EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
2025, Vol. 18, Issue 4, Article Number 6366
ISSN 1307-5543 – ejpam.com
Published by New York Business Global

Some Geometric Results for a New Subfamily of
Regular Functions in the Generalized Janowski Domain

Tamer M. Seoudy1,∗, Amnah E. Shammaky2

1 Department of Mathematics, Faculty of Science, Fayoum University, Fayoum 63514, Egypt
2 Department of Mathematics, Faculty of Science, Jazan University, Jazan 45142,
Saudi Arabia

Abstract. Let SK [P,Q;α, β] denote the subfamily of normalized regular function g(ξ) = ξ +
d2ξ

2 + d3ξ
3 + d4ξ

4 + ... in the open unit disk ∆ holding the next subordination condition:

αg (ξ) + βξg′ (ξ)

αξ + βg (ξ)
≺ 1 + [Q+ (1− γ) (P −Q)] ξ

1 +Mξ
,

where 0 ≤ α, β ≤ 1, −1 ≤ Q < P ≤ 1, 0 ≤ γ < 1 and ξ ∈ ∆. In this article, we study some
geometric properties such as convolution results, coefficient estimates, upper bounds for the initial
coefficients of the first four coefficients, and the Fekete-Szegö inequalities for this subfamily.

2020 Mathematics Subject Classifications: 30C45

Key Words and Phrases: Regular function, starlike, convolution, Fekete-Szegö problem

1. Introduction

Let us represent the family of regular (analytic) functions inside the symmetrical unit
disc ∆ = {ξ ∈ C : |ξ| < 1} by A (∆) and let H be the subfamily of A (∆) consisting of all
functions g in ∆ that has the following form:

g(ξ) = ξ +

∞∑
j=2

djξ
j (ξ ∈ ∆) . (1)

Also, let Ω be the family of functions w ∈ A (∆) satisfying w(0) = 0 and |w(ξ)| < 1 for
all ξ ∈ ∆. For h1(ξ), h2(ξ) ∈ A (∆), we say that h1 (ξ) is subordinate to h2 (ξ), written
as h1(ξ) ≺ h2(ξ) if there exists a function w(ξ) ∈ Ω such that h1(ξ) = (h2 ◦ w) (ξ) for all
ξ ∈ ∆. Moreover, if h1(ξ) is univalent function in ∆, then h1(ξ) ≺ h2(ξ) if and only if
h1(0) = h2(0) and h1(∆) ⊂ h2(∆).

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v18i4.6366

Email addresses: tms00@fayoum.edu.eg (T. M. Seoudy), aeshamakhi@jazan.edu.sa (A. E. Shammaky)

https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0)



T. M. Seoudy, A. E. Shammaky / Eur. J. Pure Appl. Math, 18 (4) (2025), 6366 2 of 15

For functions g, h ∈ H, where g is given by (1) and h has the following form:

h(ξ) = ξ +

∞∑
j=2

ejξ
j , (2)

then, the convolution of g and h is defined by

(g ∗ h) (ξ) = ξ +

∞∑
j=2

djejξ
j = (h ∗ g) (ξ). (3)

Definition 1. Let us introduce the new subfamily SK [P,Q;α, β] of H by using the sub-
ordination principle between two analytic functions as follows:

SK [P,Q, γ;α, β] =

{
g ∈ H :

αg (ξ) + βξg′ (ξ)

αξ + βg (ξ)
≺ 1 + [Q+ (1− γ) (P −Q)] ξ

1 +Mξ

}
, (4)

where 0 ≤ α, β ≤ 1, −1 ≤ Q < P ≤ 1, 0 ≤ γ < 1 and ξ ∈ ∆.

Remark 1. The function

ψ (ξ) =
1 + [Q+ (1− γ) (P −Q)] ξ

1 +Qξ

is a bilinear transformation that maps the symmetrical unit disk ∆ onto the symmetrical
disk with respect to positive real axis, which is centered at 1−[Q+(1−γ)(P−Q)]Q

1−Q2 (Q ̸= ±1)

and, with a radius (1−γ)(P−Q)
1−Q2 (Q ̸= ±1).

