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

Abstract

Let 𝐵_𝛽 be the integral operator defined by 𝐵_𝛽(𝑧)=[𝛽∫^𝑧₀𝑡^𝛽−1∏^𝑛_𝑖=1[((𝑓_𝑖(𝑡))/𝑡)^𝛾𝑖𝑝^𝜁𝑖_𝑖(𝑡)]𝑑𝑡]^1/𝛽 where each of the functions 𝑓_𝑖 and 𝑝_𝑖 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.

1. Introduction and Definitions

Let 𝒜 denote the class of the normalized functions of the form 𝑓(𝑧)=𝑧+^∞_𝑘=2𝑎_𝑘𝑧^𝑘, (1) 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 𝑧²𝑓′(𝑧)𝑓²(𝑧)−1<1−𝛿(𝑧∈𝒰). (2)

Very recently many authors studied the problem of integral operators which preserve the class 𝒮 (see, e.g., [3–15]). In this paper, we obtain new sufficient conditions for the univalence of the general integral operator 𝐵_𝛽(𝑧) defined by 𝐵_𝛽(𝑧)=𝛽^𝑧₀𝑡^𝑛𝛽−1_𝑖=1𝑓_𝑖(𝑡)𝑡^𝛾𝑖𝑝^𝜁𝑖_𝑖(𝑡)𝑑𝑡^1/𝛽, (3) where 𝑓_𝑖∈𝒜, 𝑝_𝑖∈𝒫, 𝛽∈ℂ^∗=ℂ⧵{0}, and 𝛾_𝑖,𝜁_𝑖∈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 operator 𝐵_𝛽 generalizes the following operators introduced and studied by several authors:(1)If we let 𝜁_𝑖=0, for all 𝑖=1,…,𝑛, in (3), we obtain the integral operator: 𝐼_𝛾𝑓₁,…,𝑓_𝑛(𝑧)=𝛽^𝑧₀𝑡^𝛽−1𝑓₁(𝑡)𝑡^𝛾1⋯𝑓_𝑛(𝑡)𝑡^𝛾𝑛𝑑𝑡^1/𝛽, (4) introduced and studied by D. Breaz and N. Breaz [16].(2)If we let 𝛾_𝑖=0, for all 𝑖=1,…,𝑛, in (3), we obtain the integral operator: 𝐼_𝛽𝑝₁,…,𝑝_𝑛;𝛼₁,…,𝛼_𝑛(𝑧)=^𝑧₀𝛽𝑡^𝛽−1𝑝₁(𝑡)^𝜁1⋯𝑝_𝑛(𝑡)^𝜁𝑛𝑑𝑡^1/𝛽 (5) introduced and studied by Frasin [17].(3)If we let 𝛽=1 and 𝜁_𝑖=0, for all 𝑖=1,…,𝑛, in (3), we obtain the integral operator: 𝐹_𝑛(𝑧)=^𝑧₀𝑓₁(𝑡)𝑡^𝛾1⋯𝑓_𝑛(𝑡)𝑡^𝛾𝑛𝑑𝑡, (6) 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]). Let 𝛼∈ℂ with Re(𝛼)>0. If ℎ∈𝒜 satisfies 1−|𝑧|^2Re(𝛼)(𝛼)(𝑧)Re𝑧ℎ′′ℎ′(𝑧)≤1, (7) for all 𝑧∈𝒰, then, for any complex number 𝛽 with Re(𝛽)≥Re(𝛼), the integral operator 𝐹_𝛽(𝑧)=𝛽^𝑧₀𝑡^𝛽−1ℎ′(𝑡)𝑑𝑡^1/𝛽 (8) is in the class 𝒮.

Lemma 3 (see [13]). Let 𝛽∈ℂ with Re(𝛽)>0,𝑐∈ℂ with |𝑐|≤1,𝑐≠−1. If ℎ∈𝒜 satisfies 𝑐|𝑧|^2𝛽+1−|𝑧|^2𝛽𝑧ℎ′′(𝑧)𝛽ℎ′(𝑧)≤1, (9)

for all 𝑧∈𝒰, then the integral operator 𝐹_𝛽(𝑧) defined by (8) is in the class 𝒮.

