Riemann-Stieltjes Operators between Weighted Bloch and Weighted Bergman Spaces

Transcription

1 Int. J. Contemp. Math. Sci., Vol. 2, 2007, no. 16, Riemann-Stieltjes Operators between Weighted Bloch and Weighted Bergman Spaces Ajay K. Sharma 1 School of Applied Physics and Mathematics Shri Mata Vaishno Devi University P/O Kakryal, Udhampur , India aksju 76@yahoo.com Som Datt Sharma Department of Mathematics University of Jammu, Jammu , India somdatt jammu@yahoo.co.in Abstract In this paper, Riemann-Stieltjes operators between weighted Bloch and weighted Bergman spaces are considered. We characterize boundedness and compactness of these operators using certain growth properties of holomorphic symbols. Keywords: weighted Bergman spaces, weighted Bloch spaces, Riemann- Stieltjes operator, Carleson measure 1 Introduction Let D be the open unit disk in the complex plane C. Let g : D C be a holomorphic map. Denote by H(D) the space of holomorphic functions on D. For f H(D), the Riemann-Stieltjes operator induced by g is defined by T g f(z) = z 0 f(ζ)dg(ζ) = 1 0 f(tz)zg (tz)dt, z D. The Riemann-Stieltjes operator can be viewed as a generalization of Cesaro operator defined by Tf(z) = 1 z z f(w) d(w), z D. 1 w 0 1 Supported by CSIR grant ( F.No. 9/100(100)2002 EMR-1).

2 760 Ajay K. Sharma and Som Datt Sharma Ch. Pommerenke [7] initiated the study of Riemann-Stieltjes operator on H 2, where he showed that T g is bounded on H 2 if and only if g is in BMOA. This was extended to other Hardy spaces H p, 1 p<, in [1]and [2] where compactness of T g on H p and Schatten class membership of T g on H 2, was also completely characterized in terms of the symboy g. Similar questions on weighted Bergman spaces were considered by A. Aleman and A. G. Siskakis in [3]. Recently, several authors have studied Riemann-Stieltjes operators on different spaces of analytic functions. For example, one can refer to ([5] [8] [9][10][11] and [12]) for the study of these operators on Bergman spaces, Dirichlet spaces, BMOA and VMOA and related references therein. In this paper we characterize boundedness and compactness of Riemann-Stieltjes operators between between weighted Bloch and weighted Bergman spaces. 2 Preliminaries In this section we review the basic concepts of weighted Bergman spaces A p α and weighted Bloch spaces B α and collect some essential facts that will be needed throughout the paper Weighted Bergman Spaces. Let da(z) be the area measure on D normalized so that area of D is 1. For each α ( 1, ), we set dν α (z) = (α + 1)(1 z 2 ) α da(z),z D. Then dν α is a probability measure on D. For 0 <p< the weighted Bergman space A p α is defined as ( ) 1/p A p α = {f H(D) : f p Aα = D f(z) p dν α (z) < }. Note that f A p α is a true norm only if 1 p< and in this case A p α is a Banach space. For 0 <p<1, A p α is a non-locally convex topological vector space and d(f,g) = f g p is a complete metric for it. The growth of A p α functions in the weighted Bergman spaces is essential in our study. To this end, the following sharp estimate will be useful. (see [7] p. 53.). It tells us how fast an arbitrary function from A p α grows near the boundary. Let f A p α. Then for every z in D, we have f(z) f A p α (1 z 2 ) (2+α)/p (2.1) with equality holds if and only if f is a constant multiple of the function ( 1 z 2 ) (2+α)/p k a (z) =. (2.2) (1 az) 2 It can be easily shown that k a A p α Weighted Bloch Spaces. For α>0, let B α consists of all

3 Riemann-Stieltjes operators 761 analytic functions f on D satisfying the condition z D (1 z 2 ) α f (z) <. Note that B 1 = B, the usual Bloch space. For f B α define f B α = f(0) + z D (1 z 2 ) α f (z) <. With this norm B α is a Banach space. Integrating the estimate f (z) f B α (1 z 2 ), α we obtain 1 1 f(z) f(0) z f z (tz) dt f B α 0 0 (1 t z 2 ) dt, α for all z D. In case 0 <α<1, the integral on the right is uniformly bounded by the constant 1 (1 0 t) α dt, and it follows that B α H. It is easy to check that in this case the linear space B α is an algebra. In fact, Hardy and Littlewood, have shown that for 0 <α<1, the space B α consist of all functions f analytic on D satisfying the Lipschitz condition f(z) f(w) z w 1 α, for all z, w D (see [4]). In case 1 <α<, the above estimate implies f(z) f(0) f B α α 1 1, (2.3) (1 z 2 ) α 1 while for α = 1 it is well known that the following hold ([6] and [14]). f(z) f(w) f B β(z, w) (2.4) for f B, where β(z, w) = 1 1 zw + z w log 2 1 zw z w is the Bergman metric on D. From (2.4), it follows that for f B, f(z) 1 ( log 2 f 2 ) B log. (2.5) 1 z 2 Throughout this paper we fix some positive radius 0 <r< and consider disks D(z, r) in the Bergman metric. The set D(z, r) ={w D : β(z, w) <r}, z D,

