < 2 −k 0 (hence δ 2 ≥ 2 −k 0 −1 ).Then, k 0 ≥ 2, d < δ 2 /4, and x 1 , x 2 belong to Ω ∩ B 2 −k 0 +1 (y), with y ∈ ∂Ω, for instance, the closest point in ∂Ω to x 2 (d(y, x 1 ) < δ 2 + d < 2δ 2 < 2 −k 0 +1 by the triangle inequality).For the same reason, , and combining it with the previous result and the fact that We put all the constants (everything that does not depend on r, k 0 ) together, and notice that |µ| ≤ C ′ and 2 −k 0 −1 ≤ r < 2 −k 0 .Additionally, we dismiss the term (1 − θ) k 0 −2 ≤ 1/(1 − θ) 2 as a constant in the second fraction.Simplifying, we get In either case, Observe that we have proved that For the expression to be always valid, take the maximum multiplicative constant and the minimum exponent.

The boundary Harnack in slit domains
We also consider our problem in slit domains, as introduced in [DS16, DS20].We define them in the unit ball B 1 to keep the notation uncluttered.Definition 4.1.We say Ω is a slit domain with Lipschitz constant L if Ω = B 1 \ K, with K a closed subset of the graph of a Lipschitz function g : B ′ 1 → R, with g(0) = 0: Additionally, we define the upper and lower halves of Ω, We will write Ω ± to refer to Ω + or Ω − indistinctly.
An analogous reasoning to the proof of Theorem 1.2 for slit domains yields the following result.
Theorem 4.2.Let q > n and let L be as in (1.2) or (1.3).There exist small constants c 0 > 0 and L 0 > 0 such that the following holds.
Let Ω = B 1 \ K be a slit domain as in Definition 4.1, with Lipschitz constant L < L 0 .Let u and v > 0 be L n -viscosity or weak solutions of with f and g satisfying (1.4).Additionally, assume that v(e n /2) ≥ 1, v(−e n /2) ≥ 1, and either u > 0 in B 1 \ K and max{u(e n /2), u(−e n /2)} ≤ 1, or u L p (B 1 ) ≤ 1 for some p > 0.Then, The positive constants C, c 0 , L 0 and α depend only on the dimension, q, λ, Λ, as well as p and σ, when applicable.
When both functions are positive, we recover the symmetric version of the boundary Harnack.
Let Ω = B 1 \ K be a slit domain as in Definition 4.1, with Lipschitz constant L < L 0 .Let u, v be positive L n -viscosity or weak solutions of with f and g satisfying (1.4).
Most of the proofs are identical to the one-sided theorem, because we can prove the results for each side of Γ and then put them together.There are two exceptions: Proposition 3.8 and Lemma 3.9.The proof of the proposition is even easier, taking v = u + and extending it by 0 on K, we are ready to apply Theorem 2.6 and see that v is universally bounded.
As for the lemma, we write here an adapted version and the step of the proof that needs to be changed.Lemma 4.4.Let q > n, κ > 1, and let L be as in (1.2) or (1.3).There exists L * = L * (q, n, κ, λ, Λ) such that the following holds.
Let Ω = B 1 \ K, with K ⊂ Γ, be a slit domain as in Definition 4.1 with constant L < L * .Let f such that f L q (B 1 ) ≤ c.Let u be a L n -viscosity or weak solution of for some sufficiently small ρ, ε, δ, c ∈ (0, 1), with ρ > 2δ, only depending on the dimension, κ, q, λ, Λ, as well as σ, when applicable.
Proof.The proof of the first inequality is completely analogous to the proof of Lemma 3.9.
For the second inequality, we do the same reasoning as in the one-sided case, but now, instead of picking a downwards cone C z with vertex at ∂Ω, for each x 0 such that d(x 0 , Γ) ≤ δ, we take z = x 0 − 5δ/2e n .Since Γ is a Lipschitz graph with Lipschitz constant L < 1/16, d(z, Γ) ≥ 5δ/(2 √ L 2 + 1) − δ > δ, so again z and the analogous downwards cone C z lie in the region where u ≥ ρ κ .Moreover, The rest of the proof continues analogously.

Applications to free boundary problems
5.1.C 1,α regularity of the free boundary in the obstacle problem.Consider the classical obstacle problem (1.5) in B 1 , with f ≥ τ 0 > 0, f ∈ W 1,q , and assume that 0 is a free boundary point.We will show that we can extend the proof of the C 1,α regularity of the free boundary due to Caffarelli [Caf98] to the case f ∈ W 1,q thanks to our new result.We generalize the steps of the proof in [FR20, Section 5.4].
