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Certain sufficient conditions for strongly starlikeness and convexity

Fixed Point Theory and Applications20132013:88

https://doi.org/10.1186/1687-1812-2013-88

  • Received: 16 January 2013
  • Accepted: 25 March 2013
  • Published:

Abstract

The object of the present paper is to derive some sufficient conditions for strongly starlikeness and convexity.

MSC:30C45.

Keywords

  • analytic function
  • starlike function
  • convex function
  • strongly starlike function
  • strongly convex function

1 Introduction

Let A ( n ) ( n 2 ) denote the class of functions f ( z ) of the form
f ( z ) = z + k = n a n z n
(1.1)

which are analytic in the open unit disc U = { z : | z | < 1 } . We write A = A ( 2 ) . Let S and K be the subclasses of A ( n ) consisting of all starlike functions f ( z ) in U and of all convex functions f ( z ) in U, respectively.

If f ( z ) A ( n ) satisfies
| arg ( z f ( z ) f ( z ) ) | < π 2 γ ( z U )
(1.2)
for some γ ( 0 < γ 1 ), then f ( z ) is said to be strongly starlike of order γ in U, and denoted by f ( z ) S ˜ ( γ ) . If f ( z ) A ( n ) satisfies
| arg ( 1 + z f ( z ) f ( z ) ) | < π 2 γ ( z U )
(1.3)

for some γ ( 0 < γ 1 ), then we say that f ( z ) is strongly convex of order γ in U, and we denote by K ˜ ( γ ) the class of all such functions. It is obvious that f ( z ) A ( n ) belongs to K ˜ ( γ ) if and only if z f ( z ) S ˜ ( γ ) . Further, we note that S ˜ ( 1 ) = S and K ˜ ( 1 ) = K .

The strongly starlike and convex functions have been extensively studied by several authors (see, e.g., [111]). The object of the present paper is to derive some sufficient conditions for strongly starlikeness and strongly convexity. Some previous results are extended.

For our purpose, we have to recall here the following results.

Lemma 1 (see [5])

Let a function p ( z ) = 1 + c 1 z + c 2 z 2 + be analytic in U and p ( z ) 0 ( z U ). If there exists a point z 0 U such that
| arg p ( z ) | < π 2 β ( | z | < | z 0 | )
and
| arg p ( z 0 ) | = π 2 β ( 0 < β 1 ) ,
then
z 0 p ( z 0 ) p ( z 0 ) = i k β ,
where

and p ( z 0 ) 1 / β = ± i a ( a > 0 ).

Lemma 2 (see [4])

If f ( z ) A satisfies
| f ( z ) 1 | < 20 5 ( z U ) ,

then f ( z ) S .

2 Starlikeness and convexity

Our first result is contained in the following.

Theorem 1 Let 0 < α 1 1 + 2 π 0 1 sin 1 ( 2 ρ 1 + ρ 2 ) d ρ . If f ( z ) A ( n ) ( n 2 ) satisfies
| arg f ( z ) | < π 2 α ( z U ) ,
(2.1)
then f ( z ) S ˜ ( β ) , where
β = ( 1 + 2 π 0 1 sin 1 ( 2 ρ 1 + ρ 2 ) d ρ ) α .

Proof

Note that
arg f ( z ) = arg ( z f ( z ) f ( z ) ) + arg ( f ( z ) z )
and
arg ( f ( z ) z ) = arg ( 1 z 0 z f ( t ) d t ) = arg ( 1 z 0 r f ( ρ e i θ ) e i θ d ρ ) ( z = r e i θ , t = ρ e i θ ) = arg ( 0 r f ( ρ e i θ ) d ρ ) .
(2.2)
Let
0 = ρ 0 < ρ 1 < ρ 2 < < ρ m 1 < ρ m = r ,
and
ρ j ρ j 1 = δ m ( j = 1 , 2 , , m ) .
Then, by using (2.2), we have that
| arg ( f ( z ) z ) | = | arg ( lim m j = 1 m δ m f ( ρ j e i θ ) ) | lim m j = 1 m δ m | arg f ( ρ j e i θ ) | .
Since the condition (2.1) implies that
f ( z ) ( 1 + z 1 z ) α ( z U ) ,
we obtain that
| arg ( f ( z ) z ) | lim m j = 1 m δ m | arg ( 1 + ρ j e i θ 1 ρ j e i θ ) α | < α 0 r sin 1 ( 2 ρ 1 + ρ 2 ) d ρ < α 0 1 sin 1 ( 2 ρ 1 + ρ 2 ) d ρ = π 2 α ( 2 π 0 1 sin 1 ( 2 ρ 1 + ρ 2 ) d ρ ) .
(2.3)
Furthermore, since
| arg ( z f ( z ) f ( z ) ) | | arg ( f ( z ) z ) | | arg f ( z ) | ( z U ) ,
we conclude from (2.1) and (2.3) that
| arg ( z f ( z ) f ( z ) ) | | arg f ( z ) | + | arg ( f ( z ) z ) | < π 2 α + π 2 α ( 2 π 0 1 sin 1 ( 2 ρ 1 + ρ 2 ) d ρ ) = π 2 β ,

