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Abstract and Applied Analysis
Volume 2008 (2008), Article ID 845724, 11 pages
http://dx.doi.org/10.1155/2008/845724
Research Article

Differential Subordinations Associated with Multiplier Transformations

Department of Mathematics, Faculty of Science, University of Oradea, 1 University Street, 410087 Oradea, Romania

Received 26 February 2008; Accepted 15 May 2008

Academic Editor: Ferhan Atici

Copyright © 2008 Adriana Cătaş et al. 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

The authors introduce new classes of analytic functions in the open unit disc which are defined by using multiplier transformations. The properties of these classes will be studied by using techniques involving the Briot-Bouquet differential subordinations. Also an integral transform is established.

1. Introduction and Definitions

Let be the class of analytic functions in the open unit disc𝑈=𝑧|𝑧|<1(1.1)and let [𝑎,𝑛] be the subclass of consisting of functions of the form 𝑓(𝑧)=𝑎+𝑎𝑛𝑧𝑛+𝑎𝑛+1𝑧𝑛+1+. Let 𝒜(𝑝,𝑛) denote the class of functions 𝑓(𝑧) normalized by𝑓(𝑧)=𝑧𝑝+𝑘=𝑝+𝑛𝑎𝑘𝑧𝑘𝑝,𝑛={1,2,3,}(1.2)which are analytic in the open unit disc. In particular, we set𝒜(𝑝,1)=𝒜𝑝,𝒜(1,𝑛)=𝒜(𝑛),𝒜(1,1)=𝒜=𝒜1.(1.3) If a function 𝑓(𝑧) belongs to the class 𝒜(𝑛), it has the form𝑓(𝑧)=𝑧+𝑘=𝑛+1𝑎𝑘𝑧𝑘.𝑛={1,2,3,}(1.4) For two functions 𝑓(𝑧) given by (1.4) and for 𝑔(𝑧) given by𝑔(𝑧)=𝑧+𝑘=𝑛+1𝑏𝑘𝑧𝑘(𝑛),(1.5)the Hadamard product (or convolution) (𝑓𝑔)(𝑧) is defined, as usual, by(𝑓𝑔)(𝑧)=𝑧+𝑘=𝑛+1𝑎𝑘𝑏𝑘𝑧𝑘=(𝑔𝑓)(𝑧).(1.6) If 𝑓 and 𝑔 are analytic in 𝑈, we say that 𝑓 is subordinate to 𝑔, written symbolically as𝑓𝑔or𝑓(𝑧)𝑔(𝑧)(𝑧𝑈)(1.7)if there exists a Schwarz function 𝑤(𝑧) in 𝑈 which is analytic in 𝑈 with 𝑤(0)=0 and |𝑤(𝑧)|<1 such that 𝑓(𝑧)=𝑔(𝑤(𝑧)), 𝑧𝑈.

We consider the following multiplier transformations.

Definition 1.1 (See [1]). Let 𝑓𝒜(𝑝,𝑛). For 𝛿,𝜆, 𝜆0, 𝛿0, 𝑙0, define the multiplier transformations 𝐼𝑝(𝛿,𝜆,𝑙) on 𝒜(𝑝,𝑛) by the following infinite series:𝐼𝑝(𝛿,𝜆,𝑙)𝑓(𝑧)=𝑧𝑝+𝑘=𝑝+𝑛𝑝+𝜆(𝑘𝑝)+𝑙𝑝+𝑙𝛿𝑎𝑘𝑧𝑘.(1.8)

It follows from (1.8) that𝐼𝑝(0,𝜆,𝑙)𝑓(𝑧)=𝑓(𝑧),(𝑝+𝑙)𝐼𝑝𝐼(2,𝜆,𝑙)𝑓(𝑧)=𝑝(1𝜆)+𝑙𝑝𝐼(1,𝜆,𝑙)𝑓(𝑧)+𝜆𝑧𝑝(1,𝜆,𝑙)𝑓(𝑧),𝐼𝑝𝛿1𝐼,𝜆,𝑙𝑝𝛿2,𝜆,𝑙𝑓(𝑧)=𝐼𝑝𝛿2𝐼,𝜆,𝑙𝑝𝛿1.,𝜆,𝑙𝑓(𝑧)(1.9)

