Application of Srivastava-Attiya Operator to the Generalization of Mocanu Functions

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URL:https://doi.org/10.28924/2291-8639 DOI:10.28924/2291-8639-17-2019-674

APPLICATION OF SRIVASTAVA-ATTIYA OPERATOR TO THE GENERALIZATION OF MOCANU FUNCTIONS

KHALIDA INAYAT NOOR AND SHUJAAT ALI SHAH∗

COMSATS University Islamabad, Pakistan

∗_{Corresponding author:} _{[email protected]}

Abstract. In this paper we introduce certain subclasses of analytic functions by applying Srivastava-Attiya operator. Our main purpose is to derive inclusion results by using concept of conic domain and subordination techniques. We also deduce some new as well as well-known results from our investigations.

1. _Introduction

Letχdenotes the class of analytic functionsf(z) in the open unit disk

f={z:|z|<1}such that

f(z) =z+ ∞ X n=2

anzn. (1.1)

Subordination of two functionsf and g is denoted by f ≺g and defined as f(z) =g(w(z)), wherew(z) is

schwarz function inf. LetS,S∗ andCdenotes the subclasses ofχof univalent functions, starlike functions

and convex functions respectively. For 0 ≤ δ < 1, S∗(δ) and C(δ) are the subclasses of S of functions f

satisfies;

zf0₍_z₎

f(z) ≺

1 + (1−2δ)z

1−z , (z∈f), (1.2)

Received 2019-04-24; accepted 2019-05-27; published 2019-07-01. 2010Mathematics Subject Classification. 30C45, 30C55.

Key words and phrases. Srivastava-Attiya operator; Mocanu functions; conic domains.

c

2019 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

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(zf0(z))0

f0₍_z₎ ≺

1 + (1−2δ)z

1−z , (z∈f), (1.3)

respectively. Mocanu [13] introduced the classMα ofα−convex functionsf ∈S satisfies;

(1−α)zf 0₍_z₎

f(z) +α

(zf0(z))0

f0₍_z₎

≺ 1 +z

1−z, (1.4)

whereα∈[0,1], f(_zz)f0(z)6= 0. andz∈f. We see thatM0 =S∗ andM1=C. This class is vastly studied

by several authors. See [4,15,17–19]. For k ∈[0,∞), Kanas and Wisniowska [8,9] introduced the classes

k−U CV of k-uniformly convex functions and k−ST of k-starlike functions. The analytic conditions for

these classes are given [6–9] as;

k−U CV =

f ∈S:Re

1 +zf

00₍_z₎

f0₍_z₎

> k

zf00(z)

f0₍_z₎

, (z∈f). (1.5)

k−ST =

f ∈S:Re

_zf0₍_z₎

f(z)

> k

zf0(z)

f(z) −1

, (z∈_f). (1.6)

We can rewrite the above relations easily as;

Re(p(z))> k|p(z)−1|, (1.7)

wherep(z) = 1 +zf_f000₍(_zz₎) orp(z) =

zf0(z)

f(z) . It is clear that p(f) is conic domain defined as;

Ωk={w∈C:Re(w)> k|w−1|}, (1.8)

or

Ωk =

u+iv:u > k

q

(u−1)2+v2

, (0≤k <∞). (1.9)

These conic domains are being studied by several authors. See [2,6,14,16]. Sokol and Nonukawa [23]

introduced the class defined as;

M N =

f ∈S:Re

1 +zf

00₍_z₎

f0₍_z₎

>

zf0(z)

f(z) −1

, (z∈f). (1.10)

It is obvius thatM N ⊂C. Recently S. Sivasubramanian et al. [22] extend the Sokol and Nonukawa’s work

in terms of conic domains. They introduced a new classk−M N of functionsf ∈S such that

Re

1 + zf

00₍_z₎

f0₍_z₎

> k

zf0(z)

f(z) −1

, (z∈f). (1.11)

In motivation of the work [23], A. Rasheed et al. [21], introduced an interesting class k−U Mα (0≤α≤1)

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Re

(1−α)zf 0₍_z₎

f(z) +α

(zf0(z))0

f0₍_z₎

> k

zf0(z)

f(z) −1

, (z∈f). (1.12)

Obviously, we can seek−U M1=k−M N and 1−U M1=M N.

