chapter 5 harmonic univalent and multivalent...
TRANSCRIPT
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CHAPTER 5
HARMONIC UNIVALENT AND
MULTIVALENT FUNCTIONS
CHAPTER 5
Harmonic Univalent and MultivalentFunctions
5.1 Introduction
This last Chapter is allocated for the study of harmonic holomorphic uni-
valent and meromorphic multivalent functions. A number of mathematicians
have contributed a lot in the study of harmonic functions, in particular O . P.
Ahuja and J. M. Jahangiri [2] have defined and searched a family of Noshiro-
Type complex valued harmonic functions of the form f = h+ g, where h and
g are holomorphic in the unit disc. Many interesting properties leading to
distortion theorem, extreme points, convolution condition and convex com-
bination for the family of harmonic functions have been studied by number
of several researchers. Ahuja, Jahangiri and Silverman [3] have shared a lot
about the contraction of harmonic univalent mappings. Upon expressing Tay-
lor series expansion under different conditions, these authors have obtained
good results. A comprehensive class of complex valued harmonic univalent
functions with varying arguments is introduced by Jahangiri and Silverman
[11]. T. Rosy, B. A. Stephen, K. G. Subramanian and Jahangiri [20] have
also endowed in the study of harmonic functions.
We have divided this Chapter in two sections. The first section consists
of the study of univalent harmonic function defined by Ruscheweyh deriv-
ative, we also obtain several interesting properties such as sharp coefficient
199
estimates, distortion bound, extreme points, Hadamard product and other
several results. We have also attempted for deriving applications of frac-
tional calculus operator in establishing distortion theorem. An attempt is
also made in undertaking study of multivalent harmonic meromorphic func-
tions in Section 2, we have introduced a subclass of multivalent harmonic
meromorphic functions defined in the exterior of the unit disk and obtain the
several geometric results which are routine in character.
SECTION 1
5.2 A Certain Subclass of Univalent Harmonic
Functions Defined by Ruscheweyh Derivatives
Let H be a class of all harmonic functions f = h + g that are univalent
and sense preserving in the open disk U = {z : |z| < 1} where
h(z) = z +∞∑
n=2
anzn and g(z) =
∞∑n=1
bnzn; |b1| < 1 (5.1)
normalized by f(0) = fz(0) − 1 = 0 with fz(0) denotes partial derivative of
f(z) at z = 0 and we call h and g holomorphic part and co-holomorphic part
of f respectively.
Now, we introduce a new class AJH(λ, α, k, γ) of functions f ∈ H where
h and g of the form
h(z) = z −∞∑
n=2
anzn and g(z) = −
∞∑n=1
bnzn, (5.2)
satisfying
Re
{keiξ + (1 + keiγ)
z(Dλf(z))′′
(Dλf(z))′
}≥ α, (5.3)
200
where z = reiθ, γ, ξ, α and θ are real such that 0 ≤ γ < 1, 0 ≤ α < 1,
0 ≤ ξ < 1, α < k ≤ 1, 0 ≤ r < 1 and Dλf(z) is the Ruscheweyh derivative of
f and is defined by Dλf(z) =∞∑
n=1
Bn(λ)cnzn, λ > −1,
Bn(λ) =(λ+ 1)(λ+ 2) · · · (λ+ n− 1)
(n− 1)!,
also
Dλf(z) = Dλh(z) +Dλg(z) [21] (5.4)
Further, let H be the subfamily of H consisting of harmonic functions
f = h + g where
h(z) = z −∞∑
n=2
anzn and g(z) = −
∞∑n=1
bnzn, an ≥ 0, bn ≥ 0. (5.5)
Let AJH(λ, α, k, γ) be the subclass of functions f ∈ H and (5.3) holds true.
To attempt the various properties of Harmonic convex functions of form
f = h + g and (5.5), we need the following sufficient condition studied by J.
M. Jahangiri [9].
Theorem 5.2.1 : Let, f = h+ g where
h(z) = z −∞∑
n=2
anzn and g(z) = −
∞∑n=1
bnzn.
Furthermore let;∞∑
n=1
(n(n−α)
1−α|an| + n(n+α)
1−α|bn|)≤ 2 where a1 = 1 and
0 ≤ α < 1. Then f is harmonic univalent in U and f ∈ AJH(α). AJH(α) is
the subclass of AJH(α) consisting of harmonic convex functions of order α.
In continuation of studying we get the several properties of f given by
(5.2).
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Theorem 5.2.2 : Let f = h+ g ∈ H with
∞∑n=1
(n(n(1 + k) − (1 + α))
k − α|an| +
n(n(1 + k) + (1 + α))
k − α|bn|)Bn(λ) ≤ 2
(5.6)
where a1 = 1, 0 ≤ α < 1, λ > −1, α < k ≤ 1. Then f is harmonic univalent
in U and f ∈ AJH(λ, α, k, γ).
Proof : By Theorem 5.2.1, we obtain f to be harmonic, since
∞∑n=2
n(n− α)
1 − α|an| +
∞∑n=1
n(n+ α)
1 − α|bn|
≤∞∑
n=2
(n(n(1 + k) − (1 + α))
k − α|an| +
n(n(1 + k) + (1 + α))
k − α|bn|)Bn(λ).
According to the Theorem 0.2.2, for proving f ∈ AJH(λ, α, k, γ), it suffices
to show that (5.3) holds. If
A(λ, z) = (1 + keiγ)[z2(Dλh(z))′′ + z2(Dλg(z))′′ + 2z(Dλg(z))′]
+keiξ[z(Dλh(z))′ − z(Dλg(z))′] (5.7)
and
B(λ, z) = z(Dλh(z))′ − z(Dλg(z))′, (5.8)
then we need only to show that
|A(λ, z) + (1 − α)B(λ, z)| − |A(λ, z) − (1 + α)B(λ, z)| ≥ 0. (5.9)
Substituting (5.7) and (5.8) in (5.9), we get
|A(λ, z) + (1 − α)B(λ, z)| − |A(λ, z) − (1 + α)B(λ, z)|
= |((1 − α) + keiξ)z(Dλh(z))′ + (1 + keiγ)z2(Dλh(z))′′ +
(1 + keiγ)z2(Dλg(z))′′ + [(2 + 2keiγ − keiξ) − (1 − α)]z(Dλg(z))′| −
202
|(keiξ − 1 − α)z(Dλh(z))′ + (1 + keiγ)z2(Dλh(z))′′ +
(1 + keiγ)z2(Dλg(z))′′ + (2 + 2keiγ − keiξ + 1 + α)z(Dλg(z))′|
= |(1 + keiξ − α)z +∞∑
n=2
n[(1 + keiξ − α+ (n− 1)(keiγ + 1))]anBn(λ)zn +
∞∑n=1
n[(1 + keiγ)(n− 1) + 1 + 2keiγ − keiξ + α]bnBn(λ)zn| − |(keiγ − 1 − α)z +
∞∑n=2
n[keiγ − 1 − α + (n− 1)(1 + keiγ)]anBn(λ)zn +
∞∑n=1
n[n(1 + keiγ) + 2 + α]bnBn(λ)zn| ≥ 2(k − α)|z| ×
(1 −∞∑
n=2
nn(1 + k) − (1 + α)
k − α|an|Bn(λ)|z|n−1 −
∞∑n=1
nn(1 + k) + (1 + α)
k − α|bn|Bn(λ)|z|n−1) ≥ 0 (by (5.6))
Thus f(z) ∈ AJH(λ, α, k, γ).
For sharpness, consider the harmonic functions of the form
f(z) = z +∞∑
n=2
k − α
n(n(1 + k) − (1 + α))Tnz
n +∞∑
n=1
k − α
n(n(1 + k) + (1 + α))Snz
n,
(5.10)
where∞∑
n=2
|Tn| +∞∑
n=1
|Sn| = 1. �
Theorem 5.2.3 : Let f = h + g defined by (5.5). Then the necessary and
sufficient condition for the function f to be in the class AJH(λ, α, k, γ) is that
∞∑n=1
(n(n(1 + k) − (1 + α))
k − α|an| +
n(n(1 + k) + (1 + α))
k − α|bn|)Bn(λ) ≤ 2,
(5.11)
where a1 = 1, 0 ≤ α < 1, α < k ≤ 1, λ > −1, 0 ≤ γ < 1 and
Bn(λ) = (λ+1)(λ+2)···(λ+n−1)(n−1)!
.
Proof : Since AJH(λ, α, k, γ) ⊂ AJH(λ, α, k, γ), then the “necessary” part
203
follows from Theorem 5.2.2. For the “sufficient” part, we show that (5.11)
does not hold good implies that f �∈ AJH(λ, α, k, γ).