4 762 Ajay K. Sharma and Som Datt Sharma is called hyprerbolic disk or Bergman disk of radius r about z. It is well known that D(z, r) is a Euclidean disk whose Euclidean center and Euclidean radius are (1 s 2 )z (1 s 2 z 2 ) and (1 z 2 )s (1 s 2 z 2 ), where s = tanh r (0, 1), respectively. For fixed r>0. the area of D(z, r) in D has the estimation; D(z, r) A = da(w) (1 z 2 ) 2. D(z,r) For fixed r>0, it is known that if w D(z, r), then 1 zw (1 z 2 ) and D(w, s) A C D(z, r) A. Following lemma lists additional properties of the hyperbolic disks. Lemma 2.2.[7] Fix r, 0 <r<. There exists a positive integer M and a sequence {a n } in D such that : (i) The disk D is covered by {D(a n,r)} n. (ii) Every point in D belongs to at most M sets in {D(a n, 2r)} n. (iii) If n m, then β(a n,a m ) r 2. We shall use these estimates in the proofs of the Theorems below. For general background of weighted Bergman spaces A p α and weighted Bloch spaces, one may consult [13] and [14] and the references therein. 3 Riemann-Stieltjes operators from weighted Bergman spaces A p α into weighted Bloch spaces B α In this section we characterize boundedness and compactness of Riemann- Stieltjes operators from weighted Bergman spaces A p α into weighted Bloch spaces spaces B α. The following Theorem characterizes Riemann-Stieltjes operators from weighted Bergman spaces A p β into weighted Bloch spaces Bα. Theorem 3.1. Let 1 p<, 1 <β<, α > 0 and g : D C be a holomorphic map. Then the Riemann-Stieltjes operator T g maps A p β boundedly into B α if and only if z D (1 z 2 ) (pα β 2)/p g (z) <. (3.1)

5 Riemann-Stieltjes operators 763 Proof. First pose that M = z D (1 z 2 ) (pα β 2)/p g (z) <. By (2.1), we have for all z D. Thus f(z) f A p β (1 z 2 ) (β+2)/p T g f B α = T g f(0) + z D (1 z 2 ) α (T g f) (z) = z D (1 z 2 ) α g (z)f(z) z D (1 z 2 ) (pα β 2)/p g (z) f A p β = M f A p, β hence T g maps A p β boundedly into Bα. Conversely, pose T g maps A p β boundedly into Bα. Fix a point a in D and consider the function ( 1 a 2 ) (β+2)/p f a (z) =. (1 az) 2 Then f a is a function of unit norm in A p β. Since T g maps A p β boundedly into B α, we can find a positive constant C such that T g f a B α C f a A p = C, β for all a D, hence for each point z D we have In particular, when z = a, we get whence (1 z 2 ) α f a (z)g (z) C. (1 a 2 ) α ( 1 a 2 (1 a 2 ) 2 ) (β+2)/p g (a) C, (1 a 2 ) (pα β 2)/p g (a) <C. Since a D is arbitrary, the result follows. Theorem 3.2. Let 1 p<, 1 <β<, α > 0 and g : D C be a holomorphic map. Suppose that T g maps A p β boundedly into Bα. Then T g maps A p β compactly into Bα if and only if lim (1 z 2 ) (pα β 2)/p g (z) =0. (3.2) z 1