Remark 1.2 (See [1]). For 𝑝=1, 𝑙=0, 𝜆0, 𝛿=𝑚, 𝑚0, 0={0}, the operator 𝐷𝑚𝜆=𝐼1(𝑚,𝜆,0) was introduced and studied by Al-Oboudi [2] which is reduced to the Sălăgean differential operator [3] for 𝜆=1. The operator 𝐼𝑚𝑙=𝐼1(𝑚,1,𝑙) was studied recently by Cho and Srivastava [4] and by Cho and Kim [5]. The operator 𝐼𝑚=𝐼1(𝑚,1,1) was studied by Uralegaddi and Somanatha [6], the operator 𝐷𝛿𝜆=𝐼1(𝛿,𝜆,0) was introduced by Acu and Owa [7] and the operator 𝐼𝑝(𝑚,𝑙)=𝐼𝑝(𝑚,1,𝑙) was investigated recently by Sivaprasad Kumar et al. [8].

If 𝑓 is given by (1.2), then we have𝐼𝑝(𝛿,𝜆,𝑙)𝑓(𝑧)=𝑓𝜑𝛿𝑝,𝜆,𝑙(𝑧),(1.10)where𝜑𝛿𝑝,𝜆,𝑙(𝑧)=𝑧𝑝+𝑘=𝑝+𝑛𝑝+𝜆(𝑘𝑝)+𝑙𝑝+𝑙𝛿𝑧𝑘.(1.11) In particular, we set𝐼1(𝛿,𝜆,𝑙)𝑓(𝑧)=𝐼(𝛿,𝜆,𝑙)𝑓(𝑧).(1.12) In order to prove our main results, we will make use of the following lemmas.

Lemma 1.3 (See [9]). For real or complex numbers 𝑎, 𝑏, and 𝑐(𝑐0={0,1,2,}), the following hold: 10𝑡𝑏1(1𝑡)𝑐𝑏1(1𝑡𝑧)𝑎𝑑𝑡=Γ(𝑏)Γ(𝑐𝑏)Γ(𝑐)2𝐹1(𝑎,𝑏;𝑐;𝑧)(Re𝑐>Re𝑏>0),(1.13)2𝐹1(𝑎,𝑏;𝑐;𝑧)=(1𝑧)𝑎2𝐹1𝑧𝑎,𝑐𝑏;𝑐;𝑧1,(1.14)2𝐹1(𝑎,𝑏;𝑐;𝑧)=2𝐹1(𝑏,𝑎;𝑐;𝑧),(1.15)(𝑏+1)2𝐹1(1,𝑏;𝑏+1;𝑧)=(𝑏+1)+𝑏𝑧2𝐹1(1,𝑏+1;𝑏+2;𝑧),(1.16)2𝐹1𝑎,𝑏;𝑎+𝑏+12;12=𝜋Γ(𝑎+𝑏+1)/2ΓΓ.(𝑎+1)/2(𝑏+1)/2(1.17)

Lemma 1.4 (See [10]). Let 𝛽,𝛾, 𝛽0 and let be convex in 𝑈, withRe𝛽(𝑧)+𝛾>0(𝑧𝑈).(1.18) If the function 𝑝[(0),𝑛], then𝑝(𝑧)+𝑧𝑝(𝑧)𝛽𝑝(𝑧)+𝛾(𝑧)𝑝(𝑧)(𝑧).(1.19)

