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On some sufficient conditions for univalence and starlikeness

Abstract

In this work, the conditions for univalence, starlikeness and convexity are discussed.

MSC:30C45, 30C80.

1 Introduction

We shall consider the set of all analytic functions in the open unit disc

D= { z : | z | < 1 }

on the complex plane and

A= { f H : f ( z ) = z + a 2 z 2 + } .

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 D } .

The class S α and the class K α of convex functions of order α<1

K α : = { f A : Re ( 1 + z f ( z ) f ( z ) ) > α , z D } = { 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. In particular, we denote by S 0 = S , K 0 =K the classes of starlike and convex functions, respectively.

2 Preliminaries

Lemma 2.1 Let O=0, P=α+iaα, Q=x+iax and A(0,+) be the points on the complex plane, where 0<αA/2, α<x and <a<. Then we have

| A Q | | O Q | < | A P | | O P | A α α .
(2.1)

Proof For a=0, the assertion is obvious. For a0, consider the triangles OAP and OAQ, see Figure 1. Both of them have the same angle φ at the point O. Let the first have the angle ψ at the point A and the second have the angle ψ at the point A. Then the hypothesis 0<αA/2 implies cosφcosψ. Further we have

cosφ= α | P | ,cosψ= A α | A P | .

This gives the second inequality of the assertion. The first one follows immediately from sin ψ >sinψ and

| A P | | P | = sin φ sin ψ , | A Q | | Q | = sin φ sin ψ .

 □

Figure 1
figure 1

The lines and the points.

3 Main result

Theorem 3.1 Let p(z)=1+ n = k 1 c n z n be analytic in the unit disc D and α be a positive real number 0<α1/2. Then suppose that there exists a point z 0 , | z 0 |<1 such that

Re { p ( z ) } >αfor |z|<| z 0 |
(3.1)

and

Re { p ( z 0 ) } =p( z 0 )=α.
(3.2)

Then we have

z 0 p ( z 0 ) p ( z 0 ) k(1α).
(3.3)

Proof Let us put

q(z)= p ( z ) 1 p ( z ) ,q(0)=0.

Then the function q is analytic in |z|| z 0 |<1 and from the hypothesis of Theorem 3.1 and Lemma 2.1, with A=1, we have

| q ( z ) | < 1 α α

for |z|| z 0 | and

| q ( z 0 ) | = 1 α α .

This shows that |q(z)| takes its maximum at z= z 0 on the circle |z|=| z 0 |. Then from Fukui-Sakaguchi [2] and Jack’s [3] lemmas, there exists a real number k1 such that

z 0 q ( z 0 ) q ( z 0 ) = z 0 p ( z 0 ) p ( z 0 ) 1 z 0 p ( z 0 ) p ( z 0 ) = z 0 p ( z 0 ) ( 1 α 1 1 α ) = z 0 p ( z 0 ) α ( α 1 ) k .

This shows that z 0 p ( z 0 ) is a negative real number and

z 0 p ( z 0 ) α = z 0 p ( z 0 ) p ( z 0 ) k(1α).

This completes the proof of Theorem 3.1. □

Theorem 3.1 is, in a certain sense, the supplement of Nunokawa’s lemma [4]. From Theorem 3.1 we have the following corollaries.

Corollary 3.2 Let p(z)=1+ n = 1 c n z n be analytic in the unit disc D and α be a positive real number 0<α1/2. Suppose also that for arbitrary r, 0<r<1, p satisfies the condition

min | z | r Re { p ( z ) } = min | z | r |p(z)|
(3.4)

and

Re { p ( z ) + z p ( z ) p ( z ) } >2α1for |z|<1.
(3.5)

Then we have

Re { p ( z ) } >αfor |z|<1.
(3.6)

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

Re { p ( z ) } >αfor |z|<| z 0 |

and

Re { p ( z 0 ) } =α,0<α1/2,

then from the hypothesis of Corollary 3.2, we have

Re { p ( z 0 ) } =p( z 0 )=α.

Then from Theorem 3.1, we have

z 0 p ( z 0 ) p ( z 0 ) α1,

and therefore we have

Re { p ( z 0 ) + z 0 p ( z 0 ) p ( z 0 ) } 2α1.

This contradicts the hypothesis of Corollary 3.2 and it completes the proof of Corollary 3.2. □

Corollary 3.3 Let f(z)=z+ n = 2 a n z n be analytic in the unit disc D and α be a positive real number 0<α1/2. Suppose that for arbitrary r, 0<r<1, f satisfies the condition

min | z | r Re { z f ( z ) f ( z ) } = min | z | r | z f ( z ) f ( z ) |
(3.7)

and

Re { 1 + z f ( z ) f ( z ) } >2α1for |z|<1,
(3.8)

where

1<2α10.

Then we have

Re { z f ( z ) f ( z ) } >αfor |z|<1,
(3.9)

or f is starlike of order α.

