Liang and Dong Journal of Inequalities and Applications 2014, 2014:502 R E S E A R C H Open Access New characterizations for the products of differentiation and composition operators between Bloch-type spaces Yu-Xia Liang 1 and Xing-Tang Dong 2* * Correspondence: dongxingtang@163com 2 Department of Mathematics, Tianjin University, Tianjin, 300072, PR China Full list of author information is available at the end of the article Abstract We use a brief way to give various equivalent characterizations for the boundedness and the essential norm of the operator C D m acting on Bloch-type spaces At the same time, we use this method to easily get a known characterization for the operator DC on Bloch-type spaces MSC: Primary 47B38; secondary 26A24; 32H02; 47B33 Keywords: essential norm; differentiation; composition operator; Bloch-type space 1 Introduction and preliminaries Recently there has been a considerable interest on various product-type operators see, eg [1 19]), and among them on products of composition and differentiation operators see, eg [1, 2, 4, 5, 7 12, 15 19]) One of the problems of interest is to characterize the boundedness and compactness of the composition operator C acting on Bloch-type spaces in terms of the nth power of the analytic self-mapping of the unit disk D Very recently, the first author and Zhou have given the characterizations for the boundedness and the essential norm of the products of differentiation and composition operator C D m and DC acting on Bloch-type spaces in [9, 10], respectively Inspired by [20], we present here an easier way to research the corresponding problem Moreover, by this brief method, we first give new equivalent characterizations for the boundedness and the essential norm of the operator C D m, and then we obtain the same results for the operator DC as in the paper [10] Let D denote the unit disk in the complex plane C DenoteHD) the space of all holomorphic functions on D and SD) the collection of all holomorphic self-mappings on D The composition operator C is defined by C f = f for f HD)and SD) TheBlochspaceofν-type { B ν = f HD): f Bν = νz) f z) } < is a Banach space endowed with the norm f 0) + f Bν,wheretheweightν : D R + is a continuous, strictly positive and bounded function 2014 Liang and Dong; licensee Springer This is an Open Access article distributed under the terms of the Creative Commons Attribution License http://creativecommonsorg/licenses/by/20), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited
Liangand DongJournal of Inequalities and Applications 2014, 2014:502 Page 2 of 14 For the standard weights ν α z)=1 z 2 ) α for α >0,wedenoteB ν = B α and f α = 1 z 2 ) α f z) Similarly, B α is a Banach space under the norm f B α = f 0) + f α Whenα =1,weget the classical Bloch space B We refer the readers to the book [21] for more information as regards the above spaces The weighted Banach space of analytic functions { Hν = f HD): f ν = νz) } f z) < is a Banach space endowed with the norm ν Theweightν is called radial,ifνz)=ν z ) for all z DForaweightν,theassociated weight νz)isdefinedby νz)= { f z) : f H ν, f ν 1 }) 1, z D For the standard weights ν α z)=1 z 2 ) α 0 < α < ), we have ν α z)=ν α z) We refer the interested readers to [22,p39]Inthiscase,wedenoteH ν = H ν α and f να = 1 z 2 ) α f z) Then H ν α is a Banach space under the norm f να