Hindawi Publishing Corporation Boundary Value Problems Volume 2007, Article ID 78029, 17 pages doi:10.1155/2007/78029 Research Article A Boundary Harnack Principle for Infinity-Laplacian and Some Related Results Tilak Bhattacharya Received 27 June 2006; Revised 27 October 2006; Accepted 27 October 2006 Recommended by Jos ´ e Miguel Urbano We prove a boundary comparison principle for positive infinity-harmonic functions for smooth boundaries. As consequences, we obtain (a) a doubling property for certain pos- itive infinity-harmonic functions in smooth bounded domains and the half-space, and (b) the optimality of blowup rates of Aronsson’s examples of singular solutions in cones. Copyright © 2007 Tilak Bhattacharya. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction In this work, one of our main efforts is to prove a boundary Harnack principle for positive infinity-harmonic functions on domains with smooth boundaries. This will generalize the result in [1] proven for flat boundaries. In this connection, also see [2–5]. This result will also be applied to study some special positive infinity-harmonic functions defined on such domains. One could refer to these as infinity-harmonic measures, however, b eing solutions to a nonlinear equation, these are not true measures. We derive some properties of these functions and among these would be the doubling property. A decay rate and a halving property for such functions on the half-space will also be presented. Another application will be to show optimality of Aronsson’s singular examples in cones, thus generalizing the result in [6, 7]. We now introduce notations for describing our results. Let Ω ⊂ R n , n ≥ 2, be a domain with boundary ∂Ω.Wesayu is infinity-harmonic in Ω if u solves in the sense of viscosity Δ ∞ u = n  i, j=1 D i u(x)D j u(x)D ij u(x) =0, x ∈ Ω. (1.1) For more discussion, see [8, 1, 9]. For a motivation for these problems, see [8, 10]. For 2 Boundary Value Problems r>0andx ∈ R n , B r (x) will be the open ball centered at x and has radius r.Let  A denote the closure of the set A and let χ A denote its characteristic function. Define Ω r (x) = Ω ∩ B r (x), P r (x) = ∂Ω ∩B r (x). We will assume throughout this work that ∂Ω ∈ C 2 .More precisely,wefirstdefineforeveryx ∈ ∂Ω R x to be the radius of the largest interior ball tangential to Ω at x. We will assume that R y > 0foreveryy ∈ ∂Ω and R x ≥ R y /2 > 0, x ∈ P δ y (y), for some δ y > 0. For every x ∈ ∂Ω,setν x to be the inner unit normal at x and x s = x + sν x , s>0. We will now state Theorem 1.1 which is the result about boundary Harnack principle [2, 1, 3, 4]. Theorem 1.1 (Boundary Harnack Principle). Let Ω be a domain in R n , n ≥ 2,with∂Ω satisfying the interior ball condition as stated above. Let u and v be infinity-harmonic in Ω. Suppose that y ∈ ∂Ω, 0 < 4δ ≤inf(δ y , R y /2),andu,v>0 in Ω δ y (y).Supposethatu,andv vanish continuously on P δ y (y), then there exist positive constants C, C 1 , C 2 independent of u, v,andδ, such that for every z ∈ Ω δ (y), (i) u(z) ≤ cu(y δ ), (ii) c 1 u(y δ )/v(y δ ) ≤ u(z)/v(z) ≤c 2 u(y δ )/v(y δ ). Inequality (i) is often referred to as the Carleson inequality. A proof is provided in Section 2. At this time, we are unable to determine if Theorem 1.1 also holds when Ω has Lipschitz continuous boundary. We will apply Theorem 1.1 to prove (a) the doubling property of solutions of (1.2), and (b) the optimality of blowup rates of the Aronsson singular functions in cones [6]. Let Ω be a bounded domain. Fix y ∈ ∂Ω;foreveryr>0, define Q r (y) =∂Ω \  P r (y). Consider the problem Δ ∞ u(x) =0, x ∈ Ω, u(x) = 1, x ∈ P r (y), u(x) = 0, x ∈ Q r (y). (1.2) By a solution u of (1.2), we mean that (i) u is infinity-harmonic, in