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Conditions for starlikeness of the Libera operator

Abstract

Let denote the class of functions f that are analytic in the unit disc and normalized by f(0)= f (0)1=0. In this paper some conditions are determined for starlikeness of the Libera integral operator F(z)= 2 z 0 z f(t)dt.

MSC:30C45, 30C80.

1 Introduction

Let be the class of functions analytic in the unit disk D={zC:|z|<1}, and let us denote by A n the class of functions fH with the normalization of the form

f(z)=z+ a n + 1 z n + 1 + a n + 2 z n + 2 +,zD,

with A 1 =A.

Let SS (β) denote the class of strongly starlike functions of order β, 0<β1,

SS (β)= { f A : | arg z f ( z ) f ( z ) | < β π 2 , z D } ,

which was introduced in [1] and [2], and SS (1) S is the well-known class of starlike functions in . Functions in the class

R(β)= { f A : Re { f ( z ) } > β , z D } ,

where β<1 are called functions with bounded turning. The Libera transform L:AA, L[f]=F, where

F(z)= 2 z 0 z f(t)dt,
(1.1)

is the Libera integral operator, which has been studied by several authors on different counts. In [3] Mocanu considered the problem of starlikeness of F and proved the following result.

Theorem 1.1 [3]

If f(z) is analytic and Re{ f (z)}>0 in the unit disc and if the function F is given in (1.1), then F S .

This result may be written briefly as follows:

L [ R ( 0 ) ] S = SS (1),
(1.2)

where L[R(0)]={L[f]:fR(0)}. In 1995 Mocanu [4] improved (1.2) by showing that

L [ R ( 0 ) ] SS (8/9).
(1.3)

In 2002 Miller and Mocanu [5] showed that a subcase of this last result can be sharpened to

L [ R ( 0 ) A 2 ] SS (2/3).

The problem of strongly starlikeness of L[f] for fR(0) was consider also in [6] where it is shown that

L [ R ( 0 ) A 2 ] SS (3/5).

The above inclusion relationship is equivalent to the following differential implication:

f A 2 andRe { f ( z ) } >0|arg { z F ( z ) F ( z ) } |< 3 π 10

or equivalently

F A 2 andRe { F ( z ) + 1 2 z F ( z ) } >0|arg { z F ( z ) F ( z ) } |< 3 π 10 ,

where F is given by (1.1).

In [7] Ponnusamy improved (1.2) by showing that

L [ R ( ϱ ) ] S ,ϱ=0.09032572.
(1.4)

On the order of starlikeness of convex functions was considered also in the recent paper [8].

2 Main result

In this paper we go back to the problem of starlikeness of Libera transform. We need the following lemmas.

Lemma 2.1 [[9], p.73]

Let n be a positive integer, λ>0 and let β 0 = β 0 (λ,n) be the positive root of the equation

βπ=3π/2 tan 1 (nλβ).
(2.1)

In addition, let

α=α(β,λ,n)=β+(2/π) tan 1 (nλβ)
(2.2)

for 0<β β 0 . If p(z)=1+ p n z n + p n + 1 z n + 1 + is analytic in , then

p(z)+λz p (z) ( 1 + z 1 z ) α ,zD,
(2.3)

implies the following subordination:

p(z) ( 1 + z 1 z ) β ,zD.
(2.4)

If in Lemma 2.1 we put n=1, λ=1/2, then the solution β 0 of (2.1) satisfies β 0 >1, so we may take β=1, which gives πα/2=π/2+ tan 1 (1/2)=2.03 .

Corollary 2.2 Assume that f(z) A 1 . If

|arg { F ( z ) + ( 1 / 2 ) z F ( z ) } |<π/2+ tan 1 (1/2)=2.03,zD,

then

Re { F ( z ) } >0,zD.

Note that if F(z) A 2 , then a sufficient condition for FR(0) is |arg{ f (z)}|<3π/4=2.356 ; see [[5], p.96].

Lemma 2.3 [10]

Let p(z) be of the form

p(z)=1+ n = m 1 a n z n , a m 0(zD),
(2.5)

with p(z)0 in . If there exists a point z 0 , | z 0 |<1, such that

|arg { p ( z ) } |<πα/2in |z|<| z 0 |and|arg { p ( z 0 ) } |=πα/2

for some α>0, then we have

z 0 p ( z 0 ) p ( z 0 ) =ikα,

where

km ( a 2 + 1 ) /(2a),when arg { p ( z 0 ) } =πα/2
(2.6)

and

km ( a 2 + 1 ) /(2a),when arg { p ( z 0 ) } =πα/2,
(2.7)

where

{ p ( z 0 ) } 1 / α =±ia,a>0.

