# A New Parametric Differential Operator of p-Valently Analytic Functions

Authors: Samir B. Hadid; Rabha W. Ibrahim; G. Murugusundaramoorthy
Journal: Journal of Function Spaces
 (2021)

---

## Abstract

Newly, numerous investigations are considered utilizing the idea of parametric operators (integral and differential). The objective of this effort is to formulate a new 2D-parameter differential operator (PDO) of a class of multivalent functions in the open unit disk. Consequently, we formulate the suggested operator in some interesting classes of analytic functions to study its geometric properties. The recognized class contains some recent works.


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## Body

## 1. Introduction

In analysis, a parametric differential operator (PDO) is a differential operator of a dependent variable with respect to another dependent variable that is engaged when both variables formulate on an independent third variable, typically supposed as “time.” We shall use this idea to consider the PDO of a complex variable to discuss its properties in the opinion of the geometric function theory (GFT). The field of differential operators is investigated in GFT early by the well-known Salagean differential operator and the Ruscheweyh derivative. Later, these operators are generalized by different types of parameters using a 1D-parameter fractional differential operator [1] and 2D-parameter fractional differential operator [2]. Recently, using the class of normalized functions ψ∈Σ(1)ψζ=ζ+∑n=2∞ψnζn,ζ∈Δ≔ζ∈ℂ:ζ<1.Ibrahim and Jay [3] presented PDO of the following form: for α∈0,1(2)P0ψζ=ψζ,Pαψζ=ρ1α,ζρ1α,ζ+ρ0α,ζψζ+ρ0α,ζρ1α,ζ+ρ0α,ζζψ′ζ.The functionsρ1,ρ0:0,1×Δ⟶Δ are analytic in Δ satisfying ρ1α,ζ≠−ρ0α,ζ.
(3)limα⟶0ρ1α,ζ=1,limα⟶1ρ1α,ζ=0,ρ1α,ζ≠0,∀ζ∈Δ,α∈0,1,(4)limα⟶0ρ0α,ζ=0,limα⟶1ρ0α,ζ=1,ρ0α,ζ≠0,∀ζ∈Δ,α∈0,1.More studies are given by Ibrahim and Baleanu [4, 5] using (2) to present a hybrid diff-integral operator and a quantum hybrid operator, respectively.In this effort, we generalize (2) by considering another class of analytic functions denoting by Σ℘ and constructing by
(5)ψζ=ζ℘+∑n=℘+1∞ψnζn,℘∈ℕ,which are analytic in Δ. Recently, different investigations are presented studying the geometric behavior of this class (see [6–9]).The Hadamard product [10, 11] for two functions in Σ℘ is given by the series
(6)ψ∗φζ=ζ℘+∑n=℘+1∞ψnζn∗ζ℘+∑n=℘+1∞φnζn=ζ℘+∑n=℘+1∞ψnφnζn∈Σ℘.Definition 1.
For a function
ψ
∈
Σ
℘
,
PDO is defined as follows:
(7)
Q
0
ψ
ζ
=
ψ
ζ
Q
α
ψ
ζ
=
ρ
1
α
,
ζ
ρ
1
α
,
ζ
+
ρ
0
α
,
ζ
ψ
ζ
+
ρ
0
α
,
ζ
ρ
1
α
,
ζ
+
ρ
0
α
,
ζ
ζ
℘
ψ
′
ζ
=
ζ
℘
+
∑
n
=
℘
+
1
∞
ψ
n
ρ
1
α
,
ζ
+
n
/
℘
ρ
0
α
,
ζ
ρ
1
α
,
ζ
+
ρ
0
α
,
ζ
ζ
n
,
Q
2
α
ψ
ζ
=
Q
Q
α
ψ
ζ
=
ζ
℘
+
∑
n
=
℘
+
1
∞
ψ
n
ρ
1
α
,
ζ
+
n
/
℘
ρ
0
α
,
ζ
ρ
1
α
,
ζ
+
ρ
0
α
,
ζ
2
ζ
n
,
⋮
Q
m
α
ψ
ζ
=
Q
α
Q
m
−
1
α
ψ
ζ
=
ζ
℘
+
∑
n
=
℘
+
1
∞
