# On General Integral Operator of Analytic Functions

Authors: Nasser Alkasbi; Maslina Darus
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

LetEβ be the integral operator defined by Eβz=β∫0ztβ-1f1′tα1f1t/tγ1P1ζ1t⋯fn′tαnfnt/tγnPnζntdt1/β, where each of the functions fi and Pi is, respectively, analytic functions and functions
with positive real part defined in the open unit disk for all i=1,…,n. The object of
this paper is to obtain several univalence conditions for this integral operator. Our main
results contain some interesting corollaries as special cases.


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## Body

## 1. Introduction and Definitions

LetA denote the class of functions of the form
(1)f(z)=z+∑n=2∞‍anzn,
which are analytic in the open unit disk U={z:z∈𝒞:|z|<1}:S={f∈A:f is univalent in U}. Also, let 𝒫 be the class of all functions which are analytic in U and satisfy P(0)=1, ℜ{P(z)}>0. Frasin and Darus [1] defined the family B(δ), 0≤δ<1, so that it consists of functions f∈A satisfying the condition
(2)|z2f′(z)f2(z)-1|<1-δ(z∈U).
In this paper, we obtain new sufficient conditions for the univalence of the general integral operator Eβ(z) defined by
(3)Eβ(z)=[β∫0z‍tβ-1∏i=0n‍(fi′(t))αi(fi(t)t)γiPiζi(t)dt]1/β,
where β∈𝒞, αi,γi,ζi∈𝒞, fi∈A, and Pi∈𝒫 for all i=1,2,3,…n.Here and throughout in the sequel, every multivalued functions is taken with the principal branch.Remark 1.
Note that the integral operatorEβ generalizes the following operators introduced and studied by several authors as follows.(i)
Forαi=0, where i=1,…,n, we obtain the integral operator
(4)Nβ(z)=[β∫0z‍tβ-1∏i=0n‍(fi(t)t)γiPiζi(t)dt]1/β,introduced and studied by Frasin [2].(ii)
Forαi=γi=0, i=1,…,n, we obtain the integral operator
(5)ℐβζ(Pi)(z)=[β∫0z‍tβ-1∏i=0n‍Piζi(t)dt]1/β,introduced and studied by Frasin [3].(iii)
Forζi=0, i=1,…,n, we obtain the integral operator
(6)ℐβαi,γi(f1,f2,…,fn)(z)=[β∫0z‍tβ-1∏i=0n‍(fi′(t))αi(fi(t)t)γidt]1/β,introduced and studied by Frasin [4].(iv)
Forζi=0, β=1, and i=1,…,n, we obtain the integral operator
(7)ℐαi,γi(f1,f2,…,fn)(z)=∫0z‍∏i=0n‍(fi′(t))αi(fi(t)t)γidt,introduced and studied by Frasin [5].(v)
Forζi=αi=0 and i=1,…,n, we obtain the integral operator
(8)ℐγβi(f1,f2,…,fn)(z)=[β∫0z‍tβ-1(f1(t)t)γ1⋯(fn(t)t)γndt]1/β,introduced and studied by D. Breaz and N. Breaz [6].(vi)
Forβ=1, ζi=αi=0, and i=1,…,n, we obtain the integral operator
(9)F(z)=∫0z‍(f1(t)t)γ1⋯(fn(t)t)γndt,introduced and studied by D. Breaz and N. Breaz [6].(vii)
Forβ=1, ζi=γi=0, and i=1,…,n, we obtain the integral operator
(10)Fα1,…,αn(z)=∫0z‍(f1′(t))α1⋯(fn′(t))αndt,introduced and studied by Breaz et al. [7].(viii)
Forβ=1, n=1, γ1=γ, and αi=ζi=0, we obtain the integral operator
(11)Fα,γ(z)=∫0z‍(f(t)t)γ(f′(t))αdt,studied in [8].(ix)
Forβ=1, n=1, γ1=γ, and αi=ζi=0, we obtain the integral operator
(12)Fγ(z)=∫0z‍(f(t)t)γdtstudied in [9]. In particular, for γ=1, we obtain Alexander integral operator which was introduced in [10] as follows
(13)F(z)=∫0z‍f(t)tdt.(x)
Forβ=1, n=1, αi=α, and γi=ζi=0, we obtain the integral operator(14)G(z)=∫0z‍(f′(t))αdt,studied in [11].In order to derive our main results, we have to recall here the following lemmas.Lemma 2 (see [12]).
Letη∈𝒞 with ℜ(η)>0. If h∈A satisfies
(15)1-|z|2ℜ(η)ℜ(η)|zh′′(z)h′(z)|≤1
for all z∈U, then the integral operator
(16)Fβ(z)=[β∫0z‍tβ-1(f′(t))dt]1/β
is in the class 𝒮.Lemma 3 (see [13]).
Letβ∈𝒞 with ℜ(β)>0, c∈𝒞 with |c|≤1, c≠-1. If h∈A satisfies
(17)|c|z|2β+(1+|z|2β)zh′′(z)βh′(z)|≤1
for all z∈U, then the integral operator Fβ(z) defined by (16) is in the class 𝒮.Lemma 4 (see [14]).
IfP(z)∈𝒫, then
(18)|zP′′(z)βP′(z)|≤2|z|1-|z|2.Lemma 5 (see [9]).
Iff(z)∈B(δ), then
(19)|zf′(z)f(z)-1|≤(1-δ)(1+|z|)1-|z|.
When δ=0, so f∈𝒮.Lemma 6 (see [9]).
Iff(z)∈B(δ), then
(20)|zf′′(z)f′(z)|≤(1-δ)(2+|z|)1-|z|.
Also we need the following general Schwarz lemma.Lemma 7 (see [15]).
Let the functionf be regular in the disk UR={z:|z|<R}, with |M|<R for fixed M. If f(z) has one zero with multiplicity order bigger than m for z=0, then
(21)|f(z)|≤MRm|z|m,(z∈UR).
The equality holds only if
(22)f(z)=eiθ(MRm)zm,
where θ is constant.Lemma 8 (see [16]).
Iff∈A, then
(23)|1+zf′′(z)f′(z)|≤54|zf′(z)f(z)|.