the viscosity sense, in Ω, and (ii) u assumes the values 1 and 0 continuously on P r (y)andQ r (y). More precisely, if x ∈ P r (y)andz → x, z ∈ Ω,thenu(z) → 1, and analogously for Q r (y). We show the ex- istence of bounded solutions of (1.2)inLemma 3.1.Onecouldrefertou as the nonlinear infinity-harmonic measure in Ω (although we have not shown uniqueness). Clearly, it is not a true measure. Our motivation for studying such quantities arises from the works [2–5]. In the context of boundary behavior, for instance the Fatou theorem, the works [4, 5] have studied such solutions for the linearized version of the p-Laplacian for finite p. We will show that requiring boundedness implies the maximum principle and com- parison, see Lemma 3.1.LetH ={x ∈R n : x n > 0} denote the half-space in R n .Sete n to be the unit vector along the positive x n -axis. Set T ={x ∈R n : x n = 0};fory ∈T,define P r (y) =T ∩B r (y), Q r (y) =T \  P r (y), and M u y (ρ) = sup ∂B ρ (y)∩H u. Define a solution u of Δ ∞ u(z) = 0, z ∈ H, u| P r (y) = 1, u| Q r (y) = 0, (1.3) to be infinity-harmonic in Ω, in the sense of viscosity, 0 ≤ u ≤ 1, continuous up to P r (y) and Q r (y), and lim sup ρ→∞ M u y (ρ) = 0. We will address the existence and uniqueness of such solutions in Lemma 3.4. We now state a result about the doubling property of solu- tions of (1.2)and(1.3). For r>0, set o 3r = 3re n . Tilak Bhattacharya 3 Theorem 1.2. (a) Let Ω ⊂ R n be a bounded domain. For y ∈ ∂Ω, assume that P r (y) lies on a connected component of ∂Ω.Letu r be a bounded solution of (1.2)inΩ and let r be small. Then there are positive constants c, C independent of r, such that u r (y r ) ≥ c and u 2r (z) ≤ Cu r (z), z ∈Ω \B 3r (y). (1.4) (b) Let H be the half-space in R n .Letu r o be the unique infinity-harmonic measure in H. Then there exist universal constants C 1 > 0 and 0 <C 2 < 1 such that u 2r o (z) ≤ C 1 u r o (z), z ∈H \B 3r (o), u r o  o s  ≤ C 2 u 2r o  o s  , s ≥ 3r. (1.5) Estimates in Theorem 1.2 are well known for linear equations [3] and also for the linearized version for the p-Laplacian [4, 5]. While we are able to prove the doubling property for any C 2 domain (see Lemma 3.3), it is unclear how a halving property (i.e., f (t) ≤ cf(2t), f positive, increasing, and c<1) may be proven if tr ue. In particular, it would be interesting to know if this is true when Ω is the unit ball. We now introduce notations for Theorem 1.3.Forα>0, let C α stand for the interior of the half-infinite cone in H,withapexato,thex n -axis as the axis of symmetry, and aperture 2α.For r>0, let M u (r) = sup z∈∂B r (o)∩C α u(z). We extend the result in [7] to show optimality of the Aronsson singular examples [6]. Theorem 1.3. For α>0,letC α be as described above. Let u, v be positive infinity-harmonic functions in C α . Assume that (i) both u and v vanish continuously on ∂C α \{o}, (ii) sup r>0 M u (r) =∞, sup r>0 M v (r) =∞,and(iii) lim r→∞ M u (r) = lim r→∞ M v (r) = 0. Then there exists a constant C, depending on α, u,andv such that 1 C ≤ u(z) v(z) ≤ C, z ∈ C α . (1.6) Moreover, for e very m = 1,2, 3, , if α = π/2m and ω is a direction in C π/2m ,thenforan appropriate  C =  C(ω), 1  C|z| m 2 /(2m+1) ≤ u(z) ≤  C |z| m 2 /(2m+1) , z ∈ C π/2m with z =|z|ω. (1.7) ThelastconclusioninTheorem 1.3 will follow from the works [6, 7]. While Theorem 1.3 applies to special situations, the main purpose is to understand better the blowup rates of singular solutions, and in some situations decay rates. We now state some well-known results that will be used in this work. Let u>0be infinity-harmonic in a domain Ω, suppose that a,b ∈ Ω such that the segment ab is at least η>0awayfrom∂Ω, then the following Harnack inequality holds: u(a)e |a−b|/η ≥ u(b). (1.8) 4 Boundary Value Problems Let B r (a) ⊂ Ω,ifω is a unit vector and 0 ≤t ≤ s<r,then u(a + tω) r −t ≤ u(a + sω) r −s , u(a + sω)(r −s) ≤ u(a + tω)(r −t). (1.9) We wil l re fer to (1.9) as the monotonicity property of u.For(1.8)and(1.9), see [8, 1, 11, 7, 12, 13]. Moreover, u is locally Lipschitz (C 1 if n = 2[14]) and satisfies the comparison principle [15]. Finally, we mention that it is unclear if a boundary Holder continuity of the quotient of two infinity-harmonic functions holds for smooth domains. Such a result for general Lipschitz domains would undoubtedly be quite useful. For p-harmonic functions (finite p), we direct the reader to the recent work by John Lewis and Kaj Nystrom “Boundary Behaviour for p Harmonic Functions in Lipschitz and Starlike Lipschitz Ring Domains.” We thank John Lewis for sending us this work. 2. Proof of Theorem 1.1 Our proof is an adaptation of the methods developed in [2, 1, 3]. Since Δ ∞ is t ranslation and rotation invariant, we may assume that the origin o ∈ ∂Ω.Setosc A u = sup z∈A u(z) − inf z∈A u(z) to be the oscillation function of u on the set A.RecallthatΩ r (y) = Ω ∩ B r (y), y ∈∂Ω. Step 1 (oscillation estimate near the boundary). Let u>0 be infinity-harmonic in Ω and vanishing on a neighborhood of o,in∂Ω.LetM u (r) = sup z∈Ω r (o) u(z). By the max- imum principle, M u (r) > 0andu(z) ≤ M u (r), z ∈ Ω r (o). For 0 <α≤ β, consider the function w(z) = M u (α)+[M u (β) −M u (α)](|z|−α)/(β −α), z ∈Ω β \Ω α .Clearly,u ≤ w on ∂(Ω β \Ω α ). Thus u ≤ w in Ω β \Ω α .Thus M u (γ) ≤M u (α)+  M u (β) −M u (α)  γ −α β −α , α ≤ γ ≤ β. (2.1) This implies that osc Ω r (o) u = M u (r)isconvexinr.Sinceu(o) = 0, it follows that 2osc Ω r (o) u ≤ osc Ω 2r (o) u. Step 2 (Carleson inequality). We now use the interior ball condition. Since ∂Ω ∈ C 2 , R x ≥ R o /2, x ∈ P 4δ (o), with 4δ<inf(δ y ,R o /2). For every x ∈ ∂Ω,letν x denote the unit inner normal at x,andsetx t = x + tν x ,0≤t ≤ R x .Wewillprovethatu(z) ≤Cu(o δ ), z ∈ Ω δ (o). We will adapt a device, based on the Harnack inequality, from [3]. For z ∈ Ω δ ,definex z ∈ ∂Ω to be the point nearest to z.Alsosetd(z) =|x z −z|.Thenz = x z + d(z)ν x z = (x z ) d(z) ; set z s = x z +2 s−1 d(z)ν x z , s =1,2,3, By the Harnack inequality (1.8), for z ∈ Ω 3δ (o), u(z) ≤ ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ Mu  z 2  :0<d(z) < 3δ 2 , Mu  o δ  : δ<d(z) < 3δ. (2.2) Tilak Bhattacharya 5 We take M = e 8 . We now make an observation which will be used repeatedly in what follows. If d(z) ≥ δ/2 s ,then u(z) ≤ Mu  z 2  ≤···≤ M s u  z s  ≤ M s+1 u  o δ  . (2.3) Suppose now that there is a ξ 0 ∈ Ω δ (o)suchthatu(ξ 0 ) ≥ M l+2 u(o δ ), where l = l(δ)islarge and will be determined later. Using the aforementioned observation, we obtain dist  ξ 0 ,∂Ω  ≤ δ 2 l . (2.4) Let p 0 ∈ ∂Ω be the nearest point to ξ 0 .Clearly,ξ 0 ∈ Ω 2 −l δ (p 0 ) ⊂ Ω 2δ (o). Thus, osc Ω δ2 −l (p 0 ) u ≥ u(ξ 0 ); thus by Step 1,form = 1,2,3 , osc Ω δ2 −l+m (p 0 ) u ≥ 2 m osc Ω δ2 −l (p 0 ) u ≥ 2 m u  ξ 0  , (2.5) where 2 m ≥ M 3 = e 24 . Select m = 60; thus osc Ω δ2 −l+m (p 0 ) u ≥ 2 m u(ξ 0 ) ≥ M l+5 u(o δ ). Thus there is a ξ 1 ∈ Ω δ2 −l+m (p 0 )suchthatu(ξ 1 ) ≥ M l+5 u(o δ ). Arguing as done in (2.4), we see dist(ξ 1 ,∂Ω) ≤ δ2 −l−3 . Letting p 1 ∈ ∂Ω to be closest to ξ 1 ,weseethatp 1 ∈ Ω 2δ (o). Repeat- ing our previous argument, osc Ω δ2 −l−3+m (p 1 ) u ≥ 2 m osc Ω δ2 −l−3 (p 1 ) u ≥ 2 m u  ξ 1  ≥ M l+8 u  o δ  . (2.6) Thus we may find a ξ 2 ∈ Ω δ2 −l−3+m (p 1 )suchthatu(ξ 2 ) ≥ M l+8 u(o δ ), and dist(ξ 2 ,∂Ω) ≤ δ2 −l−6 . Thus we obtain a sequence of points ξ k ∈ Ω and p k ∈ ∂Ω, k =1,2,3 ,suchthat u  ξ k  ≥ M l+2+3k u(o δ ), dist  ξ k ,∂Ω  ≤ δ2 −l−3k , ξ k ∈ Ω δ2 −l−3(k−1)+m  p k−1  . (2.7) Note that   