EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 5907 ISSN 1307-5543 – ejpam.com Published by New York Business Global Second-Order Differential Subordination with the Mittag-Leffler Operator Ebrahim Amini1,∗, Shrideh Al-Omari2, Mona Khandaqji3 1 Department of Mathematics, Payame Noor University, P. O. Box 19395-4697, Tehran, Iran 2 Department of Mathematics, Faculty of Science, Al-Balqa Applied University, Salt 11134, Jordan 3 Department of Mathematics, Applied Science Private University, Amman 11931, Jordan Abstract. In this study, we employ the generalized Mittag-Leffler function and the Komatu in- tegral operator to present a new linear operator in terms of the convolution and define related classes of admissible functions. Then, we derive several properties and characteristics of two-order differential subordinations and superordinations. Moreover, we establish sandwich-type results for a class of analytic functions on an open unit disc. Over and above, we derive various results involving univalent functions in some details. 2020 Mathematics Subject Classifications: 30C45, 30C80, 33E12, 26A33 Key Words and Phrases: p-valent function, Borel distribution, Inclusion relation, Integral operator, Convolution 1. Introduction Let A be the class of complex-valued analytic functions of the subsequent form f(ζ) = ζ + ∞∑ n=2 anζ n, (1) which are defined on the open unit disc ∆ = {ζ ∈ C; |ζ| < 1}. Let S, ST and CV denote the familiar subclasses of A of univalent, starlike and convex functions on ∆, respectively ([1, 2]). In a very recent years, various researchers have studied a number of different subclasses of univalent functions in the context of geometric function theory (see for details [3–8]). For two analytic functions f and g belonging to A, we say that the function f is subordinate to the function g (or g superordinate of function f), written as ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5907 Email addresses: eb.amini.s@pnu.ac.ir (E. Amini), shridehalomari@bau.edu.jo (S. Al-Omari), m khandakji@asu.edu.jo (M. Khandaqji) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 2 of 22 f ≺ g or f(ζ) ≺ g(ζ), if there exists a Schwartz function w where w(0) = 0, |w(ζ)| < 1 and f(ζ) = g(w(ζ)). If g is univalent in ∆, then the subordination is equivalent to say f(0) = g(0) and f(∆) ⊂ g(∆) (see [9]). Further, the convolution (or Hadamard) product of two functions f and g, where f is given by (1) and g(ζ) = ζ + ∞∑ n=2 bnζ n, is presented by Ruscheweyh [10] as f(ζ)∗g(ζ) = ζ+ ∑∞ n=2 anbnζ n. Let p(ζ) be an analytic function in ∆ and suppose that ψ(r1, r2, r3, ζ) : C3×∆ −→ C be a univalent function and p(ζ) satisfies the subsequent differential subordination ψ(p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ) ≺ φ(ζ), (2) where φ(ζ) ∈ S. Then, p(ζ) is said to be a solution of the differential subordination (2). An analytic function γ(ζ) is said to be a dominant to the solution (2), if p(ζ) ≺ γ(ζ) for all functions p(ζ) satisfies differential subordinate (2). A univalent function γ̂(ζ) that satisfies γ̂(ζ) ≺ γ(ζ) for all the subordinates γ(ζ) of (2) is called the best dominant of (2). The best dominant is unique to a rotation of ∆ (see [11]). Let p(ζ) be an analytic function in ∆ and that ϕ(r1, r2, r3, ζ) : C3×∆ −→ C be a univalent function and p(ζ) satisfies the subsequent differential superordination φ(ζ) ≺ ϕ(p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ), (3) where φ(ζ) ∈ S. Then, p(ζ) is said to be a solution of differential superordination (3). An analytic function γ(ζ) is said to be dominant of the solution (3), if γ(ζ) ≺ p(ζ) for all functions p(ζ) satisfies differential superordinate (3). A univalent function γ̂(ζ) that satisfies γ(ζ) ≺ γ̂(ζ) for