Báo cáo hóa học: "CONTINUITY OF MULTILINEAR OPERATORS ON TRIEBEL-LIZORKIN SPACES" potx

TRIEBEL-LIZORKIN SPACESLANZHE LIU Received 4 February 2006; Revised 20 September 2006; Accepted 28 September 2006 The continuity of some multilinear operators related to certain convolut

TRIEBEL-LIZORKIN SPACES

LANZHE LIU

Received 4 February 2006; Revised 20 September 2006; Accepted 28 September 2006

The continuity of some multilinear operators related to certain convolution operators on the Triebel-Lizorkin space is obtained The operators include Littlewood-Paley operator and Marcinkiewicz operator

Copyright © 2006 Lanzhe Liu This is an open access article distributed under the Cre-ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

LetT be the Calder ´on-Zygmund singular integral operator, a well-known result of

Coif-man et al (see [6]) states that the commutator [b, T]( f ) = T(b f ) − bT( f ) (where b ∈

BMO) is bounded onL p(R n) (1< p < ∞); Chanillo (see [1]) proves a similar result when

T is replaced by the fractional integral operator; in [8,9], these results on the Triebel-Lizorkin spaces and the caseb ∈Lipβ (where Lip β is the homogeneous Lipschitz space)

are obtained The main purpose of this paper is to study the continuity of some multi-linear operators related to certain convolution operators on the Triebel-Lizorkin spaces

In fact, we will obtain the continuity on the Triebel-Lizorkin spaces for the multilinear operators only under certain conditions on the size of the operators As the applications, the continuity of the multilinear operators related to the Littlewood-Paley operator and Marcinkiewicz operator on the Triebel-Lizorkin spaces are obtained

2 Notations and results

Throughout this paper,Q will denote a cube of R nwith side parallel to the axes, and for a cubeQ, let f Q = | Q | −1 

Q f (x)dx and f#(x) =supx ∈ Q | Q | −1 

Q | f (y) − f Q | d y For

1≤ r < ∞and 0≤ δ < n, let

M δ,r(f )(x) =sup

x ∈ Q

 1

| Q |1− δr/n



Q

f (y)r

d y

1/r

Hindawi Publishing Corporation

Journal of Inequalities and Applications

Volume 2006, Article ID 58473, Pages 1 11

DOI 10.1155/JIA/2006/58473

we denoteM δ,r(f ) = M r(f ) if δ =0, which is the Hardy-Littlewood maximal function whenr =1 (see [10]) Forβ > 0 and p > 1, let ˙ F p β, ∞be the homogeneous Triebel-Lizorkin space, and let the Lipschitz space ˙∧ βbe the space of functions f such that

 f  ∧˙β = sup

x, h ∈ R n,h =0

Δ[β]+1

h f (x)

whereΔk

hdenotes thekth difference operator (see [9])

We are going to study the multilinear operator as follows

Letm be a positive integer and let A be a function on R n We denote

R m+1(A; x, y) = A(x) − 

| α |≤ m

1

α! D

Definition 2.1 Let F(x, t) define on R n ×[0, +∞), denote

F t(f )(x) =



R n F(x − y, t) f (y)d y,

F A

t (f )(x) =



R n

R m+1(A; x, y)

| x − y | m F(x − y, t) f (y)d y.

(2.4)

LetH be the Hilbert space H = { h :  h  < ∞}such that, for each fixedx ∈ R n,F t(f )(x)

andF A

t(f )(x) may be viewed as a mapping from [0, + ∞) to H Then, the multilinear

operators related toF tis defined by

T A(f )(x) =F A

and also defineT( f )(x) =  F t(f )(x) 

In particular, consider the following two sublinear operators

Definition 2.2 Fix ε > 0, n > δ ≥0 Letψ be a fixed function which satisfies the following

properties:

(1)

ψ(x)dx =0;

(2)| ψ(x) | ≤ C(1 + | x |)−(n+1 − δ);

