Open Access

Subclass of univalent harmonic functions defined by dual convolution

Journal of Inequalities and Applications20132013:537

https://doi.org/10.1186/1029-242X-2013-537

Received: 6 December 2012

Accepted: 14 October 2013

Published: 12 November 2013

Abstract

In the present paper, we study a subclass of univalent harmonic functions defined by convolution and integral convolution. We obtain the basic properties such as coefficient characterization and distortion theorem, extreme points and convolution condition.

MSC:30C45, 30C50.

Keywords

harmonic functionunivalentsense-preservingintegral convolution

1 Introduction

A continuous function f = u + i v is a complex-valued harmonic function in a simply connected complex domain D C if both u and v are real harmonic in D. It was shown by Clunie and Sheil-Small [1] that such a harmonic function can be represented by f = h + g ¯ , where h and g are analytic in D. Also, a necessary and sufficient condition for f to be locally univalent and sense-preserving in D is that | h ( z ) | > | g ( z ) | (see also [24] and [5]).

Denote by S H the class of functions f that are harmonic univalent and sense-preserving in the open unit disc U = { z C : | z | < 1 } , for which f ( 0 ) = h ( 0 ) = f z ( 0 ) 1 = 0 . Then for f = h + g ¯ S H we may express the analytic functions h and g as
h ( z ) = z + n = 2 a n z n , g ( z ) = n = 1 b n z n , | b 1 | < 1 .
(1.1)

Clunie and Sheil-Small [1] investigated the class S H as well as its geometric subclasses and obtained some coefficient bounds.

Also, let S H ¯ denote the subclass of S H consisting of functions f = h + g ¯ such that the functions h and g are of the form
h ( z ) = z n = 2 | a n | z n , g ( z ) = n = 1 | b n | z n , | b 1 | < 1 .
(1.2)
Recently Kanas and Wisniowska [6] (see also Kanas and Srivastava [7]) studied the class of k-uniformly convex analytic functions, denoted by k U C V , k 0 , so that ϕ k U C V if and only if
Re { 1 + ( z ζ ) ϕ ( z ) ϕ ( z ) } 0 ( | ζ | k ; z U ) .
(1.3)
For θ R , if we let ζ = k z e i θ , then condition (1.3) can be written as
Re { 1 + ( 1 + k e i θ ) z ϕ ( z ) ϕ ( z ) } 0 .
(1.4)
Kim et al. [8] introduced and studied the class H C V ( k , α ) consisting of functions f = h + g ¯ , such that h and g are given by (1.1), and satisfying the condition
Re { 1 + ( 1 + k e i θ ) z 2 h ( z ) + 2 z g ( z ) + z 2 g ( z ) ¯ z h ( z ) z g ( z ) ¯ } α ( 0 α < 1 ; θ R ; k 0 ) .
(1.5)
Also, the class of k U S T uniformly starlike functions is defined by using (1.4) as the class of all functions ψ ( z ) = z ϕ ( z ) such that ϕ k U C V , then ψ ( z ) k U S T if and only if
Re { ( 1 + k e i θ ) z ψ ( z ) ψ ( z ) k e i θ } 0 .
(1.6)
Generalizing the class k U S T to include harmonic functions, we let HST ( k , α ) denote the class of functions f = h + g ¯ , such that h and g are given by (1.1), which satisfies the condition
Re { ( 1 + k e i θ ) z f ( z ) z f ( z ) k e i θ } α ( 0 α < 1 ; θ R ; k 0 ) .
(1.7)
Replacing h + g ¯ for f in (1.7), we have
Re { ( 1 + k e i θ ) z h ( z ) z g ( z ) ¯ h ( z ) + g ( z ) ¯ k e i θ } α ( 0 α < 1 ; θ R ; k 0 ) .
(1.8)
The convolution of two functions of the form
f ( z ) = z + n = 2 a n z n and F ( z ) = z + n = 2 A n z n
is defined as
( f F ) ( z ) = z + n = 2 a n A n z n ,
(1.9)
while the integral convolution is defined by
( f F ) ( z ) = z + n = 2 a n A n n z n .
(1.10)
From (1.9) and (1.10), we have
( f F ) ( z ) = 0 z ( f F ) ( t ) t d t .
Now we consider the subclass HST ( ϕ , ψ , k , α ) consisting of functions f = h + g ¯ , such that h and g are given by (1.1), and satisfying the condition
Re { ( 1 + k e i θ ) h ( z ) φ ( z ) g ( z ) χ ( z ) ¯ h ( z ) φ ( z ) + g ( z ) χ ( z ) ¯ k e i θ } α ( 0 α < 1 ; k 0 ; θ  real ) ,
(1.11)
where
φ ( z ) = z + n = 2 λ n z n ( λ n 0 ) and χ ( z ) = z + n = 2 μ n z n ( μ n 0 ) .
(1.12)

