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Notes on analytic functions with a bounded positive real part

Abstract

For real α and β such that 0α<1<β, we denote by S(α,β) the class of normalized analytic functions f such that α<Re{z f (z)/f(z)}<β in . We find some properties, including inclusion properties, Fekete-Szegö problem and coefficient problems of inverse functions.

MSC:30C45, 30C55.

Dedication

Dedicated to Professor Hari M Srivastava

1 Introduction

Let denote the class of analytic functions in the unit disc U={z:|z|<1} on the complex plane . Let denote the subclass of consisting of functions normalized by f(0)=0 and f (0)=1. Let denote the subclass of consisting of univalent functions. Denote by S and , the class of starlike functions and convex functions, respectively. It is well-known that K S S.

We say that f is subordinate to F in , written as fF if and only if f(z)=F(w(z)) for some Schwarz function with w(0)=0 and |w(z)|<1, zU. If F(z) is univalent in , then the subordination fF is equivalent to f(0)=F(0) and f(U)F(U).

We denote by S (A,B), the class of Janowski starlike functions, namely, the functions satisfying the subordination equation: z f (z)/f(z)(1+Az)/(1+Bz). Note that S (1,1)= S .

Now, we shall introduce the class of analytic functions used in the sequel.

Definition 1 Let α and β be real numbers such that 0α<1<β. The function fA belongs to the class S(α,β) if f satisfies the following inequality:

α<Re { z f ( z ) f ( z ) } <β(zU).

We remark that for given α, β (0α<1<β), fS(α,β) if and only if f satisfies the following two subordination equations:

z f ( z ) f ( z ) 1 + ( 1 2 α ) z 1 z and z f ( z ) f ( z ) 1 + ( 1 2 β ) z 1 z ,
(1)

since the functions (1+(12α)z)/(1z) and (1+(12β)z)/(1z) map onto the right half plane, having real part greater than α, and the left half plane, having real part smaller than β, respectively. The above class S(α,β) is introduced by Kuroki and Owa [1]. They investigated coefficient estimates for fS(α,β) and found the necessary and sufficient condition for fS(α,β) using the following subordination.

Lemma 1 (Kuroki and Owa [1])

Let fA and 0α<1<β. Then fS(α,β) if and only if

z f ( z ) f ( z ) 1+ β α π ilog ( 1 e 2 π i 1 α β α z 1 z ) (zU).

Lemma 1 means that the function p defined by

p(z)=1+ β α π ilog ( 1 e 2 π i 1 α β α z 1 z )
(2)

maps the unit disk onto the strip domain w with α<Re(w)<β. We note that the function fA, given by

f(z)=zexp { β α π i 0 z 1 t log ( 1 e 2 π i 1 α β α t 1 t ) d t } ,

is in the class S(α,β).

2 Inclusion properties

Theorem 1 For given 0α<1<β, let A and B be real numbers such that

2 α β β α B<A β 2 α β + α β α .
(3)

Then S (A,B)S(α,β).

Proof At first, we note that

1< 2 α β β α and β 2 α β + α β α <1.

For f S (A,B), we know that the following inequality holds:

1 A 1 B <Re { z f ( z ) f ( z ) } < 1 + A 1 + B (zU).

Therefore, it suffices to show that α and β satisfy the following inequalities:

α 1 A 1 B and 1 + A 1 + B β.
(4)

Using inequality (4), we can derive that

1+B 2 ( 1 α ) β α and1+A 2 β ( β α ) β α .
(5)

Also,

1B 2 ( β 1 ) β α and1A 2 α ( β a ) β α .
(6)

By the above inequalities (5) and (6), we can easily obtain the inequalities (3), so the proof of Theorem 1 is completed. □

Lemma 2 (Miller and Mocanu [2])

Let Ξ be a set in the complex plane and let b be a complex number such that Re(b)>0. Suppose that a function ψ: C 2 ×UC satisfies the condition

ψ(iρ,σ;z)Ξ

for all real ρ, σ | b i ρ | 2 /(2Re(b)) and all zU. If the function p(z) defined by p(z)=b+ b 1 z+ b 2 z 2 + is analytic in and if

ψ ( p ( z ) , z p ( z ) ; z ) Ξ,

then Re(p(z))>0 in .

