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Article

# New Applications of the Bernardi Integral Operator

by
Shigeyoshi Owa
1,* and
H. Özlem Güney
2
1
“1 Decembrie 1918” University Alba Iulia, 510009 Alba-Iulia, Romania
2
Department of Mathematics, Faculty of Science Dicle University, 21280 Diyarbakır, Turkey
*
Author to whom correspondence should be addressed.
Mathematics 2020, 8(7), 1180; https://doi.org/10.3390/math8071180
Submission received: 30 June 2020 / Revised: 15 July 2020 / Accepted: 16 July 2020 / Published: 17 July 2020
(This article belongs to the Special Issue Complex Analysis and Geometric Function Theory)

## Abstract

:
Let $A ( p , n )$ be the class of $f ( z )$ which are analytic p-valent functions in the closed unit disk $U ¯ = z ∈ C : z ≤ 1$. The expression $B − m − λ f ( z )$ is defined by using fractional integrals of order $λ$ for $f ( z ) ∈ A ( p , n ) .$ When $m = 1$ and $λ = 0 ,$ $B − 1 f ( z )$ becomes Bernardi integral operator. Using the fractional integral $B − m − λ f ( z ) ,$ the subclass $T p , n α s , β , ρ ; m , λ$ of $A ( p , n )$ is introduced. In the present paper, we discuss some interesting properties for $f ( z )$ concerning with the class $T p , n α s , β , ρ ; m , λ .$ Also, some interesting examples for our results will be considered.

