# On Integral Operator Defined by Convolution Involving Hybergeometric Functions

Authors: K. Al-Shaqsi; M. Darus
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008-02-03)

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## Abstract

Forλ>−1 and μ≥0, we consider a liner operator Iλμ on the class 𝒜 of analytic functions in the unit disk defined by the convolution (fμ)(−1)∗f(z), where fμ=(1−μ)z2F1(a,b,c;z)+μz(z2F1(a,b,c;z))', and introduce a certain new subclass of 𝒜 using this operator. Several interesting properties of these classes are obtained.


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## Body

## 1. Introduction

Let𝒜 denote the
class of functions of the form
(1.1)f(z)=z+∑k=2∞akzk,
which are
analytic in the unit disk 𝕌={z:|z|<1}.Iff(z)∈𝒜 satisfies
(1.2)|arg(zf′(z)f(z)−α)|<π2β(z∈𝕌,0≤α<1,0<β≤1),
then f(z) is said to be
strongly starlike of order β and type α in 𝕌, and denoted by S*(α,β).Iff(z)∈𝒜 satisfies
(1.3)|arg(1+zf′′(z)f′(z)−α)|<π2β(z∈𝕌,0≤α<1,0<β≤1),
then f(z) is said to be
strongly convex of order β and type α in 𝕌, and denoted by C(α,β). It is obvious that f(z)∈𝒜 belongs to C(α,β) if and if zf′(z)∈S*(α,β). Further, we note that S*(α,1)≡S*(α) and C(α,1)≡C(α) which are,
respectively, starlike and convex univalent functions of order α.LetP denote the
class of functions of the form p(z)=1+p1z+… analytic in 𝕌 which satisfy
the condition Re{P(z)}>0.For functionsf given by (1.1)
and g(z)=z+∑k=2∞bkzk, let (f*g)(z) denote the
Hadamard product (or convolution) of f(z) and g(z), defined by
(1.4)(f*g)(z)=f(z)*g(z)=z+∑k=2∞akbkzk.
If f and g are analytic in 𝕌, we say that f is subordinate
to g, written f≺g or f(z)≺g(z), if there exists a Schwarz function w in 𝕌 such that f(z)=g(w(z)) [1].Letf∈𝒜. Denote by Dλ:𝒜→𝒜 the operator defined by
(1.5)Dλ=z(1−z)λ+1*f(z)(λ>−1).
It is obvious that D0f(z)=f(z), D1f(z)=zf′(z), and
(1.6)Dδf(z)=z(zδ−1f(z))(δ)δ!(δ∈ℕ0=ℕ∪{0}).
The operator Dδf is called the δth-order
Ruscheweyh derivative of f. Recently, K. I. Noor [2] and K. I. Noor and M. A. Noor [3] defined
and studied an integral operator In:𝒜→𝒜, analogous to Dδf(z) as follows.Letfn=z/(1−z)n+1, (n∈ℕ0),  and fn(−1)(z) be defined such that
(1.7)fn(z)*fn(−1)(z)=z(1−z)2.
Then
(1.8)Inf(z)=fn(−1)(z)*f(z)=[z(1−z)n+1](−1)*f(z)(f∈𝒜).
We note that I0f(z)=zf′(z), I1f(z)=f(z). The operator In is called the
Noor integral of nth order of f (see [4, 5]), which is an important tool in defining several classes of analytic functions.
In recent years, it has been shown that Noor integral operator has fundamental
and significant applications in the geometric function theory.For real or complex numbersa, b, c other than 0,−1,−2,…, the hypergeometric series is defined by
(1.9)2F1(a,b;c;z)=∑k=0∞(a)k(b)k(c)k(1)kzk,
where (x)k is Pochhammer symbol defined by
(1.10)(x)k=Γ(x+k)Γ(x)=x(x+1)…(x+k−1)fork=1,2,3,…,x∈ℂ,(x)0=1.We note that the series (1.9) converges absolutely for all z∈𝕌 so that it
represents an analytic function in 𝕌. Also an incomplete beta function ϕ(a,c;z) is related to
