Gen. Math. Notes, Vol. 22, No. 1, May 2014, pp.17-30
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2014
www.i-csrs.org
Available free online at http://www.geman.in
A Class of Harmonic Multivalent Functions
Defined by an Integral Operator
R. Ezhilarasi1 , T.V. Sudharsan2 and K.G. Subramanian3
1,2
Department of Mathematics, SIVET College
Chennai - 600 073, India
1
E-mail: ezhilarasi2008@ymail.com
2
E-mail: tvsudharsan@rediffmail.com
3
School of Computer Sciences
Universiti Sains Malaysia
11800 Penang, Malaysia
E-mail: kgsmani1948@yahoo.com
(Received: 27-1-14 / Accepted: 12-3-14)
Abstract
A new class of harmonic multivalent functions defined by an integral operator is introduced. Coefficient inequalities, extreme points, distortion bounds,
inclusion results and closure under an integral operator for this class are obtained.
Keywords: Harmonic functions, multivalent functions, integral operator.
1
Introduction
Harmonic mappings are important in different applied fields of study [1]. Harmonic mappings in a simply connected domain D ⊆ C are univalent complex
valued harmonic functions f = u + iv where both u and v are real harmonic
in D.
Let SH denote the family of harmonic functions f = h + g [6], which are
univalent and sense-preserving in the open unit disc ∆ = {z : |z| < 1} where
h and g are analytic in D and f is normalized by f (0) = h(0) = fz (0) − 1 = 0.
Subclasses of harmonic functions have been studied by many authors (See for
18
R. Ezhilarasi et al.
example, Aouf et al. [2], Atshan and Kulkarni [3], Chandrashekar et al. [5],
Cotı̂rlă [7], Jahangiri [9, 10], Jahangiri and Ahuja [11], Jahangiri et al. [12]).
The class Hp (n) (p, n ∈ N = {1, 2, . . . }), consisting of all p-valent harmonic
functions f = h + g that are orientation preserving in ∆ was defined by Ahuja
and Jahangiri [11] where h and g are of the form
p
h(z) = z +
∞
X
ak+p−1 z
k+p−1
, g(z) =
k=2
∞
X
bk+p−1 z k+p−1 , |bp | < 1.
(1)
k=1
An integral operator I n was introduced by Salagean [14] which is given below
in a slightly modified form as stated by [7].
(i) I 0 f (z) = f (z);
(ii) I 1 f (z) = If (z) = p
Rz
0
f (t)t−1 dt;
(iii) I n f (z) = I(I n−1 f (z)), n ∈ N , f ∈ A
where A = {f ∈ H : f (z) = z + a2 z 2 + . . . } and H = H(∆), the class of
holomorphic functions in ∆.
The modified Salagean integral operator of f = h+g given by (1) is defined
[7] as
I n f (z) = I n h(z) + (−1)n I n g(z),
(2)
where
∞ X
n
p
I h(z) = z +
ak+p−1 z k+p−1 and
k
+
p
−
1
k=2
n
∞ X
p
n
bk+p−1 z k+p−1
I g(z) =
k
+
p
−
1
k=1
n
p
For 0 ≤ β < 1, 0 ≤ t ≤ 1, n ∈ N , z ∈ ∆, let Hp (n, β, t) denote the family
of harmonic functions of the form (1) such that
I n f (z)
>β
(3)
Re
(1 − t)z p + tI n+1 f (z)
where I n is defined by (2).
Let Hp (n, β, t) denote the subclass consists of harmonic functions fn =
h + gn in Hp (n, β, t) so that h and gn are of the form
