Open Access

Remarks on some starlike functions

Journal of Inequalities and Applications20132013:593

https://doi.org/10.1186/1029-242X-2013-593

Received: 12 November 2013

Accepted: 5 December 2013

Published: 30 December 2013

Abstract

Let A be the class of functions that are analytic in the unit disk D = { z C : | z | < 1 } and normalized by f ( 0 ) = f ( 0 ) 1 = 0 . In this work we investigate conditions under which | z f ( z ) / f ( z ) δ | < δ . Next we also estimate | Arg { f ( z ) / z } | , | Arg { f ( z ) / z 2 } | and | Arg { z f ( z ) / f ( z ) } | for functions of the form f ( z ) = z 2 + a 3 z 3 + in the unit disc | z | < 1 , which satisfy | f ( z ) 2 | < 2 . Furthermore, some geometric consequences of these results are given.

MSC: Primary 30C45; secondary 30C80.

Keywords

differential subordinationsNunokawa’s lemmastarlike functions of order alphaconvex functions of order alphastrongly starlike functions of order alpha

1 Introduction

Let A be the class of functions that are analytic in the unit disk D = { z C : | z | < 1 } and normalized by f ( 0 ) = f ( 0 ) 1 = 0 . The subclasses of A consisting of functions that are univalent in D , starlike with respect to the origin and convex will be denoted by S , S and C , respectively. The class S α of starlike functions of order α < 1 may be defined as
S α = { f A : Re z f ( z ) f ( z ) > α , z U } .
The class S α and the class C α of convex functions of order α < 1
K α : = { f A : Re ( 1 + z f ( z ) f ( z ) ) > α , z U } = { f A : z f S α }
were introduced by Robertson in [1]. If α [ 0 ; 1 ) , then a function in either of these sets is univalent. The convexity in one direction (it implies the univalence) of functions convex of negative order 1 / 2 was proved by Ozaki [2]. In [3] Pfaltzgraff et al. established that the constant 1 / 2 is, in a certain sense, the best possible. A lot of the other equivalent/sufficient conditions for univalence or for the starlikeness, or more, for the convexity in one direction, one can find in [3]. In this work we consider a similar problem, namely find α, β such that
Re ( 1 + z f ( z ) f ( z ) ) < α | z f ( z ) f ( z ) β | < β .

If β ( 0 , 1 ] , it implies also the starlikeness of f.

2 Preliminaries

The following lemma is a simple generalization of Nunokawa’s lemma [4], which together with the lemma from [5] has a surprising number of important applications in the theory of univalent functions.

Lemma 2.1 [6]Let p ( z ) = 1 + n = m 2 c n z n be an analytic function in D . Suppose also that there exists a point z 0 D such that
Re { p ( z ) } > 0 for  | z | < | z 0 |
and
Re { p ( z 0 ) } = 0 and p ( z 0 ) 0 .
Then we have
z 0 p ( z 0 ) p ( z 0 ) = i k ,
where k is a real number and
k m 2 ( a + 1 a ) m 2 when  Arg { p ( z 0 ) } = π 2
and
k m 2 ( a + 1 a ) m 2 when  Arg { p ( z 0 ) } = π 2 ,

where | p ( z 0 ) | = a .

3 Main results

Theorem 3.1 Assume that δ 3 / 4 and m is a positive integer such that m > 4 δ 1 . If f ( z ) = z + n = m a n z n , and z f ( z ) / f ( z ) are analytic in the unit disc D with z f ( z ) 2 δ f ( z ) , f ( z ) 0 , z D and
Re { 1 + z f ( z ) f ( z ) } < { 2 δ + ( δ 1 / 2 ) ( m 1 ) for  δ [ 3 / 4 , 1 )  and  m δ / ( 1 δ ) , m N , m 1 2 ( 2 δ 1 ) for  δ 1  and  m > 4 δ 1 , m N , m 1 2 ( 2 δ 1 ) for  δ [ 3 / 4 , 1 )  and  4 δ 1 < m < δ / ( 1 δ ) , m N ,
(3.1)
then we have
| z f ( z ) f ( z ) δ | < δ for  | z | < 1 .
Proof The function z f ( z ) / f ( z ) is analytic in D , thus we can define the function p by
z f ( z ) f ( z ) δ = δ p ( z ) + 1 2 δ p ( z ) 1 + 2 δ for  | z | < 1 ,
(3.2)

where p ( 0 ) = 1 , and p ( z ) = 1 + p m 1 z m 1 + p m z m +  , z D .

