# Certain Properties of Multivalent Functions Associated with the Dziok-Srivastava Operator

Authors: T. M. Seoudy
Journal: Journal of Mathematics
 (2013)

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## Abstract

By making use of the techniques of the differential subordination, we derive certain properties ofp-valent functions associated with the Dziok-Srivastava operator.


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## Body

## 1. Introduction

LetA(p,k) denote the class of functions of the form
(1)f(z)=zp+∑n=k∞‍an+pzn+p(p,k∈ℕ={1,2,3,…}),
which are analytic in the open unit disk  U={z∈ℂ:|z|<1}. We write A(p,1)=A(p).Suppose thatf and g are analytic in U. We say that the function f is subordinate to g in U, or g superordinate tof in U, and we write f≺g or f(z)≺g(z)(z∈U), if there exists an analytic function ω in U with ω(0)=0 and |ω(z)|<1, such that f(z)=g(ω(z))(z∈U). If g is univalent in U, then the following equivalence relationship holds true (see [1–3]):
(2)f(z)≺g(z)⟺f(0)=g(0),f(U)⊂g(U).For functionsfj∈A(p,k) given by
(3)fj(z)=zp+∑n=k∞‍an+p,jzn+p(j=1,2;p∈ℕ),
we define the Hadamard product (or convolution) of f1 and f2 by
(4)(f1*f2)(z)=zp+∑n=k∞‍an+p,1an+p,2zn+p=(f2*f1)(z).For complex parametersa1,…,aq and b1,…,bs(bj∉ℤ0-={0,-1,-2,…};j=1,…,s), the generalized hypergeometric function Fsq is defined (see [4]) by the following infinite series:
(5)qFs(a1,…,ai,…,aq;b1,…,bs;z)=∑n=0∞(a1)n⋯(aq)n(b1)n⋯(bs)nznn!(q≤s+1;q,s∈ℕ0=ℕ∪{0};z∈U),
where (θ)n is the Pochhammer symbol defined, in terms of the Gamma function Γ, by
(6)(θ)n=Γ(θ+n)Γ(θ)={1,(υ=0),θ(θ+1)⋯(θ+n-1),(υ∈ℕ).
Corresponding a function hp(a1,…,ai,…,aq;b1,…,bs;z) defined by
(7)hp(a1,…,ai,…,aq;b1,…,bs;z)=zp·qFs(a1,…,ai,…,aq;b1,…,bs;z)(z∈U),
Dziok and Srivastava [5] considered a linear operator
(8)Hp(a1,…,aq;b1,…,bs):A(p,k)→A(p,k)
defined by the following Hadamard product:
(9)Hp(a1,…,aq;b1,…,bs)f(z),=hp(a1,…,ai,…,aq;b1,…,bs;z)*f(z),(q≤s+1;q,s∈ℕ0;z∈U).
If f∈A(p,k) is given by (1), then we have
(10)Hp(a1,…,aq;b1,…,bs)f(z)=zp+∑n=k∞‍Γnan+pzn+p(z∈U),
where
(11)Γn=(a1)n⋯(aq)n(b1)n⋯(bs)n1n!,(n∈ℕ).
To make the notation simple, we write
(12)Hp,q,s(a1)f(z)=Hp(a1,…,aq;b1,…,bs)f(z).
It easily follows from (9) or (10) that
(13)z(Hp,q,s(a1)f(z))′=a1Hp,q,s(a1+1)f(z)-(a1-p)Hp,q,s(a1)f(z),(z∈U).
