# Sufficient Conditions and Applications for Carathéodory Functions

Authors: Neng Xu
Journal: Journal of Mathematics
 (2013)

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## Abstract

LetP be the class of functions of the form p(z)=1+∑n=1∞pnzn which are analytic in U. By using the method of differential subordinations, we give some sufficient conditions for Carathéodory function, that is, if p(z)∈P satisfies p(z)≠0(0<|z|<1) and p(z)m+λ(zp'(z)/(p(z))μ)≺h(z), where h(z)=((1+az)/(1+bz))m+λ(a-b)z/(1+bz)2-μ(1+az)μ, -1≤b<a≤1,λ>0, m, μ∈{0,1,2},m+μ≤2, then p(z)≺(1+az)/(1+bz). Some useful consequences of this result are also given.


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## Body

## 1. Introduction

LetA be the class of functions of the following form:
(1)f(z)=z+∑n=2∞‍anzn,
which are analytic in U={z:|z|<1}. We denote by S the subclass of A consisting of univalent functions and by S* and K the usual subclasses of S whose members are starlike (with respect to the origin) and close-to-convex, respectively. Finally, we denote by R the family of functions f∈A which satisfy the condition Ref′(z)>0(z∈U). It is well known that R⊂K.LetP be the class of functions of the following form:
(2)p(z)=1+∑n=1∞‍pnzn,
which are analytic in U. If p(z)∈P satisfies Rep(z)>0(z∈U), then we say that p(z) is the Carathéodory function. For Carathéodory functions, Nunokawa et al. [1] have shown some sufficient conditions applying the differential inequalities. In the present paper, using the method of differential subordinations, we derive certain conditions under which we have p(z)≺(1+az)/(1+bz), where -1≤b<a≤1. Our results generalize or improve some results due to [1–4].To prove our results, we need the following lemma due to Miller and Mocanu [5].Lemma 1.
Letg(z) be analytic and univalent in U and let θ(w) and φ(w) be analytic in a domain D containing g(U), with φ(w)≠0 when w∈g(U). Set
(3)Q(z)=zg′(z)φ(g(z)),h(z)=θ(g(z))+Q(z)
and suppose that (i)
Q(z)is starlike univalent in U, and(ii)
Re(zh′(z)/Q(z))=Re{θ′(g(z))/φ(g(z))+zQ′(z)/Q(z)}>0(z∈U).
Ifp(z) is analytic in U, with p(0)=g(0), p(U)⊂D and
(4)θ(p(z))+zp′(z)φ(p(z))≺θ(g(z))+zg′(z)φ(g(z))=h(z),
then p(z)≺g(z) and g(z) is the best dominant of (4).


