Double Integral Operators Concerning Starlike of Order 𝛽

Abstract

Double integral operators which were considered by S. S. Miller and P. T. Mocanu (Integral Transform. Spec. Funct. 19(2008), 591–597) are discussed. In order to show the analytic function 𝑓(𝑧) is starlike of order 𝛽 in the open unit disk 𝕌, the theory of differential subordinations for analytic functions is applied. The object of the present paper is to discuss some interesting conditions for 𝑓(𝑧) to be starlike of order 𝛽 in 𝕌 concerned with second-order differential inequalities and double integral operators.

1. Introduction, Definition and Preliminaries

Let ℋ=ℋ(𝕌) denote the class of functions 𝑓(𝑧) which are analytic in the open unit disk 𝕌={𝑧∶𝑧∈ℂand|𝑧|<1}. (1.1) For a positive integer 𝑛 and 𝑎∈ℂ, we define the following classes of analytic functions: ℋ[𝑎,𝑛]={𝑓∈ℋ∶𝑓(𝑧)=𝑎+𝑎_𝑛𝑧^𝑛+𝑎_𝑛+1𝑧^𝑛+1+⋯,𝑧∈𝕌} (1.2)

and

𝒜_𝑛={𝑓∈ℋ∶𝑓(𝑧)=𝑧+𝑎_𝑛+1𝑧^𝑛+1+𝑎_𝑛+2𝑧^𝑛+2+⋯,𝑧∈𝕌} (1.3) with 𝒜₁=𝒜.

A function 𝑓(𝑧)∈𝒜 is said to be starlike of order 𝛽 in 𝕌 if it satisfies Re(𝑧𝑓^′(𝑧)𝑓(𝑧))>𝛽(𝑧∈𝕌), (1.4) for some 𝛽(0≦𝛽<1). We denote by 𝒮^∗(𝛽) the subclass of 𝒜 consisting of all functions 𝑓(𝑧) which are starlike of order 𝛽 in 𝕌.

By the familiar principle of differential subordinations between analytic functions 𝑓(𝑧) and 𝑔(𝑧) in 𝕌, we say that 𝑓(𝑧) is subordinate to 𝑔(𝑧) in 𝕌 if there exists an analytic function 𝑤(𝑧) with 𝑤(0)=0 and |𝑤(𝑧)|<1 such that 𝑓(𝑧)=𝑔(𝑤(𝑧))(𝑧∈𝕌).

We denote this subordination by 𝑓(𝑧)≺𝑔(𝑧)(𝑧∈𝕌). (1.5) In particular, if 𝑔(𝑧) is univalent in 𝕌, then it is known that 𝑓(𝑧)≺𝑔(𝑧),(𝑧∈𝕌)⟺𝑓(0)=𝑔(0)and𝑓(𝕌)⊂𝑔(𝕌). (1.6)

To obtain some results of this paper, we need the following two lemmas concerning the differential subordinations.

Lemma 1.1 (see [1, Hallenbeck and Ruscheweyh]). Letℎ(𝑧)be a convex function with ℎ(0)=𝑎and letRe𝛾>0. If 𝑝(𝑧)∈ℋ[𝑎,𝑛] and 𝑝(𝑧)+𝑧𝑝^′(𝑧)𝛾≺ℎ(𝑧), (1.7) then 𝑝(𝑧)≺𝑞(𝑧)≺ℎ(𝑧), (1.8) where 𝛾𝑞(𝑧)=𝑛𝑧^{𝛾/𝑛𝑧}₀ℎ(𝑡)𝑡^{𝛾/𝑛−1}𝑑𝑡. (1.9) This result is sharp.

Lemma 1.2 (see [2, Al-Amiri and Mocanu]). Let 𝑛 be a positive integer, and let 𝛼 be real with0≦𝛼<𝑛. Let 𝑞(𝑧)∈ℋ with 𝑞(0)=0,𝑞^′(0)≠0 and Re𝑧𝑞^′′(𝑧)𝑞^′𝛼(𝑧)+1>𝑛. (1.10) If 𝑝∈ℋ[0,𝑛] satisfies 𝑧𝑝^′(𝑧)−𝛼𝑝(𝑧)≺𝑛𝑧𝑞^′(𝑧)−𝛼𝑞(𝑧), (1.11) then 𝑝(𝑧)≺𝑞(𝑧) and this result is sharp.

