Open Access

On a class of spiral-like functions with respect to a boundary point related to subordination

Journal of Inequalities and Applications20132013:274

https://doi.org/10.1186/1029-242X-2013-274

Received: 29 December 2012

Accepted: 1 May 2013

Published: 31 May 2013

Abstract

For μ C , φ a starlike univalent function, the class of functions f that are spiral-like with respect to a boundary point satisfying the subordination

2 μ z f ( z ) f ( z ) + 1 + z 1 z φ ( z ) , z D ,

is investigated. The integral representation, growth and distortion theorem are proved by relating these functions with Ma and Minda starlike functions. Some earlier results are shown to be a special case of the results obtained.

MSC:30C80, 30C45.

Keywords

subordination starlike with respect to boundary point spiral-like with respect to a boundary point

Dedication

Dedicated to Professor Hari M Srivastava

1 Introduction and motivation

Let D = { z : | z | < 1 } be an open unit disk of the complex plane and let A be a class of analytic functions f normalized by f ( 0 ) = 0 and f ( 0 ) = 1 . Let w 0 be an interior or a boundary point of a set D in . The set D is starlike with respect to w 0 if the line segment joining w 0 to every other point in D lies in the interior of D . If a function f A maps D onto a starlike domain with respect to origin, then f is a starlike function. The class of starlike functions with respect to origin is denoted by S . Analytically,
S : = { f A : Re z f ( z ) f ( z ) > 0 } .
Robertson [1] took a leap forward with the characterization of the class S and defined the class S b of starlike functions with respect to a boundary point. Geometrically, it is the characterization of a function f S b = { f ( z ) = 1 + d 1 z + d 2 z 2 + | f  univalent } such that f ( D ) is starlike with respect to the boundary point f ( 1 ) : = lim r 1 f ( r ) = 0 and lies in a half-plane. The analytic description given by Robertson was
S b : = { f S b : Re ( 2 z f ( z ) f ( z ) + 1 + z 1 z ) > 0 } .

This was partially proved in [1]. It was only in 1984 that the characterization was validated by Lyzzaik [2]. Todorov [3] associated this class with a functional f ( z ) / ( 1 z ) and obtained a structured formula and coefficient estimates in the year 1986. Later, Silverman and Silvia [4] gave a full description of the class of univalent functions on D , the image of which is star-shaped, with respect to a boundary point. Since then, this class of starlike functions with respect to a boundary point has gained notable interest among geometric function theorist and also other researchers. Among them, Abdullah et al. [5] studied the properties of functions in this class. The distortion results for starlike functions with respect to a boundary point were obtained in [6, 7]. The dynamical characterizations of functions starlike with respect to a boundary point can be found in [8]. In the year 2001, Lecko [9] gave another representation of starlike functions with respect to a boundary point. Also, Lecko and Lyzzaik obtained different characterizations of this class in [10].

Following the studies on the class of starlike functions, many authors extensively studied the class of spiral-like functions. For recent work on the class of spiral-like functions, see [11]. Later, there was interest towards the class of spiral-like functions with respect to a boundary point. See [1215]. Aharonov et al. [16] gave a comprehensive definition for spiral-shaped domains with respect to a boundary point.

Definition 1.1 A simply connected domain Ω C , 0 Ω , is called a spiral-shaped domain with respect to a boundary point if there is a number μ C with Re μ > 0 such that, for any point ω Ω , the curve e t μ ω , t 0 , is contained in Ω.

It was also showed in [16] (see also [17]) that each spiral-like function with respect to a boundary point is a complex power of starlike function with respect to a boundary point. In particular, if μ R in Definition 1.1, then Ω is called a star-shaped domain with respect to a boundary point. The following was proved in the same.

