EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 13, No. 4, 2020, 861-872 ISSN 1307-5543 – www.ejpam.com Published by New York Business Global Univalence of New General Integral Operator Defined by the Ruscheweyh Type q-Difference Operator Suhila Elhaddad1, Huda Aldweby2, Maslina Darus1,∗ 1 Department of Mathematical Sciences, Faculty of Science and Technology, Universiti Kebangsaan Malaysia, 43600, Bangi, Selangor, Malaysia 2 Department of Mathematics, Faculty of Science, Al-Asmarya Islamic University, Libya Abstract. In this study, by employing the Ruscheweyh type q-analogue operator we consider a new family of integral operators on the space of analytic functions. For this family, we demonstrate some sufficient conditions of univalence criteria on the class of analytical functions. 2020 Mathematics Subject Classifications: 30C45, 30C50 Key Words and Phrases: q- analogue of Ruscheweyh operator, integral operators, univalence criteria. 1. Introduction Univalence criteria for certain class of analytic functions has attracted many and some of their work can be seen widely in the literature. For example, Pascu [21], [22] studied on the univalence criterion for certain class of functions and improvement of Becker’s univalence criteria in 1985 and 1987 respectively. Then, Pescar [23] led on the generalised univalence criteria of Ahlfor’s and Becker’s. Later, Faisal and Darus [13–15] and Al-Refai and Darus [1] continued to study the same for different operators and classes. Here we are studying similar criteria for a class generated by a q-analogue of Ruscheweyh. Let A denote the class of functions of the form: f(z) = z + ∞∑ n=2 anz n, (1) which are analytic in the open unit disk U = {z ∈ C : |z|<1} and satisfy the following normalized condition: f(0) = f ′ (0)− 1 = 0. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v13i4.3817 Email addresses: P89257@siswa.ukm.edu.my (S. Elhaddad), hu.aldweby@gmail.com (H. Aldweby), maslina@ukm.edu.my (M. Darus) https://www.ejpam.com 861 c© 2020 EJPAM All rights reserved. S. Elhaddad, H. Aldweby, M. Darus / Eur. J. Pure Appl. Math, 13 (4) (2020), 861-872 862 Additionally, let S ⊂ A be the family of univalent functions in U . The Hadamard product for two analytic functions f ∈ A defined in (1) and g(z) = z + ∞∑ n=2 bnz n, is given by f(z) ∗ g(z) = z + ∞∑ n=2 anbnz n. Firstly, we will present the concepts and definitions for q-calculus which will later be applied (see [5] and [12]). Let n ∈ N, 0 −1 and [n]q! defined by (2). From the Definition 1, we note that, if q → 1 ,we have lim q−→1 Rυqf(z) = z + lim q−→1 [ ∞∑ n=2 [n+ υ − 1]q! [υ]q![n− 1]q! anz n ] = z + ∞∑ n=2 (n+ υ − 1)! (υ)!