A CLASS OF UNIVALENT HARMONIC FUNCTIONS DEFINED BY MULTIPLIER TRANSFORMATION

DEFINED BY MULTIPLIER TRANSFORMATION

RAJ KUMAR, SUSHMA GUPTA and SUKHJIT SINGH

In this paper, we define subclasses RSH(k, λ, γ) and RS_H(k, λ, γ) of univalent harmonic functions by using multiplier transformationI(k, λ). Certain coefficient conditions and extreme points for the above classes are obtained. We also discuss convolution properties of the functions from the classRS_H(k, λ, γ).

AMS 2010 Subject Classification: 30C45, 30C50.

Key words: harmonic function, univalent function, convolution.

1. INTRODUCTION

A continuous functionf(x+iy) =u(x, y) +iv(x, y) defined in a domain D ⊂ C (Complex plane) is harmonic in D if u and v are real harmonic in D. Clunie and Shiel-Small [3] showed that in a simply connected domain such functions can be written in the formf =h+g, where bothhandgare analytic.

We callhthe analytic part andg, the co-analytic part off. Letw(z) = ^g_h⁰0^(z)(z)be the dilatation off =h+g. The mappingf is sense-preserving and locally one- to-one in the open unit discE ={z:|z|<1},if and only if the Jacobian of the mapping, Jf(z) =|h⁰(z)|²− |g⁰(z)|²,is positive. So, the condition for f to be sense-preserving and locally one-to-one is that|h⁰(z)|>|g⁰(z)|or equivalently,

|w(z)|<1 inE. We denote bySH the class of harmonic, sense preserving and univalent functions in the unit disc E, normalized by the conditions f(0) = 0 andf_z(0) = 1. So, a harmonic mapping in the classS_H has the representation f =h +g,where

(1) h(z) =z+

∞

X

n=2

a_nzⁿ, g(z) =

∞

X

n=1

b_nzⁿ, |b₁|<1.

S⁰_His the subclass ofS_H whose membersf satisfy additional condition,fz(0) = b1 = 0. K_H and K_H⁰ are, respectively, subclasses ofS_H and S_H⁰ which mapE onto convex domains.

REV. ROUMAINE MATH. PURES APPL.,57(2012),4, 371-382

Convolution of two harmonic functionsf(z) =z+

∞

P

n=2

a_nzⁿ+

∞

P

n=1

b_nzⁿand F(z) =z+

∞

P

n=2

A_nzⁿ+

∞

P

n=1

B_nzⁿ is denoted byf∗F and is defined as follows (f∗F)(z) =z+

∞

X

n=2

a_nA_nzⁿ+

∞

X

n=1

b_nB_nzⁿ.

In 2001, Rosy et. al. [5] studied the subclassG_H(γ) ⊂S_H consisting of univalent harmonic functions f which satisfy the condition

Re

(1 +e^iα)zf⁰(z) f(z) −e^iα

≥γ, 0≤γ <1, α∈R.

Recently, in 2007, Yalcin et. al. [6] defined and studied the subclassRSH(k, γ) ofS_H consisting of harmonic functions f which satisfy the following condition

Re

(1 +e^iα)D^k+1f(z) D^kf(z) −e^iα

≥γ, 0≤γ <1, α∈R,

where D^kf(z) is the modified Salagean operator of harmonic univalent func- tion f introduced by Jahangiri et. al. [4] and is defined as

D^kf(z) =D^kh(z) + (−1)^kD^kg(z), k∈N. Here,

D^kh(z) =z+

∞

X

n=2

n^ka_nzⁿ and D^kg(z) =

∞

X

n=1

n^kb_nzⁿ.

