paper 07-17-2009. 81KB Jun 04 2011 12:03:06 AM

ACTA UNIVERSITATIS APULENSIS

No 17/2009

NEW SUFFICIENT CONDITIONS FOR ANALYTIC AND
UNIVALENT FUNCTIONS

Basem A. Frasin

Abstract. The object of the present paper is to obtain new sufficient conditions on f ′′′ (z) which lead to some subclasses of univalent functions defined
in the open unit disk.
2000 Mathematics Subject Classification: 30C45.
1. Introduction and definitions
Let A denote the class of functions of the form :
f (z) = z + a2 z 2 + a3 z 3 + · · · ,

(1)

which are analytic in the open unit disk U = {z : |z| < 1}. A function
f (z) ∈ A is said to be in the class S ∗ (α), the class of starlike functions of
order α, 0 ≤ α < 1, if and only if

zf ′ (z)
Re
f (z)

!

>α

(z ∈ U).

(2)

Then S ∗ = S ∗ (0) is the class of starlike functions in U. Further, if f (z) ∈ A
satisfies



zf ′ (z)  απ

,

(z ∈ U)
(3)
arg
<

f (z) 
2
for some 0 < α ≤ 1, then f (z) said to be strongly starlike function of order α
∗
∗
in U, and this class denoted by S (α). Note that S (1) = S ∗ .
Also let K(α), 0 ≤ α < 1, which consists of functions f (z) ∈ A such that
61

B.A. Frasin - New sufficient conditions for analytic and univalent functions

zf ′′ (z)
Re 1 + ′
f (z)

!

>α

(z ∈ U),

(4)

and K = K(0) is the class of convex functions in U. Further, if f (z) ∈ A
satisfies

!

zf ′′ (z)  απ

arg 1 +
<
,
(z ∈ U)
(5)


f ′ (z) 
2

for some 0 < α ≤ 1, then f (z) said to be strongly convex function of order α in
∗
U, and this class denoted by K(α). Note that K(1) = K and if zf ′ (z) ∈ S (α)
then f (z) ∈ K(α).
Furthermore, a function f (z) ∈ A is said to be in the class R(α) if and
only if
Re{f ′ (z)} > α

(0 ≤ α < 1; z ∈ U),

(6)

Also, let R(α) be the subclasses of A which satisfies
|arg f ′ (z)| <

απ

,
2

(z ∈ U)

(7)

for some 0 < α ≤ 1.
All the above mentioned classes are subclasses of univalent functions in U.
There are many results for sufficient conditions of functions f (z) which
are analytic in U to be starlike, convex, strongly starlike and strongly convex
functions have been given by several researchers. ([1], [2], [3], [4], [5], [6], [7],
[8], [9]). In this paper we will study λ such that the condition|f ′′′ (z)| ≤ λ,
z ∈ U, implies that f (z) belongs to one of the classes defined above.
In order to prove our main result, we shall need the following lemmas
obtained by Tuneski ([9]).
Lemma 1. If f (z) ∈ A satisfies
|f ′′ (z)| ≤

2(1 − α)

1−α

(z ∈ U; 0 ≤ α < 1)

(8)

then f (z) ∈ S ∗ (α).

Lemma 2. If f (z) ∈ A satisfies
|f ′′ (z)| ≤

2 sin(απ/2)
1 + sin(απ/2)
62

(z ∈ U; 0 < α ≤ 1)

(9)

B.A. Frasin - New sufficient conditions for analytic and univalent functions

∗

then f (z) ∈ S (α).

Lemma 3. If f (z) ∈ A satisfies
|f ′′ (z)| ≤

1−α
2−α

(z ∈ U; 0 ≤ α < 1)

(10)

(z ∈ U; 0 ≤ α < 1)

(11)

(z ∈ U; 0 < α ≤ 1)

(12)

then f (z) ∈ K(α).

Lemma 4. If f (z) ∈ A satisfies
|f ′′ (z)| ≤ 1 − α

then f (z) ∈ R(α).

Lemma 5. If f (z) ∈ A satisfies
|f ′′ (z)| ≤ sin

απ
2

then f (z) ∈ R(α).

