# Subclasses of Analytic Functions Defined by Generalized Hypergeometric Functions

Authors: Badr S. Alkahtani; Saima Mustafa; Teodor Bulboacă
Journal: Journal of Function Spaces
 (2016)

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## Abstract

We introduce a new subclass of analytic functions in the unit diskU, defined by using the generalized hypergeometric functions, which extends some previous well-known classes defined by different authors. Inclusion results, radius problems, and some connections with the Bernardi-Libera-Livingston integral operator are discussed.


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## Body

## 1. Introduction

LetA denote the class of functions f of the form (1)fz=z+∑k=1∞ak+1zk+1,z∈U,which are analytic in the unit disk U.Theconvolution (or theHadamard product) of two functions f and g, where f is given by (1) and g(z)=z+∑k=1∞‍bk+1zk+1, z∈U, is defined as(2)f∗gz=z+∑k=1∞ak+1bk+1zk+1,z∈U.For the complex parametersαi with βj∈C∖Z0-, where Z0-=0,-1,-2,…, i∈1,2,…,q, and j∈1,2,…,s, we define thegeneralized hypergeometric functionFsq(α1,…,αq;β1,…,βs;z) as follows (see [1, 2]): (3)Fsqα1,…,αq;β1,…,βs;z=∑k=0∞α1k⋯αqkβ1k⋯βskzkk!,z∈U,q≤s+1;q,s∈N0=N∪0;N=1,2,…,where (λ)k (or theshifted factorial) is defined by (4)λ0=1,1k=k!,k∈N0,λk=λλ+1⋯λ+k-1,k∈N.Using the functionF(α1,…,αq;β1,…,βs;z) defined by(5)Fα1,…,αq;β1,…,βs;z=z-m·Fsqα1,…,αq;β1,…,βs;z,Liu and Srivastava [3] introduced and studied the properties of the linear operator H(α1,…,αq;β1,…,βs), defined by the Hadamard (or convolution) product(6)Hα1,…,αq;β1,…,βsfz=Fα1,…,αq;β1,…,βs;z∗fz,where the function f is analytic and m-valent in the punctured unit disk U˙=U∖0 and has the form f(z)=z-m+∑k=1∞‍akzk-m. Note that linear operator H(α1,…,αq;β1,…,βs) was motivated essentially by the work of Dziok and Srivastava [4].Corresponding to the functionF(α1,…,αq;β1,…,βs;z) defined by(7)Fα1,…,αq;β1,…,βs;z=z·Fsqα1,…,αq;β1,…,βs;z,we introduce a function Fp(α1,…,αq;β1,…,βs;z) given by(8)Fα1,…,αq;β1,…,βs;z∗Fpα1,…,αq;β1,…,βs;z=z1-zp,where p>0.Analogous toH(α1,…,αq;β1,…,βs) defined by (6), we now define the linear operator Hp(α1,…,αq;β1,…,βs) on A as follows:(9)Hpα1,…,αq;β1,…,βsfz=Fpα1,…,αq;β1,…,βs;z∗fz,αj,βj∈C∖Z0-;i∈1,2,…,q;j∈1,2,…,s;p>0.For convenience, we write(10)Hp,q,sα1≔Hpα1,…,αq;β1,…,βs.Remark 1.
The linear operatorHp(α1,…,αq;β1,…,βs) includes various other linear operators which were considered in some earlier works:(i)
In particular, forp=1, q=2, and s=1, we obtain the linear operator H1(α1,α2;β1) which was defined by Hohlov [5].(ii)
