Volume 8 (2007), Issue 2, Article 59, 8 pp.
UNIVALENT HARMONIC FUNCTIONS
H.A. AL-KHARSANI AND R.A. AL-KHAL
D EPARTMENT OF M ATHEMATICS
FACULTY OF S CIENCE , G IRLS C OLLEGE
P.O. B OX 838, DAMMAM , S AUDI A RABIA
hakh73@hotmail.com
ranaab@hotmail.com
Received 26 February, 2007; accepted 20 April, 2007
Communicated by H. Silverman
A BSTRACT. A necessary and sufficient coefficient is given for functions in a class of complexvalued harmonic univalent functions using the Dziok-Srivastava operator. Distortion bounds,
extreme points, an integral operator, and a neighborhood of such functions are considered.
Key words and phrases: Harmonic functions, Dziok-Srivastava operator, Convolution, Integral operator, Distortion bounds,
Neighborhood.
2000 Mathematics Subject Classification. 30C45, 30C50.
1. I NTRODUCTION
Let U denote the open unit disc and SH denote the class of functions which are complexvalued, harmonic, univalent, sense-preserving in U normalized by f (0) = fz (0) − 1 = 0. Each
f ∈ SH can be expressed as f = h + g, where h and g are analytic in U . 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 U is that |h0 (z)| > |g 0 (z)| in U (see [3]). Thus for
f = h + g ∈ SH , we may write
∞
∞
X
X
k
(1.1)
h(z) = z +
ak z ,
g(z) =
bk z k (0 ≤ b1 < 1).
k=2
k=1
Note that SH reduces to S, the class of normalized analytic univalent functions if the co-analytic
part of f = h + g is identically zero.
For αj ∈ C (j = 1, 2, . . . , q) and βj ∈ C −{0, −1, −2, . . .} (j = 1, 2, . . . , s), the generalized
hypergeometric function is defined by
∞
X
(α1 )k · · · (αq )k z k
F
(α
,
.
.
.
,
α
;
β
,
.
.
.
,
β
;
z)
=
,
q s
1
q
1
s
(β
1 )k · · · (βs )k k!
k=0
(q ≤ s + 1; q, s ∈ N0 = {0, 1, 2, . . .}),
069-07
2
H.A. A L -K HARSANI
AND
R.A. A L -K HAL
where (a)n is the Pochhammer symbol defined by
(a)n =
Γ(a + n)
= a(a + 1) · · · (a + n − 1)
Γ(a)
for n ∈ N = {1, 2, . . .} and 1 when n = 0. Corresponding to the function
h(α1 , . . . , αq ; β1 , . . . , βs ; z) = zq Fs (α1 , . . . , αq ; β1 , . . . , βs ; z).
The Dziok-Srivastava operator [4], Hq,s (α1 , . . . , αq ; β1 , . . . , βs ) is defined by
Hq,s (α1 , . . . , αq ; β1 , . . . , βs )f (z) = h(α1 , . . . , αq ; β1 , . . . , βs ; z) ∗ f (z)
∞
X
(α1 )k−1 · · · (αq )k−1 ak z k
,
=z+
(β
1 )k−1 · · · (βs )k−1 (k − 1)!
k=2
where “∗” stands for convolution.
To make the notation simple, we write
Hq,s [α1 ]f (z) = Hq,s (α1 , . . . , αq ; β1 , . . . , βs )f (z).
We define the Dziok-Srivastava operator of the harmonic function f = h + g given by (1.1)
as
(1.2)
Hq,s [α1 ]f = Hq,s [α1 ]h + Hq,s [α1 ]g.
∗
Let SH
(α1 , β) denote the family of harmonic functions of the form (1.1) such that
∂
(arg Hq,s [α1 ]f ) ≥ β, 0 ≤ β < 1, |z| = r < 1.
∂θ
∗
For q = s + 1, α2 = β1 , . . . , αq = βs , SH
(1, β) = SH(β) [6] is the class of orientation∂
preserving harmonic univalent functions f which are starlike of order β in U , that is, ∂θ
(arg f (reiθ ) >
β.
∗
Also, SH
(n + 1, β) = RH (n, β) [7], is the class of harmonic univalent functions with
∂
n
(arg D f (z)) ≥ β, where D is the Ruscheweyh derivative (see [9]).
