Lipschitz estimates for multilinear Marcinkiewicz operators

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Lipschitz estimates for multilinear Marcinkiewicz operators

72commutators. It is well known that multilinear operators are of great interest inharmonic analysis and have been widely studied by many authors(see [2, 3, 4, 5, 7,11]). In [2], authors obtain the boundedness of multilinear singular integral operatorsgenerated by singular integrals and Lipschitz functions on L p (p > 1) and some Hardyspaces. The main purpose of this paper is to discuss the boundedness properties ofthe multilinear Marcinkiewicz operators on Hardy and Herz-Hardy spaces. Let usfirst introduce some definitions (see [8, 13, 14, 15]). Throughout this paper, M(f)will denote the Hardy-Littlewood maximal function of f, Q will denote a cube of R nwith sides parallel to the axes. Denote the Hardy spaces by H p (R n ). It is well knownthat H p (R n )(0 < p ≤ 1) has the atomic decomposition characterization(see[9]). Forβ > 0, the Lipschitz space Lip β (R n ) is the space of functions f such that||f|| Lipβ = supx, h ∈ R nh ≠ 0|f(x + h) − f(x)|/|h| β < ∞.Let B k = {x ∈ R n : |x| ≤ 2 k }, C k = B k \ B k−1 , k ∈ Z. Denote χ k =χ Ckand χ 0 =χ B0 , where χ E is the characteristic function of the set E.for k ∈ ZDefinition 1. Let 0 < p, q < ∞, α ∈ R.(1) The homogeneous Herz space is defined by˙K qα,p (R n ) = {f ∈ L q loc (Rn \ {0}) : ||f|| ˙K qα,p < ∞},where||f|| ˙K α,pq=⎡⎣∞∑k=−∞⎤2 kαp ||fχ k || p ⎦L q1/p;(2) The nonhomogeneous Herz space is defined byKqα,p (R n ) = {f ∈ L q loc (Rn ) : ||f|| Kα,p < ∞},qwhere||f|| Kα,pq=[ ∞ ∑] 1/p2 kαp ||fχ k || p L + ||fχ 0|| p q L . qk=1


73(3) The homogeneous Herz type Hardy space is defined byHα,p ˙K q (R n ) = {f ∈ S ′ (R n ) : G(f) ∈˙Kα,pq (R n )},where||f|| H ˙K α,pq= ||G(f)|| ˙K α,pq;(4) The nonhomogeneous Herz type Hardy space is defined bywhereHKqα,p (R n ) = {f ∈ S ′ (R n ) : G(f) ∈ Kq α,p (R n )},||f|| HKα,pq= ||G(f)|| Kα,pq;where G(f) is the grand maximal function of f.The Herz type Hardy spaces have the atomic decomposition characterization.Definition 2. Let α ∈ R, 1 < q < ∞. A function a(x) on R n is called a central(α, q)-atom (or a central (a, q)-atom of restrict type), if1) Suppa ⊂ B(0, r) for some r > 0 (or for some r ≥ 1),2) ||a|| L q ≤ |B(0, r)| −α/n ,3)∫ a(x)x η dx = 0 for |η| ≤ [α − n(1 − 1/q)].Lemma 1. (see [14]) Let 0 < p < ∞, 1 < q < ∞ and α ≥ n(1−1/q). A temperatedistribution f belongs to H˙Kα,pq(R n )(or HK α,p (R n )) if and only if there exist central(α, q)-atoms(or central (α, q)-atoms of restrict type)a j supported on B j = B(0, 2 j )and constants λ j , ∑ j |λ j | p < ∞ such that f = ∑ ∞j=−∞ λ j a j (or f = ∑ ∞j=0 λ j a j )in theS ′ (R n ) sense, andq||f|| H ˙K α,pq( or ||f|| HKα,p) ∼q⎛⎝ ∑ j⎞1/p|λ j | p ⎠ .✷Now we can state our results as following.