all the superordinates γ(ζ) of (3) is called the best dominant of (3). The best dominant is unique to a rotation of ∆ (see [11]). Miller et al. [12] investigated sufficient conditions on the function p, γ and ξ for which if the p(ζ) satisfying (3) then γ(ζ) ≺ p(ζ). Due to the result of Miller et al. [12], Bulboaca in [13] studied a subclass of first-order differential superordination whenever the superordination preserves operators. Moreover, Ali et al. [14] studied the sufficient condition for a function f ∈ A to satisfy γ1(ζ) ≺ ζf ′(ζ) f(ζ) ≺ γ2(ζ), where γ1(ζ) and γ2(ζ) are analytic functions with γ1(0) = γ2(0) = 1. A detailed investigation of subordination and superordination is given by many authors (see [15–17]). Komatu [18] considered the linear integral operator for σ ∈ C as follows F σ(ζ) = 2σ γ(σ) ∫ ζ 0 ( log t ζ )σ−1 f(t)dt = ζ + ∞∑ n=2 an nσ ζn. E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 3 of 22 The Pochhammer symbol, denoted by (µ)m, is defined by (µ)m = { 0, n = 0, µ ̸= 0, µ(µ+ 1)...(µ+ n− 1), n ∈ N. (4) Sharma and Jain [19] introduced theM -series as a function defined by means of the power series α pM β q ((aj)n, (bj)n, ζ) = ∞∑ n=0 (a1)n...(ap)n (b1)n...(bq)n ζn Γ(αn+ β (5) where α, β, µ ∈ C, Re(α) > 0 and (aj)n, (bj)n are the Pochhammer symbols defined by (4). The series (5) is not defined if one of the parameter aj , bs, j = 1, ..., p, s = 1, ..., q is a negative integer or zero. The series (5) is convergent for all ζ if p ≤ q (see [19]). The generalized Mittag-Leffler function is a M -series for p = q = 1, a = µ and b = 1. Thus, this function is defined by the power series [20] α 1M β 1 (µ, 1, ζ) = ∞∑ n=0 (µ)n n!Γ(αn+ β) ζn, (ζ ∈ ∆). (6) where α, β, µ ∈ C and Re(α) > 0. It is clear that the series is convergent for all ζ. A detailed investigation of analytic function by Mittag-Leffler is given by reserchers. (see [21, 22]). The normalized form of α 1M β 1 (µ, 1, ζ) can be performed as follows αE µ β (ζ) = ζ + ∞∑ n=2 Γ(β)(µ)n−1 (n− 1)!Γ(α(n− 1) + β) ζn, (ζ ∈ ∆). By making use of αE µ β , we introduce the operator σ αQ µ β : A → A, defined in terms of the convolution as σ αQ µ βf(ζ) = αE µ β (ζ) ∗ F σ(ζ) = ζ + ∞∑ n=2 Γ(β)(µ)n−1 nσ(n− 1)!Γ(α(n− 1) + β) anζ n, (ζ ∈ ∆), where α, β, µ, σ ∈ C and Re(α) > 0. The operator σ αQ µ βf(ζ) indeed satisfies the following first-order differential recurrence re- lation ζ ( σ αQ µ βf(ζ) )′ = µσαQ µ+1 β f(ζ)− (µ− 1)σαQ µ βf(ζ). (7) In this paper, we study a suitable class of admissible functions involving linear generalized Mittag-Leffler and Komatu integral operators. We also derive several sufficient conditions E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 4 of 22 of two-order differential subordinations and superordinations of analytic univalent func- tions on an open unit disc ∆. Moreover, we obtain some Sandwich-type subordination of the subsequent form: γ1(ζ) ≺ σ αQ µ βf(ζ) ≺ γ2(ζ), where γ1(ζ) and γ2(ζ) are analytic functions with γ1(0) = 0 and γ2(0) = 0. 2. Preliminaries Lemma The following lemmas are very useful in our investigation. We first recall some defini- tions. Definition 1. [12] Let ζ̃ ∈ ∆−E(f). The set of all functions f(ζ) ∈ S on ∆−E(f) such that f ′(ζ̃) ̸= 0 is denoted by H, where E(f) = {ζ̃, ζ̃ ∈ ∂∆ : lim ζ→ζ̃ f(ζ) = +∞}. Definition 2. [13] Let Ω be a subset of C and γ ∈ H. The class Ψn[Ω, γ] of admissible functions, consists of the complex-valued functions ψ : C3 × ∆ −→ C, which satisfy the following admissibility conditions: ψ(θ1, θ2, θ3; ζ) /∈ Ω, whenever θ1 = γ(ζ̃), θ2 = mζ̃γ′(ζ̃), and Re ( θ3 θ2 + 1 ) ≥ mRe [ ζ̃γ′′(ζ̃) γ′(ζ̃) + 1 ] , where ζ ∈ ∆, ζ̃ ∈ ∂∆− E(q) and m ≥ 1. Lemma 1. [23] Let Ω ⊆ C and ϕ ∈ Ψ[Ω, γ]. If p ∈ H satisfies the following condition {ϕ(p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ) : ζ ∈ ∆} ∈ Ω, then we have the following differential subordination p(ζ) ≺ γ(ζ), (ζ ∈ ∆). Definition 3. [23] Let Ω be a subset of C and γ ∈ H. The class Φ[Ω, γ] of admissible complex-valued functions ϕ : C3 × ∆ −→ C, which satisfy the following admissibility conditions: ϕ(θ1, θ2, θ3; ζ) ∈ Ω, E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 5 of 22 whenever θ1 = γ(ζ̃), θ2 = ζ̃γ′(ζ̃) m , and Re ( θ3 θ2 + 1 ) ≥ 1 m Re [ ζ̃γ′′(ζ̃) γ′(ζ̃) + 1 ] , where ζ ∈ ∆, ζ̃ ∈ ∂∆− E(q) and m ≥ n. Lemma 2. [23] Let Ω ⊆ C and ϕ ∈ Φ[Ω, γ]. If p ∈ H and ϕ(p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ) is univalent in ∆, then Ω ⊂ {ϕ(p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ) : ζ ∈ ∆}, implies that the differential subordination is as follows γ(ζ) ≺ p(ζ), (ζ ∈ ∆). Lemma 3. [2] If f ∈ A is a univalent function such that g(ζ) ≺ f(ζ). Then |g(ζ)| ≤ |f(ζ)|, for all ζ in the disc |ζ| ≤ 1 2(3− √ 5). This radius is best possible. Lemma 4. [2] If f ∈ A is a univalent function such that g(ζ) ≺ f(ζ). Then |g′(ζ)| ≤ |f ′(ζ)| for all ζ in the disc |ζ| ≤ 1 2(3− √ 8). This radius is best possible. 3. Two-order subordination result In this section, we derive a foundation result in the theory of second-order differential subordination. Furthermore, we will take several applications on the boundary of ∆. Definition 4. Let Ω be a subset of C, γ ∈ H ∩ A and µ ∈ C, (µ ̸= 0, 1). We define the class Ψ′(Ω, γ) of admissible complex valued functions ψ′ : C3 ×∆ → C such that the following admissibility conditions hold: ψ′(τ1, τ2, τ3; ζ) /∈ Ω, whenever τ1 = γ(ζ̃), τ2 = mζ̃γ′(ζ̃) + (µ− 1)γ(ζ̃) µ , and Re ( µ2τ3 − (µ− 1)τ2 µτ2 + (µ− 1)τ1 − µ+ 1 ) ≥ mRe ( ζ̃γ′′(ζ̃) γ′(ζ̃) + 1 ) , where ζ ∈ ∆, ζ̃ ∈ ∂∆− E(γ) and m ≥ 1. E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 6 of 22 Theorem 1. Let Ω be a subset of C and ψ′ ∈ Ψ′(Ω, γ). If f ∈ A satisfies{ ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) , ζ ∈ ∆ } ⊆ Ω, then we have σ αQ µ βf(ζ) ≺ γ(ζ). (8) Proof. Assume that p(ζ) = σ αQ µ βf(ζ). (9) Then, by making (7) and (9), we obtain that σ αQ µ+1 β f(ζ) = ζp′(ζ) + (µ− 1)p(ζ) µ , (ζ ∈ ∆). Moreover, a simple computation shows that σ αQ µ+2 β f(ζ) = ζ2p′′(ζ) + (2µ− 1)ζp′(ζ) + (µ− 1)2p(ζ) µ2 , (ζ ∈ ∆). Now, we define τ1 = θ1, θ2 = θ2 + (µ− 1)θ1 µ , and τ3 = θ3 + (2µ− 1)θ2 + (µ− 1)2θ1 µ2 . Further, we define the transformation h from C3 ×∆ to C as h(θ1, θ2, θ3; ζ) = ψ′(τ1, τ2, τ3; ζ) = ψ′ ( θ1, θ2 + (µ− 1)θ1 µ , θ3 + (2µ− 1)θ2 + (µ− 1)2θ1 µ2 ; ζ ) .(10) From the equations (9) to (10), we have h(p(ζ), ζp′(ζ), ζ2p′′(ζ), ζ) = ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ); ζ ) . (11) Hence, the assertion (11) becomes ψ′(p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ) /∈ Ω. We note that θ3 θ2 + 1 = µ2τ3 − (µ− 1)τ2 µτ2 − (µ− 1)τ3 − µ+ 1. Since the admissibility conditions for ψ′ ∈ Ψ′(Ω, γ) are equivalent to ψ ∈ Ψ(Ω, γ) as given in Definition 2, then, by using Lemma 1 we have p(ζ) ≺ γ(ζ), (ζ ∈ ∆). This shows that the desired differential subordination (8) is established. The result can be extended to the case Ω = h(∆) in which the complex-valued function h(ζ) is a conformal mapping of ∆ onto Ω. In this case, we write Ψ′(Ω, γ) = Ψ′(h, γ). E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 7 of 22 Theorem 2. Let ψ′ ∈ Ψ′(h, γ). If ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆ and ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) ≺ h(ζ), then, we have σ αQ µ βf(ζ) ≺ γ(ζ). Proof. Following similar proof to that of Theorem [[11], Theorem 2.3c], we can proof Theorem 2. So, it is omitted. We next consider the behaviour of γ on the boundary of ∆. The following result is an interesting consequence of Theorem 1. Theorem 3. Let 0 < ρ < 