# An Application of Differential Subordination

Authors: T. N. Shanmugam; M. P. Jeyaraman
Journal: International Journal of Mathematics and Mathematical Sciences
 (2009)

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## Abstract

We apply the general theory of differential subordination to obtain certian interesting criteria forp-valent starlikeness and strong starlikeness. Some applications of these results are also discussed.


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## Body

## 1. Introduction

Let𝒜p(p∈ℕ={1,2,3,…}) be the class of functions f(z) of the form
(1.1)f(z)=zp+∑m=1∞ap+mzp+m
which are analytic in the open unit disk Δ:={z:|z|<1}.Let𝒫 be the class of functions p(z) of the form
(1.2)p(z)=1+∑n=1∞pnzn
which are analytic in Δ. If p(z)∈𝒫 satisfies ℜp(z)>0(z∈Δ), then we say that p(z) is a Carathéodory function.With a view to recalling the principle of subordination between analytic functions, let the functionsf and g be analytic in Δ. Then we say that the function f is subordinate to g if there exists a Schwarz function w(z), analytic in Δ with
(1.3)w(0)=0,|w(z)|<1(z∈Δ),
such that
(1.4)f(z)=g(w(z))(z∈Δ).
We denote this subordination by
(1.5)f≺g,f(z)≺g(z)(z∈Δ).
In particular, if the function g is univalent in Δ, the above subordination is equivalent to
(1.6)f(0)=g(0)orf(Δ)⊂g(Δ).For-1≤b<a≤1 and 0<γ≤1, a function f∈𝒜p is said to be in the class Sp*(γ,a,b) if it satisfies
(1.7)zf′(z)f(z)≺p(1+az1+bz)γ.
Also, we write Sp*(γ,1,-1)=SSp*(γ), the class of  strongly starlike  p-valent functions  of order γ in Δ.  Sp*(1,a,b)=Sp*(a,b), the class of Janowski starlike p-valent function, Sp*(1,-1)=Sp*, the class of p-valent starlike function, and Sp*(1-2γ,1)=Sp*(γ)(0≤γ<1), the class of p-valent starlike function of order γ.For Carathéodory functions, Miller [1] obtained certain sufficient conditions applying the differential inequalities. Recently, Nunokawa et al. [2] have given some improvement of result by Miller [1]. Recently Ravichandran and Jayamala [3] studied some subordination results for Carathéodory functions. In this paper by extending the result of Ravichandran and Jayamala [3], we find sufficient conditions for the subordination p(z)≺q(z) to hold for given q(z) and criteria for p-valent starlikeness. Our results include results obtained by Nunokawa et al. [2]. We also give some criteria for p-valently starlikeness and strong starlikeness.To prove our result we need the following lemma due to Miller and Mocanu [4].Lemma 1.1 (see [4, Theorem 3.4h, page 132]).
Letq(z) be analytic and univalent in the unit disk Δ and θ(ω) and let ϕ(ω) be analytic in a domain D containing q(Δ) with ϕ(w)≠0 when w∈q(Δ). Set
(1.8)Q(z)=zq′(z)ϕ(q(z)),h(z)=θ(q(z))+Q(z).
Suppose that(i)
Q(z) is starlike univalent in Δ,(ii)
ℜ{zh′(z)/Q(z)}=ℜ{θ′(q(z))/ϕ(q(z))+zQ′(z)/Q(z)}>0 for z∈Δ.If p(z) is analytic in Δ with, p(0)=q(0), p(Δ)⊆D, and
(1.9)θ(p(z))+zp′(z)ϕ(p(z))≺θ(q(z))+zq′(z)ϕ(q(z)),
then p(z)≺q(z) and q(z) is the best dominant.


