# Fractional Hardy–Sobolev Inequalities with Magnetic Fields

Authors: Min Liu; Fengli Jiang; Zhenyu Guo
Journal: Advances in Mathematical Physics
 (2019)

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## Abstract

A fractional Hardy–Sobolev inequality with a magnetic field is studied in the present paper. Under appropriate conditions, the achievement of the best constant of the fractional magnetic Hardy–Sobolev inequality is established.


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## Body

## 1. Introduction

During the last decades, researchers paid more and more attention to the study of Sobolev inequalities and Hardy–Sobolev inequalities, including the fractional case and magnetic case, see e.g. [1–9].It is well known that the sharp constant of the embeddingD1,2RN↪L2∗RN is attained (see [10]), where 2∗≔2N/N−2 is the Sobolev critical exponent and D1,2RN={u∈L2∗RN:|∇u|∈L2RN}. That is,(1)μ≔infu∈D1,2RNu≠0∫RN∇u2dx∫RNu2∗dx2/2∗is achieved by the so-called Aubin–Talenti instanton (cf. [11, 12]) Uε,x0∈D1,2RN defined by(2)Uε,x0x≔NN−2N−2/4εε2+x−x02N−2/2,whereε>0,x0∈RN. Moreover, Uε,x0 is a positive solution of −Δu=|u|2∗−2u in RN, (3)∫RN∇Uε,x02dx=∫RNUε,x02∗dx=μN/2,and{Uε,x0:ε>0,x0∈RN} contains all positive solutions of −Δu=|u|2∗−2u in RN.For the best constant of the Hardy–Sobolev inequality(4)μt≔infu∈D1,2RNu≠0∫RN∇u2dx∫RNu2∗t/xtdx2/2∗t,where0<t<2 and 2∗t≔2N−t/N−2 is the Hardy–Sobolev critical exponent, it follows from [13] that μt is achieved by functions of the form(5)Uεx≔N−tN−2N−2/22−tεN−2/2ε2−t+x2−t−N−2/2−t,whereε>0. The function Uε is a positive solution of −Δu=|u|2∗t−2u/|x|t in RN, and moreover,(6)∫RN∇Uε2dx=∫RNUε2∗txtdx=μtN−t/2−tFor the fractional Sobolev inequality, consider the Hilbert spaceDsRN defined as Gagliardo seminorm(7)DsRN≔u∈L2s∗RN:ux−uyx−yN/2+s∈L2R2N,where0<s<1, 2s∗≔2N/N−2s is the fractional Sobolev critical exponent and the norm(8)uDs2≔cN,s2∫R2Nux−uy2x−yN+2sdxdyis induced by the scalar product(9)u,vDs≔cN,s2∫R2Nux−uyvx−vyx−yN+2sdxdy.HerecN,s is a dimensional constant precisely given by(10)cN,s=∫RN1−cosζ1ζN+2sdζ−1.Define the best constant for the fractional Sobolev inequality as(11)μs≔infu∈DsRNu≠0uDs2∫RNu2s∗dx2/2s∗.From [14–16], we see that μs is achieved by u∼x=κε2+|x−x0|−N−2s/2, that is, μs=u~Ds2/u~2s∗2. Normalizing u∼ by |u∼|2s∗, we get that u−=u∼/|u∼|2s∗ fulfills(12)μs=infu∈DsRNu2s∗=1uDs2=u¯Ds2andu− is a positive ground state solution of −Δsu=μs|u|2s∗−2u in RN (see Lemma 2.12 in [17]). Denote(13)Us,εx=ε−N−2s/2ûxε,whereu^=μs1/2s∗−2u− is a positive ground state solution of −Δsu=|u|2s∗−2u in RN and(14)ûDs2=∫RNû2s∗dx=μsN/2s.Then, Lemma 2.12 in [17] yields that Us,ε solves −Δsu=|u|2s∗−2u in RN. The fractional Hardy–Sobolev inequality(15)μs,t≔infu∈DsRNu≠0uDs2∫RNu2s∗t/xtdx2/2s∗t,was considered in [18, 19], where 2s∗t≔2N−t/N−2s is the fractional Hardy–Sobolev critical exponent. Marano and Mosconi [18] proved that μs,t is achieved by a optimizer U∈DsRN, whose asymptotic