6 764 Ajay K. Sharma and Som Datt Sharma Proof. First pose that (3.2) holds. Let {f n } be a bounded sequence in A p β that converges to zero uniformly on compact subsets of D. Let M = n f n A p <. Given ε>0, there exist an r (0, 1) such that if z >r, β then (1 z 2 ) (pα β 2)/p g (z) <ε. Thus for z D such that z >r,by (2.1) we have (1 z 2 ) α (T g f n ) (z) = (1 z 2 ) α g (z) f n (z) (1 z 2 ) α (β 2)/p g (z) f n A p β < εm, for all n. On the other hand, since f n 0 uniformly on compact subsets of D, there exist an n 0 such that if z r and n n 0, then f n (z) <ε.by Theorem 3.1, we have g B α and so we have hence N = z D (1 z 2 ) α g (z) <, (1 z 2 ) α (T g f n ) (z) = (1 z 2 ) α g (z) f n (z) The above arguments together yield < εn. T g f n B α = T g f(0) + z D (1 z 2 ) α (T g ) f n (z) (1 z 2 ) α (T g f n ) (z) + z >r (N + M)ε. (1 z 2 ) α (T g f n ) (z) Thus T g f n B α 0 as n, hence T g maps A p β compactly into Bα. Conversely, pose T g maps A p β compactly into Bα and (3.2 ) does not hold. Then there exists a positive number δ and a sequence {z n } in D such that z n 1 and (1 z n 2 ) (pα β 2)/p g (z n ) δ, for all n. For each n, consider the function f n defined as ( 1 zn 2 ) (β+2)/p f n (z) =, z D. (1 z n z) 2

7 Riemann-Stieltjes operators 765 Then the sequence {f n } is norm bounded and f n 0 uniformly on compact subsets of D, it follows that a subsequence of {T g f n } tends to 0 in B α. On the other hand, T g f n B α (1 z n 2 ) α (T g f n ) (z n ) = (1 z n 2 ) α g (z n ) f n (z n ) = (1 z n 2 ) (pα β 2)/p g (z n ) δ, which is absurd. Hence we are done. 4 Riemann-Stieltjes operators between weighted Bloch spaces B α In this section we characterize boundedness and compactness of Riemann- Stieltjes operators between weighted Bloch spaces B α. Theorem 4.1. Let α>0, β > 0 and g : D C be a holomorphic map. (i) If 0 <α<1, then T g maps B α boundedly into B β if and only if g B β. (ii) Operator T g maps B boundedly into B β if and only if z D (1 z 2 ) β g 2 (z) log (1 z 2 ) <. (iii) If α>1, then T g maps B α boundedly into B β if and only if z D (1 z 2 ) 1+β α g (z) <. Proof. First we consider the case α>1. Suppose T g maps B α boundedly into B β. Consider the function f a (z) = 1 a 2 (1 az) α, z D. Then f a B β 1+2α. Thus f a B α and M = { f a B β : a D} 1+2α. Since T g maps B α boundedly into B β, we can find a positive constant C such that T g f a B β C f a B α CM for each a D, hence for each z D, we have (1 z 2 ) β g (z) f a (z) = (1 z 2 ) β (T g f a ) (z) CM.

8 766 Ajay K. Sharma and Som Datt Sharma In particular, when z = a, we have (1 a 2 ) 1+β α g (a) CM. Since a D is arbitrary, the result follows. Conversely, pose that By (2.3), we have M = z D (1 z 2 ) 1+β α g (z) < (4.1) f(z) f(0) for all z D, independent of f B α. Since f B α (α 1)(1 z 2 ) (α 1) T g f B β = T g f(0) + z D (1 z 2 ) β (T g f) (z) z D (1 z 2 ) β f(z) f(0) g (z) + f(0) z D (1 z 2 ) β g (z) z D (1 z 2 ) β g f (z) B β + C (α 1)(1 z 2 )(α 1) z D (1 z 2 ) β g (z) ( CM + M ) f B β. (α 1) Hence T g maps B α boundedly into B β. This completes the proof of (iii). Next, we will prove (ii). Suppose T g maps B boundedly into B β. For a D, let f a (z) = log Then f a Band f a B 3. So 2 (1 az), z D. 3 T g B β T g f a B β Since a D is arbitrary, the result follows. Convesely, pose that = T g f a (0) + z D (1 z 2 ) β (T g f a ) (z) (1 a 2 ) β g (a) f a (a) = (1 a 2 ) β g 2 (a) log (1 a 2 ). M = z D (1 z 2 ) β g 2 (z) log (1 z 2 ) <.

