Weighted Composition Followed by Differentiation between Weighted Bergman Space and H on the Unit Ball 1

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1 International Journal of Mathematical Analysis Vol 9, 015, no 4, HIKARI Ltd, wwwm-hikaricom Weighted Composition Followed by Differentiation between Weighted ergman Space and H on the Unit all 1 Chao Zhang Dept of Math, Guangdong University of Education Guangzhou, , PR China Sui Huang College of Mathematics, Chongqing Normal University Chongqing , PR China Copyright c 014 Chao Zhang and Sui Huang 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 Abstract We define differentiation operator on H() by radial derivative, then we study the boundedness and compactness of products of multiplication operator, composition operator and differentiation operator acting between weighted ergman spaces and H on the unit ball Mathematics Subject Classification: 4733, 30H05 Keywords: Composition operator; Multiplication operator; Differentiation operator; Weighted ergman space 1 Supported by the Chongqing Education Commission (NO KJ13063), Chongqing Normal University ( NO13XLZ0) and Guangdong University of Education (014ARF04)

2 170 Chao Zhang and Sui Huang 1 Introduction Let D be the open unit disk in the complex plane Let = {z C n : z < 1} be the unit ball of C n, and S = its boundary We will denote by dv the normalized Lebesgue measure on Recall that for α > 1 the weighted Lebesgue measure dv α is defined by dv α (z) = c α (1 z ) α dv(z), where Γ(n α) c α = n!γ(1 + α) is a normalizing constant so that dv α is a probability measure on Let H() denotes the space of holomorphic functions on Take 1 p < Then f H() is said to be in the weighted ergman space A p α() if f p = f(z) p dv A p α α (z) < As we all know, H = {f H() : f = f(z) < } Let ϕ be an analytic self-mapping of, then the composition operator on H() is given by C ϕ f = f ϕ Recently, there have been an increasing interest in studying composition operators acting on different spaces of analytic functions, for example, see [3,5] for details about composition operators on classical spaces of analytic functions Let D be the differentiation operator defined by Df = f, f H(D) Hibschweiler and Portnoy [7] defined the linear operators DC ϕ and C ϕ D and investigated the boundedness and compactness of these operators between ergman spaces using Carleson-type measure S Ohno [10] discussed boundedness and compactness of C ϕ D between Hardy spaces Recall the multiplication operator M ψ defined by M ψ f = ψf, f H(D) A K Sharma defined [1] products of these operators in the following six ways: (M ψ C ϕ Df)(z) = ψ(z)f (), (M ψ DC ϕ f)(z) = ψ(z)(ϕ (z))f (), (C ϕ M ψ Df)(z) = ψ()f (),

3 Weighted composition followed by differentiation 171 (DM ψ C ϕ f)(z) = ψ (z)f() + ψ(z)(ϕ (z))f (), (C ϕ DM ψ f)(z) = ψ ()f() + ψ()f (), (DC ϕ M ψ f)(z) j=1 = ψ ()f()ϕ (z) + ψ()f ()ϕ (z) for z D and f H(D) There are a lot of papers researching these products, see [,4,9] Since those results focus on D, naturally, we consider similar questions on Of course, the method we used is different from the case on D For f H(), we define the differentiation operator on H() by radial derivative Recall that for z and f H(), n f f(z + rz) f(z) Rf = z j (z) = lim, r R z j r 0 r One can see that for z ϕ 1 (0), R(f ϕ)(z) = (Rf)() R Then we also have six ways of products of these operators on the unit ball: (M ψ C ϕ R)f(z) = ψ(z) (Rf)(), (C ϕ M ψ Rf)(z) = ψ() (Rf)(), (M ψ RC ϕ f)(z) ψ(z) R (Rf)() =, (C ϕ RM ψ f)(z) = (Rψ)() f() + ψ() (Rf)(), (RM ψ C ϕ f)(z) ψ(z) R (Rf)() = f() Rψ(z) +, (RC ϕ M ψ f)(z) (Rψ)() R f() + ψ() (Rf)() R = for z ϕ 1 (0) In this paper, we characterize the boundedness and compactness of RM ψ C ϕ between weighted ergman spaces and H on the unit ball, which extend the results of H Li in [4] Main results The following lemma is the Theorem 0 in [8] Lemma 1 Suppose p > 0, n+1+α > 0, then there exist a constant C > 0 (depend on p and α) such that f(z) C f A p α (1 z ) n+1+α p