## 2. Application of Differential Subordination

By making use of Lemma1.1, we first prove the following theorem.Theorem 2.1.
Let0≠α∈ℂ and λ be a positive real number. Let q(z) be convex univalent in Δ and ℜ((1-α)∖α+m(q(z))m-1)>0,m∈ℕ∖{1}. If p∈𝒫 satisfies
(2.1)(1-α)p(z)+α(p(z))m+αλzp′(z)≺h(z),
where
(2.2)h(z)=(1-α)q(z)+α(q(z))m+αλzq′(z),
then
(2.3)p(z)≺q(z),
and q(z) is the best dominant of (2.1).Proof.
Let(2.4)θ(w)=(1-α)w+αwm,ϕ(w)=αλ.
Then clearly θ(w) and ϕ(w) are analytic in ℂ and ϕ(w)≠0. Also let
(2.5)Q(z)=zq′(z)ϕ(q(z))=αλzq′(z),h(z)=θ(q(z))+Q(z)=(1-α)q(z)+α(q(z))m+αλzq′(z).
Since q(z) is convex univalent, zq′(z) is starlike univalent. Therefore Q(z) is starlike univalent in Δ, and
(2.6)ℜ(zh′(z)Q(z))=1λℜ{1-αα+m(q(z))m-1+λ(1+zq′′(z)q′(z))}>0
for z∈Δ.
From (2.1)–(2.6) we see that
(2.7)θ(p(z))+zp′(z)ϕ(p(z))≺θ(q(z))+zq′(z)ϕ(q(z))=h(z).
Therefore, by applying Lemma 1.1, we conclude that p(z)≺q(z) and q(z) is the best dominant of (2.1). The proof of the theorem is complete.By takingα as real and q(z)=((1+az)/(1+bz))γ in Theorem 2.1, we get the following corollary.Corollary 2.2.
Let-1≤b<a≤1, m∈ℕ∖{1}, 0<γ≤1/(m-1), λ be real number such that λ>0 and 0<α≤1. If p∈𝒫 satisfies
(2.8)(1-α)p(z)+α(p(z))m+αλzp′(z)≺h(z),
where
(2.9)h(z)=(1-α)(1+az1+bz)γ+α(1+az1+bz)mγ+αλγ(a-b)z(1+az)1-γ(1+bz)1+γ,
then
(2.10)p(z)≺(1+az1+bz)γ,
and ((1+az)/(1+bz))γ is the best dominant of (2.8).Corollary 2.3.
Let-1≤b<a≤1, λ>0. If f∈𝒜p satisfies f(z)≠0 in 0<|z|<1 and
(2.11)zf′(z)pf(z)[1-α+αp(1-λp)zf′(z)f(z)+αλ(1+zf′′(z)f′(z))]≺h(z),
where
(2.12)h(z)=(1-α)(1+az1+bz)γ+α(1+az1+bz)2γ+αλγ(a-b)z(1+az)1-γ(1+bz)1+γ,
then
(2.13)zf′(z)pf(z)≺(1+az1+bz)γ.Proof.
Letp(z)=zf′(z)/pf(z), then p∈𝒫 and (2.11) can be written as
(2.14)(1-α)p(z)+αp2(z)+αλzp′(z)≺(1-α)(1+az1+bz)γ+α(1+az1+bz)2γ+αλγ(a-b)z(1+az)1-γ(1+bz)1+γ.
Taking m=2 in Corollary 2.2 and using (2.14), we have
(2.15)zf′(z)pf(z)≺(1+az1+bz)γ.By takingp=λ=γ=a=1 and b=-1 in Corollary 2.3, we get the following result of Padmanabhan [5].Corollary 2.4.
Letf∈𝒜 and
(2.16)zf′(z)f(z)+αz2f′′(z)f′(z)≺2α(z2+2z)+1-z2(1-z)2(0<α≤1),
then
(2.17)ℜ(zf′(z)f(z))>0.Theorem 2.5.
Letα,β,ξ,η∈ℂ and η≠0. Let q(z) be convex univalent in Δ and satisfy
(2.18)ℜ[1η(β+2ξq(z))]>0.
If p∈𝒫 satisfies
(2.19)α+βp(z)+ξp2(z)+ηzp′(z)≺α+βq(z)+ξq2(z)+ηzq′(z)=h(z),
