SpringerOpen Newsletter

Receive periodic news and updates relating to SpringerOpen.

Open Access Research

New subclasses of analytic functions

Basem Aref Frasin

Author Affiliations

Faculty of Science, Department of Mathematics, Al al-Bayt University, P.O. Box 130095, Mafraq, Jordan

Journal of Inequalities and Applications 2012, 2012:24  doi:10.1186/1029-242X-2012-24

The electronic version of this article is the complete one and can be found online at: http://www.journalofinequalitiesandapplications.com/content/2012/1/24


Received:13 July 2011
Accepted:9 February 2012
Published:9 February 2012

© 2012 Frasin; licensee Springer.

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

For analytic functions f (z) in the open unit disk U , subclasses T ( β 1 , β 2 , β 3 : λ ) , P ( θ , α ) , and K ( θ , α ) are introduced. The object of the present article is to discuss some interesting properties of functions f (z) associated with classes T ( β 1 , β 2 , β 3 : λ ) , P ( θ , α ) , and K ( θ , α ) .

Mathematics Subject Classification (2010): 30C45.

Keywords:
analytic; univalent functions; Cauchy-Schwarz inequality

1. Introduction and Definitions

Let A denotes the class of the normalized functions of the form

f ( z ) = z + n = 2 a n z n , (1.1)

which are analytic in the open unit disk U = { z : | z | < 1 } . Also, a function f (z) belonging to A is said to be convex of order α if it satisfies

Re 1 + z f ( z ) f ( z ) > α ( z U ) (1.2)

for some α(0 ≤ α < 1). We denote by K ( α ) the subclass of A consisting of functions which are convex of order α in U (see, [1,2]). Further, a function f (z) belonging to A is said to be in the class P ( α ) iff

Re ( z f ( z ) + f ( z ) ) > α , ( z U ) . (1.3)

for some α(0 ≤ α < 1).

For analytic functions f (z), Uyanik and Owa [3], obtained some interesting properties for analytic functions in the subclass A ( β 1 , β 2 , β 3 ; λ ) defined by

β 1 z f ( z ) z + β 2 z f ( z ) z + β 3 z f ( z ) z λ ( β 1 , β 2 , β 3 ; λ > 0 ; z U ) ,

associated with close-to-convex functions and starlike functions of order α.

In this article, we define the following subclass of analytic functions.

Definition 1.1. A function f (z) belonging to A is said to be in the class T ( β 1 , β 2 , β 3 ; λ ) , if it satisfies

β 1 z f ( z ) + β 2 z 2 f ( z ) + β 3 z 3 f ( 4 ) ( z ) λ ( z U ) , (1.4)

for some complex numbers β1, β2, β3, and for some real λ > 0.

Example 1.2. Let us consider the function fγ (z), γ ∈ ℝ, given by

f γ ( z ) = z ( 1 + z ) γ .

Then, we observe that

| β 1 z f γ ( z ) + β 2 z 2 f γ ( z ) + β 3 z 3 f γ ( 4 ) ( z ) | = | n = 2 n ( n 1 ) ( γ n 1 ) ( β 1 + ( n 2 ) β 2 + ( n 2 ) ( n 3 ) β 3 ) z n 1 | ,

where

γ n - 1 = γ ( γ - 1 ) ( γ - 2 ) ( γ - n + 2 ) ( n - 1 ) ! .

Therefore, if γ = 1, then

β 1 z f 1 ( z ) + β 2 z 2 f 1 ( z ) + β 3 z 3 f 1 ( 4 ) ( z ) = 2 β 1 z 2 β 1 .

This implies that f 1 ( z ) T ( β 1 , β 2 , β 3 ; λ ) for λ ≥ 2 |β1|. If γ = 2, then

β 1 z f 2 ( z ) + β 2 z 2 f 2 ( z ) + β 3 z 3 f 2 ( 4 ) ( z ) = 4 β 1 z + 6 ( β 1 + β 2 ) z 2 10 β 1 + 6 β 2 .