9 Riemann-Stieltjes operators 767 By (2.5), for f B α, we have T g f B β = T g f(0) + z D (1 z 2 ) β (T g f) (z) z D (1 z 2 ) β g 1 (z) log 2 log 2 (1 z 2 ) f B α = 1 log 2 M f B, hence T g maps B boundedly into B β. This completes the proof of (ii). Finally we will prove (i). First pose that T g maps B α boundedly into B β, then g = g(0) + T g 1 B β. Conversely, pose that g B β. Then Thus, if f B α, then f(z) f B α(1 + (1 z 2 ) 1 α ), α 1,z D. T g f B β z D (1 z 2 ) β (T g f) (z) z D ((1 z 2 ) β +(1 z 2 ) β+1 α ) g (z) f B α C z D ((1 z 2 ) β ) g (z) f B α C g B β f B α. As a result, T g maps B α boundedly into B β. Lemma 4.2.[6] Let 0 <α<1 and let T be a bounded linear operator from B α into a normed linear space X. Then T is compact if and only if Tf n X 0, whenever {f n } is a bounded sequence in B α that converges to zero uniformly on D. Theorem 4.3. Let α>0, β > 0 and g : D C be a holomorphic map. Suppose that T g maps B α boundedly into B β. (i) If 0 <α<1, then T g maps B α compactly into B β. (ii) Operator T g maps B compactly into B β if and only if lim (1 z 1 z 2 ) β g 2 (z) log (1 z 2 ) =0. (iii) If α>1, then T g maps B α compactly into B β if and only if lim (1 z 1 z 2 ) 1+β α g (z) =0.

10 768 Ajay K. Sharma and Som Datt Sharma Proof. First we consider the case α>1. To prove that the condition in (iii) is sufficient for compactness of the operator T g from B α into B β, it is enough to show that if {f n } is a bounded sequence in B α that converges to zero uniformly on compact subsets of D, then lim n T g f n B β =0. Let M = n f n B α <. Given ε>0, there exists an r (0, 1) such that, if z >r,then (1 z 2 ) 1+β α g (z) <ε. By (2.3), we have f n (z) f n (0) f n B α (α 1)(1 z 2 ) (α 1) for all z D. Since T g f n B β = T g f n (0) + z D (1 z 2 ) β (T g f n ) (z) z D (1 z 2 ) β f n (z) f n (0) g (z) + f n (0) z D (1 z 2 ) β g (z) (1 z 2 ) β g (z) f n (z) f n (0) + f n (0) (1 z 2 ) β g (z) + (1 z 2 ) β g (z) f n (z) f n (0) + f n (0) (1 z 2 ) β g (z) z >r z >r (1 z 2 ) β g (z) f n (z) f n (0) + f n (0) (1 z 2 ) β g (z) + (1 z 2 ) β g (z) f B α z >r (α 1)(1 z 2 ) + f n(0) (1 z 2 ) β g (z) (α 1) z >r ε(4 g B β + M α 1 + M) as n n 0. Thus, T g maps B α compactly into B β. Conversely, pose that T g maps B α compactly into B β and (iii ) does not hold. Then there exists a positive number δ and a sequence {z n } in D such that z n 1 and (1 z n 2 ) 1+β α g (z n ) δ, for all n. For each n, let f n (z) = 1 z n 2 (1 z n z), α z D. Then the sequence f n is norm bounded and f n 0 uniformly on compact subsets of D. Hence there exists a subsequence of {T g f n } which tends to 0 in B β. On the other hand, T g f n B β (1 z n 2 ) β (T g f n ) (z n ) = (1 z n 2 ) β g (z n ) f n (z n ) = (1 z n 2 ) 1+β α g (z n ) δ,

11 Riemann-Stieltjes operators 769 which is absurd. Hence we are done. Next, we will prove (ii). Let {f n } is a bounded sequence in B that converges to zero uniformly on compact subsets of D. Let M = n f n B <. Given ε>0, there exists an r (0, 1) such that, if z >r,then By (2.5), we have (1 z 2 ) β g (z) log 2 (1 z 2 ) <ε. f n (z) 1 log 2 f 2 n B log (1 z 2 ) for all z D, f B α. Also, by Theorem 4.1, we have g B β and so, for the above ε, we can find n 0 N such that T g f n B β = T g f n (0) + z D (1 z 2 ) β g (z) f n (z) (1 z 2 ) β g (z) f n (z) + (1 z 2 ) β g (z) f n (z) ε( g B β + M) for n n 0. Thus T g maps B compactly into B β. Conversely, pose T g maps B compactly into B β and condition in (ii ) does not hold. Then there exists a positive number δ and a sequence {z n } in D such that z n 1 and for all n. For each n, let (1 z n 2 ) β g (z n ) log z >r 2 (1 z n 2 ) δ, 2 f n (z) = log (1 z n z), z D. Then the sequence f n is norm bounded and f n 0 uniformly on compact subsets of D. By the compactness of T g we can find a subsequence of {T g f n } which tends to 0 in B β. On the other hand, T g f n B β (1 z n 2 ) β (T g f n ) (z n ) = (1 z n 2 ) β g (z n ) f n (z n ) = (1 z n 2 ) β g 2 (z n ) log (1 z n 2 ) δ, which is absurd. We are done. Finally, we will prove (i). Suppose that n f n B α M and f n 0 uniformly on D. Then z D (1 z 2 ) β g (z) f n (z) f n (z) g B β 0 as n. z 61

