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Volume 2007, Article ID 78029, 17 pagesdoi:10.1155/2007/78029 Research Article A Boundary Harnack Principle for Infinity-Laplacian and Some Related Results Tilak Bhattacharya Received 27

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, being 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 LetΩ⊂ R n,n ≥2, be a domain with boundary∂Ω We say u 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 i j u(x) =0, x ∈ Ω. (1.1) For more discussion, see [8,1,9] For a motivation for these problems, see [8,10] For

r > 0 and x ∈ R n,B r(x) will be the open ball centered at x and has radius r LetA denote

the closure of the setA and let χ Adenote its characteristic function DefineΩr(x)=Ω∩

B r(x), Pr(x)= ∂Ω ∩ B r(x) We will assume throughout this work that ∂Ω∈ C2 More precisely, we first define for everyx ∈ ∂Ω R x to be the radius of the largest interior ball tangential toΩ at x We will assume that R y > 0 for every y ∈ ∂Ω and R x ≥ R y /2 > 0,

x ∈ P δ y(y), for some δy > 0 For every x ∈ ∂Ω, set ν x to be the inner unit normal atx

andx s = x + sν x,s > 0 We will now stateTheorem 1.1which is the result about boundary Harnack principle [2,1,3,4]

Theorem 1.1 (Boundary Harnack Principle) Let Ω be a domain inRn, 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(δy,R y /2), and u,v > 0 in Ω δ y(y) Suppose that u, and v

vanish continuously on P δ y(y), then there exist positive constants C, C1,C2independent of

u, v, and δ, such that for every z ∈Ωδ(y),

(i)u(z) ≤ cu(y δ ),

(ii)c1u(y δ)/v(yδ)≤ u(z)/v(z) ≤ c2u(y δ)/v(yδ ).

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 ifTheorem 1.1 also holds whenΩ has Lipschitz continuous boundary We will applyTheorem 1.1to 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] LetΩ be a bounded domain Fix y ∈ ∂Ω; for every r > 0,

defineQ 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 solutionu 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) and Qr(y) More precisely,

ifx ∈ P r(y) and z→ x, z ∈ Ω, then u(z) →1, and analogously forQ r(y) We show the ex-istence of bounded solutions of (1.2) inLemma 3.1 One could refer tou as the nonlinear

infinity-harmonic measure inΩ (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, seeLemma 3.1 LetH = { x ∈ R n:x n > 0 }denote the half-space inRn Sete nto

be the unit vector along the positivex n-axis SetT = { x ∈ R n:x n =0}; fory ∈ T, define

P r(y)= T ∩ B r(y), Qr(y)= T \  P r(y), and Mu

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 inΩ, in the sense of viscosity, 0≤ u ≤1, continuous up toP r(y) andQ r(y), and limsupρ →∞ M u

y(ρ)=0 We will address the existence and uniqueness of such solutions inLemma 3.4 We now state a result about the doubling property of solu-tions of (1.2) and (1.3) Forr > 0, set o3r =3ren

Theorem 1.2 (a) LetΩ⊂ R n be a bounded domain For y ∈ ∂Ω, assume that P r(y) lies on

a connected component of ∂Ω Let u r be a bounded solution of ( 1.2 ) in Ω and let r be small Then there are positive constants c, C independent of r, such that u r(yr)≥ c and

u2r(z)≤ Cu r(z), z ∈Ω\ B3r(y) (1.4)

(b) Let H be the half-space inRn Let u r

o be the unique infinity-harmonic measure in H Then there exist universal constants C1> 0 and 0 < C2< 1 such that

u2r

o (z)≤ C1u r

o(z), z ∈ H \ B3r(o), u r

o



o s

≤ C2u2r o



o s

, s ≥3r (1.5)

Estimates inTheorem 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 anyC2domain (seeLemma 3.3), it is unclear how a halving property (i.e.,

f (t) ≤ c f (2t), f positive, increasing, and c < 1) may be proven if true In particular, it

would be interesting to know if this is true whenΩ is the unit ball We now introduce notations forTheorem 1.3 For α > 0, let C α stand for the interior of the half-infinite cone inH, with apex at o, the x n-axis as the axis of symmetry, and aperture 2α For

r > 0, let M u(r)=supz ∈ ∂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, let C 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) supr>0 M u(r)= ∞ , sup r>0 M v(r)= ∞ , and (iii) lim r →∞ M u(r)=limr →∞ M v(r)= 0 Then there exists a constant C, depending on α, u, and v such that

1

C ≤ u(z)

