Gen. Math. Notes, Vol. 24, No. 1, September 2014, pp.109-126
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2014
www.i-csrs.org
Available free online at http://www.geman.in
On Some Subclasses of Analytic Functions With
Negative Coefficients using a Convolution Approach
E.A. Oyekan1 , T.O. Opoola2 and A.T. Ademola3
1
Department of Mathematical Sciences
Ondo State University of Science and Technology (OSUSTECH)
P.M.B. 353 Okitipupa, Ondo State, Nigeria
E-mail: shalomfa@yahoo.com; ea.oyekan@osustech.edu.ng
2
Department of Mathematics
University of Ilorin, Ilorin, Nigeria
E-mail: opoolato@unilorin.edu.ng
3
Department of Mathematics
University of Ibadan, Ibadan, Nigeria
E-mail: ademola672000@yahoo.com
(Received: 29-8-13 / Accepted: 14-7-14)
Abstract
In this work, we introduce and investigate new subclasses
Em,n (Φ, Ψ; A, B, α, β, γ) and Ẽm,n (Φ, Ψ; A, B, α, β, γ)
of analytic functions by making use of the familiar convolution structure of analytic functions whose Taylor-Maclaurin coefficients from the second onwards
are all negative. In particular, we derive the coefficient inequalities and some
other interesting properties for functions belonging to these subclasses. Our
results generalize some earlier known results.
Keywords: Analytic functions, coefficient estimates, distortion inequalities, Littlewood subordination theorem, integral means.
110
1
E.A. Oyekan et al.
Introduction and Definitions
Let A denote the class of functions of the form
f (z) = z +
∞
X
ak z k ,
(1)
k=2
which are analytic in the open unit disk U = {z ∈ C : |z| < 1}.
Also, let S be the subclass of A, consisting of analytic and univalent functions
in U.
We denote by
S ∗ (α) and K(α)
(0 ≤ α < 1)
the class of starlike functions of order α in U and the class of convex functions
of order α in U, respectively.
Further denote by Em,n (Φ, Ψ; A, B, α, β), the class of functions f (z) ∈ A which
are analytic in U and satisfy therein the condition
m
D (f ∗ Φ)(z)
Dm (f ∗ Φ)(z)
≺ (1 − α) 1 + Az + α (z ∈ U ), (2)
−
β
−
1
Dn (f ∗ Ψ)(z)
Dn (f ∗ Ψ)(z)
1 + Bz
where ∗ and ≺ denotes convolution(or Hadamard product) and subordination
respectively, (f ∗ Ψ)(z) 6= 0, A and B are arbitrarily fixed numbers such that
−1 ≤ B < A ≤ 1 and −1 ≤ B < 0 and 0 ≤ α < 1, β ≤ 0 and m ≥ n
(m, n ∈ N0 ).
The class Em,n (Φ, Ψ; A, B, α, β) was introduced and studied by Srivastava et
al.[3] and the class is due to the class Em,n (Φ, Ψ; A, B, α) which was earlier
introduced by Eker and Seker [14].
Note that
∞
X
Dn f (z) = z +
k n ak z k , (n ∈ N0 = N ∪ {0}),
(3)
k=2
n
where D is the usual Salagean operator (see [1]) and
Φ(z) = z +
∞
X
λk z k ,
Ψ(z) = z +
k=2
∞
X
µk z k
(4)
k=2
which are also analytic in U with λk ≥ 0, µk ≥ 0 and λk ≥ µk .
Definition 1.1. (Hadamard Product(or Convolution)) If f (z) and g(z)
are analytic in U, where g(z) is given by
g(z) = z +
∞
X
k=2
bk z k
(5)
111
On Some Subclasses of Analytic Functions With...
then, their Hadamard product(or convolution), f ∗ g is the function
(f ∗ g)(z) = z +
∞
X
ak bk z k = (g ∗ f )(z).
(6)
k=2
The function f ∗ g is also analytic in U.
Definition 1.2. (Subordination Principle) Let f (z) and g(z) be analytic in
the unit disk U. Then g(z) is said to be subordinate to f (z) in U and we write
g(z) ≺ f (z),
z ∈ U,
if there exists a Schwarz function w(z), analytic in U with w(0) = 0, |w(z)| < 1
such that
g(z) = f (w(z)), z ∈ U.
(7)
In particular, if the function f (z) is univalent in U, then g(z) is subordinate
to f (z) if
g(0) = f (0), g(u) ⊂ f (u).
