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A simple proof of the generalized Leibniz rule on bounded Euclidean domains

Quoc-Hung Nguyen, Yannick Sire and Juan-Luis Vázquez ORCID logo
From the journal Forum Mathematicum

Abstract

This paper is devoted to a simple proof of the generalized Leibniz rule in bounded domains. The operators under consideration are the so-called spectral Laplacian and the restricted Laplacian. Equations involving such operators have lately been considered by Constantin and Ignatova in the framework of the SQG equation [P. Constantin and M. Ignatova, Critical SQG in bounded domains, Ann. PDE 2 2016, 2, Article ID 8] in bounded domains, and by two of the authors [Q.-H. Nguyen and J. L. Vázquez, Porous medium equation with nonlocal pressure in a bounded domain, Comm. Partial Differential Equations 43 2018, 10, 1502–1539] in the framework of the porous medium with nonlocal pressure in bounded domains. We will use the estimates in this work in a forthcoming paper on the study of porous medium equations with pressure given by Riesz-type potentials.

MSC 2010: 42B37; 35J25

Communicated by Christopher D. Sogge


Funding statement: The first author is supported by the Shanghai Tech University startup fund. The third author was partially funded by Project PGC2018-098440-B-I00 from MICINN, of the Spanish Government. Also, he partially performed as an honorary professor at University Complutense de Madrid.

A Appendix

In this appendix, we provide two results supporting our conjecture on the failure of the usual form of the Leibniz rule in the case of the restricted Laplacian. Our purpose is to relate a weighted (by a suitable power of the distance function) Lp-norm for p=1,2 of the function to the Lp-norm of its fractional Laplacian. We would like to make in particular three comments:

  1. By the very definition of the restricted Laplacian, since the functions are supported on Ω, the Leibniz rule reduces to estimate the integrals

    d(f(x)-f(y))(g(x)-g(y))|x-y|d+2α𝑑y.

    Since we are interested in estimating L2-norms in Ω, one is led to consider quantities of the type

    Ω×Ω(f(x)-f(y))(g(x)-g(y))|x-y|d+2α𝑑x𝑑y,Ω×Ωc(f(x)-f(y))(g(x)-g(y))|x-y|d+2α𝑑x𝑑y.
  2. It is by now well-known that smooth functions that are compactly supported in Ω and have finite Hα semi-norm behave like dist(x,Ω)α close to the boundary of Ω.

  3. Finally, notice that there are two different ways to define a semi-norm (even in d) in W˙α,p(d), namely

    d×d|f(x)-f(y)|p|x-y|d+pα𝑑x𝑑yandd|(-Δ)α2f(x)|p𝑑x.

    In the case of the whole space d and p=2, these latter norms are equivalent. Actually, according to [24], depending on p, these spaces are ordered for every α(0,1) in bounded domains and they are still equivalent for p=2.

According to the previous remarks, if one seeks for a counter-example, one would need to understand how the L2-norm in d of the commutator behaves with respect to its L2-norm in Ω. The following computations show that the boundary behavior plays a crucial role.

Let α(0,12). Let uεCc(B1(0)) be a cut-off function such that uε=1 in B1-2ε, uε=0 in B1-εc, and |uε|Cε, 0uε1. We easily get first

B1uε(x)2(1-|x|2)2α𝑑x1for all ε(0,110).

and for any 0<α<α0<12,

B1B1|uε(x)-uε(y)|2|x-y|d+2α𝑑x𝑑ys0ε1-2α0.

Indeed,

B1B1|uε(x)-uε(y)|2|x-y|d+2α𝑑x𝑑y=2B1B1-3εB1B1-3ε|uε(x)-uε(y)|2|x-y|d+2α𝑑x𝑑y+2B1B1-2εB1-3ε|1-uε(y)|2|x-y|d+2α𝑑x𝑑y
ε-2α0B1B1-3εB1B1-3ε1|x-y|d-2(α0-α)𝑑x𝑑y+B1B1-2εB1-3ε1|x-y|d+2α𝑑x𝑑y
ε-2α0B1B1-3ε𝑑y+B1B1-2εε-2α𝑑y
ε1-2α0+ε1-2αε1-2α0.

On the other hand, for any xB1,

(A.1)(-Δ)α2uεL2(B1)1.

Indeed, for xB1,

|(-Δ)α2uε(x)-B1uε(x)-uε(y)|x-y|d+α𝑑y|=|B1cuε(x)|x-y|d+α𝑑y||uε(x)|(1-|x|2)α.

So,

(A.2)B1|(-Δ)α2uε(x)-B1uε(x)-uε(y)|x-y|d+α𝑑y|2𝑑x=|B1cuε(x)|x-y|d+α𝑑y|B1|uε(x)|2(1-|x|2)2α𝑑x1.

Moreover, for α<α1<α2<12,

B1|B1uε(x)-uε(y)|x-y|d+α𝑑y|2𝑑xα1B1B1|uε(x)-uε(y)|2|x-y|d+2α1𝑑y𝑑xα1,α2ε1-2α2,

where the second relation follows by (2.2). Combining this with (A.2) yields (A.1). As a consequence, we have

B1uε(x)2(1-|x|2)2α𝑑x(-Δ)α2uεL2(B1)21,

but

B1B1|uε(x)-uε(y)|2|x-y|d+2α𝑑x𝑑y0as ε0.

This is an explicit example that in a bounded domain for α<12 the Hardy inequality in L2 does not hold.

However, the analogous result in L1 does hold.

Theorem A.1.

Let uCc(Ω) be a solution to

du(x)-u(y)|x-y|d+2α𝑑y=f(x)

in Ω, where Ω is a smooth bounded domain. Then

(A.3)Ω|u(x)|dist(x,Ω)2α𝑑x+uW2α-δ,1(d)Ω,δfL1(Ω)

for any δ(0,α4)

Proof.

Using Tε(u(x))=sign(u(x))min{ε,|u(x)|} as test function, we obtain

dd|u(x)-u(y)||Tε(u(x))-Tε(u(y))||x-y|d+2α𝑑x𝑑y=2Tε(u(x))f(x)𝑑x.

This implies

ΩcΩ|u(x)||Tε(u(x))||x-y|d+2α𝑑x𝑑y=ΩcΩ|u(x)-u(y)||Tε(u(x))-Tε(u(y))||x-y|d+2α𝑑x𝑑yεfL1(Ω).

So,

ΩcΩ|u(x)|ε-1Tε(|u(x)|)|x-y|d+2α𝑑x𝑑yfL1(Ω).

Letting ε0 yields

ΩcΩ|u(x)||x-y|d+2α𝑑x𝑑yfL1(Ω).

Since

Ωc1|x-y|d+2αdydist(x,Ω)-2α,

we have

(A.4)Ω|u(x)|dist(x,Ω)2α𝑑xfL1(Ω).

Moreover, we can write

du(x)-u(y)|x-y|d+2α𝑑y=𝟏xΩf(x)+𝟏xΩΩ-u(y)|x-y|d+2α𝑑y:=g(x)

for any xd. Thus, by the standard regularity theory, we have

uW2α-δ,1(d)Ω,δgL1(d),

for any δ(0,α2). Since

gL1(d)fL1(Ω)+ΩcΩ|u(y)||x-y|d+2α𝑑y𝑑xfL1(Ω)+Ω|u(y)|dist(y,Ω)2α𝑑yfL1(Ω),

where the second relation follows by (A.4), we get (A.3). ∎

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Received: 2020-08-17
Revised: 2021-08-23
Published Online: 2021-09-25
Published in Print: 2021-11-01

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