# Starlike and Convex Properties for

Authors: Oh Sang Kwon; Nak Eun Cho
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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## Abstract

The purpose of the present paper is to give some characterizations for a (Gaussian)
hypergeometric function to be in various subclasses of starlike and convex functions.
We also consider an integral operator related to the hypergeometric function.


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## Body

## 1. Introduction

Let𝒯 be the class consisting of functions of the
form (1.1)f(z)=z−∑n=2∞anzn,an≥0, that are analytic and
univalent in the open unit disk 𝕌={z:|z|<1}.
Let 𝒯*(α) and 𝒞(α) denote the subclasses of 𝒯 consisting of starlike and convex functions of
order α(0≤α<1),
respectively [1].Recently, Bharati et al. [2] introduced the following
subclasses of starlike and convex functions.Definition 1.1.
A functionf of the form (1.1) is in 𝒮p𝒯(α,β) if it satisfies the condition (1.2)Re{zf′(z)f(z)}≥α|zf′(z)f(z)−1|+β,α≥0,0≤β<1, and f∈𝒰𝒞𝒯(α,β) if and only if zf′∈𝒮p𝒯(α,β).Definition 1.2.
A functionf of the form (1.1) is in 𝒫𝒯(α) if it satisfies the condition (1.3)Re{zf′(z)f(z)}+α≥|zf′(z)f(z)−α|,0<α<∞, and f∈𝒞𝒫𝒯(α) if and only if zf′∈𝒫𝒯(α).Bharati et al. [2] showed that 𝒮p𝒯(α,β)=𝒯*((α+β)/(1+α)),𝒰𝒞𝒯(α,β)=𝒞((α+β)/(1+α)),𝒫𝒯(α)=𝒯*(1−α)(0<α≤1),
and 𝒞𝒫𝒯(α)=𝒞(1−α)(0<α≤1).
In particular, we note that 𝒰𝒞𝒯(1,0) is the class of uniformly convex functions
given by Goodman [3] (also see [4–6]).LetF(a,b;c;z) be the (Gaussian) hypergeometric function
defined by (1.4)F(a,b;c;z)=∑n=0∞(a)n(b)n(c)n(1)nzn, where c≠0,−1,−2,…,
and (λ)n is the Pochhammer symbol defined by (1.5)(λ)n={1ifn=0,λ(λ+1)⋯(λ+n−1)ifn∈ℕ={1,2,…}. We note that F(a,b;c;1) converges for Re(c−a−b)>0 and is related to the Gamma function by (1.6)F(a,b;c;1)=Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b). Silverman [7] gave necessary
and sufficient conditions for zF(a,b;c;z) to be in 𝒯*(α) and 𝒞(α),
and also examined a linear operator acting on hypergeometric functions. For the
other interesting developments for zF(a,b;c;z) in connection with various subclasses of
univalent functions, the readers can refer to the works of Carlson and Shaffer
[8], Merkes and Scott [9], and Ruscheweyh and Singh [10].In the present
paper, we determine necessary and sufficient conditions forzF(a,b;c;z) to be in 𝒮p𝒯(α,β),𝒰𝒞𝒯(α,β),𝒫𝒯(α), and 𝒞𝒫𝒯(α).
Furthermore, we consider an integral operator related to the hypergeometric
function.