(3)| ψ(x + y) − ψ(x) | ≤ C | y | ε(1 +| x |)−(n+1+ε − δ)when 2| y | < | x |

The multilinear Littlewood-Paley operator is defined by

g A

δ(f )(x) =

∞ 0

F A

t (f )(x) 2dt

t

1/2

where

F t A(f )(x) =



R n

R m+1(A; x, y)

| x − y | m ψ t(x − y) f (y)d y (2.7)

andψ t(x) = t − n+δ ψ(x/t) for t > 0 Denote that F t(f ) = ψ t ∗ f , and also define that

g δ(f )(x) =

∞ 0

F t(f )(x) 2dt

t

1/2

which is the Littlewood-Paleyg function when δ =0 (see [11])

LetH be the space H = { h :  h  =(∞

0 | h(t) |2dt/t)1/2 < ∞}, then, for each fixedx ∈ R n,

F t A(f )(x) may be viewed as a mapping from [0, + ∞) toH, and it is clear that

g δ(f )(x) =F t(f )(x), g A

δ(f )(x) =F A

t (f )(x). (2.9)

Definition 2.3 Let 0 ≤ δ < n, 0 < γ ≤1 andΩ be homogeneous of degree zero on R n

such that

S n −1Ω(x)dσ(x)=0 Assume thatΩ∈Lipγ(S n −1), that is, there exists a con-stantM > 0 such that for any x, y ∈ S n −1,| Ω(x) − Ω(y) | ≤ M | x − y | γ The multilinear Marcinkiewicz operator is defined by

μ A δ(f )(x) =

∞ 0

F A

t (f )(x) 2dt

t3

1/2

where

F A

t (f )(x) =



| x − y |≤ t

Ω(x − y)

| x − y | n −1− δ

R m+1(A; x, y)

| x − y | m f (y)d y; (2.11) denote

F t(f )(x) =



| x − y |≤ t

Ω(x − y)

| x − y | n −1− δ f (y)d y, (2.12) and also define that

μ δ(f )(x) =

∞ 0

F t(f )(x) 2dt

t3

1/2

which is the Marcinkiewicz operator whenδ =0 (see [12])

LetH be the space H = { h :  h  =(∞

0 | h(t) |2dt/t3)1/2 < ∞} Then, it is clear that

μ δ(f )(x) =F t(f )(x), μ A

δ(f )(x) =F A

t (f )(x). (2.14)

It is clear that Definitions2.2and 2.3are the particular examples ofDefinition 2.1 Note that whenm =0,T Ais just the commutator ofF t andA, while when m > 0, it is

nontrivial generalizations of the commutators It is well known that multilinear oper-ators are of great interest in harmonic analysis and have been widely studied by many authors (see [2–5,7]) The main purpose of this paper is to study the continuity for the multilinear operators on the Triebel-Lizorkin spaces We will prove the following theo-rems inSection 3

Theorem 2.4 Let g δ A be the multilinear Littlewood-Paley operator as in Definition 2.2 If

0< β < min(1, ε) and D α A ∈ ∧˙β for | α | = m, then

(a)g A

δ maps L p(R n ) continuously into ˙ F q β, ∞(R n ), for 1 < p < n/δ and 1/q =1/ p − δ/n;

(b)g δ A maps L p(R n ) continuously into L q(R n ) for 1 < p < n/(δ + β) and 1/ p −1/q =

(δ + β)/n.

Theorem 2.5 Let μ A

δ be the multilinear Marcinkiewiz operator as in Definition 2.3 If 0 <

β < min(1/2, γ) and D α A ∈ ∧˙β for | α | = m, then

(a)μ A δ maps L p(R n ) continuously into ˙ F q β, ∞(R n ) for 1 < p < n/δ and 1/q =1/ p − δ/n,

(b)μ A

δ maps L p(R n ) continuously into L q(R n ) for 1 < p < n/(δ + β) and 1/ p −1/q =

(δ + β)/n.

3 Main theorem and proof

We first prove a general theorem

Theorem 3.1 (main theorem) Let 0 ≤ δ < n, 0 < β < 1, and D α A ∈ ∧˙β for | α | = m Sup-pose F t , T, and T A are the same as in Definition 2.1, if T is bounded from L p(R n ) to L q(R n)

for 1 < p < n/δ and 1/q =1/ p − δ/n, and T satisfies the following size condition:

F A

t (f )(x) − F A

t (f )

| α |= m

D α A

˙

∧ β | Q | β/n M δ,1 f (x) (3.1)

for any cube Q with supp f ⊂(2Q) c and x ∈ Q, then

(a)T A is bounded from L p(R n ) to ˙ F q β, ∞(R n ) for 1 < p < n/δ and 1/q =1/ p − δ/n,

(b)T A is bounded from L p(R n ) to L q(R n ) for 1 < p < n/(δ + β) and 1/q =1/ p −(δ + β)/n.