ψ
n
ρ
1
α
,
ζ
+
n
/
℘
ρ
0
α
,
ζ
ρ
1
α
,
ζ
+
ρ
0
α
,
ζ
m
ζ
n
,
Q
m
α
ψ
ζ
=
ζ
℘
+
∑
n
=
℘
+
1
∞
ψ
n
Λ
n
m
ζ
n
,
ζ
∈
Δ
,
℘
∈
ℕ
,
α
∈
0
,
1
,
m
∈
ℕ
,
where(8)Λn=ρ1α,ζ+n/℘ρ0α,ζρ1α,ζ+ρ0α,ζ.
ρ
1 and ρ0 are defined in (3) and (4), respectively.Remark 2.
(i)
It is clear thatQmαψζ∈Σ℘, and it is a generalization of (2) (℘=1)
(ii)
The integral operator that corresponds toQmαψζ is
(9)
L
m
α
ψ
ζ
=
ζ
℘
+
∑
n
=
℘
+
1
∞
ψ
n
Λ
n
m
ζ
n
.
ζ
∈
Δ
,
℘
∈
ℕ
,
α
∈
0
,
1
,where
(10)Qmα∗Lmαψζ=Lmα∗Qmαψζ=ψζ.Moreover, we have the following property:Proposition 3(semigroup property).
Consider the PDO; then forψ and φ∈Σ℘(11)Qmαaψζ+bφζ=aQmαψζ+bQmαφζ,a,b∈ℝ.Proof.
Letm=1; the definition of Qmν implies
(12)Qαaψζ+bφζ=ρ1α,ζρ1α,ζ+ρ0α,ζaψζ+bφζ+ρ0α,ζρ1α,ζ+ρ0α,ζζ℘aψζ+bφζ′=aρ1α,ζρ1α,ζ+ρ0α,ζψζ+ρ0α,ζρ1α,ζ+ρ0α,ζζ℘ψ′ζ+bρ1α,ζρ1α,ζ+ρ0α,ζφζ+ρ0α,ζρ1α,ζ+ρ0α,ζζ℘φ′ζ=aQαψζ+bQαφζ.
Hence, for allm, we have the desired assertion.Our study is about the following class:Definition 4.
A function
ψ
∈
Σ
℘
is called in the class
Σ
℘
α
σ
,
p
if it satisfies the inequality
(13)
1
−
σ
ζ
℘
Q
m
α
ψ
ζ
+
σ
℘
ζ
℘
−
1
Q
m
α
ψ
ζ
′
≺
p
ζ
=
μ
ζ
+
1
ν
ζ
+
1
,
(14)
ζ
∈
Δ
,
α
,
σ
∈
0
,
1
,
−
1
≤
ν
<
μ
≤
1
,
℘
∈
ℕ
,where the symbol ≺ presents the subordination symbol [12] and p is convex univalent in Δ.For example(15)pζ=μζ+1νζ+1=Yμ,νζ,which is univalent convex in Δ, and it is the extreme function in the set
(16)P≔p∈Δ:pζ=1+∑n=1∞piζn.Define a functionalΨ:Δ⟶Δ, as follows:
(17)Ψζ≔1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′,(18)=1+∑n=℘+1∞Ψn+℘ψnζn−℘,ζ∈Δ,where
(19)Ψn+℘=1+n−1℘σΛnm.Shortly, by Definition4 we say
(20)Ψζ≺Yμ,νζ≔μζ+1νζ+1,ζ∈Δ.Our aim is to study the operator formulaΨ. We recall the following results:Lemma 5 (see[12]).
Let two analytic functions
f
ζ
and
g
ζ
be convex univalent defined in
Δ
such that
f
0
=
g
0
.
Moreover, for a constant
c
≠
0
,
R
c
≥
0
,
the subordination
(21)
f
ζ
+
1
/
c
f
′
ζ
≺
g
ζimplies
(22)fζ≺gζ.Lemma 6 (see[12]).
Define the general class of holomorphic functions
(23)
ℍ
a
,
n
=
h
:
h
ζ
=
a
+
a
n
ζ
n
+
a
n
+
1
ζ
n
+
1
+
⋯
,where a∈ℂ and n is a positive integer. If c∈ℝ, then
(24)Rhζ+cζh′ζ>0⇒Rhζ>0.
Moreover, ifc>0 and h∈ℍ1,n, then there are fixed numbers ℓ1>0 and ℓ2>0 with the inequality
(25)hζ+cζh′ζ≺1+ζ1−ζℓ1,hζ≺1+ζ1−ζℓ2.Lemma 7 (see[13]).
Let
ℏ
,
p
∈
ℍ
a
,
n
, where
p
is convex univalent in
Δ
and for
k
1
,
k
2
∈
ℂ
,
k
2
≠
0
;
then
(26)
k
1
ℏ
ζ
+
k
2
ζ
ℏ
′
ζ
≺
k
1
p
ζ
+
k
2
ζ
p
′
ζ
⟶
ℏ
ζ
≺
p
ζ
.Lemma 8 (see[14]).
Let
h
,
p
∈
ℍ
a
,
n
, where
p
is convex univalent in
Δ
such that
h
ζ
+
k
ζ
h
′
ζ
is univalent; then
(27)
p
ζ
+
k
ζ
p
′
ζ
≺
h
ζ
+
k
ζ
h
′
ζ
⟶
p
ζ
≺
h
ζ
.Lemma 9 (see[15]).
Let
ℏ
,
y
̈
,
g
∈
ℍ
a
,
n
, and
g
is convex univalent in
Δ
such that
ℏ
≺
g
and
y
≺
g
; then
(28)
k
ℏ
+
1
−
k
y
≺
g
,
k
∈
0
,
1
.


## 2. The Results

In this section, we illustrate our main results concerning the classΣ℘ασ,p for some special pζ,ζ∈Δ.


