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Article

New Class of Specific Functions with Fractional Derivatives

by
Hatun Özlem Güney
1 and
Shigeyoshi Owa
2,*
1
Department of Mathematics, Faculty of Science, Dicle University, 21280 Diyarbakır, Turkey
2
“1 Decembrie 1918” University Alba Iulia, 510009 Alba-Iulia, Romania
*
Author to whom correspondence should be addressed.
Axioms 2025, 14(8), 608; https://doi.org/10.3390/axioms14080608
Submission received: 25 June 2025 / Revised: 31 July 2025 / Accepted: 31 July 2025 / Published: 5 August 2025
(This article belongs to the Special Issue Recent Advances in Complex Analysis and Related Topics)

Abstract

Let A n be the class of specific analytic functions f ( z ) = z + k = 1 a 1 + k n z 1 + k n ( n N = { 1 , 2 , 3 , } ) in the open unit disk U . For f ( z ) A n , fractional derivatives D z λ f ( z ) and D z j + λ f ( z ) ( 0 λ < 1 , j N ) are defined by using Gamma functions. Applying such fractional derivatives, we introduce a new subclass A n ( j , λ , α , β ) of A n . In this paper, we establish sufficient conditions for f ( z ) for A n ( j , λ , α , β ) , coefficient inequalities for | a 1 + 1 n | and | a 1 + k n | ( k = 2 , 3 , 4 , ) of f ( z ) A n ( j , λ , α , β ) , and some interesting argument properties of fractional derivatives for f ( z ) A n through an example.

1. Introduction

Let A be the class of functions f ( z ) that are analytic in the open unit disk U = { z C : | z | < 1 } and normalized by the conditions f ( 0 ) = 0 and f ( 0 ) = 1 . For this class, we introduce a subclass A n of A consisting of specific functions f ( z ) given by
f ( z ) = z + k = 1 a 1 + k n z 1 + k n ( n N = { 1 , 2 , 3 , } ) .
This class was introduced by Güney, Breaz and Owa [1]. They considered f ( z ) A n given by
f ( z ) = z 1 z n 2 n ( 1 α ) = z + k = 1 = 1 k ( + 2 n ( 1 α ) 1 ) k ! z 1 + k n ( n N = { 1 , 2 , 3 , } )
for 0 α < 1 and showed that
R e z f ( z ) f ( z ) = R e 1 + ( 1 2 α ) z n 1 z n > α ( z U ) .
For f ( z ) A n , we define the fractional integral of order λ as follows:
D z λ f ( z ) = 1 Γ ( λ ) 0 z f ( t ) ( z t ) 1 λ d t ( λ > 0 ) ,
where Γ ( λ ) is the Gamma function.
The fractional derivative of order λ ( 0 λ < 1 ) for f ( z ) A n is defined by
D z λ f ( z ) = 1 Γ ( 1 λ ) d d z 0 z f ( t ) ( z t ) λ d t .
Further, D z j + λ f ( z ) ( j N ) is defined by
D z j + λ f ( z ) = d j d z j D z λ f ( z ) .
The definitions D z λ f ( z ) , D z λ f ( z ) and D z j + λ f ( z ) were given by Owa [2,3]. With the above definitions, we know that
D z 0 f ( z ) = f ( z ) ,
D z j f ( z ) = f ( j ) ( z )
and
D z j + λ f ( z ) = D z λ f ( j ) ( z ) .
Thus, we have
D z λ f ( z ) = 1 Γ ( 2 λ ) z 1 λ + k = 1 Γ k n + 2 Γ k n + 2 λ a 1 + k n z 1 + k n λ
and
D z j + λ f ( z ) = = 1 j ( 2 λ ) Γ ( 2 λ ) + k = 1 Γ k n + 2 Γ k n + 2 λ = 1 j k n + 2 λ a 1 + k n z k n + 1 λ j
for j N . With the above the fractional derivative, we introduce the subclass A n ( j , λ , α , β ) of A n as follows:
R e e i α z D z j + λ f ( z ) ( 2 λ j ) D z j 1 + λ f ( z ) > β cos α ( z U )
with some real α ( | α | < π 2 ) and β ( 0 β < 1 ) , where j N and 0 λ < 1 .
If j = 1 ,   λ = 0 and α = 0 in (1), then we have
R e z f ( z ) f ( z ) > β ( z U )
for the class A n ( 1 , 0 , 0 , β ) . Functions f ( z ) A n in the class A n ( 1 , 0 , 0 , β ) were discussed by Güney, Breaz and Owa [1]. Let us consider a function f ( z ) A n given by
D z j 1 + λ f ( z ) = z 2 λ j Γ ( 2 λ ) 1 z n t
with
t = 2 n ( 2 λ j ) ( 1 β ) e i α cos α ( j N ) .
Then, f ( z ) satisfies
D z j + λ f ( z ) D z j 1 + λ f ( z ) = ( 2 λ j ) + t n z n 1 z n
and
R e e i α z D z j + λ f ( z ) ( 2 λ j ) D z j 1 + λ f ( z ) = cos α + 2 ( 1 β ) cos α R e z n 1 z n > cos α ( 1 β ) cos α = β cos α ( z U ) .
Therefore, f ( z ) A n given by (2) belongs to the class A n ( j , λ , α , β ) . Accordingly, it is very important to provide example functions f ( z ) for the new class A n ( j , λ , α , β ) .

