Communications on Applied Nonlinear Analysis 

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Certain Properties on Univalent Functions Related to New Linear and 

Integral Operators 

 

Lect. Dr . Bassim Kareem Mihsin 

General Directorate of Education in Karbala, Iraq 

E- mail: bassim_kareem@karbala.edu.iq 

basmk3756@gmail.com 

 

Article History: 

Received: 06-04-2024 

Revised: 28-05-2024 

Accepted: 12-06-2024 

Abstract:  

We are implementing the two new operators, 𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏) and 𝔩𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞 𝑓(𝓏) say linear and 

integral operators respectively, of analaytic functions in open unit disk Ʋ , to pedimenting 

new results for superordination and subordination. We conclude several sandwich-type 

results are the master goal for this paper. 

Keywords: Analytic  functions ,multivalent functions, Hadamart product , Differential 

subordination , Superordination , Sandwich theorems , dominant , Subordinant .   

 

1. Introduction  

Suppose that ℬ to be class functions intailing the following function:  

 𝑓(𝓏) = 𝒶 + 𝒶𝔫 𝓏𝔫 + 𝒶𝔫+1𝓏𝔫+1 + ⋯ (𝒶 ∈ ₵, (𝔫 ∈ ℕ = {1,2, … }; 𝓏 ∈ Ʋ),                     (1,1)  

where 𝑓(𝓏) be analytic to open unit disk Ʋ =  {𝓏 ∈  ₵ ∶  |𝓏| <  1 } . Assume 𝑀[𝒶, 𝔫] be subclass of 

the function 𝑓 ∈ 𝒢  . For 𝒶 ∈ ₵ and 𝔫  to be positive integer number , 

If   𝑓 ∈ 𝒢 defined by (1.1) and 𝑔 ∈ 𝒢 is given by formula: 

𝑓(𝓏) = 𝓏 + ∑ 𝒶𝔫𝓏𝔫∞
𝔫=2  , 𝑔(𝓏) = 𝓏 + ∑ 𝑏𝔫𝓏𝔫∞

𝔫=2  

By ussing convolution of 𝑓 𝑎𝑛𝑑 𝑔 to get    

(𝑓 ∗ 𝑔)(𝓏) = 𝓏 + ∑ 𝒶𝔫𝑏𝔫𝓏𝔫∞
𝔫=2 = (𝑔 ∗ 𝑓)(𝓏) . 

If the functions  𝑓 𝑎𝑛𝑑 𝑔  be analytic functions in  , so  𝑓  be subordinate to  𝑔 in Ʋ , for that we can 

say  𝑓 (𝓏)  ≺  𝑔 (𝓏)  , if existing a Schwarz function ѡ(𝓏) to be analytic within Ʋ to satisfy the 

conditions that |ѡ(𝓏)| <  1 ( 𝓏 ∈  Ʋ) and ѡ(0)  =  0 and where 𝑓 (𝓏)  =  𝑔 (ѡ(𝓏)), ( 𝓏 ∈  Ʋ ) . As 

additional ,to that if 𝑔 be univalent function in Ʋ, so we satisfy the  equivalence relation link  (see 

[15],[16]and[18])    

𝑓 (Ʋ)  ⊂   𝑔(Ʋ), (𝓏 ∈  Ʋ) and 𝑓(𝓏)   ≺ 𝑔(𝓏) ⇔  𝑓 (0)  =  𝑔(0). 

Definition (1.1)[15]: Assume that ȴ(𝑧)  to be  univalent in Ʋ and Ə:₵3 × Ʋ ⇢  ₵ . Let 𝒫(𝓏) be 

analytic in Ʋ to satisfy the  following differential subordination of second – order: 

 Ə (𝒫(𝓏), 𝓏(𝒫(𝓏))
′
, 𝓏2(𝒫(𝓏))

′′
; 𝓏) ≺ ȴ(𝓏),                                         (1.2) 



Communications on Applied Nonlinear Analysis 

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therefore the equation (1.2) of differential subordination have the solution 𝒫(𝓏) . The formula (1.2) 

represent the solution of differential subordination which has  ϥ(𝓏) as a dominant , or more, 

additional to that to be simply dominant, if  𝒫(𝓏) ≺ ϥ(𝓏) to all 𝒫(𝓏) for that will  satisfy (1.2).If 

ϥ̃(𝓏) is univalent dominant which satisfys ϥ̃(𝓏) ≺ ϥ(𝓏) for every dominant ϥ(𝓏) for (1.2) ,  so the 

formula (1.2) will be satisfied by the best dominant.    