1 and h(ζ), γ(ζ) ∈ S satisfy the conditions γρ(ζ) = γ(ρζ) and hρ(ζ) = h(ρζ). Let ψ′ : C3 ×∆ −→ C satisfy one of the subsequent conditions: (i) ψ′ ∈ Ψ′(h, γρ), (ii) there exist ρ0 ∈ (0, 1) such that ψ′ ∈ Ψ′(hρ, γρ), for all ρ ∈ (ρ0, 1). If ψ′ ∈ Ψ′(h, γ), ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is analytic in ∆ and ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) ≺ h(ζ), then we have σ αQ µ βf(ζ) ≺ γ(ζ). Proof. By following the proof Theorem [[11], Theorem 2.3d], we can proof Theorem 3. So, it has been omitted. Theorem 4. Let k ∈ {2, 3, 4, ...}, 0 < ρ < 1, h ∈ S and ψ′ : C3 ×∆ −→ C. Suppose that the differential equation ψ′ ( σ αQ µ βf(ζ), kζ k−1σ αQ µ+1 β f(ζ), k2ζ2(k−1)σ αQ µ+2 β f(ζ); ζ ) = h(ζ) (12) has a solution γ(ζ) with γ(0) = 0 and one of the subsequent conditions is satisfied: (i) γ ∈ H and ψ′ ∈ Ψ′(h, γ), (ii) γ ∈ S and ψ′ ∈ Ψ′(h, γρ), or (iii) γ ∈ S and there exists ρ0 ∈ (0, 1) such that ψ′ ∈ Ψ′(hρ, γρ) for all ρ ∈ (0, 1). E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 8 of 22 If p(ζ) = σ αQ µ βf(ζ k) (13) and ψ′(p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ) ∈ A such that ψ′(p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ) ≺ h(ζ), (14) then p(ζ) ≺ γ(ζ) and γ is the best dominant. Proof. Because of Theorems 2 and 3, we deduce that γ is dominant (14). From (7) and (13), we obtain that kζk−1σ αQ µ+1 β f(ζk) = ζp′(ζ) + (µ− 1)kζk−1p(ζ) µ . (15) Moreover, a simple computation shows that k2ζ2(k−1)σ αQ µ+2 β f(ζk) = ζ2p′′(ζ) + ( 1− k + 2(µ− 1)kζk+1 ) ζp′(ζ) + (µ− 1)2k2ζ2k−1p(ζ) µ2 .(16) Similar to the proof of the Theorem 1, we define the transformation h : C3 ×∆ → C as follows h ( p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ ) = ψ′ ( σ αQ µ βf(ζ k), kζk−1σ αQ µ+1 β f(ζk), k2ζ2(k−1)σ αQ µ+2 β f(ζk); ζ ) . Therefore, from (12) we obtain that h ( p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ ) = γ(ζk) ≺ γ(ζ). Since p(∆) = γ(∆), we conclude that γ is the best dominant. This completes the proof of Theorem 4. In this particular case, we define the function γ2 : ∆ −→ C as follows γ2(ζ) = z − β1z 2, |β1| < 1. (17) Now, we introduce and investigate the class Ψ′(∆, γ2) consisting of all admissible functions. Definition 5. Let γ2(ζ) be given by (22) and µ ∈ C, (µ ̸= 0, 1). Then, we define the class of admissible functions Ψ′(∆, γ2) to be the set of all functions ψ′ : C3×∆ −→ C satisfying the following admissibility conditions: Ψ′(τ ′1, τ ′ 2, τ ′ 3, ζ) /∈ ∆ where τ ′1 = z − β1z 2, τ ′2 = meiθ(1− 2β1e iθ) + (µ− 1)(eiθ − 2β1e 2iθ) µ , E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 9 of 22 τ ′3 = L+ (2µ− 1)meiθ(1− 2β1e iθ) + (µ− 1)2(eiθ − β1e iθ µ2 , such that Re { Le−iθ 1− 2β1eiθ } ≥ m2 2β(1− cos θ) 1 + 2β(1− 2 cos θ) , where ζ ∈ ∆, θ ∈ R, |β1| < 1 and m ≥ 1. Theorem 5. Let γ2(ζ) be given by (22), ψ′ ∈ Ψ′(∆, γ2) and µ ∈ C, (µ ̸= 0, 1). If f ∈ A satisfies { ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) , ζ ∈ ∆ } ∈ ∆, then, we have σ αQ µ βf(ζ) ≺ z − β1z 2, (|β1| < 1). Proof. Similar to the proof of Theorem 1, we can proof Theorem 5 Theorem 6. Let h(ζ) be a conformal mapping from ∆ onto ∆, γ2(ζ) be given by (22), ψ′ ∈ Ψ′(∆, γ2), µ ∈ C, (µ ̸= 0, 1) and ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆. If ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) ≺ h(ζ), then, we have ∣∣∣σαQµ βf(ζ) ∣∣∣ ≤ 1 + |β1|, (18) for all ζ in the disc |ζ| ≤ 1 2(3− √ 5) and |β1| < 1. This radius is best possible. Proof. In view of Theorem 2, we obtain that σ αQ µ βf(ζ) ≺ γ2(ζ). Now, by applying Lemma 3, we get∣∣∣σαQµ βf(ζ) ∣∣∣ ≤ |γ2(ζ)|, for all ζ in the disc |ζ| ≤ 1 2(3− √ 5). From the Maximum-Modulus Principle, we have |γ2(ζ)| ≤ 1 + |β1|. This establishes inequality (34). By using Lemma 3 we conclude that this