then
(2.20)p(z)≺q(z),
and q(z) is the best dominant of (2.19)Proof.
By settingθ(w):=α+βw+ξw2 and ϕ(w):=η it can be easily  observed that  θ(w) and  ϕ(w) are  analytic  in ℂ and  that ϕ(w)≠0(w∈ℂ∖{0}).
Also, by letting
(2.21)Q(z)=zq′(z)ϕ(q(z))=ηzq′(z),h(z)=θ(q(z))+Q(z)=α+βq(z)+ξq2(z)+ηzq′(z),
we find that Q(z) is starlike univalent in Δ and that
(2.22)ℜ(zh′(z)Q(z))=ℜ[1η(β+2ξq(z))+(1+zq′′(z)q′(z))]>0.
The differential subordination
(2.23)α+βp(z)+ξp2(z)+ηzp′(z)≺α+βq(z)+ξq2(z)+ηzq′(z)
becomes
(2.24)θ(p(z))+zp′(z)ϕ(p(z))≺θ(q(z))+zq′(z)ϕ(q(z)).
Now, the result follows as an application of Lemma 1.1.Theorem 2.6.
Letα,β,ξ,η, and δ be complex numbers, δ≠0. Let 0≠q(z) be univalent in Δ and satisfy the following conditions for z∈Δ: (1)
letQ(z)=δzq′(z)/q(z) be starlike,(2)
ℜ{(β/δ)q(z)+(2ξ/δ)q2(z)-(η/δq(z))+zQ′(z)/Q(z)}>0.If p∈𝒫 satisfies
(2.25)α+βp(z)+ξ(p(z))2+ηp(z)+δzp′(z)p(z)≺α+βq(z)+ξ(q(z))2+ηq(z)+δzq′(z)q(z),
then
(2.26)p(z)≺q(z),
and q(z) is the best dominant.Proof.
The proof of this theorem is much akin to the proof of  Theorem2.5 and hence can be omitted.Remark 2.7.
By takingα=β=0, ξ=(λ/μ)μ>0, λ>-μ/2, η=1, and q(z)=(1+z)/(1-z) in Theorem 2.5 we get the result of Nunokawa et al. [2] which was proved by a different method.Remark 2.8.
For the choices ofα=β=0 in Theorem 2.5, we get the result of [3, Theorem  1, page  192] and for α=ξ=η=0 in Theorem 2.6 we get the result of [3, Theorem  2, page  194].Corollary 2.9.
Let-1≤b<a≤1, 0<γ≤1 and λ>0. If f∈𝒜p satisfies f(z)f′(z)≠0 in 0<|z|<1, then
(2.27)(1-λ)zf′(z)f(z)+λ(1+zf′′(z)f′(z))≺p(1+az1+bz)γ+λγ(a-b)z(1+az)(1+bz)
implies
(2.28)f∈Sp*(γ,a,b).
Also,
(2.29)1+zf′′(z)f′(z)-zf′(z)f(z)≺γ(a-b)z(1+az)(1+bz)
implies
(2.30)f∈Sp*(γ,a,b).Proof.
By takingα=ξ=η=0, β=p/λ, δ=1, p(z)=zf′(z)/pf(z), and q(z)=((1+az)/(1+bz))γ in Theorem 2.6, we get the first part.
Proof of the second part follows, by settingα=β=ξ=η=0, δ=1, p(z)=zf′(z)/pf(z), and q(z)=((1+az)/(1+bz))γ.Forα=ξ=0, β=1, p(z)=zf′(z)/f(z), and q(z)=(1+az)/(1-z), -1<a≤1 in Theorem 2.5, we have the following result.Corollary 2.10.
Iff∈𝒜 satisfies f(z)≠0, z∈Δ and
(2.31)zf′(z)f(z)[(1-ηzf′(z)f(z))+η(1+zf′′(z)f′(z))]≺h(z),
where
(2.32)h(z)=1+az1-z+η(1+a)z(1-z)2,
then
(2.33)zf′(z)f(z)≺1+az1-z.
