MATEMATIQKI VESNIK
originalni nauqni rad
research paper
65, 4 (2013), 555–564
December 2013
ON A NEW CLASS OF HARMONIC UNIVALENT FUNCTIONS
Waggas Galib Atshan and Abbas Kareem Wanas
Abstract. We define a new class of harmonic univalent functions of the form f = h + g in
the open unit disk U . Also we study some properties of this class, like coefficient bounds, extreme
points, convex combination, distortion bounds, integral operator and convolution property.
1. Introduction
A continuous function f = u + iv is a complex valued harmonic function in
a complex domain C, if both u and v are real harmonic in C. In any simply
connected domain D ⊂ C, we can write f = h + g, where h and g are analytic in
D. We call h the analytic part and g the co-analytic part of f . A necessary and
sufficient condition for f to be locally univalent and sense-preserving in D is that
|h0 (z)| > |g 0 (z)| in D (see Clunie and Sheil-Small [2]).
Denote by MH the class of functions f = h + g that are harmonic univalent
and sense preserving in the unit disk U = {z ∈ C : |z| < 1}. So f = h + g ∈ MH
is normalized by f (0) = fz (0) − 1 = 0. For f = h + g ∈ MH , we may express the
analytic functions h and g as
∞
∞
P
P
h(z) = z +
an z n , g(z) =
bn z n , |b1 | < 1.
(1.1)
n=2
n=1
Also denote by WH the subclass of MH containing all functions f = h + g where h
and g are given by
∞
∞
P
P
h(z) = z −
an z n , g(z) = −
bn z n , (an ≥ 0, bn ≥ 0, |b1 | < 1).
(1.2)
n=2
n=1
We denote by KMH (γ, α, β) the class of all functions of the form (1.1) that satisfy
the condition
¯
¯
½
¾
¯ zf 00 (z) + f 0 (z)
¯
zf 00 (z) + f 0 (z)
¯
Re
>β¯ 0
− 1¯¯ + α,
(1.3)
f 0 (z) + γzf 00 (z)
f (z) + γzf 00 (z)
2010 AMS Subject Classification: 30C45, 30C50.
Keywords and phrases: Harmonic function; extreme points; convex combination; integral
operator.
555
556
W. G. Atshan, A. K. Wanas
where 0 ≤ α < 1, β ≥ 0, 0 ≤ γ < 1 and z ∈ U .
Let KWH (γ, α, β) be the subclass of KMH (γ, α, β), where
KWH (γ, α, β) = WH ∩ KMH (γ, α, β).
When β = 0, the class reduces to KWH (γ, α, 0) = C(γ, α), which was studied by
Mostafa [6] for analytic part.
Such type of study was carried out by various authors for another classes,
like Frasin [4], Yalcin [9], Gencel and Yalcin [5], Cotirla [3], Porwal et al. [7] and
Seker [8].
2. Coefficient bounds
First we determine the sufficient condition for f = h + g to be in the class
KMH (γ, α, β).
Theorem 2.1. Let h + g with h and g given by (1.1). If
∞
P
n=2
n{(n − 1)[β(1 − γ) − αγ] + n − α}(|an | + |bn |) ≤ (1 − α)(1 − |b1 |),
(2.1)
where 0 ≤ α < 1, β ≥ 0, 0 ≤ γ < 1, then f is harmonic univalent in U and
f ∈ KMH (γ, α, β).
Proof. For proving f ∈ KMH (γ, α, β), we must show that (1.3) holds true. It
is sufficient to show that
¾
½
zf 00 (z) + f 0 (z)
iθ
iθ
(1
+
βe
)
−
βe
> α (−π ≤ θ ≤ π),
Re
f 0 (z) + γzf 00 (z)
or equivalently
½
¾
(1 + βeiθ )(zf 00 (z) + f 0 (z)) − βeiθ (f 0 (z) + γzf 00 (z))
Re
> α.
f 0 (z) + γzf 00 (z)
(2.2)
If we put
A(z) = (1 + βeiθ )(zf 00 (z) + f 0 (z)) − βeiθ (f 0 (z) + γzf 00 (z))
and B(z) = f 0 (z) + γzf 00 (z), we only need to prove that
|A(z) + (1 − α)B(z)| − |A(z) − (1 + α)B(z)| ≥ 0.