Lemma 1.5 (See [11]). Let 𝜇 be a positive measure on the unit interval 𝐼=[0,1]. Let 𝑔(𝑡,𝑧) be a function analytic in [0,1]×𝑈, for each 𝑡𝐼 and integrable in 𝑡, for each 𝑧𝑈 and for almost all 𝑡𝐼. Suppose also thatRe𝑔(𝑡,𝑧)>0(𝑧𝑈;𝑡𝐼),(1.20)𝑔(𝑡,𝑟) is real for real 𝑟 and1Re1𝑔(𝑡,𝑧).𝑔(𝑡,𝑟)|𝑧|𝑟<1,𝑡𝐼(1.21) If𝑔(𝑧)=𝐼𝑔(𝑡,𝑧)𝑑𝜇(𝑡),(1.22)then1Re1𝑔(𝑧).𝑔(𝑟)|𝑧|𝑟<1(1.23)

Lemma 1.6 (See [12]). Let 𝜓(𝑧) be univalent in the unit disc 𝑈 and let 𝑣 and 𝜑 be analytic in a domain 𝐷𝜓(𝑈) with 𝜑(𝑤)0, when 𝑤𝜓(𝑈). Set𝑄(𝑧)=𝑧𝜓(𝑧)𝜑𝜓(𝑧),(𝑧)=𝑣𝜓(𝑧)+𝑄(𝑧).(1.24) Suppose that
(1)𝑄(𝑧) is starlike in 𝑈 and(2)Re(𝑧(𝑧)/𝑄(𝑧))>0 for 𝑧𝑈. If 𝑞(𝑧) is analytic in 𝑈, with 𝑞(0)=𝜓(0), 𝑞(𝑈)𝐷, and𝑣𝑞(𝑧)+𝑧𝑞(𝑧)𝜑𝑞(𝑧)𝑣𝜓(𝑧)+𝑧𝜓(𝑧)𝜑𝜓(𝑧),(1.25)then 𝑞(𝑧)𝜓(𝑧) and 𝜓(𝑧) is the best dominant.

Lemma 1.7 (See [12, Theorem 3.3d]). Let 𝛽,𝛾,𝐴, with Re[𝛽+𝛾]>0 and let 𝐵[1,0] satisfy eitherRe𝛽(1+𝐴𝐵)+𝛾(1+𝐵2)||𝛽𝐴+𝛽𝐵+𝐵(𝛾+||,𝛾)(1.26)when 𝐵(1,0], or𝛽(1+𝐴)>0,Re𝛽(1𝐴)+2𝛾0,(1.27)when 𝐵=1. If 𝑝[1,𝑛] satisfies𝑝(𝑧)+𝑧𝑝(𝑧)𝛽𝑝(𝑧)+𝛾1+𝐴𝑧,1+𝐵𝑧(1.28)then𝑝(𝑧)𝑞(𝑧)=𝑞𝑛(𝑧)1+𝐴𝑧,1+𝐵𝑧(1.29)where 𝑞𝑛 is the univalent solution of the differential equation𝑞(𝑧)+𝑛𝑧𝑞(𝑧)=𝛽𝑞(𝑧)+𝛾1+𝐴𝑧.1+𝐵𝑧(1.30)In addition, the function 𝑞𝑛, is the best (1,𝑛)-dominant and the function 𝑞𝑛 is given by𝑞(𝑧)=𝛽+𝛾𝛽𝑘(𝑧)𝐾(𝑧)𝛽/𝑛𝛾𝛽=𝑧𝐾(𝑧)=1𝐾(𝑧)𝛾𝛽𝑔(𝑧)𝛽,(1.31)where𝑘(𝑧)=𝑧exp𝑧0((𝑡)1)𝑡1𝑑𝑡,𝐾(𝑧)=𝛽+𝛾𝑛𝑧𝛾/𝑛𝑧0𝑘𝛽/𝑛(𝑡)𝑡(𝛾/𝑛)1𝑑𝑡𝑛/𝛽,(1.32)and the univalent function 𝑔 is given by1𝑔(𝑧)=𝑛10𝑘(𝑡𝑧)𝑘(𝑧)𝛽/𝑛𝑡(𝛾/𝑛)1𝑑𝑡.(1.33)

Now we define new classes of analytic functions by using the multiplier transformations 𝐼(𝑚,𝜆,𝑙) defined by (1.8) as follows.