By making use of these lemmas, Miller and Mocanu [3] have investigated some second-order differential inequality that implies starlikeness and deduced the following lemma.

Lemma 1.3. Let 𝑓(𝑧)∈𝒜_𝑛 and let 0≦𝛼<𝑛. If 𝑓(𝑧) satisfies |𝑧𝑓^′′(𝑧)−𝛼(𝑓^′(𝑧)−1)|<𝑛−𝛼, (1.12) then 𝑓(𝑧) is starlike.

Furthermore, by using Lemma 1.3, Miller and Mocanu [3] obtained some result concerning the double integral starlike operator as follows.

Lemma 1.4. Let 0≦𝛼<𝑛 and let 𝑔(𝑧)∈ℋ. If 𝑔(𝑧) satisfies |𝑔(𝑧)|≦𝑛−𝛼, (1.13) then the function 𝑓(𝑧) given by 𝑓(𝑧)=𝑧+𝑧^𝑛+11₀𝑔(𝑟𝑠𝑧)𝑟^{𝑛−𝛼−1}𝑠^𝑛𝑑𝑟𝑑𝑠 (1.14) is starlike.

2. Some Second-Order Differential Inequalities for Starlike of Order 𝜷

In this section, we deduced some conditions concerning the second-order differential inequality to show that 𝑓(𝑧) is starlike of order 𝛽 in 𝕌.

Theorem 2.1. Let 𝑓(𝑧)∈𝒜_𝑛 and let 0≦𝛼<𝑛 and 0≦𝛽<1. If 𝑓(𝑧) satisfies |𝑧𝑓^′′(𝑧)−𝛼(𝑓^′(𝑧)−1)|<(𝑛+1)(1−𝛽)(𝑛−𝛼),𝑛+1−𝛽 (2.1) then 𝑓(𝑧) is starlike of order 𝛽 in 𝕌.

Proof. We can rewrite the inequality (2.1) in terms of subordination as 𝑧𝑓^′′(𝑧)−𝛼(𝑓^′(𝑧)−1)≺(𝑛+1)(1−𝛽)(𝑛−𝛼)𝑛+1−𝛽𝑧. (2.2) If we set 𝑃(𝑧)=𝑓^′(𝑧)−(1+𝛼)𝑓(𝑧)𝑧=−𝛼+(𝑛−𝛼)𝑎_𝑛+1𝑧^𝑛+(𝑛+1−𝛼)𝑎_𝑛+2𝑧^𝑛+1+⋯∈ℋ[−𝛼,𝑛], (2.3) then the subordination (2.2) becomes 𝑃(𝑧)+𝑧𝑃^′(𝑧)≺−𝛼+(𝑛+1)(1−𝛽)(𝑛−𝛼)𝑛+1−𝛽𝑧. (2.4) Applying Lemma 1.1 to this first-order differential subordination, we obtain 1𝑃(𝑧)≺𝑛𝑧^1/𝑛𝑧₀{−𝛼+(𝑛+1)(1−𝛽)(𝑛−𝛼)𝑛+1−𝛽𝑡}𝑡^1/𝑛−1𝑑𝑡=−𝛼+(1−𝛽)(𝑛−𝛼)𝑛+1−𝛽𝑧, (2.5) or equivalently 𝑓^′(𝑧)−(1+𝛼)𝑓(𝑧)𝑧≺−𝛼+(1−𝛽)(𝑛−𝛼)𝑛+1−𝛽𝑧. (2.6) If we consider 𝑞(𝑧)=1−𝛽𝑛+1−𝛽𝑧(𝑞(0)=0,𝑞^′(0)=1−𝛽𝑛+1−𝛽≠0) (2.7) and 𝑝(𝑧)=𝑓(𝑧)𝑧−1=𝑎_𝑛+1𝑧^𝑛+𝑎_𝑛+2𝑧^𝑛+1+⋯∈ℋ[0,𝑛], (2.8) then the subordination (2.6) can be written as 𝑧𝑝^′(𝑧)−𝛼𝑝(𝑧)≺(1−𝛽)(𝑛−𝛼)𝑛+1−𝛽𝑧=𝑛𝑧𝑞^′(𝑧)−𝛼𝑞(𝑧). (2.9) Since 0≦𝛼<𝑛 and the function 𝑞(𝑧) satisfies Re𝑧𝑞^′′(𝑧)𝑞^′𝛼(𝑧)+1=1>𝑛, (2.10) using Lemma 1.2, we obtain the subordination 𝑝(𝑧)≺𝑞(𝑧), or 𝑓(𝑧)𝑧−1≺1−𝛽𝑛+1−𝛽𝑧. (2.11) It follows from the subordination (2.6) that |𝑓^′(𝑧)−(1+𝛼)𝑓(𝑧)𝑧|<𝛼+(1−𝛽)(𝑛−𝛼)=𝑛+1−𝛽𝑛(1+𝛼−𝛽),𝑛+1−𝛽 (2.12) while from the subordination (2.11) that |𝑓(𝑧)𝑧|>1−1−𝛽=𝑛𝑛+1−𝛽.𝑛+1−𝛽 (2.13) Combining these last two inequalities, we see that 𝑛|𝑛+1−𝛽𝑧𝑓^′(𝑧)𝑓(𝑧)−(1+𝛼)|<|𝑓(𝑧)𝑧||𝑧𝑓^′(𝑧)𝑓(𝑧)−(1+𝛼)|=|𝑓^′(𝑧)−(1+𝛼)𝑓(𝑧)𝑧|<𝑛(1+𝛼−𝛽),𝑛+1−𝛽 (2.14) which simplifies to |𝑧𝑓^′(𝑧)𝑓(𝑧)−(1+𝛼)|<𝑛(1+𝛼−𝛽)⋅𝑛+1−𝛽𝑛+1−𝛽𝑛=1+𝛼−𝛽. (2.15) This gives us that Re𝑧𝑓^′(𝑧)𝑓(𝑧)>(1+𝛼)−(1+𝛼−𝛽)=𝛽, (2.16) which proves that 𝑓(𝑧) is starlike of order 𝛽 in 𝕌.