But
¯
³P
∞
∞
P
¯
|A(z) + (1 − α)B(z)| = ¯(1 + βeiθ )
n(n − 1)an z n−1 +
n(n − 1)bn (z)n−1
n=2
+1+
∞
P
n=2
+
∞
P
n=1
nan z n−1 +
nbn (z)n−1 +
∞
P
n=1
∞
P
n=2
´
³
n=1
nbn (z)n−1 − βeiθ 1 +
γn(n − 1)an z n−1 +
∞
P
n=1
∞
P
n=2
nan z n−1
γn(n − 1)bn (z)n−1
´
³
∞
∞
∞
P
P
P
+ (1 − α) 1 +
nan z n−1 +
nbn (z)n−1 +
γn(n − 1)an z n−1
n=2
n=1
n=2
557
On a new class of harmonic univalent functions
+
∞
P
n=1
´¯
¯
γn(n − 1)bn (z)n−1 ¯
¯
∞
¡
¢
P
¯
= ¯(2 − α) +
n (n − 1)(βeiθ (1 − γ) + (1 − α)γ) + n + 1 − α an z n−1
n=2
+
¯
¢
¯
n (n − 1)(βeiθ (1 − γ) + (1 − α)γ) + n + 1 − α bn (z)n−1 ¯.
¡
∞
P
n=1
Also
¯
³P
∞
∞
P
¯
|A(z) − (1 + α)B(z)| = ¯(1 + βeiθ )
n(n − 1)an z n−1 +
n(n − 1)bn (z)n−1
n=2
+1+
∞
P
n=2
+
∞
P
n=1
nan z n−1 +
nbn (z)n−1 +
∞
P
n=1
∞
P
n=2
´
³
n=1
nbn (z)n−1 − βeiθ 1 +
γn(n − 1)an z n−1 +
∞
P
n=1
∞
P
n=2
nan z n−1
γn(n − 1)bn (z)n−1
´
³
∞
∞
∞
P
P
P
− (1 + α) 1 +
nan z n−1 +
γn(n − 1)an z n−1
nbn (z)n−1 +
n=2
+
∞
P
n=1
´¯
¯
γn(n − 1)bn (z)n−1 ¯
n=1
n=2
¯
∞
¡
¢
P
¯
=¯−α+
n (n − 1)(βeiθ (1 − γ) − (1 + α)γ) + n − (1 + α) an z n−1
n=2
+
¯
¢
¯
n (n − 1)(βeiθ (1 − γ) − (1 + α)γ) + n − (1 + α) bn (z)n−1 ¯.
¡
∞
P
n=1
Then
|A(z) + (1 − α)B(z)| − |A(z) − (1 + α)B(z)|
∞
P
≥ 2(1 − α) −
2n((n − 1)(β(1 − γ) − αγ) + n − α)|an ||z|n−1
n=2
−
∞
P
n=1
2n((n − 1)(β(1 − γ) − αγ) + n − α)|bn ||z|n−1
n
∞
P
> 2 (1 − α) −
n((n − 1)(β(1 − γ) − αγ) + n − α)|an |
n=2
−
∞
P
n=1
o
n((n − 1)(β(1 − γ) − αγ) + n − α)|bn | > 0.
