A Subclass of Harmonic Functions Associated with a Convolution Structure

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Vol. 4, No.2, 2013, 182-191

ISSN: 2279-087X (P), 2279-0888(online) Published on 20 November 2013

www.researchmathsci.org

182

A Subclass of Harmonic Functions Associated with a

Convolution Structure

R. Ezhilarasi and T.V. Sudharsan

Department of Mathematics, SIVET College, Chennai − 600 073, India Email : [email protected], [email protected]

Received 20 October 2013; accepted 5 November 2013

Abstract. In this paper, we introduce a certain subclass of harmonic univalent functions associated with convolution. We also derive the coefficient conditions, extreme points, distortion bounds, convolution conditions and convex combination for this class.

Keywords: Harmonic, univalent, starlike, convex, convolution AMS Mathematics Subject Classification (2010): 30C45, 30C50 1. Introduction

In 1984, Clunie and Sheil-Small [3] investigated the class SH, consisting of complex valued harmonic sense-preserving univalent functions f =h+g in a simply connected domain D ⊆ C defined on the open unit disc ∆ = {z : |z| < 1} and normalized by f(0) = fz(0) − 1 = 0 with h and g given by,

1. b , z b g(z) ;

z a z

h(z) ₁

1 n

n n 2

n n

n = <

+

=

∑

∞

= ∞

=

(1.1)

Clunie and Sheil-Small [3] also considered the geometric subclasses of SH and obtained some coefficient bounds. Since then, there have been several related papers on SH and its subclasses. For a brief history on Harmonic univalent functions, see (Duren [5], Ahuja [1], Jahangiri et al. [8] and Ponnusamy and Rasila [10, 11]).

In 1999, Jahangiri [7] defined the class TH(α) consisting of functions f =h+g such that h and g are of the form

∑

∞

= ∞

=

= −

=

1 n

n n 2

n n

n z ; g(z) b z

a z

h(z) (1.2)

which satisfy the condition

1. r z 1,

α

0

α, )) f(re (arg

θ

i ≥ ≤ < = < ∂

∂

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183

γ

ke ]]

ψ)(z) * (g )(z) *

λ[(h

λ)z [(1 z

] (z) )

ψ

* z(g (z) ) * )[z(h ke (1

Re iα

α i

≥    

 

− +

+ − ′

′ −

′ +

φ φ

(1.3)

for all real α, (z) z λ z ,

2 n

n n

∑

∞

=

+ =

φ

∑

∞

=

+ =

2 n

n nz

µ z

ψ(z) are analytic with conditions,

λn≥ 0, µn≥ 0, 0 ≤λ≤ 1, (z re ),

θ

z = iθ

∂ ∂ =

′ 0 ≤ r < 1, 0 ≤ θ < 2π and 0 ≤γ < 1. The

operator * denotes the Hadamard product or convolution of two power series.

We further let S*H(φ,ψ,λ,γ,k) denote the subclass of S ( ,ψ,λ,γ,k) *

H φ consisting of functions f =h+g∈SH such that h and g are of the form (1.2).

Remark 1.1.

When α = 0, 

     + =    

 

−

− 2

γ 1 T γ,1 ,1, z 1

z , z 1

z

S*_H _H [7]

In the present paper, we obtain the coefficient inequality, extreme points, distortion bounds, convolution conditions and convex combination for the functions of the class

k)

γ,

λ,

ψ, , (

S*H φ .

2. Coefficient Bounds

We begin with a sufficient condition for functions in S*_H(φ,ψ,λ,γ,k).

Theorem 2.1. Let the function f =h+g be so that h and g are given by (1.1). Furthermore, let

1 b

γ

1

γ)]µ λ(k k) [n(1 a

γ

1

γ)]λ λ(k k) [n(1

1 n

n n

2 n

n

n ≤

   

 

− + + + +

   

 

− + −

+

_∑

∑

∞

= ∞

=

(2.1)

where 0 ≤γ < 1, 0 ≤λ≤ 1, k ≥ 0, α real and if

. γ)]µ λ(k k) [n(1 γ)]λ λ(k k) [n(1 γ)

n(1− ≤ + − + n ≤ + + + n

Then f is sense preserving, harmonic univalent mapping in ∆ and for ,

γ

1

γ

1

λ

+ − =

f ∈ S*_H(φ,ψ,λ,γ,k).