We further consider the subclass HST ¯ ( ϕ , χ , k , α ) of HST ( ϕ , χ , k , α ) for h and g given by (1.2).

We note that
  1. (i)

    HST ¯ ( ϕ , χ , 0 , α ) = H S ¯ ( ϕ , χ , α ) (see Dixit et al. [9]);

     
  2. (ii)

    HST ¯ ( z ( 1 z ) 2 , z ( 1 z ) 2 , 1 , α ) = G H ¯ ( α ) (see Rosy et al. [10]);

     
  3. (iii)

    HST ¯ ( z + z 2 ( 1 z ) 3 , z + z 2 ( 1 z ) 3 , k , α ) = H ¯ C V ( k , α ) (see Kim et al. [8]);

     
  4. (iv)

    HST ¯ ( z ( 1 z ) 2 , z ( 1 z ) 2 , 0 , α ) = T H ( α ) (see Jahangiri [3], see also Joshi and Darus [11]);

     
  5. (v)

    HST ¯ ( z + z 2 ( 1 z ) 3 , z + z 2 ( 1 z ) 3 , 0 , α ) = C H ( α ) (see Jahangiri [3], see also Joshi and Darus [11]).

     

In this paper, we extend the results of the above classes to the classes HST ( ϕ , χ , k , α ) and HST ¯ ( ϕ , χ , k , α ) , we also obtain some basic properties for the class HST ¯ ( ϕ , χ , k , α ) .

2 Coefficient characterization and distortion theorem

Unless otherwise mentioned, we assume throughout this paper that φ ( z ) and χ ( z ) are given by (1.12), 0 α < 1 , k 0 and θ is real. We begin with a sufficient condition for functions in the class HST ( ϕ , χ , k , α ) .

Theorem 1 Let f = h + g ¯ be such that h and g are given by (1.1). Furthermore, let
n = 2 λ n n ( ( 1 + k ) n ( k + α ) ( 1 α ) ) | a n | + n = 1 μ n n ( ( 1 + k ) n + ( k + α ) ( 1 α ) ) | b n | 1 ,
(2.1)
where
n 2 ( 1 α ) λ n [ ( 1 + k ) n ( k + α ) ] and n 2 ( 1 α ) μ n [ ( 1 + k ) n + ( k + α ) ] for n 2 .

Then f is sense-preserving, harmonic univalent in U and f HST ( ϕ , χ , k , α ) .

