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volume 6, issue 2, article 56, 2005.

Received 14 May, 2005;

accepted 19 May, 2005.

Communicated by:L.-E. Persson

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Journal of Inequalities in Pure and Applied Mathematics

SHARP CONSTANTS FOR SOME INEQUALITIES CONNECTED TO THEP-LAPLACE OPERATOR

JOHAN BYSTRÖM

Luleå University of Technology SE-97187 Luleå, Sweden EMail:johanb@math.ltu.se

c

2000Victoria University ISSN (electronic): 1443-5756 158-05

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Sharp Constants for Some Inequalities Connected to the

p-Laplace Operator Johan Byström

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J. Ineq. Pure and Appl. Math. 6(2) Art. 56, 2005

http://jipam.vu.edu.au

Abstract

In this paper we investigate a set of structure conditions used in the existence theory of differential equations. More specific, we find best constants for the corresponding inequalities in the special case when the differential operator is thep-Laplace operator.

2000 Mathematics Subject Classification:26D20, 35A05, 35J60.

Key words:p-Poisson, p-Laplace, Inequalities, Sharp constants, Structure conditions.

1. Introduction

When dealing with certain nonlinear boundary value problems of the kind ( −div (A(x,∇u)) =f onΩ⊂Rⁿ,

u∈H₀^1,p(Ω), 1< p <∞,

it is common to assume that the function A : Ω×Rⁿ → Rⁿsatisfies suitable continuity and monotonicity conditions in order to prove existence and unique- ness of solutions, see e.g. the books [6], [9], [11] and [12]. For C₁^∗ and C₂^∗ finite and positive constants, a popular set of such structure conditions are the following:

|A(x, ξ₁)−A(x, ξ₂)| ≤C₁^∗(|ξ₁|+|ξ₂|)^p−1−α|ξ₁−ξ₂|^α, hA(x, ξ₁)−A(x, ξ₂), ξ₁−ξ₂i ≥C₂^∗(|ξ₁|+|ξ₂|)^p−β|ξ₁−ξ₂|^β,

where0 ≤ α ≤ min (1, p−1)andmax (p,2)≤ β < ∞. See for instance the articles [1], [2], [3], [4], [7], [8] and [10], where these conditions (or related variants) are used in the theory of homogenization. It is well known that the corresponding function

A(x,∇u) = |∇u|^p−2∇u

for thep-Poisson equation satisfies these conditions, see e.g. [12], but the best possible constantsC₁^∗andC₂^∗ are in general not known. In this article we prove that the best constantsC₁ andC₂for the inequalities

|ξ1|^p−2ξ1− |ξ2|^p−2ξ2

≤C1(|ξ1|+|ξ2|)^p−1−α|ξ1−ξ2|^α, |ξ₁|^p−2ξ₁− |ξ₂|^p−2ξ₂, ξ₁−ξ₂

≥C₂(|ξ₁|+|ξ₂|)^p−β|ξ₁−ξ₂|^β,

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are

C₁ = max 1,2^2−p,(p−1) 2^2−p , C₂ = min 2^2−p,(p−1) 2^2−p

, see Figure1.

0 0.5 1 1.5

2

1 2 3 4 5 6 7 8

p C

2 1

C

Figure 1: The constantsC₁ andC₂plotted for different values ofp.

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2. Main Results

Leth·,·idenote the Euclidean scalar product onRⁿand letpbe a real constant, 1 < p < ∞. Moreover, we will assume that|ξ1| ≥ |ξ2| > 0,which poses no restriction due to symmetry reasons. The main results of this paper are collected in the following two theorems:

Theorem 2.1. Letξ₁, ξ₂ ∈Rⁿand assume that the constantαsatisfies 0≤α≤min (1, p−1).

Then it holds that

|ξ₁|^p−2ξ₁− |ξ₂|^p−2ξ₂

≤C₁(|ξ₁|+|ξ₂|)^p−1−α|ξ₁−ξ₂|^α, with equality if and only if

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ξ₁ =−ξ₂, for1< p <2,

∀ξ₁, ξ₂ ∈Rⁿ, forp= 2,

ξ₁ =ξ₂, for2< p <3andα= 1, ξ₁ =kξ₂,1≤k <∞, forp= 3,

ξ₁ =kξ₂whenk→ ∞, for3< p <∞.

