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U N I V E R S I T A T I S M A R I A E C U R I E - S K Ł O D O W S K A L U B L I N – P O L O N I A

VOL. LXVI, NO. 1, 2012 SECTIO A 63–73

ALI MUHAMMAD

On inclusion relationships

of certain subclasses of meromorphic functions involving integral operator

Abstract. In this paper, we introduce some subclasses of meromorphic func- tions in the punctured unit disc. Several inclusion relationships and some other interesting properties of these classes are discussed.

1. Introduction. Let M denote the class of functions f (z) of the form

(1.1) f (z) = 1

z+

X

k=0

ak zk, which are analytic in the punctured open unit disc

E = {z : z ∈ C and 0 < |z| < 1} = E\{0}.

If f (z) is given by (1.1) and g(z) is given by

(1.2) g(z) = 1

z+

X

k=0

bkzk,

2000 Mathematics Subject Classification. 30C45, 30C50.

Key words and phrases. Meromorphic functions, functions with bounded boundary and bounded radius rotation, quasi-convex functions, close-to-convex functions, general- ized hypergeometric functions, functions with positive real part, Hadamard product (or convolution), linear operators.

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we define the Hadamard product (or convolution) of f (z) and g(z) by (1.3) (f ? g) (z) = 1

z+

X

k=0

akbkzk= (g ? f ) (z) (z ∈ E) .

Let Pk(ρ) be the class of functions p(z) analytic in E with p(0) = 1 and

(1.4)

Z

0

Re p(z) − ρ 1 − ρ

dθ ≤ kπ, z = re,

where k > 2 and 0 ≤ ρ < 1. This class was introduced by Padmanbhan et al. in [16]. We note that Pk(0) = Pk, see [17], P2(ρ) = P (ρ), the class of analytic functions with positive real part greater than ρ and P2(0) = P, the class of functions with positive real part. From (1.4) we can easily deduce that p(z) ∈ Pk(ρ) if and only if, there exists p1(z), p2(z) ∈ P (ρ) such that for z ∈ E,

(1.5) p(z) = k

4 +1 2



p1(z) − k 4 − 1

2

 p2(z).

In recent years, several families of integral operators and differential opera- tors were introduced using Hadamard product (or convolution). For exam- ple, we choose to mention the Rushcheweyh derivative [18], the Carlson–

Shaffer operator [1], the Dziok–Srivastava operator [4], the Noor integral operator [14], also see [3, 5, 6, 11]. Motivated by the work of N. E. Cho and K. I. Noor [2, 9], we introduce a family of integral operators defined on the space of meromorphic functions in the class M. By using these integral operators, we define several subclasses of meromorphic functions and in- vestigate various inclusion relationships and some other properties for the meromorphic function classes introduced here.

For complex parameters α1, . . . , αqand β1, . . . , βsj∈ C\Z0, j = 1, . . . , s;

Z0 = {0, −1, −2, . . . }) we now define the function φ(α1, . . . , αq; β1, . . . , βs; z) by

φ(α1, . . . , αq; β1, . . . , βs; z) = 1 z+

X

k=0

1)k+1. . . (αq)k+11)k+1. . . (βs)k+1(k + 1)!zk, (q ≤ s + 1; s ∈ N0 = N ∪ {0}; N = {1, 2, . . . }; z ∈ E),

where (v)k is the Pochhammer symbol (or shifted factorial) defined in (terms of the Gamma function) by

(v)k = Γ(v + k) Γ(v) =

(1 if k = 0 and v ∈ C\{0}

v(v + 1) . . . (v + k − 1) if k ∈ N and v ∈ C.

Now we introduce the following operator

Iµp1, . . . , αq, β1, . . . , βs) : M −→ M

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as follows:

Let Fµ,p(z) = z1 +P k=0

k+µ+1 µ

p

zk, p ∈ N0, µ 6= 0 and let Fµ,p−1(z) be defined such that

Fµ,p(z) ∗ Fµ,p−1(z) = φ(α1, . . . , αq; β1, . . . , βs; z).

Then

(1.6) Iµp1, . . . , αq, β1, . . . , βs)f (z) = Fµ,p−1(z) ∗ f (z).

