RIEMANN-STIELTJES OPERATORS FROM F(p, q,s) SPACES TO α-BLOCH SPACES ON THE UNIT BALL SONGXIAO LI Received December 2005; Accepted 19 April 2006 Let H(B) denote the space of all holomorphic functions on the unit ball B Cn We investigate the following integral operators: Tg ( f )(z) = f (tz) g(tz)(dt/t), Lg ( f )(z) = f (tz)g(tz)(dt/t), f ¾ H(B), z ¾ B, where g ¾ H(B), and h(z) = n=1 z j (∂h/∂z j )(z) j is the radial derivative of h The operator Tg can be considered as an extension of the Ces` ro operator on the unit disk The boundedness of two classes of Riemann-Stieltjes a operators from general function space F(p, q,s), which includes Hardy space, Bergman space, Q p space, BMOA space, and Bloch space, to α-Bloch space Ꮾα in the unit ball is discussed in this paper Copyright © 2006 Songxiao Li This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited Introduction Let z = (z1 , ,zn ) and w = (w1 , ,wn ) be points in the complex vector space Cn and ¯ ¯ z,w = z1 w1 + ¡¡¡ + zn wn (1.1) Let dv stand for the normalized Lebesgue measure on Cn For a holomorphic function f we denote Öf = ∂f ∂f , , ∂z1 ∂zn (1.2) Let H(B) denote the class of all holomorphic functions on the unit ball Let f (z) = stand for the radial derivative of f ¾ H(B) [21] It is easy to see that, if f ¾ H(B), f (z) = α aα zα , where α is a multiindex, then n j =1 z j (∂ f /∂z j )(z) α aα z α f (z) = α Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2006, Article ID 27874, Pages 1–14 DOI 10.1155/JIA/2006/27874 (1.3) Riemann-Stieltjes operators from F(p, q,s) to Ꮾα ¾ H(B) such that The α-Bloch space Ꮾα (B) = Ꮾα , α > 0, is the space of all f bα ( f ) = sup   z z¾B f (z) < ½ α (1.4) On Ꮾα the norm is introduced by f Ꮾα = f (0) + bα ( f ) (1.5) With this norm Ꮾα is a Banach space If α = 1, we denote Ꮾα simply by Ꮾ o For a,z ¾ B, a = 0, let ϕa denote the Mă bius transformation of B taking to a defined by ϕa (z) = a   Pa (z)     z Qa (z) ,   z,a (1.6) where Pa (z) is the projection of z onto the one dimensional subspace of Cn spanned by a and Qa (z) = z   Pa (z) which satisfies (see [21]) ϕa Ỉ ϕa = id, ϕa (0) = a,   ϕa (z) ϕa (a) = 0, Let < p, s < ½,  n   < q < ½ A function f F(p, q,s)(B) (see [19, 20]) if f p F(p,q,s) = f (0) p + sup a¾B B Ư f (z) = 1  a 1  z   z,a (1.7) ¾ H(B) is said to belong to F(p, q,s) = 1  z p g (z,a)dv(z) < ½, q s (1.8) where g(z,a) = log ϕa (z)  1 is Green’s function for B with logarithmic singularity at a We call F(p, q,s) general function space because we can get many function spaces, such as BMOA space, Q p space (see [9]), Bergman space, Hardy space, Bloch space, if we take special parameters of p, q,s in the unit disk setting, see [20] If q + s  1, then F(p, q,s) is the space of constant functions For an analytic function f (z) on the unit disk D with Taylor expansion f (z) = ½ a zn , the Ces` ro operator acting on f is a n =0 n Ꮿ f (z) = ½ n =0 n ak z n n + k =0 (1.9) The integral form of Ꮿ is Ꮿ( f )(z) = z z f (ζ) 1 ζ dζ = z z f (ζ) ln 1 ζ ¼ dζ, (1.10) Songxiao Li taking simply as a path of the segment joining and z, we have Ꮿ( f )(z) = f (tz) ln ¼ 1 ζ ζ =tz dt (1.11) The following operator: zᏯ( f )(z) = z f (ζ) dζ, 1 ζ (1.12) is closely related to the previous operator and on many spaces the boundedness of these two operators is equivalent It is well known that Ces` ro operator acts as a bounded linear a operator on various analytic function spaces (see [4, 8, 11–13, 16] and the references therein) Suppose that g ¾ H(D), the operator Jg f (z) = z f (ξ)dg(ξ) = f (tz)zg ¼ (tz)dt = z f ()g ẳ ()d, z ắ D, (1.13) where f ¾ H(D), was introduced in [10] where Pommerenke showed that Jg is a bounded operator on the Hardy space H (D) if and only if g ¾ BMOA The operator Jg acting on various function spaces have been studied recently in [1–3, 14, 17, 18] Another operator was recently defined in [18], as follows: Ig f (z) = z f ¼ (ξ)g(ξ)dξ (1.14) The above operators Jg , Ig can be naturally extended to the unit ball Suppose that g : B C1 is a holomorphic map of the unit ball, for a holomorphic function f , define Tg f (z) = f (tz) dg(tz) = dt f (tz) g(tz) dt , t z ¾ B (1.15) This operator is called Riemann-Stieltjes operator (or extended-Ces` ro operator) It was a introduced in [5], and studied in [5–7, 15, 17] Here, we extend operator Ig for the case of holomorphic functions on the unit ball as follows: Lg f (z) = f (tz)g(tz) dt , t z ¾ B (1.16) To the best of our knowledge operator Lg on the unit ball is introduced in the present paper for the first time The purpose of this paper is to study the boundedness of the two Riemann-Stieltjes operators Tg , Lg from F(p, q,s) to α-Bloch space The corollaries of our results generalized the former results and some results are new even in the unit disk setting 4 Riemann-Stieltjes operators from F(p, q,s) to Ꮾα In this paper, constants are denoted by C, they are positive and may differ from one occurrence to the other a b means that there is a positive constant C such that a Cb Moreover, if both a b and b a hold, then one says that a b Tg ,Lg : F(p, q,s) Ꮾα In order to prove our results, we need some auxiliary results which are incorporated in the following lemmas The first one is an analogy of the following one-dimensional result: z ¼ f (ζ)g ¼ (ζ)dζ z = f (z)g ¼ (z), f ¼ (ζ)g(ζ)dζ ¼ = f ¼ (z)g(z) (2.1) Lemma 2.1 [5] For every f ,g ¾ H(B), it holds that Tg ( f ) (z) = f (z) g(z), Lg ( f ) (z) = g has the expansion Proof Assume that the holomorphic function f Tg ( f ) (z) = α aα (tz)α f (z)g(z) dt = t α aα α z = α α aα z aα z α , (2.2) α Then (2.3) α which is what we wanted to prove The proof of the second formula is similar and will be omitted The following lemma can be found in [19] Lemma 2.2 For < p, s < ½, Ꮾ(n+1+q)/ p and  n   < q < ½, q + s >  1, if f ¾ F(p, q,s), then f ¾ f Ꮾ(n+1+q)/ p C f F(p,q,s) (2.4) The following lemma can be found in [15] Lemma 2.3 If f ¾ Ꮾα , then f (z) ⎧ ⎪ f (0) + f ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ C ⎪ f (0) + f ⎪ ⎪ ⎪ ⎪ ⎪ f (0) + ⎪ ⎩ < α < 1; Ꮾα , Ꮾα log f 1  z Ꮾα , 1  z , α 1 α = 1, (2.5) α > 1, for some C independent of f 2.1 Case p < n + + q In this section we consider the case p < n + + q Our first result is the following theorem Songxiao Li Theorem 2.4 Let g be a holomorphic function on B, < p, s < ½,  n   < q < ½, q + s >  1, n + + q pα, p < n + + q Then Tg : F(p, q,s) Ꮾα is bounded if and only if α+1 (n+1+q)/ p sup z zắB g(z) < ẵ (2.6) Moreover, the following relationship: Tg F(p,q,s) Ꮾα α+1 (q+n+1)/ p sup   z z¾B g(z) (2.7) holds Proof For f ,g ¾ H(B), note that Tg f (0) = 0, by Lemmas 2.1, 2.2, and 2.3, Ꮾα = sup   z α = sup   z Tg f α Tg f (z) z¾B z¾B f (z) C f Ꮾ(n+1+q)/ p sup z¾B C f F(p,q,s) sup z¾B g(z) 1  z 1  z α+1 (n+1+q)/ p (2.8) g(z) α+1 (n+1+q)/ p g(z) Therefore (2.6) implies that Tg : F(p, q,s) Ꮾα is bounded Conversely, suppose Tg : F(p, q,s) Ꮾα is bounded For w ¾ B, let fw (z) = 1  w   z,w (n+1+q)/ p (2.9) It is easy to see that fw (w) = 1  w fw (w) , (n+1+q)/ p 1 1  w w (n+1+q)/ p (2.10) If w = then fw obviously belongs to F(p, q,s) From [19] we know that fw ¾ F(p, q,s), moreover there is a positive constant K such that supw¾B fw F(p,q,s) K Therefore 1  z α fw (z) g(z) =   z α Tg fw (z) Tg fw Ꮾα K Tg F(p,q,s) Ꮾα , (2.11) for every z,w ¾ B From this and (2.10), we get 1  w α+1 (n+1+q)/ p g(w) =   w α fw (w) g(w) K Tg F(p,q,s) Ꮾα , (2.12) from which (2.6) follows From the above proof, we see that (2.7) holds 6 Riemann-Stieltjes operators from F(p, q,s) to Ꮾα Theorem 2.5 Let g be a holomorphic function on B, < p, s < ½,  n   < q < ½, q + s >  1, n + + q pα, p < n + + q Then Lg : F(p, q,s) Ꮾα is bounded if and only if sup   z zắB (n+1+q)/ p g(z) < ẵ (2.13) Moreover, the following relationship: Lg sup   z z¾B F(p,q,s) Ꮾα α (n+1+q)/ p g(z) (2.14) holds Proof Assume that (2.13) holds Let f (z) ¾ F(p, q,s) sup   z z¾B (n+1+q)/ p Ꮾ(n+1+q)/ p , then f (z) < ½ (2.15) Therefore by Lemmas 2.1 and 2.2 we have Ꮾα = sup   z α = sup   z Lg f α Lg f (z) z¾B f (z) z¾B sup   z z¾B (n+1+q)/ p C f Ꮾ(n+1+q)/ p sup z¾B C f F(p,q,s) sup z¾B 1  1  z g(z) f (z) sup   z z¾B α (n+1+q)/ p z2 g(z) α (n+1+q)/ p α (n+1+q)/ p g(z) (2.16) g(z) Here we used the fact Lg f (0) = It follows that Lg is bounded Conversely, if Lg : F(p, q,s) Ꮾα is bounded Let β(z,w) denote the Bergman metric between two points z and w in B It is well known that β(z,w) = + ϕz (w) log   ϕz (w) (2.17) For a ¾ B and r > the set D(a,r) = z ¾ B : β(a,z) < r , a ¾ B, (2.18) is a Bergman metric ball at a with radius r It is well known that (see [21]) 1  a   a,z n+1 2(n+1) 1  z2 n+1 1  a2 n+1 D(a,r) (2.19) Songxiao Li when z ¾ D(a,r) For w ¾ B, let fw (z) be defined by (2.9), then by (2.10) and (2.19) we have 1  w  2(n+1+q)/ p g(w) fw (w)g(w) C w 1  C n+1 D(w,r) 1  C w 2 w C w 1  D(w,r) n+1 fw (z) g(z) Lg f w 1  z dv(z) 1  z 2α+n+1 sup 1  z z¾D(w,r) 2α fw (z) D(w,r) 2 g(z) dv(z) 1  z 2α 2α fw (z) dv(z) 2α g(z) (2.20) 2 Ꮾα , that is, 1  w 2α 2(n+1+q)/ p g(w) w C Lg f w Ꮾα CK Lg F(p,q,s) Ꮾα (2.21) Taking supremum in the last inequality over the set 1/2 w < and noticing that by the maximum modulus principle there is a positive constant C independent of g ¾ H(B) such that 1  w sup w 1/2 α (q+n+1)/ p g(w) C sup 1/2 w Then the operator Tg : A2 Ꮾα is bounded if and only if sup   z z¾B Lg : A2 α (n+1)/2 g(z) < ½ (2.24) α (n+1)/2 1 g(z) < ½ (2.25) Ꮾα is bounded if and only if sup   z z¾B Riemann-Stieltjes operators from F(p, q,s) to Ꮾα