We introduce the following example for Theorem 2.1.

Example 2.2. For the function 𝑓(𝑧)=𝑧+((1−𝛽)/(𝑛+1−𝛽))𝑧^𝑛+1(0≦𝛽<1), we have |𝑧𝑓^′′(𝑧)−𝛼(𝑓^′(𝑧)−1)|=|𝑛(𝑛+1)(1−𝛽)𝑧𝑛+1−𝛽^𝑛−𝛼(𝑛+1)(1−𝛽)𝑧𝑛+1−𝛽^𝑛|=|(𝑛+1)(1−𝛽)(𝑛−𝛼)𝑛+1−𝛽||𝑧|^𝑛<(𝑛+1)(1−𝛽)(𝑛−𝛼).𝑛+1−𝛽 (2.17) Furthermore, we see that Re𝑧𝑓^′(𝑧)𝑓(𝑧)=Re(1+((𝑛+1)(1−𝛽)/(𝑛+1−𝛽))𝑧^𝑛1+((1−𝛽)/(𝑛+1−𝛽))𝑧^𝑛)>1−(𝑛+1)(1−𝛽)/(𝑛+1−𝛽)1−((1−𝛽)/(𝑛+1−𝛽))=𝛽(𝑧∈𝕌). (2.18)

Remark 2.3. Letting 𝛽=0 in Theorem 2.1, we obtain Lemma 1.3 given by Miller and Mocanu [3].

Also, setting 𝛼=0 in Theorem 2.1, we have the following.

Corollary 2.4. Let𝑓(𝑧)∈𝒜_𝑛 and let0≦𝛽<1. If 𝑓(𝑧) satisfies |𝑧𝑓^′′(𝑧)|<𝑛(𝑛+1)(1−𝛽),𝑛+1−𝛽 (2.19) then 𝑓(𝑧) is starlike of order 𝛽 in 𝕌.

For the case 𝑛=1 in the above corollary, we find the following.

Remark 2.5. For 𝑓(𝑧)∈𝒜, we have that |𝑧𝑓^′′(𝑧)|<2(1−𝛽)2−𝛽⇒𝑓(𝑧)∈𝒮^∗(𝛽). (2.20) The case 𝛽=0 was first discussed by Obradović [4].

