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Axioms 2018, 7(2), 27; doi:10.3390/axioms7020027

Article
Subordination Properties for Multivalent Functions Associated with a Generalized Fractional Differintegral Operator
1
Department of Mathematics, Faculty of Science, Menofia University, Shebin Elkom 32511, Egypt
2
Department of Mathematics, Faculty of Science, Mansoura University, Mansoura 35516, Egypt
*
Author to whom correspondence should be addressed.
Received: 19 January 2018 / Accepted: 17 April 2018 / Published: 24 April 2018

## Abstract

:
Using of the principle of subordination, we investigate some subordination and convolution properties for classes of multivalent functions under certain assumptions on the parameters involved, which are defined by a generalized fractional differintegral operator under certain assumptions on the parameters involved.
Keywords:
differential subordination; p-valent functions; generalized fractional differintegral operator
JEL Classification:
30C45; 30C50

## 1. Introduction and Definitions

Denote by $A ( p )$ the class of analytic and p-valent functions of the form:
$f ( z ) = z p + ∑ n = 1 ∞ a p + n z p + n ( p ∈ N = { 1 , 2 , … } ; z ∈ U = { z ∈ C : z < 1 } ) .$
For functions $f , g$ analytic in $U$, f is subordinate to g, written $f ( z ) ≺ g ( z )$ if there exists a function w, analytic in $U$ with $w ( 0 ) = 0$ and $w ( z ) < 1 ,$ such that $f ( z ) = g ( w ( z ) ) , z ∈ U .$ If g is univalent in $U ,$ then (see [1,2]):
$f ( z ) ≺ g ( z ) ⇔ f ( 0 ) = g ( 0 ) and f ( U ) ⊂ g ( U ) .$
If $φ ( z )$ is analytic in $U$ and satisfies:
$H ( φ ( z ) , z φ ′ ( z ) ) ≺ h ( z ) ,$
then $φ$ is a solution of (2). The univalent function q is called dominant, if $φ ( z ) ≺ q ( z )$ for all $φ$. A dominant $q ˜$ is called the best dominant, if $q ˜ ( z ) ≺ q ( z )$ for all dominants q.
Let $2 F 1 ( a , b ; c ; z )$ $c ≠ 0 , − 1 , − 2 , ⋯$ be the well-known (Gaussian) hypergeometric function defined by:
$2 F 1 ( a , b ; c ; z ) : = ∑ n = 0 ∞ ( a ) n ( b ) n ( c ) n ( 1 ) n z n , z ∈ U ,$
where:
$( λ ) n : = Γ λ + n Γ λ .$
We will recall some definitions that will be used in our paper.
Definition 1.
For $f ( z ) ∈ A ( p ) ,$ the fractional integral and fractional derivative operators of order λ are defined by Owa [3] (see also [4]) as:
$D z − λ f ( z ) : = 1 Γ ( λ ) ∫ 0 z f ( ζ ) ( z − ζ ) 1 − λ d ζ ( λ > 0 ) ,$
$D z λ f ( z ) : = 1 Γ ( 1 − λ ) d d z ∫ 0 z f ( ζ ) ( z − ζ ) λ d ζ ( 0 ≤ λ < 1 ) ,$
where f is an analytic function in a simply-connected region of the complex z-plane containing the origin, and the multiplicity of $( z − ζ ) λ − 1 ( ( z − ζ ) − λ )$ is removed by requiring $log ( z − ζ )$ to be real when $z − ζ > 0$.
Definition 2.
For $f ( z ) ∈ A ( p )$ and in terms of $2 F 1 ,$ the generalized fractional integral and generalized fractional derivative operators defined by Srivastava et al. [5] (see also [6]) as:
$I 0 , z λ , μ , η f ( z ) : = z − λ − μ Γ ( λ ) ∫ 0 z ( z − ζ ) λ − 1 f ( ζ ) 2 F 1 μ + λ , − η ; λ ; 1 − ζ z d ζ λ > 0 , μ , η ∈ R ,$
$J 0 , z λ , μ , η f ( z ) : = d d z z λ − μ ∫ 0 z ( z − ζ ) − λ f ( ζ ) 2 F 1 μ − λ , 1 − η ; 1 − λ ; 1 − ζ z d ζ Γ ( 1 − λ ) ( 0 ≤ λ < 1 ) , d n d z n J 0 , z λ − n , μ , η f ( z ) ( n ≤ λ < n + 1 ; n ∈ N ) ,$
