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Submitted on 4 Apr 2017

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non-smooth domains

Konstantinos Gkikas, Laurent Véron

To cite this version:

Konstantinos Gkikas, Laurent Véron. The spherical p-harmonic eigenvalue problem in non-smooth

domains. Journal of Functional Analysis, Elsevier, 2018, 274, pp.1155-1176. �hal-01501604�

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in non-smooth domains

Konstantinos Gkikas

Laurent Véron

Abstract

We prove the existence of p-harmonic functions under the formu(r, σ) =r−βω(σ)in any coneCS generated by a spherical domainSand vanishing on∂CS. We prove the uniqueness of the exponentβ and of the normalized functionωunder a Lipschitz condition onS.

2010 Mathematics Subject Classification.35J72; 35J92.

Key words.p-Laplacian operator; polar sets; Harnack inequality; boundary Harnack inequality;p-Martin boundary.

Contents

1 Introduction 1

2 Existence 2

2.1 Estimates . . . 2

2.2 Approximations from inside . . . 5

2.3 Approximations from outside . . . 9

3 Uniqueness 10 3.1 Uniqueness of exponentβ . . . 10

3.2 Uniqueness of eigenfunction . . . 11

3.2.1 The convex case . . . 11

3.2.2 Proof of Theorem E . . . 12

1 Introduction

Letp >1,Sa domain of the unit sphereSN−1ofRN andCS :={(r, σ) :r > 0, σ∈S}the positive cone generated byS. If one looks forp-harmonic functions inCS under the formu(x) = u(r, σ) = r−βω(σ)vanishing on∂CS\ {0}, thenωsatisfies thesphericalp-harmonic eigenvalue problemonS

−div0

β2ω2+|∇0ω|2p−22

0ω

= (p−1)β(β−β0)

β2ω2+|∇0ω|2p−22

ω inS

ω= 0 in∂S

(1.1)

[email protected]

[email protected]

1

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withβ0 = Np−1−p and werediv0 and∇0 denote the divergence operator and the covariant gradient on SN−1endowed with the metric induced by its isometric inbedding intoRN. Separable solutions play a key role for describing the boundary behaviour and the singularities of solutions of a large variety of quasilinear equations. WhenN = 2the equation is completely integrable and has been solved by Kroll in the regular caseβ <0and Kichenassamy and Véron in the the singular caseβ >0. In higher dimension, Tolksdorff [15] proved the following:

Theorem AIfS is a smooth spherical domain, there exist two couples(βS, ωS)and (βS0, ωS0)where βS >0andβS0 <0,ωS andωS0 are positiveC2(S)-functions vanishing on∂Swhich solve(1.1)with (β, ω) = (βS, ωS) or(β, ω) = (βS0, ω0S). FurthermoreβS andβS0 are unique, andωS and ωS0 are unique up to an homothety.

A more general and transparent proof has been obtained by Porretta and Véron [13], but always in the case of a smooth spherical domain. The aim of this article is to extend Theorem A to a general spherical domain. If we consider an increasing sequence of smooth domains{Sk}such thatSk ⊂Sk ⊂Sk+1

and∪kSk=Swe prove the following:

Theorem BAssume thatScis not polar. Then the sequence of theβSk>0from Theorem A is decreasing and converges toβS >0. There existsωS ∈W01,p(S)∩L(S)weak solution of(1.1)withβ =βS. FurthermoreβS>0is the largest exponentβsuch that(1.1)admits a positive solutionωS ∈W01,p(S).

Under a mild assumption onSit is possible to approximate it by a decreasing sequence of smooth domainsSk0 such thatSk0 ⊂S0k ⊂Sk−10 and∩kSk0 =S

Theorem CAssume thatS =

o

S. Then the sequenceβS0

k > 0is increasing and converges toβˆS >0 and there existsωˆS ∈W01,p(S)∩L(S)weak solution of(1.1)withβ = ˆβS. FurthermoreβˆS is the smallest exponentβsuch that(1.1)admits a positive solutionωS∈W01,p(S).

We prove the uniqueness of the exponentβ, under a Lipschitz assumption onS.

Theorem D Assume thatS is a Lipschitz domain, then βS = ˆβS and if ω and ω0 are two positive solutions of(1.1)inW01,p(S), there exists a constantc >0such thatc−1ω0 ≤ω≤cω0.