Our starting point will be the existence of a regular blow-up.We will also take for granted the following nondegeneracy condition: if x 0 ∈ {u > 0}, sup Br(x 0 ) u ≥ cr 2 , which follows easily from the fact f ≥ τ 0 > 0; see [FR20, Proposition 5.9].
Proposition 5.1.Let u be a solution of (1.5), with f ∈ W 1,n and f ≥ τ 0 > 0. Assume that 0 is a regular free boundary point as in Definition 1.5.
Then, for every L 0 > 0 there exists r > 0 such that the free boundary is the graph of a Lipschitz function in B r with Lipschitz constant L < L 0 .
We will denote u r (x) := u(rx) r 2 .Observe that the blow-up hypothesis implies that for all ε > 0, there exists r 0 such that To prove that the free boundary is Lipschitz, we will use the interior and exterior cone conditions, and to do this we will prove that ∂ ν u r ≥ 0, with ν a unit vector, when ν • e > c(L), where c(L) is the positive constant that ensures that the cone {x ∈ R n : x • e = |x|c(L)} has Lipschitz constant L. We need a positivity lemma.
Proof.First, it is clear that w > 0 in Ω \ N δ .Suppose there exists x 0 ∈ B 1/2 ∩ N δ such that w(x 0 ) < 0. We will arrive at a contradiction using the maximum princple, combined with the ABP estimate, with the function Consider the set Ω ∩ B 1/4 (x 0 ).On ∂Ω, u r = 0, hence v ≥ 0. On Notice that u r C 1 is uniformly bounded as r → 0. Hence, choosing η small enough, the second inequality implies v ≥ M/2.For the first inequality, choosing now small enough ε and δ, we obtain v ≥ η/(64n).
Using the lemma, we prove that there is an arbitrarily wide cone of directions where ∂ ν u r ≥ 0, for small r > 0.
Proof of Proposition 5.1.We only need to check that, for any ν ∈ S n−1 such that ν • e > c(L), the hypotheses of the lemma hold.By construction, we only need to check that (5.1) holds for a small enough r > 0.
Let δ = ε 1/8 .By the blow-up, there exists r > 0 such that Moreover, u r = 0 in {x • e < −δ 2 }, as we prove by contradiction from the nondegeneracy.Suppose u r (y) > 0 for some y such that y • e < −δ.Then, sup B δ 2 (y) u r ≥ cδ 4 = cε 1/2 , but since B δ 2 (y) ⊂ {x • e < 0}, u r < ε, which cannot happen if ε is small enough.Hence, the free boundary is contained in the strip {|x • e| < δ 2 }.Now, let a unit ν such that ν • e > c(L).The lower bound for ∂ ν u r in N δ only takes into account the convergence of the blow-up, On the other hand, if z ∈ Ω \ N δ , since the free boundary is at a distance lower than δ 2 from the hyperplane {x • e = 0}, z • e > δ − δ 2 .Hence, where M > 0 provided that ε is small enough.
In the interior of the cone, there are no free boundary points because u is 0 in a neighbourhood of all points.This is the interior cone.To check the exterior cone condition, suppose there is another free boundary point x 1 in the set B r ∩ {x 0 + tν : ν • e > c(L), t ∈ R + }.Then, by applying the interior cone condition to x 1 , we get that x 0 cannot be a free boundary point, a contradiction.This proves that, in B r , the free boundary is a Lipschitz graph with constant L in the direction e.Now we can use our new boundary Harnack inequality to prove the C 1,α regularity of the free boundary at regular points à la Caffarelli.To do this, we must ask the right hand side to belong to W 1,q with q > n, which is slightly more restrictive and implies that f is Hölder continuous.
Proof of Corollary 1.6.As it is customary in this kind of proof, we will use the boundary Harnack with the derivatives of u r .Let L = L 0 (q, n, 1, 1)/2 with the L 0 defined in Corollary 1.4.From Proposition 5.1, there exists r > 0 such that the free boundary is a Lipschitz graph with constant L in B r .Assume without loss of generality that the direction of the graph is e = e n , and that L < 1.
