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FRACTIONAL INTEGRAL OPERATOR FORL1VECTOR FIELDS AND ITS APPLICATIONS

Zhibing Zhang

School of Mathematics and Physics, Anhui University of Technology,

Ma’anshan 243032, P. R. China

e-mail: [email protected]

(Received 18 March 2017; accepted 27 October 2017)

This paper studies fractional integral operator for vector fields in weightedL1. Using the

esti-mates on fractional integral operator and Stein-Weiss inequalities, we can give a new proof for a

class of Caffarelli-Kohn-Nirenberg inequalities and establish new div-curl inequalities for vector

fields.

Key words : Stein-Weiss inequality; fractional integral operator;L1vector fields; div-curl in-equalities.

1. INTRODUCTION

For0< λ < n, the fractional integral operatorIλis defined by

Iλf(x) = (∆)−λ2f(x) =Cn,λ

Z

Rn

f(y) |x−y|n−λdy,

which is also called the Riesz potential. It is well-known thatIλ is a bounded linear operator from

Lp(Rn)toLq(Rn), where1/q= 1/p−λ/nand1< p < n/λ, which is so-called Hardy-Littlewood-Sobolev inequality. Stein and Weiss [15] established a doubly weighted generalization as follows.

Theorem 1.1 — [15]. Let 0 < λ < n, 1 < p < ∞, α < pn0, β < nq, α+β 0, and

1

q = 1p +α+nβ−λ. Ifp≤q <∞, then there exists a constantC, independent off, such that °

°

°|x|−βIλf ° ° °

Lq ≤Ck|x|

αfk Lp.

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symmetric functions ifp >1, Theorem 1.1 holds forα+β (n−1)(1q 1p); ifp = 1, Theorem 1.1 holds forα+β >(n−1)(1q 1).

As we are interested in the extreme casep= 1of Stein-Weiss inequality, we turn to the end-point estimates forL1 vector fields, which was pioneered by Bourgain and Brezis [1]. For any vector-valued functionf ∈L1(Rn,Rn), letu= (∆)−1f = Γf be the Newtonian potential off, where Γis defined by

Γ(x) =

  

 

1

2π ln|x|, n= 2,

1

|Sn−1|(n2)|x|n−2, n≥3.

Forn≥2, Bourgain and Brezis [2, 3] proved that

k∇ukLn0(Rn) ≤C(kfkL1(Rn)+kdivfkW˙2,n0(Rn)), n0=n/(n−1). (1.1)

Maz’ya [13] established the weighted inequalities related to (1.1) as follows:

(1) Let1≤q < n0,β = 1−n(11q)andf ∈L1(Rn,Rn)satisfyingdivf = 0. Then it holds

that °

° ° °|x|uβ

° ° ° °

Lq(Rn)

≤CkfkL1(Rn). (1.2)

(2) Let1 < q < n0, β = 1−n(11q),f L1(Rn,Rn) and(∆)−1div f L1(Rn,Rn). Then it holds that

° ° ° °|x|uβ

° ° ° °

Lq(Rn)

≤C(kfkL1(Rn)+

°

°(∆)−1divf°°

L1(Rn)). (1.3)

Soon after, Bousquet and Mironescu [4] gave a short proof of Maz’ya’s result with improvements. They found that the inequality (1.3) holds also forq = 1. In fact, using Leray decomposition and a similar trick used in the proof of Theorem 1.2, we see that the inequality (1.2) and the result of Bousquet and Mironescu are equivalent, see Remark 3.6. Inspired by these inequalities, we try to extend the range of exponents that Theorem 1.1 holds for weighted vector fields. We introduce a more general fractional integral operatorTλdefined by

Tλf(x) =K∗f(x) =

Z

Rn

K(x−y)f(y)dy,

where the kernelK(x)satisfies

(i)|K(x)| ≤C|x|λ−n, if|x| 6= 0; (ii)|K(x−y)−K(x)| ≤C |y|

|x|n+1−λ, if|y| ≤

|x|

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Our main result is that

Theorem 1.2 — Letn 2,0 < λ < n,α <1,β < nq+β > 0, 1q = 1 + α+nβ−λ. Suppose thatK(x)satisfies the conditions in (1.4). If1≤q <∞, then

° °

°|x|−βTλf ° ° °

Lq ≤C(k|x|

αfk L1 +

°

°|x|α(∆)−1divf°°

L1). (1.5)

This paper is organized as follows. In Section 2, we give some notations that have appeared in the context. Section 3 shows the proof of Theorem 1.2. In Section 4, applying the new inequality and Stein-Weiss inequality, we give a new proof to Hardy inequality as well as a class of Caffarelli-Kohn-Nirenberg inequalities and obtain new div-curl inequalities for vector fields.