where $f ( z )$ is an analytic function in a simply-connected region of the complex $z −$ plane containing the origin with the order $f ( z ) = O ( z ε ) , z → 0$ when $ε > max { 0 , μ − η } − 1$, and the multiplicity of $( z − ζ ) λ − 1$ $( ( z − ζ ) − λ )$ is removed by requiring $log ( z − ζ )$ to be real when $z − ζ > 0$.
We note that:
$I 0 , z λ , − λ , η f ( z ) = D z − λ f ( z ) ( λ > 0 ) and J 0 , z λ , λ , η f ( z ) = D z λ f ( z ) ( 0 ≤ λ < 1 ) ,$
where $D z − λ f ( z )$ and $D z λ f ( z )$ are the fractional integral and fractional derivative operators studied by Owa [3].
$S 0 , z λ , μ , η , p f ( z ) = Γ ( p + 1 − μ ) Γ ( p + 1 − λ + η ) Γ ( p + 1 ) Γ ( p + 1 − μ + η ) z μ J 0 , z λ , μ , η f ( z ) ( 0 ≤ λ < η + p + 1 ; z ∈ U ) , Γ ( p + 1 − μ ) Γ ( p + 1 − λ + η ) Γ ( p + 1 ) Γ ( p + 1 − μ + η ) z μ I 0 , z − λ , μ , η f ( z ) ( − ∞ < λ < 0 ; z ∈ U ) .$
For $f ( z ) ∈ A ( p )$, we have:
$S 0 , z λ , μ , η , p f ( z ) = z 3 p F 2 ( 1 , 1 + p , 1 + p + η − μ ; 1 + p − μ , 1 + p + η − λ ; z ) ∗ f ( z ) = z p + ∑ n = 1 ∞ ( p + 1 ) n ( p + 1 − μ + η ) n ( p + 1 − μ ) n ( p + 1 − λ + η ) n a p + n z p + n ( p ∈ N ; μ , η ∈ R ; μ < p + 1 ; − ∞ < λ < η + p + 1 ) ,$
where “∗” stands for convolution of two power series, and $q F s ( q ≤ s + 1 ; q , s ∈ N 0 = N ∪ { 0 } )$ is the well-known generalized hypergeometric function.
Let:
$G p , η , μ λ ( z ) = z p + ∑ n = 1 ∞ ( p + 1 ) n ( p + 1 − μ + η ) n ( p + 1 − μ ) n ( p + 1 − λ + η ) n z p + n ( p ∈ N ; μ , η ∈ R ; μ < p + 1 ; − ∞ < λ < η + p + 1 ) .$
and:
$G p , η , μ λ ( z ) ∗ G p , η , μ λ ( z ) − 1 = z p ( 1 − z ) δ + p ( δ > − p ; z ∈ U ) .$
Tang et al. [9] (see also [10,11,12,13,14,15]) defined the operator $H p , η , μ λ , δ : A ( p ) → A ( p ) ,$ where:
$H p , η , μ λ , δ f ( z ) = z p + ∑ n = 1 ∞ ( δ + p ) n ( p + 1 − μ ) n ( p + 1 − λ + η ) n ( 1 ) n ( p + 1 ) n ( p + 1 − μ + η ) n a p + n z p + n ( p ∈ N , δ > − p , μ , η ∈ R , μ < p + 1 , − ∞ < λ < η + p + 1 ) .$
It is easy to verify that:
$z H p , η , μ λ , δ f ( z ) ′ = ( δ + p ) H p , η , μ λ , δ + 1 f ( z ) − δ H p , η , μ λ , δ f ( z ) ,$
and:
$z H p , η , μ λ + 1 , δ f ( z ) ′ = ( p + η − λ ) H p , η , μ λ , δ f ( z ) − ( η − λ ) H p , η , μ λ + 1 , δ f ( z ) .$
By using the operator $H p , η , μ λ , δ$, we introduce the following class.
Definition 3.
For $A , B ( − 1 ≤ B < A ≤ 1 )$, $f ∈ A ( p )$ is in the class $T p , η , μ λ , δ ( A , B )$ if
$( H p , η , μ λ , δ f ( z ) ) ′ p z p − 1 ≺ 1 + A z 1 + B z ( z ∈ U ; p ∈ N ) ,$
which is equivalent to:
$( H p , η , μ λ , δ f ( z ) ) ′ p z p − 1 − 1 B ( H p , η , μ λ , δ f ( z ) ) ′ p z p − 1 − A < 1 ( z ∈ U ) .$
For convenience, we write $T p , η , μ λ , δ 1 − 2 ξ p , − 1 = T p , η , μ λ , δ ( ξ ) ( 0 ≤ ξ < p )$, which satisfies the inequality:
$ℜ ( H p , η , μ λ , δ f ( z ) ) ′ z p − 1 > ξ ( 0 ≤ ξ < p ) .$
In this paper, we investigate some subordination and convolution properties for classes of multivalent functions, which are defined by a generalized fractional differintegral operator. The theory of subordination received great attention, particularly in many subclasses of univalent and multivalent functions (see, for example, [13,15,16,17]).