(i) Substituting α = γ = 0 in (4), we get the subfamily

S [P,Q] =

{
g ∈ H :

ξg′ (ξ)

g (ξ)
≺ 1 + Pξ

1 +Qξ

}
,

which was introduced by Janowski [1, 2]. Furthermore, this subfamily has been
studied by the many authors (see [3–13]). By taking β = γ = 0 in (4), we obtain
the next subfamily

K [P,Q] =

{
g ∈ H :

g (ξ)

ξ
≺ 1 + Pξ

1 +Qξ

}
;

(ii) Putting P = 1 and Q = −1 in (4), we obtain the next subfamily

SK [γ;α, β] =

{
g ∈ H : ℜ

{
αg (ξ) + βξg′ (ξ)

αξ + βg (ξ)

}
> γ

}
, (5)

by taking α = 0 in (5), we obtain the next subfamily

S (γ) =

{
g ∈ H : ℜ

{
ξg′ (ξ)

g (ξ)

}
> γ

}
,



T. M. Seoudy, A. E. Shammaky / Eur. J. Pure Appl. Math, 18 (4) (2025), 6366 3 of 15

where S (γ) is the family of starlike of order γ in ∆ (see [14–19]) and by putting
β = 0 in (5), we obtain the next subfamily

K (γ) =

{
g ∈ H : ℜ

{
g (ξ)

ξ

}
> γ

}
;

(iii) Substituting the value of P = ρ and Q = −ρ with 0 < ρ ≤ 1 in (4), we get the
following subfamily

SK (γ, ρ;α, β) =

g ∈ H :

∣∣∣∣∣∣
αg(ξ)+βξg′(ξ)

αξ+βg(ξ) − 1

αg(ξ)+βξg′(ξ)
αξ+βg(ξ) + 1− 2γ

∣∣∣∣∣∣ < ρ

 (6)

by substituting α = 0 in (6), we have

S (γ, ρ) =

g ∈ H :

∣∣∣∣∣∣
ξg′(ξ)
g(ξ) − 1

ξg′(ξ)
g(ξ) + 1− 2γ

∣∣∣∣∣∣ < ρ

 ,

where the subfamily S (γ, ρ) was introduced in [20] and also, by putting α = 0 in
(6), we get the following subfamily

K (γ, ρ) =

g ∈ H :

∣∣∣∣∣∣
g(ξ)
ξ − 1

g(ξ)
ξ + 1− 2γ

∣∣∣∣∣∣ < ρ

 .

In order to obtain the results of this study, we will mention the next definition and the
next lemmas.

Definition 2. The family of regular functions ϕ(ξ) in ∆ with the positive real part in ∆,
and of the form

ϕ(ξ) = 1 +

∞∑
j=1

δjξ
j (ξ ∈ ∆) (7)

is denoted by Φ, that calls the well-known Carathéodory family.

Lemma 1. If ϕ given by (7) belongs to Φ, then

|δj | ≤ 2, for j ∈ N = {1, 2, 3, ...} , (8)

for u, v are complex numbers we have∣∣δ2 − uδ21
∣∣ ≤ 2max{1; |2u− 1|} (9)

and
|δi+j − vδiδj | ≤ 2max{1; |1− 2v|}. (10)



T. M. Seoudy, A. E. Shammaky / Eur. J. Pure Appl. Math, 18 (4) (2025), 6366 4 of 15

The inequality (8) is Carathéodory’s result (see [21, 22]), while the inequality (9) may
be detected in [23], and the inequality (10) is from [24].

Lemma 2. [25] If ϕ given by (7) belongs to Φ, then

∣∣δ2 − uδ21
∣∣ ≤


−4u+ 2 if u ≤ 0,
2 if 0 ≤ u ≤ 1,
4u− 2 if u ≥ 1.

(11)

Also the upper bound (11) can be improved as follows for 0 ≤ u ≤ 1
2 , we have∣∣δ2 − uδ21

∣∣+ u |δ1|2 ≤ 2,

and for 1
2 ≤ u ≤ 1, we have ∣∣δ2 − uδ21

∣∣+ (1− u) |δ1|2 ≤ 2.

In the following sections, for the function subfamily SK [P,Q;α, β] and its special sub-
families S [P,Q] and K [P,Q], we study the well-known results, like convolution properties,
necessary and sufficient conditions, coefficient estimates, and Fekete-Szegö inequalities.

2. Convolution Properties

Unless otherwise stated, we suppose throughout this article that 0 ≤ θ < 2π; 0 ≤
α, β ≤ 1; α+ β ̸= 0; −1 ≤ Q < P ≤ 1; 0 ≤ γ < 1 and g ∈ H has the series form (1).