Motivated by above papers, we define the subclass RSH(k, λ, γ) of class SH

consisting of harmonic functions f of the form (1) which satisfy the condition

(2) Re

(1 +e^iα)I(k+ 1, λ)f(z) I(k, λ)f(z) −e^iα

≥γ, 0≤γ <1, α∈R, where I(k, λ) is the modified multiplier transformation defined as follows

I(k, λ)f(z) =I(k, λ)h(z) + (−1)^kI(k, λ)g(z), 0≤λ≤1, k∈N0 =N∪ {0}.Here,

I(k, λ)h(z) =z+

∞

X

n=2

n+λ 1 +λ

k

a_nzⁿ and I(k, λ)g(z) =

∞

X

n=1

n+λ 1 +λ

k

b_nzⁿ. The multiplier transformation I(k, γ) for analytic functionf given by

I(k, γ)f(z) =z+

∞

X

n=2

n+γ 1 +γ

k

a_nzⁿ,

was introduced by Cho and Kim [2].RS_H⁰(k, λ, γ) is the corresponding subclass of S_H⁰ satisfying condition (2).

For k∈ N0, let f_k =h+g_k be a univalent harmonic function, where h and g_k are of the form

(3) h(z) =z−

∞

X

n=2

|a_n|zⁿ, g_k(z) = (−1)^k

∞

X

n=1

|b_n|zⁿ.

We denote by RS_H(k, λ, γ), the class of functions of the type f_k = h+g_k, which satisfy the condition (2).

In Section 2 of the present paper, we obtain sufficient coefficient condi- tions for a harmonic function to be in RS_H(k, λ, γ) or RS_H(k, λ, γ).Extreme points for the class RS_H(k, λ, γ) are obtained in Section 3. In the last sec- tion, we investigate the properties of convolution of the function in the class RSH(k, λ, γ) with other harmonic functions.

2. COEFFICIENT CONDITIONS

Theorem 1. Let a function f =h+g∈S_H given by (1) satisfy

∞

X

n=1

n+λ 1 +λ

k

2(n+λ)−(1 +λ)(1 +γ) 1 +λ

|a_n|+

(4)

+

2(n+λ) + (1 +λ)(1 +γ) 1 +λ

|b_n|

≤2(1−γ),

0≤γ <1, α∈R, 0≤λ≤1, k∈N0 =N∪ {0}. Then,f ∈RS_H(k, λ, γ).

Proof. For n ∈ N, 0 ≤ γ < 1 and 0 ≤ λ ≤ 1, we have the following inequalities

n <

n+λ 1 +λ

k

2(n+λ)−(1 +λ)(1 +γ) (1 +λ)(1−γ)

(5)

<

n+λ 1 +λ

k

2(n+λ) + (1 +λ)(1 +γ) (1 +λ)(1−γ)

, (6)

for k= 0,1,2,3. . ..

Now, we shall prove thatf is sense-preserving and univalent inE. Since

|z|<1, so

|h⁰(z)| ≥1−

∞

X

n=2

n|a_n| |z|ⁿ⁻¹ >1−

∞

X

n=2

n|a_n|

≥1−

∞

X

n=2

n+λ 1 +λ

k

2(n+λ)−(1 +λ)(1 +γ) (1 +λ)(1−γ)

|a_n|

(by inequality (5))

≥

∞

X

n=1

n+λ 1 +λ

k

2(n+λ) + (1 +λ)(1 +γ) (1 +λ)(1−γ)

|b_n|

(using condition (4) as|a₁|= 1)

>

∞

X

n=1

n|b_n|z||ⁿ⁻¹ (by inequality (6))>|g⁰(z)|.

This shows thatf is sense-preserving inE. Next, we prove the univalence of f. Forz16=z2 ∈E,

f(z₁)−f(z₂) h(z1)−h(z2)

≥1−

g(z₁)−g(z₂) h(z1)−h(z2)

>1−

∞

P

n=1

n|b_n| 1−

∞

P

n=2

n|a_n|

>1−

∞

P

n=1

n+λ 1+λ

kn2(n+λ)+(1+λ)(1+γ) (1+λ)(1−γ)

o|b_n|

1−

∞

P

n=2

n+λ 1+λ

kn2(n+λ)−(1+λ)(1+γ) (1+λ)(1−γ)

o|a_n|

>0 (using condition (4)).