2. Conditions of f ( z ) for the Class A n ( j , λ , α , β )

We first consider the following condition on f ( z ) for the class A n ( j , λ , α , β ) .
Theorem 1. 
If f ( z ) A n satisfies
z D z j + λ f ( z ) ( 2 λ j ) D z j 1 + λ f ( z ) 1 < ( 1 β ) cos α ( z U )
for some real α ( | α | < π 2 ) ,   0 β < 1 ,   0 λ < 1 and j N , then f ( z ) A n ( j , λ , α , β ) .
Proof. 
We define a function w ( z ) as follows:
z D z j + λ f ( z ) ( 2 λ j ) D z j 1 + λ f ( z ) 1 = ( 1 β ) cos α w ( z ) ( z U )
for f ( z ) A n which satisfies (3). Then, we see that w ( z ) is analytic in U , that w ( 0 ) = 0 and that | w ( z ) | < 1 ( z U ) . For such a function w ( z ) , we have
R e e i α z D z j + λ f ( z ) ( 2 λ j ) D z j 1 + λ f ( z ) = R e ( e i α ( 1 + ( 1 β ) cos α w ( z ) ) ) = cos α + ( 1 β ) cos α R e ( e i α w ( z ) ) cos α ( 1 β ) cos α | e i α w ( z ) | > β cos α ( z U ) .
This shows us that f ( z ) A n ( j , λ , α , β ) .
If we consider j = 1 and λ = 0 in Theorem 1, then we have the following corollary.
Corollary 1. 
If f ( z ) A n satisfies
z f ( z ) f ( z ) 1 < ( 1 β ) cos α ( z U )
for some real α ( | α | < π 2 ) and 0 β < 1 , then
R e e i α z f ( z ) f ( z ) > β cos α ( z U ) .
Next, we derive the following theorem.
Theorem 2. 
If f ( z ) A n satisfies
Γ ( 1 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ 1 λ + k n ( 1 β ) sec α a 1 + k n 1
for some real α ( | α | < π 2 ) ,   0 β < 1 and 0 λ < 1 , then f ( z ) A n ( 1 , λ , α , β ) . If f ( z ) A n satisfies
Γ ( 2 λ ) = 1 j ( 2 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ 2 λ j + k n ( 1 β ) sec α × = 1 j 1 k n + 2 λ a 1 + k n 1
for some real α ( | α | < π 2 ) ,   0 β < 1 ,   0 λ < 1 and j = 2 , 3 , , then f ( z ) A n ( 1 , λ , α , β ) .
Proof. 
For f ( z ) A n and j 2 , we have that
z D z j + λ f ( z ) ( 2 λ j ) D z j 1 + λ f ( z ) = = 1 j ( 2 λ ) Γ ( 2 λ ) z 2 λ j + k = 1 Γ k n + 2 Γ k n + 2 λ = 1 j k n + 2 λ a 1 + k n z k n + 2 λ j ( 2 λ j ) = 1 j 1 ( 2 λ ) Γ ( 2 λ ) z 2 λ j + k = 1 Γ k n + 2 Γ k n + 2 λ = 1 j 1 k n + 2 λ a 1 + k n z k n + 2 λ j = k = 1 Γ k n + 2 Γ k n + 2 λ k n = 1 j 1 k n + 2 λ a 1 + k n z k n + 2 λ j .
It follows from the above that
z D z j + λ f ( z ) ( 2 λ j ) D z j 1 + λ f ( z ) 1 = Γ ( 2 λ ) = 1 j ( 2 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ k n = 1 j 1 k n + 2 λ a 1 + k n z k n 1 + Γ ( 2 λ ) = 1 j ( 2 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ = 1 j 1 k n + 2 λ a 1 + k n z k n < Γ ( 2 λ ) = 1 j ( 2 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ k n = 1 j 1 k n + 2 λ a 1 + k n 1 Γ ( 2 λ ) = 1 j 1 ( 2 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ = 1 j 1 k n + 2 λ a 1 + k n
for z U . Thus, we know by Theorem 1 that if f ( z ) satisfies
Γ ( 2 λ ) = 1 j ( 2 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ k n = 1 j 1 k n + 2 λ a 1 + k n ( 1 β ) cos α 1 Γ ( 2 λ ) = 1 j 1 ( 2 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ = 1 j 1 k n + 2 λ a 1 + k n ,
then f ( z ) A n ( j , λ , α , β ) . Finally, we see that the inequality (6) implies (5).
If j = 1 , we also have that
z D z 1 + λ f ( z ) ( 1 λ ) D z λ f ( z ) 1 < Γ ( 1 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ k n a 1 + k n 1 Γ ( 2 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ a 1 + k n .
Noting that if f ( z ) satisfies
Γ ( 1 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ k n a 1 + k n ( 1 β ) cos α 1 Γ ( 2 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ a 1 + k n ,
we obtain the inequality in (4). Further, if we consider a function f ( z ) A n given by
f ( z ) = z + k = 1 c 1 + k n z 1 + k n
with
c 1 + k n = Γ k n + 2 λ e i θ Γ ( 1 λ ) Γ k n + 2 1 λ + k n ( 1 β ) sec α k ( k + 1 ) ( 0 θ < 2 π ) ,
then f ( z ) satisfies
Γ ( 1 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ 1 λ + k n ( 1 β ) sec α c 1 + k n = k = 1 1 k ( k + 1 ) = k = 1 1 k 1 k + 1 = 1 .
If we take a function f ( z ) A n given by
f ( z ) = z + k = 1 c 1 + k n z 1 + k n
with
c 1 + k n = = 1 j ( 2 λ ) Γ k n + 2 λ e i θ Γ ( 2 λ ) Γ k n + 2 2 λ j + k n ( 1 β ) sec α × 1 = 1 j 1 k n + 2 λ k ( k + 1 ) ( 0 θ < 2 π ) ,
then f ( z ) satisfies the equality in (5). □