Tg : H Ꮾα is bounded if and only if sup   z z¾B Lg : H 2 α n/2 g(z) < ½ (2.26) α n/2 1 g(z) < ½ (2.27) Ꮾα is bounded if and only if sup   z z¾B 2.2 Case p > n + + q Theorem 2.8 Let g be a holomorphic function on B, < p, s < ½,  n   < q < ½, q + s >  1, α 0, n + + q pα, p > n + + q Then Tg : F(p, q,s) Ꮾα is bounded if and only if g ¾ Ꮾα Moreover, the following relationship: Tg F(p,q,s) Ꮾα sup   z z¾B α g(z) (2.28) holds Proof Since f ¾ F(p, q,s) Tg f Ꮾ(n+1+q)/ p , by Lemmas 2.1, 2.2, and 2.3, Ꮾα = sup   z α z¾B C f F(p,q,s) sup z ¾B f (z) g(z) 1  z (2.29) α g(z) Therefore g ¾ Ꮾα implies that Tg : F(p, q,s) Ꮾα is bounded Conversely, suppose Tg : F(p, q,s) Ꮾα is bounded For w ¾ B, let fw (z) = 1  w (p+n+1+q)/ p   z,w 2(n+1+q)/ p   1  w   z,w (n+1+q)/ p + (2.30) Then it is easy to see that fw (w) = 1, fw (z) C 1  w   z,w (n+1+q+p)/ p , fw (w) 1  w w (n+1+q)/ p (2.31) Songxiao Li By [19], we know that fw ¾ F(p, q,s), moreover there exists a constant L such that supz¾B fw F(p,q,s) L Hence g(w) 2 = fw (w) g(w) 1  C C w n+1 D(w,r) 1  C w fw (z) D(w,r) n+1 D(w,r) g(z) dv(z) 1  z 2α+n+1 sup 1  z z¾D(w,r) C 1  w 2α 1  w α Tg fw fw (z) 1  z 2α 2 2α fw (z) g(z) dv(z) 1  z dv(z) 2α g(z) 2 Ꮾα , (2.32) that is, g(w) C Tg fw CL Tg Ꮾα (2.33) F(p,q,s) Ꮾα for every w ¾ B The result follows Theorem 2.9 Let g be a holomorphic function on B, < p, s < ½,  n   < q < ½, q + s >  1, α 0, n + + q pα, p > n + + q Then Lg : F(p, q,s) Ꮾα is bounded if and only if sup   z z¾B α (n+1+q)/ p g(z) < ½ (2.34) Moreover, the following relationship: Lg sup   z z¾B F(p,q,s) Ꮾα α (n+1+q)/ p g(z) (2.35) holds Proof Suppose (2.34) holds Let f (z) ¾ F(p, q,s) sup   z z¾B (n+1+q)/ p Ꮾ(n+1+q)/ p , then f (z) < ½ (2.36) Hence Lg f Ꮾα = sup   z α f (z) z¾B sup   z z¾B (n+1+q)/ p C f Ꮾ(n+1+q)/ p sup z¾B C f F(p,q,s) sup z¾B It follows that Lg is bounded 1  1  z g(z) f (z) sup   z z¾B α (n+1+q)/ p z2 g(z) α (n+1+q)/ p g(z) α (n+1+q)/ p g(z) (2.37) 10 Riemann-Stieltjes operators from F(p, q,s) to Ꮾα Ꮾα is bounded, for w ¾ B, let fw (z) be defined by (2.30) Conversely, if Lg : F(p, q,s) Then by (2.31), 1  w  2(n+1+q)/ p fw (w)g(w) 1  C C w n+1 w 1  C w n+1 fw (z) D(w,r) 2 2α fw (z) D(w,r) g(z) dv(z) 1  z 2α+n+1 sup 1  z z¾D(w,r) D(w,r) C 1  w g(w) 1  z 2α 2 g(z) dv(z) 2α 1  z fw (z) dv(z) 2α g(z) (2.38) 2 Ꮾα , Lg f w that is, 1  w 2α 2(n+1+q)/ p g(w) w C Lg f w Ꮾα CL Lg F(p,q,s) Ꮾα (2.39) Similarly to the proof of Theorem 2.5, we get the desired result 2.3 Case p = n + + q Theorem 2.10 Let g be a holomorphic function on B, < p, s < ½,  n   < q < ½, q + s >  1, s > n, α 1, p = n + + q Then Tg : F(p, q,s) Ꮾα is bounded if and only if sup   z z¾B α log 1  z g(z) < ½ (2.40) Moreover the following relationship: Tg F(p,q,s) Ꮾα sup   z z ¾B α log 1  z g(z) (2.41) holds Proof Since f ¾ F(p, q,s) Ꮾ, by Lemmas 2.1, 2.2, and 2.3, Ꮾα = sup   z α = sup   z Tg f α z¾B z¾B C f F(p,q,s) sup z¾B Tg f (z) f (z) 1  z g(z) α log (2.42) 1  z g(z) Therefore (2.40) implies that Tg is a bounded operator from F(p, q,s) to Ꮾα Songxiao Li 11 Conversely, suppose Tg is a bounded operator from F(p, q,s) to Ꮾα For w ¾ B, let fw (z) = log 1  z,w (2.43) Then by [19] we see