Proof. First we note that f is locally univalent and sense-preserving in ∆. This is because

. (z) g r b n b

n

b

µ γ

1

γ)]

λ(k k) [n(1

a

λ γ

1

γ)]

λ(k k) [n(1 1

a n 1 r a n 1 (z) h

1 n

1 n n 1

n n 1 n

n n 2

n

n n 2 n

n 2

n

1 n n

′ ≥ >

≥

− + + + ≥

− + − + −

≥

− > −

≥ ′

∑

∞

=

− ∞

= ∞

=

∞

= ∞

=

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184

To show that f is univalent in ∆, we show that f(z1) ≠ f(z2) whenever z1≠ z2. Suppose z1, z2∈ ∆ so that z1≠ z2. Since the unit disc ∆ is simply connected and convex, we have z(t) = (1−t)z1 + tz2 in D, where 0 ≤ t ≤ 1. Then we write

∫

− ′ + − ′

= −

1

0

1 2 1

2 1

2) f(z ) [(z z )h(z(t)) (z z )g(z(t))]dt f(z

On dividing throughout by z2− z1≠ 0 and taking only the real parts we obtain

∫



  

 

′ − − + ′ = −

− 1

0 2 1

1 2

1

2 _g_(z(t)) _dt

z z

) z (z (z(t)) h Re z

z

) f(z ) f(z Re

>

∫

′ − ′ 1

0

dt ] (z(t)) g (z(t)) h

[Re (2.2)

On the other hand

(2.1). by 0

b µ γ

1

γ)] λ(k k) [n(1

a λ γ

1

γ)] λ(k k) [n(1 1

b n a

n 1

b n (z) h Re (z) g (z) h Re

1 n

n n 2

n

n n 1

n n 2

n n

1 n

n

≥

− + + + −

− + − + −

≥

− −

≥

− ′ ≥ ′ − ′

∑

∞

= ∞

=

∞

= ∞

=

∞

=

Therefore this together with inequality (2.2) implies the univalence of f(z). Next we show that f ∈ S*H(φ,ψ,λ,γ,k).

From (1.3)

γ. B(z) A(z) Re ]]

ψ)(z) * (g )(z) * λ[(h λ)z [(1 z

]]] ψ)(z) * (g )(z) * λ[(h λ)z [(1 z [ ke

] (z) ) ψ * z(g (z) ) * )[z(h ke (1

Re

α i

≥       =       

      

+ +

− ′

+ +

− ′ −

′ −

′ +

φ φ φ

where

]]]

ψ)(z) * (g )(z) *

λ[(h

λ)z [(1 z [ ke ] (z) )

ψ

* z(g (z) ) * )[z(h ke (1

A(z)= + iα φ ′ − ′ − iα ′ − + φ +

and B(z)=z′[(1−λ)z+λ[(h*φ)(z)+(g*ψ)(z)]].

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185

∑

∞ = ∞ = ∞ = ∞ = ∞ = ∞ = ∞ = ∞ = ∞ = ∞ = + + + − + − + − − ≥ + + + + − − + + − − − + + − − − − + − + − − ≥ + + + + − − + + − + − − + + − − − − + − + + − = + + − ′ + − + + − ′ − ′ − ′ + − + + − ′ − + + + − ′ − ′ − ′ + = + − − − + 1 n n n n 2 n n n n 1 n n n n α i 2 n n n n α i 1 n n n n α i 2 n n n n α i 1 n n n n α i 2 n n n n α i 1 n n n n α i 2 n n n n α i α i α i α i α i z b k)]µ λ(γ k) [n(1 2 z a k)]λ λ(γ k) [n(1 2 z γ) 2(1 z b µ λ) (n ke γ)λ) (1 (n z a λ λ) (n ke γ)λ) (1 (n z γ z b µ λ) (n ke γ)λ) (1 (n z a λ λ) (n ke γ)) λ(1 (n z γ) (2 z b λ)]µ (n ke γ)λ) (1 [(n z a λ)]λ (n ke γ)λ) (1 [(n z γ z b λ)]µ (n ke γ)λ) (1 [(n z a λ)]λ (n ke γ)) λ(1 [(n γ)z (2 ]]] ψ)(z) * (g φ)(z) * λ[(h λ)z [(1 z γ)[ (1 ]]] ψ)(z) * (g φ)(z) * λ[(h λ)z [(1 z [ ke ] (z) ) ψ * z(g (z) ) φ * )[z(h ke (1 ]]] ψ)(z) * (g φ)(z) * λ[(h λ)z [(1 z γ)[ (1 ]]] ψ)(z) * (g φ)(z) * λ[(h λ)z [(1 z [ ke ] (z) ) ψ * z(g (z) ) φ * )[z(h ke (1 γ)B(z) (1 A(z) γ)B(z) (1 A(z) 0 b µ γ 1 k)] λ(γ k) [n(1 a λ γ 1 k)] λ(γ k) [n(1 1 z γ) (1 2 z b µ γ 1 k)] λ(γ k) [n(1 z a λ γ 1 k)] λ(γ k) [n(1 1 z γ) (1 2 1 n n n 2 n n n 1 n 1 n n n 2 n 1 n n n ≥               − + + + − − + − + − − ≥               − + + + − − + − + − − ≥

