Differential inequalities for certain analytic functions

Computers and Mathematics with Applications 56 (2008) 2908–2914

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Differential inequalities for certain analytic functions Mamoru Nunokawa a , Nak Eun Cho b,∗ , Oh Sang Kwon c , Shigeyoshi Owa d , Hitoshi Saitoh e a

Gunma University, Hoshikuki 798-8, Chuou-Ward, Chiba 260-0808, Japan

b

Department of Applied Mathematics, Pukyong National University, Busan 608-737, Republic of Korea

c

Department of Mathematics, Kyungsung University, Busan 608-736, Republic of Korea

d

Department of Mathematics, Kinki University, Higashi-Osaka, Osaka 577-8502, Japan

e

Department of Mathematics, Gunma College of Technology, Toriba, Maebashi, Gunma 371-8530, Japan

article

info

Article history: Received 5 July 2007 Received in revised form 22 April 2008 Accepted 10 September 2008 Keywords: Analytic function Univalent function Starlike function of order α Convex function Subordination

a b s t r a c t The purpose of the present paper is to derive some differential inequalities for certain analytic functions in the open unit disk by appealing to a result obtained by the first author. Relevant connections of the results presented here with those obtained in earlier works are also pointed out. Crown Copyright © 2008 Published by Elsevier Ltd. All rights reserved.

1. Introduction Let A be the class of functions of the form f (z ) = z +

∞ X

an z n

n =2

which are analytic in the open unit disk U = {z ∈ C : |z | < 1}. A function f (z ) ∈ A is said to be starlike of order α in U if and only if it satisfies

 Re

zf 0 (z ) f (z )



> α (0 ≤ α < 1; z ∈ U).

We denote by S ∗ (α) the subclass of A consisting of all functions f (z ) which are starlike of order α in U and in particular, S ∗ (0) ≡ S ∗ . A function f (z ) ∈ A is said to be convex in U if and only if it satisfies

 1 + Re

zf 00 (z ) f 0 (z )



> 0 (z ∈ U).

∗

Corresponding author. E-mail addresses: [email protected] (M. Nunokawa), [email protected] (N.E. Cho), [email protected] (O.S. Kwon), [email protected] (S. Owa), [email protected] (H. Saitoh). 0898-1221/$ – see front matter Crown Copyright © 2008 Published by Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2008.09.007

M. Nunokawa et al. / Computers and Mathematics with Applications 56 (2008) 2908–2914

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If f (z ) and g (z ) are analytic functions in U, then we say that f (z ) is subordinate to g (z ), written as f (z ) ≺ g (z ), if there is a function w(z ) analytic in U, with w(0) = 0 and |w(z )| < 1 for z ∈ U, and such that f (z ) = g (w(z )) for z ∈ U. If g (z ) is univalent in U, then f (z ) ≺ g (z ) if and only if f (0) = g (0) and f (U) ⊂ g (U). Now we introduce the following theorems obtained by some authors. Theorem A ([1]). If f (z ) ∈ A satisfies

 Re

zf 0 (z )

zf 00 (z )

 1+

f (z )



f 0 (z )

> 0 (z ∈ U),

then f (z ) ∈ S ∗ . Theorem B ([2]). If f (z ) ∈ A satisfies

 Re

zf 0 (z ) f (z )

zf 00 (z )

 1+

f 0 (z )



> 0 (z ∈ U),

then f (z ) ∈ S ∗ (1/2). Theorem C ([3]). Let f (z ) ∈ A with f (z )f 0 (z )/z 6= 0 for z ∈ U. If

   zf 00 (z ) zf 0 (z ) +α 1+ 0 > 0 (0 ≤ α < 1; z ∈ U), Re (1 − α) f (z ) f (z ) then f (z ) ∈ S ∗ . Theorem D ([4]). Let h(z ) be a convex function for z ∈ U. If f (z ) ∈ A satisfies