ξ k −o   ≤ k−1  i=1   ξ i+1 −ξ i   +   ξ 0 −o   ≤ δ  1+2 k−1  i=0 2 −l−3i+m  . (2.8) Choose l ≥ 70, then |ξ k −o|≤2δ. Noting that u vanishes continuously on ∂Ω and letting k →∞in (2.7) result in a contradiction. Thus the Carleson inequality in Theorem 1.1 follows. Step 3 (bounds near the boundary). We first derive a lower bound in terms of the distance to the boundary. For every z ∈ Ω δ (o), let x z and d(z)beasinStep 2.Notethatd(z) ≤|z − o|≤δ.Thusx z ∈ Ω 2δ (o). Call ζ z = x z + δν x z ,observethatζ z ∈ Ω 3δ (o). By monotonicity (1.9) and the interior ball condition, we have u(z) d(z) ≥ u  ζ z  δ ≥ e −6 u  o δ  δ , (2.9) since |ζ z −o δ |≤|x z + δν x z −δν o |≤4δ. Let z ∈ Ω δ (o). As noted previously, x z ∈ Ω 2δ (o)andΩ δ (x z ) ⊂ Ω 3δ (o). Note that z ∈ Ω δ (x z ). Set μ z = sup Ω δ (x z ) u, then by comparison u(ξ) ≤ μ z |ξ −x z |/δ, ξ ∈ Ω δ (x z ). Thus 6 Boundary Value Problems u(z) ≤ μ z d(z)/δ. By the Carleson inequality, μ z ≤ Cu(ζ z ). Note that |x δ − o δ |=|x z + δν x −δν o |≤4δ. By the Harnack inequality, u(ζ z ) ≤ e 4 u(o δ ). Thus there is universal  C, such that u(z) d(z) ≤  C u  o δ  δ , z ∈ Ω δ (o). (2.10) If u, v are two positive infinity-harmonic functions in Ω 4δ (o), then by (2.9)and(2.10), there exist universal constants C 1 and C 2 such that C 1 u  o δ  v  o δ  ≤ u(z) v(z) ≤ C 2 u  o δ  v  o δ  , z ∈ Ω δ (o). (2.11) This proves Theorem 1.1. Remark 2.1. We comment that the distance function d(z) = dist(z,∂Ω), z ∈ Ω,isC 2 and infinity-harmonic near ∂Ω. Also the oscillation estimate in Step 1 continues to hold for Lipschitz boundaries. One could show a Carleson inequality by following the ideas in [2]. 3. Proof of Theorem 1.2 In this section, we will assume that Ω is a bounded C 2 domain. For y ∈∂Ω and r>0, re- call the definitions of P r (y)andQ r (y). Note that both P r (y)andQ r (y) are relatively open in ∂Ω.Letu be a solution of (1.2). As in Section 2,forx ∈ ∂Ω, ν x and x t = x + tν x , t>0, are as defined in Section 2. We will assume that Ω is bounded but we can extend our arguments to the case of the half-space H.Wewillalwaystakeu to be bounded in this section. This w ill imply the maximum principle. At this time, it is not clear whether un- bounded solutions to (1.2) exist. Let C y be the connected component of ∂Ω that contains y.InLemma 3.1, we assume that B r (y) ∩∂Ω = B r (y) ∩C y . Lemma 3.1. Let Ω ∈ C 2 be a bounded domain. Let y ∈∂Ω and r>0. The following holds. (i) There exists a solution u of the problem in (1.2) such that 0 <u<1 in Ω. (ii) If v is any bounded solution of (1.2), then 0 <v<1 in Ω. (iii) There are a maximal solution u r y and a minimal solution u r y ,inΩ such that if v is any bounded solution of (1.2), then u r y ≤ v ≤ u r y . (iv) If t<r, then u t y ≤ lim ρ↑r u ρ y =  u r y ≤ u r y = lim r↓r u r y . Moreover, u r y satisfies the following compar ison principle: if ω, w ∈ C(  Ω) are infinity- harmonic, and ω ≤ u r y ≤ w on ∂Ω, then ω ≤ u r y ≤ w in Ω. Proof. Fix y ∈ ∂Ω and r>0. We have broken up our proof into five steps. We first start with the existence of bounded solutions. Step 1 (existence). We use the existence results proven in [8, 15], for Lipschitz bound- ary data. Let η>0besmall.SetI r (y) = ∂B r (y) ∩ ∂Ω,andfort>0, set S t = P r (y) ∪ (  x∈I r (y) B t (x) ∩ ∂Ω). The set S t is obtained by appending a t-band to P r (y). For l = 1,2,3, ,let f l be such that (i) f l ∈ C(∂Ω), (ii) f l (x) = 1, x ∈ P r (y), Tilak Bhattacharya 7 (iii) f l (x) = 0, x ∈ ∂Ω \S η/l , (iv) f l (x) = (η/l) −dist(x,P r (y))/(η/l), x ∈ S η/l . Now let u l ∈ C(  Ω) be the unique viscosity solution of the problem Δ ∞ u l (z) = 0, z ∈ Ω, u l | ∂Ω = f l . (3.1) Clearly, 0 <u l < 1inΩ.Since f l ≥ f l+1 , by comparison, there is a function