Lemma 4 (see [19]). If 𝑝(𝑧)∈𝒫, then we have 𝑧𝑝′(𝑧)𝑝(𝑧)<2|𝑧|1−|𝑧|²,(𝑧∈𝒰). (10)

Lemma 5 (see [20]). If 𝑓(𝑧)∈ℬ(𝛿), then 𝑧𝑓′(𝑧)𝑓(𝑧)<(|𝑧−11−𝛿)(1+|)|𝑧|().1−𝑧∈𝒰 (11) Also, we need the following general Schwarz Lemma.

Lemma 6 (see [21]). Let the function 𝑓 be regular in the disk 𝒰_𝑅={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 𝑓(𝑧≤𝑀)𝑅^𝑚|𝑧|^𝑚𝑧∈𝒰_𝑅. (12)
The equality can hold only if 𝑓(𝑧)=𝑒^𝑖𝜃𝑀𝑅^𝑚𝑧^𝑚, (13) where 𝜃 is constant.

2. Univalence Conditions for the Operator 𝐵_𝛽

We first prove the following theorem.

Theorem 7. Let 𝑓(𝑧)∈ℬ(𝛿_𝑖); 0≤𝛿_𝑖<1 and 𝑝_𝑖(𝑧)∈𝒫 for all 𝑖=1,…,𝑛. Let 𝛼∈ℂ with Re(𝛼)=𝑎>0. If ^𝑛_𝑖=1𝛾_𝑖1−𝛿_𝑖+𝜁_𝑖𝑎≤min2;14, (14) then the integral operator 𝐵_𝛽(𝑧) defined by (3) is in the class 𝒮.

Proof. Define the regular function ℎ(𝑧) by ℎ(𝑧)=^𝑧₀^𝑛_𝑖=1𝑓_𝑖(𝑡)𝑡^𝛾𝑖𝑝^𝜁𝑖_𝑖(𝑡)𝑑𝑡. (15)
Then it is easy to see that ℎ^′(𝑧)=^𝑛_𝑖=1𝑓_𝑖(𝑧)𝑧^𝛾𝑖𝑝^𝜁𝑖_𝑖(𝑧), (16) and ℎ(0)=ℎ′(0)−1=0. Differentiating both sides of (16) logarithmically, we obtain 𝑧ℎ′′(𝑧)(𝑧)=ℎ′^𝑛_𝑖=1𝛾_𝑖𝑧𝑓_𝑖′(𝑧)𝑓_𝑖(𝑧)−1+^𝑛_𝑖=1𝜁_𝑖𝑧𝑝_𝑖′(𝑧)𝑝_𝑖(𝑧). (17)
Thus, we have 𝑧ℎ′′(𝑧)(𝑧)≤ℎ′^𝑛_𝑖=1𝛾_𝑖𝑧𝑓^′_𝑖(𝑧)𝑓_𝑖(𝑧)−1+^𝑛_𝑖=1𝜁_𝑖𝑧𝑝_𝑖′(𝑧)𝑝_𝑖(𝑧). (18)
Since 𝑓_𝑖(𝑧)∈ℬ(𝛿_𝑖) and 𝑝_𝑖(𝑧)∈𝒫 for all 𝑖=1,…,𝑛, from (18), (11), and (10), we obtain 𝑧ℎ′′(𝑧)(𝑧)≤ℎ′^𝑛_𝑖=1𝛾_𝑖1−𝛿_𝑖1+|𝑧||𝑧|+1−^𝑛_𝑖=1𝜁_𝑖2|𝑧|1−|𝑧|²≤^𝑛_𝑖=1𝛾_𝑖1−𝛿_𝑖2|𝑧|+1−^𝑛_𝑖=1𝜁_𝑖2|𝑧|.1− (19)
Multiply both sides of (19) by (1−|𝑧|^2Re(𝛼))/Re(𝛼), we get 1−|𝑧|^2Re(𝛼)(𝛼)(𝑧)Re𝑧ℎ′′(𝑧)≤|𝑧|ℎ′1−^2Re(𝛼)|𝑧|21−(𝛼)×Re_𝑛𝑖=1𝛾_𝑖1−𝛿_𝑖+𝜁_𝑖, (20) for all 𝑧∈𝒰.
Let us denote |𝑧|=𝑥, 𝑥∈[0,1), Re(𝛼)=𝑎>0, and Φ(𝑥)=(1−𝑥^2𝑎)/(1−𝑥). It is easy to prove that Φ(𝑥)≤11,if0<𝑎<212𝑎,if2<𝑎<∞. (21)
From (20), (21), and the hypothesis (14), we have 1−|𝑧|^2𝑎𝑎𝑧ℎ^′′(𝑧)ℎ^′(𝑧)≤2𝑎^𝑛_𝑖∑₌₁𝛾_𝑖1−𝛿_𝑖+𝜁_𝑖1,if0<𝑎<24∑^𝑛_𝑖=1𝛾_𝑖1−𝛿_𝑖+𝜁_𝑖1,if2<𝑎<∞.≤1, (22)
for all 𝑧∈𝒰. Applying Lemma 2 for the function ℎ(𝑧), we prove that 𝐵_𝛽(𝑧)∈𝒮.