2. NOTATIONS ANDDEFINITIONS

LetΩRn. The notationD(Ω,Rn)denotes the space ofn-dimensional vector-valued functions that are infinitely differentiable and have compact supports inΩ. LetD0(Ω,Rn)denote the dual space of

D(Ω,Rn). The spaceHp(div,Ω)is defined by

Hp(div,Ω) ={v∈Lp(Ω,Rn) : divv∈Lp(Ω)}

and is provided with the norm

kvkHp(div,Ω) =kvkLp(Ω)+kdivvkLp(Ω).

We denote by Hp0(div,Ω) the closure of C∞

c (Ω,Rn) in Hp(div,Ω). It is well-known that

W1,p(Rn) = W1,p

0 (Rn). Proposition 3.1 below implies that Hp(div,Rn) has the same property, i.e.,Hp(div,Rn) =Hp0(div,Rn).

We recall the definition of curl operator. It is defined as a matrix of ordern≥2. We denote the elements of the matrixCURLvby

CURLijv= ∂vi

∂xj

−∂vj ∂xi

, i, j = 1,2,· · · , n, for anyv= (v1, v2,· · · , vn)∈ D0(Ω,Rn).

For a matrixA= (aij)∈ D0(Ω,Rn

2

), wherei, j= 1,2,· · · , n, we define its divergence by

divA= (

n X

j=1

∂a1j ∂xj ,

n X

j=1

∂a2j ∂xj ,· · ·,

n X

j=1

∂anj ∂xj ).

For anyv= (v1, v2,· · ·, vn)∈ D0(Ω,Rn), it holds that

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Throughout this paper, bold typeface will indicate vector or matrix quantities; normal typeface will be used for vector and matrix components and for scalars. To simplify the notations, we write

k · kLp instead ofk · kLp(Rn).

3. PROOFS OFTHEOREM1.2

Proposition 3.1 — Let1≤p <∞. ThenHp(div,Rn) =Hp0(div,Rn).

PROOF : It suffices to show that Cc(Rn,Rn) is dense in Hp(div,Rn). Assume that v Hp(div,Rn). Setvε=η

ε∗v, whereηis the standard mollifier. We have thatvε∈C∞(Rn,Rn)and

vεvinHp(div,Rn)asε0. Letζ C

c (B2(0))be a cut-off function such that0≤ζ 1and

ζ 1inB1(0). Setvkε(x) =vε(x)ζ(xk), thenvεk Cc∞(Rn,Rn). By the definition of divergence

operator, we get

divvεk(x) = divvε(x)ζ(x

k) +

1

kv

ε(x)·(ζ)(x

k).

Therefore, for any fixedε, we have

kdivvεkdivvεkpLp≤C

ÃZ

|x|>k

|divvε|pdx+1

kkv

εkp Lp

!

0ask→ ∞

and

kvεkvεkpLp

Z

|x|>k

|vε|pdx→0ask→ ∞.

By the above two inequalities and using the fact thatvε→vinHp(div,Rn)asε→0, we know that for anyε >0, there existsk=k(ε)such thatvεkvinHp(div,Rn)asε→0. 2

Since Cc(Rn,Rn) is dense inH1(div,Rn), by the divergence theorem and an approximation argument, we obtain the following corollary.

Corollary 3.2 — For anyvH1(div,Rn), it holds that

Z

Rndivvdx= 0.

Letρ0 ∈Cc∞(R+)be a cut-off function such that0≤ρ0 1and

ρ0(t) =

 

1, t≤ 14,

0, t≥ 1

2.

We denoteρ(y, x) =ρ0

µ

|y| |x|

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Lemma 3.3 — Letn≥2,f ∈L1loc(Rn,Rn)withdivf = 0. Then we have ¯ ¯ ¯ ¯ Z Rn

ρ(y, x)f(y)dy

¯ ¯ ¯ ¯≤C

Z

|y|≤|x2|

|y|

|x||f(y)|dy.

PROOF: For any|x| 6= 0, we haveyiρ(y, x)f(y)H1(div,Rn)and

div(yiρ(y, x)f(y)) =y(yiρ(y, x))·f(y) +yiρ(y, x)divf =y(yiρ(y, x))·f(y).