Definition (1.2)([15]𝑎𝑙𝑠𝑜 𝑠𝑒𝑒[16]) ∶ Assume that Ə:₵3 × Ʋ ⇢  ₵ and assume the function ȴ(𝓏) 

which be analytic in Ʋ. If Ə (𝒫(𝓏), 𝓏(𝒫(𝓏))
′
, 𝓏2(𝒫(𝓏))

′′
; 𝓏) and the univalent function 𝒫(𝓏) 

within Ʋ where 𝒫(𝓏) satisfy  the following differential Superordination of second–order  : 

ȴ(𝓏) ≺ Ə (𝒫(𝓏), 𝓏(𝒫(𝓏))
′
, 𝓏2(𝒫(𝓏))

′′
; 𝓏),                                              (1.3) 

then the equation (1.3) have the differential superordination solution of (1.3)  say 𝒫(𝓏). Equation 

(1.3) leads to the solution of subordinant ϥ(𝓏), where  must be  analytic function or we can say that 

subordinant will be more  simple when ϥ(𝓏) ≺ 𝒫(𝓏) for every 𝒫(𝓏) halds (1.3). The function ϥ̃(𝓏) 

be univalent subordinant which satisfy ϥ(𝓏) ≺ ϥ̃(𝓏) to  all subordinants ϥ(𝓏) of (1.3) ,will be best 

subordinant . 

In [16], Miller and Macanu have obtained sufficient conditions to functions ȴ, ϥ 𝑎𝑛𝑑 Ə for where the 

implicationts given by:  

ϥ(𝓏)  ≺ 𝒫(𝓏) ⇒ ȴ(𝓏) ≺ Ə(𝒫(𝓏), 𝓏𝒫′(𝓏), 𝓏2𝒫′′(𝓏); 𝓏)                                       (1.4)                

by taking the results ( see [1,2,4,5,6,7,,8,919] ) for obtaining sufficient conditions to normalized 

analytic functions for satisfing: 

ϥ1(𝓏) ≺
𝓏𝑓′(𝓏)

𝑓(𝓏)
≺ ϥ2(𝓏) , 

such that the two  tbe univalent function in Ʋ sayϥ1(𝓏)𝑎𝑛𝑑 ϥ2(𝓏) where ϥ1(0) = ϥ2(0) = 1 . 

Addition to that, El-Ashwah and Aouf  [23] ,Ali et al. [3] ,Atshan and Hadi [6] , Atshan and Ali [4] 

,and Gochhayat [25] derived some superordination and subordination results to analytic functions in  

Ʋ . Recently ,Al-Ameedee et al. [1] ,Atshan et al. [4,5] and Gochhayat [24] got sandwich results to 

some classes of analytic functions  

The function 𝜙𝕥(𝓏, 𝕥, 𝕤) define the following series:  

𝜙𝕥(𝓏, 𝕥, 𝕤) = ∑
𝓏𝑛

(
𝕣+𝑛𝕤

𝕤
)

𝕥 
∞
𝑛=0  . 

Where 𝓏 ∈ Ʋ, 𝕣 ∈ ₵ ∖ 𝑧0
− = {0, −1, −2, … }, 𝕥 ∈ ₵, 𝑅𝑒(𝕥) > 1, 𝓏 ∈ 𝜕Ʋ, 𝕤 ∈ 𝑁 ∖ {1}. 

The following normalized function 𝒥𝕣,𝕤
𝕥 (𝓏) defined as: 

𝒥𝕣,𝕤
𝕥 (𝓏) = (

𝕣

𝕤
+ 1)

𝕥

[𝜙𝕥(𝓏, 𝕥, 𝕤) − (
𝕣

𝕤
)

−𝕥

] = 𝓏 + ∑ (
𝕣+𝕤

𝕣+𝓃𝕤
)

𝕥

 ∞
𝑛=0 𝒶𝑛𝓏𝑛, 𝓏 ∈ Ʋ  ,        (1. 5)  

see more [22]. 



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Swamy [21] defined linear operator ℱ𝕦,𝕧
𝕞 𝑓(𝓏) for  𝕞 ∈ ℕ0 = ℕ ∪ {0}, 𝕦 ∈ 𝑅  , 𝕧 ≥ 0 , 𝕦 + 𝕧 > 0  

and define by: 

ℱ𝕦,𝕧
𝕞 𝑓(𝓏) = 𝓏 + ∑ (

𝕦+𝔫𝕧

𝕦+𝕧
)

𝕞
∞
𝒏=𝟐 𝒶𝑛𝓏𝑛 .                                                                   (1.6) 

See [10,11,12,13,14,17,20,21].  