One notes that ifh(z)=u+iv, then h(Δ) is the exterior of the parabola given by
(2.34)v2=-(1+a)η[u-2-2a-η(1+a)4]
with its vertex as ((2-2a-η(1+a)/4),0) (see [5, 6]).By takingη=a=1 in Corollary 2.10, we obtain the following.Corollary 2.11.
Iff∈𝒜 satisfies f(z)≠0, z∈Δ, and
(2.35)zf′(z)f(z)[2-zf′(z)f(z)+zf′′(z)f′(z)]≺1+2z-z2(1-z)2,
then
(2.36)zf′(z)f(z)≺1+z1-z.
Region h(Δ) has been shown shaded in Figure 1.Figure 1
η=a=1.Lettingα=β=0, ξ=η=1, p(z)=zf′(z)/f(z), and q(z)=(1+(1-2γ)z)/(1-z) in Theorem 2.5, we get the following.Corollary 2.12.
Iff∈𝒜 satisfies f(z)≠0, 0<|z|<1, and
(2.37)zf′(z)f(z)(1+zf′′(z)f′(z))≺h(z),
where
(2.38)h(z)=(1-2γ)2z2+2(2-3γ)z+1(1-z)2
for some γ(0≤γ<1), then
(2.39)ℜ(zf′(z)f(z))>γ.For the univalent functionh(z) given by (2.38), One now finds the image h(Δ) of the unit disk Δ.Leth=u+iv, where u and v are real. One has
(2.40)u=-(2-3γ)+(1+2γ2-2γ)cosθ(1-cosθ),v=2γ(1-γ)sinθ1-cosθ.
Elimination of θ yields
(2.41)v2=-8γ2(1-γ)3-2γ[u-2γ2+γ-12].
Therefore, one concludes that
(2.42)h(Δ)={w=u+iv;v2>-8γ2(1-γ)3-2γ[u-2γ2+γ-12]},
which properly contains the half plane ℜw>(2γ2+γ-1)/2.Corollary 2.13.
Let-1≤b<a≤1 and ℜβ≥0. If f∈𝒜p satisfies f′(z)≠0 in 0<|z|<1 and
(2.43)(1-β)f(z)zf′(z)+f(z)f′′(z)(f′(z))2≺h(z),
where
(2.44)h(z)=b(pb-βa)z2+((2p+1-β)b-(1+β)a)z+p-βp(1+bz)2,
then
(2.45)f∈Sp*(b,a).Proof.
If we letp(z)=pf(z)/zf′(z), then p∈𝒫 and (2.43) can be expressed as
(2.46)βp(z)+zp′(z)≺β(1+az1+bz)+(a-b)z(1+bz)2.
Hence, by taking α=ξ=0, η=1, q(z)=(1+az)/(1+bz) and ℜβ≥0 in Theorem 2.5, we have p(z)≺(1+az)/(1+bz). So, f(z)∈Sp*(b,a).Settingp=1 and b=-1 in Corollary 2.13, we get the following corollay.Corollary 2.14.
Let-1<a≤1 and ℜβ≥0. If f∈𝒜 satisfies f′(z)≠0 in 0<|z|<1 and
(2.47)(1-β)f(z)zf(z)+f(z)f′′(z)(f′(z))2≺h(z),
where
(2.48)h(z)=(1+βa)z2+((β-3)-(1+β)a)z+1-β(1-z)2,
then
(2.49)f(z)zf′(z)≺1+az1-z.Remark 2.15.
For the functionh(z) given by (2.48), we have
(2.50)h(Δ)={w=u+iv;v2>a0[u-b0]},
which properly contains the half plane ℜw>b0, where
(2.51)a0=(1+a)β2,b0=5+a+2β(a-1)4.By puttingp=a=β=1 and b=-1 in Corollary 2.13, we get the following result of Tuneski [7].Corollary 2.16.
Iff(z)∈𝒜 and
(2.52)f(z)f′′(z)(f′(z))2≺2z(z-2)(1-z)2,
then
(2.53)ℜ(f(z)zf′(z))>0.Remark 2.17.
By putting0=b<a≤1, p=1, and β=0 in Corollary 2.13, we get the result obtained by Singh [8], which refines the result of Silverman [9].Corollary 2.18.
