1 Introduction

Let Σ denote the class of functions of the form:

f ( z ) = 1 z + n = 0 a n z n ,
(1.1)

which are analytic in the punctured open unit disk

U * = { z : 0 < z < 1 } =U\ { 0 } ,
(1.2)

where U is the open unit disk U= { z : | z | < 1 } ..

We say that a function f ∈ Σ is meromorphic starlike of order δ (0 ≤ δ < 1), and belongs to the class Σ*(δ), if it satisfies the inequality:

- z f ( z ) f ( z ) >δ.
(1.3)

A function f ∈ Σ is a meromorphic convex function of order δ (0 ≤ δ < 1), if f satisfies the following inequality:

- 1 + z f ( z ) f ( z ) >δ,
(1.4)

and we denote this class by Σ k (δ).

For f ∈ Σ, Wang et al. [1] and Nehari and Netanyahu [2] introduced and studied the subclass Σ N (β) of Σ consisting of functions f(z) satisfying

- z f ( z ) f ( z ) + 1 <β ( β > 1 , z U ) .
(1.5)

Let A denote the class of functions f of the form f ( z ) =z+ n = 2 a n z n , which are analytic in the open unit disk U.

Analogous to several subclasses [310] of analytic functions of A, we define the following subclasses of Σ.

Definition 1.1 Let a function f ∈ Σ be analytic in U * . Then f is in the class Σ b ( δ ) if, and only if, f satisfies

1 - 1 b z f ( z ) f ( z ) + 1 >δ,
(1.6)

where b\ { 0 } , 0 ≤ δ < 1.

Definition 1.2 Let a function f ∈ Σ be analytic in U * . Then f is in the class ΣK b (δ) if, and only if, f satisfies

1 - 1 b z f ( z ) f ( z ) + 2 >δ,
(1.7)

where bC\{0}, 0 ≤ δ < 1. We note that f ∈ Σ K b (δ) if, and only if, -z f Σ b ( δ ) .

Furthermore, the classes

Σ 1 ( δ ) Σ ( δ ) ,Σ K 1 ( δ ) Σ k ( δ )

are the classes of meromorphic starlike functions of order δ and meromorphic convex functions of order δ in U * , respectively. Moreover, the classes

Σ 1 ( 0 ) Σ ( 0 ) ,Σ K 1 ( 0 ) Σ k ( 0 )

are familiar classes of starlike and convex functions in U * , respectively.

Definition 1.3 Let a function f ∈ Σ be analytic in U * . Then f is in the class Σ U ( α , δ , b ) if, and only if, f satisfies

1 - 1 b z f ( z ) f ( z ) + 1 >α 1 b z f ( z ) f ( z ) + 1 +δ,
(1.8)

where α ≥ 0, δ ∈ [-1,1), α + δ ≥ 0, b\ { 0 } .

Definition 1.4 Let a function f ∈ Σ be analytic in U * . Then f is in the class ΣKU ( α , δ , b ) if, and only if, f satisfies

1 - 1 b z f ( z ) f ( z ) + 2 >α 1 b z f ( z ) f ( z ) + 2 +δ,
(1.9)

where α ≥ 0, δ ∈ [-1,1), α + δ ≥ 0, b\ { 0 } .

We note that fΣKU ( α , δ , b ) if, and only if, -z f Σ U ( α , δ , b ) .

For α = 0, we have

Σ U ( 0 , δ , b ) Σ b ( δ ) ,ΣKU ( 0 , δ , b ) Σ K b ( δ ) .

Definition 1.5 Let a function f ∈ Σ be analytic in U * . Then f is in the class Σ UH ( α , b ) if, and only if, f satisfies

1 - 1 b z f ( z ) f ( z ) + 1 - 2 α ( 2 - 1 ) < 2 1 - 1 b z f ( z ) f ( z ) + 1 + 2 α ( 2 - 1 ) ,
(1.10)

where α > 0, b\ { 0 } .

Definition 1.6 Let a function f ∈ Σ be analytic in U * . Then f is in the class ΣKUH ( α , b ) if, and only if, f satisfies

1 - 1 b z f ( z ) f ( z ) + 2 - 2 α ( 2 - 1 ) < 2 1 - 1 b z f ( z ) f ( z ) + 2 + 2 α ( 2 - 1 ) .
(1.11)

where α > 0, b\ { 0 } .

We note that fΣKUH ( α , b ) if, and only if, -z f Σ UH ( α , b ) .