12 770 Ajay K. Sharma and Som Datt Sharma It follows from Lemma 4.2 that T g maps B α compactly into B β. Lemma 4.4. Let 1 p<q< and 1 <α<. Then the injection map from A q α into Ap α is compact. Proof Let {f n } n=1 be a bounded sequence in A q α, and let M = n N f n A q α. By 2.1, {f n : n N} is a normal family, hence we can find a subsequence {f nk } of {f n } that converges uniformly on compact subsets of D to an analytic function f. By Fatou s lemma, D f(z) p ν α (z) lim inf k D f nk(z) p ν α (z) M q <. Since p<q,it follows that f A q α. We claim that {f nk} converges to f in A p α. Let ε>0 and let Ω be an arbitrary compact subset of D. Now ( D\Ω (f n k f)(z) p ν α (z) D\Ω ) p/q ( (f n k f)(z) q ν α (z) D\Ω ) 1 p/q ν α (z) ( ) p/q(να D (f nk f)(z) q ν α (z) (D \ Ω)) 1 p/q ) p/q(να (2 q D (f nk(z) q + f(z) q )ν α (z) (D \ Ω)) 1 p/q 2 p ( f n k p + A q f p )(ν α A α(d \ Ω)) 1 p/q q α 2 p+1 M p (ν α (D \ Ω)) 1 p/q, where in the first line we have used Holder s inequality and the elementary inequalities (x + y) a 2 a (x a + y b ), (x + y) b (x b + y b ) which holds when x, y 0, and 0 <b<1. By choosing the compact set Ω so that D \ Ω has sufficiently small area, we obtain (f n k f)(z) p ν α (z) <ε/2 D\Ω for k large enough. On the other hand, since f n f uniformly on Ω we can choose k large enough so that (f n k f)(z) p ν α (z) <ε/2. Ω Thus {f n k} converges to f in A p α. Hence the injection map is compact. Corollary 4.5. Let 1 p<, 1 <β<, α>0 and g : D C be holomorphic. Then T g maps B α compactly into A p β if and only if T g maps B α boundedly into B β. Proof. Suppose T g maps B α boundedly into B β, thus also into the large space

13 Riemann-Stieltjes operators 771 A p β. Since convergence in either space implies uniform convergence on compact sets, it follows from closed graph theorem that T g maps B α boundedly into A p β. In order to show that T g maps B α compactly into A p β, choose any q such that q>pand factorize T g through the intermediate space A q β : B α e T g A q β I A p β, where T g is the Riemann-Stieltjes operator from B α to A q β, and I is the injection map. Since I is compact and T g is bounded, so T g maps B α compactly into A p β. References [1] A. Aleman and A. G. Siskakis, An integral operator on H p and Hardy inequality, J. Anal. Math. 85 (2001), [2] A. Aleman and A. G. Siskakis, An integral operator on H p, Complex variables. 28 (1995), (1997), [3] A. Aleman and A. G. Siskakis, Integral operators on Bergman spaces, Indiana Univ. Math. J. 46 (1997), [4] P. L. Duren, Theory of H p spaces, Academic Press, New York, [5] B.D. MacCluer and R. Zhao, Vanishing logarithmic Carleson measures, Illinois J. Math. 46 (2002), [6] S. Ohno, K. Stroethoff and R. Zhao, Weighted composition operators between Bloch-type spaces, Rocky Mountain J. Math., 33 (2003), [7] C. Pommerenke, Schlichte Funktionen und analytische Funktionen von beschrankter mittlerer Oszillation, Comment. Math. Helv. 52 (1977), [8] A. G. Siskakis and R. Zhao, A Volterra type operator on spaces of analytic functions, Contemp. Math. 232 (1999), [9] J. Xiao, Cesaro operators on Hardy, BMOA and Bloch spaces, Arch. Math. 68 (1997), [10] J. Xiao, The -problem for multipliers of the Sobolev space, Acta. Sci Math. (Szeged) 69 (2003), [11] J. Xiao, Riemann-Stieltjes operators on weighted Bloch and Bergman spaces of the unit ball, J. London Math. Soc. 70 (2004),

14 772 Ajay K. Sharma and Som Datt Sharma [12] R. Zhao, On logarithmic Carleson measures, Contemp. Math. 232 (1999), [13] K. Zhu, Operator theory in function spaces, Marcel Dekker, New York, [14] K. Zhu, Bloch type spaces of analytic functions, Rocky Mountain J. Math., 23 (1993), Received: October 5, 2006