Then it follows that
1 + z f ( z ) f ( z ) = 2 δ p ( z ) p ( z ) 1 + 2 δ + 2 δ 1 p ( z ) 1 + 2 δ z p ( z ) p ( z ) .
(3.3)
If there exists a point z 0 D such that
| z f ( z ) f ( z ) δ | < δ for  | z | < | z 0 |
and
| z 0 f ( z 0 ) f ( z 0 ) δ | = δ ,
then by (3.2)
Re { p ( z ) } > 0 for  | z | < | z 0 |
and
Re { p ( z 0 ) } = 0
and p ( z 0 ) 0 by (3.3). Then applying Lemma 2.1, we have
z 0 p ( z 0 ) p ( z 0 ) = i k ,
where
k ( m 1 ) ( a 2 + 1 ) 2 a when  Arg { p ( z 0 ) } = π 2
(3.4)
and
k ( m 1 ) ( a 2 + 1 ) 2 a when  Arg { p ( z 0 ) } = π 2 ,
and where p ( z 0 ) = ± i a and 0 < a . For the case Arg { p ( z 0 ) } = π / 2 , p ( z 0 ) = i a and 0 < a it follows from (3.3) that
Re { 1 + z 0 f ( z 0 ) f ( z 0 ) } = Re 2 δ i a i a 1 + 2 δ + Re ( 2 δ 1 ) i k i a 1 + 2 δ = 2 a 2 δ a 2 + ( 2 δ 1 ) 2 + ( 2 δ 1 ) a k a 2 + ( 2 δ 1 ) 2 = 2 a 2 δ + ( 2 δ 1 ) a k a 2 + ( 2 δ 1 ) 2 .
Therefore, we have from (3.4)
Re { 1 + z 0 f ( z 0 ) f ( z 0 ) } 2 a 2 δ + ( 2 δ 1 ) a m 1 2 a 2 + 1 a a 2 + ( 2 δ 1 ) 2 = 4 a 2 δ + ( 2 δ 1 ) ( m 1 ) ( a 2 + 1 ) 2 ( a 2 + ( 2 δ 1 ) 2 ) = 2 δ + ( δ 1 / 2 ) ( m 1 ) + 2 δ ( 2 δ 1 ) ( m 1 ) ( 1 δ ) ( 2 δ 1 ) a 2 + ( 2 δ 1 ) 2 for  a > 0 .
In the last expression, the numerator ( m 1 ) ( 1 δ ) ( 2 δ 1 ) is nonnegative if and only if δ [ 3 / 4 , 1 ) and m δ / ( 1 δ ) but this expression tends to 0+ when a . Therefore, in this case we have
Re { 1 + z 0 f ( z 0 ) f ( z 0 ) } 2 δ + ( δ 1 / 2 ) ( m 1 ) for  δ [ 3 / 4 , 1 )  and  m > δ / ( 1 δ ) .
(3.5)
Furthermore, the numerator ( m 1 ) ( 1 δ ) ( 2 δ 1 ) is negative if and only if δ 1 and m N or δ [ 3 / 4 , 1 ) and m < δ / ( 1 δ ) . In this case the quotient decreases when a 0 + . Therefore, in this case we have
Re { 1 + z 0 f ( z 0 ) f ( z 0 ) } 2 δ + ( δ 1 / 2 ) ( m 1 ) + 2 δ { ( m 1 ) ( 1 δ ) ( 2 δ 1 ) } 2 δ 1 = m 1 2 ( 2 δ 1 ) .
(3.6)
We have assumed that m > 4 δ 1 to have the right-hand side in (3.1) greater to 1. So in this case we have
4 δ 1 < m < δ 1 δ for  δ [ 3 / 4 , 1 ) .
(3.7)
Therefore, we can write (3.6) in the form
Re { 1 + z 0 f ( z 0 ) f ( z 0 ) } m 1 2 ( 2 δ 1 ) { either for δ 1  and  m > 4 δ 1 , m N , or for δ [ 3 / 4 , 1 )  and  4 δ 1 < m < δ / ( 1 δ ) .
(3.8)
Inequalities (3.5) and (3.8) contradict the hypothesis of Theorem 3.1, and therefore we have
Re { p ( z ) } > 0 for  | z | < 1 .
(3.9)
Furthermore, from (3.2) and (3.9) we obtain
| z f ( z ) f ( z ) δ | = | δ p ( z ) + 1 2 δ p ( z ) 1 + 2 δ | < δ for  | z | < 1 .
(3.10)