Definition (1.3): Suppose 𝑓 ∈ 𝒢, 𝓏 ∈ Ʋ, 𝕣 ∈ ₵ ∖ 𝑧0
− = {0, −1, −2, … }, 𝕥 ∈ ₵, , 𝕞 ∈ ℕ0 = ℕ ∪

{0}, 𝕦 ∈ 𝑅  , 𝕤 ∈ 𝑁 ∖ {1}, 𝕧 ≥ 0 , 𝕦 + 𝕧 > 0 , 𝑅𝑒(𝕥) > 1 𝑤ℎ𝑒𝑟𝑒 𝓏 ∈ 𝜕Ʋ and we define new operator:  

𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏): 𝒢 → 𝒢 ,   where    

𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏) = 𝒥𝕣,𝕤

𝕥 (𝓏) ∗ ℱ𝕦,𝕧
𝕞 𝑓(𝓏) 

𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏) = 𝔃 + ∑ (

𝕣+𝕤

𝕣+𝕤𝔫
)

𝕥

(
𝕦+𝖓𝕧

𝕦+𝕧
)

𝕞
∞
𝒏=𝟐 𝒶𝑛𝓏𝑛.                                            (1.7)   

We have from (1.7) that : 

(𝕦 + 𝕧)𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏) = 𝕦𝒯𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞 𝑓(𝓏) − 𝕧𝓏 (𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏))

′

, 𝕧 > 0 .             (1.8)   

We observe that 𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏): 𝒢 → 𝒢 is an integral operator and for f given by (1.2) we have: 

𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏) = 𝔃 + ∑ (

𝕣+𝕤

𝕣+𝕤𝔫
)

𝕥

(
𝕦+𝕧

𝕦+𝔫𝕧
)

𝕞
∞
𝒏=𝟐 𝒶𝑛𝓏𝑛  , 𝓏 ∈ Ʋ .                      (1.9) 

It follows form (1.9) that: 

(𝕦 + 𝕧)𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏) = 𝕦𝔩𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞+1 𝑓(𝓏) + 𝕧𝓏 (𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏))

′

 .              (1.10) 

Now,in this work has been dedicated to derive several superordination, subordination  and sandwich 

results of differential containing new operators 𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏) and 𝔩𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞 𝑓(𝓏). 

 

2. Preliminaries 

Constructing our major results ,some following lemmas will be needed with its references , (see also 

[23]) . 

Definition (2.1)[15]: called by 𝚀  which represent all 𝑓 functions,they must be analytic and one–to–

one on  Ʋ ∖ 𝐸(𝑓), where   𝐸(𝑓) = { 𝜁 ∈ 𝜕Ʋ: 𝑙𝑖𝑚
𝓏→𝜁

𝑓(𝓏) = ∞} and Ʋ̅ = Ʋ ∪ {𝓏 ∈ 𝜕Ʋ},  in addition to 

that  𝑓′(𝜁) ≠ 0 of 𝜁 ∈ 𝜕Ʋ\𝐸(𝑓). More that, we consider a subclass of 𝚀 to 𝑓(0) = 𝒶 is called 

𝚀(𝒶) 𝑤ℎ𝑒𝑟𝑒 𝚀(1) = 𝚀1𝑎𝑛𝑑 𝚀(0) = 𝚀0.   

Lemma  (2.2) [15]: Assume the function ϥ(𝓏) be univalent and convex in Ʋ assume 𝛽 ∈   ₵ ∕ {0} 

𝛼 ∈ ₵,  , 𝛽 ≠ 0  , and suppose  

𝑅𝑒 {1 +
𝓏ϥ" (𝓏)

ϥ′(𝓏)
} > max {0, −𝑅𝑒 (

α

β
)} . 

If  𝑃(𝓏) is analytic function in  Ʋ, and 



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𝛼𝒫 (𝓏) +  𝛽𝓏(𝒫(𝓏))
′

≺  𝛼ϥ (𝓏) +  𝛽𝓏(ϥ(𝓏))
′
 ,                                                   (2.1) 

hence ϥ(𝓏)  will be best dominant and 𝒫(𝓏)  ≺  ϥ(𝓏)  .  

Lemma (2.3) [15]: Let ϥ(𝓏) is belongs in Ʋ with 𝑞(0)  =  1,where ϥ(𝓏) is convex and univalent. Let 

𝑅𝑒 (𝛽)  >  0 and 𝛽 ∈  ₵ . If 𝒫(𝓏) ∈ 𝑀[ϥ(0), 1]  ∩  𝚀  and  𝒫(𝓏) +  𝛽𝓏(𝒫(𝓏))
′
   is univalent in Ʋ, 

then ϥ(𝓏)  +  𝛽𝓏(ϥ(𝓏))′   ≺  𝒫(𝓏)  +  𝛽𝓏(𝒫(𝓏))′ , which implies that ϥ(𝓏) ≺  𝒫(𝓏) 𝑎𝑛𝑑 ϥ(𝓏) will 

be best subordinant . 