3. Coefficient Inequalities of f ( z ) for A n ( j , λ , α , β )

We would like to discuss coefficient inequalities of f ( z ) for the class A n ( j , λ , α , β ) . We introduce the lemma established by Carathéodory [4].
Lemma 1. 
If a function p ( z ) given by
p ( z ) = 1 + k = 1 c k z k
is analytic in U and R e p ( z ) > α ( z U ) with 0 α < 1 , then
| c k | 2 ( 1 α ) ( z U ) .
The equality in (7) is satisfied by p ( z ) such that
p ( z ) = 1 + ( 1 2 α ) z 1 z = 1 + 2 ( 1 α ) k = 1 z k .
With the above lemma, we see the following lemma.
Lemma 2. 
If a function p ( z ) given by
p ( z ) = 1 + k = 1 c k n z k n ( n N )
is analytic in U and R e p ( z ) > α ( z U ) with 0 α < 1 , then
| c k n | 2 ( 1 α ) ( k N ) .
The equality in (8) is satisfied by p ( z ) such that
p ( z ) = 1 + ( 1 2 α ) z n 1 z n = 1 + 2 ( 1 α ) k = 1 z k n .
Applying Lemma 2, we prove the following theorem.
Theorem 3. 
If f ( z ) A n ( j , λ , α , β ) , then
a 1 + 1 n 2 n e i α β cos α Γ 1 n + 2 λ = 1 j | 2 λ | Γ ( 2 λ ) Γ 1 n + 2 = 1 j 1 1 n + 2 λ
and
a 1 + k n 2 n e i α β cos α Γ k n + 2 λ = 1 j | 2 λ | k Γ ( 2 λ ) Γ k n + 2 = 1 j 1 k n + 2 λ × m = 1 k 1 1 + 2 n m e i α β cos α | 2 λ j |
for k = 2 , 3 , 4 , , where j = 2 , 3 , 4 , . The equalities in (9) and (10) are satisfied by f ( z ) A n such that
D z j 1 + λ f ( z ) = z 1 z n t 2 λ j
with
t = 2 n β cos α e i α 1 .
Proof. 
Let us consider a function p ( z ) as follows:
p ( z ) = e i α z D z j + λ f ( z ) ( 2 λ j ) D z j 1 + λ f ( z ) β cos α e i α β cos α = 1 + k = 1 c k n z k n
for f ( z ) A n ( j , λ , α , β ) . Then, we know that p ( z ) is analytic in U ,   p ( 0 ) = 1 and that R e p ( z ) > 0 ( z U ) . Thus, Lemma 2 implies that
c k n 2 ( k N )
and the equality in (12) is satisfied by
p ( z ) = 1 + z n 1 z n .
By (11), we see
e i α z D z j + λ f ( z ) β ( 2 λ j ) cos α D z j 1 + λ f ( z ) = ( 2 λ j ) e i α β cos α p ( z ) D z j 1 + λ f ( z ) .
It is clear that
e i α z D z j + λ f ( z ) β ( 2 λ j ) cos α D z j 1 + λ f ( z ) = e i α β cos α Γ ( 2 λ ) = 1 j ( 2 λ ) z 2 λ j + k = 1 e i α k n + 2 λ β ( 2 λ j ) Γ k n + 2 Γ k n + 2 λ × = 1 j 1 k n + 2 λ a 1 + k n z k n + 2 λ j .
Further, we calculate
( 2 λ j ) e i α β cos α p ( z ) D z j 1 + λ f ( z ) = ( 2 λ j ) e i α β cos α 1 + k = 1 c k n z k n × = 1 j 1 ( 2 λ ) Γ ( 2 λ ) z 2 λ j + k = 1 Γ k n + 2 Γ k n + 2 λ = 1 j 1 k n + 2 λ a 1 + k n z k n + 1 λ j = ( 2 λ j ) ( e i α β cos α ) = 1 j 1 ( 2 λ ) Γ ( 2 λ ) z 2 λ j + c 1 n = 1 j 1 ( 2 λ ) Γ ( 2 λ ) + Γ 1 n + 2 Γ 1 n + 2 λ = 1 j 1 1 n + 2 λ a 1 + 1 n z 1 n + 2 λ j + c 2 n = 1 j 1 ( 2 λ ) Γ ( 2 λ ) + c 1 n Γ 2 n + 2 Γ 2 n + 2 λ = 1 j 1 1 n + 2 λ a 1 + 1 n + Γ 2 n + 2 Γ 2 n + 2 λ = 1 j 1 2 n + 2 λ a 1 + 2 n z 2 n + 2 λ j + c 3 n = 1 j 1 ( 2 λ ) Γ ( 2 λ ) + c 2 n Γ 1 n + 2 Γ 1 n + 2 λ = 1 j 1 2 n + 2 λ a 1 + 1 n + c 1 n Γ 