### 2.1. General Properties

Theorem 10.
Suppose that
ψ
∈
Σ
℘
α
σ
,
p
. If
R
Ψ
ζ
>
0
,
then the coefficient bounds of
Ψ
satisfy the inequality
(29)
∣
Ψ
n
∣
2
≤
∫
0
2
π
∣
e
−
i
n
θ
∣
d
M
θ
,where dM is a probability measure. Also, if
(30)ReiχΨζ>0,ζ∈Δ,χ∈ℝ,then ψ∈Σ℘ασ,νζ+1/νζ+1, that is
(31)Ψζ≈μζ+1νζ+1,ζ∈Δ.Proof.
By the assumption, we have(32)RΨζ=R1+∑n=℘+1∞Ψnζn>0.
Thus, the Carathéodory positivist technique yields(33)∣ψn∣≤2∫02π∣e−inθ∣dMθ,where dM is a probability measure. In addition, if
(34)ReiχΨζ>0,ζ∈Δ,χ∈ℝ,then according to Theorem 1.6 in [10] and for fixed χ∈ℝ, we have
(35)Ψζ≈pζ=μζ+1νζ+1,ζ∈Δ.
Hence,ψ∈Σ℘ασ,νζ+1/νζ+1.
The next results show the sufficient and necessary conditions for the sandwich behavior of the functionalΨζ=1−σ/ζ℘Qmαψζ+σ/℘ζ℘−1Qmαψζ′.Theorem 11.
Let the following assumptions hold
(36)
σ
ζ
Q
m
α
ψ
ζ
′
′
+
−
2
σ
℘
+
2
σ
+
℘
Q
m
α
ψ
ζ
′
+
σ
−
1
℘
−
1
℘
Q
m
α
ψ
ζ
℘
ζ
℘
−
1
≺
p
2
ζ
+
ζ
p
2
′
ζ
,where p20=1 and convex in Δ. Moreover, let Ψζ be univalent in Δ such that Ψ∈ℍp10,1∩ℚ, where ℚ represents the set of all injection analytic functions f with limζ∈∂Δf≠∞ and
(37)p1ζ+ζp1′ζ≺σζQmαψζ′′+−2σ℘+2σ+℘Qmαψζ′+σ−1℘−1℘Qmαψζ℘ζ℘−1.
Then(38)p1ζ≺Ψζ≺p2ζ,and p1ζ is the best subdominant, and p2ζ is the best dominant.Proof.
Since(39)Ψζ+ζΨ′ζ=σζQmαψζ′′+−2σ℘+2σ+℘Qmαψζ′+σ−1℘−1℘Qmαψζ℘ζ℘−1,then we obtain the next double inequality
(40)p1ζ+ζp1′ζ≺Ψζ+ζΨ′ζ≺p2ζ+ζp2′ζ.
Thus, Lemmas7 and 8 imply the desired assertion.Theorem 12.
Let
p
be a univalent convex function in
Δ
such that
p
0
=
0
and
(41)
Q
m
α
ψ
ζ
≺
p
ζ
,
L
m
α
ψ
ζ
≺
p
ζ
.
Then(42)Amαψζ≔kQmαψζ+1−kLmαψζ≺pζ,k∈0,1.Proof.
By the definition ofQmαψζ and Lmαψζ, clearly we have Amαψζ∈Σ℘. Hence, a direct application of Lemma 9, we obtain the result.


