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

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

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.

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

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