2. Main Results

Definition 2.1. Let 1𝐵<𝐴1, 𝜆>0, 𝑙0, 𝑚{0}. A function 𝑓𝒜(𝑛) is said to be in the class 𝑆(𝑚,𝜆,𝑙;𝐴,𝐵) if it satisfies the following subordination:𝑙+1𝜆𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧)𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)1𝜆+𝑙𝑙+11+𝐴𝑧1+𝐵𝑧(𝑧𝑈).(2.1)

Remark 2.2. We note that𝑆(0,1,0;12𝛼,1)𝑆(𝛼),𝑆(1,1,0;12𝛼,1)𝐾(𝛼),(2.2)where 𝑆(𝛼) and 𝐾(𝛼)(0𝛼<1) denote the subclasses of functions in 𝒜 which are, respectively, starlike of order 𝛼 and convex of order 𝛼 in 𝑈. Also we have the class𝑆(𝑚,𝜆,0;𝐴,𝐵)𝑆𝑚𝜆(𝐴,𝐵)(2.3)studied by Patel [13].

Let 𝜙(𝑧) be analytic in 𝑈 and 𝜙(0)=1. We introduce the following definition.

Definition 2.3. A function 𝑓𝒜(𝑛) is said to be in the class 𝒜(𝑚,𝜆,𝑙,𝑛;𝜙) if it satisfies the following subordination:𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧)𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)𝜙(𝑧),(𝑧𝑈).(2.4)

Remark 2.4. We note that the classes 𝒜(𝑚,1,𝑙,𝑛;𝜙) were investigated recently by Sivaprasad Kumar et al. [8].

Theorem 2.5. Let 1𝐵<𝐴1, 𝑙0, 𝜆>0, and(1𝐵)(1𝜆+𝑙)+𝜆(1𝐴)>0.(2.5) (i) Then𝑆(𝑚+1,𝜆,𝑙;𝐴,𝐵)𝑆(𝑚,𝜆,𝑙;𝐴,𝐵).(2.6) Further, for 𝑓𝑆(𝑚+1,𝜆,𝑙;𝐴,𝐵), the following hold: 𝑙+1𝜆𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧)𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)1𝜆+𝑙𝑙+1𝑞(𝑧)1+𝐴𝑧,1+𝐵𝑧(2.7)where1𝑞(𝑧)=(1/𝑛)10𝑠(𝑙+1)/𝜆𝑛1(1+𝐵𝑧𝑠)/(1+𝐵𝑧)(1/𝑛)(𝐴/𝐵1)𝑑𝑠1𝜆+𝑙𝜆1if𝐵0,(1/𝑛)10𝑠(𝑙+1)/𝜆𝑛1exp𝐴𝑧(𝑠1)/𝑛𝑑𝑠1𝜆+𝑙𝜆if𝐵=0,(2.8)and 𝑞 is the best dominant of (2.7).
(ii) Furthermore, in addition to (2.5), one consider the inequality𝐵𝐴𝑙+1+𝜆(𝑛1)𝜆,(2.9)where 1𝐵<0, then𝑆(𝑚+1,𝜆,𝑙;𝐴,𝐵)𝑆(𝑚,𝜆,𝑙;12𝜂,1),(2.10)where𝜂=2𝐹111,𝑛𝐴1𝐵,𝑙+1𝐵𝜆𝑛+1,𝐵11(1𝜆+𝑙)/𝜆.(2.11) The result is the best possible.