radius is best possible. Hence, the proof of Theorem 6 is completed. Plots of the suggested function γ2(ζ) = ζ − β1ζ 2 in the unit disc ∆ are illustrated in figure 1(a). The parameter was β1 = 1 2 . By putting ψ′(τ ′1, τ ′ 2, τ ′ 3, ζ) = τ ′2 in Theorem (6) we obtain the following corollary. E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 10 of 22 Figure 1: (a)γ2(ζ) = ζ − βζ2 for β = 1 2 , (b)γ′ 1(ζ) = 1− 2βζ for β = 1 2 Corollary 1. Let h(ζ) be a conformal mapping of ∆ onto ∆, γ2(ζ) be given by (22), ψ′ ∈ Ψ′(∆, γ2), µ ∈ C, (µ ̸= 0, 1) and ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆. If σ αQ µ+1 β f(ζ) ≺ h(ζ), then we have ∣∣∣σαQµ βf(ζ) ∣∣∣ ≤ 1 + |β1|, for all ζ in the disc |ζ| ≤ 1 2(3− √ 5) and |β1| < 1. This radius is best possible. Theorem 7. Let h(ζ) be a conformal mapping of ∆ onto ∆, γ2(ζ) be given by (22), ψ′ ∈ Ψ′(∆, γ2), µ ∈ C, (µ ̸= 0, 1) and ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆. If ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) ≺ h(ζ), then we have ∣∣∣∣1ζ [µσαQµ+1 β f(ζ)− (µ− 1)σαQ µ βf(ζ) ]∣∣∣∣ ≤ 1 + 2|β1|, (19) for all ζ in the disc |ζ| ≤ 1 2(3− √ 8) and |β1 < 1. This radius is best possible. Proof. In view of Theorem 2, we obtain that σ αQ µ βf(ζ) ≺ γ2(ζ). Now, by applying Lemma 4, we get∣∣∣∣(σαQµ βf(ζ) )′∣∣∣∣ ≤ |γ′2(ζ)|, (20) E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 11 of 22 for all ζ in the disc |ζ| ≤ 1 2(3− √ 8). From the Maximum-Modulus Principle, we have |γ′2(ζ)| ≤ 1 + 2|β1|. (21) From the inequalities (7), (21) and (20), we establish ineqality (35). By using Lemma 4 we conclude that this radius is best possible. Hence, the proof of Theorem 7 is completed. Plots of the suggested function γ′2(ζ) = 1 − 2β1ζ in the unit disc ∆ are illustrated in figure 1(b). The parameter was β1 = 1 2 . Similarly, putting ψ′(τ ′1, τ ′ 2, τ ′ 3, ζ) = τ ′2 in Theorem (6) leads to the following corollary. Corollary 2. Let h(ζ) be a conformal mapping of ∆ onto ∆, γ2(ζ) be given by (22), ψ′ ∈ Ψ′(∆, γ2), µ ∈ C, (µ ̸= 0, 1) and ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆. If σ αQ µ+1 β f(ζ) ≺ h(ζ), then we have ∣∣∣∣1ζ [µσαQµ+1 β f(ζ)− (µ− 1)σαQ µ βf(ζ) ]∣∣∣∣ ≤ 1 + 2|β1|, for all ζ in the disc |ζ| ≤ 1 2(3− √ 8) and |β1 < 1. This radius is best possible. [rgb]1.00,0.00,0.00For another example, we define the function γ3 : ∆ −→ ∆ as follow γ3(ζ) = −2ζ 1 + ζ . (22) Now, we introduce and investigate the class of Ψ′(∆, γ3) consisting of admissible functions. Definition 6. Let γ3(ζ) that is given by (22) and µ ∈ C, (µ ̸= 0, 1). Then, we define the class of admissible functions Ψ′(∆, γ3) to be the set of all function ψ′ : C3 ×∆ −→ C satisfying the following admissibility conditions: Ψ′(τ11 , τ 1 2 , τ 1 3 , ζ) /∈ ∆ where τ11 = γ3(ζ), τ12 = ζγ′3(ζ) (m− (µ− 1)(1 + ζ)) , τ13 = ζγ′′3 (ζ) [ ζ − (2µ− 1)(1 + ζ) 2 − (µ− 1)2(1 + ζ)2 2 ] . such that Re ( µ2τ13 − (µ− 1)τ12 µτ12 + (µ− 1)τ11 − µ+ 1 ) ≥ 0 where ζ ∈ ∆, ζ ∈ ∂∆− E(q) and m ≥ 1. E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 12 of 22 Theorem 8. Let γ3(ζ) be given by (22), ψ′ ∈ Ψ′(∆, γ3) and µ ∈ C, (µ ̸= 0, 1). If f ∈ A satisfies { ψ′ ( σ αQ µ βf(v), σ αQ µ+1 β f(v), σαQ µ+2 β f(v), v ) , v ∈ ∆ } ⊆ ∆. Then, we have σ αQ µ βf(ζ) ≺ −2ζ 1 + ζ . Proof. Similar to the proof of Theorem 1, we can proof of Theorem 8 Plots of the suggested function γ3(ζ) = −2ζ 1+ζ in the unit disc ∆ are illustrated in figure 2. Figure 2: γ3(ζ) = −2ζ 1+ζ 4. Two-order superordination result In this section, we investigate the following new class of admissible functions which yield a result of two-order differential superordination for the operator Qµ βf(ζ). Definition 7. Let Ω be a subset of C, γ ∈ H∩A and µ ∈ C, (µ ̸= 0, 1). We define the set Φ′(Ω, γ) of admissible complex valued functions