For SD), u HD), the weighted composition operator uc is defined by uc f )=u f ), f HD) As for u 1, the weighted composition operator is the usual composition operator C When is the identity mapping I, the operator uc I is the multiplication operator M u The differentiation operator D is defined by Df = f, f HD) The products of differentiation and composition operators DC and C D m are defined, respectively, as follows: DC f z)=f z) ) z), C D m f = f m), f HD), m N The essential norm of a continuouslinear operator T between two normed linear spaces X and Y is its distance from the compact operators That is, T e,x Y = inf { T K : K is compact }, where denotes the operator norm Notice that T e,x Y =0ifandonlyifT is compact, so the estimate on T e,x Y will lead to the condition for the operator T to be compact
Liangand DongJournal of Inequalities and Applications 2014, 2014:502 Page 3 of 14 Throughout this paper, C will denote a positive constant, the exact value of which will vary from one appearance to the next The notations A B, A B, A B mean that there maybe different positive constants C such that B/C A CB, A CB, A CB For convenience of the reader we list the results related with our conclusions in this paper Theorem A [9,Theorem1] Let 0<α, <, m be a nonnegative integer and be a holomorphic self-map of the unit disk D Then C D m : B α B is bounded if and only if n α 1 C D m I n z) n N Theorem B [9,Theorem2] Let 0<α, <, m be a nonnegative integer and be a holomorphic self-map of the unit disk D Suppose that C D m : B α B is bounded Then the estimate for the essential norm of C D m : B α B is C D m e lim n α 1 C D m I n z), where I n z)=z n, z D, n N Theorem C [10, Theorem23] Let 0<α, <, and SD) Then DC : B α B is bounded if and only if n α I < and n α J n 1 ) Theorem D [10, Theorem35] Let 0<α, < and SD) Suppose that DC : B α B is bounded Then the estimate for the essential norm of DC : B α B is { } A max 3 2, B DC α+1 2 α+1 e A + B), 3α +4) e where A := 2α+1) )α+1 lim n α I n ) and B := e 2α )α lim n α J n 1 ) The definitions of I n ) and J n 1 ) can be found in Section 4 We would like to point out that the first author and Zhou got the above four theorems by using complex calculations and intricate discussions In this paper, we will use a brief way to give other equivalent characterizations for the boundedness and the essential norm of C D m : B α B on the unit disk in Section 3 In addition, using this method we will show new proofs of Theorem C and Theorem D in Section 4 2 Lemmas In this section we quote some lemmas for our further application The first lemma is a well-knowncharacterizationforb α 0 < α < ) Lemma 21 For f HD), m N and α >0Then f B α m 1 f α f j) 0) + 1 z 2 ) α+m 1 f m) z) j=0
Liangand DongJournal of Inequalities and Applications 2014, 2014:502 Page 4 of 14 So for f B α, the above lemma implies that f Hν α and more general f m+1) Hν α+m Therefore, theories of the weighted composition operator uc : Hν Hw play a key role in the proof of our main results Here we list some lemmas which will be used later Lemma 22 [23, Proposition31] Let ν and w be weights Then the weighted composition operator uc : Hν Hw is bounded if and only if wz) uz) νz)) Moreover, the following holds: uc H ν H w = wz) uz) νz)) Lemma 23 [23, Theorem 44] Let ν and w be radial, non-increasing weights tending to zero at the