4 17 Chao Zhang and Sui Huang for all f A p α, z Recall that for a holomorphic function f in we write ( f f(z) = (z),, f ) (z) z 1 z n and call f(z) the gradient of f at z Then we give the following lemma Lemma Let p > 0, α > 1, then there exist a constant C > 0 such that Rf(z) C z f A α () (1 z ) n+1+α +1 for all f A α() and z Proof For a f in A α(), from the Exercise 359 in [3] we know that f A α () = f(0) + f(z) (1 z ) α+ dv(z) On the other hand, Then by Lemma 1, f A α () = f(z) y the proof of Lemma,14 in [6] we have then we draw the conclusion f(z) (1 z ) α dv(z) C f A α () (1 z ) n+1+α +1 Rf(z) z f(z), Lemma 3 Suppose p > 0, α > 1, ϕ : be analytic ψ H(), set T = RM ψ C ϕ : A α() H Then T is compact if and only if T is bounded and for any bounded sequence {f k } in A α() which converges to zero uniformly on compact subsets of, T f k 0 as k y standard arguments from Proposition 311 in [3], this lemma follows Theorem 4 Assume that p > 0, α > 1, ϕ : be analytic, ψ H() Then RM ψ C ϕ : A α() H is bounded if and only if ψ(z) R (1 ) n+1+α +1 Rψ(z) (1 ) n+1+α <, (1) < ()

5 Weighted composition followed by differentiation 173 Proof Assume that RM ψ C ϕ : A α() H is bounded for every a, let ( a z )( 1 a ) n+1+α f a (z) = 1 az (1 az) Then a change of variables yields that f a A = f α() a (z) (1 z ) α dv(z) = a z (1 a ) α (1 z ) α 1 a n+1 dv(z) 1 az 1 az α (1 az) α 1 a n+1 dv(z) C (1 az) Obviously, f a A α(), a f a A α () < C It is easy to see that Rf a (z) = which imjplies that Then we have RM ψ C ϕ f ( a z ) (n α)(1 a ) n+1+α az 1 az (1 az) n++α ( 1 a ) n+1+α a(a z) z(1 az) + (1 az) (1 az) (Rf )() = f () = 0, (1 ) n+1+α +1 f () Rψ(z) + ψ(z) R (Rf )() ψ(z) R (1 ) n+1+α +1 Then (1) can be obtained Next we assume ( 1 a ) n+1+α g a (z) = (1 az) Similarly as above, g a A α(), a g a A α () < C On the other hand, which implies that, RM ψ C ϕ g g () = (Rg )() = 1 (1 ) n+1+α (n α) (1 ) n+1+α, +1 g () Rψ(z) + ψ(z) R (Rg )()

6 174 Chao Zhang and Sui Huang Rψ(z) (1 ) n+1+α (n α) ψ(z) R (1 ) n+1+α +1 Since ϕ is an analytic self-map of, and (1), we get () For the converse, assume (1) and () hold y the Lemma 1 and Lemma, (RM ψ C ϕ f)(z) = f() Rψ(z) + C Rψ(z) f A α () (1 ) n+1+α ψ(z) R (Rf)() + From the assumption, RM ψ C ϕ : A α() H is bounded C ψ(z) R f A α () (1 ) n+1+α +1 Theorem 5 Assume that p > 0, α > 1, ϕ : be analytic, ψ H() Then RM ψ C ϕ : A α() H is compact if and only if 1 ψ(z) R <, (3) 1 Rψ(z) <, (4) ψ(z) R (1 ) n+1+α +1 Rψ(z) (1 ) n+1+α = 0, (5) = 0 (6) Proof Assume that RM ψ C ϕ : A α() H is compact From Lemma 3, we know that RM ψ C ϕ is bounded and by taking f(z) = 1, it follows that Rψ(z) < y taking the tunction f(z) = z, we have that ψ(z) R + Rψ(z) < From above and ϕ < 1, we get ψ(z) R < If we pose (5) does not hold, then there exist a positive number δ (0, 1) and a sequence {z n } n N in, and ϕ(z n ) 1 as n such that ϕ(z n) 1 ψ(z n ) Rϕ(z n ) (1 ϕ(z n ) ) δ n+1+α +1 for all n N Next consider funtion ( ϕ(zn ) z )( 1 ϕ(zn ) ) n+1+α f n (z) =, z 1 ϕ(z n )z (1 ϕ(z n )z) From Theorem 4 we know f n A α(), f n 0 uniformly on compact subsets of From Lemma 3, it follows that a sequence {RM ψ C ϕ f n } tends to 0 in H