## 1. Introduction

Let $A ( p , n )$ be the class of functions $f ( z )$ of the form
$f ( z ) = z p + ∑ k = p + n ∞ a k z k , n ∈ N = { 1 , 2 , 3 , … }$
that are analytic p-valent functions in the closed unit disk $U ¯ = z ∈ C : z ≤ 1$. For functions $f ( z ) ∈ A ( p , n ) ,$ we consider
$B − 1 f ( z ) = p + γ z γ ∫ 0 z t γ − 1 f ( t ) d t = z p + ∑ k = p + n ∞ p + γ k + γ a k z k , γ ∈ N .$
If $p = 1 ,$ for $f ( z ) ∈ A ( 1 , n )$
$B − 1 f ( z ) = 1 + γ z γ ∫ 0 z t γ − 1 f ( t ) d t = z + ∑ k = n + 1 ∞ 1 + γ k + γ a k z k , γ ∈ N$
is considered by Bernardi [1]. Therefore, $B − 1 f ( z )$ in (3) is said to be the Bernardi integral operator. Further, if $p = 1$ and $γ = 1 ,$ for $f ( z ) ∈ A ( 1 , n )$
$L − 1 f ( z ) = 2 z ∫ 0 z f ( t ) d t = z + ∑ k = n + 1 ∞ 2 k + 1 a k z k$
is defined by Libera [2]. Therefore, $L − 1 f ( z )$ in (4) is called the Libera integral operator.
For $B − 1 f ( z )$ in (2), we consider
$B − 2 f ( z ) = B − 1 B − 1 f ( z ) = z p + ∑ k = p + n ∞ p + γ k + γ 2 a k z k$
and
$B − m f ( z ) = B − 1 B − m + 1 f ( z ) = z p + ∑ k = p + n ∞ p + γ k + γ m a k z k$
with $m ∈ N$ and $B 0 f ( z ) = f ( z ) .$
From the various definitions of fractional calculus of $f ( z ) ∈ A ( p , n )$ (that is, fractional integrals and fractional derivatives) given in the literature, we would like to recall here the following definitions for fractional calculus which were used by Owa [3] and Owa and Srivastava [4].
Definition 1.
The fractional integral of order λ for $f ( z ) ∈ A ( p , n )$ is defined by
$D z − λ f ( z ) = 1 Γ ( λ ) ∫ 0 z f ( t ) ( z − t ) 1 − λ d t , λ > 0$
where the multiplicity of $( z − t ) λ − 1$ is removed by requiring $l o g ( z − t )$ to be real when $z − t > 0$ and Γ is the Gamma function.
With the above definitions, we know that
$D z − λ f ( z ) = Γ ( p + 1 ) Γ ( p + 1 + λ ) z p + λ + ∑ k = p + n ∞ Γ ( k + 1 ) Γ ( k + 1 + λ ) a k z k + λ$
for $λ > 0$ and $f ( z ) ∈ A ( p , n ) .$ Using the fractional integral operator over $A ( p , n ) ,$ we consider
$B − λ f ( z ) = Γ ( p + γ + λ ) Γ ( p + γ ) z 1 − γ − λ D z − λ z γ − 1 f ( z ) = z p + ∑ k = p + n ∞ Γ ( k + γ ) Γ ( p + γ + λ ) Γ ( p + γ ) Γ ( k + γ + λ ) a k z k ,$