Hence,f is univalent in E. In order to prove that f ∈RS_H(k, λ, γ) we have to show that

Re

(1 +e^iα)I(k+ 1, λ)f(z) I(k, λ)f(z) −e^iα

≥γ, 0≤γ <1, α∈R.

We know that Re(w)≥γ if and only if |1−γ+w| ≥ |1 +γ−w|. So, it suffices to show that

(1−γ)I(k, λ)f(z) + (1 +e^iα)I(k+ 1, λ)f(z)−e^iαI(k, λ)f(z)|−

− |(1 +γ)I(k, λ)f(z)−(1 +e^iα)I(k+ 1, λ)f(z) +e^iαI(k, λ)f(z) ≥0.

Now,

(1−γ)I(k, λ)f(z) + (1 +e^iα)I(k+ 1, λ)f(z)−e^iαI(k, λ)f(z)|−

− |(1 +γ)I(k, λ)f(z)−(1 +e^iα)I(k+ 1, λ)f(z) +e^iαI(k, λ)f(z) =

=

(2−γ)z+

∞

X

n=2

1−γ−e^iα+

n+λ 1 +λ

+

n+λ 1 +λ

e^iα n+λ 1 +λ

k

a_nzⁿ−

−(−1)^k

∞

X

n=1

n+λ 1 +λ

+

n+λ 1 +λ

e^iα−1 +γ+e^iα n+λ 1 +λ

k

b_nzⁿ

−

−

γz−

∞

X

n=2

n+λ 1 +λ

+

n+λ 1 +λ

e^iα−1−γ−e^iα n+λ 1 +λ

k

a_nzⁿ+ + (−1)^k

∞

X

n=1

n+λ 1 +λ

+

n+λ 1 +λ

+ 1 +γ+e^iα n+λ 1 +λ

k

b_nzⁿ

≥

≥(2−γ)|z| −

∞

X

n=2

n+λ 1 +λ

k 2

n+λ 1 +λ

−γ

|a_n| |z|ⁿ−

−

∞

X

n=1

n+λ 1 +λ

k 2

n+λ 1 +λ

+γ

|b_n| |z|ⁿ−

−γ|z| −

∞

X

n=2

n+λ 1 +λ

k 2

n+λ 1 +λ

−γ−2

|a_n| |z|ⁿ−

−

∞

X

n=1

n+λ 1 +λ

k 2

n+λ 1 +λ

+γ+ 2

|b_n| |z|ⁿ=

= 2(1−γ)|z| −2

∞

X

n=2

n+λ 1 +λ

k 2

n+λ 1 +λ

−γ−1

|a_n| |z|ⁿ−

−2

∞

X

n=1

n+λ 1 +λ

k 2

n+λ 1 +λ

+γ+ 1

|b_n| |z|ⁿ=

= 2(1−γ)|z|

"

1−

∞

X

n=2

n+λ 1 +λ

k2(n+λ)−(1 +γ)(1 +λ)

(1 +λ)(1−γ) |a_n| |z|ⁿ⁻¹−

−

∞

X

n=1

n+λ 1 +λ

k

2(n+λ) + (1 +γ)(1 +λ)

(1 +λ)(1−γ) |b_n| |z|ⁿ⁻¹

#

>

>2(1−γ)

"

1−

∞

X

n=2

n+λ 1 +λ

k

2(n+λ)−(1 +γ)(1 +λ) (1 +λ)(1−γ) |a_n|−

−

∞

X

n=1

n+λ 1 +λ

k

2(n+λ) + (1 +γ)(1 +λ) (1 +λ)(1−γ) |b_n|

#

≥0,

in view of condition (4). The proof of the theorem is now complete.

In the next result, we show that the condition (4) is also necessary for the function RS_H(k, λ, γ).

Theorem 2. Letf_k =h+g_k be given by (3). Then f_k∈RS_H(k, λ, γ) if and only if

∞

X

n=1

n+λ 1 +λ

k

2(n+λ)−(1 +λ)(1 +γ) 1 +λ

|a_n|+ +

2(n+λ) + (1 +λ)(1 +γ) 1 +λ

|b_n|

≤2(1−γ).

