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The second eigenvalue of the fractional p-Laplacian

  • Lorenzo Brasco EMAIL logo and Enea Parini
Published/Copyright: August 29, 2015

Abstract

We consider the eigenvalue problem for the fractional p-Laplacian in an open bounded, possibly disconnected set Ωn, under homogeneous Dirichlet boundary conditions. After discussing some regularity issues for eigenfunctions, we show that the second eigenvalue λ2(Ω) is well-defined, and we characterize it by means of several equivalent variational formulations. In particular, we extend the mountain pass characterization of Cuesta, De Figueiredo and Gossez to the nonlocal and nonlinear setting. Finally, we consider the minimization problem

inf{λ2(Ω):|Ω|=c}.

We prove that, differently from the local case, an optimal shape does not exist, even among disconnected sets. A minimizing sequence is given by the union of two disjoint balls of volume c/2 whose mutual distance tends to infinity.

MSC 2010: 35P30; 47J10; 35R09

Communicated by Juha Kinnunen


A Some useful inequalities I

We will repeatedly use that for 1<p< the real function

Jp(t):=|t|p-2t

is monotone increasing.

Lemma A.1

Lemma A.1 (Towards subsolutions)

Let 1<p< and f:RR be a C1 convex function. Then

(A.1)Jp(a-b)[AJp(f(a))-BJp(f(b))]Jp(f(a)-f(b))(A-B)

for every a,bR and every A,B0.

Proof.

By convexity of f we have

(A.2)f(a)-f(b)f(a)(a-b)andf(a)-f(b)f(b)(a-b).

By writing the left-hand side of (A.1) as

Jp(a-b)[AJp(f(a))-BJp(f(b))]=Jp(f(a)(a-b))A-Jp(f(b)(a-b))B,

we can get the conclusion by simply using (A.2) and the monotonicity of Jp. ∎

The next pointwise inequality generalizes a similar estimate in [4, Appendix C].

Lemma A.2

Lemma A.2 (Towards Moser’s iteration)

Let 1<p< and let g:RR be an increasing function. We define

G(t)=0tg(τ)1p𝑑τ,t.

Then we have

(A.3)Jp(a-b)(g(a)-g(b))|G(a)-G(b)|p.

Proof.

We first observe that we can suppose a>b without loss of generality. Then

Jp(a-b)(g(a)-g(b))=(a-b)p-1bag(τ)𝑑τ=(a-b)p-1baG(τ)p𝑑τ(baG(τ)𝑑τ)p,

thanks to Jensen inequality. ∎

Remark A.3

With the same proof one can show that if g is decreasing, then

|a-b|p-2(a-b)(g(b)-g(a))|H(a)-H(b)|p,

where

H(t)=0t-g(τ)1pdτ.
Lemma A.4

Let β1. Then for every a,b0 we have

(A.4)|a-b|p(aβ-1+bβ-1)(max{1,(3-β)})|a-b|p-2(a-b)(aβ-bβ).

Proof.

We first observe that (A.4) is trivially true for a=b, thus let us consider ab. It is not restrictive to assume that a>b: then (A.4) is equivalent to

(1-t)p(1+tβ-1)C(1-t)p-1(1-tβ)for 0t<1,

that is

(A.5)(1-t)(1+tβ-1)C(1-tβ).

By observing that

(1-t)(1+tβ-1)=(1-tβ)+tβ-1-t,

and remembering that 0t<1, we easily get the conclusion for β2, since tβ-1-t0 in this case. If on the contrary 1<β<2, then by concavity of the function ττβ-1 we have

tβ-1-t=(tβ-1-1)-(t-1)(β-1)(t-1)-(t-1)(2-β)(1-tβ).

This finally shows (A.4) for 1<β<2 as well. The case β=1 is evident. ∎

Lemma A.5

Let p1. Then

(1β)1pβ+p-1p1for every β>0.

Proof.

For p=1 there is nothing to prove, thus let us assume that p>1. The result follows from the convexity of the function ttp, which implies

β-1p(β1p-1).