We recall a Hurwitz-Lerch Zeta function Φ(s, b;z) [25] defined by

Φ(s, a;z) = ∞ X n=2

zn

(n+b)s, (1.13)

(b C\Z−0 ;s Cwhen|z|<1;Re(s)>1 when|z|= 1),

where_Cand_Z−₀ denotes the set of complex numbers and the set of negative integers respectively.

Srivastava and Attiya [24] introduced the linear operatorJs,b:χ→χdefined in terms of the convolution

( or Hadamard product ), by

Js,bf(z) =Gs,b(z)∗f(z), (1.14)

where

Gs,b(z) = (1 +b)s[Φ(s, b;z)−bs], (1.15)

withz f, b C\Z−0 ands C. Therefore, using (1.13) to (1.15), we have

Js,bf(z) =z+ ∞ X n=2

_{1 +}_b

n+b

s

anzn, (1.16)

wherez _f,b _C\_Z−₀ ands _C.

The srivastava-Attiya operator generalizes the integral operators introduced by Alexandar [1], Libera [10],

Bernardi [3] and Jung et al. [5].

In 2007, Raducanu and Srivastava [20] introduced and studied the classS_s,b∗ (δ) of functionsf χsatisfies

Js,bf(z) S∗(δ).

Now by using concepts of conic domains and Srivastava-Attiya integral operator, we introduce new classes

as following.

Definition 1.1. Let k∈[0,∞)andα, β∈[0,1]. Thenf ∈k−U M(α, β)if and only if

Re

(1−α)zf 0₍_z₎

f(z) +α

(zf0(z))0

f0₍_z₎

> k

(1−β)zf 0₍_z₎

f(z) +β

(zf0(z))0

f0₍_z₎ −1

, (z∈f).

Some of the special cases are given below and we refer to [8,9,21–23].

Special cases:

(i) Forβ= 0, the classk−U M(α, β) reduces to the classk−U Mα. See [21].

(ii) Forα= 1 and β= 0, the classk−U M(α, β) reduces to the classk−M N. See [22].

(iii) Forα= 1,β= 0 and k= 1, the classk−U M(α, β) reduces to the classM N. See [23].

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(v) Forα= 0 andβ = 0 , the classk−U M(α, β) reduces to the classk−ST. See [8].

Definition 1.2. Let α, β [0,1],k [0,∞),b C\Z−0 ands C. Then

f k−U Ms

b(α, β)if and only if Js,bf(z) k−U M(α, β).

Clearly, fors= 0 the classesk−U Ms

b(α, β) andk−U M(α, β) coincides.

2. _{Preliminaries}

Lemma 2.1. [12] Let ~be an analytic function on fexcept for at most one pole on ∂fand univalent on f,℘ be an analytic function in fwith ℘(0) =~(0) and℘(z)6=℘(0), z ∈f. If ℘is not subordinate to ~,

then there exist pointsz0∈f,ξ0∈∂fandε≥1 for which

℘(|z|<|z0|)⊂~(f), ℘(z0) =~(ξ0), z0℘0(z0) =εξ0℘0(ξ0).

Lemma 2.2. [6] Iff ∈S∗(α)for someα∈1 2,0

, then

Re

_f₍_z₎

z

> 1

3−2α.

Lemma 2.3. [6] IfRepf0₍_z₎_{> α}_{for some}_α_∈1 2,0

, then

Re

_f₍_z₎

z

>2α 2_{+ 1}

3 .