Lemma 2.4 [[9], p.75], [11]

Let p(z)=1+ n = 1 a n z n be analytic in the unit disc . If

p(z)+z p (z) 1 + z 1 z ,zD,

then

p(z)q(z)= 2 z log 1 1 z 1

and

|arg { p ( z ) } |< θ 0 = max | z | = 1 |arg { q ( z ) } |=0.9110,zD,
(2.8)

where θ 0 lies between 0.911621904 and 0.911621907.

Theorem 2.5 Let q(z) be analytic in and suppose that

|arg { q ( z ) } |< β π 2 ,zD

for some β(0,1]. If p(z) is analytic and p(z)0 in with p(0)=1 and such that

|arg { q ( z ) ( z p ( z ) + p 2 ( z ) + p ( z ) ) } |< tan 1 β,zD,
(2.9)

then we have

|arg { p ( z ) } |< β π 2 ,zD.

Proof If there exists a point z 0 , | z 0 |<1, for which

|arg { p ( z ) } |<πβ/2 ( | z | < | z 0 | )

and

|arg { p ( z 0 ) } |=πβ/2,p( z 0 )= ( ± i a ) β ,

then from Nunokawa’s Lemma 2.3, we have

z 0 p ( z 0 ) p ( z 0 ) =ikβ,

where

k a 2 + 1 2 a 1,when arg { p ( z 0 ) } =πβ/2

and

k a 2 + 1 2 a 1,when arg { p ( z 0 ) } =πβ/2.

For the case arg{p( z 0 )}=βπ/2, we have

| arg { q ( z 0 ) [ z 0 p ( z 0 ) + p 2 ( z 0 ) + p ( z 0 ) ] } | = | arg { q ( z 0 ) p ( z 0 ) [ 1 + p ( z 0 ) + z 0 p ( z 0 ) / p ( z 0 ) ] } | = | π β 2 + arg { q ( z 0 ) } + arg { 1 + p ( z 0 ) + z 0 p ( z 0 ) / p ( z 0 ) } | = | π β 2 + arg { q ( z 0 ) } + tan 1 { β k + a β sin ( π β / 2 ) 1 + a β cos ( π β / 2 ) } | ,
(2.10)

where p( z 0 )= ( i a ) β , 0<a and

k a 2 + 1 2 a 1.

Let us put

g(a)= k β + a β sin ( π β / 2 ) 1 + a β cos ( π β / 2 ) ,0<a,

then it is easy to see that

g(a) β + a β sin ( π β / 2 ) 1 + a β cos ( π β / 2 ) ,0<a.
(2.11)

Putting

h(x)= β + x sin ( π β / 2 ) 1 + x cos ( π β / 2 ) ,0x,

we have

h (x)= sin ( π β / 2 ) β cos ( π β / 2 ) ( 1 + x cos ( π β / 2 ) ) 2 >0,0x,

because tan(πβ/2)>β. Therefore, for x>0 we get h(x)>h(0)=β, so from (2.11) we have

g(a)>β,

and so

tan 1 { k β + a β sin ( π β / 2 ) 1 + a β cos ( π β / 2 ) } > tan 1 β,0<a.

Therefore, we have the following inequality from (2.10):

| arg { q ( z 0 ) [ z 0 p ( z 0 ) + p 2 ( z 0 ) + p ( z 0 ) ] } | π β 2 + tan 1 k + a β sin ( π β / 2 ) 1 + a β cos ( π β / 2 ) | arg { q ( z 0 ) } |
(2.12)
> tan 1 β.
(2.13)

This contradicts the hypothesis and for the case arg{p( z 0 )}=βπ/2, applying the same method as above, we also have (2.12). This is also a contradiction and it completes the proof. □

Corollary 2.6 Assume that

|arg { f ( z ) } |< tan 1 β,zD
(2.14)

and

|arg { F ( z ) / z } |< β π 2 ,zD
(2.15)

for some β(0,1], where F(z) is given in (1.1). Then we have

|arg { z F ( z ) F ( z ) } |< β π 2 ,zD,

hence F(z) is strongly starlike of order β.

Proof If we set

p(z)= z F ( z ) F ( z ) ,

then

f (z)= F (z)+ 1 2 z F (z)= 1 2 ( F ( z ) z ) ( z p ( z ) + p 2 ( z ) + p ( z ) ) .

