RESEARC H Open Access Weighted differentiation composition operators from weighted bergman space to nth weighted space on the unit disk Liang Zhang and Hong-Gang Zeng * * Correspondence: zhgng@tju.edu. cn Department of Mathematics, Tianjin University, Tianjin 300072, People’s Republic of China Abstract This paper characterizes the boundedness and compactness of the weighted differentiation composition operator from weighted Bergman space to nth weigh ted space on the unit disk of ≤. 2000 Mathematics Subject Classification: Primary: 47B38; Secondary: 32A37, 32H02, 47G10, 47B33. Keywords: weighted differentiation composition operators, weighted Bergman space, nth weighted space, boundedness, compactness 1. Introduction Let D be the open unit disk in the complex space ≤,dA the Lebesegue measure on D normalized so that A ( D ) = 1 . Let H ( D ) be the space of all analytic functions on D . Let a >-1, p>0. f is said to belong to the weighted Bergman space, denoted by A p α (= A p α (D) ) ,if f ∈ H ( D ) and  f  p =(α +1)  D | f ( z ) | p (1 −|z | 2 ) α dA < ∞ . When 0 <p<1, it is complete metric space; when p ≥1, it is a Banach space. Let μ(z)(weight) be a positive continuous function o n D and n Î N 0 .Thenth weighted space on the unit disk, denoted by ω ( n ) μ (D ) , consists of all f ∈ H ( D ) such that b ω (n) μ (D)  f  =sup z ∈ D μ ( z ) | f ( n) ( z ) | < ∞ . For n = 0, the space beco mes the weighted-type space H ∞ μ (D ) ;forn = 1, the Bloch- type space B μ (D ) ; and for n = 2, the Zygmud-type Z μ (D ) . For more details about these spaces, we recommend the readers to ([1,2]). The expression b ω (n) μ (D) (f ) defines a semi-norm on the nth weighted space ω (n) μ (D ) , while the natural norm is given by  f  ω (n) μ (D) = n− 1  j =0    f ( j ) ( 0 )    + b ω (n) μ (D)  f  . Zhang and Zeng Journal of Inequalities and Applications 2011, 2011:65 http://www.journalofinequalitiesandapplications.com/content/2011/1/65 © 2011 Zhang a nd Zeng; licensee Springer. T his is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecomm ons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproductio n in any medium, provided the original work is prop erly cited. With this norm, ω (n) μ (D ) becomes a Banach space. The little nth weighted space, denoted by ω (n) μ ,0 (D ) , is a closed subspace of ω (n) μ (D ) , consisting of those f for which lim | z | →1 μ ( z ) | f ( n) ( z ) | =0 . Let  beanon-constantanalyticself-mapof D , u( z ) ∈ H ( D ) ,andm Î N.The weighted differentiation composition operator D m ϕ , u is defined by D m ϕ ,u f ( z ) = u ( z ) f ( m) ( ϕ ( z )) , for z ∈ D , f ∈ H ( D ) .Ifm =1,u (z)=’ (z), then D m ϕ ,u = DC ϕ ;ifletm =1,u (z)=1, then D m ϕ ,u = C ϕ D . Recently, there have been some interests in study ing some parti cular cases of opera- tors, such as DC  , C  D and D m ϕ , u , between different function spaces. From those stu- dies, they gave some sufficient and necessary conditions for these operators to be bounded and compact. Concerning these results, we also re commend the interested readers to ([3-9]). In this paper, we characterize the boundedness and compactness of the operator D