where $0 ≤ λ ≤ 1 .$ If $λ = 0$ in (9), then $B 0 f ( z ) = f ( z )$ and if $λ = 1$ in (9), then we see that
$B − 1 f ( z ) = p + γ z γ ∫ 0 z t γ − 1 f ( t ) d t .$
With the operator $B − λ f ( z )$ given by (9), we know
$B − m − λ f ( z ) = B − m B − λ f ( z ) = z p + ∑ k = p + n ∞ p + γ k + γ m Γ ( k + γ ) Γ ( p + γ + λ ) Γ ( p + γ ) Γ ( k + γ + λ ) a k z k ,$
where $0 ≤ λ ≤ 1$ and $m ∈ N .$ The operator $B − m − λ f ( z )$ is a generalization of the Bernardi integral operator $B − 1 f ( z )$. From the definition of $B − m − λ f ( z ) ,$ we know that
$B − m − λ f ( z ) = B − m B − λ f ( z ) = B − λ B − m f ( z ) .$
From s different boundary points $z l ( l = 1 , 2 , 3 , … , s )$ with $| z l | = 1 ,$ we consider
$α s = 1 s ∑ l = 1 s B − m − λ f ( z l ) z l p ,$
where $α s ∈ e i β B − m − λ f ( U ) ,$$α s ≠ 1 ,$ $− π 2 ≤ β ≤ π 2$ and $U = { z ∈ C : | z | < 1 }$ is the open unit disk. For such $α s ,$ if $f ( z ) ∈ A ( p , n )$ satisfies
$e i β B − m − λ f ( z ) z p − α s e i β − α s − 1 < ρ , z ∈ U$
for some real $ρ > 0 ,$ we say that the function $f ( z )$ belongs to the class $T p , n α s , β , ρ ; m , λ .$
It is clear that a function $f ( z ) ∈ A ( p , n )$ belongs to the class $T p , n α s , β , ρ ; m , λ$ provided that the condition
$B − m − λ f ( z ) z p − 1 < ρ e i β − α s , z ∈ U ,$
is satisfied. If we consider the function $f ( z ) ∈ A ( p , n )$ given by
$f ( z ) = z p + p + n + γ p + γ m Γ p + γ Γ p + n + γ + λ Γ p + n + γ Γ p + γ + λ ρ ( e i β − α s ) z p + n$
then $f ( z )$ satisfies
$B − m − λ f ( z ) z p − 1 = ρ e i β − α s | z | m < ρ e i β − α s , z ∈ U .$
Therefore, $f ( z )$ given by (16) is in the class $T p , n α s , β , ρ ; m , λ .$
Discussing our problems for $f ( z ) ∈ T p , n α s , β , ρ ; m , λ ,$ we have to recall here the following lemma due to Miller and Mocanu [5,6] (refining the old one in Jack [7].)
Lemma 1.
Let the function $w ( z )$ given by
$w ( z ) = a n z n + a n + 1 z n + 1 + a n + 2 z n + 2 + … , n ∈ N$
be analytic in $U$ with $w ( 0 ) = 0 .$ If $| w ( z ) |$ attains its maximum value on the circle $| z | = r$ at a point $z 0 , ( 0 < | z 0 | < 1 )$ then there exists a real number $k ≥ n$ such that
$z 0 w ′ ( z 0 ) w ( z 0 ) = k$
and
$R e 1 + z 0 w ″ ( z 0 ) w ′ ( z 0 ) ≥ k .$