Letting 𝑛=1, 𝛿₁=𝛿, 𝛾₁=𝛾, 𝜁₁=𝜁, and 𝑓₁=𝑓 in Theorem 7, we obtain the following corollary.

Corollary 8. Let 𝑓(𝑧)∈ℬ(𝛿); 0≤𝛿<1 and 𝑝(𝑧)∈𝒫. Also, let 𝛼,𝛾,𝜁∈ℂ with Re(𝛼)=𝑎>0. If 𝛾)+𝜁𝑎(1−𝛿≤min2;14, (23) then the integral operator 𝐵_{𝛽,𝛾,𝜁}(𝑧) defined by 𝐵_{𝛽,𝛾,𝜁}(𝑧)=𝛽^𝑧₀𝑡^𝛽−1𝑓(𝑡)𝑡^𝛾𝑝^𝜁(𝑡)𝑑𝑡^1/𝛽 (24) is in the class 𝒮.

If we set 𝛿=0 in Corollary 8, we have the following.

Corollary 9. Let 𝑓(𝑧)∈𝒮 and 𝑝(𝑧)∈𝓅. Also, let 𝛼,𝛾,𝜁∈ℂ with Re(𝛼)=𝑎>0. If 𝛾+𝜁𝑎≤min2;14, (25) then the integral operator 𝐵_{𝛽,𝛾,𝜁}(𝑧) defined by (24) is in the class 𝒮.

Next, we prove the following theorem.

Theorem 10. Let 𝑓_𝑖(𝑧)∈𝒜 satisfies Re(𝑓_𝑖(𝑧)/𝑧)>0, and 𝛾_𝑖+𝜁_𝑖𝑧𝑓^′_𝑖(𝑧)𝑓_𝑖(𝑧)−1≤(2𝑎+1)^{(2𝑎+1)/2𝑎}2𝑛, (26)for all 𝑖=1,…,𝑛, where 𝛾_𝑖,𝜁_𝑖,𝛼∈ℂ with Re(𝛼)=𝑎>0, then the integral operator 𝐵_𝛽(𝑧) defined by (3) is in the class 𝒮.

Proof. Suppose that Re(𝑓_𝑖(𝑧)/𝑧)>0for all 𝑖=1,…,𝑛. Thus, we have 𝑓_𝑖(𝑧)𝑧=𝑝_𝑖(𝑧), (27) where 𝑝_𝑖(𝑧)∈𝒫 for all 𝑖=1,…,𝑛. Differentiating both sides of (27) logarithmically, we obtain 𝑧𝑓^′_𝑖(𝑧)𝑓_𝑖(𝑧)−1=𝑧𝑝^′_𝑖(𝑧)𝑝_𝑖(𝑧),(𝑖=1,…,𝑛). (28)
Define the regular function ℎ(𝑧) as in (15). Thus from (28) and (17), we have 𝑧ℎ^′′(𝑧)ℎ^′(𝑧)=^𝑛_𝑖=1𝛾_𝑖+𝜁_𝑖𝑧𝑓′_𝑖(𝑧)𝑓_𝑖(𝑧)−1,𝑧∈𝒰. (29)
Form the hypothesis (26) and (29), we immediately have 𝑧ℎ^′′(𝑧)ℎ^′(𝑧)≤(2𝑎+1)^{(2𝑎+1)/2𝑎}2, (30) for all 𝑧∈𝒰. Applying Lemma 6, we obtain 𝑧ℎ^′′(𝑧)ℎ^′(𝑧)≤(2𝑎+1)^{(2𝑎+1)/2𝑎}2|𝑧|,𝑧∈𝒰. (31)
Thus from (29) and (31) we have, 1−|𝑧|^2𝑎𝑎𝑧ℎ^′′(𝑧)ℎ^′(𝑧)≤|𝑧|1−|𝑧|^2𝑎𝑎⋅(2𝑎+1)^{(2𝑎+1)/2𝑎}2, (32) for all 𝑧∈𝒰. Let us denote |𝑧|=𝑥, 𝑥∈[0,1], Re(𝛼)=𝑎>0, and Ψ(𝑥)=𝑥(1−𝑥^2𝑎). It is easy to prove that the maximum is attained at the point 𝑥=1/(2𝑎+1)^1/2𝑎, and thus we have Ψ(𝑥)≤2𝑎(2𝑎+1)^{(2𝑎+1)/2𝑎}. (33)
In view of this inequality and (32), we obtain 1−|𝑧|^2𝑎𝑎𝑧ℎ′′(𝑧)(𝑧)().ℎ′≤1𝑧∈𝒰 (34)
Applying Lemma 2 for the function ℎ(𝑧), we prove that 𝐵_𝛽(𝑧)∈𝒮.