Let us also introduce the following families of new subclasses Σ F 1 ( δ , b ) , Σ F 2 ( α , δ , b ) , and Σ F 3 ( α , b ) as follows.

Definition 1.7 Let a function f ∈ Σ be analytic in U * . Then f is in the class Σ F 1 ( δ , b ) , if, and only if, f satisfies

1 - 1 b z z f ( z ) + 3 f ( z ) z f ( z ) + 2 f ( z ) + 1 >δ.
(1.12)

where b\ { 0 } , 0 ≤ δ < 1.

We note that fΣ F 1 ( δ , b ) if, and only if, z f ( z ) +2f ( z ) Σ b ( δ ) .

Definition 1.8 Let a function f ∈ Σ be analytic in U * . Then f is in the class Σ F 2 ( α , δ , b ) if, and only if, f satisfies

1 - 1 b z z f ( z ) + 3 f ( z ) z f ( z ) + 2 f ( z ) + 1 >α 1 b z z f ( z ) + 3 f ( z ) z f ( z ) + 2 f ( z ) + 1 +δ,
(1.13)

where α ≥ 0, δ ∈ [-1,1), α + δ ≥ 0, b\ { 0 } .

We note that fΣ F 2 ( α , δ , b ) if, and only if, z f ( z ) +2f ( z ) Σ U ( α , δ , b ) .

Definition 1.9 Let a function f ∈ Σ be analytic in U * . Then f is in the class Σ F 3 ( α , b ) if, and only if, f satisfies

1 - 1 b z z f ( z ) + 3 f ( z ) z f ( z ) + 2 f ( z ) + 1 - 2 α ( 2 - 1 ) < 2 1 - 1 b z z f ( z ) + 3 f ( z ) z f ( z ) + 2 f ( z ) + 1 + 2 α ( 2 - 1 ) ,
(1.14)

where α > 0, b\ { 0 } .

We note that fΣ F 3 ( α , b ) if, and only if, z f ( z ) +2f ( z ) Σ UH ( α , b ) .

Recently, many authors introduced and studied various integral operators of analytic and univalent functions in the open unit disk U [1121].

Most recently, Mohammed and Darus [22, 23] introduced the following two general integral operators of meromorphic functions Σ:

H n ( z ) = 1 z 2 0 z u f 1 ( u ) γ 1 u f n ( u ) γ n du,
(1.15)

and

H γ 1 , , γ n ( z ) = 1 z 2 0 z - u 2 f 1 ( u ) γ 1 - u 2 f n ( u ) γ n du.
(1.16)

Goyal and Prajapat [24] obtained the following results for f ∈ Σ to be in the class Σ*(δ), 0 ≤ δ < 1.

Corollary 1.1 If f ∈ Σ satisfies the following inequality

z f ( z ) f ( z ) - 2 z f ( z ) f ( z ) < 1 - δ 3 - δ 2 - δ ,0δ<1,
(1.17)

then f ∈ Σ*(δ).

Corollary 1.2 If f ∈ Σ satisfies the following inequality

z f ( z ) f ( z ) - z f ( z ) f ( z ) + 1 < 1 2 ,
(1.18)

then f ∈ Σ*.

Corollary 1.3 If f ∈ Σ satisfies the following inequality

z f ( z ) f ( z ) 2 z f ( z ) f ( z ) - z f ( z ) f ( z ) - 1 >- 1 2 ,
(1.19)

then f ∈ Σ*.

In this article, we derive several properties for the integral operators H n ( z ) and H γ 1 , , γ n ( z ) of the subclasses given by (1.5) and Definitions 1.1 to 1.6.

2 Some properties for H n ( z )

In this section, we investigate some properties for the integral operator H n ( z ) defined by (1.15) of the subclasses given by (1.5), Definitions 1.1, 1.3, and 1.5.

Theorem 2.1 For i ∈ {1,..., n}, let γ i > 0, f i ∈ Σ and

z f i ( z ) f i ( z ) - 2 z f i ( z ) f i ( z ) < 1 - δ 3 - δ 2 - δ , 0 δ < 1 .
(2.1)

If

i = 1 n γ i > 2 1 - δ ,
(2.2)

then H n ( z ) Σ N ( β ) , where β > 1.