Now, a function f ∈ AJH(λ, α, k, γ) if and only if
Re
{keiξ + (1 + keiγ)
z(Dλf(z))′′
(Dλf(z))′
}≥ α.
Therefore,
Re{[(k − α)|z| −∞∑
n=2
n(n(1 + k) − (1 + α))|an|Bn(λ)|z|n −∞∑
n=1
n(n(1 + k) + (1 + α))|bn|Bn(λ)|z|n]/[z −∞∑
n=2
n|an|Bn(λ)|z|n +
∞∑n=1
n|bn|Bn(λ)|z|n]} ≥ 0.
The last inequality must hold for all z, |z| = r < 1.
Choosing the values of z on the positive real axis where 0 < |z| = r < 1
we must have,
[(k − α) −∞∑
n=2
n(n(1 + k) − (1 + α))|an|Bn(λ)rn−1 −∞∑
n=1
n(n(1 + k) +
(1 + α))|bn|Bn(λ)rn−1]/[1 −∞∑
n=2
n|an|Bn(λ)rn−1 +
∞∑n=1
n|bn|Bn(λ)rn−1]
≥ 0 (5.12)
We note that if the condition (5.11) does not hold, then the numerator in
(5.12) when r goes to 1 is negative. This is a contradiction for f(z) ∈AJH(λ, α, k, γ) and the proof is complete. �
In the next theorem we obtain the extreme points of the closed convex
hulls of AJH(λ, α, k, γ) denoted by clcoAJH(λ, α, k, γ).
204
Theorem 5.2.4 : The function f(z) ∈ clco AJH(λ, α, k, γ) if and only if
f(z) =
∞∑n=1
(Tnhn(z) + Sngn(z)), (5.13)
where h1(z) = z,
hn(z) = z − k − α
n(n(1 + k) − (1 + α))Bn(λ)zn, n = 2, 3, 4, · · ·
gn(z) = z − k − α
n(n(1 + k) + (1 + α))Bn(λ)zn, n = 1, 2, 3, 4, · · ·
∞∑n=1
(Tn + Sn) = 1, Tn ≥ 0 and Sn ≥ 0.
In particular the extreme points of AJH(λ, α, k, γ) are {hn} and {gn}.
Proof : Let f be written as (5.13). Then we have
f(z) =∞∑
n=1
(Tn + Sn)z −∞∑
n=2
k − α
n(n(1 + k) − (1 + α))Bn(λ)Tnz
n
−∞∑
n=1
k − α
n(n(1 + k) + (1 + α))Bn(λ)Snz
n = z −∞∑
n=2
anzn −
∞∑n=1
bnzn.
Therefore,
∞∑n=2
n(n(1 + k) − (1 + α))Bn(λ)
k − α|an| +
∞∑n=1
n(n(1 + k) + (1 + α))Bn(λ)
k − α|bn|
=
∞∑n=2
Tn +
∞∑n=1
Sn = 1 − T1 ≤ 1.
Then f ∈ clco AJH(λ, α, k, γ).
Conversely, assume that f ∈ clco AJH(λ, α, k, γ). Putting
Tn =n(n(1 + k) − (1 + α))Bn(λ)
k − α|an|, n = 2, 3, · · ·
Sn =n(n(1 + k) + (1 + α))Bn(λ)
k − α|bn|, n = 1, 2, · · ·
205
and
T1 = 1 −∞∑
n=2
Tn −∞∑
n=1
Sn,
then∞∑
n=1
(Tn + Sn) = 1, 0 ≤ Tn ≤ 1 (n = 2, 3, · · ·), 0 ≤ Sn ≤ 1 (n = 1, 2, · · ·).Thus, by simple calculations we get
f(z) =∞∑
n=1
(Tnhn(z) + Sngn(z))
and the proof is complete. �
Theorem 5.2.5 : Let f ∈ AJH(λ, α, k, γ). Then
|f(z)| ≤ (1+|b1|)r+ 1
2B2(λ)
[k − α
(1 + 2k − α)− 2 + k + α
(1 + 2k − α)|b1|]r2, |z| = r < 1
and
|f(z)| ≥ (1−|b1|)r− 1
2B2(λ)
[k − α
(1 + 2k − α)− 2 + k + α
(1 + 2k − α)|b1|]r2, |z| = r < 1.
(5.14)
Proof : We have
|f(z)| ≤ (1 + |b1|)r +∞∑
n=2
(|an| + |bn|)rn
≤ (1 + |b1|)r +∞∑
n=2
(|an| + |bn|)r2
= (1 + |b1|)r +k − α
2B2(λ)(1 + 2k − α)
∞∑n=2
[2(1 + 2k − α)
k − α|an| +
2(1 + 2k − α)
k − α|bn|]Bn(λ)r2
≤ (1 + |b1)r +k − α
2B2(λ)(1 + 2k − α)
∞∑n=2
(n(n(1 + k) − (1 + α))
k − α|an| +
n(n(1 + k) + (1 + α))
k − α|bn|)Bn(λ)r2
≤ (1 + |b1|)r +1
2B2(λ)(
k − α
1 + 2k − α− 2 + k + α
1 + 2k − α|b1|)r2.
206
The next inequality can be proved by using similar arguments. This completes
the proof of theorem. �
Now we define the convolution of two harmonic functions. [see 16].
If f(z) and g(z) be given by
f(z) = z −∞∑
n=2
|an|zn −∞∑
n=1
|bn|zn, g(z) = z −∞∑
n=2
|cn|zn −∞∑
n=1
|dn|zn.
Then the Hadamard product of f(z) and g(z) defined by
(f ∗ g)(z) = f(z) ∗ g(z) = z −∞∑
n=2
|an||cn|zn −∞∑
n=1
|bn||dn|zn. (5.15)
Theorem 5.2.6 : Let f(z) ∈ AJH(λ, α, k, γ) and g(z) ∈ AJH(λ, β, k, γ).
Then for 0 ≤ β ≤ α < 1, we have
(f ∗ g)(z) ∈ AJH(λ, α, k, γ) ⊂ AJH(λ, β, k, γ).
Proof : Since f(z) ∈ AJH(λ, α, k, γ) and g(z) ∈ AJH(λ, β, k, γ), then both
of them satisfy (5.11) and since (|cn| ≤ 1, |dn| ≤ 1), then we have
∞∑n=2
n(n(1 + k) − (1 + α))k − α
|ancn|Bn(λ) +∞∑
n=1
n(n(1 + k) + (1 + α))k − α
|bndn|Bn(λ)
≤∞∑
n=1
n(n(1 + k) − (1 + α))k − α
|an|Bn(λ) +∞∑
n=1
n(n(1 + k) + (1 + α))k − α
|bn|Bn(λ).
The right hand side of the last inequality is bounded above by 1, then
f ∗ g ∈ AJH(λ, α, k, γ) ⊂ AJH(λ, β, k, γ).
The proof of this theorem is complete. �
Theorem 5.2.7 : The class AJH(λ, α, k, γ) is closed under convex combina-
tion.
207
Proof : Suppose fi ∈ AJH(λ, α, k, γ)(i = 1, 2, · · ·) are defined by
fi(z) = z −∞∑
n=2
|ai,n|zn −∞∑
n=1
|bi,n|zn,
from (5.11), we have
∞∑n=1
(n(n(1 + k) − (1 + α))
k − α|ai,n| +
n(n(1 + k) + (1 + α))
k − α|bi,n|
)Bn(λ) ≤ 2.
For∞∑i=1
si = 1, 0 ≤ si ≤ 1, we may write the convex combination of fi as
∞∑i=1
sifi(z) = z −∞∑
n=2
( ∞∑i=1
si|ai,n|)zn −
∞∑n=1
( ∞∑i=1
si|bi,n|)zn.
Thus
∞∑n=1
[n(n(1 + k) − (1 + α))
k − α
( ∞∑i=1
si|ai,n|)
+n(n(1 + k) + (1 + α))
k − α
( ∞∑i=1
si|bi,n|)]
Bn(λ)
=∞∑
i=1
[[ ∞∑n=1
n(n(1 + k) − (1 + α))k − α
|ai,n| + n(n(1 + k) + (1 + α))k − α
|bi,n|]
Bn(λ)
]si
≤ 2∞∑
i=1
si = 2,
then∞∑i=1
sifi(z) ∈ AJH(λ, α, kγ). The proof is complete. �
Next, we introduce a class ASH(λ, α, k, γ) of functions f ∈ H of the form
(5.1)satisfying
Re
{(1 + keiγ)
z2(Dλh(z))′′ + z2(Dλg(z))′′ + 2z(Dλg(z))′
z(Dλh(z))′ − z(Dλg(z))′+ 1
}≥ α (5.16)
where z = reiθ, γ, α, and θ are real such that 0 ≤ γ < 1, 0 ≤ α < 1, 0 ≤ k <
∞, 0 ≤ r < 1 and Dλf(z) is the Ruscheweyh derivative of f .