Theorem 1.1 Let f be an analytic function with f ( 0 ) = 1 , f ( 1 ) = 0 , and let it be a spiral-like function with respect to a boundary point. Then there exists a number μ Ω : = { λ C : | λ 1 | 1 , λ 0 } such that
Re ( 2 μ z f ( z ) f ( z ) + 1 + z 1 z ) > 0 .
(1.1)

Conversely, if f is a univalent function with f ( 0 ) = 1 and f ( 1 ) = 0 satisfies (1.1) for some μ Ω , then f is a spiral-like function with respect to a boundary point.

Elin [18] then considered the class of spiral-like functions of order β ( 0 < β 1 ) with respect to a boundary point and obtained interesting results including the distortion and covering theorems.

On the other hand, Ma and Minda [19] gave a unified presentation of the class starlike using the method of subordination. For two functions h and g in A , the function h is subordinate to g, written
h ( z ) g ( z ) , z D ,
if there exists a function w A , with w ( 0 ) = 0 and | w ( z ) | < 1 , such that h ( z ) = g ( w ( z ) ) . In particular, if the function g is univalent in D , then h ( z ) g ( z ) is equivalent to h ( 0 ) = g ( 0 ) and h ( D ) g ( D ) . A function h A is starlike if z h ( z ) / h ( z ) is subordinated to ( 1 + z ) / ( 1 z ) . Ma and Minda [19] introduced the class
S ( φ ) = { h A : z h ( z ) h ( z ) φ ( z ) } ,

where φ is an analytic function with a positive real part in D , φ ( D ) is symmetric with respect to the real axis and starlike with respect to φ ( 0 ) = 1 and φ ( 0 ) > 0 . A function f S ( φ ) is called Ma and Minda starlike (with respect to φ). The class S ( β ) consisting of starlike functions of order β, 0 β < 1 and the class S ( A , B ) of Janowski starlike functions are special cases of S ( φ ) when φ ( z ) : = ( 1 + ( 1 2 β ) z ) / ( 1 z ) and φ ( z ) : = ( 1 + A z ) / ( 1 + B z ) for 1 B < A 1 , respectively.

In the same direction and motivated mainly by [18] and [19], we consider the following class.

Definition 1.2 Let f S b , f ( 0 ) = 1 and μ Ω : = { λ C : | λ 1 | 1 , λ 0 } . Also, let φ be an analytic function with a positive real part D , let φ ( D ) be symmetric with respect to the real axis and starlike with respect to φ ( 0 ) = 1 and φ ( 0 ) > 0 . The function f S b ( μ , φ ) if the subordination
2 μ z f ( z ) f ( z ) + 1 + z 1 z φ ( z ) , z D ,
(1.2)

holds.

For φ ( z ) = ( 1 + A z ) / ( 1 + B z ) ( 1 B < A 1 ), denote the class S b ( μ , φ ) by S b ( μ , A , B ) . For 0 β < 1 , A = 1 2 β and B = 1 , denote S b ( μ , A , B ) by S b ( μ , β ) .

The class S b ( μ , φ ) defined by subordination is investigated to obtain representation, estimates for f and f and subordination conditions. We obtained some interesting result in a wider context and our approach is mainly based on [19].

2 Representation for the class S b ( μ , φ )

The following result provides an integral representation of functions belonging to the class S b ( μ , φ ) .

Theorem 2.1 The function f S b ( μ , φ ) if and only if there exists p satisfying p φ such that
f ( z ) = ( 1 z ) μ exp ( μ 2 0 z p ( ζ ) 1 ζ d ζ ) .
Proof Let f S b ( μ , φ ) . Then define p : D C by
p ( z ) = 2 μ z f ( z ) f ( z ) + 1 + z 1 z .
Then f S b ( μ , φ ) implies that p φ . Rewriting the above equation as
2 μ f ( z ) f ( z ) + 2 1 z = p ( z ) 1 z
and integrating from 0 to z, it follows that
log ( f ( z ) 2 μ ( 1 z ) 2 ) = 0 z p ( ζ ) 1 ζ d ζ .
An exponentiation gives
f ( z ) 2 μ = ( 1 z ) 2 exp ( 0 z p ( ζ ) 1 ζ d ζ ) .