ϕ′ : C3 ×∆ → C such that the subsequent admissibility conditions hold: ϕ′(τ1, τ2, τ3; ζ) ∈ Ω, whenever τ1 = γ(ζ̃), τ2 = ζ̃γ′(ζ̃)/m1 + µγ(ζ̃) µ− 1 , E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 13 of 22 and Re ( µ2τ3 − (µ− 1)τ2 µτ2 + (µ− 1)τ1 − µ+ 1 ) ≥ 1 m1 Re ( ζ̃γ′′(ζ̃) γ′(ζ̃) + 1 ) , where ζ ∈ ∆, ζ̃ ∈ ∂∆− E(γ) and m1 ≥ 1. Theorem 9. Let Ω be a subset of C and ϕ′ ∈ Φ′(Ω, γ). If f ∈ A satisfies Ω ⊆ { ϕ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) , ζ ∈ ∆ } , then we have γ(ζ) ≺ σ αQ µ βf(ζ). (23) Proof. Assume that p(ζ) = σ αQ µ βf(ζ). (24) Similar to the proof of Theorem 1, we can obtain the transformation h1 : C3 ×∆ → C as follows h1(p(ζ), ζp ′(ζ), ζ2p′′(ζ); ζ) = ϕ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) . (25) From equations (23) and (25), we obtain Ω ⊆ {h1(p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ)}. Because of (25), we conclude that the admissibility condition ϕ′ ∈ Φ′(Ω, γ) and the ad- missibility condition for ϕ in Definition 3 is equivalent. Thus, by using Lemma 2, we conclude that γ(ζ) ≺ p(ζ). (26) Hence, the differential subordination (26) is equivalent to (23). This completes the proof of Theorem 9. The result can be extended to the case Ω = h(∆) in which the complex-valued function h(z) is a conformal mapping of ∆ onto Ω. In this function, we write Φ′(Ω, γ) = Φ′(h, γ). Theorem 10. Let ϕ′ ∈ Φ′(h, γ). If ϕ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆ and h(ζ) ≺ ϕ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) , then we have γ(ζ) ≺ σ αQ µ βf(ζ). E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 14 of 22 Proof. Similar to the proof of Theorem 2, we can prove Theorem 10. Theorem 11. Let 0 < ρ < 1 and h(ζ), γ(ζ) ∈ S satisfy the conditions γρ(ζ) = γ(ρζ) and hρ(ζ) = h(ρζ). Let ϕ′ : C3 ×∆ −→ C satisfy one of the following conditions: (i) ϕ′ ∈ Φ′(h, γρ), (ii) there exist ρ0 ∈ (0, 1) such that ϕ′ ∈ Φ′(hρ, γρ), for all ρ ∈ (ρ0, 1). If ϕ′ ∈ Φ′(h, γ), ϕ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is analytic in ∆ and h(ζ) ≺ ϕ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) . Then, we have γ(ζ) ≺ σ αQ µ βf(ζ). Proof. Similar to the proof of Theorem 3, we can prove Theorem 11. Theorem 12. Let k ∈ {2, 3, 4, ...}, 0 < ρ < 1, h ∈ S and ϕ′ : C3×∆ −→ C. Suppose that the differential equation ϕ′ ( σ αQ µ βf(ζ), kζ k−1σ αQ µ+1 β f(ζ), k2ζ2(k−1)σ αQ µ+2 β f(ζ); ζ ) = h(ζ), (27) has a solution γ(ζ) with γ(0) = 0 and one of the following conditions is satisfied: (i) γ ∈ H and ϕ′ ∈ Φ′(h, γ), (ii) γ ∈ S and ϕ′ ∈ Φ′(h, γρ), or (iii) γ ∈ S and there exists ρ0 ∈ (0, 1) such that ϕ′ ∈ Φ′(hρ, γρ) for all ρ ∈ (0, 1). If p(ζ) = τ αQ µ βf(ζ k), and ϕ′(p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ) ∈ A such that h(ζ) ≺ ϕ′(p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ), (28) then γ(ζ) ≺ p(ζ) and γ is the best dominant. E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 15 of 22 Proof. In view of the Theorems 10 and 11, we deduce that γ(ζ) is a dominant (28). By following similar proof to the proof of Theorem (9) and using the assertions (15) and (16), we define the transformation h1 : C×∆ → C as follows h1 ( p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ ) = ϕ′ ( σ αQ µ βf(ζ k), kζk−1σ αQ µ+1 β f(ζk), k2ζ2(k−1)σ αQ µ+2 β f(ζk); ζ ) . Therefore from (27), we obtain that γ(ζ) ≺ γ(ζk) = h1 ( p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ ) . Since p(∆) = γ(∆), we conclude that γ(ζ) is the best dominant. This completes the proof of Theorem 12. In this particular case, we define the function γ1 : ∆ −→ C as follows γ1(ζ) = ζeλζ , 0 < λ ≤ 1. (29) In what follows, we introduce the class Φ′(∆, γ1) of admissible functions. Definition 8. Let γ1(ζ) be given by (29) and µ ∈ C, (µ ̸= 0, 1). We