Further, let H be the subfamily of H consisting of harmonic functions
f = h + g where
h(z) = z −∞∑
n=2
|an|zn and g(z) = −∞∑
n=1
|bn|zn, |b1| < 1.
208
Let ASH(λ, α, k, γ) be the subclass of functions f ∈ H and (5.16) holds true.
Special cases were studied by Kanas and Wisniowska [13], Kanas and
Srivastava [12] and Kim et. al. [16].
Now we obtain conditions for f(z) ∈ ASH(λ, α, k, γ) and f(z) ∈ ASH(λ, α, k, γ).
Theorem 5.2.8 : Let f = h+ g ∈ H of the form (5.1) with
∞∑n=1
(n(n(1 + k) − k − α)
1 − α|an| +
n(n(1 + k) + k + α)
1 − α|bn|)Bn(λ) ≤ 2
(5.17)
where a1 = 1, 0 ≤ α < 1, λ > −1, 0 ≤ k < ∞. Then f is harmonic univalent
in U and f ∈ ASH(λ, α, k, γ).
Proof : It is clear that
∞∑n=2
n(n− α)
1 − α|an| +
∞∑n=1
n(n + α))
1 − α|bn|
≤∞∑
n=2
(n(n(1 + k) − k − α)
1 − α|an| +
n(n(1 + k) + k + α))
1 − α|bn|)Bn(λ),
where 0 ≤ α < 1, 0 ≤ k <∞, λ > −1. Thus, by Theorem 5.2.1 f is harmonic.
Now by using Theorem 0.2.2 for proving f ∈ ASH(λ, α, k, γ), it suffices
to show that (5.16) holds. If we put
A(λ, z) = (1 + keiγ)[z2(Dλh(z))′′ + z2(Dλg(z))′′ + 2z(Dλg(z))′]
+[z(Dλh(z))′ − z(Dλg(z))′]
and
B(λ, z) = z(Dλh(z))′ − z(Dλg(z))′,
we want to show that
|A(λ, z) + (1 − α)B(λ, z)| − |A(λ, z) − (1 + α)B(λ, z)| ≥ 0.
209
But
|A(λ, z) + (1 − α)B(λ, z)| − |A(λ, z) − (1 + α)B(λ, z)|
= |(1 + keiγ)[z2(Dλh(z))′′ + z2(Dλg(z))′′ + 2z(Dλg(z))′] +
[z(Dλh(z))′ − z(Dλg(z))′] + (1 − α)[z(Dλh(z))′ − z(Dλg(z))′| −
|(1 + keiγ)[z2(Dλh(z))′′ + z2(Dλg(z))′′ + 2z(Dλg(z))′] +
[z(Dλh(z))′ − z(Dλg(z))′] − (1 + α)[z(Dλ(h(z))′ − z(Dλg(z))′]|
= |(2 − α)z +∞∑
n=2
n(n+ 1 − α + k(n− 1)eiγ)Bn(λ)anzn +
∞∑n=1
n(n− 1 + α + k(n+ 1)eiγ)Bn(λ)bnzn
≥ (2 − α)|z| −∞∑
n=2
n(n(k + 1) + 1 − k − α]Bn(λ)|an||z|n −∞∑
n=1
n(n(k + 1) − 1 + k + α)Bn(λ)|bn||z|n −
α|z| −∞∑
n=2
n[(n(k + 1) − 1 − k − α]Bn(λ)|an||z|n −∞∑
n=1
n(n(k + 1) + 1 + k + α)Bn(λ)|bn||z|n
≥ 2(1 − α)|z|[
1 −∞∑
n=2
n(nk + n− k − α)
1 − αBn(λ)|an| −
∞∑n=1
n(nk + n + k + α)
1 − αBn(λ)|bn|
]≥ 0 (by (5.17)
This completes the proof of theorem.
The harmonic function
f(z) = z +
∞∑n=2
1 − α
n[n(k + 1) − k − α]Tnz
n +
∞∑n=1
1 − α
n[n(k + 1) + k + α]Snz
n
(5.18)
210
where∞∑
n=2
|Tn| +∞∑
n=1
|Sn| = 1, shows the sharpness of (5.17). �
Theorem 5.2.9 : Let f(= h + g) ∈ H. Then the necessary and suffi-
cient condition for the function f to be in the class ASH(λ, α, k, γ) where
ASH(λ, α, k, γ) is the subclass of ASH(λ, α, k, γ), is that
∞∑n=1
(n(n(k + 1) − k − α)
1 − α|an| +
n(n(1 + k) + k + α)
1 − α|bn|)Bn(λ) ≤ 2
(5.19)
where a1 = 1, 0 ≤ α < 1, 0 ≤ k <∞, λ > −1, 0 ≤ γ < 1 and
Bn(λ) = (λ+1)(λ+2)···(λ+n−1)(n−1)!
.
Proof : Since ASH(λ, α, k, γ) ⊂ ASH(λ, α, k, γ), then the “necessary” part
follows from Theorem 5.2.8. For the “sufficient” part, we show that (5.19)
does not hold good implies that f �∈ ASH(λ, α, k, γ).
Now, a function f ∈ ASH(λ, α, k, γ) if and only if
Re
{1 + (1 + keiγ)
z(Dλf(z))′′
(Dλf(z))′
}≥ α.
Therefore,
Re
(1 − α)|z| −∞∑
n=2n(n(1 + k) − k − α)|an|Bn(λ)|z|n −
∞∑n=1
n(n(1 + k) + k + α))|bn|Bn(λ)|z|n
z −∞∑
n=2n|an|Bn(λ)|z|n +
∞∑n=1
n|bn|Bn(λ)|z|n
≥ 0.
The last inequality must hold for all z, |z| = r < 1.
Choosing the values of z on the positive real axis where 0 < |z| = r < 1
we must have,
(1 − α) −∞∑
n=2n(n(1 + k) − k − α)|an|Bn(λ)rn−1 −
∞∑n=1
n(n(1 + k) + k + α)|bn|Bn(λ)rn−1
1 −∞∑
n=2n|an|Bn(λ)rn−1 +
∞∑n=1
n|bn|Bn(λ)rn−1
≥ 0.
(5.20)
211
We note that if the condition (5.19) does not hold, then the numerator in
(5.20) when r goes to 1, is negative. This is a contradiction for
f(z) ∈ ASH(λ, α, k, γ) and this completes the proof of theorem. �
Theorem 5.2.10 : f ∈ clco ASH(λ, α, k, γ) if and only if
f(z) =∞∑
n=1
(Tnhn(z) + Sngn(z)) (5.21)
where, h1(z) = z,
hn(z) = z − 1 − α
n(n(1 + k) − k − α)Bn(λ)zn, n = 2, 3, 4, · · ·
gn(z) = z − 1 − α
n(n(1 + k) + k + α))Bn(λ)zn, n = 1, 2, 3, 4, · · ·
∞∑n=1
(Tn + Sn) = 1, Tn ≥ 0 and Sn ≥ 0.
In particular the extreme points of ASH(λ, α, k, γ) are {hn} and {gn}.
Proof : Let f be written as (5.21), then we have
f(z) =
∞∑n=1
(Tn + Sn)z −∞∑
n=2
1 − α
n(n(1 + k) − k − α))Bn(λ)Tnz
n
−∞∑
n=1
1 − α
n(n(1 + k) + k + α))Bn(λ)Snz
n = z −∞∑
n=2
anzn −
∞∑n=1
bnzn.
Therefore,
∞∑n=2
n(n(1 + k) − k − α)Bn(λ)
1 − α|an| +
∞∑n=1
n(n(1 + k) + k + α)Bn(λ)
1 − α|bn|
=∞∑
n=2
Tn +∞∑
n=1
Sn = 1 − Tn ≤ 1,
and so f ∈ clco ASH(λ, α, k, γ).
Conversely, suppose that f ∈ clco ASH(λ, α, k, γ). Setting
Tn =n(n(1 + k) − k − α))Bn(λ)
2 + k − α|an|, n = 2, 3, · · ·
212
and
Sn =n(n(k + 1) + k + α))Bn(λ)
2 + k − α|bn|, n = 1, 2, · · ·
where,∞∑
n=1
[Tn + Sn] = 1, then we obtain by simple calculations
f(z) =
∞∑n=1
(Tnhn(z) + Sngn(z)).