Theorem 1. The function g ∈ SK [P,Q;α, β] if and only if

1

ξ

g (ξ) ∗ ξ −
(
α+(α+β)R

α+β

)
ξ2 +

(
αR
α+β

)
ξ3

(1− ξ)2

 ̸= 0, (12)

where R is defined by

R = R (θ, P,Q, γ) =
e−iθ +Q

(1− γ) (P −Q)
+ 1. (13)

Proof. It is clear to check the next:

g (ξ) = g (ξ) ∗ ξ

1− ξ
,

ξg′ (ξ) = g (ξ) ∗ ξ

(1− ξ)2
.

 (14)

To show that (12) is true, we will write (4) as follows

αg (ξ) + βξg′ (ξ)

αξ + βg (ξ)
=

1 + [Q+ (1− γ) (P −Q)]w (ξ)

1 +Qw (ξ)
, (15)



T. M. Seoudy, A. E. Shammaky / Eur. J. Pure Appl. Math, 18 (4) (2025), 6366 5 of 15

where w (ξ) ∈ Ω, hence we have

αg (ξ) + βξg′ (ξ)

αξ + βg (ξ)
̸= 1 + [Q+ (1− γ) (P −Q)] eiθ

1 +Qeiθ
,

which is equivalent to

1

ξ

{(
1 +Qeiθ

) [
αg (ξ) + βξg′ (ξ)

]
−

(
1 + [Q+ (1− γ) (P −Q)] eiθ

)
[αξ + βg (ξ)]

}
̸= 0.

(16)
By using (14) in (16), we get

1

ξ

 g (ξ) ∗
(

αξ
1−ξ +

βξ

(1−ξ)2

) (
1 +Qeiθ

)
−g (ξ) ∗

(
αξ + βξ

1−ξ

) (
1 + [Q+ (1− γ) (P −Q)] eiθ

)


=
1

ξ

 g (ξ) ∗
(
(α+β)ξ−αξ2

(1−ξ)2

) (
1 +Qeiθ

)
−g (ξ) ∗

(
(α+β)ξ−(2α+β)ξ2+αξ3

(1−ξ)2

) (
1 + [Q+ (1− γ) (P −Q)] eiθ

)


= − (α+β)(1−γ)(P−Q)eiθ

ξ

g (ξ) ∗
ξ −

α+(α+β)

(
e−iθ+Q

(1−γ)(P−Q)
+1

)
α+β ξ2 +

α

(
e−iθ+Q

(1−γ)(P−Q)
+1

)
α+β ξ3

(1− ξ)2


̸= 0

which explains the necessary condition (12) for this theorem.
Reversely, assume that g ∈ H satisfies (12). Since the first part has been proven,

assumption (12) is equivalent to (16), so we get that

αg (ξ) + βξg′ (ξ)

αξ + βg (ξ)
̸= 1 + [Q+ (1− γ) (P −Q)] eiθ

1 +Qeiθ
, (17)

if we assume to φ (ξ) = αg(ξ)+βξg′(ξ)
αξ+βg(ξ) and ψ (ξ) = 1+[Q+(1−γ)(P−Q)]ξ

1+Qξ , the relationship

(17) indicates that φ (∆) ∩ ψ (∂∆) = ∅. Therefore, the simply connected domain φ (∆)
lies inside a connected component of C\ψ (∂∆). Thus, using the fact that the function
ψ(ξ) is univalent and φ (0) = ψ (0), it follows that φ (ξ) ≺ ψ (ξ), which means that
g ∈ SK [P,Q, γ;α, β]. This finishes Theorem 1.

Letting P = 1 and Q = −1 in Theorem 1, we get the following result.

Corollary 1. The function g ∈ SK [γ;α, β] if and only if

1

ξ

g (ξ) ∗ ξ −
(
α+(α+β)T

α+β

)
ξ2 +

(
αT
α+β

)
ξ3

(1− ξ)2

 ̸= 0,

where T is given by

T = T (θ, γ) =
e−iθ − 1

2 (1− γ)
+ 1. (18)



T. M. Seoudy, A. E. Shammaky / Eur. J. Pure Appl. Math, 18 (4) (2025), 6366 6 of 15

Putting α = γ = 0 in Theorem 1, we have the next corollary.