define the class Φ′(∆, γ1) of admissible functions ϕ′ : C3 ×∆ −→ C, which satisfy the following admissi- bility conditions: Φ′(τ ′′1 , τ ′′ 2 τ ′′ 3 , ζ) ∈ Ω, whenever τ ′′1 = ζ̄eλζ̄ , τ ′′2 = ζ̄eλζ̄ 1 + λζ̄ +mµ m(µ− 1) , τ ′′3 = L+ ζ̄eλζ̄ [µ(2µ− 1)(1 + λζ̄ +mµ)−m(µ− 1)2] mµ2(µ− 1) , such that Re { L ζ̄eλζ̄(1 + λζ̄ +mµ) } ≥ µ m2(µ− 1) Re { λ 1 + λζ̄ + λ+ 1 } , where ζ ∈ ∆, ζ̃ ∈ ∂∆− E(q) and m ≥ 1. Theorem 13. Let γ1(ζ) be given by (29), ϕ′ ∈ Φ′(∆, γ2) and µ ∈ C, (µ ̸= 0, 1). If f ∈ A satisfies ∆ ⊆ { ϕ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) , ζ ∈ ∆ } , then we have ζeλζ ≺ σ αQ µ βf(ζ). E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 16 of 22 Proof. Similar proof to the proof of Theorem 9, we can proof Theorem 13. Theorem 14. Let h(ζ) be a conformal mapping of ∆ onto C, γ1(ζ) be given by (29), ϕ′ ∈ Φ′(∆, γ2), µ ∈ C, (µ ̸= 0, 1) and ϕ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆ . If h(ζ) ≺ ϕ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) , then we have 1 eλ ≤ ∣∣∣σαQµ βf(ζ) ∣∣∣ , (30) for all ζ in the disc |ζ| ≤ 1 2(3− √ 5) and 0 < λ ≤ 1. This radius is best possible. Proof. In view of Theorem 10, we obtain that γ1(ζ) ≺ σ αQ µ βf(ζ). Now, by applying Lemma 3, we get |γ1(ζ)| ≤ ∣∣∣σαQµ βf(ζ) ∣∣∣ , for all ζ in the disc |ζ| ≤ 1 2(3− √ 5). From the Maximum-Modulus Principle, we have 1 eλ ≤ |γ1(ζ)|. This establish inequality (30). By using Lemma 3 we conclude that this radius is best possible. Hence, the proof of Theorem 14 is completed. Plots of the suggested function γ1(ζ) = ζeλζ in the unit disc ∆ are illustrated in figure 2(c). The parameter was λ = 1 4 . Putting ψ′(τ ′1, τ ′ 2, τ ′ 3, ζ) = τ ′2 in Theorem (14) yields the following corollary. Corollary 3. Let h(ζ) be a conformal mapping of ∆ onto C, γ1(ζ) be given by (29), ϕ′ ∈ Φ′(∆, γ1), µ ∈ C, (µ ̸= 0, 1) and ϕ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆ . If h(ζ) ≺ σ αQ µ+1 β f(ζ), then we have 1 eλ ≤ ∣∣∣σαQµ βf(ζ) ∣∣∣ , for all ζ in the disc |ζ| ≤ 1 2(3− √ 5) and 0 < λ ≤ 1. This radius is best possible. E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 17 of 22 Theorem 15. Let h(ζ) be a conformal mapping of ∆ onto C, γ1(ζ) be given by (29), ϕ′ ∈ Φ′(∆, γ1), µ ∈ C, (µ ̸= 0, 1) and ϕ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆. If h(ζ) ≺ ϕ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) , then we have 1− λ eλ ≤ ∣∣∣∣1ζ [µσαQµ+1 β f(ζ)− (µ− 1)σαQ µ βf(ζ) ]∣∣∣∣ , (31) for all ζ in the disc |ζ| ≤ 1 2(3− √ 8) and |β1 < 1. This radius is best possible. Proof. In view of Theorem 10, we obtain that γ1(ζ) ≺ σ αQ µ βf(ζ). Now, by applying Lemma 4, we get |γ′1(ζ)| ≤ ∣∣∣∣(σαQµ βf(ζ) )′∣∣∣∣ , (32) for all ζ in the disc |ζ| ≤ 1 2(3− √ 8). From the Maximum-Modulus Principle, we have 1− λ eλ ≤ |γ′1(ζ)|. (33) From the inequalities (7), (33) and (32), we establish inequality (31). By using Lemma 4 we conclude that this radius is best possible. Hence, the proof of Theorem 15 is completed. Plots of the suggested function γ′1(ζ) = eλζ + λζeλζ in the unit disc ∆ are illustrated in figure 2(d). The parameter was λ = 1 4 . Similarly, putting ϕ′(τ ′1, τ ′ 2, τ ′ 3, ζ) = τ ′2 in the Theorem (15) leads to the following corollary. Corollary 4. Let h(ζ) be a conformal mapping of ∆ onto C, γ1(ζ) be given by (22), ϕ′ ∈ Φ′(∆, γ1), µ ∈ C, (µ ̸= 0, 1) and ϕ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆. If h(ζ) ≺ σ αQ µ+1 β f(ζ), then we have 1− λ eλ ≤ ∣∣∣∣1ζ [µσαQµ+1 β f(ζ)− (µ− 1)σαQ µ βf(ζ) ]∣∣∣∣ , for all ζ in the disc |ζ| ≤ 1 2(3− √ 8) and 0 < λ < 1. This radius is best possible. E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 18 of 22 Figure 3: (c)γ1(ζ) = ζeλζ for λ = 1 4 , (d)γ′ 1(ζ) = eλζ + ζ 4 eλζ for λ = 1 4 5. Result on Sandwich Theorems In this section, we