7 Weighted composition followed by differentiation 175 On the other hand, (RM ψ C ϕ f n )(z n ) f n (ϕ(z n )) Rψ(z n ) + ψ(z n) Rϕ(z n ) (Rf n )(ϕ(z n )) ϕ(z n ) C ψ(z n ) Rϕ(z n ) (1 ϕ(z n ) ) n+1+α +1 Which is absurd, so (5) holds Now we consider the other function ( 1 ϕ(zk ) ) n+1+α g k (z) =, z (1 ϕ(z k )z) As the proof of Theorem 4, g k A α(), g k 0 uniformly on compact subsets of From Lemma 3, we can see that the sequence RM ψ C ϕ g k 0 in H (RM ψ C ϕ g k )(z k ) g k (ϕ(z k )) Rψ(z k ) + ψ(z k) Rϕ(z k ) (Rg k )(ϕ(z k )) ϕ(z k ) C ψ(z k ) Rϕ(z k ) (1 ϕ(z k ) ) Rψ(z k ) n+1+α +1 (1 ϕ(z k ) ) n+1+α Thus lim ϕ(z k ) 1 ψ(z k ) Rϕ(z k ) (1 ϕ(z k ) ) = lim n+1+α +1 ϕ(z k ) 1 Rψ(z k ) (1 ϕ(z k ) ) n+1+α From the above proof, we get (6) Conversely, for any bounded sequence {f n } A α() with {f n } 0 uniformly on compact subsets of, by Lemma 3, it is enough to prove {RM ψ C ϕ g k } tends to 0 in H From (5) and (6), given every ε > 0, there exits a δ (0, 1) such that when δ < < 1, ψ(z) R (1 ) < ε; Rψ(z) < ε n+1+α +1 (1 ) n+1+α On the other hand, since {f n } 0 uniformly on compact subsets of, if we set h n (z) = Rfn(z), then Cauchy, s estimates give that {h z n } 0 uniformly on compact subsets of, thus there exists an N 0 N, for every n > N 0, δ, f n () < ε and h n () < ε Thus, from (3) and (4) we have f n () Rψ(z) + ψ(z) R (Rf n)() δ ε ψ(z) R + Rψ(z) Cε δ δ

8 176 Chao Zhang and Sui Huang Thus RM ψ C ϕ f n = f n () Rψ(z) + ψ(z) R (Rf n)() f n () Rψ(z) + ψ(z) R (Rf n)() δ< <1 + f n () Rψ(z) + ψ(z) R (Rf n)() δ Cε + ε So RM ψ C ϕ : A α() H is compact References [1] A K Sharma, Products of multiplication, composition and differention between weighted ergman-nevanlinna and loch-tyoe spacers, Turk J Marth, 35 (011), [] A K Sharma, S D Sharma and S Kumar, Weighted composition followed by differentiation between ergman spaces, International Mathematical Forum, (007), [3] C C Cowen, D MacCluer, Composition Operators on Spaces of Analytic Functions, CRC Press, oca Raton, 1995 [4] H Li, C Wang, X Zhang, Weighted composition followed by differentiation between weighted ergman space and H, Int Journal of Math Analysis, 5 (011), [5] J H Shapiro, Composition Operators and Classical Function Theory, Springer-Verlag, New York, [6] K Zhu, Spaces of holomorphic functions in the unit ball, Springer-Verlag, New York, 004 [7] R A Hibschweiler, N Portnoy, Composition followed by differentiation between ergman and Hardy spaces, Rock Mountain Journal of Mathematics, 35 (005), [8] R Zhao and K Zhu, Theory of ergman spaces in the unit ball of C n, Mém Soc Math Fr, 008, 115, vi+103 pages [9] S Kumar, K J Singh, Weighted composition operators on weighted ergman spaces, Extracta Mathematicae, (007), [10] S Ohno, Products of composition and differention between Hardy spaces, ull Austral Math Soc 73 (006), Received: November 17, 014; Published: January 1, 015