∑

∞ = ∞ = ∞ = − ∞ = −

The coefficient bound (2.1) is sharp for the function

∑

∞ = ∞ = − + + + + − + − + + = 1 n n n 2 n n n

n µ y

γ 1 k)] λ(γ k) [n(1 z x λ γ 1 k)] λ(γ k) [n(1 z

f(z) zn (2.4)

where x y 1.

1 n

n 2

n

n +

∑

=

∑

∞

= ∞

=

□

Next we show that the above sufficient condition is also necessary for functions in

k) γ, λ, ψ, , (

S*_H φ .

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186

1 b µ γ

1

k)] λ(γ k) [n(1 a

λ γ

1

k)] λ(γ k) [n(1

1 n

n n 2

n

n ≤

− + + + +

− + −

+

_∑

∑

∞

= ∞

=

(2.5)

where 0 ≤γ < 1, 0 ≤λ≤ 1, k ≥ 0, α real and

n n [n(1 k) λ(k γ)]µ

γ)]λ λ(k k) [n(1 γ)

n(1− ≤ + − + ≤ + + +

and |bn| > |an|, for every n ≥ 2.

Proof. Since S*H(φ,ψ,λ,γ,k) ⊂ S ( ,ψ,λ,γ,k) *

H φ , we need only to prove the only if part of the theorem. For functions f(z) of the form (1.2), we note that the condition

γ

ke ]]

ψ)(z) * (g )(z) *

λ[(h

λ)z [(1 z

] (z) )

ψ

* z(g (z) ) * )[z(h ke (1

Re iα

α i

≥   

  

− +

+ − ′

′ −

′ +

φ φ

is equivalent to

γ ]]

ψ)(z) * (g )(z) * λ[(h λ)z [(1 z

]]] ψ)(z) * (g )(z) * λ[(h λ)z [(1 z [ ke

] (z) ) ψ * z(g (z) ) * )[z(h ke (1

Re

α i

≥  

  

 

 

  

 

+ +

− ′

+ +

− ′ −

′ −

′ +

φ φ φ

,

or

0

z b µ λ z

a λ λ z

z b λ)]µ (n ke γ) λ [(n

z a λ)]λ (n ke γ) λ [(n γ)z

(1

Re

1 n

n n n 2

n

n n n 1 n

n n n α

i 2

n

n n n α

i

≥

    

     

 

    

     

 

+ −

+ +

+ −

− +

− −

−

∑

∞

= ∞

=

(2.6)

The above required condition (2.6) must hold for all real values of α, |z| = r < 1.

Upon choosing α = 0 and the value of z on the positive real axis where 0 ≤ z = r < 1, we must have for bn ≥an, for every n ≥ 2

0

r b µ λ r

a λ λ 1

r b µ k)] λ(γ k)) [n(1

r a λ k)] λ(γ k) [n(1 γ)

(1

1 n

1 n n n 2

n

1 n n n 1 n

1 n n n 2

n

1 n n n

≥

    

     

 

    

     

 

+ −

+ + + −

+ − + −

−

∑

∞

=

− ∞

=

− ∞

=

− ∞

=

−

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187

If the condition (2.5) does not hold, then the numerator in (2.7) is negative for r sufficiently close to 1. Hence there exist z0 = r0 in (0, 1) for which the quotient in (2.7) is

negative. This contradicts the required condition for f ∈ S*H(φ,ψ,λ,γ,k) and so the proof

is complete. □

3. Distortion Bounds

In next theorem, we determine distortion bounds for functions in S*H(φ,ψ,λ,γ,k).