The constantC₁is sharp and given by

C₁ = max 2^2−p,(p−1) 2^2−p,1 .

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Theorem 2.2. Letξ₁, ξ₂ ∈Rⁿand assume that the constantβsatisfies max (p,2)≤β <∞.

Then it holds that

|ξ₁|^p−2ξ₁− |ξ₂|^p−2ξ₂, ξ₁−ξ₂

≥C₂(|ξ₁|+|ξ₂|)^p−β|ξ₁−ξ₂|^β, with equality if and only if

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ξ1 =ξ2, for1< p <2andβ = 2,

∀ξ₁, ξ₂ ∈Rⁿ, forp= 2, ξ₁ =−ξ₂, for2< p <∞.

The constantC₂is sharp and given by

C2 = min 2^2−p,(p−1) 2^2−p .

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3. Some Auxiliary Lemmas

In this section we will prove the four inequalities |ξ₁|^p−2ξ₁− |ξ₂|^p−2ξ₂

≤c₁|ξ₁−ξ₂|^p−1, 1< p≤2, |ξ₁|^p−2ξ₁− |ξ₂|^p−2ξ₂, ξ₁−ξ₂

≥c₂(|ξ₁|+|ξ₂|)^p−2|ξ₁−ξ₂|², 1< p ≤2, |ξ₁|^p−2ξ₁ − |ξ₂|^p−2ξ₂

≤c₁(|ξ₁|+|ξ₂|)^p−2|ξ₁−ξ₂|, 2≤p <∞, |ξ1|^p−2ξ1− |ξ2|^p−2ξ2, ξ1−ξ2

≥c2|ξ1−ξ2|^p, 2≤p <∞.

Note that, by symmetry, we can assume that|ξ₁| ≥ |ξ₂|>0. By putting η₁ = _|ξ^ξ¹

1|, |η₁|= 1, η₂ = _|ξ^ξ²

2|, |η₂|= 1, γ =hη₁, η₂i, −1≤γ ≤1, k = ^|ξ_|ξ¹^|

2| ≥1,

we see that the four inequalities above are in turn equivalent with k^p−1η1−η2

≤c1|kη1−η2|^p−1, 1< p≤2, (3.1)

k^p−1η₁−η₂, kη₁−η₂

≥c₂(k+ 1)^p−2|kη₁−η₂|², 1< p ≤2, (3.2)

k^p−1η₁−η₂

≤c₁(k+ 1)^p−2|kη₁ −η₂|, 2≤p <∞, (3.3)

k^p−1η₁−η₂, kη₁−η₂

≥c₂|kη₁−η₂|^p, 2≤p < ∞.

(3.4)

Before proving these inequalities, we need one lemma.

Lemma 3.1. Letk ≥1andp >1. Then the function h(k) = (3−p) 1−k^p−1

+ (p−1) k−k^p−2

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satisfiesh(1) = 0.Whenk >1, h(k)is positive and strictly increasing forp∈ (1,2)∪(3,∞),and negative and strictly decreasing forp∈(2,3).Moreover, h(k)≡0forp= 2orp= 3.

Proof. We easily see thath(1) = 0.Two differentiations yield h⁰(k) = (p−1) (p−3)k^p−2+ 1−(p−2)k^p−3

, h⁰⁰(k) = (p−1) (p−2) (p−3)k^p−3

1− 1

k

,

with h⁰(1) = 0 and h⁰⁰(1) = 0. When p ∈ (1,2) ∪(3,∞), we have that h⁰⁰(k) > 0for k > 1which implies thath⁰(k) > 0fork > 1, which in turn implies that h(k) > 0fork > 1.When p ∈ (2,3),a similar reasoning gives that h⁰(k) < 0 and h(k) < 0 for k > 1. Finally, the lemma is proved by observing thath(k)≡0forp= 2orp= 3.