## 2. Univalence Conditions for the OperatorEβ

We first prove the following theorem.Theorem 9.
Letf∈B(δi), 0≤δi<1, Pi(z)∈𝒫 for all i=1,…,n and η∈𝒞 with ℜ(η)=a>0. If
(24)∑i=1n‍[(3|αi|+2|γi|)(1-δi)+2|ζi|]<min{a;12},
then the integral operator Eβ defined by (3) is in the class 𝒮.Proof.
Define the regular functionh(z) by
(25)h(z)=∫0z‍∏i=0n‍[(fi′(t))αi(fi(t)t)γiPiζi(t)]dt.
Then it is easy to see that
(26)h′(z)=∏i=0n‍[(fi′(t))αi(fi(t)t)γiPiζi(t)],
and h(0)=1-h′(0)=0. Differentiating both sides of (26) logarithmically, we obtain
(27)zh′′(z)h′(z)=∑i=1n‍αi(zfi′′(z)fi′(z))+∑i=1n‍γi(zfi′(z)fi(z)-1)+∑i=1n‍ζi(zPi′(z)Pi(z)).
Thus, we have
(28)|zh′′(z)h′(z)|≤∑i=1n‍|αi||zfi′′(z)fi′(z)|+∑i=1n‍|γi||zfi′(z)fi(z)-1|+∑i=1n‍|ζi||zPi′(z)Pi(z)|.
Since fi∈B(δi), Pi(z)∈𝒫 for all i=1,…,n, from (28), (18), (19), and (20), we obtain
(29)|zh′′(z)h′(z)|≤∑i=1n‍|αi|(1-δi)(2+|z|1-|z|)+∑i=1n‍|γi|(1-δi)(1+|z|1-|z|)+∑i=1n‍|ζi|(2|z|1-|z|2)≤∑i=1n‍|αi|(1-δi)(31-|z|)+∑i=1n‍|γi|(1-δi)(21-|z|)+∑i=1n‍|ζi|(21-|z|).
Multiplying both sides of (29) by (1-|z|2ℜ(η))/ℜ(η), we get
(30)1-|z|2ℜ(η)ℜ(η)|zh′′(z)h′(z)|≤1-|z|2ℜ(η)(1-|z|)ℜ(η)×∑i=1n‍[(3|αi|+2|γi|)(1-δi)+2|ζi|],
for all z∈U.
Let us denote|z|=x, x∈[0,1), ℜ(η)=a>0, and Φ(x)=(1-x2a)/(1-x). It is easy to prove that
(31)Φ(x)≤{1,0<a<1,2a,12<a<∞.
From (31), (30), and the hypotheses (24), we have
(32)1-|z|2aa|zh′′(z)h′(z)|≤{1a∑i=1n‍[(3|αi|+2|γi|)(1-δi)+2|ζi|],0<a<1,2∑i=1n‍[(3|αi|+2|γi|)(1-δi)+2|ζi|],12<a<∞≤1,
for all z∈U. Applying Lemma 2 for the function h(z), we prove that Eβ(z)∈𝒮.Lettingn=1, δ1=δ, α1=α, γ1=γ, and ζ1=ζ in Theorem 9, we obtain the following corollary.Corollary 10.
Letf∈B(δ), 0≤δ<1, P(z)∈𝒫, and all η,γ,ζ∈𝒞 with ℜ(η)=a>0. If
(33)(3|α|+2|γ|)(1-δ)+2|ζ|<min{a;12},
and then the integral operator Nβα,γ,ζ defined by
(34)Nβα,γ,ζ(z)=[β∫0z‍tβ-1(f′(t))α(f(t)t)γPζ(t)dt]1/β
is in the class 𝒮.If we setδ=0 in Corollary 10, we have the following.Corollary 11.
Letf∈𝒮, P(z)∈𝒫 and all η,γ,ζ∈𝒞 with ℜ(η)=a>0. If
(35)3|α|+2|γ|+2|ζ|<min{a;12},