Theorem 3.1. Let f ∈ S_H*(φ,ψ,λ,γ,k) and

n n 2

γ)]µ λ(k k) [n(1

γ)]λ λ(k k) [n(1

γ)]λ λ(k k) [2(1

+ + + ≤

+ − + ≤ + − +

for n ≥ 2. Then we have,

1 r z , r b

γ)]λ λ(k k) [2(1

γ)]µ λ(k k [1

γ)]λ λ(k k) [2(1

γ

1 )r

b (1

f(z) ₁ 2

2 1

2

1  = <

  

 

+ − +

+ + + − +

− +

+ ≤

(3.1) and

1 r z , r b

γ)]λ λ(k k) [2(1

γ)]µ λ(k k [1

γ)]λ λ(k k) [2(1

γ

1 )r

b (1

f(z) ₁ 2

2 1

2

1  = <

  

 

+ − +

+ + + − +

− +

− −

− ≥

(3.2) Proof. We only prove the first inequality. The proof for the second inequality is similar. Let f ∈ S*_H(φ,ψ,λ,γ,k). Taking the absolute value of f we obtain

∑

∞

=

∞

= ∞

=

   

 

− + − + + −

+ − +

− +

+ =

+ +

+ ≤

+ +

+ ≤

2 n

2 n 2 n

2

2 1

2 n

2 n n 1

2 n

n n n 1

r b

γ

1

γ)]λ λ(k k) [2(1 a

γ

1

γ)]λ λ(k k) [2(1

γ

1 r

) b (1

r ) b a ( r ) b (1

r ) b a ( r ) b (1 f(z)

_∑

∞

= 

 

 

− + + + + −

+ − +

− +

+ ≤

2 n

2 n n n

n 2 1

r b

γ

1

γ)]µ λ(k k) [n(1 a

γ

1

γ)]λ λ(k k) [n(1

γ)]λ λ(k k) [2(1

γ

1 r

) b (1

b r ,

γ

1

γ)]µ λ(k k [1 1

γ)]λ λ(k k) [2(1

γ

1 r

) b

(1 1 ₁ 2

2

1 

  

 

− + + + − +

− +

+ ≤

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188

b r .

γ)]λ λ(k k) [2(1

γ)]µ λ(k k [1

γ)]λ λ(k k) [2(1

γ

1 r

) b

(1 1 2

2 1

2

1 

  

 

+ − +

+ + + − +

− +

+ ≤

4. Extreme Points

Now we obtain the extreme points of S*_H(φ,ψ,λ,γ,k).

Theorem 4.1. Let h1(z) = z,

, z γ)]λ λ(k k) [n(1

γ) (1 z

(z)

h n

n

n ₊ ₋ ₊

− −

= n ≥ 2 and

, z γ)]µ λ(k k) [n(1

γ) (1 z

(z)

g n

n

n ₊ ₊ ₊

− +

= n ≥ 1.

Then f ∈ S*H(φ,ψ,λ,γ,k) if and only if it can be expressed as

, ) g y h (x f(z)

1 n

n n n n

∑

∞

=

+

= (4.1)

where xn ≥ 0, yn ≥ 0, (x y ) 1 1

n

n n+ =

∑

∞

=

. In particular, the extreme points of

k)

γ,

λ,

ψ, , (

S*H φ are {hn} and {gn}.

Proof. Suppose

∑

∞

= ∞

=

∞

= ∞

=

+ + +

− +

+ − +

− −

=

+ + +

− +

+ − +

− −

+ =

1 n

n n n 2

n

n n n

1 n

n n n 2

n

n n n 1

n

n n 1 n

n n n n

z y

γ)]µ λ(k k) [n(1

γ) (1 z

x

γ)]λ λ(k k) [n(1

γ) (1 z

z y

γ)]µ λ(k k) [n(1

γ) (1 z

x

γ)]λ λ(k k) [n(1

γ) (1 z

) y (x

) g y h (x f(z)

Then from (2.5) we have

1 x 1 y x

y

γ)]µ λ(k k) [n(1

γ) (1

γ)]µ λ(k k) [n(1

x

γ)]λ λ(k k) [n(1

γ) (1

γ)]λ λ(k k) [n(1

b

µ γ

1

k)]

λ(γ

k) [n(1 a

λ γ

1

k)]

λ(γ

k) [n(1

1 1

n n 2

n n 1 n

n n n

2 n

n n n

1 n

n n 2

n

n n

≤ − = +

=

   

 

+ + +

− −

+ + + +

   

 

+ − +

− −

+ − + =

− + + + +

− + − +

∑

∞

= ∞

=

∞

= ∞

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189 and so f ∈ S*_H(φ,ψ,λ,γ,k).

Conversely, if f ∈ S*H(φ,ψ,λ,γ,k), then

n n

γ)]λ λ(k k) [n(1

γ) (1 a

+ − +

− ≤

and

. γ)] λ(k k) [n(1

γ) (1 b

n

n ₊ ₊ ₊ _µ

− ≤

Setting

, a γ)

(1

γ)]λ λ(k k) [n(1

x n _n

n ₋

+ − +

= n ≥ 2 and b ,

γ) (1

γ)]µ λ(k k) [n(1

y n _n

n ₋

+ + +

= n ≥ 1.