## 2. Preliminaries

To prove our main results, we shall need the following lemmas.
Lemma 1.
[18]. Let h be an analytic and convex (univalent) function in $U$ with $h ( 0 ) = 1$. Additionally, let φ given by:
$ϕ ( z ) = 1 + c n z n + c n + 1 z n + 1 + …$
be analytic in $U$. If:
$ϕ ( z ) + z ϕ ′ ( z ) σ ≺ h ( z ) ( ℜ ( σ ) ⩾ 0 ; σ ≠ 0 ) ,$
then:
$ϕ ( z ) ≺ ψ ( z ) = σ n z − σ n ∫ 0 z t σ n − 1 h ( t ) d t ≺ h ( z ) ,$
and ψ is the best dominant of (6).
Denote by $P ( ς )$ the class of functions $Φ$ given by:
$Φ ( z ) = 1 + c 1 z + c 1 z 2 + … ,$
which are analytic in $U$ and satisfy the following inequality:
$ℜ Φ ( z ) > ς ( 0 ≤ ς < 1 ) .$
Using the well-known growth theorem for the Carathéodory functions (cf., e.g., [19]), we may easily deduce the following result:
Lemma 2.
[19]. If $Φ ∈$ $P ( ς )$. Then
$ℜ Φ ( ς ) ⩾ 2 ς − 1 + 2 ( 1 − ς ) 1 + z ( 0 ≤ ς < 1 ) .$
Lemma 3.
[20]. For $0 ≤ ς 1 , ς 2 < 1 ,$
$P ( ς 1 ) ∗ P ( ς 2 ) ⊂ P ( ς 3 ) ( ς 3 = 1 − 2 ( 1 − ς 1 ) ( 1 − ς 2 ) ) .$
The result is the best possible.
Lemma 4.
[21]$.$ Let ϕ be such that $φ ( 0 ) = 1$ and $φ ( z ) ≠ 0$ and $A , B ∈ C ,$ with $A ≠ B , B ≤ 1 , ν ∈ C * .$
$( i )$ If $ν ( A − B ) B − 1 ≤ 1$ or $ν ( A − B ) B + 1 ≤ 1 , B ≠ 0$ and $φ ( z )$ satisfies:
$1 + z φ ′ ( z ) ν φ ( z ) ≺ 1 + A z 1 + B z ,$
then:
$φ ( z ) ≺ ( 1 + B z ) ν A − B B$
and this is the best dominant.
$( i i )$ If $B = 0$ and $ν A < π$ and if $φ$ satisfies:
$1 + z φ ′ ( z ) ν φ ( z ) ≺ 1 + A z ,$
then:
$φ ( z ) ≺ e ν A z ,$
and this is the best dominant.
Lemma 5.
[2]. Let $Ω ⊂ C$, $b ∈ C ,$ $ℜ b > 0$ and $ψ : C 2 × U → C$ satisfy $ψ i x , y ; z ∉ Ω$ for all $x , y ≤ − b − i x 2 2 ℜ b$ and all $z ∈ U$. If $p ( z ) = 1 + p 1 z + p 2 z 2 + … ,$ is analytic in $U$ and if:
$ψ p ( z ) , z p ′ z ; z ∈ Ω ,$
then $ℜ p ( z ) > 0$ in $U$.
Lemma 6.
[22]. Let $ψ z$ be analytic in $U$ with $ψ ( 0 ) = 1$ and $ψ ( z ) ≠ 0$ for all $z .$ If there exist two points $z 1 , z 2 ∈ U$ such that:
$− π 2 ρ 1 = arg { ψ ( z 1 ) } < arg { ψ ( z ) } < π 2 ρ 2 = arg { ψ ( z 2 ) } ,$
for some $ρ 1$ and $ρ 2 ρ 1 , ρ 2 > 0$ and for all $z z < z 1 = z 2 ,$ then:
$z 1 ψ ′ ( z 1 ) ψ ( z 1 ) = − i ρ 1 + ρ 2 2 κ a n d z 2 ψ ′ ( z 2 ) ψ ( z 2 ) = i ρ 1 + ρ 2 2 κ ,$
where:
$κ ≥ 1 − a 1 + a a n d a = i tan ρ 2 − ρ 1 ρ 2 + ρ 1 .$