Proof First we note that f is locally univalent and sense-preserving in U. This is because
| h ( z ) | 1 n = 2 n | a n | r n 1 > 1 n = 2 n | a n | 1 n = 2 λ n n ( ( 1 + k ) n ( k + α ) ( 1 α ) ) | a n | n = 1 μ n n ( ( 1 + k ) n + ( k + α ) ( 1 α ) ) | b n | n = 1 n | b n | n = 1 n | b n | r k 1 > | g ( z ) | .
To show that f is univalent in U, suppose z 1 , z 2 U so that z 1 z 2 , then
| f ( z 1 ) f ( z 2 ) h ( z 1 ) h ( z 2 ) | 1 | g ( z 1 ) g ( z 2 ) h ( z 1 ) h ( z 2 ) | = 1 | n = 1 b n ( z 1 n z 2 n ) ( z 1 z 2 ) + n = 2 a n ( z 1 n z 2 n ) | 1 n = 1 n | b n | 1 n = 2 n | a n | > 1 n = 1 μ n n ( ( 1 + k ) n + ( k + α ) ( 1 α ) ) | b n | 1 n = 2 λ n n ( ( 1 + k ) n ( k + α ) ( 1 α ) ) | a n | 0 .
Now, we prove that f HST ( ϕ , ψ , k , α ) , by definition, we only need to show that if (2.1) holds, then condition (1.11) is satisfied. From (1.11), it suffices to show that
Re { ( 1 + k e i θ ) ( h ( z ) φ ( z ) g ( z ) χ ( z ) ¯ ) ( k e i θ + α ) ( h ( z ) φ ( z ) + g ( z ) χ ( z ) ¯ ) h ( z ) φ ( z ) + g ( z ) χ ( z ) ¯ } 0 .
(2.2)
Substituting for h, g, φ and χ in (2.2) and dividing by ( 1 α ) z , we obtain Re A ( z ) B ( z ) 0 , where
A ( z ) = 1 + n = 2 λ n n ( 1 + k e i θ ) n ( k e i θ + α ) ( 1 α ) a n z n 1 ( z ¯ z ) n = 1 μ n n ( 1 + k e i θ ) n + ( k e i θ + α ) ( 1 α ) b n z ¯ n 1
and
B ( z ) = 1 + n = 2 λ n n a n z n 1 + ( z ¯ z ) n = 1 μ n n b n z ¯ n 1 .
Using the fact that Re ( w ) 0 if and only if | 1 + w | | 1 w | in U, it suffices to show that | A ( z ) + B ( z ) | | A ( z ) B ( z ) | 0 . Substituting for A ( z ) and B ( z ) gives
| A ( z ) + B ( z ) | | A ( z ) B ( z ) | = | 2 + n = 2 λ n n ( 1 + k e i θ ) n ( k e i θ + 2 α 1 ) ( 1 α ) a n z n 1 ( z ¯ z ) n = 1 μ n n ( 1 + k e i θ ) n + ( k e i θ + 2 α 1 ) ( 1 α ) b n z ¯ n 1 | | n = 2 λ n n ( 1 + k e i θ ) n ( 1 + k e i θ ) ( 1 α ) a n z n 1 ( z ¯ z ) n = 1 μ n n ( 1 + k e i θ ) n + ( 1 + k e i θ ) ( 1 α ) b n z ¯ n 1 | 2 n = 2 λ n n ( 1 + k ) n ( k + 2 α 1 ) ( 1 α ) | a n | | z | n 1 n = 1 μ n n ( 1 + k ) n + ( k + 2 α 1 ) ( 1 α ) | b n | | z | n 1 n = 2 λ n n ( 1 + k ) n ( 1 + k ) ( 1 α ) | a n + 1 | | z | n 1 n = 1 μ n n ( 1 + k ) n + ( 1 + k ) ( 1 α ) | b n | | z | n 1 2 { 1 n = 2 λ n n ( 1 + k ) n ( k + α ) ( 1 α ) | a n | n = 1 μ n n ( 1 + k ) n + ( k + α ) ( 1 α ) | b n | } 0 by (2.1) .
The harmonic functions
f ( z ) = z + n = 2 n λ n ( 1 α ) ( 1 + k ) n ( k + α ) x n z n + n = 1 n μ n ( 1 α ) ( 1 + k ) n + ( k + α ) y ¯ n z ¯ n ,
(2.3)
where n = 2 | x n | + n = 1 | y n | = 1 , show that the coefficient bound given by (2.1) is sharp. The functions of the form (2.3) are in the class HST ( ϕ , χ , k , α ) because
n = 2 [ λ n n ( 1 + k ) n ( k + α ) ( 1 α ) | a n | + n = 1 μ n n ( 1 + k ) n + ( k + α ) ( 1 α ) | b n | ] = n = 2 | x n | + n = 1 | y n | = 1 .

This completes the proof of Theorem 1. □

In the following theorem, it is shown that condition (2.1) is also necessary for functions f = h + g ¯ , where h and g are given by (1.2).