Proof. SinceRS_H(k, λ, γ)⊂RS_H(k, λ, γ) therefore, “if” part of the Theo- rem 2 follows from Theorem 1. We only need to prove the “only if” part.

Letf_k=h+g_k∈RS_H(k, λ, γ) then, the condition (2) is equivalent to Re

(1 +e^iα)I(k+ 1, λ)f(z)

I(k, λ)f(z) −e^iα−γ

≥0, 0≤γ <1, α∈R.

After substitutingI(k+ 1, λ)f(z) andI(k, λ)f(z) in the above inequality we get.

Re









(1−γ)z−

∞

P

n=2

hn+λ 1+λ

−γ+

n+λ 1+λ −1

e^iαi

n+λ 1+λ

k

|a_n|zⁿ

I(k, λ)f(z) −

−

(−1)^2k

∞

P

n=1

hn+λ 1+λ

+γ+

n+λ 1+λ + 1

i n+λ 1+λ

k

|b_n|zⁿ

I(k, λ)f(z)









≥0.

Re









(1−γ)−

∞

P

n=2

hn+λ 1+λ

−γ+

n+λ 1+λ −1

e^iαi

n+λ 1+λ

k

|a_n|zⁿ⁻¹

1−

∞

P

n=2

n+λ 1+λ

k

|a_n|zⁿ⁻¹+^z_z(−1)^2k

∞

P

n=1

n+λ 1+λ

k

|b_n|zⁿ⁻¹

−

−

z z(−1)^2k

∞

P

n=1

hn+λ 1+λ

+γ+

n+λ

1+λ + 1i

n+λ 1+λ

k

|b_n|zⁿ⁻¹

1−

∞

P

n=2

n+λ 1+λ

k

|a_n|zⁿ⁻¹+^z_z(−1)^2k

∞

P

n=1

n+λ 1+λ

k

|b_n|zⁿ⁻¹









≥0.

The above condition holds for all values ofz,|z|=r <1. Choosing the values of z on the +ve real axis, where 0≤z=r <1, we have

Re









(1−γ)−

∞

P

n=2

n+λ 1+λ −γ

e^iα

n+λ 1+λ

^k

|an|rⁿ⁻¹−

∞

P

n=1

n+λ

1+λ +γ ^n+λ_1+λ^k

|bn|rⁿ⁻¹ 1−

∞

P

n=2

n+λ 1+λ

k

|an|rⁿ⁻¹+

∞

P

n=1

n+λ 1+λ

k

|bn|rⁿ⁻¹

−

e^iαn ^∞ P

n=2

n+λ 1+λ −1

e^iα

n+λ 1+λ

k

|an|rⁿ⁻¹+

∞

P

n=1

n+λ

1+λ+ 1 ^n+λ_1+λk

|bn|rⁿ⁻¹o

1−

∞

P

n=2

n+λ 1+λ

^k

|an|rⁿ⁻¹+

∞

P

n=1

n+λ 1+λ

^k

|bn|rⁿ⁻¹









≥0.

Since Re(−e^iα)≥ −1, the above inequality reduces to (1−γ)−

∞

P

n=2

2^n+λ_1+λ−γ−1

n+λ 1+λ

^k

|an|rⁿ⁻¹−

∞

P

n=1

2^n+λ_1+λ + 1 +γ ^n+λ_1+λ^k

|bn|rⁿ⁻¹

1− P^∞

n=2

n+λ 1+λ

k

|a_n|rⁿ⁻¹+

∞

P

n=1

n+λ 1+λ

k

|b_n|rⁿ⁻¹

≥0.

(7)

In case the condition (4) fails then, the numerator of (7) and hence, the expression in the left hand side of (7) becomes negative for r sufficiently close to 1. This contradicts the condition for RS_H(k, λ, γ).

Hence the result is proved.

Theorem 3. Members of class RS_H(k, λ, γ) map unit disk E onto a starlike domain.