3. Subordination Results: 

Theorem (3.1): Let ϥ(𝓏) is belongs in Ʋ with ϥ(0)  =  1,where ϥ(𝓏) is convex and univalent.    Let 

𝜉 ∈ ₵
∗
,𝜇, 𝕧 > 0, 𝕦 real number such that 𝕦 + 𝕧 > 0 andSuppose that ϥ(𝓏) satisfies : 

𝑅𝑒 {1 +
𝑧ϥ " (𝓏)

ϥ′(𝓏)
} > max {0, 𝑅𝑒 (

𝜇(𝕦+𝕧)

𝜉𝕧
)}  .                                                            (3.1) 

Let  𝑓 ∈ 𝒢 holds the subordination 

𝔇(𝕞, 𝜉, 𝜇, 𝕦, 𝕧) ≺ ϥ(𝓏) +
𝜉𝕧

𝜇(𝕦+𝕧)
𝓏ϥ′(𝓏),                                                              (3.2) 

Where 

𝔇(𝕞, 𝜉, 𝜇, 𝕦, 𝕧) = (1 − 𝜉) (
𝒯𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞 𝑓(𝓏)

𝓏
)

𝜇

+ 𝜉 (
𝒯𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞 𝑓(𝓏)

𝓏
)

𝜇

+ (
𝒯𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞+1 𝑓(𝓏)

𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏)

) ,         (3.3) 

   

then  

  (
𝒯𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞 𝑓(𝓏)

𝓏
)

𝜇

≺ ϥ(𝓏),                                                                                        (3.4) 

and the equation (3.2) have the best dominant say ϥ . 

Proof: Put  

𝒮(𝓏)  = (
𝒯𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞 𝑓(𝓏)

𝓏
)

𝜇

, 𝓏 ∈ Ʋ.                                                                        (3.5) 

Hence differentiate  (3.5) logarithmically according to 𝓏, and taking identity (1.8) in resultant 

equation , to get 

𝓏 𝒮′(𝓏)

𝒮(𝓏)
=  𝜇 (

(𝕦+𝕧)

𝕧
) (

𝓏𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏) 

𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏) 

− 1),and which can be written as 

𝕧

𝜇(𝕦 + 𝕧)
𝓏𝒮′(𝓏) = (

𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏)

𝓏
)

𝜇

(
𝒯𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞+1 𝑓(𝓏) 

𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏) 

− 1) 

Thus the equation of subordination which is represented by (3.2) be equivalent by  

𝒮(𝓏) +
𝜉𝕧

𝜇(𝕦+𝕧)
𝓏𝒮′(𝓏) ≺ ϥ(𝓏) +

𝜉𝕧

𝜇(𝕦+𝕧)
𝓏ϥ′(𝓏) . 

By apply  Lemma (2.2)2.1  when  𝜎 =
𝜉𝕧

𝜇(𝕦+𝕧)
   , the proof of theorem(3.1) is complete. 



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Now, in theorem above,we put , ϥ(𝓏) =  
1+𝐴𝓏

1+𝐵𝓏
 ,so we get the following corollary. 

Corollary (3.1): Assume𝜉, 𝐴, 𝐵 ∈ ₵ , 𝐴 ≠ 𝐵, |𝐵| < 1, 𝜇 > 0, 𝑅𝑒(𝜉) > 0 𝑎𝑛𝑑 𝕦 real number such that 

𝕦 + 𝕧 > 0 ,if  𝑓 ∈ 𝒢 satisfy the recent subordination case :   

  𝔇(𝕞, 𝜉, 𝜇, 𝕦, 𝕧) ≺ (
1+𝐴𝓏

1+𝐵𝓏
) +

𝜉𝕧

𝜇(𝕦+𝕧)
(

(𝐴+𝐵)𝓏

(1+𝐵𝓏)2) , such that 𝔇(𝕞, 𝜉, 𝜇, 𝕦, 𝕧) known by (3.3) ,then      

(
𝒯𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞 𝑓(𝓏)

𝓏
)

𝜇

≺
1 + 𝐴𝓏

1 + 𝐵𝓏
  , 

and will be best  dominant say  
1+𝐴𝓏

1+𝐵𝓏
  . 

Now ,put 𝕞 = 0  in the Theorem above ,to get a new result.  

Corollary (3.2): Let ϥ(𝓏) is belongs in Ʋ with ϥ(0)  =  1,where ϥ(𝓏) is univalent. Let 𝜉 ∈ ₵
∗
,𝜇, 𝕧 >

0, 𝕦 real number such that 𝕦 + 𝕧 > 0 andSuppose that ϥ(𝓏) satisfies : 

Let  𝑓 ∈ 𝒢 holds the subordination 

𝔇1(0, 𝜉, 𝜇, 𝕦, 𝕧) ≺ ϥ(𝓏) +
𝜉𝕧

𝜇(𝕦 + 𝕧)
𝓏ϥ′(𝓏), 

Where 

𝔇1(0, 𝜉, 𝜇, 𝕦, 𝕧) = (1 − 𝜉) (
𝒯𝕣,𝕤,𝕥,

0 𝑓(𝓏)

𝓏
)

𝜇

+ 𝜉 (
𝒯𝕣,𝕤,𝕥,,

0 𝑓(𝓏)

𝓏
)

𝜇

+ (
𝒯𝕣,𝕤,𝕥,,𝕦,𝕧

1 𝑓(𝓏)

𝑓(𝓏)
) , 

then  

(
𝒯𝕣,𝕤,𝕥,

0 𝑓(𝓏)

𝓏
)

𝜇

≺ ϥ(𝓏), 

and the equation (3.2) have the best dominant say ϥ .  

