PRODUCTS OF MULTIPLICATION, COMPOSITION AND DIFFERENTIATION OPERATORS FROM MIXED-NORM SPACES TO WEIGHTED-TYPE SPACES. Fang Zhang and Yongmin Liu

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1 TAIWANESE JOURNAL OF MATHEMATICS Vol. 18, No. 6, pp , December 2014 DOI: /tjm This paper is available online at PRODUCTS OF MULTIPLICATION, COMPOSITION AND DIFFERENTIATION OPERATORS FROM MIXED-NORM SPACES TO WEIGHTED-TYPE SPACES Fang Zhang Yongmin Liu Abstract. Let ϕ be an analytic self-map of the unit disk D, H(D) the space of analytic functions on D ψ 1,ψ 2 H(D). Recently Stević co-workers defined the following operator T ψ1,ψ 2,ϕf(z) =ψ 1 (z)f(ϕ(z)) + ψ 2 (z)f (ϕ(z)), f H(D). The boundedness compactness of the operators T ψ1,ψ 2,ϕ from mixed-norm spaces to weighted-type spaces are investigated in this paper. 1. INTRODUCTION Let H(D) denote the space of all analytic functions in the open unit disc D of the complex plane C. A positive continuous function φ on the interval [0,1) is called normal if there exist positive numbers a, b, 0 <a<b t 0 [0, 1), such that φ(t) (1 t 2 ) a φ(t) (1 t 2 ) a φ(t) is decreasing for t 0 t<1 lim t 1 (1 t 2 ) =0; a (see, e.g., [20]). is increasing for t 0 t<1 lim t 1 φ(t) (1 t 2 ) a = For 0 <p<, 0 <q< a normal function φ, the mixed-norm space H(p, q, φ) is the space of analytic functions on the unit disk D such that ( 1 f p,q,φ = 0 M p q (f, r)φp (r) 1 r rdr ) 1/p, Received January 14, 2014, accepted April 14, Communicated by Alexer Vasiliev Mathematics Subject Classification: 47B38, 47B33, 46E14, 30H10. Key words phrases: Multiplication operator, Differentiation operator, Composition operator, Mixednorm space, Weighted-type space. The authors are supported by the Natural Science Foundation of China (No ) the Priority Academic Program Development of Jiangsu Higher Education Institutions. 1927

2 1928 Fang Zhang Yongmin Liu where the integral means M p (f, r) are defined by M p (f, r) = ( 1 2π 1/p, f(re iθ ) dθ) p 0 r<1. 2π 0 For 1 p<, H(p, q, φ) equipped with the norm p,q,φ is a Banach space. When 0 <p<1, p,q,φ is a quasinorm on H(p, q, φ), H(p, q, φ) is a Fréchet space but not a Banach space. If 0 <p= q<, thenh(p, p, φ) is the Bergman-type space H(p, p, φ)={f H(D): D f(z) pφp ( z ) da(z) < }, 1 z where da(z) denotes the normalized Lebesgue area measure on the unit disk D such that A(D) =1. Note that if φ(r) =(1 r) (α+1)/p,thenh(p, p, φ) is the weighted Bergman space A p α defined for 0 <p< α> 1, as the space of all f H(D) such that f p A p = f(z) p (1 z 2 ) α da(z) < α D (see, e.g., [4]). Let μ be a positive continuous function on D (weight). The weighted-type space Hμ (D) =Hμ consists of all f H(D) such that f H μ =sup μ(z) f(z) <. It is known that Hμ is a Banach space. Let Hμ,0 denote the subspace of H μ consisting of those f Hμ such that sup μ(z) f(z) = 0. This space is called the little z 1 weighted-type space. For some results on weighted-type spaces see, e.g.[6] the related references therein. Denote by S(D) the set of analytic self-maps of D. Forϕ S(D) the composition operator C ϕ is defined by C ϕ f = f ϕ, f H(D). It is interesting to provide a function theoretic characterization for ϕ inducing a bounded or compact composition operator on various spaces. It is well known that the composition operator is bounded on Hardy space, the Bergman space the Bloch space. The composition operator was studied extensively by many people, see, for example, [1, 19, 31] references therein. For ψ H(D), the multiplication operator M ψ is defined by M ψ f = ψ f, f H(D).