74Theorem 1. Let 0 < β ≤ 1, max(n/(n + β), n/(n + γ), n/(n + 1/2)) < p ≤ 1 and1/p − 1/q = β/n. If D α A ∈ Lip β (R n ) for |α| = m. Then µ A Ω, µ A S and µ A λ all mapH p (R n ) continuously into L q (R n ).Theorem 2. Let 0 < β < min(1/2, γ). If D α A ∈ Lip β (R n ) for |α| = m. Then˜µ A Ω, ˜µ A S and ˜µ A λ all map H n/(n+β) (R n ) continuously into L 1 (R n ). ✷Theorem 3. Let 0 < β < min(1/2, γ). If D α A ∈ Lip β (R n ) for |α| = m. Thenµ A Ω, µ A S and µ A λ all map H n/(n+β) (R n ) continuously into weak L 1 (R n ). ✷Theorem 4. Let 0 < β ≤ 1, 0 < p < ∞, 1 < q 1 , q 2 < ∞, 1/q 1 − 1/q 2 = β/nand n(1 − 1/q 1 ) ≤ α < min(n(1 − 1/q 1 ) + β, n(1 − 1/q 1 ) + γ, n(1 − 1/q 1 ) + 1/2). IfD α A ∈ Lip β (R n ) for |α| = m. Then µ A Ω, µ A S and µ A α,pλ all map H ˙K q 1(R n ) continuouslyα,pinto ˙K q 2(R n ). ✷Remark. Theorem 4 also holds for the nonhomogeneous Herz type Hardy space.✷3. SOME LEMMASWe begin with some preliminary lemmas.Lemma 2. (see [5]) Let A be a function on R n and D α A ∈ L q (R n ) for |α| = mand some q > n. Then|R m (A; x, y)| ≤ C|x − y| m ∑|α|=m(1| ˜Q(x, y)|∫1/q|D α A(z)| dz) q ,˜Q(x,y)where ˜Q(x, y) is the cube centered at x and having side length 5 √ n|x − y|.Lemma 3. Let 0 < β ≤ 1, 1 < p < n/β, 1/q = 1/p − β/n and D α A ∈ Lip β (R n )for |α| = m. Then µ A Ω, µ A S and µ A λ all map L p (R n ) continuously into L q (R n ).Proof. By Minkowski inequality, for µ A Ω, we haveµ A Ω(f)(x) ≤∫(|Ω(x − y)||R m+1 (A; x, y)| ∫ ∞|f(y)|R n |x − y| m+n−1≤∫|R m+1 (A; x, y)|C|f(y)|dy;R n |x − y| m+n|x−y|dtt 3 ) 1/2dy✷


75For µ A S , note that |x − z| ≤ 2t, |y − z| ≥ |x − z| − t ≥ |x − z| − 3t when |x − y| ≤ t,|y − z| ≤ t, we obtain≤≤≤≤µ A S (f)(x)⎡∫ ∫ ∫⎣R n |x−y|≤t( |Ω(y − z)||Rm+1 (A; x, z)||f(z)|∫[|R m+1 (A; x, z)||f(z)| ∫ ∫CR n |x − z| m |x−y|≤t∫[|R m+1 (A; x, z)||f(z)| ∫ ∞CR n |x − z| m+3/2 |x−z|/2∫|R m+1 (A; x, z)|C|f(z)|dz;R n |x − z| m+n) ⎤2χ|y − z| n−1 |x − z| m Γ(z) (y, t) dydt ⎦t n+3χ Γ(z) (y, t)t −n−3(|x − z| − 3t) 2n−2 dydt ] 1/2dzdt(|x − z| − 3t) 2n−2 ] 1/2dzFor µ A λ , note that |x − z| ≤ t(1 + 2 k+1 ) ≤ 2 k+2 t, |y − z| ≥ |x − z| − 2 k+3 t when|x − y| ≤ 2 k+1 t and |y − z| ≤ t, we haveµ A λ (f)(x)⎡∫ ∫ ∫ () nλ ( ) ⎤2t≤ ⎣|Ω(y − z)||Rm+1 (A; x, z)||f(z)|χR n R n+1 t + |x − y| |y − z| n−1 |x − z| m Γ(z) (y, t) dydt1/2⎦ dzt n+3 +∫|R m+1 (A; x, z)||f(z)|≤ CR n |x − z| m⎡∫ ∞ ∫ () ⎤nλ 1/2tχ× ⎣Γ(z) (y, t) dydt⎦ dz|x−y|≤t t + |x − y| (|x − z| − 3t) 2n−2 t n+3+C0∫R n |R m+1 (A; x, z)||f(z)||x − z| m⎡∫ ∞ ∞∑∫(× ⎣0k=02 k t


