# On a Class of Analytic Functions Defined by an Integral Operator

Authors: Pravati Sahoo; Saumya Singh
Journal: Journal of Mathematics
 (2013)

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## Abstract

We define a new subclassM(λ,μ,α,β)
by using an integral operator Qμλf(z). We find a coefficient inequality and using that we derive many sharp results. These results generalize many results which are existing in the literature.


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## Body

## 1. Introduction and Preliminaries

Let𝒜 denote the class of functions of the form
(1)f(z)=z+∑k=2∞akzk,
which are analytic in the unit disc Δ={z∈ℂ:|z|<1}. For f(z)∈𝒜 and for λ>0, μ>-1, the integral operator is defined by
(2)Qμλf(z)=(λ+μμ)λzμ∫0ztμ-1(1-tz)λ-1f(t)dt,=z+∑n=1∞(λ+1)n(λ+μ+1)nan+1zn+1(z∈Δ),
where (λ)n is the Pochhammer symbol given by
(3)(λ)n={1,n=0,λ(λ+1)(λ+2)⋯(λ+n-1),n∈ℕ,(λ+μμ)=Γ(λ+1)Γ(λ+μ+1)(μ+1)=(λλ-μ).Forλ>0, μ>0, Qμλf(z) is defined by Komatu in [1, 2]. Here,
(4)Q00f(z)=f(z),Q0nf(z)=f(z),n∈ℕ,z(Qμλf(z))′=(λ+μ)Qμλ-1f(z)m(Qμλf(z))m-(λ+μ-1)Qμλf(z),forλ>1,μ>-1,Re{Qμλf(z)}′=λ(λλ-μ)∫01(1-t)λ-1tμf′(zt)dt,(λ>0,μ>-1).The operatorQ1λf(z) is the Bernardi operator [3, 4]. In fact the operator Qμλf(z) is related rather closely to the Beta or Euler transformation. Moreover, for μ>-1, the operator Qμλf(z) was used by Owa and Srivastava [5–8].For-1≤α<1, β≥0, we define a class of all analytic functions involving the integral operator, Qμλ, by
(5)M(λ,μ;α,β)={f∈𝒜:Re{(λ+1)Qμλ+1f(z)Qμλf(z)-λ}>β|(λ+1)Qμλ+1f(z)Qμλf(z)-(λ+1)|+α,z∈Δ}.The aim of this paper is to study the classM(λ,μ;α,β) and find the similar type of results proved by Frasin in [9], where the author has defined similar type of class involving the operator
(6)ℒ(a,c)f(z)=z+∑n=1∞(a)n(c)nan+1zn+1,(z∈Δ).
Also in [10], the authors studied the similar type class involving the well-known Salagean operator.


## 2. Definitions and Lemmas

Definition 1.
Letg(z) be analytic and univalent in Δ. If f(z) is analytic in Δ, f(0)=g(0), and f(Δ)⊂g(Δ), then we say that the function f(z) is subordinate to g(z) in Δ, and we write f(z)≺g(z).Definition 2 (subordinating factor sequence).
A sequence{bn}n=1∞ of complex numbers is called a subordinating sequence if, whenever f(z) is analytic, univalent, and convex in Δ, we have the subordination given by
(7)∑n=2∞bnanzn≺f(z),(z∈Δ,a1=1).Lemma 3 (see [11]).
A sequence{bn}n=1∞ is a subordinating factor sequence if and only if
(8)Re{1+2∑n=1∞bnzn}>0,(z∈Δ).Lemma 4.
If(9)∑n=1∞γn(λ,μ;α,β)|an+1|≤1-α,
where -1≤α<1;β>0, n≥1 and
(10)γn(λ,μ;α,β)=(λ+1)n(λ+μ+1)n+1×{(1+β)μn+(1-α)(λ+μ+n+1)},
then f(z)∈M(λ,μ;α,β).Proof.
It is sufficient to show that(11)β|(λ+1)Qμλ+1f(z)Qμλf(z)-(λ+1)|m-Re{(λ+1)Qμλ+1f(z)Qμλf(z)-λ}≤1-α.
Now, we have(12)β|(λ+1)Qμλ+1f(z)Qμλf(z)-(λ+1)|-Re{(λ+1)Qμλ+1f(z)Qμλf(z)-λ}≤(1+β)m×∑n=1∞‍((λ+1)n/(λ+μ+1)n)((μn)/(λ+μ+n+1))|an+1|1-∑n=1∞‍((λ+1)n/(λ+μ+1)n)|an+1|.
The above expansion is bounded by1-α if
(13)∑n=1∞(λ+1)n(λ+μ+1)n+1m×{(1+β)μn+(1-α)(λ+μ+n+1)}≤1-α,
hence the proof follows from (10).LetM*(λ,μ;α,β) denote the class of functions f(z)∈𝒜 whose coefficients satisfy the condition (10). So M(λ,μ;α,β)⊂M*(λ,μ;α,β).