To prove the theorem, we need the following lemmas

Lemma 3.2 (see [9]) For 0 < β < 1, 1 < p < ∞ ,

 f  F˙β, p ∞ ≈

sup

Q

1

| Q |1+β/n



Q

f (x) − f Qdx

L p

≈

sup

·∈ Q

inf

c

1

| Q |1+β/n



Q

f (x) − cdx

L p

(3.2)

Lemma 3.3 (see [9]) For 0 < β < 1, 1 ≤ p ≤ ∞ ,

 f  ∧˙β ≈sup

Q

1

| Q |1+β/n



Q

f (x) − f Qdx

≈sup

Q

1

| Q | β/n

 1

| Q |



Q

f (x) − f Qp

dx

1/ p

.

(3.3)

Lemma 3.4 (see [1,2]) Suppose that 1 ≤ r < p < n/δ and 1/q =1/ p − δ/n Then

M δ,r(f )

Lemma 3.5 (see [5]) Let A be a function on R n and D α A ∈ L q(R n ) for | α | = m and some

q > n Then

R m(A; x, y)  ≤ C | x − y | m 

| α |= m

 1

 Q(x, y)



Q(x,y)

D α A(z)q

dz

1/q

, (3.5)

where Q(x, y) is the cube centered at x and has side length 5 √ n | x − y | .

Proof of Theorem 3.1 (main theorem) Fix a cube Q = Q(x0,l) and x∈ Q Let Q=5√ nQ and A(x) = A(x) −| α |= m(1/α!)(D α A) Qx α, then R m(A; x, y) = R m(A; x, y) and D α A=

D α A −(D α A) Qfor| α | = m We write, for f1= f χ Qand f2= f χ R n \ Q,

F t A(f )(x) =



R n

R m+1 A; x, y

| x − y | m F(x − y, t) f (y)d y

=



R n

R m+1 A; x, y

| x − y | m F(x − y, t) f2(y)d y

+



R n

R m A; x, y

| x − y | m F(x − y, t) f1(y)d y

| α |= m

1

α!



R n

F(x − y, t)(x − y) α

| x − y | m D α A(y) f 1(y)d y,

(3.6)

then

T A(f )(x) − T A

t (f )(x) − F A

t

f2 x0

≤

F t

R

m A; x, ·

| x − ·| m f1

 (x)



| α |= m

1

α!



F t

(x − ·)α

| x − ·| m D α A f 1(x)



+F A

t

f2 (x) − F A

t

f2 x0 A(x) + B(x) + C(x),

(3.7)

thus,

1

| Q |1+β/n



Q

T A(f )(x) − T A(f )

x0 dx

| Q |1+β/n



Q A(x)dx + 1

| Q |1+β/n



Q B(x)dx

| Q |1+β/n



Q C(x)dx : = I + II + III.

(3.8)

Now, let us estimateI, II, and III, respectively First, for x ∈ Q and y ∈ Q, using Lemmas

3.3and3.5, we get

R m A; x, y C | x − y | m 

| α |= m

sup

x ∈ Q

D α A(x) − D α A Q

≤ C | x − y | m | Q | β/n 

| α |= m

D α A˙

∧ β,

(3.9)

thus, takingr, s such that 1 ≤ r < p and 1/s =1/r − δ/n, by the (L r,L s) boundedness ofT

and Holder’ inequality, we obtain

I ≤ C 

| α |= m

D α A˙

∧ β

1

| Q |



Q

T

f1 (x)dx ≤ C 

| α |= m

D α A˙

∧ βT

f1 L s | Q | −1/s

≤ C 

| α |= m

D α A

˙

∧ βf1

L r | Q | −1/s ≤ C 

| α |= m

D α A

˙

∧ β M δ,r(f )(x).