For i = 1, . . ., n − 1, consider the functions They are both solutions of ∆w j = g j , with g 1 (x) = r∂ i f (rx), g 2 (x) = r∂ n f (rx).Moreover, w 2 is positive.To be able to use the boundary Harnack, we need to see that the right hand is small.Indeed, taking r → 0, g j L q (B 1 ) ≤ r∇f (rx) L q (B 1 ) = 2r 1−n/q ∇f L q (Br) → 0.
Thus, we can normalize w j dividing by w j (e n /2) and the right hand side still converges to 0 in norm.
Let Ω r = {u r > 0}.By the boundary Harnack with right hand side, Theorem 1.2, The unit normal vector to any level set {u r = t}, t > 0, is, by components, As this expression is C 0,α up to the boundary, this proves the normal vector to the free boundary is C 0,α , and by a simple calculation it follows that the free boundary is C 1,α .5.2.C 1,α regularity of the free boundary in the fully nonlinear obstacle problem.Consider the fully nonlinear obstacle problem in the general version (1.7), under the assumptions in Corollary 1.7.
Our starting point will be the existence of a regular blow-up in the sense of Definition 1.5, i.e., there exists r k ↓ 0 such that for some γ > 0 and e ∈ S n−1 .We also need the classical nondegeneracy condition: if x 0 ∈ {u > 0}, sup From here, we will extend the proof of [IM16] to the case where f ∈ W 1,q (and not necessarily Lipschitz), and we will also prove C 1,α regularity of the free boundary.
Our first step is an analogue to [IM16, Lemma 3.7] for the case f ∈ W 1,n .
The proof is the same as in our source, except for the final step.We provide it here for the convenience of the reader.
Remark 5.4.For this lemma, the case q = n, i.e., when f ∈ W 1,n and F ∈ W 1,n with respect to the second variable, is also true.
Proof.Let x ∈ {u > 0} and ∂ 1 F (M, x) denote the subdifferential of F at the point (M, x) with respect to the first variable.Since F is convex in M, then ∂ 1 F (M, x) = ∅.Consider a measurable function P M with P M (x) ∈ ∂ 1 F (M, x).Since f ∈ C α , by interior regularity estimates u ∈ C 2,α loc ({u > 0}), and thus we can define the measurable coefficients Since F is convex in the first and F (0, x) ≡ 0, then for any unit vector ν, Suppose there exists y 0 ∈ Ω ∩ B 1/2 such that C 0 ∂ ν u(y 0 ) − u(y 0 ) < 0.Then, we consider the auxiliary function Then, Hence, by the ABP estimate, since R ∈ L n (B 1 ), By the scaling of the L n norm, the second term in the sum is bounded by C R L n (B r/4 (ry 0 )) → 0 as r → 0. On the other hand, w ≡ 0 on ∂Ω, and Therefore, choosing ε and r small enough we reach w > 0 in Ω ∩ B 1/4 (y 0 ), a contradiction.Now, as we show next, by the C 1 convergence of the blow-up we can fulfill the sufficient conditions in Lemma 5.3, and prove that the free boundary is Lipschitz at regular points, analogously to Proposition 5.1.Then, applying the boundary Harnack inequality, we can improve the regularity up to C 1,α .
Proof of Corollary 1.7.Let be the blow-up at 0. Let s ∈ (0, 1).Then, for any direction ν ∈ S n−1 such that ν • e ≥ s/2.From the C 1 convergence of the blow-up, there exists r k such that for some sufficiently small ρ > 0.
In particular, this shows that the free boundary fulfills the interior and exterior cone conditions in B ρ and therefore it is Lipschitz, with Lipschitz constant L(s), that satisfies L(s) → 0 as s → 0. Now, assume without loss of generality e = e n .For i = 1, . . ., n − 1, consider the functions w 1 = ∂ i u ρ and w 2 = ∂ n u ρ .
Notice that w 2 ≥ 0. Since F is C 1 with respect to D 2 u and F (D 2 u, x) ∈ W 1,q , then u ∈ W 3,q and we can commute the third derivatives as follows, Here, Lw = Tr(A(x)w), with A(x) = (F ij (D 2 u ρ , ρx)) ij .Hence, w 1 and w 2 are both solutions of To be able to use the boundary Harnack, we need to show that the right hand is small.Indeed, taking ρ → 0, Finally, by the blow-up convergence, w j (e n /2) > γ/2 − ε > γ/4, w j (e n /2) < γ/2 + ε < γ.