Moreover, puttingp=1, q=2, s=1, and α2=1, we obtain the well-known Carlson-Shaffer operator L(α1,β1)=H1(α,1;β1) (see [6, 7]).From definitions (8) and (9), using notation (10), it is easy to prove the differentiation formula(11)zHp,q,sλfz′=pHp+1,q,sλfz-p-1Hp,q,sλfz.We note that the operatorHp,q,s(λ) is closely related to the Choi-Saigo [8] operator for analytic functions, which includes the integral studied by Liu [9] and K. I. Noor and M. A. Noor [10].LetPkρ be the class of functions h, analytic in the unit disk U, satisfying the condition h(0)=1 and(12)∫02πRe⁡hz-ρ1-ρdθ≤kπ,where z=reiθ, k≥2, and 0≤ρ<1. This class was introduced in [11], and as a special case we note that the class Pk(0) was defined by Pinchuk in [12]. Moreover, P(ρ)≔P2(ρ) is the class of analytic functions with the real part greater than ρ.Remark 2.
(i) Like in [13, 14], it can easily be seen that the function h, analytic in U, with h(0)=1, belongs to Pk(ρ) if and only if there exists the functions h1,h2∈P(ρ) such that(13)hz=k4+12h1z-k4-12h2z,z∈U.
(ii) It is known from [15] that the class Pk(ρ) is a convex set.We will assume throughout our discussions, unless otherwise stated, thatαi,βj∈C∖Z0-, i∈1,2,…,q, j∈1,2,…,s, with q≤s+1, p>0, and all the powers representthe principal branches; that is, log⁡1=0.Using the linear operatorHp,q,s(λ), we will define the following classes of analytic functions.Definition 3.
Let0≤ρ<1, c>0, and let b be a complex number such that Re⁡b≥0. A function f∈A is said to be in the class Hp,q,sb(λ,ρ,c,δ,k) if and only if(14)1-bHp,q,sλfzHp,q,sλgzc+bHp+1,q,sλfzHp+1,q,sλgzHp,q,sλfzHp,q,sλgzc-1∈Pkρ,where k≥2 and g∈A satisfies the condition(15)βz=Hp+1,q,sλgzHp,q,sλgz∈Pδ,0≤δ<1.Remark 4.
From the above definition, the following subclasses ofA emerge as special cases:(i)
Forp=q=s=1, k=2, b=0, c=1, g(z)=z, and λ=1, we have(16)H1,1,101,ρ,1,0,2=f∈A:fzz∈Pρ,0≤ρ<1,and this class was studied by Chen [16].(ii)
Whenp=q=s=1, g(z)=z, and λ=1, then H1,1,1b(1,ρ,c,δ,k) reduces to the class studied by Noor [13].(iii)
Forp=q=s=1, b=1, k=2, g(z)=z, and λ=1, we obtain the class(17)H1,1,111,ρ,c,0,2=f∈A:f′zfzzc-1∈Pρ,0≤ρ<1,which was studied by Ponnusamy and Karunakaran [17].We will use the following lemmas to prove our main results.Lemma 5 (see [18]).
Ifh is an analytic function in U, with h(0)=1, and if λ1 is a complex number satisfying Re⁡λ1≥0, λ1≠0, then(18)Re⁡hz+λ1zh′z>ρ,z∈U,0≤ρ<1implies(19)Re⁡hz>γ,z∈U,where γ is given by(20)γ=ρ+1-ρ2γ1-1,γ1=∫011+tRe⁡λ1-1dt,and γ1 is an increasing function of Re⁡λ1, and 1/2≤γ1<1. The estimate is sharp in the sense that the bound cannot be improved.Lemma 6 (see [19]).
Letu=u1+iu2, v=v1+iv2, and let ψ(u,v) be a complex-valued function satisfying the following conditions: (i)
ψ(u,v) is continuous in a domain D⊂C2.(ii)
(1,0)∈D and Re⁡ψ(1,0)>0.(iii)
Re⁡ψ(iu2,v1)≤0, whenever (iu2,v1)∈D and v1≤-(1/2)1+u22.
Ifp(z)=1+∑m=1∞‍cmzm is an analytic function in U, such that (p(z),zp′(z))∈D for all z∈U, and Re⁡ψ(p(z),zp′(z))>0 for all z∈U, then Re⁡p(z)>0 for all z∈U.In this paper, we investigate several properties of the classHp,q,sb(λ,ρ,c,δ,k), like inclusion results and radius problems; moreover, a connection with the Bernardi-Libera-Livingston integral operator is also discussed.