## 2. Properties of Functions Concerning with the Class Tp,n (αs, β, ρ; m, λ)

We begin with a sufficient condition on a function $f ( z ) ∈ A ( p , n )$ which makes it a member of $T p , n α s , β , ρ ; m , λ .$
Theorem 1.
If $f ( z ) ∈ A ( p , n )$ satisfies
$B − m − λ + 1 f ( z ) B − m − λ f ( z ) − 1 < e i β − α s n ρ ( p + γ ) ( 1 + e i β − α s ρ ) , z ∈ U$
for some $α s$ given by (13) with $α s ≠ 1$ such that $z g ∈ ∂ U ( g = 1 , 2 , 3 , … , s ) ,$ and for some real $ρ > 1 ,$ then
$B − m − λ f ( z ) z p − 1 < ρ e i β − α s , z ∈ U$
that is, $f ( z ) ∈ T p , n α s , β , ρ ; m , λ .$
Proof.
We introduce the function $w ( z )$ defined by
$w ( z ) = e i β B − m − λ f ( z ) z p − α s e i β − α s − 1 = e i β e i β − α s ∑ k = p + n ∞ p + γ k + γ m Γ ( k + γ ) Γ ( p + γ + λ ) Γ ( p + γ ) Γ ( k + γ + λ ) a k z k − p .$
Then, $w ( z )$ is analytic in $U$ with $w ( 0 ) = 0$ and
$B − m − λ f ( z ) z p = 1 + ( 1 − e − i β α s ) w ( z ) .$
Noting that
$B − m − λ + 1 f ( z ) = γ p + γ B − m − λ f ( z ) + 1 p + γ z ( B − m − λ f ( z ) ) ′ ,$
we obtain that
$B − m − λ + 1 f ( z ) B − m − λ f ( z ) − 1 = ( 1 − e − i β α s ) z w ′ ( z ) ( p + γ ) ( 1 + ( 1 − e − i β α s ) w ( z ) )$
and that
$B − m − λ + 1 f ( z ) B − m − λ f ( z ) − 1 = 1 p + γ ( 1 − e − i β α s ) z w ′ ( z ) 1 + ( 1 − e − i β α s ) w ( z ) < | e i β − α s | n ρ ( p + γ ) ( 1 + | e i β − α s | ρ ) .$
by employing (21). Assume, to arrive at a contradiction, that there exists a point $z 0 , ( 0 < | z 0 | < 1 )$ such that
$max { | w ( z ) | ; | z | ≤ | z 0 | } = | w ( z 0 ) | = ρ > 1 .$
Then, we can write that $w ( z 0 ) = ρ e i θ , ( 0 ≤ θ ≤ 2 π )$ and $z 0 w ′ ( z 0 ) = k w ( z 0 ) , ( k ≥ n )$ by Lemma 1. For such a point $z 0 ∈ U ,$ $f ( z )$ satisfies
$B − m − λ + 1 f ( z 0 ) B − m − λ f ( z 0 ) − 1 = 1 p + γ ( 1 − e − i β α s ) z 0 w ′ ( z 0 ) 1 + ( 1 − e − i β α s ) w ( z 0 )$
$= 1 p + γ ( 1 − e − i β α s ) k ρ 1 + ( 1 − e − i β α s ) ρ e i θ$
$≥ | 1 − e − i β α s | n ρ ( p + γ ) ( 1 + | 1 − e − i β α s | ρ )$
$= | e i β − α s | n ρ ( p + γ ) ( 1 + | e i β − α s | ρ ) .$
Since this contradicts our condition (21), we see that there is no $z 0 , ( 0 < | z 0 | < 1 )$ such that $| w ( z 0 ) | = ρ > 1 .$ This shows us that
$| w ( z ) | = e i β B − m − λ f ( z ) z p − 1 e i β − α s < ρ , z ∈ U ,$
that is, that
$B − m − λ f ( z ) z p − 1 < ρ e i β − α s , z ∈ U .$
This completes the proof of the theorem.  □
Example 1.
We consider a function $f ( z ) ∈ A ( p , n )$ given by
$f ( z ) = z p + a p + n z p + n , z ∈ U$
with $0 < | a p + n | < 1 2 Q$, where
$Q = p + γ p + n + γ m Γ ( p + n + γ ) Γ ( p + γ + λ ) Γ ( p + γ ) Γ ( p + n + γ + λ ) .$
For such $f ( z )$, we have