The proof of Theorem C is based upon a sharp form of boundary Harnack inequality proved in [10],

lnωω(σ011))−lnωω(σ022))

≤c11−σ2|α ∀σ1, σ2∈S, (1.2) for somec1 =c1(N, p, S)>0andα∈(0,1). Actually we have a stronger result, much more delicate to obtain.

Theorem ELetSbe a Lipschitz subdomain ofSN−1. Then two positive solutions of(1.1)inW01,p(S) are proportional.

The proof is based upon a non trivial adaptation of a series of deep results of Lewis and Nyström [10] concerning thep-Martin boundary of domains.

AcknowledgementsThis article has been prepared with the support of the collaboration programs ECOS C14E08.

2 Existence

2.1 Estimates

Through this article we assume thatScis not polar, or equivalently that it has positivecS1,pN−1-capacity.

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Lemma 2.1. Assumep >1. Then any solutionω∈W01,p(S)of(1.1)satisfies

kωkCγ(S)≤c1kωkLp(S), (2.1)

ifp > N−1whereγ= 1−Np−1 ifp > N−1and

kωkL(S)≤c1kωkLp(S), (2.2)

if1< p≤N−1, wherec1>0depends onp,N,β.

Proof. Multiplying the equation byωand using Hölder’s inequality, we derive (i)

Z

S

β2ω2+|∇0ω|2p2

dS≤(β(pβ−(p−1)β0))p2 Z

S

|ω|pdS if p≥2, (ii)

Z

S

β2ω2+|∇0ω|2p2

dS≤βp−1(pβ−(p−1)β0) Z

S

|ω|pdS if 1< p <2.

(2.3)

Notice that these inequalities hold for allp >1. Ifp > N−1 (2.1)follows by Morrey’inequality. Here after we assume1< p≤N−1. Letα≥1andk >0. Thenζ = min{|ω|, k}α−1ωis an admissible test function, hence

1- Ifp≥2, Z

S

β2ω2+|∇0ω|2p−22

h∇0ω.∇0ζidS= (p−1)β(β−β0) Z

S

β2ω2+|∇0ω|2p−22 ωζdS

≤c2

Z

S

|∇0ω|p−2ω2min{|ω|, k}α−1dS+c2βp Z

S

|ω|pmin{|ω|, k}α−1dS

≤c2 Z

S

|ω|pmin{|ω|, k}α−1dS

p−2p Z

S

|∇0ω|pmin{|ω|, k}α−1dS p2

+c2βp Z

S

|ω|pmin{|ω|, k}α−1dS,

(2.4)

wherec2=c2(N, p, β)>0. Since Z

S

β2ω2+|∇0ω|2p−22

h∇0ω.∇0ζidS≥c3(p) Z

S

|∇0ω|pmin{|ω|, k}α−1dS, it implies that there existsc4=c4(N, p, β)such that

Z

S

|∇0ω|pmin{|ω|, k}α−1dS≤c4 Z

S

|ω|pmin{|ω|, k}α−1dS, (2.5) which yields

Z

S

|∇0j(ω)|pdS≤c4

Z

S

|j(ω)|pdS, (2.6)

wherej(ω) = min{|ω|, k}α−1p ω.

2- If1< p <2, then Z

S

β2ω2+|∇0ω|2p−22

h∇0ω.∇0ζidS= Z

S

β2ω2+|∇0ω|2p−22

|∇0ω|2min{|ω|, k}α−1dS

+ (α−1) Z

S∩{|ω|<k}

β2ω2+|∇0ω|2p−22

|∇0ω|2|ω|α−1dS.