 0  arg f (z ) ≤ π α (0 ≤ α ≤ 1; z ∈ U), h0 (z ) 2 then

  arg f (z2 ) − f (z1 ) ≤ π α (0 ≤ α ≤ 1; z ∈ U). h(z ) − h(z ) 2 2

1

Theorem E ([5]). If f (z ) ∈ A satisfies Re{f 0 (z ) + α zf 00 (z )} > 0 (α ≥ 0; z ∈ U), then Re{f 0 (z )} > 0 for z ∈ U. We note that Chichra [5] supposed the condition Re{α} ≥ 0 in the hypothesis of Theorem E, but the proof was done only for the case α ≥ 0. In the present paper, we extend or modify all theorems mentioned above using the result given by the first author [6]. 2. Main results In proving our main results, we need the following lemma. Lemma 1 ([6]). Let p(z ) be analytic for z ∈ U with p(0) = 1 and p(z ) 6= 0 for z ∈ U. Suppose also that there exists a point z0 ∈ U such that

| arg{p(z )}| <

π 2

α for |z | < |z0 |

(2.1)

and

| arg{p(z0 )}| =

π 2

α,

(2.2)

where 0 < α ≤ 1. Then z0 p0 (z0 ) p(z0 )

= iα k,

(2.3)

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M. Nunokawa et al. / Computers and Mathematics with Applications 56 (2008) 2908–2914

where k≥

1

 a+

2

1



when arg{p(z0 )} =

a

π 2

α

(2.4)

and k≤−

1

 a+

2

1



when arg{p(z0 )} = −

a

π 2

α,

(2.5)

where p(z0 )1/α = ±ia

(a > 0).

(2.6)

Theorem 1. Let f (z ) ∈ A with f (z ) 6= 0 for z ∈ U \ {0} and let Q (z ) be defined by Q (z ) =

zf 0 (z )

 1+

f (z )

zf 00 (z )



(z ∈ U).

f 0 (z )

If Q (z ) 6= m < − 12 and Q (z ) 6= 0 for z ∈ U, then f (z ) ∈ S ∗ . Proof. Let us put zf 0 (z )

p(z ) =

(z ∈ U).

f (z )

Then from the assumption of Theorem 1, p(z ) is analytic in U with p(0) = 1 and p(z ) 6= 0 for z ∈ U. By a simple calculation, we have Q (z ) =

zf 0 (z )



f (z )

1+

zf 00 (z ) f 0 (z )



  zp0 (z ) = p(z ) p(z ) + . p(z )

If there exists a point z0 ∈ U such that conditions (2.1) and (2.2) are satisfied, then by using Lemma 1 for α = 1, we obtain (2.3) under restrictions (2.4), (2.5) and (2.6). For the case arg{p(z0 )} = π /2 and p(z0 ) = ia (a > 0), we have Q (z0 ) = −a2 − ak 1 1 ≤ − (1 + 3a2 ) < − . 2 2 This shows that Q (z0 ) is a real number and less than −1/2, which evidently contradicts the assumption of Theorem 1. For the case arg{p(z0 )} = −π /2 and p(z0 ) = −ia (a > 0), we also have Q (z0 ) = −a2 + ak 1 1 ≤ − (1 + 3a2 ) < − . 2 2 This also is a contradiction to the assumption of Theorem 1. Therefore we complete the proof of Theorem 1. Theorem 2. If f (z ) ∈ A with f (z ) 6= 0 for z ∈ U \ {0} satisfies

 0   00 arg zf (z ) 1 + zf (z ) < π − Tan−1 f (z ) 0 0 f (z ) f (z ) zf (z ) and zf 0 (z ) f (z )

6=

1 2

(z ∈ U),

then

 Re

zf 0 (z ) f (z )



>

or, equivalently, f (z ) ∈ S ∗ (1/2).

1 2

(z ∈ U)

(z ∈ U)



M. Nunokawa et al. / Computers and Mathematics with Applications 56 (2008) 2908–2914

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Proof. Let us put Q (z ) =

zf 0 (z )

 1+

f (z )

zf 00 (z )



f 0 (z )

and zf 0 (z )

=

f (z )

1 2

(p(z ) + 1).

Then from the assumption of Theorem 2, p(z ) is analytic in U with p(0) = 1 and p(z ) 6= 0 for z ∈ U. If there exists a point z0 ∈ U such that conditions (2.1) and (2.2) are satisfied, then by using Lemma 1 for α = 1, we obtain (2.3) under restrictions (2.4)–(2.6). For the case arg{p(z0 )} = π /2 and p(z0 ) = ia (a > 0), we have Q (z0 ) =

= =

z0 f 0 (z0 )

 1+

f (z0 ) 1 2 1 4

(ia + 1)



1 2

z0 f 00 (z0 )



f 0 (z0 )

(ia + 1) +

z0 p0 (z0 )

ia



p(z0 ) (ia + 1)

(1 − a − 2ak + i2a), 2

which implies that 1 Re{Q (z0 )} ≤ − a2 2 and Im{Q (z0 )} =

1 2

a.

