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Radius properties for analytic and p-valently starlike functions

Journal of Inequalities and Applications20112011:107

https://doi.org/10.1186/1029-242X-2011-107

Received: 28 June 2011

Accepted: 7 November 2011

Published: 7 November 2011

Abstract

Let A p be the class of functions f(z) which are analytic in the open unit disk U and satisfy z p f ( z ) 0 ( z U ) . Also, let S p * ( α ) denotes the subclass of A p consisting of f(z) which are p-valently starlike of order α(0 α < p). A new subclass U p ( λ ) of A p is introduced by

z 2 z p - 1 f ( z ) - 1 z λ ( z U )

for some real λ > 0. The object of the present paper is to consider some radius properties for f ( z ) S p * ( α ) such that δ - p f ( δ z ) U p ( λ ) .

2010 Mathematics Subject Classification: Primary 30C45.

Keywords

Analyticradius problemCauchy-Schwarz inequalityp-valently starlike of order α.

1 Introduction

Let A p be the class of functions f(z) of the form
f ( z ) = z p + n = p + 1 a n z n ( p = 1 , 2 , 3 , )
(1.1)
which are analytic in the open unit disk U = { z : | z | < 1 } and satisfy
z p f ( z ) = 1 + n = p + 1 b n z n - p 0 ( z U ) .
(1.2)
For f ( z ) A p , we say that f(z) belongs to the class U p ( λ ) if it satisfies
z 2 z p - 1 f ( z ) - 1 z λ ( z U )
(1.3)

for some real number λ > 0.

Let us consider a function f δ (z) given by
f δ ( z ) = z p ( 1 - z ) δ ( δ ) .
(1.4)
Then, we can write that
f δ ( z ) = z p 1 + n = 1 a n z n
with
a n = ( - 1 ) n δ n
and
z 2 z p - 1 f δ ( z ) - 1 z = n = 1 ( n - 1 ) a n z n < n = 1 ( n - 1 ) | a n | .
Thus, if δ = 2, then
z 2 z p - 1 f 2 ( z ) - 1 z < 1 .

This shows that f 2 ( z ) U p ( λ ) for λ 1.

If δ = 3, then we have that
z 2 z p - 1 f 3 ( z ) - 1 z < 5

Which shows that f ( z ) U p ( λ ) for λ 5.

Further, if δ = 4, then
z 2 z p - 1 f 4 ( z ) - 1 z < 1 1

which shows that f ( z ) U p ( λ ) for λ 11.

If p = 1, then f ( z ) U 1 ( λ ) is defined by
z 2 1 f ( z ) - 1 z λ ( z U )
(1.5)
for some real number λ > 0. Note that (1.5) is equivalent to
f ( z ) z f ( z ) 2 - 1 λ ( z U ) .

Therefore, this class U 1 ( λ ) was considered by Obradović and Ponnusamy [1]. Further-more, this class was extended as the class U ( β 1 , β 2 ; λ ) by Shimoda et al. [2].

Let S p * ( α ) denotes the subclass of A p consisting of f(z) which satisfy
R e z f ( z ) f ( z ) > α ( z U )
(1.6)

for some real α (0 α < p).

A function f ( z ) S p * ( α ) is said to be p-valently starlike of order α in U (cf. Robertson [3]).

2 Coefficient inequalities

For f ( z ) A p , we consider the sufficient condition for f(z) to be in the class U p ( λ ) .

Lemma 1 If f ( z ) A p satisfies
n = p + 2 ( n - p - 1 ) | b n | λ ,
(2.1)

then f ( z ) U 1 ( λ ) .

Proof We note that
z 2 z p - 1 f ( z ) - 1 z = n = p + 1 ( n - p - 1 ) b n z n - p < n = p + 1 ( n - p - 1 ) | b n | .
Therefore, if
n = p + 1 ( n - p - 1 ) | b n | = n = p + 2 ( n - p - 1 ) | b n | λ ,

then f ( z ) U p ( λ ) .

Example 1 If we consider a function f ( z ) A p given by
z p f ( z ) = 1 + b p + 1 z + n = p + 2 λ e i φ ( n - p ) ( n - p - 1 ) 2 z n - p 0 ( z U )
with
b n = λ e i φ ( n - p ) ( n - p - 1 ) 2 ( λ > 0 , φ )
for n p + 2, then we see that
n = p + 2 ( n - p - 1 ) | b n | = n = p + 2 λ e i φ ( n - p ) ( n - p - 1 ) < λ n = p + 2 1 n - p - 1 - 1 n - p = λ .
Thus, this function f(z) satisfies the inequality (2.1). Also, we see that
| z 2 ( z p 1 f ( x ) 1 z ) | = | n = p + 2 λ e i φ n p 1 ) ( n p ) z n p | < λ n = p + 2 ( 1 n p 1 1 n p ) = λ .

Therefore, we say that f ( z ) U p ( λ ) .

Next, we discuss the necessary condition for the class S p * ( α ) .

Lemma 2 If f ( z ) S p * ( α ) satisfies
z p f ( z ) = 1 + n = p + 1 b n z n - p 0 ( z U )
with b n = |b n | ei(n-p)θ(n = p + 1, p + 2, p + 3,...), then
n = p + 1 ( n + α - 2 p ) | b n | p - α .
Proof Let us define the function F(z) by
F ( z ) = z p f ( z ) = 1 + n = p + 1 b n z n - p .
It follows that
R e z f ( z ) f ( z ) = R e p - z F ( z ) F ( z ) = R e p - n = p + 1 ( n - 2 p ) b n z n - p 1 + n = p + 1 b n z n - p = R e p - n = p + 1 ( n - 2 p ) | b n | e i ( n - p ) θ z n - p 1 + n = p + 1 | b n | e i ( n - p ) θ z n - p > α
for z U . Letting z = |z| e -iθ , we have that
p - n = p + 1 ( n - 2 p ) | b n | | z | n - p 1 + n = p + 1 | b n | | z | n - p > α ( z U ) .
If we take |z| → 1-, we obtain that
p - n = p + 1 ( n - 2 p ) | b n | 1 + n = p + 1 | b n | α
which implies that
n = p + 1 ( n + α - 2 p ) | b n | p - α .

