# General Integral Operator of Analytic Functions Involving Functions with Positive Real Part

Authors: B. A. Frasin
Journal: Journal of Mathematics
 (2013)

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## Abstract

LetBβ be the integral operator defined by Bβ(z)=[β∫0ztβ-1∏i=1n[((fi(t))/t)γipiζi(t)]dt]1/βwhere each of the functions fi and pi are, 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 the normalized functions of the form
(1)f(z)=z+∑k=2∞akzk,
which are analytic in the open unit disk 𝒰={z∈ℂ:|z|<1}. Further, by  𝒮  we shall denote the class of all functions in 𝒜 which are univalent in 𝒰. Also, let 𝒫 be the class of all functions which are analytic in  𝒰  and satisfy p(0)=1,Re{p(z)}>0.Frasin and Darus [1] (see also [2]) 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).Very recently many authors studied the problem of integral operators which preserve the class𝒮 (see, e.g., [3–15]). In this paper, we obtain new sufficient conditions for the univalence of the general integral operator Bβ(z) defined by
(3)Bβ(z)=[β∫0ztβ-1∏i=1n[(fi(t)t)γipiζi(t)]dt]1/β,
where  fi∈𝒜, pi∈𝒫, β∈ℂ*=ℂ∖{0}, and γi,ζi∈ℂ  for all i=1,…,n.Here and throughout in the sequel, every many-valued function is taken with the principal branch.Remark 1.
Note that the integral operatorBβ generalizes the following operators introduced and studied by several authors:(1)
If we letζi=0, for all  i=1,…,n,  in (3), we obtain the integral operator:
(4)Iγ(f1,…,fn)(z)={β∫0ztβ-1(f1(t)t)γ1⋯(fn(t)t)γndt}1/β,
introduced and studied by D. Breaz and N. Breaz [16].(2)
If we letγi=0,  for all  i=1,…,n,  in (3), we obtain the integral operator:
(5)Iβ(p1,…,pn;α1,…,αn)(z)={∫0zβtβ-1(p1(t))ζ1⋯(pn(t))ζndt}1/β
introduced and studied by Frasin [17].(3)
If we letβ=1  and  ζi=0, for all i=1,…,n, in (3), we obtain the integral operator:
(6)Fn(z)=∫0z(f1(t)t)γ1⋯(fn(t)t)γndt,
introduced and studied by D. Breaz and N. Breaz [16].In order to derive our main results, we have to recall here the following lemmas.Lemma 2 (see [18]).
Letα∈ℂ  with Re(α)>0. If  h∈𝒜  satisfies
(7)1-|z|2Re(α)Re(α)|zh′′(z)h′(z)|≤1,
for all  z∈𝒰, then, for any complex number  β with Re(β)≥Re(α), the integral operator
(8)Fβ(z)={β∫0ztβ-1h′(t)dt}1/β
is in the class 𝒮.Lemma 3 (see [13]).
Letβ∈ℂ  with  Re(β)>0,c∈ℂ  with |c|≤1,c≠-1.  If h∈𝒜  satisfies
(9)|c|z|2β+(1-|z|2β)zh′′(z)βh′(z)|≤1,for allz∈𝒰, then the integral operator  Fβ(z)  defined by (8) is in the class  𝒮.Lemma 4 (see [19]).
Ifp(z)∈𝒫, then we have
(10)|zp′(z)p(z)|<2|z|1-|z|2,(z∈𝒰).Lemma 5 (see [20]).
Iff(z)∈ℬ(δ), then
(11)|zf′(z)f(z)-1|<(1-δ)(1+|z|)1-|z|(z∈𝒰).
Also, we need the following general Schwarz Lemma.Lemma 6 (see [21]).
Let the functionf be regular in the disk  𝒰R={z:|z|<R}, with |f(z)|<M  for fixed M. If f(z) has one zero with multiplicity order bigger than  m  for z=0, then
(12)|f(z)|≤MRm|z|m(z∈𝒰R).
The equality can hold only if(13)f(z)=eiθ(MRm)zm,
where  θ  is constant.