Theorem 2 Let fA, 1/2α<1 and Re{z f (z)/f(z)}>α in . Then

Re { f ( z ) z } >γ(α):= 1 3 2 α (zU).
(7)

Proof Write γ(α):=γ and note that 1 2 γ<1 for 1 2 α<1. Let p be defined by

p(z)= 1 1 γ ( z f ( z ) f ( z ) γ ) .

Then p is analytic in , p(0)=1 and

z f ( z ) f ( z ) =1+ ( 1 γ ) z p ( z ) ( 1 γ ) p ( z ) + γ =ψ ( p ( z ) , z p ( z ) ) ,

where

ψ(r,s)=1+ ( 1 γ ) s ( 1 γ ) r + γ .
(8)

Also,

{ ψ ( p ( z ) , z p ( z ) ) : z U } { w C : Re ( w ) > α } := Ω α .

Now for all real ρ, σ 1 2 (1+ ρ 2 ),

Re ( ψ ( i ρ , σ ) ) = Re ( 1 + ( 1 γ ) σ ( 1 γ ) i ρ + γ ) = 1 + γ ( 1 γ ) σ γ 2 + ( 1 γ ) 2 ρ 2 1 1 2 γ ( 1 γ ) 1 + ρ 2 γ 2 + ( 1 γ ) 2 ρ 2 .

Now, we let

g(ρ)= 1 + ρ 2 γ 2 + ( 1 γ ) 2 ρ 2 .
(9)

Then

g (ρ)= 2 ( 2 γ 1 ) ρ ( γ 2 + ( 1 γ ) 2 ρ 2 ) 2 ,

hence g (ρ)=0 occurs at only ρ=0 and g satisfies

g(0)= 1 γ 2

and

lim ρ g(ρ)= 1 ( 1 γ ) 2 .

Since 1/2γ<1, we have

1 γ 2 g(ρ)< 1 ( 1 γ ) 2 ,

hence we get

Re ( ψ ( i ρ , σ ) ) 1 1 2 γ(1γ) 1 γ 2 = 3 γ 1 2 γ =α.

This shows that Re{ψ(iρ,σ)} Ω α . By Lemma 2, we get Re(p(z))>0 in , and this shows that inequality (7) holds and the proof of Theorem 2 is completed. □

Theorem 3 Let fA, 1<β<3/2 and Re{z f (z)/f(z)}<β in . Then

Re { f ( z ) z } <δ(β):= 1 3 2 β (zU).
(10)

Proof Note that δ:=δ(β)= 1 3 2 β >1 for β>1. Let p be defined by

p(z)= 1 1 δ ( z f ( z ) f ( z ) δ ) .

Then p is analytic in , p(0)=1 and

z f ( z ) f ( z ) =ψ ( p ( z ) , z p ( z ) ) ,

where ψ is given in (8). Also

{ ψ ( p ( z ) , z p ( z ) ) : z U } { w C : Re ( w ) < β } := Ω β .

Now, for all real ρ, σ 1 2 (1+ ρ 2 ),

Re ( ψ ( i ρ , σ ) ) 1 1 2 δ(1δ)g(ρ),

where g(ρ) is given (9). Since

1 ( 1 δ ) 2 <g(ρ) 1 δ 2

for all δ>1, we have

Re ( ψ ( i ρ , σ ) ) 3 δ 1 2 δ =β.