The harmonic univalent function
∞
P
xn
zn
f (z) = z +
n((n
−
1)(β(1
−
γ) − αγ) + n − α)
n=2
∞
P
yn
+
(z)n ,
(2.3)
n=1 n((n − 1)(β(1 − γ) − αγ) + n − α)
P∞
P∞
where n=2 |xn |+ n=1 |yn | = 1−α, show that the coefficient bound given by (2.1)
is sharp. The functions of the form (2.3) are in the class KMH (γ, α, β), because
∞
P
|xn |
n((n − 1)(β(1 − γ) − αγ) + n − α)
n((n − 1)(β(1 − γ) − αγ) + n − α)
n=2
558
W. G. Atshan, A. K. Wanas
+
∞
P
n=1
=
∞
P
n=2
|yn |
n((n − 1)(β(1 − γ) − αγ) + n − α) n((n−1)(β(1−γ)−αγ)+n−α)
|xn | +
∞
P
n=1
|yn | = 1 − α.
The restriction placed in Theorem 2.1 on the moduli of the coefficients of f = h + g
enables us to conclude for arbitrary rotation of the coefficients of f that the resulting
functions would still be harmonic univalent and f ∈ KMH (γ, α, β).
In the following theorem, it is shown that the condition (2.1) is also necessary
for functions in KWH (γ, α, β).
Theorem 2.2. Let f = h + g with h and g be given by (1.2). Then f ∈
KWH (γ, α, β) if and only if
∞
P
n((n − 1)(β(1 − γ) − αγ) + n − α)an
n=2
∞
P
+
n=1
n((n − 1)(β(1 − γ) − αγ) + n − α)bn ≤ 1 − α
(2.4)
where 0 ≤ α < 1, β ≥ 0, 0 ≤ γ < 1.
Proof. Since KWH (γ, α, β) ⊂ KMH (γ, α, β), we only need to prove the “only
if” part of the theorem. Assume that f ∈ KWH (γ, α, β). Then by (1.3), we have
¾
½
zf 00 (z) + f 0 (z)
iθ
iθ
(1 + βe ) − βe
> α.
Re
f 0 (z) + γzf 00 (z)
This is equivalent to
¾
½
(1 + βeiθ )(zf 00 (z) + f 0 (z)) − βeiθ (f 0 (z) + γzf 00 (z))
Re
f 0 (z) + γzf 00 (z)
n
∞
P
= Re (1 − α) −
n((n − 1)(βeiθ (1 − γ) − αγ) + n − α)an z n−1
n=2
−
∞
P
n=1
n((n − 1)(βeiθ (1 − γ) − αγ) + n − α)bn (z)n−1
o
n
o−1
∞
∞
P
P
× 1−
n(1 + γ(n − 1))an z n−1 −
n(1 + γ(n − 1))bn (z)n−1
≥ 0.
n=2
(2.5)
n=1
The above required condition (2.5) must hold for all values of z in U . Upon choosing
the values of z on the positive real axis where 0 ≤ z = r < 1, we must have
n
hP
∞
Re (1 − α) −
(n(n − α) − n(n − 1)αγ)an rn−1 +
n=2
∞
P
− βe
iθ
hP
∞
n=2
(n(n − α) − n(n − 1)αγ)bn rn−1
i
n=1
n(n − 1)(1 − γ)an rn−1 +
∞
P
n=1
n(n − 1)(1 − γ)bn rn−1
io
o−1
n
∞
∞
P
P
× 1−
n(1 + γ(n − 1))an rn−1 −
n(1 + γ(n − 1))bn rn−1
≥ 0.
n=2
n=1
On a new class of harmonic univalent functions
559
Since Re(−eiθ ) ≥ −|eiθ | = −1, and letting r → 1− , the above inequality reduces to
1−α−
1−
∞
P
n=2
∞
P
n=2
n((n − 1)(β(1 − γ) − αγ) + n − α)an
∞
P
n(1 + γ(n − 1))an −
n=1
∞
P
−
1−
n=1
∞
P
n=2
n(1 + γ(n − 1))bn
n((n − 1)(β(1 − γ) − αγ) + n − α)bn
n(1 + γ(n − 1))an −
∞
P
n=1
≥ 0.
n(1 + γ(n − 1))bn
This gives (2.4) and the proof is complete.