## 2. Results

To establish
our main results, we need the following lemmas due to Bharati et al. [2].Lemma 2.1.
(i) A
functionf of the form (1.1) is in 𝒮p𝒯(α,β) if and only if it satisfies (2.1)∑n=2∞(n(1+α)−(α+β))an≤1−β.
(ii) A functionf of the form (1.1) is in 𝒰𝒞𝒯(α,β) if and only if it satisfies (2.2)∑n=2∞n(n(1+α)−(α+β))an≤1−β.Lemma 2.2.
(i) A functionf of the form (1.1) is in 𝒫𝒯(α) if and only if it satisfies (2.3)∑n=2∞(n−1+α)an≤α.
(ii) A functionf of the form (1.1) is in 𝒞𝒫𝒯(α) if and only if it satisfies (2.4)∑n=2∞n(n−1+α)an≤α.Theorem 2.3.
(i) Ifa,b>−1,c>0, and ab<0,
then zF(a,b;c;z) is in 𝒮p𝒯(α,β) if and only if (2.5)c≥a+b+1−(1+α)ab(1−β).
(ii) Ifa,b>0 and c>a+b+1,
then F1(a,b;c;z)=z(2−F(a,b;c;z)) is in 𝒮p𝒯(α,β) if and only if (2.6)Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)(1+(1+α)ab(1−β)(c−a−b−1))≤2.Proof.
(i) Since(2.7)zF(a,b;c;z)=z+abc∑n=2∞(a+1)n−2(b+1)n−2(c+1)n−2(1)n−1zn=z−|abc|∑n=2∞(a+1)n−2(b+1)n−2(c+1)n−2(1)n−1zn, according to (i) of Lemma 2.1,
we must show that (2.8)∑n=2∞(n(1+α)−(α+β))(a+1)n−2(b+1)n−2(c+1)n−2(1)n−1≤|cab|(1−β). Noting that (λ)n=λ(λ+1)n−1 and then applying (1.6), we
have (2.9)∑n=0∞((n+2)(1+α)−(α+β))(a+1)n(b+1)n(c+1)n(1)n+1=(1+α)∑n=0∞(a+1)n(b+1)n(c+1)n(1)n+(1−β)cab∑n=1∞(a)n(b)n(c)n(1)n=(1+α)Γ(c+1)Γ(c−a−b−1)Γ(c−a)Γ(c−b)+(1−β)cab(Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)−1). Hence, (2.8) is equivalent
to (2.10)Γ(c+1)Γ(c−a−b−1)Γ(c−a)Γ(c−b)(1+α+(1−β)c−a−b−1ab)≤(1−β)(c|ab|+cab)=0. Thus, (2.10) is valid if and only
if 1+α+(1−β)(c−a−b−1)/(ab)≤0 or, equivalently, c≥a+b+1−(1+α)ab/(1−β).
(ii) Since(2.11)F1(a,b;c;z)=z−∑n=2∞(a)n−1(b)n−1(c)n−1(1)n−1zn, by (i) of Lemma 2.1, we need
only to show that (2.12)∑n=2∞(n(1+α)−(α+β))(a)n−1(b)n−1(c)n−1(1)n−1≤1−β. Now, (2.13)∑n=2∞(n(1+α)−(α+β))(a)n−1(b)n−1(c)n−1(1)n−1=(1+α)∑n=1∞n(a)n(b)n(c)n(1)n(1−β)∑n=1∞(a)n(b)n(c)n(1)n=(1+α)abc∑n=1∞(a+1)n−1(b+1)n−1(c+1)n−1(1)n−1+(1−β)∑n=1∞(a)n(b)n(c)n(1)n=Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)((1+α)abc−a−b−1+1−β)−(1−β). But this last expression is
bounded above by 1−β if and only if (2.6) holds.Theorem 2.4.
(i) Ifa,b>−1,ab<0, and c>a+b+2,
then zF(a,b;c;z) is in 𝒰𝒞𝒯(α,β) if and only if (2.14)(1+α)(a)2(b)2+(3+2α−β)ab(c−a−b−2)+(1−β)(c−a−b−2)2≥0.
(ii) Ifa,b>0 and c>a+b+2,
then F1(a,b;c;z)=z(2−F(a,b;c;z)) is in 𝒰𝒞𝒯(α,β) if and only if (2.15)Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)((1+α)(a)2(b)2(1−β)(c−a−b−2)2+(3+2α−β1−β)(abc−a−b−1)+1)≤2.Proof.
(i) SincezF has the form (2.7), we see from (ii) of Lemma