From (1.6) it can be easily seen

(1.7)

Iµp1, . . . αq, β1, . . . βs)f (z)

= 1 z +

X

k=0

 µ

k + µ + 1

p

1)k+1. . . (αq)k+1

1)k+1. . . (βs)k+1(k + 1)! akzk. For conveniences, we shall henceforth denote

(1.8) Iµp1, . . . αq, β1, . . . βs)f (z) = Iµp1, β1)f (z).

For the choices of the parameters p = 0, q = 2, s = 1, the operator Iµp1, β1)f (z) is reduced to an operator by N. E. Cho and K. I. Noor [2] and K. I. Noor [9] and when p = 0, q = 2, s = 1, α1 = λ, α2 = 1, β1 = (n + 1), the operator Iµp1, β1)f (z) is reduced to an operator recently introduced by S.-M. Yuan et al. in [20].

It can be easily verified from the above definition of the operator Iµp1, β1) that

(1.9) z(Iµp+11, β1)f (z))0= µIµp1, β1)f (z) − (µ + 1)Iµp+11, β1)f (z) and

(1.10) z(Iµp1, β1)f (z))0= α1Iµp1+ 1, β1)f (z) − (α1+ 1)Iµp1, β1)f (z).

By using the operator Iµp1, β1), we now introduce the following subclasses of meromorphic functions:

Definition 1.1 ([9]). A function f ∈ M is said to belong to the class M Rk(γ) for z ∈ E, 0 ≤ γ < 1, k ≥ 2, if and only if

−zf0(z)

f (z) ∈ Pk(γ),

and f ∈ M Vk(γ), for z ∈ E, 0 ≤ γ < 1, k ≥ 2, if and only if

−(zf (z))0

f0(z) ∈ Pk(γ).

We call f ∈ M Rk(γ) a meromorphic function with bounded radius rotation of order γ and f ∈ M Vk(γ) a meromorphic function with bounded boundary rotation.

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Definition 1.2. Let f ∈ M, 0 ≤ γ < 1, k ≥ 2, z ∈ E. Then f ∈ M Rk,µp1, β1, γ) if and only if Iµp1, β1)f ∈ M Rk(γ).

Also

f ∈ M Vk,µp1, β1, γ) if and only if Iµp1, β1)f ∈ M Vk(γ), z ∈ E. We observe that, for z ∈ E,

f ∈ M Vk,µp1, β1, γ) ⇔ −zf0 ∈ M Rpk,µ1, β1, γ).

Definition 1.3. Let λ ≥ 0, f ∈ M, p ∈ N0, 0 ≤ γ, ρ < 1, µ > 0 and z ∈ E. Then f ∈ Bλ,pk,µ1, β1, γ, ρ), if and only if there exists a function g ∈ M V2,µp1, β1, γ), such that



(1 − λ)(Iµp1, β1)f (z))0 (Iµp1, β1)g(z))0 + λ



−(z(Iµp1, β1)f (z))0)0 (Iµp1, β1)g(z))0



∈ Pk(ρ).

In particular, for λ = 0 = p, k = q = µ = 2 and s = 1, we obtain the class of meromorphic close-to-convex function, see [7], see also K. I. Noor [9]. For λ = 1, p = 0, k = q = µ = 2, s = 1, we have the class of meromorphic quasi-convex functions defined for z ∈ E. We note that the class C of quasi-convex univalent functions, analytic in E, was first introduced and studied in [12], see also [13, 15].

In order to establish our main results, we need the following lemma, which is properly known as the Miller–Mocanu Lemma.

Lemma 1.1 ([8]). Let u = u1+ iu2, v = v1+ iv2 and Ψ (u, v) be a complex valued function satisfying the conditions:

(i) Ψ (u, v) is continuous in a domain D ⊂ C2, (ii) (1, 0) ∈ D and Re Ψ (1, 0) > 0,

(iii) Re Ψ (iu2, v1) ≤ 0, whenever (iu2, v1) ∈ D and v1 ≤ −12 1 + u22 . If h (z) = 1 + c1z + c2z2 + · · · is a function analytic in E such that (h(z), zh0(z)) ∈ D and Re(Ψ (h(z), zh0(z)) > 0 for z ∈ E, then Re h(z) > 0 in E.

2. Main results.

Theorem 2.1. Let Re α1 > 0, µ > 0 and 0 ≤ γ < 1. Then

M Rpk,µ1+ 1, β1, γ) ⊂ M Rk,µp1, β1, ρ) ⊂ M Rp+1k,µ1, β1, η).

