Poc. Indian Acad. Sci. (Math. Sci. Vol. 24, No. 2, May 24, pp. 93 23. c Indian Academy of Sciences oundedness fo Macinkiewicz integals associated with Schödinge opeatos WENHUA GAO and LIN TANG 2 School of Applied Mathematics, eijing Nomal Univesity Zhuhai, Zhuhai 5985, People s Republic of China 2 LMAM, School of Mathematical Sciences, Peking Univesity, eijing 87, People s Republic of China E-mail: gaowenhua@sina.com; tanglin@math.pku.edu.cn MS eceived 8 Octobe 22 Abstact. Let L = V be a Schödinge opeato, whee is the Laplacian on R n, while nonnegative potential V belongs to the evese Hölde class. In this pape, we will show that Macinkiewicz integal associated with Schödinge opeato is bounded on MO L, and fom HL (Rn to L (R n. Keywods. Macinkiewicz integal; Schödinge opeato. 2 Mathematics Classification. 422, 4225, 4235.. Intoduction Let R n,n 3bethen-dimensional Euclidean space and S n be the unit sphee in R n equipped with nomalized Lebesgue measue dσ = dσ(x.let L s (S n, s be a homogeneous function of degee zeo on R n and satisfy (xdσ(x=. (. S n The Macinkiewicz integal opeato μ is defined by ( μf (x = (x y f(y 2 /2 n. (.2 The above opeato was intoduced by Stein in [8] as an extension of the notion of Macinkiewicz function fom one dimension to highe dimensions. Meanwhile, Stein [8] showed that if Lip α (S n fo some <α, then μ is a bounded opeato on L p (R n fo <p 2, and a bounded mapping fom L (R n to weak L (R n. enedek et al. [] showed that if is continuously diffeentiable in x =, then (.2 above holds fo <p<. Recently, Ding et al. [2] poved that the Macinkiewicz function μ is bounded fom H (R n (Ha space; see [6] to L (R n. On the othe hand, the stu of Schödinge opeato L = V ecently attacted much attention. In paticula, Shen [7] consideed L p estimates fo Schödinge opeatos L with cetain potentials which include Schödinge Riesz tansfoms R L j = 93
94 Wenhua Gao and Lin Tang x j L 2, j =,...,n. Then, Dziubanński and Zienkiewicz [4] intoduced the Ha type space HL (Rn associated with the Schödinge opeato L, which is lage than the classical Ha space H (R n. The dual space of HL (Rn is the MO type space MO L investigated by Dziubanński et al. [5]. Recently, Dong and Liu [3] poved that the Schödinge Riesz tansfoms Rj L ae bounded on MO L(R n. Simila to the classical Macinkiewicz function μ, we define the Macinkiewicz functions μ L j associated with the Schödinge opeato L by ( μ L j f(x= Kj L 2 /2 (x, yf (y, whee Kj L(x, y = K j L(x, y and K j L(x, y is the kenel of RL j = x j L 2,j=,...,n. In paticula, when V =, K j (x, y = K j (x, y = (x y j/ x y and x y n K j (x, y is the kenel of R j = x j 2,j=,...,n. In this pape, we wite K j (x, y = K j (x, y and ( μ j f(x= K j (x, yf (y 2 /2. Obviously, μ j ae classical Macinkiewicz functions. Theefoe, it will be an inteesting thing to stu the popety of μ L j. The main pupose of this pape is to show that Macinkiewicz integals associated with Schödinge opeatos ae bounded on MO L, and fom HL (Rn to L (R n. To state ou esults, let us fist intoduce some notations. Note that a nonnegative locally L q integal function V(x on R n is said to belong to q ( <q< if thee exists C>suchthat the evese Hölde inequality ( /q ( V (y q V(y (.3 (x, (x, (x, (x, holds fo evey x R n and <<, whee(x, denotes the ball centeed at x with adius. It is woth pointing out that the q class is that, if V q fo some q>, then thee exists ɛ>, which depends only on n and the constant C in (.3, such that V qɛ. Thoughout this pape, we always assume that V n. Let ρ(x be the auxiliay function defined by { } ρ(x = sup > : n 2 V(y. (x, Function f is an element of MO L if thee exists C>suchthat sup sup f f C and sup sup x R n <ρ(x (x, x R n ρ(x f C, (x, whee f = f(y. Let f MO L (R n be the smallest C in the inequalities above. It is easy to veify that f MO(R n 2 f MOL (R n fo all f MO L (R n.
