IJMMSInternational Journal of Mathematics and Mathematical Sciences1687-04250161-1712Hindawi Publishing Corporation68543510.1155/2009/685435685435Research ArticleAn Application of Differential SubordinationShanmugamT. N.1JeyaramanM. P.2GovilNarendra Kumar1Department of MathematicsCollege of EngineeringAnna UniversityChennai 600025Indiaannauniv.edu2Department of MathematicsEaswari Engineering CollegeChennai 600089Indiasrmeaswari.ac.in200923112009200927072009081020092009Copyright © 2009This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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

1. Introduction

Let 𝒜p(p={1,2,3,}) be the class of functions f(z) of the form f(z)=zp+m=1ap+mzp+m which are analytic in the open unit disk Δ:={z:|z|<1}.

Let 𝒫 be the class of functions p(z) of the form p(z)=1+n=1pnzn which are analytic in Δ. If p(z)𝒫 satisfies p(z)>0  (zΔ), then we say that p(z) is a Carathéodory function.

With a view to recalling the principle of subordination between analytic functions, let the functions f 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 w(0)=0,|w(z)|<1(zΔ), such that f(z)=g(w(z))(zΔ). We denote this subordination by fg,f(z)g(z)(zΔ). In particular, if the function g is univalent in Δ, the above subordination is equivalent to f(0)=g(0)orf(Δ)g(Δ).

For -1b<a1 and 0<γ1, a function f𝒜p is said to be in the class Sp*(γ,a,b) if it satisfies 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)(0γ<1), the class of p-valent starlike function of order γ.

For Carathéodory functions, Miller  obtained certain sufficient conditions applying the differential inequalities. Recently, Nunokawa et al.  have given some improvement of result by Miller . Recently Ravichandran and Jayamala  studied some subordination results for Carathéodory functions. In this paper by extending the result of Ravichandran and Jayamala , 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. . 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 .

Lemma 1.1 (see [<xref ref-type="bibr" rid="B3">4</xref>, Theorem 3.4h, page 132]).

Let q(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 wq(Δ). Set Q(z)=zq(z)ϕ(q(z)),h(z)=θ(q(z))+Q(z). Suppose that

Q(z) is starlike univalent in Δ,

{zh(z)/Q(z)}={θ(q(z))/ϕ(q(z))+zQ(z)/Q(z)}>0 for zΔ.

If p(z) is analytic in Δ with, p(0)=q(0), p(Δ)D, and 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 Lemma 1.1, we first prove the following theorem.

Theorem 2.1.

Let 0α and λ be a positive real number. Let q(z) be convex univalent in Δ and ((1-α)α+m(q(z))m-1)>0,m{1}. If p𝒫 satisfies (1-α)p(z)+α(p(z))m+αλzp(z)h(z), where h(z)=(1-α)q(z)+α(q(z))m+αλzq(z), then p(z)q(z), and q(z) is the best dominant of (2.1).

Proof.

Let θ(w)=(1-α)w+αwm,ϕ(w)=αλ. Then clearly θ(w) and ϕ(w) are analytic in and ϕ(w)0. Also let 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 (zh(z)Q(z))=1λ{1-αα+m(q(z))m-1+λ(1+zq′′(z)q(z))}>0 for zΔ.

From (2.1)–(2.6) we see that

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 taking α as real and q(z)=((1+az)/(1+bz))γ in Theorem 2.1, we get the following corollary.

Corollary 2.2.

Let -1b<a1, m{1}, 0<γ1/(m-1), λ be real number such that λ>0 and 0<α1. If p𝒫 satisfies (1-α)p(z)+α(p(z))m+αλzp(z)h(z), where 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 p(z)(1+az1+bz)γ, and ((1+az)/(1+bz))γ is the best dominant of (2.8).

Corollary 2.3.

Let -1b<a1, λ>0. If f𝒜p satisfies f(z)0 in 0<|z|<1 and zf(z)pf(z)[1-α+αp(1-λp)zf(z)f(z)+αλ(1+zf′′(z)f(z))]h(z), where 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 zf(z)pf(z)(1+az1+bz)γ.

Proof.