h(z) = z p −
∞
X
ak+p−1 z k+p−1 and gn (z) = (−1)n−1
k=2
where ak+p−1 , bk+p−1 ≥ 0 and |bp | < 1.
∞
X
k=1
bk+p−1 z k+p−1
(4)
19
A Class of Harmonic Multivalent Functions...
Remark 1.1 The class Hp (n, β, t) reduces to the class Hp (n, β) [7] and to
the class Hp (n + 1, n, β, 0) [8], when t = 1.
Coefficient inequalities, extreme points, distortion bounds, inclusion results
and closure under an integral operator for functions in the class Hp (n, β, t) are
obtained.
2
Main Results
A sufficient coefficient condition for harmonic functions belonging to the class
Hp (n, β, t) is now derived.
Theorem 2.1 Let f = h + g be given by (1). If
∞
X
φ(n, p, k, β, t)|ak+p−1 | +
k=2
∞
X
ψ(n, p, k, β, t)|bk+p−1 | ≤ 1
(5)
k=1
where
p
k+p−1
n h
i
p
1 − βt k+p−1
p
k+p−1
1−β
n h
i
p
1 + βt k+p−1
φ(n, p, k, β, t) =
ψ(n, p, k, β, t) =
1−β
,
0 ≤ β < 1, 0 ≤ t ≤ 1, n ∈ N . Then f ∈ Hp (n, β, t).
Proof. To show that f ∈ Hp (n, β, t) according to the condition (3), we only
need to show that if (5) holds, then
Re
I n f (z)
(1 − t)z p + tI n+1 f (z)
= Re
A(z)
≥β
B(z)
where z = reiθ , 0 ≤ θ ≤ 2π, 0 ≤ r < 1 and 0 ≤ β < 1.
Note that A(z) = I n f (z) and
B(z) = (1 − t)z p + tI n+1 f (z).
Using the fact that Re w ≥ β if and only if |1 − β + w| ≥ |1 + β − w|, it suffices
to show that
|A(z) + (1 − β)B(z)| − |A(z) − (1 + β)B(z)| ≥ 0
(6)
20
R. Ezhilarasi et al.
Substituting A(z) and B(z) in (6) we obtain
|A(z) + (1 − β)B(z)| − |A(z) − (1 + β)B(z)|
= I n f (z) + (1 − β)[(1 − t)z p + tI n+1 f (z)]
− I n f (z) − (1 + β)[(1 − t)z p + tI n+1 f (z)]
n
n
∞ ∞ X
p
p
p X
k+p−1
n
= z +
bk+p−1 z k+p−1
ak+p−1 z
+ (−1)
k
+
p
−
1
k
+
p
−
1
k=2
k=1
"
n+1
∞
X
p
p
p
ak+p−1 z k+p−1
+ (1 − β) (1 − t)z + tz + t
k
+
p
−
1
k=2
#
n+1
∞
X
p
+t(−1)n+1
bk+p−1 z k+p−1 k
+
p
−
1
k=1
n
n
∞ ∞ X
p
p
p X
k+p−1
n
− z +
bk+p−1 z k+p−1
ak+p−1 z
+ (−1)
k
+
p
−
1
k
+
p
−
1
k=2
k=1
"
n+1
∞
X
p
p
p
− (1 + β) (1 − t)z + tz + t
ak+p−1 z k+p−1
k+p−1
k=2
#
∞
n+1
X
p
n+1
k+p−1
bk+p−1 z
+t(−1)
k+p−1
k=1
∞
n
X
p
p
p
= (2 − β)z +
1 + (1 − β)t
ak+p−1 z k+p−1
k+p−1
k+p−1
k=2
∞
n
X
p
p
n+1
k+p−1
bk+p−1 z
−(−1)
1 − (1 − β)t
k+p−1
k+p−1
k=1
n ∞ X
p
p
p
− −βz +
1 − (1 + β)t
ak+p−1 z k+p−1
k
+
p
−
1
k
+
p
−
1
k=2
n ∞
X
p
p
n+1
k+p−1
−(−1)
1 + (1 + β)t
bk+p−1 z
k+p−1
k+p−1
k=1
∞
n
X
p
p
p
1 + (1 − β)t
|ak+p−1 ||z|k+p−1
≥ (2 − β)|z| −
k+p−1
k+p−1
k=2
∞
n
X
p
p
−
1 − (1 − β)t
|bk+p−1 ||z|k+p−1
k
+
p
−
1
k
+
p
−
1
k=1
n ∞ X
p
p
p
1 − (1 + β)t
|ak+p−1 ||z|k+p−1
− β|z| −
k
+
p
−
1
k
+
p
−
1
k=2
n ∞
X
p
p
−
1 + (1 + β)t
|bk+p−1 ||z|k+p−1
k+p−1
k+p−1
k=1
21
A Class of Harmonic Multivalent Functions...
∞ X
n p
p
≥ 2(1 − β)|z| −
1 + (1 − β)t
k+p−1
k+p−1
k=2
p
|ak+p−1 ||z|k+p−1
+1 − (1 + β)t
k+p−1
n ∞ X
p
p
−
1 − (1 − β)t
k+p−1
k+p−1
k=1
p
+1 + (1 + β)t
|bk+p−1 ||z|k+p−1
k+p−1
n ∞
X
p
p
p
2
≥ 2(1 − β)|z| −
1 − βt
|ak+p−1 ||z|k+p−1
k+p−1
k+p−1
k=2
∞
n
X
p
p
−
2
1 + βt
|bk+p−1 ||z|k+p−1
k
+
p
−
1
k
+
p
−
1
k=1
p