For the case Arg { p ( z 0 ) } = π / 2 , p ( z 0 ) = i a and 0 < a , applying the same method as above, we also have (3.9). Therefore, we get (3.10), which completes the proof of Theorem 3.1. □

Substituting δ = 1 in Theorem 3.1 leads to the following corollary.

Corollary 3.2 If f ( z ) = z + n = m a n z n is analytic in the unit disc D and
Re { 1 + z f ( z ) f ( z ) } < m 1 2 ,
then we have
| z f ( z ) f ( z ) 1 | < 1 for  | z | < 1 .

Substituting δ = 3 / 4 , m = 3 in Theorem 3.1 gives the following corollary.

Corollary 3.3 If f ( z ) = z + n = 3 a n z n is analytic in the unit disc D and
Re { 1 + z f ( z ) f ( z ) } < 13 8 ,
then we have
| z f ( z ) f ( z ) 3 4 | < 3 4 for  | z | < 1 .

Substituting δ = 4 / 5 , m = 3 in Theorem 3.1 gives the following corollary.

Corollary 3.4 If f ( z ) = z + n = 3 a n z n is analytic in the unit disc D and
Re { 1 + z f ( z ) f ( z ) } < 5 2 ,
then we have
| z f ( z ) f ( z ) 4 5 | < 4 5 for  | z | < 1 .
As a supplement to the above results recall here the known result [[7], p.61] that if f ( z ) = z + n = 2 a n z n is analytic in the unit disc D and
Re { 1 + z f ( z ) f ( z ) } R 1 , 1 ( z ) = 1 + z 1 z + 2 z 1 z 2 ,
then
Re z f ( z ) f ( z ) > 0 for  | z | < 1 .

Note that the open door function R 1 , 1 ( z ) maps D onto the complex plane with slits along the half-lines Re { w } = 0 , and | Im { w } | 3 . Next we give the bounds for | Arg { z f ( z ) / f ( z ) } | .

Theorem 3.5 Let f ( z ) = z 2 + n = 3 a n z n be analytic in the unit disc D . If
| f ( z ) 2 | < 2 for  | z | < 1 ,
(3.11)
then
| f ( z ) z 2 | < 1 for  | z | < 1 .
(3.12)
Proof By the Schwarz lemma we have
| f ( t e i φ ) 2 | 2 t , t [ 0 , 1 ) .
Let z = r e i φ , r [ 0 , 1 ) , and let φ be fixed. Using this we obtain
| f ( z ) z 2 | = | f ( z ) 2 z | | z | = | 0 z ( f ( u ) 2 ) d u | | z | = | 0 r ( f ( t e i φ ) 2 ) d ( t e i φ ) | | r e i φ | = | 0 r e i φ ( f ( t e i φ ) 2 ) d t | | r e i φ | 0 r | e i φ ( f ( t e i φ ) 2 ) | d t | r e i φ | 0 r 2 t d t r = r 2 r = r < 1 .

Therefore, we obtain (3.12). □

For the function f ( z ) = z 3 / 3 + z 2 , condition (3.11) is satisfied while (3.12) becomes | z | < 1 in the unit disc, which shows that the constant 1 in (3.12) cannot be replaced by a smaller one. A simple geometric observation yields the following corollary.

Corollary 3.6 Let f ( z ) = z 2 + n = 3 a n z n be analytic in the unit disc D . If
| f ( z ) 2 | < 2 for  | z | < 1 ,
(3.13)
then
| Arg { f ( z ) z } | < π 6 for  | z | < 1 .
(3.14)

Using the same method as in the proof of Theorem 3.5, we can obtain the following result.