∂θ
∗
We also let VH (α1 , β) = SH
(α1 , β) ∩ VH , where VH [5], the class of harmonic functions f
of the form (1.1) and there exists φ so that, mod 2π,
(1.3)
(1.4)
arg(ak ) + (k − 1)φ = π,
arg(bk ) + (k − 1)φ = 0
k ≥ 2.
Jahangiri and Silverman [5] gave the sufficient and necessary conditions for functions of the
form (1.1) to be in VH (β), where 0 ≤ β < 1.
Note for q = s + 1, α1 = 1, α2 = β1 , . . . , αq = βs and the co-analytic part of f = h + g
being zero, the class VH (α1 , β) reduces to the class studied in [10].
In this paper, we will give a sufficient condition for f = h + g given by (1.1) to be in
∗
SH
(α1 , β) and it is shown that this condition is also necessary for functions in VH (α1 , β). Distortion theorems, extreme points, integral operators and neighborhoods of such functions are
considered.
2. M AIN R ESULTS
In our first theorem, we introduce a sufficient coefficient bound for harmonic functions in
∗
SH
(α1 , β).
Theorem 2.1. Let f = h + g be given by (1.1). If
∞
X
1
k−β
k+β
1+β
(2.1)
|ak | +
|bk | Γ(α1 , k) ≤ 1 −
|b1 |,
(k − 1)! 1 − β
1−β
1−β
k=2
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 59, 8 pp.
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U NIVALENT H ARMONIC F UNCTIONS
3
1 )k−1 ···(αq )k−1 ∗
where a1 = 1, 0 ≤ β < 1 and Γ(α1 , k) = (α
, then f ∈ SH
(α1 , β).
(β1 )k−1 ···(βs )k−1 ∗
Proof. To prove that f ∈ SH
(α1 , β), we only need to show that if (2.1) holds, then the required
condition (1.3) is satisfied. For (1.3), we can write
(
)
∂
z(Hq,s [α1 ]h(z))0 − z(Hq,s [α1 ]g(z))0
(arg Hq,s [α1 ]f (z)) = Re
∂θ
Hq,s [α1 ]h + Hq,s [α1 ]g
= Re
A(z)
.
B(z)
Using the fact that Re ω ≥ β if and only if |1 − β + ω| ≥ |1 + β − ω|, it suffices to show that
|A(z) + (1 − β)B(z)| − |A(z) − (1 + β)B(z)| ≥ 0.
(2.2)
Substituting for A(z) and B(z) in (2.1) yields
(2.3)
|A(z) + (1 − β)B(z)| − |A(z) − (1 + β)B(z)|
∞
X
k+1−β
≥ (2 − β)|z| −
Γ(α1 , k)|ak ||z|k
(k − 1)!
k=2
−
∞
X
k−1+β
k=1
−
(k − 1)!
∞
X
k−1−β
k=2
(k − 1)!
(
Γ(α1 , k)|bk ||z|k − β|z|
k
Γ(α1 , k)|ak ||z| −
∞
X
k+1+β
k=1
(k − 1)!
Γ(α1 , k)|bk ||z|k
k−β
Γ(α1 , k)|ak |
(1
−
β)(k
−
1)!
k=2
)
∞
X
k+β
Γ(α1 , k)|bk |
−
(1 − β)(k − 1)!
k=1
(
1+β
= 2(1 − β)|z| 1 −
|b1 |
1−β
"∞
#)
X
1
k−β
k+β
−
|ak | +
|bk | Γ(α1 , k)
.
(k − 1)! 1 − β
1−β
k=2
≥ 2(1 − β)|z| 1 −
X
∗
The last expression is non-negative by (2.1) and so f ∈ SH
(α1 , β).
Now, we obtain the necessary and sufficient conditions for f = h + g given by (1.4).
Theorem 2.2. Let f = h + g be given by (1.4). Then f ∈ VH (α1 , β) if and only if
∞
X
1
k−β
k+β
1+β
(2.4)
|ak | +
|bk | Γ(α1 , k) ≤ 1 −
|b1 |,
(k − 1)! 1 − β
1−β
1−β
k=2
1 )k−1 ···(αq )k−1 where a1 = 1, 0 ≤ β < 1 and Γ(α1 , k) = (α
.
(β1 )k−1 ···(βs )k−1 ∗
Proof. Since VH (α1 , β) ⊂ SH
(α1 , β), we only need to prove the “only if” part of the theorem.
∂
(arg Hq,s [α1 ]f (z)) ≥ β
To this end, for functions f ∈ VH (α1 , β), we notice that the condition ∂θ
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 59, 8 pp.