(n− 1)! anz n = Rυf(z), where Rυf(z) is Ruscheweyh operator that was presented in [24] and has been examined by many authors, for instance [19] and [26]. In fact, the q-derivative type of Ruscheweyh operator has been studied recently by Hussain et.al [17], Aldweby and Darus [3] for dif- ferent properties. Other type of q-derivative can be seen in [16]. S. Elhaddad, H. Aldweby, M. Darus / Eur. J. Pure Appl. Math, 13 (4) (2020), 861-872 863 Definition 2. A function f ∈ A is said to be in the class Bυ(q, ϑ) if it is satisfying the condition ∣∣∣∣∣z2 ( Rυqf(z) )′[ Rυqf(z) ]2 − 1 ∣∣∣∣∣<ϑ, (z ∈ U, 0 < ϑ ≤ 1), (4) where Rυqf(z) is the operator defined by (3). Note that, B0(q → 1, ϑ) = B(ϑ), where the analytic and univalent functions class B(ϑ) was presented and studied in [11]. Using the operatorRυqf(z), we now introduce the general integral operator as following: Definition 3. Let m ∈ N ∪ {0}, let γ1, γ2, ..., γn, |q|<1 and % ∈ C \ {0,−1, ...}, then the integral operator Iγn,%(υ, q, z) : A → A is defined by Iγn,%(υ, q, z) = ( % ∫ z 0 t%−1 m∏ n=1 (Rυqfn(t) t ) 1 γn dt ) 1 % , (5) where fn ∈ A. Remark 1. Interestingly, the integral operator Iγn,%(υ, q, z) generalizes a number of op- erators that have been implemented and studied by several authors, for instance • For υ = 0 and γ1, ..., γm = σ , we get the following operator Iσ,%(z) = ( % ∫ z 0 t%−1 m∏ n=1 ( fn(t) t ) 1 σ dt ) 1 % , (6) that considered by Breaz and Breaz [7]. • For υ = 0,m = 1, γn = 1 σn , % = 1, σ1 = 1, σ2 = ... = σm = 0 and f1 = f2 = ... = fm = f ∈ S, we have the following integral operator developed and studied by Alexander [4], I(z) = ∫ z 0 f(t) t dt. (7) • For υ = 0, % = 1 and γn = 1 σn , we obtain the following integral operator introduced by Breaz and Breaz [6], f(z) = ∫ z 0 [ f1(t) t ]σ1 ... [ fm(t) t ]σm dt. (8) • For υ = 0, γn = 1 σ − 1 and % = m(σ − 1) + 1, we have the integral operator: Gm,σ(z) = ( [m(σ − 1) + 1] ∫ z o (f1(t)) σ−1...(fm(t))σ−1dt ) 1 m(σ−1)+1 , (9) studied by Breaz et al. [9]. S. Elhaddad, H. Aldweby, M. Darus / Eur. J. Pure Appl. Math, 13 (4) (2020), 861-872 864 • For υ = 0,m = 1, γn = 1 an , % = 1, σ1 = σ, σ2 = ... = σm = 0 and f1 = f2 = ... = fm = f ∈ S, we obtain the integral operator: Iσ(z) = ∫ z 0 [ f(t) t ]σ dt, (10) introduced by Miller and Mocanu [18]. • For υ = 0, γn = 1 σ − 1 , % = σ and f1 = f2 = ... = fm = f ∈ A where σ ∈ C and <(σ)>0,we obtain the following operator: Gσ(z) = ( σ ∫ z 0 (f(t))σ−1dt ) 1 σ , (11) studied and introduced by Pescar [23]. • For υ = 1, q → 1, γn = 1 σ − 1 and % = 1 +m(σ− 1),we get the integral operator that Selvaraj and Karthikeyan [25] introduced Gσ(z) = ( [m(σ − 1) + 1] ∫ z o tm(σ−1) ( f ′ 1(t) )σ−1 ... ( f ′ m(t) )σ−1 dt ) 1 1+m(σ−1) . (12) • For υ = 1, q → 1, γn = 1 σ and % = 1, we obtain the following integral operator: Gσ(z) = ∫ z o ( f ′ 1(t) )σ ... ( f ′ m(t) )σ dt, (13) studied and introduced by Breaz and Güney [10]. 