Remark 1 If we take p = 1 in Lemmas 1 and 2, then we have that

(i) f ( z ) A 1 , n = 2 ( n - 2 ) | b n | λ f ( z ) U 1 ( λ )

and

(ii) f ( z ) S * ( α ) , | b n | = | b n | e i ( n - 1 ) θ n = 2 ( n + α - 2 ) | b n | 1 - α .

3 Radius problems

Our main result for the radius problem is contained in

Theorem 1 Let f ( z ) S p * ( α ) (p - 1 α < p) with
z p f ( z ) = 1 + n = p + 1 b n z n - p 0 ( z U ) .
and b n = | b n | ei(n-p)θ(n = p + 1, p + 2, p + 3, ...). If δ ( | δ | < 1 ) , then 1 δ p f ( δ z ) belongs to the class U p ( λ ) for 0 < | δ | | δ 0 ( λ ) | , where |δ0(λ)| is the smallest positive root of the equation
| δ | 2 1 - α - ( 1 - | δ | 2 ) λ = 0 ,
(3.1)
that is,
| δ 0 ( λ ) | = λ λ + 1 - α .
(3.2)
Proof Since
f ( δ z ) = δ p z p + n = p + 1 a n δ n z n ,
we have that
z p 1 δ p f ( δ z ) = 1 + n = p + 1 b n δ n - p z n - p .
In view of Lemma 1, we have to show that
n = p + 2 ( n - p - 1 ) | b n | | δ | n - p λ .
Note that f ( z ) S p * ( α ) satisfies
| b n | p - α n + α - 2 p < 1 ( p - 1 α < p ) .
Applying Cauchy-Schwarz inequality, we obtain that
n = p + 2 ( n - p - 1 ) | b n | | δ | n - p n = p + 2 ( n - p - 1 ) | b n | 2 1 2 n = p + 2 ( n - p - 1 ) | δ | 2 ( n - p ) 1 2 n = p + 2 ( n - p - 1 ) | δ | 2 ( n - p ) 1 2 p - α .
Let |δ|2 = x. Then, we have that
n = p + 2 ( n - p - 1 ) x n - p = x 2 n = p + 2 ( n - p - 1 ) x n - p - 2 = x 2 n = p + 2 x n - p - 1 = x 2 n = 1 x n - 1 = x 2 ( 1 - x ) 2 .
This gives us that
n = p + 2 ( n - p - 1 ) | b n | | δ | n - p | δ | 2 p - α 1 - | δ | 2 .
Let us define the function h(|δ|) by
h ( | δ | ) = | δ | 2 p - α - ( 1 - | δ | 2 ) λ .
Then, h (|δ|) satisfies h (0) = < 0 and h ( 1 ) = p - α > 0 . Indeed, we have that h (|δ0(λ) |) = 0 for
0 < | δ 0 ( λ ) | = λ λ + p - α < 1 .

This completes the proof of the theorem.

Corollary 1 Let f ( z ) S 1 * ( α ) ( 0 α < 1 ) with
z f ( z ) = 1 + n = 2 b n z n - 1 0 ( z U )
and b n = |b n | ei(n-1)θ(n = 2, 3, 4,...). If δ (|δ| < 1), then 1 δ f ( δ z ) belongs to the class U 1 ( λ ) for 0 < | δ | | δ 0 ( λ ) | , where |δ0(λ)| is the smallest positive root of the equation
| δ | 2 1 - α - ( 1 - | δ | 2 ) λ = 0 ,
that is,
| δ 0 ( λ ) | = λ λ + 1 - α .
Remark 2 In view of (3.2), we define the function g(λ) by
g ( λ ) = | δ 0 ( λ ) | = λ λ + p - α .
Then, we have that
g ( λ ) = 1 2 p - α λ ( λ + p - α ) 3 > 0

for λ > 0. Therefore, |δ0(λ)| given by (3.2) is increasing for λ > 0.

Remark 3 If we put α = p - 1 2 in Theorem 1, then
| δ 0 ( λ ) | = 2 λ 2 λ + 2 .
Therefore, if we consider λ = 1 2 , then we see that
δ 0 1 2 = 1 1 + 2 = 0 . 6 4 3 5 9
and if we make λ = 5, then we have that
| δ 0 ( 5 ) | = 1 0 1 0 + 2 = 0 . 9 3 6 0 0

Declarations

Authors’ Affiliations

(1)
Department of Mathematics, Kazim Karabekir Faculty of Education, Atatürk University, Erzurum, Turkey
(2)
Department of Mathematics, Kinki University, Higashi-Osaka, Osaka, Japan

References

  1. Obradović M, Ponnusamy S: Radius properties for subclasses of univalent functions. Analysis 2005, 25: 183–188. 10.1524/anly.2005.25.3.183MATHGoogle Scholar
  2. Shimoda Y, Hayami T, Owa S: Notes on radius properties of certain univalent functions. Acta Univ Apul 2009, 377–383. (Special Issue)Google Scholar
  3. Robertson MS: On the theory of univalent functions. Ann Math 1936, 37: 374–408. 10.2307/1968451View ArticleGoogle Scholar

Copyright

© Uyanik and Owa; licensee Springer. 2011

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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