3 Products of Multiplication, Composition Differentiation Operators 1929 Given ψ H(D) ϕ S(D), the weighted composition operator with symbols ψ ϕ is defined as the linear operator on H(D) given by (ψc ϕ f)(z) =ψ(z)f(ϕ(z)) = (M ψ C ϕ f)(z), f H(D), z D. Special cases for ψ(z) =1 ϕ(z) =z, z D, are the composition operator C ϕ the multiplication operator M ψ. For some recent articles on weighted composition operators on some H -type spaces, see, for example, [7, 9, 17, 21, 22, 23] references therein. Let D be the differentiation operator, it is defined by Df = f, f H(D). The differentiation operator is typically unbounded on many analytic function spaces. Products of concrete linear operators between spaces of holomorphic functions have been the object of study for recent several years, see, e.g. [3, 5, 8, 10, 11, 12, 14, 15, 18, 25, 27, 30, 32] the related references therein. The products of composition operator differentiation operator DC ϕ C ϕ D are defined respectively as follows DC ϕ f = f (ϕ)ϕ, f H(D) C ϕ Df = f ϕ, f H(D). They have been recently studied, for example, in [5, 8, 10, 11, 12, 14, 18, 25, 27, 29, 30] (see also the related references therein). Ohon in [18] devoted most of the paper to finding necessary sufficient conditions for C ϕ D to be bounded as well as for C ϕ D to be compact on the Hardy space H 2. The operator DC ϕ was studied for the first time in [5], where the boundedness compactness of DC ϕ between Bergman Hardy spaces are investigated. Li Stević in [8, 10, 12] studied the boundedness compactness of the operator DC ϕ between Bloch type space, weighted Bergman space A p α Bloch type space B β, mixed-norm space α-bloch space B α as well as the space of bounded analytic functions the Bloch-type space. Liu Yu in [15] studied the boundedness compactness of the operator DC ϕ from H Bloch spaces to Zygmund spaces. Yang in [28] studied the same problems for operators C ϕ D DC ϕ from Q K (p, q) space to B μ B μ,0. The products of differentiation operator multiplication operator, denoted by DM ϕ, is defined as follows DM ψ f = ψ f + ψ f, f H(D). Stević in [26] studied the boundedness compactness of the products of differentiation multiplication operators DM ψ from mixed-norm spaces to weighted-type

