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Article

New Results on Differential Subordination and Superordination for Multivalent Functions Involving New Symmetric Operator

by
Abdul Rahman S. Juma
1,
Nihad Hameed Shehab
2,
Daniel Breaz
3,
Luminiţa-Ioana Cotîrlă
4,*,
Maslina Darus
5 and
Alin Danciu
6
1
Department of Mathematics, College of Education for Pure Sciences, University of Anbar, Anbar 31001, Iraq
2
Mathematic, General Directorate of Education in Anbar Governorate, Anbar 31001, Iraq
3
Department of Mathematics, “1 Decembrie 1918” University of Alba Iulia, 510009 Alba Iulia, Romania
4
Department of Mathematics, Technical University of Cluj-Napoca, 400114 Cluj-Napoca, Romania
5
School of Mathematical Sciences, Faculty of Sciences and Technology, Universiti Kebangsaan Malaysia, Bangi 43600, Selangor, Malaysia
6
Department of Mathematics, Babes Bolyai University, 400084 Cluj-Napoca, Romania
*
Author to whom correspondence should be addressed.
Symmetry 2024, 16(10), 1326; https://doi.org/10.3390/sym16101326
Submission received: 18 August 2024 / Revised: 3 October 2024 / Accepted: 5 October 2024 / Published: 8 October 2024

Abstract

:
This article aims to significantly advance geometric function theory by providing a valuable contribution to analytic and multivalent functions. It focuses on differential subordination and superordination, which characterize the interactions between analytic functions. To achieve our goal, we employ a method that relies on the characteristics of differential subordination and superordination. As one of the latest advancements in this field, this technique is able to derive several results about differential subordination and superordination for multivalent functions defined by the new operator M λ , p m v , ρ ; η F ξ within the open unit disk A . Additionally, by employing the technique, the differential sandwich outcome is achieved. Therefore, this work presents crucial exceptional instances that follow the results. The findings of this paper can be applied to a wide range of mathematical and engineering problems, including system identification, orthogonal polynomials, fluid dynamics, signal processing, antenna technology, and approximation theory. Furthermore, this work significantly advances the knowledge and understanding of the analytical functions of the unit and its interactive higher relations. The characteristics and consequences of differential subordination theory are symmetric to those of differential superordination theory. By combining them, sandwich-type theorems can be derived.

