Sufficient conditions for Carathéodory functions

Computers and Mathematics with Applications 59 (2010) 2067–2073

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Sufficient conditions for Carathéodory functions In Hwa Kim, Nak Eun Cho ∗ Department of Applied Mathematics, Pukyong National University, Busan 608-737, Republic of Korea

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Article history: Received 23 June 2009 Received in revised form 9 December 2009 Accepted 9 December 2009

The purpose of the present paper is to derive some sufficient conditions for Carathéodory functions in the open unit disk by using Miller and Mocanu’s lemma. Several special cases are considered as the corollaries of main results. © 2009 Elsevier Ltd. All rights reserved.

Keywords: Analytic function Carathéodory function Subordination Univalent function Starlike function

1. Introduction Let P be the class of functions p of the form p(z ) =

∞ X

pn z n

n=0

which are analytic in the open unit disk U = {z ∈ C : |z | < 1}. If p in P satisfies Re{p(z )} > 0

(z ∈ U),

then we say that p is the Carathéodory function. Let A denote the class of all functions f analytic in the open unit disk U = {z : |z | < 1} with the usual normalization f (0) = f 0 (0) − 1 = 0. A function f ∈ A is said to be starlike in U if and only if

 Re

zf 0 (z ) f (z )



> 0 (z ∈ U).

We denote by S ∗ the subclass of A consisting of all such functions. Let f and F be members of A. The function f is said to be subordinate to F if there exists a function w analytic in U, with

w(0) = 0 and |w(z )| < 1 (z ∈ U), such that f (z ) = F (w(z ))

(z ∈ U).

We note (cf. [1,2]) that, if the function F is univalent in U, then f is subordinate to F if and only if f (0) = F (0) and

∗

f (U) ⊂ F (U).

Corresponding author. E-mail addresses: [email protected] (I.H. Kim), [email protected] (N.E. Cho).

0898-1221/$ – see front matter © 2009 Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2009.12.012

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Denote by Q the class of functions q that are analytic and injective on U \ E (q), where

  E (q) = ζ ∈ ∂ U : lim q(z ) = ∞ z →ζ

and are such that q0 (ζ ) 6= 0 (ζ ∈ ∂ U \ E (q)). Further, let the subclass of Q for which q(0) = a be denoted by Q(a). Various sufficient conditions for Carathéodory functions were studied by many authors (see [3–7]), which have been used widely on the space of analytic and univalent functions in U (see [8–12]). In the present paper, we investigate some sufficient conditions for Carathéodory functions by using Miller and Mocanu’s lemma [1]. Moreover, we consider several applications as special cases of main results presented here. 2. Main results To prove our main results, we need the following lemma due to Miller and Mocanu [1, p. 24]. Lemma 1. Let q ∈ Q(a) and let p(z ) = a + an z n + · · ·

(n ≥ 1)

be analytic function in U with p(0) 6= a. If p is not subordinate to q, then there exist points z0 ∈ U

and ξ0 ∈ ∂ U \ E (q)

for which (i) p(z0 ) = q(ξ0 ) and (ii) z0 p0 (z0 ) = mξ0 q0 (ξ0 ) (m ≥ n ≥ 1). With the help of Lemma 1, we now derive the following theorem. Theorem 1. Let P :U→C with Re{P (z )} ≥ Im{P (z )} tan α ≥ 0 (0 ≤ α < π /2). If p is an analytic function in U with p(0) = 1 and Re p(z ) + P (z )zp0 (z ) >





1  2A

(cos α + 2A) sin2 α − A2 cos α ,

where A = Re{P (z )} cos α − Im{P (z )} sin α

(0 ≤ α < π /2; z ∈ U).

(1)

then

|arg{p(z )}| <

π 2

− α (0 ≤ α < π /2; z ∈ U).

Proof. First, let us define the functions q and h1 , respectively, by putting q(z ) = eiα p(z )

(q(z ) 6= eiα ; 0 ≤ α < π /2; z ∈ U)

(2)

and h1 ( z ) =

eiα + eiα z 1−z

(0 ≤ α < π /2; z ∈ U).