4 1930 Fang Zhang Yongmin Liu spaces. Liu Yu in [14] studied the operators DM ψ from H to Zygmund spaces. Yu Liu in [30] investigated the same problems for operators DM ψ from mixednorm spaces to Bloch-type spaces. Zhu in [32] completely characterized the boundedness compactness of linear operators which are obtained by taking products of differentiation, composition multiplication operators which act from Bergman type spaces to Bers spaces. Kumar Singh investigated the same problem for operators DC ϕ M ψ acting on A p α used the Carleson-type conditions. They also found the essential norm estimates of M ψ DC ϕ in the spirit of the work by Čučković Zhao [2]. The products of composition, multiplication differentiation operators can be defined in following six ways (1) (M ψ C ϕ Df)(z) =ψ(z)f (ϕ(z)); (M ψ DC ϕ f)(z) =ψ(z)ϕ (z)f (ϕ(z)); (C ϕ M ψ Df)(z) =ψ(ϕ(z))f (ϕ(z)); (DM ψ C ϕ f)(z) =ψ (z)f(ϕ(z)) + ψ(z)ϕ (z)f (ϕ(z)); (C ϕ DM ψ f)(z) =ψ (ϕ(z))f(ϕ(z)) + ψ(ϕ(z))f (ϕ(z)); (DC ϕ M ψ f)(z) =ψ (ϕ(z))ϕ (z)f(ϕ(z)) + ψ(ϕ(z))ϕ (z)f (ϕ(z)); for z D f H(D). It is interesting to provide a function theoretic characterization of ψ ϕ when the six above operators become bounded or compact operators between spaces of analytic functions in the unit disk, the polydisk the unit ball. Note that the operator M ψ C ϕ D induces many known operators. If ψ(z) =1,then M ψ C ϕ D = C ϕ D. When ψ(z) =ϕ (z), then we get the operator DC ϕ. If we put ϕ(z)=z, thenm ψ C ϕ D = M ψ D, that is, the product of differentiation operator. Also note that M ψ DC ϕ = M ψϕ C ϕ D C ϕ M ψ D = M ψ ϕ C ϕ D. Thus the corresponding characterizations of boundedness compactness of M ψ DC ϕ C ϕ M ψ D can be obtained by replacing ψ, respectively by ψϕ ψ ϕ in the results stated for M ψ C ϕ D. Let ψ 1,ψ 2 H(D) ϕ be a holomorphic self-map of D. The products of multiplication composition differentiation operators are defined as follows T ψ1,ψ 2,ϕf(z) =ψ 1 (z)f(ϕ(z)) + ψ 2 (z)f (ϕ(z)), f H(D). The operator T ψ1,ψ 2,ϕ was studied by Stević co-workers for the first time in [27], where the boundedness compactness of T ψ1,ψ 2,ϕ between Bergman spaces are investigated. It is clear that all products of composition, multiplication differentiation operators in (1.1) can be obtained from the operator T ψ1,ψ 2,ϕ by fixing ψ 1 ψ 2.More specifically we have M ψ C ϕ D = T 0,ψ,ϕ, M ψ DC ϕ = T 0,ψϕ,ϕ, C ϕ M ψ D = T 0,ψ ϕ,ϕ,

5 Products of Multiplication, Composition Differentiation Operators 1931 DM ψ C ϕ = T ψ,ψϕ,ϕ, C ϕ DM ψ = T ψ ϕ,ψϕ,ϕ,dc ϕ M ψ = T (ψ ϕ)ϕ,(ψ ϕ)ϕ,ϕ. Motivated by the results [26, 27], we consider the boundedness compactness of the operator T ψ1,ψ 2,ϕ from mixed-norm space H(p, q, φ) to weighted-type space H μ. Throughout this article, the letter C denotes a positive constant which may vary at each occurrence but it is independent of the essential variables. 2. SOME LEMMAS Lemma 2.1. [24]. Assume that p, q (0, ), φ is normal f H(p, q, φ). Then for each n N 0, there is a positive constant C independent of f such that f p,q,φ f (n) (z) C φ( z )(1 z 2 ) 1/q+n, z D. By stard arguments (see [16, 21]) the following lemmas follows. Lemma 2.2. Assume that ψ 1,ψ 2 H(D) ϕ S(D).Then T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ is compact if only if T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ is bounded for any bounded sequence f k in H(p, q, φ) which converges to zero uniformly on compact subsets of D as k, we have T ψ1,ψ 2,ϕf k H μ 0 as k. Lemma 2.3. AclosedsetK in Hμ,0 is compact if only if K is bounded satisfies lim sup μ(z) f(z) =0. z 1 f K 3. MAIN RESULTS AND PROOFS Theorem 3.1. Assume that ψ 1,ψ 2 H(D) ϕ S(D). Then T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ is bounded if only if (1) sup φ( ϕ(z) )(1 ϕ(z) 2 <, μ(z) ψ 2 (z) (2) sup φ( ϕ(z) )(1 ϕ(z) 2 <. +1 Proof. Suppose that T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ is bounded, i.e., there exists a constant C such that T ψ1,ψ 2,ϕf H μ C f p,q,φ.forafixedw D, set f w (z) = (1 w 2 ) b+1 φ( w ) 1 ( (1 wz) α 2α(1 w 2 ) (α + 1)(1 wz) α+1 + α(1 w 2 ) 2 (α + 2)(1 wz) α+2 ),