3. Extreme points
Theorem 3.1. Let f be given by (1.2). Then f ∈ KWH (γ, α, β) if and only if
f can be expressed as
f (z) =
∞
P
n=1
(µn hn (z) + δn gn (z)), (z ∈ U )
where h1 (z) = z,
hn (z) = z −
1−α
z n , (n = 2, 3, · · · )
n((n − 1)(β(1 − γ) − αγ) + n − α)
and
gn (z) = z −
P∞
n=1 (µn
1−α
(z)n , (n = 1, 2, · · · ),
n((n − 1)(β(1 − γ) − αγ) + n − α)
+ δn ) = 1, (µn ≥ 0, δn ≥ 0).
In particular, the extreme points of KWH (γ, α, β) are {hn } and {gn }.
Proof. Assume that f can be expressed by (3.1). Then, we have
f (z) =
∞
P
(µn hn (z) + δn gn (z))
n=1
∞
P
∞
P
1−α
µn z n
n=1
n=2 n((n − 1)(β(1 − γ) − αγ) + n − α)
∞
P
1−α
−
δn (z)n
n((n
−
1)(β(1
−
γ) − αγ) + n − α)
n=1
∞
P
1−α
=z−
µn z n
n((n
−
1)(β(1
−
γ)
−
αγ)
+
n
−
α)
n=2
∞
P
1−α
−
δn (z)n .
n=1 n((n − 1)(β(1 − γ) − αγ) + n − α)
=
(µn + δn )z −
(3.1)
560
W. G. Atshan, A. K. Wanas
Therefore
∞
P
n((n − 1)(β(1 − γ) − αγ) + n − α)
1−α
µn
n((n − 1)(β(1 − γ) − αγ) + n − α)
n=2
∞
P
1−α
+
n((n − 1)(β(1 − γ) − αγ) + n − α)
δn
n((n
−
1)(β(1
−
γ) − αγ) + n − α)
n=1
¶
µ∞
P
= (1 − α)
(µn + δn ) − µ1 = (1 − α)(1 − µ1 ) ≤ 1 − α.
n=1
So f ∈ KWH (γ, α, β).
Conversely, let f ∈ KWH (γ, α, β), by putting
n((n − 1)(β(1 − γ) − αγ) + n − α)
µn =
an , (n = 2, 3, · · · )
1−α
and
n((n − 1)(β(1 − γ) − αγ) + n − α)
δn =
bn , (n = 1, 2, · · · ).
1−α
∞
P
P∞
We define µ1 = 1 −
µn − n=1 δn .
n=2
Then note that 0 ≤ µn ≤ 1 (n = 2, 3, · · · ), 0 ≤ δn ≤ 1E (n = 1, 2, · · · ). Hence
∞
∞
P
P
f (z) = z −
an z n −
bn (z)n
n=2
∞
P
n=1
1−α
µn z n
n=2 n((n − 1)(β(1 − γ) − αγ) + n − α)
∞
P
1−α
−
δn (z)n
n=1 n((n − 1)(β(1 − γ) − αγ) + n − α)
∞
∞
P
P
=z−
(z − hn (z))µn −
(z − gn (z))δn
n=2
n=1
µ
¶
∞
∞
∞
∞
P
P
P
P
= 1−
µn −
δn z +
µn hn (z) +
δn gn (z)
=z−
n=2
n=2
= µ1 h1 (z) +
∞
P
n=2
n=2
µn hn (z) +
∞
P
n=1
n=1
δn gn (z) =
∞
P
(µn hn (z) + δn gn (z)),
n=1
that is the required representation.
4. Convex combination
Theorem 4.1. The class KWH (γ, α, β) is closed under convex combinations.
Proof. For j = 1, 2, 3, · · · , let fj ∈ KWH (γ, α, β), where fj is given by
∞
∞
P
P
fj (z) = z −
αn,j z n −
bn,j (z)n .
n=2
n=1
Then by (2.4), we have
∞
P
n((n − 1)(β(1 − γ) − αγ) + n − α)αn,j
n=2
+
∞
P
n=1
n((b − 1)(β(1 − γ) − αγ) + n − α)bn,j ≤ 1 − α.