Then we see that q and h1 are analytic in U with q(0) = h1 (0) = eiα ∈ C and h1 (U) = {w : Re{w} > 0}. Now we suppose that q is not subordinate to h1 . Then by Lemma 1, there exist points z0 ∈ U

and ξ0 ∈ ∂ U \ {1}

(3)

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2069

such that q(z0 ) = h1 (ξ0 ) = iρ

(ρ ∈ R) and z0 q0 (z0 ) = mξ0 h01 (ξ0 ) (m ≥ 1).

(4)

Here we note that 1 ξ 0 = h− 1 [q(z0 )] =

q(z0 ) − eiα

ξ0 h01 (ξ0 ) = −

and

q(z0 ) + eiα

ρ 2 − 2ρ sin α + 1 ≡ σ1 . 2 cos α

(5)

Using Eqs. (2)–(5) and letting B = Re{P (z0 )} cos α + Im{P (z0 )} sin α

(0 ≤ α < π /2),

we obtain Re p(z0 ) + P (z0 )z0 p0 (z0 ) = Re e−iα q(z0 ) + P (z0 )e−iα z0 q0 (z0 )









= = = ≤

Re e−iα h1 (ξ0 ) + P (z0 )e−iα mξ0 h01 (ξ0 )

=

2 cos α





Re {(cos α − i sin α)iρ + P (z0 )(cos α − i sin α)mσ1 }

ρ sin α + mσ1 {Re{P (z0 ) cos α} + Im{P (z0 ) sin α}} ρ sin α + σ1 B  2 1 = −ρ B + 2ρ sin α(cos α + B) − B 2 cos α 1

g (ρ).

(6)

Then, by a simple calculation, we see that the function g (ρ) in (6) takes the maximum value at ρ ∗ given by

ρ∗ =

sin α(cos α + B) B

.

Hence, we have 1 g (ρ ∗ ) 2 cos α 1  = (cos α + 2B) sin2 α − B2 cos α 2B 1  ≤ (cos α + 2A) sin2 α − A2 cos α , 2A

Re p(z0 ) + P (z0 )z0 p0 (z0 ) ≤





where A is given by (1). This evidently contradicts the assumption of Theorem 1. Therefore we obtain Re{eiα p(z )} > 0 (0 ≤ α < π /2; z ∈ U).

(7)

Next, let us put r (z ) = e−iα p(z )

(0 ≤ α < π /2; z ∈ U)

(8)

and h2 (z ) =

e−iα + e−iα z 1−z

(0 ≤ α < π /2; z ∈ U).

(9)

Then we see that the functions r and h2 are analytic in U with r (0) = h2 (0) = e−iα ∈ C

and h2 (U) = {w : Re{w} > 0} = h1 (U).

If we suppose that r is not subordinate to h2 , then by Lemma 1, there exist points z0 ∈ U

and ξ0 ∈ ∂ U \ {1}

such that r (z0 ) = h2 (ξ0 ) = iρ

(ρ ∈ R) and z0 r 0 (z0 ) = mξ0 h02 (ξ0 ) (m ≥ 1).

(10)

We also note that 1 ξ 0 = h− 2 [r (z0 )] =

r (z0 ) − e−iα r (z0 ) + e−iα

and ξ0 h02 (ξ0 ) = −

ρ 2 + 2ρ sin α + 1 ≡ σ2 . 2 cos α

(11)

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Then applying Eqs. (8)–(11), we get Re p(z0 ) + P (z0 )z0 p0 (z0 ) = Re eiα q(z0 ) + P (z0 )eiα z0 q0 (z0 )







= = = ≤ = ≡ ≤ =



Re eiα h1 (ξ0 ) + P (z0 )eiα mξ0 h01 (ξ0 )





Re {(cos α + i sin α)iρ + P (z0 )(cos α + i sin α)mσ2 }

−ρ sin α + mσ2 {Re{P (z0 )} cos α − Im{P (z0 )} sin α} −ρ sin α + σ2 A  2 1 − ρ A + 2ρ sin α(cos α + A) + A 2 cos α g (ρ) g (− sin α(cos α + A)/A). 1  (cos α + 2A) sin2 α − A2 cos α , 2A

where A is given by (1). This also contradicts the assumption of Theorem 1. Therefore we obtain Re{e−iα p(z )} > 0 (0 ≤ α < π /2; z ∈ U).