Corollary 2. The function g ∈ S [P,Q] if and only if

1

ξ

{
g (ξ) ∗ ξ −Rξ2

(1− ξ)2

}
̸= 0,

where R is given by (13).

Putting β = γ = 0 in Theorem 1, we get the next.

Corollary 3. The function g ∈ K [P,Q] if and only if

1

ξ

{
g (ξ) ∗ ξ −Rξ2

1− ξ

}
̸= 0,

where R is given by (13).

Theorem 2. The function g ∈ SK [P,Q, γ;α, β] if and only if

1−
∞∑
j=2

α
(
e−iθ +Q

)
+ β

[
(j − 1)

(
e−iθ +Q

)
− (1− γ) (P −Q)

]
(α+ β) (1− γ) (P −Q)

djξ
j−1 ̸= 0. (19)

Proof. From Theorem 1, we find that g(ξ) ∈ SK [P,Q, γ;α, β] if and only if

1

ξ

g(ξ) ∗ ξ − α+(α+β)R
α+β ξ2 + αR

α+β ξ
3

(1− ξ)2

 ̸= 0 (20)

for all R given by (13). The left hand side of (20) can be written as

1

ξ

g(ξ) ∗
 αR

α+ β
ξ +

(
α+(β−α)R

α+β

)
ξ

1− ξ
+

(
β(1−R)
α+β

)
ξ

(1− ξ)2


=

1

ξ

{(
αR

α+ β

)
ξ +

(
α+ (β − α)R

α+ β

)
g (ξ) +

(
β (1−R)

α+ β

)
ξg′ (ξ)

}
= 1−

∞∑
j=2

α (R− 1) + β (j (R− 1)−R)

α+ β
djξ

j−1

= 1−
∞∑
j=2

α
(
e−iθ +Q

)
+ β

[
(j − 1)

(
e−iθ +Q

)
− (1− γ) (P −Q)

]
(α+ β) (1− γ) (P −Q)

djξ
j−1.

Thus, the proof of Theorem 2 is completed.

Putting P = 1 andQ = −1 in Theorem 2, we obtain the following result for SK [γ;α, β].



T. M. Seoudy, A. E. Shammaky / Eur. J. Pure Appl. Math, 18 (4) (2025), 6366 7 of 15

Corollary 4. The function g ∈ SK [γ;α, β] if and only if

1−
∞∑
j=2

α
(
e−iθ − 1

)
+ β

[
(j − 1)

(
e−iθ − 1

)
− 2 (1− γ)

]
2 (α+ β) (1− γ)

djξ
j−1 ̸= 0.

Putting α = γ = 0 in Theorem 2, we obtain the following result for S [P,Q].

Corollary 5. The function g ∈ S [P,Q] if and only if

1−
∞∑
j=2

(j − 1)
(
e−iθ +Q

)
− P +Q

P −Q
djξ

j−1 ̸= 0.

Putting β = γ = 0 in Theorem 2, we obtain the following result for K [P,Q].

Corollary 6. The function g ∈ K [P,Q] if and only if

1−
∞∑
j=2

e−iθ +Q

P −Q
djξ

j−1 ̸= 0.

Theorem 3. If the function g(ξ) satisfy the inequality

∞∑
j=2

{α (1−Q) + β [(j − 1) (1−Q) + (1− γ) (P −Q)]} |dj | ≤ (α+ β) (P −Q) , (21)

then g ∈ SK [P,Q, γ;α, β].

Proof. Note that∣∣∣∣∣∣1−
∞∑
j=2

α
(
e−iθ +Q

)
+ β

[
(j − 1)

(
e−iθ +Q

)
− (1− γ) (P −Q)

]
(α+ β) (1− γ) (P −Q)

djξ
j−1

∣∣∣∣∣∣

≥ 1−

∣∣∣∣∣∣
∞∑
j=2

α
(
e−iθ +Q

)
+ (β)

[
(j − 1)

(
e−iθ +Q

)
− (1− γ) (P −Q)

]
(α+ β) (1− γ) (P −Q)

djξ
j−1

∣∣∣∣∣∣
≥ 1−

∞∑
j=2

α (1−Q) + β [(j − 1) (1−Q) + (1− γ) (P −Q)]

(α+ β) (1− γ) (P −Q)
|dj | > 0.

Thus, the inequality (21) holds and our result follows from Theorem 3.

Putting P = 1 andQ = −1 in Theorem 3, we obtain the following result for SK [γ;α, β].