## 3. Properties Involving $H p , η , μ λ , δ$

Unless otherwise mentioned, we assume throughout this paper that $p ∈ N , δ > − p , μ , η ∈ R , μ < p + 1 , − ∞ < λ < η + p + 1 ,$ $− 1 ≤ B < A ≤ 1 , θ > 0$, and the powers are considered principal ones.
Theorem 1.
Let $f ∈ A ( p )$ satisfy:
$( 1 − θ ) H p , η , μ λ , δ f ( z ) ′ p z p − 1 + θ H p , η , μ λ , δ + 1 f ( z ) ′ p z p − 1 ≺ 1 + A z 1 + B z .$
Then:
$ℜ H p , η , μ λ , δ f ( z ) ′ p z p − 1 1 τ > δ + p θ ∫ 0 1 u δ + p θ − 1 1 − A u 1 − B u d u 1 τ , τ ≥ 1 .$
The estimate in (13) is sharp.
Proof.
Let:
$ϕ ( z ) = H p , η , μ λ , δ f ( z ) ′ p z p − 1 ( z ∈ U ) .$
Then, φ is analytic in $U$. After some computations, we get:
$( 1 − θ ) H p , η , μ λ , δ f ( z ) ′ p z p − 1 + θ H p , η , μ λ , δ + 1 f ( z ) ′ p z p − 1 = ϕ ( z ) + θ z ϕ ′ ( z ) δ + p ≺ 1 + A z 1 + B z .$
Now, by using Lemma 1, we deduce that:
$H p , η , μ λ , δ f ( z ) ′ p z p − 1 ≺ δ + p θ z − δ + p θ ∫ 0 z t δ + p θ − 1 1 + A t 1 + B t d t ,$
or, equivalently,
$H p , η , μ λ , δ f ( z ) ′ p z p − 1 = δ + p θ ∫ 0 1 u δ + p θ − 1 1 + A u w ( z ) 1 + B u w ( z ) d u ,$
and so:
$ℜ H p , η , μ λ , δ f ( z ) ′ p z p − 1 > δ + p θ ∫ 0 1 u δ + p θ − 1 1 − A u 1 − B u d u .$
Since:
$ℜ χ 1 τ ≥ ℜ χ 1 τ χ ∈ C , ℜ χ ≥ 0 , τ ≥ 1 .$
The inequality (13) now follows from (16) and (17). To prove that the result is sharp, let:
$H p , η , μ λ , δ f ( z ) ′ p z p − 1 = δ + p θ ∫ 0 1 u δ + p θ − 1 1 + A u z 1 + B u z d u .$
Now, for $f ( z )$ defined by (18), we have:
$( 1 − θ ) H p , η , μ λ , δ f ( z ) ′ p z p − 1 + θ H p , η , μ λ , δ + 1 f ( z ) ′ p z p − 1 = 1 + A z 1 + B z z ∈ U ,$
Letting $z → − 1 ,$ we obtain:
$H p , η , μ λ , δ f ( z ) ′ p z p − 1 → δ + p θ ∫ 0 1 u δ + p θ − 1 1 − A u 1 − B u d u ,$
which ends our proof. ☐
Putting $θ = 1$ and using Lemma 1 for Equation (15) in Theorem 1, we obtain the following example.
Example 1.
Let the function $f ( z ) ∈ A ( p ) .$ Then, following containment property holds,
$T p , μ , η λ , δ + 1 ( A , B ) ⊂ T p , μ , η λ , δ ( A , B ) .$
Using (4) instead of (3) in Theorem 1, one can prove the following theorem.
Theorem 2.
Let $f ∈ A ( p )$ satisfy
$( 1 − θ ) H p , η , μ λ + 1 , δ f ( z ) ′ p z p − 1 + θ H p , η , μ λ , δ f ( z ) ′ p z p − 1 ≺ 1 + A z 1 + B z .$
Then:
$ℜ H p , η , μ λ + 1 , δ f ( z ) ′ p z p − 1 1 τ > p + η − λ θ ∫ 0 1 u p + η − λ θ − 1 1 − A u 1 − B u d u 1 τ , τ ≥ 1 .$