m ϕ , u from A p α to nth weighted space. For the case of D m ϕ ,u : A p α → ω (n ) μ , we have the following results: Theorem 1. Assume that p >0, a >-1, n, m Î N, μ is a weight on D ,  is a non- constant analytic self-map of D , and u ∈ H ( D ) . Then, (1a) D m ϕ,u : A p α → ω (n ) μ is bounded if and only if for each k Î {0, 1, , n} sup z∈D μ(z)    n l=k C l n u ( n− l ) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ ( l − k +1 ) (z)     1 −   ϕ(z)   2  k+m+ α+2 p < ∞ . (1) (1b ) D m ϕ ,u : A p α → ω (n ) μ is compact if and only if D m ϕ ,u : A p α → ω (n ) μ is bounded and for each k Î {0, 1, , n} lim | ϕ(z) | →1 μ(z)    n l=k C l n u ( n−l ) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ ( l−k+1 ) ( z )     1 −   ϕ(z)   2  k+m+ α+2 p =0. (2) For the case of D m ϕ ,u : A p α → ω (n) μ , 0 , our main results are the following: Theorem 2. Assume that p >0,a >-1,n, m Î N, μ is a weight on D ,  is a non- constant analytic self-map of D , and u ∈ H ( D ) . Then, (2a) D m ϕ ,u : A p α → ω (n) μ , 0 is bounded if and only if D m ϕ ,u : A p α → ω (n ) μ is bounded and for each k Î {0, 1, , n}, lim | z | →1 μ(z)      n  l = k C l n u (n−l) ( z ) B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)       =0 . (3) (2b) D m ϕ ,u : A p α → ω (n) μ , 0 is compact if and only if D m ϕ ,u : A p α → ω (n) μ , 0 is bounded and for each k Î {0, 1, , n}, Zhang and Zeng Journal of Inequalities and Applications 2011, 2011:65 http://www.journalofinequalitiesandapplications.com/content/2011/1/65 Page 2 of 10 lim | z | →1 μ(z)    n l=k C l n u ( n− l ) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ ( l − k +1 ) (z)     1 −   ϕ(z)   2  k+m+ α+2 p =0 . (4) The organization of this paper is as follows: we give some lemmas in Section 2, and then prove Theorem 1 in Section 3 and Theorem 2 in Section 4, respectively. Throughout this paper, we will use the symbol C to denote a finite positive number, and it may differ from one occurrence to the other. 2. Some Lemmas Lemma1isadirectconsequenceofthewell-knownestimatein([10],Proposition 1.4.10). Hence, we omit its proof. Lemma 1. Assume that p >0,a >-1,n Î N, n >0,and w ∈ D . Then the function g w,n (z)=  1 − | w | 2  n ( 1 − wz ) n+ 2+α p belongs to A p α . Moreover, sup w ∈ D   g w,n   A p α < ∞ . The next lemma comes from ([11]). Lemma 2. Assume that p >0,a >-1,n Î N, and z ∈ D . Then, there is a positive constant C independent of f such that   f ( n) (z)   ≤ C   f   A p α  1 − | z | 2  n+ 2+α p . Lemma 3. Let p >0,a > -1, m Î N, a = m +1+ α+2 p and D n+1 =         11··· 1 aa+1 ··· a + n ···  n−1 j =0  a + j   n−1 j =0  a + j +1  ···  n−1 j =0  a + j + n          . Then, D n+1 =  n j =1 j ! . Proof.With a = m +1+ α+2 p and replacing n by n + 1 in ([12], Lemm a 2.3), the lemma easily follows. □ The next lemma can be found in ([7], Lemma 4). Lemma 4. Assume n Î N, g, u ∈ H ( D ) and  is an analytic self-map of D . Then,  u(z)g(ϕ( z ))  (n) =  n k=0 g (k)  ϕ(z)   n l=k C l n u (n−l) ( z ) B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)  , where B l,k (ϕ  (z), ϕ  (z), , ϕ (l−k+1) )=  k 1 , ,k l l! k 1 ! k l ! l  j =1  ϕ (j) (z) j!  k j , (5) and the sum in (5) is overall non-neg ative integers k 1 , , k l satisfying k 1 +k 2 + +k 1 = k and k 1 +2k 2 + + lk l = l. By a proof in a standard way ([1], Proposition 3.11), we can get the next lemma. Zhang and Zeng Journal of Inequalities and Applications 2011, 2011:65 http://www.journalofinequalitiesandapplications.com/content/2011/1/65 Page 3 of 10 Lemma 5. Suppose u ∈ H ( D ) , p >0,a >-1,n, m Î N and  is a non-constant analytic self-map of D . Then the operator D m ϕ ,u : A p α → ω (n ) μ is compact if and only if D m ϕ ,u : A p α → ω (n ) μ is bounded and for any bounded sequence { f k } kÎN in A p α which converges to zero uniformly on compact subsets of D as k ® ∞, we have D m ϕ ,u f k → 0 in ω ( n ) μ as k ® ∞. Lemma 6. Suppose n Î N and μ is a radial weight such that lim |z|®1 μ (z)=0.A closed set K in ω (n) μ , 0 is compact if and only if it is bounded and satisfies lim | z | →1 sup f ∈K μ (z)    f (n) (z)    =0 . Proof. The proof of this Lemma is followed by standard arguments similar to those outlined in ([13]). We omit the details. □ 3. The Proof of Theorem 1 (1a) Boundedness of D m ϕ , u . We will prove the sufficiency first. Suppose that the conditions in (1) hold. Then, for any f ∈ A p α , from Lemma 2 and Lemma 4, we obtain μ(z)     D m ϕ,u f  (n) (z)    = μ(z)      n  k=0 f (k+m)  ϕ(z)  n  l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)       ≤ μ(z) n  k=0    f (k+m)  ϕ(z)          n  l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)       ≤ C   f   A p α n  k=0 μ(z)     n  l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)       1 −   ϕ(z)   2  k+m+ α+2 p . (6) For z = 0 and every d Î {0, 1, , n - 1},     D m ϕ,u f  (d) (0)    =      d  k=0 f (k+m) (ϕ(0)) d  l=k C l d u (d−l) (0)B l,k  ϕ  (0), ϕ  (0), ϕ (l−k+1) (0)       ≤ C   f   A p α d  k=0      d  l=k C l d u (d−l) (0)B l,k  ϕ  (0), ϕ  (0), , ϕ (l−k+1) (0)        1 − | ϕ ( 0 ) | 2  k+m+ α+2 p (7) From (1), (6), and (7), we know that D m ϕ ,u : A p α → ω (n ) μ is bounded. Conversely, suppose that D m ϕ ,u : A p α → ω (n ) μ is bounded. Then, ther e exists a constant C such that   D m ϕ,u f   ω (n) μ ≤ C   f   A p α , for all f ∈ A p α . For a fixed w ∈ D , and constants c 1 , c 2 , , c n +1 , set g w ( z ) = n+1  j=1 c j  j + 2+α p  j + 2+α p +1   j + 2+α p + m − 1   1 − | w | 2  j ( 1 − wz ) j+ 2+α p . (8) Zhang and Zeng Journal of Inequalities and Applications 2011, 2011:65 http://www.journalofinequalitiesandapplications.com/content/2011/1/65 Page 4 of 10 Applying Lemma 1 and triangle inequality, it is easy to get that g w ∈ A p α for every w ∈ D . Moreover, we have that sup w ∈ D   g w   A p α < ∞ . (9) Now we show that for each s Î {m, m + 1, , m + n}, there are constants c 1 , c 2 , , c n+1 , such that, g (s) w (w)= w s  1−|w | 2  s+ α+2 p , g (t) w (w)=0, t ∈{m, , m + n}\{s } (10) Indeed, by differentiating function g w for each s Î {m, m + 1, , m + n}, the system in (10) becomes c 1 + c 2 + ···+ c n+1 =0 (m +1+ α+2 p )c 1 +(m +2+ α+2 p )c 2 + ···+(m + n +1+ α+2 p )c n+1 =0 (m +1+ α+2 p ) (s + α+2 p )c 1 + ···+(m + n +1+ α+2 p )  n + s + α+2 p  c n+1 =1 (m +1+ α+2 p ) (m + n + α+2 p )c 1 + ···+(m + n +1+ α+2 p )  m +2n + α+2 p  c n+1 = 0 (11) By Lemma 3, the determinant of system (11) is different from zero, which implies the statement. For each k Î {0, 1, 2, , n}, we choose the corresponding family of func- tions that satisfy (10) with s = m + k and denote it by g w,k . For each fixed k Î {0, 1, , n}, the boundedness of the operator D m ϕ ,u : A p α → ω (n ) μ , along with Lemma 4 and (8) implies that for each (w) ≠ 0, μ(w)   ϕ(w)   m+k    n l=k C l n u (n−l) (w)B l,k  ϕ  (w), ϕ  (w), , ϕ (l−k+1) (w)     1−|ϕ(w) | 2  k+m+ α+2 p ≤ C sup w∈ D   D m ϕ,u (g ϕ(w),k )   ω (n) μ ≤ C   D m ϕ,u   A p α →ω (n) μ . (12) From (12), it follows that for each k Î {0, 1, , n}, sup | ϕ(z) | > 1 2 μ(z)    n l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)    (1 −|ϕ(z ) | 2 ) k+m+ α+2 p ≤ C   D m ϕ,u   A p α →ω (n) μ . (13) Now we use the test functions h k ( z ) = z k+m , k =0,1, , n . For each k Î N, it is easy to get that h k ∈ A p α ,  h k  A p α ≤ 1 . (14) By applying Lemma 4 to the h 0 (z)=z m , we get (D m ϕ,u h 0 ) (n) (z) = h (m) 0 (ϕ(z)) n  l=0 C l n u (n−l) (z)B l,0  ϕ  (z), ϕ  (z), , ϕ (l+1) (z)  = m! n  l = 0 C l n u (n−l) ( z ) B l,0  ϕ  (z), ϕ  (z), , ϕ (l+1) (z)  , Zhang and Zeng Journal of Inequalities and Applications 2011, 2011:65 http://www.journalofinequalitiesandapplications.com/content/2011/1/65 Page 5 of 10 which along with boundedness of the operator D m ϕ,u : A p α → ω (n ) μ implies that m!sup z∈D μ(z)      n  l = 0 C l n u (n−l) (z)B l,0  ϕ  (z), ϕ  (z), , ϕ (l+1) (z)       ≤ C   D m ϕ,u   A p α →ω (n) μ . (15) Assume now that we have proved the following inequalities sup z∈D μ(z)       n  l=j C l n u (n−l) (z)B l,j  ϕ  (z), ϕ  (z), , ϕ (l−j+1) (z)        ≤ C   D m ϕ,u   A p α →ω (n) μ . (16) for j Î {0, 1, , k - 1}, k ≤ n. Apply Lemma 4 to the h k (z)=z m+k , and knowing that z (s) ≡ 0fors >m + k and the boundedness of the operator D m ϕ ,u : A p α → ω (n ) μ , we get    (D m ϕ,u h k ) (n) (z)    ≥       k−1  j=0 (m + k) (k − j +1)(ϕ(z)) (k−j) n  l=j C l n u (n−l) (z)B l,j  ϕ  (z), , ϕ (l−j+1) (z)        −      (m + k)! n  l = k C l n u (n−l) (z)B l,k  ϕ  (z), , ϕ (l−k+1) (z)       . (17) Using hypothesis (16), we can know that sup z∈D μ(z)      n  l = k C l n u (n−l) (z) B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)       ≤ C   D m ϕ,u   A p α →ω (n) μ . (18) for each k Î {0, 1, , n}. Then, for each k Î {0, 1, , n}, sup | ϕ(z) | ≤ 1 2 μ(z)    n l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)     1−|ϕ(z) | 2  k+m+ α+2 p ≤ C sup z∈D μ(z)      n  l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), ϕ (l−k+1) (z)       . ≤ C   D m ϕ,u   A p α →ω (n) μ (19) From (13) and (19), we know that (1) holds. (1b) Compactness of D m ϕ , u . Suppose D m ϕ ,u : A p α → ω (n ) μ is bounded and (2) holds. Then, by (1a), (1) holds. Let  f i  i ∈N beasequencein A p α ,suchthat, sup i ∈ N   f i   A p α ≤ M and f i converges to 0 uniformly on compact subsets of D as i ® ∞. By the assumption, for any ε > 0, there is a δ Î (0, 1), such that, for each k Î {0, 1, , n} and δ <|(z)|<1, μ(z)    n l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)     1−|ϕ(z) | 2  k+m+ α+2 p <ε . (20) Zhang and Zeng Journal