(12)

Hence combining inequalities (7) and (12), we conclude that

|arg{p(z )}| <

π 2

− α (0 ≤ α < π /2; z ∈ U).

Therefore we complete the proof of Theorem 1.



If we take P (z ) ≡ β (β > 0) in Theorem 1, the we have the following corollary. Corollary 1. Let p be analytic function in U with p(0) = 1. If Re p(z ) + β zp0 (z ) >





1  2(1 + β) sin2 α − β 2

2β

(0 ≤ α < π /2; β > 0; z ∈ U),

then

|arg{p(z )}| <

π 2

− α (0 ≤ α < π /2; z ∈ U).

Remark. If we let α = 0 in Corollary 1, then we have the result obtained by Nunokawa et al. [13]. If we take P (z ) ≡ 1 in Theorem 1 or β = 1 in Corollary 1, then we have the following result. Corollary 2. Let p be analytic function in U with p(0) = 1. If Re p(z ) + zp0 (z ) > 2 sin2 α −





1 2

(0 ≤ α < π /2; z ∈ U),

then

|arg{p(z )}| <

π 2

− α (0 ≤ α < π /2; z ∈ U).

We can list some special cases of Corollary 2 as follows. Corollary 3. Let f ∈ A. Then (i) Re f 0 (z ) > − 21 (z ∈ U) H⇒ Re

n

f (z ) z

o

> 0 (z ∈ U), n o f (z ) (ii) Re f 0 (z ) > 0 (z ∈ U) H⇒ arg z < π3 (z ∈ U), 





(iii)

 Re

zf 0 (z ) f (z )

 2+

zf 00 (z ) f 0 (z )

−

zf 0 (z ) f (z )



>−

1

(z ∈ U)  0  zf (z ) H⇒ Re > 0 (z ∈ U). f (z ) 2

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2071

Letting P (z ) =

g (z )

(f ∈ A) and α = 0,

zg 0 (z )

we have the following result. Corollary 4. Let f ∈ A and g ∈ S ∗ .

 Re

f 0 (z )



g 0 (z )

1

> − Re

g (z )





zg 0 (z )

2

(z ∈ U) H⇒ Re



f (z )



g (z )

> 0 (z ∈ U).

Theorem 2. Let p be nonzero analytic function in U with p(0) = 1. If



δ1 (α) < Im p(z ) +

zp0 (z )



p(z )

< δ2 (α) (0 ≤ α < π /2; z ∈ U),

where

√

2 cos2 α + 1 + sin α

δ1 (α) = −

cos α

√ and

δ2 (α) =

2 cos2 α + 1 − sin α cos α

,

then

|arg{p(z )}| <

π 2

− α (0 ≤ α < π /2; z ∈ U).

Proof. We define the functions q and h1 by (2) and (3), respectively. If q is not subordinate to h1 , then there exist points z0 ∈ U

and ξ0 ∈ ∂ U \ {1}

satisfying (4). By using Eqs. (2)–(5), we have



Im p(z0 ) +

z0 p0 (z0 )



p(z0 )

  z0 q0 (z0 ) = Im e−iα q(z0 ) + q(z0 )   m ξ0 h01 (ξ0 ) −iα = Im e h(ξ0 ) + h1 (ξ0 ) mσ1 = ρ cos α − (ρ ∈ R \ {0}). ρ

For the case ρ > 0, since σ1 < 0 and m ≥ 1, we obtain

ρ cos α −

mσ1

ρ

σ1 ρ  2 ρ (2 cos2 α + 1) − 2ρ sin α + 1

≥ ρ cos α − =

1 2ρ cos α

≡ g (ρ).