3. _{Main Results}

Theorem 3.1. Let k∈[0,∞)andα, β∈[0,1]. Also, letpbe a function analytic in the unit disk such that

p(0) = 1. If

Re

p(z) +αzp

0₍_z₎

p(z)

−k

p(z)−1 +βzp

0₍_z₎

p(z)

>0,

then

p(z)≺ 1 + (1−2γ)z

1−z :=h(z),

whereγ=γ(k, α, β) is given by

γ(k, α, β) =1 4

"s

(α−2k+βk)2

(1 +k)2 +

8 (α+βk) (1 +k) −

(α−2k+βk) (1 +k)

#

. (3.1)

Proof. We may assume thatγ≥1

2 since the conditionRe

p(z) +zp_p₍0_z(z₎)>0

implies at least Re(p(z))> 1

2. (See [11]). Suppose now, on the contrary that p⊀h. Then, by Lemma 2.1,

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p(z0) =γ+ix, z0p0(z0) =my, where y≤ −

(1−γ)2₊_x2

2 (1−γ) , (x, y∈R).

Using these relations, we have

Re

p(z0) +α

z0p0(z0) p(z0)

−k

p(z0)−1 +β

z0p0(z0) p(z0)

>0,

or

0< Re

p(z0) +α

z0p0(z0) p(z0)

−k

p(z0)−1 +β

z0p0(z0) p(z0)

=Re

γ+ix+α my

γ+ix

−k

γ+ix−1 +β my

γ+ix

=γ+ αmyγ

γ2₊_x2 −k

(γ+ix)2−(γ+ix) +βmy γ+ix

≤γ− αγ 2(1−γ)

(1−γ)2+x2 γ2₊_x2

−k

q

(X+Y x2₎2₊_{T x}2

γ2₊_x2 =R(x),

where X = (2γ+β₂)(1−γ),Y = 2(1₂₍₁−γ_−γ)+₎β andT = (2γ−1)2. The functionR(x) is even in regard ofx. Now

we have to show that R(x) has maximum value at x= 0 when α, β∈[0,1] and γ∈1 2,1

. We can easily

check

R0(x) =−x

 

αγ(2γ−1) (1−γ) (γ2₊_x2₎−k

  

2Y(X+Y x2) +T −2 q

(X+Y x2₎2₊_{T x}2

γ2₊_x2

    .

ThenR0(x) = 0, if and only if,x= 0. Since α, β∈[0,1] andγ∈1 2,1

. So one can see

R00(x) =−

_α₍₂_γ₋₁₎

γ(1−γ) −

k

2 n

(2(1−γ) +β) (2γ+β) + 2 (2γ−1)2o

<0.

ThusR(x) has maximum value atx= 0, that is

R(x)≤R(0) =γ−αγ(1−γ) 2γ −

k(1−γ) (2γ+β) 2γ = 0,

forγ=γ(k, α, β) is given by (3.1), which contradicts the assumption. Hence

p(z)≺ 1 + (1−2γ)z

1−z :=h(z),

whereγ=γ(k, α, β) is given by (3.1).

Theorem 3.2. Let α, β [0,1],k [0,∞),b _C\_Z−₀ and s _C. Then

k−U M_bs(α, β)⊂S_s,b∗ (γ),

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Proof. Letf ∈k−U Ms

b(α, β). Then, by Definition 1.2,Js,bf(z) k−U M(α, β), that is

Re

"

(1−α)z(Js,bf(z)) 0

Js,bf(z)

+α z(Js,bf(z))

00

(Js,bf(z))0 #

> k

(1−β)z(Js,bf(z)) 0

Js,bf(z)

+β z(Js,bf(z))

00

(Js,bf(z))0 −1

, (z∈_f).

Puttingp(z) =z(Js,bf(z))0

Js,bf(z) , we have

Re

p(z) +αzp

0₍_z₎

p(z)

−k

p(z)−1 +βzp

0₍_z₎

p(z)

>0.

Our required result follows easily by using Theorem 3.1.

Whens= 0, then we have the following new result for classk−U M(α, β)

Theorem 3.3. Let k∈[0,∞)andα, β∈[0,1]. Then

k−U M(α, β)⊂S∗(γ),

The proof is straight forward by puttingp(z) =zf_f₍0(_zz₎) and using Theorem 3.1.

Whenk= 0, then we have the following result for a class 0−U M(α, β) =Mα,introduced by Mocanu [13].