= 2(1 − β)|z|p 1 −
n h
i

p
p
∞
X
1
−
βt
k+p−1
k+p−1

+
k=2
∞
X
p
k+p−1
n h
1−β
k=1

≥ 2(1 − β) 1 −
1−β
i
p
1 + βt k+p−1
n h
i

p
p
∞
X
1
−
βt
k+p−1
k+p−1

+
1−β
k=2
n h
i
p
p
∞
1
+
βt
X
k+p−1
k+p−1
1−β
k=1
|ak+p−1 ||z|k−1


k−1 
|bk+p−1 ||z|

|ak+p−1 |


|bk+p−1 | 

≥ 0, by (5).
This completes the proof.
The harmonic univalent functions
p
f (z) = z +
∞
X
k=2
where n ∈ N and
∞
X
1
1
k+p−1
xk z
+
yk z k+p−1
φ(n, p, k, β, t)
ψ(n,
p,
k,
β,
t)
k=1
∞
X
k=2
by (5) is sharp.
|xk |+
∞
X
k=1
(7)
|yk | = 1, shows that the coefficient bound given
22
R. Ezhilarasi et al.
This is because
∞
X
φ(n, p, k, β, t)|ak+p−1 | +
k=2
=
∞
X
=
ψ(n, p, k, β, t)|bk+p−1 |
k=1
φ(n, p, k, β, t)
k=2
∞
X
∞
X
|Xk | +
∞
X
k=2
∞
X
1
1
|Xk | +
|Yk |
ψ(n, p, k, β, t)
φ(n, p, k, β, t)
ψ(n,
p,
k,
β,
t)
k=1
|Yk | = 1.
k=1
We now show that the condition (5) is also necessary for functions fn =
h + gn , where h and gn are of the form (4).
Theorem 2.2 Let fn = h + gn be given by (4). Then fn ∈ Hp (n, β, t) if
and only if
∞
X
k=2
φ(n, p, k, β, t)ak+p−1 +
∞
X
ψ(n, p, k, β, t)bk+p−1 ≤ 1.
(8)
k=1
where 0 ≤ β < 1, 0 ≤ t ≤ 1, n ∈ N , with bk+p−1 > ak+p−1 , for every k ≥ 2.
Proof. We only need to prove the “only if” part of the theorem because
Hp (n, β, t) ⊂ Hp (n, β, t). To this end, for functions fn of the form (4), we
notice that the condition
I n f (z)
Re
>β
(1 − t)z p + tI n+1 f (z)
is equivalent to