76∫|R m+1 (A; x, z)|= C|f(z)|dz.R n |x − z| m+nThus, the lemma follows from [2].✷4. PROOFS OF THEOREMSProof of Theorem 1. To be simply, we denote that T A = µ A Ω or µ A Sor µ A λ . Itsuffices to show that there exists a constant C > 0 such that for every H p -atom a,||T A (a)|| L q ≤ C.Let a be a H p -atom, that is that a supported on a cube Q = Q(x 0 , r), ||a|| L ∞ ≤ |Q| −1/pand ∫ a(x)x η dx = 0 for |η| ≤ [n(1/p − 1)]. We write∫( ∫ )[T A (a)(x)] q dx = + [TR n 2Q∫(2Q) A (a)(x)] q dx = I + II.cFor I, taking 1 < p 1 < n/β and q 1 such that 1/p 1 −1/q 1 = β/n, by Holder’s inequalityand the (L p 1, L q 1)-boundedness of T A (see Lemma 3), we see thatI ≤ C||T A (a)|| q L q 1|2Q| 1−q/q 1≤ C||a|| q L p 1 |Q| 1−q/q 1≤ C.To obtain the estimate of II, we need to estimate T A (a)(x) for x ∈ (2Q) c . Let˜Q = 5 √ nQ and Ã(x) = A(x) − ∑ |α|=m 1 α! (Dα A) ˜Qx α . Then R m (A; x, y) = R m (Ã; x, y)and D α Ã(y) = D α A(y) − (D α A) Q .For µ A Ω, we have, by the vanishing moment of a,≤|Ft A (a)(x)|∫ ∣ ∣∣∣∣ Ω(x − y)|R n |x − y| − Ω(x − x ∣0) ∣∣∣∣χ n+m−1 |x − x 0 | n+m−1 Γ(x) (y, t)|R m (Ã; x, y)||a(y)|dy∫χ Γ(x) (y, t)|Ω(x − x 0 )|+|RR n |x − x 0 | n+m−1 m (Ã; x, y) − Rm(Ã; x, x 0)||a(y)|dy∫+(χ∣ Γ(x) (y, t) − χ Γ(x) (x 0 , t)) Ω(x − x 0)R m (Ã; x, x 0)a(y)dyR n |x − x 0 | n+m−1 ∣+ ∑1∫Ω(x − y)(x − y) α D α A(y)a(y)dyα! ∣ Γ(x) |x − y| n+m−1 ∣|α|=m= II 1 + II 2 + II 3 + II 4 .


77we getFor II 1 , by Lemma 2 and the following inequality, for b ∈ Lip β (R n ),|b(x) − b Q | ≤ 1 ∫||b|| Lipβ |x − y| β dy ≤ ||b|| Lipβ (|x − x 0 | + r) β ,|Q| Q∑|R m (Ã; x, y)| ≤ ||D α A|| Lipβ (|x − y| + r) m+β ,|α|=mon the other hand, by the following inequality (see [16]):Ω(x − y)∣|x − y| − Ω(x − x ∣ (0) ∣∣∣∣ r≤n+m−1 |x − x 0 | n+m−1 |x − x 0 | + r γ )n+m |x − x 0 | n+m+γ−1and note that |x − y| ∼ |x − x 0 | for y ∈ Q and x ∈ R n \ Q, we obtain, similar to theproof of Lemma 3,≤≤(∫ ∞) 1/2|II 1 | 2 dt/t 30(rC∫R n |x − x 0 | + r γ ) ∑||D α A|| n+m+1 |x − x 0 | n+m+γ Lipβ |x − x 0 | m+β |a(y)|dy|α|=mC ∑( )|Q|||D α β/n+1−1/pA|| Lipβ + |Q|γ/n+1−1/p.|x − x|α|=m0 | n |x − x 0 | n+γ−βFor II 2 , by the following equality (see [5]):we obtainR m (Ã; x, y) − R m(Ã; x, x 0) = ∑|η|