Corollary 3.1. Let f ∈Mα. Thenf ∈S∗(γ), where

γ(α) =−α+ √

α2_{+ 8}_α

4 . (3.2)

Whenβ= 0, then we have the following result, proved in [21].

Corollary 3.2. Let f ∈k−U M(α,0) =k−U Mα. Thenf ∈S∗(γ), where

γ(α, k) = (2ϑ−η) + q

(2ϑ−η)2+ 8η

4 , (3.3)

whereϑ= _k₊₁k , η=α_k+₊₁k.

Whenα= 1, β= 0, then we have the following result, proved in [22].

Corollary 3.3. Let f ∈k−U M(1,0) =k−M N. Thenf ∈S∗(γ), where

γ(k) =1 4

 

s

₁₋₂_k 1 +k

2 + 8

(1 +k)−

₁₋₂_k 1 +k



. (3.4)

Whenα= 1, β= 0, andk= 1, then we have the following result, proved in [22].

Corollary 3.4. Let f ∈1−U M(1,0) =M N. Then f ∈S∗(γ), whereγ'0.6403.

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Corollary 3.5. Let f ∈k−U M(1,1) =k−U CV. Thenf ∈S∗(γ), where

γ(k) = 1 4

 

s ₁₋_k

1 +k

2 + 8−

₁₋_k 1 +k



. (3.5)

Theorem 3.4. Let f ∈k−U Ms

b(α, β). Then

Js,bf(z)

z ≺

1 + (1−2η)z

1−z ,

whereη= ₃₋1₂_γ andγ=γ(k, α, β) is given by (3.1).

Proof. Letf ∈k−U Ms

b(α, β). Then by Theorem 3.2 we have

z(Js,bf(z))0

Js,bf(z)

≺1 + (1−2γ)z 1−z ,

whereγ=γ(k, α, β) is given by (3.1). Using Lemma 2.2, we get

Js,bf(z)

z ≺

1 + (1−2η)z

1−z ,

whereη= ₃₋1₂_γ.

Whens= 0, then one can prove the following result by using Theorem 3.3 together with Lemma 2.2.

Theorem 3.5. Let f ∈k−U M(α, β). Then

f(z)

z ≺

1 + (1−2η)z

1−z ,

whereη= ₃₋1₂_γ andγ=γ(k, α, β) is given by (3.1).

Corollary 3.6. Let f ∈k−U M(1,0) =k−M N. Then

f(z)

z ≺

1 + (1−2η)z

1−z ,

whereη= ₃₋1₂_γ andγ=γ(k)is given by (3.4).

Whenα= 1, β= 0 andk= 1, then we have the following result, proved in [22].

Corollary 3.7. Let f ∈1−U M(1,0) =M N. Then

f(z)

z ≺

1 + (1−2η)z

1−z , where η'0.58159.

Whenα=β= 1, then we have the following result, proved in [6].

Corollary 3.8. Let f ∈k−U M(1,1) =k−U CV. Then

f(z)

z ≺

1 + (1−2η)z

1−z ,

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Whenα=β=k= 1, then we have the following result, proved in [6].

Corollary 3.9. Let f ∈1−U M(1,1) = 1−U CV. Then

Re

_f₍_z₎

z

>0.6289.

Theorem 3.6. Let α, β [0,1],k [0,∞),b _C\Z−0 and s C. If

Re

q

(Js,bf(z))0+α

z(Js,bf(z))00

2 (Js,bf(z))0

> k

(Js,bf(z))0+β

z(Js,bf(z))00

2 (Js,bf(z))0 −1

,

then

q

(Js,bf(z)) 0

≺ 1 + (1−2γ)z 1−z ⇒

Js,bf(z)

z ≺

1 + (1−2η)z

1−z

whereη= 2γ2₃+1 andγ=γ(k, α, β)is given by (3.1).

Proof. If we putp(z) = q

(Js,bf(z)) 0

, then

zp0(z)

p(z) =

z(Js,bf(z))00

2 (Js,bf(z))0

.

The proof follows easily by using Theorem 3.1 along with Lemma 2.3.

We can deduce the following result from Theorem 3.6 by choosings= 0.

