# Lipschitz estimates for multilinear Marcinkiewicz operators 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 , 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). 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 ) 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 ) 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 .✷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 ):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 ):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 , 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 , 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 S. Chanillo, A note on commutators, Indiana Univ. Math. J. 31 (1982), 7-16. W. G. Chen, A Besov estimate for a class of multilinear singular integrals, ActaMath. Sinica, 16 (2000), 613-626. J. Cohen, A sharp estimate for a multilinear singular integral on R n , IndianaUniv. Math. J., 30 (1981), 693-702. J. Cohen and J. Gosselin, On multilinear singular integral operators on R n , StudiaMath., 72 (1982), 199-223.

84 J. Cohen and J. Gosselin, A BMO estimate for multilinear singular integral operators,Illinois J. Math., 30 (1986), 445-465. R. Coifman, R.Rochberg and G.Weiss, Factorization theorems for Hardy spacesin several variables, Ann. of Math. 103 (1976), 611-635. Y. Ding and S. Z. Lu, Weighted boundedness for a class rough multilinear operators,Acta Math. Sinica, 17 (2001), 517-526. J. Garcia-Cuerva and M. J. L. Herrero, A theory of Hardy spaces associated toHerz spaces, Proc. London Math. Soc., 69 (1994), 605-628. J. Garcia-Cuerva and J. L. Rubio de Francia, Weighted norm inequalities andrelated topics, North-Holland Math.16, Amsterdam, 1985. S. Janson, Mean oscillation and commutators of singular integral operators, Ark.Math., 16 (1978), 263-270. L. Z. Liu,Boundedness for Multilinear Marcinkiewicz Operators on certain hardySpaces, Inter. J. of Math. and Math. Sci., 2 (2003), 87-96. S. Z. Lu, Q. Wu and D. C. Yang, Boundedness of commutators on Hardy typespaces, Sci.in China(ser.A), 45 (2002), 984-997. S. Z. Lu and D. C. Yang, The decomposition of the weighted Herz spaces and itsapplications, Sci. in China (ser.A), 38 (1995), 147-158. S. Z. Lu and D. C. Yang, The weighted Herz type Hardy spaces and its applications,Sci. in China(ser.A), 38 (1995), 662-673. M. Paluszynski, Characterization of the Besov spaces via the commutator operatorof Coifman, Rochberg and Weiss, Indiana Univ. Math. J., 44 (1995), 1-17. A. Torchinsky and S. Wang, A note on the Marcinkiewicz integral, Colloq. Math.60/61 (1990), 235-243.

More magazines by this user
Similar magazines