Thus, we can normalize w j dividing by w j (e n /2) and the right hand side still converges to 0 in norm.
Let Ω ρ = {u ρ > 0}.By the boundary Harnack with right hand side, Theorem 1.2, The unit normal vector to any level set {u ρ = t}, t > 0, is, by components, As this expression is C 0,α up to the boundary, this proves the normal vector to the free boundary is C 0,α , and it follows that the free boundary is C 1,α .5.3.C 1,α regularity of the free boundary in the fully nonlinear thin obstacle problem.Recall the fully nonlinear thin obstacle problem (1.9), under the assumptions in Corollary 1.9.
We will denote u = v − ϕ.In this case, we know the following.

Proposition 5.5 ([RS16]
). Assume that 0 is a regular free boundary point for (1.9), where F is uniformly elliptic, convex and F (0) = 0, and ϕ ∈ C 1,1 .Then, there exists e ∈ S n−1 ∩ {x n = 0} such that for any L > 0 there exists r > 0 for which In particular, the free boundary is Lipschitz in B r , with Lipschitz constant L. Now, using our new boundary Harnack in slit domains, Theorem 4.2, on top of this proposition, we derive the C 1,α regularity of the free boundary at regular points.
For i = 1, . . ., n − 2, consider the functions Since F ∈ C 1 and F (D 2 v) = 0, then v ∈ W 3,p for all p < ∞ and we can commute the third derivatives as follows, Moreover, w 2 is positive.Then, using that v = u + ϕ, where Lw = Tr(AD 2 w), A = (F ij • D 2 u) ij .Now, we will check that, after a scaling, w 2 (e n /2) ≥ 1 and the right hand side becomes arbitrarily small.Define w2 (x) = w 2 (sx) s and φ(x) = ϕ(sx) s 2 .Now, we check that the right hand side is as small as required.Indeed, letting s k → 0, The right hand side becomes arbitrarily small in the equation for w 1 analogously.Then, since 0 is a regular free boundary point, by the convergence of the blow-up, w2 (e n−1 /2) → ∞ for a sequence of values {s k } → 0. Now, by the interior Harnack inequality combined with the ABP estimate, since w2 ≥ 0 in Ω and the distance between the segment joining e n−1 /2 and ±e n /2 and the contact set is positive and larger than some constant c(L, n) only depending on the Lipschitz constant of the free boundary and the dimension, for some positive c 1 and c 2 only depending on the dimension, L, λ and Λ.Therefore, letting s k → 0, w2 (±e n /2) ≥ 1.If w1 L 1 > 1, we normalize it (notice that this step can only make the right hand side smaller).Thus, by the boundary Harnack inequality with right hand side for slit domains, Theorem 4.2, w 1 /w 2 ∈ C 0,α in Ω ∩ B s/2 .Thus, ∂ i u/∂ n−1 u ∈ C 0,α .Now, the unit normal vector to any level set in the thin space {x n = 0} ∩ {u = t} with t > 0 is, by components, Then, letting t → 0 + , we recover that the normal vector to the free boundary is C 0,α , and hence the free boundary is a C 1,α graph.

Sharpness of the results
We construct two examples that show that: • Without the smallness assumption on the Lipschitz constant of the domain, Theorem 1.2 fails.• For q = n, Theorem 1.2 fails.
• For divergence form operators, if the coefficients are only bounded and measurable, Theorem 1.2 fails.As a first observation, see [AS19], take for instance Ω = {x n > 0} ⊂ R n , and let u(x) = x n , v(x) = x 2 n .These functions are normalized in the sense that u(e n ) = v(e n ) = 1, and vanish contiuously on ∂Ω.Even in a flat domain, a function with a too large Laplacian, |∆v| = 2, will never be comparable to a harmonic function near the boundary.Hence, the right hand side of the equation must be small, otherwise the result fails.
The following example in two dimensions shows that if we ask ∆v to be small in L q norm, for any q > n there is a cone narrow enough such that we can find harmonic functions that are not comparable with v.Moreover, if we consider a fixed cone, there exists q > n such that the L q boundedness of the right hand side is not enough to have a boundary Harnack.If q = n, such counterexamples are valid for any cone.Proposition 6.1.Let L > 0, q > 0, and assume π 2 arctan(1/L) + 2 q > 2. (6.1) Then, for every δ > 0, there exists a cone Ω ⊂ R 2 with Lipschitz constant L, and positive functions u, v that vanish continuously on ∂Ω such that u(0, 1) = v(0, 1) = 1, ∆u = 0 and ∆v L q (Ω) < δ, but sup In particular, Theorem 1.2 fails for q = n.