(3.10) Secondly, using the following inequality (see [9]):

 D α A − D α A Q f χ Q

L r ≤ C | Q |1/s+β/nD α A

˙

∧ β M δ,r(f )(x), (3.11) and similar to the proof ofI, we gain

II ≤ C 

| α |= m

D α A

˙

∧ β M δ,r(f )(x). (3.12) ForIII, using the size condition of T, we have

III ≤ C 

| α |= m

D α A

˙

∧ β M δ,1(f )( x). (3.13)

We now put these estimates together; and taking the supremum over allQ such that

x ∈ Q, and using Lemmas3.2and3.4, we obtain

T A(f )

˙

F β, q ∞ ≤ C 

| α |= m

D α A

˙

This completes the proof of (a)

(b) By same argument as in proof of (a), we have

1

| Q |



Q

T A(f )(x) − T A

f2 x0 dx

≤ C 

| α |= m

D α A˙

∧ β

M δ+β,r(f ) + M δ+β,1(f ) , (3.15)

thus,

T A(f ) #≤ C 

| α |= m

D α A

˙

∧ β

M δ+β,r(f ) + M δ+β,1(f ) (3.16)

Now, usingLemma 3.4, we gain

T A(f )

L q ≤ C T A(f ) #

L q

≤ C 

| α |= m

D α A

˙

∧ β M δ+β,r(f )

L q+M δ+β,1(f )

L q ≤ C  f  L p (3.17)

To prove Theorems2.4and2.5, sinceg δandμ δare all bounded fromL p(R n) toL q(R n) for 1< p < n/δ and 1/q =1/ p − δ/n (see [11,12]), it suffices to verify that g A

δ andμ A δ

satisfy the size condition inTheorem 3.1 (main theorem).

Suppose suppf ⊂(2Q) c and x ∈ Q = Q(x0,l) Note that | x0− y | ≈ | x − y |for y ∈

(2Q) c

Forg δ A, we write

F t A(f )(x) − F t A(f )

x0

=



R n \ Q

ψ

t(x − y)

| x − y | m − ψ t

x0− y

x0− ym



R m A; x, y f (y)d y

+



R n \ Q

ψ t

x0− y f (y)

x0− ym



R m A; x, y − R m A; x0,y d y

| α |= m

1

α!



R n \ Q

ψ

t(x − y)(x − y) α

| x − y | m − ψ t

x0− y x0− y α

x0− ym



D α A(y) f (y)d y

= I1+I2+I3.

(3.18)

By the condition onψ, we obtain

I1 ≤C



R n \ Q

x − x0

x0− ym+1R m A; x, y f (y)∞

0

tdt

t +x0− y 2(n+1 − δ)

1/2

d y

+C



R n \ Q

x − x0ε

x0− ymR m A; x, y f (y)∞

0

tdt

t +x0− y 2(n+1+ε − δ)

1/2

d y

≤ C 

| α |= m

D α A˙

∧ β | Q | β/n∞

k =0



2k+1 Q\2k+1 Q

x − x0

x0− yn+1 − δ+ x − x0ε

x0− yn+ε − δ f (y)d y

≤ C 

| α |= m

D α A

˙

∧ β | Q | β/n

∞



k =1

2− k+ 2− kε

 1

2k Q  1− δ/n



2k Q f (y)d y

≤ C 

| α |= m

D α A

˙

∧ β | Q | β/n M δ,1(f )(x).

(3.19)

ForI2, by the formula (see [5]):

R m A; x, y − R m A; x0,y = 

| η | <m

1

η! R m −| η |

D η A; x, x 0 (x − y) η (3.20)

andLemma 3.5, we get

R m A; x, y − R m A; x0,y C 

| α |= m

D α A˙

∧ β | Q | β/nx − x0x0− ym −1

, (3.21)

thus, similar to the proof ofI1,

I2 ≤C



R n \ Q

R m A; x, y − R m A; x0,y

x0− ym+n − δ f (y)d y

≤ C 

| α |= m

D α A

˙

∧ β | Q | β/n∞

k =0



2k+1 Q\2k Q

x − x0

x0− yn+1 − δf (y)d y

≤ C 

| α |= m

D α A

˙

∧ β | Q | β/n M δ,1(f )(x).