Proof. A function f =h+g∈ RS_H(k, λ, γ) maps the unit disk E onto starlike domain if, Ren_zh0(z)−zg⁰(z)

h(z)+g(z)

o

>0, for all z∈E.

We know that Re(w)>0 if and only if |1 +w|>|1−w|. So, it suffices to show that,

|h(z) +g(z) +zh⁰(z)−zg⁰(z)| − |h(z) +g(z)−zh⁰(z) +zg⁰(z)|>0.

Now,

h(z) +g(z) +zh⁰(z)−zg⁰(z)| − |h(z) +g(z)−zh⁰(z) +zg⁰(z) =

=

2z−

∞

X

n=2

(n+ 1)|a_n|zⁿ+

∞

X

n=1

(n−1)|b_n|zⁿ

−

−

∞

X

n=2

(n−1)|a_n|zⁿ+

∞

X

n=1

(n+ 1)|b_n|zⁿ

≥

≥2|z|

"

1−

∞

X

n=2

n|a_n| |z|ⁿ⁻¹−

∞

X

n=1

n|b_n| |z|ⁿ⁻¹

#

≥

≥2|z|

"

1−

∞

X

n=2

n|a_n| −

∞

X

n=1

n|b_n| |z|

#

>

>2|z|

"

1−

∞

X

n=2

n+λ 1 +λ

k 2(n+λ)−(1 +γ)(1 +λ) (1 +λ)(1−γ) |a_n|−

−

∞

X

n=1

n+λ 1 +λ

k 2(n+λ) + (1 +γ)(1 +λ) (1 +λ)(1−γ) |b_n|

#

>0,

in view of condition (4) and Theorem 2. Hence it completes the proof.

3. EXTREME POINTS

In the next theorem, we obtain the extreme points for the class RS_H(k, λ, γ).

Theorem 4. Let fk = h+gk where h and gk are given by (3). Then f_k ∈RS_H(k, λ, γ) if and only if

f_k(z) =

∞

X

n=1

(X_nh_n(z) +Y_ng_kn(z)), where h₁(z) =z,

h_n(z) =z− (1−γ)(1 +λ)^(k+1)

[2(n+λ)−(1 +λ)(1 +γ)](n+λ)^kzⁿ, n= 2,3,4, . . . (8)

g_kn=z+ (−1)^k (1−γ)(1 +λ)^(k+1)

[2(n+λ) + (1 +λ)(1 +γ)](n+λ)^kzⁿ, n= 1,2,3, . . . ,

∞

X

n=1

(Xn+Yn) = 1,

X_n ≥ 0, Y_N ≥ 0. In particular, the extreme points of RS_H(k, r, λ) are {h_n} and {g_kn}.

Proof. Substituting forhn(z) and gkn(z) in fk, we have fk(z) =

∞

X

n=1

(Xnhn(z) +Yngkn(z)) =

=

∞

X

n=1

(X_n+Y_n)z−

∞

X

n=2

(1−γ)(1 +λ)^(k+1)

[2(n+λ)−(1 +λ)(1 +γ)](n+λ)^kX_nzⁿ+ +(−1)^k

∞

X

n=1

(1−γ)(1 +λ)^(k+1)

[2(n+λ) + (1 +λ)(1 +γ)](n+λ)^kY_nzⁿ. Comparing it with (3), we get

|a_n|= (1−γ)(1 +λ)^(k+1)

[2(n+λ)−(1 +λ)(1 +γ)](n+λ)^kXn, and

|b_n|= (−1)^k

∞

X

n=1

(1−γ)(1 +λ)^(k+1)

[2(n+λ) + (1 +λ)(1 +γ)](n+λ)^kY_n. Now,

∞

X

n=2

n+λ 1 +λ

k

2(n+λ)−(1 +λ)(1 +γ) (1−γ)(1 +λ)

|a_n|+

+

∞

X

n=1

2(n+λ) + (1 +λ)(1 +γ) (1−γ)(1 +λ)

|b_n|

#

=

∞

X

n=2

X_n+

∞

X

n=1

Y_n= 1−X₁ ≤1.