(See for details Duren[10])
In view of (1), (4) and definition 1.1 we note that
(f ∗ Φ)(z) = z +
∞
X
k
(f ∗ Ψ)(z) = z +
ak λk z ,
∞
X
ak µ k z k ,
(8)
k=2
k=2
such that by using Binomial expansion on (8) we have
∞
∞
X
X
(f ∗Φ)γ (z) = z γ +
ak (γ)λk (γ)z k+γ−1 , (f ∗Ψ)(z) = z γ +
ak (γ)µk (γ)z k+γ−1 ,
k=2
k=2
(9)
Now, let
γ
γ
h(z) = (f ∗ Φ) (z) = z +
∞
X
ak (γ)λk (γ)z k+γ−1
k=2
and
γ
q(z) = (f ∗ Ψ)(z) = z +
∞
X
ak (γ)µk (γ)z k+γ−1
k=2
γ ≥ 0.
Then a function f ∈ A is said to be in the class Em,n (Φ, Ψ; A, B, α, β, γ) if and
only if
m
D h(z)
Dm h(z)
≺ (1 − α) 1 + Az + α (z ∈ U ),
−
β
−
1
(10)
n
n
D q(z)
D q(z)
1 + Bz
112
E.A. Oyekan et al.
where ≺ denotes subordination as earlier defined, q(z) 6= 0, A and B are
arbitrarily fixed numbers such −1 ≤ B < A ≤ 1 and − 1 ≤ B < 0 and
0 ≤ α < 1, β ≤ 0 and m ≥ n (m, n ∈ N0 ).
In other words, f ∈ Em,n (Φ, Ψ; A, B, α, β, γ) if and only if there exists an
analytic function ω(z) satisfying
ω(0) = 0 and |ω(z)| < 1
(z ∈ U )
such that
m
D h(z)
Dm h(z)
≺ (1 − α) 1 + Aω(z) + α (z ∈ U ),
−
β
−
1
n
n
D q(z)
D q(z)
1 + Bω(z)
(11)
The condition (11) is equivalent to the following inequality:
Dm h(z)
Dn q(z)
m h(z)
− β| DDn q(z)
− 1| − 1
< 1
m h(z)
m h(z)
(A − B)(1 − α) − B DDn q(z)
− β| DDn q(z)
− 1| − 1
(z ∈ U ). (12)
Let τ denote the subclass of A whose Taylor-Maclaurin expansion about z = 0
can be expressed in the following form:
f (z) = z −
∞
X
ak z k
(ak ≥ 0).
(13)
k=2
We shall denote Ẽm,n (Φ, Ψ; A, B, α, β, γ) the subclass of functions in
Em,n (Φ, Ψ; A, B, α, β, γ) that has their non-zero Taylor-Maclaurin coefficients,
from the second term onwards, all negative.
Thus we can write
Ẽm,n (Φ, Ψ; A, B, α, β, γ) = Em,n (Φ, Ψ; A, B, α, β, γ) ∩ τ.
(14)
It is easy to check that various known or new subclasses of τ referred to above
can be represented in terms of Ẽm,n (Φ, Ψ; A, B, α, β, γ) for suitable choices of
the function Φ and Ψ (and the parameters m, n, A, B, β and γ).
For example, we have the following relationship with known classes of functions:
z
z
,
; 1, −1, α, 0, 1) = S ∗ (α)
Ẽ0,0 (
2
(1 − z) 1 − z
and
z
z + z2
Ẽ0,0 (
,
; 1, −1, α, 0, 1) = k(α)
3
(1 − z) (1 − z)2
which were studied by Silverman[7];
Ẽ0,0 (
z
, z; 1, −1, α, 0, 1) = P∗ (α),
2
(1 − z)
113
On Some Subclasses of Analytic Functions With...
which was studied by Bhoosnurmath and swamy[13] and Gupta and Jain[15],
Ẽ0,0 (
z + (1 − 2α)z 2
z
,
; 1, −1, α, 0, 1) = R(α),
3−2α
(1 − z)
(1 − z)2−2α
which was studiedby Silverman and Silivia[6].
2
Main Results
We begin by proving the following results.