Next, by making use of Theorem 2.1, we obtain the following result concerning the double, integral operator for starlike of order 𝛽.

Theorem 2.6. Let a function𝑔(𝑧)∈ℋ satisfy |𝑔(𝑧)|≦(𝑛+1)(1−𝛽)(𝑛−𝛼),𝑛+1−𝛽 (2.21) for some 0≦𝛼<𝑛 and 0≦𝛽<1. Then the function 𝑓(𝑧) given by 𝑓(𝑧)=𝑧+𝑧^𝑛+11₀𝑔(𝑟𝑠𝑧)𝑟^{𝑛−𝛼−1}𝑠^𝑛𝑑𝑟𝑑𝑠 (2.22) is starlike of order 𝛽 in 𝕌.

Proof. We first consider the function 𝑓(𝑧)∈𝒜_𝑛 satisfying the differential equation 𝑧𝑓^′′(𝑧)−𝛼(𝑓^′(𝑧)−1)=𝑧^𝑛𝑔(𝑧). (2.23) Then, it is clear that |𝑧𝑓^′′(𝑧)−𝛼(𝑓^′(𝑧)−1)|=|𝑧|^𝑛|𝑔(𝑧)|<(𝑛+1)(1−𝛽)(𝑛−𝛼)𝑛+1−𝛽(𝑧∈𝕌). (2.24) Thus, from Theorem 2.1, we see that the solution of the differential equation (2.23) must be starlike of order 𝛽. The solution of (2.23) can be obtained in two integrations. If we set 𝜑(𝑧)=𝑓^′(𝑧)−1, then the equation (2.23) can be simplified to 𝑧𝜑^′(𝑧)−𝛼𝜑(𝑧)=𝑧^𝑛𝑔(𝑧), (2.25) which has the solution 𝜑(𝑧) given by 𝜑(𝑧)=𝑧^𝛼𝑧₀𝑔(𝜁)𝜁^{𝑛−𝛼−1}𝑑𝜁=𝑧^𝑛1₀𝑔(𝑟𝑧)𝑟^{𝑛−𝛼−1}𝑑𝑟. (2.26) Since 𝜑(𝑧)=𝑓^′(𝑧)−1, we have 𝑓^′(𝑧)−1=𝑧^𝑛1₀𝑔(𝑟𝑧)𝑟^{𝑛−𝛼−1}𝑑𝑟, (2.27) that is, 𝑓(𝑧)=𝑧+^𝑧₀𝜁^𝑛(¹₀𝑔(𝑟𝜁)𝑟^{𝑛−𝛼−1}𝑑𝑟)𝑑𝜁=𝑧+𝑧^𝑛+11₀𝑔(𝑟𝑠𝑧)𝑟^{𝑛−𝛼−1}𝑠^𝑛𝑑𝑟𝑑𝑠. (2.28)

Remark 2.7. Taking 𝛽=0 in Theorem 2.6, we find Lemma 1.4 given by Miller and Mocanu [3]. However, (1.14) and (2.22) are double integral operators of the same form.

Moreover, making 𝑛=1 and 𝛼=0 in Theorem 2.6, we have the following.

Corollary 2.8. If𝑔(𝑧)∈ℋ and |𝑔(𝑧)|≦2(1−𝛽)2−𝛽(𝑧∈𝕌) (2.29) for some 0≦𝛽<1, then 𝑓(𝑧)=𝑧+𝑧²¹₀𝑔(𝑟𝑠𝑧)𝑠𝑑𝑟𝑑𝑠∈𝒮^∗(𝛽). (2.30)

As examples of Corollary 2.8, we get the following.