1. Introduction and Definitions

Geometric function theory is a branch of complex analysis concerned with, and investigating, the geometric characteristics of analytic functions. The study of the theory of analytic univalent and multivalent functions is a cornerstone of complex analysis, captivating researchers due to its elegant geometric and abundant research prospects.
The study of univalent functions constitutes a fundamental pillar of complex analysis for both single and many variables. This concept traces its roots to the novel works of Koebe in 1907 [1], Gronwall in 1914, and Bieberbach in 1916. Their contributions, including Koebe’s fundamental publication, Gronwall’s area theorem, and Bieberbach’s second coefficient estimate, have been considered the groundwork for the subsequent development in this field [2]. Univalent function theory had become a distinct discipline at that point. Sanford S. Miller and Petru T. Mocanu proposed the notion of differential subordinations. Miller and Mocanu first introduced the concept of differential subordination in their book published in 2000 [3]. Also, in 2003, they developed the concept of differential superordination as a complementary concept [4].
In 2009, the idea of strong differential superordination was introduced, expanding on the twin notions of differential subordination and differential superordination [4]. Additionally, the authors in [5] provided initial instances of strong differential subordinations and superordinations for analytic functions. Numerous scholars have explored second-order differential superordination and subordination [6,7]. To investigate the characteristics of subordination and superordination, Ibrahim et al. [8,9], in 2015, introduced a novel operator by employing a convolution technique between the Carlson–Shaffer operator and a fractional integral operator. In 2021, Lupas and Oros [10] examined the concept of subordination and its characteristics by utilizing the fractional integral of the confluent Hypergeometric function. The subsequent year witnessed the publication of many papers investigating the features of subordination and subordination (such as [11,12]). Renowned mathematicians, such as Zayed and Bulboacă [13], Attiy et al. [14], Oros and Oros [15], Reem and Kassim [16], Abdulnabi et al. [17], and others have recently carried out considerable advanced investigations based on subordinate techniques (see further details in [9,18,19,20,21,22,23,24,25]).
The family Ξ A denotes the class of all analytic functions in the open unit disk
A = { ξ : ξ < 1   a n d   ξ C }
and Ξ [ a ,   n ] is the subclass of Ξ A which consists of the form functions
F ξ = a + a n ξ n + a n + 1 ξ n + 1 + . . . ,       a C , ξ A , n N .
Let A p be the class of all multivalent in open unit disk A functions of the form
F ξ = ξ p + n = 1 + p a n ξ n ,       ξ A , p N .  
Furthermore, we denote by A = A 1 the class of analytic functions in A and normalized with F ( 0 ) = 0 , F ( 0 ) = 1 . Also, consider S to be the class of univalent function in A . Let S ( ϱ ) , C ( ϱ ) and K be the subclasses of A such that
If F S * : R e ξ F ( ξ ) F ( ξ ) > ϱ , ξ A , 0 < ϱ < 1 .
Then F is starlike function of order ϱ .
If F C : R e 1 + ξ F ξ F ξ > ϱ , ξ A , 0 < ϱ < 1 .
Then F is convex function of order ϱ .
If F K : R e F 1 ξ T ξ > 0 : T C , ξ A .
Then F is close-to-convex function of order ϱ .
Particularly in complex analysis and geometric function theory, symmetry refers to a function’s property of remaining invariant when its variables are substituted with an equal or balanced number [2]. In complex functions, symmetry is when F ( ξ ) is a complex function, it may be said to be symmetric if and only if F ( ξ ) = F ( ξ ), ξ A . Stated otherwise, the function’s values at ξ and ξ are identical. The function is symmetric around the origin in the complex plane because of this characteristic.
Conversely, the symmetry derivatives for the function F ( ξ ) may be altered by this feature when F ξ and F ( ξ ) are symmetric. While F ξ may not change, it is common to see that F ( ξ ) changes in sign.