## 2. Main Results

Theorem 7.
Ifb≥0, then Hp,q,sb(λ,ρ,c,δ,k)⊂Hp,q,s0(λ,ρ,c,δ,k).Proof.
Let an arbitrary functionf∈Hp,q,sb(λ,ρ,c,δ,k), and denote(21)hz≔Hp,q,sλfzHp,q,sλgzc,where h is analytic in U, with h(0)=1, and g satisfies condition (15). From part (i) of Remark 2, we have that f∈Hp,q,s0(λ,ρ,c,δ,k), if and only if(22)hz=k4+12h1z-k4-12h2z,where h1,h2∈P(ρ).
Using the differentiation formula (11) together with (15), after an elementary computation, we obtain(23)1-bHp,q,sλfzHp,q,sλgzc+bHp+1,q,sλfzHp+1,q,sλgzHp,q,sλfzHp,q,sλgzc-1=hz+bzh′zpcβz,where β is given by (15).
Now, using the representation formula (22), we have(24)hz+bzh′zpcβz=k4+12h1z+bzh1′zpcβz-k4-12h2z+bzh2′zpcβz.
Sincef∈Hp,q,sb(λ,ρ,c,δ,k), from relations (23) and (24), it follows that(25)hiz+bzhi′zpcβz∈Pρ,i=1,2,and using the substitutionHi(z)≔hi(z)-ρ, i=1,2, the above relation becomes(26)Hiz+bzHi′zpcβz∈P0,i=1,2.
To prove our result, we need to show that (26) implies Hi∈P(0), i=1,2. We will define the functional ψ(u,v) by taking u=Hi(z), and v=zHi′(z), and thus we have(27)ψu,v=u+bvpcβz.
It is easy to see that the first two conditions of Lemma6 are satisfied; hence, we proceed to verify condition (iii). Since β∈P(δ), that is, Re⁡β(z)>δ for all z∈U, 0≤δ<1, it follows that(28)Re⁡ψiu2,v1=Re⁡bv1pcβz≤-b1+u22δ2pcβz2≤0,whenever v1≤-(1/2)1+u22. Using Lemma 6, we conclude that Hi∈P(0), for i=1,2, which completes our proof.Theorem 8.
If0≤b1<b2, then Hp,q,sb2(λ,ρ,c,δ,k)⊂Hp,q,sb1(λ,ρ,c,δ,k).Proof.
If we consider an arbitrary functionf∈Hp,q,sb2(λ,ρ,c,δ,k), then φ2∈Pk(ρ), where(29)φ2z≔1-b2Hp,q,sλfzHp,q,sλgzc+b2Hp+1,q,sλfzHp+1,q,sλgzHp,q,sλfzHp,q,sλgzc-1.
According to Theorem7, we have(30)φ1z≔Hp,q,sλfzHp,q,sλgzc∈Pkρ,and a simple computation shows that(31)1-b1Hp,q,sλfzHp,q,sλgzc+b1Hp+1,q,sλfzHp+1,q,sλgzHp,q,sλfzHp,q,sλgzc-1=1-b1b2φ1z+b1b2φ2z.Since the class Pk(ρ) is a convex set (see (ii) from Remark 2), it follows that the right-hand side of (31) belongs to Pk(ρ) for 0≤b1<b2, which implies that f∈Hp,q,sb1(λ,ρ,c,δ,k).Now, let us define the operatorJc:A→A by(32)Jcfz=c+1zc∫0ztc-1ftdtc>-1.For c∈N, the operator Jc was introduced by Bernardi [20], while the special case J1 was previously studied by Libera [21] and Livingston [22].Theorem 9.
Iff∈A, Jc(f) is given by (32), and b∈C with Re⁡b>0, then(33)1-bHp,q,sλJcfzz+bHp,q,sλfzz∈Pkρimplies that(34)Hp,q,sλJcfzz∈Pkγ,where γ is given by (20), with λ1=b/(c+1).Proof.