boundary of D Suppose uc : Hν Hw is bounded Then uc e,h ν H w lim r 1 z) >r wz) uz) νz)) Lemma 24 [24, Theorem 24] Let ν and w be radial, non-increasing weights tending to zero at the boundary of D Then a) uc : Hν Hw is bounded if and only if u n w < n 0 z n ν with the norm comparable to the above remum b) uc e,h ν Hw = lim u n w z n ν Lemma 25 [22, Lemma 21] For α >0,we have lim n +1) α z n να = 2α e )α The following criterion for compactness follows from an easy modification of [25, Proposition 311] Hence we omit the details Lemma 26 Let 0<α, < and T be a linear operator from B α to B Then T : B α B is compact if and only if T : B α B is bounded and for any bounded sequence {f k } k N in B α which converges to zero uniformly on compact subsets of D, Tf k B 0 as k 3 Boundedness and essential norm of C D m In this section, we give other equivalent characterizations for the boundedness and the essential norm of the operator C D m : B α B with 0 < α, Theorem 31 Let 0<α, <, m N, and SD) Then the following statements are equivalent: a) C D m : B α B is bounded
Liangand DongJournal of Inequalities and Applications 2014, 2014:502 Page 5 of 14 b) 1 z 2 ) z) 31) 1 z) 2 ) α+m c) n α+m n 1 ν 32) Proof a) b) Suppose that C D m : B α B is bounded Choose f 1 z)=z m+1 and f w z)= 1 w) 2 1 w)z) α, w D It is easy to verify that f 1 B α and f w B α for w DBy C D m f f α for f B α,we obtain and 1 z 2 ) z) <, 1 z 2 ) z) z) m+1 1 z) 2 ) α+m Then it follows that and z) 1 2 z) > 1 2 1 z 2 ) z) 1 z) 2 ) α+m 1 z 2 ) z) <, 1 z 2 ) z) 1 z 2 ) z) z) m+1 1 z) 2 ) α+m 1 z) 2 ) α+m That is, b) holds b) c) From Lemma 22, the condition b) is a necessary and sufficient condition for the boundedness of weighted composition operator C : H ν α+m H ν Furtherby Lemma 24a) and Lemma 25, the boundedness of the weighted composition operator C : H ν α+m H ν is equivalent to the following: n 1 ν z n 1 να+m n α+m n 1 ν = n α+m z n 1 να+m n α+m n 1 ν b) a) Suppose b) holds For every f B α, then it follows from Lemma 21 that C D m f = 1 z 2 ) f m+1) z) ) z) 1 z 2 ) z) 1 z) 2 ) α+m f B α
Liangand DongJournal of Inequalities and Applications 2014, 2014:502 Page 6 of 14 f B α Moreover, C D m f 0) = f m) 0)) Thus C 1 0) 2 ) α+m 1 D m f B <, and hence a) holds Remark 32 1) The relation a) b) was essentially proved in a very general result in [18] For convenience of the reader, we sketch the proof in [18] 2) One can easily see that n α 1 C D m I n z) = n α 1 nn 1) n m) n m 1 ν n N n m+1 = k 1k + m) α k + m 1) k k 1 ν k 1 k α+m k 1 ν = n α+m n 1 ν Therefore, the characterizations for the boundedness of the operator C D m in Theorem 31 are equivalentto thatin TheoremA As an application of Theorem 31, we present an example of the bounded operator C D m, according to either 31)or32) Example 33 Let z)=z 2 for z D and = α + m Thenwestudytheboundednessof C D m : B α B α+m Firstly,by31), it is clear that 1 z 2 ) α+m z) = 1 z) 2 ) α+m Secondly, by 32)weobtain 1 z 2 ) α+m 2z 1 z 4 ) α+m n α+m n 1 ν = n α+m 2zz 2n 1) ν = n α+m n α+m 1 z 2 ) α+m 2zz 2n 1) 1 x) α+m x n 2 1 x [0,1[ = n α+m 1 n 1 ) α+m 2 n 1 ) n 1 2 2 + n 1 + n 1 2 2 = n + n 1 2 ) α+m n 1 ) n 1 2 2 + n 1 2 From each of these conditions, one sees that C D m : B α B α+m is bounded Next we estimate the essential norm of the operator C D m : B α B for all 0 < α, < Denote B α = {f B α : f 0) = 0}LetD m+1 : B α Hν α+m be defined by D m+1 f = f m+1) z) Then we have D m+1 f νm+α f B α for f B α Sincef m+1) Hν α+m when f B α,and further by the equality C D m f ) = f m+1) ) for all f B α, it follows that C D m e, B α B C e,h να+m H ν 33)