The desired result follows from this. The converse follows easily. □

3 Estimates for f and f in the class S b ( μ , φ )

Theorem 3.1 Let h φ be an analytic function with h φ ( 0 ) = 0 , h φ ( 0 ) = 1 satisfying the equation z h φ ( z ) / h φ ( z ) = φ ( z ) . If f S b ( μ , φ ) , then
h φ ( r ) r | 1 z | 2 | f ( z ) 2 μ | h φ ( r ) r | 1 z | 2 , | z | = r .
(3.1)
Proof Define the function h A by
h ( z ) = z ( 1 z ) 2 f ( z ) 2 μ , z D .
(3.2)
Since f is univalent and f ( 1 ) : = lim r 1 f ( r ) = 0 , it is clear that f ( z ) 0 in D . Therefore, the function h is well defined and analytic in D . A computation shows that
z h ( z ) h ( z ) = 2 μ z f ( z ) f ( z ) + 1 + z 1 z .
(3.3)
Hence we have the relation f S b ( μ , φ ) if and only if h S ( φ ) . Ma and Minda [[19], Corollary 1′] have shown that for h S ( φ ) ,
h φ ( r ) | h ( z ) | h φ ( r ) , | z | = r .
Using this inequality for h in (3.2) gives
h φ ( r ) | z ( 1 z ) 2 f ( z ) 2 μ | h φ ( r ) , | z | = r

and hence the desired result follows. □

If S b ( μ , A , B ) and hence
h φ ( z ) = { z ( 1 + B z ) A B B , B 0 , z exp ( A z ) , B = 0 ,
then
| 1 z | 2 ( 1 B r ) A B B | f ( z ) 2 μ | | 1 z | 2 ( 1 + B r ) A B B for  B 0 , | 1 z | 2 exp ( A r ) | f ( z ) 2 μ | | 1 z | 2 exp ( A r ) for  B = 0 .
If S b ( μ , β ) and
h φ ( z ) = z ( 1 z ) 2 2 β ,
then
| 1 z | 2 ( 1 + r ) 2 2 β | f ( z ) 2 μ | | 1 z | 2 ( 1 r ) 2 2 β .
In particular, for 0 μ R , the inequality reduces to the following inequality [18]:
| 1 z | μ ( 1 + r ) μ ( 1 β ) | f ( z ) | | 1 z | μ ( 1 r ) μ ( 1 β ) .
Theorem 3.2 Let φ ( z ) = z h φ ( z ) / h φ ( z ) and f S b ( μ , φ ) . Then, for | z | = r ,
| arg f ( z ) 1 μ ( 1 z ) | 1 2 max | z | = r arg h φ ( z ) z .
For 0 μ R ,
| arg f ( z ) ( 1 z ) μ | | μ | 2 max | z | = r arg h φ ( z ) z .
Proof For a function h S ( φ ) , in the paper [[19], Corollary 3′] it is shown that
| arg h ( z ) z | max | z | = r arg h φ ( z ) z , | z | = r .