Here, we review the subordination principles between the two analytic functions in A , F ( ξ ), and T ( ξ ). If there is an analytic function t ( ξ ) in A , with t ( 0 ) = 0 and t ( ξ ) < 1 , ξ   A , such that F ( ξ ) = T ( t ( ξ )), then F ( ξ ) is subordinate to T ( ξ ) , written F ( ξ ) T ( ξ ) , ξ   A . Furthermore, if T is univalent in A , then F ( ξ ) T ξ is equivalent to T 0 = F 0   and F ( A ) ⸦ T ( A ).
The convolution operation (also known as Hadamard product) is a famous mathematical procedure credited to Hadamard and represented by the symbol *. This concept describes an intriguing new way to develop convolution operators and special functions. The formulation is as follows:
For F , T A p , where F ξ is defined by (2) and T ( ξ ) is given by
T ( ξ ) = ξ p + n = 1 + p b n ξ n , ξ A ,
the convolution (Hadamard product) of the function F    and  T , written as F ξ T ξ , yields a new analytic function stated as [2]
F ξ T ξ = ξ p + n = 1 + p a n b n ξ n , ξ A .
For   λ > 0 ,   m N 0 = N 0 and F A p , Goyal et al. [26] introduced the following differential operator:
D λ 0 F ξ = F ξ , D λ 1 F ξ = 1 λ F ξ + λ p ξ F ξ = D λ F ξ .
In general,
D λ m F ξ = ξ p + n = 1 + p 1 + n p 1 λ m a n ξ n , ξ A .
Among the fundamental principles and typical applications of fractional calculus (see for examples [27,28]), the Riemann–Liouville fractional integral operator of order α   C , ( R e ( α ) > 0 ) is a highly utilized operator. It is defined as follows:
( Ι 0 + α F ) x = 1 Γ α 0 x x μ α 1 F μ d μ ,       x > 0   ; R e α > 0 ,
utilizing the well-known Gamma function Γ ( α ) (Euler’s). As an intriguing substitute for the Riemann–Liouville operator Ι 0 + α , consider the Erdelyi–Kober fractional integral operator of order α   C ( R e ( α ) > 0 ) , which is defined as follows:
( Ι 0 + , ϱ , η α F ) x = ϱ x ϱ ( ϱ + η ) Γ α 0 x x μ ϱ 1 + η 1 x ϱ μ ϱ ϱ 1 F μ d μ ,
where x > 0 ; R e α > 0 .
This corresponds roughly to (4) when ϱ 1 = η = 0 , since
Ι 0 + , ϱ , η α F x = x α Ι 0 + α F x ,   x > 0   ; R e α > 0 ,
Here, we examine [29] the integral operator   J p v , ρ ; μ F ( ξ ) , primarily inspired by the particular situation of the expression (5) when x = ϱ = 1 ,   η = v 1 , and α = ρ v and F A p .
J p v , ρ ; η F ξ = Γ ( ρ + η p ) Γ ( v + η p ) Γ ( ρ v   ) 0 x 1 μ ρ v 1 μ v 1 F ξ μ η d μ ,
where v , ρ R , η > 0 , ρ > v > η p , p N and ξ A .
Applying the Eulerian Beta-function integral to evaluate (Euler’s) Gamma function as follows:
K α , β = { 1 μ β 1 μ α 1 d μ                 min {   R e α ,   R e β > 0 Γ α Γ β Γ α + β                                                                                         α , β C \ Z 0   ) ,
Ekram et al. [30] introduced the differintegral operator J p v , ρ ; η F ( ξ ) : A p A p defined by
J p v , ρ ; η F ( ξ ) = ξ p + Γ ( ρ + p η ) Γ ( ρ + p η ) n = 1 + p Γ ( ρ + n η ) Γ ( ρ + n η ) a n ξ n .  
where v , ρ R , η > 0 , ρ > v > η p , p N and ξ A .
If ρ > v implies that J p v , ρ ; η F ξ = F ξ .
Definition 1. 
Suppose  m N 0 = N 0 , λ > 0 , F A p ,   v , ρ R , η > 0 ,   ρ > v > η p ,   p N  and ξ A , we definition the new operator  M λ , p m v , ρ ; η F ( ξ ) = : A p A p ,    where
M λ , p m v , ρ ; η F ( ξ ) = D λ m F ξ J p v , ρ ; η F ( ξ )
M λ , p m v , ρ ; η F ( ξ ) = ξ p + Γ ( ρ + p η ) Γ ( v + p η ) n = 1 + p 1 + n p 1 λ m Γ ( v + n η ) Γ ( ρ + n η ) a n ξ n .
It is readily verified from (6) that
ξ ( M λ , p m v , ρ ; η F ( ξ ) ) = ( λ v η + p ) ( M λ , p m v + 1 , ρ ; η F ( ξ ) ) λ v η ( M λ , p m v , ρ ; η F ( ξ ) )  
Remark 1. 
Special cases of operator  M λ , p m v , ρ ; η F ξ  are indicated below.
1. 
When  P = 1   a n d   ρ = v    the operator  M λ , p m v , ρ ; η F ξ    reduce to  D λ m F ξ  that introduced by Al-Oboudi [31].
2. 
When  P = 1 ,   λ = 1 ,   m = m   a n d   ρ = v    the operator  M λ , p m v , ρ ; η F ξ    reduce to  D 1 m F ξ  that introduced by Salagean operator [32].
3. 
When  m = 0   a n d   λ = 0    the operator  M λ , p m v , ρ ; η F ξ    reduce to  J p v , ρ ; η F ( ξ )  that introduced by Ekram et al. [30].
4. 
When  m = m ,   λ = λ   a n d   ρ = v    the operator  M λ , p m v , ρ ; η F ξ    reduce to  D λ m F ξ  that introduced by Goyal et al. [26].
5. 
When  p = 1 ,   m = 0 ,   ρ > v   a n d   λ = 0 , the operator  M λ , p m v , ρ ; η F ξ  reduce to  Ι ~ v , ρ ; η F ξ  that introduced by Ŕand and Sharma [33].
6. 