(13)

Since the function g in (13) takes the minimum value at ρ ∗ given by

ρ∗ = √

1 2 cos2 α + 1

,

we have

ρ cos α −

mσ1

ρ

≥ g (ρ ∗ ) √ =

2 cos2 α + 1 − sin α

= δ2 (α),

cos α

,

which is a contradiction to the assumption of Theorem 2. For the case ρ < 0, we put

ρ = −ρ 0 (ρ 0 > 0).

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Then using the same method mentioned above, we obtain

ρ cos α −

mσ1

≤ −ρ 0 cos α +

ρ

1

=

20 cos α ≡ g (ρ 0 )



σ1 ρ0

(ρ 0 )2 (2 cos2 α + 1) + 2ρ 0 sin α + 1

 p  ≤ g 1/ 2 cos2 α + 1 √ 2 cos2 α + 1 + sin α =− , cos α = δ1 (α), which also contradicts the assumption of Theorem 2. Hence we have Re{eiα p(z )} > 0 (0 ≤ α < π /2; z ∈ U).

(14)

Next, considering the functions r and h2 , respectively, defined by (8) and (9) and using a similar method as the above, we obtain Re{e−iα p(z )} > 0 (0 ≤ α < π /2; z ∈ U). Therefore making use of (14) and (15), we have the conclusion of Theorem 2.

(15) 

Taking p(z ) =

zf 0 (z )

(f ∈ A) and α = 0

f (z )

in Theorem 2, we have the following corollary. Corollary 5. Let f ∈ A. Then

  00 √ Im 1 + zf (z ) < 3 (z ∈ U) H⇒ f ∈ S ∗ . f 0 (z ) Theorem 3. Let p be nonzero analytic function in U with p(0) = 1. If

0 p(z ) + zp (z ) − 1 < 3 |p(z )| cos α (0 ≤ α < π /2; z ∈ U), 2 p(z ) then

|arg {p(z )}| <

π 2

− α (0 ≤ α < π /2; z ∈ U).

Proof. Let q(z ) =

eiα p(z )

(0 ≤ α < π /2; z ∈ U)

and let h1 be the function as in the proof of Theorem 1. If q is not subordinate to h1 , then there exist points and ξ0 ∈ ∂ U \ {1}

z0 ∈ U

satisfying (4). By using Eqs. (2)–(5) and (16), we have

p(z0 ) +

z0 p0 (z0 ) p(z0 )

|p(z0 )|

− 1

= e−iα q(z0 ) + e−iα z0 q0 (z0 ) − 1 = e−iα h1 (ξ0 ) + e−iα mξ0 h01 (ξ0 ) − 1 = |(ρ mσ1 sin α cos α − 1) + i(ρ cos α − mσ1 sin α)|  1/2 = (ρ − sin α)2 + (mσ1 − cos α)2 1/2  ≥ (ρ − sin α)2 + (|σ1 | + cos α)2

(16)

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( (ρ − sin α)2 +

= ≥

3 2



(ρ − sin α)2 3 + cos α 2 cos α 2

2073

2 )1/2

cos α,

which is a contradiction to the assumption of Theorem 3. Hence we have ei α

 Re



p(z )

> 0 (0 ≤ α < π /2; z ∈ U).

(17)

Next, we consider the function r defined by r (z ) =

e−iα p(z )

(0 ≤ α < π /2; z ∈ U)

and the function h2 defined by (9). Using a similar method as the above, we obtain

 Re

e− i α p(z )



> 0 (0 ≤ α < π /2; z ∈ U).

Therefore by virtue of (17) and (18), we have Theorem 3.

(18) 

If we take p(z ) =

zf 0 (z ) f (z )

and

α=0

in Theorem 3, we have the following result. Corollary 6. Let f ∈ A with f (z )/z 6= 0 in U. Then

00 zf (z ) 3 zf 0 (z ) < f 0 (z ) 2 f (z )

(z ∈ U) H⇒ f ∈ S ∗ .

Acknowledgements The authors would like to express their gratitude to the referees for many valuable advices regarding a previous version of this paper. This work was supported by the Korea Research Foundation Grant funded by the Korean Government (MOEHRD, Basic Research Promotion Fund) (KRF-2008-313-C00035). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13]

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