Proof On successive differentiation of H n ( z ) , which is defined in (1.5), we get

z 2 H n ( z ) +2z H n ( z ) = z f 1 ( z ) γ 1 z f n ( z ) γ n ,
(2.3)

and

z 2 H n ( z ) + 4 z H n ( z ) + 2 H n ( z ) = i = 1 n γ i z f i ( z ) + f i ( z ) z f i ( z ) z f 1 ( z ) γ 1 z f n ( z ) γ n .
(2.4)

Then from (2.3) and (2.4), we obtain

z 2 H n ( z ) + 4 z H n ( z ) + 2 H n ( z ) z 2 H n ( z ) + 2 z H n ( z ) = i = 1 n γ i f i ( z ) f i ( z ) + 1 z .
(2.5)

By multiplying (2.5) with z yield

z 2 H n ( z ) + 4 z H n ( z ) + 2 H n ( z ) z H n ( z ) + 2 H n ( z ) = i = 1 n γ i z f i ( z ) f i ( z ) + 1 .
(2.6)

This is equivalent to

z 2 H n ( z ) + 3 z H n ( z ) z H n ( z ) + 2 H n ( z ) +1= i = 1 n γ i z f i ( z ) f i ( z ) + 1 .
(2.7)

Therefore, we have

- z H n ( z ) H n ( z ) + 1 - 2 1 + 2 H n ( z ) z H n ( z ) =- i = 1 n γ i z f i ( z ) f i ( z ) + 1 +1.
(2.8)

Then, we easily get

- z H n ( z ) H n ( z ) + 1 = 2 H n ( z ) z H n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 1 + 1 + i = 1 n γ i - z f i ( z ) f i ( z ) + 3 - i = 1 n γ i .
(2.9)

Taking real parts of both sides of (2.9), we obtain

- z H n ( z ) H n ( z ) + 1 = 2 H n ( z ) z H n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 1 + 1 + i = 1 n γ i - z f i ( z ) f i ( z ) + 3 - i = 1 n γ i 2 H n ( z ) z H n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 1 + 1 + i = 1 n γ i - z f i ( z ) f i ( z ) + 3 - i = 1 n γ i .
(2.10)

Let

β = 2 H n ( z ) z H n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 1 + 1 + i = 1 n γ i - z f i ( z ) f i ( z ) + 3 - i = 1 n γ i .
(2.11)

Since 2 H n ( z ) z H n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 1 + 1 >0, applying Corollary 1.1, we have

β>δ i = 1 n γ i +3- i = 1 n γ i =3- 1 - δ i = 1 n γ i .
(2.12)

Then, by the hypothesis (2.2), we have β > 1. Therefore, H n ( z ) Σ N ( β ) , where β > 1. This completes the proof.   □

Letting δ = 0 in Theorem 2.1, we have

Corollary 2.2 For i ∈ {1,..., n}, let γ i > 0, f i ∈ Σ and

z f i ( z ) f i ( z ) - 2 z f i ( z ) f i ( z ) < 3 2 .
(2.13)

If

i = 1 n γ i >2,
(2.14)

then H n ( z ) Σ N ( β ) , where β > 1.

Making use of (2.11), Corollary 1.2 and Corollary 1.3, one can prove the following results.

Theorem 2.3 For i ∈ {1,..., n}, let γ i > 0, f i ∈ Σ and

z f i ( z ) f i ( z ) - z f i ( z ) f i ( z ) + 1 < 1 2 .
(2.15)

If

i = 1 n γ i >2,
(2.16)

then H n ( z ) Σ N ( β ) , where β > 1.

Theorem 2.4 For i ∈ {1,..., n}, let γ i > 0, f i ∈ Σ and

z f i ( z ) f i ( z ) 2 z f i ( z ) f i ( z ) - z f i ( z ) f i ( z ) - 1 >- 1 2 .
(2.17)

If

i = 1 n γ i >2,
(2.18)

then H n ( z ) Σ N ( β ) , where β > 1.

Theorem 2.5 For i ∈ {1,..., n}, let γ i > 0 and f i Σ b ( δ i ) (0 ≤ δ < 1 and b\ { 0 } ).

If

0< i = 1 n γ i 1 - δ i 1,
(2.19)

then H n ( z ) is in the class Σ F 1 ( μ , b ) ,μ=1- i = 1 n γ i 1 - δ i .

