Let P 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.

1. Introduction

Let A 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.

Let P 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.

Let g(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

If p(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 ResultsTheorem 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.

Let β>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.

Let a=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 Corollary 3 was also proved by Nunokawa et al. [1] using another method.

Taking a=1, b=-1, λ=β/α, α>0, β>0, and m=μ=1 in the theorem, we have the following.

Corollary 5.

Let α>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.

Letting f(z)∈A, p(z)=f′(z), α=1 in Corollary 5, we have the following.

Corollary 7.

If f(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.

Let f(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.

Letting f(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).

For 0≤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].

Letting f(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.

If f(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 function h4(z), we now find the image h4(U) of the unit disk U.

Let h4(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].

Acknowledgments

This work was partially supported by the National Natural Science Foundation of China (Grant no. 11171045). The author would like to thank the referees for their careful reading and for making some valuable comments which have essentially improved the presentation of this paper.

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