Moreover, for every m =1, 2, 3, , if α = π/2m and ω is a direction in C π/2m , then for an appropriate C=  C(ω),

1



C | z | m2/(2m+1) ≤ u(z) ≤ C

| z | m2/(2m+1), z ∈ C π/2m with z = | z | ω. (1.7) The last conclusion inTheorem 1.3will follow from the works [6,7] WhileTheorem 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 Letu > 0 be

infinity-harmonic in a domainΩ, suppose that a,b ∈ Ω such that the segment ab is at

leastη > 0 away from ∂Ω, then the following Harnack inequality holds:

u(a)e | a − b | /η ≥ u(b). (1.8)

LetB r(a)⊂ Ω, if ω 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 will refer to (1.9) as the monotonicity property ofu For (1.8) and (1.9), see [8,1,11,

7,12,13] Moreover,u is locally Lipschitz (C1ifn =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 Forp-harmonic functions (finite p), we direct the reader to the recent work by John Lewis and Kaj Nystrom “Boundary

Behaviour forp 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] SinceΔ∞is translation and rotation invariant, we may assume that the origino ∈ ∂Ω Set osc A u =supz ∈ A u(z) −

infz ∈ A u(z) to be the oscillation function of u on the set A Recall that Ω 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 ofo, in ∂Ω Let M u(r)=supz ∈Ωr(o) u(z) By the

max-imum principle,M u(r) > 0 and u(z)≤ M u(r), z∈Ωr(o) For 0 < α≤ β, consider the

functionw(z) = M u(α) + [Mu(β)− M u(α)](| z | − α)/(β − α), z ∈Ωβ \Ωα Clearly,u ≤ w

on∂(Ω β \Ωα) Thusu ≤ w in Ω β \Ωα 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 oscΩr(o) u = M u(r) is convex in r Since u(o)=0, it follows that 2oscΩr(o) u

≤oscΩ2r(o) u.

Step 2 (Carleson inequality) We now use the interior ball condition Since ∂Ω ∈ C2,R x ≥

R o /2, x ∈ P4δ(o), with 4δ < inf(δy,Ro /2) For every x ∈ ∂Ω, let ν x denote the unit inner normal atx, and set x t = x + tν x, 0≤ t ≤ R x We will prove thatu(z) ≤ Cu(o δ),z ∈Ωδ(o)

We will adapt a device, based on the Harnack inequality, from [3] Forz ∈Ωδ, definex z ∈

∂Ω to be the point nearest to z Also set d(z) = | x z − z | Thenz = x z+d(z)ν x z =(xz)d(z); setz s = x z+ 2s −1d(z)ν x z,s =1, 2, 3, By the Harnack inequality (1.8), forz ∈Ω3δ(o),

u(z) ≤

⎧

⎪

⎨

⎪

⎩

Mu

z2 

: 0< d(z) <3δ

2 ,

Mu

o δ

:δ < d(z) < 3δ.

(2.2)

We takeM = e8 We now make an observation which will be used repeatedly in what follows Ifd(z) ≥ δ/2 s, then

u(z) ≤ Mu

z2 

z s

o δ

Suppose now that there is aξ0∈Ωδ(o) such that u(ξ0)≥ M l+2 u(o δ), wherel = l(δ) is large

and will be determined later Using the aforementioned observation, we obtain

dist

ξ0,∂Ω

Letp0∈ ∂Ω be the nearest point to ξ0 Clearly,ξ0∈Ω2− l δ(p0)⊂Ω2δ(o) Thus, oscΩδ2 − l(p0 )u

≥ u(ξ0); thus byStep 1, form =1, 2, 3 ,

oscΩδ2 − l+m(p0 )u ≥2moscΩδ2 − l(p0 )u ≥2m u

ξ0



where 2m ≥ M3= e24 Select m =60; thus oscΩδ2 − l+m(p0 )u ≥2m u(ξ0)≥ M l+5 u(o δ) Thus there is aξ1∈Ωδ2 − l+m(p0) such thatu(ξ1)≥ M l+5 u(o δ) Arguing as done in (2.4), we see dist(ξ1,∂Ω)≤ δ2 − l −3 Lettingp1∈ ∂Ω to be closest to ξ1, we see thatp1∈Ω2δ(o) Repeat-ing our previous argument,

oscΩδ2 − l −3+m(p1 )u ≥2moscΩδ2 − l −3 (p1 )u ≥2m u

ξ1 

o δ

Thus we may find aξ2∈Ωδ2 − l −3+m(p1) such thatu(ξ2)≥ M l+8 u(o δ), and dist(ξ2,∂Ω)≤