Let p(z)=zf(z)/pf(z), then p𝒫 and (2.11) can be written as (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 zf(z)pf(z)(1+az1+bz)γ.

By taking p=λ=γ=a=1 and b=-1 in Corollary 2.3, we get the following result of Padmanabhan .

Corollary 2.4.

Let f𝒜 and zf(z)f(z)+αz2f′′(z)f(z)2α(z2+2z)+1-z2(1-z)2(0<α1), then (zf(z)f(z))>0.

Theorem 2.5.

Let α,β,ξ,η and η0. Let q(z) be convex univalent in Δ and satisfy [1η(β+2ξq(z))]>0. If p𝒫 satisfies α+βp(z)+ξp2(z)+ηzp(z)α+βq(z)+ξq2(z)+ηzq(z)=h(z), then p(z)q(z), and q(z) is the best dominant of (2.19)

Proof.

By setting θ(w):=α+βw+ξw2 and ϕ(w):=η it can be easily observed that θ(w) and ϕ(w) are analytic in and that ϕ(w)0  (w{0}).

Also, by letting

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 (zh(z)Q(z))=[1η(β+2ξq(z))+(1+zq′′(z)q(z))]>0. The differential subordination α+βp(z)+ξp2(z)+ηzp(z)α+βq(z)+ξq2(z)+ηzq(z) becomes 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.

Let α,β,ξ,η, and δ be complex numbers, δ0. Let 0q(z) be univalent in Δ and satisfy the following conditions for zΔ:

let Q(z)=δzq(z)/q(z) be starlike,

{(β/δ)q(z)+(2ξ/δ)q2(z)-(η/δq(z))+zQ(z)/Q(z)}>0.

If p𝒫 satisfies 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 p(z)q(z), and q(z) is the best dominant.

Proof.

The proof of this theorem is much akin to the proof of Theorem 2.5 and hence can be omitted.

Remark 2.7.

By taking α=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.  which was proved by a different method.

Remark 2.8.

For the choices of α=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 -1b<a1, 0<γ1 and λ>0. If f𝒜p satisfies f(z)f(z)0 in 0<|z|<1, then (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 fSp*(γ,a,b). Also, 1+zf′′(z)f(z)-zf(z)f(z)γ(a-b)z(1+az)(1+bz) implies fSp*(γ,a,b).

Proof.

By taking α=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 setting α=β=ξ=η=0, δ=1, p(z)=zf(z)/pf(z), and q(z)=((1+az)/(1+bz))γ.

For α=ξ=0, β=1, p(z)=zf(z)/f(z), and q(z)=(1+az)/(1-z), -1<a1 in Theorem 2.5, we have the following result.

Corollary 2.10.

If f𝒜 satisfies f(z)0, zΔ and zf(z)f(z)[(1-ηzf(z)f(z))+η(1+zf′′(z)f(z))]h(z), where h(z)=1+az1-z+η(1+a)z(1-z)2, then zf(z)f(z)1+az1-z.

One notes that if h(z)=u+iv, then h(Δ) is the exterior of the parabola given by

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 taking η=a=1 in Corollary 2.10, we obtain the following.

Corollary 2.11.

If f𝒜 satisfies f(z)0, zΔ, and zf(z)f(z)[2-zf(z)f(z)+zf′′(z)f(z)]1+2z-z2(1-z)2, then zf(z)f(z)1+z1-z. Region h(Δ) has been shown shaded in Figure 1.

η=a=1.

Letting α=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.

If f𝒜 satisfies f(z)0, 0<|z|<1, and zf(z)f(z)(1+zf′′(z)f(z))h(z), where h(z)=(1-2γ)2z2+2(2-3γ)z+1(1-z)2 for some γ(0γ<1), then (zf(z)f(z))>γ.

For the univalent function h(z) given by (2.38), One now finds the image h(Δ) of the unit disk Δ.

Let h=u+iv, where u and v are real. One has u=-(2-3γ)+(1+2γ2-2γ)cosθ(1-cosθ),v=2γ(1-γ)sinθ1-cosθ. Elimination of θ yields v2=-8γ2(1-γ)3-2γ[u-2γ2+γ-12]. Therefore, one concludes that h(Δ)={w=u+iv;v2>-8γ2(1-γ)3-2γ[u-2γ2+γ-12]}, which properly contains the half plane w>(2γ2+γ-1)/2.