Therefore, f 2 ( z ) T ( β 1 , β 2 , β 3 ; λ ) for λ ≥ 10 |β1| + 6 |β2|. Further, if γ = 3; then we have

β 1 z f 3 ( z ) + β 2 z 2 f 3 ( z ) + β 3 z 3 f 3 ( 4 ) ( z ) = 6 β 1 z + 18 ( β 1 + β 2 ) z 2 + 12 ( β 1 + 2 β 2 + 2 β 3 ) z 3 36 β 1 + 42 β 2 + 24 β 3 .

Thus, f 3 ( z ) T ( β 1 , β 2 , β 3 ; λ ) for λ ≥ 36 |β1| + 42 |β2| + 24 |β3|.

Now, let A θ denotes the subclass of A consisting of functions f (z) with

a n = a n e i ( ( n - 1 ) θ + π ) ( n = 2 , 3 , . . . ) .

Also, we introduce the subclasses P ( θ , α ) and K ( θ , α ) of A θ as follows:

P ( θ , α ) = A θ P ( α ) and K ( θ , α ) = A θ K ( α ) .

2. Properties of the class T ( β 1 , β 2 , β 3 ; λ )

We first prove

Theorem 2.1. If f ( z ) A satisfies

n = 2 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n λ (2.1)

for some complex numbers β1, β2, β3 and for some real λ > 0, then f ( z ) T ( β 1 , β 2 , β 3 ; λ ) .

Proof. We observe that

β 1 z f 3 ( z ) + β 2 z 2 f 3 ( z ) + β 3 z 3 f 3 ( 4 ) ( z ) = n = 2 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n z n - 1 n = 2 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n z n - 1 < n = 2 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n .

Therefore, if f (z) satisfies the inequality (2.1), then f ( z ) T ( β 1 , β 2 , β 3 ; λ ) .

Next, we prove

Theorem 2.2. if f ( z ) T ( β 1 , β 2 , β 3 ; λ ) with arg β1 = arg β2 = arg β3 = ϕ and an = |an|ei((n-1)θ-ϕ)(n = 2, 3,...), then we have

n = 2 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n λ .

Proof. For f ( z ) T ( β 1 , β 2 , β 3 ; λ ) , we see that

β 1 z f 3 ( z ) + β 2 z 2 f 3 ( z ) + β 3 z 3 f 3 ( 4 ) ( z ) = n = 2 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n z n - 1 = n = 2 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n e i ( n - 1 ) θ z n - 1 λ .

for all z U . Let us consider a point z U such that z = |z| e-.

Then we have

n = 2 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n z n - 1 λ .

Letting |z| → 1-, we obtain

n = 2 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n λ .

Corollary 2.3. If f ( z ) T ( β 1 , β 2 , β 3 ; λ ) with arg β1 = arg β2 = arg β3 = ϕ and an = |an| ei((n-1)θ-ϕ) (n = 2, 3,...), then we have

a n λ n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) ( n = 2 , 3 , . . . ) .

Example 2.4. Let us consider the function f ( z ) T ( β 1 , β 2 , β 3 ; λ ) with arg β1 = arg β2 = arg β3 = ϕ and

a n = λ e i ( ( n - 1 ) θ - ϕ ) n 2 ( n - 1 ) 2 ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) ( n = 2 , 3 . . . ) .

Then, we see that

n = 2 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n = λ n = 2 1 n ( n - 1 ) = λ n = 2 1 n - 1 - 1 n = λ .

Corollary 2.5. If f ( z ) T ( β 1 , β 2 , β 3 ; λ ) with arg β1 = arg β2 = arg β3 = ϕ and an = |an| ei((n-1)θ-ϕ) (n = 2, 3,...), then we have

z - n = 2 j a n z n - A j z j + 1 f ( z ) z + n = 2 j a n z n + A j z j + 1

with

A j = ( λ n = 2 j n ( n 1 ) ( | β 1 | + ( n 2 ) | β 2 | + ( n 2 ) ( n 3 ) | β 3 | ) | a n | ) j ( j + 1 ) ( | β 1 | ) + ( j 1 ) | β 2 | + ( j 1 ) ( j 2 ) | β 3 | )

and

1 - n = 2 j a n z n - 1 - B j z j f ( z ) 1 + n = 2 j a n z n - 1 + B j z j

with

B j = λ - n = 2 j n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n j ( β 1 + ( j - 1 ) β 2 + ( j - 1 ) ( j - 2 ) β 3 )

Proof. In view of Theorem 2.1, we know that

n = j + 1 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n λ - n = 2 j n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n .