### 2.2. Inclusion Properties

In this part, we deal with the inclusion properties.Theorem 13.
For
σ
2
≤
σ
1
<
0
and
ψ
∈
Σ
℘
,
then
(43)
Σ
℘
α
σ
2
,
p
⊂
Σ
℘
α
σ
1
,
p
.Proof.
Letψ∈Σ℘ασ2,p. Define the analytic function in Δ, as follows:
(44)ϕζ=ζ−℘Qmαψζ,satisfying ϕ0=1. A computation gives
(45)1−σ2ζ℘Qmαψζ+σ2℘ζ℘−1Qmαψζ′=ϕζ+σ2℘ζϕ′ζ.
Consequently, we get the inequality(46)ϕζ+σ2℘ζϕ′ζ≺μζ+1νζ+1.
Applying Lemma5 with σ2/℘>0 gives
(47)ϕζ≺μζ+1νζ+1.
Since0<σ1/σ2<1 and Yμ,νζ is convex univalent in Δ, we arrive at the inequality
(48)1−σ1ζ℘Qmαψζ+σ1℘ζ℘−1Qmαψζ′=1−σ1ϕζ+σ1℘ζ℘−1Qmαψζ′=1−σ1ϕζ+σ1℘ζϕ′ζ+℘ϕζ,=1−σ1ϕζ+σ1℘ζϕ′ζ+℘ϕζ+σ1σ2ϕζ−σ1σ2ϕζ=σ1σ21−σ2ϕζ+σ2℘ζϕ′ζ+℘ϕζ+1−σ1σ2ϕζ=σ1σ21−σ2ζ℘Qmαψζ+σ2℘ζ℘−1Qmαψζ′+1−σ1σ2ϕζ≺μζ+1νζ+1=Yμ,νζ.
Hence, by Definition4, we conclude that ψ∈Σ℘ασ1,p.Theorem 14.
Let
(49)
Ψ
ζ
=
1
−
σ
ζ
℘
Q
m
α
ψ
ζ
+
σ
℘
ζ
℘
−
1
Q
m
α
ψ
ζ
′
.
Then(50)Qmαψζ′ζ℘ℏ1+Qmαψζζ℘−1ℏ1+1+℘ℏ2+ℏ2+ℏ2ζ2−℘Qmαψζ′′≺1+ζ1−ζℓ1⇒Ψζ≺1+ζ1−ζℓ2,where
(51)ℓ1>0,ℓ2>0,ℏ1=1−σ,ℏ2=σ℘,℘>0.Proof.
A calculation implies that(52)Ψζ+ζΨ′ζ=1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′+ζ1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′=Qmαψζ′ζ℘ℏ1+Qmαψζζ℘−1ℏ1+1+℘ℏ2+ℏ2+ℏ2ζ2−℘Qmαψζ′′≺1+ζ1−ζℓ1.
According to Lemma6 joining the value c=1, we get
(53)Ψζ≺1+ζ1−ζℓ2.Corollary 15.
Let
Ψ
ζ
be assumed as in Theorem
14
. If the subordination
(54)
Q
m
α
ψ
ζ
′
ζ
℘
ℏ
1
+
Q
m
α
ψ
ζ
ζ
℘
−
1
ℏ
1
+
1
+
℘
ℏ
2
+
ℏ
2
+
ℏ
2
ζ
2
−
℘
Q
m
α
ψ
ζ
′
′
≺
1
+
ζ
1
−
ζ
,where ℓ1>0,ℓ2>0,ℏ1=1−σ,ℏ2=σ/℘,℘>0 holds, then ψ∈Σ℘ασ,1+ζ/1−ζ.Proof.
Taking,ℓ1=ℓ2=1 in Theorem 14 implies that Ψζ≺1+ζ/1−ζ. Consequently, we have ψ∈Σ℘ασ,1+ζ/1−ζ.Theorem 16.
Let
ψ
∈
Σ
℘
α
σ
,
p
and
f
∈
Σ
℘
.
If
(55)
R
Q
m
α
ψ
ζ
ζ
℘
>
1
2
,then ψ×f∈Σ℘ασ,p.Proof.
A convolution product indicates that(56)1−σζ℘Qmαψζ×fζ+σ℘ζ℘−1Qmαψζ×fζ′=1−σQmαψζζ℘×Qmαfζζ℘+σ℘Qmαfζ′ζ℘−1×Qmαfζζ℘=1−σQmαψζζ℘+σ℘ζ℘−1Qmαfζ′×Qmαfζζ℘=Ψζ×Qmαfζζ℘,where Ψζ≺Yμ,νζ. In view of real inequality (55), we get that Qmαfζ/ζ℘ has the Herglotz integral formula [11].
(57)Qmαfζζ℘=∫∣τ∣=1dςτ1−τζ,where dς conforms the probability measure on the unit circle ∣τ∣=1 and
(58)∫∣τ∣=1dςτ=1.
But,Yμ,νζ is convex in Δ; then we have
(59)1−σζ℘Qmαψζ∗fζ+σ℘ζ℘−1Qmαψζ∗fζ′=Θζ∗Qmαfζζ℘=∫τ=1Θτζdςτ≺Yμ,νζ.
Thus,ψ×f∈Σ℘ασ,p.