of Inequalities and Applications 2011, 2011:65 http://www.journalofinequalitiesandapplications.com/content/2011/1/65 Page 6 of 10 From Lemma 2 and (20), we have sup i∈N μ(z)      n  k=0 f (k+m) i  ϕ(z)  n  l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ (l−k+1) (z)       ≤ C sup i∈N   f i   A p α n  k=0 sup z∈D μ(z)     n  l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)       1−|ϕ(z) | 2  k+m+ α+2 p ≤ CM ( n +1 ) ε. (21) If |(z)|≤ r, then by Cauchy’s estimate and (19), we have sup μ(z)      n  k=0 f (k+m) i (ϕ(z)) n  l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ (l−k+1) (z)       ≤ C n  k = 0 sup |ϕ(z)|≤r    f (m+k) i (ϕ(z))    → 0, (i →∞) (22) For j = 0, 1, , n - 1, we have    (D m ϕ,u f i ) (j) (0)    → 0, (i →∞ ) (23) Applying (21), (22), and (23), we know that   D m ϕ,u f i   ω (n) μ → 0, (i →∞ ) .FromLemma 5, D m ϕ,u : A p α → ω (n ) μ is compact. Conversely, suppose that D m ϕ ,u : A p α → ω (n ) μ is compact, then D m ϕ ,u : A p α → ω (n ) μ is bounded. Let (z i ) iÎ N be a seq uence in D such that |(z i )| ® 1, i ® ∞.Ifsucha sequence does not exist, then the condition in (2) is e asily satisfied. Now, assume that when |||| ∞ = 1 and (2) does not hold, then there is k Î {0, 1, , n} and δ > 0 such that μ(z i )    n l=k C l n u (n−l) (z i )B l,k  ϕ  (z i ), ϕ  (z i ), ϕ (l−k+1) (z i )     1−|ϕ(z i ) | 2  k+m+ α+2 p ≥ δ . Let g i (z)=g ϕ (z i ),k (z), i ∈ N , k Î 0, 1, , n be as in Theorem 1. Then, sup i ∈ N   g i   A p α ≤ M and g i ® 0 uniformly on compact subsets of D as i ® ∞. By the assumption and Lemma 5, we have that for k Î {0, 1, , n} lim i →∞   D m ϕ,u g ϕ(z i ),k   ω (n) μ =0 . (24) On the other hand, from (12), we obtain   D m ϕ,u g ϕ(z i ),k   ω (n) μ ≥ μ(z i ) | ϕ(z i ) | k+m    n l=k C l n u (n−l) (z i )B l,k  ϕ  (z i ), ϕ  (z i ), ϕ (l−k+1) (z i )     1−|ϕ(z i ) | 2  k+m+ α+2 p > δ 3 (25) for large enough i. From (24) and (25), this is a contradiction. So, (2) holds. Now the proof of Theorem 1 is completed. Zhang and Zeng Journal of Inequalities and Applications 2011, 2011:65 http://www.journalofinequalitiesandapplications.com/content/2011/1/65 Page 7 of 10 4. The Proof of Theorem 2 (2a) Boundedness of D m ϕ , u . First, suppose that D m ϕ ,u : A p α → ω (n ) μ is bounded and (3) holds. For each polynomial p (z), we obtain |p (m+k) (z)| ≤ C p , z Î D, C p is s constant depending on p. And μ(z)    (D m ϕ,u p) (n) (z)    = μ(z)      n  k=0 p (k+m) (ϕ(z)) n  l=k C l n u (n−l) (z)B l,k  ϕ  (z), , ϕ (l−k+1) (z)       ≤ C p n  k=0 | μ(z) |      n  l=k C l n u (n−l) (z)B l,k  ϕ  (z), , ϕ (l−k+1) (z)       → 0, | z | → 1. (26) From (26), we have that for each polynomial p(z), D m ϕ ,u p ∈ ω (n) μ , 0 . Since the set of poly- nomials is d ense in A p α , we have that for each f ∈ A p α ,thereisasequenceofpolyno- mials ( p k ) k∈ N ,suchthat  f − p k  A p α → 0 as k ® ∞ . From the boundness of D m ϕ ,u : A p α → ω (n ) μ , we have that   D m ϕ,u f − D m ϕ,u p k   ω (n) μ ≤   D m ϕ,u   A p α →ω (n) μ   f − p k   A p α → 0, k →∞ . Then, D m ϕ ,u f ∈ ω (n) μ , 0 , from which the boundedness of D m ϕ ,u : A p α → ω (n) μ , 0 follows. Conversely, suppose that D m ϕ ,u : A