By adding p on both sides, we get the conclusion. ∎

B Some useful inequalities II

We still use the notation Jp(t)=|t|p-2t.

Lemma B.1

Let 1<p< and U,VR such that UV0. We define the following function:

g(t)=|U-tV|p+Jp(U-V)V|t|p,t.

Then we have

g(t)g(1)=Jp(U-V)U,t.

Proof.

Let us start observing that if U=0, then we have

g(t)=0for every t.

In the same manner, if V=0, then we have

g(t)=|U|pfor every t.

In both cases, the conclusion trivially holds true.

Thus we can suppose that UV0. Then we have

g(t)=-pJp(U-tV)V+pJp(U-V)VJp(t)=pV[Jp(tU-tV)-Jp(U-tV)].

We distinguish two cases.

  1. Case V<0 and U>0. Then we have

    g(t)0t(U-V)U-tVt1.

    This implies that t=1 is a global maximum point for the function g.

  2. Case V>0 and U<0. We now have

    g(t)0t(U-V)U-tVt1,

    since now U<0. Again, we get that t=1 is a global maximum point.

In both cases, we get the desired conclusion. ∎

Lemma B.2

Let 1<p<. Then for every a,bR such that ab0, we have

(B.1)Jp(a-b)a{|a|p-(p-1)|a-b|p-2ab,if 1<p2,|a|p-(p-1)|a|p-2ab,if p>2.

Proof.

We start with some elementary considerations. First of all, there is no loss of generality in supposing a0 and b0. Then we notice that the function Jp on [0,) is convex for p>2 and concave for 1<p2. Thus

(B.2)Jp(x)+Jp(y)(y-x)Jp(y)for 1<p2,0xy,
(B.3)Jp(x)+Jp(x)(y-x)Jp(y)for p>2,0xy.

We now come to the proof of (B.1), starting with the case 1<p2. By using (B.2) with the choices

y=a-bandx=a,

we get

Jp(a-b)Jp(a)-Jp(a-b)b=|a|p-2a-(p-1)|a-b|p-2b,

and multiplying by a0 we conclude. This ends the proof in the case 1<p2.

The case p>2 is handled in a similar manner, by using (B.3) instead of (B.2). ∎

Lemma B.3

Let 1<p<. Then there exists a constant cp>0 such that for every a,bR we have

(B.4)|a-b|p|a|p+|b|p+cp(|a|2+|b|2)p-22|ab|.

Proof.

We first suppose ab0, without loss of generality we can suppose that a,b0 and ab. Then we have

|a-b|p=(a-b)pap|a|p+|b|p,

thus (B.4) is proved.

Let us consider now the case ab0. Without loss of generality, we can suppose that a0 and b0. Then (B.4) is equivalent to

(a+τ)pap+τp+cp(a2+τ2)p-22aτ,a,τ0.

Of course, this is easily seen to be true if a=0, so let us take a>0 and divide the previous by ap. Then we are reduced to show that

(1+m)p1+mp+cp(1+m2)p-22m,

that is

supm>0(1+m)p-1-mpm(1+m2)p-22<+.

To this end, it is sufficient to observe that

limm0+(1+m)p-1-mpm(1+m2)p-22=pandlimm+(1+m)p-1-mpm(1+m2)p-22=p.

This concludes the proof. ∎

Finally, we recall the following classical inequalities. For the proofs the reader is referred to [22, Section 10].

Lemma B.4

For 1<p2, we have

(B.5)(|b|2+|a|2)2-p2(Jp(b)-Jp(a))(b-a)(p-1)|b-a|2,a,b.

For 2<p< we have

(B.6)(Jp(b)-Jp(a))(b-a)22-p|b-a|p,a,b.

Acknowledgements

Part of this work has been conducted during the conference “Journées d’Analyse Appliquée Nice-Toulon-Marseille” held in Porquerolles in May 2014. Organizers and hosting institutions are gratefully acknowledged.

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Received: 2015-2-20
Revised: 2015-7-5
Accepted: 2015-7-21
Published Online: 2015-8-29
Published in Print: 2016-10-1

© 2016 by De Gruyter

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