employ the results obtained in sections 3 and 4 and derive the sandwich-type theorem. Theorem 16. Let Ω be a subset of C and φ ∈ Φ′(Ω, γ1) ⋂ Ψ′(Ω, γ2). If f ∈ A satisfies{ φ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) , ζ ∈ ∆ } = Ω, then we have γ1(ζ) ≺ σ αQ µ βf(ζ) ≺ γ2(ζ). Proof. We can combine the Theorems 1 and 9 and obtain the Theorem 16. Theorem 17. Let h1 and h2 be two conformal mapping of ∆ onto Ω, γ1 and γ2 be two analytic functions in ∆ with γ1(0) = γ2(0) = 0 and φ ∈ Φ′(h, γ1) ⋂ Ψ′(h, γ2). If f ∈ A, σ αQ µ βf(ζ) ∈ A ⋂ H and φ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆, then γ1(ζ) ≺ φ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) ≺ γ2(ζ), implies the following subordination γ1(ζ) ≺ σ αQ µ βf(ζ) ≺ γ2(ζ). Proof. We can combine Theorems 2 and 10 and obtain Theorem 17. Theorem 18. Let 0 < ρ < 1, h1, h2 be two conformal mapping of ∆ onto Ω satisfying the conditions h1ρ(ζ) = h1(ρζ) and h2ρ(ζ) = h2(ρζ). Let γ1 and γ2 be two analytic functions in ∆ with γ1(0) = γ2(0) = 0 satisfying the conditions, γ1ρ(ζ) = γ1(ρζ) and one of the following conditions: E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 19 of 22 (i) φ ∈ Φ′(h, γρ) ⋂ Ψ′(h, γρ), or (ii) there exist ρ0 ∈ (0, 1) such that φ ∈ Φ′(hρ, γρ) ⋂ Ψ′(hρ, γρ), for all ρ ∈ (ρ0, 1). If φ ∈ Φ′(h, γ) ⋂ Ψ′(h, γ), φ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆ and h1(ζ) ≺ φ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) ≺ h2(ζ), then, we have γ1(ζ) ≺ σ αQ µ βf(ζ) ≺ γ2(ζ). Proof. We can combine Theorems 3 and 11, and then obtain Theorem 18. Theorem 19. Let k ∈ {2, 3, 4, ...}, 0 < ρ < 1 and h1, h2 be two conformal mapping of ∆ onto Ω satisfying the conditions h1ρ(ζ) = h1(ρζ) and h2ρ(ζ) = h2(ρζ). Let γ1 and γ2 be two analytic functions in ∆ with γ1(0) = γ2(0) = 0 satisfying the conditions, γ1ρ(ζ) = γ1(ρζ). Suppose that the differential equation φ ( σ αQ µ βf(ζ), kζ k−1σ αQ µ+1 β f(ζ), k2ζ2(k−1)σ αQ µ+2 β f(ζ); ζ ) = h1(ζ), has a solution γ1(ζ) and φ ( σ αQ µ βf(ζ), kζ k−1σ αQ µ+1 β f(ζ), k2ζ2(k−1)σ αQ µ+2 β f(ζ); ζ ) = h2(ζ), has a solution γ2(ζ) and one of the following conditions is satisfied: (i) γ1, γ2 ∈ H and φ ∈ Ψ′(h1, γ1) ⋂ Φ′(h2, γ2), (ii) γ1, γ2 ∈ S and φ ∈ Ψ′(h1, γ1ρ) ⋂ Φ′(h2, γ2ρ), or (ii) γ1, γ2 ∈ S and there exists ρ0 ∈ (0, 1) such that φ ∈ Ψ′(h1ρ, γ1ρ) ⋂ Φ′(h2ρ, γ2ρ) for all ρ ∈ (0, 1). If p(ζ) = σ αQ µ βf(ζ k), and φ(p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ) ∈ A such that h1(ζ) ≺ φ(p(ζ), ζp′(ζ), ζ2p′′(ζ); ζ) ≺ h2(ζ), then γ1(ζ) ≺ p(ζ) ≺ γ2(ζ) and γ1, γ2 are the best dominant. Proof. We can combine Theorems 4 and 12 and obtain Theorem 19. E. Amini, S. Al-Omari, M. Khandaqji / Eur. J. Pure Appl. Math, 18 (2) (2025), 5907 20 of 22 Theorem 20. Let h1(ζ) and h2(ζ) be two conformal mapping of ∆ onto C, γ2(ζ) that is given by (22), γ1(ζ) that is given by (29), φ ∈ Ψ′(∆, γ2) ⋂ Φ′(∆, γ1), µ ∈ C, (µ ̸= 0, 1) and φ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆. If h1(ζ) ≺ ψ′ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) ≺ h2(ζ), then we have 1 eλ ≤ ∣∣∣σαQµ βf(ζ) ∣∣∣ ≤ 1 + |β1|, (34) for all ζ in the disc |ζ| ≤ 1 2(3− √ 5), 0 < λ < 1 and |β1| < 1. This radius is best possible. Proof. We can combine Theorems 6 and 14 and obtain Theorem 20. Theorem 21. Let h1(ζ) and h2(ζ) be two conformal mapping of ∆ onto C, γ2(ζ) given by (22), γ1(ζ) is given by (29), φ ∈ Ψ′(∆, γ2) ⋂ Φ′(∆, γ1), µ ∈ C, (µ ̸= 0, 1) and φ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) is univalent in ∆. If h1(ζ) ≺ φ ( σ αQ µ βf(ζ), σ αQ µ+1 β f(ζ), σαQ µ+2 β f(ζ), ζ ) ≺ h2(ζ), then we have 1− λ eλ ≤ ∣∣∣∣1ζ [µσαQµ+1 β f(ζ)− (µ− 1)σαQ µ βf(ζ) ]∣∣∣∣ ≤ 1 + 2|β1|, (35) for all ζ in the disc |ζ| ≤ 1 2(3− √ 8), 0 < λ < 1 and |β1 < 1. 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