Gauss hypergeometric function z2F1(a,b;c;z) as
(1.11)ϕ(a,c;z)=z2F1(1,a;c;z),
and we note that ϕ(a,1;z)=z/(1−z)a, where ϕ(2,1;z) is Koebe
function. Using ϕ(a,c;z), a convolution
operator [6], was defined by Carlson and Shaferr. Furthermore, Hohlov [7]
introduced a convolution operator using F12(a,b;c;z).N. Shukla and P. Shukla [8] studied the mapping properties
of a function fμ to be as given in
(1.12)fμ(a,b,c)(z)=(1−μ)z2F1(a,b,c;z)+μz(z2F1(a,b,c;z))′(μ≥0),
and investigated the geometric properties of an integral operator of the form
(1.13)I(z)=∫0zfμ(t)tdt.Kim and Shon [9] considered linear operator Lμ:𝒜→𝒜 defined by Lμ(a,b,c)f(z)=fμ(a,b,c)(z)*f(z).We now introduce a function(fμ)(−1) given by
(1.14)fμ(a,b,c)(z)*(fμ(a,b,c)(z))(−1)=z(1−z)λ+1(μ≥0,λ>−1),
and obtain the following linear operator:
(1.15)Iμλ(a,b,c)f(z)=(fμ(a,b,c)(z))(−1)*f(z).
The operator Iμλ is known as the
generalized integral operator. For μ=0 in (1.14), Iλ(a,b;c)f(z):=Iμλ(a,b;c)f(z), which was introduced by K. I. Noor [10].Now we find the explicit form of the function(fμ)(−1). It is well known that λ>−1(1.16)z(1−z)λ+1=∑k=0∞(λ+1)kk!zk+1(z∈𝕌).
Putting (1.9) and (1.16) in (1.14), we get
(1.17)∑k=0∞(μk+1)(a)k(b)k(c)k(1)kzk+1*(fμ)(−1)=∑k=0∞(λ+1)kk!zk+1.
Therefore, the
function (fμ)(−1) has the
following form:
(1.18)(fμ(a,b,c)(z))(−1)=∑k=0∞(λ+1)k(c)k(μk+1)(a)k(b)kzk+1(z∈𝕌).Now we note that(1.19)Iμλ(a,b,c)f(z)=z+∑k=1∞(λ+1)k(c)k(μk+1)(a)k(b)kak+1zk+1.From (1.19), we note that
(1.20)I0λ(a,λ+1,a)f(z)=f(z),I01(a,1,a)f(z)=zf′(z).Also it can easily be verified that(1.21)z(Iμλ(a,b,c)f(z))′=(λ+1)Iμλ+1(a,b,c)f(z)−λIμλ(a,b,c)f(z),(1.22)z(Iμλ(a+1,b,c)f(z))′=aIμλ(a,b,c)f(z)−(a−1)Iμλ(a+1,b,c)f(z).
Now we introduce the following classes in term of the new operator Iμλ(a,b,c). For λ>−1, μ≥0, 0≤α<1, and 0<β≤1, let 𝒮μλ(a,b,c;α,β) be the class of
functions f∈𝒜 satisfying
(1.23)|arg(z(Iμλ(a,b,c)f(z))′Iμλ(a,b,c)f(z)−α)|<π2β(z∈𝕌).
Observe that Iμλ(a,b,c)f(z)∈S*(α,β) and z(Iμλ(a,b,c)f(z))′/Iμλ(a,b,c)f(z)≠α. Also, for λ>−1, μ≥0, 0≤α<1, and 0<β≤1, let 𝒞μλ(a,b,c;α,β) be the class of
functions f∈𝒜 satisfying
(1.24)|arg(1+z(Iμλ(a,b,c)f(z))′′(Iμλ(a,b,c)f(z))′−α)|<π2β(z∈𝕌).
Observe that Iμλ(a,b,c)f(z)∈C(α,β) and 1+z(Iμλ(a,b,c)f(z))′′/(Iμλ(a,b,c)f(z))′≠α.Clearly,f∈𝒞μλ(a,b,c;α,β) if and only if zf′(z)∈𝒮μλ(a,b,c;α,β).Note that𝒮0λ(a,λ+1,a;α,β)≡S*(α,β),   𝒮0λ(a,λ+1,a;α,1)≡S*(α), C0λ(a,λ+1,a;α,β)≡C(α,β), and C0λ(a,λ+1,a;α,1)≡C(α).Finally, let𝒦μλ(a,b,c;α,β,γ;A,B) be the class of
functions f∈𝒜 satisfying
(1.25)|arg(z(Iμλ(a,b,c)f(z))′Iμλ(a,b,c)g(z)−α)|<π2β(z∈𝕌)
for λ>−1, μ≥0, 0≤α<1, 0<β≤1, and g∈𝒬μλ(a,b,c;γ;A,B), where
(1.26)𝒬μλ(a,b,c;γ;A,B)={g∈𝒜:11−γ(z(Iμλ(a,b,c)g(z))′Iμλ(a,b,c)g(z)−γ)≺1+Az1+Bz}(z∈𝕌;0≤γ<1,−1≤B<A≤1).
We also note that 𝒦0λ(a,λ+1,a;α,1,γ;1,−1) and 𝒦0λ(a,1,a;α,1,γ;1,−1) are the classes
of quasiconvex and close-to-convex functions of order α and type γ, respectively, introduced and studied by Noor and
Alkhorasani [11] and Silverman [12].