Example 2.9. For the function 𝑔(𝑧)=2(1−𝛽)/(2−𝛽)(0≦𝛽<1), we find 𝑓(𝑧)=𝑧+1−𝛽𝑧2−𝛽²∈𝒮^∗(𝛽), (2.31) because Re𝑧𝑓^′(𝑧)𝑓(𝑧)=Re(1+(2(1−𝛽)/(2−𝛽))𝑧1+((1−𝛽)/(2−𝛽))𝑧)>1−(2(1−𝛽)/(2−𝛽))1−((1−𝛽)/(2−𝛽))=𝛽(𝑧∈𝕌). (2.32)

Example 2.10. For the function 𝑔(𝑧)=(2(1−𝛽)/(2−𝛽))𝑧(0≦𝛽<1), we have 𝑓(𝑧)=𝑧+1−𝛽𝑧3(2−𝛽)³. (2.33) Then, we see that Re𝑧𝑓^′(𝑧)𝑓(𝑧)=Re(1+((1−𝛽)/(2−𝛽))𝑧²1+((1−𝛽)/3(2−𝛽))𝑧²3)>5−2𝛽>𝛽,(𝑧∈𝕌), (2.34) that is, 𝑓(𝑧)=𝑧+1−𝛽𝑧3(2−𝛽)³∈𝒮^∗(35−2𝛽)⊂𝒮^∗(𝛽). (2.35)

3. Other Result for Starlikeness of Order 𝜷

To obtain that 𝑓(𝑧) is starlike of order 𝛽 in Theorem 2.1, we showed that |𝑧𝑓^′(𝑧)𝑓(𝑧)−(1+𝛼)|<1+𝛼−𝛽(𝑧∈𝕌). (3.1) In this section, to obtain that 𝑓(𝑧) is starlike of order 𝛽, we consider some second-order differential inequality concerning the order 𝛽 and show the following inequality: |𝑓(𝑧)𝑧𝑓^′−1(𝑧)12𝛽|<2𝛽(𝑧∈𝕌) (3.2) for some 0<𝛽<1.

Remark 3.1. Foe some 𝛽(0<𝛽<1), we see that |𝑓(𝑧)𝑧𝑓^′−1(𝑧)12𝛽|<2𝛽⟺Re𝑧𝑓^′(𝑧)𝑓(𝑧)>𝛽. (3.3)

Now, we consider the following theorem.

Theorem 3.2. Let 𝑓(𝑧)∈𝒜_𝑛 and let0<𝛽<1. If 𝑓(𝑧) satisfies |𝑧𝑓^′′(𝑧)+(1−2𝛽)(𝑓^′(𝑧)−1)|<𝛽(𝑛+1)(𝑛+1−2𝛽)1𝑛+1−𝛽(0<𝛽<2)(1−𝛽)(𝑛+1)(𝑛+1−2𝛽)1𝑛+1−𝛽,(2≦𝛽<1), (3.4) then |𝑓(𝑧)𝑧𝑓^′−1(𝑧)12𝛽|<2𝛽(𝑧∈𝕌), (3.5) or 𝑓(𝑧) is starlike of order 𝛽 in 𝕌.