Theorem 3.7. Let k∈[0,∞)andα, β∈[0,1]. If

Re

p

f0₍_z_{) +}_αzf 00₍_z₎

2f0₍_z₎

> k

p

f0₍_z_{) +}_βzf 00₍_z₎

2f0₍_z₎ −1

,

then

p

f0₍_z₎_≺ 1 + (1−2γ)z 1−z ⇒

f(z)

z ≺

1 + (1−2η)z

1−z

whereη= 2γ2₃+1 andγ=γ(k, α, β)is given by (3.1).

Corollary 3.10. If

Re

p

f0₍_z_{) +}zf 00₍_z₎

2f0₍_z₎

> k

p

f0₍_z₎₋₁ ,

then

Repf0₍_z₎_{> γ}_⇒_Re _f₍_z₎

z

> η,

whereη= 2γ2₃+1 andγ=γ(k)is given by (3.4).

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Corollary 3.11. If

Re

p

f0₍_z_{) +}zf 00₍_z₎

2f0₍_z₎ > p

f0₍_z₎₋₁ , then

Repf0₍_z₎_{> γ}_'₀_.₆₄_⇒_Re

f(z)

z

> η'0.60.

Fork= 0, we have the following result, refer to [22].

Corollary 3.12. If

Re

p

f0₍_z_{) +}zf 00₍_z₎

2f0₍_z₎

>0,

then

Repf0₍_z₎_{> γ}_'₀_.₆₄_⇒_Re

f(z)

z

> η'0.60.

Theorem 3.8. Let α, β [0,1],k [0,∞),b C\Z−0 and s C. If

Re

_J s,bf(z)

z +α

_z₍_J

s,bf(z))0

Js,bf(z) −1 > k

Js,bf(z)

z +β

_z₍_J

s,bf(z))0

Js,bf(z) −1 −1 , then

Js,bf(z)

z ≺

1 + (1−2γ)z

1−z ,

The proof follows easily by substitutingp(z) =Js,bf(z)

z in Theorem 3.1.

Fors= 0, we can easily deduce the following result.

Re

_f₍_z₎

z +α

_zf0₍_z₎

f(z) −1 > k

f(z)

z +β

_zf0₍_z₎

f(z) −1 −1 , then

f(z)

z ≺

1 + (1−2γ)z

1−z ,

Corollary 3.13. If

Re

_zf0₍_z₎

f(z) +

f(z)

z −1

> k

f(z)

z −1

⇒f(z)

z ≺

1 + (1−2γ)z

1−z

whereγ=γ(k)is given by (3.4).

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Corollary 3.14. If

Re

_zf0₍_z₎

f(z) +

f(z)

z −1

>

f(z)

z −1

⇒Re

_f₍_z₎

z

> γ'0.64.

Whenα= 1, β= 0 andk= 0, then we have following result.

Corollary 3.15. If

Re

_zf0₍_z₎

f(z) +

f(z)

z −1

>0⇒Re

_f₍_z₎

z

> 1

2.

If we substitute p(z) = (Js,bf(z))0 in Theorem 3.1, then we have the following result.

Theorem 3.10. Let α, β [0,1],k [0,∞),b _C\Z−0 ands C. If

Re

(Js,bf(z)) 0

+αz(Js,bf(z))

00

(Js,bf(z))0

> k

(Js,bf(z)) 0

+βz(Js,bf(z))

00

(Js,bf(z))0 −1

,

then

Re (Js,bf(z)) 0

> γ

Fors= 0, we have the following result.

Re

f0(z) +αzf

00₍_z₎

f0₍_z₎

> k

f0(z) +βzf

00₍_z₎

f0₍_z₎ −1

,

then

Re(f0(z))> γ

Corollary 3.16. If

Re

f0(z) +zf 00₍_z₎

f0₍_z₎

> k|f0(z)−1| ⇒Re(f0(z))> γ'0.64.

Whenα= 1, β=k= 0, then we have the following result.

Corollary 3.17. If Rehf0₍_z_{) +}zf00(z)

f0₍_z₎

i

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