p α → ω (n) μ , 0 is bounded. It is clear that D m ϕ ,u : A p α → ω (n ) μ is bounded. Then, taking the test functions h k (z)=z m+k for each k Î {0, 1, , n}, we obtain D m ϕ ,u z m+k ∈ ω (n) μ , 0 . By the proof of Theorem 1, for each k Î {0, 1, , n}, lim | z | →1 μ ( z )      n  l = k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), ϕ (l−k+1) (z)       =0 . (2a) is completed. (2b) Compactness of D m ϕ , u . First, assume that D m ϕ ,u : A p α → ω (n) μ , 0 is compact, so it is bounded and (3) holds. Hence, if |||| ∞ <1, μ(z)    n l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)     1 −   ϕ(z)   2  k+m+ α +2 p ≤ μ(z)    n l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)     1−ϕ  2 ∞  k+m+ α+2 p → 0, ( | z | → 1 ) (27) Zhang and Zeng Journal of Inequalities and Applications 2011, 2011:65 http://www.journalofinequalitiesandapplications.com/content/2011/1/65 Page 8 of 10 If |||| ∞ =1,since D m ϕ,u : A p α → ω (n ) μ is compact too and (2) holds, then for all ε >0, there is an r Î (0, 1), such that when r <| (z) | < 1, for k Î {0, 1, , n}, μ(z)    n l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)     1−|ϕ(z) | 2  k+m+ α+2 p <ε . (28) By (3), we know there is a δ Î (0, 1), such that δ <|z| < 1, for k Î {0, 1, , n}, μ(z)      n  l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)       <ε ( 1 − r 2 ) k+m+ α+2 p . (29) Then, when δ <|z| < 1 and r <| (z) | < 1 for k Î {0, 1, , n}, we get μ(z)    n l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)    (1 −   ϕ(z)   2 ) k+m+ α+2 p <ε . (30) In addition, when | (z)|≤ r and δ <|z| < 1, we have μ(z)    n l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)     1−|ϕ(z) | 2  k+m+ α+2 p < μ ( z )    n l=k C l n u (n−l) (z)B l,k  ϕ  (z), ϕ  (z), , ϕ (l−k+1) (z)    ( 1 − r 2 ) k+m+ α+2 p <ε . (31) Combining (30) and (31), we know (4) holds. Conversely, assum e D m ϕ ,u : A p α → ω ( n ) μ , 0 is bounded and (4) holds. Taking the supre- mum in (6) for all f in the unit ball of A p α , and using the condition (4), we have lim | z | →1 sup  f  A p α ≤1 μ(z)    (D m ϕ,u f ) (n) (z)    = 0 , from which by Lemma 6, the compactness of D m ϕ ,u : A p α → ω ( n ) μ , 0 follows. Now the proof of Theorem 2 is finished. Acknowledgements We are grateful to the referee(s) for many helpful comments on the manuscript. Hong-Gang Zeng is supported in part by the National Natural Science Foundation of China (Grand Nos. 10971153, 10671141). Authors’ contributions LZ found the question and drafted the manuscript. HGZ joined in the discussion about the question and revised the paper. All authors read and approved the final manuscript Competing interests The authors declare that they have no competing interests. Received: 29 January 2011 Accepted: 21 September 2011 Published: 21 September 2011 References 1. Cowen, CC, Maccluer, BD: Composition Operators on Spaces of Analytic Functions. CRC Press, Boca Raton, FL (1995) 2. Zhu, K: Operator Theory in Function Spaces. Marcel Dekker, New York (1990) 3. Hibschweiler, RA, Portnoy, N: Composition followed by differentiation between Bergman and Hardy spaces. The Rocky Mountain Journal of Mathematics. 