If we let q(z)=F(z)/z, then by (2.14) and (2.15) the assumptions of Theorem 2.5 are satisfied. Therefore,

|arg { p ( z ) } |< β π 2 ,zD.

 □

Theorem 2.7 Let q(z) be analytic in , with q(0)=1 and satisfy

Re { z q ( z ) + q ( z ) } >0,zD.

If p(z) is analytic in , with p(0)=1 and if

|arg { q ( z ) ( z p ( z ) + p 2 ( z ) + p ( z ) ) } |< 5 π 6 θ 0 =1.706,zD,

where θ 0 is given in (2.8), then we have

Re { p ( z ) } >0,zD.

Proof By Lemma 2.4, we have

|arg { q ( z ) } |< θ 0 =0.911,zD.
(2.16)

If there exists a point z 0 , | z 0 |<1, such that

|arg { p ( z ) } |<π/2 ( | z | < | z 0 | )

and

|arg { p ( z 0 ) } |=π/2,p( z 0 )=±ia,0<a,

then from Nunokawa’s Lemma 2.3, we have

z 0 p ( z 0 ) p ( z 0 ) =ik,

where

k a 2 + 1 2 a 1,when arg { p ( z 0 ) } =π/2

and

k a 2 + 1 2 a 1,when arg { p ( z 0 ) } =π/2.

For the case arg{p( z 0 )}=π/2, we have

arg { 1 + i a + i k } arg { 1 + i a + i a 2 + 1 2 a } = tan 1 Im { 1 + i a + i a 2 + 1 2 a } Re { 1 + i a + i a 2 + 1 2 a } = tan 1 { 3 a 2 + 1 2 a } tan 1 { 3 } = π 3 .

Therefore, for the case arg{p( z 0 )}=π/2, we have

π 3 arg{1+ia+ik}< π 2 .

Moreover, by (2.16)

arg { q ( z 0 ) } < θ 0 .

Therefore, we can write

| arg { q ( z 0 ) ( z 0 p ( z 0 ) + p 2 ( z 0 ) + p ( z 0 ) ) } | = | arg { p ( z 0 ) [ 1 + p ( z 0 ) + z 0 p ( z 0 ) / p ( z 0 ) ] q ( z 0 ) } | | arg { p ( z 0 ) ( 1 + i a + i k ) } | | arg { q ( z 0 ) } | π 2 + π 3 | arg { q ( z 0 ) } | 5 π 6 θ 0 .
(2.17)

This contradicts the hypothesis and for the case arg{p( z 0 )}=π/2, applying the same method as above, we have

|arg { q ( z 0 ) ( z 0 p ( z 0 ) + p 2 ( z 0 ) + p ( z 0 ) ) } | 5 π 6 θ 0 .

This is also a contradiction and it completes the proof. □

Corollary 2.8 Assume that

|arg { f ( z ) } |< 5 π 6 θ 0 =1.706,zD,
(2.18)

then we have

Re { z F ( z ) F ( z ) } >0,zD,
(2.19)

where F(z) is Libera integral given in (1.1).

Proof Because

f (z)= F (z)+ 1 2 z F (z),

by Corollary 2.2 and by (2.18) we obtain

Re { F ( z ) } >0,zD.

Therefore, if we let q(z)=F(z)/z, then

Re { z q ( z ) + q ( z ) } =Re { F ( z ) } >0,zD.

If we set

p(z)= z F ( z ) F ( z ) ,

then

f (z)= F (z)+ 1 2 z F (z)= 1 2 ( F ( z ) z ) ( z p ( z ) + p 2 ( z ) + p ( z ) ) .

The assumptions of Theorem 2.7 are satisfied. Therefore, (2.19) holds. □

Corollary 2.8 is an extension of Mocanu’s result (1.2) from the paper [3] because in (2.18) we have |arg{ f (z)}|<1.706 , while in (1.2) we have the stronger assumption that |arg{ f (z)}|<π/2=1.57 .

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Acknowledgements

The authors would like to express their thanks to the referees for valuable advice regarding a previous version of this paper. This research was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (No. 2011-0007037).

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Nunokawa, M., Sokół, J., Cho, N.E. et al. Conditions for starlikeness of the Libera operator. J Inequal Appl 2014, 135 (2014). https://doi.org/10.1186/1029-242X-2014-135

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Keywords

  • Nunokawa’s lemma
  • starlike functions
  • strongly starlike functions