This completes the proof of theorem. �
Theorem 5.2.11 : Let f ∈ ASH(λ, α, k, γ). Then
|f(z)| ≤ (1 + |b1|)r +1
2B2(λ)
[1 − α
(2 + k − α)− 1 + 2k + α
(2 + k − α)|b1|]r2, |z| = r < 1
and
|f(z)| ≥ (1−|b1|)r− 1
2B2(λ)
[1 − α
(2 + k − α)− 1 + 2k + α
(2 + k − α)|b1|]r2, |z| = r < 1.
(5.22)
Proof : We have
|f(z)| ≤ (1 + |b1|)r +
∞∑n=2
(|an| + |bn|)rn
≤ (1 + |b1|)r +
∞∑n=2
(|an| + |bn|)r2
= (1 + |b1|)r +1 − α
2(2 + k − α)B2(λ)
∞∑n=2
[2(2 + k − α)
1 − α|an| +
2(2 + k − α)
1 − α|bn|]Bn(λ)r2
≤ (1 + |b1|)r +1
2B2(λ)
(1 − α
2 + k − α− 1 + 2k + α
2 + k − α|b1|)r2.
Also,
f(z) ≥ (1 − |b1|)r −∞∑
n=2
(|an| + |bn|)rn
≥ (1 − |b1|)r −∞∑
n=2
(|an| + |bn|)r2
213
= (1 − |b1|)r − 1 − α
2(2 + k − α)B2(λ)
∞∑n=2
[2(2 + k − α)
1 − α|an| +
2(2 + k − α)
1 − α|bn|]Bn(λ)r2
≥ (1 − |b1|)r − 1 − α
2(2 + k − α)B2(λ)
∞∑n=2
[(n(n(1 + k) − k − α)
1 − α|an| +
n(n(1 + k) + k + α)
1 − α|bn|]Bn(λ)r2
≥ (1 − |b1|)r − 1
2B2(λ)[
1 − α
2 + k − α− 1 + 2k + α
2 + k − α|b1|]r2.
This completes the proof of theorem. �
Theorem 5.2.12 : Let f(z) ∈ ASH(λ, α, k, γ) and g(z) ∈ ASH(λ, β, k, γ).
Then for 0 ≤ β ≤ α < 1, we have
(f ∗ g)(z) ∈ ASH(λ, α, k, γ) ⊂ ASH(λ, β, k, γ).
Proof : We have
(f ∗ g)(z) = z −∞∑
n=2
|an||cn|zn −∞∑
n=1
|bn||dn|zn.
By noting that |cn| ≤ 1 and |dn| ≤ 1, the theorem follows easily by using the
condition (5.19).
The proof of this theorem is complete. �
In the next theorem we show that ASH(λ, α, k, γ) is closed under convex
combination of its members.
Theorem 5.2.13 : Let for i = 1, 2, 3, · · · the function
fi(z) = z −∞∑
n=2
|ai,n| −∞∑
n=1
|bi,n|zn
belong to ASH(λ, α, k, γ). Then∞∑
n=1
µifi(z) belongs to ASH(λ, α, k, γ), where
∞∑i=1
µi = 1, 0 ≤ µi < 1, i = 1, 2, · · · .
214
Proof : Since fi ∈ ASH(λ, α, k, γ)(i = 1, 2, · · ·), then from (5.19), we have
∞∑n=1
(n(n(1 + k) − k − α)
1 − α|ai,n| +
n(n(1 + k) + k + α)
1 − α|bi,n|
)Bn(λ) ≤ 2.
For∞∑i=1
µi = 1, 0 ≤ µi ≤ 1, we may write the convex combination of fi as
∞∑i=1
µifi(z) = z −∞∑
n=2
( ∞∑i=1
µi|ai,n|)zn −
∞∑n=1
( ∞∑i=1
µi|bi,n|)zn.
Thus
∞∑n=1
[n(n(1 + k) − k − α)
1 − α
( ∞∑i=1
µi|ai,n|)
+n(n(1 + k) + k + α)
1 − α×
( ∞∑i=1
µi|bi,n|)]
Bn(λ) =∞∑i=1
[[ ∞∑n=1
n(n(1 + k) − k − α)
1 − α|ai,n| +
n(n(1 + k) + k + α)
1 − α|bi,n|
]Bn(λ)
]µi ≤ 2
∞∑i=1
µi = 2,
then∞∑i=1
µifi(z) ∈ ASH(λ, α, kγ). This completes the proof of theorem. �
Now, we introduce a new class MAH(α, β, γ) of harmonic univalent func-
tion defined in the open unit disk. Jahangiri [10] defined a class MAH(α)
consisting of harmonic starlike functions f = h + g which are of order α
(0 ≤ α < 1), where
h(z) = z −∞∑
n=2
|an|zn, g(z) =
∞∑n=1
|bn|zn (5.23)
and which satisfy the condition that
∂
∂θ(arg f(reiθ)) ≥ α, 0 ≤ α < 1, |z| = r < 1. (5.24)
It was proved in [10] that if f = h + g where h, g are given by (5.1) and if
∞∑n=1
(n− α
1 − α|an| +
n + α
1 − α|bn|)
≤ 2, (0 ≤ α < 1, a1 = 1), (5.25)
215
then f is harmonic, univalent and starlike of order α in U . The condition in
(5.25) is shown to be necessary also if h and g are of the form (5.23). Avci
and Zlotkiewicz in [4] showed that if∞∑
n=2
n(|an|+|bn|) ≤ 1, then f ∈ MAH(0).
Silverman in [22] proved that the last condition is also necessary if f = h+ g
has negative coefficients.
Here we define a new class MAH(α, β, γ) of harmonic univalent functions
as follows:
A function f(= h+g) ∈ MAH(α) is said to be in the class MAH(α, β, γ)
if the holomorphic functions h and g satisfy the condition
�{1 + αz2h′′(z) − βzh′(z) + βg(z))} > 1 − |γ|, (5.26)
where 0 ≤ β ≤ α, α ≥ 0, γ ∈ C and z ∈ U .Here we obtain sufficient conditions for a function f to be in the class
MAH(α, β, γ) and derive distortion theorems by using fractional calculus
operators. We also obtain the radii of starlikeness, close-to-convexity and
convexity of functions belonging to the above class.
We begin by proving the following:
Theorem 5.2.14 : Let f = h + g (h and g being given by (5.23)). If
f ∈ MAH(α, β, γ), then
∞∑n=2
[n(α(n− 1) − β)|an| − β
1 − 3α
n + α
]≤ |γ|, (5.27)
where a1 = b1 = 1, 0 < α ≤ 1/3, 0 ≤ β < α and γ ∈ C. The result is sharp.
Proof : Assume that f ∈ MAH(α, β, γ). Using (5.26), we get
Re
{1 −
∞∑n=2
αn(n− 1)|an|zn +
∞∑n=2
βn|an|zn +
∞∑n=2
β|bn|zn
}> 1 − |γ|.
216
Choosing z real and letting z → 1−, we obtain
1 −[ ∞∑
n=2
n(α(n− 1) − β)|an| −∞∑
n=2
β|bn|]≥ 1 − |γ|
which implies∞∑
n=2
n(α(n− 1) − β)|an| − β|bn| ≤ |γ|. (5.28)
Since f ∈ MAH(α), therefore, it follows that
∞∑n=1
n + α
1 − α|bn| ≤ 2
and from the above inequality we infer that
|bn| ≤(
1 − 3α
n+ α
)(n ≥ 2),
and consequently
β|bn| ≤ β1 − 3α
n + α. (5.29)
The assertion (5.27) of Theorem 5.2.14 now follows upon combining (5.28)
and (5.29). �
Deduction 5.2.1 : If the function f = h + g where h and g are given by
(5.23) belong to the class MAH(α, β, γ), then
|an| ≤ (n+ α)|γ| + β(1 − 3α)
n(α(n− 1) − β)(n+ α)(n ≥ 2, 0 < α ≤ 1/3, 0 ≤ β < α, γ ∈ C).
(5.30)
The result is sharp for the functions h(z) and g(z), respectively, given by
h(z) = z − (n+ α)|γ| + β(1 − 3α)
n(α(n− 1) − β)(n+ α)zn (n ≥ 2), (5.31)
and
g(z) = z +β(1 − 3α)
n+ αzn (n ≥ 2). (5.32)
217
Making use of the operators in Definition 0.1.20, we establish the following
distortion theorems.
Theorem 5.2.15 : If the function f = h+ g (h and g being given by (5.23))
belong to the class MAH(α, β, γ), then
|D−λz h(z)| ≥ |z|1+λ
Γ(2 + λ)
{1 − (2 + α)|γ| + β(1 − 3α)
(α− β)(2 + α)(2 + λ)|z|}
(5.33)
and
|D−λz h(z)| ≤ |z|1+λ
Γ(2 + λ)
{1 +
(2 + α)|γ| + β(1 − 3α)
(α− β)(2 + α)(2 + λ)|z|}, (5.34)
for λ > 0, 0 < α ≤ 1/3, 0 ≤ β < α, γ ∈ C and z ∈ U . The results are sharp.