then the integral operator Nβα,γ,ζ defined by (34) is in the class 𝒮.Next, we prove the following theorem.Theorem 12.
Letαi,γi,ζi∈𝒞 for all i=1,…,n and each fi∈A satisfies condition (15) with ℜ(fi(z)/z)>0 and
(36)∑i=1n‍[29|αi|+16|γi+ζi|]≤{4ℜ(η),ifℜ(η)∈(0,1),4,ifℜ(η)∈[1,∞)
and then, for any complex number σ with ℜ(σ)≥ℜ(η)>0, the integral operator Nβ defined by (4) is in the class 𝒮.Proof.
Suppose thatℜ(fi(z)/z) for all i=1,…,n. Thus we have
(37)fi(z)z=Pi(z),
where Pi∈𝒫 for all i=1,…,n. Differentiating both sides of (37) logarithmically, we obtain
(38)zfi′(z)fiz-1=zPi′(z)Pi(z).
Define the regular function h(z) as in (25). Thus from (27) we have
(39)zh′′(z)h′(z)=∑i=1n‍αi(zfi′′(z)fi′(z))+∑i=1n‍(γi+ζi)(zfi′(z)fi(z)-1)
and so
(40)zh′′(z)h′(z)=∑i=1n‍αi(zfi′′(z)fi′(z)+1)-∑n=i0‍αi+∑i=1n‍(γi+ζi)(zfi′(z)fi(z)-1).
From Lemma 8, it follows that
(41)|zh′′(z)h′(z)|≤54∑i=1n‍|αi||zfi′(z)fi(z)|+∑i=1n‍|γi+ζi||zfi′(z)fi(z)-1|+∑n=i0‍|αi|≤54∑i=1n‍|αi|[|zfi′(z)fi(z)-1|+1]+∑i=1n‍|γi+ζi||zfi′(z)fi(z)-1|+∑n=i0‍|αi|≤∑i=1n‍[(54|αi|+|γi+ζi|)|zfi′(z)fi(z)-1|]+94∑i=1n‍|αi|≤∑i=1n‍[(54|αi|+|γi+ζi|)(|zfi′(z)fi(z)|+1)]+94∑i=1n‍|αi|.
Multiplying both sides of (41) by (1-|z|2ℜ(η))/ℜ(η), from Lemma 5 with δ=0, we get
(42)1-|z|2ℜ(η)ℜ(η)|zh′′(z)h′(z)|≤1-|z|2ℜ(η)ℜ(η)∑i=1n‍(54|αi|+|γi+ζi|)(21-|z|)+9(1-|z|2ℜ(η))∑i=1n|αi|4ℜ(η).
Suppose that ℜ(η)∈(0,1). Define a function Φ:(0,1)→R by
(43)Φ(x)=1-a2a(0<a<1).
Then Φ is an increasing function and consequently, for |z|=a; z∈U, we obtain
(44)1-|z|2ℜ(η)<1-|z|2.
We thus find from (42) and (44) that
(45)1-|z|2ℜ(η)ℜ(η)|zh′′(z)h′(z)|≤∑i=1n(5|αi|+4|γi+ζi|)ℜ(η)+9∑i=1n|αi|4ℜ(η)=∑i=1n(29|αi|+19|γi+ζi|)4ℜ(η).
Using the hypotheses (36) for ℜ(η)∈(0,1), we readily get
(46)1-|z|2ℜ(η)ℜ(η)|zh′′(z)h′(z)|≤1.
Now if ℜ(η)∈[1,∞), we define a function Ψ:[1,∞)→R by
(47)Ψ(x)=1-a2xx,(0<a<1).
We observe that the function Ψ is decreasing and consequently, for |z|=a;z∈U, we have
(48)1-|z|2ℜ(η)ℜ(η)<1-|z|2,
for all z∈U. It follows from (40) and (42) that