6 1932 Fang Zhang Yongmin Liu where the constant b is from the definition of the normality of the function φ α =1/q + b +1. A straightforward calculation show that f w (z) =(1 w 2 ) b+1 w α ( φ( w ) (1 wz) α+1 2α(1 w 2 ) (1 wz) α+2 + α(1 w 2 ) 2 ), (1 wz) α+3 2 f w (w) = (α +1)(α +2)φ( w )(1 w 2, f w(w) =0, sup f w p,q,φ C (see [24, 26]). Hence, w D C T ψ1,ψ 2,ϕf ϕ(w) H μ μ(w) ψ 1 (w)f ϕ(w) (ϕ(w)) + ψ 2 (w)f ϕ(w) (ϕ(w)) 2 = μ(w) ψ 1 (w) (α +1)(α +2)φ( ϕ(w) )(1 ϕ(w) 2, for every w D. Therefore sup φ( ϕ(z) )(1 ϕ(z) 2 <. For a fixed w D. Set (1 w 2 ) t+1 g w (z) = φ( w )(1 wz+t+1, where the constant t is from the definition of the normality of the function φ. straightforward calculation show that g w(z) = (t +1+1/q)(1 w 2 ) t+1 w φ( w )(1 wz+t+2, 1 g w (w) = φ( w )(1 w 2, sup g w p,q,φ C (see [13]). Hence, w D A C T ψ1,ψ 2,ϕg ϕ(w) H μ μ(w) ψ 1 (w)g ϕ(w) (ϕ(w)) + ψ 2 (w)g ϕ(w) (ϕ(w)) μ(w) ψ 2 (w) g ϕ(w) (ϕ(w)) μ(w) ψ 1(w) g ϕ(w) (ϕ(w)) = μ(w) ψ 2 (w) (t +1+1/q)(1 ϕ(w) 2 ) t+1 ϕ(w) φ( ϕ(w) )(1 ϕ(w) 2 +t+2 1 μ(w) ψ 1 (w) φ( ϕ(w) )(1 ϕ(w) 2 = (t +1+1/q)μ(w) ψ 2(w) ϕ(w) φ( ϕ(w) )(1 ϕ(w) 2 +1 μ(w) ψ 1 (w) φ( ϕ(w) )(1 ϕ(w) 2

7 Products of Multiplication, Composition Differentiation Operators 1933 for every w D. Therefore, (t +1+1/q)μ(w) ψ 2 (w) ϕ(w) μ(w) ψ 1 (w) φ( ϕ(w) )(1 ϕ(w) 2 +1 C + φ( ϕ(w) )(1 ϕ(w) 2. From (1), we get μ(z) ψ 2 (z) ϕ(z) (3) sup φ( ϕ(z) )(1 ϕ(z) 2 <. +1 From (3), we have (4) sup ϕ(z) > 1 2 μ(z) ψ 2 (z) φ( ϕ(z) )(1 ϕ(z) 2 +1 sup ϕ(z) > 1 2 Since f(z) =1,g(z)=z H(p, q, φ), it follows that 2μ(z) ψ 2 (z) ϕ(z) φ( ϕ(z) )(1 ϕ(z) 2 <. ) 1/q+1 sup T ψ1,ψ 2,ϕf H μ C sup μ(z) ψ 1 (z)ϕ(z)+ψ 2 (z) T ψ1,ψ 2,ϕg H μ C. It is easy to see that μ(w) ψ 2 (w) T ψ1,ψ 2,ϕg H μ + μ(w) ψ 1 (w)ϕ(w) C for every w D. From this the fact φ is normal we obtain (5) sup ϕ(z) 1 2 μ(z) ψ 2 (z) φ( ϕ(z) )(1 ϕ(z) 2 +1 C sup ϕ(z) 1 2 μ(z) ψ 2 (z) <. Combining (4) (5), we get (2) as desired. For the converse, suppose that (1) (2) hold. For any f H(p, q, φ), by Lemma 2.1, we have μ(z) ψ 1 (z)f(ϕ(z)) + ψ 2 (z)f (ϕ(z)) f p,q,φ φ( ϕ(z) )(1 ϕ(z) 2 + μ(z) ψ 2 (z) f p,q,φ φ( ϕ(z) )(1 ϕ(z) Therefore, T ψ1,ψ 2,ϕ : H(p, q, φ) H μ is bounded. The proof of the theorem is complete. Theorem 3.2. Assume that ψ 1,ψ 2 H(D) ϕ S(D). Then T ψ1,ψ 2,ϕ : H(p, q, φ) H μ is compact if only if T ψ1,ψ 2,ϕ : H(p, q, φ) H μ is bounded,