78≤≤≤∫|R mC(Ã; x, x ∣∫∫0)||a(y)| ∣∣∣χR n |x − x 0 | n+m−1 Γ(x) (y, t)dt/t 3 −C∫R n |R m (Ã; x, x 0)||a(y)||x 0 − y| 1/2|x − x 0 | n+m+1/2 dyC ∑|α|=m|Q| 1+1/(2n)−1/p||D α A|| Lipβ|x − x 0 | ;n+1/2−βχ Γ(x) (x 0 , t)dt/t 3 ∣ ∣∣∣ 1/2dySimilarly,≤(∫ ∞) 1/2|II 4 | 2 dt/t 30C ∑( )|Q|||D α β/n+1−1/pA|| Lipβ + |Q|1+γ/n−1/p |Q|1+1/(2n)−1/p+ .|x − x|α|=m0 | n |x − x 0 |n+γ−β|x − x 0 | n+1/2−βFor µ A S , we write≤|Ft A (a)(x, y)|∫ ∣ ∣∣∣∣ Ω(y − z)R n |y − z| n−1 |x − z| − Ω(y − x ∣0) ∣∣∣∣χ m |y − x 0 | n+m−1 Γ(y) (z, t)|R m (Ã; x, z)||a(z)|dz∫χ Γ(y) (x 0 , t)|Ω(y − x 0 )|+|RR n |y − x 0 | n+m−1 m (Ã; x, z) − Rm(Ã; x, x 0)||a(z)|dz∫+(χ∣ Γ(y) (z, t) − χ Γ(y) (x 0 , t)) Ω(y − x 0)R m (Ã; x, x 0)a(z)dzR n |y − x 0 | n+m−1 ∣+ ∑1∫Ω(y − z)(x − z) α D α Ã(z)a(z)dzα! ∣ Γ(y) |y − z| n−1 |x − z| m ∣|α|=m= II 1 + II 2 + II 3 + II 4 .We obtain, similar to the proof of Lemma 3 and µ A Ω,[ ∫ ∫ ] 1/2|II 1 | 2 dydt/t n+3Γ(x)≤≤C∫R n (C ∑|α|=mr|x − x 0 | n+m+1 + r γ|x − x 0 | n+m+γ ) ∑|α|=m( )|Q|||D α β/n+1−1/pA|| Lipβ + |Q|γ/n+1−1/p|x − x 0 | n |x − x 0 | n+γ−β||D α A|| Lipβ |x − x 0 | m+β |a(y)|dy


79and≤≤[ ∫ ∫ ] 1/2|II 2 | 2 dydt/t n+3Γ(x)C ∑|α|=mC ∑|α|=m||D α A|| Lipβ∫⎛⎝ ∑R n|η|