When  m = 0 ,   λ = 0 ,   v = β ,   ρ = β + 1 ,   ρ > v  and  η = 1 , the operator  M λ , p m v , ρ ; η F ξ  reduce to  J p β F ( ξ ) β > p  that introduced by Saitoh et al. [34].
7. 
When  m = 0 ,   λ = 0 ,   v = β ,   ρ = α + β δ + 1 ,   ρ > v  and  η = 1 , the operator  M λ , p m v , ρ ; η F ξ  reduce to  N β , p α , δ F ( ξ ) β > p , δ > 0 ,   α δ 1  that introduced by Aouf et al. [35].
8. 
When  m = 0 ,   λ = 0 ,   v = β ,   ρ = α + β ,   ρ > v  and  η = 1 , the operator  M λ , p m v , ρ ; η F ξ  reduce to  Q β , p α F ( ξ ) β > p ,   α 0  that introduced by Liu and Owa [36].
9. 
When  p = 1 ,   m = 0 ,   λ = 0 ,   v = β ,   ρ = α + β δ + 1 ,   ρ > v  and  η = 1 , the operator  M λ , p m v , ρ ; η F ξ  reduce to  N β α F ( ξ ) β > 1 , δ > 0 ,   α δ 1  that introduced by Jung et al. [37].
10. 
When  p = 1 ,   m = 0 ,   λ = 0 ,   v = α 1 ,   ρ = β 1 ,   ρ > v  and  η = 1 , the operator  M λ , p m v , ρ ; η F ξ  reduce to  L ( α , β ) F ( ξ ) β , α C \ Z 0 ,   Z 0 = { 0 , 1 , 2 , }  that introduced by Carlson and Shaffer [38].
11. 
When  p = 1 ,   m = 0 ,   λ = 0 ,   v = v 1 ,   ρ = j ,   ρ > v  and  η = 1 , the operator  M λ , p m v , ρ ; η F ξ  reduce to  I v , j F ( ξ ) v > 0 ,   j 1  that introduced by Choi et al. [39].
12. 
When  P = 1 ,   m = 0 ,   λ = 0 ,   v = α ,   ρ = 0  and  η = 1 , the operator  M λ , p m v , ρ ; η F ξ  reduce to  D F ( ξ ) α > 1  that introduced by Ruscheweyh [40].
13. 
When  m = 0 ,   λ = 0 ,   v = 1 ,   ρ = κ ,   ρ > v  and  η = 1 , the operator  M λ , p m v , ρ ; η F ξ  reduce to  I κ F ( ξ ) κ N 0 F ξ  that introduced by Noor [41].
14. 
When  P = 1 ,   m = 0 ,   λ = 0 ,   v = β ,   ρ = β + 1  and  η = 1 , the operator  M λ , p m v , ρ ; η F ξ  reduce to  J β F ( ξ )    that introduced by Bernadi [42].
15. 
When  P = 1 ,   m = 0 ,   λ = 0 ,   v = 1 ,   ρ = 2  and  η = 1 , the operator  M λ , p m v , ρ ; η F ξ  reduce to  J F ( ξ )    that introduced by Libera [43].
Definition 2. 
[44] Let  ψ  : C 3 × A C , and h ( ξ )  be univalent in  A . If p ( ξ ) is analytic in  A  that fulfill the second-order differential subordination
ψ p ξ , ξ p ξ ,   ξ 2 p ξ ; ξ h ( ξ )
then p ξ is a solution of differential subordination in (8).
Definition 3. 
[44] Let    h  ( ξ )  be univalent in  A ,   and  ψ 1 : C 3 × A C . If p ( ξ ) and ψ 1 p ξ , ξ p ξ ,   ξ 2 p ξ ; ξ are univalent in  A   and p ξ that fulfill the second-order differential superordination
h ( ξ ) ψ 1 p ξ , ξ p ξ ,   ξ 2 p ξ ; ξ ,
then p ξ is a solution of differential superordination in (9).
Definition 4. 
[44] Let  Q  be the families of functions  F   which are injective and analytic on   A ¯ E ( F ) , when E F = ς A : lim ξ ς F ξ = , and F ξ 0 for ς   A E F .
Lemma 1. 
[4] Let  Σ and ϑ   be holomorphic in a domain  D  and let   p 1 ξ be univalent function in  A ,  p 1 ( A ) D ,   with ϑ ( ξ ) 0 when ξ p 1 ( A ). Set
O ξ = ξ   p 1 ξ ϑ ( p 1 ξ )  and   ξ = Σ ( p 1 ξ + O ξ . Assume that
i   O ξ  is starlike in  A .
i i   R e ξ ξ O ξ > 0 ,    ξ   A . If  p 2 ξ  is holomorphic in  A   with  p 2 0 = p 1 0 ,   p 2 ( A ) D  and
Σ p 2 ξ + ξ p 2 ξ ϑ p 2 ξ Σ p 1 ξ + ξ p 1 ξ ϑ p 1 ξ ,
then
p 2 ξ   p 1 ξ
Lemma 2 
[44] Let  p 1 ξ  be convex in  A   and β 1 C , β 2 C with R e 1 + p 1 ξ p 1 ξ > max 0 , R e β 1 β 2 . If p 2 ξ is holomorphic in  A   and
β 1 p 2 ξ + β 2 ξ p 2 ξ β 1 p 1 ξ + β 2 ξ p 1 ξ ,
then
p 2 ξ   p 1 ξ
Lemma 3. 
[4] Let p 1 ξ  be convex univalent in  A   and let and let    and ϑ be holomorphic in a domain   D , p 1 ( A ) D .  Suppose that
i   ξ p 1 ξ ϑ ( p 1 ξ  is starlike univalent in  A .
i i   R e p 1 ξ ϑ p 1 ξ > 0 ,   ξ A .
If  p 2 ξ A p 1 0 , 1   Q , with   p 2 ( A ) D ,  Σ p 2 ξ + ξ p 2 ξ ϑ ( p 2 ξ  is univalent in  A  and
Σ p 1 ξ + ξ p 1 ξ ϑ p 1 ξ Σ p 2 ξ + ξ p 2 ξ ϑ p 2 ξ ,
then
p 1 ξ   p 2 ξ
Lemma 4. 
[4] Let  p 1 ξ  be convex in  A   and R e β > 0 . If p 2 ξ A p 1 0 , 1   Q ,    p 2 ξ + β ξ p 2 ξ   is univalent in  A   and  p 1 ξ + β ξ p 1 ξ p 2 ξ + β ξ p 2 ξ , then p 1 ξ p 2 ξ .
The new operator, which is symmetric in the unit disk A , is used in this article. We use the results of multivalent functions applying operator M λ , p m v , ρ ; η F ( ξ ) to derive many differential subordinations and superordinations. Following that, a few sandwich-type outcomes are also provided.