Proof. Setting𝑧𝜒(𝑧)=𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)(𝑧𝑈),(2.12)we note that 𝜒(𝑧) is analytic in 𝑈 and𝜒(𝑧)=1+𝑎𝑛𝑧𝑛+𝑎𝑛+1𝑧𝑛+1+.(2.13) Using the identity(𝑙+1)𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧)=(1𝜆+𝑙)𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)+𝜆𝑧𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)(2.14)in definition of 𝜒(𝑧) and carrying out logarithmic differentiation in the resulting equation, one obtains𝑧𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧)𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧)=𝜒(𝑧)+𝑧𝜒(𝑧).𝜒(𝑧)+(1𝜆+𝑙)/𝜆(2.15) Since 𝑓𝑆(𝑚+1,𝜆,𝑙;𝐴,𝐵), we get𝜒(𝑧)+𝑧𝜒(𝑧)𝜒(𝑧)+(1𝜆+𝑙)/𝜆1+𝐴𝑧.1+𝐵𝑧(2.16) By applying Lemma 1.4, we obtain that𝜒(𝑧)1+𝐴𝑧.1+𝐵𝑧(2.17) Hence we have shown the inclusion (2.6). Also, making use of Lemma 1.7 with 𝛽=1 and 𝛾=(1𝜆+𝑙)/𝜆, 𝑞 is the best dominant of (2.7) and 𝑞 is defined by (2.8). This proves part (i) of Theorem 2.5.
To establish (2.10), we need to show thatinf|𝑧|<1Re𝑞(𝑧)=𝑞(1).(2.18)The proof of the assertion (2.18) will be deduced on the same lines as in [14] making use of Lemma 1.5. If we set 𝑎=(1/𝑛)(1𝐴/𝐵),𝑏=(𝑙+1)/𝜆𝑛,𝑐=(𝑙+1)/𝜆𝑛+1, 𝐵<0, then by using (1.13), (1.14), and (1.15), we find from (2.8) that1𝑞(𝑧)=(1/𝑛)𝑄(𝑧)1𝜆+𝑙𝜆,(2.19)where𝑄(𝑧)=(1+𝐵𝑧)𝑎10𝑠𝑏1(1+𝐵𝑠𝑧)𝑎=𝑑𝑠Γ(𝑏)Γ(𝑐)2𝐹11,𝑎;𝑐;𝐵𝑧.𝐵𝑧+1(2.20)By using (1.13), the above equality yields𝑄(𝑧)=10𝑔(𝑠,𝑧)𝑑𝜇(𝑠),(2.21)where𝑔(𝑠,𝑧)=1+𝐵𝑧,1+(1𝑠)𝐵𝑧𝑑𝜇(𝑠)=Γ(𝑐)𝑠Γ(𝑎)Γ(𝑐𝑎)𝑎1(1𝑠)𝑐𝑎1𝑑𝑠(2.22)is a positive measure on the closed interval [0,1].
For 1𝐵<1, we note that Re𝑔(𝑠,𝑧)>0, 𝑔(𝑠,𝑟) is real for 0𝑟<1 and 𝑠[0,1] and1Re𝑔(𝑠,𝑧)1(1𝑠)𝐵𝑟=11𝐵𝑟𝑔(𝑠,𝑟),|𝑧|𝑟<1.(2.23)Therefore, by using Lemma 1.5, one obtains1Re1𝑄(𝑧)𝑄(𝑟),|𝑧|𝑟<1(2.24)which, upon letting 𝑟11, yields1Re1𝑄(𝑧).𝑄(1)(2.25)Now, the assertion (2.18) follows by using Lemma 1.5. The result is the best possible and 𝑞 is the best dominant of (2.7). This completes the proof of Theorem 2.5.

Taking 𝑚=0, 𝑛=1, 𝜆=1, 𝑙=0, 𝐴=12𝛼, 𝐵=1 in Theorem 2.5, we get the following result due to MacGregor [15].

Corollary 2.6. For 0𝛼<1, one obtains𝐾(𝛼)𝑆𝜂1,(2.26)where𝜂1=2𝐹111,2(1𝛼),2,21=12𝛼22(1𝛼)122𝛼11,𝛼2,112ln2,𝛼=2.(2.27) The result is the best possible.