80For µ A λ , similarly,µ A λ (a)(x) ≤ C ∑≤ThusII∞∑∫k=1⎛≤ C ⎝ ∑≤ C2 k+1 Q\2 k Q|α|=m⎛⎝ ∑|α|=m|α|=m[ |Q|||D α β/n+1−1/pA|| Lipβ + |Q|γ/n+1−1/p|x − x 0 | n |x − x 0 |[T A (a)(x)] q dx⎞||D α A|| Lipβ⎠⎞||D α A|| Lipβ⎠q ∞ ∑qk=1,|x − x 0 | n+1/2−β ]n+γ−β+|Q|1+1/(2n)−1/p[2 kqn(1/p−(n+β)/n) + 2 kqn(1/p−(n+γ)/n) + 2 kqn(1/p−(n+1/2)/n)]which together with the estimate for I yields the desired result. This finishes theproof of Theorem 1.✷.Proof of Theorem 2. We only give the proof of ˜µ A Ω and omit the proof of ˜µ A Sand ˜µ A λ for their similarity. It suffices to show that there exists a constant C > 0 suchthat for every H n/(n+β) -atom a supported on Q = Q(x 0 , r), we haveWe write∫R n ˜µ A Ω(a)(x)dx =For J 1 , by the following equality||˜µ A Ω(a)|| L 1 ≤ C.Q m+1 (A; x, y) = R m+1 (A; x, y) − ∑[ ∫ ]+ ˜µ2Q∫(2Q) A Ω(a)(x)dx := J 1 + J 2 .c|α|=mwe have, similar to the proof of Lemma 3,˜µ A Ω(a)(x) ≤ µ A Ω(a)(x) + C ∑ ∫|α|=m1α! (x − y)α (D α A(x) − D α A(y)),R n |D α A(x) − D α A(y)||x − y| n |a(y)|dy,thus, ˜µ A Ω is (L p , L q )-bounded by Lemma 3 and [10], where 1 < p < n/β and 1/q =1/p − β/n. We see thatJ 1 ≤ C||˜µ A Ω(a)|| L q|2Q| 1−1/q ≤ C||a|| L p|Q| 1−1/q ≤ C|Q| 1+1/p−1/q−(n+β)/n .


81To obtain the estimate of J 2 , we denote that Ã(x) = A(x) − ∑ |α|=m 1 α! (Dα A) 2Q x α .Then Q m (A; x, y) = Q m (Ã; x, y).We write, by the vanishing moment of a andQ m+1 (A; x, y) = R m (A; x, y) − ∑ |α|=m 1 α! (x − y)α D α A(x), for x ∈ (2Q) c ,==˜F t A (a)(x)∫Ω(x − y)R m (Ã; x, y)a(y)dy − ∑Γ(x) |x − y| n+m−1∫1 Ω(x − y)D α Ã(x)(x − y) αa(y)dyα! Γ(x) |x − y||α|=mn+m−1− χ Γ(x)(x 0 , t)Ω(x − x 0 )R m (Ã; x, x ]0)|x − y| n+m−1 |x − x 0 | n+m−1∫ [ χ Γ(x) (y, t)Ω(x − y)R m (Ã; x, y)− ∑|α|=m1α!∫ [ χ Γ(x) (y, t)Ω(x − y)(x − y) α− χ Γ(x)(x 0 , t)Ω(x − x 0 )(x − x 0 ) α ]|x − y| n+m−1 |x − x 0 | m×D α Ã(x)a(y)dy,thus, similar to the proof of Theorem 1, we obtain, for x ∈ (2Q) ca(y)dy|˜µ A Ω(a)(x)|≤ C|Q| −β/n ∑([||D α |Q| 1/nA|| Lipβ|x − x|α|=m0 | + |Q| 1/(2n)n+1−β |x − x 0 | + |Q| γ/n )n+1/2−β |x − x 0 | n+γ−β( |Q|+|D α 1/nÃ(x)||x − x 0 | + |Q| 1/(2n))|Q|γ/n+ ],n+1 |x − x 0 |n+1/2−β|x − x 0 | n+γso that,J 2 ≤ C ∑|α|=m||D α A|| Lipβ∞ ∑k=1[2 k(β−1) + 2 k(β−1/2) + 2 k(β−γ)] ≤ C,which together with the estimate for J yields the desired result. This finishes theproof of Theorem 2.Proof of Theorem 3. We only give the proof of µ A Ω. By the following equalityR m+1 (A; x, y) = Q m+1 (A; x, y) + ∑|α|=mand similar to the proof of Lemma 3, we getµ A Ω(f)(x) ≤ ˜µ A Ω(f)(x) + C ∑|α|=m∫1α! (x − y)α (D α A(x) − D α A(y))R n |D α A(x) − D α A(y)||x − y| n |f(y)|dy,✷