By Corollary 3.2, we get Z

Rn

div(yiρ(y, x)f(y))dy= 0.

Thus Z

R(y, x)fi(y) +

yi

|y||x|ρ

0 0(

|y| |x|)

n X

j=1

yjfj(y)dy= 0.

So we get ¯

¯ ¯ ¯ Z

R(y, x)fi(y)dy

¯ ¯ ¯ ¯≤C

Z

|y|≤|x2|

|y|

|x||f(y)|dy.2

PROOF OF THEOREM 1.2 : First, we claim that Theorem 1.2 is equivalent to the following statement.

Claim 3.4 : Letn≥2,0< λ < n,α <1,β < nq,α+β >0, 1q = 1 +α+βn−λ anddivf = 0. If

1≤q <∞, then °

°

°|x|−βTλf

° ° °

Lq ≤Ck|x|

αfk

L1. (3.1)

It is easy to see that Claim 3.4 is a special case of Theorem 1.2. We only need to derive Theorem 1.2 from Claim 3.4. We decomposefasf =f+(∆)−1divf− ∇(∆)−1divf, which is called the Leray decomposition off. f+(∆)−1divf is the divergence-free part while(∆)−1divf is the curl-free part. Sincediv (f+(∆)−1divf) = 0, by Claim 3.4, we obtain

° °

°|x|−βTλ(f+(∆)−1divf) ° ° °

Lq ≤C

°

°|x|α(f+(∆)−1divf)°° L1

≤C(k|x|αfkL1 +

°

°|x|α(∆)−1divf°°

L1).

(3.2)

On the other hand, for 1 i < j n, set(gi, gj) = (∂xj(∆)−1div f,−∂xi(∆)−1div f),

gk= 0ifk6=i, j. Then we havedivg= 0. By Claim 3.4, we obtain °

° °

°|x|−βTλ(∂x i(∆)

−1divf)

° ° ° ° Lq + ° ° °

°|x|−βTλ(∂x j(∆)

−1divf)

° ° ° ° Lq 2 ° °

°|x|−βTλg

° ° °

Lq ≤Ck|x|

αgk L1 ≤C

°

°|x|α(∆)−1divf°°

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Hence, we have

° °

°|x|−βTλ((∆)−1divf) ° ° °

Lq ≤C

°

°|x|α(∆)−1divf°°

L1. (3.3)

Therefore, the inequality (3.1) follows from the inequalities (3.2) and (3.3).

Hence, we only need to prove the case ofdivf = 0. The benefit of this observation is that there is no need to deal with the termRRnyiρ(y, x)divf(y)dyin Lemma 3.3 forfis a divergence-free vector

field, which is different from [4].

The proof of Claim 3.4 consists of the following steps.

Step 1 : We writeTλf(x) =J1(x) +J2(x), where

J1(x) =

Z

Rn

ρ(y, x)K(x−y)f(y)dy, J2(x) =

Z

Rn

(1−ρ(y, x))K(x−y)f(y)dy.

By the condition (i) in (1.4) and generalized Minkowski’s inequality, we have

° °

°|x|−βJ2(x)

° ° °

Lq ≤C

° ° ° ° °|x|

−β Z

|y|≥|x4|

|f(y)| |x−y|n−λdy

° ° ° ° °

Lq

≤C

Z

Rn

ÃZ

|x|≤4|y|

|x|−βq |x−y|(n−λ)qdx

!1 q

|f(y)|dy.

Since

Z

|x|≤4|y|

|x|−βq

|x−y|(n−λ)qdx= Z

|x|≤|y2|

|x|−βq

|x−y|(n−λ)qdx+ Z

|y|

2≤|x|≤4|y|

|x|−βq

|x−y|(n−λ)qdx

≤C

Ã

|y|(λ−n)q

Z

|x|≤|y2|

|x|−βqdx+|y|−βq

Z

|x−y|≤5|y|

|x−y|(λ−n)qdx

!