Remark 3.2. The special casep = 2is trivial, with equality (c₁ =c₂ ≡ 1) for allξ_i ∈Rⁿin all four inequalities (3.1) – (3.4). Hence this case will be omitted in all the proofs below.

Lemma 3.3. Let1< p <2andξ₁, ξ₂ ∈Rⁿ.Then |ξ₁|^p−2ξ₁− |ξ₂|^p−2ξ₂

≤c₁|ξ₁−ξ₂|^p−1,

with equality if and only ifξ₁ =−ξ₂.The constantc₁ = 2^2−p is sharp.

Proof. We want to prove (3.1) fork ≥1.By squaring and puttingγ =hη₁, η₂i, we see that this is equivalent with proving

k^2(p−1)+ 1−2k^p−1γ ≤c²₁ k²+ 1−2kγp−1

,

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where−1≤γ ≤1.Now construct f₁(k, γ) = k^2(p−1)+ 1−2k^p−1γ

(k²+ 1−2kγ)^p−1 = (k^p−1−1)² + 2k^p−1(1−γ) (k−1)²+ 2k(1−γ)^p−1 . Then

f₁(k, γ)<∞.

Moreover,

∂f₁

∂γ =− 2k

(1−k^p−2) (k^p−1) + (2−p)

2k^p−1(1−γ) + (k^p−1−1)² (k²+ 1−2kγ)^p

<0.

Hence we attain the maximum for f1(k, γ) (and thus also for p

f1(k, γ)) on the borderγ =−1.We therefore examine

g₁(k) = p

f₁(k,−1) = k^p−1+ 1 (k+ 1)^p−1. We have

g₁⁰ (k) =− p−1

(k+ 1)^p 1−k^p−2

≤0,

with equality if and only ifk = 1.The smallest possible constantc₁ for which inequality (3.1) will always hold is the maximum value ofg₁(k),which is thus attained fork = 1.Hence

c₁ =g₁(1) = 2^2−p.

This constant is attained fork = 1andγ =−1,that is, whenξ1 =−ξ2.

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Lemma 3.4. Let1< p <2andξ₁, ξ₂ ∈Rⁿ.Then |ξ₁|^p−2ξ₁− |ξ₂|^p−2ξ₂, ξ₁−ξ₂

≥c₂(|ξ₁|+|ξ₂|)^p−2|ξ₁−ξ₂|², with equality if and only ifξ1 =ξ2.The constantc2 = (p−1) 2^2−p is sharp.

Proof. We want to prove (3.2) fork ≥ 1.By puttingγ = hη1, η2i,we see that this is equivalent with proving

k^p+ 1− k^p−2 + 1

kγ ≥c₂(k+ 1)^p−2 k²+ 1−2kγ , where−1≤γ ≤1.Now construct

f₂(k, γ) = k^p + 1−(k^p−2+ 1)kγ (k+ 1)^p−2(k²+ 1−2kγ)

= (k^p−1−1) (k−1) + (k^p−1 +k) (1−γ) (k+ 1)^p−2 (k−1)²+ 2k(1−γ) . Then

f₂(k, γ)>0.

Moreover,

∂f₂

∂γ =− k(1−k^p−2) (k²−1)

(k+ 1)^p−2(k²+ 1−2kγ)² ≤0,

with equality for k = 1. Hence we attain the minimum for f₂(k, γ) on the borderγ = 1.We therefore examine

g₂(k) =f₂(k,1) = k^p−1−1 (k−1) (k+ 1)^p−2.