82from Theorem 1 and 2 with [10], we obtain|{x ∈ R n : µ A Ω(f)(x) > η}|≤≤|{x ∈ R n : ˜µ A Ω(f)(x) > η/2}|⎧⎫⎨+∣⎩ x ∈ ∑∫|D α A(x) − D α A(y)|⎬Rn :|f(y)|dy > CηR n |x − y| n ⎭∣|α|=mCη −1 ||f|| H n/(n+β).This completes the proof of Theorem 3.✷Proof of Theorem 4. We only give the proof of µ A Ω. Let f ∈ Hα,p ˙K q 1(R n ) andf(x) = ∑ ∞j=−∞ λ j a j (x) be the atomic decomposition for f as in Lemma 1. We write≤||µ A Ω(f)|| α,pp˙K q∞∑k=−∞= L 1 + L 2 .2 kαp ⎛⎝k−3 ∑j=−∞⎞|λ j |||µ A Ω(a j )χ k || L q 2⎠p+∞∑k=−∞2 kαp ⎛⎝∞∑j=k−2For L 2 , by the (L q 1, L q 2) boundedness of µ A Ω(see Lemma 3), we haveL 2 ≤ C≤≤⎧⎪⎨⎪⎩C∞∑k=−∞2 kαp ⎛⎝∞∑j=k−2⎞|λ j |||a j || L q 1⎠C ∑ ∞j=−∞ |λ j | p ( ∑ j+2k=−∞ 2(k−j)αp) , 0 < p ≤ 1p⎞|λ j |||µ A Ω(a j )χ k || L q 2⎠C ∑ ∞j=−∞ |λ j | ( p ∑ ( j+2k=−∞ 2(k−j)αp/2) ∑ ) j+2p/p/2 ′k=−∞ 2(k−j)αp′ , p > 1∞∑|λ j | p ≤ C||f|| p Hj=−∞˙Kα,pq 1.For L 1 , similar to the proof of Theorem 1, we have, for x ∈ C k , j ≤ k − 3,p≤≤µ A Ω(a j )(x)( |Bj | β/nC|x| n+ |B j| 1/(2n)|x| n+1/2−β + |B j| γ/n|x| n+γ−β ) ∫|a j (y)|dyC ( 2 j(β+n(1−1/q 1)−α) |x| −n+2 j(1/2+n(1−1/q 1)−α) |x| β−n−1/2 + 2 j(γ+n(1−1/q 1)−α) |x| β−n−γ) ,


83thus||µ A Ω(a j )χ k || L q 2≤ C2 −kα ( 2 (j−k)(β+n(1−1/q 1)−α) + 2 (j−k)(1/2+n(1−1/q 1)−α) + 2 (j−k)(γ+n(1−1/q 1)−α) )and≤ CL 1∞∑k=−∞( k−3 ∑j=−∞|λ j | ( 2 (j−k)(β+n(1−1/q 1)−α)+2 (j−k)(1/2+n(1−1/q1)−α) + 2 )) (j−k)(γ+n(1−1/q p1)−α)⎧C ∑ ∞j=−∞ |λ j | p ∑ (∞k=j+3 2(j−k)(β+n(1−1/q 1 )−α)⎪⎨+2 (j−k)(1/2+n(1−1/q1)−α) + 2 ) (j−k)(γ+n(1−1/q p1)−α) , 0 < p ≤ 1≤C ∑ ∞j=−∞ |λ j | p ∑ (∞k=j+3 2(j−k)p(β+n(1−1/q 1 )−α)/2⎪⎩+2 (j−k)p(1/2+n(1−1/q1)−α)/2 + 2 ) (j−k)p(γ+n(1−1/q 1)−α)/2, p > 1∞∑≤ C |λ j | p≤j=−∞C||f|| p H˙Kα,pq 1.This finishes the proof of Theorem 4.✷References[1] S. Chanillo, A note on commutators, Indiana Univ. Math. J. 31 (1982), 7-16.[2] W. G. Chen, A Besov estimate for a class of multilinear singular integrals, ActaMath. Sinica, 16 (2000), 613-626.[3] J. Cohen, A sharp estimate for a multilinear singular integral on R n , IndianaUniv. Math. J., 30 (1981), 693-702.[4] J. Cohen and J. Gosselin, On multilinear singular integral operators on R n , StudiaMath., 72 (1982), 199-223.


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