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4
H.A. A L -K HARSANI
AND
R.A. A L -K HAL
is equivalent to
∂
(arg Hq,s [α1 ]f (z)) − β = Re
∂θ
(
z(Hq,s [α1 ]h(z))0 − z(Hq,s [α1 ]g(z))0
Hq,s [α1 ]h(z) + Hq,s [α1 ]g(z)
)
−β
≥ 0.
That is,
"
Re
(2.5)
(1 − β)z +
P∞
z+
P∞ k+β
k−β
k
k
k=2 (k−1)! Γ(α1 , k)|ak |z −
k=1 (k−1)! Γ(α1 , k)|bk |z
P∞
P∞
k
k +
Γ(α
,
k)|a
|z
1
k
k=2
k=1 Γ(α1 , k)|bk |z
#
≥ 0.
The above condition must hold for all values of z in U . Upon choosing φ according to (1.4), we
must have
P k−β
k+β
(1 − β) − (1 + β)|b1 | − ∞
|a
|
+
|b
|
Γ(α1 , k)rk−1
k
k
k=2 (k−1)!
(k−1)!
P
(2.6)
≥ 0.
k−1
1 + |b1 | + ∞
k=2 (|ak | + |bk |) Γ(α1 , k)r
If condition (2.4) does not hold then the numerator in (2.6) is negative for r sufficiently close to
1. Hence there exist z0 = r0 in (0, 1) for which the quotient of (2.6) is negative. This contradicts
the fact that f ∈ VH (α1 , β) and so the proof is complete.
The following theorem gives the distortion bounds for functions in VH (α1 , β) which yields a
covering result for this class.
Theorem 2.3. If f ∈ VH (α1 , β), then
and
1
|f (z)| ≤ (1 + |b1 |)r +
Γ(α1 , 2)
1−β 1+β
−
|b1 | r2
2−β 2−β
|z| = r < 1
1
|f (z)| ≥ (1 + |b1 |)r −
Γ(α1 , 2)
1−β 1+β
−
|b1 | r2
2−β 2+β
|z| = r < 1.
Proof. We will only prove the right hand inequality. The proof for the left hand inequality is
similar.
Let f ∈ VH (α1 , β). Taking the absolute value of f , we obtain
∞
∞
X
X
|f (z)| ≤ (1 + |b1 |)r +
(|ak | + |bk |)rk ≤ (1 + |b1 |)r +
(|ak | + |bk |)r2 .
k=2
k=2
That is,
∞ X
1−β
2−β
2−β
|f (z)| ≤ (1 + |b1 |)r +
|ak | +
|bk | Γ(α1 , 2)r2
Γ(α1 , 2)(2 − β) k=2 1 − β
1−β
1−β
1+β
≤ (1 + |b1 |)r +
1−
|b1 | r2
Γ(α1 , 2)(2 − β)
1−β
1
1−β 1+β
≤ (1 + |b1 |)r +
−
|b1 | r2 .
Γ(α1 , 2) 2 − β 2 − β
Corollary 2.4. Let f be of the form (1.1) so that f ∈ VH (α1 , β). Then
n
2Γ(α1 , 2) − 1 − (Γ(α1 , 2) − 1)β
(2.7)
ω : |ω| <
(2 − β)Γ(α1 , 2)
o
2Γ(α1 , 2) − 1 − (Γ(α1 , 2) − 1)β
−
|b1 | ⊂ f (U ).
(2 + β)Γ(α1 , 2)
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 59, 8 pp.
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U NIVALENT H ARMONIC F UNCTIONS
5
Next, we examine the extreme points for VH (α1 , β) and determine extreme points of VH (α1 , β).
Theorem 2.5. Set
λk =
(1 − β)(k − 1)!
(k − β)Γ(α1 , k)
µk =
and
(1 − β)(k − 1)!
.
(k + β)Γ(α1 , k)
For b1 fixed, the extreme points for VH (α1 , β) are
o
n
(2.8)
z + λk xz k + b1 z ∪ z + b1 z + µk xz k ,
where k ≥ 2 and |x| = 1 − |b1 |.