(2.7)

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Since Z

S

β2ω2+|∇0ω|2p−22

|∇0ω|2min{|ω|, k}α−1dS= Z

S

β2ω2+|∇0ω|2p2

min{|ω|, k}α−1dS

−β2 Z

S

β2ω2+|∇0ω|2p−22

min{|ω|, k}α−1ω2dS

≥ Z

S

|∇0ω|pmin{|ω|, k}α−1dS−β2 Z

S

β2ω2+|∇0ω|2p−22

min{|ω|, k}α−1ω2dS, we derive

Z

S

|∇0ω|pmin{|ω|, k}α−1dS≤βp−1(pβ−(p−1)β0) Z

S

|ω|pmin{|ω|, k}α−1dS, (2.8) which leads to(2.6). Lettingk→ ∞we infer by Fatou’s lemma,

Z

S

0|ω|α−1p +1

p

dS≤c4 Z

S

|ω|α−1+pdS. (2.9)

Ifp < N−1we derive from Sobolev inequality and puttingq=α−1 +pands= N−1−pN−1 >1 Z

S

|ω|sqdS 1s

≤c5 Z

S

|ω|qdS, (2.10)

andc5>0depends onN,pandβ. Iterating this estimate by Moser’s method we derive(2.10).

Ifp=N−1we have for1≤m < p−1andm=m(N−1)N−1−m c6

Z

S

|ω|(α−1p +1)mdS mpm

≤ Z

S

0|ω|α−1p +1

m

dS mp

≤ |S|mp−1c4

Z

S

|ω|α−1+pdS, andc6=c6(N, p), hence

Z

S

|ω|tqdS 1t

≤c5

Z

S

|ω|qdS, (2.11)

witht=p(Nm(N−1−m)−1) =N−1−mm . The proof follows again by Moser’s iterative scheme.

Proposition 2.2. LetS1andS2be two subdomains ofSN−1such thatS1⊂S1⊂S2andS2not polar.

Letβj >0, j=1,2, such that there exist positive solutionsωj∈W01,p(Sj)solutions of

−div0

βj2ω2j+|∇0ωj|2p−22

0ωj

= (p−1)βjj−β0)

βj2ωj2+|∇0ωj|2p−22

ωj inSj

ωj= 0 in∂Sj.

(2.12) Thenβ1≥β2.

Proof. Setuj(r, σ) =r−βjωj(σ),CSj = (0,∞)×SJ and assumeβ1 < β2. By Harnack inequality ω2≥c >0onS1, thus

u2(r, σ)≥cr−β2 a.e. inCS1. For >0there existr>0such that

u2(x)≥u1(x) ∀x∈CS1∩Br.

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Letδ >0, there existsRδ >0such that

u1(x)≤δ ∀x∈CS1∩BRcδ. Henceζ = (u1−u2−δ)+ ∈W01,p(QrS,Rδ

1 ), whereQrS,Rδ

1 = {x ∈ CS1 : r <|x| < Rδ}. This implies

0 = Z

Qr,RδS

1

|∇u1|p−2∇u1− |∇(u1)|p−2∇(u1).∇ζ dx

= Z

Qr,RδS

1 ∩{u1−u2≥δ}

|∇u1|p−2∇u1− |∇(u1)|p−2∇(u1).∇(u1−u2) dx.

Therefore∇(u1−u2−δ)+= 0a.e. inQrS,Rδ

1 , which leads tou1−u2≤δin the same set. Letting δ →0 yieldsRδ → ∞, thus we obtainu1 ≤u2inCS1 ∩Bcr henceu1 ≤ 0inCS1, contradiction.

2.2 Approximations from inside

Proof of Theorem B. Let{Sk}be an increasing sequence of smooth domains such thatSk⊂Sk ⊂Sk+1. We denote by{(βSk, ωk)}the corresponding sequence of solutions of(1.1)withβ=βSkandω=ωk. The sequence{βSk}is uniquely determined by [15], it admits a limitβ:=βS, and theωkare the unique positive solutions such that

Z

Sk

k|dS= 1.

Ifp≥2, we have Z

Sk

|∇0ωk|pdS≤ Z

Sk

β2Skω2k+|∇0ωk|2p−22

|∇0ωk|2dS

= (p−1)βSkSk−β0) Z

Sk

β2S

kω2k+|∇0ωk|2p−22 ωk2dS

≤2

(p−4)+

2 (p−1)βSkSk−β0) Z

Sk

βSp−2

k ωkp+|∇0ωk|p−2ωk2 dS

≤c7(N, p, βSk) Z

Sk

ωpkdS+1 2 Z

Sk

|∇0ωk|pdS.

SinceβSk≤β1, we derive

Z

Sk

|∇0ωk|pdS≤c8, (2.13)

from the normalization assumption withc8= 2c7(N, p, β1).