n ∞ h
X

p
p

p

(1 − β)z −
1 − βt
ak+p−1 z k+p−1


k+p−1
k+p−1


k=2

n 
∞ i

X

p
p
2n−1
k+p−1


+(−1)
1
+
βt
b
z
k+p−1


k+p−1
k+p−1
k=1
Re
n+1
∞ h
X

p

p

z −t
ak+p−1 z k+p−1



k
+
p
−
1

k=2

n+1

∞ i

X

p

2n
k+p−1

+t(−1)
bk+p−1 z


k+p−1
k=1
≥0































(9)
23
A Class of Harmonic Multivalent Functions...
We observe that the above required condition (9) must hold for all values of z
in ∆. Choosing the values of z on the positive real axis where 0 ≤ z = r < 1,
we have for bk+p−1 > ak+p−1 , for every k ≥ 2,
∞ h
X
(1 − β) −
n p
p
1 − βt
ak+p−1 rk−1
k
+
p
−
1
k
+
p
−
1
k=2
n ∞ i
X
p
p
k−1
−
1 + βt
bk+p−1 r
k+p−1
k+p−1
k=1
≥ 0 (10)
n+1
∞ h
X
p
k−1
1−t
ak+p−1 r
k
+
p
−
1
k=2
n+1
∞ i
X
p
+t
bk+p−1 rk−1
k+p−1
k=1
If the condition (8) does not hold, then the expression in (10) is negative
for r sufficiently close to 1. Hence there exist z0 = r0 in (0, 1) for which
the quotient in (10) is negative. This contradicts the required condition for
fn ∈ Hp (n, β, t) and this completes the proof.
The extreme points of closed convex hull of Hp (n, β, t), denoted by clco Hp (n, β, t)
is now determined.
Theorem 2.3 Let fn be given by (4). Then fn ∈ Hp (n, β, t) if and only if
∞
X
fn (z) =
[xk+p−1 hk+p−1 (z) + yk+p−1 gnk+p−1 (z)],
k=1
1
z k+p−1 , k = 2, 3, . . . ,
φ(n, p, k, β, t)
1
and gnk+p−1 (z) = z p + (−1)n−1
z k+p−1 , k = 1, 2, 3, . . . .
ψ(n, p, k, β, t)
∞
∞
X
X
xk+p−1 ≥ 0, yk+p−1 ≥ 0, xp = 1 −
xk+p−1 −
yk+p−1 .
where hp (z) = z p , hk+p−1 (z) = z p −
k=2
k=1
In particular, the extreme point of Hp (n, β, t) are {hk+p−1 } and {gnk+p−1 }.
24
R. Ezhilarasi et al.
Proof. Suppose
∞
X
fn (z) =
[xk+p−1 hk+p−1 (z) + yk+p−1 gnk+p−1 (z)]
=
k=1
∞
X
(xk+p−1 + yk+p−1 )z p −
k=1
k=2
+ (−1)n−1
∞
X
k=1
= zp −
∞
X
∞
X
k=2
1
xk+p−1 z k+p−1
φ(n, p, k, β, t)
1
yk+p−1 z k+p−1
ψ(n, p, k, β, t)
1
xk+p−1 z k+p−1
φ(n, p, k, β, t)
n−1
+ (−1)
∞
X
k=1
1
yk+p−1 z k+p−1
ψ(n, p, k, β, t)
Then
∞
X
φ(n, p, k, β, t)|ak+p−1 | +
k=2
∞
X
ψ(n, p, k, β, t)|bk+p−1 |
k=1
∞
X
1
φ(n, p, k, β, t)
xk+p−1
=
φ(n,
p,
k,
β,
t)
k=2
∞
X
1
ψ(n, p, k, β, t)
+
yk+p−1
ψ(n,
p,
k,
β,
t)
k=1
=
∞
X
k=2
xk+p−1 +
∞
X
yk+p−1 = 1 − xp ≤ 1.
k=1
and so fn (z) ∈ clco Hp (n, β, t).
Conversely, if fn (z) ∈ clco Hp (n, β, t). Letting
xp = 1 −
∞
X
k=2
xk+p−1 −
∞
X
yk+p−1 .
k=1
Set
xk+p−1 = φ(n, p, k, β, t)ak+p−1 , k = 2, 3, . . . and
yk+p−1 = ψ(n, p, k, β, t)bk+p−1 , k = 1, 2, . . . .
25
A Class of Harmonic Multivalent Functions...
The required representations is obtained as
∞
X
fn (z) = z p −