T. M. Seoudy, A. E. Shammaky / Eur. J. Pure Appl. Math, 18 (4) (2025), 6366 8 of 15

Corollary 7. If the function g(ξ) satisfy the inequality

∞∑
j=2

[α+ β (j − γ)] |dj | ≤ (α+ β) (1− γ) ,

then g ∈ SK [γ;α, β].

Putting α = γ = 0 in Theorem 3, we obtain the following result for S [P,Q].

Corollary 8. If g(ξ) satisfy the inequality

∞∑
j=2

[(j − 1) (1−Q) + P −Q] |dj | ≤ P −Q,

then g ∈ S [P,Q].

Putting β = γ = 0 in Theorem 3, we obtain the following result for K [P,Q].

Corollary 9. If g(ξ) satisfy the inequality

∞∑
j=2

(1−Q) |dj | ≤ P −Q,

then g ∈ K [P,Q].

3. Initial Coefficient Estimates and Fekete–Szegő Problems

The Fekete-Szegő problems are a major area of research in complex analysis concerning
the maximum value of functionals of the form

∣∣d3 − µd22
∣∣ over families of univalent functions

found by Fekete and Szegő [26] (see also, [27–32]). In the following theorems, we determine
the upper bounds for the coefficients |d2|, |d3|, |d4| and the Fekete–Szegő functional for
the subfamily SK [P,Q, γ;α, β].

Theorem 4. If g ∈ SK [P,Q, γ;α, β], then

|d2| ≤ (1− γ) (P −Q) ,

|d3| ≤ (α+ β) (1− γ) (P −Q)

α+ 2β
max

{
1;

∣∣∣∣Q− β (1− γ) (P −Q)

α+ β

∣∣∣∣} ,
|d4| ≤ (α+β)(1−γ)(P−Q)

(α+3β)

 max
{
1;
∣∣∣ (2α+3β)β(1−γ)(P−Q)

(α+β)(α+2β) − 1− 2Q
∣∣∣}

+
∣∣∣1 +Q− β(1−γ)(P−Q)

α+β

∣∣∣ ∣∣∣1 +Q− β(1−γ)(P−Q)
(α+2β)

∣∣∣
 .



T. M. Seoudy, A. E. Shammaky / Eur. J. Pure Appl. Math, 18 (4) (2025), 6366 9 of 15

Proof. If g ∈ SK [P,Q, γ;α, β], then there exists a function w (ξ) ∈ Ω in ∆ such that

αg (ξ) + βξg′ (ξ)

αξ + βg (ξ)
=

1 + [Q+ (1− γ) (P −Q)]w (ξ)

1 +Qw (ξ)
= J (w (ξ)) . (22)

Since g (ξ) is given by (1), it follows that

αg (ξ) + βξg′ (ξ)

αξ + βg (ξ)
= 1 + d2ξ +

[
α+2β
α+β d3 −

β
α+βd

2
2

]
ξ2

+
[
α+3β
α+β d4 −

(2α+3β)β

(α+β)2
d2d3 +

β2

(α+β)2
d32

]
ξ3 + ... . (23)

Define the function ϕ (ξ) by

ϕ (ξ) =
1 + w (ξ)

1− w (ξ)
= 1 + δ1ξ + δ2ξ

2 + δ3ξ
3 + ... ,

since w (ξ) ∈ Ω , we see that ϕ ∈ Φ and

w (ξ) =
ϕ (ξ)− 1

ϕ (ξ) + 1
=

δ1ξ + δ2ξ
2 + δ3ξ

3 + ...

2 + δ1ξ + δ2ξ2 + δ3ξ3 + ...
. (24)

According to (24), we have

J (w (ξ)) =
1 + [Q+ (1− γ) (P −Q)]w (ξ)

1 +Qw (ξ)
= 1+ (1−γ)(P−Q)

2 δ1ξ+
(1−γ)(P−Q)

2

[
δ2 − (1+Q)

2 δ21

]
ξ2

+
(1− γ) (P −Q)

2

[
δ3 − (1 +Q) δ1δ2 +

(1 +Q)2

4
δ31

]
ξ3 + ... . (25)

Equating the corresponding coefficients of (23) and (25), we obtain

d2 =
(1− γ) (P −Q) δ1

2
, (26)

d3 =
(α+ β) (1− γ) (P −Q)

2 (α+ 2β)