δ2 − l −6 Thus we obtain a sequence of pointsξ k ∈ Ω and p k ∈ ∂Ω, k =1, 2, 3 , such that

u

ξ k



≥ M l+2+3k u(o δ), dist

ξ k,∂Ω

≤ δ2 − l −3k, ξ k ∈Ωδ2 − l −3(k −1)+m

p k −1

. (2.7)

Note that

ξ k − o ≤ k−1

i =1

ξ i+1 − ξ i + ξ0− o ≤ δ 1 + 2

k−1

i =0

2− l −3i+m



Choosel ≥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 inTheorem 1.1

follows

Step 3 (bounds near the boundary) We first derive a lower bound in terms of the distance

to the boundary For everyz ∈Ωδ(o), let xzandd(z) be as inStep 2 Note thatd(z) ≤ | z −

o | ≤ δ Thus x z ∈Ω2δ(o) Call ζz = x z+δν x z, observe thatζ z ∈Ω3δ(o) By monotonicity (1.9) and the interior ball condition, we have

u(z) d(z) ≥ u



ζ z

δ ≥ e −6u

o δ

since| ζ z − o δ | ≤ | x z+δν x z − δν o | ≤4δ

Letz ∈Ωδ(o) As noted previously, xz ∈Ω2δ(o) and Ωδ(xz)⊂Ω3δ(o) Note that z∈

Ωδ(xz) Setμ z =supΩ(x)u, then by comparison u(ξ) ≤ μ z | ξ − x z | /δ, ξ ∈Ωδ(xz) Thus

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(ζz)≤ e4u(o δ) Thus there is universalC,

such that

u(z) d(z) ≤  C u



o δ

Ifu, v are two positive infinity-harmonic functions in Ω4δ(o), then by (2.9) and (2.10), there exist universal constantsC1andC2such that

C1u

o δ



v

o δ  ≤ u(z)

v(z) ≤ C2u

o δ



v

o δ

, z ∈Ωδ(o) (2.11) This provesTheorem 1.1

Remark 2.1 We comment that the distance function d(z) =dist(z,∂Ω), z∈ Ω, is C2and infinity-harmonic near∂Ω Also the oscillation estimate inStep 1continues 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 thatΩ is a bounded C2domain Fory ∈ ∂Ω and r > 0,

re-call the definitions ofP r(y) and Qr(y) Note that both Pr(y) and Qr(y) are relatively open

in∂Ω Let u be a solution of (1.2) As inSection 2, forx ∈ ∂Ω, ν xandx t = x + tν x,t > 0,

are as defined inSection 2 We will assume thatΩ is bounded but we can extend our arguments to the case of the half-spaceH We will always take u to be bounded in this

section This will imply the maximum principle At this time, it is not clear whether un-bounded solutions to (1.2) exist LetC ybe the connected component of∂Ω that contains

y InLemma 3.1, we assume thatB r(y)∩ ∂Ω = B r(y)∩ C y

Lemma 3.1 LetΩ∈ C2be 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 ur

y , in Ω such that if v is any bounded solution of ( 1.2 ), then ur

y ≤ v ≤ u r

y (iv) If t < r, then u t

y ≤limρ ↑ r u ρ y =  u r

y ≤ u r

y =limr↓ r ur

y Moreover, u r

y satisfies the following comparison 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 > 0 be small Set I r(y)= ∂B r(y)∩ ∂Ω, and for t > 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 lbe such that

(i) f l ∈ C(∂Ω),

(ii) f l(x)=1,x ∈ P r(y),

(iii) f l(x)=0,x ∈ ∂Ω \ S η/l,

(iv) f l(x)=(η/l)−dist(x,Pr(y))/(η/l), x∈ S η/l

Now letu l ∈ C(Ω) be the unique viscosity solution of the problem

Δ∞ul(z)=0, z ∈ Ω, u l | ∂Ω = f l (3.1) Clearly, 0< u l < 1 in Ω Since f l ≥ f l+1, by comparison, there is a functionu η such that

u l ↓ u η inΩ We first show that if x ∈ P r(y) and z→ x ∈ P r(y), z∈ Ω, then u η(z)→1 Consider the setΩδ(x), where δ=infξ ∈ Q r(y) | x − ξ | /2 For z ∈Ωδ(x), set w(z)=1− | z −

x | /δ By comparison, for every l, w ≤ u l ≤1 inΩδ(x) Thus 1− | z − x | /δ ≤ u η(z)≤1,z ∈