Theorem 2 Let f = h + g ¯ be such that h and g are given by (1.2). Then f HST ¯ ( ϕ , χ , k , α ) if and only if
n = 2 λ n n ( ( 1 + k ) n ( k + α ) ( 1 α ) ) | a n | + n = 1 μ n n ( ( 1 + k ) n + ( k + α ) ( 1 α ) ) | b n | 1 .
(2.4)
Proof Since HST ¯ ( ϕ , χ , k , α ) HST ( ϕ , χ , k , α ) , we only need to prove the ‘only if’ part of the theorem. To this end, we notice that the necessary and sufficient condition for f HST ¯ ( ϕ , χ , k , α ) is that
Re { ( 1 + k e i θ ) h ( z ) φ ( z ) g ( z ) χ ( z ) ¯ h ( z ) φ ( z ) + g ( z ) χ ( z ) ¯ k e i θ } α .
This is equivalent to
Re { ( 1 + k e i θ ) ( h ( z ) φ ( z ) g ( z ) χ ( z ) ¯ ) ( k e i θ + α ) ( h ( z ) φ ( z ) + g ( z ) χ ( z ) ¯ ) h ( z ) φ ( z ) + g ( z ) χ ( z ) ¯ } > 0 ,
which implies that
Re { ( 1 α ) z n = 2 λ n n [ ( 1 + k e i θ ) n ( k e i θ + α ) ] | a n | z n z n = 2 λ n n | a n | z n + n = 1 μ n n | b n | z ¯ n n = 1 μ n n [ ( 1 + k e i θ ) n + ( k e i θ + α ) ] | b n | z ¯ n z n = 2 λ n n | a n | z n + n = 1 μ n n | b n | z ¯ n } = Re { ( 1 α ) n = 2 λ n n [ ( 1 + k e i θ ) n ( k e i θ + α ) ] | a n | z n 1 1 n = 2 λ n n | a n | z n 1 + ( z ¯ z ) n = 1 μ n n | b n | z ¯ n 1 ( z ¯ z ) n = 1 μ n n [ ( 1 + k e i θ ) n + ( k e i θ + α ) ] | b n | z ¯ n 1 1 n = 2 λ n n | a n | z n + ( z ¯ z ) n = 1 μ n n | b n | z ¯ n 1 } > 0 ,
(2.5)
since Re ( e i θ ) | e i θ | = 1 , the required condition (2.5) is equivalent to
{ 1 n = 2 λ n n ( 1 + k ) n ( k + α ) ( 1 α ) | a n | r n 1 1 n = 2 λ n n | a n | r n 1 + n = 1 μ n n | b n | r n 1 n = 1 μ n n ( 1 + k ) n + ( k + α ) ( 1 α ) | b n | r n 1 1 n = 2 λ n n | a n | r n 1 + n = 1 μ n n | b n | r n 1 } 0 .
(2.6)

If condition (2.4) does not hold, then the numerator in (2.6) is negative for z = r sufficiently close to 1. Hence there exists z 0 = r 0 in ( 0 , 1 ) for which the quotient in (2.6) is negative. This contradicts the required condition for f HST ¯ ( ϕ , χ , k , α ) , and so the proof of Theorem 2 is completed. □

Theorem 3 Let f HST ¯ ( ϕ , χ , k , α ) . Then, for | z | = r < 1 , | b 1 | < 1 α 2 k + α + 1 and
D n λ n n , E n μ n n for n 2 and C = min { D 2 , E 2 } ,
(2.7)
we have
| f ( z ) | ( 1 + | b 1 | ) r + { ( 1 α ) C ( 2 + k α ) 2 k + 1 + α C ( 2 + k α ) | b 1 | } r 2
and
| f ( z ) | ( 1 | b 1 | ) r { ( 1 α ) C ( 2 + k α ) 2 k + 1 + α C ( 2 + k α ) | b 1 | } r 2 .

The results are sharp.

Proof We prove the left-hand side inequality for | f | . The proof for the right-hand side inequality can be done by using similar arguments.