$B − m − λ + 1 f ( z ) B − m − λ f ( z ) = z p + p + n + γ p + γ Q a p + n z p + n z p + Q a p + n z p + n = 1 + p + n + γ p + γ Q a p + n z n 1 + Q a p + n z n$
that is, that
$B − m − λ + 1 f ( z ) B − m − λ f ( z ) − 1 = n p + γ Q a p + n z n 1 + Q a p + n z n < n p + γ Q | a p + n | 1 − Q | a p + n | , z ∈ U .$
Now, we consider five boundary points such that
$z 1 = e − i a r g ( a p + n ) n$
$z 2 = e i π − 6 a r g ( a p + n ) 6 n$
$z 3 = e i π − 4 a r g ( a p + n ) 4 n$
$z 4 = e i π − 3 a r g ( a p + n ) 3 n$
and
$z 5 = e i π − 2 a r g ( a p + n ) 2 n .$
For these five boundary points, we know that
$B − m − λ f ( z 1 ) z 1 p = 1 + Q a p + n e − i a r g ( a p + n ) = 1 + Q | a p + n | ,$
$B − m − λ f ( z 2 ) z 2 p = 1 + Q a p + n e i π 6 − a r g ( a p + n ) = 1 + 3 + i 2 Q | a p + n | ,$
$B − m − λ f ( z 3 ) z 3 p = 1 + Q a p + n e i π 4 − a r g ( a p + n ) = 1 + 2 ( 1 + i ) 2 Q | a p + n | ,$
$B − m − λ f ( z 4 ) z 4 p = 1 + Q a p + n e i π 3 − a r g ( a p + n ) = 1 + 1 + 3 i 2 Q | a p + n | ,$
and
$B − m − λ f ( z 5 ) z 5 p = 1 + Q a p + n e i π 2 − a r g ( a p + n ) = 1 + i Q | a p + n | .$
Thus $α 5$ is given by
$α 5 = 1 5 ∑ l = 1 5 B − m − λ f ( z l ) z l p = 1 + ( 3 + 2 + 3 ) ( 1 + i ) 10 Q | a p + n | .$
This gives us that
$1 − e − i β α 5 = 2 ( 3 + 2 + 3 ) 10 Q | a p + n |$
with $β = 0 .$ For such $α 5$ and $β ,$ we take $ρ > 1$ with
$n p + γ Q | a p + n | 1 − Q | a p + n | ≤ | e i β − α 5 | n ρ ( p + γ ) ( 1 + | e i β − α 5 | ρ ) .$
It follows from the above that
$ρ ≥ 10 2 ( 3 + 2 + 3 ) ( 1 − 2 Q | a p + n | ) > 10 2 ( 3 + 2 + 3 ) > 1 .$
For such $α 5$ and $ρ > 1$, $f ( z )$ satisfies
$B − m − λ f ( z ) z p − 1 < Q | a p + n | ≤ ρ | e i β − α 5 | , z ∈ U .$
Our next result reads as follows.
Theorem 2.
If $f ( z ) ∈ A ( p , n )$ satisfies
$B − m − λ + 1 f ( z ) B − m − λ f ( z ) − 1 B − m − λ f ( z ) z p − 1 < e i β − α s 2 n ρ 2 ( p + γ ) ( 1 + e i β − α s ρ ) , z ∈ U$
for some $α s$ defined by (13) with $α s ≠ 1$ such that $z g ∈ ∂ U ( g = 1 , 2 , 3 , … s ) ,$ and for some real $ρ > 1 ,$ then
$B − m − λ f ( z ) z p − 1 < ρ e i β − α s , z ∈ U$
that is, $f ( z ) ∈ T p , n α s , β , ρ ; m , λ .$
Proof.
Define a function $w ( z )$ by (23). Using (24) and (26), we have
$B − m − λ + 1 f ( z ) B − m − λ f ( z ) − 1 B − m − λ f ( z ) z p − 1 = 1 − e − i β α s 2 z w ( z ) w ′ ( z ) ( p + γ ) ( 1 + 1 − e − i β α s w ( z ) ) .$
We suppose that there exists a point $z 0 , ( 0 < | z 0 | < 1 )$ such that
$max { | w ( z ) | ; | z | ≤ | z 0 | } = | w ( z 0 ) | = ρ > 1 .$
Then, Lemma 1 leads us that $w ( z 0 ) = ρ e i θ , ( 0 ≤ θ ≤ 2 π )$ and $z 0 w ′ ( z 0 ) = k w ( z 0 ) , ( k ≥ n ) .$ It follows from the above that
$B − m − λ + 1 f ( z 0 ) B − m − λ f ( z 0 ) − 1 B − m − λ f ( z 0 ) z 0 p − 1 = 1 − e − i β α s 2 z 0 w ( z 0 ) w ′ ( z 0 ) ( p + γ ) ( 1 + 1 − e − i β α s w ( z 0 ) )$