As a consequence of Theorem 3 in [10], we obtain the following lemma:
Lemma 6. If f (z) ∈ A, satisfies
q

|f ′ (z) + αzf ′′ (z) − 1| < (1 + α) (α − 1)/(α2 + 3α)

(z ∈ U)

(13)

where α > 1, then f (z) ∈ K(1/2).
2.Main Result
Theorem 1. If f (z) = z + a2 z 2 + a3 z 3 + · · · ∈ A satisfies
|f ′′′ (z)| ≤

2(1 − α)
− 2 |a2 |
1−α

then f (z) ∈ S ∗ (α).

Proof. Noting that

63

(z ∈ U; 0 ≤ α < 1)

(14)

B.A. Frasin - New sufficient conditions for analytic and univalent functions

|f ′′ (z)| − 2 |a2 | ≤ |f ′′ (z) − 2a2 |
=


 z

Z


 f ′′′ (σ)dσ 





0

≤

Z|z|
0

and thus,



 ′′′ iθ 
f (te ) dt

Z|z|

 ′′′ iθ 
f (te ) dt + 2 |a2 |
|f (z)| ≤
′′

0

≤

! Z|z|
2(1 − α)
dt + 2 |a2 |
− 2 |a2 |
1−α
0

!

2(1 − α)
− 2 |a2 | |z| + 2 |a2 |
1−α
2(1 − α)
.
≤
1−α

≤

Then by Lemma 1, we have f (z) ∈ S ∗ (α).

Corollary 1. Let f (z) ∈ A with f ′′ (0) = 0 . Then
(i) |f ′′′ (z)| ≤ 2 implies f (z) ∈ S ∗ ;
(ii) |f ′′′ (z)| ≤ 4/5 implies f (z) ∈ S ∗ (1/3);
(iii) |f ′′′ (z)| ≤ 2/3 implies f (z) ∈ S ∗ (1/2); and
(iv) |f ′′′ (z)| ≤ 1/2 implies f (z) ∈ S ∗ (2/3).

Applying the same method as in the proof of Theorem 1 and using Lemmas
2-5 instead of Lemma 1, we obtain the following theorems, respectively.
Theorem 2. If f (z) = z + a2 z 2 + a3 z 3 + · · · ∈ A satisfies
|f ′′′ (z)| ≤

2 sin(απ/2)
− 2 |a2 |
1 + sin(απ/2)

∗

then f (z) ∈ S (α).
64

(z ∈ U; 0 < α ≤ 1)

(15)

B.A. Frasin - New sufficient conditions for analytic and univalent functions
Corollary 2. Let f (z) ∈ A with f ′′ (0) = 0 . Then
(i) |f ′′′ (z)| ≤ 1 implies f (z) ∈ S ∗ ;
∗
(ii) |f ′′′ (z)| ≤ 2/3√implies f (z) ∈ S (1/3);.
∗
(iii) |f ′′′ (z)| ≤ 2( √2 − 1) = 0.8284 . . . implies f (z) ∈ S (1/2); and
∗
(iii) |f ′′′ (z)| ≤ 2(2 3 − 3) = 0.9282 . . . implies f (z) ∈ S (2/3).
Theorem 3. If f (z) = z + a2 z 2 + a3 z 3 + · · · ∈ A satisfies
|f ′′′ (z)| ≤

1−α
− 2 |a2 |
2−α

(z ∈ U; 0 ≤ α < 1)

(16)

then f (z) ∈ K(α).

Corollary 3. Let f (z) ∈ A with f ′′ (0) = 0 . Then
(i) |f ′′′ (z)| ≤ 1/2 implies f (z) ∈ K;.
(ii) |f ′′′ (z)| ≤ 2/5 implies f (z) ∈ K(1/3);.
(iii) |f ′′′ (z)| ≤ 1/3 implies f (z) ∈ K(1/2); and
(iv) |f ′′′ (z)| ≤ 1/4 implies f (z) ∈ K(2/3).

Theorem 4. If f (z) = z + a2 z 2 + a3 z 3 + · · · ∈ A satisfies
|f ′′′ (z)| ≤ 1 − α − 2 |a2 |

(z ∈ U; 0 ≤ α < 1)

(17)

then f (z) ∈ R(α).