It should be remarked that the linear operator Hp,q,s(a1) is a generalization of many other linear operators considered earlier. In particular, for f∈A(p) we have the following observations:(i)
H1,2,1(a,b;c)f(z)=(Ica,b)f(z)(a,b∈ℂ;c∉ℤ0-), where the linear operator Ica,b was investigated by Hohlov [6];(ii)
Hp,2,1(n+p,1;1)f(z)=Dn+p-1f(z)(n∈ℕ;n>-p), where the linear operator Dn+p-1 was studied by Goel and Sohi [7]. In the case when p=1, Dnf(z) is the Ruscheweyh derivative of f(z) (see [8]);(iii)
Hp,2,1(μ+p,1;μ+p+1)f(z)=Jp,δ(f)(z)=((p+δ)/zδ)∫0z‍tδ-1f(t)dt(δ>-p), where Jp,δ is the generalized Bernardi-Libera-Livingston integral operator (see [9]);(iv)
Hp,2,1(p+1,1;p+1-λ)f(z)=Ωz(λ,p)f(z)=(Γ(p+1-λ)/Γ(p+1))zλDzλf(z) (-∞≤λ<p+1;z∈U), where Dzλf(z) is the fractional integral of f of order -λ when -∞≤λ<0 and fractional derivative of f of order λ when 0≤λ<p+1. The extended fractional differintegral operator Dz(λ,p) was introduced and studied by Patel and Mishra [10]. The fractional differential operator Ωz(λ,p) with 0≤λ<1 was investigated by Srivastava and Aouf [11]. The operator Ωz(λ,1)=Ωzλ was introduced by Owa and Srivastava [12] (see also [13–15]).(v)
Hp,2,1(a,1;c)f(z) = Lp(a,c)f(z)(a∈ℝ;c∈ℝ∖ℤ0-), where the linear operator Lp(a,c) was studied by Saitoh [16] which yields the operator L(a,c) introduced by Carlson and Shaffer [17] for p=1;(vi)
H1,2,1(μ,1;λ+1)f(z) = Iλ,μf(z)(λ>-1;μ>0), where Iλ,μ is the Choi-Saigo-Srivastava operator [9] which is closely related to the Carlson-Shaffer [17] operator L(μ,λ+1)f(z);(vii)
Hp,2,1(p+1,1;n+p)f(z) = In,pf(z)(n∈ℤ;n>-p), where the operator In,p was considered by Liu and Noor [18];(viii)
Hp,2,1(λ+p,c;a)f(z) = Ipλ(a,c)f(z)(a,c∈ℝ∖ℤ0-;λ>-p), where Ipλ(a,c) is the Cho-Kwon-Srivastava operator [19].In recent years, many interesting subclasses of analytic functions, associated with the Dziok-Srivastava operatorHp,q,s(a1) and its many special cases, were investigated by, for example, Dziok and Srivastava [5, 20], Gangadharan et al. [21], Liu and Noor [18], Liu [22], Liu and Srivastava [23], and others (see also [19, 24–26]). In the present paper, we shall use the method based upon the differential subordination to derive inclusion relationships and other interesting properties and characteristics of the Dziok-Srivastava operator Hp,q,s(a1).