2 n + 2 Γ 2 n + 2 λ = 1 j 1 2 n + 2 λ a 1 + 2 n + Γ 3 n + 2 Γ 3 n + 2 λ = 1 j 1 3 n + 2 λ a 1 + 3 n z 3 n + 2 λ j + + c k n = 1 j 1 ( 2 λ ) Γ ( 2 λ ) + c k 1 n Γ 1 n + 2 Γ 1 n + 2 λ = 1 j 1 1 n + 2 λ a 1 + 1 n + c k 2 n Γ 2 n + 2 Γ 2 n + 2 λ = 1 j 1 2 n + 2 λ a 1 + 2 n + + c 2 n Γ k 2 n + 2 Γ k 2 n + 2 λ = 1 j 1 k 2 n + 2 λ a 1 + k 2 n + c 1 n Γ k 1 n + 2 Γ k 1 n + 2 λ = 1 j 1 k 1 n + 2 λ a 1 + k 1 n + Γ k n + 2 Γ k n + 2 λ = 1 j 1 k n + 2 λ a 1 + k n z k n + 2 λ j + .
Therefore, for k = 1 , we have
e i α Γ 1 n + 2 n Γ 1 n + 2 λ = 1 j 1 1 n + 2 λ a 1 + 1 n = c 1 n ( 2 λ j ) e i α β cos α = 1 j 1 ( 2 λ ) Γ ( 2 λ ) .
Using | c 1 n | 2 for p ( z ) , we have
a 1 + 1 n 2 n | e i α β cos α | Γ 1 n + 2 λ = 1 j 1 | 2 λ | Γ ( 2 λ ) Γ 1 n + 2 = 1 j 1 1 n + 2 λ .
This proves the coefficient inequality (9) for a 1 + 1 n .
Considering the coefficient for z 2 n + 2 λ j in (13) and (14), we prove
2 e i α Γ 2 n + 2 n Γ 2 n + 2 λ = 1 j 1 2 n + 2 λ a 1 + 2 n = ( 2 λ j ) e i α β cos α c 2 n = 1 j 1 ( 2 λ ) Γ ( 2 λ ) + c 1 n Γ 1 n + 2 Γ 1 n + 2 λ = 1 j 1 1 n + 2 λ a 1 + 1 n .
Therefore, using (8) and (15), it follows that
a 1 + 2 n 2 n | e i α β cos α | Γ 2 n + 2 λ = 1 j 1 | 2 λ | 2 Γ ( 2 λ ) Γ 2 n + 2 = 1 j 1 2 n + 2 λ × 1 + 2 n | 2 λ j | e i α β cos α .
This shows the coefficient inequality for a 1 + 2 n in (10).
Furthermore, we have from (13) and (14) that
k e i α Γ k n + 2 n Γ k n + 2 λ = 1 j 1 k n + 2 λ a 1 + k n = ( 2 λ j ) e i α β cos α c k n = 1 j 1 ( 2 λ ) Γ ( 2 λ ) + c k 1 n Γ 1 n + 2 Γ 1 n + 2 λ = 1 j 1 1 n + 2 λ a 1 + 1 n + c k 2 n Γ 2 n + 2 Γ 2 n + 2 λ = 1 j 1 2 n + 2 λ a 1 + 2 n + + c 1 n Γ k 1 n + 2 Γ k 1 n + 2 λ = 1 j 1 k 1 n + 2 λ a 1 + k 1 n .
Thus, by mathematical induction, we also prove that
a 1 + k n 2 n | e i α β cos α | Γ k n + 2 λ = 1 j | 2 λ | k Γ ( 2 λ ) Γ k n + 2 = 1 j 1 k n + 2 λ × m = 1 k 1 1 + 2 n m e i α β cos α | 2 λ |
for k = 2 , 3 , 4 , .
Finally, considering a function p ( z ) given by
p ( z ) = 1 + z n 1 z n = 1 + 2 k = 1 z k n ,
Equation (11) implies that
z D z j + λ f ( z ) D z j 1 + λ f ( z ) = ( 2 λ j ) e i α β cos α + ( 1 e i α β cos α ) 1 + z n 1 z n = ( 2 λ j ) 1 + ( 1 2 e i α β cos α ) z n 1 z n
and that
D z j + λ f ( z ) D z j 1 + λ f ( z ) = ( 2 λ j ) 1 z + 2 ( 1 e i α β cos α ) z n z ( 1 z n ) .
This gives us
log D z j 1 + λ f ( z ) = ( 2 λ j ) ( log z ) + 2 n ( β cos α e i α 1 ) log ( 1 z n ) .
Thus, we obtain the following:
D z j 1 + λ f ( z ) = z ( 1 z n ) t 2 λ j
with
t = 2 n ( β cos α e i α 1 ) .
Taking λ = 0 in Theorem 3, we see the following corollary.
Corollary 2. 
If f ( z ) A n ( j , 0 , α , β ) , then
a 1 + 1 n 2 n e i α β cos α = 1 j | 2 | = 1 j 1 1 n + 2
and
a 1 + k n 2 n e i α β cos α = 1 j | 2 | k = 1 j 1 k n + 2 × m = 1 k 1 1 + 2 n m e i α β cos α | 2 j |
for k = 2 , 3 , 4 , , where j = 2 , 3 , 4 , .