Proof. We prove the first part of the Theorem 2.1 and the second part follows by using similar techniques. Let

f ∈ M Rpk,µ1+ 1, β1, γ), z ∈ E and set

(2.1) −(Iµp1, β1)f (z))0 (Iµp1, β1)f (z)) = k

4 +1 2



h1(z) − k 4 − 1

2



h2(z) = H(z).

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Simple computation together with (2.1) and (1.10) yields (2.2) −(Iµp1+ 1, β1)f (z))0

(Iµp1+ 1, β1)f (z)) =



H(z) + zH0(z)

−H(z) + α1+ 1



∈ Pk(γ), z ∈ E.

Let

Φα1(z) = 1 α1+ 1

"

1 z+

X

k=0

zk

#

+ α1 α1+ 1

"

1 z +

X

k=0

kzk

# , then

(2.3)

H(z) ∗ zΦα1(z) =



H(z) + zH0(z)

−H(z) + α1+ 1



= k 4 +1

2

 

h1(z) + zh01(z)

−h1(z) + α1+ 1



− k 4 −1

2

 

h2(z) + zh02(z)

−h2(z) + α1+ 1

 . Since f ∈ M Rpk,µ1+ 1, β1, γ), it follows from (2.2) and (2.3) that



hi(z) + zh0i(z)

−hi(z) + α1+ 1



∈ P (γ), i = 1, 2, z ∈ E.

Let hi(z) = (1 − ρ)pi(z) + ρ. Then



(1 − ρ)pi(z) + ρ − γ + (1 − ρ)zp0i(z)

−(1 − ρ)pi(z) − ρ + α1+ 1



∈ P, z ∈ E.

We shall show that pi(z) ∈ P, i = 1, 2.

We form the functional Ψ(u, v) by taking u = u1 + iu2 = pi(z), v = v1 + iv2 = zp0i(z). The first two conditions of Lemma 1.1 can be easily verified. We need to verify condition (iii) as follows:

Ψ(u, v) =



(1 − ρ)u + ρ − γ + (1 − ρ)v

−(1 − ρ)u − ρ + α1+ 1

 , implies that

Re Ψ(iu2, v1) = ρ − γ + (1 − ρ)(α1+ 1 − ρ)v1

(1 − ρ)2u22+ (−ρ + α1+ 1)2. By taking v1 ≤ −12(1 + u22), we have

Re Ψ(iu2, v1) ≤ A + Bu22 2C , where

A = 2(ρ − γ)(α1+ 1 − ρ)2− (1 − ρ)(α1+ 1 − ρ), B = 2(ρ − γ)(1 − ρ)2− (1 − ρ)(α1+ 1 − ρ), C = (α1+ 1 − ρ)2+ (1 − ρ)2u22 > 0.

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We note that Re Ψ(iu2, v1) ≤ 0 if and only if A ≤ 0 and B ≤ 0. From A ≤ 0, we obtain

(2.4) ρ = 1

4 n

(3 + 2α1+ 2γ) −p

(3 + 2α1+ 2γ)2− 8o , and B ≤ 0 gives us 0 ≤ ρ < 1.

Now using Lemma 1.1, we see that pi(z) ∈ P for z ∈ E, i = 1, 2 and hence f ∈ M Rpk,µ1, β1, ρ) with ρ given by (2.4). 

In particular, we note that ρ = 1

4



(3 + 2α1) − q

(12α1+ 4α21) + 1

 . Theorem 2.2. Let Re α1, µ > 0. Then

M Vk,µp1+ 1, β1, γ) ⊂ M Vk,µp1, β1, ρ) ⊂ M Vk,µp+11, β1, η).

Proof. We observe that

f (z) ∈ M Vk,µp1+ 1, β1, γ) ⇔ −zf0(z) ∈ M Rpk,µ1+ 1, β1, γ)

⇒ −zf0(z) ∈ M Rpk,µ1, β1, ρ)

⇔ f (z) ∈ M Vk,µp1, β1, ρ), where ρ is given by (2.4).

The second part can be proved by means of similar arguments.  Theorem 2.3. Let Re α1, µ > 0.Then

Bλ,pk,µ1+ 1, β1, γ1, ρ1) ⊂ Bk,µλ,p1, β1, γ2, ρ2) ⊂ Bk,µλ,p+11, β1, γ3, ρ3), where γi= γii, µ), i = 1, 2, 3 are given in the proof.