which shows that f ( z ) S ˜ ( β ) . □

Theorem 2 Let 0 < α 1 . If f ( z ) A ( n ) ( n 2 ) satisfies
| arg ( f ( z ) + z f ( z ) ) | < π 2 α ( α 1 + 2 π tan 1 α 1 ) ( z U ) ,
(2.4)
then f ( z ) K ˜ ( α ) , where α 1 = 0.3834 is the root of the equation
2 α 1 + 2 π tan 1 α 1 = 1 .

Proof

Note that
arg ( f ( z ) + z f ( z ) ) = arg f ( z ) + arg ( 1 + z f ( z ) f ( z ) ) .
If there exists a point z 0 U such that
| arg f ( z ) | < π 2 α 1 α ( | z | < | z 0 | )
and
| arg f ( z 0 ) | = π 2 α 1 α ,
then by Lemma 1, we have
z 0 f ( z 0 ) f ( z 0 ) = i α 1 α k .
Therefore, if arg f ( z 0 ) = π 2 α 1 α , then we have
arg f ( z 0 ) + arg ( 1 + z 0 f ( z 0 ) f ( z 0 ) ) = π 2 α 1 α + arg ( 1 + i α 1 α k ) = π 2 α 1 α + tan 1 ( α 1 α k ) π 2 α 1 α + α tan 1 α 1 = π 2 α ( α 1 + 2 π tan 1 α 1 ) ,
which contradicts (2.4). If arg f ( z 0 ) = π 2 α 1 α , then applying the same method for the previous case, we also have
arg f ( z 0 ) + arg ( 1 + z 0 f ( z 0 ) f ( z 0 ) ) π 2 α ( α 1 + 2 π tan 1 α 1 ) ,
which contradicts (2.4). Therefore, there exists no z 0 U such that | arg f ( z 0 ) | = π 2 α 1 α . This implies that
| arg f ( z ) | < π 2 α 1 α ( z U ) .
Furthermore, since
| arg ( 1 + z f ( z ) f ( z ) ) | | arg f ( z ) | | arg ( f ( z ) + z f ( z ) ) | < π 2 α ( α 1 + 2 π tan 1 α 1 ) ( z U ) ,
we conclude that
| arg ( 1 + z f ( z ) f ( z ) ) | < π 2 α ( 2 α 1 + 2 π tan 1 α 1 ) = π 2 α ( z U ) ,

which shows that f ( z ) K ˜ ( α ) . □

Theorem 3 If f ( z ) = z + a n z n + A ( n ) ( n 2 ) satisfies
| f ( n ) ( z ) | 20 5 ( z U ) ,
(2.5)

then f ( z ) S .

Proof From (2.5), one can see that
Noting that
| f ( z ) 1 | = | 0 z f ( t ) d t | 0 | z | | f ( t ) | | d t | 20 5 | z | < 20 5 ( z U ) .

By Lemma 2, we have f ( z ) S . □

Theorem 4 If f ( z ) = z + a n z n + A ( n ) ( n 2 ) satisfies
| f ( n ) ( z ) | 5 5 ( z U ) ,
(2.6)

then f ( z ) K .

Proof

By using the same method as in the proof of Theorem 3, we have
| f ( z ) | 5 5 ( z U ) .
It follows that
| ( z f ( z ) ) 1 | = | f ( z ) + z f ( z ) 1 | | f ( z ) 1 | + | z f ( z ) | | 0 z f ( t ) d t | + | z f ( z ) | 0 | z | | f ( t ) | | d t | + 5 5 | z | 2 5 5 | z | < 20 5 ( z U ) .

Therefore, using Lemma 2, we see that z f ( z ) S , or f ( z ) K . □

Declarations

Acknowledgements

Dedicated to Professor Hari M Srivastava.

We would like to express sincere thanks to the referees for careful reading and suggestions which helped us to improve the paper.

Authors’ Affiliations

(1)
Department of Mathematics, Maanshan Teacher’s College, Maanshan, 243000, China
(2)
Department of Mathematics, Yangzhou University, Yangzhou, 225002, China

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Copyright

© Tao and Liu; licensee Springer 2013

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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