2.1
A Set of Coefficient Inequalities
Theorem 2.1. If f (z) ∈ A satisfies the following inequality:
∞
X
{(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
k=2
+ (A − B)(1 − α)(γ + k − 1)n µk (γ)}|ak (γ)|
≤ (A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
(15)
where
(λk (γ) ≥ µk (γ) ≥ 0; 0 ≤ α < 1; β ≥ 0; m ≥ n
(m, n ∈ N0 ))
and ak (γ), λk (γ), µk (γ) are the coefficients ak , λk , µk depending on γ, then f (z) ∈
Em,n (Φ, Ψ; A, B, α, β, γ).
Proof. Let the condition (15) hold true. Then it suffice to show that
Dm h(z)
Dm h(z)
−
β|
− 1| − 1
Dn q(z)
Dn q(z)
< 1 (z ∈ U ). (16)
m
D h(z)
Dm h(z)
(A − B)(1 − α) − B
−
β|
−
1|
−
1
Dn q(z)
Dn q(z)
114
E.A. Oyekan et al.
Thus we have that
m
D h(z) − Dn q(z) − βeiθ |Dm h(z) − Dn q(z)|
n
m
n
iθ
m
n
− (A − B)(1 − α)D q(z) − B[D h(z) − D q(z) − βe |D h(z) − D q(z)|]
∞
X
= |(γ − γ )z +
[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]ak (γ)z γ+K−1
m
n
γ
k=2
− βeiθ |(γ m − γ n )z γ +
∞
X
[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]ak (γ)z γ+K−1 ||
k=2
m γ
− |(A − B)(1 − α)γ z + (A − B)(1 − α)
∞
X
(γ + k − 1)n ak (γ)µk (γ)z γ+k−1
k=2
− B{(γ m − γ n )z γ +
∞
X
[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]ak (γ)z γ+k−1
k=2
∞
X
− βeiθ |(γ m − γ n z γ ) +
[(γ + k − 1)m λk − (γ + k − 1)n µk (γ)]ak (γ)z γ+k−1 |]|
k=2
≤ (γ m − γ n ) +
∞
X
[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]|aγ |
k=2
m
n
+ β(γ − γ ) + β
∞
X
|z|r |z|k
|z|
[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]|ak (γ)|
k=2
|z|r |z|k
|z|
− (A − B)(1 − α)γ m |z|γ
∞
X
|z|r |z|k
+ (A − B)(1 − α)
(γ + k − 1)n |ak (γ)|µk (γ)
|z|
k=2
m
n
r
+ |B|(γ − γ )|z| + |B|
∞
X
[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]|ak (γ)|
k=2
+ |B|β(γ m − γ n ) + |B|β
m
∞
X
[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]|ak (γ)|
k=2
n
≤ (1 − B)(1 + β)(γ − γ ) − (A − B)(1 − β)γ m
+
∞
X
{(1 − B)(1 + β)[(γ + k − 1)m λk (γ)
k=2
− (γ + k − 1)n µk (γ)] + (A − B)(1 − α)(γ + k − 1)n µk (γ)}|ak (γ)| ≤ 0,
|z|r |z|k
|z|
|z|r |z|k
|z|
115
On Some Subclasses of Analytic Functions With...
which implies that
∞
X
{(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
k=2
+ (A − B)(1 − α)(γ + k − 1)n µk (γ)}|ak (γ)|
≤ (A − B)(1 − β)γ m + (1 − B)(1 + β)(γ m − γ n )
This completes the proof of Theorem 2.1
In theorem 2.2 below, it is shown that the condition (15) is also necessary for
functions f (z) of the form (13) to be in the class Ẽm,n (Φ, Ψ; A, B, α, β, γ).
Theorem 2.2. Let f (z) ∈ τ, then f (z) ∈ Ẽm,n (Φ, Ψ; A, B, α, β, γ) if and
only if
∞
X
{(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
k=2
+ (A − B)(1 − α)(γ + k − 1)n µk (γ)}|ak (γ)|
≤ (A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
(17)
where
(λk (γ) ≥ µk (γ) ≥ 0; 0 ≤ α < 1; β ≥ 0; m ≥ n
(m, n ∈ N0 ))
and ak (γ), λk (γ), µk (γ) are the coefficients ak , λk , µk depending on γ.