4. Argument Properties of Fractional Derivatives

For analytic functions f ( z ) and g ( z ) in U , f ( z ) is said to be subordinate to g ( z ) in U , written f ( z ) g ( z ) ( z U ) , if there exists a function w ( z ) analytic in U with w ( 0 ) = 0 and | w ( z ) | < 1 ( z U ) , such that f ( z ) = g ( w ( z ) ) ( z U ) . If g ( z ) is univalent in U , then f ( z ) g ( z ) ( z U ) if and only if f ( 0 ) = g ( 0 ) and f ( U ) g ( U ) [5]. For subordinations, Miller and Mocanu [6] gave the following lemma.
Lemma 3. 
Let β 0 = 1.21872 be the solution of β π = 3 2 π T a n 1 β and α = α ( β ) = β + 2 T a n 1 β π for 0 < β < β 0 . If g ( z ) is analytic in U with g ( 0 ) = 1 , then
g ( z ) + z g ( z ) 1 + z 1 z α ( z U )
gives
g ( z ) 1 + z 1 z β ( z U ) .
It follows that the subordination given in (16) implies that
| arg ( g ( z ) + z g ( z ) ) | < π 2 α ( z U )
and the subordination (17) implies that
| arg g ( z ) | < π 2 β ( z U ) .
We also introduce the next lemma, which was established by Nunokawa [7].
Lemma 4. 
Let g ( z ) be analytic in U with g ( 0 ) = 1 and g ( z ) 0 ( z U ) . If there exists a point z 0 U such that
| arg g ( z ) | < π 2 α ( | z | < | z 0 | )
and
| arg g ( z 0 ) | = π 2 α
for some real α > 0 , then we have
z 0 g ( z 0 ) g ( z 0 ) = i k α ,
where
k 1 2 a + 1 a ( f o r arg g ( z 0 ) = π 2 α ) ,
k 1 2 a + 1 a ( f o r arg g ( z 0 ) = π 2 α ) ,
and
g ( z 0 ) 1 α = ± i a ( a > 0 ) .
Now we have the following theorem.
Theorem 4. 
If f ( z ) A n satisfies
arg D z j + λ f ( z ) z 1 j λ z D z j + 1 + λ f ( z ) D z j + λ + j + λ < π 2 α ( z U )
for some α > 0 ,   0 λ < 1 and j N , then
arg D z j + λ f ( z ) z 1 j λ < π 2 α β ( z U ) ,
where
β = 2 π 0 1 S i n 1 2 ρ 1 + ρ 2 d ρ = 0.55872876 .
Proof. 
For f ( z ) A n satisfying (20), we define the function g ( z ) as follows:
g ( z ) = Γ ( 2 λ ) D z j + λ f ( z ) = 1 j ( 2 λ ) z 1 j λ = 1 + Γ ( 2 λ ) = 1 j ( 2 λ ) k = 1 Γ k n + 2 Γ k n + 2 λ = 1 j k n + 2 λ a 1 + k n z k n .
Then, g ( z ) is analytic in U with g ( 0 ) = 1 and g ( z ) 0 ( z U ) , as follows from (20). For the above-defined g ( z ) , we see that
arg g ( z ) = arg 1 z ( z g ( z ) ) = arg 1 z 0 z ( t g ( t ) ) d t = arg 1 z 0 r d d ρ ( ρ e i θ g ( ρ e i θ ) ) e i θ d ρ = arg 0 r g ( ρ e i θ ) + ρ e i θ d d ρ g ( ρ e i θ ) d ρ ,
where z = r e i θ and t = ρ e i θ . Let
0 = ρ 0 < ρ 1 < ρ 2 < < ρ n 1 < ρ n = r , = 1 n ρ = ρ
and
ρ ρ 1 = δ ( = 1 , 2 , 3 , , n ) .
Then, we have
| arg g ( z ) | = arg lim n = 1 n δ ( g ( ρ e i θ ) + ρ e i θ g ( ρ e i θ ) ) lim n = 1 n δ arg ( g ( ρ e i θ ) + ρ e i θ g ( ρ e i θ ) ) .
Since
| arg g ( z ) + z g ( z ) | = arg D z j + λ f ( z ) z 1 j λ z D z j + 1 + λ f ( z ) D z j + λ f ( z ) + j + λ < π 2 α
by (20), using Lemma 4, we obtain the following:
| arg g ( z ) | = arg D z j + λ f ( z ) z 1 j λ lim n = 1 n δ arg 1 + ρ e i θ 1 ρ e i θ α = α 0 r S i n 1 2 ρ 1 + ρ 2 d ρ < α 0 1 S i n 1 2 ρ 1 + ρ 2 d ρ = π 2 α β .
This completes the proof of the theorem. □