2. Subordination Results

In this section, we use the differential subordination process to study several convexity criteria of the new operator.
Theorem 1. 
Let  k ( ξ )  be convex univalent in  A , with k 0 = 1 , and suppose the symmetry condition  k ξ = k ( ξ )   for all ξ A . Assume the function  k ( ξ )   satisfies the condition
R e 1 + k ξ k ξ > max ξ A 0 , R e c 1 c 2 .
where  c 1 > 0 , c 2 0 ,   c 2 C   and ξ A .
If  F A , satisfies the subordination
1 c 2 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1                                   + c 2 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) c 1 M λ , p m v , ρ ; η F ( ξ ) ξ p c 1                                   k ξ + c 2 c 1 ξ k ξ ,
then
M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 k ξ .
Proof. 
Consider
q ξ = M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 ,
then
q ξ = c 1 M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 ξ p M λ , p m v , ρ ; η F ( ξ )
ξ p M λ , p m v , ρ ; η F ( ξ ) p ξ p 1 M λ , p m v , ρ ; η F ( ξ ) ξ p 2 = c 1 M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 M λ , p m v , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) p ξ .
We have
ξ q ξ q ξ = c 1 ξ M λ , p m v , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) p .
By using (5), we get
ξ q ξ q ξ = c 1 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + p M λ , p m v , ρ ; η F ( ξ ) λ v + η p η M λ , p m v , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) p = c 1 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) M λ , p m v + 1 , ρ ; η F ( ξ ) 1 ,
c 2 ξ q ξ c 1 = c 2 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 c 2 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 .
And
By using the hypothesis, and apply Lemma 2, when β 1 = 1 and β 2 = c 2 c 1 , then
q ξ + c 2 c 1 ξ q ξ k ξ + c 2 c 1 ξ k ξ
By using the symmetry condition k ξ = k ξ for all ξ A , and the subordination principle, we get
q ξ k ξ .  
Thus
M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 k ξ .  
Corollary 1. 
Let  k ( ξ )  be convex univalent in  A , with k 0 = 1 , and suppose the symmetry condition  k ξ = k ( ξ )   for all ξ A . Assume the function   k ( ξ )   satisfies the condition
R e 1 + k ξ k ξ > max ξ A 0 , R e c 1 c 2 .
where  c 1 > 0 , c 2 0 , c 2 C   and ξ A .
If  F A , satisfies the subordination
1 c 2 v + η p η J p v , ρ ; η F ( ξ ) ξ p c 1                                   + c 2 v + η p η J p v + 1 , ρ ; η F ( ξ ) J p v , ρ ; η F ( ξ ) c 1 J p v , ρ ; η F ( ξ ) ξ p c 1                                   k ξ + c 2 c 1 ξ k ξ ,
then
J p v , ρ ; η F ( ξ ) ξ p c 1 k ξ .
Theorem 2. 
Let  k ( ξ )  be convex univalent in  A , with k 0 = 1 , and suppose the symmetry condition   k ξ = k ( ξ )   for all ξ A . Assume the function  k ( ξ )   satisfies the condition
R e 1 + k ξ k ξ > sup ξ A 0 , R e ( c 1 ) R e ( c 2 ) k ξ k ξ .
where  c 1 > 0 , 0 c 2 C  and ξ A .
If  F A , satisfies the subordination
1 c 2 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1                                   + c 2 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) c 1 M λ , p m v , ρ ; η F ( ξ ) ξ p c 1                                   k ξ + c 2 c 1 ξ k ξ ,
then
M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 k ξ .
Proof. 
Consider
q ξ = M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 ,
then
q ξ = c 1 M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 1 ξ p M λ , p m v , ρ ; η F ξ p ξ p 1 M λ , p m v , ρ ; η F ξ ξ p 2 = c 1 q ξ M λ , p m v , ρ ; η F ξ ξ p M λ , p m v , ρ ; η F ξ ξ M λ , p m v , ρ ; η F ξ .
Thus,
ξ q ξ q ξ = c 1 ξ M λ , p m v , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) p .
Applying the hypothesis, we get
ξ q ξ q ξ = c 1 λ v + η p η . M λ , p m v + 1 , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) p .
Hence,
c 2 ξ q ξ c 1 = c 2 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) c 2 p c 1 .
By using the condition of subordination, we obtain
1 c 2 λ v + η p η q ξ + c 2 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) q ξ k ξ + c 2 c 1 ξ k ξ ,
Now, applying the convexity condition, we get
R e 1 + k ξ k ξ > sup ξ A 0 , R e c 1 c 2 k ξ k ξ .
By using the symmetry condition k ξ = k ξ for all ξ A , and the subordination principle, and applying Lemma 2, when β 1 = 1 and β 2 = c 2 c 1 , we conclude that,
q ( ξ ) k ξ
then
M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 k ξ .
Corollary 2. 
Let  k ( ξ )  be convex univalent in  A , with k 0 = 1 ,   c 1 > 0 , 0 c 2 C  and suppose
R e 1 + k ξ k ξ > max ξ A 0 , R e ( c 1 ) c 2 .
If  F A , satisfies the subordination
M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 k ξ + c 2 c 1 ξ k ξ .
Then the special case in the following are holds.
Example 1. 
If we choose  c 1 = 1 ,   c 2 = 1 , v = ρ ,   m = 0 ,   p = 0    and λ = 1 , then the condition clarify to
M λ , p m v , ρ ; η F ( ξ ) ξ p k ξ ξ k ξ .
In a special case, we use the above conditions to get
F ( ξ ) 1 k ξ ξ k ξ .
Proof. 
Consider F ξ = ξ 1 ξ 2   the transformed of this function becomes