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By Lemma3.1we have that

g₂⁰ (k) = (3−p) (1−k^p−1) + (p−1) (k−k^p−2) (k−1)²(k+ 1)^p−1 ≥0,

with equality if and only if k = 1.The largest possible constant c₂ for which inequality (3.2) always will hold is the minimum value ofg₂(k),which thus is attained fork = 1.Hence

c2 = lim

k→1g2(k) = lim

k→1

k^p−1 −1

(k−1) (k+ 1)^p−2 = (p−1) 2^2−p. This constant is attained fork = 1andγ = 1,that is, whenξ₁ =ξ₂. Lemma 3.5. Let2< p <∞andξ₁, ξ₂ ∈Rⁿ.Then

|ξ₁|^p−2ξ₁− |ξ₂|^p−2ξ₂

≤c₁(|ξ₁|+|ξ₂|)^p−2|ξ₁−ξ₂|, with equality if and only if

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ξ1 =ξ2, for2< p <3, ξ₁ =kξ₂when1≤k < ∞, forp= 3, ξ₁ =kξ₂whenk→ ∞, for3< p <∞.

The constantc₁ is sharp, wherec₁ = (p−1) 2^2−pfor2< p <3andc₁ = 1for 3≤p < ∞.

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Proof. We want to prove (3.3) fork ≥1.By squaring and puttingγ =hη₁, η₂i, we see that this is equivalent with proving

k^2(p−1)+ 1−2k^p−1γ ≤c²₁(k+ 1)^2(p−2) k²+ 1−2kγ , where−1≤γ ≤1.Now construct

f₃(k, γ) = k^2(p−1)+ 1−2k^p−1γ (k+ 1)^2(p−2)(k² + 1−2kγ)

= (k^p−1−1)²+ 2k^p−1(1−γ) (k+ 1)^2(p−2) (k−1)²+ 2k(1−γ). Then

f₃(k, γ)<∞.

Moreover,

∂f₃

∂γ = 2k(k^p−2 −1) (k^p−1)

(k+ 1)^2(p−2)(k²+ 1−2kγ) ≥0,

with equality for k = 1.Hence we attain the maximum forf₃(k, γ)(and thus also forp

f₃(k, γ)) on the borderγ = 1.We therefore examine

g3(k) =p

f3(k,1) = k^p−1 −1 (k−1) (k+ 1)^p−2. First we note that

g₃(k)≡1

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whenp = 3,implying thatc₁ = 1with equality for all ξ₁ = kξ₂,1 ≤k < ∞.

Moreover, we have that

g₃⁰ (k) = (3−p) (1−k^p−1) + (p−1) (k−k^p−2) (k−1)²(k+ 1)^p−1 .

By Lemma 3.1 it follows thatg₃(k) ≤ 0for 2 < p < 3 with equality if and only ifk = 1. The smallest possible constantc₁ for which inequality (3.3) will always hold is the maximum value ofg₃(k),which thus is attained fork = 1.

Hence c₁ = lim

k→1g₃(k) = lim

k→1

k^p−1−1

(k−1) (k+ 1)^p−2 = (p−1) 2^2−p, for2< p <3.

This constant is attained fork = 1andγ = 1,that is, whenξ₁ =ξ₂.

Again using Lemma 3.1, we see that g₃(k) ≥ 0 for 3 < p < ∞, with equality if and only if k = 1. The smallest possible constant c₁ for which inequality (3.3) will always hold is the maximum value ofg₃(k),which thus is attained whenk → ∞.Hence

c1 = lim

k→∞g3(k) = lim

k→∞

k^p−1−1

(k−1) (k+ 1)^p−2 = 1, for3< p <∞.

This constant is attained when k → ∞ and γ = 1, that is, when ξ1 = kξ2, k → ∞.

Lemma 3.6. Let2< p <∞andξ₁, ξ₂ ∈Rⁿ.Then |ξ₁|^p−2ξ₁− |ξ₂|^p−2ξ₂, ξ₁−ξ₂

≥c₂|ξ₁−ξ₂|^p, with equality if and only ifξ1 =−ξ2.The constantc2 = 2^2−p is sharp.

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Proof. We want to prove (3.4) fork ≥1.By squaring and puttingγ =hη₁, η₂i, we see that this is equivalent to proving

k^p+ 1− k^p−2 + 1 kγ2

≥c²₂ k²+ 1−2kγp

, where−1≤γ ≤1.Now construct

f₄(k, γ) = (k^p+ 1−(k^p−2+ 1)kγ)² (k²+ 1−2kγ)^p

= ((k^p−1−1) (k−1) + (k^p−1+k) (1−γ))² (k−1)² + 2k(1−γ)p . Then

f₄(k, γ)>0.