{
δ2 −

1

2

[
1 +Q− β (1− γ) (P −Q)

α+ β

]
δ21

}
(27)

d4 =
(α+β)(1−γ)(P−Q)

2(α+3β)


δ3 −

[
1 +Q− (2α+3β)β(1−γ)(P−Q)

2(α+β)(α+2β)

]
δ1δ2

+

{(
1+Q
2

)2
− (2α+3β)β(1+Q)(1−γ)(P−Q)

4(α+β)(α+2β) + β2(P−Q)2(1−γ)2

4(α+β)(α+2β)

}
δ31


(28)

Using (26), we have

|d2| =
(1− γ) (P −Q)

2
|δ1| ,

and from (8), we have |δ1| ≤ 2, therefore

|d2| ≤ (1− γ) (P −Q) .



T. M. Seoudy, A. E. Shammaky / Eur. J. Pure Appl. Math, 18 (4) (2025), 6366 10 of 15

The relation (27) yields to

|d3| =
(α+ β) (1− γ) (P −Q)

2 (α+ 2β)

∣∣∣∣δ2 − 1

2

[
1 +Q− β (1− γ) (P −Q)

α+ β

]
δ21

∣∣∣∣
and according to (9), we get

|d3| ≤ (α+ β) (1− γ) (P −Q)

2 (α+ 2β)
.2.max

{
1;

∣∣∣∣2.12
[
1 +Q− β (1− γ) (P −Q)

α+ β

]
− 1

∣∣∣∣}
=

(α+ β) (1− γ) (P −Q)

α+ 2β
max

{
1;

∣∣∣∣Q− β (1− γ) (P −Q)

α+ β

∣∣∣∣} .
The equality (28) leads to

|d4| = (α+β)(1−γ)(P−Q)
2(α+3β)

∣∣∣∣∣∣∣
δ3 −

[
1 +Q− (2α+3β)β(1−γ)(P−Q)

2(α+β)(α+2β)

]
δ1δ2

+

{(
1+Q
2

)2
− (2α+3β)β(1+Q)(1−γ)(P−Q)

4(α+β)(α+2β) + β2(P−Q)2(1−γ)2

4(α+β)(α+2β)

}
δ31

∣∣∣∣∣∣∣
and using the triangle inequality we have

|d4| ≤ (α+β)(1−γ)(P−Q)
2(α+3β)


∣∣∣δ3 − [

1 +Q− (2α+3β)β(1−γ)(P−Q)
2(α+β)(α+2β)

]
δ1δ2

∣∣∣
+

∣∣∣∣(1+Q
2

)2
− (2α+3β)β(1+Q)(1−γ)(P−Q)

4(α+β)(α+2β) + β2(P−Q)2(1−γ)2

4(α+β)(α+2β)

∣∣∣∣ |δ1|3
 ,

from (8) and (10), the above inequality implies that

|d4| ≤ (α+β)(1−γ)(P−Q)
2(α+3β)

 max
{
1;
∣∣∣ (2α+3β)β(1−γ)(P−Q)

(α+β)(α+2β) − 1− 2Q
∣∣∣}

+
∣∣∣1 +Q− β(1−γ)(P−Q)

α+β

∣∣∣ ∣∣∣1 +Q− β(1−γ)(P−Q)
α+2β

∣∣∣
 .

This completes the proof of Theorem 4.

Putting P = 1 and Q = −1 in Theorem 4, we get the following.

Corollary 10. If g ∈ SK [γ;α, β], then

|d2| ≤ 2 (1− γ) ,

|d3| ≤ 2 [2β (1− γ) + α+ β] (1− γ)

α+ 2β
,

|d4| ≤ 2 [2β (1− γ) + α+ β] [2β (1− γ) + α+ 2β] (1− γ)

(α+ 2β) (α+ 3β)
.

Putting α = γ = 0 in Theorem 4, we get the following.

Corollary 11. If g ∈ S [P,Q], then

|d2| ≤ P −Q,

|d3| ≤ P −Q

2
max {1; |2Q− P |} ,

|d4| ≤ P−Q
3

[
max

{
1;
∣∣1 + 7

2Q− 3
2P

∣∣}+ |1 + 2Q− P |
∣∣∣∣1 + 3

2
Q− 1

2
P

∣∣∣∣] .