M λ , p m v , ρ ; η F ξ ξ p ξ 1 ξ 2 .
Now we will compare this simplified form:
ξ 1 ξ 2 k ξ ξ k ξ .
Putting k ξ = 1 + ξ 1 ξ .
We derive both sides with respect to   ξ , we get
k ξ = 2 1 ξ 2 .
Therefore,
k ξ ξ k ξ = 1 + ξ 1 ξ 2 ξ 1 ξ 2 .
To put it simply,
1 + ξ 1 ξ 2 ξ 1 ξ 2 = 1 ξ ξ 2 1 ξ 2 .
Thus,
ξ 1 ξ 2 1 ξ ξ 2 1 ξ 2 .
Hence,
F ξ k ξ ξ k ξ ,
Theorem 3. 
Let  k  be convex univalent in  A ,   k 0 = 1 , and k ξ 0 for each ξ A , and assume that  k  satisfies
R e p + ξ t σ ξ p c 2 + ξ ε σ + 1 ξ p c 2 ξ + σ 1 ξ k ξ k ξ + ξ k ξ k ξ > 0 ,  
where  σ ,   ε , t C , c 1 > 0 , 0 c 2 C  and ξ A .
Suppose that  ξ p k ξ σ 1 k ξ  is starlike univalent in  A .
If  F A , satisfies the subordination
N p , m , λ , v , η , ε , c 1 , c 2 ; ξ t + ε k ξ k ξ σ + c 2 k ξ σ 1 k ξ
where
N p , m , λ , v , η , ε , c 1 , c 2 ; ξ = t λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 σ + ε λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 σ + 1 + c 2 c 1 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 σ ξ λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) p ,
then
λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 k ξ
Proof. 
Let H β = ( t + ε β ) β σ and L β = c 2 ( β ) σ 1 , 0 β C , when H β and L β are analytic in C . Then, we get
G ξ = ξ k ξ L k ξ = c 2 ξ p ( k ξ ) σ 1 k ξ ,
and
y ξ = H k ξ + G ξ = t + ε k ξ k ξ σ + c 2 ξ p k ξ σ 1 k ξ .
Since ξ p k ξ σ 1 k ξ starlike. Then G ξ is starlike in A , and
R e y ξ G ξ = R e p + ξ t σ ξ p c 2 + ξ ε σ + 1 ξ p c 2 ξ + σ 1 ξ k ξ k ξ + ξ k ξ k ξ > 0 .
Also, consider
q ξ = λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 ,
then
q ξ = c 1 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1                           λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) 1 ξ ,
we get
t λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 σ + ε λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 σ + 1 = t q ξ σ + ε q ξ σ q ξ = t + ε q ξ q ξ σ
Since
c 1 ξ λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) p = ξ q ξ q ξ ,
that
c 2 c 1 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 σ ξ λ v + η p η M λ , p m v + 1 , ρ ; η F ξ + 1 λ v + η p η M λ , p m v , ρ ; η F ξ λ v + η p η M λ , p m v + 1 , ρ ; η F ξ + 1 λ v + η p η M λ , p m v , ρ ; η F ξ p = c 2   ξ q ξ σ 1 q ξ .
From (10) we get t + ε q ξ q ξ σ + c 2 ξ q ξ σ 1 q ξ
t + ε k ξ k ξ σ + c 2 k ξ σ 1 k ξ ,
by Lemma 1, we obtain q ξ k ξ .
Corollary 3. 
Let  k  be convex univalent in  A ,   k 0 = 1 , and k ξ 0 for each ξ A , and assume that  k  satisfies
R e p + ξ t σ ξ p c 2 + ξ ε σ + 1 ξ p c 2 ξ + σ 1 ξ k ξ k ξ + ξ k ξ k ξ > 0 ,
where  σ ,   ε , t C , c 1 > 0 , 0 c 2 C  and  ξ A .
Suppose that  ξ p k ξ σ 1 k ξ  is starlike univalent in  A .  If  F A , satisfies the subordination
N ( σ , t , ε , v , η , c 1 , c 2 ; ξ ) t + ε k ξ k ξ σ + c 2 k ξ σ 1 k ξ
then
v + η p η J p v + 1 , ρ ; η F ( ξ ) + 1 v + η p η J p v , ρ ; η F ( ξ ) ξ p c 1 k ξ .
Theorem 4. 
Let  k  be convex univalent in  A ,   k 0 = 1 , and k ξ 0 for each ξ A , and assume that  k  satisfies
R e ξ p t σ k ξ σ + ! k ξ + ξ k ξ k ξ + ξ ε σ + 2 k ξ σ + 2 k ξ + ξ k ξ k ξ + σ 2 ξ 2 k ξ k ξ 2 > 0 ,  
where  σ ,   ε , t C , c 1 > 0 , 0 c 2 C  and  ξ A .
Suppose that  ξ p k ξ σ 2 k ξ  is starlike univalent in  A .  If  F A , satisfies the subordination
N p , m , λ , v , η , ε , c 1 , c 2 ; ξ t + ε k ξ k ξ σ + 1 + c 2 k ξ σ k ξ + ξ k ξ k ξ
where
N p , m , λ , v , η , ε , c 1 , c 2 ; ξ                          = t λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) c 1 σ                          + ε λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) c 1 σ + 1                          + c 2 c 1 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) c 1 σ                          × ξ λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) p ,  
then
λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) c 1 k ξ
Proof. 
Let H β = ( t + ε β ) β σ and L β = c 2 ( β ) σ 1 , 0 β C , when H β and L β are analytic in C . Then, we get
G ξ = ξ k ξ L k ξ = c 2 ξ p k ξ σ 1 k ξ + ξ k ξ k ξ ,
and
y ξ = H k ξ + G ξ = t + ε k ξ k ξ σ + c 2 ξ p k ξ σ 1 k ξ + ξ k ξ k ξ .
Since ξ p k ξ σ 2 k ξ is starlike. Then G ξ is starlike in A , and
R e y ξ G ξ = R e ξ p t σ k ξ σ + ! k ξ + ξ k ξ k ξ + ξ ε σ + 2 k ξ σ + 2 k ξ + ξ k ξ k ξ + σ 2 ξ 2 k ξ k ξ 2 > 0 ,
Also, we defined
q ξ = λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) c 1 .
Now, taking the derivative of the function q ξ , getting
q ξ = c 1 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) c 1 × ξ λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) 1 ξ ,
Thus, we get
t q ξ σ + ε q ξ σ + 1 = t + ε q ξ q ξ σ
Since
ξ q ξ q ξ = c 1 ξ λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) 1 ξ ,
that
c 2 c 1 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 σ ξ λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) p = c 2   ξ q ξ σ 1 q ξ .
From (12) we get t + ε q ξ q ξ σ + c 2 ξ q ξ σ 1 q ξ
t + ε k ξ k ξ σ + c 2 k ξ σ 1 k ξ ,
by Lemma 1, we obtain
q ξ k ξ .