u η such that u l ↓ u η in Ω. We first show that if x ∈ P r (y)andz → x ∈ P r (y), z ∈ Ω,thenu η (z) → 1. Consider the set Ω δ (x), where δ = inf ξ∈Q r (y) |x −ξ|/2. For z ∈Ω δ (x), set w(z) = 1 −|z − x|/δ. By comparison, for every l, w ≤u l ≤ 1inΩ δ (x). Thus 1 −|z −x|/δ ≤ u η (z) ≤ 1, z ∈ Ω δ (x). We see that lim z→x u η (z) = 1. For x ∈ Q r (y)andδ = inf ξ∈P r (y) |ξ −x|/2, it is clear that for l large, u l (z) ≤|z −x|/δ, z ∈ B δ (x) . Thus u η (z) → 0asz → x. Moreover, the limit function u η does not depend on the width η of the appended band S η .Anargument based on comparison shows easily that for any η 1 , η 2 > 0, u η 1 = u η 2 .Setu = u η .Our next step is to show that u is a viscosity solution in Ω. We first show that u is locally Lipschitz in Ω. To see this, take x 1 ∈ Ω and t>0suchthatB 4t (x 1 ) ⊂ Ω. Select x 2 ∈ B t (x 1 ); set μ l = sup B 4t (x 1 ) u l . Applying monontonicity (1.9)inB t (x 1 ), we have for every l,(μ l − u l (x 1 ))/t ≤ (μ l −u l (x 2 ))/(t −|x 1 −x 2 |). Rearranging terms (see [1, Lemma 3.6], also see [12]), noting that u l (x 1 ), u l (x 2 ) ≥ 0andμ l ≤ μ 1 ≤ 1, we obtain |u l (x 2 ) −u l (x 1 )|/|x 1 − x 2 |≤1/t. Fixing x 1 , x 2 and letting l →∞,weobtainthatu is locally Lipschitz. Fix ξ ∈ Ω and for 0 ≤ t<dist(ξ, pΩ), set M l (t) = sup B t (ξ) u l , m l = inf B t (ξ) u l , M(t) = sup B t (ξ) u,and m(t) = inf B t (ξ) u. Using that (i) u k ≤ u j ≤ u l when l<j<k, (ii) M l is convex and m l is concave in t, it follows that for a<c<band z ∈ ∂B c (ξ), b −c b −a m k (a)+ c −a b −a m k (b) ≤ m k (c) ≤ u j (z) ≤ M l (c) ≤ b −c b −a M l (a)+ c −a b −a M l (b). (3.2) Now in (3.2)firstlettingk →∞,replacingu j (z)byu(z), and then letting l →∞,weobtain that M(t)isconvexandm(t) is concave. This implies that u is a viscosity solution [8]. Part (i) now follows. A proof also could be worked by showing cone comparison. Throughout the rest of the proof, u will stand for the solution constructed in Step 1. Step 2 (comparison). We now prove an easy comparison result for u.Let f ∈ C(∂Ω)and let u f ∈ C(  Ω) be the unique infinity-harmonic function with boundary values f .Let f ≤ χ P r (y) . Using comparison, we see that for every l, u f ≤ u l in Ω.Thusu f ≤ u in Ω. Now let f ≥ χ P r (y) ,setε>0. Since f ≥ 1inP r (y), there exists a δ>0suchthat f + ε ≥ 1 in B δ (x) ∩∂Ω,foreveryx ∈ ∂Ω ∩∂B r (y). Take l large so that η/l ≤ δ/2. By comparison, u l ≤ u f + ε in Ω.Thuswehaveu ≤u f in Ω. Step 3 (maximum principle). We now prove part (ii). Let v be any bounded solution of (1.2). We will adapt an argument used in [11]. We observe that there is an R 0 > 0such that for x ∈ P 2r (y), R x ≥ R 0 , and consequently,  x∈P 2r (y) B R 0 /4 (x R 0 /4 ) ⊂ Ω.Inwhatfollows we take the quantities σ, η<R 0 /10. We exploit the special geometry of P r (y) to achieve our proof. Set I r (y) = ∂Ω ∩∂B r (y); for every x ∈ I r (y)andσ>0, define m x (σ) = inf ∂B σ (x)∩Ω v and M x (σ) = sup ∂B σ (x)∩Ω v.Clearly,m x (σ) ≤ 0andM x (σ) ≥ 1. We claim that M x is convex 8 Boundary Value Problems and m x is concave in σ. To see this, take z ∈ Ω with 0 <a≤|z −x|≤b.Setw(z) = m x (a)+ [m x (b) −m x (a)](|z −x|−a)/(b − a). Clearly, w ≤ 0. By comparison, w ≤ v in (B b (x) \ B a (x)) ∩ Ω.Thusm x (σ)isconcaveinσ, and one can show analogously that M x (σ)is convex. Define m y (σ) = inf x∈I r (y) m x (σ)andM y (σ) = sup x∈I r (y) M x (σ), then for σ>0, (i) M y (σ) ≥ 1isconvex, m y (σ) ≤ 0isconcaveinσ, (ii) m y (σ) ≤ v(z) ≤M y (σ), z ∈ Ω \∪ x∈I r (y) B σ (x), (iii) M y (σ) ↑, m y (σ) ↓ as σ ↓0. (3.3) Note that v = 0or1on∂Ω \  x∈I r (y) B σ (x). Thus (3.3)(i) follows easily. Now using (3.3)(i) and comparison in the set Ω \  x∈I r (y) B σ (x)yields(3.3)(ii). Clearly, M y (σ)(m y (σ)) is the supremum (infimum) of v on Ω \  x∈I r (y) B σ (x). The conclusion in (3.3)(iii) follows by observing that  x∈I r (y) B σ 1 (x) ⊂  x∈I r (y) B σ 2 (x), when σ 1 >σ 2 .By(3.3), the quanti- ties M(0) = lim σ→0 M y (σ)andm(0) = lim σ→0 m y (σ) exist. By our assumptions, −∞ < m(0) ≤ v ≤ M(0) < ∞. We show that m(0) = 0. Assume instead that m(0) < 0. Recall that v is continuous up to Q r (y)andP r (y). For x ∈ ∂Ω,letρ(x) = dist(x,P r (y)) and ρ(x) = dist(x,Q r (y)). For x ∈ Q r (y), define w x (z) = m(0)|z −x|/ρ(x) in the set Ω ρ(x) (x). By comparison w x ≤ v in Ω ρ(x) (x), and v ≥ m(0)/2, in Ω ρ(x)/2 (x). For x ∈ P r (y), define ω x (z) = 1+(m(0) −1)|z −x|/ ρ(x)inΩ ρ(x) (x). Then v ≥ ω x in Ω ρ(x) (x)andv ≥m(0)/2, in Ω ρ(x)/2 (x). Let η>0besmall.SetA η ={x ∈ ∂Ω : ρ(x) ≥ η} and B η ={x ∈ P r (y): ρ(x) ≥ η}. We now apply the above observations to obtain v(z) ≥ ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ m(0) 2 : z ∈ Ω ρ(x)/2 (x), x ∈A η , m(0) 2 : z ∈ Ω ρ(x)/2 (x), x ∈B η . (3.4) Set S =  η>0  x∈A η Ω ρ(x)/2 (x)andT =  η>0  x∈B η Ω ρ(x)/2 (x), and call G y = Ω \(S ∪T). For l = 1,2,3 ,letz l ∈ Ω be such that v(z l ) ≤ 7m(0)/8andv(z l ) → m(0), as l →∞.By (3.4), z l ∈ Ω \G y , and by the maximum principle, dist(z l ,I r (y)) →0. In the discussion that follows, we will assume that n>2. Recalling that I r (y)= ∂B r (y) ∩ ∂Ω, it follows that I r (y) is smooth. For every l,letx l ∈ I r (y) be the closest point to z l and d l =|x l −z l |. Note that the seg ment x l z l is normal to I r (y). Since x l ∈ ∂B r (y), yx l ⊥ ∂B r (y), and so yx l ⊥ I r (y). Let T l be the hyperplane tangential to ∂Ω at x l ,andlet Π l be the 2-dimensional plane containing the segments yx l and yz l .ThusΠ l ⊥ I r (y)atx l and ν x l lies in Π l .NotethatΠ l ⊥ T l and I r (y)istangentialtoT l at x l .CallJ l = ∂Ω ∩Π l , observe that the curve J l ⊥ I r (y)atx l . It is easy to see that if x ∈ J l is close to x l , then (i) ρ(x) =|x −x l | if x ∈ P r (y), and (ii) ρ(x) =|x −x l | if x ∈ Q r (y). Now consider the set C l = Π l ∩∂B d l (x l ) \G y . As noted above z l ∈ C l , moreover one can find α l ∈ C l such that v(α l ) = 3m(0)/4. We will apply the Harnack inequality in C l to obtain a contradiction. In (3.4), take η = d l and we observe the following. Since ∂Ω ∈ C 2 and x l ’s lie in a com- pact set, it follows that for q ∈ C l , dist(q,∂Ω) ≈ dist(q,T l ) = O(d l ), as d l → 0. In other words, dist(q,∂Ω) has a lower bound of the order of d l . We show this as follows. First note that since ∂Ω ∈ C 2 , it permits a local parametrization near x l ,wherex n = ν x l , x n = 0 is T l ,andx n = φ(x 1 , ,x n−1 ) describes ∂Ω. Clearly, dist(q,∂Ω) ≤|q −x l |=d l .Wewill Tilak Bhattacharya 9 show that (a) dist(q,∂Ω) ≥ dist(q,T l )+O(d 2 l ) and (b) dist(q,T l ) ≈ O(d l ), uniformly in l. (a) Let (i) q ∂Ω be the point on ∂Ω closest to q, (ii) let q T l be the point on T l closest to q, (iii) q int the point of intersection of the line, containing the seg ment qq ∂Ω ,andT l , and (iv) let q T l ∂Ω be the point on T L closest to q ∂Ω .Clearly,|q −q T l |≤|q −x l |=d l and q ∂Ω ∈ B 2d l (x l ). Since ∂Ω ∈ C 2 , |q ∂Ω −q T l ∂Ω |=O(d 2 l ). If |q −q ∂Ω |≥|q −q int |,then|q − q ∂Ω |≥|q − q int |+ dist(q ∂Ω ,T l ) =|q − q int |+ O(d 2 l ) ≥|q − q T l |+ O(d 2 l ). Let |q − q ∂Ω | < |q −q int |.If|q −q ∂Ω |≥|q −q T l |, then we are done. Otherwise, |q −q ∂Ω |+ |q ∂Ω −q T l ∂Ω |≥ | q −q T l ∂Ω |≥|q −q T l |.Thus|q −q ∂Ω |≥|q −q T l |+ O(d 2 l ). (b) We now estimate |q − q T l |.Letp l = J l ∩ ∂B d l (x l ), then |q − p l |≥d l /2. See the paragraph preceding proof of (a). Note that dist(p l ,T l ) = O(d 2 l ), since ∂Ω ∈ C 2 .Ifp l − x l ,ν x l ≥0, then dist(q,T l ) ≥ d l /3. If p l −x l ,ν x l  < 0, it again follows