2.1 that our conclusion is equivalent to (2.16)∑n=2∞n(n(1+α)−(α+β))(a+1)n−2(b+1)n−2(c+1)n−2(1)n−1≤c|ab|(1−β). Writing (n+2)((n+2)(1+α)−(α+β))=(1+α)(n+1)2+(2+α−β)(n+1)+(1−β),
we see that (2.17)∑n=0∞(n+2)((n+2)(1+α)−(α+β))(a+1)n(b+1)n(c+1)n(1)n+1=(1+α)∑n=0∞(n+1)(a+1)n(b+1)n(c+1)n(1)n+(2+α−β)∑n=0∞(a+1)n(b+1)n(c+1)n(1)n+(1−β)∑n=0∞(a+1)n(b+1)n(c+1)n(1)n+1=(1+α)(a+1)(b+1)c+1∑n=0∞(a+2)n(b+2)n(c+2)n(1)n+(3+2α−β)∑n=0∞(a+1)n(b+1)n(c+1)n(1)n+(1−β)cab∑n=1∞(a)n(b)n(c)n(1)n=Γ(c+1)Γ(c−a−b−2)Γ(c−a)Γ(c−b)((1+α)(a+1)(b+1)+(3+2α−β)(c−a−b−2)+1−βab(c−a−b−2)2)−(1−β)cab. This last expression is bounded
above by (1−β)c/|ab| if and only if (2.18)(1+α)(a+1)(b+1)+(3+2α−β)(c−a−b−2)+1−βab(c−a−b−2)2≤0, which is equivalent to (2.14).
(ii) In view of (ii) of Lemma2.1, we need only to
show that (2.19)∑n=2∞n(n(1+α)−(α+β))(a)n−1(b)n−1(c)n−1(1)n−1≤1−β. Now, (2.20)∑n=0∞(n+2)((n+2)(1+α)−(α+β))(a)n+1(b)n+1(c)n+1(1)n+1=(1+α)∑n=0∞(n+2)2(a)n+1(b)n+1(c)n+1(1)n+1−(α+β)∑n=0∞(n+2)(a)n+1(b)n+1(c)n+1(1)n+1. Writing n+2=(n+1)+1,
we have (2.21)∑n=0∞(n+2)(a)n+1(b)n+1(c)n+1(1)n+1=∑n=0∞(a)n+1(b)n+1(c)n+1(1)n+∑n=0∞(a)n+1(b)n+1(c)n+1(1)n+1,∑n=0∞(n+2)2(a)n+1(b)n+1(c)n+1(1)n+1=∑n=0∞(n+1)(a)n+1(b)n+1(c)n+1(1)n+2∑n=0∞(a)n+1(b)n+1(c)n+1(1)n+∑n=0∞(a)n+1(b)n+1(c)n+1(1)n+1=∑n=1∞(a)n+1(b)n+1(c)n+1(1)n−1+3∑n=0∞(a)n+1(b)n+1(c)n+1(1)n+∑n=1∞(a)n(b)n(c)n(1)n. Substituting (2.21) into the right-hand side of (2.20), we obtain (2.22)(1+α)∑n=0∞(a)n+2(b)n+2(c)n+2(1)n+(3+2α−β)∑n=0∞(a)n+1(b)n+1(c)n+1(1)n+(1−β)∑n=0∞(a)n+1(b)n+1(c)n+1(1)n+1. Since (a)n+k=(a)k(a+k)n,
we write (2.22) as (2.23)(1+α)(a)2(b)2(c)2Γ(c+2)Γ(c−a−b−2)Γ(c−a)Γ(c−b)+(3+2α−β)abcΓ(c+1)Γ(c−a−b−1)Γ(c−a)Γ(c−b)+(1−β)(Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)−1). By simplification, we see that
the last expression is bounded above by 1−β if and only if (2.15) holds.Theorem 2.5.
(i) Ifa,b>−1,c>0, and ab<0,
then zF(a,b;c;z) is in 𝒫𝒯(α) if and only if (2.24)c≥a+b+1−abα.
(ii) Ifa,b>0 and c>a+b+1,
then F1(a,b;c;z)=z(2−F(a,b;c;z)) is in 𝒫𝒯(α) if and only if (2.25)Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)(1+abα(c−a−b−1))≤2.Proof.
(i) Since(2.26)zF(a,b;c;z)=z+abc∑n=2∞(a+1)n−2(b+1)n−2(c+1)n−2(1)n−1zn=z−|ab|c∑n=2∞(a+1)n−2(b+1)n−2(c+1)n−2(1)n−1zn, according to (i) of Lemma 2.2,
we must show that (2.27)∑n=2∞(n−1+α)(a+1)n−2(b+1)n−2(c+1)n−2(1)n−1≤c|ab|α. Noting that (λ)n=λ(λ+1)n−1 and then applying (1.6), we