T. M. Seoudy, A. E. Shammaky / Eur. J. Pure Appl. Math, 18 (4) (2025), 6366 11 of 15

Putting β = γ = 0 in Theorem 4, we get the next.

Corollary 12. If g ∈ K [P,Q], then

|d2| ≤ P −Q,

|d3| ≤ P −Q,

|d4| ≤ (P −Q)
[
max {1; |1 + 2Q|}+ (1 +Q)2

]
.

Theorem 5. If g ∈ SK [P,Q, γ;α, β], then∣∣d3 − µd22
∣∣ ≤ (α+ β) (1− γ) (P −Q)

α+ 2β
max

{
1;

∣∣∣∣Q− β(1−γ)(P−Q)
α+β

(
1− α+ 2β

β
µ

)∣∣∣∣} .
(29)

Proof. If g ∈ SK [P,Q, γ;α, β], then from (26) and (27) we get

d3 − µd22 =
(α+ β) (1− γ) (P −Q)

2 (α+ 2β)

{
δ2 − uδ21

}
, (30)

where

u =
1

2

[
1 +Q− β (1− γ) (P −Q)

α+ β

(
1− α+ 2β

β
µ

)]
. (31)

Applying (9) to (30), it follows that∣∣d3 − µd22
∣∣ ≤ (α+β)(1−γ)(P−Q)

2(α+2β) .2max

{
1;

∣∣∣∣2.12 [
1 +Q− β(1−γ)(P−Q)

α+β

(
1− α+2β

β µ
)]

− 1

∣∣∣∣}
= (α+β)(1−γ)(P−Q)

α+2β max
{
1;
∣∣∣Q− β(1−γ)(P−Q)

α+β

(
1− α+2β

β µ
)∣∣∣} .

This completes the proof of Theorem 5.

Putting P = 1 and Q = −1 in Theorem 5, we get the following.

Corollary 13. If g ∈ SK [γ;α, β], then∣∣d3 − µd22
∣∣ ≤ 2 (α+ β) (1− γ)

α+ 2β
max

{
1;

∣∣∣∣Q− 2β (1− γ)

α+ β

(
1− α+ 2β

β
µ

)∣∣∣∣} .
Putting α = γ = 0 in Theorem 5, we get the following.

Corollary 14. If g ∈ S [P,Q], then∣∣d3 − µd22
∣∣ ≤ P−Q

2 max {1; |Q− (P −Q) (1− 2µ)|} .

Putting β = γ = 0 in Theorem 5, we get the following.

Corollary 15. If g ∈ K [P,Q], then∣∣d3 − µd22
∣∣ ≤ P −Q.



T. M. Seoudy, A. E. Shammaky / Eur. J. Pure Appl. Math, 18 (4) (2025), 6366 12 of 15

Theorem 6. Let

σ1 =
β (1− γ) (P −Q)− (α+ β) (1 +Q)

(α+ 2β) (1− γ) (P −Q)
, σ2 =

β (1− γ) (P −Q) + (α+ β) (1−Q)

(α+ 2β) (1− γ) (P −Q)
,

σ3 =
β (1− γ) (P −Q)− (α+ β)Q

(α+ 2β) (1− γ) (P −Q)
.

If g ∈ SK [P,Q, γ;α, β], then

∣∣d3 − µd22
∣∣ ≤


− (α+β)(1−γ)(P−Q)

(α+2β)

[
Q− β(1−γ)(P−Q)

α+β

(
1− α+2β

β µ
)]

(µ ≤ σ1) ,
(α+β)(1−γ)(P−Q)

(α+2β) (σ1 ≤ µ ≤ σ2) ,
(α+β)(1−γ)(P−Q)

(α+2β)

[
Q− β(1−γ)(P−Q)

α+β

(
1− α+2β

β µ
)]

(µ ≥ σ2) .

Further, if σ1 ≤ µ ≤ σ3, then∣∣d3 − µd22
∣∣+ α+ β

α+ 2β

[
1 +Q

(1− γ) (P −Q)
− β

α+ β

(
1− α+ 2β

β
µ

)]
|d2|2 ≤

(α+ β) (1− γ) (P −Q)

α+ 2β
.

If σ3 ≤ µ ≤ σ2, then∣∣d3 − µd22
∣∣+ α+ β

α+ 2β

[
1−Q

(1− γ) (P −Q)
+

β

α+ β

(
1− α+ 2β

β
µ

)]
|d2|2 ≤

(α+ β) (1− γ) (P −Q)

α+ 2β
.