## 2. Main Results

In order to give our results, we need the following lemmas.Lemma 2.1 (see [13]).
Letβ, γ be complex numbers. Let ϕ(z) be convex univalent in 𝕌 with ϕ(0)=1 and Re[βϕ(z)+γ]>0, z∈𝕌 and q∈𝒜 with q(z)≺ϕ(z), z∈𝕌. If p∈P is analytic in 𝕌, then
(2.1)p(z)+zp′(z)βq(z)+γ≺ϕ(z)(z∈𝕌)
implies
(2.2)p(z)≺ϕ(z)(z∈𝕌).Lemma 2.2 (see [14]).
Letδ, η be complex
numbers. Let ϕ(z) be convex
univalent in 𝕌 with ϕ(0)=1 and Re[δϕ(z)+η]>0, z∈𝕌. If p∈P is analytic in 𝕌, then
(2.3)p(z)+zp′(z)δp(z)+η≺ϕ(z)(z∈𝕌)
implies
(2.4)p(z)≺ϕ(z)(z∈𝕌).Lemma 2.3 (see [15]).
Letϕ(z) be convex
univalent in 𝕌 and let E≥0. Suppose B(z) is analytic in 𝕌 with ReB(z)≥E, If g∈P is analytic in 𝕌, then
(2.5)Ez2g′′(z)+B(z)zg′(z)+g(z)≺ϕ(z)(z∈𝕌)
implies
(2.6)g(z)≺ϕ(z)(z∈𝕌).Theorem 2.4.
Letϕ(z) be convex univalent in 𝕌 with ϕ(0)=1 and Reϕ(z)≥0. If f(z)∈𝒜 satisfies the condition
(2.7)11−γ(z(Iμλ+1(a,b,c)f(z))′Iμλ+1(a,b,c)f(z)−γ)≺ϕ(z)(z∈𝕌),
then
(2.8)11−γ(z(Iμλ(a,b,c)f(z))′Iμλ(a,b,c)f(z)−γ)≺ϕ(z)(z∈𝕌)
for λ>−1, μ≥0, and 0≤γ<1.Proof.
Let(2.9)p(z)=11−γ(z(Iμλ(a,b,c)f(z))′Iμλ(a,b,c)f(z)−γ),
where p∈P. By using (1.21) in (2.9) and then differentiating,
we get
(2.10)11−γ(z(Iμλ+1(a,b,c)f(z))′Iμλ+1(a,b,c)f(z)−γ)=p(z)+zp′(z)(λ+1)q(z),
where q(z)=Iμλ+1(a,b,c)f(z)/Iμλ(a,b,c)f(z) and q(z)≺ϕ(z). Hence by applying Lemma 2.1, we obtain the required result.Theorem 2.5.
Letϕ(z) be convex univalent in 𝕌 with ϕ(0)=1 and Reϕ(z)≥0. If f(z)∈𝒜 satisfies the condition
(2.11)11−γ(z(Iμλ(a,b,c)f(z))′Iμλ(a,b,c)f(z)−γ)≺ϕ(z)(z∈𝕌),
then
(2.12)11−γ(z(Iμλ(a+1,b,c)f(z))′Iμλ(a+1,b,c)f(z)−γ)≺ϕ(z)(z∈𝕌)
for λ>−1, μ≥0, and 0≤γ<1.Proof.
By using the same technique in the proof of Theorem2.4 and using (1.22) and
applying Lemma 2.2, we obtain the required result.Takingϕ(z)=(1+Az)/(1+Bz)(−1≤B<A≤1) in Theorem 2.4
and in Theorem 2.5, we have the following.Corollary 2.6.
It holds that(2.13)𝒬μλ+1(a,b,c;γ;A,B)⊂Qμλ(a,b,c;γ;A,B),𝒬μλ(a,b,c;γ;A,B)⊂Qμλ(a+1,b,c;γ;A,B)
for λ>−1, μ≥0, 0≤γ<1, and Rea>1−γ.Also, by takingϕ(z)=((1+z)/(1−z))β(0<β≤1) in Theorem 2.4
and in Theorem 2.5, we have the following.Corollary 2.7.