ak+p−1 z k+p−1 + (−1)n−1
k=2
k=2
n−1
+ (−1)
= zp −
"
= 1−
+
=
k=2
∞
X
1
xk+p−1 z k+p−1
φ(n, p, k, β, t)
∞
X
k=1
∞
X
p
1
yk+p−1 z k+p−1
ψ(n, p, k, β, t)
[z − hk+p−1 (z)]xk+p−1 −
k=2
∞
X
k=2
∞
X
bk+p−1 z k+p−1
k=1
∞
X
= zp −
∞
X
∞
X
[z p − gk+p−1 (z)]yk+p−1
k=1
xk+p−1 −
∞
X
#
yk+p−1 z p
k=1
xk+p−1 hk+p−1 (z) +
∞
X
yk+p−1 gnk+p−1 (z)
k=1
[xk+p−1 hk+p−1 (z) + yk+p−1 gnk+p−1 (z)]
k=1
We now obtain the distortion bounds for functions in Hp (n, β, t).
Theorem 2.4 Let fn ∈ Hp (n, β, t). Then for |z| = r < 1 we have
|fn (z)| ≤ (1 + bp )rp + {θ(n, p, k, β, t) − Ω(n, p, k, β, t)bp }rn+p+1
and
|fn (z)| ≥ (1 − bp )rp − {θ(n, p, k, β, t) − Ω(n, p, k, β, t)bp }rn+p+1
where
θ(n, p, k, β, t) = 1 − βt
n h
i
p
p
1
−
βt
p+1
p+1
Ω(n, p, k, β, t) = 1 + βt
n h
i
p
p
1
−
βt
p+1
p+1
We prove the right hand side inequality for |fn |. The proof for the left hand
inequality is similar. Let fn ∈ Hp (n, β, t) taking the absolute value of fn then
26
R. Ezhilarasi et al.
by Theorem 2.2, we obtain:
∞
∞
X
p X
|fn (z)| = z −
bk+p−1 z k+p−1 ak+p−1 z k+p−1 + (−1)n−1
k=2
≤ rp +
∞
X
k=1
ak+p−1 rk+p−1 +
k=2
∞
X
bk+p−1 rk+p−1
k=1
= r p + bp r p +
∞
X
(ak+p−1 + bk+p−1 )rk+p−1
k=2
≤ r p + bp r p +
∞
X
(ak+p−1 + bk+p−1 )rp+1
k=2
p
∞
X
p
k=2
n+p+1
= (1 + bp )r + θ(n, p, k, β, t)
1
(ak+p−1 + bk+p−1 )rp+1
θ(n, p, k, β, t)
≤ (1 + bp )r + θ(n, p, k, β, t)r
"∞
#
X
φ(n, p, k, β, t)ak+p−1 + ψ(n, p, k, β, t)bk+p−1
k=2
p
≤ (1 + bp )r + {θ(n, p, k, β, t) − Ω(n, p, k, β, t)bp }rn+p+1 .
3
Closure Property of the Class Hp(n, β, t)
In the next two theorems, we prove that the class Hp (n, β, t) is invariant under
convolution and convex combinations of its members.
The convolution of two harmonic functions,
∞
X
p
fn (z) = z −
ak+p−1 z
k+p−1
n−1
+ (−1)
k=2
and
p
Fn (z) = z −
∞
X
k=2
∞
X
bk+p−1 z k+p−1
(11)
Bk+p−1 z k+p−1
(12)
k=1
Ak+p−1 z
k+p−1
+ (−1)
n−1
∞
X
k=1
is defined as
(fn ∗ Fn )(z) = fn (z) ∗ Fn (z)
∞
∞
X
X
p
k+p−1
n−1
=z −
ak+p−1 Ak+p−1 z
+ (−1)
bk+p−1 Bk+p−1 z k+p−1
k=2
k=1
(13)
Using this definition, we first show that the class Hp (n, β, t) is closed under
convolution.
27
A Class of Harmonic Multivalent Functions...
Theorem 3.1 For 0 ≤ α ≤ β < 1, 0 ≤ t ≤ 1, let fn ∈ Hp (n, β, t) and
Fn ∈ Hp (n, α, t). Then
fn ∗ Fn ∈ Hp (n, β, t) ⊂ Hp (n, α, t).
p
Proof. Let fn (z) = z −
∞
X
ak+p−1 z
k=2
∞
X
Hp (n, β, t) and Fn (z) = z p −
k+p−1
+ (−1)
n−1
∞
X
bk+p−1 z k+p−1 be in
k=1
∞
X
n−1
Ak+p−1 z k+p−1 + (−1)
k=2
Bk+p−1 z k+p−1 be
k=1
in Hp (n, α, t).
Then the convolution fn ∗ Fn is given by (13). We wish to show that the