If1< p <2, we have Z

Sk

|∇0ωk|pdS≤ Z

Sk

βS2

kωk2+|∇0ωk|2p2 dS

≤βSk(pβSk+ (p−1)β0) Z

Sk

β2S

kω2k+|∇0ωk|2p−22 ωk2dS

≤βkp−1(pβSk+ (p−1)β0) Z

Sk

ωkpdS,

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and we obtain(2.13)withc81p−1(pβ1+ (p−1)β0).

Next we extend ωk by 0 inSkc. Then there existsω ∈ W01,p(S) such that ωk * ω weakly in W01,p(S), up to subsequence that we still denote{ωk}, andωk→ωinLp(S).

Step 1: We claim that∇0ωkconverges to∇0ωlocally inLp(S).

Leta∈Sandr >0such thatB4r(a)⊂S. Then fork≥k0,B2r(a)⊂Sk. Letζ∈C0(B2r(a))such that0≤ζ≤1,ζ= 1inBr(a). For test function we chooseηk =ζ(ω−ωk), then

Z

Sk

β2Skω2k+|∇0ωk|2p−22

h∇0ωk.∇0ηkidS= (p−1)βSkSk−β0) Z

Sk

βS2kω2k+|∇0ωk|2p−22

ωkηkdS.

By the above inequality, we have Z

B2r(a)

β2ω2+|∇0ω|2p−22

0ω− βS2

kω2k+|∇0ωk|2p−22

0ωk.∇0ηk

dS

= Z

B2r(a)

β2ω2+|∇0ω|2p−22

h∇0ω.∇0ηkidS

−(p−1)βSkSk−β0) Z

Sk

βS2

kωk2+|∇0ωk|2p−22

ωkηkdS.

Using the weak convergence of the gradient, we have

k→∞lim Z

B2r(a)

β2ω2+|∇0ω|2p−22

h∇0ω.∇0ηkidS= 0.

Sinceωkis uniformly bounded inW01,p(S)andωk→ωinLp(S), we have

k→∞lim Z

B2r(a)

βS2

kωk2+|∇0ωk|2p−22

ωkηkdS= 0, and

k→∞lim Z

B2r(a)

(ω−ωk)

β2ω2+|∇0ω|2p−22

0ω−

βS2kωk2+|∇0ωk|2p−22

0ωk.∇0ζ

dS= 0.

Combining the above relations we infer

k→∞lim Z

B2r(a)

ζ

β2ω2+|∇0ω|2p−22

0ω−

β2Skω2k+|∇0ωk|2p−22

0ωk.∇0(ω−ωk)

dS= 0.

(2.14) Next we write

Z

B2r(a)

ζ

β2ω2+|∇0ω|2p−22

0ω− βS2

kωk2+|∇0ωk|2p−22

0ωk.∇0(ω−ωk)

dS

=1 2 Z

B2r(a)

ζ

β2ω2+|∇0ω|2p−22 +

βS2

kωk2+|∇0ωk|2p−22

|∇0(ω−ωk)|2dS

+1 2 Z

B2r(a)

ζ

β2ω2+|∇0ω|2p−22

βS2kωk2+|∇0ωk|2p−22

×

|∇0ω|22ω2−β2S

kω2k− |∇0ωk|2 dS

−1 2 Z

B2r(a)

ζ

β2ω2+|∇0ω|2p−22

− βS2

kωk2+|∇0ωk|2p−22

β2ω2−β2S

kωk2 dS.

(2.15)

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Ifp≥2, we have from(2.4), Z

B2r(a)

ζ

β2ω2+|∇0ω|2p−22

0ω− βS2

kωk2+|∇0ωk|2p−22

0ωk.∇0(ω−ωk)

dS

≥1 2 Z

B2r(a)

ζ

|∇0ω|p−2+|∇0ωk|p−2

|∇0(ω−ωk)|2dS

−1 2 Z

B2r(a)

ζ

β2ω2+|∇0ω|2p−22

− βS2

kωk2+|∇0ωk|2p−22

β2ω2−β2S

kωk2 dS

≥min{2−1,22−p} Z

B2r(a)

ζ|∇0(ω−ωk)|pdS

−1 2 Z

B2r(a)

ζ

β2ω2+|∇0ω|2p−22

− βS2

kωk2+|∇0ωk|2p−22

β2ω2−β2S

kωk2 dS.