Proof : Applying (0.12) to (5.23), we find that
D−λz h(z) =
1
Γ(2 + λ)z1+λ −
∞∑n=2
Γ(n+ 1)
Γ(n+ 1 + λ)|an|zn+λ
which yields
Γ(2 + λ)z−λD−λz f(z) = z −
∞∑n=2
Γ(n + 1)Γ(2 + λ)
Γ(n + 1 + λ)|an|zn.
We observe that the function ϕ(n) = Γ(n+1)Γ(2+λ)Γ(n+1+λ)
(n ≥ 2) is a decreasing
function of n, and this implies that
0 < ϕ(n) ≤ ϕ(2) =2
2 + λ(n ≥ 2).
In view of Theorem 5.2.14, we conclude that
|Γ(2 + λ)z−λD−λz h(z)| ≥ |z| − ϕ(2)|z|2
∞∑n=2
|an|
≥ |z| − (2 + α)|γ| + β(1 − 3α)
(α− β)(2 + α)(2 + λ)|z|2
218
which gives the desired assertion (5.33) of Theorem 5.2.15. Similarly, it follows
that
|Γ(2 + λ)z−λD−λz h(z)| ≤ |z| + ϕ(2)|z|2
∞∑n=2
|an|
≤ |z| +(2 + α)|γ| + β(1 − 3α)
(α− β)(2 + α)(2 + λ)|z|2.
This yields the second assertion (5.34) of Theorem 5.2.15. The sharpness of
the inequalities (5.32) and (5.33) are achieved when h(z) is given by
D−λz h(z) =
z1+λ
Γ(2 + λ)
{1 − (2 + α)|γ| + β(1 − 3α)
2(α− β)(2 + α)(2 + λ)z
}. �
Following similar steps as used in proving Theorem 5.2.15, wherein, the
fractional derivative operator (0.13) is applied to (5.23), we can prove the
following.
Theorem 5.2.16 : If the function f = h+ g (h and g being given by (5.23))
belong to the class MAH(α, β, γ), then
|Dλzh(z)| ≥ |z|1−λ
Γ(2 − λ)
{1 − (2 + α)|γ| + β(1 − 3α)
(α− β)(2 + α)(2 − λ)|z|}
(5.35)
and
|Dλzh(z)| ≤ |z|1−λ
Γ(2 − λ)
{1 +
(2 + α)|γ| + β(1 − 3α)
(α− β)(2 + α)(2 − λ)|z|}, (5.36)
where 0 ≤ λ < 1, 0 < α ≤ 1/3, 0 ≤ β < α, γ ∈ C and z ∈ U . The results are
sharp for the function given by
Dλzh(z) =
z1−λ
Γ(2 − λ)
{1 − (2 + α)|γ| + β(1 − 3α)
(α− β)(2 + α)(2 − λ)z
}. (5.37)
Letting λ→ 0 in Theorem 5.2.15, we obtain
219
Corollary 5.2.2 : If the function f = h+ g ( h and g being given by (5.23))
belong to the class MAH(α, β, γ), then
r− (2 + α)|γ| + β(1 − 3α)
2(α− β)(2 + α)r2 ≤ |h(z)| ≤ r+
(2 + α)|γ| + β(1 − 3α)
2(α− β)(2 + α)r2, (5.38)
where 0 < α ≤ 1/3, 0 ≤ β < α, γ ∈ C and |z| = r < 1. The result is sharp.
Also, by letting λ→ 1 in Theorem 5.2.16, we get
Corollary 5.2.3 : If the function f = h+ g ( h and g being given by (5.23))
belong to the class MAH(α, β, γ), then
1− (2 + α)|γ| + β(1 − 3α)
(α− β)(2 + α)r ≤ |h′(z)| ≤ 1+
(2 + α)|γ| + β(1 − 3α)
(α− β)(2 + α)r (5.39)
where 0 < α ≤ 1/3, 0 ≤ β < α, γ ∈ C and |z| = r < 1. The result is sharp.
We claim that the above results are entirely new.
Now, we begin by recalling the fractional integral operator Iλ,ξ,δ0,z and the
fractional derivative operator J λ,ξ,δ0,z of Saigo type as follows [23](see also [5]
and [19]):
The fractional integral operator of order λ for a function h(z) is defined by
Iλ,ξ,δ0,z h(z) =
z−λ−ξ
Γ(λ)
∫ z
0
(z − t)λ−12F1(λ+ ξ,−δ;λ; 1 − t
z)h(t)dt, (5.40)
where λ > 0, k > max{0, ξ − δ} − 1. The function h(z) is holomorphic
in a simply-connected region of the z-plane containing the origin with the
order h(z) = O(|z|k), z → 0, and the multiplicity of (z − t)λ−1 is removed by
requiring log(z − t) to be real when (z − t) > 0.
The fractional derivative operator of a function h(z) of order λ(0 ≤ λ < 1)
is defined by
J λ,ξ,δ0,z h(z) =
1Γ(1 − λ)
d
dz
{zλ−ξ
∫ z
0
(z − t)−λ2F1
(ξ − λ, 1 − δ; 1 − λ; 1 − t
z
)h(t)dt
}(0 ≤ λ < 1) (5.41)
220
which holds under similar constraints as mentioned for the operator Iλ,ξ,δ0,z
defined above by (5.40).
We observe the following relationships of the operators (5.40) and (5.41):
Iλ,−λ,δ0,z h(z) = D−λ
z h(z) and Jλ,λ,δ0,z h(z) = Dλ
zh(z).
Corresponding to the fractional derivative operator (5.40), we make use
of the operator ∆λ,ξ,δ0,z which is defined by ([5])
∆λ,ξ,δ0,z h(z) =
Γ(2 − ξ)Γ(2 − λ+ ξ)
Γ(2 − ξ + δ)zξJ λ,ξ,δ
0,z h(z). (5.42)
(0 ≤ λ < 1, ξ < 2, δ > 0)
In view of (5.23) and (5.41) we obtain the series representation
∆λ,ξ,δ0,z h(z) = z −
∞∑n=2
(2 − ξ + δ)n−1(2)n−1
(2 − ξ)n−1(2 − λ+ δ)n−1|an|zn (5.43)
Theorem 5.2.17 : Let the function f = h + g such that h and g are given
by (5.23). If f ∈ MAH(α, β, γ), then
|∆ν,σ,s0,z h(z)| ≥ |z| − |z|2 (2 − σ + s)[(2 + α)|γ| + β(1 − 3α)]
(2 − σ)(2 − ν + s)(α− β)(2 + α)(5.44)
and
|∆ν,σ,s0,z h(z)| ≤ |z| + |z|2 (2 − σ + s)[(2 + α)|γ| + β(1 − 3α)]
(2 − σ)(2 − ν + s)(α− β)(2 + α)(5.45)
where 0 < α ≤ 13, 0 ≤ β < α, γ ∈ C, 0 ≤ ν < 1, σ < 2, s ∈ IR+, provided that
σ(ν − s)
ν≤ 3.
The results are sharp.
221
Proof : From (5.27) of Theorem 2.2.14, we obtain
∞∑n=2
|an| ≤ (2 + α)|γ| + β(1 − 3α)
2(α− β)(2 + α),
and in view of (5.43) the operator ∆ν,σ,s0,z applied to h(z) gives
∆ν,σ,s0,z h(z) = z −
∞∑n=2
(2 − σ + s)n−1(2)n−1
(2 − σ)n−1(2 − ν + s)n−1|an|zn. (5.46)
Under the assumptions stated with Theorem 5.2.17, we observe that the func-
tion
ψ(n) =(2 − σ + s)n−1(2)n−1
(2 − σ)n−1(2 − ν + s)n−1
(n ≥ 2)
is non-increasing and, therefore, we get
0 < ψ(n) ≤ ψ(2) =(2 − σ + s)(2)
(2 − σ)(2 − ν + s).
Consequently, we find that
|∆ν,σ,s0,z h(z)| ≥ |z| − |z|2ψ(2)
∞∑n=2
|an|
≥ |z| − |z|2 (2 − σ + s)[(2 + α)|γ| + β(1 − 3α)]
(2 − σ)(2 − ν + s)(α− β)(2 + α)
and
|∆ν,σ,s0,z h(z)| ≤ |z| + |z|2 (2 − σ + s)[(2 + α)|γ| + β(1 − 3α)]
(2 − σ)(2 − ν + s)(α− β)(2 + α).