Liangand DongJournal of Inequalities and Applications 2014, 2014:502 Page 7 of 14 Thus we only need to estimate C e,h να+m Hν for the upper bound of the essential norm of C D m It is obvious that every compact operator T K B α, B )canbeextended to a compact operator K KB α, B ) In fact, for every f B α, f f 0) B α,andwecan define Kf ):=Tf f 0)) + f 0), which is a compact operator from B α to B,duetoKf k ) has convergent subsequence when {f k } is a bounded sequence In the following lemma we will use the compact operator K r defined on the space B α by K r f z)=f rz) Lemma 34 If 0<α, < and C D m is a bounded operator from B α to B, then C D m e, B α B = C D m e,b α B Proof Although the proof is similar to [20, Lemma31],wewillgiveallthedetailsfor convenience of the reader It is obvious that C D m e, B α B C D m e,b α B Conversely, let T KB α, B ) be given Choose an increasing sequence r n ) n in 0, 1) converging to 1 We denote by A theclosedsubspaceofb α consisting of all constant functions Then we have Hence C D m T B α B = inf T KB α,b ) C D m f ) Tf ) B f B α 1 f B α 1 + f B α 1 C D m f f 0) ) T B α f f 0) ) B C D m f 0) ) T f 0) ) B C D m g) T B α g) B + C D m h) T A h) B g B α h A C D m T B α B inf T KB α,b ) + inf T KB α,b ) Since C D m : B α B is bounded, it follows that C D m T B α B α B C D m T A B A C D m e, B α B + lim C D m I K rn ) A B lim C D m I K rn ) A B C lim I K r n A B =0 Thus we obtain C D m e, B α B C D m e,b α B Theproofisfinished Thus by Lemma 34 and 33) it follows that C D m e,b α B C e,h να+m H ν 34)
Liangand DongJournal of Inequalities and Applications 2014, 2014:502 Page 8 of 14 Theorem 35 Let 0<α, <, m N, and SD) Suppose that C D m : B α B is bounded Then C D m e,b α B lim n α+m n 1 ν lim z) 1 1 z 2 ) z) 35) 1 z) 2 ) α+m Proof If <1,thenby[26, Lemma 31], the operator uc : B α Hμ is compact The boundedness compactness) of C D m : B α B is equivalent to the boundedness compactness) of C : B α+m H Inthiscase,allitemsin35) are zero If =1,sinceC D m : B α B is bounded, then the boundedness of C : Hν α+m Hν follows from the proof in Theorem 31Thusby34), Lemma 24b), and Lemma 25, C D m e,b α B n 1 ν C e,h να+m Hν = lim z n 1 να+m = lim n α+m n 1 ν lim n α+m z n 1 να+m nα+m n 1 ν Since =1,wemaychooseasequence{z k } k N D such that z k ) 1ask Define f k z)= 1 z k) 2 1 z k )z) α, k N It is easy to show that f k B α and converges to zero uniformly on the compact subsets of D as k Moreover, f m+1) k zk ) ) = αα +1) α + m)z k)) m+1 1 z k ) 2 ) α+m Then for every compact operator T : B α lim Tf k =0Thus B, by Lemma 26, it follows that C D m T B α B lim C D m f k ) lim Tf k Consequently, = lim lim C D m f k ) 1 zk 2) f m+1) zk ) ) z k ) lim 1 zk 2) z k ) z k ) m+1 1 z k ) 2 ) α+m = lim 1 z k 2 ) z k ) 1 z k ) 2 ) α+m C D m 1 z 2 ) z) e,b α B lim z) 1 1 z) 2 ) α+m k