Let f HST ¯ ( ϕ , χ , k , α ) , then we have
| f ( z ) | = | z n = 2 | a n | z n + n = 1 | b n | z ¯ n | r | b 1 | r n = 2 ( | a n | + | b n | ) r 2 r | b 1 | r ( 1 α ) C ( 2 + k α ) n = 2 C ( ( 1 + k ) n ( k + α ) ) ( 1 α ) ( | a n | + | b n | ) r 2 r | b 1 | r ( 1 α ) C ( 2 + k α ) n = 2 { C ( ( 1 + k ) n ( k + α ) ) ( 1 α ) | a n | + C ( ( 1 + k ) n + ( k + α ) ) ( 1 α ) | b n | } r 2 ( 1 | b 1 | ) r ( 1 α ) C ( 2 + k α ) { 1 2 k + 1 + α ( 1 α ) | b 1 | } r 2 ( 1 | b 1 | ) r { ( 1 α ) C ( 2 + k α ) 2 k + 1 + α C ( 2 + k α ) | b 1 | } r 2 .
The bounds given in Theorem 3 are respectively attained for the following functions:
f ( z ) = z + | b 1 | z ¯ + ( ( 1 α ) C ( 2 + k α ) 2 k + 1 + α C ( 2 + k α ) | b 1 | ) z ¯ 2
and
f ( z ) = ( 1 | b 1 | ) z ( ( 1 α ) C ( 2 + k α ) 2 k + 1 + α C ( 2 + k α ) | b 1 | ) z 2 .

 □

The following covering result follows from the left side inequality in Theorem 3.

Corollary 1 Let f HST ¯ ( ϕ , χ , k , α ) , then for | b 1 | < 1 α 2 k + α + 1 the set
{ w : | w | < 1 ( 1 α ) C ( 2 + k α ) ( 1 2 k + 1 + α C ( 2 + k α ) ) | b 1 | }

is included in f ( U ) , where C is given by (2.7).

3 Extreme points

Our next theorem is on the extreme points of convex hulls of the class HST ¯ ( ϕ , χ , k , α ) , denoted by c l c o HST ¯ ( ϕ , χ , k , α ) .

Theorem 4 Let f = h + g ¯ be such that h and g are given by (1.2). Then f c l c o HST ¯ ( ϕ , χ , k , α ) if and only if f can be expressed as
f ( z ) = n = 1 [ X n h n ( z ) + Y n g n ( z ) ] ,
(3.1)
where
h 1 ( z ) = z , h n ( z ) = z n ( 1 α ) λ n ( ( 1 + k ) n ( k + α ) ) z n ( n 2 ) , g n ( z ) = z + n ( 1 α ) μ n ( ( 1 + k ) n + ( k + α ) ) z ¯ n ( n 1 ) , X n 0 , Y n 0 , n = 1 [ X n + Y n ] = 1 .

In particular, the extreme points of the class HST ¯ ( ϕ , χ , k , α ) are { h n } and { g n } , respectively.

Proof For functions f ( z ) of the form (3.1), we have
f ( z ) = n = 1 [ X n + Y n ] z n = 2 n ( 1 α ) λ n ( ( 1 + k ) n ( k + α ) ) X n z n + n = 1 n ( 1 α ) μ n ( ( 1 + k ) n + ( k + α ) ) Y n z ¯ n .
Then
n = 2 λ n ( ( 1 + k ) n ( k + α ) ) n ( 1 α ) ( n ( 1 α ) λ n ( ( 1 + k ) n ( k + α ) ) ) X n + n = 1 μ n ( ( 1 + k ) n + ( k + α ) ) n ( 1 α ) ( n ( 1 α ) μ n ( ( 1 + k ) n + ( k + α ) ) ) Y n = n = 2 X n + n = 1 Y n = 1 X 1 1 ,
and so f ( z ) c l c o HST ¯ ( ϕ , χ , k , α ) . Conversely, suppose that f ( z ) c l c o HST ¯ ( ϕ , χ , k , α ) . Set
X n = λ n ( ( 1 + k ) n ( k + α ) ) n ( 1 α ) | a n | ( n 2 )
and
Y n = μ n ( ( 1 + k ) n + ( k + α ) ) n ( 1 α ) | b n | ( n 1 ) ,

then note that by Theorem 2, 0 X n 1 ( n 2 ) and 0 Y n 1 ( n 1 ).

Consequently, we obtain
f ( z ) = n = 1 [ X n h n ( z ) + Y n g n ( z ) ] .

Using Theorem 2, it is easily seen that the class HST ¯ ( ϕ , χ , k , α ) is convex and closed and so c l c o HST ¯ ( ϕ , χ , k , α ) = HST ¯ ( ϕ , χ , k , α ) . □

4 Convolution result

For harmonic functions of the form
f ( z ) = z n = 2 | a n | z n + n = 1 | b n | z ¯ n
(4.1)
and
G ( z ) = z n = 2 A n z n + n = 1 B n z ¯ n ( A n , B n 0 ) ,
(4.2)
we define the convolution of two harmonic functions f and G as
( f G ) ( z ) = f ( z ) G ( z ) = z n = 2 a n A n z n + n = 1 b n B n z ¯ n .