The inequalities (5.44) and (5.45) are sharp, and the equalities are attained
for the function h(z) given by
h(z) = z − (2 + α)|γ| + β(1 − 3α)
2(α− β)(2 + α)z2. (5.47) �
Evidently, in the special case when σ = ν, Theorem 5.2.17 would correspond
to Theorem 5.2.16. In a similar manner, one can obtain the distortion in-
equalities involving the Saigo type fractional integral operator (5.40).
222
We merely state here the following results (Theorems 5.2.18 to 5.2.20)
giving the radii of starlikeness, close to convexity and convexity for the holo-
morphic part of the function f belonging to the class MAH(α, β, γ). The
proofs of these results are omitted here as these results can be established by
following [23] (see also [19]).
Theorem 5.2.18 : Let f ∈ MAH(α, β, γ) where f = h + g with h and g
given by (5.23), then h(z) is starlike of order ρ (0 ≤ ρ < 1) in |z| < r1 where
r1 = infn
[2(α− β)(2 + α)(1 − ρ)
[(2 + α)|γ| + β(1 − 3α)](n− ρ)
] 1n−1
, (n ≥ 2). (5.48)
Theorem 5.2.19 : Let f ∈ MAH(α, β, γ) where f = h + g with h and g
given by (5.23), then h(z) is close-to-convex of order ρ (0 ≤ ρ < 1) in |z| < r2
where
r2 = infn
[(1 − ρ)(α− β)(2 + α)
(2 + α)|γ| + β(1 − 3α)
] 1n−1
, n ≥ 2. (5.49)
Theorem 5.2.20 : Let f ∈ MAH(α, β, γ) where f = h+ g with h and g are
given by (5.23), then h(z) is convex of order ρ (0 ≤ ρ < 1) in |z| < r3 where
r3 = infn
[(1 − ρ)(α− β)(2 + α)
[(2 + α)|γ| + β(1 − 3α)](n− ρ)
] 1n−1
, n ≥ 2. (5.50)
223
SECTION 2
5.3 On a Certain Class of Multivalently Harmonic
Meromorphic Functions
In this Section we research a certain subclass of multivalent harmonic
meromorphic functions. This class with other subclasses has been researched
by O. P. Ahuja and J. M. Jahangiri [1], T. Rosy, B. A. Stephen, K. G.
Subramanian and J. M. Jahangiri [20].
W. Hengartner and G. Schober [7], considered harmonic sense preserving
univalent mappings defined on U = {z : |z| > 1} that map ∞ to ∞ and
represented by
f(z) = h(z) + g(z) + A log |z| where
h(z) = αz +
∞∑n=0
anz−n, g(z) = βz +
∞∑n=1
bnz−n
are holomorphic in U and |α| > |β| ≥ 0, A ∈ C, further fz
fzis holomorphic
and∣∣∣fz
fz
∣∣∣ < 1.
Now, let us denote the family Σp(H) consisting of all harmonic sense-
preserving multivalent meromorphic mapping
f(z) = h(z) + g(z) (5.51)
where
h(z) = zp +
∞∑n=1
an+p−1z−(n+p−1), g(z) =
∞∑n=1
bn+p−1z−(n+p−1), |bp| < 1, |z| > 1.
(5.52)
For 0 ≤ α < 2(1 − k), 0 ≤ λ ≤ 1, 12≤ k < 1, 0 ≤ θ < 1, 0 ≤ ξ < 1 and
z = reiβ ;
224
1 < r < ∞; β, ξ, θ and α are real, we introduce the subclass AOH(α, λ, k, p)
consisting of all functions f satisfying
Re
{(1 + eiθ)
zf ′(z)/f(z)
λzf ′(z)/f(z) + (1 − pλ)− pk(1 + eiξ)
}≥ pα. (5.53)
Also the subclass of multivalent meromorphic harmonic functions with
f = zp +∞∑
n=1
|an+p−1|z−(n+p−1) −∞∑
n=1
|bn+p−1|z−(n+p−1), |bp| < 1 (5.54)
that satisfies (5.53) is denoted by AOH(α, λ, k, p).
The particular case λ = 0, p = 1 recently is investigated by [15] for uni-
valent harmonic functions defined in U = {z : |z| < 1}.
In order to study the various properties of AOH(α, λ, k, p) we need the
following result due to O.P. Ahuja and J. M. Jahangiri [1].
Theorem 5.3.1 : Let f = h+ g with h and g given by (5.52). If
∞∑n=1
(n + p− 1)(|an+p−1| + |bn+p−1|) ≤ p (5.55)
then f is harmonic, sense preserving and multivalent in U and f ∈ Σp(H).
Theorem 5.3.2 : Let f = h+g where g and h given by (5.52). Furthermore,
let
∞∑n=1
([2(n + p− 1) − p(α+ 2k)(λn+ 2pλ− λ− 1)]|an+p−1|
+[2(n+ p− 1) − p(α + 2k)(λn− λ+ 1)]|bn+p−1|) ≤ p(2 − 2k − α) , (5.56)
where 0 ≤ α < 2(1−k), 0 ≤ λ ≤ 1 and 12≤ k < 1. Then f is sense preserving
multivalent meromorphic harmonic functions with f ∈ AOH(α, λ, k, p).
Proof : If the inequality (5.56) holds for coefficients of f = h + g, then by
(5.55), f is sense preserving and harmonic multivalent in U . Now we want to
225
show that f ∈ AOH(α, λ, k, p). According to (5.53), we have
Re
{[zh′(z) − zg′(z)](1 + eiθ)
λ(zh′(z) − zg′(z)) + (1 − pλ)(h(z) + g(z))− pk(1 + eiξ)
}≥ pα,
where z = reiγ , 0 ≤ γ ≤ 2π, 0 ≤ r < 1, 0 ≤ α < 2(1 − k), 0 ≤ θ < 1,
0 ≤ ξ < 1, 12≤ k < 1, and 0 ≤ λ ≤ 1.
Let N(λ, z) = (1 + eiθ)(zh′(z) − zg′(z)) − pk(1 + eiξ)[λ(zh′(z) − zg′(z)) +
(1 − pλ)(h(z) + g(z))] and
M(λ, z) = λ(zh′(z) − zg′(z)) + (1 − pλ)(h(z) + g(z)).
By using the fact Re w > pα if and only if |p(1 − α) + w| > |p(1 + α) − w|,it is enough to show that
|p(1 − α)M(λ, z) +N(λ, z)| − |N(λ, z) − p(1 + α)M(λ, z)| ≥ 0.
Therefore
|N(λ, z) + p(1 − α)M(λ, z)| = |(1 + eiθ)(zh′(z) − zg(z)′) − pk(1 + eiξ) ×
[λ(zh′(z) − zg′(z) + (1 − pλ)(h(z) + g(z))] + p(1 − α)[λ(zh′(z) − zg′(z)) +
(1 − pλ)(h(z) + g(z))]| = |[(1 + eiθ) − pλk(1 + eiξ) + pλ(1 − α)]zh′(z) −
[(1 − pλ)pk(1 + eiξ) − p(1 − α)(1 − pλ)]h(z) − [(1 + eiθ) − λpk(1 + eiξ) +
p(1 − α)λ]zg′(z) − [pk((1 + eiξ)(1 − pλ) − p(1 − α)(1 − pλ)]g(z)|
= |[(1 + eiθ) − pλk(1 + eiξ) + pλ(1 − α)][pzp −∞∑
n=1
(n+ p− 1)an+p−1z−(n+p−1)] −
[(1 − pλ)pk(1 + eiξ) − p(1 − α)(1 − pλ)][zp +∞∑
n=1
an+p−1z−(n+p−1)] −
[(1 + eiθ) − pλk(1 + eiξ) + p(1 − α)λ][−∞∑
n=1
(n + p− 1)bn+p−1z−(n+p−1)] −
226
[pk(1 + eiξ)(1 − pλ) − p(1 − α)(1 − pλ)]∞∑
n=1
bn+p−1z−(n+p−1)
≥ (3p− 2kp− pα)|z|p −∞∑
n=1
[2(n+ p− 1) + p(1 − α− 2k) ×
(λn+ 2pλ− λ− 1)]|an+p−1||z|−(n+p−1) −∞∑
n=1
[2(n + p− 1) + p(1 − α− 2k)(λn− λ+ 1)]|bn+p−1||z|−(n+p−1),
also we have
|N(λ, z) − p(1 + α)M(λ, z)| = |(1 + eiθ)(zh′(z) − zg′(z) − pk(1 + eiξ) ×
[λ(zh′(z) − zg′(z)) + (1 − pλ)(h(z) + g(z))] −
p(1 + α)[λ(zh′(z) − zg′(z)) + (1 − pλ)(h(z) + g(z))]|
= |[(1 + eiθ) − pkλ(1 + eiξ) − pλ(1 + α)]zh′(z) − [(1 − pλ)pk(1 + eiξ) +
p(1 + α)(1 − pλ)]h(z) − [(1 + eiθ) − λpk(1 + eiξ) − p(1 + α)λ]zg′(z) −
[pk(1 + eiξ)(1 − pλ) + p(1 + α)(1 − pλ)]g(z)|
≥ (pα + 2pk − p)|z|p +∞∑
n=1
[2(n+ p− 1) − p(1 + α + 2k) ×
(λn+ 2λp− λ− 1)]|an−p−1||z|−(n+p−1) +∞∑
n=1
[2(n+ p− 1) − p(2k + 1 + α)(λn− λ+ 1)]|bn+p−1||z|−(n+p−1).