## 3. Main Results

By using the technique used earlier by Attiya [12] and Singh [13], we state and prove the following theorem.Theorem 5.
Let the functionf(z) defined by (1) be in the class M*(λ,μ;α,β), where -1≤α<1;β≥0;λ>0;μ>-1. Also let 𝒦 denote the class of functions f(z)∈𝒜 which are convex and univalent in Δ. Then,
(14)γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)](f*g)(z)≺g(z),(15)Re(f(z))>-1-α+γ2(λ,μ;α,β)γ2(λ,μ;α,β).
The constant (1-α+γ2(λ,μ;α,β))/γ2(λ,μ;α,β) is the best estimate.Proof.
f
(
z
)
∈
M
*
(
λ
,
μ
;
α
,
β
) and let g(z)=z+∑n=1∞cn+1zn+1∈𝒦. Then,
(16)γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)](f*g)(z)m=γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)](z+∑n=1∞‍an+1cn+1zn+1).
Thus by the Definition2 and (14) will hold if the sequence
(17){γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)]ak}k=1∞
is a subordinating factor sequence, with a1=1, in view of Lemma 3, this will be the case if and only if
(18)Re{1+2∑k=n+1∞‍γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)]akzk}>0,nnnnnnnnnnnnnnnnn(n∈ℕ0=ℕ∪{0},z∈Δ).
Now(19)Re{1+γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)]zm+1[1-α+γ2(λ,μ;α,β)]∑n=1∞‍γn(λ,μ;α,β)an+1zn+1}.
Case I (-1<μ<0). From (19), we obtain
(20)Re{1+γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)]z+1[1-α+γ2(λ,μ;α,β)]×∑n=1∞γn(λ,μ;α,β)an+1zn+1}≥1-{γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)]z-1[1-α+γ2(λ,μ;α,β)]×∑n=1∞‍γn(λ,μ;α,β)an+1zn+1}.
Sinceγn(λ,μ;α,β) is an increasing function of n(n≥1), so
(21)1-{γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)]r-1[1-α+γ2(λ,μ;α,β)]×∑n=1∞γn(λ,μ;α,β)an+1rn+1}>1-{γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)]r-1-α[1-α+γ2(λ,μ;α,β)]r}>0,mmmmmmmmmmmmmfor|z|=r<1.
Case II(μ>0). From (19), we obtain
(22)Re{1+γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)]z+1[1-α+γ2(λ,μ;α,β)]m×∑n=1∞‍γn(λ,μ;α,β)an+1zn+1}≥1-{γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)]z+1[1-α+γ2(λ,μ;α,β)]×∑n=1∞‍γn(λ,μ;α,β)an+1zn+1}.
Sinceγn(λ,μ;α,β) is decreasing function of n(n≥1), so
(23)1-{γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)]r+1[1-α+γ2(λ,μ;α,β)]m×∑n=1∞γn(λ,μ;α,β)an+1rn+1}>1-{γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)]rmmmm+(1-α)r[1-α+γ2(λ,μ;α,β)]}>0,mmmmmmmmmmmmmfor|z|=r<1.
Thus, (18) holds true in Δ. This prove the inequality (14). The inequality (15) follows by taking the convex function g(z)=z/(1-z)=z+∑n=1∞‍zn+1 in (14). To prove the sharpness of the constant, we suppose that the function f0(z)∈M*(λ,μ;α,β) given by
(24)f0(z)=z-1-αγ2(λ,μ;α,β)z2,(-1≤α<1;β≥0),
from (14), we have
(25)γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)]f0(z)≺z1-z.
After a simple calculation, we get
(26)min{Re(γ2(λ,μ;α,β)2[1-α+γ2(λ,μ;α,β)]f0(z))}=-12.Corollary 6.
Let the functionf(z) defined by (1) be in the class M*(1,1;α,β) and satisfy the condition
(27)∑n=1∞2(n+3)(n+2)m×{(1+β)n+(1-α)(n+3)}≤1-α,
then
(28)2β-5α+72(17-15α+2β)(f*g)(z)≺g(z),Re(f(z))>-17-15α+2β2β-5α+7.
The constant(2β-5α+7)/2(17-15α+2β) is the best estimate.Corollary 7.
Let the functionf(z) defined by (1) be in the class M*(1,1;0,0) and satisfy the condition
(29)∑n=1∞2(2n+3)(n+3)(n+2)≤1,
then
(30)734(f*g)(z)≺g(z),Re(f(z))>-177.
The constant7/34 is the best estimate.Corollary 8.
Let the functionf(z) defined by (1) be in the class M*(2,1;α,β) and satisfy the condition
(31)∑n=1∞‍3(n+3)(n+4){(1+β)n+(1-α)(n+4)}≤1-α,
then
(32)4+β-3α18-16α+2β(f*g)(z)≺g(z),Re(f(z))>-9-8α+β4+β-3α.
The constant(4+β-3α)/(18-16α+2β) is the best estimate.



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