## 2. Main Results

Unless otherwise mentioned, we assume throughout the sequel thatai>0; ai∉ℤ0-(i=1,…,q); α>0; μ>0 and -1≤B<A≤1.LetP[k] denote the class of functions of the form
(14)φ(z)=1+ckzk+ck+1zk+1+⋯
that are analytic in U, we write P[1]=P. In our present investigation, we shall require the following lemmas.Lemma 1 (see [2]).
Leth be analytic and convex (univalent) in U with h(0)=1 and φ∈P[k]. If
(15)φ(z)+zφ′(z)γ≺h(z),
then, for γ≠0 and ℜ(γ)≥0,
(16)φ(z)≺q(z)=γkz-γ/k∫0z‍tγ/k-1h(t)dt≺h(z),
and q is the best dominant.Lemma 2 (see [1]).
LetD be a set in the complex plane ℂ and b be a complex number satisfying ℜ(b)>0. Suppose that the function Ψ:ℂ2×U→ℂ satisfies the condition Ψ(ix,y)∉D for all real x,y≤-|b-ix|/2ℜ(b) and for all z∈U. If the functions φ∈P and ℜ{Ψ(φ(z),zφ′(z);z)}∈D, then ℜ{φ(z)}>0 in U.Lemma 3 (see [27]).
Letϕ be analytic in U with ϕ(0)=1 and ϕ(z)≠0 for all z∈U. If there exist two points z1,z2∈U such that
(17)-π2δ1=arg{ϕ(z1)}<arg{ϕ(z)}<arg{ϕ(z2)}=π2δ2
for some δ1 and δ2(δ1,δ2>0) and for all z(|z|<|z1|=|z2|), then
(18)z1ϕ′(z1)ϕ(z1)=-i(δ1+δ22m),z2ϕ′(z2)ϕ(z2)=-i(δ1+δ22m),
where
(19)m≥1-|b|1+|b|,b=itan(δ2-δ1δ2+δ1).Theorem 4.
Letm≥1, γ>0. Let f∈A(k,p), then
(20)ℜ{(Hp,q,s(a1+1)f(z))(j)(Hp,q,s(a1)f(z))(j)}<a1+γa1(z∈U;0≤j<p),
implies
(21)ℜ{((Hp,q,s(a1+1)f(z))(j)zp-j)-1/2γm}>2-1/m(z∈U;0≤j<p).
The bound 2-1/m is the best possible.Proof.
It easily follows from (13) that
(22)z(Hp,q,s(a1)f(z))(j+1)=a1(Hp,q,s(a1+1)f(z))(j)-(a1-p+j)(Hp,q,s(a1)f(z))(j)(z∈U;0≤j<p).
From (20) and (22), we have
(23)ℜ{z(Hp,q,s(a1+1)f(z))(j+1)(Hp,q,s(a1)f(z))(j)}<γ+p-j(z∈U;0≤j<p).
That is,
(24)-12γ(z(Hp,q,s(a1+1)f(z))(j+1)(Hp,q,s(a1)f(z))(j)-p+j)≺z1-z(z∈U).
Let
(25)φ(z)=((p-j)!p!(Hp,q,s(a1)f(z))(j)zp-j)-1/2γ(z∈U),
then (24) may be written as
(26)z(logφ(z))′≺z(log11-z)′.
By using a well-known result (see  [28]) to (26) we obtain that
(27)φ(z)≺11-z,
or, equivalently,
(28)((p-j)!p!(Hp,q,s(a1)f(z))(j)zp-j)-1/2γm=(11-ω(z))1/m,
where ω is analytic in U, ω(0)=0 and |ω(z)|<1 for z∈U. Since ℜ(t1/m)≥(ℜ(t))1/m for ℜ(t)>0 and m≥1, (28) yields
(29)ℜ((p-j)!p!(Hp,q,s(a1)f(z))(j)zp-j)-1/2γm≥(ℜ(11-ω(z)))1/m≥2-1/m(z∈U).
To see that the bound 2-1/m cannot be increased, we consider the function
(30)g(z)=zp+p!(p-j)!∑n=1∞‍(-2γ)n(n+p-j)!n!(n+p)!Γnzn+p,(z∈U).
Since
(31)(p-j)!p!(Hp,q,s(a1)g(z))(j)zp-j=(1-z)-2γ,
we easily have that g satisfies (20) and
(32)ℜ((p-j)!p!(Hp,q,s(a1)g(z))(j)zp-j)-1/2γm→2-1/m
as ℜ(z)=z→1-. This completes the proof of Theorem 4.Theorem 5.
Letα≥0, γ>1. If f∈A(p) satisfies the following inequality
(33)ℜ{(1-α)(Hp,q,s(a1+1)f(z))(j)(Hp,q,s(a1)f(z))(j)+α(Hp,q,s(a1+2)f(z))(j)(Hp,q,s(a1+1)f(z))(j)}<γ(0≤j<p;z∈U),
then
(34)ℜ{(Hp,q,s(a1+1)f(z))(j)(Hp,q,s(a1)f(z))(j)}<β(0≤j<p;z∈U),
where β∈(1,∞) is the positive root of the equation
