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Subclass of univalent harmonic functions defined by dual convolution

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.

1 Introduction

A continuous function f=u+iv is a complex-valued harmonic function in a simply connected complex domain DC 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={zC:|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 kUCV, k0, so that ϕkUCV if and only if

Re { 1 + ( z ζ ) ϕ ( z ) ϕ ( z ) } 0 ( | ζ | k ; z U ) .
(1.3)

For θR, if we let ζ=kz 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 HCV(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(0α<1;θR;k0).
(1.5)

Also, the class of kUST uniformly starlike functions is defined by using (1.4) as the class of all functions ψ(z)=z ϕ (z) such that ϕkUCV, then ψ(z)kUST 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 kUST 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(0α<1;θR;k0).
(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(0α<1;θR;k0).
(1.8)

The convolution of two functions of the form

f(z)=z+ n = 2 a n z n andF(z)=z+ n = 2 A n z n

is defined as

(fF)(z)=z+ n = 2 a n A n z n ,
(1.9)

while the integral convolution is defined by

(fF)(z)=z+ n = 2 a n A n n z n .
(1.10)

From (1.9) and (1.10), we have

(fF)(z)= 0 z ( f F ) ( t ) t dt.

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(0α<1;k0;θ real),
(1.11)

where

φ(z)=z+ n = 2 λ n z n ( λ n 0)andχ(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 ¯ CV(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), 0α<1, k0 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 fHST(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 fHST(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 (1α)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||1w| 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 forn2andC=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 clco HST ¯ (ϕ,χ,k,α).

Theorem 4 Let f=h+ g ¯ be such that h and g are given by (1.2). Then fclco 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)clco HST ¯ (ϕ,χ,k,α). Conversely, suppose that f(z)clco HST ¯ (ϕ,χ,k,α). Set

X n = λ n ( ( 1 + k ) n ( k + α ) ) n ( 1 α ) | a n |(n2)

and

Y n = μ n ( ( 1 + k ) n + ( k + α ) ) n ( 1 α ) | b n |(n1),

then note that by Theorem 2, 0 X n 1 (n2) and 0 Y n 1 (n1).

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 clco 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

(fG)(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 0α<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].

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

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El-Ashwah, R.M. Subclass of univalent harmonic functions defined by dual convolution. J Inequal Appl 2013, 537 (2013). https://doi.org/10.1186/1029-242X-2013-537

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Keywords

  • harmonic function
  • univalent
  • sense-preserving
  • integral convolution