35(3), 843–855 (2005). doi:10.1216/rmjm/1181069709 4. Li, S, Stević, S: Composition followed by differentiation between Bloch-type spaces. J Comput Anal Appl. 9(2), 195–205 (2007) Zhang and Zeng Journal of Inequalities and Applications 2011, 2011:65 http://www.journalofinequalitiesandapplications.com/content/2011/1/65 Page 9 of 10 5. Ohno, S: Products of composition and diffferentiation between Hardy spaces. Bull Aust Math Soc. 73(2), 235–243 (2006). doi:10.1017/S0004972700038818 6. Stević, S: Products of composition and differentiation operators on the weighted Bergman space. Bull Belg Math Soc Simon Stevin. 16(4), 623–635 (2009) 7. Stević, S: Weighted differentiation composition operators from H ∞ and Bloch spaces to nth weighted-type spaces on the unit disk. Appl Math Comput. 216(12), 3634–3641 (2010). doi:10.1016/j.amc.2010.05.014 8. Yang, WF: Products of composition and differentiation operators from Q k (p, q) spaces to Bloch-Type Spaces. Abst Appl Anal2009, 14. Article ID 741920 9. Zhou, XL: Products of differentiation, composition and multiplication from Bergman type spaces to Bers type spaces. Int Trans Special Funct. 18(3), 223–231 (2007). doi:10.1080/10652460701210250 10. Rudin, W: Function Theory in the Unit ball of C n , vol. 241 of Fundamental Principles of Mathematical Science. Springer, New York, USA (1980) 11. Flett, TM: The dual of an inequality and Littlewood and some related inequalities. J Math Anal Appl. 38, 746–765 (1972). doi:10.1016/0022-247X(72)90081-9 12. Stević, S: composition operators from the weighted Bergman space to the nth weighted-type space on the unit disk. Hindawi Publ Corp Discret Dyn Nat Soc2009, 11. Article ID 742019 13. Madigan, K, Matheson, A: Compact composition operators on the Bloch Space. Trans Am Math Soc. 347(7), 2679–2687 (1995). doi:10.2307/2154848 doi:10.1186/1029-242X-2011-65 Cite this article as: Zhang and Zeng: Weighted differentiation composition operators from weighted bergman space to nth weighted space on the unit disk. Journal of Inequalities and Applications 2011 2011:65. Submit your manuscript to a journal and benefi t from: 7 Convenient online submission 7 Rigorous peer review 7 Immediate publication on acceptance 7 Open access: articles freely available online 7 High visibility within the fi eld 7 Retaining the copyright to your article Submit your next manuscript at 7 springeropen.com Zhang and Zeng Journal of Inequalities and Applications 2011, 2011:65 http://www.journalofinequalitiesandapplications.com/content/2011/1/65 Page 10 of 10 . RESEARC H Open Access Weighted differentiation composition operators from weighted bergman space to nth weighted space on the unit disk Liang Zhang and Hong-Gang Zeng * * Correspondence: zhgng@tju.edu. cn Department. of composition and differentiation operators on the weighted Bergman space. Bull Belg Math Soc Simon Stevin. 16(4), 623–635 (2009) 7. Stević, S: Weighted differentiation composition operators from. differentiation composition operators, weighted Bergman space, nth weighted space, boundedness, compactness 1. Introduction Let D be the open unit disk in the complex space ≤,dA the Lebesegue measure on D normalized