The result then follows easily as the relation (3.3) holds. □

Corollary 3.1 If f S b ( μ , A , B ) , then for | z | = r ,
| arg f ( z ) 1 μ ( 1 z ) | A B 2 B max | z | = r arg ( 1 + B z ) for B 0
and
| arg f ( z ) 1 μ ( 1 z ) | 1 2 max | z | = r arg exp ( A z ) for B = 0 .
Corollary 3.2 If f S b ( μ , β ) , then for | z | = r
| arg f ( z ) 1 μ ( 1 z ) | ( 1 β ) max | z | = r arg 1 ( 1 z ) .
Theorem 3.3 Let φ ( z ) = z h φ ( z ) / h φ ( z ) and
min | z | = r | φ ( z ) | = φ ( r ) and max | z | = r | φ ( z ) | = φ ( r ) .
(3.4)
Also, let
H φ 1 = | μ | | 1 z | μ 2 r ( h φ ( r ) r ) μ 2 ( | 1 + z 1 z | + φ ( r ) )
and
H φ 2 = | μ | | 1 z | μ 2 r ( h φ ( r ) r ) μ 2 ( | 1 + z 1 z | + φ ( r ) ) .
For μ R , if f S b ( μ , φ ) then
H φ 1 | f ( z ) | H φ 2 .
Proof By Definition 1.2, for f S b ( μ , φ ) , we have
2 μ z f ( z ) f ( z ) + 1 + z 1 z φ ( z ) , z D .
When (3.4) holds, the above subordination indicates that
φ ( r ) | 2 μ z f ( z ) f ( z ) + 1 + z 1 z | φ ( r ) , | z | = r .
This shows that
| 1 + z 1 z | + φ ( r ) | 2 μ z f ( z ) f ( z ) | | 1 + z 1 z | + φ ( r )
or
| μ | 2 r ( | 1 + z 1 z | + φ ( r ) ) | f ( z ) f ( z ) | | μ | 2 r ( | 1 + z 1 z | + φ ( r ) ) .
(3.5)
For μ R , Theorem 3.1 gives
| 1 z | μ ( h φ ( r ) r ) μ 2 | f ( z ) | | 1 z | μ ( h φ ( r ) r ) μ 2 .
(3.6)

Combining (3.5) and (3.6), the desired results follows. □

We have the following corollaries as (3.4) holds.

Corollary 3.3 Let φ ( z ) = z h φ ( z ) / h φ ( z ) . For B 0 , let
H φ 1 = | μ | | 1 z | μ 2 r ( 1 B r ) μ ( A B ) 2 B ( | 1 + z 1 z | + 1 A r 1 B r )
and
H φ 2 = | μ | | 1 z | μ 2 r ( 1 + B r ) μ ( A B ) 2 B ( | 1 + z 1 z | + 1 + A r 1 + B r ) .
For B = 0 , let
H φ 1 = | μ | | 1 z | μ 2 r exp ( μ A r 2 ) ( | 1 + z 1 z | r exp ( A r ) )
and
H φ 2 = | μ | | 1 z | μ 2 r exp ( μ A r 2 ) ( | 1 + z 1 z | + r exp ( A r ) ) .
For μ R , if f S b ( μ , A , B ) then
H φ 1 | f ( z ) | H φ 2 .
Corollary 3.4 Let φ ( z ) = z h φ ( z ) / h φ ( z ) ,
H φ 1 = | μ | | 1 z | μ 2 r ( 1 + r ) μ ( 1 β ) ( | 1 + z 1 z | + 1 ( 1 2 β ) r 1 + r )
and
H φ 2 = | μ | | 1 z | μ 2 r ( 1 r ) μ ( 1 β ) ( | 1 + z 1 z | + 1 + ( 1 2 β ) r 1 r ) .
For μ R , if f S b ( μ , β ) then
H φ 1 | f ( z ) | H φ 2 .

4 Necessary and sufficient condition

Theorem 4.1 Let φ be a convex univalent function defined on D . The function f S b ( μ , φ ) if and only if for all | s | 1 , | t | 1 ,
s t ( 1 t z 1 s z ) 2 ( f ( s z ) f ( t z ) ) 2 μ h φ ( s z ) h φ ( t z ) ,

where h φ ( z ) = z exp ( 0 z ( ( φ ( ζ ) 1 ) / ζ ) d ζ ) .

Proof Ruscheweyh [[20], Theorem 1] showed that for φ a convex univalent function, F as in the hypothesis and h A
z h ( z ) h ( z ) φ ( z )
if and only if for all | s | 1 , | t | 1 ,
h ( s z ) h ( t z ) h φ ( s z ) h φ ( t z ) .
(4.1)
From the relation (3.3), we know that f S b ( μ , φ ) if and only if h S ( φ ) . Substituting (3.2) in (4.1), we have
s z ( 1 s z ) 2 f ( s z ) 2 μ t z ( 1 t z ) 2 f ( t z ) 2 μ h φ ( s z ) h φ ( t z )

and hence the desired result follows. □

The following corollaries hold for φ ( z ) = 1 + A z 1 + B z is convex univalent on D .

