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

NEW SUFFICIENT CONDITIONS FOR ANALYTIC AND UNIVALENT FUNCTIONS

Basem A. Frasin

Abstract. _{The object of the present paper is to obtain new sufficient} con-ditions 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+a2z

2

+a3z 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

Re zf ′_(z) f(z)

!

> α (z _{∈ U}). (2)

Then _S∗ _{= S}∗_{(0) is the class of starlike functions in U. Further, if f(z) ∈ A}

satisfies

argzf ′_(z) f(z)

< απ

2 , (z∈ U) (3)

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∗.

Re 1 + zf ′′_(z) f′_(z)

!

> α (z _{∈ U}), (4)

and _K = _K(0) is the class of convex functions in _U. Further, if f(z) _{∈ A}

satisfies

arg 1 + zf ′′_(z) f′_(z)

!

< απ

2 , (z ∈ U) (5)

for some 0< α_≤1,thenf(z) said to be strongly convex function of orderαin U, and this class denoted by_K(α). Note that _K(1) =_Kand 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

|argf′(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 thatf(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)

then f(z)_{∈ S} (α).

Lemma 3. If f(z)_{∈ A} satisfies

|f′′(z)_{| ≤} 1−α

2₋α (z ∈ U; 0≤α <1) (10)

then f(z)_{∈ K}(α).

Lemma 4. If f(z)_{∈ A} satisfies

|f′′(z)_{| ≤}1₋α (z _{∈ U}; 0_≤α <1) (11)

then f(z)_{∈ R}(α).

Lemma 5. If f(z)_{∈ A} satisfies

|f′′(z)_{| ≤}sinαπ

2 (z ∈ U; 0< α≤1) (12)

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

|f′(z) +αzf′′(z)₋1_|<(1 +α)q(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+a2z 2

+a3z 3

+_{· · · ∈ A} satisfies

|f′′′(z)_{| ≤} 2(1−α)

1₋α −2|a2| (z ∈ U; 0≤α <1) (14)

then f(z)_{∈ S}∗_(α).

|f′′(z)_{| −}2_|a2| ≤ |f′′(z)−2a2| = z Z 0

f′′′(σ)dσ

≤ |z_|

Z

0

f

′′′_(teiθ

) dt

and thus,

|f′′(z)_{| ≤} |z|

Z

0

f

′′′_(teiθ

)

dt+ 2|a2|

≤ 2(1₁ −α)

−α −2|a2|

! |z| Z

0

dt+ 2_|a2|

≤ 2(1−α)

1₋α −2|a2|

!

|z_|+ 2_|a2|

≤ 2(1₁ −α) −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+a2z 2

+a3z 3

+_{· · · ∈ A} satisfies

|f′′′(z)_{| ≤} 2 sin(απ/2)

1 + sin(απ/2)−2|a2| (z ∈ U; 0< α≤1) (15)

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+a2z

2

+a3z 3

+_{· · · ∈ A} satisfies

|f′′′(z)_{| ≤} 1−α

2₋α −2|a2| (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+a2z 2

+a3z 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 impliesRef(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+a2z

2

+a3z 3

+_{· · · ∈ A} satisfies

|f′′′(z)_{| ≤}sinαπ

2 −2|a2| (z ∈ U; 0< α≤1) (18)

then f(z)_{∈ R}(α).

Corollary 5. Let f(z)_{∈ A} with f′′_{(0) = 0 . Then}

(i)_|f′′′_{(z)| ≤1/2 implies f(z)∈ R(1/3);}

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

0

f′′(t)dt

+α_|zf′′(z)_|

≤ |z_|

Z

0

|f′′(t)dt_|+αq(α₋1)/(α2

+ 3α)_|z_| ≤ (1 +α)q(α₋1)/(α2

+ 3α)_|z_| < (1 +α)q(α₋1)/(α2_{+ 3α),}

Using Lemma 6, we have f(z)_{∈ K}(1/2).

References

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Author:

Basem A. Frasin

Department of Mathematics Al al-Bayt University

P.O. Box: 130095 Mafraq, Jordan