Liangand DongJournal of Inequalities and Applications 2014, 2014:502 Page 9 of 14 is bounded, then applying Lemma 23, Lem- Since the operator C : Hν α+m Hν ma 24b), and Lemma 25,weget 1 z 2 ) z) lim z) 1 1 z) 2 ) C e,h α+m να+m Hν = lim n 1 ν z n 1 να+m lim n α+m n 1 ν Thus Hence lim n α+m n 1 ν C D m e,b α B 1 z 2 ) z) lim z) 1 1 z) 2 ) α+m lim n α+m n 1 ν C D m e,b α B lim n α+m n 1 ν 1 z 2 ) z) lim z) 1 1 z) 2 ) α+m This completes the proof Remark 36 1) The relation C D m e,b α B lim 1 z 2 ) z) z) 1 canbeprovedsimilarlyto 1 z) 2 ) α+m [26, Theorem 32] Here we give a complete proof for the reader s convenience 2) Similar to Remark 32,onecanget lim n α+m n 1 ν lim n α 1 C D m I n z) Therefore, the characterizations for the essential norms of the operator C D m in Theorem 35 areequivalentto thatin TheoremB The following corollary is an immediate consequence of Theorem 35 Corollary 37 Let 0<α, <, m N, and SD) Then the following statements are equivalent: a) C D m : B α B is compact b) C D m : B α B is bounded and lim z) 1 1 z 2 ) z) =0 1 z) 2 ) α+m c) C D m : B α B is bounded and lim n α+m n 1 ν =0
Liang and Dong Journal of Inequalities and Applications 2014, 2014:502 Page 10 of 14 4 Boundedness and essential norm of DC In this section, the corresponding problems for the operator DC : B α B are considered Let u HD), then for every f HD), define I u f z)= z 0 f ζ )uζ ) dζ, J u f z)= z 0 f ζ )u ζ ) dζ Then it follows that I n ) z)= z By an easy calculation, one can get n ) ζ ) ζ ) dζ, J n 1 ) z z)= n 1 ζ ) ζ ) dζ 0 0 I n ) z) ) = nz) n 1 z) ) 2 41) and J n 1 ) z) ) = z) n 1 z) 42) In 2007, S Li and S Stević gave the following characterizations for the boundedness and compactness of the operator DC : B α B Lemma 41 Let α, >0and SD) Then the following statements hold: a) [4, Theorem 1] DC : B α B is bounded if and only if z) 2 1 z 2 ) z) 1 z 2 ) < and 43) 1 z) 2 ) α+1 1 z) 2 ) α b) [4,Theorem2]DC : B α B is compact if and only if DC : B α B is bounded, z) 2 1 z 2 ) z) 1 z 2 ) lim =0 and lim =0 z) 1 1 z) 2 ) α+1 z) 1 1 z) 2 ) α First,wewillgiveabriefproofofTheoremC as regards the bounded operator DC : B α B for all 0 < α, Theorem 42 Let 0<α, < and SD) Then DC : B α B is bounded if and only if n α I < and n α J n 1 ) Proof Lemma 41 shows that DC maps B α boundedly into B if and only if 43)holdsOn the other hand, Lemma 22 shows that 43) holds if and only if the weighted composition operators ) 2 C maps H ν α+1 boundedly into H ν and C maps H ν α boundedly into H ν, and hence it follows from Lemma 24a) that 43)isequivalentto ) 2 n 1 ν n 1 ν < and z n 1 να+1 z n 1 να
Liang and Dong Journal of Inequalities and Applications 2014, 2014:502 Page 11 of 14 Using Lemma 25, 41) and42), then the boundedness of DC : B α B is equivalent to and ) 2 n 1 ν n α+1 n α+1 z n 1 να+1 ) 2 n 1 ν n α+1 = n α I < n α n 1 ν n α z n 1 να n α n 1 ν = n α J n 1 ) This completes the proof Now,wegiveanewproofofTheoremD about the essential norm of DC : B α B for 0<α, Wedenote B α = {f B α : f 0) = 0}LetD α : B α Hν α and S α : B α Hν α+1 be the first-order derivative operator and the second-order derivative operator, respectively That is, D α f )=f, S α f )=f ByLemma21we have D α f να = f B α and S α f να+1 f B α for f B α For f B α, by Lemma 21, f Hν α+1,andf Hν α Then by the equation DC f ) = f ) ) 2 + f ), it follows that DC e, B α B ) 2 C e,h να+1 H ν + C e,h να H ν 44) Moreover, every compact operator T K B α, B ) can be extended to a compact operator K KB α, B ) Then similar to Lemma 34,onecaneasilyget DC e, B α B = DC e,b α B Thus combining the above equation with 44), we obtain DC e,b α B ) 2 C e,h να+1 H ν + C e,h να H ν 45) According to 45), we only need to estimate the right two essential norms for the upper bound of the essential norm of DC : B α B Theorem 43 Let 0<α, < and SD) Suppose that DC : B α B is bounded Then DC e,b α B max {lim n α I, lim n α J n 1 ) }