Now ,put 𝕦 = 𝕧 = 1  in the Theorem above ,to get a new result. 

Corollary (3.3): Let ϥ(𝓏) is belongs in Ʋ with ϥ(0)  =  1,where ϥ(𝓏) is univalent. Let 𝜉 ∈ ₵
∗
,𝜇 > 0,  

and Suppose that (3.1) holds. Let  𝑓 ∈ 𝒢 holds the subordination 

𝔇2(𝕞, 𝜉, 𝜇, 1,1) ≺ ϥ(𝓏) +
𝜉𝕧

𝜇(𝕦 + 𝕧)
𝓏ϥ′(𝓏), 

Where 

𝔇2(𝕞, 𝜉, 𝜇, 1,1) = (1 − 𝜉) (
𝒯𝕣,𝕤,𝕥,,1,1

𝕞 𝑓(𝓏)

𝓏
)

𝜇

+ 𝜉 (
𝒯𝕣,𝕤,𝕥,,1,1

𝕞 𝑓(𝓏)

𝓏
)

𝜇

+ (
𝒯𝕣,𝕤,𝕥,,1,1

𝕞+1 𝑓(𝓏)

𝒯𝕣,𝕤,𝕥,,1,1
𝕞 𝑓(𝓏)

) , 

then  

(
𝒯𝕣,𝕤,𝕥,,1,1

𝕞 𝑓(𝓏)

𝓏
)

𝜇

≺ ϥ(𝓏), 

and the equation (3.2) have the best dominant say ϥ .    



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By the same way we can take Theorem (3.1), to prove the following theorems by using the 

identity(1.10). 

Theorem (3.2): Let ϥ(𝓏) is belongs in Ʋ with ϥ(0)  =  1,where ϥ(𝓏) is convex and univalent. Let 

𝜉 ∈ ₵
∗
,𝕧 > 0, 𝕦 real number such that 𝕦 + 𝕧 > 0 andSuppose that ϥ(𝓏) satisfies : 

𝑅𝑒 {1 +
𝑧ϥ " (𝓏)

ϥ′(𝓏)
} > max {0, −𝑅𝑒 (

𝜇

𝜉
)}  .                                                      (3.6)                                                                                   

Let  𝑓 ∈ 𝒢 holds the subordination 

𝔏(𝜇, 𝕞, 𝕦, 𝕧, 𝜉) ≺ ϥ(𝓏) +
𝜉

𝜇
𝓏ϥ′(𝓏),                                                     (3.7)           

where 

 𝔏(𝜇, 𝕞, 𝕦, 𝕧, 𝜉) = (1 − 𝜉) (
𝕦+𝕧

𝕧
) (

𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏)

𝓏
)

𝜇

+ 𝜉 (
𝕦+𝕧

𝕧
) (

𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏)

𝓏
)

𝜇

+ (
𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏)

𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏)

),             ( 3.8)       

   

then  

  (
𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏)

𝓏
)

𝜇

≺ ϥ(𝓏),                                                                (3.9) 

and the equation (3.7) have the best dominant say ϥ . 

Proof: Put  

𝒮(𝓏)  = (
𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏)

𝓏
)

𝜇

, 𝓏 ∈ Ʋ.                                                       (3.10) 

Hence differentiate  (3.10) logarithmically according to 𝓏, and taking identity (1.10) in resultant 

equation , to get 

𝓏 𝒮′(𝓏)

𝒮(𝓏)
=  𝜇 (

(𝕦+𝕧)

𝕧
) (

𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏) 

𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏) 

− 1),and which can be written as 

𝕧

𝜇(𝕦+𝕧)
𝓏𝒮′(𝓏) = (

𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏)

𝓏
)

𝜇

(
𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏) 

𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏) 

− 1)  . 

Thus the equation of subordination which is represented by (3.7) be equivalent by  

𝒮(𝓏) +
𝜉

𝜇
𝓏𝒮′(𝓏) ≺ ϥ(𝓏) +

𝜉

𝜇
𝓏ϥ′(𝓏) . 