Further, we note that

j ( j + 1 ) ( β 1 + ( j - 1 ) β 2 + ( j - 1 ) ( j - 2 ) β 3 ) n = j + 1 a n n = j + 1 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n ,

which is equivalent to

n = j + 1 a n λ - n = 2 j n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n j ( j + 1 ) ( β 1 + ( j - 1 ) β 2 + ( j - 1 ) ( j - 2 ) β 3 ) = A j .

Thus, we have

f ( z ) z + n = 2 j a n z n + n = j + 1 a n z n z + n = 2 j a n z n + A j z j + 1

and

f ( z ) z - n = 2 j a n z n - n = j + 1 a n z n z - n = 2 j a n z n - A j z j + 1 .

Next, we observe that

j ( β 1 + ( j - 1 ) β 2 + ( j - 1 ) ( j - 2 ) β 3 ) n = j + 1 n a n n = j + 1 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n λ n = 2 j n ( n - 1 ) ( β 1 + ( n - 1 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n ,

which implies that

n = j + 1 n a n λ - n = 2 j n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) a n j ( β 1 + ( j - 1 ) β 2 + ( j - 1 ) ( j - 2 ) β 3 ) = B j .

Therefore, we obtain that

f ( z ) 1 + n = 2 j n a n z n - 1 + n = j + 1 n a n z n - 1 1 + n = 2 j a n z n - 1 + B j z j

and

f ( z ) 1 - n = 2 j n a n z n - 1 - n = j + 1 n a n z n - 1 1 - n = 2 j a n z n - 1 - B j z j .

3. Radius problem for the class P ( θ , α )

To obtain the radius problem for the class P ( θ , α ) , we need the following lemma.

Lemma 3.1. If f ( z ) P ( θ , α ) , then

n = 2 n 2 a n 1 - α . (3.1)

Proof. Let f ( z ) P ( θ , α ) . Then, we have

Re { ( z f ( z ) + f ( z ) ) } = Re 1 + n = 2 n 2 a n z n - 1 = Re 1 + n = 2 n 2 a n e i ( ( n - 1 ) θ + π ) z n - 1 = Re 1 - n = 2 n 2 a n e i ( ( n - 1 ) θ ) z n - 1 > α

for all z U . Let us consider a point z U such that z = |z| e-.

Then we have

1 - n = 2 n 2 a n z n - 1 > α

Letting |z| → 1-, we obtain the inequality (3.1).

Corollary 3.2. If f ( z ) P ( θ , α ) , then

a n 1 - α n 2 ( n = 2 , 3 , . . . ) .

Remark 3.3. By Lemma 3.1, we observe that if f ( z ) P ( θ , α ) , then

n = 2 n ( n - 1 ) a n n = 2 n 2 a n 1 - α .

Applying Theorem 2.1 and Lemma 3.1, we derive

Theorem 3.4. Let f ( z ) P ( θ , α ) , and δ ∈ ℂ (0 < |δ| < 1). Then the function 1 δ f ( δ z ) T ( β 1 , β 2 , β 3 ; λ ) for (0 < |δ| ≤ |δ0(λ)|, where |δ0(λ)| is the smallest positive root of the equation

| β 1 | | δ | 2 ( 1 α ) ( 1 | δ | 2 ) 3 / 2 + | β 2 | | δ | 2 ( 6 + 18 | δ | 2 ) ( 1 α 2 | a 2 | 2 ) ( 1 | δ | 2 ) 5 / 2 + | β 3 | 4 3 | δ | 3 ( 1 + 8 | δ | 2 + 6 | δ | 4 ) ( 1 α 2 ) | a 2 | 2 6 | a 3 | 2 ) ( 1 | δ | 2 ) 7 / 2 = λ (3.2)

in 0 < |δ| < 1.