Corollary 2.13.

Let -1b<a1 and     β0. If f𝒜p satisfies f(z)0 in 0<|z|<1 and (1-β)f(z)zf(z)+f(z)f′′(z)(f(z))2h(z), where h(z)=b(pb-βa)z2+((2p+1-β)b-(1+β)a)z+p-βp(1+bz)2, then fSp*(b,a).

Proof.

If we let p(z)=pf(z)/zf(z), then p𝒫 and (2.43) can be expressed as 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   β0 in Theorem 2.5, we have p(z)(1+az)/(1+bz). So, f(z)Sp*(b,a).

Setting p=1 and b=-1 in Corollary 2.13, we get the following corollay.

Corollary 2.14.

Let -1<a1 and   β0. If f𝒜 satisfies f(z)0 in 0<|z|<1 and (1-β)f(z)zf(z)+f(z)f′′(z)(f(z))2h(z), where h(z)=(1+βa)z2+((β-3)-(1+β)a)z+1-β(1-z)2, then f(z)zf(z)1+az1-z.

Remark 2.15.

For the function h(z) given by (2.48), we have h(Δ)={w=u+iv;v2>a0[u-b0]}, which properly contains the half plane w>b0, where a0=(1+a)β2,b0=5+a+2β(a-1)4.

By putting p=a=β=1 and b=-1 in Corollary 2.13, we get the following result of Tuneski .

Corollary 2.16.

If f(z)𝒜 and f(z)f′′(z)(f(z))22z(z-2)(1-z)2, then (f(z)zf(z))>0.

Remark 2.17.

By putting 0=b<a1, p=1, and β=0 in Corollary 2.13, we get the result obtained by Singh , which refines the result of Silverman .

Corollary 2.18.

Let 0η and q(z) be convex univalent in Δ with q(0)=1 and satisfy (2.18).

Let f𝒜p and ψ(z):=α+βp(f(z)zp)μ+ξp2(f(z)zp)2μ+ημ(f(z)zp)μ[zf(z)pf(z)-1]. If ψ(z)α+βq(z)+ξq2(z)+ηzq(z), then 1p(f(z)zp)μq(z), and q(z) is the best dominant.

Proof.

By taking p(z)=(1/p)(f(z)/zp)μ in Theorem 2.5, we have the above corollary.

Corollary 2.19.

Let 0λ and q(z) be convex univalent in Δ with q(0)=1 and satisfy (μλ)>0.

If f𝒜 satisfies (1-λ)(f(z)z)μ+λf(z)(f(z)z)μ-1q(z)+λμzq(z), then (f(z)z)μq(z).

If f𝒜 satisfies f(z)(f(z)z)μ-1-(f(z)z)μ1μzq(z), then (f(z)z)μq(z), and q(z) is the best dominant.

Proof.

Proof of the first part follows from Corollary 2.18, by taking β=p=1, α=ξ=0, and η=λ/μ.

The proof of the second part follows from Corollary 2.18, by taking α=β=ξ=0, p=1, and η=1/μ.

By taking λ=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 .

Corollary 2.20.

Let f𝒜, then for a positive integer n, one has that {(1-n)(f(z)z)n+nf(z)(f(z)z)n-1}>β implies (f(z)z)nA+(1-A)(-1-2zlog(1-z)), and A+(1-A)[-1-(2/z)log(1-z)] is the best dominant.

Remark 2.21.

By taking μ=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 .

By taking β=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 .

Corollary 2.22.

Let f𝒜 and zf(z)f(z)1+Az, then (f(z)z)μeμAz, and eμAz is the best dominant.

We remark here that q(z)=eμAz is univalent if and only if |μA|<π.

Remark 2.23.

For a special case when p(z)=(1/p)(f(z)/zp)μ, q(z)=1/(1-z)2b where b{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 .

Corollary 2.24.

If f𝒜 satisfies (1+λ)(zf(z))μ-λf(z)(zf(z))μ+1q(z)+λμzq(z), then (zf(z))μq(z), and q is the best dominant.

Proof.

By taking p(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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