Theorem 3.7 Let f ( z ) = z 2 + n = 3 a n z n be analytic in the unit disc D . If
| f ( z ) 2 | < 2 for  | z | < 1 ,
(3.15)
then
| f ( z ) z 2 1 | < 1 3 for  | z | < 1 .
(3.16)

For the function f ( z ) = z 3 / 3 + z 2 , condition (3.15) is satisfied while (3.16) becomes | z / 3 | < 1 / 3 in the unit disc, which shows that the constant 1 / 3 in (3.16) cannot be replaced by a smaller one. A simple geometric observation yields the following corollary.

Corollary 3.8 Let f ( z ) = z 2 + n = 3 a n z n be analytic in the unit disc D . If
| f ( z ) 2 | < 2 for  | z | < 1 ,
(3.17)
then
| Arg { f ( z ) z 2 } | < sin 1 1 3 for  | z | < 1 .
(3.18)

Using Corollaries 3.6 and 3.8 together, we obtain the next one.

Corollary 3.9 Let f ( z ) = z 2 + n = 3 a n z n be analytic in the unit disc D . If
| f ( z ) 2 | < 2 for  | z | < 1 ,
(3.19)
then
| Arg { z f ( z ) f ( z ) } | < π 6 + sin 1 1 3 0.8634 for  | z | < 1 .
(3.20)
Proof From (3.12) and from (3.16), we have
| Arg { z f ( z ) f ( z ) } | = | Arg { f ( z ) z z 2 f ( z ) } | | Arg { f ( z ) z } | + | Arg { f ( z ) z 2 } | < π 6 + sin 1 1 3 0.8634 .

 □

Recall the class SS ( β ) of strongly starlike functions of order β, 0 < β 1 ,
SS ( β ) : = { f A : | Arg z f ( z ) f ( z ) | < β π 2 , z U } ,

which was introduced in [8] and [9]. Therefore, Corollary 3.9 says that if f satisfies the assumptions, then it is 2-valently strongly starlike of order at least 0.8634.

Declarations

Authors’ Affiliations

(1)
University of Gunma
(2)
Department of Mathematics, Rzeszów University of Technology

References

  1. Robertson MS: On the theory of univalent functions. Ann. Math. 1936, 37: 374–408. 10.2307/1968451View ArticleGoogle Scholar
  2. Ozaki S: On the theory of multivalent functions. Sci. Rep. Tokyo Bunrika Daigaku 1941, 4: 45–86.MathSciNetGoogle Scholar
  3. Pfaltzgraff JA, Reade MO, Umezawa T: Sufficient conditions for univalence. Ann. Fac. Sci. Kinshasa Zaïre Sect. Math.-Phys. 1976, 2: 94–100.MathSciNetGoogle Scholar
  4. Nunokawa M: On properties of non-Carathéodory functions. Proc. Jpn. Acad. Ser. A 1992, 68(6):152–153. 10.3792/pjaa.68.152MathSciNetView ArticleGoogle Scholar
  5. Nunokawa M: On the order of strongly starlikeness of strongly convex functions. Proc. Jpn. Acad. Ser. A 1993, 69(7):234–237. 10.3792/pjaa.69.234MathSciNetView ArticleGoogle Scholar
  6. Nunokawa, M, Sokół, J: New conditions for starlikeness and strongly starlikeness of order alpha (submitted)Google Scholar
  7. Miller SS, Mocanu PT Series of Monographs and Textbooks in Pure and Applied Mathematics 225. In Differential Subordinations: Theory and Applications. Dekker, New York; 2000.Google Scholar
  8. Stankiewicz J: Quelques problèmes extrêmaux dans les classes des fonctions α -angulairement étoilées. Ann. Univ. Mariae Curie-Skłodowska, Sect. A 1966, 20: 59–75.MathSciNetGoogle Scholar
  9. Brannan DA, Kirwan WE: On some classes of bounded univalent functions. J. Lond. Math. Soc. 1969, 1(2):431–443.MathSciNetView ArticleGoogle Scholar

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© Nunokawa and Sokół; licensee Springer. 2013

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