The result is sharp.
Putting $θ = 1$ in Theorem 2, we obtain the following example.
Example 2.
Let the function $f ( z ) ∈ A ( p ) .$ Then, following inclusion property holds
$T p , μ , η λ , δ ( A , B ) ⊂ T p , μ , η λ + 1 , δ ( A , B ) .$
For a function $f ∈ A ( p ) ,$ the generalized Bernardi–Libera–Livingston integeral operator $F p , γ$ is defined by (see [23]):
$F p , γ f ( z ) = γ + p z p ∫ 0 z t γ − 1 f ( t ) d t = z p + ∑ k = 1 ∞ γ + p γ + p + k z p + k ∗ f ( z ) ( γ > − p ) = z p 3 F 2 1 , 1 , γ + p ; 1 , γ + p + 1 ; z ∗ f ( z ) .$
Lemma 7.
If $f ∈ A p ,$ prove that:
(i)
$H p , η , μ λ , δ F p , γ f = F p , γ H p , η , μ λ , δ f ,$
(ii)
$z H p , η , μ λ , δ F p , γ f ( z ) ′ = ( p + γ ) H p , η , μ λ , δ f ( z ) − γ H p , η , μ λ , δ F p , γ f ( z ) .$
Proof.
Since
$H p , η , μ λ , δ F p , γ f = z p 3 F 2 δ + p , p + 1 − μ , p + 1 − λ + η ; p + 1 , p + 1 − μ + η ; z ∗ F p , γ f = z p 3 F 2 δ + p , p + 1 − μ , p + 1 − λ + η ; p + 1 , p + 1 − μ + η ; z ∗ z p 3 F 2 1 , 1 , γ + p ; 1 , γ + p + 1 ; z ∗ f ( z ) ,$
and:
$F p , γ H p , η , μ λ , δ f = z p 3 F 2 1 , 1 , γ + p ; 1 , γ + p + 1 ; z ∗ H p , η , μ λ , δ f = z p 3 F 2 1 , 1 , γ + p ; 1 , γ + p + 1 ; z ∗ z p 3 F 2 δ + p , p + 1 − μ , p + 1 − λ + η ; p + 1 , p + 1 − μ + η ; z ∗ f ( z ) .$
Now, the first part of this lemma follows. Furthermore,
$z F p , γ f ( z ) ′ = ( p + γ ) f ( z ) − γ F p , γ f ( z ) .$
If we replace $f ( z )$ by $H p , η , μ λ , δ f ( z )$ and using the first part of this lemma, we get (21). ☐
Theorem 3.
Suppose that $p + γ > 0 , f ∈$ $T p , η , μ λ , δ ( A , B )$ and $F p , γ$ defined by (20). Then:
$ℜ H p , η , μ λ , δ F p , γ f ( z ) ′ p z p − 1 1 τ > p + γ ∫ 0 1 u p + γ − 1 1 − A u 1 − B u d u 1 τ , τ ≥ 1 .$
The result is sharp.
Proof.
Let:
$ϕ ( z ) = H p , η , μ λ , δ F p , γ f ( z ) ′ p z p − 1 ( z ∈ U ) .$
Then, φ is analytic in $U$. After some calculations, we have:
$( H p , η , μ λ , δ f ( z ) ) ′ p z p − 1 = ϕ ( z ) + z ϕ ′ ( z ) p + γ ≺ 1 + A z 1 + B z .$
Employing the same technique that was used in proving Theorem 1, the remaining part of the theorem can be proven. ☐
Theorem 4.
Let $− 1 ≤ B i < A i ≤ 1 ( i = 1 , 2 ) .$ If each of the functions $f i ∈ A ( p )$ satisfies:
$( 1 − θ ) H p , η , μ λ , δ f i ( z ) z p + θ H p , η , μ λ , δ + 1 f i ( z ) z p ≺ 1 + A i z 1 + B i z ( i = 1 , 2 ) ,$