Proof. We prove the first inclusion of this result and the other part follows along similar lines.

Let f ∈ Bk,µλ,p1 + 1, β1, γ1, ρ1). Then by Definition 1.3, there exists a function g ∈ M V2,µp1+ 1, β1, γ1) such that

(2.5)



(1 − λ)(Iµp1+ 1, β1)f (z))0 (Iµp1+ 1, β1)g(z))0 + λ



−(z(Iµp1+ 1, β1)f (z))0)0 (Iµp1+ 1, β1)g(z))0



∈ Pk1).

Set

(2.6) h(z) =



(1 − λ)(Iµp1, β1)f (z))0 (Iµp1, β1)g(z))0 + λ



−(z(Iµp1, β1)f (z))0)0 (Iµp1, β1)g(z))0



, where h(z) is an analytic function in E with h(0) = 1.

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Now, g ∈ M V2,µp1+ 1, β1, γ1) ⊂ M V2,µp1, β1, γ2), where γ2 is given by the equation

(2.7) 2γ22+ (3 + 2α1− 2γ12− {2γ1(1 + α1) + 1} = 0.

Therefore,

q(z) = −(zIµp1, β1)g(z))0

(Iµp1, β1)g(z))0 ∈ P (γ2), z ∈ E.

By using (1.10), (2.5), (2.6) and (2.7), we have (2.8)



h(z) + λzh0(z)

−q(z) + α1+ 1



∈ Pk1), q(z) ∈ P (γ2), z ∈ E.

With

h(z) = k 4 +1

2



[(1 − ρ2)h1(z) + ρ2] − k 4 −1

2



[(1 − ρ2)h2(z) + ρ2], (2.8) can be written as

 k 4 +1

2

 

(1 − ρ2)h1(z) + ρ2+(1 − ρ2)λzh01(z)

−q(z) + α1+ 1



− k 4 −1

2

 

(1 − ρ2)h2(z) + ρ2+(1 − ρ2)λzh02(z)

−q(z) + α1+ 1

 , where



(1 − ρ2)hi(z) + ρ2+(1 − ρ2)λzh0i(z)

−q(z) + α1+ 1



∈ P (ρ1), z ∈ E, i = 1, 2.

That is



(1 − ρ2)hi(z) + ρ2− ρ1+(1 − ρ2)λzh0i(z)

−q(z) + α1+ 1



∈ P, z ∈ E, i = 1, 2.

We form the functional Ψ(u, v) by choosing u = u1 + iu2 = hi(z), v = v1+ iv2= zh0i(z), and

Ψ(u, v) =



(1 − ρ2)u + ρ2− ρ1+ (1 − ρ2)λv

−q(z) + α1+ 1



, (q = q1+ iq2).

The first two conditions of Lemma 1.1 are clearly satisfied. We verify (iii), with v1 ≤ −12(1 + u22) as follows:

Re Ψ(iu2, v1) = ρ2− ρ1+ Re λ(1 − ρ2)v1{(−q1+ α1+ 1) + iq2} (−q1+ α1+ 1)2+ q22



≤ 2(ρ2− ρ1) |−q + α1+ 1|2− λ(1 − ρ2){(−q1+ α1+ 1)(1 + u22) 2 |−q + α1+ 1|2

= A + Bu22 2C ≤ 0,

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if A ≤ 0 and B ≤ 0, where

A = 2(ρ2− ρ1) |−q + α1+ 1|2− λ(1 − ρ2){(−q1+ α1+ 1), B = −λ(1 − ρ2){(−q1+ α1+ 1) ≤ 0,

C = |−q + α1+ 1|2 > 0.

From A ≤ 0, we obtain

ρ2 = 2ρ1|−q + α1+ 1|2+ λ Re(−q(z) + α1+ 1) 2 |−q + α1+ 1|2+ λ Re(−q(z) + α1+ 1) .

Hence, using Lemma 1.1, it follows that h(z), defined by (2.6), belongs to Pk2) and thus f ∈ Bk,µλ,p1, β1, γ2, ρ2) for z ∈ E. This completes the proof of the first part. The second part of this result can be obtained by using similar techniques and the relation (1.9).  Theorem 2.4. Let Re α1, µ > 0. Then

(i) Bk,µλ,p1, β1, γ, ρ) ⊂ Bk,µ0,p1, β1, γ, ρ4).