oundedness fo Macinkiewicz integals 95 A function f L (R n is said to be in H L (Rn if the semigoup maximal function M L f belongs to L (R n. The nom of such a function is defined by f H L (R n = ML f L (R n, whee M L f(x= sup s> Ts L f(x and {T L s } s> ={e sl } s> is a C contaction semigoup geneated by Schödinge opeato L = V ;see[4]. The authos [5] have poven that MO L (R n is the dual space of HL (Rn. The main esults in this pape ae stated as follows. Theoem. The opeatos μ L j ae bounded on Lp (R n fo <p<, and bounded fom L (R n to weak L (R n. Theoem 2. The opeatos μ L j ae bounded on MO L(R n. Theoem 3. The opeatos μ L j ae bounded fom H L (Rn to L (R n. Thoughout this pape, C denotes the constants that ae independent of the main paametes involved but whose value may diffe fom line to line. y A, we mean that thee exists a constant C>suchthat/C A/. 2. Poof of the main theoems Fist, we need the following lemmas. Lemma [7]. Thee exists l > such that ( C l ρ(y ( ρ(x ρ(x l /(l. ρ(x In paticula, ρ(x ρ(y if <Cρ(x. Lemma 2[7]. Fo any l>, thee exists C l > such that and Kj L (x, y l ( ρ(x l n Kj L (x, y K j (x, y C ρ(x n 2. Poof of Theoem. It suffices to show that μ L j f(x μ j f(x CMf(x, a.e.x R n, (2. whee M denotes the standad Ha Littlewood opeato.
96 Wenhua Gao and Lin Tang Fix x R n and let = ρ(x. Notice that ( μ L j f(x Kj L 2 /2 (x, yf (y ( Kj L 2 /2 (x, yf (y x y ( Kj L 2 /2 (x, yf (y < ( [Kj L (x, y K j (x, y]f(y ( K j (x, yf (y 2 /2 ( Kj L 2 /2 (x, yf (y x y ( Kj L 2 /2 (x, yf (y < := E E 2 E 3 E 4. Fo E, by Lemma 2, we have ( E Obviously, E 2 μ j f(x. Fo E 3, using Lemma 2 again, we get ( E 3 x y f(y f(y n n 2 2 /2 Mf(x. 2 /2 Mf(x. /2 It emains to estimate E 4. Fom Lemma 2, we obtain ( E 4 f(y < n [log 2 t/] (2 k n k= x y 2 k ( ([log 2 t/]mf (x ( t /2 Mf (x2 Mf(x. Thus, Theoem is poved. /2 f(y /2 2 /2
oundedness fo Macinkiewicz integals 97 Poof of Theoem 2. Let f MO L (R n and fix a ball = (x,. We conside two cases ρ(x and <ρ(x.if ρ(x, wite f = fχ fχ ( c = f f 2, whee = (x, 2 and χ S is the chaacteistic function of set S. y Theoem, we know that μ L j ae bounded on L2 (R n.wethenhave ( /2 f (x dx f(x 2 dx ( /2 f(x 2 dx f MOL. Let x. Fom Lemma, ρ(x. y Lemma 2, we get ( μ L j f 2(x = Kj L (x, yf 2(y ( ( ρ(x ρ(x ( ρ(x /2 Kj L 2 /2 (x, yf (y [log 2 t/] k= f(y n (2 k n ([log2 t/] /2 x y 2 k f(y /2 f MOL 2 ρ(x f MOL f MOL. (2.2 Hence f 2(x dx f MOL. If <ρ(x,weset f = fχ fχ ( c = f f 2, whee = (x, 2ρ(x. Note that ρ(x ρ(x fo any x. Simila to (2.2, we have f 2 (x dx f MO L. It emains to pove that thee exists a suitable constant C such that f (x C dx f MOL. (2.3 /2
98 Wenhua Gao and Lin Tang The left of (2.3 is dominated by f (x μ j f (x dx μ j f (x C dx. Let x and x,k = (x, 2 2 k ρ(x, k =,,... Note that ρ(x ρ(x.let f( = f(y and f( = f(y. It is easy to see that f( x, f( x, C f MOL. Note that f( x,k f( x,k f( x,k f( x,k C f MO, so f( x,k f( x,k C(k f MOL. Fom this, we have x,k f(x dx (k x,k f MOL. (2.4 y Lemma 2 and (2.4, we obtain f (x μ j f ( (x ( 4ρ(x x y <min{t,4ρ(x } x y <t ( C 4ρ(x ( 4ρ(x ρ(x C ( 4ρ(x 4ρ(x C 4ρ(x f MOL. Kj L (x, y K j (x, y f(y Kj L (x, y K j (x, y f(y x y <2ρ(x x y <t x y <2ρ(x ρ(x 2 k= 2 k= /2 Kj L (x, y K j (x, y f(y f(y f(y (2 k t 2 n (2 k ρ(x n n 2 n x y <2 k t x y <2 k ρ(x /2 /2 f(y f(y /2 /2 /2 /2