### 2.3. Fekete-Szegö Inequality

In this section, we obtain the Fekete-Szegö relation coefficient estimates for the classΣ℘ασ,p. Let Ω be the class of functions of the form
(60)ϖζ=1+ϖ1ζ+ϖ2ζ2+⋯,in the open unit disk Δ satisfying ∣ϖz∣<1. To prove our results, we need the following lemma.Lemma 17 (see[16]).
If
ϖ
∈
Ω
, then for any complex number
ρ
(61)
∣
ϖ
2
−
ρ
ϖ
1
2
∣
≤
max
1
,
∣
ρ
∣
.
The result is sharp for the functions given byϖζ=ζ or ϖζ=ζ2.Theorem 18.
Let the function
ψ
be formulated by ((5)). Then,ψ∈Σ℘ασ,pand(62)ψ℘+2−ρψ℘+12≤μ−ν℘℘+℘+1σΛ℘+2mmax1,∣ℵ∣,where
(63)ℵ=ν+ρμ−ν℘+℘+1σΛ℘+2m℘1+σ2Λ2m℘+1.Proof.
Sinceψ∈Σ℘ασ,p, we have
(64)1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′≺pζ=μζ+1νζ+1.
In addition, there is a Schwarz functionϖζ=1+ϖ1ζ+ϖ2ζ2+⋯ in Ω such that
(65)1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′≺pwζpwζ=1+μwζ1+νwζ=1+μ−νϖ1ζ+μ−νϖ2−νμ−νϖ12ζ2+⋯.
Now by (18), we have
(66)1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′=1+∑n=℘+1∞1+n−1℘σΛnmψnζn−℘,ζ∈Δ,where Λnm is given by (19). Equating the coefficients of ζ and ζ2, we get
(67)1+σΛ℘+1mψ℘+1=μ−νϖ1,(68)1+℘+1℘σΛ℘+2mψ℘+2=μ−νϖ2−νμ−νϖ12,(69)℘+℘+1σ℘Λ℘+2mψ℘+2=μ−νϖ2−νμ−νϖ12.
From (67) and (69), we get
(70)ψ℘+1=μ−νϖ11+σΛ℘+1m,ψ℘+2=μ−ν℘℘+℘+1σΛ℘+2mϖ2−νμ−ν℘℘+℘+1σΛ℘+2mϖ12.
For anyρ∈ℂ, we get
(71)ψ℘+2−ρψ℘+12=μ−ν℘℘+℘+1σΛ℘+2mϖ2−νμ−ν℘℘+℘+1σΛ℘+2mϖ12−μ−νϖ11+σΛ℘+1m2=μ−ν℘℘+℘+1σΛ℘+2mϖ2−ν+ρμ−ν℘+℘+1σΛ℘+2m℘1+σ2Λ℘+12mϖ12=μ−ν℘℘+℘+1σΛ℘+2mϖ2−ℵϖ12,where
(72)ℵ=ν+ρμ−ν℘+℘+1σΛ℘+2m℘1+σ2Λ2m℘+1.
By applying Lemma17, we get
(73)ψ℘+2−ρψ℘+12≤μ−ν℘℘+℘+1σΛ℘+2mmax1,ℵ.
The result is sharp for the function(74)1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′=pζ2=μζ2+1νζ2+1,or
(75)1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′=pζ=μζ+1νζ+1.Remark 19.
By fixing
ρ
=
1
in Theorem
18
, we get
(76)
ψ
℘
+
2
−
ψ
℘
+
1
2
≤
μ
−
ν
℘
℘
+
℘
+
1
σ
Λ
℘
+
2
m
max
1
,
∣
ℵ
∣
,where
(77)ℵ=ν+μ−ν℘+℘+1σΛ℘+2m℘1+σ2Λ2m℘+1.
From Definition4, a function ψ∈Σ℘ is said to be in the class Σ℘ασ,p if it satisfies the inequality (13); then we have
(78)Ψζ−1μ−νΨζ<1,where
(79)Ψζ=1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′is as given in (17).Now, we obtain coefficient estimates forf∈Σ℘ασ,p.Theorem 20.
Let the function
ψ
be defined by ((5)). Then,ψ∈Σ℘ασ,pif(80)∑n=p+1∞Ψn,℘1+νψn≤μ−ν,(81)Ψn,℘=1+n−1℘σΛnm,where Λnm is given by (8).Proof.
Supposeψ satisfies (80). Then, for ∣ζ∣=r<1(82)Ψζ−1−μ−νΨζ=∑n=p+1∞Ψn,℘ψnζn−p−βμ−ν+ν∑n=p+1∞Ψn,℘∣ψnζn−p≤∑n=p+1∞Ψn,℘ψn−μ−ν+∑n=p+1∞Ψn,℘νψn=∑n=p+1∞Ψn,℘1+νψn−μ−ν≤0.