Moreover,

∂f₄

∂γ = 2k((p−2)A(k) +B(k))A(k) (k²+ 1−2kγ)^p+1 >0, where

A(k) = k^p−1−1

(k−1) + k^p−1+k

(1−γ), B(k) = k^p−2−1

k²−1 .

Hence we attain the minimum forf₄(k, γ)(and thus also forp

f₄(k, γ)) on the borderγ =−1.We therefore examine

g₄(k) = p

f₄(k,−1) = k^p−1+ 1 (k+ 1)^p−1.

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We have

g₄⁰ (k) = (p−1) (k^p−2−1) (k+ 1)^p ≥0,

with equality if and only if k = 1.The largest possible constant c₂ for which inequality (3.4) will always hold is the minimum value ofg₄(k),which thus is attained fork = 1.Hence

c₂ =g₄(1) = 2^2−p.

This constant is attained fork = 1andγ =−1,that is, whenξ₁ =−ξ₂.

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4. Proof of the Main Theorems

Proof of Theorem2.1. Let1< p <2.Then the condition0≤α≤min (1, p−1)

=p−1implies thatp−1−α≥0.From Lemma3.3it follows that |ξ₁|^p−2ξ₁− |ξ₂|^p−2ξ₂

≤c₁|ξ₁ −ξ₂|^p−1−α|ξ₁−ξ₂|^α

≤c₁(|ξ₁|+|ξ₂|)^p−1−α|ξ₁−ξ₂|^α,

with c₁ = 2^2−p, and we have equality when ξ₁ = −ξ₂.Now let 2 < p < ∞.

Then0≤ α≤ min (1, p−1) = 1implies that1−α ≥ 0.From Lemma3.5it follows that

|ξ₁|^p−2ξ₁− |ξ₂|^p−2ξ₂

≤c₁(|ξ1|+|ξ2|)^p−1−α

(|ξ₁|+|ξ₂|)^1−α |ξ₁−ξ₂|

≤c₁(|ξ₁|+|ξ₂|)^p−1−α|ξ₁−ξ₂|^α, with

a) c₁ = (p−1) 2^2−p,equality forξ₁ =ξ₂ whenα = 1for2< p <3, b) c1 = 1,equality forξ1 =kξ2whenk→ ∞for3< p <∞.

The case p = 2 is trivial and the case p = 3 has equality for ξ₁ = kξ₂, 1 ≤ k < ∞, both cases with constantc1 = 1.The theorem follows by taking these two inequalities together.

Proof of Theorem2.2. Let 1 < p < 2. Then the condition 2 = max (p,2) ≤ β <∞implies thatβ−2≥0.From Lemma3.4it follows that

|ξ1|^p−2ξ1− |ξ2|^p−2ξ2, ξ1−ξ2

≥c2(|ξ1|+|ξ2|)^p−β(|ξ1|+|ξ2|)^β−2|ξ1 −ξ2|²

≥c₂(|ξ₁|+|ξ₂|)^p−β|ξ₁−ξ₂|^β,

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withc₂ = (p−1) 2^2−p and equality forξ₁ =ξ₂ whenβ = 2.Now let2< p <

∞.Thenp= max (p,2)≤β <∞implies thatβ−p≥0.From Lemma3.6it follows that

|ξ₁|^p−2ξ₁− |ξ₂|^p−2ξ₂, ξ₁−ξ₂

≥c₂|ξ₁−ξ₂|^p−β|ξ₁−ξ₂|^β

≥c₂(|ξ₁|+|ξ₂|)^p−β|ξ₁−ξ₂|^β,

with c₂ = 2^2−p and equality for ξ₁ = −ξ₂. The case p = 2 is trivial, with constant c2 = 1. The theorem is proven by taking these two inequalities together.

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[9] S. FU ˇCIK AND A. KUFNER, Nonlinear Differential Equations, Elsevier Scientific, New York, 1980.

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