This shows that Re{ψ(iρ,σ)} Ω β . By Lemma 2, we get Re(p(z))>0 in , and this is equivalent to

Re { f ( z ) z } <δ(zU),

and the proof of Theorem 3 is completed. □

Combining the above Theorems 2 and 3, we can obtain the following result:

Theorem 4 Let fA, 1/2α<1<β<3/2 and α<Re{z f (z)/f(z)}<β in . Then

γ(α)<Re { f ( z ) z } <δ(β)(zU),

where γ(α) and δ(β) is given (7) and (10).

3 Some coefficient problems

In this section, we investigate coefficient problems for functions in the class S(α,β). In [1], Kuroki and Owa investigated the coefficient of the function p given by (2); the function p can be written as

p(z)=1+ n = 1 B n z n ,

where

B n = 2 ( β α ) n π sin ( 1 α β α π ) .

We denote by S σ (α,β) the class of bi-univalent functions f such that fS(α,β) and the inverse function f 1 S(α,β). Srivastava et al. investigated the estimates on the initial coefficient for certain subclasses of analytic and bi-univalent functions in [3, 4]. Ali et al. have studied similar problems in [5].

In theorem, we shall solve the Fekete-Szegö problem for fS(α,β). We need the following lemma:

Lemma 3 (Keogh and Merkers [6])

Let p(z)=1+ c 1 z+ c 2 z 2 + be a function with positive real part in . Then, for any complex number ν,

| c 2 ν c 1 2 | 2max { 1 , | 1 2 ν | } .

The following result holds for the coefficient of fS(α,β).

Theorem 5 Let 0α<1<β and let the function f given by f(z)=z+ n = 2 a n z n be in the class S(α,β). Then, for a complex number μ,

| a 3 μ a 2 2 | β α 2 π 2 2 cos ( 1 α β α 2 π ) max { 1 ; | 1 2 + ( 1 2 μ ) β α π i + ( 1 2 ( 1 2 μ ) β α π i ) e 2 π i 1 α β α | } .
(11)

Proof Let us consider a function q given by q(z)=z f (z)/f(z). Then, since fS(α,β), we have q(z)p(z), where

p(z)=1+ β α π ilog ( 1 e 2 π i 1 α β α z 1 z ) .

Let

h(z)= 1 + p 1 ( q ( z ) ) 1 p 1 ( q ( z ) ) =1+ h 1 z+ h 2 z 2 +(zU).

Then h is analytic and has positive real part in the open unit disk . We also have

q(z)=p ( h ( z ) 1 h ( z ) + 1 ) (zU).
(12)

We find from equation (12) that

a 2 = 1 2 B 1 h 1

and

a 3 = 1 4 B 1 h 2 1 8 B 1 h 1 2 + 1 8 B 2 h 1 2 + 1 8 B 1 2 h 1 2 ,

which imply that

a 3 μ a 2 2 = 1 4 B 1 ( h 2 ν h 1 2 ) ,

where

ν= 1 2 ( 1 B 2 B 1 B 1 + 2 μ B 1 ) .

Applying Lemma 3, we can obtain

| a 3 μ a 2 2 | = 1 4 | B 1 | | h 2 ν h 1 2 | 1 2 max { 1 ; | 1 2 ν | } .
(13)

And substituting

B 1 = β α π i ( 1 e 2 π i 1 α β α )
(14)

and

B 2 = β α 2 π i ( 1 e 4 π i 1 α β α )
(15)

in (13), we can obtain the result as asserted. □

Using the above Theorem 5, we can get the following result.

Corollary 1 Let the function f, given by f(z)=z+ n = 2 a n z n , be in the class S(α,β). Also let the function f 1 , defined by

f 1 ( f ( z ) ) =z=f ( f 1 ( z ) ) ,
(16)

be the inverse of f. If

f 1 (w)=w+ n = 2 b n w n ( | w | < r 0 ; r 0 > 1 4 ) ,
(17)

then

| b 2 | 2 ( β α ) π sin ( 1 α β α π )

and

| b 3 | β α 2 π 2 2 cos ( 1 α β α 2 π ) max { 1 ; | 1 2 3 β α π i + ( 1 2 + 3 β α π i ) e 2 π i 1 α β α | } .