## 2.1. General Properties

Theorem 10.
Suppose that
ψ
∈
Σ
℘
α
σ
,
p
. If
R
Ψ
ζ
>
0
,
then the coefficient bounds of
Ψ
satisfy the inequality
(29)
∣
Ψ
n
∣
2
≤
∫
0
2
π
∣
e
−
i
n
θ
∣
d
M
θ
,where dM is a probability measure. Also, if
(30)ReiχΨζ>0,ζ∈Δ,χ∈ℝ,then ψ∈Σ℘ασ,νζ+1/νζ+1, that is
(31)Ψζ≈μζ+1νζ+1,ζ∈Δ.Proof.
By the assumption, we have(32)RΨζ=R1+∑n=℘+1∞Ψnζn>0.
Thus, the Carathéodory positivist technique yields(33)∣ψn∣≤2∫02π∣e−inθ∣dMθ,where dM is a probability measure. In addition, if
(34)ReiχΨζ>0,ζ∈Δ,χ∈ℝ,then according to Theorem 1.6 in [10] and for fixed χ∈ℝ, we have
(35)Ψζ≈pζ=μζ+1νζ+1,ζ∈Δ.
Hence,ψ∈Σ℘ασ,νζ+1/νζ+1.
The next results show the sufficient and necessary conditions for the sandwich behavior of the functionalΨζ=1−σ/ζ℘Qmαψζ+σ/℘ζ℘−1Qmαψζ′.Theorem 11.
Let the following assumptions hold
(36)
σ
ζ
Q
m
α
ψ
ζ
′
′
+
−
2
σ
℘
+
2
σ
+
℘
Q
m
α
ψ
ζ
′
+
σ
−
1
℘
−
1
℘
Q
m
α
ψ
ζ
℘
ζ
℘
−
1
≺
p
2
ζ
+
ζ
p
2
′
ζ
,where p20=1 and convex in Δ. Moreover, let Ψζ be univalent in Δ such that Ψ∈ℍp10,1∩ℚ, where ℚ represents the set of all injection analytic functions f with limζ∈∂Δf≠∞ and
(37)p1ζ+ζp1′ζ≺σζQmαψζ′′+−2σ℘+2σ+℘Qmαψζ′+σ−1℘−1℘Qmαψζ℘ζ℘−1.
Then(38)p1ζ≺Ψζ≺p2ζ,and p1ζ is the best subdominant, and p2ζ is the best dominant.Proof.
Since(39)Ψζ+ζΨ′ζ=σζQmαψζ′′+−2σ℘+2σ+℘Qmαψζ′+σ−1℘−1℘Qmαψζ℘ζ℘−1,then we obtain the next double inequality
(40)p1ζ+ζp1′ζ≺Ψζ+ζΨ′ζ≺p2ζ+ζp2′ζ.
Thus, Lemmas7 and 8 imply the desired assertion.Theorem 12.
Let
p
be a univalent convex function in
Δ
such that
p
0
=
0
and
(41)
Q
m
α
ψ
ζ
≺
p
ζ
,
L
m
α
ψ
ζ
≺
p
ζ
.
Then(42)Amαψζ≔kQmαψζ+1−kLmαψζ≺pζ,k∈0,1.Proof.
By the definition ofQmαψζ and Lmαψζ, clearly we have Amαψζ∈Σ℘. Hence, a direct application of Lemma 9, we obtain the result.


## 2.2. Inclusion Properties

In this part, we deal with the inclusion properties.Theorem 13.
For
σ
2
≤
σ
1
<
0
and
ψ
∈
Σ
℘
,
then
(43)
Σ
℘
α
σ
2
,
p
⊂
Σ
℘
α
σ
1
,
p
.Proof.
Letψ∈Σ℘ασ2,p. Define the analytic function in Δ, as follows:
(44)ϕζ=ζ−℘Qmαψζ,satisfying ϕ0=1. A computation gives
(45)1−σ2ζ℘Qmαψζ+σ2℘ζ℘−1Qmαψζ′=ϕζ+σ2℘ζϕ′ζ.
Consequently, we get the inequality(46)ϕζ+σ2℘ζϕ′ζ≺μζ+1νζ+1.
Applying Lemma5 with σ2/℘>0 gives
(47)ϕζ≺μζ+1νζ+1.
Since0<σ1/σ2<1 and Yμ,νζ is convex univalent in Δ, we arrive at the inequality
(48)1−σ1ζ℘Qmαψζ+σ1℘ζ℘−1Qmαψζ′=1−σ1ϕζ+σ1℘ζ℘−1Qmαψζ′=1−σ1ϕζ+σ1℘ζϕ′ζ+℘ϕζ,=1−σ1ϕζ+σ1℘ζϕ′ζ+℘ϕζ+σ1σ2ϕζ−σ1σ2ϕζ=σ1σ21−σ2ϕζ+σ2℘ζϕ′ζ+℘ϕζ+1−σ1σ2ϕζ=σ1σ21−σ2ζ℘Qmαψζ+σ2℘ζ℘−1Qmαψζ′+1−σ1σ2ϕζ≺μζ+1νζ+1=Yμ,νζ.
Hence, by Definition4, we conclude that ψ∈Σ℘ασ1,p.Theorem 14.
Let
(49)
Ψ
ζ
=
1
−
σ
ζ
℘
Q
m
α
ψ
ζ
+
σ
℘
ζ
℘
−
1
Q
m
α
ψ
ζ
′
.
Then(50)Qmαψζ′ζ℘ℏ1+Qmαψζζ℘−1ℏ1+1+℘ℏ2+ℏ2+ℏ2ζ2−℘Qmαψζ′′≺1+ζ1−ζℓ1⇒Ψζ≺1+ζ1−ζℓ2,where
(51)ℓ1>0,ℓ2>0,ℏ1=1−σ,ℏ2=σ℘,℘>0.Proof.
A calculation implies that(52)Ψζ+ζΨ′ζ=1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′+ζ1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′=Qmαψζ′ζ℘ℏ1+Qmαψζζ℘−1ℏ1+1+℘ℏ2+ℏ2+ℏ2ζ2−℘Qmαψζ′′≺1+ζ1−ζℓ1.
According to Lemma6 joining the value c=1, we get
(53)Ψζ≺1+ζ1−ζℓ2.Corollary 15.
Let
Ψ
ζ
be assumed as in Theorem
14
. If the subordination
(54)
Q
m
α
ψ
ζ
′
ζ
℘
ℏ
1
+
Q
m
α
ψ
ζ
ζ
℘
−
1
ℏ
1
+
1
+
℘
ℏ
2
+
ℏ
2
+
ℏ
2
ζ
2
−
℘
Q
m
α
ψ
ζ
′
′
≺
1
+
ζ
1
−
ζ
,where ℓ1>0,ℓ2>0,ℏ1=1−σ,ℏ2=σ/℘,℘>0 holds, then ψ∈Σ℘ασ,1+ζ/1−ζ.Proof.
Taking,ℓ1=ℓ2=1 in Theorem 14 implies that Ψζ≺1+ζ/1−ζ. Consequently, we have ψ∈Σ℘ασ,1+ζ/1−ζ.Theorem 16.
Let
ψ
∈
Σ
℘
α
σ
,
p
and
f
∈
Σ
℘
.
If
(55)
R
Q
m
α
ψ
ζ
ζ
℘
>
1
2
,then ψ×f∈Σ℘ασ,p.Proof.
A convolution product indicates that(56)1−σζ℘Qmαψζ×fζ+σ℘ζ℘−1Qmαψζ×fζ′=1−σQmαψζζ℘×Qmαfζζ℘+σ℘Qmαfζ′ζ℘−1×Qmαfζζ℘=1−σQmαψζζ℘+σ℘ζ℘−1Qmαfζ′×Qmαfζζ℘=Ψζ×Qmαfζζ℘,where Ψζ≺Yμ,νζ. In view of real inequality (55), we get that Qmαfζ/ζ℘ has the Herglotz integral formula [11].
(57)Qmαfζζ℘=∫∣τ∣=1dςτ1−τζ,where dς conforms the probability measure on the unit circle ∣τ∣=1 and
(58)∫∣τ∣=1dςτ=1.
But,Yμ,νζ is convex in Δ; then we have
(59)1−σζ℘Qmαψζ∗fζ+σ℘ζ℘−1Qmαψζ∗fζ′=Θζ∗Qmαfζζ℘=∫τ=1Θτζdςτ≺Yμ,νζ.
Thus,ψ×f∈Σ℘ασ,p.