By apply  Lemma (2.2) when  𝜎 =
𝜉

𝜇
   , the proof of theorem(3.2) is complete  therefore ,we put 𝕧 =

1  in above theorem to obtain the results below.   

Corollary(3.4): Let ϥ(𝓏) is belongs in Ʋ with ϥ(0)  =  1,where ϥ(𝓏) is univalent . Let 𝜉 ∈ ₵
∗
,𝜇 >

0, 𝕦 real number such that 𝕦 + 𝕧 > 0 andSuppose that ϥ(𝓏) satisfies : 

𝑅𝑒 {1 +
𝑧ϥ " (𝓏)

ϥ′(𝓏)
} > max {0, −𝑅𝑒 (

𝜇(𝕦+𝕧)

𝜉𝕧
)}  .                                    (3.11) 

Let  𝑓 ∈ 𝒢 holds the subordination 



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𝔏1(𝜇, 𝕞, 𝕦, 𝕧, 𝜉) ≺ ϥ(𝓏) +
𝜉

𝜇
𝓏ϥ′(𝓏),                                                         (3.12)           

Where 

𝔏1(𝜇, 𝕞, 𝕦, 𝕧, 𝜉) = (1 − 𝜉) (
𝕦+𝕧

𝕧
) (

𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏)

𝓏
)

𝜇

+ 𝜉 (
𝕦+𝕧

𝕧
) (

𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏)

𝓏
)

𝜇

+ (
𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏)

𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏)

)          (3.13)                                    

   

then  

  (
𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏)

𝓏
)

𝜇

≺ ϥ(𝓏),                                                                    (3.14) 

and the equation (3.7) have the best dominant say ϥ . 

4. Superordination Results : 

Theorem (4.1) : Assume that the function ϥ are univalent and convex in Ʋ 𝑤𝑖𝑡ℎ ϥ(0) =  1, 𝑅𝑒(𝜉) >

0, 𝜉 ∈ ₵, 𝜇, 𝕧 > 0, 𝕦 ∈ 𝑅 such that 𝕦 + 𝕧 > 0. if  𝑓 ∈ 𝒢 , where   

(
𝒯𝕣,𝕤,𝕥,,1,1

𝕞 𝑓(𝓏)

𝓏
)

𝜇

∈ 𝑀[ϥ(0), 1] ∩ 𝚀   .                                                              (4.1) 

If 𝔇(𝕞, 𝜉, 𝜇, 𝕦, 𝕧)is univalent function in Ʋ  as defined by (3.3),and satisfies the superordination case 

below; 

  ϥ(𝓏) +   
𝜉𝕧

𝜇(𝕦+𝕧)
 𝓏 ϥ′(𝓏) ≺ 𝔇(𝕞, 𝜉, 𝜇, 𝕦, 𝕧),                                                    (4.2)           

then  ϥ(z) ≺ (
𝒯𝕣,𝕤,𝕥,,1,1

𝕞 𝑓(𝓏)

𝓏
)

𝜇

,                                                                 (4.3)  

and ϥ(z) will be best subordination .  

Proof: Put  

𝒮(𝓏)  = (
𝒯𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞 𝑓(𝓏)

𝓏
)

𝜇

, 𝓏 ∈ Ʋ.                                                        (3.4) 

Hence differentiate  (3.4) logarithmically according to 𝓏, and taking identity (1.8) in resultant 

equation , to get 

𝓏 𝒮′(𝓏)

𝒮(𝓏)
=  𝜇 (

(𝕦+𝕧)

𝕧
) (

𝓏𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏) 

𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏) 

− 1),and which can be written as 

𝕧

𝜇(𝕦 + 𝕧)
𝓏𝒮′(𝓏) = (

𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏)

𝓏
)

𝜇

(
𝒯𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞+1 𝑓(𝓏) 

𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏) 

− 1) 

Thus the equation of subordination which is represented by (4.2) be equivalent by  

𝒮(𝓏) +
𝜉𝕧

𝜇(𝕦+𝕧)
𝓏𝒮′(𝓏) ≺ 𝑝(𝓏) +

𝜉𝕧

𝜇(𝕦+𝕧)
𝓏𝑝′(𝓏) . 

By apply  Lemma (2.2), the proof of theorem(4.1) is complete. 

Now, in theorem above,we put , 𝕞 =  0 ,so we get the following corollary. 