Liang and Dong Journal of Inequalities and Applications 2014, 2014:502 Page 12 of 14 Proof By Lemma 41a) and Lemma 22, the boundedness of DC : B α B is equivalent to ) 2 C : Hν α+1 Hν and C : Hν α Hν are bounded weighted composition operators The upper estimate From Lemma 24b) and Lemma 25,weobtain ) 2 ) 2 n 1 ν Ce,H να+1 Hν = lim z n 1 να+1 = lim n α+1 ) 2 n 1 ν n α+1 z n 1 να+1 C e,h να H ν lim = lim n α+1 ) 2 n 1 ν = lim n α I, n 1 ν z n 1 να = lim n 1 ν n α z n 1 να n α lim n α n 1 ν = lim n α J n 1 ) Then it follows from 45)that DC e,b α B max {lim n α I, lim n α J n 1 ) } The lower estimate Let {z k } k N be a sequence in D such that z k ) 1ask Define f k z)= 1 z k) 2 1 z k )z) α α α +1 g k z)= 1 z k) 2 1 z k )z) α α α +2 1 z k ) 2 ) 2 1 z k )z), α+1 1 z k ) 2 ) 2 1 z k )z) α+1 We can easily show both f k and g k belong to B α and converge to zero uniformly on the compact subsets of D as k Moreover, f k zk ) ) =0, f zk ) ) = g k zk ) ) = k αz k )) 2 1 z k ) 2 ) α+1 ; αz k ) α + 2)1 z k ) 2 ) α, g k zk ) ) =0 Then for every compact operator T : B α B, by Lemma 26 we obtain DC T B α B lim DC f k ) lim 1 zk 2) α z k )) 2 z k )) 2 1 z k ) 2 ) α+1, DC T B α B lim DC g k ) lim 1 zk 2) α z k )z k ) α + 2)1 z k ) 2 ) α
Liang and Dong Journal of Inequalities and Applications 2014, 2014:502 Page 13 of 14 Since the weighted composition operators ) 2 C : Hν α+1 Hν and C : Hν α Hν are bounded Then applying Lemma 23, Lemma 24b), and Lemma 25, it follows that DC e,b α B lim 1 z 2 ) z)) 2 1 z) 2 ) α+1 z) 1 ) 2 C e,hν α+1 H ν = lim ) 2 n 1 ν z n 1 να+1 Hence lim n α+1 ) 2 n 1 ν = lim n α I, DC e,b α B lim 1 z 2 ) z) 1 z) 2 ) α z) 1 C e,h να H ν = lim lim DC e,b α B max {lim n 1 ν z n 1 να n α n 1 ν = lim n α J n 1 ) n α I, lim n α J n 1 ) } This completes the proof The following result is an immediate consequence of Theorem 43 and Lemma 41b) Corollary 44 Let α, >0and SD) Then the following statements are equivalent: a) DC : B α B is compact b) DC : B α B is bounded, lim n α I =0 and lim n α J n 1 ) =0 Competing interests The authors declare that they have no competing interests Authors contributions All authors conceived and drafted the manuscript, and read and approved the final manuscript Author details 1 School of Mathematical Sciences, Tianjin Normal University, Tianjin, 300387, PR China 2 Department of Mathematics, Tianjin University, Tianjin, 300072, PR China Acknowledgements The authors would like to thank the Journal Editorial Office and the referees for the useful comments and suggestions, which improved the presentation of this article This work was ported in part by the National Natural Science Foundation of China Grant Nos 11201331; 11301373; 11401431) Received: 29 September 2014 Accepted: 28 November 2014 Published: 12 Dec 2014 References 1 Hibschweiler, RA, Portnoy, N: Composition followed by differentiation between Bergman and Hardy spaces Rocky Mt J Math 353), 843-855 2005) 2 Hyvärinen, O, Nieminen, I: Weighted composition followed by differentiation between Bloch-type spaces Rev Mat Complut 2014) doi:101007/s13163-013-0138-y
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