(4.1)
561
On a new class of harmonic univalent functions
For
P∞
j=1 tj
= 1, 0 ≤ tj ≤ 1, the convex combination of fj ’s may be written as
´
´
∞
∞ ³ P
∞
∞ ³ P
∞
P
P
P
tj fj (z) = z −
tj an,j z n −
tj bn,j (z)n .
n=2
j=1
n=1
j=1
j=1
Then by (4.1), we have
∞
P
n=2
³P
´
∞
n((n − 1)(β(1 − γ) − αγ) + n − α)
tj an,j
j=1
+
∞
P
n=1
=
∞
P
tj
j=1
∞
P
³P
´
∞
n((n − 1)(β(1 − γ) − αγ) + n − α)
tj bn,j
j=1
nP
∞
n=2
+
n=1
n((n − 1)(β(1 − γ) − αγ) + n − α)an,j
o
n((n − 1)(β(1 − γ) − αγ) + n − α)bn,j
Therefore
∞
P
j=1
≤
∞
P
j=1
tj (1 − α) = 1 − α.
tj fj (z) ∈ KWH (γ, α, β).
This completes the proof.
5. Distortion bounds
Theorem 5.1. Let f ∈ KWH (γ, α, β). Then for |z| = r < 1, we have
|f (z)| ≥ (1 − b1 )r −
(1 − α)(1 − b1 )
r2
2(β(1 − γ) − α(1 + γ) + 2)
(5.1)
|f (z)| ≤ (1 + b1 )r +
(1 − α)(1 − b1 )
r2 .
2(β(1 − γ) − α(1 + γ) + 2)
(5.2)
and
Proof. Assume that f ∈ KWH (γ, α, β). Then by (2.4), we get
¯
¯
∞
∞
∞
P
P
P
¯
¯
|f (z)| = ¯z −
an z n −
(an + bn )rn
bn (z)n ¯ ≥ (1 − b1 )r −
n=2
≥ (1 − b1 )r −
= (1 − b1 )r −
×
∞
P
n=2
∞
P
n=2
(an + bn )r
2
n=2
1
2(β(1 − γ) − α(1 + γ) + 2)
2(β(1 − γ) − α(1 + γ) + 2)(an + bn )r2
≥ (1 − b1 )r −
×
n=2
n=1
∞
P
1
2(β(1 − γ) − α(1 + γ) + 2)
n((n − 1)(β(1 − γ) − αγ) + n − α)(an + bn )r2
562
W. G. Atshan, A. K. Wanas
1
[(1 − α) − (1 − α)b1 ]r2
2(β(1 − γ) − α(1 + γ) + 2)
(1 − α)(1 − b1 )
= (1 − b1 )r −
r2 .
2(β(1 − γ) − α(1 + γ) + 2)
≥ (1 − b1 )r −
Relation (5.2) can be proved by using similar statements. So the proof is complete.
6. Integral operator
Definition 6.1. [1] The Bernardi operator is defined by
Z
c + 1 z c−1
Lc (k(z)) =
ε k(ε) dε, c ∈ N = {1, 2, · · · }.
zc
0
P∞
If k(z) = z + n=2 en z n , then
∞ c+1
P
en z n .
L(c(k(z)) = z +
n=2 c + n
(6.1)
(6.2)
Remark 6.1. If f = h + g, where
∞
∞
P
P
h(z) = z −
an z n , g(z) = −
bn z n (an ≥ 0, bn ≥ 0),
n=2
n=1
then
Lc (f (z)) = Lc (h(z)) + Lc (g(z)).
Theorem 6.1.
KWH (γ, α, β).