Proof Putting

p(z)= z f ( z ) f ( z ) ,

it follows that

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

Then from Corollary 3.2, we have (3.9). □

Theorem 3.4 Let p(z)=1+ n = k 1 c n z n be analytic in the unit disc D and α be a positive real number 1/2<α<1. Then suppose that there exists a point z 0 D such that

Re { p ( z ) } >αfor |z|<| z 0 |
(3.10)

and

Re { p ( z 0 ) } =p( z 0 )=α.
(3.11)

Then we have

z 0 p ( z 0 ) p ( z 0 ) =Re { z 0 p ( z 0 ) p ( z 0 ) } k ( 2 α ) 2 .
(3.12)

Proof Let us put

q(z)= 2 p ( z ) p ( z ) 1,q(0)=0.

Then from Lemma 2.1, with A=2, we have that |q(z)+1| takes its maximum value at z= z 0 on the circle |z|=| z 0 | or

max | z | = | z 0 | | q ( z ) + 1 | = max | z | = | z 0 | | 2 p ( z ) p ( z ) |= 2 α α .

Applying Jack [3], Miller-Mocanu [5] and Fukui-Sakaguchi’s [2] lemmas, there exists a real number mk such that

z 0 q ( z 0 ) q ( z 0 ) = z 0 p ( z 0 ) 2 p ( z 0 ) z 0 p ( z 0 ) p ( z 0 ) = z 0 p ( z 0 ) ( 1 2 α + 1 α ) = 2 z 0 p ( z 0 ) α ( 2 α ) = z 0 p ( z 0 ) p ( z 0 ) ( 2 2 α ) m k .

This shows that

z 0 p ( z 0 ) p ( z 0 ) k ( 2 α ) 2 .

This completes the proof of Theorem 3.4. □

Corollary 3.5 Let f(z)=z+ n = 2 a n z n be analytic in the unit disc D and α be a positive real number 1/2<α<1. Suppose that for arbitrary r, 0<r<1, f satisfies the condition

min | z | r Re { z f ( z ) f ( z ) } = min | z | r | z f ( z ) f ( z ) |
(3.13)

and

Re { 1 + z f ( z ) f ( z ) } > 3 α 2 2 for |z|<1.
(3.14)

Then we have

Re { z f ( z ) f ( z ) } >αfor |z|<1.
(3.15)

Proof Applying the same method as in the proof of Corollary 3.3 and in the proof of Lemma 2.1, we can obtain Corollary 3.5. □

Corollary 3.6 Let F(z)=1/z+ n = 1 b n z n be analytic and not vanishing in the punctured unit disc 0<|z|<1 and let α be a positive real number 0<α<1/2. Suppose also that for arbitrary r, 0<r<1, F satisfies the following condition:

min | z | r Re { z F ( z ) F ( z ) } = min | z | r | z F ( z ) F ( z ) |
(3.16)

and

Re { 1 + z F ( z ) F ( z ) } <2αfor |z|<1.
(3.17)

Then we have

Re { z F ( z ) F ( z ) } >αfor |z|<1.
(3.18)

Proof Putting

p(z)= z F ( z ) F ( z ) = z ( 1 / z 2 + n = 1 n b n z n 1 ) 1 / z + n = 1 b n z n ,

then we have

p(z)=1+ n = 2 c n z n ,p(0)=1

and it follows that

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

If there exists a point z 0 D such that

Re { p ( z ) } >αfor |z|<| z 0 |

and

Re { p ( z 0 ) } =p( z 0 )=α,

then from the hypothesis, we have p( z 0 )=α. Applying Theorem 3.1, we have

z 0 p ( z 0 ) p ( z 0 ) 2(1α)

and therefore we have

Re { p ( z 0 ) z 0 p ( z 0 ) p ( z 0 ) } = Re { 1 + z 0 F ( z 0 ) F ( z 0 ) } α + 2 ( 1 α ) = 2 α .

This contradicts the hypothesis and therefore we have

Re { p ( z ) } =Re { z F ( z ) F ( z ) } >αfor |z|<1,

and this shows that F is meromorphic starlike of order α in the punctured unit disc 0<|z|<1. □

We note the following interesting result which was published in a minor journal and so it was not well known in the public of univalent function theory but is strongly connected with the previous Corollaries 3.3, 3.5 and 3.6.

Lemma 3.7 ([6])

Let f(z)=z+ a 2 z 2 + be an analytic function in |z|<1 with f(z) f (z)/z0 in |z|<1. Then, for each α, 1/2<α<0, there exists a function f which satisfies

1+Re z f ( z ) f ( z ) >αin |z|<1,
(3.19)

but f is not starlike in |z|<1.

In [6] the authors pointed out that the function

f(z)= ( 1 z ) 2 α 1 1 1 2 α
(3.20)

satisfies the above conditions.

Lemma 3.8 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 ) } >0for |z|<| z 0 |

and

Re { p ( z 0 ) } =0andp( z 0 )0.

Then we have

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

where k is a real number and

k m 2 ( a + 1 a ) m2when argp( z 0 )= π 2

and

k m 2 ( a + 1 a ) m2when argp( z 0 )= π 2 ,

where

p( z 0 )=±iaanda>0.

Proof Let us put

ϕ(z)= 1 p ( z ) 1 + p ( z ) ,|z|<1.