(2.16) Sinceωk →ωinLp(S),βSk→βandωk, ωare uniformly bounded inW01,p(S), we derive

Z

B2r(a)

ζ

β2ω2+|∇0ω|2p−22

− βS2

kω2k+|∇0ωk|2p−22

β2ω2−βS2

kωk2

dS→0 ask→ ∞. Jointly with(2.14)we infer that

k→∞lim Z

Br(a)

|∇0(ω−ωk)|pdS= 0. (2.17)

If1< p <2, then Z

B2r(a)

ζ

β2ω2+|∇0ω|2p−22

0ω− β2S

kω2k+|∇0ωk|2p−22

0ωk.∇0(ω−ωk)

dS

= Z

B2r(a)

ζ

β2S

kω2k+|∇0ω|2p−22

0ω− βS2

kωk2+|∇0ωk|2p−22

0ωk.∇0(ω−ωk)

dS

+ Z

B2r(a)

ζ

β2ω2+|∇0ω|2p−22

− βS2

kωk2+|∇0ω|2p−22

0ω.∇0(ω−ωk)

dS.

(2.18) Up to extracting a subsequence, we have thatωk →ω a.e. inS and that there existsΦ∈L1(S)such that

k|p+|ω|p≤Φ a.e. inS and ∀k≥1. (2.19) Since

β2Skω2k+|∇0ω|2p−22

|∇ω| ≤

βS2kω2k+|∇0ω|2p−12

≤βSp−1

k ωp−1k +|∇0ω|p−1, and

β2ω2+|∇0ω|2p−22

|∇ω| ≤βp−1ωp−1+|∇0ω|p−1, we derive that

β2ω2+|∇0ω|2p−22

βS2kω2k+|∇0ω|2p−22

|∇0ω| ≤2

βp−1Φp−1+|∇0ω|p−1 , which implies that

ζ

β2ω2+|∇0ω|2p−22

βS2kωk2+|∇0ω|2p−22

0ω→0 inLp0(S)

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wherep0is the conjugate ofp, and finally Z

B2r(a)

ζh

β2ω2+|∇0ω|2p−22

− β2S

kω2k+|∇0ω|2p−22

0ω.∇0(ω−ωk)idS→0 ask→ ∞.

(2.20) For the last term on the right-hand side of(2.18), we have, forγ∈R+andA,B∈RN,

γ+|B|2p−22 B−

γ+|A|2p−22 A=

Z 1 0

d dt

γ+|tB+ (1−t)A|2p−22

(tB+ (1−t)A)

dt

= Z 1

0

γ+|tB+ (1−t)A|2p−22 dt

(B−A)

+ (p−2) Z 1

0

γ+|tB+ (1−t)A|2p−42

htB+ (1−t)A.B−Ai(tB+ (1−t)A)dt.

This implies h

γ+|B|2p−22 B−

γ+|A|2p−22

A.B−Ai= Z 1

0

γ+|tB+ (1−t)A|2p−22 dt

|B−A|2

+ (p−2) Z 1

0

γ+|tB+ (1−t)A|2p−42

htB+ (1−t)A.B−Ai2dt.

We observe that Z 1

0

γ+|tB+ (1−t)A|2p−42

htB+ (1−t)A.B−Ai2dt

≤ |B−A|2 Z 1

0

γ+|tB+ (1−t)A|2p−22 dt, and since1< p <2, we finally obtain

h

γ+|B|2p−22 B−

γ+|A|2p−22

A.B−Ai

≥(p−1) Z 1

0

γ+|tB+ (1−t)A|2p−22 dt

|B−A|2

≥(p−1)|B−A|2

γ+ 1 +|B|2+|A|2p−22 .

(2.21)

We plug this estimate into(2.18)withγ=β2kωk2,A=∇0ωandB=∇0ωk, then Z

B2r(a)

ζh β2S

kω2k+|∇0ω|2p−22

0ω− βS2

kωk2+|∇0ωk|2p−22

0ωk.∇0(ω−ωk)idS

≥ Z

B2r(a)

ζ|∇0(ω−ωk)|2

βk2ω2k+ 1 +|∇0ωk|2+|∇0ω|2p−22 dS.