(3.22)

ForI3, similar to the proof ofI1, we obtain

I3 ≤C 

| α |= m



R n \ Q

x − x0

x0− yn+1 − δ + x − x0ε

x0− yn+ε − δ f (y)D α A(y) d y

≤ C 

| α |= m

D α A

˙

∧ β | Q | β/n∞

k =1

2k(β −1)+ 2k(β − ε) M δ,1(f )(x)

≤ C 

| α |= m

D α A

˙

∧ β | Q | β/n M δ,1(f )(x)

(3.23)

so that

F A

t (f )(x) − F t A(f )

| α |= m

D α A

˙

∧ β | Q | β/n M δ,1(f )(x). (3.24)

Forμ A δ, we write

F A

t (f )(x) − F A

t(f )

x0

≤

∞

0



| x − y |≤ t, | x0− y | >t

Ω(x − y)R m A; x, y

| x − y | m+n −1− δ f (y)d y2dt

t3

1/2

+

∞

0



| x − y | >t, | x0− y |≤ t

Ω

x0− y R m A; x0,y

x0− ym+n −1− δ f (y)d y2dt

t3

1/2

+

∞

0



| x − y |≤ t, | x0− y |≤ t





Ω(x − y)R m(A; x, y)

| x − y | m+n −1− δ

−Ω x0− y R m A; x0,y

x0− ym+n −1− δ





 f (y)d y2dt

t3

1/2

+C 

| α |= m

∞ 0



|

x − y |≤ t

Ω(x − y)(x − y) α

| x − y | m+n −1− δ −



| x0− y |≤ t

Ω x0− y x0− y α

x0− ym+n −1− δ

× D α A(y) f (y)d y 





2

dt

t3

1/2

:= J1+J2+J3+J4.

(3.25)

Then

J1≤ C



R n \ Q

f (y)R m A; x, y

| x − y | m+n −1− δ



| x − y |≤ t< | x0− y |

dt

t3

1/2

d y

≤ C



R n \ Q

f (y)R m A; x, y

| x − y | m+n −1− δ

x0− x 1/2

| x − y |3/2 d y

≤ C 

| α |= m

D α A˙

∧ β | Q | β/n∞

k =1

2− k/22k Q1 1− δ/n



2k Q

f (y)d y

≤ C 

| α |= m

D α A

˙

∧ β | Q | β/n M δ,1(f )(x),

(3.26)

similarly, we haveJ2≤ C

| α |= m  D α A  ∧˙β | Q | β/n M δ,1(f )(x).

ForJ3, by the following inequality (see [12]):





 Ω(x − y)

| x − y | m+n −1− δ − Ω x0− y

x0− ym+n −1− δ





 ≤ C

x − x0

x0− ym+n − δ+ x − x0γ

x0− ym+n −1− δ+γ ,

(3.27)

we gain

J3≤ C 

| α |= m

D α A˙

∧ β | Q | β/n



R n \ Q

x − x0

x0− yn − δ + x − x0γ

x0− yn −1− δ+γ

×



| x0− y |≤ t, | x − y |≤ t

dt

t3

1/2f (y)d y

≤ C 

| α |= m

D α A˙

∧ β | Q | β/n∞

k =1

2− k+ 2− γk M δ,1(f )(x)

≤ C 

| α |= m

D α A

˙

∧ β | Q | β/n M δ,1(f )(x).

(3.28)

ForJ4, similar to the proof ofJ1,J2, andJ3, we obtain

J4≤ C 

| α |= m



R n \ Q

x − x0

x0− yn+1 − δ+ x − x0 1/2

x0− yn+1/2 − δ+ x − x0γ

x0− yn+γ − δ

×D α A(y) f (y)d y

≤ C 

| α |= m

D α A˙

∧ β | Q | β/n∞

k =1

2k(β −1)+ 2k(β −1/2)+ 2k(β − γ) 2k1Q 



2k Q

f (y)d y

≤ C 

| α |= m

D α A

˙

∧ β | Q | β/n M δ,1(f )(x).

(3.29) These yield the desired results

Acknowledgment

The author would like to express his gratitude to the referee for his comments and sug-gestions

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˙

This completes the proof of (a)

(b) By same argument as in proof of (a), we have

1

| Q |... δ,r(f )(x). (3.12) ForIII, using the size condition of T, we have

III ≤ C 

|... I1+I2+I3.

(3.18)

By the condition on< i>ψ, we obtain

I1 ≤C