Remark 6.2.Since arctan(1/L) ∈ (0, π/2), the first term in the condition (6.1) is always greater than 1, hence, if q ≤ 2 there are always counterexamples to the boundary Harnack with right hand side bounded in L q .On the other hand, if L > 1, arctan(1/L) < π/4, and the condition is fulfilled for all q > 0 and q = ∞.
The limiting case L = 0, q = 2 (or q = n in higher dimensions) corresponds to domains that are locally a half-space.We have not considered this particular case in our setting.The existence of such example shows that, to have a boundary Harnack inequality for equations with a right hand side, we need the Lipschitz constant of the boundary to be sufficiently small, and also the right hand side to be small compared to the values of the function.It also shows a trade-off between the maximum possible slope of the boundary and the exponent of the L q boundedness of the right hand side.
On the other hand, it is impossible to have a boundary Harnack for equations with right hand side in Lipschitz domains with narrow corners, even less in Hölder domains or more general domains, under the reasonable hypothesis ∆u = f ∈ L ∞ , with f L ∞ small.However, the boundary Harnack holds for divergence form operators in Lipschitz domains with big Lipschitz constants when the right hand side vanishes as a big enough power of the distance [AS19].It is likely that we could prove the same for non-divergence form operators, but we will not pursue this because it cannot be used in the context of free boundary problems.
The following example is based in a counterexample to the Hopf lemma for divergence operators with discontinuous coefficients [Naz12], and shows that the boundary Harnack for equations with a right hand side fails in this setting.Proposition 6.3.There exists L in divergence form with discontinuous coefficients and positive functions u, v in {y > 0} ⊂ R 2 that vanish continuously at {y = 0} such that u(1, 1) = v(1, 1) = 1, Lu = 0 in {y > 0} and Lv L ∞ (B + 1 ) < δ, for any given δ > 0, but sup In particular, Theorem 1.2 fails if the divergence form operator has discontinuous coefficients.
It is easy to check that L is uniformly elliptic and that u(x, y) = y 3 + 18|x|y 2 + 72x 2 y 91 is a solution of Lu = 0. Now we will define v as a perturbation of u, in a similar way as in Proposition 6.1.We will use that the coefficients A(x, y) are constant in the positive quadrant.
Now, we construct v as the following infinite sum, that converges uniformly.
Since the supports of ψ 2 −k are disjoint, v ≥ 0. On the other hand, as k 0 → ∞.Hence, we can choose k 0 big enough so that Lv L ∞ (B + 1 ) < δ.To end, for k ≥ k 0 , u(2 We can use this result to prove a generalized Hopf lemma for the solutions of non-divergence equations with small right hand side. Corollary 7.2.Let q > n and L in non-divergence form as in (1.2).There exist small c 0 > 0 and L 0 > 0 such that the following holds.
Let Ω be a Lipschitz domain as in Definition 1.1, with Lipschitz constant L < L 0 .Suppose further that ∂Ω is a C 1,Dini graph.Let v be a solution of in the L n -viscosity, with v > 0 in Ω ∩ B 1 and f L q (B 1 ) ≤ c 0 v(e n /2).
Then, for any l = (l 1 , . . ., l n ) ∈ R n with |l| = 1 and l n > 0, v(rl) ≥ cl n v(e n /2)r, r ∈ (0, δ), where c, c 0 and δ are positive and depend only on the dimension, λ, Λ and the modulus of continuity of the domain.

1. 4 .
Applications to obstacle problems.The boundary Harnack inequality is the technical tool that allows us to prove C 1,α regularity of the free boundary once we know it is Lipschitz in the classical obstacle problem with constant right hand side [FR20, Section 5.6] and in the thin obstacle problem with zero obstacle [Fer20, Section 5].The functions to which we apply the boundary Harnack are derivatives of the solution to the free boundary problem.Hence, if the original free boundary problem is the classical obstacle problem, ∆u = f χ {u>0} u ≥ 0, (1.5) the derivatives of u are solutions of ∆(∂ ν u) = ∂ ν f in {u > 0} ∂ ν u = 0 on ∂{u > 0}, and we can apply the boundary Harnack of Allen and Shahgholian if f ∈ W 1,∞