oundedness fo Macinkiewicz integals 99 Thus f (x μ j f (x dx f MO L. We next need only to show μ j f (x C dx f MOL. Let k = (x, 2 k ρ(x, k =,,...,k,wheek satisfies 2 k ρ(x < 2 k ρ(x. Note that μ j f = μ j (f f( k.set f f( k = (f f( k χ k (f f( k χ \ k f( k χ ( c = f, f,2 f,3. y the L 2 (R n boundedness of μ j,wehave ( μ j f, (x dx ( k /2 μ j f, (x 2 dx f MO. k f f( k 2 dx /2 Now conside the tem f,2.next,givenx, we estimate I := μ j f,2 (x μ j f,2 (x. It is easy to see that ( I ( ( := I I 2 I 3. < x y x y t< x y x y t, K j (x, yf,2 (y /2 K j (x,yf,2 (y /2 (K j (x, y K j (x, yf,2 (y /2 Since the estimates fo I and I 2 follow along simila lines, we only conside I.Since x y, by Minkowski s inequality, we have I R n /2 ( f,2 (y n x y t< x y R n f,2 (y n/2 /2
2 Wenhua Gao and Lin Tang k k= 2 (k k /2 k k ( f(y f( k f( k f( k k (k k 2 (k k/2 f MO f MO. k= Fo I 3, again by Minkowski s inequality, we get ( f,2 (y I 3 R n n f,2 (y n R n k k= 2 (k k k k x y t, k (k k 2 (k k f MO k= f MO. /2 ( f(y f( k f( k f( k Thus μ j f,2 (x μ j f,2 (x dx f MOL. Fo the thid tem, simila to above aguments, by (2.4, we have μ j f,3 (x μ j f,3 (x ( f(k ( f(k ( f(k x y t< x y < x y x y t, /2 2 K j (x, yχ ( c(y f(k ( /2 x y >ρ(x x y >ρ(x n (k 2 k /2 f MOL 2 /2 K j (x,yχ ( c(y 2 /2 K j (x, yχ ( c(y (K j (x, y n/2 f MOL.
oundedness fo Macinkiewicz integals 2 Theefoe μ j f,3 (x μ j f,3 (x dx f MOL. Thus, the poof of Theoem 2 is completed. Poof of Theoem 3. Fom the atom decomposition theoy (see [4], we only need to pove that thee exists a constant C>such that fo any HL atom a(x, μl j a L (R n C. A function a(x is said to be HL atom if it satisfies the following conditions: (i supp a (x,, (ii a 2 /2 and (iii if <ρ(x,then (x, a(xdx =. Let supp a (x,and = (x, 2.Wethenhave μ L j a L (R n = μ L j a(xdx μ L ( c j a(xdx = II II 2. Fo II,usingL 2 boundedness of μ j,weget ( /2 μ L j a(xdx μ L j a(x 2 dx /2. To estimate II 2, we conside two cases fo. If ρ(x, note that x y x x fo x ( c and y. y Lemma, we obtain ( II 2 K L 2 /2 ( c j (x, ya(y dx ( ρ(x2(l x y n 2l a(y 2 /2 ( c dx ( ρ(x 2 x x n 2l 2l a(y 2 /2 ( c dx ( x x n ρ(x /2 ( c dx ( ρ(x 2. (2.5 If <ρ(x,wehave ( II 2 = Kj L 2 /2 (x, ya(y dx 2< x x <2ρ(x ( 2ρ(x x x := II 2, II 2,2. Kj L 2 /2 (x, ya(y dx
22 Wenhua Gao and Lin Tang Simila to the poof of (2.5, we have II 2,2. Fo II,,wehave II 2, 2< x x <2ρ(x ( ρ(x ( 2< x x <2ρ(x 2< x x <2ρ(x := III III 2 III 3. ρ(x ( ρ(x Fo III,sinceρ(x ρ(x, we obtain ( ρ(x III ρ(x 2< x x <2ρ(x ρ(x ρ(x /2 2< x x <2ρ(x ρ(x /2 Fo III 2,weget III 2 2< x x <2ρ(x 2< x x <2ρ(x 2< x x <2ρ(x ρ(x (Kj L (x, y K j (x, ya(y 2 /2 dx Kj L 2 /2 (x, ya(y dx K j (x, ya(y 2 /2 dx a(y x x (n 2 dx. ( ρ(x ( ρ(x 2< x x <2ρ(x n 2 2 /2 dx a(y x x n 2 /2 dx Kj L 2 /2 (x, y a(y dx a(y 2 /2 x x n dx x x (n dx. Fom the aguments in pages 78 79 in [2], we have ( III 3 K j (x, ya(y 2 /2 dx. 2< x x Thus, Theoem 3 is poved.
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