have (2.28)∑n=0∞(n+1+α)(a+1)n(b+1)n(c+1)n(1)n+1=∑n=0∞(a+1)n(b+1)n(c+1)n(1)n+αcab∑n=1∞(a)n(b)n(c)n(1)n=Γ(c+1)Γ(c−a−b−1)Γ(c−a)Γ(c−b)+αcab(Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)−1). Hence, (2.27) is equivalent
to (2.29)Γ(c+1)Γ(c−a−b−1)Γ(c−a)Γ(c−b)(1+αc−a−b−1ab)≤α(c|ab|−cab)=0. Thus, (2.29) is valid if and only if 1+α(c−a−b−1)/ab≤0 or, equivalently, c≥a+b+1−ab/α.
(ii) Since(2.30)F1(a,b;c;z)=z−∑n=2∞(a)n−1(b)n−1(c)n−1(1)n−1zn, by (i) of Lemma 2.2, we need
only to show that (2.31)∑n=2∞(n−1+α)(a)n−1(b)n−1(c)n−1(1)n−1≤α. Now, (2.32)∑n=2∞(n−1+α)(a)n−1(b)n−1(c)n−1(1)n−1=∑n=1∞n(a)n(b)n(c)n(1)n+α∑n=1∞(a)n(b)n(c)n(1)n=abc∑n=1∞(a+1)n−1(b+1)n−1(c+1)n−1(1)n−1+α∑n=1∞(a)n(b)n(c)n(1)n=Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)(abc−a−b−1+α)−α. But this last expression is
bounded above by α if and only if (2.25) holds.Theorem 2.6.
(i) Ifa,b>−1,ab<0, and c>a+b+2,
then zF(a,b;c;z) is in 𝒞𝒫𝒯(α) if and only if (2.33)(a)2(b)2+(2+α)ab(c−a−b−2)+α(c−a−b−2)2≥0.
(ii) Ifa,b>0 and c>a+b+2,
then F1(a,b;c;z)=z(2−F(a,b;c;z)) is in 𝒞𝒫𝒯(α) if and only if (2.34)Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)((a)2(b)2α(c−a−b−2)2+(2+αα)(abc−a−b−1)+1)≤2.Proof.
(i) SincezF has the form (2.26), we see from (ii) of Lemma
2.2 that our conclusion is equivalent to (2.35)∑n=2∞n(n−1+α)(a+1)n−2(b+1)n−2(c+1)n−2(1)n−1≤c|ab|α. Writing (n+2)(n+1+α)=(n+1)2+(1+α)(n+1)+α,
we see that (2.36)∑n=0∞(n+2)(n+1+α)(a+1)n(b+1)n(c+1)n(1)n+1=∑n=0∞(n+1)(a+1)n(b+1)n(c+1)n(1)n+(1+α)∑n=0∞(a+1)n(b+1)n(c+1)n(1)n+α∑n=0∞(a+1)n(b+1)n(c+1)n(1)n+1=(a+1)(b+1)c+1∑n=0∞(a+2)n(b+2)n(c+2)n(1)n+(2+α)∑n=0∞(a+1)n(b+1)n(c+1)n(1)n+αcab∑n=1∞(a)n(b)n(c)n(1)n=Γ(c+1)Γ(c−a−b−2)Γ(c−a)Γ(c−b)((a+1)(b+1)+(2+α)(c−a−b−2)+αab(c−a−b)2)−αcab. This last expression is bounded
above by αc/|ab| if and only if (a+1)(b+1)+(2+α)(c−a−b−2)+(α/ab)(c−a−b−1)2≤0,
which is equivalent to (2.33).
(ii) In view of (ii) of Lemma2.2, we need only to
show that (2.37)∑n=2∞n(n−1+α)(a)n−1(b)n−1(c)n−1(1)n−1≤α. Now, (2.38)∑n=0∞(n+2)(n+2−(1−α))(a)n+1(b)n+1(c)n+1(1)n+1=∑n=0∞(n+2)2(a)n+1(b)n+1(c)n+1(1)n+1−(1−α)∑n=0∞(n+2)(a)n+1(b)n+1(c)n+1(1)n+1. Substituting (2.21) into the right-hand side of (2.38), we obtain (2.39)∑n=0∞(a)n+2(b)n+2(c)n+2(1)n+(2+α)∑n=0∞(a)n+1(b)n+1(c)n+1(1)n+α∑n=1∞(a)n(b)n(c)n(1)n. Since (a)n+k=(a)k(a+k)n,
we may write (2.39) as (2.40)(a)2(b)2(c)2Γ(c+2)Γ(c−a−b−2)Γ(c−a)Γ(c−b)+(2+α)abcΓ(c+1)Γ(c−a−b−1)Γ(c−a)Γ(c−b)+α(Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)−1). By simplification, we see that
the last expression is bounded above by α if and only if (2.34) holds.