Differentiating relation (32), we have(35)zJcfz′=c+1fz-cJcfz,and using definition (9), this implies(36)zHp,q,sλJcfz′=c+1Hp,q,sλfz-cHp,q,sλJcfz.If we let(37)hz≔Hp,q,sλJcfzz,according to part (i) of Remark 2, we need to prove that h is of the form(38)hz=k4+12h1z-k4-12h2z,where h1,h2∈P(γ).
Using (36), from the above relation, we have (39)1-bHp,q,sλJcfzz+bHp,q,sλfzz=hz+bzh′zc+1=k4+12h1z+bzh1′zc+1-k4-12h2z+bzh2′zc+1∈Pkρ.Thus, from part (i) of Remark 2, it follows that(40)hiz+bzhi′zc+1∈Pρ,i=1,2,and from Lemma 5, we conclude that hi∈P(γ), i=1,2, with γ given by (20) and λ1=b/(c+1).The next result deals with the converse of Theorem7.Theorem 10.
Letb>0 and 0<δ<1. If f∈Hp,q,s0(λ,ρ,c,δ,k), then f∈Hp,q,sb(λ,ρ,c,δ,k) for z<R, where(41)R=min⁡-b+b2+p2c2δ2pcδ;pcρδ-b+pcρδ-b2+p2c2δ21-ρ2pcδ1-ρ.Proof.
For arbitraryf∈Hp,q,s0(λ,ρ,c,δ,k), let us define the function h as in (43). Thus, it follows that(42)Hp,q,sλfzHp,q,sλgzc=hz∈Pkρ,and g∈A satisfies the condition(43)βz=Hp+1,q,sλgzHp,q,sλgz∈Pδ.
From part (i) of Remark2, we have that (42) holds if and only if(44)hz=k4+12h1z-k4-12h2z,where h1,h2∈P(ρ).
Using the above representation formula, similar to the proof of Theorem7, we deduce that (45)1-bHp,q,sλfzHp,q,sλgzc+bHp+1,q,sλfzHp+1,q,sλgzHp,q,sλfzHp,q,sλgzc-1=k4+12h1z+bzh1′zpcβz-k4-12h2z+bzh2′zpcβz,and substituting Hi(z)≔hi(z)-ρ, i=1,2, we finally obtain (46)1-bHp,q,sλfzHp,q,sλgzc+bHp+1,q,sλfzHp+1,q,sλgzHp,q,sλfzHp,q,sλgzc-1=k4+12H1z+ρ+bzH1′zpcβz-k4-12H2z+ρ+bzH2′zpcβz,where H1,H2∈P(0).
To prove our result, we need to determine the value ofR, such that(47)Re⁡Hiz+ρ+bzHi′zpcβz>0,forz<R,i=1,2,wheneverH1,H2∈P(0).
Using the well-known estimates for the classP(0) [23], that is, (48)Hi′z≤2Re⁡Hiz1-r2,z≤r<1,i=1,2,Re⁡Hiz≥1-r1+r,z≤r<1,i=1,2,and according to (43), we obtain(49)Re⁡Hiz+ρ+bzHi′zpcβz≥ρ+Re⁡Hiz-bpczHi′zβz≥ρ+Re⁡Hiz-bpczHi′zδ≥ρ+Re⁡Hiz1-2bpcδr1-r2,for all z≤r<1 and i=1,2.
A simple computation shows that1-2b/pcδr/1-r2≥0 (0≤r<1) if and only if(50)r∈0,-b+b2+p2c2δ2pcδ.Assuming that (50) holds, from (49), we deduce that(51)Re⁡Hiz+ρ+bzHi′zpcβz≥ρ+1-r1+r1-2bpcδr1-r2,z≤r<1,for i=1,2. It is easy to see that the right-hand side of the above inequality is greater than or equal to zero if and only if(52)r∈0,min⁡1;pcρδ-b+pcρδ-b2+p2c2δ21-ρ2pcδ1-ρ,and combining this with (50), we obtain our result.Remark 11.
We note the following special case obtained from the above result: forρ=0, formula (41) reduces to(53)R=-b+b2+p2c2δ2pcδ.



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