## 2. Univalence Conditions for the OperatorBβ

We first prove the following theorem.Theorem 7.
Letf(z)∈ℬ(δi); 0≤δi<1  and pi(z)∈𝒫  for all i=1,…,n. Let  α∈ℂ  with Re(α)=a>0.  If
(14)∑i=1n‍[|γi|(1-δi)+|ζi|]≤min{a2;14},
then the integral operator Bβ(z)  defined by (3) is in the class 𝒮.Proof.
Define the regular functionh(z)  by
(15)h(z)=∫0z∏i=1n[(fi(t)t)γipiζi(t)]dt.
Then it is easy to see that(16)h′(z)=∏i=1n[(fi(z)z)γipiζi(z)],
and  h(0)=h′(0)-1=0. Differentiating both sides of (16) logarithmically, we obtain
(17)zh′′(z)h′(z)=∑i=1nγi(zfi′(z)fi(z)-1)+∑i=1nζizpi′(z)pi(z).
Thus, we have(18)|zh′′(z)h′(z)|≤∑i=1n|γi||zfi′(z)fi(z)-1|+∑i=1n|ζi||zpi′(z)pi(z)|.
Sincefi(z)∈ℬ(δi) and  pi(z)∈𝒫  for all i=1,…,n, from (18), (11), and (10), we obtain
(19)|zh′′(z)h′(z)|≤∑i=1n|γi|(1-δi)(1+|z|1-|z|)+∑i=1n|ζi|(2|z|1-|z|2)≤∑i=1n|γi|(1-δi)(21-|z|)+∑i=1n|ζi|(21-|z|).
Multiply both sides of (19) by  (1-|z|2Re(α))/Re(α), we get
(20)1-|z|2Re(α)Re(α)|zh′′ࢉ(z)h′ࢉ(z)|≤1-|z|2Re(α)1-|z|2Re(α)×∑i=1n[|γi|(1-δi)+|ζi|],
for all  z∈𝒰.
Let us denote|z|=x, x∈[0,1), Re(α)=a>0, and Φ(x)=(1-x2a)/(1-x).  It is easy to prove that
(21)Φࢉ(x)≤{1,if0<a<122a,if12<a<∞.
From (20), (21), and the hypothesis (14), we have
(22)1-|z|2aa|zh′′(z)h′(z)|≤{2a∑i=1n[|γi|(1-δi)+|ζi|],if0<a<124∑i=1n[|γi|(1-δi)+|ζi|],if12<a<∞.≤1,
for allz∈𝒰. Applying Lemma 2 for the function  h(z), we prove that  Bβ(z)∈𝒮.Lettingn=1, δ1=δ, γ1=γ, ζ1=ζ, and f1=f in Theorem 7, we obtain the following corollary.Corollary 8.
Letf(z)∈ℬ(δ); 0≤δ<1  and  p(z)∈𝒫. Also, let α,γ,ζ∈ℂ  with Re(α)=a>0.  If
(23)|γ|(1-δ)+|ζ|≤min{a2;14},
then the integral operator Bβ,γ,ζ(z) defined by
(24)Bβ,γ,ζ(z)=[β∫0ztβ-1(f(t)t)γpζ(t)dt]1/β
is in the class 𝒮.If we setδ=0  in Corollary 8, we have the following.Corollary 9.
Letf(z)∈𝒮 and  p(z)∈𝓅.  Also, let  α,γ,ζ∈ℂ with Re(α)=a>0.  If
(25)|γ|+|ζ|≤min{a2;14},
then the integral operator Bβ,γ,ζ(z) defined by (24) is in the class  𝒮.Next, we prove the following theorem.Theorem 10.
Letfi(z)∈𝒜  satisfies  Re(fi(z)/z)>0,  and
(26)|(γi+ζi)(zfi′(z)fi(z)-1)|≤(2a+1)(2a+1)/2a2n,for all i=1,…,n, where  γi,ζi,α∈ℂ  with Re(α)=a>0, then the integral operator Bβ(z)  defined by (3) is in the class  𝒮.Proof.