Proof. (i) For the case 0<𝛽<1/2, the inequality (3.4) can be written as follows: 𝑧𝑓^′′(𝑧)+(1−2𝛽)(𝑓^′(𝑧)−1)≺𝛽(𝑛+1)(𝑛+1−2𝛽)𝑛+1−𝛽𝑧. (3.6) If we set 𝑃(𝑧)=𝑓^′(𝑧)−2𝛽𝑓(𝑧)𝑧=(1−2𝛽)+(𝑛+1−2𝛽)𝑎_𝑛+1𝑧^𝑛+(𝑛+2−2𝛽)𝑎_𝑛+2𝑧^𝑛+1+⋯∈ℋ[1−2𝛽,𝑛], (3.7) then the subordination (3.6) becomes 𝑃(𝑧)+𝑧𝑃^′(𝑧)≺(1−2𝛽)+𝛽(𝑛+1)(𝑛+1−2𝛽)𝑛+1−𝛽𝑧. (3.8) Applying Lemma 1.1 as well as the proof of Theorem 2.1, we obtain that 1𝑃(𝑧)≺𝑛𝑧^1/𝑛𝑧₀{(1−2𝛽)+𝛽(𝑛+1)(𝑛+1−2𝛽)𝑛+1−𝛽𝑡}𝑡^1/n−1𝑑𝑡=(1−2𝛽)+𝛽(𝑛+1−2𝛽)𝑛+1−𝛽𝑧, (3.9) or equivalently, that 𝑓^′(𝑧)−2𝛽𝑓(𝑧)𝑧≺(1−2𝛽)+𝛽(𝑛+1−2𝛽)𝑛+1−𝛽𝑧. (3.10) Also, if the function 𝑝(𝑧) is defined by 𝑝(𝑧)=𝑓^′(𝑧)−1=(𝑛+1)𝑎_𝑛+1𝑧^𝑛+(𝑛+2)𝑎_𝑛+2𝑧^𝑛+1+⋯∈ℋ[0,𝑛], (3.11) then the subordination (3.6) becomes 𝑧𝑝^′(𝑧)+(1−2𝛽)𝑝(𝑧)≺𝛽(𝑛+1)(𝑛+1−2𝛽)𝑛+1−𝛽𝑧, (3.12) namely, 𝑝(𝑧)+𝑧𝑝^′(𝑧)≺1−2𝛽𝛽(𝑛+1)(𝑛+1−2𝛽)(1−2𝛽)(𝑛+1−𝛽)𝑧. (3.13) Since 0<1−2𝛽<1, we can use Lemma 1.1 as 𝛾=1−2𝛽 and obtain 𝑝(𝑧)≺1−2𝛽𝑛𝑧^{(1−2𝛽)/𝑛𝑧}₀𝛽(𝑛+1)(𝑛+1−2𝛽)𝑡(1−2𝛽)(𝑛+1−𝛽)^{(1−2𝛽)/𝑛}𝑑𝑡=𝛽(𝑛+1)𝑛+1−𝛽𝑧, (3.14) that is, 𝑓^′(𝑧)−1≺𝛽(𝑛+1)𝑛+1−𝛽𝑧. (3.15) From the subordination (3.10), we find |𝑓^′(𝑧)−2𝛽𝑓(𝑧)𝑧|<(1−2𝛽)+𝛽(𝑛+1−2𝛽)=𝑛+1−𝛽𝑛(1−𝛽)+(1−2𝛽),𝑛+1−𝛽 (3.16) while, from the subordination (3.15), we get |𝑓^′(𝑧)|>1−𝛽(𝑛+1)=𝑛+1−𝛽𝑛(1−𝛽)+(1−2𝛽).𝑛+1−𝛽 (3.17) By combining these last two inequalities, we obtain that 𝑛(1−𝛽)+(1−2𝛽)|𝑛+1−𝛽𝑓(𝑧)𝑧𝑓^′−1(𝑧)2𝛽|<|𝑓^′(𝑧)||𝑓(𝑧)𝑧𝑓^′−1(𝑧)|=12𝛽2𝛽|𝑓^′(𝑧)−2𝛽𝑓(𝑧)𝑧|<𝑛(1−𝛽)+(1−2𝛽)2𝛽(𝑛+1−𝛽)(𝑧∈𝕌), (3.18) which simplifies to |𝑓(𝑧)𝑧𝑓^′−1(𝑧)12𝛽|<2𝛽(𝑧∈𝕌). (3.19) (ii) For the case 1/2≦𝛽<1, we can rewrite (3.4) in terms of the subordination as 𝑧𝑓^′′(𝑧)+(1−2𝛽)(𝑓^′(𝑧)−1)≺(1−𝛽)(𝑛+1)(𝑛+1−2𝛽)𝑛+1−𝛽𝑧. (3.20) Applying Lemma 1.1 in similar to case (i), we can obtain that 𝑓(𝑧)−2𝛽𝑓(𝑧)𝑧≺(1−2𝛽)+(1−𝛽)(𝑛+1−2𝛽)𝑛+1−𝛽𝑧. (3.21) Furthermore, if we set 𝑞(𝑧)=(1−𝛽)(𝑛+1)𝑧𝑛+1−𝛽and𝑝(𝑧)=𝑓^′(𝑧)−1∈ℋ[0,𝑛], (3.22) then the subordination (3.20) can be written as 𝑧𝑝^′(𝑧)−(2𝛽−1)𝑝(𝑧)≺𝑛𝑧𝑞^′(𝑧)−(2𝛽−1)𝑞(𝑧). (3.23) Since 0≦2𝛽−1<1≦𝑛 and the function 𝑞(𝑧) satisfies 𝑞(0)=0,𝑞^′(0)=(1−𝛽)(𝑛+1)≠𝑛+1−𝛽andRe𝑧𝑓^′′(𝑧)𝑓^′(𝑧)+1=1>2𝛽−1𝑛, (3.24) we can use Lemma 1.2 and obtain the following fact: 𝑧𝑝^′(𝑧)−(2𝛽−1)𝑝(𝑧)≺𝑛𝑧𝑞^′(𝑧)−(2𝛽−1)𝑞(𝑧)⇒𝑝(𝑧)≺𝑞(𝑧). (3.25) This implies that 𝑓^′(𝑧)−1≺(1−𝛽)(𝑛+1)𝑛+1−𝛽𝑧. (3.26) From subordinations (3.21) and (3.26), we find |𝑓^′(𝑧)−2𝛽𝑓(𝑧)𝑧|<|(1−2𝛽)−(1−𝛽)(𝑛+1−2𝛽)𝑛+1−𝛽|=𝛽𝑛𝑛+1−𝛽(𝑧∈𝕌) (3.27) and |𝑓^′(𝑧)|>1−(1−𝛽)(𝑛+1)=𝑛+1−𝛽𝛽𝑛𝑛+1−𝛽(𝑧∈𝕌). (3.28) Therefore, we obtain 𝛽𝑛|𝑛+1−𝛽𝑓(𝑧)𝑧𝑓^′−1(𝑧)2𝛽|<|𝑓^′(𝑧)||𝑓(𝑧)𝑧𝑓^′−1(𝑧)|=12𝛽2𝛽|𝑓^′(𝑧)−2𝛽𝑓(𝑧)𝑧|<𝛽𝑛2𝛽(𝑛+1−𝛽)(𝑧∈𝕌), (3.29) which simplifies to |𝑓(𝑧)𝑧𝑓^′−1(𝑧)12𝛽|<2𝛽(𝑧∈𝕌). (3.30) This completes the proof of this theorem.