Proof From (2.7), we have

z z H n ( z ) + 3 H n ( z ) z H n ( z ) + 2 H n ( z ) +1= i = 1 n γ i z f i ( z ) f i ( z ) + 1 .
(2.20)

Equivalently, (2.20) can be written as

1- 1 b z z H n ( z ) + 3 H n ( z ) z H n ( z ) + 2 H n ( z ) + 1 = i = 1 n γ i 1 - 1 b z f i ( z ) f i ( z ) + 1 +1- i = 1 n γ i .
(2.21)

Taking the real part of both terms of the last expression, we have

1 - 1 b z z H n ( z ) + 3 H n ( z ) z H n ( z ) + 2 H n ( z ) + 1 = i = 1 n γ i 1 - 1 b z f i ( z ) f i ( z ) + 1 +1- i = 1 n γ i .
(2.22)

Since f i Σ b ( δ i ) , hence

1 - 1 b z z H n ( z ) + 3 H n ( z ) z H n ( z ) + 2 H n ( z ) + 1 > i = 1 n γ i δ i +1- i = 1 n γ i .
(2.23)

so that

1 - 1 b z z H n ( z ) + 3 H n ( z ) z H n ( z ) + 2 H n ( z ) + 1 >1- i = 1 n γ i 1 - δ i .
(2.24)

Then H n ( z ) Σ F 1 ( μ , b ) ,μ=1- i = 1 n γ i 1 - δ i .

Now, adopting the techniques used by Breaz et al. [11] and Bulut [21], we prove the following two theorems.

Theorem 2.6 For i ∈ {1,..., n}, let γ i > 0 and f i Σ U ( α , δ , b ) (α ≥ 0, δ ∈ [-1,1), α + δ ≥ 0 and b\ { 0 } ). If

i = 1 n γ i 1,
(2.25)

then H n ( z ) is in the class Σ F 2 ( α , δ , b ) .

Proof Since f i Σ U ( α , δ , b ) , it follows from Definition 1.3 that

1 - 1 b z f i ( z ) f i ( z ) + 1 >α 1 b z f i ( z ) f i ( z ) + 1 +δ.
(2.26)

Considering Definition 1.8 and with the help of (2.21), we obtain

1 - 1 b z z H n ( z ) + 3 H n ( z ) z H n ( z ) + 2 H n ( z ) + 1 - α 1 b z z H n ( z ) + 3 H n ( z ) z H n ( z ) + 2 H n ( z ) + 1 - δ = 1 - i = 1 n γ i + i = 1 n γ i 1 - 1 b z f i ( z ) f i ( z ) + 1 - α i = 1 n γ i 1 b z f i ( z ) f i ( z ) + 1 - δ 1 - i = 1 n γ i + i = 1 n γ i 1 - 1 b z f i ( z ) f i ( z ) + 1 - α i = 1 n γ i 1 b z f i ( z ) f i ( z ) + 1 - δ > 1 - i = 1 n γ i + i = 1 n γ i α 1 b z f i ( z ) f ( z ) + 1 + δ - α i = 1 n γ i 1 b z f i ( z ) f i ( z ) + 1 - δ = 1 - δ 1 - i = 1 n γ i 0 .
(2.27)

This completes the proof.   □

Theorem 2.7 For i ∈ {1,..., n}, let γ i > 0 and f i Σ UH ( α , b ) (α > 0 and b\ { 0 } ). If

i = 1 n γ i 1,
(2.28)

then H n ( z ) is in the class Σ F 3 ( α , b ) .

Proof Since f i Σ UH ( α , b ) , it follows from Definition 1.5 that

2 1 - 1 b z f i ( z ) f i ( z ) + 1 +2α ( 2 - 1 ) - 1 - 1 b z f i ( z ) f i ( z ) + 1 - 2 α ( 2 - 1 ) >0.
(2.29)