(2.22)

Setφ(.) =βk2ω2k+ 1 +|∇0ωk|2+|∇0ω|2, then Z

Br(a)

|∇0ω− ∇0ωk|pdS= Z

Br(a)

|∇0ω− ∇0ωk|pφp(p−2)4 φp(p−2)4 dS

≤ Z

Br(a)

|∇0ω− ∇0ωk|2φp−22 dS

!p2 Z

Br(a)

φp2dS

!2−p2 .

(2.23)

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Jointly with(2.14)and(2.22)we conclude that(2.17). Step 1 follows by a standard covering argument.

Step 2: We claim thatωkconverges toωinW01,p(S).

Up to a subsequence that we denote again by{k}, we can assume thatωk →ωand∇0ωk → ∇0ωa.e.

inS. Letζ ∈C0(S), then there existsk∈Nsuch that the supportKofζis a compact subset ofSk

for allk≥k. If1< p <2,

βS2kωk2+|∇0ωk|2p−22

|∇0ωk| ≤ |∇0ωk|p−1,

which bounded inLp0(K), then uniformly integrable inKand by Vitali’s convergence theorem βS2

kω2k+|∇0ωk|2p−22

0ωk

β2ω2+|∇0ω|2p−22

0ω, inL1loc(S). Similarly

β2Skωk2+|∇0ωk|2p−22

ωk

β2ω2+|∇0ω|2p−22 ω, inL1loc(S). Ifp≥2

βS2kωk2+|∇0ωk|2p−22

|∇0ωk| ≤c

k|p−1+|∇0ωk|p−1 ,

and we conclude again by Vitali’s convergence theorem that the previous convergences hold. Since Z

Sk

β2S

kω2k+|∇0ωk|2p−22

h∇0ωk.∇0ζidS= (p−1)βSkSk−β0) Z

Sk

β2S

kω2k+|∇0ωk|2p−22 ωkζdS

we conclude thatωis a weak solution of(1.1)withβ=βS.

2.3 Approximations from outside

Proof of Theorem C. SinceSchas a non-empty interior, the existence of a sequence{ω0k}corresponding to solutions of(1.1)inSk0 withβ =βSk0 is the consequence of [13]. The fact that{βSk0}is increasing follows from Proposition 2.2. We denote byβˆ:= ˆβSits limit, and it is smaller or equal toβS. Estimates (2.4)are valid withS0kk0 andβSk0 instead ofS,ωandβ. If we extendω0kby0inSk0cthese estimates are valid with SN−1 instead ofS0k. Then up to a subsequence the existsω ∈ W1,p(SN−1) and a subsequence stil denoted by{k}such thatωk0 * ωweakly inW1,p(SN−1), strongly inLp(SN−1)and a.e. inSN−1. Furthermore, as in the proof of Theorem A, for any compact setK ⊂S,∇0ωk0 → ∇0ω0 inLp(K). This is sufficient to assert thatωis a weak solution of

−div0

βˆ2ω02+|∇0ω0|2p−22

0ω0

= (p−1) ˆβ( ˆβ−β0)

βˆ2ω2+|∇0ω0|2p−22

ω0 inS.

Moreoverω0bS0

k belongs toW01,p(Sk0)for allk. Sinceωk0 = 0inSkc and converges a.e. toω, this last function vanishes a.e. in∪kSkc = (∩kSk)c =Sc. Thereforeωvanishes a.e. inScand since it is quasi continuous, it vanishes,(1−p)- quasi everywhere inSc. From Netrusov’s theorem (see [1, Th 10.1.1]- (iii)) there exists a sequence{ηn} ⊂ C0(S)which converges toω inW1,p(S), thusω ∈ W01,p(S).