behavior is(16)Ux≃x2s−Nasx→+∞.For the magnetic Hardy–Sobolev inequality, we regard the range of function asC, that is, u:RN→C,N≥3, A=A1,…,AN:RN→RN is a magnetic vector potential. Setting −ΔA≔−i∇+A2,∇A≔∇+iA, and(17)DA1,2RN,C≔u∈L2∗RN,C:∇Au∈L2RN,then−ΔAu=−Δu−iudivA−iA·∇u+|A|2u and DA1,2RN,C is the Hilbert space obtained as the closures of Cc∞RN,C with respect to scaler product(18)Re∫RN∇Au·∇Av¯dx,where the bar denotes complex conjugation. Define(19)μt,A≔infu∈DA1,2RNu≠0∫RN∇Au2dx∫RNu2∗t/xtdx2/2∗t,where2∗t≔2N−t/N−2 is the Hardy–Sobolev critical exponent. Then, by Theorem 1.1 in [20], we see that if A∈LlocNRN,RN, then μt,A is attained by some u∈DA1,2RN,C\{0} if and only if curlA≡0, where curl A is the usual curl operator for N=3 and the N×N skew-symmetric matrix with entries ajk=∂jAk−∂kAj for N≥4.In our paper, we consider a fractional Hardy–Sobolev inequality with a magnetic field. To show our question, we first introduce the fractional magnetic Sobolev spaceDAsRN,C, which is the completion of Cc∞RN,C with respect to the so-called fractional magnetic Gagliardo seminorm ·DAs given by(20)uDAs2≔cN,s2∫R2Ne−ix−y·∫01A1−θx+θydθux−uy2x−yN+2sdxdy,wherecN,s is given by (10), u:RN→C,0<s<1,N≥3, and A=A1,…,AN:RN→RN is a magnetic vector potential. The scalar product in DAsRN,C is(21)u,vDAs≔cN,s2Re∫R2Ne−ix−y·∫01A1−θx+θydθux−uyx−yN+2s·e−ix−y·∫01A1−θx+θydθvx−vy¯dxdy.Although·DAs is a seminorm, by fractional magnetic Sobolev embeddings (see Lemma 3.5 in [21]), we can view ·DAs as a norm ‖·‖DAs≔·DAs in the space DAsRN,C. As in Propositions 2.1 and 2.2 in [21], we can verify that DAsRN,C is a Hilbert space. LpRN,dx/|x|t denotes the space of Lp-integrable functions with respect to the measure dx/|x|t, endowed with norm(22)up,t≔∫RNupxtdx1/p.The aim of the present paper is to investigate the following fractional magnetic Hardy–Sobolev inequality(23)μs,t,A≔infu∈DAsRNu≠0uDAs2u2s∗t,t2,where0<s<1,0<t<2s and 2s∗t≔2N−t/N−2 is the fractional Hardy–Sobolev critical exponent. Problem (23) relates to the fractional magnetic Laplacian defined by(24)−ΔAsux=cN,slimε→0+∫Bεcxux−eix−y·∫01A1−θx+θydθuyx−yN+2sdy,x∈RN.ForN=3 and with mid-point prescription, the fractional magnetic Laplacian was studied in [21]. In particular, d’Avenia and Squassina [21] considered the operator(25)−ΔAsux=cN,slimε→0+∫Bεcxux−eix−y·Ax+y/2uyx−yN+2sdy,x∈R3.Obviously, (24) can be regarded as an extension of the above-mentioned operator involving mid-point prescription. The fractional magnetic Laplacian −ΔAs:DAsRN,C→DA−sRN,C can also be defined by duality as(26)−ΔAsu,v≔cN,s2Re∫R2Ne−ix−y·∫01A1−θx+θydθux−uyx−yN+2s·e−ix−y·∫01A1−θx+θydθvx−vy¯dxdy=cN,s2Re∫R2Nux−eix−y·∫01A1−θx+θydθuyx−yN+2s·vx−eix−y·∫01A1−θx+θydθvy¯dxdy.Denote(27)J=u∈DAsRN,C:u2s∗t,t=1.Then (23) can be characterized as:(28)μs,t,A=infu∈JuDAs2.Our main result is:Theorem 1.
IfA:RN→RN is a continuous function with locally bounded gradient, then μs,t,A is achieved by a nonzero element Us,t,A∈DAsRN,C.