Corollary 4. Let f (z) ∈ A with f ′′ (0) = 0 . Then
(i) |f ′′′ (z)| ≤ 1 implies Re f (z) > 0;
(i) |f ′′′ (z)| ≤ 2/3 implies f (z) ∈ R(1/3);.
(ii) |f ′′′ (z)| ≤ 1/2 implies f (z) ∈ R(1/2); and
(iii) |f ′′′ (z)| ≤ 1/3 implies f (z) ∈ R(2/3).

Theorem 5. If f (z) = z + a2 z 2 + a3 z 3 + · · · ∈ A satisfies
|f ′′′ (z)| ≤ sin

απ
− 2 |a2 |
2

then f (z) ∈ R(α).

(z ∈ U; 0 < α ≤ 1)

Corollary 5. Let f (z) ∈ A with f ′′ (0) = 0 . Then
(i) |f ′′′ (z)| ≤ 1/2
√ implies f (z) ∈ R(1/3);
′′′
(ii) |f (z)| ≤ √2/2 = 0.7071 . . . implies f (z) ∈ R(1/2); and
(iii) |f ′′′ (z)| ≤ 3/2 = 0.8660 . . . implies f (z) ∈ R(2/3).
Finally, we prove

65

(18)

B.A. Frasin - New sufficient conditions for analytic and univalent functions
Theorem 6. If f (z) ∈ A, satisfies
|f ′′ (z)| <

q

(α − 1)/(α2 + 3α)

(z ∈ U)

(19)

where α > 1, then f (z) ∈ K(1/2).
Proof. It follows that

|f ′ (z) + αzf ′′ (z) − 1| ≤ |f ′ (z) − 1| + α |zf ′′ (z)|

 z

Z



≤  f ′′ (t)dt + α |zf ′′ (z)|


0

≤

Z|z|
0

q

|f ′′ (t)dt| + α (α − 1)/(α2 + 3α) |z|
q

≤ (1 + α) (α − 1)/(α2 + 3α) |z|
q

< (1 + α) (α − 1)/(α2 + 3α),
Using Lemma 6, we have f (z) ∈ K(1/2).
References
[1] B.A. Frasin, New Sufficient Conditions for Strongly Starlikeness and
Strongly Convex Functions, Kyungpook Math. J. 46(2006), 131-137.
[2] P.T. Mocanu, Some starlikeness conditions for analytic functions, Rev.
Roumaine Math. Pures. Appl. 33 (1988), 117-124.
[3] P.T. Mocanu, Two simple conditions for starlikeness, Mathematica (Cluj).
34 (57) (1992), 175-181.
[4] P.T. Mocanu, Some simple criteria for for starlikeness and convexity,
Libertas Mathematica, 13 (1993), 27-40.
[5] M. Nunokawa, On the order of strongly starlikeness of strongly convex
functions, Proc. Japan. Acad., Sec.A 68 (1993), 234-237.
[6] M. Nunokawa, S. Owa, Y. Polatoğlu, M. Cağlar and E. Yavuz, Some sufficient conditions for starlikeness and convexity, International Short Joint Research Workshop Study on Geometric Univalent Function Theory, May 16-18,
2007, Research Institute for Mathematical Sciences, Kyoto University (RIMS)
at Kyoto, Japan.
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B.A. Frasin - New sufficient conditions for analytic and univalent functions
[7] M. Obradovič, Simple sufficients conditions for starlikeness, Matematicki Vesnik. 49 (1997), 241-244.
[8] G. Oros, A condition for convexity, Gen. Math.Vol. 8, No. 3-4
(2000),15-21.
[9] S. N. Tuneski, On simple sufficient conditions for univalence, Mathematica Bohemica, Vol. 126, No. 1, (2001), 229-236.
[10] S. Ponnusamy and S. Rajasekaran, New sufficient conditions for starlike and univalent functions, Soochow J. Math. 21 (2) (1995), 193-201.
Author:
Basem A. Frasin
Department of Mathematics
Al al-Bayt University
P.O. Box: 130095
Mafraq, Jordan
email:bafrasin@yahoo.com

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