Taking j = 1 and λ = 0 in Theorem 4, we have the following corollary.
Corollary 3. 
If f ( z ) A n satisfies
arg f ( z ) + z f ( z ) < π 2 α ( z U ) ,
for some α > 0 , then
arg ( f ( z ) < π 2 α β ( z U ) ,
where β is given by (21).
Remark 1. 
If we take α = 3 2 , then we have
α β = 3 π 1 2 π log 2 = 0.533546701 .
Next, applying Lemma 4, as developed by Nunokawa [7], we prove the following theorem.
Theorem 5. 
If f ( z ) A n satisfies
arg z D z j + 1 + λ f ( z ) D z j + λ f ( z ) + Γ ( 2 λ ) D z j + λ f ( z ) = 1 j ( 2 λ ) z 1 j λ + j + λ 1 < T a n 1 β ( α ) ( z U )
for some α ( 0 < α < 1 ) ,   0 λ < 1 and j N , then
arg D z j + λ f ( z ) z 1 j λ < π 2 α ( z U ) ,
where
β ( α ) = α 2 1 + α 1 α 1 α 2 + 1 α 1 + α 1 + α 2 sec π 2 α + tan π 2 α .
Proof. 
We consider a function g ( z ) given by (22) for f ( z ) A n . Then, g ( z ) is analytic in U with g ( 0 ) = 1 . For the above-defined g ( z ) , we see that
z g ( z ) g ( z ) + g ( z ) = z D z j + 1 + λ f ( z ) D z j + λ f ( z ) + Γ ( 2 λ ) D z j + λ f ( z ) = 1 j ( 2 λ ) z 1 j λ + j + λ 1 .
We suppose that there exists a point z 0 U such that
| arg g ( z ) | = arg D z j + λ f ( z ) z 1 j λ < π 2 α ( | z | < | z 0 | )
and
| arg g ( z 0 ) | = arg D z j + λ f ( z 0 ) z 0 1 j λ = π 2 α .
Then, Lemma 4 gives us the following:
z 0 g ( z 0 ) g ( z 0 ) = z 0 D z j + 1 + λ f ( z 0 ) D z j + λ f ( z 0 ) + j + λ 1 = i k α ,
where k is given by (18) and (19). If arg g ( z 0 ) = π 2 α , then g ( z 0 ) = a α e i π 2 α and
arg z 0 g ( z 0 ) g ( z 0 ) + g ( z 0 ) = arg z 0 D z j + 1 + λ f ( z 0 ) D z j + λ f ( z 0 ) + Γ ( 2 λ ) D z j + λ f ( z 0 ) = 1 j ( 2 λ ) z 0 1 j λ + j + λ 1 = arg i k α + a α e i π 2 α = arg a α cos π 2 α + i k α + a α sin π 2 α arg a α cos π 2 α + i α 2 a + 1 a + a α sin π 2 α = T a n 1 α 2 a 1 α + a 1 α sec π 2 α + tan π 2 α T a n 1 α 2 1 + α 1 α 1 α 2 + 1 α 1 + α 1 + α 2 sec π 2 α + tan π 2 α = T a n 1 β ( α ) .
This contradicts the condition (23) of the theorem. If arg g ( z 0 ) = π 2 α , then we also have that
arg z 0 g ( z 0 ) g ( z 0 ) + g ( z 0 ) = arg z 0 D z j + 1 + λ f ( z 0 ) D z j + λ f ( z 0 ) + Γ ( 2 λ ) D z j + λ f ( z 0 ) = 1 j ( 2 λ ) z 0 1 j λ + j + λ 1 T a n 1 α 2 1 + α 1 α 1 α 2 + 1 α 1 + α 1 + α 2 sec π 2 α + tan π 2 α = T a n 1 β ( α ) .
This also contradicts the condition (23) of the theorem for g ( z 0 ) = π 2 α . Therefore, we say that there is no z 0 U such that
| arg g ( z ) | = arg D z j + λ f ( z ) z 1 j λ < π 2 α ( | z | < | z 0 | )
and
| arg g ( z 0 ) | = arg D z j + λ f ( z 0 ) z 0 1 j λ = π 2 α .
This implies that
| arg g ( z ) | = arg D z j + λ f ( z ) z 1 j λ < π 2 α ( z U ) .
Taking j = 1 and λ = 0 in Theorem 5, we have the following corollary.
Corollary 4. 
If f ( z ) A n satisfies
arg z f ( z ) f ( z ) + f ( z ) < T a n 1 β ( α ) ( z U ) ,
for some real α ( 0 < α < 1 ) , then
arg f ( z ) < π 2 α ( z U ) ,
where β ( α ) is given by (24).
Next, we consider f ( z ) A n given by