Proof. Any function f ∈ VH (α1 , β) may be expressed as
∞
X
X
f (z) = z +
|ak |eiγk |z k + b1 z +
|bk |eiδk z k ,
k=2
k=2
where the coefficients satisfy the inequality (2.1). Set
h1 (z) = z, g1 (z) = b1 z, hk (z) = z + λk eiγk z k
gk = b1 z + µk eiδk z k ,
Writing Xk =
|ak |
, Yk
λk
=
|bk |
,
µk
for
k = 2, 3, . . .
k = 2, 3, . . . and
X1 = 1 −
∞
X
Xk ;
k=2
we have
f (z) =
∞
X
Y1 = 1 −
∞
X
Yk ,
k=2
(Xk hk (z) + Yk gk (z)).
k=1
In particular, setting
f1 (z) = z + b1 z
and
fk (z) = z + λk xz k + b1 z + µk yz k
(k ≥ 2, |x| + |y| = 1 − |b1 |),
we see that the extreme points of VH (α1 , β) are contained in {fk (z)}.
To see that f1 is not an extreme point, note that f1 may be written as
1
1
f1 (z) = {f1 (z) + λ2 (1 − |b1 |)z 2 } + {f1 (z) − λ2 (1 − |b1 |)z 2 },
2
2
a convex linear combination of functions in VH (α1 , β). If both |x| 6= 0 and |y| 6= 0, we will
show that it can also be expressed as a convex linear combination of functions in VH (α1 , β).
. Set
Without loss of generality, assume |x| ≥ |y|. Choose > 0 small enough so that < |x|
|y|
x A = 1 + and B = 1 − y . We then see that both
t1 (z) = z + λk Axz k + b1 z + µk yBz k
and
t2 (z) = z + λk (2 − A)xz k + b1 z + µk y(2 − B)z k
are in VH (α1 , β) and note that
1
fn (z) = {t1 (z) + t2 (z)}.
2
The extremal coefficient bound shows that the functions of the form (2.8) are the extreme points
for VH (α1 , β) and so the proof is complete.
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 59, 8 pp.
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6
H.A. A L -K HARSANI
AND
R.A. A L -K HAL
For q = s + 1, α2 = β1 , . . . , αq = βs , α1 = n + 1, Theorems 2.1 to 2.5 give Theorems 1, 2,
3 and 4 in [7].
Now, we will examine the closure properties of the class VH (α1 , β) under the generalized
Bernardi-Libera-Livingston integral operator Lc (f ) which is defined by
Z
c + 1 z c−1
Lc (f (z)) =
t f (t)dt,
c > −1.
zc 0
Theorem 2.6. Let f ∈ VH (α1 , β). Then Lc (f (z)) belongs to the class VH (α1 , β).
Proof. From the representation of Lc (f (z)), it follows that
Z
c + 1 z c−1
Lc (f (z)) =
t {h(t) + g(t)}dt
zc 0

!
Z
Z z
∞
X
c + 1  z c−1
k
=
t
t+
ak t dt +
tc−1
zc
0
0
k=2
=z+
∞
X
Ak z k +
k=2
∞
X
∞
X
! 
bk tk dt
k=1
Bk z k ,
k=1
where
Ak =
c+1
ak ,
c+k
Bk =
c+1
bk .
c+k
Therefore,
∞
X
k=2
1
(k − 1)!
(k − β)(c + 1)
(k + β)(c + 1)
|ak | +
|bk | Γ(α1 , k)
(1 − β)(c + k)
(1 − β)(c + k)
∞
X
1
k−β
k+β
≤
|ak | +
|bk | Γ(α1 , k)
(k
−
1)!
1
−
β
1
−
β
k=2
≤1−
1+β
b1 .
1−β
Since f ∈ VH (α1 , β), therefore by Theorem 2.2, Lc (f (z)) ∈ VH (α1 , β).
The next theorem gives a sufficient coefficient bound for functions in S ∗ (α1 , β).
∗
Theorem 2.7. f ∈ SH
(α1 , β) if and only if
2(1 − β)z + (ξ − 1 + 2β)z 2
Hq,s [α1 ]h(z) ∗
(1 − z)2
2(ξ + β)z − (ξ − 1 + 2β)z 2
6= 0,
+ Hq,s [α1 ]g ∗
(1 − z)2
|ξ| = 1,
z ∈ U.
∗
Proof. From (1.3), f ∈ SH
(α1 , β) if and only if for z = reiθ in U , we have
i
∂
∂ h
(arg(Hq,s [α1 ]f (reiθ )) =
arg Hq,s [α1 ]h(reiθ ) + Hq,s [α1 ]g(reiθ ) ≥ β.