that fw ¾ F(p, q,s) and fw (w) = log , 1  w w fw (w) 1  w Moreover there is a positive constant M such that supw¾B f log 1  w g(w) = fw (w) g(w) 1  C C w n+1 1  C w 2α (2.44) M Hence F(p,q,s) C 1  w D(w,r) n+1 fw (z) Tg fw fw (z) D(w,r) g(z) dv(z) 1  z 2α+n+1 sup 1  z z¾D(w,r) D(w,r) 1  z 2α 2 g(z) dv(z) 2α fw (z) 1  z dv(z) 2α g(z) (2.45) 2 Ꮾα , that is, 1  w α log 1  w g(w) C Tg fw Ꮾα CM Tg F(p,q,s) Ꮾα (2.46) The result follows Theorem 2.11 Let g be a holomorphic function on B, < p, s < ½,  n   < q < ½, q + s >  1, s > n, α 1, p = n + + q Then Lg : F(p, q,s) Ꮾα is bounded if and only if sup   z z¾B α 1 g(z) < ½ (2.47) Moreover the following relationship: Lg holds F(p,q,s) Ꮾα sup   z z¾B α 1 g(z) (2.48) 12 Riemann-Stieltjes operators from F(p, q,s) to Ꮾα Proof Suppose (2.47) holds Let f (z) ắ F(p, q,s) ẵ By Lemmas 2.1 and 2.2 we have Lg f Ꮾα = sup   z α z ¾B f (z) 1  z C f 1  w  2 g(w) w fw (w)g(w) 1  C C w D(w,r) C 1  w n+1 Ꮾ sup z¾B C f It follows that Lg is bounded Conversely, if Lg : F(p, q,s) then by (2.44) we have F(p,q,s) sup z ¾B f (z) < g(z) g(z) 1  z (2.49) α 1 g(z) Ꮾα is bounded For w ¾ B, let fw (z) be defined by (2.43), C w 1  n+1 fw (z) D(w,r) Lg f w fw (z) D(w,r) g(z) 1  z dv(z) 1  z 2α+n+1 sup 1  z z¾D(w,r) 2α Ꮾ, then supz¾B (1   z ) 2α 2 2α g(z) dv(z) 1  z fw (z) dv(z) 2α g(z) (2.50) 2 Ꮾα , that is, 1  w 2α 2 g(w) w C Lg f w Ꮾα (2.51) Similarly to the proof of Theorem 2.5, we get the desired result Similarly to the proof of Theorems 2.10 and 2.11, we can obtain the following results We omit the details Corollary 2.12 Let g be a holomorphic function on B, < p < ½, and α Q p Ꮾα is bounded if and only if sup   z z¾B Lg : Q p α log 1  z g(z) < ½ Then Tg : (2.52) Ꮾα is bounded if and only if sup   z z¾B α 1 g(z) < ½ Corollary 2.13 Let g be a holomorphic function on B Then Lg : Ꮾ only if g ắ H ẵ Especially, we have the following known result (see [6, 15, 17]) (2.53) Ꮾ is bounded if and Songxiao Li Corollary 2.14 Let g be a holomorphic function on B Then Tg : Ꮾ and only if sup   z z¾B log 1  z 13 Ꮾ is bounded if g(z) < ½ (2.54) Acknowledgment This author is supported partly by the NNSF of China (no 10371051, 10371069) References [1] A Aleman and J A Cima, An integral operator on H p and Hardy’s 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New York, 2005 Songxiao Li: Department of Mathematics, JiaYing University, 514015, Meizhou, GuangDong, China Current address: Department of Mathematics, Shantou University, 515063, Shantou, GuangDong, China E-mail addresses: lsx@mail.zjxu.edu.cn; jyulsx@163.com ... even in the unit disk setting 4 Riemann-Stieltjes operators from F(p, q,s) to Ꮾα In this paper, constants are denoted by C, they are positive and may differ from one occurrence to the other a... operator from F(p, q,s) to Ꮾα Songxiao Li 11 Conversely, suppose Tg is a bounded operator from F(p, q,s) to Ꮾα For w ¾ B, let fw (z) = log 1  z,w (2.43) Then by [19] we see that fw ¾ F(p, q,s). .. in the unit disk setting, see [20] If q + s  1, then F(p, q,s) is the space of constant functions For an analytic function f (z) on the unit disk D with Taylor expansion f (z) = ½ a zn , the