Letting 𝑛=1, 𝛾₁=𝛾, 𝜁₁=𝜁, and 𝑓₁=𝑓 in Theorem 10, we have the following corollary.

Corollary 11. Let 𝑓(𝑧)∈𝒜 satisfies Re(𝑓(𝑧))/𝑧>0, and (𝑧)𝛾+𝜁𝑧𝑓′𝑓(𝑧)≤()−12𝑎+1^{(2𝑎+1)/2𝑎}2, (35) where 𝛾,𝜁,𝛼∈ℂ with Re(𝛼)=𝑎>0, then the integral operator 𝐵_{𝛽,𝛾,𝜁}(𝑧) defined by (24) is in the class 𝒮.

Using Lemma 3, we derive the following theorem.

Theorem 12. Suppose that each of the functions 𝑓_𝑖(𝑧)∈𝒜 satisfies Re((𝑓_𝑖(𝑧))/𝑧)>0, and 𝛾_𝑖+𝜁_𝑖𝑧𝑓^′_𝑖(𝑧)𝑓_𝑖(𝑧)≤𝛽−1𝑛(1−|𝑐|), (36)
for all 𝑖=1,…,𝑛, where 𝛾_𝑖,𝜁_𝑖,𝛽∈ℂ(Re(𝛽)>0) and 𝑐∈ℂ(|𝑐|<1), then the integral operator 𝐵_𝛽(𝑧) defined by (3) is in the class 𝒮.

Proof. From (29), we have𝑐|𝑧|^2𝛽+1−|𝑧|^2𝛽𝑧ℎ^′′(𝑧)𝛽ℎ^′(𝑧)=𝑐|𝑧|^2𝛽+1−|𝑧|^2𝛽𝛽^𝑛_𝑖=1𝛾_𝑖+𝜁_𝑖𝑧𝑓^′_𝑖(𝑧)𝑓_𝑖(𝑧)≤|𝑐|+1−1𝛽^𝑛_𝑖=1𝛾_𝑖+𝜁_𝑖𝑧𝑓^′_𝑖(𝑧)𝑓_𝑖(𝑧)−1. (37)
Now by using the hypothesis (36), we obtain 𝑐|𝑧|^2𝛽+1−|𝑧|^2𝛽𝑧ℎ′′(𝑧)𝛽ℎ′(𝑧)≤1. (38)
Finally, by applying Lemma 3, we conclude that 𝐵_𝛽(𝑧)∈𝒮.

Letting 𝑛=1, 𝛾₁=𝛾, 𝜁₁=𝜁, and 𝑓₁=𝑓 in Theorem 12, we have the following corollary.

Corollary 13. Suppose that the functions 𝑓(𝑧)∈𝒜 satisfy Re(𝑓(𝑧)/𝑧)>0, and 𝛾+𝜁𝑧𝑓^′(𝑧)𝑓(𝑧)−1≤𝛽(1−|𝑐|), (39) where 𝛾, 𝜁, 𝛽∈ℂ(Re(𝛽)>0) and 𝑐∈ℂ(|𝑐|<1), then the integral operator 𝐵_{𝛽,𝛾,𝜁}(𝑧) defined by (24) is in the class 𝒮.

Acknowledgment

The author would like to thank the referee for his helpful comments and suggestions.