## 2.3. Fekete-Szegö Inequality

In this section, we obtain the Fekete-Szegö relation coefficient estimates for the classΣ℘ασ,p. Let Ω be the class of functions of the form
(60)ϖζ=1+ϖ1ζ+ϖ2ζ2+⋯,in the open unit disk Δ satisfying ∣ϖz∣<1. To prove our results, we need the following lemma.Lemma 17 (see[16]).
If
ϖ
∈
Ω
, then for any complex number
ρ
(61)
∣
ϖ
2
−
ρ
ϖ
1
2
∣
≤
max
1
,
∣
ρ
∣
.
The result is sharp for the functions given byϖζ=ζ or ϖζ=ζ2.Theorem 18.
Let the function
ψ
be formulated by ((5)). Then,ψ∈Σ℘ασ,pand(62)ψ℘+2−ρψ℘+12≤μ−ν℘℘+℘+1σΛ℘+2mmax1,∣ℵ∣,where
(63)ℵ=ν+ρμ−ν℘+℘+1σΛ℘+2m℘1+σ2Λ2m℘+1.Proof.
Sinceψ∈Σ℘ασ,p, we have
(64)1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′≺pζ=μζ+1νζ+1.
In addition, there is a Schwarz functionϖζ=1+ϖ1ζ+ϖ2ζ2+⋯ in Ω such that
(65)1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′≺pwζpwζ=1+μwζ1+νwζ=1+μ−νϖ1ζ+μ−νϖ2−νμ−νϖ12ζ2+⋯.
Now by (18), we have
(66)1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′=1+∑n=℘+1∞1+n−1℘σΛnmψnζn−℘,ζ∈Δ,where Λnm is given by (19). Equating the coefficients of ζ and ζ2, we get
(67)1+σΛ℘+1mψ℘+1=μ−νϖ1,(68)1+℘+1℘σΛ℘+2mψ℘+2=μ−νϖ2−νμ−νϖ12,(69)℘+℘+1σ℘Λ℘+2mψ℘+2=μ−νϖ2−νμ−νϖ12.
From (67) and (69), we get
(70)ψ℘+1=μ−νϖ11+σΛ℘+1m,ψ℘+2=μ−ν℘℘+℘+1σΛ℘+2mϖ2−νμ−ν℘℘+℘+1σΛ℘+2mϖ12.
For anyρ∈ℂ, we get
(71)ψ℘+2−ρψ℘+12=μ−ν℘℘+℘+1σΛ℘+2mϖ2−νμ−ν℘℘+℘+1σΛ℘+2mϖ12−μ−νϖ11+σΛ℘+1m2=μ−ν℘℘+℘+1σΛ℘+2mϖ2−ν+ρμ−ν℘+℘+1σΛ℘+2m℘1+σ2Λ℘+12mϖ12=μ−ν℘℘+℘+1σΛ℘+2mϖ2−ℵϖ12,where
(72)ℵ=ν+ρμ−ν℘+℘+1σΛ℘+2m℘1+σ2Λ2m℘+1.
By applying Lemma17, we get
(73)ψ℘+2−ρψ℘+12≤μ−ν℘℘+℘+1σΛ℘+2mmax1,ℵ.
The result is sharp for the function(74)1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′=pζ2=μζ2+1νζ2+1,or
(75)1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′=pζ=μζ+1νζ+1.Remark 19.
By fixing
ρ
=
1
in Theorem
18
, we get
(76)
ψ
℘
+
2
−
ψ
℘
+
1
2
≤
μ
−
ν
℘
℘
+
℘
+
1
σ
Λ
℘
+
2
m
max
1
,
∣
ℵ
∣
,where
(77)ℵ=ν+μ−ν℘+℘+1σΛ℘+2m℘1+σ2Λ2m℘+1.
From Definition4, a function ψ∈Σ℘ is said to be in the class Σ℘ασ,p if it satisfies the inequality (13); then we have
(78)Ψζ−1μ−νΨζ<1,where
(79)Ψζ=1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′is as given in (17).Now, we obtain coefficient estimates forf∈Σ℘ασ,p.Theorem 20.
Let the function
ψ
be defined by ((5)). Then,ψ∈Σ℘ασ,pif(80)∑n=p+1∞Ψn,℘1+νψn≤μ−ν,(81)Ψn,℘=1+n−1℘σΛnm,where Λnm is given by (8).Proof.
Supposeψ satisfies (80). Then, for ∣ζ∣=r<1(82)Ψζ−1−μ−νΨζ=∑n=p+1∞Ψn,℘ψnζn−p−βμ−ν+ν∑n=p+1∞Ψn,℘∣ψnζn−p≤∑n=p+1∞Ψn,℘ψn−μ−ν+∑n=p+1∞Ψn,℘νψn=∑n=p+1∞Ψn,℘1+νψn−μ−ν≤0.