## 2. Main Results

Theorem 2.
Let-1≤b<a≤1, λ>0, m,μ∈{0,1,2}, and m+μ≤2. If p(z)∈P satisfies p(z)≠0(0<|z|<1) and
(5)p(z)m+λzp′(z)(p(z))μ≺h(z),
where
(6)h(z)=(1+az1+bz)m+λ(a-b)z(1+bz)2-μ(1+az)μ,
then p(z)≺(1+az)/(1+bz) and (1+az)/(1+bz) is the best dominant of (5).Proof.
Let-1≤b<a≤1, λ>0, m,μ∈{0,1,2},
(7)D={Cμ=0,C∖{0}μ=1,2,
and choose
(8)g(z)=1+az1+bz,θ(w)=wm,φ(w)=λwμ.
Then g(z) is analytic and univalent in U, g(0)=p(0)=1, p(U)⊂D, θ(w) and φ(w) satisfy the conditions of the lemma. The function
(9)Q(z)=zg′(z)φ(g(z))=λ(a-b)z(1+bz)2-μ(1+az)μ
is univalent and starlike in U because
(10)RezQ′(z)Q(z)=1-(2-μ)Rebz1+bz-μReaz1+az=-1+(2-μ)Re11+bz+μRe11+az>-1+(2-μ)11+|b|+μ11+|a|≥0(z∈U).
Further, we have
(11)θ(g(z))+Q(z)=(1+az1+bz)m+λ(a-b)z(1+bz)2-μ(1+az)μ=h(z),zh′(z)Q(z)=mλ(1+az1+bz)m+μ-1+zQ′(z)Q(z)(z∈U).
Since m+μ≤2, from (11) it is easy to know that
(12)Rezh′(z)Q(z)>0(z∈U).
Hence the function h(z) is close-to-convex and univalent in U. Now it follows from (5)–(12) that
(13)θ(p(z))+zp′(z)φ(p(z))≺θ(g(z))+zg′(z)φ(g(z))=h(z).
Therefore, by virtue of the lemma, we conclude that p(z)≺g(z) and g(z) is the best dominant of (5). The proof of the theorem is complete.Making use of the theorem, we can obtain a number of interesting results.Corollary 3.
Letβ>0, α>0. If p(z)∈P satisfies
(14)(p(z))2+βαzp′(z)≺α+2(1+β)z+αz2α(1-z)2=h1(z)(z∈U),
then  Rep(z)>0.Proof.
Leta=1, b=-1, λ=β/α, β>0, α>0, m=2, and μ=0 in the theorem, then we have
(15)g(z)=1+z1-z,θ(w)=w2,φ(w)=βα,Q(z)=2βzα(1-z)2
satisfy the conditions of the lemma. Note that
(16)Rezh1′(z)Q(z)=(2αβ+1)Re1+z1-z>0(z∈U).
Hence, similar to the proof of the theorem, we conclude that p(z)≺g(z), that is, Rep(z)>0.Remark 4.
Note that Corollary3 was also proved by Nunokawa et al. [1] using another method.
Takinga=1, b=-1, λ=β/α, α>0, β>0, and m=μ=1 in the theorem, we have the following.Corollary 5.
Letα>0, β>0, p(z)∈P satisfies p(z)≠0(0<|z|<1), and
(17)αp(z)+βzp′(z)p(z)≺α1+z1-z+2βz1-z2=h2(z)(z∈U),
then Rep(z)>0.Remark 6.
Note that the function(18)w=h2(z)=α1+z1-z+2βz1-z2(α>0,β>0)
maps U onto the w-plane slit along the half-lines Rew=0, Imw≥β(2α+β) and Rew=0, Imw≤-β(2α+β). Hence Corollary 5 coincides with the result obtained by Nunokawa et al. [1, Theorem 2] using another method.
Lettingf(z)∈A, p(z)=f′(z), α=1 in Corollary 5, we have the following.Corollary 7.
Iff(z)∈A, f′(z)≠0(0<|z|<1), and
(19)f′(z)+βzf′′(z)f′(z)≠ib(z∈U),
where β>0 is real and |b|≥β(2+β), then f(z)∈R.Remark 8.
Lewandowski et al. [3] proved if f(z)∈A, f′(z)≠0(0<|z|<1), β>0, and
(20)Re{f′(z)+βzf′′(z)f′(z)}>0(z∈U),
then f(z)∈R. We see that Corollary 7 improves this result.Corollary 9.
Letf(z)∈A with f(z)f′(z)≠0(0<|z|<1). If
(21)(zf(z))′′f′(z)-2zf′(z)f(z)≺(1-δ)z1+(1-δ)z=h3(z)
for some δ(0≤δ<1), then f(z)∈S and
(22)|z2f′(z)f2(z)-1|<1-δ(z∈U).Proof.
Lettingf(z)∈A, f(z)f′(z)≠0(0<|z|<1), then p(z)=z2f′(z)/f2(z)∈P,
(23)zp′(z)p(z)=(zf(z))′′f′(z)-2zf′(z)f(z).
Taking m=0, μ=λ=1 and a=1-δ(0≤δ<1), b=0 in the theorem, it follows from (5), (6), and (21) that |(z2f′(z))/(f2(z))-1|<1-δ(z∈U), which implies that f(z)∈S (see [6]).Remark 10.
Frassin and Darus [2] have shown that if f(z)∈A and
(24)|(zf(z))′′f′(z)-2zf′(z)f(z)|<1-δ2-δ(z∈U)
for some δ(0≤δ<1), then |(z2f′(z)/f2(z))-1|<1-δ(z∈U).
For0≤δ<1, the function
(25)h3(z)=(1-δ)z1-(1-δ)z
is analytic and convex univalent in U. Since
(26)h3(U)={w:|w+(1-δ)2δ(2-δ)|<1-δδ(2-δ)}(0<δ<1),h3(U)={w:Rew<12}(δ=0),
the disk |w|<(1-δ)/(2-δ) is properly contained in h3(U). Therefore, in view of (21), we see that Corollary 9 is better than the result given in [2].
Lettingf(z)∈A, p(z)=f′(z)∈P, m=1, μ=0, a=1-2α, 0≤α<1, and b=-1 in the theorem, we have the following.Corollary 11.
Iff(z)∈A, λ>0, 0≤α<1, and
(27)f′(z)+λzf′′(z)≺1+(1-2α)z1-z+λ2(1-α)z(1-z)2=h4(z)(z∈U),
then Ref′(z)>α.
For the univalent functionh4(z), we now find the image h4(U) of the unit disk U.
Leth4(eiθ)=u+iv, where u and v are real. We have
(28)u=α-(1-α)λ1-cosθ,v=(1-α)sinθ1-cosθ.
Elimination of θ yields
(29)v2=2α(1-α)-(1-α)2λλ-2(1-α)λu.
Therefore, we conclude that
(30)h4(U)={w=u+iv:v2>-2(1-α)λhghh×(u+(1-α)λ2-α)}.Remark 12.
Nunokawa and Hoshino [4] have proved that if f(z)∈A, λ>0 and
(31)Re{f′(z)+λzf′′(z)}>-λ2,(z∈U),
then Ref′(z)>0. For α=0, h4(U) properly contains the half plane Rew>-λ/2. Hence Corollary 11 with α=0 is better than the result given in [4].



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