So,f_k∈RS_H(k, λ, γ).

Conversely, letfk ∈RS_H(k, λ, γ). Set Xn=

n+λ 1 +λ

k

2(n+λ)−(1 +λ)(1 +γ) (1−γ)(1 +λ)

|a_n|, n= 2,3,4, . . . and

(9)

Y_n=

n+λ 1 +λ

k

2(n+λ) + (1 +λ)(1 +γ) (1−γ)(1 +λ)

|b_n|, n= 1,2,3, . . . , where P∞

n=1(X_n+Y_n) = 1.Since f_k∈RS_H(k, λ, γ), therefore f_k=z−

∞

X

n=2

|a_n|zⁿ+ (−1)^k

∞

X

n=1

|b_n|zⁿ. Substituting values of |a_n|and |b_n|from (6), we get

f_k=

∞

X

n=1

(X_nh_n(z) +Y_ng_kn(z)),

where hn(z) and gkn are given by (5). Hence, it completes the proof.

Remark 1. Ahuja et. al. [1] proved that a functionf =h+gwherehand g are given by (1) is said to be in the multiplier family F_H{c_n, d_n}, if there exist sequencescnand dn of positive real numbers such that

∞

X

n=2

cn|a_n|+

∞

X

n=1

dn|b_n| ≤1, d1|b₁|<1.

Further, if n ≤cn and n ≤ dn then, F_H{c_n, dn} ⊂ S_H.If we set k= 2p (an even number) in Theorem 1 and 2 we obtain,

cn=

∞

X

n=1

n+λ 1 +λ

2p

2(n+λ)−(1 +λ)(1 +γ) (1 +λ)(1−γ)

, d_n=

∞

X

n=1

n+λ 1 +λ

2p

2(n+λ) + (1 +λ)(1 +γ) (1 +λ)(1−γ)

.

It is pertinent to mention here that our result does not follow from that of Ahuja et. al. [1], where K is an odd number.

4. CONVOLUTION PROPERTIES

Theorem 5. Let f ∈ RS_H(k, λ, γ1) and F ∈ RS_H(k, λ, γ2) where 0 ≤ γ₁ ≤γ₂ <1, then f∗F ∈RS_H(k, λ, γ₂)⊂RS_H(k, λ, γ₁).

Proof. Letf(z) =z−P∞

n=2|a_n|zⁿ+P∞

n=1|b_n|zⁿbe inRS_H(k, λ, γ1) and F(z) =z−P∞

n=2|A_n|zⁿ+P∞

n=1|B_n|zⁿbe inRS_H(k, λ, γ₂) satisfy the coeffi- cient condition (4) and also |A_n| ≤1 and |B_n| ≤1. Now, for the coefficients of f ∗F, we can write

∞

X

n=1

n+λ 1 +λ

k

2(n+λ)−(1 +λ)(1 +γ2) (1−γ₂)(1 +λ)

|a_nAn|+

+

2(n+λ) + (1 +λ)(1 +γ2) (1−γ₂)(1 +λ)

|b_nBn|

≤

≤

∞

X

n=1

n+λ 1 +λ

k

2(n+λ)−(1 +λ)(1 +γ₂) (1−γ₂)(1 +λ)

|a_n|+

+

2(n+λ) + (1 +λ)(1 +γ₂) (1−γ2)(1 +λ)

|b_n|

≤

≤

∞

X

n=1

n+λ 1 +λ

k

2(n+λ)−(1 +λ)(1 +γ₁) (1−γ1)(1 +λ)

|a_n|+

+

2(n+λ) + (1 +λ)(1 +γ₁) (1−γ1)(1 +λ)

|b_n|

≤2.

Hence,f∗F ∈RS_H(k, λ, γ2)⊂RS_H(k, λ, γ1).

Remark 2. By setting k=λ= 0 in Theorems 1, 2, 3 and 4 we get the corresponding results proved by Rosy et. al. [5] and by taking λ= 0 in these theorems, we obtain the corresponding results of Yalcin et. al. [6].