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3 Uniqueness

3.1 Uniqueness of exponent β

Proof of Theorem D. IfS is Lipschitz,CS is also Lipschitz. We fixz ∈ S ≈ SN−1∩∂CS and we apply [10, Th 2] inGz =CS ∩B1

2(z)to two separablep-harmonic functionsu(r, σ) = r−βω(σ)and u0(r, σ) =r−β0ω0(σ). There existγ∈(0,12),c10>0andα∈(0,1)such that

ln u(y1)

u0(y1)−ln u(y2) u0(y2)

≤c10|y1−y2|α ∀y1, y2∈CS∩Bγ(z). (3.24) Assume|y1|=|y2|= 1, then

ln ω(y1)

ω0(y1)−ln ω(y2) ω0(y2)

≤c10|y1−y2|α ∀y1, y2∈S∩Bγ(z). (3.25) We denote by`(x, y)the geodesic distance onSN−1and by`(x, K)the geodesic distance from a point x∈SN−1to a subsetK. Since the setSγ ={σ∈S :`(σ, ∂S)≤γ2}can be covered by a finite number of balls with center on∂S, we infer that

ln ω(y1)

ω0(y1)−ln ω(y2) ω0(y2)

≤c11 ∀y1, y2∈Sγ. (3.26) InS\Sγ

2

we can use Harnack inequality to obtain

−c12≤lnω(y1)

ω(y2) ≤c12 ∀y1, y2∈S\Sγ

2

s.t.`(y1, y2)≤ γ2. (3.27) Hence there exists a constantc13>0such that(3.27)holds for anyy1, y2∈S\Sγ

2

, withc12replaced byc13. Furthermore ω0 satisfies the same inequality in S\Sγ

2

. Combining the two inequalities we obtain

−2c13≤lnω(y1)

ω(y2)−lnω0(y1)

ω0(y2)≤2c13 ∀y1, y2∈S\Sγ

2

. (3.28)

Combining this estimate with(3.25)we derive that it holds for ally1, y2∈S. This implies e−2c13 ω(y2)

ω0(y2) ≤ ω(y1)

ω0(y1) ≤e2c13ω(y2)

ω0(y2) ∀y1, y2∈S. (3.29) Assume now that there exist two exponentsβ > β0 > 0 such thatr−βω(.)andr−β0ω0(.)arep- harmonic and positive in the coneCSand vanishes on∂CS. Putθ=ββ0,η=ωand

T(η) =−div0

β2η2+|∇0η|2p−22

0η

−(p−1)β(β−β0)

β2η2+|∇0η|2p−22 η, then

T(η) =−θp−2

β02ω02+|∇0ω0|2p−22

(β−β002+ (p−1)θ(θ−1)|∇0ω0|2

≤0.

Up to multiplyingω0 byλ, we can assume thatη ≤ ωand that the graphs of η andω are tangent in S. Sinceω0 ≤ cω,η = o(ω)near∂S. Hence there existsσ0 ∈ S such thatω(σ0) = η(σ0)and the

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coincidence set ofηandωis a compact subset ofS. We putw=ω−η, since∇ω(σ0) =∇η(σ0)we proceed as in [14, Th 4.1] (see also [4] in the flat case) and derive thatwsatisfies, in a system of local coordinates(σ1, ..., σN−1)nearσ0,

Lw:=−X

`,j

∂σ`

Aj,`

∂w

∂σj

+X

j

Cj

∂w

∂σ`+Cw≥0,

where the matrix(Aj,`)is smooth, symmetric and positive nearσ0and theCjandCare bounded. Hence wis locally zero. By a standard argument of connectedness, this implies that the zero set ofwmust be

empty, contradiction. Henceβ=β0.

3.2 Uniqueness of eigenfunction

The proof is based upon a delicate adaptation of the characterisation of thep-Martin boundary obtained in [10], but we first give a proof in the convex case.

3.2.1 The convex case

Theorem 3.1. AssumeS is a convex spherical subdomain. Then two positive solutions of(1.1)are proportional.

Proof. We recall that a domainSis (geodesically) convex if a minimal geodesic joining two points of S is contained in S. IfS ⊂ SN−1 is convex, the cone CS is convex too. SinceS is convex, it is Lipschitz and by Theorem D,βS = ˆβS :=β. Letωandω0be two positive solutions of(1.1)satisfying supSω= supSω0 = 1. We denote byuω(x) =|x|−βω(.)anduω0(x) =|x|−βω0(.)the corresponding separablep-harmonic functions defined inCS. If0< a < b, we setCSa,b =CS∩(Bb\Ba). Then for 0< <1we denote byuthe unique function which satisfies

−∆pu= 0 inCS,1 u=−βω inCS∩∂B

u= 0 in (CS∩∂B1)∪ ∂CS∩(B1\B) .