then:
$( 1 − θ ) H p , η , μ λ , δ F ( z ) z p + θ H p , η , μ λ , δ + 1 F ( z ) z p ≺ 1 + ( 1 − 2 ϱ ) z 1 − z ,$
where:
$F ( z ) = H p , η , μ λ , δ ( f 1 ∗ f 2 ) ( z )$
and:
$ϱ = 1 − 4 ( A 1 − B 1 ) ( A 2 − B 2 ) ( 1 − B 1 ) ( 1 − B 2 ) 1 − 1 2 2 F 1 1 , 1 ; δ + p θ + 1 ; 1 2 .$
The result is possible when $B 1 = B 2 = − 1 .$
Proof.
Suppose that $f i ∈ A ( p ) ( i = 1 , 2 )$ satisfy the condition (25). Setting:
$p i ( z ) = ( 1 − θ ) H p , η , μ λ , δ f i ( z ) z p + θ H p , η , μ λ , δ + 1 f i ( z ) z p ( i = 1 , 2 ) ,$
we have:
$p i ( z ) ∈ P ( ς i ) ς i = 1 − A i 1 − B i , i = 1 , 2 .$
Thus, by making use of the identity (3) in (29), we get:
$H p , η , μ λ , δ f i ( z ) = δ + p θ z p − δ + p θ ∫ 0 z t δ + p θ − 1 p i ( t ) d t ( i = 1 , 2 ) ,$
which, in view of F given by (27) and (30), yields:
$H p , η , μ λ , δ F ( z ) = δ + p θ z p − δ + p θ ∫ 0 z t δ + p θ − 1 F ( t ) d t ,$
where:
$F ( z ) = ( 1 − θ ) H p , η , μ λ , δ F ( z ) z p + θ H p , η , μ λ , δ + 1 F ( z ) z p = δ + p θ z − δ + p θ ∫ 0 z t δ + p θ − 1 ( p 1 ∗ p 2 ) ( t ) d t .$
Since $p i ( z ) ∈ P ( ς i ) ( i = 1 , 2 ) ,$ it follows from Lemma 3 that:
$( p 1 ∗ p 2 ) ( z ) ∈ P ( ς 3 ) ( ς 3 = 1 − 2 ( 1 − ς 1 ) ( 1 − ς 2 ) ) .$
Now, by using (33) in (32) and then appealing to Lemma 2, we have:
$ℜ F ( z ) = δ + p θ ∫ 0 1 u δ + p θ − 1 ℜ ( p 1 ∗ p 2 ) ( u z ) d u ⩾ δ + p θ ∫ 0 1 u δ + p θ − 1 2 ς 3 − 1 + 2 ( 1 − ς 3 ) 1 + u z d u > δ + p θ ∫ 0 1 u δ + p θ − 1 2 ς 3 − 1 + 2 ( 1 − ς 3 ) 1 + u d u = 1 − 4 ( A 1 − B 1 ) ( A 2 − B 2 ) ( 1 − B 1 ) ( 1 − B 2 ) 1 − δ + p θ ∫ 0 1 u δ + p θ − 1 ( 1 + u ) − 1 d u = 1 − 4 ( A 1 − B 1 ) ( A 2 − B 2 ) ( 1 − B 1 ) ( 1 − B 2 ) 1 − 1 2 2 F 1 1 , 1 ; δ + p θ + 1 ; 1 2 = ϱ .$
When $B 1 = B 2 = − 1 ,$ we consider the functions $f i ( z ) ∈ A ( p ) ( i = 1 , 2 )$, which satisfy (25), are defined by:
$H p , η , μ λ , δ f i ( z ) = δ + p θ z p − δ + p θ ∫ 0 z t δ + p θ − 1 1 + A i t 1 − t d t ( i = 1 , 2 ) .$
Thus, it follows from (32) that:
$F ( z ) = δ + p θ ∫ 0 1 u δ + p θ − 1 1 − ( 1 + A 1 ) ( 1 + A 2 ) + ( 1 + A 1 ) ( 1 + A 2 ) ( 1 − u z ) d u$
$= 1 − ( 1 + A 1 ) ( 1 + A 2 ) + ( 1 + A 1 ) ( 1 + A 2 ) ( 1 − z ) − 1 2 F 1 1 , 1 ; δ + p θ + 1 ; z z − 1$
$→ 1 − ( 1 + A 1 ) ( 1 + A 2 ) + 1 2 ( 1 + A 1 ) ( 1 + A 2 ) 2 F 1 1 , 1 ; δ + p θ + 1 ; 1 2 as z → − 1 ,$