Hence we obtain that arg{Q (z0 )} ≥ π − Tan−1

  1 a

−1 f (z0 ) . = π − Tan 0 z f (z ) 0

0

This contradicts the assumption of Theorem 2. For the case arg{p(z0 )} = −π /2, applying the same method as the above, we also have a contradiction to the assumption of Theorem 2. Therefore the proof of Theorem 2 is completed.  Theorem 3. If f (z ) ∈ A with f (z ) 6= 0 for z ∈ U \ {0} satisfies

  zf 00 (z ) +β 1+ 0 6= ±iγ (z ∈ U), f (z ) f (z ) √ where 0 < β or β < −2 and β(β + 2) ≤ γ , then f (z ) ∈ S ∗ . (1 − β)

zf 0 (z )

Proof. Let us put p(z ) =

zf 0 (z ) f (z )

(z ∈ U).

Then p(z ) is analytic in U with p(0) = 1. Applying the same method as in the proof of [7, Theorem 2], and using the assumption of Theorem 3, we can easily see that p(z ) 6= 0 (z ∈ U). Then by a simple calculation, we have

(1 − β)

zf 0 (z ) f (z )



+β 1+

zf 00 (z ) f 0 (z )



= p(z ) + β

zp0 (z ) p(z )

≡ Q (z ).

If there exists a point z0 ∈ U such that conditions (2.1) and (2.2) are satisfied, then by using Lemma 1 for α = 1, we obtain (2.3) under restrictions (2.4)–(2.6).

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M. Nunokawa et al. / Computers and Mathematics with Applications 56 (2008) 2908–2914

For the case arg{p(z0 )} = π /2, p(z0 ) = ia (a > 0) and β > 0, we have

(1 − β)

z0 f 0 (z0 )



+β 1+

f (z0 )

z0 f 00 (z0 )



f 0 (z0 )

= p(z0 ) + β

z0 p0 (z0 ) p(z0 )

= i(a + β k) = Q (z0 ). Then from Lemma 1, we obtain

≥



1

Im{Q (z0 )} ≥

2

p

β

(2 + β)a +



a

(2 + β)β.

This contradicts the assumption of Theorem 3. For the case arg{p(z0 )} = −π /2, p(z0 ) = −ia (a > 0) and β > 0, applying the same method as the above, we have Im{Q (z0 )} ≤ − (2 + β)β.

p

This also contradicts the assumption of Theorem 3. The remaining part of the proof is similar to that of the first part and so we omit the details involved.



Remark 1. We see easily that Theorems 1–3 are more extended results in comparison with Theorems A–C, respectively. Theorem 4. Let f (z ) ∈ A with f (z ) 6= 0 for z ∈ U \ {0} and let g (z ) ∈ A be convex for z ∈ U. If

 0  αβ arg f (z ) < π α + Sin−1 √ α − Sin−1 √ 0 g (z ) 2 1 + α2 1 + α2

(0 < α < 1; 0 < β < 1; z ∈ U)

and zg 0 (z )

≺

g (z )

1 1 − βz

(0 < β < 1; z ∈ U),

then

  arg f (z ) < π α (0 < α < 1; z ∈ U). g (z ) 2 Proof. Let us put p(z ) =

f (z ) g (z )

,

(z ∈ U).

Then p(z ) is analytic in U with p(0) = 1 and p(z ) 6= 0 for z ∈ U. It follows that f 0 (z ) g 0 (z )

g (z ) = p(z ) + 0 p0 (z ) g (z )   zp0 (z ) g (z ) = p(z ) 1 + p(z ) zg 0 (z ) ≡ Q (z ).

If there exists a point z0 ∈ U such that conditions (2.1) and (2.2) are satisfied, then by using Lemma 1, we obtain (2.3) under restrictions (2.4)–(2.6). For the case arg{p(z0 )} = (π /2)α , we have



arg{Q (z0 )} = arg{p(z0 )} + arg 1 + iα k

g (z0 )



z0 g 0 (z0 )

.