## 3. An Application

In this section, we consider the suggested classΣ℘ασ,1+ζ/1−ζ for all α∈0,1.Theorem 21.
Consider the class of analytic functions
Σ
℘
α
σ
,
1
+
ζ
/
1
−
ζ
.
Then, the solution of the differential equation corresponds to this class is
(83)
Q
m
α
ψ
ζ
=
c
1
ζ
σ
−
1
℘
σ
+
ζ
℘
2
℘
ζ
2
F
1
1
,
σ
+
℘
/
σ
,
℘
/
σ
+
2
,
ζ
+
1
σ
+
℘
,where 2F1a,b,c;ζ represents the hypergeometric function.Proof.
Suppose thatψ∈Σ℘ασ,1+ζ/1−ζ. Then, it satisfies the differential equation
(84)1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′=ωζ+11−ωζ,where ω0=0 and ∣ω∣<1. This leads to the solution
(85)Qmαψζ=ζσ−1℘/σ∫0ζ−℘z℘/σ−1ωz+1σωz−1dz.
To find the upper solution, we letωζ=ζ. Thus, we have the differential equation
(86)1−σζ℘Qmαψζ+σ℘ζ℘−1Qmαψζ′=ζ+11−ζ.
Rewrite the above equation as follows:(87)Qmαψζ′+℘1−σσζQmαψζ=℘ζ℘−1σ1+ζ1−ζ.
Multiplying the above equation by the functional(88)Tζ=exp∫℘σ+ζ−σζ−1σζζ−1dζ,we obtain
(89)ζ℘1/σ−1Qmαψζ′−Qmαψζ℘ζ℘1/σ−1−1σ+ζ−σζ−1σ1−ζ=℘ζ℘/σ−1σ1+ζ1−ζ.
Hence, it follows the solution (26).Example 1.
For
(i)
℘
=
1
,
σ
=
0.5
,
c
1
=
0
, the solution is
(90)
Q
m
α
ψ
ζ
=
−
ζ
−
4
ζ
+
log
1
−
ζ
ζ
(ii)
℘
=
1
,
σ
=
0.25
,
c
1
=
0; the solution becomes
(91)
Q
m
α
ψ
ζ
=
−
ζ
−
4
2
ζ
3
+
3
ζ
2
+
6
ζ
+
6
log
1
−
ζ
3
ζ
3
(iii)
℘
=
2
,
σ
=
0.5
,
c
1
=
0; then the solution is given by the formula
(92)
Q
m
α
ψ
ζ
=
−
ζ
2
−
4
2
ζ
3
+
3
ζ
2
+
6
ζ
+
6
log
1
−
ζ
3
ζ
2
.


## 4. Conclusion

Commencing overhead, we formulated a new parametric differential operator for a certain class of multivalently analytic functions. We investigated some geometric conducts of the operator connecting with the Janowski function, which is convex univalent in the open unit disk. As an application, we presented the formula of the suggested class involving the operator. For future works, one can generalize the suggested fractional operator using various classes of analytic functions such as meromorphic and harmonic functions.



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*Source: 6708137-2021-10-31.xml*