It holds that(2.14)𝒮μλ+1(a,b,c;γ,β)⊂𝒮μλ(a,b,c;γ,β),𝒮μλ(a,b,c;γ,β)⊂𝒮μλ(a+1,b,c;γ,β)
for λ>−1, μ≥0, 0≤γ<1, and Rea>1−β.Corollary 2.8.
Forλ>−1, μ≥0, 0≤γ<1, and Rea>1−β, one has
(2.15)𝒞μλ+1(a,b,c;γ,β)⊂𝒞μλ(a,b,c;γ,β),𝒞μλ(a,b,c;γ,β)⊂𝒞μλ(a+1,b,c;γ,β).Proof.
We will proof the first relation and by the same method we can proof the second
relation(2.16)f(z)∈𝒞μλ+1(a,b,c;γ,β)⟺zf′(z)∈𝒮μλ+1(a,b,c;γ,β)⟺zf′(z)∈𝒮μλ(a,b,c;γ,β)⟺Iμλ(a,b,c)(zf′(z))∈S*(γ,β)⟺z(Iμλ(a,b,c)f(z))′∈S*(γ,β)⟺Iμλ(a,b,c)f(z)∈C(γ,β)⟺f(z)∈𝒞μλ(a,b,c;γ,β).Theorem 2.9.
Letϕ(z) be convex
univalent in 𝕌 with ϕ(0)=1 and Reϕ(z)≥0. If f(z)∈𝒜 satisfies the condition
(2.17)11−γ(z(Iμλ(a,b,c)f(z))′Iμλ(a,b,c)f(z)−γ)≺ϕ(z)(0≤γ<1;z∈𝕌),
then
(2.18)11−γ(z(Iμλ(a,b,c)F(z))′Iμλ(a,b,c)F(z)−γ)≺ϕ(z)(0≤γ<1;z∈𝕌),
where F be the integral operator defined by
(2.19)F(z)=c+1zc∫0ztc−1f(t)dt(c>−1).Proof.
From (2.19), we have
(2.20)z(Iμλ(a,b,c)F(z))′=(c+1)Iμλ(a,b,c)f(z)−cIμλ(a,b,c)F(z).
Now, let
(2.21)p(z)=11−γ(z(Iμλ(a,b,c)F(z))′Iμλ(a,b,c)F(z)−γ),
where p∈P. Then by using (2.20), we get
(2.22)(1−γ)p(z)+c+γ=(c+1)Iμλ(a,b,c)f(z)Iμλ(a,b,c)F(z).
Differentiating both sides of (2.22) logarithmically, we obtain
(2.23)p(z)+zp′(z)c+γ+(1−γ)p(z)=11−γ(z(Iμλ(a,b,c)f(z))′Iμλ(a,b,c)f(z)−γ).
Then, by Lemma 2.2, we obtain that
(2.24)11−γ(z(Iμλ(a,b,c)f(z))′Iμλ(a,b,c)f(z)−γ)≺ϕ(z)(0≤γ<1;z∈𝕌).Now, by lettingϕ(z)=(1+Az)/(1+Bz)(−1≤B<A≤1) in Theorem 2.9, we have the following.Corollary 2.10.
Forλ>−1, μ≥0, c>−γ, and 0≤γ<1. If f∈𝒬μλ(a,b,c;γ;A,B), then F∈𝒬μλ(a,b,c;γ;A,B), where F given by (2.19).Also, by takingϕ(z)=((1+z)/(1−z))β(0<β≤1) in Theorem 2.9, we have the following.Corollary 2.11.
Forλ>−1, μ≥0, c>−β, 0<β≤1, and 0≤γ<1. If f∈𝒮μλ(a,b,c;γ,β), then F∈𝒮μλ(a,b,c;γ,β).Corollary 2.12.
Forλ>−1, μ≥0, c>−β, 0<β≤1, and 0≤γ<1. If f∈𝒞μλ(a,b,c;γ,β), then F∈𝒞μλ(a,b,c;γ,β).Proof.
It holds that(2.25)f(z)∈𝒞μλ(a,b,c;γ,β)⟺F(zf′(z))∈𝒮μλ(a,b,c;γ,β)⟺z(F(z))′∈𝒮μλ(a,b,c;γ,β)⟺F(z)∈𝒞μλ(a,b,c;γ,β).Theorem 2.13.
Letf∈𝒜. Then
(2.26)𝒦μλ+1(a,b,c;α,β,γ;A,B)⊂𝒦μλ(a,b,c;α,β,γ;A,B)
for Rea>1−β, 0<β≤1, 0≤α<1, 0≤γ<1,  and −1≤B<A≤1.Proof.
Letf∈𝒦μλ+1(a,b,c;α,β,γ;A,B), then by the definition, we can write