$= e i β − α s 2 ρ 2 k ( p + γ ) 1 + ( 1 − e − i β α s ) ρ e i θ$
$≥ e i β − α s 2 n ρ 2 ( p + γ ) ( 1 + e i β − α s ρ ) .$
This contradicts our condition (51) for $f ( z )$. Therefore, there is no $z 0 , ( 0 < | z 0 | < 1 )$ such that $| w ( z 0 ) | = ρ > 1 .$ This means that
$B − m − λ f ( z ) z p − 1 < ρ | e i β − α s | , z ∈ U .$
□
Example 2.
Consider a function $f ( z )$ given by (32) with $0 < | a p + n | < 1 Q$, where Q is given by (33). For this function $f ( z ) ,$ we have
$B − m − λ + 1 f ( z ) B − m − λ f ( z ) − 1 B − m − λ f ( z ) z p − 1 = n Q 2 a p + n 2 z 2 n ( p + γ ) ( 1 + Q a p + n z n )$
$< n Q 2 | a p + n | 2 ( p + γ ) ( 1 − Q | a p + n | ) , z ∈ U .$
Consider five boundary points $z 1 , z 2 , z 3 , z 4$ and $z 5$ in Example 1. Then, we have
$1 − e − i β α 5 = 2 ( 3 + 2 + 3 ) 10 Q | a p + n |$
with $β = 0 .$ With such $α 5$ and $β ,$ we take $ρ > 1$ by
$n Q 2 | a p + n | 2 ( p + γ ) ( 1 − Q | a p + n | ) ≤ e i β − α 5 2 n ρ 2 ( p + γ ) ( 1 + e i β − α 5 ρ ) .$
Then, this ρ satisfies
$ρ ≥ 10 2 ( 3 + 2 + 3 ) Q | a p + n | > 1 .$
For such $α 5$ and ρ, we know that
$B − m − λ f ( z ) z p − 1 < ρ | e i β − α 5 | , z ∈ U .$
Next, we derive the following result.
Theorem 3.
If $f ( z ) ∈ A ( p , n )$ satisfies
$B − m − λ + g f ( z ) z p − 1 < ρ | e i β − α s | p + n + γ p + γ , z ∈ U$
for some $α s$ defined by (13) with $α s ≠ 1 , g = 1 , 2 , 3 , … m ,$ and for some real $ρ > 1 ,$ then
$B − m − λ + g − 1 f ( z ) z p − 1 < ρ e i β − α s , z ∈ U .$
Proof.
Define the function $w ( z )$ by
$w ( z ) = e i β B − m − λ + g − 1 f ( z ) z p − α s e i β − α s − 1$
$= e i β e i β − α s ∑ k = p + n ∞ p + γ k + γ m − g Γ ( k + γ ) Γ ( p + γ + λ ) Γ ( p + γ ) Γ ( k + γ + λ ) a k z k − p .$
It follows from the above that
$B − m − λ + g − 1 f ( z ) = z p + ( 1 − e − i β α s ) z p w ( z ) .$
By the definition of $B − m − λ + g f ( z ) ,$ we know that
$B − m − λ + g f ( z ) = z 1 − γ p + γ z γ B − m − λ + g − 1 f ( z ) ′ = z p 1 + ( 1 − e − i β α s ) w ( z ) 1 + z w ′ ( z ) ( p + γ ) w ( z ) .$
Our condition implies that
$B − m − λ + g f ( z ) z p − 1 = ( 1 − e − i β α s ) w ( z ) 1 + z w ′ ( z ) ( p + γ ) w ( z ) < ρ e i β − α s p + n + γ p + γ$
for all $z ∈ U .$ Suppose that there exists a point $z 0 , ( 0 < | z 0 | < 1 )$ such that
$max { | w ( z ) | ; | z | ≤ | z 0 | } = | w ( z 0 ) | = ρ > 1 .$
Then, Lemma 1 says that $w ( z 0 ) = ρ e i θ , ( 0 ≤ θ ≤ 2 π )$ and $z 0 w ′ ( z 0 ) = k w ( z 0 ) , ( k ≥ n )$. Therefore, we have
$B − m − λ + g f ( z 0 ) z 0 p − 1 = ( 1 − e − i β α s ) w ( z 0 ) 1 + z 0 w ′ ( z 0 ) ( p + γ ) w ( z 0 )$
$= ρ e i β − α s 1 + k p + γ ≥ ρ e i β − α s p + n + γ p + γ$
which contradicts the inequality (67). This means that there is no $z 0 ∈ U$ such that $| w ( z 0 | = ρ > 1 .$ Thus we know that