Using this definition, we show that the class HST ¯ ( ϕ , χ , k , α ) is closed under convolution.

Theorem 5 For 0 α < 1 , let f HST ¯ ( ϕ , χ , k , α ) and G HST ¯ ( ϕ , χ , k , α ) . Then f ( z ) G ( z ) HST ¯ ( ϕ , χ , k , α ) .

Proof Let the functions f ( z ) defined by (4.1) be in the class HST ¯ ( ϕ , χ , k , α ) , and let the functions G ( z ) defined by (4.2) be in the class HST ¯ ( ϕ , χ , k , α ) . Obviously, the coefficients of f and G must satisfy a condition similar to inequality (2.4). So, for the coefficients of f ( z ) G ( z ) , we can write
n = 2 λ n n ( 1 + k ) n ( k + α ) ( 1 α ) | a n | A n + n = 1 μ n n ( 1 + k ) n + ( k + α ) ( 1 α ) | b n | B n n = 2 [ λ n n ( 1 + k ) n ( k + α ) ( 1 α ) | a n | + n = 1 μ n n ( 1 + k ) n + ( k + α ) ( 1 α ) | b n | ] ,

the right-hand side of this inequality is bounded by 1 because f HST ¯ ( ϕ , χ , k , α ) . Then f ( z ) G ( z ) HST ¯ ( ϕ , χ , k , α ) . □

Finally, we show that HST ¯ ( ϕ , χ , k , α ) is closed under convex combinations of its members.

Theorem 6 The class HST ¯ ( ϕ , χ , k , α ) is closed under convex linear combination.

Proof For i = 1 , 2 , 3 ,  , let f i HST ¯ ( ϕ , χ , k , α ) , where the functions f i are given by
f i ( z ) = z n = 2 | a n , i | z n + n = 1 | b n , i | z ¯ n .
For i = 1 t i = 1 ; 0 t i 1 , the convex linear combination of f i may be written as
i = 1 t i f i ( z ) = z n = 2 ( i = 1 t i | a n , i | ) z n + n = 1 ( i = 1 t i | b n , i | ) z ¯ n ,
then by (2.4) we have
n = 2 λ n n ( 1 + k ) n ( k + α ) ( 1 α ) i = 1 t i | a n , i | + n = 1 μ n n ( 1 + k ) n + ( k + α ) ( 1 α ) i = 1 t i | b n , i | = i = 1 t i { n = 2 [ λ n n ( 1 + k ) n ( k + α ) ( 1 α ) | a n , i | + n = 1 μ n n ( 1 + k ) n + ( k + α ) ( 1 α ) | b n , i | ] } i = 1 t i = 1 .

This condition is required by (2.4) and so i = 1 t i f i ( z ) HST ¯ ( ϕ , χ , k , α ) . This completes the proof of Theorem 6. □

Remarks
  1. (i)

    Putting k = 0 in our results, we obtain the results obtained by Dixit et al. [9];

     
  2. (ii)

    Putting φ ( z ) = χ ( z ) = z ( 1 z ) 2 and k = 1 in our results, we obtain the results obtained by Rosy et al. [10];

     
  3. (iii)

    Putting φ ( z ) = χ ( z ) = z + z 2 ( 1 z ) 3 in our results, we obtain the results obtained by Kim et al. [8];

     
  4. (iv)

    Putting φ ( z ) = χ ( z ) = z ( 1 z ) 2 and k = 0 in our results, we obtain the results obtained by Jahangiri [3];

     
  5. (v)

    Putting φ ( z ) = χ ( z ) = z + z 2 ( 1 z ) 3 and k = 0 in our results, we obtain the results obtained by Jahangiri [2].

     

Declarations

Acknowledgements

The author would like to express her sincere gratitude to Springer Open Accounts Team for their kind help.

Authors’ Affiliations

(1)
Department of Mathematics, Faculty of Science, Damietta University

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© El-Ashwah; licensee Springer. 2013

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