Proof. For f ( z ) P ( θ , α ) , we see that

1 δ f ( δ z ) = z + n = 2 δ n - 1 a n z n

and

n = 2 n ( n - 1 ) a n 2 1 - α .

Thus, to show that 1 δ f ( δ z ) T ( β 1 , β 2 , β 3 ; λ ) , from Theorem 2.1, it is sufficient to prove that

n = 2 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) δ n - 1 a n λ .

Applying Cauchy-Schwarz inequality, we note that

n = 2 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) δ n - 1 a n β 1 δ n = 2 n ( n - 1 ) δ 2 n 1 2 n = 2 n ( n - 1 ) a n 2 1 2 + β 2 δ n = 3 n ( n - 1 ) ( n - 2 ) 2 δ 2 n 1 2 n = 3 n ( n - 1 ) a n 2 1 2 + β 3 δ n = 4 n ( n - 1 ) ( n - 2 ) 2 ( n - 3 ) 2 δ 2 n 1 2 n = 4 n ( n - 1 ) a n 2 1 2 β 1 δ n = 2 n ( n - 1 ) δ 2 n 1 2 1 - α + β 2 δ n = 3 n ( n - 1 ) ( n - 2 ) 2 δ 2 n 1 2 1 - α - 2 a 2 2 + β 3 δ n = 4 n ( n - 1 ) ( n - 2 ) 2 ( n - 3 ) 2 δ 2 n 1 2 1 - α - 2 α 2 2 - 6 a 3 2 . (3.3)

We note that

n = 0 x n = 1 1 - x , ( x < 1 ) ,

thus, we have

n = 2 n ( n - 1 ) x n = 2 x 2 ( 1 - x ) 3 . (3.4)

Since

n = 3 ( n - 2 ) x n - 1 = x 2 n = 3 ( n - 2 ) x n - 3 = x 2 n = 3 x n - 2 = x 2 ( 1 - x ) 2 ,

we see that

n = 3 ( n - 1 ) ( n - 2 ) 2 x n = x 3 x 2 ( 1 - x ) 2 = 2 x 3 + 4 x 4 ( 1 - x ) 4 .

and thus, we obtain

n = 3 n ( n - 1 ) ( n - 2 ) 2 x n = 6 x 3 + 18 x 4 ( 1 - x ) 5 . (3.5)

Furthermore, we have

n = 4 ( n - 1 ) ( n - 2 ) 2 ( n - 3 ) 2 x n = x 4 n = 4 ( n - 1 ) ( n - 2 ) 2 ( n - 3 ) 2 x n - 4 = x 4 n = 4 ( n - 2 ) ( n - 3 ) x n - 1 ,

but

n = 4 ( n - 2 ) ( n - 3 ) x n - 1 = x 3 n = 4 ( n - 2 ) ( n - 3 ) x n - 4 = 2 x 3 ( 1 - x ) 3

thus, we have

n = 4 ( n - 1 ) ( n - 2 ) 2 ( n - 3 ) 2 x n = 12 x 4 + 72 x 5 + 36 x 6 ( 1 - x ) 6 ,

which yields

n = 4 n ( n - 1 ) ( n - 2 ) 2 ( n - 3 ) 2 x n = 48 x 4 ( 1 + 8 x + 6 x 2 ) ( 1 - x ) 7 . (3.6)

Therefore, from (3.3)-(3.6) with |δ|2 = x, we obtain

n = 2 n ( n - 1 ) ( β 1 + ( n - 2 ) β 2 + ( n - 2 ) ( n - 3 ) β 3 ) δ n - 1 a n β 1 δ 2 ( 1 - α ) ( 1 - δ 2 ) 3 / 2 + β 2 δ 2 ( 6 + 18 δ 2 ) ( 1 - α - 2 a 2 2 ) ( 1 - δ 2 ) 5 / 2 β 3 4 3 δ 3 ( 1 + 8 δ 2 + 6 δ 4 ) ( 1 - α - 2 a 2 2 - 6 a 3 2 ) ( 1 - δ 2 ) 7 / 2