that dist(q,T l ) ≥ d l /3. We now apply the Harnack inequality, employing the above estimate for (q, ∂Ω), to see that for some c>0 independent of d l ,  v  z l  − m(0)  ≥ e −c  v  α l  − m(0)  ≥ e −c |m(0)| 4 . (3.5) Letting l →∞,weget0≥|m(0)|/4. Thus m(0) = 0. To show that M(0) =1, we work with function 1 −u and in place of m(0), we take 1 −M(0). Arguing analogously, one may now show that M(0) = 1. When n = 2, I r (y) reduces to two points and one may again adapt the above argument to obtain part (ii). Step 4 (maximal solution u r y ). Our next goal is to show that u ≥ v,wherev is any bounded solution of (1.2). Recall that for x ∈ ∂Ω, ν x is the unit inner normal to ∂Ω at x and x s = x + sν x .Since∂Ω ∈ C 2 and is bounded, there exists a δ>0suchthatforevery x ∈ ∂Ω, R x ≥ δ.Letε>0, small, with ε ≤ min(1/10 4 ,δ 2 /10 4 ,r 2 /10 4 ). For every x ∈ ∂Ω, set Ω ε ={x ∈Ω : dist(x,∂Ω) ≥ ε}.Then∂Ω ε ={x ε : x ∈ ∂Ω}. We w ill estimate u and v on ∂Ω ε . To this end, set P ε ={x ε : x ∈ P r (y)} and Q ε ={x ε : x ∈ Q r (y)}.Notethat Q ε = ∂Ω ε \  P ε , dist(∂Ω,∂Ω ε ) = ε, Ω ε ↑ Ω,asε ↓ 0. (3.6) For z ∈ ∂Ω ε ,letz ε be the nearest point on ∂Ω.Clearly,z = (z ε ) ε .Ifz ∈ Q ε ,thenu(z ε ) = 0, and if z ∈ P ε ,thenu(z ε ) = 1. Set N ε =  z ∈ Q ε : dist  z ε ,P r (y)  ≥ √ ε  , O ε =  z ∈ P ε : dist  z ε ,Q r (y)  ≥ √ ε  . (3.7) For v, we use comparison as follows. For x ∈ Q r (y) with dist(x,P r (y)) ≥ √ ε, v(ξ) ≤ | ξ −x| √ ε , ξ ∈ B √ ε (x) ∩Ω, implying that 0 <v(x ε ) ≤ √ ε. (3.8) Similarly, for x ∈ P r (y) with dist(x,Q r (y)) ≥ √ ε, 1 −v(ξ) ≤ | ξ −x| √ ε , ξ ∈ B √ ε (x) ∩Ω, implying that 1 − √ ε ≤ v(x ε ) ≤ 1. (3.9) 10 Boundary Value Problems Thus from (3.8)and(3.9), we obtain that 0 <v ≤ √ ε on N ε ,1− √ ε ≤ v<1onO ε . (3.10) Note that (3.10) is satisfied by any solution of (1.2), and in particular holds also for u. However, we will work with u l instead. Fix η>0, and recall from Step 1 that for l = 1,2, 3, , f l (x) = (η/l) −dist  x, P r (y)  (η/l) , x ∈ S η/l , f l (x) = 0, x ∈ ∂Ω \S η/l . (3.11) For ease of presentation, set j = 4l/η. We will work with l’s such that j √ ε<1. For x ∈∂Ω with dist(x,P r (y)) ≤3 √ ε,weseethat u l (x) = 1, x ∈ P r (y), u l (x) ≥ (η/l) −3 √ ε (η/l) ≥ 1 − j √ ε, x ∈ P r (y). (3.12) We now use comparison in B √ ε (x) ∩Ω, with dist(x,P r (y)) ≤2 √ ε.Setw x (z) = j √ ε +(1− j √ ε)|z −x|/ √ ε.Clearly,w x ≥ 1 −u l in B √ ε (x) ∩Ω. Using (3.9) and noting that u l ≥ u,we have for x ∈ ∂Ω, (i) u l (x ε ) ≥ 1 − √ ε, x ∈ P r (y), with dist(x,Q r (y)) ≥ √ ε, (ii) u l (x ε ) ≥ (1 − j √ ε)(1 − √ ε), with dist(x,P r (y)) ≤2 √ ε. Call J ε ={x ε : x ∈ ∂Ω and dist(x,P r (y)) ≤ 2 √ ε}.From(3.7), J ε ⊃ O ε , J ε ∩N ε = ∅,and u l (x ε ) ≥ (1 − j √ ε)(1 − √ ε), x ∈ J ε . Using (3.8)and(3.9), we see that u l +2j √ ε ≥ v on ∂Ω ε . By comparison, u l +2j √ ε ≥ v in Ω ε . Letting ε → 0, we obtain u l ≥ v in Ω.Now letting l →∞,weseethatu ≥v in Ω. From here on, we call u r y = u and refer to it as the maximal solution of (1.2); clearly, v ≤ u r y . Step 5 (minimal solution u r y ). It is clear from Step 1 that for r 1 <r 2 , u r 1 y ≤ u r 2 y (working with the corresponding u l ’s). Note that u r y is locally Lipschitz but uniformly so in r.Set u r y = sup t<r u t y . Using Step 1, u r y is a solution of (1.2)andu r y ≤ u r y . The comparison prin- ciple in Step 2 also holds. We now show that u r y ≤ v,wherev is any solution of (1.2). We do this by showing that u t y ≤ v whenever t<r.Fixt<r; we proceed as in Step 4.Let δ>0beasinStep 4.Letd>0, small, such that 0 <d ≤ min (δ 2 /10 4 ,r 2 /10 4 ,(r −t) 2 /100); set Ω d ={z ∈ Ω : dist(z,∂Ω) ≥ d}.AsinStep 4,defineP d,s ={x d : x ∈ P s (y)},wheres is either r or t.NowdefineQ d,s analogously. Then (3.6)holds.Forz ∈ ∂Ω d ,recallthat z d ∈ ∂Ω is such that |z −z d |=d.Nowforeachs =r,t, define the sets N d,s and O d,s , both subsets of Ω d , along the lines of (3.7). Using (3.8), (3.9), and (3.10), we obtain (i) 0 <u t y (ξ) ≤ √ d, ξ ∈ N d,t , (ii) 1 − √ d ≤ u t y (ξ) ≤1, ξ ∈ O d,t , (iii) 0 <v(ξ) ≤ √ d, ξ ∈ N d,r , (iv) 1 − √ d ≤ v(x d ) ≤ 1, x ∈ O d,r . Clearly, O d,r ⊃ O d,t , N d,t ⊃ N d,r ,andforsmalld, N d,t ∩O d,r = ∅.Thusv +2 √ d ≥ u t y on ∂Ω d . Using comparison in Ω d and taking d → 0, we obtain v ≥ u t y . The claim now follows. Call u r y the minimal solution. By Step 4 and arguments presented here, we have the first [...]... E Fabes, N Garofalo, S Mar´n-Malave, and S Salsa, “Fatou theorems for some nonlinear elliptic ı equations,” Revista Matem´ tica Iberoamericana, vol 4, no 2, pp 227–251, 1988 a [5] J J Manfredi and A Weitsman, “On the Fatou theorem for p-harmonic functions,” Communications in Partial Differential Equations, vol 13, no 6, pp 651–668, 1988 [6] G Aronsson, “Construction of singular solutions to the p-harmonic... 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Electronic Journal of Differential Equations, vol 2002, no 101, pp 1–22, 2002 [2] P Bauman, “Positive solutions of elliptic equations in nondivergence form and their adjoints,” Arkiv f¨r Matematik, vol 22, no 2, pp 153–173, 1984 o [3] L Caffarelli, E Fabes, S Mortola, and S Salsa, Boundary behavior of nonnegative solutions of elliptic operators in divergence form,” Indiana University Mathematics Journal, vol... del Seminario Matematico Universit` e Politecnico di Torino, pp 15–68 (1991), 1989, special issue a [11] T Bhattacharya, “On the behaviour of ∞-harmonic functions near isolated points,” Nonlinear Analysis Theory, Methods & Applications, vol 58, no 3-4, pp 333–349, 2004 [12] T Bhattacharya, “On the behaviour of ∞-harmonic functions on some special unbounded domains,” Pacific Journal of Mathematics, vol... the appendix in [11, 7] Note that wm (|x|,θ) > 0, −π/2m ≤ θ ≤ π/2m, and wm (±π/2m) = 0 We now have the desired conclusion by using Step 1 Acknowledgments We thank the referees for careful reading of the manuscript Their various suggestions have improved the presentation and the clarity of this work References [1] T Bhattacharya, “On the properties of ∞-harmonic functions and an application to capacitary... modified boundary data fl Since ur decreases in l l l, it follows that ur → ur , where now ur solves (3.21) Let v be any solution of (3.21) y y l We show that ur ≥ v in H Let ε > 0 be small We adapt Step 4 in Lemma 3.1 and use y Tilak Bhattacharya 13 the comparison lemma [7, Lemma 2.2] We work with ur and estimate both ur and v on l l √ the set {x ∈ H : xn = ε} Let r0 > 100 ε be such that 0 ≤ sup(M... 1.1(ii), the Harnack inequality, and following the proof of Theorem 1.1 in [7], we see that there are universal constants C1 , C2 such that C1 ur (x) ur ot ur (x) ≤ ≤ C2 , v(x) v(x) v ot x ∈ At , t ≥ 2r (3.25) 14 Boundary Value Problems Set Γ(t) = supAt ur /v, γ(t) = inf At ur /v, t ≥ 2r We now proceed as in [7, Corollary 2.3] to see that there is a universal constant C3 such that for t ≥ 2r and 2r ≤ t1 . Hindawi Publishing Corporation Boundary Value Problems Volume 2007, Article ID 78029, 17 pages doi:10.1155/2007/78029 Research Article A Boundary Harnack Principle for Infinity-Laplacian and Some. normal at x and x s = x + sν x , s>0. We will now state Theorem 1.1 which is the result about boundary Harnack principle [2, 1, 3, 4]. Theorem 1.1 (Boundary Harnack Principle) . Let Ω be a domain. u>0be infinity-harmonic in a domain Ω, suppose that a, b ∈ Ω such that the segment ab is at least η>0awayfrom∂Ω, then the following Harnack inequality holds: u (a) e |a b|/η ≥ u(b). (1.8) 4 Boundary Value