(2.27)11−α(z(Iμλ+1(a,b,c)f(z))′Iμλ+1(a,b,c)g(z)−α)≺(1+z1−z)β(z∈𝕌)
for some g∈Qμλ+1(a,b,c;γ;A,B).
Lettingh(z)=z(Iμλ(a,b,c)f(z))′/Iμλ(a,b,c)g(z) and H(z)=z(Iμλ(a,b,c)g(z))′/Iμλ(a,b,c)g(z), we observe that h(z), H(z)∈P(z). Now by Corollary 2.6, g∈Qμλ(a,b,c;γ;A,B) and so ReH(z)>γ. Also, note that
(2.28)z(Iμλ(a,b,c)f(z))′=(Iμλ(a,b,c)g(z))h(z).
Differentiating
both sides in (2.28) yields
(2.29)z(Iμλ(a,b,c)f(z))′Iμλ(a,b,c)g(z)=z(Iμλ(a,b,c)g(z))′Iμλ+1(a,b,c)g(z)h(z)+zh′(z)=H(z)h(z)+zh′(z).
Now by using
the identity (1.21), we obtain
(2.30)z(Iμλ+1(a,b,c)f(z))′Iμλ+1(a,b,c)g(z)=Iμλ+1(a,b,c)(zf′(z))Iμλ+1(a,b,c)g(z)=z(Iμλ(a,b,c)(zf′(z)))′+λIμλ(a,b,c)(zf′(z))z(Iμλ(a,b,c)g(z))′+λIμλ(a,b,c)g(z)=z(Iμλ(a,b,c)(zf′(z)))′/Iμλ(a,b,c)g(z)+λ(Iμλ(a,b,c)(zf′(z))/Iμλ(a,b,c)g(z))z(Iμλ(a,b,c)g(z))′/Iμλ(a,b,c)g(z)+λ=H(z)h(z)+zh′(z)+λh(z)H(z)+λ=h(z)+zh′(z)H(z)+λ.
From (2.27),
(2.28), and (2.30), we conclude that
(2.31)11−α(h(z)+zh′(z)H(z)+λ−α)≺(1+z1−z)β.
Letting E=0 and B(z)=(1/(1−α))(1/(H(z)+λ)), we obtain
(2.32)Re[B(z)]=11−α1|H(z)+λ|2Re[H(z)+λ]>0.
The above
inequality satisfies the conditions required by Lemma 2.3. Hence ϕ(z)≺((1+z)/(1−z))β and so the proof is complete.Theorem 2.14.
Letf∈𝒜. Then
(2.33)𝒦μλ(a,b,c;α,β,γ;A,B)⊂𝒦μλ(a+1,b,c;α,β,γ;A,B)
for Rea>1−β, 0<β≤1, 0≤α<1, 0≤γ<1, and −1≤B<A≤1.Proof.
By using the same technique as in the proof of Theorem2.13, we get
(2.34)11−α(h(z)+zh′(z)H(z)+(α−1)−α)≺(1+z1−z)β.
By letting E=0 and B(z)=(1/(1−α))(1/(H(z)+(a−1))), we obtain
(2.35)Re[B(z)]=11−α1|H(z)+(a−1)|2Re[H(z)+(a−1)]>0.
Then, by applying Lemma 2.3, we obtain the required result.Theorem 2.15.
Letc>−β, 0<β≤1, 0≤α<1, 0≤γ<1, and −1≤B<A≤1. If f∈𝒦μλ(a,b,c;α,β,γ;A,B), then F∈𝒦μλ(a,b,c;α,β,γ;A,B), where F is given by (2.19).Proof.
Also, by using the same technique as in the proof of Theorem2.13, we get
(2.36)11−α(zh′(z)H(z)+c+h(z)−α)≺(1+z1−z)β.
By letting E=0 and B(z)=(1/(1−α))(1/(H(z)+c)), we obtain
(2.37)Re[B(z)]=11−α1|H(z)+c|2Re[H(z)+c]>0.
Then, applying Lemma 2.3, we obtain the required result.



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