Let0≠η and q(z) be convex univalent in Δ with q(0)=1 and satisfy (2.18).
Letf∈𝒜p and
(2.54)ψ(z):=α+βp(f(z)zp)μ+ξp2(f(z)zp)2μ+ημ(f(z)zp)μ[zf′(z)pf(z)-1].
If
(2.55)ψ(z)≺α+βq(z)+ξq2(z)+ηzq′(z),
then
(2.56)1p(f(z)zp)μ≺q(z),
and q(z) is the best dominant.Proof.
By takingp(z)=(1/p)(f(z)/zp)μ in Theorem 2.5, we have the above corollary.Corollary 2.19.
Let0≠λ∈ℂ and q(z) be convex univalent in Δ with q(0)=1 and satisfy
(2.57)ℜ(μλ)>0.(i)
Iff∈𝒜 satisfies
(2.58)(1-λ)(f(z)z)μ+λf′(z)(f(z)z)μ-1≺q(z)+λμzq′(z),
then
(2.59)(f(z)z)μ≺q(z).(ii)
Iff∈𝒜 satisfies
(2.60)f′(z)(f(z)z)μ-1-(f(z)z)μ≺1μzq′(z),
then
(2.61)(f(z)z)μ≺q(z),
and q(z) is the best dominant.Proof.
Proof of the first part follows from  Corollary2.18,  by taking  β=p=1, α=ξ=0, and η=λ/μ.
The proof of the second part follows from Corollary2.18, by taking α=β=ξ=0, p=1, and η=1/μ.By takingλ=μ=n where n is a positive integer and q(z)=A+(1-A)[-1-(2/z)log(1-z)] in the first part of Corollary 2.19, we get the following result of Ponnusamy [10].Corollary 2.20.
Letf∈𝒜, then for a positive integer n, one has that
(2.62)ℜ{(1-n)(f(z)z)n+nf′(z)(f(z)z)n-1}>β
implies
(2.63)(f(z)z)n≺A+(1-A)(-1-2zlog(1-z)),
and A+(1-A)[-1-(2/z)log(1-z)] is the best dominant.Remark 2.21.
By takingμ=1 and q(z)=1+(A/(1+δ))z in Corollary 2.19 and μ=λ=1 and q(z)=A/B+(1-A/B)(log(1+Bz)/Bz) we get the result of Ponnusamy and Juneja [11].By takingβ=ξ=η=0, α=p=1, δ=1/μ, p(z)=(1/p)(f(z)/zp)μ, and q(z)=eμAz in Theorem 2.5, we get the following result obtained by Owa and Obradović [12].Corollary 2.22.
Letf∈𝒜 and
(2.64)zf′(z)f(z)≺1+Az,
then
(2.65)(f(z)z)μ≺eμAz,
and eμAz is the best dominant.We remark here thatq(z)=eμAz is univalent if and only if |μA|<π.Remark 2.23.
For a special case whenp(z)=(1/p)(f(z)/zp)μ, q(z)=1/(1-z)2b where b∈ℂ∖{0} and β=ξ=η=0, α=μ=p=1, and δ=1/b in Theorem 2.6, we have the result obtained by Srivastava and Lashin [13].Corollary 2.24.
Iff∈𝒜 satisfies
(2.66)(1+λ)(zf(z))μ-λf′(z)(zf(z))μ+1≺q(z)+λμzq′(z),
then
(2.67)(zf(z))μ≺q(z),
and q is the best dominant.Proof.
By takingp(z)=(1/p)(zp/f(z))μ and α=ξ=0, β=p=1 and η=λ/μ in Theorem 2.5, we get the previous corollary.



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