In this paper, we investigate a subordination property and the coefficient inequality for the class M(1, b), The lower bound is also provided for the real part of functions belonging to the class M(1, b).
Let A denote the class of function f ( z ) analytic in the open unit disk U = { z ∈ ℂ : | z | < 1 } and let S be the subclass of A consisting of functions univalent in U and have the form
f ( z ) = z + ∑ k = 2 ∞ a k z k , (1.1)
The class of convex functions of order α in U, denoted as K ( α ) is given by
K ( α ) = { f ∈ S : R e ( 1 + z f ″ ( z ) f ′ ( z ) ) > α , 0 ≤ α < 1, z ∈ U }
Definition 1.1. The Hadamard product or convolution f ∗ g of the func- tion f ( z ) and g ( z ) , where f ( z ) is as defined in (1.1) and the function g ( z ) is given by
g ( z ) = z + ∑ k = 2 ∞ b k z k ,
is defined as:
( f ∗ g ) ( z ) = z + ∑ k = 2 ∞ a k b k z k = ( g ∗ f ) ( z ) , (1.2)
Definition 1.2. Let f ( z ) and g ( z ) be analytic in the unit disk U . Then f ( z ) is said to be subordination to g ( z ) in U and written as:
f ( z ) ≺ g ( z ) , z ∈ U
if there exist a Schwarz function ω ( z ) , analytic in U with ω ( 0 ) = 0 , | ω ( z ) | < 1 such that
f ( z ) = g ( ω ( z ) ) , z ∈ U (1.3)
In particular, if the function g ( z ) is univalent in U, then f ( z ) is said to be subordinate to g ( z ) if
f ( 0 ) = g ( 0 ) , f ( u ) ⊂ g ( u ) (1.4)
Definition 1.3. The sequence { c k } k = 1 ∞ of complex numbers is said to be a subordinating factor sequence of the function f ( z ) if whenever f ( z ) in the form (1.1) is analytic, univalent and convex in the unit disk U , the subordination is given by
∑ k = 1 ∞ a k c k z k ≺ f ( z ) , z ∈ U , a 1 = 1
We have the following theorem:
Theorem 1.1. (Wilf [
R e { 1 + 2 ∑ k = 1 ∞ c k z k } > 0 , z ∈ U (1.5)
Definition 1.4. A function P ∈ A which is normalized by P ( 0 ) = 1 is said to be in P ( 1, b ) if
| P ( z ) − 1 | < b , b > 0 , z ∈ U .
The class P ( 1, b ) was studied by Janwoski [
( z f ′ ( z ) f ( z ) ) ∈ P ( 1 , 1 − α ) , 0 ≤ α < 1
Then f ( z ) is starlike of order α in U and if
( 1 + z f ″ ( z ) f ′ ( z ) ) ∈ P ( 1 , 1 − α ) , 0 ≤ α < 1
Then f ( z ) is convex of order α in U.
Let F ( 1, b ) be the subclass of P ( 1,1 − α ) consisting of functions P ( f ) such that
P ( f ) = z f ′ ( z ) f ( z ) ( 1 + z f ″ ( z ) f ′ ( z ) ) (1.6)
we have the following theorem
Theorem 1.2. [
∑ k = 2 ∞ ( k 2 + b − 1 ) | a k | < b , b > 0
then P ( f ) ∈ F ( 1, b ) , 0 < b < 0.935449 .
It is natural to consider the class
M ( 1 , b ) = { f ∈ A : ∑ k = 2 ∞ ( k 2 + b − 1 ) | a k | < b , b > 0 }
0 < b < 0.935449
Remark 1.1. [
∑ k = 2 ∞ ( k − α ) | a k | < ∑ k = 2 ∞ ( k 2 − α ) | a k |
Our main focus in this work is to provide a subordination results for functions belonging to the class M (1,b)
Let f ( z ) ∈ M ( 1, b ) , then
3 + b 2 ( 3 + 2 b ) ( f ∗ g ) ( z ) ≺ g ( z ) (2.1)
where 0 < b < 0.935449 and g ( z ) is convex function.
Proof:
Let
f ( z ) ∈ M (1,b)
and suppose that
g ( z ) = z + ∑ b k z k ∈ C (α)
that is g ( z ) is a convex function of order α .