which evidently ends the proof. ☐
Theorem 5.
Let $υ ∈ C ∗$, and let $A , B ∈ C$ with $A ≠ B$ and $B ≤ 1 .$ Suppose that:
$υ ( δ + p ) ( A − B ) B − 1 ≤ 1 o r υ ( δ + p ) ( A − B ) B + 1 ≤ 1 i f B ≠ 0 , υ ( δ + p ) A ≤ π i f B = 0 .$
If $f ∈ A ( p )$ with $H p , η , μ λ , δ f ( z ) ≠ 0$ for all $z ∈ U ∗ = U ∖ { 0 } ,$ then:
$H p , η , μ λ , δ + 1 f ( z ) H p , η , μ λ , δ f ( z ) ≺ 1 + A z 1 + B z ,$
implies:
$H p , η , μ λ , δ f ( z ) z p υ ≺ q ( z ) ,$
where:
$q ( z ) = ( 1 + B z ) υ ( δ + p ) ( A − B ) / B i f B ≠ 0 , e υ ( δ + p ) A z i f B = 0 ,$
is the best dominant.
Proof.
Putting:
$Δ ( z ) = H p , η , μ λ , δ f ( z ) z p υ ( z ∈ U ) .$
Then, $Δ$ is analytic in $U$, $Δ ( 0 ) = 1$ and $Δ ( z ) ≠ 0$ for all $z ∈ U .$ Taking the logarithmic derivatives on both sides of (34) and using (3), we have:
$1 + z Δ ′ ( z ) υ ( δ + p ) Δ ( z ) = H p , η , μ λ , δ + 1 f ( z ) H p , η , μ λ , δ f ( z ) ≺ 1 + A z 1 + B z .$
Now, the assertions of Theorem 5 follow by Lemma 4. ☐
Theorem 6.
Let $0 ≤ α ≤ 1 , ζ > 1 .$ If $f ( z ) ∈ A p$ satisfies:
$ℜ 1 − α H p , η , μ λ , δ + 2 f ( z ) ′ H p , η , μ λ , δ + 1 f ( z ) ′ + α H p , η , μ λ , δ + 1 f ( z ) ′ H p , η , μ λ , δ f ( z ) ′ < ζ ,$
then:
$ℜ H p , η , μ λ , δ + 1 f ( z ) H p , η , μ λ , δ f ( z ) < β ,$
where $β ∈ 1 , ∞$ is the positive root of the equation:
$2 δ + p + α β 2 − 2 ζ δ + p + 1 − 1 − α β − 1 − α = 0 .$
Proof.
Let:
$H p , η , μ λ , δ + 1 f ( z ) H p , η , μ λ , δ f ( z ) = β + 1 − β φ ( z ) .$
Then, φ is analytic in $U$, $φ ( 0 ) = 1$ and $φ ( z ) ≠ 0$ for all $z ∈ U .$ Taking the logarithmic derivatives on both sides of (37) and using the identity (3), we have:
$δ + p + 1 H p , η , μ λ , δ + 2 f ( z ) ′ H p , η , μ λ , δ + 1 f ( z ) ′ − δ + p H p , η , μ λ , δ + 1 f ( z ) ′ H p , η , μ λ , δ f ( z ) ′ = 1 + 1 − β z φ ′ ( z ) β + 1 − β φ ( z ) ,$
and so:
$1 − α H p , η , μ λ , δ + 2 f ( z ) ′ H p , η , μ λ , δ + 1 f ( z ) ′ + α H p , η , μ λ , δ + 1 f ( z ) ′ H p , η , μ λ , δ f ( z ) ′ = α β + 1 − α δ + p β δ + p + 1$
$+ 1 − β α + 1 − α δ + p δ + p + 1 φ ( z ) + 1 − α 1 − β β + 1 − β φ ( z ) δ + p + 1 z φ ′ z .$
Let:
$Ψ r , s ; z = α β + 1 − α δ + p β δ + p + 1 + 1 − β α + 1 − α δ + p δ + p + 1 r + 1 − α 1 − β β + 1 − β φ ( z ) δ + p + 1 s ,$
and:
$Ω = w ∈ C : ℜ ( w ) < ζ .$
Then, for $x , y ≤ − 1 + x 2 2 ,$ we have:
$ℜ Ψ i x , y ; z = α β + 1 − α δ + p β δ + p + 1 + 1 − α 1 − β β y β 2 + 1 − β 2 x 2 δ + p + 1 ≥ α β + 1 − α δ + p β δ + p + 1 − 1 − α 1 − β 2 β δ + p + 1 = ζ ,$
where $β$ is the positive root of Equation (36). Suppose that:
$R β = 2 δ + p + α β 2 − 2 ζ δ + p + 1 − 1 − α β − 1 − α = 0 .$
For $β = 0 , R 0 = − 1 − α ≤ 0$ and for $β = 1 , R 1 = 2 δ + p 1 − ζ + 2 α − ζ ≤ 0 .$ This proves that $β ∈ 1 , ∞ .$ Thus, for $z ∈ U$, $Ψ i x , y ; z ∉ Ω$, and so, we obtain the required result by an application of Lemma 5. ☐
Theorem 7.
Suppose that $0 < ε 1 , ε 2 ≤ 1 .$ If:
$− π 2 ε 1 < arg ( 1 − θ ) H p , η , μ λ , δ f ( z ) ′ p z p − 1 + θ H p , η , μ λ , δ + 1 f ( z ) ′ p z p − 1 < π 2 ε 2 ,$
then:
$− π 2 ξ 1 < arg H p , η , μ λ , δ f ( z ) ′ p z p − 1 < π 2 ξ 2 ,$
where:
$ε 1 = ξ 1 + 2 π arctan ξ 1 + ξ 2 θ 2 δ + p 1 − a 1 + a , ε 2 = ξ 2 + 2 π arctan ξ 1 + ξ 2 θ 2 δ + p 1 − a 1 + a .$
Proof.
Let:
$ϕ ( z ) = H p , η , μ λ , δ f ( z ) ′ p z p − 1 ( z ∈ U ) .$
Then, from Theorem 1, we have:
$( 1 − θ ) H p , η , μ λ , δ f ( z ) ′ p z p − 1 + θ H p , η , μ λ , δ + 1 f ( z ) ′ p z p − 1 = ϕ ( z ) + θ z ϕ ′ ( z ) δ + p .$
Let $U ( z )$ be the function that maps $U$ onto the domain:
$w ∈ C : − π 2 ε 1 < arg ( w ) < π 2 ε 2 ,$
with $U ( 0 ) = 1 ,$ then:
$ϕ ( z ) + θ z ϕ ′ ( z ) δ + p ≺ U ( z ) .$
Assume that $z 1 , z 2$ are two points in $U$ such that the condition (9) is satisfied, then by Lemma 6, we obtain (10) under the constraint (11). Therefore,
$arg δ + p ϕ ( z 1 ) + θ z 1 ϕ ′ ( z 1 ) = arg ϕ ( z 1 ) δ + p + θ z 1 ϕ ′ ( z 1 ) ϕ ( z 1 ) = arg ϕ ( z 1 ) + arg δ + p + θ z 1 ϕ ′ ( z 1 ) ϕ ( z 1 ) = − π 2 ξ 1 + arg δ + p − i θ ξ 1 + ξ 2 κ 2 = − π 2 ξ 1 − arctan ξ 1 + ξ 2 θ κ 2 δ + p ≤ − π 2 ξ 1 − arctan ξ 1 + ξ 2 θ 2 δ + p 1 − a 1 + a ,$
and:
$arg δ + p ϕ ( z 2 ) + θ z 2 ϕ ′ ( z 2 ) ≥ π 2 ξ 2 + arctan ξ 1 + ξ 2 θ 2 δ + p 1 − a 1 + a .$
which contradicts the assumption (38). This evidently completes the proof of Theorem 7. ☐

## Acknowledgments

The authors would like to thank all referees for their valuable comments which led to the improvement of this paper.

## Author Contributions

All the authors read and approved the final manuscript as a consequence of the authors meetings.

## Conflicts of Interest

The authors declare no conflict of interest.

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