=C|y|n−βq+(λ−n)q =C|y|αq,

here we requiren−βq >0andn+ (λ−n)q >0, then we get

° °

°|x|−βJ2(x)

° ° °

Lq ≤C

Z

Rn

|y|α|f(y)|dy. (3.4)

Step 2 : We writeJ1(x) =J11(x) +J12(x), where

J11(x) =

Z

Rn

ρ(y, x)(K(x−y)−K(x))f(y)dy, J12(x) =

Z

Rn

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Thus by generalized Minkowski’s inequality and the condition (ii) in (1.4), we obtain

° °

°|x|−βJ11(x)

° ° °

Lq

Z

Rn

µZ

Rn

(ρ(y, x)|x|−β|K(x−y)−K(x)|)qdx

¶1 q

|f(y)|dy

≤C

Z

Rn

ÃZ

|x|≥2|y|

|x|(λ−n−1−β)qdx

!1 q

|y||f(y)|dy

=C

Z

Rn(|y|

(α−1)q)1q|y||f(y)|dy=C

Z

Rn|y|

α|f(y)|dy,

(3.5)

here we requiren+ (λ−n−1−β)q <0, i.e.,α <1.

Step 3 : At last we deal with the termJ12(x). By the condition (i) in (1.4), we have

|J12(x)| ≤C 1

|x|n−λ ¯ ¯ ¯ ¯ Z Rn

ρ(y, x)f(y)dy

¯ ¯ ¯ ¯.

Using Lemma 3.3 and generalized Minkowski’s inequality, we get

° °

°|x|−βJ12(x)

° ° °

Lq ≤C

° ° ° ° °|x|

λ−n−β Z

|y|≤|x2|

|y|

|x||f(y)|dy

° ° ° ° ° Lq ≤C Z Rn ÃZ

|x|≥2|y|

|x|(λ−n−β−1)qdx

!1 q

|y||f(y)|dy

=C

Z

Rn

|y|α|f(y)|dy,

(3.6)

here we requiren+ (λ−n−β−1)q <0.

Combining the inequalities (3.4), (3.5) and (3.6), we obtain the result. 2

Remark 3.5 : Letu= (∆)−1f. Forn≥3, using Theorem 1.2 withK(x) =|x|2−n, we have

° ° ° °|xu|β

° ° ° °

Lq

≤C(kfkL1 +k∇(∆)−1divfkL1),

where1≤q < n−2n ,β = 2 +n(1q1). If we setK(x) = |xx|jn in Theorem 1.2, then we can see that our result generalizes the inequality (1.3) in a doubly weighted form.

Remark 3.6 : Applying inequality (1.2) togas the proof of Theorem 1.2, we can get

° ° °

°(∆)

−1(∆)−1divf

|x|β

° ° ° °

Lq

≤C°°(∆)−1divf°°L1,

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4. APPLICATIONS

There are many proofs to Hardy inequality [6, p. 111]. Here we give a new proof of Hardy inequality by the theory of singular integrals. If1< p < n, we can use Theorem 1.1 to prove Hardy inequality

° ° ° °|ux|

° ° ° °

Lp

≤Ck∇ukLpfor anyu∈Cc(Rn).

In fact, for anyu∈Cc(Rn), we have the equality (see [8, Lemma 7.14])

u(x) = 1 |Sn−1|

Z

Rn

(x−y)· ∇u(y) |x−y|n dy.

Ifp >1, by Theorem 1.1, we have

° ° ° °|ux|

° ° ° ° Lp 1 |Sn−1|

° ° ° °|1x|

Z

Rn

|∇u(y)| |x−y|n−1dy

° ° ° °

Lp

≤Ck∇ukLp.

But Theorem 1.1 doesn’t work whenp = 1. Our theorem make it possible to prove the case

p= 1. In view of (2.1), for any functionf ∈Cc(Rn,Rn), we have

k|x|αfkL1 +

°

°|x|α(∆)−1divf°°

L1 2(k|x|αfkL1 +

°

°|x|αdiv(∆)−1CURLf°°

L1).

Therefore, for divergence-free or curl-free smooth vector fields with compact support, the term

°

°(∆)−1divf°°

L1 can be removed in (1.5). Ifp= 1, since

(∆)−1div∇u=−∇uorCURL∇u=0,

settingK(x) = |xx|jn in Theorem 1.2, we get

° ° ° °|ux|

° ° ° ° L1 = 1

|Sn−1|

° ° ° °|x1|

Z

Rn

(x−y)· ∇u(y) |x−y|n dy

° ° ° ° L1 1

|Sn−1|

n X j=1 ° ° ° °|x1|

Z

Rn

xj −yj

|x−y|n∇u(y)dy ° ° ° °

L1

≤Ck∇ukL1.