Corollary 4.1 The function f S b ( μ , A , B ) if and only if for all | s | 1 , | t | 1 ,
( 1 t z 1 s z ) μ ( f ( s z ) f ( t z ) ) ( 1 + B s z 1 + B t z ) μ ( A B ) 2 B for B 0 , ( 1 t z 1 s z ) μ ( f ( s z ) f ( t z ) ) exp ( μ A z ( s t ) 2 ) for B = 0 .

Let 0 β < 1 , A = 1 2 β and B = 1 in Corollary 4.1 and hence we have the result.

Corollary 4.2 [18]

The function f S b ( μ , β ) if and only if for all | s | 1 , | t | 1 ,
( 1 t z 1 s z ) μ f ( s z ) f ( t z ) ( 1 t z 1 s z ) μ ( 1 β ) .

Theorem 4.2 as well as Corollaries 4.3 and 4.4 below are respectively special cases of Theorem 4.1 and Corollaries 4.1 and 4.2 when s = 1 and t = 0 . However, we prove the below without the convexity assumption on φ.

Theorem 4.2 If f S b ( μ , φ ) , then
f ( z ) 2 μ ( 1 z ) 2 h φ ( z ) z ,

where h φ ( z ) = z exp ( 0 z ( ( φ ( ζ ) 1 ) / ζ ) d ζ ) .

Proof Clearly z h φ ( z ) / h φ ( z ) = φ ( z ) . If h S ( φ ) , then
z h ( z ) h ( z ) z h φ ( z ) h φ ( z ) .
Therefore by [[19], Theorem 1′]
h ( z ) z h φ ( z ) z .

Let h ( z ) be defined as in (3.2) and hence we arrive at the desired conclusion. □

Corollary 4.3 If f S b ( μ , A , B ) then
f ( z ) ( 1 z ) μ ( 1 + B z ) μ ( A B ) 2 B for B 0
and
f ( z ) ( 1 z ) μ exp ( μ A z 2 ) for B = 0 .

When 0 β < 1 , A = 1 2 β and B = 1 , the above corollary reduces to the following result.

Corollary 4.4 [18]

If f S b ( μ , β ) then
f ( z ) ( 1 z ) μ 1 ( 1 z ) μ ( 1 β ) .

5 Coefficient estimate for f S b ( φ )

In particular, when μ = 1 , (1.2) becomes
2 z f ( z ) f ( z ) + 1 + z 1 z φ ( z ) , z D .

We denote the class satisfying the above subordination as S b ( φ ) .