(6.3)
If f ∈ KWH (γ, α, β), then Lc (f ) (c ∈ N) is also in
Proof. By (6.2) and (6.3), we get
³
´
∞
∞
P
P
Lc (f (z)) = Lc z −
an z n −
bn (z)n
n=2
n=1
∞ c+1
∞ c+1
P
P
=z−
an z n −
bn (z)n .
n=2 c + n
n=1 c + n
Since f ∈ KWH (γ, α, β), then by Theorem 2.2, we have
∞ n(n − 1)(β(1 − γ) − αγ) + n − α)
P
an
1−α
n=2
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)
P
+
bn ≤ 1.
1−α
n=1
Since c ∈ N = {1, 2, · · · }, then
c+1
c+n
≤ 1, therefore
µ
¶
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)
P
c+1
an
1−α
c+n
n=2
µ
¶
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)
P
c+1
bn
+
1−α
c+n
n=1
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)
P
≤
an
1−α
n=2
On a new class of harmonic univalent functions
+
563
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)
P
bn ≤ 1,
1−α
n=1
and this gives the result.
7. Convolution property
For our next theorem, we need to define the convolution of two harmonic
functions. For harmonic functions of the form
∞
∞
P
P
f (z) = z −
an z n −
bn (z)n ,
n=2
and
F (z) = z −
∞
P
n=2
n=1
cn z n −
∞
P
n=1
dn (z)n ,
we define the convolution of two harmonic functions f and F as
∞
∞
P
P
(f ∗ F )(z) = f (z) ∗ F (z) = z −
an cn z n −
bn dn (z)n .
n=2
n=1
Theorem 7.1. For 0 ≤ η ≤ α < 1, let f ∈ KWH (γ, α, β) and F ∈
KWH (γ, η, β). Then
f ∗ F ∈ KWH (γ, α, β) ⊂ KWH (γ, η, β).
Proof. It is easy to see that KWH (γ, α, β) ⊂ KWH (γ, η, β).
Since f ∈ KWH (γ, α, β) and F ∈ KWH (γ, η, β), then by Theorem 2.2, we
have
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)
P
an
1−α
n=2
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)
P
+
bn ≤ 1 (7.1)
1−α
n=1
and
∞ n((n − 1)(β(1 − γ) − ηγ) + n − η)
P
cn
1−η
n=2
∞ n((n − 1)(β(1 − γ) − ηγ) + n − η)
P
+
dn ≤ 1.
1−η
n=1
From (7.2), we get the following inequalities
1−η
cn <
(n = 2, 3, · · · ), c1 = 1
n((n − 1)(β(1 − γ) − ηγ) + n − η)
1−η
(n = 2, 3, · · · ), 0 ≤ d1 < 1.
dn <
n((n − 1)(β(1 − γ) − ηγ) + n − η)
Therefore
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)
P
an cn
1−α
n=2
(7.2)
564
W. G. Atshan, A. K. Wanas
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)
P
bn dn
1−α
n=1
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)
P
= b1 d1 +
an cn
1−α
n=2
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)
P
+
bn dn
1−α
n=2
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)(1 − η)
P
< b1 +
an
n=2 n((n − 1)(β(1 − γ) − ηγ) + n − η)(1 − α)
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)(1 − η)
P
+
bn
n=2 n((n − 1)(β(1 − γ) − ηγ) + n − η)(1 − α)
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)(1 − η)
P
an
=
n=2 n((n − 1)(β(1 − γ) − ηγ) + n − η)(1 − α)
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)(1 − η)
P
+
bn
n=1 n((n − 1)(β(1 − γ) − ηγ) + n − η)(1 − α)
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)
P
≤
an
1−α
n=2
∞ n((n − 1)(β(1 − γ) − αγ) + n − α)
P
+
bn ≤ 1.
1−α
n=1
+
Then f ∗ F ∈ KWH (γ, α, β) ⊂ KWH (γ, η, β), and the proof is complete.
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(received 20.01.2012; in revised form 14.07.2012; available online 01.10.2012)
Department of Mathematics, College of Computer Science and Mathematics, University of AlQadisiya, Diwaniya, Iraq
E-mail: waggashnd@gmail.com, waggas hnd@yahoo.com, k.abbaswmk@yahoo.com