Proof Relations (16) and (17) give

b 2 = a 2 and b 3 =2 a 2 2 a 3 .

Thus, we can get the estimate for | b 2 | by

| b 2 |=| a 2 | 2 ( β α ) π sin ( 1 α β α π ) ,

immediately. An application of Theorem 5 (with μ=2) gives the estimates for | b 3 |, hence the proof of Corollary 1 is completed. □

Next, we shall estimate on some initial coefficient for the bi-univalent functions f S σ (α,β).

Theorem 6 Let f be given by f(z)=z+ n = 2 a n z n be in the class S σ (α,β). Then

| a 2 | | B 1 | | B 1 | | B 1 2 + B 1 B 2 |
(18)

and

| a 3 || B 1 |+| B 2 B 1 |,
(19)

where B 1 and B 2 are given by (14) and (15).

Proof If f S σ (α,β), then fS(α,β) and gS(α,β), where g= f 1 . Hence

Q(z):= z f ( z ) f ( z ) p(z)andL(z):= z g ( z ) g ( z ) p(z),

where p(z) is given by (2). Let

h(z)= 1 + p 1 ( Q ( z ) ) 1 p 1 ( Q ( z ) ) =1+ h 1 z+ h 2 z 2 +

and

k(z)= 1 + p 1 ( L ( z ) ) 1 p 1 ( L ( z ) ) =1+ k 1 z+ k 2 z 2 +.

Then h and k are analytic and have positive real part in . Also, we have

Q(z)=p ( h ( z ) 1 h ( z ) + 1 ) andL(z)=p ( k ( z ) 1 k ( z ) + 1 ) .

By suitably comparing coefficients, we get

a 2 = 1 2 B 1 h 1 ,
(20)
2 a 3 a 2 2 = 1 2 B 1 h 2 1 4 B 1 h 1 2 + 1 4 B 2 h 1 2 ,
(21)
a 2 = 1 2 B 1 k 1
(22)

and

3 a 2 2 2 a 3 = 1 2 B 1 k 2 1 4 B 1 k 1 2 + 1 4 B 2 k 1 2 ,
(23)

where B 1 and B 2 are given by (14) and (15), respectively. Now, considering (20) and (22), we get

h 1 = k 1 .
(24)

Also, from (21), (22), (23) and (24), we find that

a 2 2 = B 1 3 ( h 2 + k 2 ) 4 ( B 1 2 + B 1 B 2 ) .
(25)

Therefore, we have

| a 2 2 | | B 1 | 3 4 | B 1 2 + B 1 B 2 | ( | h 2 | + | k 2 | ) | B 1 | 3 | B 1 2 + B 1 B 2 | .

This gives the bound on | a 2 | as asserted in (18). Now, further computations from (21), (23), (24) and (25) lead to

a 3 = 1 8 ( B 1 ( h 2 + 3 k 2 ) + 2 h 1 2 ( B 2 B 1 ) ) .

This equation, together with the well-known estimates:

| h 1 |2,| h 2 |2and| k 2 |2

lead us to inequality (19). Therefore, the proof of Theorem 6 is completed. □

References

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Acknowledgements

The research was supported by Kyungsung University Research Grants in 2013.

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Correspondence to Oh Sang Kwon.

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Authors’ contributions

The corresponding author, OSK carried out the subclasses of analytic functions studies and conceived of the study. YJS participated in the sequence alignment and drafted the manuscript. All authors read and approved the final manuscript.

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Sim, Y.J., Kwon, O.S. Notes on analytic functions with a bounded positive real part. J Inequal Appl 2013, 370 (2013). https://doi.org/10.1186/1029-242X-2013-370

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Keywords

  • starlike functions
  • bounded positive real part
  • Fekete-Szegö problem
  • bi-univalent functions