∂θ
∂θ
Therefore, we must have
#)
(
"
1
z(Hq,s [α1 ]h(z))0 − z(Hq,s [α1 ]g(z))0
−β
≥ 0.
Re
1−β
Hq,s [α1 ]h(z) + Hq,s [α1 ]g(z)
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 59, 8 pp.
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U NIVALENT H ARMONIC F UNCTIONS
Since
1
1−β
"
z(Hq,s [α1 ]h(z))0 − z(Hq,s [α1 ]g(z))0
7
#
− β = 1 at z = 0,
Hq,s [α1 ]h(z) + Hq,s [α1 ]g(z)
the above required condition is equivalent to
"
#
z(Hq,s [α1 ]h(z))0 − z(Hq,s [α1 ]g(z))0
1
ξ−1
(2.9)
− β 6=
,
1−β
ξ+1
Hq,s [α1 ]h(z) + Hq,s [α1 ]g(z)
|ξ| = 1, ξ 6= −1, 0 < |z| < 1.
By a simple algebraic manipulation, inequality (2.9) yields
0 6= (ξ + 1)[z(Hq,s [α1 ]h(z))0 − z(Hq,s [α1 ]g(z))0 ]
− (ξ − 1 + 2β)[Hq,s [α1 ]h(z) + Hq,s [α1 ]g(z)]
(ξ + 1)z ξ − 1 + 2β
= Hq,s [α1 ]h(z) ∗
−
(1 − z)2
1−z
"
#
(ξ + 1)z (ξ − 1 + 2β)z
− Hq,s [α1 ]g(z) ∗
+
(1 − z)2
1−z
2(1 − β)z + (ξ − 1 + 2β)z 2
= Hq,s [α1 ]h(z) ∗
(1 − z)2
"
#
2
2(ξ + β)z − (ξ − 1 + 2β)z
+ Hq,s [α1 ]g(z) ∗
,
(1 − z)2
which is the condition required by Theorem 2.7.
Finally, for f given by (1.1), the δ-neighborhood of f is the set
(
∞
∞
X
X
Ak z k +
Nδ (f ) = F = z +
Bk z k :
k=2
k=1
∞
X
)
k(|ak − Ak | + |bk − Bk |) + |b1 − B1 | ≤ δ
k=2
(see [1] [8]). In our case, let us define the generalized δ-neighborhood of f to be the set
(
∞
X
Γ(α1 , k)
N (f ) = F :
[(k − β)|ak − Ak | + (k + β)|bk − Bk |]
(k
−
1)!
k=2
)
+ (1 + β)|b1 − B1 | ≤ (1 − β)δ .
Theorem 2.8. Let f be given by (1.1). If f satisfies the conditions
∞
∞
X
X
k(k − β)
k(k + β)
(2.10)
|ak |Γ(α1 , k) +
|bk |Γ(α1 , k) ≤ 1 − β, 0 ≤ β < 1
(k − 1)!
(k − 1)!
k=2
k=1
and
1−β
δ≤
2−β
1+β
1−
|b1 | ,
1−β
∗
then N (f ) ⊂ SH
(α1 , β).
J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 59, 8 pp.
http://jipam.vu.edu.au/
8
H.A. A L -K HARSANI
AND
R.A. A L -K HAL
Proof. Let f satisfy (2.10) and
F (z) = z + B1 z +
∞ X
Ak z k + Bk z k
k=2
belong to N (f ). We have
∞
X
Γ(α1 , k)
(1 + β)|B1 | +
((k − β)|Ak | + (k + β)|Bk |)
(k − 1)!
k=2
≤ (1 + β)|B1 − b1 | + (1 + β)|b1 | +
∞
X
Γ(α1 , k)
k=2
+
∞
X
Γ(α1 , k)
k=2
(k − 1)!
(k − 1)!
[(k − β)|Ak − ak | + (k + β)|Bk − bk |]
[(k − β)|ak | + (k + β)|bk |]
≤ (1 − β)δ + (1 + β)|b1 | +
1 X Γ(α1 , k)
k
[(k − β)|ak | + (k + β)|bk |]
2 − β k=2 (k − 1)!
≤ (1 − β)δ + (1 + β)|b1 | +
1
[(1 − β) − (1 + β)|b1 |]
2−β
≤ 1 − β.
Hence, for
1−β
1+β
δ≤
1−
|b1 | ,
2−β
1−β
∗
we have F (z) ∈ SH
(α1 , β).
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