By definition (1.1) we have
3 + b 2 ( 3 + 2 b ) ( f ∗ g ) ( z ) = 3 + b 2 ( 3 + 2 b ) ( z + ∑ k = 2 ∞ a k b k z k ) = ∑ k = 1 ∞ 3 + b 2 ( 3 + 2 b ) a k b k z k , a 1 = 1 , b 1 = 1 (2.2)
Hence, by Definition 1.3…to show subordination (2.1) is by establishing that
{ 3 + b 2 ( 3 + 2 b ) a k } k = 1 ∞ (2.3)
is a subordinating factor sequence with a 1 = 1 . By Theorem 1.1, it is sufficient to show that
R e { 1 + 2 ∑ k = 1 ∞ 3 + b 2 ( 3 + 2 b ) a k z k } > 0, z ∈ U (2.4)
Now,
R e { 1 + 2 ∑ k = 1 ∞ 3 + b 2 ( 3 + 2 b ) a k z k } = R e { 1 + 3 + b 3 + 2 b z + ∑ k = 2 ∞ 3 + b 3 + 2 b a k z k } > R e { 1 − 3 + b 3 + 2 b r − 3 + b 3 + 2 b ∑ k = 2 ∞ | a k | r k } > R e { 1 − 3 + b 3 + 2 b r − 1 3 + 2 b ∑ k = 2 ∞ ( k 2 − b + 1 ) | a k | r k } > R e { 1 − 3 + b 3 + 2 b r − b r 3 + 2 b } = 1 − r > 0
Since ( | z | = r < 1 ), therefore we obtain
R e { 1 + 2 ∑ k = 1 ∞ 3 + b 2 ( 3 + 2 b ) a k z k } > 0, z ∈ U
which by Theorem 1.1 shows that 3 + b 2 ( 3 + 2 b ) a k is a subordinating factor, hence, we have established Equation (2.5).
Given f ( z ) ∈ M ( 1, b ) , then
R e f ( z ) > − 3 + 2 b 3 + b (2.6)
The constant factor 3 + 2 b 3 + b cannot be replaced by a larger one.
Proof:
Let
g ( z ) = z 1 − z
which is a convex function, Equation (2.1) becomes
3 + b 2 ( 3 + 2 b ) f ( z ) ∗ z 1 − z ≺ z 1 − z
Since
R e ( z 1 − z ) > − 1 2 , | z | = r (2.7)
This implies
R e { 3 + b 2 ( 3 + 2 b ) f ( z ) ∗ z 1 − z } > − 1 2 (2.8)
Therefore, we have
R e ( f ( z ) ) > − 3 + 2 b 3 + b
which is Equation (2.6).
Now to show that sharpness of the constant factor
3 + b 3 + 2 b
We consider the function
f 1 ( z ) = z ( 3 + b ) + b z 2 3 + b (2.9)
Applying Equation (2.1) with g ( z ) = z 1 − z and f ( z ) = f 1 ( z ) , we have
z ( 3 + b ) + b z 2 2 ( 3 + b ) ≺ z 1 − z (2.10)
Using the fact that
| R e ( z ) | ≤ | z | (2.11)
We now show that the
m i n z ∈ U { R e ( z ( 3 + b ) + b z 2 2 ( 3 + b ) ) } = − 1 2 (2.12)
we have
| R e ( z ( 3 + b ) + b z 2 2 ( 3 + b ) ) | ≤ | z ( 3 + b ) + b z 2 2 ( 3 + b ) | ≤ | z | | ( 3 + b ) + b z | | 2 ( 3 + b ) | ≤ | ( 3 + b ) + b z | 2 ( 3 + b ) ≤ ( 3 + b ) + b 2 ( 3 + 2 b ) ≤ 3 + 2 b 2 ( 3 + 2 b ) = 1 2 , ( | z | = 1 )
This implies that
| R e ( z ( 3 + b ) + b z 2 2 ( 3 + b ) ) | ≤ 1 2
and therefore
− 1 2 ≤ R e ( z ( 3 + b ) + b z 2 2 ( 3 + b ) ) ≤ 1 2
Hence, we have that
m i n z ∈ U { R e ( z ( 3 + b ) + b z 2 2 ( 3 + b ) ) } = − 1 2
That is
m i n z ∈ U { R e 3 + b 2 ( 3 + 2 b ) ( f 1 ∗ g ( z ) ) } = − 1 2
which shows the Equation (2.12).
Let
f ( z ) = z + ∑ k = 2 ∞ a k z k ∈ M ( 1 , b ) , 0 < b < 0.935449
then | a k | ≤ 1 2 .
Proof:
Let
f ( z ) = z + ∑ k = 2 ∞ a k z k ∈ M ( 1 , b )
then by definition of the class M ( I , b ) ,
∑ k = 2 ∞ ( k 2 + b − 1 ) | a k | ≤ b , 0 < b < 0.935449
we have that
k 2 + b − 1 b − k > 0
which gives that
∑ k = 2 ∞ k | a k | ≤ k 2 + b − 1 b | a k | ≤ 1
i . e ∑ k = 2 ∞ k | a k | ≤ 1
hence
2 ∑ | a k | ≤ 1
| a k | ≤ 1 2
Bello, R.A. (2017) On a Subordination Result of a Subclass of Analytic Functions. Advances in Pure Mathematics, 7, 641-646. https://doi.org/10.4236/apm.2017.711038