8 1934 Fang Zhang Yongmin Liu (6) lim ϕ(z) 1 φ( ϕ(z) )(1 ϕ(z) 2 =0, μ(z) ψ 2 (z) (7) lim ϕ(z) 1 φ( ϕ(z) )(1 ϕ(z) 2 =0. +1 Proof. Suppose that T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ is compact. Then let {z k} k N be a sequence in D such that ϕ(z k ) 1 as k. We can use the test functions in Theorem 3.2. Let f k (z) =f ϕ(zk )(z). We have 2 1 f k (ϕ(z k )) = (α +1)(α +2) φ( ϕ(z k ) )(1 ϕ(z k ) 2, f k (ϕ(z k)) = 0 sup f k p,q,φ C. For z = r<1, using the fact that φ is k N normal, we have C f k (z) (1 r+1 (1 ϕ(z k) ) 0 (k ), that is, f k converges to 0 uniformly on compact subsets of D, using the compactness of T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ Lemma 2.2, we obtain 2 1 μ(z k ) ψ 1 (z k ) (α +1)(α +2) φ( ϕ(z k ) )(1 ϕ(z k ) 2 T ψ 1,ψ 2,ϕf k H μ 0, as k.fromthis, ϕ(z k ) 1, it follows that lim k consequently (6) holds. In order to prove (7), choose μ(z k ) ψ 1 (z k ) φ( ϕ(z k ) )(1 ϕ(z k ) 2 =0, g k (z) =g ϕ(zk )(z). We have g k (ϕ(z k )) = 1 φ( ϕ(z k ) )(1 ϕ(z k ) 2, g k (ϕ(z k)) = (t +1+1/q)(1 ϕ(z k) 2 ) t+1 ϕ(z k ) φ( ϕ(z k ) )(1 ϕ(z k ) 2 ) 2+t+1/q sup g k p,q,φ C, k N

9 Products of Multiplication, Composition Differentiation Operators 1935 g k converges to 0 uniformly on compact subsets of D. The lemma 2.2 implies that It follows that lim T ψ 1,ψ 2,ϕg k H k μ =0. (t +1+1/q)μ(z k ) ψ 2 (z k ) ϕ(z k ) φ( ϕ(z k ) )(1 ϕ(z k ) 2 +1 T ψ1,ψ 2,ϕg k H μ + 0, μ(z k ) ψ 1 (z k ) φ( ϕ(z k ) )(1 ϕ(z k ) 2 as k.fromthis, ϕ(z k ) 1, it follows that μ(z k ) ψ 2 (z k ) lim k φ( ϕ(z k ) )(1 ϕ(z k ) 2 =0, +1 consequently (7) holds. Conversely, assume that T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ is bounded the conditions (6) (7) hold. For any bounded sequence {f k } in H(p, q, φ) with f k 0 uniformly on compact subsets of D. To establish the assertion, it suffices, in view of Lemma 2.2, to show that T ψ1,ψ 2,ϕf k H μ 0, as k. We assume that f k p,q,φ 1. From (6) (7), there exists a δ (0, 1), whenδ< ϕ(z) < 1, wehave (8) φ( ϕ(z) )(1 ϕ(z) 2 + μ(z) ψ 2 (z) φ( ϕ(z) )(1 ϕ(z) 2 <ε. ) 1/q+1 From the proof of Theorem 3.1, we see that sup C. sup μ(z) ψ 2 (z) C. Since f k 0 uniformly on compact subsets of D, Cauchy s estimate gives that f k converges to 0 uniformly on compact subsets of D, there exists a K 0 N such that k>k 0 implies that (9) sup μ(z) ψ 1 (z)f k (ϕ(z)) + sup μ(z) ψ 2 (z)f k(ϕ(z)) <Cε. ϕ(z) δ ϕ(z) δ From (8), (9) Lemma 2.1, we have T ψ1,ψ 2,ϕf k H μ =supμ(z) ψ 1 (z)f k (ϕ(z)) + ψ 2 (z)f k(ϕ(z)) μ(z) ψ 1 (z)f k (ϕ(z)) + sup sup ϕ(z) δ ϕ(z) δ μ(z) ψ 2 (z)f k (ϕ(z)) + sup ( ϕ(z) >δ φ( ϕ(z) )(1 ϕ(z) 2 + μ(z) ψ 2 (z) φ( ϕ(z) )(1 ϕ(z) 2 +1 ) < (C +1)ε,