(ii) Bk,µλ1,p1, β1, γ, ρ) ⊂ Bk,µλ2,p1, β1, γ, ρ), for 0 ≤ λ2 < λ1. Proof. (i). Let

h(z) = (Iµp1, β1)f (z))0 (Iµp1, β1)g(z))0, h(z) is analytic in E and h(0) = 1. Then

(2.10)



(1 − λ)(Iµp1, β1)f (z))0 (Iµp1, β1)g(z))0 + λ



−(z(Iµp1, β1)f (z))0)0 (Iµp1, β1)g(z))0



= h(z) + λ zh0(z)

−h0(z), where

h0(z) = −(z(Iµp1, β1)f (z))0)0

(Iµp1, β1)g(z))0 ∈ P (γ).

Since f ∈ Bk,µλ,p1, β1, γ, ρ), it follows that



h(z) + λ zh0(z)

−h0(z)



∈ Pk(ρ), h0 ∈ P (γ), for z ∈ E.

Let

h(z) = k 4 +1

2



h1(z) − k 4 −1

2

 h2(z).

Thus (2.10) implies that



hi(z) + λzh0i(z)

−h0(z)



∈ P (ρ), z ∈ E, i = 1, 2.

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and using similar techniques, together with Lemma 1.1, it follows that hi(z) ∈ P (ρ4), i = 1, 2, where

ρ4 = 2ρ |h0(z)|2+ λ Re h0(z) 2 |h0(z)|2+ λ Re h0(z) .

Therefore h(z) ∈ Pk4), and f ∈ Bk,µ0,p1, β1, γ, ρ4), for z ∈ E. In partic- ular, it can be shown that hi(z) ∈ P (ρ), i = 1, 2. Consequently h ∈ Pk(ρ) and f ∈ Bk,µ0,p1, β1, γ, ρ) in E.

For λ2 = 0, we have part (i). Therefore, we let λ2 > 0 and f ∈ Bk,µλ1,p1, β1, γ, ρ). There exist two functions H1(z), H2(z) ∈ Pk(ρ) such that



(1 − λ1)(Iµp1+ 1, β1)f (z))0 (Iµp1+ 1, β1)g(z))0 + λ1



−(z(Iµp1+ 1, β1)f (z))0)0 (Iµp1+ 1, β1)g(z))0



= H1(z) (Iµp1+ 1, β1)f (z))0

(Iµp1+ 1, β1)g(z))0 = H2(z), where g(z) ∈ M V2,µp1, β1, γ).

Now

(2.11)



(1 − λ2)(Iµp1+ 1, β1)f (z))0 (Iµp1+ 1, β1)g(z))02



−(z(Iµp1+ 1, β1)f (z))0)0 (Iµp1+ 1, β1)g(z))0



= λ2 λ1

H1(z) +

 1 −λ2

λ1

 H2(z).

Since the class Pk(ρ) is convex, see [10], it follows that the right hand side of (2.11) belongs to Pk(ρ) and this shows that f ∈ Bλk,µ2,p1, β1, γ, ρ) for

z ∈ E. This completes the proof. 

Inclusion properties involving the integral operator Fc. Consider the operator Fc, defined by

(2.12) Fc(f )(z) = c zc+1

z

Z

0

tcf (t)dt (f ∈ M; c > 0).

From the Definition of Fc defined by (2.12), we observe that

(2.13) z((Iµp1, β1)Fcf (z))0 = c(Iµp1, β1)f (z) − (c + 1)(Iµp1, β1)Fcf (z).

Using (2.12), (2.13) with similar arguments as used earlier, we can prove the following theorem.

Theorem 2.5. Let f ∈ M Rpk,µ1, β1, γ) or f ∈ M Vk,µp1, β1, γ) or f ∈ Bk,µλ,p1, β1, γ, ρ), for z ∈ E. Then Fc(f ) defined by (2.12) is also in the same class for z ∈ E.

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Acknowledgement. I am thankful for the valuable suggestions of referee which improved this paper.

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Ali Muhammad

Department of Basic Sciences

University of Engineering and Technology Peshawar

Pakistan

e-mail: ali7887@gmail.com Received March 25, 2011

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