Thus,
|p(1 − α)M(λ, z) +N(λ, z)| − |N(λ, z) − p(1 + α)M(λ, z)|
≥ 2p(2 − 2k − α)|z|p − 2∞∑
n=1
[2(n+ p− 1) − p(α + 2k) ×
(λn+ 2λp− λ− 1)]|an+p−1||z|−(n+p−1) −
2∞∑
n=1
[2(n+ p− 1) − p(α + 2k)(λn− λ+ 1)]|bn−p+1||z|−(n+p−1)
≥ 0 (by (5.56)).
227
So f(z) ∈ AOH(α, λ, k, p). �
Theorem 5.3.3 : Let f = h + g, where h and g have the form given by
(5.52). Then f ∈ AOH(α, λ, k, p) if and only if
∞∑n=1
([2(n+ p− 1) − p(α + 2k)(λn+ 2pλ− λ− 1)]|an+p−1| + [2(n+ p− 1)
−p(α + 2k)(λn− λ+ 1)]|bn+p−1|) / (p(2 − 2k − α)) ≤ 1 (5.57)
Proof : The “if” part is clear, since AOH(α, λ, k, p) ⊆ AOH(α, λ, k, p), for “
only if” part we show that f �∈ AOH(α, λ, k, p) if the inequality (5.57) does
not hold. So, we must show that
Re
{(zh′(z) − zg′(z)
λ(zh′(z) − zg′(z)) + (1 − pλ)(h(z) + g(z))
)(1 + eiθ) − pk(1 + eiξ) − pα)
}
= Re
{C(z)D(z)
}≥ 0,
where
C(z) = [zh′(z)−zg′(z)](1+eiθ)−p(k(1+eiξ)+α)[λ(zh′(z)−zg′(z))+(1−pλ)(h(z)+g(z))],
also
D(z) = λ(zh′(z) − zg′(z)) + (1 − pλ)(h(z) + g(z)),
then
C(z) = p(2 − 2k − α)|z|p −∞∑
n=1
[2(n + p− 1) − p(α + 2k)(λn+ 2λp− λ− 1] ×
|an+p−1||z|−(n+p−1) + [2(n+ p− 1) − p(α+ 2k)(λn− λ+ 1)]|bn+p−1||z|−(n+p−1)
and
D(z) = zp−∞∑
n=1
(λn+2pλ−λ−1)|an+p−1||z|−(n+p−1)+∞∑
n=1
(λn−λ+1)|bn+p−1||z|−(n+p−1).
228
Upon choosing the values of z on the positive real axis, where |z| = r < 1,
then we must show that
[(2 − 2k − α) −∞∑
n=1
[(2(n+ p− 1) − p(α + 2k)(λn+ 2λp− λ− 1)]|an+p−1| +
[2(n+ p− 1) − p(α + 2k)(λn− λ+ 1)]|bn+p−1|]/[1 −∞∑
n=1
(λn+ 2pλ− λ− 1) ×
|an+p−1|r−(n+2p−1) +
∞∑n=1
(λn− λ+ 1)|bn+p−1|r−(n+2p−1)] ≥ 0. (5.58)
We note that the last inequality is negative for r sufficiently close to 1, then
the inequality (5.57) does not hold, therefore Re{
C(z)D(z)
}is negative. This
contradicts the required condition for f ∈ AOH(α, λ, k, p). This completes
the proof of theorem. �
Next we obtain the distortion bounds and extreme points.
Theorem 5.3.4 : Let f(z) ∈ AOH(α, λ, k, p). Then
rp − p(2 − 2k − α)r−p ≤ |f(z)| ≤ rp + p(2 − 2k − α)r−p.
Proof : Let f ∈ AOH(α, λ, k, p), then for |z| = r > 1 we have
|f(z)| =
∣∣∣∣∣zp +
∞∑n=1
an+p−1|z|−(n+p−1) −∞∑
n=1
bn+p−1z−(n+p−1)
∣∣∣∣∣≤ rp +
∞∑n=1
(an+p−1 + bn+p−1)r−(n+p−1) ≤ rp + r−p
∞∑n=1
(|an+p−1| + |bn+p−1|)
≤ rp + r−p∞∑
n=1
[[2(n + p− 1) − p(α + 2k)(λn+ 2pλ− λ− 1)]|an+p−1| ×
r−(n+2p−1) + [2(n + p− 1) − p(α+ 2k)(λn− λ+ 1)|bn+p−1|r−(n+2p−1)]]
≤ rp + p(2 − 2k − α)r−p.
The left hand inequality can be proved by using similar arguments. This
completes the proof of theorem. �
229
Theorem 5.3.5 : The function
f(z) = h(z) + g(z) ∈ AOH(α, λ, k, p)
if and only if
f(z) =
∞∑n=0
(Sn+p−1hn+p−1(z) + Tn+p−1gn+p−1(z)), z ∈ U , p ≥ 1 (5.59)
where
hp−1(z) = zp, hn+p−1(z) = zp+p(2 − 2k − α)
2(n+ p− 1) − p(α+ 2k)(λn+ 2λp− λ− 1)z(n+p−1),
gp−1(z) = zp, gn+p−1(z) = zp− p(2 − 2k − α)
2(n+ p− 1) − p(α + 2k)(λn− λ+ 1)z−(n+p−1)
for (n ≥ 1),∞∑
n=0
(Sn+p−1 + Tn+p−1) = 1, Sn+p−1 ≥ 0 and Tn+p−1 ≥ 0. In
particular the extreme points of AOH(α, λ, k, p) are {hn+p−1} and {gn+p−1}.
Proof : Suppose that f can be written of the form (5.59), then
f(z) = Sp−1hp−1(z) + Tp−1gp−1(z) +
∞∑n=1
Sn+p−1 ×(zp +
p(2 − 2k − α)
2(n+ p− 1) − p(α + 2k)(λn− 2λp− λ− 1)z−(n+p−1)
)+
∞∑n=1
Tn+p−1
(zp − p(2 − 2k − α)
2(n+ p− 1) − p(α + 2k)(λn− λ+ 1)z−(n+p−1)
)
=
∞∑n=0
(Sn+p−1 + Tn+p−1)zp +
∞∑n=1
[p(2 − 2k − α)
2(n+ p− 1) − p(α + 2k)(λn+ 2λp− λ− 1)Sn+p−1z
−(n+p−1) −
p(2 − 2k − α)
2(n+ p− 1) − p(α + 2k)(λn− λ+ 1)Tn+p−1z
−(n+p−1)
].
Since
∞∑n=1
((2(n+ p− 1) − p(α + 2k)(λn+ 2λp− λ− 1))×
230
(p(2 − 2k − α)
2(n+ p− 1) − p(α + 2k)(λn+ 2λp− λ− 1)Sn+p−1
)+
(2(n+ p− 1) − p(α+ 2k)(λn− λ+ 1)) ×(p(2 − 2k − α)
2(n+ p− 1) − p(α + 2k)(λn− λ+ 1)Tn+p−1
))
= p(2 − 2k − α)
∞∑n=1
(Sn+p−1 + Tn+p−1) ≤ p(2 − 2k − α).
So by Theorem 5.3.3, f ∈ AOH(α, λ, k, p).
Conversely, if f ∈ AOH(α, λ, k, p), then by Theorem 5.3.3 we have
∞∑n=1
1
p(2 − 2k − α)[(2(n+ p− 1) − p(α + 2k)(λn+ 2λp− λ− 1))|an+p−1| +
(2(n+ p− 1) − p(α+ 2k)(λn− λ+ 1))|bn+p−1|] ≤ 1.