10 1936 Fang Zhang Yongmin Liu when k>k 0. By using Lemma 3.2, it follows that the operator T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ is compact. Theorem 3.3. Assume that ψ 1,ψ 2 H(D) ϕ S(D). Then T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ,0 is bounded if only if T ψ 1,ψ 2,ϕ : H(p, q, φ) Hμ is bounded, (10) lim z 1 μ(z) ψ 1(z) =0, (11) lim z 1 μ(z) ψ 2(z) =0, Proof. Suppose that T ψ1,ψ 2,ϕ : H(p, q, φ) H μ,0 is bounded. Then it is clear that T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ is bounded. Taking the functions f(z) =1 f(z) =z, respectively, we obtain lim μ(z) ψ 1(z) =0 z 1 Since lim μ(z) ψ 1(z)ϕ(z)+ψ 2 (z) =0. z 1 μ(z) ψ 1 (z)ϕ(z)+ψ 2 (z) μ(z) ψ 2 (z) μ(z) ψ 1 (z)ϕ(z), μ(z) ψ 2 (z) μ(z) ψ 1 (z)ϕ(z) + μ(z) ψ 1 (z)ϕ(z)+ψ 2 (z), we get lim μ(z) ψ 2(z) =0. z 1 Conversely, assume that T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ (10), (11) hold. For each polynomial p(z), weget is bounded the conditions (12) μ(z) (T ψ1,ψ 2,ϕ)p(z) = μ(z) ψ 1 (z)p(ϕ(z)) + ψ 2 (z)p (ϕ(z)). p(ϕ(z)) < sup p (ϕ(z)) <, from (12) it follows that T ψ1,ψ 2,ϕp. From the set of all polynomials is dense in H(p, q, φ), we have that for every Since sup H μ,0 f H(p, q, φ), there is a sequence of polynomials {p k } k N such that f p k p,q,φ 0 as k. Hence T ψ1,ψ 2,ϕf T ψ1,ψ 2,ϕp k H μ T ψ1,ψ 2,ϕ f p k p,q,φ 0

11 Products of Multiplication, Composition Differentiation Operators 1937 as k, by using the boundedness of the operator T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ. Since Hμ,0 is a closed subset of H μ,weobtaint ψ1,ψ 2,ϕ(H(p, q, φ)) Hμ,0.Therefore T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ,0 is bounded. Theorem 3.4. Assume that ψ 1,ψ 2 H(D) ϕ S(D). Then T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ,0 is compact if only if (13) lim z 1 φ( ϕ(z) )(1 ϕ(z) 2 =0, μ(z) ψ 2 (z) (14) lim z 1 φ( ϕ(z) )(1 ϕ(z) 2 =0. +1 Proof. Assume that conditions (13) (14) hold. Then it is clear that (1) (2) hold. Hence T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ is bounded by Theorem 3.1. Since μ(z) T ψ1,ψ 2,ϕf(z) = μ(z) ψ 1 (z)f(ϕ(z)) + ψ 2 (z)f (ϕ(z)) f p,q,φ φ( ϕ(z) )(1 ϕ(z) 2 + μ(z) ψ 2 (z) f p,q,φ φ( ϕ(z) )(1 ϕ(z) 2 +1, Taking the supremum in above inequality over all f H(p, q, φ) such that f p,q,φ 1 letting z 1, yields lim sup z 1 f p,q,φ 1 μ(z) T ψ1,ψ 2,ϕf(z) =0. Hence, by Lemma 2.3 we see that the operator T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ,0 is compact. Now assume that T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ,0 is compact. Then T ψ 1,ψ 2,ϕ : H(p, q, φ) Hμ,0 is bounded, by taking the function f(z) =1, it follows that (16) sup =0. z 1 Since T ψ1,ψ 2,ϕ : H(p, q, φ) Hμ,0 is compact, then T ψ 1,ψ 2,ϕ : H(p, q, φ) Hμ compact, by Theorem 3.2 we have lim ϕ(z) 1 φ( ϕ(z) )(1 ϕ(z) 2 =0. is It follows that for every ε>0, there exists δ (0, 1) such that (17) φ( ϕ(z) )(1 ϕ(z) 2 <ε,