Proof. Using Lemma 2 to (30) and (31), we can derive our results, which are confirmed
by Theorem 6.

Taking P = 1 and Q = −1 in Theorem 6, we derive the following.

Corollary 16. Let

σ4 =
β

α+ 2β
, σ5 =

β (1− γ) + α+ β

(α+ 2β) (1− γ)
, σ6 =

2β (1− γ) + α+ β

2 (α+ 2β) (1− γ)
.

If g ∈ SK [γ;α, β], then

∣∣d3 − µd22
∣∣ ≤


2(α+β)(1−γ)

α+2β

[
1 + 2β(1−γ)

α+β

(
1− α+2β

β µ
)]

(µ ≤ σ4) ,
2(α+β)(1−γ)

α+2β (σ4 ≤ µ ≤ σ5) ,

−2(α+β)(1−γ)
α+2β

[
1 + 2β(1−γ)

α+β

(
1− α+2β

β µ
)]

(µ ≥ σ5) .

Further, if σ4 ≤ µ ≤ σ6, then∣∣d3 − µd22
∣∣− β

α+ 2β

(
1− α+ 2β

β
µ

)
|d2|2 ≤

2 (α+ β) (1− γ)

(α+ 2β)
.

If σ3 ≤ µ ≤ σ2, then∣∣d3 − µd22
∣∣+ α+ β

α+ 2β

[
1

1− γ
+

β

α+ β

(
1− α+ 2β

β
µ

)]
|d2|2 ≤

2 (α+ β) (1− γ)

(α+ 2β)
.



T. M. Seoudy, A. E. Shammaky / Eur. J. Pure Appl. Math, 18 (4) (2025), 6366 13 of 15

Putting α = γ = 0 in Theorem 6, we get the following.

Corollary 17. Let

σ10 =
P − 2Q− 1

2 (P −Q)
, σ11 =

P − 2Q+ 1

2 (P −Q)
, σ12 =

P − 2Q

2 (P −Q)
.

If g ∈ S [P,Q], then

∣∣d3 − µd22
∣∣ ≤


− (P−Q)[Q−(P−Q)(1−2µ)]

2 (µ ≤ σ10) ,
P−Q
2 (σ10 ≤ µ ≤ σ11) ,

(P−Q)[Q−(P−Q)(1−2µ)]
2 (µ ≥ σ11) .

Further, if σ10 ≤ µ ≤ σ12, then∣∣d3 − µd22
∣∣+ 1

2

(
1+Q
P−Q − 1 + 2µ

)
|d2|2 ≤ P−Q

2 .

If σ12 ≤ µ ≤ σ11, then ∣∣d3 − µd22
∣∣+ 1

2

(
1−Q
P−Q + 1− 2µ

)
|d2|2 ≤ P−Q

2 .

Putting β = γ = 0 in Theorem 5, we obtain the following.

Corollary 18. Let

σ13 = − 1 +Q

P −Q
, σ14 =

1−Q

P −Q
, σ15 = − Q

P −Q
.

If g ∈ K [P,Q], then

∣∣d3 − µd22
∣∣ ≤


− (P −Q)Q (µ ≤ σ13) ,
P −Q (σ13 ≤ µ ≤ σ14) ,
(P −Q)Q (µ ≥ σ14) .

Further, if σ13 ≤ µ ≤ σ15, then∣∣d3 − µd22
∣∣+ 1+Q

P−Q |d2|2 ≤ P −Q.

If σ15 ≤ µ ≤ σ14, then ∣∣d3 − µd22
∣∣+ 1−Q

P−Q |d2|2 ≤ P −Q.

4. Conclusion

This paper mainly focuses on finding the convolution results and coefficient esti-
mates for a new subfamily SK [P,Q, γ;α, β] of analytic functions linked to the generalized
Janowski domain. We also studied the upper bounds of the first four coefficients and the
Fekete-Szegö inequalities for the functions in this subfamily. We note that the results of
this study naturally include many of the known results for these subfamilies, which are
listed in the Introduction section. For future studies, we can derive important geometric
properties by introducing the same subfamily in the case of multivalent (or meromorphic)
regular functions and applying the same techniques used in this paper.



T. M. Seoudy, A. E. Shammaky / Eur. J. Pure Appl. Math, 18 (4) (2025), 6366 14 of 15

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