(35)2(a1-α+1)x2+(3α-2γα-2γ)x-α=0.Proof.
Let(36)φ(z)=1β-1[β-(Hp,q,s(a1+1)f(z))(j)(Hp,q,s(a1)f(z))(j)](z∈U),
then φ(z) is analytic in U and φ(0)=1. Differentiating (36) and using (22), we obtain that
(37)(1-α)(Hp,q,s(a1+1)f(z))(j)(Hp,q,s(a1)f(z))(j)+α(Hp,q,s(a1+2)f(z))(j)(Hp,q,s(a1+1)f(z))(j)=β-α(β-1)a1+1-(a1-α+1)(β-1)a1+1φ(z)-α(β-1)a1+1zφ′(z)β-(β-1)φ(z)=ψ(φ(z),zφ′(z)),
where
(38)ψ(r,s)=β-α(β-1)a1+1-(a1-α+1)(β-1)a1+1r-α(β-1)a1+1sβ-(β-1)r.
Using (33) and (38), we have
(39){ψ(φ(z),zφ′(z)):z∈U}⊂D={z∈ℂ:ℜ(z)<γ}.
Now for all real x,y≤-(1+x2)/2, we have
(40)ℜ{ψ(ix,y)}=β-α(β-1)a1+1-α(β-1)a1+1βyβ2+(β-1)2x2≥β-α(β-1)a1+1+αβ(β-1)2(a1+1)1+x2β2+(β-1)2x2≥β-α(β-1)a1+1+α(β-1)2β(a1+1)=β-α(β-1)(2β-1)2β(a1+1)=γ,
where β is the positive root of (35).
Note that forα≥0,  γ>1, a1>0 and
(41)h(x)=2(a1-α+1)x2+(3α-2γα-2γ)x-α,
we have h(0)=-α≤0 and h(1)=2a1(1-γ)-2γ<0. This shows β∈(0,+∞). Hence for each z∈U, ψ(ix,y)∉Ω. By Lemma 2, we get ℜ{φ(z)}>0(z∈U), and this proves (34).Theorem 6.
Suppose that0≤j<p; α>0 and 0<δ1, δ2≤1. If Fα given by
(42)Fα(z)=(1-α-αa1+αp)Hp,q,s(a1)f(z)+αa1Hp,q,s(a1+1)f(z)
satisfies
(43)-π2δ1<arg{Fα(j)(z)zp-j}<π2δ2(z∈U),
then
(44)-π2η1<arg{(Hp,q,s(a1)f(z))(j)zp-j}<π2η2(z∈U),
where η1 and η2 are the solution of the equations:
(45)δ1=η1+2πarctan[α(η1+η2)2(1-α+αp)(1-|b|1+|b|)],δ2=η2+2πarctan[α(η1+η2)2(1-α+αp)(1-|b|1+|b|)],
where b is given by (19).Proof.
Using (42) and the identity (22), it follows that
(46)Fα(j)(z)=(1-α+αj)(Hp,q,s(a1)f(z))(j)+αz(Hp,q,s(a1)f(z))(j+1),
for 0≤j<p. Putting
(47)φ(z)=(p-j)!p!(Hp,q,s(a1)f(z))(j)zp-j(z∈U).
On differentiating (47) followed by a simple calculation, we get
(48)Fα(j)(z)zp-j=p!(1-α+αp)(p-j)!×{φ(z)+α1-α+αpzφ′(z)}(z∈U).
Let h be the function which maps U onto the angular domain {w∈ℂ:-(π/2)δ1<arg{w}<(π/2)δ2} with h(0)=1. By using (43) in (48), we get
(49)φ(z)+α1-α+αpzφ′(z)≺h(z).
Further, an application of Lemma 1 yields ℜ{φ(z)}>0 in U and hence φ(z)≠0 for z∈U.
Suppose there exist two pointsz1,z2∈U such that the condition (28) is satisfied. Then by Lemma 3, we obtain (18) under the constraint (19). Therefore, we have
(50)arg{(1-α+αp)φ(z1)+αzφ′(z1)}=arg{φ(z1)}+arg{(1-α+αp)+αz1φ′(z1)φ(z1)}=-π2η1+arg{(1-α+αp)-iα(η1+η2)2m}=-π2η1-arctan{α(η1+η2)2(1-α+αp)m}≤-π2η1-arctan{α(η1+η2)2(1-α+αp)(1-|b|1+|b|)},arg{(1-α+αp)φ(z2)+αzφ′(z2)}≥-π2η2-arctan{α(η1+η2)2(1-α+αp)(1-|b|1+|b|)},
which contradicts the assumption (43). This proves the assertion (44) of the Theorem 6.
Forδ1=δ2=δ, Theorem 6 reduces to the following corollary.Corollary 7.
Suppose that0≤j<p and α>0. If Fα defined by (42) satisfies
(51)|arg{Fα(j)(z)zp-j}|<π2δ(0<δ≤1;z∈U),
then
(52)|arg{(Hp,q,s(a1)f(z))(j)zp-j}|<π2η(z∈U),
where η(0<η≤1) is the solution of the equation:
(53)δ=η+2πarctan(αη1-α+αp).



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