Ωδ(x) We see that limz → x u η(z)=1 Forx ∈ Q r(y) and δ=infξ ∈ P r(y) | ξ − x | /2, it is clear

that forl large, u l(z)≤ | z − x | /δ, z ∈ B δ(x) Thus uη(z)→0 asz → x Moreover, the limit

functionu η does not depend on the widthη of the appended band S η An argument based on comparison shows easily that for any η1, η2> 0, u η1= u η2 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 inΩ To see this, take x1∈ Ω and t > 0 such that B4t(x1)⊂ Ω Select x2∈ B t(x1); setμ l = sup B4t(x1 )u l Applying monontonicity (1.9) inB t(x1), we have for everyl, (μ l −

u l(x1))/t≤(μl − u l(x2))/(t− | x1− x2|) Rearranging terms (see [1, Lemma 3.6], also see [12]), noting that u l(x1), u l(x2)≥0 andμ l ≤ μ1≤1, we obtain| u l(x2)− u l(x1)| / | x1−

x2| ≤1/t Fixing x1,x2and lettingl → ∞, we obtain thatu is locally Lipschitz Fix ξ ∈Ω and for 0≤ t < dist(ξ, pΩ), set M l(t)=supB t(ξ) u l, m l =infB t(ξ) u l,M(t) =supB t(ξ) u, and m(t) =infB t(ξ) u Using that (i) u k ≤ u j ≤ u l whenl < j < k, (ii) M l is convex andm l is concave int, it follows that for a < c < b and 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) first lettingk → ∞, replacingu j(z) by u(z), and then letting l→ ∞, we obtain thatM(t) is convex and m(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 inStep 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 everyl, u f ≤ u lin Ω Thus u f ≤ u in Ω.

Now let f ≥ χ P r(y), setε > 0 Since f ≥1 inP r(y), there exists a δ > 0 such that f + ε≥1

inB δ(x)∩ ∂Ω, for every x ∈ ∂Ω ∩ ∂B r(y) Take l large so that η/l≤ δ/2 By comparison,

u l ≤ u f+ε in Ω Thus we have u ≤ u f inΩ

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 anR0> 0 such

that forx ∈ P2r(y), Rx ≥ R0, and consequently,

x ∈ P2r(y) B R0/4(xR0/4)⊂Ω In what follows

we take the quantitiesσ, η < R0/10 We exploit the special geometry of P r(y) to achieve our proof

SetI r(y)= ∂Ω ∩ ∂B r(y); for every x∈ I r(y) and σ > 0, define mx(σ)=inf∂B σ(x) ∩Ωv

andM x(σ)=sup∂B (x) ∩Ωv Clearly, m x(σ)≤0 andM x(σ)≥1 We claim thatM xis convex

andm xis concave inσ To see this, take z ∈ Ω with 0 < a ≤ | z − x | ≤ b Set w(z) = m x(a) + [mx(b)− m x(a)](| z − x | − a)/(b − a) Clearly, w ≤0 By comparison,w ≤ v in (B b(x)\

B a(x))∩ Ω Thus m x(σ) is concave in σ, and one can show analogously that Mx(σ) is convex Definem y(σ)=infx ∈ I r(y) m x(σ) and My(σ)=supx ∈ I r(y) M x(σ), then for σ > 0,

(i)M y(σ)≥1 is convex, m y(σ)≤0 is concave inσ,

(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)↓ asσ ↓0

(3.3)

Note thatv =0 or 1 on∂Ω \x ∈ I r(y) B σ(x) Thus (3.3)(i) follows easily Now using (3.3)(i) and comparison in the setΩ\x ∈ I r(y) B σ(x) yields (3.3)(ii) Clearly,M y(σ)(my(σ)) is the supremum (infimum) ofv on Ω \x ∈ I r(y) B σ(x) The conclusion in (3.3)(iii) follows

by observing that 

x ∈ I r(y) B σ1(x)⊂x ∈ I r(y) B σ2(x), when σ1> σ2 By (3.3), the quanti-tiesM(0) =limσ →0M y(σ) and m(0)=limσ →0m 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) and Pr(y) For x∈ ∂Ω, let ρ(x) =dist(x,Pr(y)) and



ρ(x) =dist(x,Qr(y)) For x∈ Q r(y), define wx(z)= m(0) | z − x | /ρ(x) in the set Ω ρ(x)(x)

By comparisonw 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) in Ω ρ(x) (x) Then v≥ ω xinΩρ(x) (x) and v≥ m(0)/2,

in Ωρ(x)/2 (x) Let η > 0 be small Set Aη = { 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 ∈Ωρ(x)/2(x), x∈ A η,

m(0)

2 :z ∈Ωρ(x)/2 (x), x∈ B η

(3.4)

SetS =η>0x ∈ A ηΩρ(x)/2(x) and T=η>0x ∈ B ηΩρ(x)/2 (x), and call Gy =Ω\(S∪ T).