Now, let us consider the complex number δ (0 < |δ| < 1) such that

β 1 δ 2 ( 1 - α ) ( 1 - δ 2 ) 3 / 2 + β 2 δ 2 ( 6 + 18 δ 2 ) ( 1 - α - 2 a 2 2 ) ( 1 - δ 2 ) 5 / 2 β 3 4 3 δ 3 ( 1 + 8 δ 2 + 6 δ 4 ) ( 1 - α - 2 a 2 2 - 6 a 3 2 ) ( 1 - δ 2 ) 7 / 2 = λ .

If we define the function h(|δ|) by

h ( δ ) = β 1 δ ( 1 - δ 2 ) 2 2 ( 1 - α ) + β 2 δ 2 ( 1 - δ 2 ) ( 6 + 18 δ 2 ) ( 1 - α - 2 a 2 2 ) + 4 3 β 3 δ 3 ( 1 + 8 δ 2 + 6 δ 4 ) ( 1 - α - 2 a 2 2 - 6 a 3 2 ) - λ ( 1 - δ 2 ) 7 / 2 ,

then we have h(0) = -λ < 0 and h ( 1 ) = 12 5 β 3 1 - α - 2 a 2 2 - 6 a 3 2 > 0 . This means that there exists some δ0 such that h(|δ0|) = 0 (0 < |δ0| < 1). This completes the proof of the theorem.

4. Radius problem for the class K ( θ , α )

For the class K ( θ , α ) , we prove the following lemma.

Lemma 4.1. If f ( z ) K ( θ , α ) , then

n = 2 n ( n - α ) a n 1 - α . (4.1)

Proof. Let f ( z ) K ( θ , α ) . Then, we have

Re 1 + z f ( z ) f ( z ) = Re 1 + n = 2 n 2 a n z n - 1 1 + n = 2 n a n z n - 1 = Re 1 - n = 2 n 2 a n e i ( n - 1 ) θ z n - 1 1 - n = 2 n a n e i ( n - 1 ) θ z n - 1 > α

for all z U . Let us consider a point z U . such that z = |z|e-.

Then we have

1 - n = 2 n 2 a n z n - 1 1 - n = 2 n a n z n - 1 > α

Letting |z| → 1-, we obtain the inequality (4.1).

Corollary 4.2. If f ( z ) K ( θ , α ) , then

a n 1 - α n ( n - α ) ( n = 2 , 3 , ) .

Remark 4.3. If f ( z ) K ( θ , α ) , then

n = 2 n ( n - 1 ) a n n = 2 n ( n - α ) a n 1 - α .

Applying Theorem 2.1, Lemma 4.1 and using the same technique as in the proof of Theorem 3.4, we derive

Theorem 4.4. Let f ( z ) K ( θ , α ) , and δ ∈ ℂ (0 < |δ| < 1). Then the function 1 δ f ( δ z ) T ( β 1 , β 2 , β 3 ; λ ) for (0 < |δ| ≤ |δ0(λ)|, where |δ0(λ)| is the smallest positive root of the equation

β 1 δ 2 ( 1 - α ) ( 1 - δ 2 ) 3 / 2 + β 2 δ 2 ( 6 + 18 δ 2 ) ( 1 - α - 2 a 2 2 ) ( 1 - δ 2 ) 5 / 2 + β 3 4 3 δ 3 ( 1 + 8 δ 2 + 6 δ 4 ) ( 1 - α - 2 α 2 2 - 6 a 3 2 ) ( 1 - | δ | 2 ) 7 / 2 = λ (4.2)

in 0 < |δ| < 1.

Competing interests

The author declares that they have no competing interests.

References

  1. Duren, PL: Univalent Functions. Springer-Verlag, Berlin (1983)

  2. Goodman, AW: Univalent Functions. Mariner, Tampa (1983)

  3. Uyanik, N, Owa, S: New extensions for classes of analytic functions associated with close-to-convex and starlike of order α. Math Comput Model. 54, 359–366 (2011). Publisher Full Text OpenURL