Theorem 2.7. Let 𝜓(𝑧) be univalent in 𝑈 with 𝜓(0)=1, Re𝜓(𝑧)>0, and let 𝑧𝜓(𝑧)/𝜓(𝑧) be starlike in 𝑈. Let 𝜙(𝑧) be defined by𝜆𝜙(𝑧)=𝑙+1𝑙+1𝜆𝜓(𝑧)+𝑧𝜓(𝑧).𝜓(𝑧)(2.28)Then𝒜(𝑚+1,𝜆,𝑙,𝑛;𝜙)𝒜(𝑚,𝜆,𝑙,𝑛;𝜓).(2.29)

Proof. Setting𝑞(𝑧)=𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧),𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)(2.30)we note that 𝑞(𝑧) is analytic in 𝑈.
By a simple computation, we observe from (2.30) that𝑧𝑞(𝑧)=𝑧𝑞(𝑧)𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧)𝑧𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧)𝐼(𝑚,𝜆,𝑙)𝑓(𝑧).𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)(2.31) Making use of the identity (2.14), one obtains from (2.31)𝐼(𝑚+2,𝜆,𝑙)𝑓(𝑧)=𝜆𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧)𝑙+1𝑙+1𝜆𝑞(𝑧)+𝑧𝑞(𝑧).𝑞(𝑧)(2.32) By the hypothesis of Theorem 2.7 that 𝑓 belongs to the class 𝒜(𝑚+1,𝜆,𝑙,𝑛;𝜙) and in view of (2.32), we have𝜆𝑙+1𝑙+1𝜆𝑞(𝑧)+𝑧𝑞(𝑧)𝜆𝑞(𝑧)𝑙+1𝑙+1𝜆𝜓(𝑧)+𝑧𝜓(𝑧).𝜓(𝑧)(2.33) If we let𝑄(𝑧)=𝑧𝜓(𝑧)𝜑𝜓(𝑧),(2.34)where𝜑=𝜆𝜓(𝑧)1𝑙+1,𝜓(𝑧)(𝑧)=𝜓(𝑧)+𝑄(𝑧)(2.35)and since 𝑄(𝑧) is starlike, our theorem is an immediate consequence of Lemma 1.6.

Theorem 2.8. Let 𝜓 be univalent in 𝑈, 𝜓(0)=1 and let 𝛾 be a complex number. Suppose that
(1)Re[𝜆(𝛾+1)(𝑙+1)+(𝑙+1)𝜓(𝑧)]>0 and(2)𝑧𝜓(𝑧)/(𝜆(𝛾+1)(𝑙+1)+(𝑙+1)𝜓(𝑧)) is starlike in U.
Let the function 𝐹(𝑧) be defined by𝐹(𝑧)=𝛾+1𝑧𝛾𝑧0𝑡𝛾1𝑓(𝑡)𝑑𝑡(2.36)and the function(𝑧)=𝜓(𝑧)+𝜆𝑧𝜓(𝑧),𝜆(𝛾+1)(𝑙+1)+(𝑙+1)𝜓(𝑧)(2.37)then 𝑓𝒜(𝑚,𝜆,𝑙,𝑛;) implies 𝐹𝒜(𝑚,𝜆,𝑙,𝑛;𝜓).