12 1938 Fang Zhang Yongmin Liu when δ< ϕ(z) < 1. Using (16) we see that there exists τ (0, 1) such that (18) <ε inf t [0,δ] φ(t)(1 t2, when τ< z < 1. Therefore, when τ< z < 1 δ< ϕ(z) < 1, by (17) we have (19) φ( ϕ(z) )(1 ϕ(z) 2 <ε, On the other h, when δ < ϕ(z) < 1 ϕ(z) δ, by (18) we obtain (20) φ( ϕ(z) )(1 ϕ(z) 2 inf φ(t)(1 t [0,δ] t2 ) 1/q <ε. From (19) (20), we obtain (13), as desired. Similarly, the result (14) holds. This completes the proof of the theorem. ACKNOWLEDGMENTS The authors are grateful to the anonymous referees for each of the in-depth extensive suggestions. This has led to a significantly improved article. REFERENCES 1. C. Cowen B. MacCluer, Composition Operators on Spaces of Analytic Functions, Stud. Adv. Math., Boca Raton: CRC Press, Ž. Čučkovič Z. Zhao, Weighted composition operators between different weighted Bergman spaces diffferent Hardy spaces, (English summary), Illinois J. Math. (Electronic), 51 (2007), Y. Liu Y. Yu, Weighted differentiation composition operators from mixted-norm to Zygmund spaces, Numer. Funct. Anal. Optim., 31 (2010), H. Hedenmalm, B. Korenblum K. Zhu, Theory of Bergman Spaces, Graduate Text in Mathematics, Springer, New York, R. Hibschweiler N. Portnoy, Composition followed by differentiation between Bergman Hardy spaces, Rocky Mountain J. Math., 35 (2005), Z. Jiang S. Stevič, Compact differences of weighted composition operators from weighted Bergman spaces to weighted-type spaces, Appl. Math. Comput., 217 (2010), S. Li S. Stevič, Weighted composition operators from α-bloch space to H on the polydisk, Numer. Funct. Anal. Optim., 28 (2007),

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14 1940 Fang Zhang Yongmin Liu 27. S. Stevič, A. Sharma A. Bhat, Products of multiplication composition differentiation operators on weighted Bergman space, Appl. Math. Comput., 217 (2011), W. Yang, Products of composition differentiation operators from Q K (p, q) spaces to Bloch-type spaces, Abstr. Appl. Anal., (2009), Art. ID , 2009, 14 pp. 29. W. Yang, Generalized weighted composition operators from the F (p, q, s) space to the Bloch-type space, Appl. Math. Comput., 218 (2012), Y. Yu Y. Liu, The product of differentiation operator multiplication operator from the mixed-norm to Bloch-type space, Acta Math. Sci. Ser. A Chin. Ed., 32 (2012), (in Chinese). 31. K. Zhu, Operator Theory in Function Spaces, New York Basel: Marcel Dekker Inc, X. Zhu, Products of differentiation, composition multiplication from Bergman type spaces to Bers type spaces, Integral Transforms Spec. Funct., 18 (2007), Fang Zhang Department Applied Mathematics Changzhou University Changzhou P. R. China fangzhang188@163.com Yongmin Liu Department of Mathematics Jiangsu Normal University Xuzhou P. R. China