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Corollary (4.1) : Let ϥ(𝓏) is belongs in Ʋ with ϥ(0)  =  1,where ϥ(𝓏) is convex and 

univalent, 𝑅𝑒(𝜉) > 0, 𝜉 ∈ ₵, 𝜇, 𝕧 > 0, 𝕦 ∈ 𝑅. if  𝑓 ∈ 𝒢 , where   

(
𝒯𝕣,𝕤,𝕥 

𝕞 𝑓(𝓏)

𝓏
)

𝜇

∈ 𝑀[ϥ(0), 1] ∩ 𝚀   .                                                          (4.5) 

If 𝔇1(0, 𝜉, 𝜇, 𝕦, 𝕧)is univalent function as defined by (3.3),and satisfies the superordination case 

below; 

  ϥ(𝓏) +   
𝜉𝕧

𝜇(𝕦+𝕧)
 𝓏 ϥ′(𝓏) ≺ 𝔇1(0, 𝜉, 𝜇, 𝕦, 𝕧),                                                  (4.6)           

then  ϥ(z) ≺ (
𝒯𝕣,𝕤,𝕥

𝕞 𝑓(𝓏)

𝓏
)

𝜇

,                                                                    (4.7)  

and ϥ(z) will be best subordination .  

Now, in theorem above,we put , 𝕧 =  1 ,so to obtain corollary below. 

Corollary (4.2) :Let ϥ(𝓏) is belongs in Ʋ with ϥ(0)  =  1,where ϥ(𝓏) is convex and 

univalent, 𝑅𝑒(𝜉) > 0, 𝜉 ∈ ₵, 𝜇 > 0,such that 𝕦 + 𝕧 > 0. if  𝑓 ∈ 𝒢 , where   

(
𝒯𝕣,𝕤,𝕥,,𝕦,1

𝕞 𝑓(𝓏)

𝓏
)

𝜇

∈ 𝑀[ϥ(0), 1] ∩ 𝚀   .                                                        (4.8) 

If 𝔇(𝕞, 𝜉, 𝜇, 𝕦, 1)is univalent function as defined by (3.3),and satisfies the superordination case 

below; 

  ϥ(𝓏) +   
𝜉𝕧

𝜇(𝕦+𝕧)
 𝓏 ϥ′(𝓏) ≺ 𝔇3(𝕞, 𝜉, 𝜇, 𝕦, 1),                                               (4.9)    

where 

𝔇3(𝕞, 𝜉, 𝜇, 𝕦, 1) = (1 − 𝜉) (
𝒯𝕣,𝕤,𝕥,,𝕦1

𝕞 𝑓(𝓏)

𝓏
)

𝜇

+ 𝜉 (
𝒯𝕣,𝕤,𝕥,,𝕦,1

𝕞 𝑓(𝓏)

𝓏
)

𝜇

+ (
𝒯𝕣,𝕤,𝕥,,𝕦,1

𝕞+1 𝑓(𝓏)

𝒯𝕣,𝕤,𝕥,,𝕦,1
𝕞 𝑓(𝓏)

)          

then  ϥ(z) ≺ (
𝒯𝕣,𝕤,𝕥,,𝕦,1

𝕞 𝑓(𝓏)

𝓏
)

𝜇

,                                                           (4.10)  

and ϥ(z) will be best subordination . 

Theorem (4.2) : Assume that the function ϥ are univalent and convex in Ʋ 𝑤𝑖𝑡ℎ ϥ(0) =  1, 𝑅𝑒(𝜉) >

0, 𝜉 ∈ ₵, 𝜇, 𝕧 > 0,such that 𝕦 + 𝕧 > 0. if  𝑓 ∈ 𝒢 , where   

(
𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏)

𝓏
)

𝜇

∈ 𝑀[ϥ(0), 1] ∩ 𝚀   .                                            (4.11) 

If 𝔏(𝜇, 𝕞, 𝕦, 𝕧, 𝜉)is univalent function as defined by (3.8),and satisfies the superordination case 

below; 

  ϥ(𝓏) +   
𝜉

𝜇
 𝓏 ϥ′(𝓏) ≺ 𝔏(𝜇, 𝕞, 𝕦, 𝕧, 𝜉),                                               (4.12)           

then                                             ϥ(z) ≺ (
𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞+1 𝑓(𝓏)

𝓏
)

𝜇

,                                                                  (4.13)  

and ϥ(z) will be best subordination .  



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thus by taking Lemma (2.3) ,we obtain result that required . 

Now, in theorem above,we put , 𝕧 =  1 ,so to obtain corollary below. 

Corollary (4.3) :Let ϥ(𝓏) is belongs in Ʋ with ϥ(0)  =  1,where ϥ(𝓏) is convex and 

univalent, 𝑅𝑒(𝜉) > 0, 𝜉 ∈ ₵, 𝜇 > 0, 𝕦 𝑏𝑒 real number. if  𝑓 ∈ 𝒢 , where   

(
𝔩𝕣,𝕤,𝕥,,𝕦,1
𝕞+1 𝑓(𝓏)

𝓏
)

𝜇

∈ 𝑀[ϥ(0), 1] ∩ 𝚀   .                                        (4.14) 

If 𝔏1(𝜇, 𝕞, 𝕦, 1, 𝜉)is univalent function as defined by (3.8),and satisfies the superordination case 

below; 

  ϥ(𝓏) +   
𝜉

𝜇
 𝓏 ϥ′(𝓏) ≺ 𝔏1(𝜇, 𝕞, 𝕦, 1, 𝜉),                                               (4.15)           

then                                                    ϥ(z) ≺ (
𝔩𝕣,𝕤,𝕥,,𝕦,1
𝕞+1 𝑓(𝓏)

𝓏
)

𝜇

,                                                           (4.16)  

and ϥ(z) will be best subordination .  

thus by taking Lemma (2.3) ,we obtain result that required . 