Proof. From the definition of 𝐹(𝑧) and(𝛾+1)𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)=𝑙+1𝜆𝐼(𝑚+1,𝜆,𝑙)𝐹(𝑧)+𝛾1𝜆+𝑙𝜆𝐼(𝑚,𝜆,𝑙)𝐹(𝑧),(2.38)if we let𝑞(𝑧)=𝐼(𝑚+1,𝜆,𝑙)𝐹(𝑧),𝐼(𝑚,𝜆,𝑙)𝐹(𝑧)(2.39)then we note that 𝑞(𝑧) is analytic in 𝑈. Using (2.38) and (2.39), one obtains(𝛾+1)𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)𝐼(𝑚,𝜆,𝑙)𝐹(𝑧)=𝛾1𝜆+𝑙𝜆+𝑙+1𝜆𝑞(𝑧).(2.40) Differentiating this equality, we obtain𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧)𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)=𝑞(𝑧)+𝜆𝑧𝑞(𝑧).𝜆(𝛾+1)(𝑙+1)+(𝑙+1)𝑞(𝑧)(2.41) For 𝑓𝒜(𝑚,𝜆,𝑙,𝑛;), we have from (2.41)𝑞(𝑧)+𝜆𝑧𝑞(𝑧)𝜆(𝛾+1)(𝑙+1)+(𝑙+1)𝑞(𝑧)𝜓(𝑧)+𝜆𝑧𝜓(𝑧).𝜆(𝛾+1)(𝑙+1)+(𝑙+1)𝜓(𝑧)(2.42) If we let𝑄(𝑧)=𝑧𝜓(𝑧)𝜑𝜓(𝑧),(2.43)where𝜑=𝜆𝜓(𝑧),𝑣𝜆(𝛾+1)(𝑙+1)+(𝑙+1)𝜓(𝑧)(𝑧)=𝑣𝜓(𝑧)+𝑄(𝑧),𝜓(𝑧)=𝜓(𝑧)(2.44)and since 𝑄(𝑧) is starlike in 𝑈, our theorem is an immediate consequence of Lemma 1.6.

Theorem 2.9. Let 𝑓(𝑧)𝒜(𝑛). Then 𝑓 belongs to the class 𝒜(𝑚,𝜆,𝑙,𝑛;𝜒) if and only if 𝐹(𝑧) defined by𝐹(𝑧)=𝑙+1𝑧(1𝜆+𝑙)/𝜆𝑧0𝑡(1𝜆+𝑙)/𝜆1𝑓(𝑡)𝑑𝑡(2.45)belongs to the class 𝒜(𝑚+1,𝜆,𝑙,𝑛;𝜒).

Proof. From the definition of 𝐹(𝑧), we have1𝜆+𝑙𝜆𝐹(𝑧)+𝑧𝐹(𝑧)=(𝑙+1)𝑓(𝑧).(2.46) By convoluting (2.46) with the function𝑢(𝑚,𝜆,𝑙,𝑛;𝑧)=𝑧+𝑘=𝑛+11+𝜆(𝑘1)+𝑙𝑙+1𝑚𝑧𝑘(2.47)and using a convolution property𝑧(𝑓𝑔)(𝑧)=𝑓(𝑧)𝑧𝑔(𝑧),(2.48)one obtains(𝑙+1)𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)=1𝜆+𝑙𝜆𝐼(𝑚,𝜆,𝑙)𝐹(𝑧)+𝑧𝑢(𝑚,𝜆,𝑙,𝑛;𝑧)𝐹(𝑧),(2.49)that is,(𝑙+1)𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)=1𝜆+𝑙𝜆𝐼(𝑚,𝜆,𝑙)𝐹(𝑧)+𝑧𝐼(𝑚,𝜆,𝑙)𝐹(𝑧).(2.50) By using identity (2.14), we get1𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)=𝜆𝐼(𝑚+1,𝜆,𝑙)𝐹(𝑧).(2.51) Also, we obtain(𝑙+1)𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧)=(1𝜆+𝑙)𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)+𝜆𝑧𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)=1𝜆+𝑙𝜆𝐼(𝑚+1,𝜆,𝑙)𝐹(𝑧)+𝑧𝐼(𝑚+1,𝜆,𝑙)𝐹(𝑧)=𝑙+1𝜆𝐼(𝑚+2,𝜆,𝑙)𝐹(𝑧).(2.52) From (2.51) and (2.52), we get𝐼(𝑚+2,𝜆,𝑙)𝐹(𝑧)=𝐼(𝑚+1,𝜆,𝑙)𝐹(𝑧)𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧).𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)(2.53) By the hypothesis of Theorem 2.9 that𝐼(𝑚+1,𝜆,𝑙)𝑓(𝑧)𝐼(𝑚,𝜆,𝑙)𝑓(𝑧)𝜒(𝑧)(2.54)and using (2.53), the desired result follows at once.

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