Then we have ϕ(0)= ϕ (0)== ϕ ( m 1 ) (0)=0, |ϕ(z)|<1, for |z|<| z 0 | and |ϕ( z 0 )|=1. From [2, 3] and [5], we obtain

z 0 ϕ ( z 0 ) ϕ ( z 0 ) = 2 z 0 p ( z 0 ) 1 p 2 ( z 0 ) = 2 z 0 p ( z 0 ) 1 + | p ( z 0 ) | 2 m.

This shows that

z 0 p ( z 0 ) m 2 ( 1 + | p ( z 0 ) | 2 )

and z 0 p ( z 0 ) is a negative real number. For the case argp( z 0 )=π/2, p( z 0 )=ia and 0<a, we have

Re { z 0 p ( z 0 ) p ( z 0 ) } =Re { z 0 p ( z 0 ) i a } =Re { i z 0 p ( z 0 ) a } =0

and

Im { z 0 p ( z 0 ) p ( z 0 ) } =Im { i z 0 p ( z 0 ) a } m 2 ( 1 + | p ( z 0 ) | 2 a ) = m 2 ( a + 1 a ) m.

For the case argp( z 0 )=π/2, p( z 0 )=ia and 0<a, applying the same method as above, we have

Re { z 0 p ( z 0 ) p ( z 0 ) } =0

and

Im { z 0 p ( z 0 ) p ( z 0 ) } m 2 ( a + 1 a ) m.

This completes the proof. □

Theorem 3.9 Let f(z)=z+ n = m + 1 a n z n be analytic in the unit disc D. Suppose also that

Re { 1 + z f ( z ) f ( z ) } < 8 + m 6 for |z|<1
(3.21)

where 1m. If z f (z)/f(z) is analytic in D and omits 4/3, then f is starlike in the unit disc D.

Proof Let us put

z f ( z ) f ( z ) = 4 p ( z ) 1 + 3 p ( z ) ,

where p(0)=1, p (0)= p (0)== p ( m 1 ) (0)=0. Then it follows that

1+ z f ( z ) f ( z ) = z f ( z ) f ( z ) + z p ( z ) p ( z ) 3 z p ( z ) 1 + 3 p ( z ) .

If there exists a point z 0 D such that

Re { p ( z ) } >0for |z|<| z 0 |

and

Re { p ( z 0 ) } =0

and p( z 0 )0, because p( z 0 )=0 contradicts the hypothesis (3.21), then from Lemma 3.8, we have

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

where k is a real number and m|k|. For the case argp( z 0 )=π/2, p( z 0 )=ia and 0<a, we have

1 + z 0 f ( z 0 ) f ( z 0 ) = z 0 f ( z 0 ) f ( z 0 ) + z 0 p ( z 0 ) p ( z 0 ) 3 z 0 p ( z 0 ) p ( z 0 ) p ( z 0 ) 1 + 3 p ( z 0 ) = 4 i a 1 + 3 i a + i k 3 i k i a 1 + 3 i a = 4 i a ( 1 3 i a ) 1 + 9 a 2 + i k + a k ( 1 3 i a ) 1 + 9 a 2 .

Therefore, we have

Re { 1 + z 0 f ( z 0 ) f ( z 0 ) } = 12 a 2 1 + 9 a 2 + 3 a k 1 + 9 a 2 12 a 2 1 + 9 a 2 + 3 m ( 1 + a 2 ) 2 ( 1 + 9 a 2 ) = 24 a 2 + 3 m a 2 + 3 m 2 ( 1 + 9 a 2 ) .

Putting

h(x)= ( 8 + m ) x 2 + m 1 + 9 x 2 ,x>0,

we have

h (x)= 16 ( 1 m ) x ( 1 + 9 x 2 ) 2 <0,x>0.

This shows that

Re { 1 + z 0 f ( z 0 ) f ( z 0 ) } lim a 24 a 2 + 3 m a 2 + 3 m 2 ( 1 + 9 a 2 ) = 24 + 3 m 18 .
(3.22)

This contradicts the hypothesis (3.21).

For the case argp( z 0 )=π/2, p( z 0 )=ia and 0<a, applying the same method as above, we also have (3.22). This is also a contradiction and therefore it completes the proof of Theorem 3.9. □

Remark 3.10 Singh and Singh obtained in [7] that if f(z)=z+ a 2 z 2 + is analytic in D and

Re { 1 + z f ( z ) f ( z ) } < 3 2 for |z|<1,
(3.23)

then f is starlike in D. Earlier, Ozaki [8] proved the univalence of f in D under the same assumption (3.23).

For m=1, the inequality (3.21) becomes (3.23), so Theorem 3.9 is a generalization of the above result.

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Acknowledgements

The authors sincerely thank the referees for pointing out a few corrections in the original draft of the paper.

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Sokół, J., Nunokawa, M. On some sufficient conditions for univalence and starlikeness. J Inequal Appl 2012, 282 (2012). https://doi.org/10.1186/1029-242X-2012-282

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Keywords

  • strongly starlike functions
  • convex functions of order alpha
  • Jack’s lemma
  • Nunokawa’s lemma
  • Umezawa condition
  • univalence criteria