Considering this inequality and (2.21), we obtain

2 1 - 1 b z z H n ( z ) + 3 H n ( z ) z H n ( z ) + 2 H n ( z ) + 1 + 2 α 2 - 1 - 1 - 1 b z z H n ( z ) + 3 H n ( z ) z H n ( z ) + 2 H n ( z ) + 1 - 2 α 2 - 1 = 2 1 - i = 1 n γ i 1 b z f i ( z ) f i ( z ) + 1 + 2 α 2 - 1 - 1 - i = 1 n γ i 1 b z f i ( z ) f i ( z ) + 1 - 2 α 2 - 1 = 2 - 2 i = 1 n γ i 1 b z f i ( z ) f i ( z ) + 1 + 2 α 2 - 1 - 1 - i = 1 n γ i 1 b z f i ( z ) f i ( z ) + 1 - 2 α 2 - 1 = 2 + 2 i = 1 n γ i 1 - 1 b z f i ( z ) f i ( z ) + 1 - 2 i = 1 n γ i + 2 α 2 - 1 - 1 + i = 1 n γ i 1 - 1 b z f i ( z ) f i ( z ) + 1 - 2 α 2 - 1 - i = 1 n γ i + 2 α 2 - 1 i = 1 n γ i - 2 α 2 - 1 = 2 1 - i = 1 n γ i + 2 α 2 - 1 + 2 i = 1 n γ i 1 - 1 b z f i ( z ) f i ( z ) + 1 - 1 - 2 α 2 - 1 1 - i = 1 n γ i + i = 1 n γ i 1 - 1 b z f i ( z ) f i ( z ) + 1 - 2 α 2 - 1 2 1 - i = 1 n γ i + 2 α 2 - 1 + 2 i = 1 n γ i 1 - 1 b z f i ( z ) f i ( z ) + 1 - i = 1 n γ i 1 - 1 b z f i ( z ) f i ( z ) + 1 - 2 α 2 - 1 - 1 - 2 α 2 - 1 1 - i = 1 n γ i = i = 1 n γ i 2 1 - 1 b z f i ( z ) f i ( z ) + 1 + 2 α 2 - 1 - 1 - 1 b z f i ( z ) f i ( z ) + 1 - 2 α 2 - 1 + 2 1 - i = 1 n γ i + 2 α 2 - 1 - 2 α 2 - 1 i = 1 n γ i - 1 - 2 α 2 - 1 1 - i = 1 n γ i > 2 + 2 α 2 - 1 - 1 - 2 α 2 - 1 1 - i = 1 n γ i > 1 - i = 1 n γ i min 2 - 1 1 + 4 α , 2 + 1 0 .

This completes the proof.   □

3 Some properties for H γ 1 , , γ n ( z )

In this section, we investigate some properties for the integral operator H γ 1 , , γ n ( z ) defined by (1.16) of subclasses given by (1.5), Definitions 1.2, 1.4, and 1.6.

Theorem 3.1 For i ∈ {1,..., n}, let γ i > 0, f i ∈ Σ and

i = 1 n γ i > 2 1 - δ , 0 δ < 1 .
(3.1)

If f i Σ k ( δ ) , then H γ 1 , , γ n ( z ) Σ N ( β ) , β > 1.

Proof On successive differentiation of H γ 1 , , γ n ( z ) , which is defined in (1.16), we have

2z H γ 1 , , γ n ( z ) + z 2 H γ 1 , , γ n ( z ) = - z 2 f 1 ( z ) γ 1 - z 2 f n ( z ) γ n ,
(3.2)

and

z 2 H γ 1 , , γ n ( z ) + 4 z H γ 1 , , γ n ( z ) + 2 H γ 1 , , γ n ( z ) = i = 1 n γ i f i ( z ) f i ( z ) + 2 z - z 2 f 1 ( z ) γ 1 - z 2 f n ( z ) γ n .
(3.3)

Then from (3.2) and (3.3), we obtain

z 2 H γ 1 , , γ n ( z ) + 4 z H γ 1 , , γ n ( z ) + 2 H γ 1 , , γ n ( z ) z 2 H γ 1 , , γ n ( z ) + 2 z H γ 1 , , γ n ( z ) = i = 1 n γ i f i ( z ) f i ( z ) + 2 z .
(3.4)

By multiplying (3.4) with z yields,

z 2 H γ 1 , , γ n ( z ) + 4 z H γ 1 , , γ n ( z ) + 2 H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) + 2 H γ 1 , , γ n ( z ) = i = 1 n γ i z f i ( z ) f i ( z ) + 2 .
(3.5)

that is equivalent to

z 2 H γ 1 , , γ n ( z ) + 3 z H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) + 2 H γ 1 , , γ n ( z ) +1= i = 1 n γ i z f i ( z ) f i ( z ) + 2 .
(3.6)

Therefore, we have

- z H γ 1 , , γ n ( z ) H γ 1 , , γ n ( z ) + 1 - 2 1 + 2 H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) =- i = 1 n γ i z f i ( z ) f i ( z ) + 2 +1.
(3.7)

so that

- z H γ 1 , , γ n ( z ) H γ 1 , , γ n ( z ) + 1 = 2 H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 2 + 1 + i = 1 n γ i - z f i ( z ) f i ( z ) + 1 + 3 - i = 1 n γ i .
(3.8)