coefficients of fn ∗ Fn satisfy the required condition given in Theorem 2.2. For
Fn ∈ Hp (n, α, t), we note that Ak+p−1 < 1 and Bk+p−1 < 1. Now, for the
convolution function fn ∗ Fn , we obtain
∞
X
φ(n, p, k, α, t)ak+p−1 Ak+p−1 +
k=2
≤
≤
∞
X
ψ(n, p, k, α, t)bk+p−1 Bk+p−1
k=1
∞
X
φ(n, p, k, α, t)ak+p−1 +
∞
X
k=2
k=1
∞
X
∞
X
φ(n, p, k, β, t)ak+p−1 +
k=2
ψ(n, p, k, α, t)bk+p−1
ψ(n, p, k, β, t)bk+p−1
k=1
≤ 1,
since 0 ≤ α ≤ β < 1 and fn ∈ Hp (n, β, t).
Now, we show that Hp (n, β, t) is closed under convex combination of its
members.
Theorem 3.2 The class Hp (n, β, t) is closed under convex combination.
Proof. For i = 1, 2, 3, . . . . Suppose fni ∈ Hp (n, β, t), where fni is given by
fni (z) = z p −
∞
X
ai,k+p−1 z k+p−1 + (−1)n−1
k=2
∞
X
bi,k+p−1 z k+p−1 .
k=1
Then by (8)
∞
X
k=2
φ(n, p, k, β, t)ai,k+p−1 +
∞
X
k=1
ψ(n, p, k, β, t)bi,k+p−1 ≤ 1
(14)
28
R. Ezhilarasi et al.
For
∞
X
ti = 1, 0 ≤ ti ≤ 1, the convex combination of fni may be written as
i=1
∞
X
p
ti fni (z) = z −
i=1
n−1
+ (−1)
∞
∞
X
X
∞
X
k=2
∞
X
k=1
i=1
!
ti ai,k+p−1 z k+p−1
i=1
!
ti bi,k+p−1 z k+p−1
Using the inequality (14), we obtain
!
!
∞
∞
∞
∞
X
X
X
X
φ(n, p, k, β, t)
ti ai,k+p−1 +
ψ(n, p, k, β, t)
ti bi,k+p−1
i=1
k=2
=
∞
X
i=1
≤
∞
X
ti
∞
X
k=1
∞
X
φ(n, p, k, β, t)ai,k+p−1 +
i=1
!
ψ(n, p, k, β, t)bi,k+p−1
k=1
k=2
ti = 1,
i=1
which is the required coefficient condition.
Finally, we examine the closure property of the class Hp (n, β, t) under the
generalized Bernardi-Libera-Livingston integral operator (see [4, 13]) Lc (f )
which is defined by,
Z
c + p z c−1
t f (t)dt, c > −1.
Lc (f ) =
zc 0
Theorem 3.3 Let fn (z) ∈ Hp (n, β, t). Then
Lc (f (z)) ∈ Hp (n, β, t).
Proof. From the representation of Lc (fn (z)), it follows that
"
#
Z
∞
∞
X
c + p z c−1 p X
Lc (fn (z)) =
t
t −
ak+p−1 tk+p−1 + (−1)n−1
bk+p−1 tk+p−1 dt
zc 0
k=2
k=1
∞
X
p
=z −
k=2
n−1
+ (−1)
c+p
ak+p−1 z k+p−1
c+p+k−1
∞
X
k=1
= zp −
∞
X
k=2
c+p
bk+p−1 z k+p−1
c+p+k−1
Xk+p−1 z k+p−1 + (−1)n−1
∞
X
k=1
Yk+p−1 z k+p−1
A Class of Harmonic Multivalent Functions...
29
where
c+p
ak+p−1 and
c+p+k−1
c+p
=
bk+p−1 .
c+p+k−1
Xk+p−1 =
Yk+p−1
Hence
∞
X
c+p
c+p
ak+p−1 +
bk+p−1
ψ(n, p, k, β, t)
φ(n, p, k, β, t)
c
+
p
+
k
−
1
c
+
p
+
k
−
1
k=1
k=2
∞
X
≤
∞
X
φ(n, p, k, β, t)ak+p−1 +
k=2
∞
X
ψ(n, p, k, β, t)bk+p−1
k=1
≤ 1 by (8).
Hence by Theorem 2.2, Lc (fn (z)) ∈ Hp (n, β, t).
Acknowledgements: The authors are grateful to the reviewer for useful
comments.
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