Then note that by Theorem 2.2, 0 ≤ xn≤ 1, (n ≥ 2) and 0 ≤ yn ≤ 1, (n ≥ 1). We define

0, y x

1 x

1 n

n 2

n n

1 = −

∑

−

∑

≥

∞

= ∞

=

by Theorem 2.2.

Consequently, we can see that f(z) can be expressed in the form (4.1). This completes the

proof of the Theorem 4.1. □

5. Convolution and Convex Combination

In this section, we show that the class S*H(φ,ψ,λ,γ,k) is invariant under convolution and convex combination of its members. The convolution of two harmonic functions

∑

∞

= ∞

=

+ −

=

1 n

n n 2

n n

nz b z a

z f(z)

and

, z B z

A z F(z)

1 n

n n 2

n n

n

∑

∞

= ∞

=

+ −

=

is defined as

. z B b z

A a z F(z) * f(z) F)(z) * (f

1 n

n n n 2

n

n n

n

∑

∞

= ∞

=

+ −

= =

Theorem 5.1. If f ∈ S*H(φ,ψ,λ,γ,k) and F ∈ S ( ,ψ,λ,γ,k) *

H φ then

f * F ∈ S*H(φ,ψ,λ,γ,k).

Proof. Let

∑

∞

= ∞

=

+ −

=

1 n

n n 2

n

n z b z

a z

f(z) and

∑

∞

= ∞

=

+ −

=

1 n

n n 2

n

n z B z

A z

F(z) be in

k)

γ,

λ,

ψ, , (

S*H φ , then by Theorem 2.2 we have

1 b

γ) (1

γ)]µ λ(k k) [n(1 a

γ) (1

γ)]λ λ(k k) [n(1

1 n

n n

2 n

n

n _≤

− + + + +

− + −

+

_∑

∑

∞

= ∞

=

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190 and

1. B

γ) (1

γ)]µ λ(k k) [n(1 A

γ) (1

γ)]λ λ(k k) [n(1

1 n

n n

2 n

n

n _≤

− + + + +

− + −

+

_∑

∑

∞

= ∞

=

(5.2)

So for the coefficients of f * F we can write

1.

b γ)

(1

γ)]µ λ(k k) [n(1 a

γ) (1

γ)]λ λ(k k) [n(1

B b γ)

(1

γ)]µ λ(k k) [n(1 A

a γ)

(1

γ)]λ λ(k k) [n(1

1 n

n n

2 n

n n

1 n

n n n

2 n

n n n

≤

− + + + +

− + − + ≤

− + + + +

− + − +

∑

∞

= ∞

=

∞

= ∞

=

Thus f * F ∈ S*H(φ,ψ,λ,γ,k).

Finally, we prove that S*H(φ,ψ,λ,γ,k) is closed under convex combinations of its members.

Theorem 5.2. The class S*_H(φ,ψ,λ,γ,k) where fi is given by . z b z

a z (z) f

1 n

n i n, 2

n

n i n,

i

∑

∞

= ∞

=

+ −

= (5.3)

is closed under convex combination.

Proof. For i = 1, 2, ..., suppose fi(z) ∈ S ( ,ψ,λ,γ,k) *

H φ where fi is given by (5.3). Then by (2.5),

1. b

γ) (1

γ)]µ λ(k k) [n(1 a

γ) (1

γ)]λ λ(k k) [n(1

1 n

i n, n

2 n

i n,

n ≤

− + + + +

− + −

+

_∑

∑

∞

= ∞

=

(5.4)

For t 1,

1 i

i =

∑

∞

=

0 ≤ ti≤ 1, the convex combination of fi(z) may be written as

. z b t z

a t z

(z) f

t n

1 n i1

n,i i n

2 n i1

n,i i 1

i i

i

∑ ∑

∑

∞

= ∞

= 

   

   +     

   − =

Then by (5.4), we obtain

1 t

b

γ

1

γ)]µ λ(k k) [n(1 a

γ

1

γ)]λ λ(k k) [n(1 t

b t

γ

1

γ)]µ λ(k k) [n(1 a

t

γ

1

γ)]λ λ(k k) [n(1

1 i

i

1 n

i n, n

1 i n 2

i n, n i

1

n i 1

i n, i n

2

n i1

i n, i n

= ≤

   −

+ + + +

  

− + − + =

   

   

− + + + +

   

− + − +

∑

∑ ∑

∑

∞

=

∞

= ∞

=

∞

=

∞

= ∞

=

∞

=

and so by Theorem 2.2, we have

∑

∈

∞

=1 i

i if (z)

(10)

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