Theorem 5.1 Let φ ( z ) = 1 + B 1 z + B 2 z 2 +  . If f S b ( φ ) , then the coefficients d 1 , d 2 , d 3 satisfy the following inequalities:
| d 1 | B 1 2 + 1 , | d 2 | B 1 4 max { 1 , | B 2 B 1 + B 1 2 | } + B 1 2 , | d 3 | B 1 6 H ( 6 B 1 2 + 16 B 2 8 B 1 , B 1 3 + 6 B 1 B 2 + 8 B 3 8 B 1 ) + B 1 4 max { 1 , | B 2 B 1 + B 1 2 | } ,
where H ( q 1 , q 2 ) a is as defined in [21] (see also [[22], Lemma 3]) and
| d 2 ν d 1 2 | { B 1 4 ( B 2 B 1 ( 2 ν 1 ) B 1 2 ) + ( 2 ν + 1 ) B 1 2 + 2 ν , ν σ 1 , B 1 4 + ( 2 ν + 1 ) B 1 2 + 2 ν , σ 1 ν σ 2 , B 1 4 ( ( 2 ν 1 ) B 1 2 B 2 B 1 ) + ( 2 ν + 1 ) B 1 2 + 2 ν , ν σ 2 ,
where
σ 1 = 1 B 1 ( B 2 B 1 1 ) + 1 2 , σ 2 = 1 B 1 ( B 2 B 1 + 1 ) + 1 2 .
Proof Define the function g ( z ) = 1 + g 1 z + g 2 z 2 + by
g ( z ) = f ( z ) ( 1 z ) , z D .
Then a computation shows that
2 z g ( z ) g ( z ) + 1 = 2 z f ( z ) f ( z ) + 1 + z 1 z .
Since f S b ( φ ) , we have
2 z g ( z ) g ( z ) + 1 φ ( z ) ,
or there is an analytic function w ( z ) = w 1 z + w 2 z 2 + such that
2 z g ( z ) g ( z ) + 1 = φ ( w ( z ) ) .
Since
2 z g ( z ) g ( z ) + 1 = 1 + 2 g 1 z + ( 2 g 1 2 + 4 g 2 ) z 2 + ( 2 g 1 3 6 g 1 g 2 + 6 g 3 ) z 3 +
and
φ ( w ( z ) ) = 1 + B 1 w 1 z + ( B 2 w 1 2 + B 1 w 2 ) z 2 + ( B 3 w 1 3 + 2 B 2 w 1 w 2 + b 1 w 3 ) z 3 + ,
we see that
g 1 = B 1 w 1 2 , g 2 = B 1 4 ( w 2 + ( B 2 B 1 + B 1 2 ) w 1 2 ) , g 3 = B 1 6 ( w 3 + ( 6 B 1 2 + 16 B 2 8 B 1 ) w 1 w 2 + ( B 1 3 + 6 B 1 B 2 + 8 B 3 8 B 1 ) w 1 3 ) .
In view of the well-known inequality | w 1 | 1 , we have
| g 1 | B 1 2 .
Applying [[23], inequality 7, p.10] and [[22], Lemma 3] (see also [21]), we get
| g 2 | B 1 4 max { 1 , | B 2 B 1 + B 1 2 | }
and
| g 3 | B 1 6 H ( 6 B 1 2 + 16 B 2 8 B 1 , B 1 3 + 6 B 1 B 2 + 8 B 3 8 B 1 ) ,
respectively. Also, we see that applying [[22], Lemma 1] (see also [19]) to inequality
g 2 ν g 1 2 = B 1 4 ( w 2 ( ( 2 ν 1 ) B 1 2 B 2 B 1 ) w 1 2 )
yields
| g 2 ν g 1 2 | { B 1 4 ( B 2 B 1 ( 2 ν 1 ) B 1 2 ) , ν σ 1 , B 1 4 , σ 1 ν σ 2 , B 1 4 ( ( 2 ν 1 ) B 1 2 B 2 B 1 ) , ν σ 2
for σ 1 and σ 2 as in the hypothesis. Todorov in [3] shows that for
g ( z ) = 1 + 1 g n z n ,
the coefficient
g n = 1 + d 1 + d 2 + + d n ,

and hence from the above relation the desired results are obtained. □

Corollary 5.1 When φ ( z ) = ( 1 + z ) / ( 1 z ) , our results coincide with [[3], Corollary 2.3].

Remark 5.1 All the results for the special case when μ = 1 or the class starlike with respect to a boundary point defined by subordination were presented at the 8th International Symposium on GFTA, 27-31 August 2012, Ohrid, Republic of Macedonia and thereafter published as [24].

Endnote

The expression for H is too lengthy to be reproduced here. See [21] or [22] for the full expression.

Declarations

Acknowledgements

The work here is partially supported by MOHE:LRGS/TD/2011/UKM/ICT/03/02.

Authors’ Affiliations

(1)
Faculty of Science and Technology, School of Mathematical Sciences, Universiti Kebangsaan Malaysia

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© Haji Mohd and Darus; licensee Springer 2013

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