Taking the real parts of both terms of the last expression, we obtain

- z H γ 1 , , γ n ( z ) H γ 1 , , γ n ( z ) + 1 = 2 H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 2 + 1 + i = 1 n γ i - z f i ( z ) f i ( z ) + 1 + 3 - i = 1 n γ i 2 H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 2 + 1 + i = 1 n γ i - z f i ( z ) f i ( z ) + 1 + 3 - i = 1 n γ i .
(3.9)

Let

β= 2 H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 2 + 1 + i = 1 n γ i - z f i ( z ) f i ( z ) + 1 +3- i = 1 n γ i .
(3.10)

Since 2 H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 2 + 1 >0, f i ∈ Σ k (δ), we get

β>3- ( 1 - δ ) i = 1 n γ i ,
(3.11)

which, in light of the hypothesis (3.1), yields β > 1.

Therefore, H γ 1 , , γ n ( z ) Σ N ( β ) , β > 1. This completes the proof.   □

Theorem 3.2 For i ∈ {1,..., n}, let γ i > 0, f i ∈ Σ and

1< i = 1 n γ i <2.
(3.12)

If H γ 1 , , γ n ( z ) Σ ( δ ) , then H γ 1 , , γ n ( z ) Σ N ( β ) , β > 1.

Proof It follows from (3.7) that

- z H γ 1 , , γ n ( z ) H γ 1 , , γ n ( z ) + 1 = 1 + 2 H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 1 + 2 i = 1 n γ i - 1 - H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) + 3 - i = 1 n γ i .
(3.13)

Taking the real parts of both terms of the last expression, we obtain

- z H γ 1 , , γ n ( z ) H γ 1 , , γ n ( z ) + 1 = 1 + 2 H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 1 + 2 i = 1 n γ i - 1 - H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) + 3 - i = 1 n γ i .
(3.14)

Thus, we have

- z H γ 1 , , γ n ( z ) H γ 1 , , γ n ( z ) + 1 = 1 + 2 H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 1 + 2 i = 1 n γ i - 1 - 1 z H γ 1 , , γ n ( z ) H γ 1 , , γ n ( z ) + 3 - i = 1 n γ i 1 + 2 H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 1 + 2 i = 1 n γ i - 1 - z H γ 1 , , γ n ( z ) H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) H γ 1 , , γ n ( z ) 2 + 3 - i = 1 n γ i .
(3.15)

Let

β = 1 + 2 H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 1 + 2 i = 1 n γ i - 1 - z H γ 1 , , γ n ( z ) H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) H γ 1 , , γ n ( z ) 2 + 3 - i = 1 n γ i .
(3.16)

Since 1 + 2 H γ 1 , , γ n ( z ) z H γ 1 , , γ n ( z ) - i = 1 n γ i z f i ( z ) f i ( z ) + 1 >0 and H γ 1 , , γ n ( z ) Σ ( δ ) , we have

β>2 i = 1 n γ i - 1 δ z H γ 1 , , γ n ( z ) H γ 1 , , γ n ( z ) 2 +3- i = 1 n γ i .>3- i = 1 n γ i .
(3.17)

Then, by the hypothesis (3.12), we see that β > 1. Therefore, H γ 1 , , γ n ( z ) Σ N ( β ) , β > 1. This completes the proof.   □

Now, using the method given in the proofs of Theorems 2.5, 2.6, and 2.7, one can prove the following results:

Theorem 3.3 For i ∈ {1,..., n}, let γ i > 0 f i ∈ ΣK b (δ i ) (0 ≤ δ < 1 and b\ { 0 } )). If

0< i = 1 n γ i 1 - δ i 1,
(3.18)

then H γ 1 , , γ n ( z ) is in the class Σ F 1 ( μ , b ) ,μ=1- i = 1 n γ i 1 - δ i .

Theorem 3.4 For i ∈ {1,..., n}, let γ i > 0 and f i ΣKU ( α , δ , b ) (α ≥ 0, δ ∈ [-1,1), α + δ ≥ 0 and b\ { 0 } )). If

i = 1 n γ i 1,
(3.19)

then H γ 1 , , γ n ( z ) is in the class Σ F 2 ( α , δ , b ) .

Theorem 3.5 For i ∈ {1,..., n}, let γ i > 0 and f i ΣKUH ( α , b ) (α ≥ 0, and b\ { 0 } )). If

i = 1 n γ i 1,
(3.20)

then H γ 1 , , γ n ( z ) is in the class Σ F 3 ( α , δ , b ) .