(3.30)

Then

(uω−1)+≤u≤uω inCS,1. (3.31)

Furthermore7→ uis increasing. When ↓0,u ↑u0whereu0is positive andp-harmonic inCS1,0, vanishes on∂CS1,0\ {0}and satisfies(3.30)with= 0. In particular

r→0limrβu0(r, σ) =ω(σ) locally uniformly inS. (3.32) We construct the same approximationu0inCS,1withω0instead ofω. Mutadis mutandis(3.31)holds andu0↑u00which is positive andp-harmonic inCS1, satisfies

(uω0−1)+≤u00≤uω0 inCS1,0, and thus

r→0limrβu00(r, σ) =ω0(σ) locally uniformly inS. (3.33) However, by [10, Th 4]u0andu00are proportional. Combined with(3.32),(3.33)it implies the claim.

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3.2.2 Proof of Theorem E

In what follows we borrow most of our construction from [10] that we adapt to the case of an infinite cone a make explicit for the sake of completeness. The nextnondegeneracy propertyof positivep-harmonic functions is proved in [10, Lemma 4.28].

Proposition 3.2. Let Ω ⊂ RN be a bounded Lipschitz domain and 1 < p < ∞. Then there exist constantsρ >0,c14, c15>0depending respectively onΩ(forρ), andp,N and the Lipschitz normM of∂Ω(forc14andc15) with the property that for anyw∈∂Ωand any positivep-harmonic functionu inΩ, continuous inΩ∩B(w)and vanishing on∂Ω∩Bρ(w), one can findξ∈SN−1, independent of u, such that

c−114 u(y)

dist(y, ∂Ω)≤ h∇u(y), ξi ≤ |∇u(y)| ≤c14

u(y)

dist(y, ∂Ω), (3.34) for ally∈CS∩Bρ|w|

c15

(w).

IfΩis replaced by a coneCS, the nondegeneracy property still holds uniformly on∂CS\ {0}.

Corollary 3.3. Let1< p <∞,S⊂SN−1is a Lipschitz domain andCSthe cone generated byS.

(i) Then there exist constantsρ < 12,c14, c15>0depending respectively onS(forρ), andp,Nand the Lipschitz normMof∂Sanddiam(S)(forc14andc15) with the property that for anyw∈∂CSand any positivep-harmonic functionuinCS, continuous inCS∩B2ρ|w|(w)and vanishing on∂CS∩Bρ|w|(w) continuous, one can findξ∈SN−1, independent ofu, such that

c−114 u(y)

dist(y, ∂CS)≤ h∇u(y), ξi ≤ |∇u(y)| ≤c14 u(y)

dist(y, ∂CS), (3.35) for ally∈B ρ

c15

(w)∩CS.

(ii) Then there exist positive constantsκandc16, c17depending onS(forκ), andp,Nand the Lipschitz normM of∂Sanddiam(S)(forc16, c17such that for anya >0and any positivep-harmonic function uinCSa vanishing on∂CS∩Bac, there holds

c−116 u(y)

dist(y, ∂CS) ≤ |∇u(y)| ≤c16

u(y)

dist(y, ∂CS) ∀y∈CSc17a s.t.dist(y, ∂CS)≤κ|y|. (3.36) Letω, ω0 ∈W01,p(S)∩C(S)be positive solutions(1.1). Sinceωω0 is bounded from above and from below inS by positive constants, we can assume, as in the proof of Theorem D, thatω ≥ω0 inSand that the graphs ofωandω0are tangent. hence, ifω6=ω0, thenω > ω0 inSand there exists a sequence {σn}converging toσ0∈∂Sasn→ ∞such that

n→∞lim ω0n)

ω(σn) = 1.

We defineδ1= sup{δ >0 :δω < ω0}. Fort∈(δ1,1), we set φt= sup{ω0, tω} and ψt= inf

t δ1

ω0, ω

(3.37) We also set

vφt(r, σ) =r−βφt(σ) and vψt(r, σ) =r−βψt(σ) ∀(r, σ)∈(0,∞)×S. (3.38)

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