Suppose thatRe(fi(z)/z)>0for alli=1,…,n. Thus, we have
(27)fi(z)z=pi(z),
where pi(z)∈𝒫 for all i=1,…,n. Differentiating both sides of (27) logarithmically, we obtain
(28)zfi′(z)fi(z)-1=zpi′(z)pi(z),(i=1,…,n).
Define the regular functionh(z)  as in (15). Thus from (28) and (17), we have
(29)zh′′(z)h′(z)=∑i=1n[γi+ζi](zf′i(z)fi(z)-1),z∈𝒰.
Form the hypothesis (26) and (29), we immediately have
(30)|zh′′(z)h′(z)|≤(2a+1)(2a+1)/2a2,
for all z∈𝒰. Applying Lemma 6, we obtain
(31)|zh′′(z)h′(z)|≤(2a+1)(2a+1)/2a2|z|,z∈𝒰.
Thus from (29) and (31) we have,
(32)1-|z|2aa|zh′′(z)h′(z)|≤|z|(1-|z|2a)a·(2a+1)(2a+1)/2a2,
for all z∈𝒰. Let us denote |z|=x, x∈[0,1], Re(α)=a>0, and Ψ(x)=x(1-x2a). It is easy to prove that the maximum is attained at the point x=1/(2a+1)1/2a, and thus we have
(33)Ψ(x)≤2a(2a+1)(2a+1)/2a.
In view of this inequality and (32), we obtain
(34)1-|z|2aa|zh′′(z)h′(z)|≤1(z∈𝒰).
Applying Lemma2 for the function  h(z), we prove that  Bβ(z)∈𝒮.Lettingn=1, γ1=γ, ζ1=ζ, and f1=f in Theorem 10, we have the following corollary.Corollary 11.
Letf(z)∈𝒜 satisfies Re(f(z))/z>0, and
(35)|(γ+ζ)(zf′(z)f(z)-1)|≤(2a+1)(2a+1)/2a2,
where γ,ζ,α∈ℂ with Re(α)=a>0, then the integral operator Bβ,γ,ζ(z) defined by (24) is in the class 𝒮.Using Lemma3, we derive the following theorem.Theorem 12.
Suppose that each of the functionsfi(z)∈𝒜 satisfies Re((fi(z))/z)>0, and
(36)|(γi+ζi)(zfi′(z)fi(z)-1)|≤|β|n(1-|c|),
for alli=1,…,n, where γi,ζi,β∈ℂ(Re(β)>0) and c∈ℂ(|c|<1), then the integral operator Bβ(z)  defined by (3) is in the class 𝒮.Proof.
From (29), we have(37)|c|z|2β+(1-|z|2β)zh′′(z)βh′(z)|=|c|z|2β+(1-|z|2β)β∑i=1n(γi+ζi)(zfi′(z)fi(z)-1)|≤|c|+1|β|∑i=1n|γi+ζi||zfi′(z)fi(z)-1|.
Now by using the hypothesis (36), we obtain
(38)|c|z|2β+(1-|z|2β)zh′′(z)βh′(z)|≤1.
Finally, by applying Lemma3, we conclude that Bβ(z)∈𝒮.Lettingn=1, γ1=γ, ζ1=ζ, and f1=f in Theorem 12, we have the following corollary.Corollary 13.
Suppose that the functionsf(z)∈𝒜 satisfy Re(f(z)/z)>0, and
(39)|(γ+ζ)(zf′(z)f(z)-1)|≤|β|(1-|c|),
where γ, ζ, β∈ℂ(Re(β)>0) and c∈ℂ(|c|<1), then the integral operator Bβ,γ,ζ(z) defined by (24) is in the class  𝒮.



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*Source: 260127-2013-01-08.xml*