Forl =1, 2, 3 , let z l ∈ Ω be such that v(z l)≤7m(0)/8 and v(zl)→ m(0), as l → ∞ By (3.4),z l ∈Ω\ G y, and by the maximum principle, dist(zl,Ir(y))→0

In the discussion that follows, we will assume thatn > 2 Recalling that I r(y)= ∂B r(y)∩

∂Ω, it follows that I r(y) is smooth For every l, let xl ∈ I r(y) be the closest point to

z l andd l = | x l − z l | Note that the segment x l z l is normal toI r(y) Since xl ∈ ∂B r(y),

yx l ⊥ ∂B r(y), and so yxl ⊥ I r(y) Let Tlbe the hyperplane tangential to∂Ω at x l, and let

Πlbe the 2-dimensional plane containing the segmentsyx landyz l ThusΠl ⊥ I r(y) at xl

andν x l lies inΠl Note thatΠl ⊥ T landI r(y) is tangential to Tlatx l CallJ l = ∂Ω ∩Πl, observe that the curveJ l ⊥ I r(y) at xl It is easy to see that ifx ∈ J lis close tox l, then (i)

ρ(x) = | x − x l |ifx ∈ P r(y), and (ii)ρ(x) = | x − x l |ifx ∈ Q r(y) Now consider the set

C l =Πl ∩ ∂B d l(xl)\ G y As noted abovez l ∈ C l, moreover one can findα l ∈ C lsuch that

v(α l)=3m(0)/4 We will apply the Harnack inequality in Clto obtain a contradiction

In (3.4), takeη = d l and we observe the following Since∂Ω ∈ C2andx l’s lie in a com-pact set, it follows that for q ∈ C l, dist(q,∂Ω)≈dist(q,Tl)= O(d l), asd l →0 In other words, dist(q,∂Ω) has a lower bound of the order of dl We show this as follows First note that since∂Ω ∈ C2, it permits a local parametrization nearx l, wherex n = ν x l,x n =0

isT l, andx n = φ(x1, ,x n −1) describes∂Ω Clearly, dist(q,∂Ω) ≤ | q − x l | = d l We will

show that (a) dist(q,∂Ω)≥dist(q,Tl) +O(d2

l) and (b) dist(q,Tl)≈ 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 onT l closest

toq, (iii) qintthe point of intersection of the line, containing the segmentqq ∂Ω, andT l, and (iv) letq T l

∂Ω be the point on T Lclosest toq ∂Ω Clearly,| q − q T l | ≤ | q − x l | = d l and

q ∂Ω ∈ B2d l(xl) Since∂Ω ∈ C2,| q ∂Ω − q T l

∂Ω | = O(d2

l) If| q − q ∂Ω | ≥ | q − qint|, then| q −

l)≥ | q − q T l |+O(d2

l) Let | q − q ∂Ω | <

∂Ω | ≥

| q − q T l

∂Ω | ≥ | q − q T l | Thus| q − q ∂Ω | ≥ | q − q T l |+O(d2

l)

(b) We now estimate | q − q T l | Let p l = J l ∩ ∂B d l(xl), then | q − p l | ≥ d l /2 See the

paragraph preceding proof of (a) Note that dist(p l,Tl)= O(d2

l), since∂Ω ∈ C2 If p l −

x l,ν x l  ≥0, then dist(q,Tl)≥ 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 somec > 0 independent of d l,



v

z l

≥ e − c

v

α l

≥ e − c | m(0) |

Lettingl → ∞, we get 0≥ | m(0) | /4 Thus m(0) =0 To show thatM(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 thatM(0) =1 Whenn =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, where v is any

bounded solution of (1.2) Recall that forx ∈ ∂Ω, ν x is the unit inner normal to∂Ω at

x and x s = x + sν x Since∂Ω ∈ C2and is bounded, there exists aδ > 0 such that for every

x ∈ ∂Ω, R x ≥ δ Let ε > 0, small, with ε ≤min(1/104,δ2/104,r2/104) For everyx ∈ ∂Ω,

setΩε = { x ∈ Ω : dist(x,∂Ω) ≥ ε } Then∂Ω ε = { x ε:x ∈ ∂Ω } We will estimateu and v on