## 2. Proof of the Main Result

To prove Theorem1, we need the following Lemma, which is obtained in [22].Lemma 1 (Diamagnetic inequality).
For anyu∈DAsRN,C, we have(29)ux−ux≤e−ix−y·∫01A1−θx+θydθux−uyfora.e.x,y∈RN
and(30)uDs≤uDAs,
which means|u|∈DsRN,R.By the fractional Hardy–Sobolev inequality (15) and Lemma 1, we have the following lemma.Lemma 2 (Fractional magnetic Sobolev embeddings).
The embedding(31)DAsRN,C→L2s∗tRN,C
is continuous.Proof of Theorem1.
Since the best constantμs,t of fractional Hardy–Sobolev inequality (15) is achievable, we only need to show that(32)μs,t,A=μs,t.
In fact, for anyε>0, there exists u∈Cc∞RN,R such that(33)uDs2≤μs,t+εandu2s∗t,t=1.
Similarly to Lemma 4.6 in [21], for any ϵ>0, consider the scaling(34)uϵx≔ϵ2s−N/2uxϵ,x∈RN.
Substitutingx=ϵX and y=ϵY, we derive that(35)uϵDAs2=cN,s2∫R2Ne−ix−y·∫01A1−θx+θydθuϵx−uϵy2x−yN+2sdxdy=cN,s2∫R2Nϵ2s−Ne−iϵX−Y·∫01Aϵ1−θX+θYdθuX−uY2ϵN+2sX−YN+2sϵ2NdXdY=cN,s2∫R2Ne−iϵx−y·∫01Aϵ1−θx+θydθux−uy2x−yN+2sdxdy
and the following invariance of scaling:(36)uϵDs2=cN,s2∫R2Nuϵx−uϵy2x−yN+2sdxdy=cN,s2∫R2Nϵ2s−NuϵX−uϵY2ϵN+2sX−YN+2sϵ2NdXdY=uDs2,
(37)
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Noticing that|u−v|2=|u|2+|v|2−2Reuv for u,v∈C, we have(38)uϵDAs2−uDs2=cN,s2∫R2Ne−iϵx−y·∫01Aϵ1−θx+θydθux−uy2x−yN+2sdxdy−cN,s2∫R2Nux−uy2x−yN+2sdxdy=cN,s2∫R2N2Re1−e−iϵx−y·∫01Aϵ1−θx+θydθuxuyx−yN+2sdxdy=cN,s∫R2N1−cosϵx−y·∫01Aϵ1−θx+θydθx−yN+2suxuydxdy≕cN,s∫R2NΥϵx,ydxdy=cN,s∫K×KΥϵx,ydxdy,
whereK is compact support of u. Obviously, Υϵx,y→0 a.e. in R2N as ϵ→0. Since A is locally bounded, for x,y∈K, we have(39)1−cosϵx−y·∫01Aϵ1−θx+θydθ≤Cx−y2.
Then, there existsC>0 such that for x,y∈K,(40)Υϵx,y≤Cx−yN−2+2s,ifx−y<1,Cx−yN+2s,ifx−y≥1.
Define(41)mx,y≔Cx−yN−2+2s,ifx−y<1,Cx−yN+2s,ifx−y≥1.
Then,|Υϵx,y|≤mx,y for x,y∈K. Since(42)∫K×Kmx,ydxdy=∫K×K∩x−y<1mx,ydxdy+∫K×K∩x−y≥1mx,ydxdy≤C∫z<11zN−2+2sdz+C∫z≥11zN+2sdz=C∫011rN−2+2srN−1dr+C∫1+∞1rN+2srN−1dr<+∞,
whereC may be different from line to line, we get that m∈L1K×K. By Lebesgue’s Dominated Convergence Theorem, we see that limϵ→0‖uϵ‖DAs2=‖u‖D0s2. Then, it follows from (33) that(43)μs,t,A≤limϵ→0uϵDAs2=uDs2≤μs,t+ε,
which means thatμs,t,A≤μs,t. The opposite inequality holds also because of Lemma 1. Thus, (32) holds, which completes the proof of Theorem 1.



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*Source: 6595961-2019-12-12.xml*