We introduce a new subclass of analytic functions in the unit disk U, 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.

1. Introduction

Let A 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.

The convolution (or the Hadamard 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 parameters αi with βj∈C∖Z0-, where Z0-=0,-1,-2,…, i∈1,2,…,q, and j∈1,2,…,s, we define the generalized hypergeometric function Fsq(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 the shifted factorial) is defined by (4)λ0=1,1k=k!,k∈N0,λk=λλ+1⋯λ+k-1,k∈N.

Using the function F(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 function F(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 to H(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 operator Hp(α1,…,αq;β1,…,βs) includes various other linear operators which were considered in some earlier works:

In particular, for p=1, q=2, and s=1, we obtain the linear operator H1(α1,α2;β1) which was defined by Hohlov [5].

Moreover, putting p=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 operator Hp,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].

Let Pkρ be the class of functions h, analytic in the unit disk U, satisfying the condition h(0)=1 and(12)∫02πRehz-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, that αi,βj∈C∖Z0-, i∈1,2,…,q, j∈1,2,…,s, with q≤s+1, p>0, and all the powers represent the principal branches; that is, log1=0.

Using the linear operator Hp,q,s(λ), we will define the following classes of analytic functions.

Definition 3.

Let 0≤ρ<1, c>0, and let b be a complex number such that Reb≥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 of A emerge as special cases:

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].

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].

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 [<xref ref-type="bibr" rid="B20">18</xref>]).

If h is an analytic function in U, with h(0)=1, and if λ1 is a complex number satisfying Reλ1≥0, λ1≠0, then(18)Rehz+λ1zh′z>ρ,z∈U,0≤ρ<1implies(19)Rehz>γ,z∈U,where γ is given by(20)γ=ρ+1-ρ2γ1-1,γ1=∫011+tReλ1-1dt,and γ1 is an increasing function of Reλ1, and 1/2≤γ1<1. The estimate is sharp in the sense that the bound cannot be improved.

Lemma 6 (see [<xref ref-type="bibr" rid="B12">19</xref>]).

Let u=u1+iu2, v=v1+iv2, and let ψ(u,v) be a complex-valued function satisfying the following conditions:

ψ(u,v) is continuous in a domain D⊂C2.

(1,0)∈D and Reψ(1,0)>0.

Reψ(iu2,v1)≤0, whenever (iu2,v1)∈D and v1≤-(1/2)1+u22.

If p(z)=1+∑m=1∞cmzm is an analytic function in U, such that (p(z),zp′(z))∈D for all z∈U, and Reψ(p(z),zp′(z))>0 for all z∈U, then Rep(z)>0 for all z∈U.

In this paper, we investigate several properties of the class Hp,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 ResultsTheorem 7.

If b≥0, then Hp,q,sb(λ,ρ,c,δ,k)⊂Hp,q,s0(λ,ρ,c,δ,k).

Proof.

Let an arbitrary function f∈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.

Since f∈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 Lemma 6 are satisfied; hence, we proceed to verify condition (iii). Since β∈P(δ), that is, Reβ(z)>δ for all z∈U, 0≤δ<1, it follows that(28)Reψiu2,v1=Rebv1pcβz≤-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.

If 0≤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 function f∈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 Theorem 7, 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 operator Jc: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.

If f∈A, Jc(f) is given by (32), and b∈C with Reb>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 Theorem 7.

Theorem 10.

Let b>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 arbitrary f∈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 Remark 2, 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 Theorem 7, 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 of R, such that(47)ReHiz+ρ+bzHi′zpcβz>0,for z<R,i=1,2,wheneverH1,H2∈P(0).

Using the well-known estimates for the class P(0) [23], that is, (48)Hi′z≤2ReHiz1-r2,z≤r<1,i=1,2,ReHiz≥1-r1+r,z≤r<1,i=1,2,and according to (43), we obtain(49)ReHiz+ρ+bzHi′zpcβz≥ρ+ReHiz-bpczHi′zβz≥ρ+ReHiz-bpczHi′zδ≥ρ+ReHiz1-2bpcδr1-r2,for all z≤r<1 and i=1,2.

A simple computation shows that 1-2b/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)ReHiz+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,min1;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: for ρ=0, formula (41) reduces to(53)R=-b+b2+p2c2δ2pcδ.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The work of the third author was entirely supported by the grant given by Babeş-Bolyai University, dedicated for Supporting the Excellence Research 2015. The first author, Badr S. Alkahtani, is grateful to King Saud University, Deanship of Scientific Research, College of Science Research Center, for supporting this project.

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