Γ ( 2 λ ) z j + λ D z j + λ f ( z ) = 1 j ( 2 λ ) = 0 z 1 + t n 1 t n α d t ( z U )
with 0 α < 1 , 0 λ < 1 and j N . Then f ( z ) satisfies
arg Γ ( 2 λ ) z j + λ D z j + λ f ( z ) = 1 j ( 2 λ ) = arg z j + λ ( j + λ ) D z j + λ f ( z ) z + D z j + 1 + λ f ( z ) = α arg 1 + z n 1 z n < π 2 α ( z U ) .
To discuss such an argument problem, we need the result devised by Fejér and Riesz [8] (or by Tsuji [9]).
Lemma 5. 
Let a function f ( z ) be analytic in | z | 1 . Then, f ( z ) satisfies
1 1 | f ( z ) | α | d z | 1 2 | z | = 1 | f ( z ) | α | d z | ( α > 0 ) ,
where the above integral on the left-hand side is considered along the real axis.
Remark 2. 
When we make the change of variables in Lemma 5, Equation (25) becomes
r r | f ( ρ e i θ ) | α d ρ r 2 0 2 π | f ( r e i θ ) | α d θ .
Further, Gwynme [10] gave the following lemma.
Lemma 6. 
Let f ( z ) be a complex valued harmonic function defined on a neighborhood of a closed disc of radius one and center the origin in the complex plane. Then
f ( r e i θ ) = 1 2 π 0 2 π f ( e i ρ ) 1 r 2 1 + r 2 2 r cos ( θ ρ ) d ρ
and
1 2 π 0 2 π f ( e i ρ ) 1 r 2 1 + r 2 2 r cos ρ d ρ = 1 .
Now, we derive the following theorem.
Theorem 6. 
If f ( z ) A n satisfies
z D z j + 1 + λ f ( z ) D z j + λ f ( z ) α 2 ( z U )
for some α ( 0 α < 1 ) , then
arg z j 1 + λ D z j + λ f ( z ) < π 2 α ( z U ) ,
where j N and 0 λ < 1 .
Proof. 
We note that
log Γ ( 2 λ ) z j 1 + λ D z j + λ f ( z ) = 1 j ( 2 λ ) = log Γ ( 2 λ ) z j 1 + λ D z j + λ f ( z ) = 1 j ( 2 λ ) + i arg ( z j 1 + λ D z j + λ f ( z ) )
and
log Γ ( 2 λ ) z j 1 + λ D z j + λ f ( z ) = 1 j ( 2 λ ) = 0 z log Γ ( 2 λ ) t j 1 + λ D z j + λ f ( t ) = 1 j ( 2 λ ) d t = 0 z j 1 + λ t + D z j + 1 + λ f ( t ) D z j + λ f ( t ) d t .
It follows from (28) and (29) that
arg z j 1 + λ D z j + λ f ( z ) = I m 0 z j 1 + λ t + D z j + 1 + λ f ( t ) D z j + λ f ( t ) d t I m 0 r j 1 + λ ρ e i θ + D z j + 1 + λ f ( ρ e i θ ) D z j + λ f ( ρ e i θ ) e i θ d ρ = 0 r I m j 1 + λ ρ + e i θ D z j + 1 + λ f ( ρ e i θ ) D z j + λ f ( ρ e i θ ) d ρ r r D z j + 1 + λ f ( ρ e i θ ) D z j + λ f ( ρ e i θ ) d ρ ,
where z = r e i θ ( 0 θ < 2 π ) , 0 r < 1 and 0 ρ r . Using (26) for α = 1 , we see that
arg z j 1 + λ D z j + λ f ( z ) r 2 0 2 π D z j + 1 + λ f ( r e i θ ) D z j + λ f ( r e i θ ) d θ = 1 2 0 2 π r e i θ D z j + 1 + λ f ( r e i θ ) D z j + λ f ( r e i θ ) d θ .
Therefore, if f ( z ) satisfies (27), then f ( z ) satisfies
arg z j 1 + λ D z j + λ f ( z ) < π 2 α ( z U ) .
Taking j = 1 and λ = 0 in Theorem 6, we have the following corollary.
Corollary 5. 
If f ( z ) A n satisfies
z f ( z ) f ( z ) α 2 ( z U )
for some α ( 0 α < 1 ) , then
arg f ( z ) < π 2 α ( z U ) .
Further, on applying Lemma 6, we arrive at the following theorem.
Theorem 7. 
If f ( z ) A n satisfies
z D z j + 1 + λ f ( z ) D z j + λ f ( z ) < α 2 R e 1 + β z n 1 z n ( z U )
for some α ( 0 α < 1 ) and β ( β 1 ) , then
arg z j 1 + λ D z j + λ f ( z ) < π 2 α ( z U ) ,