Making use of Theorem 3.2, we obtain the following result concerning the double integral operator for starlike of order 𝛽.

Theorem 3.3. If𝑔(𝑧)∈ℋ satisfies |𝑔(𝑧)|≦𝛽(𝑛+1)(𝑛+1−2𝛽)1𝑛+1−𝛽(0<𝛽<2)(1−𝛽)(𝑛+1)(𝑛+1−2𝛽)(1𝑛+1−𝛽2≦𝛽<1), (3.31) for some 0<𝛽<1, then 𝑓(𝑧)=𝑧+𝑧^𝑛+11₀𝑔(𝑟𝑠𝑧)𝑟^𝑛−2𝛽𝑠^𝑛𝑑𝑟𝑑𝑠, (3.32) is a starlike function of order 𝛽.

This theorem can be shown as well as the proof of Theorem 2.6.

Making 𝑛=1 in Theorem 3.3, we have

Corollary 3.4. If 𝑔(𝑧)∈ℋ satisfies |𝑔(𝑧)|≦4𝛽(1−𝛽)12−𝛽(0<𝛽<2)4(1−𝛽)²(12−𝛽2≦𝛽<1), (3.33) for some 0<𝛽<1, then 𝑓(𝑧)=𝑧+𝑧²¹₀𝑔(𝑟𝑠𝑧)𝑟^1−2𝛽𝑠𝑑𝑟𝑑𝑠∈𝒮^∗(𝛽). (3.34)

Let us consider an example for Corollary 3.4.

Example 3.5. If we consider 𝑔(𝑧)=4𝛽(1−𝛽)/(2−𝛽)(0<𝛽<1/2), then we find 𝛽𝑓(𝑧)=𝑧+𝑧2−𝛽²∈𝒮^∗(2−3𝛽2(1−𝛽))⊂𝒮^∗(𝛽), (3.35) because Re𝑧𝑓^′(𝑧)𝑓(𝑧)=Re(1+(2𝛽/(2−𝛽))𝑧1+(𝛽/(2−𝛽))𝑧)>(2−3𝛽)/2(1−𝛽)>𝛽(𝑧∈𝕌). (3.36) Further, if we take 𝑔(𝑧)=4(1−𝛽)²/(2−𝛽)(1/2≦𝛽<1), then we see that 𝑓(𝑧)=𝑧+1−𝛽𝑧2−𝛽²∈𝒮^∗(𝛽). (3.37)