Moreover, by the same idea, we can give a new proof of a class of Caffarelli-Kohn-Nirenberg inequalities (see [5]) as follows:

(1) Letn≥2. Ifα <1,β < nq,0< α+β≤1and1q = 1 +α+βn−1, then

° ° ° °|xu|β

° ° ° °

Lq

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(2) If1< p <∞,α < pn0,β < nq,α+β≥0andq1 = 1p +α+nβ−1,p≤q <∞, then °

° ° °|xu|β

° ° ° °

Lq

≤Ck|x|α∇ukLp. (4.2)

Takeα= 0in the inequality (4.1), we get

° ° ° °|xu|β

° ° ° °

Lq

≤Ck∇ukL1,

wheren≥2,β = 1−n(11q)and1≤q < n0.

Another application is to establish some new weighted div-curl inequalities. By the way, we point out that div-curl inequalities involvingL1norm have been studied by [2, 10, 11, 14, 16, 17] and the references therein.

Theorem 4.1 — Letu∈Cc(Rn,Rn).

(1) Letn≥3,α <1,β < nq,0< α+β 1and 1q = 1 + α+nβ−1. Ifdivu= 0, then

° ° ° °|xu|β

° ° ° °

Lq

≤Ck|x|αCURLukL1.

(2) Letn≥2,1< p <∞,α < pn0,β < nq+β≥0and1q = 1p+α+nβ−1. Ifp≤q <∞, then °

° ° °|xu|β

° ° ° °

Lq

≤C(k|x|αdivukLp+k|x|αCURLukLp).

PROOF: By Green’s representation formula and the identity (2.1), we obtain

u(x) =

Z

Rn

Γ(x−y)(∆u)(y)dy

=

Z

Rn

Γ(x−y)(div CURLu− ∇(divu))(y)dy

=

Z

Rn

yΓ(x−y)·CURLu(y) +yΓ(x−y)divu(y)dy

= 1

|Sn−1|

Z

Rn

x−y

|x−y|n·CURLu(y) +

x−y

|x−y|ndivu(y)dy,

(4.3)

where the dot product between a vector v and a matrix A = (A1,A2,· · ·,An)T is defined by

v·A= (v·A1,v·A2,· · · ,v·An).

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Next we turn to the casep = 1anddiv u = 0. For1 i < j < k n, set(fi, fj, fk) =

( ∂yi,

∂yj,

∂yk)×(ui, uj, uk),fl = 0ifl 6=i, j, k. Then we havedivf = 0. Applying Theorem 1.2 tof, then we obtain

° ° ° °|x1|β

Z

Rn

xm−ym

|x−y|n(

∂ui

∂yj

∂uj

∂yi)(y)dy ° ° ° °

Lq

≤Ck|x|αfkL1 ≤Ck|x|αCURLukL1,

for any1≤m≤n. Then using (4.3) withdivu= 0, we get the first inequality. 2

We can give another proof to the second inequality in Theorem 4.1. We first state the following two facts:

(1) For anyu∈Cc(Rn,Rn), it holds that

u=−∇(∆)−1div CURLu− ∇(∆)−1∇(divu),

(2) If1< p <∞and−np < α < pn0, then|x|αpis in the classAp(see [12, Proposition 1.4.4]).

Using the above two results and the idea of proof given in [9], we can generalize [9, Lemma 2.4] to any dimensionn≥2:

If1< p <∞and−np < α < pn0, then for anyu∈Cc(Rn,Rn), we have the following weighted

inequality for div-curl-grad operators

k|x|α∇ukLp ≤C(k|x|αdivukLp+k|x|αCURLukLp). (4.4)

By Caffarelli-Kohn-Nirenberg inequality (4.2) and the above inequality, we can also obtain the second conclusion of Theorem 4.1. However, this method is not applicable for the casep= 1.

ACKNOWLEDGEMENT

First, I would like to express my gratitude to my supervisor Prof. Xingbin Pan for guidance and constant encouragement. I also would like to thank Dr. Xingfei Xiang for introducing me some inequalities involvingL1-norm, Deliang Chen, Yong Zeng and Dr. Huyuan Chen for useful discus-sions and suggestions. The work was partly supported by the National Natural Science Foundation of China grant no. 11171111 and by Outstanding Doctoral Dissertation Cultivation Plan of Action (PY2015038).

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17. X. F. Xiang and Z. B. Zhang, Hardy-type inequalities for vector fields with vanishing tangential

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