∂Ω ε To this end, setP ε = { x ε:x ∈ P r(y)}andQ ε = { x ε:x ∈ Q r(y)} Note that

Q ε = ∂Ω ε \  P ε, dist(∂Ω,∂Ωε)= ε, Ωε ↑ Ω, as ε ↓0 (3.6) Forz ∈ ∂Ω ε, letz εbe the nearest point on∂Ω Clearly, z =(zε)ε Ifz ∈ Q ε, thenu(z ε)=0, and ifz ∈ P ε, thenu(z ε)=1 Set

N ε =z ∈ Q ε: dist

z ε,Pr(y)

, O ε =z ∈ P ε: dist

z ε,Qr(y)

. (3.7) Forv, we use comparison as follows For x ∈ Q r(y) with dist(x,Pr(y))≥ √ ε,

v(ξ) ≤ | ξ √ − x |

ε , ξ ∈ B √ ε(x)∩ Ω, implying that 0 < v(x ε)≤ √ ε. (3.8) Similarly, forx ∈ P r(y) with dist(x,Qr(y))≥ √ ε,

1− v(ξ) ≤ | ξ √ − x |

ε , ξ ∈ B √ ε(x)∩Ω, implying that 1− √ ε ≤ v(x ε)≤1 (3.9)

Thus from (3.8) and (3.9), we obtain that

0< v ≤ √ ε on N ε, 1− √ ε ≤ v < 1 on O 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, setj =4l/η We will work with l’s such that j√

ε < 1 For x ∈ ∂Ω

with dist(x,Pr(y))≤3√

ε, we see that

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 inB √ ε(x)∩ Ω, with dist(x,P r(y))≤2√

ε Set w x(z)= j √

ε + (1 −

j √

ε) | z − x | / √

ε Clearly, w x ≥1− u linB √ ε(x)∩Ω Using (3.9) and noting thatu l ≥ u, we

have forx ∈ ∂Ω,

(i)u l(xε)≥1− √ ε, x ∈ P r(y), with dist(x,Qr(y))≥ √ ε,

(ii)u l(xε)≥(1− j √

ε)(1 − √ ε), with dist(x,P r(y))≤2√

ε.

CallJ 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(xε)≥(1− j √

ε)(1 − √ ε), x ∈ J ε Using (3.8) and (3.9), we see thatu l+ 2j√

ε ≥ v on

∂Ω ε By comparison, u l+ 2j √

ε ≥ v in Ω ε Lettingε →0, we obtainu l ≥ v in Ω Now

lettingl → ∞, we see thatu ≥ v in Ω.

From here on, we callu r

y = u and refer to it as the maximal solution of (1.2); clearly,

v ≤ u r

y

Step 5 (minimal solution ur

y) It is clear fromStep 1that forr1< r2,u r1

y ≤ u r2

y (working with the correspondingu l’s) Note thatu r

yis locally Lipschitz but uniformly so inr Set



u r

y =supt<r u t

y UsingStep 1,ur

yis a solution of (1.2) andur

y ≤ u r

y The comparison prin-ciple in Step 2also holds We now show thatur

y ≤ v, where v is any solution of (1.2)

We do this by showing thatu t

y ≤ v whenever t < r Fix t < r; we proceed as inStep 4 Let

δ > 0 be as inStep 4 Letd > 0, small, such that 0 < d ≤min (δ2/104,r2/104, (r− t)2/100);

setΩd = { z ∈ Ω : dist(z,∂Ω) ≥ d } As inStep 4, defineP d,s = { x d:x ∈ P s(y)}, wheres

is eitherr or t Now define Q d,sanalogously Then (3.6) holds Forz ∈ ∂Ω d, recall that

z d ∈ ∂Ω is such that | z − z d | = d Now for each s = r,t, define the sets N d,sandO d,s, both subsets ofΩd, 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, and for smalld, N d,t ∩ O d,r =∅ Thusv + 2 √

d ≥ u t

y on

∂Ω d Using comparison inΩdand takingd →0, we obtainv ≥ u t

y The claim now follows Callur

ythe minimal solution ByStep 4and arguments presented here, we have the first

and< i>m xis concave inσ To see this, take z ∈ Ω with < a ≤ | z −... ≤ | q − x l | = d l We will

show... ∈Ωδ(xz) Thus

u(z) ≤