5. Sandwich  Results 

Joining Theorems (3.1) and (4.1) , in order to get sandwich theorem (5.1) . 

Theorem (5.1) : Assume the two convex functions in Ʋ sayϥ1(𝓏) 𝑎𝑛𝑑 ϥ2(𝓏)  together  with ϥ1(0) =

ϥ2(0) = 1.Suppose that 𝑅𝑒(𝜉) > 0, 𝜉 ∈ ₵, 𝜇, 𝕧 > 0, 𝕦 be real number such that 𝕦 + 𝕧 > 0. .I𝑓 𝑓 ∈

𝒢 ,where 

(
𝒯𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞 𝑓(𝓏)

𝓏
)

𝜇

 ∈ 𝑀[ϥ(0), 1] ∩ 𝚀 , 

and 𝔇(𝕞, 𝜉, 𝜇, 𝕦, 𝕧) which is given by (3.3) be univalent function and holds 

ϥ1(𝓏) +
𝜉𝕧

𝜇(𝕦+𝕧)
𝓏ϥ′

1
(𝓏) ≺ 𝔇(𝕞, 𝜉, 𝜇, 𝕦, 𝕧) ≺ ϥ2(𝓏) +

𝜉𝕧

𝜇(𝕦+𝕧)
𝓏ϥ′

2
(𝓏),          (5.1) 

implies ϥ1(𝓏) ≺  (
𝒯𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞 𝑓(𝓏)

𝓏
)

𝜇

≺ ϥ2(𝓏),with ϥ1(𝓏)  best subordinant and ϥ2(𝓏) best dominant (5.1) 

respectively. 

Joining Theorems (3.2) and (4.2) , in order to get sandwich theorem (5.2) . 

Theorem (5.2) : Assume the two univalent convex functions in Ʋ sayϥ1(𝓏) 𝑎𝑛𝑑 ϥ2(𝓏)  together  

with ϥ1(0) = ϥ2(0) = 1. Suppose that 𝑅𝑒(𝜉) > 0, 𝜉 ∈ ₵, 𝜇, 𝕧 > 0, 𝕦  real number such that 𝕦 +

𝕧 > 0.  

.I𝑓 𝑓 ∈ 𝒢 ,where(
𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏)

𝓏
)

𝜇

∈  𝑀[ϥ(0),1] ∩ 𝚀 ,   

Assume univalent function in Ʋ say 𝔏(𝜇, 𝕞, 𝕦, 𝕧, 𝜉),  

 then 

 ϥ1(𝓏) +
𝜉

𝜇
ϥ2

1
(𝓏) ≺  𝔏(𝜇, 𝕞, 𝕦, 𝕧, 𝜉) ≺  ϥ2(𝓏) +

𝜉

𝜇
𝓏ϥ2

′ (𝓏) .                                (5.2)                                               



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implies  ϥ1(𝓏) ≺ (
𝔩𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏)

𝓏
)

𝜇

≺ ϥ2(𝓏)  , and with ϥ1(𝓏)  best subordinant and ϥ2(𝓏) best dominant 

(5.2) respectively . 

6.  Conclusions 

The main aim of our present work is dedecated to give a new results connected by new operators, 

𝒯𝕣,𝕤,𝕥,,𝕦,𝕧
𝕞 𝑓(𝓏) and 𝔩𝕣,𝕤,𝕥,,𝕦,𝕧

𝕞 𝑓(𝓏) say linear and integral operators respectively for univalent function in 

open unit disc Ʋ ,by using differential superordinations and subordinations. The introduced results 

have properties of differential subordinations which are analogous to differential superordination 

properties in sandwich theorm,in supplement to that the results of this work include with a new ideas, 

which can be applied on analytic and multivalent functions theory. 

Refrences 

[1] S. A. Al- Ameedee ,W. G. Atshan and F. A. Al-Maamori, On sandwich results of univalent functions defined by 

a linear operator, Journal of Interdisciplinary Mathematics, 23(4)(2020), 803-809 . 

[2] S. A. Al-Ameedee, W. G. Atshan and F. A. Al-Maamori ,Some new results of differential subordiantions for 

higher-order derivatives of multivalent functions,  Journal of Physics : Conference Series ,1804 (2021) 012111 

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