Putting Sn+p−1 = 2(n+p−1)−p(α+2k)(λn+2λp−λ−1)p(2−2k−α)
, Tn+p−1 = 2(n+p−1)−p(α+2k)(λn−λ+1)p(2−2k−α)
,
0 ≤ Sp−1 ≤ 1 and Tp−1 = 1−Sp−1−∞∑
n=1
(Sn+p−1 +Tn+p−1) we get the required
result. �
Definition 5.3.1 : Let
f(z) = zp +
∞∑n=1
|an+p−1|z−(n+p−1) −∞∑
n=1
|bn+p−1|z−(n+p−1) (5.60)
g(z) = zp +
∞∑n=1
|cn+p−1|z−(n+p−1) −∞∑
n=1
|dn+p−1|z−(n+p−1). (5.61)
be in AOH(α, λ, k, p), for real number β, we define the β-convolution of f and
g as follows.
(f ⊗β g)(z) = z +
∞∑n=1
|an+p−1cn+p−1|(n+ p− 1)β
z−(n+p−1) +
∞∑n=1
|bn+p−1dn+p−1|(n + p− 1)β
z−(n+p−1).
The 0-convolution of f and g is the familiar Hadamard product, also the
1-convolution of f and g is named integral convolution and is defined by
(f ⊗1 g)(z) = z +
∞∑n=1
|an+p−1cn+p−1|n+ p− 1
z−(n+p−1) +
∞∑n=1
|bn+p−1dn+p−1|n+ p− 1
z−(n+p−1).
231
Theorem 5.3.6 : Let f(z), g(z) defined as (5.60), (5.61) respectively be in
AOH(α, λ, k, p). Then the β-convolution of f and g where β ≥ p(2−2k−α)2p−p(α+2k)(2λp−1)
belongs to AOH(α, λ, k, p), if furthermore of β,
then, we have
β ≥ maxn �=1
{(log(n+ p− 1))−1 log
2 − 2k − α
2 − (α + 2k)(2λp− 1),
(log(n+ p− 1))−1 log2 − 2k − α
2 − (α + 2k)
}.
Proof : By assumption, we have f, g ∈ AOH(α, λ, k, p), therefore
∞∑n=1
2(n+ p− 1) − p(α+ 2k)(λn+ 2pλ− λ− 1)
p(2 − 2k − α)|an+p−1| +
2(n+ p− 1) − p(α + 2k)(λn− λ+ 1)
p(2 − 2k − α)|bn+p−1| ≤ 1 (5.62)
∞∑n=1
2(n+ p− 1) − p(α+ 2k)(λn+ 2pλ− λ− 1)
p(2 − 2k − α)|cn+p−1| +
2(n+ p− 1) − p(α + 2k)(λn− λ+ 1)
p(2 − 2k − α)|dn+p−1| ≤ 1. (5.63)
Also we have
|cn+p−1| ≤ p(2 − 2k − α)
2(n+ p− 1) − p(α+ 2k)(λn+ 2pλ− λ− 1),
|dn+p−1| ≤ p(2 − 2k − α)
2(n+ p− 1) − p(α+ 2k)(λn− λ+ 1),
we have to show that
∞∑n=1
2(n+ p− 1) − p(α + 2k)(λn+ 2pλ− λ− 1)
p(2 − 2k − α)
|an+p−1cn+p−1|(n+ p− 1)β
+
[2(n+ p− 1) − p(α+ 2k)(λn− λ+ 1)]
p(2 − 2k − α)
|bn+p−1dn+p−1|(n+ p− 1)β
≤ 1 (5.64)
For this purpose, we have
∞∑n=1
2(n+ p− 1) − p(α + 2k)(λn+ 2pλ− λ− 1)
p(2 − 2k − α)
|an+p−1cn+p−1|(n+ p− 1)β
+
232
2(n+ p− 1) − p(2 + 2k)(λn− λ+ 1)|bn+p−1dn+p−1|p(2 − 2k − α)(n+ p− 1)β
≤∞∑
n=1
|an+p−1| + |bn+p−1|(n+ p− 1)β
.
Therefore the inequality in (5.64) holds true if
(n+ p− 1)β ≥ p(2 − 2k − α)
2(n+ p− 1) − p(α + 2k)(λn+ 2pλ− λ− 1),
(n+ p− 1)β ≥ p(2 − 2k − α)
2(n+ p− 1) − p(2 + 2α)(λn− λ+ 1)
holds true
or
β ≥ maxn�=1
{(log(n + p − 1))−1 log
2 − 2k − α
2 − (α + 2k)(2λp − 1), (log(n + p − 1))−1 log
2 − 2k − α
2 − (α + 2k)
}.
�
Theorem 5.3.7 : Suppose 0 ≤ α1 ≤ α2 < 2(1 − k), 12≤ k < 1 and
f ∈ AOH(α2, λ, k, p), g ∈ AOH(α1, λ, k, p), then
f ∗ g ∈ AOH(α2, λ, k, p) ⊂ AOH(α1, λ, k, p).
Proof : By assumption it is clear that AOH(α2, λ, k, p) ⊂ AOH(α1, λ, k, p).
Further,
∞∑n=1
([2(n+ p− 1) − p(α2 + 2k)(λn+ 2pλ− λ− 1)]|an+p−1| +
[2(n+ p− 1) − p(α2 + 2k)(λn− λ+ 1)]|bn+p−1|)/(p(2 − 2k + α2)) ≤ 1
∞∑n=1
{[2(n+ p− 1) − p(α1 + 2k)(λn+ 2pλ− λ− 1)]|cn+p−1| +
[2(n+ p− 1) − p(α1 + 2k)(λn− λ+ 1)|dn+p−1]}/{p(2 − 2k + α)} ≤ 1
233
and consequently, we have
|cn+p−1| ≤ p(2 − 2k − α1)
2(n+ p− 1) − p(α1 + 2k)(λn+ npλ− λ− 1),
|dn+p−1| ≤ p(2 − 2k − α1)
2(n+ p− 1) − p(α1 + 2k)(λn− λ+ 1).
Therefore,
∞∑n=1
([2(n + p− 1) − p(α2 + 2k)(λn+ npλ− λ− 1)]|an+p−1cn+p−1| +
[2(n+ p− 1) − p(α2 + 2k)(λn− λ+ 1)|bn+p−1dn+p−1|)/(p(2 − 2k − α2))
≤ [2(n+ p− 1) − p(α2 + 2k)(λn+ npλ− λ− 1)][p(2 − 2k + α1)
(p(2 − 2k − α2)[2(n+ p− 1) − p(α1 + 2k)(λn+ npλ− λ− 1)]|an+p−1| +
∞∑n=1
[2(n+ p− 1) − b(α2 + 2k)(λn− λ− 1)](p(2 + 2k + α1)
p(2 − 2k − α2)[2(n+ p− 1) − p(α1 + 2k)(λn− λ− 1)]|bn+p−1| ≤ 1,
then by Theorem 5.3.6, we have f ∗ g ∈ AOH(α2, λ, k, p). �
Theorem 5.3.8 : Let f ∈ AOH(α, λ, k, p). Then f is closed under convex
combination.
Proof : Let fj(z) ∈ AOH(α, λ, k, p) for j ≥ 1 be defined by the following
form
fj(z) = zp +
∞∑n=1
an+p−1,jz−(n+p−1) −
∞∑n=1
bn+p−1,jz−(n+p−1),
(an+p−1,j ≥ 0 and bn+p−1,j ≥ 0).
Therefore by Theorem 5.3.3, we have
∞∑n=1
[(2(n+ p− 1) − p(α + 2k)(λn+ 2λp− λ− 1)
p(2 − 2k − α)an+p−1,j +
2(n+ p− 1) − p(α + 2k)(λn− λ+ 1)
p(2 − 2k − α)bn+p−1,j
]≤ 1. (5.65)
Then we can write for∞∑
j=1
Sj = 1, 0 ≤ Sj ≤ 1, the convex combination of fj
234
as
∞∑n=1
Sjfj(z) = zp+
∞∑n=1
(
∞∑j=1
Sjan+p−1,j)z−(n+p−1)−
∞∑n=1
(
∞∑j=1
Sjbn+p−1,j)z−(n+p−1).
From (5.65) we get
∞∑n=1
[2(n+ p− 1) − p(α + 2k)(λn+ 2λp− λ− 1)
p(2 − 2k − α)(
∞∑j=1
Sjan+p−1,j) +
2(n+ p− 1) − p(α + 2k)(λn− λ+ 1)
p(2 − 2k − α)(
∞∑j=1
Sjbn+p−1,j)
]
=∞∑
j=1
Sj
( ∞∑n=1
[2(n+ p− 1) − p(α + 2k)(λn+ 2λp− λ− 1)
p(2 − 2k − α)an+p−1,j +
2(n + p− 1) − p(α+ 2k)(λn− λ+ 1)
p(2 − 2k − α)bn+p−1,j
])≤
∞∑j=1
Sj = 1.
Therefore∞∑
j=1
Sjfj(z) ∈ AOH(α, λ, k, p). This completes the proof. �
235
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