## 3. An Integral Operator

In the next
theorems, we obtain similar-type results in connection with a particular
integral operatorG(a,b;c;z) acting on F(a,b;c;z) as follows: (3.1)G(a,b;c;z)=∫0zF(a,b;c;t)dt.Theorem 3.1.
Leta,b>−1,ab<0, and c>max{0,a+b}.
Then,
(i)
G(a,b;c;z) defined by (3.1) is in 𝒮p𝒯(α,β) if and only if (3.2)Γ(c+1)Γ(c−a−b)Γ(c−a)Γ(c−b)((1+α)ab−(α+β)(c−a−b)(a−1)2(b−1)2)+(α+β)(c−1)2(a−1)2(b−1)2≤0;(ii)
G(a,b;c;z) defined by (3.1) is in 𝒫𝒯(α) if and only if (3.3)Γ(c+1)Γ(c−a−b)Γ(c−a)Γ(c−b)(1ab+(α−1)(c−a−b)(a−1)2(b−1)2)−(α−1)(c−a)2(a−1)2(b−1)2≤0.Proof.
(i) Since(3.4)G(a,b;c;z)=z−|ab|c∑n=2∞(a+1)n−2(b+1)n−2(c+1)n−2(1)nzn, by (i) of Lemma 2.1, we need
only to show that (3.5)∑n=2∞(n(1+α)−(α+β))(a+1)n−2(b+1)n−2(c+1)n−2(1)n≤(1−β)c|ab|. Now, (3.6)∑n=0∞((n+2)(1+α)−(α+β))(a+1)n(b+1)n(c+1)n(1)n+2=(1+α)∑n=0∞(a+1)n(b+1)n(c+1)n(1)n+1−(α+β)cab∑n=1∞(a)n(b)n(c)n(1)n+1=Γ(c+1)Γ(c−a−b)Γ(c−a)Γ(c−b)(1+αab−(α+β)(c−a−b)(a−1)2(b−1)2)+(α+β)(c−1)2(a−1)2(b−1)2−(1−β)cab≤(1−β)c|ab|, which is equivalent to (3.2).
(ii) According to (i) of Lemma2.2, it is sufficient
to show that (3.7)∑n=2∞(n−1+α)(a+1)n−2(b+1)n−2(c+1)n−2(1)n≤αc|ab|. Now, (3.8)∑n=0∞(n+1+α)(a+1)n(b+1)n(c+1)n(1)n+2=∑n=0∞(a+1)n(b+1)n(c+1)n(1)n+1+(α−1)∑n=0∞(a+1)n(b+1)n(c+1)n(1)n+2=cab(Γ(c)Γ(c−a−b)Γ(c−a)Γ(c−b)−1)+(α−1)cab((c−1)(a−1)(b−1)(Γ(c−1)Γ(c−a−b+1)Γ(c−a)Γ(c−b)−1)−1)=Γ(c+1)Γ(c−a−b)Γ(c−a)Γ(c−b)(1ab+(α−1)(c−a−b)(a−1)2(b−1)2)−(α−1)(c−1)2(a−1)2(b−1)2−αcab≤αc|ab|, which is equivalent to (3.3).
Now, we observe
thatG(a,b;c;z)∈𝒰𝒞𝒯(α,β)(𝒞𝒫𝒯(α)) if and only if zF(a,b;c;z)∈𝒮p𝒯(α,β)(𝒫𝒯(α)). Thus, any result of functions belonging to the class 𝒮p𝒯(α,β)(𝒫𝒯(α)) about zF leads to that of functions belonging to the
class 𝒰𝒞𝒯(α,β)(𝒞𝒫𝒯(α)).
Hence, we obtain the following analogues to Theorems 2.3 and 2.5.Theorem 3.2.
Leta,b>−1,ab<0, and c>a+b+2.
Then,
(i)
G(a,b;c;z) defined by (3.1) is in 𝒰𝒞𝒯(α,β) if and only if (3.9)c≥a+b+1−(1+α)ab(1−β);(ii)
G(a,b;c;z) defined by (3.1) is in 𝒞𝒫𝒯(α) if and only if (3.10)c≥a+b+1−abα.



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