$| w ( z ) | = e i β B − m − λ + g − 1 f ( z ) z p − α s e i β − α s − 1 = e i β e i β − α s B − m − λ + g − 1 f ( z ) z p − 1 < ρ .$
This completes the proof of the theorem.  □
Theorem 3 implies the following one.
Theorem 4.
If $f ( z ) ∈ A ( p , n )$ satisfies
$B − m − λ + g f ( z ) z p − 1 < ρ e i β − α s p + n + γ p + γ g , z ∈ U$
for some $α s$ given by (13) with $α s ≠ 1 , g = 1 , 2 , 3 , … , m ,$ and for some real $ρ > 1 ,$ then
$B − m − λ f ( z ) z p − 1 < ρ e i β − α s , z ∈ U$
or, equivalently, $f ( z ) ∈ T p , n α s , β , ρ ; m , λ .$
Proof.
By means of Theorem 3, we see that if $f ( z )$ satisfies the inequality (71), then
$B − m − λ + g − 1 f ( z ) z p − 1 < ρ e i β − α s p + n + γ p + γ g − 1 , z ∈ U$
Similarly, we have
$B − m − λ + g − 2 f ( z ) z p − 1 < ρ e i β − α s p + n + γ p + γ g − 2 , z ∈ U .$
Continuing this consideration, we obtain that
$B − m − λ f ( z ) z p − 1 < ρ e i β − α s , z ∈ U .$
□
Example 3.
Consider the function
$f ( z ) = z p + a p + n z p + n , z ∈ U$
which satisfies
$B − m − λ + g f ( z ) = z p + Γ ( p + n + γ ) Γ ( p + γ + λ ) Γ ( p + γ ) Γ ( p + n + γ + λ ) p + γ p + n + γ m − g a p + n z p + n .$
It follows from (77) that
$B − m − λ + g f ( z ) z p − 1 = Γ ( p + n + γ ) Γ ( p + γ + λ ) Γ ( p + γ ) Γ ( p + n + γ + λ ) p + γ p + n + γ m − g | a p + n | | z | n$
$< Q p + n + γ p + γ g | a p + n | , z ∈ U ,$
where Q is given by (33). Now, we consider the five boundary points $z 1 , z 2 , z 3 , z 4$ and $z 5$ as in Example 1. Then we see
$| e i β − α 5 | = 2 ( 3 + 2 + 3 ) 10 Q | a p + n |$
where $β = 0 .$ With the above relation (79), we consider $ρ > 1$ such that
$Q p + n + γ p + γ g | a p + n | ≤ ρ | e i β − α 5 | ,$
that is, ρ satisfies
$ρ ≥ 10 2 ( 3 + 2 + 3 ) p + n + γ p + γ g > 1 .$
Thus, we have that
$B − m − λ f ( z ) z p − 1 ≤ Q | a p + n |$
$≤ ρ | e i β − α 5 | p + γ p + n + γ g$
$< ρ | e i β − α 5 | , z ∈ U .$
Remark 1.
If we take $γ = 1$ in the results of this section, then these results correspond to applications of the Libera integral operator as introduced by Libera [2].
Let us write that
$B − m − λ f ( z ) = L − m − λ f ( z ) = z p + ∑ k = p + n ∞ p + 1 k + 1 m Γ ( p + 1 + λ ) k ! Γ ( k + 1 + λ ) p ! a k z k$
for $γ = 1$ in (11). Then Theorem 1 says that if $f ( z ) ∈ A ( p , n )$ satisfies
$L − m − λ + 1 f ( z ) L − m − λ f ( z ) − 1 < | e i β − α s | n ρ ( p + 1 ) ( 1 + | e i β − α s | ρ ) , z ∈ U ,$
for some $α s$ given by
$α s = 1 s ∑ l = 1 s L − m − λ f ( z l ) z l p , z l ∈ U ¯ ,$
where $− π 2 ≤ β ≤ π 2 ,$ and for some real $ρ > 1 ,$ then
$L − m − λ f ( z ) z p − 1 < | e i β − α s | ρ , z ∈ U .$
For another result, we consider again the Libera integral operator with $γ = 1 .$