Since from the assumption of Theorem 4, g (z ) zg 0 (z )

≺ 1 − β z (z ∈ U),

we obtain that arg{Q (z0 )} ≥

=

π 2

π 2

α + min arg{1 + iα k(1 − β z )} |z |=|z0 |

α + Sin−1 √

α

1+α

2

− Sin−1 √

αβ

1 + α2

.

M. Nunokawa et al. / Computers and Mathematics with Applications 56 (2008) 2908–2914

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This contradicts the assumption of Theorem 4. For the case arg{p(z0 )} = −(π /2)α , using similar arguments to those used in the proof of the first part, we have a contradiction to the assumption of Theorem 4. The proof of Theorem 4 is thus completed.  Remark 2. If we suppose

 0  αβ π arg f (z ) < π α + Sin−1 √ α − Sin−1 √ < 2 2 g 0 (z ) 2 2 1+α 1+α

(0 < α < 1; 0 < β < 1; z ∈ U)

in Theorem 4, then from Theorem D, we have

  arg f (z ) < π g (z ) 2

(z ∈ U).

This shows that f (z ) 6= 0 (z ∈ U \ {0}), and so we do not need the above condition in the assumption of Theorem 4. Theorem 5. If f (z ) ∈ A satisfies Re{f 0 (z ) + β zf 00 (z )} > 0 (β > 0; z ∈ U), then

| arg{f 0 (z )}| <

π 2

α0 (z ∈ U),

where α0 (0 < α0 < 1) is the positive root of the equation

α0 +

2

π

Tan−1 βα0 = 1.

Proof. Let us put p(z ) = f 0 (z ),

(z ∈ U).

Then from Theorem E, we have p(z ) 6= 0 (p(0) = 1; z ∈ U). If there exists a point z0 ∈ U such that conditions (2.1) and (2.2) are satisfied, then by using Lemma 1 for α = α0 , we obtain (2.3) under restrictions (2.4)–(2.6). For the case arg{p(z0 )} = π2 α0 , we see that arg{f 0 (z0 ) + β zf 00 (z0 )} = arg{p(z0 ) + β zp0 (z0 )}

=

π

2

α0 + arg{1 + iβα0 k}

π

α0 + arg{1 + iβα0 }   π 2 −1 = α0 + Tan βα0 . 2 π ≥

2

This contradicts the assumption of Theorem 5. For the case arg{p(z0 )} = − π2 α0 , applying the same method as the above, we also have a contradiction to the assumption of Theorem 5. Therefore we complete the proof of Theorem 5.  Theorem 6. Let f (z ) ∈ A with f 0 (z ) 6= 0 for z ∈ U. If

| arg{f 0 (z ) + β zf 00 (z )}| < π − Tan−1 then Re{f 0 (z )} > 0 (z ∈ U).

1

β

(β > 0; z ∈ U),

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M. Nunokawa et al. / Computers and Mathematics with Applications 56 (2008) 2908–2914

Proof. Let us put p(z ) = f 0 (z )

(z ∈ U).

If there exists a point z0 ∈ U such that

| arg{p(z )}| <

π

for |z | < |z0 |

2

and

| arg{p(z0 )}| =

π 2

,

then from [8, p. 152], we have 1 z0 p0 (z0 ) ≤ − (1 + |p(z0 )|2 ). 2 For the case arg{p(z0 )} = π /2 and p(z0 ) = ia (a > 0), we have arg{f 0 (z0 ) + β z0 f 00 (z0 )} = arg{p(z0 ) + β z0 p0 (z0 )}

  1 ≥ arg ia − β(1 + a2 ) 2

= π − Tan−1

2a

β(1 + a2 ) 1

≥ π − Tan−1 . β This is a contradiction to the assumption of Theorem 6. The remaining part of the proof is much akin to that of the first part and so the details may be omitted.



Remark 3. We note that Theorems 5 and 6 are comparable with Theorem E. Acknowledgments The authors would like to express their gratitude to the referees for many valuable advices regarding a previous version of this paper. References [1] [2] [3] [4] [5] [6] [7] [8]

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