3. Superordinations Results

In this section, we use the differential superordination process to study several convexity criteria of the new operator.
Theorem 5. 
Let  k ξ  be convex in  A , with k 0 = 1 ,     c 1 > 0 ,  R e ( c 2 > 0 ) .  If F A ,
M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 Ξ q 0 , 1 Q
and let
1 c 2 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 + c 2 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) c 1 M λ , p m v , ρ ; η F ( ξ ) ξ p c 1
be univalent in A  and satisfies the superordination.
k ξ + c 2 c 1 ξ k ξ                                1 c 2 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1                                + c 2 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) c 1 M λ , p m v , ρ ; η F ( ξ ) ξ p c 1
then
k ξ M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 .
Proof. 
Consider
q ξ = M λ , p m v , ρ ; η F ξ ξ p c 1 ,
then
q ξ = c 1 M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 1 ξ p M λ , p m v , ρ ; η F ( ξ ) ξ p 2 p ξ p 1 M λ , p m v , ρ ; η F ( ξ ) ξ ξ p 2 .
We have
q ξ q ξ = c 1 M λ , p m v , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) 1 ξ ,
with the same steps of Theorem 1, and using the hypothesis, we get
k ξ + c 2 c 1 ξ k ξ k ξ + c 2 c 1 ξ k ξ .
Applying Lemma 4, we get
k ξ M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 .
Corollary 4. 
Let  k ξ  be convex in  A , with k 0 = 1 , c 1 > 0 , R e c 2 > 0 ,
If F A ,
J p n ( θ , λ ) F ξ ξ p c 1 Ξ q 0 , 1 Q
and let
1 c 2 v + η p η J p v , ρ ; η F ( ξ ) ξ p c 1 + c 2 p + λ θ J p v + 1 , ρ ; η F ( ξ ) J p v , ρ ; η F ( ξ ) c 1 J p v , ρ ; η F ( ξ ) ξ p c 1
be univalent in A and satisfies the superordination.
k ξ + c 2 c 1 ξ k ξ 1 c 2 v + η p η J p v , ρ ; η F ( ξ ) ξ p c 1 + c 2 v + η p η J p v + 1 , ρ ; η F ( ξ ) J p v , ρ ; η F ( ξ ) c 1 J p v , ρ ; η F ( ξ ) ξ p c 1
then
k ξ J p v , ρ ; η F ξ ξ p c 1 .
Theorem 6. 
Let  k ( ξ )  be convex univalent in  A , with k 0 = 1 , R e c 1 > 0 ,   R e c 2 > 0 . If F A , satisfies the superordination
M λ , p m v , ρ ; η F ξ ξ p c 1 Ξ q 0 , 1 Q ,
and
1 c 2 λ v + η p η M λ , p m v , ρ ; η F ξ ξ p c 1 + c 2 λ v + η p η M λ , p m v + 1 , ρ ; η F ξ M λ , p m v , ρ ; η F ξ c 1 M λ , p m v , ρ ; η F ξ ξ p c 1 ,
be univalent in A  and satisfies the superordination
k ξ + c 2 c 1 ξ k ξ                               1 c 2 λ v + η p η M λ , p m v , ρ ; η F ξ ξ p c 1                               + c 2 λ v + η p η M λ , p m v + 1 , ρ ; η F ξ M λ , p m v , ρ ; η F ξ c 1 M λ , p m v , ρ ; η F ξ ξ p c 1 ,
then
k ξ M λ , p m v , ρ ; η F ξ ξ p c 1 .
Proof. 
We defined the function q ξ as the following:
q ξ = M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 ,
we derive the above function, we get
q ξ = c 1 M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 1 ξ p M λ , p m v , ρ ; η F ξ p ξ p 1 M λ , p m v , ρ ; η F ξ ξ p 2 = c 1 q ξ M λ , p m v , ρ ; η F ξ ξ p M λ , p m v , ρ ; η F ξ ξ M λ , p m v , ρ ; η F ξ .
Thus,
ξ q ξ q ξ = c 1 ξ M λ , p m v , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) p .
with the same steps of Theorem 2 and using the hypothesis, we get
k ξ + c 2 c 1 ξ k ξ 1 c 2 λ v + η p η q ξ + c 2 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) q ξ ,
applying Lemma 4, when β 1 = 1 and β 2 = c 2 c 1 , we conclude that,
q ( ξ ) k ξ
then
M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 k ξ .
Corollary 5. 
Let  k ( ξ )  be convex univalent in  A , with k 0 = 1 ,   c 1 > 0 , 0 c 2 C and suppose
R e 1 + k ξ k ξ > max ξ A 0 , R e ( c 1 ) c 2 .
If  F A , satisfies the superordination
k ξ + c 2 c 1 ξ k ξ M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 .
Then the special case in the following are holds.
Example 2. 
If we choose  c 1 = 1 ,   c 2 = 1 , v = ρ ,   m = 0 ,   p = 0    and λ = 1 , then the condition clarify to
k ξ ξ k ξ M λ , p m v , ρ ; η F ( ξ ) ξ p .
In a special case, we use the above conditions to get
k ξ ξ k ξ F ( ξ ) 1 .
Proof. 
Consider F ξ = ξ 1 ξ 2 the transformed of this function becomes
ξ 1 ξ 2 M λ , p m v , ρ ; η F ξ ξ p .
Now we will compare this simplified form:
k ξ ξ k ξ ξ 1 ξ 2 .
with the same steps of Corollary 2 and using the hypothesis, we get
k ξ ξ k ξ F ξ
Theorem 7. 
Let  k  be convex univalent in  A ,   k 0 = 1 , and k ξ 0 for each ξ A , and assume that  k  satisfies
R e t σ c 2 k ξ + ε σ + 1 c 2 k ξ k ξ > 0
where  σ ,   ε , t C , 0 c 2 C ,  ξ A  and ξ k ξ σ 1 k ξ is starlike univalent in  A .
If  F A , satisfies the condition
λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 Ξ k 0 , 1   Q ,
and
N p , n , λ , v , η , ε , c 1 , c 2 ; ξ  is univalent in  A ,
If   t + ε   k ξ k ξ σ + c 2 k ξ σ 1 k ξ N σ , t , ε , v , λ , c 1 , c 2 ; ξ ,  then
k ξ λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 .
Proof. 
Let H β = ( t + ε β ) β σ and L β = c 2 ( β ) σ 1 , 0 β C , when H β is analytic in C and L β 0 is analytic in C 0 . Then we get
G ξ = ξ p k ξ L k ξ = c 2 ξ p k ξ σ 1 k ξ .
Since ξ p k ξ σ 1 k ξ starlike, then G ξ is starlike in A , and
R e H k ξ L k ξ = R e t + ε k ξ k ξ σ c 2 k ξ σ 1 = R e t σ c 2 k ξ + ε σ + 1 c 2 k ξ k ξ > 0 .
Now let
q ξ = λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 .
From (10) we get
t + ε k ξ k ξ σ + c 2 k ξ σ 1 k ξ t + ε q ξ q ξ σ + c 2   ξ q ξ σ 1 q ξ .
By Lemma 3 we get k ξ q ξ .
Corollary 6. 
Let  k  be convex univalent in  A ,   k 0 = 1 , and k ξ 0 for each ξ A , and assume that  k  satisfies
R e t σ c 2 k ξ + ε σ + 1 c 2 k ξ k ξ > 0 ,
where  σ ,   ε , t C , 0 c 2 C ,  ξ A  and ξ k ξ σ 1 k ξ is starlike univalent in  A .
If  F A p , satisfies the condition
v + η p η J p v + 1 , ρ ; η F ( ξ + 1 v + η p η J p v , ρ ; η F ( ξ ξ p c 1 Ξ k 0 , 1   Q ,
and
N p , n , λ , θ , ε , c 1 , c 2 ; ξ is univalent in A .
If  t + ε k ξ k ξ σ + c 2 k ξ σ 1 k ξ N σ , t , ε , h μ , μ , c 1 , c 2 ; ξ ,  then
k ξ v + η p η J p v + 1 , ρ ; η F ( ξ + 1 v + η p η J p v , ρ ; η F ( ξ ξ p c 1 .
Theorem 8. 
Let  k  be convex univalent in  A ,   k 0 = 1 , and k ξ 0 for each ξ A , and assume that  k  satisfies
R e ξ p t σ k ξ σ + ! k ξ + ξ k ξ k ξ + ξ ε σ + 2 k ξ σ + 2 k ξ + ξ k ξ k ξ + σ 2 ξ 2 k ξ k ξ 2 > 0 ,  
where  σ ,   ε , t C , c 1 > 0 , 0 c 2 C  and ξ A .
Suppose that  ξ p k ξ σ 2 k ξ  is starlike univalent in  A . If  F A , satisfies the condition
λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) c 1 Ξ k 0 , 1   Q ,
and N p , m , λ , v , η , ε , c 1 , c 2 ; ξ is univalent in  A .  If
t + ε   k ξ k ξ σ + 1 + c 2 k ξ σ k ξ + ξ k ξ k ξ N p , m , λ , v , η , ε , c 1 , c 2 ; ξ ,
then
k ξ λ v + η p η M λ , p m v + 1 , ρ ; η F ξ + 1 λ v + η p η M λ , p m v , ρ ; η F ξ c 1 .
Proof. 
Let H β = ( t + ε β ) β σ and L β = c 2 ( β ) σ 1 , 0 β C , when H β and L β are analytic in C . Then, we get
G ξ = ξ k ξ L k ξ = c 2 ξ p k ξ σ 1 k ξ + ξ k ξ k ξ ,
and
y ξ = H k ξ + G ξ = t + ε k ξ k ξ σ + c 2 ξ p k ξ σ 1 k ξ + ξ k ξ k ξ .
Since ξ p k ξ σ 2 k ξ is starlike. Then G ξ is starlike in A , and
R e y ξ G ξ = R e ξ p t σ k ξ σ + ! k ξ + ξ k ξ k ξ + ξ ε σ + 2 k ξ σ + 2 k ξ + ξ k ξ k ξ + σ 2 ξ 2 k ξ k ξ 2 > 0 ,
Also, we defined
q ξ = λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) c 1 .
From (14) we get
t + ε k ξ k ξ σ + c 2 k ξ σ 1 k ξ t + ε q ξ q ξ σ + c 2   ξ q ξ σ 1 q ξ
by Lemma 3, we obtain
k ξ q ξ .