## 3. Application of Carathéodory Lemma

In this section, we will apply Carathéodory Lemma for coefficients of functions $f ( z ) ∈ A ( p , n ) .$
In 1907, Carathéodory [8] gave the following result.
Lemma 2.
Let a function $g ( z )$ given by
$g ( z ) = 1 + ∑ k = 1 ∞ c k z k$
be analytic in $U$ and $R e g ( z ) > 0 , z ∈ U .$ Then $g ( z )$ satisfies
$| c k | ≤ 2 , ( k = 1 , 2 , 3 , … ) .$
The inequality (88) is sharp for each $k .$
Applying the above lemma, we derive the following thorem.
Theorem 5.
If $f ( z ) ∈ A ( p , n )$ is in the class $T p , n α s , β , ρ ; m , λ ,$ then
$| a k | ≤ 2 | e i β − α s | ρ R , ( k = p + n , p + n + 1 , … ) ,$
where
$R = p + γ k + γ m Γ ( k + γ ) Γ ( p + γ + λ ) Γ ( p + γ ) Γ ( k + γ + λ ) .$
The result is sharp for $f ( z )$ given by
$B − m − λ f ( z ) = z p e i θ − ( 1 + 2 | e i β − α s | ρ ) z e i θ − z .$
Proof.
For $f ( z ) ∈ T p , n α s , β , ρ ; m , λ ,$ we see that
$B − m − λ f ( z ) z p − 1 < | e i β − α s | ρ , z ∈ U .$
If we define a function $g ( z )$ with $f ( z ) ∈ T p , n α s , β , ρ ; m , λ$ by
$g ( z ) = B − m − λ f ( z ) z p − ( 1 − | e i β − α s | ρ ) | e i β − α s | ρ , z ∈ U ,$
then $g ( z )$ is analytic in $U$ with $g ( 0 ) = 1$ and $R e g ( z ) > 0 , z ∈ U .$ Also, $g ( z )$ has the following power series expansion:
$g ( z ) = 1 + ∑ k = p + n ∞ R | e i β − α s | ρ a k z k − p .$
Therefore, by applying Lemma 2 to $g ( z ) ,$ we obtain
$R | e i β − α s | ρ | a k | ≤ 2 , ( k = p + n , p + n + 1 , … ) .$
This shows the coefficient inequalities (89). Note that
$g ( z ) = e i θ + z e i θ − z = 1 + ∑ k = 1 ∞ 2 e i θ z k = 1 + ∑ k = 1 ∞ c k z k$
is analytic in $U ,$ $g ( 0 ) = 1 ,$ $R e g ( z ) > 0 , ( z ∈ U )$ and $| c k | = 2 , ( k = 1 , 2 , 3 , … ) .$ Therefore, considering $f ( z )$ such that
$g ( z ) = B − m − λ f ( z ) z p − ( 1 − | e i β − α s | ρ ) | e i β − α s | ρ = e i θ + z e i θ − z ,$
we have
$B − m − λ f ( z ) = z p e i θ − ( 1 + 2 | e i β − α s | ρ ) z e i θ − z .$
This completes the proof of the theorem.  □
Remark 2.
If we take $γ = 1$ in Theorem 5, then we get the following result for the Libera integral operator.
If $f ( z ) ∈ A ( p , n )$ satisfies
$L − m − λ f ( z ) z p − 1 < | e i β − α s | ρ , z ∈ U ,$
then
$| a k | ≤ 2 | e i β − α s | ρ R 0 , ( k = p + n , p + n + 1 , … ) ,$
where
$R 0 = p + 1 k + 1 m Γ ( p + 1 + λ ) k ! Γ ( k + 1 + λ ) p ! .$
The result is sharp for $f ( z )$ given by
$L − m − λ f ( z ) = z p e i θ − ( 1 + 2 | e i β − α s | ρ ) z e i θ − z .$
Finally, we derive
Theorem 6.
If $f ( z ) ∈ A ( p , n )$ satisfies
$∑ k = p + n ∞ R | a k | ≤ | e i β − α s | ρ ,$
then $f ( z ) ∈ T p , n α s , β , ρ ; m , λ ,$ where R is given by (90).
Proof.
For $f ( z ) ∈ A ( p , n ) ,$ we consider
$B − m − λ f ( z ) z p − 1 = ∑ k = p + n ∞ R a k z k$
$< ∑ k = p + n ∞ R | a k | ≤ | e i β − α s | ρ , z ∈ U .$
Therefore, if $f ( z ) ∈ A ( p , n )$ satisfies (102), then we know $f ( z ) ∈ T p , n α s , β , ρ ; m , λ .$ □
Remark 3.
Letting $γ = 1$ in Theorem 6, we have the result concerning with the Libera integral operator $L − m − λ f ( z ) .$

## Author Contributions

Conceptualization, S.O.; Investigation, S.O. and H.Ö.G.; Methodology, S.O.; Writing—original draft, S.O.; Writing—review and editing, H.Ö.G. All authors have read and agreed to the published version of the manuscript.

## Funding

This research received no external funding.

## Acknowledgments

The authors would like to give our thanks to T. Bulboaca, Department of Mathematics, Faculty of Mathematics and Computer Sciences, Babes-Bolyai University, Romania for his kind support for our paper and also the reviewers for valuable remarks and suggestions in order to revise and improve of our paper.

## Conflicts of Interest

The authors declare no conflict of interest.

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Owa, S.; Güney, H.Ö. New Applications of the Bernardi Integral Operator. Mathematics 2020, 8, 1180. https://doi.org/10.3390/math8071180

AMA Style

Owa S, Güney HÖ. New Applications of the Bernardi Integral Operator. Mathematics. 2020; 8(7):1180. https://doi.org/10.3390/math8071180

Chicago/Turabian Style

Owa, Shigeyoshi, and H. Özlem Güney. 2020. "New Applications of the Bernardi Integral Operator" Mathematics 8, no. 7: 1180. https://doi.org/10.3390/math8071180

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