2. Preliminaries In order to prove our main results, we need to recall the following. Lemma 1. (see [21] and [22]) Let % ∈ C with <(%)>0. If f ∈ A satisfies 1− |z|2<(%) <(%) ∣∣∣∣∣zf ′′ (z) f ′(z) ∣∣∣∣∣ ≤ 1, z ∈ U, then the operator f%(z) = { % ∫ z 0 t%−1f ′ (t)dt } 1 % , is belonging to S. S. Elhaddad, H. Aldweby, M. Darus / Eur. J. Pure Appl. Math, 13 (4) (2020), 861-872 865 Lemma 2. (see [23]) Let c ∈ C with |c| ≤ 1, c 6= −1, % ∈ C with <(%)>0. If f ∈ A satisfies ∣∣∣∣∣c|z|2% + (1− |z|2%)zf ′′ (z) %f ′(z) ∣∣∣∣∣ ≤ 1, z ∈ U, then the operator f%(z) = { % ∫ z 0 t%−1f ′ (t)dt } 1 % , is belonging to S. Lemma 3. (see [20]) (Generalized Schwarz Lemma) Let f ∈ A within UR = {z : |z| m for z = 0, thus |f(z)| ≤ N Rm |z|m, (z ∈ UR). Equality can only be achieved if f(z) = eiθ ( N Rm ) zm, where θ is constant. 3. Main Results In this part, by utilizing the above lemmas, we find the univalence of this integral operator defined by Ruscheweyh type q-analogue. Theorem 1. Let f1, ..., fm ∈ A and %, γ1, ..., γm ∈ C. Let N ≥ 1 with 1 <(%) m∑ n=1 [(1 + ϑn)N + 1] |γn| ≤ 1. (14) If f1, ..., fm ∈ Bυ(q, ϑn), 0 < ϑn ≤ 1, n = 1, ...,m and |Rυqfn(z)| ≤ N, (z ∈ U), then the function Iγn,%(υ, q, z) given by (5) is univalent. Proof. From the definition of the operator Rυqf(z) we have Rυqf(z) z = z + ∑∞ n=2 [n+υ−1]q ! [υ]q ![n−1]q !anz n z = 1 + ∞∑ n=2 [n+ υ − 1]q! [υ]q![n− 1]q! anz n−1, S. Elhaddad, H. Aldweby, M. Darus / Eur. J. Pure Appl. Math, 13 (4) (2020), 861-872 866 then Rυqf(z) z 6= 0, (z ∈ U), and for z = 0 and n = 1, ...,m, we have(Rυqf1(z) z ) 1 γ1 ... (Rυqfm(z) z ) 1 γm = 1. Define the function f(z) = ∫ z 0 m∏ n=1 (Rυqfn(t) t ) 1 γn dt, (15) then we have f(0) = 0 and f ′(0) = 1. Therefore f ′ (z) = m∏ n=1 (Rυqfn(z) z ) 1 γn . (16) The equality (16) implies ln f ′ (z) = m∑ n=1 1 γn ( ln Rυqfn(z) z ) . Or equivalently ln f ′ (z) = m∑ n=1 1 γn ( lnRυqfn(z)− lnz ) . By differentiating the above equality, we have zf ′′ (z) f ′(z) = m∑ n=1 1 γn ( z ( Rυqfn(z) )′ Rυqfn(z) − 1 ) . (17) From (17), we have∣∣∣∣∣zf ′′ (z) f ′(z) ∣∣∣∣∣ ≤ m∑ n=1 1 |γn| (∣∣∣∣∣z ( Rυqfn(z) )′ Rυqfn(z) ∣∣∣∣∣+ 1 ) = m∑ n=1 1 |γn| (∣∣∣∣∣z2 ( Rυqfn(z) )′ [Rυqfn(z)]2 ∣∣∣∣∣ ∣∣∣∣Rυqfn(z) z ∣∣∣∣+ 1 ) . (18) From the hypothesis, we have |Rυqfn(z)| ≤ N , fn ∈ Bυ(q, ϑn), (n = 1, ...,m, z ∈ U), then by using lemma 3, we get that |Rυqfn(z)| ≤ N |z| , (n = 1, ...,m, z ∈ U). From (18), we get∣∣∣∣∣zf ′′ (z) f ′(z) ∣∣∣∣∣ ≤ m∑ n=1 1 |γn| (∣∣∣∣∣z2 ( Rυqfn(z) )′ [Rυqfn(z)]2 ∣∣∣∣∣N + 1 ) S. Elhaddad, H. Aldweby, M. Darus / Eur. J. Pure Appl. Math, 13 (4) (2020), 861-872 867 ≤ m∑ n=1 1 |γn| (∣∣∣∣∣z2 ( Rυqfn(z) )′ [Rυqfn(z)]2 − 1 ∣∣∣∣∣N +N + 1 ) ≤ m∑ n=1 1 |γn| (ϑnN +N + 1) = m∑ n=1 (1 + ϑn)N + 1 |γn| , which easily shows that 1− |z|2<(%) <(%) ∣∣∣∣∣zf ′′ (z) f ′(z) ∣∣∣∣∣ = 1− |z|2<(%) <(%) ∣∣∣∣∣ m∑ n=1 1 γn ( z ( Rυqfn(z) )′ Rυqfn(z) − 1 )∣∣∣∣∣ ≤ 1 <(%) m∑ n=1 (1 + ϑn)N + 1 |γn| , since 1 <(%) ∑m n=1 [(1 + ϑn)N + 1] |γn| ≤ 1. Using Lemma 1 , we obtain that the integral Iγn,%(υ, q, z) given by (5) is univalent. Setting N = 1, υ = 0, γn = 1 σ − 1 , and % = m(σ − 1) + 1 in Theorem 1, we get Corollary 1. [8] Let f1, ..., fm ∈ A and σ ∈ C with |σ − 1| ≤ <(σ) 3m , if ∣∣∣∣∣ z2f ′ k(z) (fn(z))2 − 1 ∣∣∣∣∣<1, (z ∈ U), then the function Gm,σ(z) defined by (9) is univalent. Setting N = 1, υ = 0, γn = 1 σ − 1 , f1 = ... = fm = f ∈ A and % = σ where σ ∈ C in Theorem 1, we get Corollary 2. Let f ∈ A and σ ∈ C with |σ − 1| ≤ <(σ) 3 , if ∣∣∣∣∣z2f ′ (z) (f(z))2 − 1 ∣∣∣∣∣<1, (z ∈ U), then the function Gσ(z) defined by (11) is univalent. S. Elhaddad, H. Aldweby, M. Darus / Eur. J. Pure Appl. Math, 13 (4) (2020), 861-872 868 Next, we prove Theorem 2. Let f1, ..., fm ∈ A, γ1, ..., γm ∈ C and % ∈ C with <(%) > ∑m n=1 [(1+ϑn)N+1] |γn| . Let c ∈ C and N ≥ 1 with |c| ≤ 1− 1 <(%) m∑ n=1 [(1 + ϑn)N + 1] |γn| . If f1, ..., fm ∈ Bυ(q, ϑn), 0 < ϑn ≤ 1, n = 1, ...,m and |Rυqfn(z)| ≤ N, (z ∈ U), then the function Iγn,%(υ, q, z) given by (5) is univalent. Proof. Following the proof of Theorem 1, we get zf ′′ (z) f ′(z) = m∑ n=1 1 γn ( z ( Rυqfn(z) )′ Rυqfn(z) − 1 ) . Then we have∣∣∣∣∣c|z|2% + (1− |z|2%)zf ′′ (z) %f ′(z) ∣∣∣∣∣ = ∣∣∣∣∣c|z|2% + (1− |z|2%)1 % m∑ n=1 1 γn ( z ( Rυqfn(z) )′ Rυqfn(z) − 1 )∣∣∣∣∣ ≤ |c|+ 1 |%| m∑ n=1 1 |γn| (∣∣∣∣∣z2 ( Rυqfn(z) )′ [Rυqfn(z)]2 ∣∣∣∣∣ ∣∣Rυqfn(z) ∣∣ |z| + 1 ) . Now directly from the proof of Theorem 1, we have∣∣∣∣∣c|z|2% + (1− |z|2%)zf ′′ (z) %f ′(z) ∣∣∣∣∣ ≤ |c|+ 1 |%| m∑ n=1 [(1 + ϑn)N + 1] |γn| ≤ |c|+ 1 <(%) m∑ n=1 [(1 + ϑn)N + 1] |γn| , since |c| ≤ 1− 1 % ∑m n=1 [(1 + ϑn)N + 1] |γn| , thus we have ∣∣∣∣∣c|z|2% + (1− |z|2%)zf ′′ (z) %f ′(z) ∣∣∣∣∣ ≤ 1, (z ∈ U). Using Lemma 2 for the function f(z) we obtain that the integral operator Iγn,%(υ, q, z) given by (5) is univalent. S. Elhaddad, H. Aldweby, M. Darus / Eur. J. Pure Appl. Math, 13 (4) (2020), 861-872 869 Corollary 3. Let f1, ..., fm ∈ A, γ ∈ C and % ∈ C with <(%) > m[(1+ϑn)N+1] |γ| . Let N ≥ 1 with |c| ≤ 1− 1 <(%) m[(1 + ϑn)N + 1] |γ| , (c ∈ C). If for all n = 1, ..,m, fn ∈ Bυ(q, ϑn), 0 < ϑn ≤ 1, and |Rυqfn(z)| ≤ N, (z ∈ U). Then the integral operator Iγn,%(υ, q, z) = ( % ∫ z 0 t%−1 m∏ n=1 (Rυqfn(t) t ) 1 γ dt ) 1 % , is univalent. Proof. In Theorem 2, we consider γ1 = γ2 = ... = γm = γ. Corollary 4. Let f1, ..., fm ∈ A, γn ∈ C and % ∈ C with <(%) > ∑m n=1 [ϑn+2] |γn| . Let c ∈ C with |c| ≤ 1− 1 <(%) m∑ n=1 [ϑn + 2] |γn| . If for all n = 1, ..,m, fn ∈ Bυ(q, ϑn), 0 < ϑn ≤ 1, and |Rυqfn(z)| ≤ 1, (z ∈ U), then the function Iγn,%(υ, q, z) given by (5) is univalent. Proof. In Theorem 2, we consider N = 1. Setting υ = 0, γn = 1 σ − 1 , and % = m(σ − 1) + 1 where σ ∈ R in Theorem 2, we have Corollary 5. 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