4. Sandwich Results

The two sandwich theorems that follow are obtained by merging the aforementioned ideas.
Theorem 9. 
Let  k 1   a n d   k 2  be convex univalent in  A , with k 1 0 = k 2 0 = 1 , R e   ( c 2 ) > 0 and
R e 1 + q ξ q ξ > max 0 , R e c 1 c 2 ,
where c 1 > 0 , c 2 0 ,   c 2 C . If F A   and
M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 Ξ 1,1   Q ,
and
  1 c 2 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 + c 2 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) c 1 M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 ,
is univalent in A satisfies
k 1 ξ + c 2 c 1 ξ k 1 ξ 1 c 2 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 + c 2 λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) M λ , p m v , ρ ; η F ( ξ ) c 1 M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 k 2 ξ + c 2 c 1 ξ k 2 ξ ,
then
k 1 ξ M λ , p m v + 1 , ρ ; η F ( ξ ) ξ p c 1 k 2 ξ .
Theorem 10. 
Let  k 1   a n d   k 2  be convex univalent in  A , with k 1 0 = k 2 0 = 1 , and let F A p , satisfies the condition:
λ v + η p η M λ , p m v + 1 , ρ ; η F ξ + 1 λ v + η p η M λ , p m v , ρ ; η F ξ ξ p c 1 Ξ 1,1 Q ,
and
N p , n , λ , v , η , ε , c 1 , c 2 ; ξ ,  is univalent in  A .
If  t + ε   k 1 ξ k 1 ξ σ + c 2 k 1 ξ σ 1 k 1 ξ N p , n , λ , v , η , ε , c 1 , c 2 ; ξ
t + ε   k 2 ξ k 2 ξ σ + c 2 k 2 ξ σ 1 k 2 ξ ,
then
k 1 ξ λ v + η p η M λ , p m v + 1 , ρ ; η F ( ξ ) + 1 λ v + η p η M λ , p m v , ρ ; η F ( ξ ) ξ p c 1 k 2 ξ .

5. Conclusions

In this article, we applied higher-order derivatives we uncovered (derived) in significant relationships between differential subordination and superordination for the new operator M λ , p m v , ρ ; η F ( ξ ) of multivalent functions that are analytic in A . Following the theoretical analysis, particular cases have been evaluated to demonstrate the outcomes of differential subordination. A sandwich-type approach was utilized to achieve differential outcomes. The novel results have the potential to assist mathematicians in exploring further specific outcomes in the geometric function theory domain. Accordingly, our results represent a significant contribution to this field. Furthermore, investigating the symmetry properties of the functions could lead to deriving solutions with certain characteristics for an equation or inequality. By examining inequalities in the context of differential subordination and superordination, exploring the symmetry features of new operator might provide intriguing outcomes. As a part of our future work, we will further investigate the symmetry properties of various functions within the framework of differential subordination and superordination. This investigation could include different types of holomorphic functions, such as harmonic and meromorphic functions, and extend their application to fuzzy differential subordination and superordination.

Author Contributions

The idea was proposed by N.H.S. and improved by N.H.S., A.R.S.J., D.B., L.-I.C., M.D. and A.D. The author, N.H.S., wrote and completed all the calculations. The authors, A.R.S.J., D.B., L.-I.C., M.D. and A.D., checked all the results. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Data Availability Statement

Data is contained within the article.

Conflicts of Interest

The authors declare no conflicts of interest.

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MDPI and ACS Style

Juma, A.R.S.; Hameed Shehab, N.; Breaz, D.; Cotîrlă, L.-I.; Darus, M.; Danciu, A. New Results on Differential Subordination and Superordination for Multivalent Functions Involving New Symmetric Operator. Symmetry 2024, 16, 1326. https://doi.org/10.3390/sym16101326

AMA Style

Juma ARS, Hameed Shehab N, Breaz D, Cotîrlă L-I, Darus M, Danciu A. New Results on Differential Subordination and Superordination for Multivalent Functions Involving New Symmetric Operator. Symmetry. 2024; 16(10):1326. https://doi.org/10.3390/sym16101326

Chicago/Turabian Style

Juma, Abdul Rahman S., Nihad Hameed Shehab, Daniel Breaz, Luminiţa-Ioana Cotîrlă, Maslina Darus, and Alin Danciu. 2024. "New Results on Differential Subordination and Superordination for Multivalent Functions Involving New Symmetric Operator" Symmetry 16, no. 10: 1326. https://doi.org/10.3390/sym16101326

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