In order to prove our next result, we require the following lemma of Clunie and Shiel-Small [3].

Lemma 1. If F ∈K_H⁰ is given by F(z) =z+

∞

P

n=2

Anzⁿ+Bnzⁿ , then

|A_n| ≤ n+ 1

2 and |B_n| ≤ n−1 2 . Theorem 6. If f ∈ RS⁰

H(k, λ, γ) and F ∈ K_H⁰, then, for 0 ≤ λ ≤ 1, f ∗F ∈RS⁰

H(k−1, λ, γ).

Proof. Sincef ∈RS_H⁰(k, λ, γ), therefore,

∞

X

n=2

n+λ 1 +λ

k

2(n+λ)−(1 +λ)(1 +γ) 1 +λ

|a_n|+ +

2(n+λ) + (1 +λ)(1 +γ) 1 +λ

|b_n|

≤(1−γ).

We know that

(f∗F)(z) =z+

∞

X

n=2

a_nA_nzⁿ+b_nB_nzⁿ . Now,

∞

X

n=2

n+λ 1 +λ

(k−1)

2(n+λ)−(1 +λ)(1 +γ) 1 +λ

|a_nAn|+ +

2(n+λ) + (1 +λ)(1 +γ) 1 +λ

|b_nB_n|

=

=

∞

X

n=2

n+λ 1 +λ

(k−1)

2(n+λ)−(1 +λ)(1 +γ) 1 +λ

|a_n| |A_n|+ +

2(n+λ) + (1 +λ)(1 +γ) 1 +λ

|b_n| |B_n|

≤

≤

∞

X

n=2

n+λ 1 +λ

(k−1)

2(n+λ)−(1 +λ)(1 +γ) 1 +λ

n+ 1 2

|a_n|+ +

2(n+λ) + (1 +λ)(1 +γ) 1 +λ

n−1 2

|b_n|

(using Lemma 1)≤

≤

∞

X

n=2

n+λ 1 +λ

(k−1)

2(n+λ)−(1 +λ)(1 +γ) 1 +λ

n+λ 1 +λ

|a_n|+ +

2(n+λ) + (1 +λ)(1 +γ) 1 +λ

n+λ 1 +λ

|b_n|

≤

≤

∞

X

n=2

n+λ 1 +λ

k

2(n+λ)−(1 +λ)(1 +γ) 1 +λ

|a_n|+ +

2(n+λ) + (1 +λ)(1 +γ) 1 +λ

|b_n|

≤1−γ.

Hence,f∗F ∈RS⁰_H(k−1, λ, γ).

Acknowledgements.First author is thankful to Council of Scientific and Industrial Research, New Delhi, for financial support in the form of Junior Research Fellowship (grant no. 09/797/0006/2010 EMR-1).

REFERENCES

[1] O.P. Ahuja and J.M. Jahangiri, Certain multipliers of univalent harmonic functions.

Appl. Math. Lett.18(2005), 1319–1324.

[2] N.E. Cho and T.H. Kim, Multiplier transformation and strongly close-to-convex func- tions. Bull. Korean Math. Soc.40(2003), 399–410.

[3] J. Clunie and Shiel-Small, Harmonic univalent functions. Ann. Acad. Sci. Fenn. Math.

9(1984), 3–25.

[4] J.M. Jahangiri, G. Murugusundaramoorthy and K. Vijaya,Salagean-type harmonic uni- valent function. Southwest J. Pure Appl. Math.2(2002), 77–82.

[5] T. Rosy, B.A. Stephen and K.G. Subramanian,Goodman-Ronning type harmonic univa- lent functions. Kyungpook Math. J.41(2001), 45–54.

[6] S. Yalcin, M. Ozturk and M. Yamankaradenz,On the subclass of salagean-type harmonic univalent functions. J. Inequal. Pure Appl. Math.8(2007),2.

Received 19 January 2012 Sant Longowal Institute of Engineering and Technology Department of Mathematics Longowal-148106(Punjab), India

[email protected] [email protected]

sukhjit [email protected]