(49)1-|z|2ℜ(η)ℜ(η)|zh′′(z)h′(z)|≤∑i=1n‍(294|αi|+4|γi+ζi|).
Using once again the hypotheses (36) when ℜ(η)∈[1,∞), we easily get
(50)1-|z|2ℜ(η)ℜ(η)|zh′′(z)h′(z)|≤1.
Finally by applying Lemma 2, we conclude that the integral operator Nβ defined by (4) is in the class 𝒮.Lettingn=1, δ1=δ, α1=α, γ1=γ, and ζ1=ζ in Theorem 12, we obtain the following corollary.Corollary 13.
Letα,γ,ζ∈𝒞, and f∈A satisfies condition (15). If
(51)29|α|+16|γ+ζ|≤{4ℜ(η),ifℜ(η)∈(0,1),4,ifℜ(η)∈[1,∞)
then, for any complex number σ with ℜ(σ)≥ℜ(η)>0, the integral operator Nβα,γ,ζ defined by (4) is in the class 𝒮.Using Lemma3, we derive the following theorem.Theorem 14.
Letαi,γi,ζi,β∈𝒞 for all i=1,…,n, ℜ(β)>0, c∈𝒞(|c|≤1) and each fi∈A satisfies condition (15). If
(52)∑i=1n‍[29|αi|+16|γi+ζi|]≤{4β(1-|c|),ifℜ(η)∈(0,1),4(1-|c|),ifℜ(η)∈[1,∞)
then, for any complex number σ with ℜ(σ)≥ℜ(η)>0, the integral operator Nβ defined by (4) is in the class 𝒮.Proof.
From (40), we have
(53)|c|z|2β+(1-|z|2β)zh′′(z)βh′(z)|=|∑i=1n‍(γi+ζi)(zfi′(z)fi(z)-1)c|z|2β+1-|z|2ββ×[∑i=1n‍αi(zfi′′(z)fi′(z)+1)-∑i=1n‍αi+∑i=1n‍(γi+ζi)(zfi′(z)fi(z)-1)]|≤|c|+|1-|z|2||β|×[∑i=1n‍(54|αi|+|γi+ζi|)(|zfi′(z)fi(z)|+1)+94∑i=1n‍|αi|]≤|c|+|1-|z|2||β|×[∑i=1n‍(54|αi|+|γi+ζi|)(21-|z|)+94∑i=1n‍|αi|].
Suppose that β∈(0,1). Define a function Φ:(0,1)→R by
(54)Φ(x)=1-a2x(0<a<1).
Then Φ is an increasing function and consequently for |z|=a; z∈U, we obtain
(55)1-|z|2β<1-|z|2.
We thus find from (53) that
(56)|c|z|2β+(1-|z|2β)zh′′(z)βh′(z)|≤∑i=1n(29|αi|+16|γi+ζi|)4|β|.
Using the hypotheses (52) for β∈(0,1), we readily get
(57)|c|z|2β+(1-|z|2β)zh′′(z)βh′(z)|≤1.
Now if β∈[1,∞), we define a function Ψ:[1,∞)→R by
(58)Ψ(x)=1-a2xx,(0<a<1).
We observe that the function Ψ is decreasing and consequently for |z|=a; z∈U, and using once again the hypotheses (36) when ℜ(η)∈[1,∞), we easily get
(59)|c|z|2β+(1-|z|2β)zh′′(z)βh′(z)|≤1.
Finally, by applying Lemma 3, we conclude that Nβ∈𝒮.



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