where j N and 0 λ < 1 .
Proof. 
From (30), we see that
arg z j 1 + λ D z j + λ f ( z ) < 1 2 0 2 π r e i θ D z j + 1 + λ f ( r e i θ ) D z j + λ f ( r e i θ ) d θ
for f ( z ) A n . It follows that (31) gives
arg z j 1 + λ D z j + λ f ( z ) < α 4 0 2 π R e 1 + β r n e i θ n 1 r n e i θ n d θ = α 4 0 2 π 1 + ( β 1 ) r n cos θ n β ( r n ) 2 1 2 r n cos θ n ( r n ) 2 d θ = α 4 0 2 π 1 β 2 + 1 + β 2 1 ( r n ) 2 1 + ( r n ) 2 2 r n cos θ n d θ .
Applying Lemma 6, we obtain
0 2 π 1 ( r n ) 2 1 + ( r n ) 2 2 r n cos θ n d θ = n 0 2 π n 1 ( r n ) 2 1 + ( r n ) 2 2 r n cos ρ d ρ 2 π .
Therefore, (32) becomes
arg z j 1 + λ D z j + λ f ( z ) < π 2 α ( z U ) .
Setting j = 1 and λ = 0 in Theorem 7, we obtain the following corollary.
Corollary 6. 
If f ( z ) A n satisfies
z f ( z ) f ( z ) < α 2 R e 1 + β z n 1 z n ( z U )
for some α ( 0 α < 1 ) and β ( β 1 ) , then
arg f ( z ) < π 2 α ( z U ) .
Example 1. 
We consider a function f ( z ) A n given by
Γ ( 2 λ ) z j 1 + λ D z j + λ f ( z ) = 1 j ( 2 λ ) = 2 2 z n 3 α ( z U )
with 0 < α < 1 . If we write that
w ( z ) = 2 2 z n ( z U ) ,
w ( z ) satisfies
w ( z ) 4 3 < 2 3 ( z U )
and
arg w ( z ) = arg 2 2 z n < π 6 ( z U ) .
Therefore, we have
arg z j 1 + λ D z j + λ f ( z ) = 3 α arg 2 2 z n < π 2 α ( z U ) .
For such a function f ( z ) , we obtain
z D z j + 1 + λ f ( z ) D z j + λ f ( z ) = 3 α n z n 2 z n ( j 1 + λ ) < 3 α n z n 2 z n + ( j 1 + λ ) < 3 α n + ( j 1 + λ ) ( z U ) .
If we consider some β such that
β n 12 n 4 ( j 1 + λ ) α ,
then
z D z j + 1 + λ f ( z ) D z j + λ f ( z ) < 3 α n + ( j 1 + λ ) α ( 1 β ) 4 < α 2 R e 1 + β z n 1 z n ( z U ) .

Author Contributions

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

Funding

This research received no external funding.

Data Availability Statement

The manuscript has no associated data.

Acknowledgments

The authors would like to thank the editor and reviewers for their valuable comments and suggestions which helped us to improve the content of this paper.

Conflicts of Interest

The authors declare no conflicts of interest.

References

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Güney, H.Ö.; Owa, S. New Class of Specific Functions with Fractional Derivatives. Axioms 2025, 14, 608. https://doi.org/10.3390/axioms14080608

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Güney HÖ, Owa S. New Class of Specific Functions with Fractional Derivatives. Axioms. 2025; 14(8):608. https://doi.org/10.3390/axioms14080608

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Güney, Hatun Özlem, and Shigeyoshi Owa. 2025. "New Class of Specific Functions with Fractional Derivatives" Axioms 14, no. 8: 608. https://doi.org/10.3390/axioms14080608

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Güney, H. Ö., & Owa, S. (2025). New Class of Specific Functions with Fractional Derivatives. Axioms, 14(8), 608. https://doi.org/10.3390/axioms14080608

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