Proof. Since Ẽm,n (Φ, Ψ; A, B, α, β, γ) ⊂ Em,n (Φ, Ψ; A, B, α, β, γ),
we only need to prove the only if part of Theorem 2.2. For functions f (z) ∈ τ,
we can write
(Dm h(z)/Dn q(z)) − (β|Dm h(z)/Dn q(z) − 1|) − 1
(A − B)(1 − α) − B[(Dm h(z)/Dn q(z)) − β|(Dm h(z)/Dn q(z) − 1|) − 1] m
n
iθ
m
n
D
h(z)
−
D
q(z)
−
βe
|D
h(z)
−
D
q(z)|
=
= (A − B)(1 − α)Dn q(z) − B[Dm h(z) − Dn q(z) − βeiθ |Dm h(z) − Dn q(z)| (γ m − γ n )z γ
+ k=2 [(γ + k − 1) λk (γ) − (γ + k − 1)n µk (γ)]ak (γ)z γ+k−1
+βeiθ |(γ m − γ n )z γ
P∞
+ k=2 [(γ + k − 1)m λk (γ) − (γ P
+ k − 1)n µk (γ)]ak (γ)z γ+k−1 |
γ n
n
γ+k−1
(A − B)(1 P
− α)z γ + (A − B)(1 − α) ∞
k=2 (γ + k − 1) µk (γ)]ak (γ)z
∞
m
n
γ+k−1
+B P
k=2 [(γ + k − 1) λk (γ) − (γ + k − 1) µk (γ)]ak (γ)z
∞
+Bβeiθ | k=2 [(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]ak (γ)z γ+k−1 |
P∞
< 1.
m
116
E.A. Oyekan et al.
Since Re(z) ≤ |z| < 1 (z ∈ U, ) we thus find that
P
m
n
γ+k−1
(γ m − γ n )z γ + ∞
k=2 [(γ + k − 1) λk (γ) − (γ + k − 1) µk (γ)]ak (γ)z
iθ
m
n γ
+βe |(γ − γ )z
P∞
m
+ k=2 [(γ + k − 1) λk (γ) − (γ P
+ k − 1)n µk (γ)]ak (γ)z γ+k−1 |
Re
n
γ+k−1
(A − B)(1 P
− α)z γ γ n + (A − B)(1 − α) ∞
k=2 (γ + k − 1) µk (γ)]ak (γ)z
∞
m
n
γ+k−1
+B P
k=2 [(γ + k − 1) λk (γ) − (γ + k − 1) µk (γ)]ak (γ)z
iθ
m
n
γ+k−1
+Bβe | ∞
|
k=2 [(γ + k − 1) λk (γ) − (γ + k − 1) µk (γ)]ak (γ)z
< 1.
If we now choose z to be real and let z → 1− , we have
∞
X
{(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
k=2
+ (A − B)(1 − α)(γ + k − 1)n µk (γ)}ak (γ)
≤ (A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
(18)
which is equivalent to equation (17).
Remark 2.3. Taking different choices for the function Φ(z) and Ψ(z) as
stated in section 1 when γ = 1, Theorem 2.2 leads us to the necessary and
sufficient conditions for a function f to be in each of the following classes:
S ∗ (α),
K(α),
P∗ (α) and R[α].
Remark 2.4. If we set γ = 1 in Theorem 2.1 and 2.2 above, we obtain
the result given in [9].
Remark 2.5. If we set m = n = β = 0 and γ = 1 in Theorem
2.1 and 2.2 above, we obtain the results given in [9].
Moreover, for m ≥ n (m, n ∈ N0 ), β = 0 and γ = 1, our results will
coincide with those presented in [14].
2.2
Distortion Theories Involving Fractional Calculus
In this section, we shall prove several distortion theorems for functions belonging to the general class Ẽm,n (Φ, Ψ; A, B, α, β, γ). Each of these theorems would
involve certain operators of fractional calculus (i.e, fractional integrals and fractional derivatives), which are defined as follows(see for details,[12, 11, 2, 5, 4]):
Definition 2.6. The fractional integral of order δ is defined, for a function
f, by
Z z
f (ξ)
1
−δ
dξ
(δ > 0, )
(19)
Dz f (z) =
Γ(δ) 0 (z − ξ)1−δ
117
On Some Subclasses of Analytic Functions With...
where f is an analytic function in a simply-connected region of the complex
z-plane containing the origin, and the multiplicity of (z − ξ)ξ−1 is removed by
requiring log(z − ξ) to be real when z − ξ > 0.
Definition 2.7. The fractional derivative of order δ is defined, for a function f, by
Z z
1
d
f (ξ)
δ
Dz f (z) =
dξ
(0 ≤ δ < 1, )
(20)
Γ(1 − δ) dz 0 (z − ξ)δ
where f is constrained, and the multiplicity of (z − ξ)ξ is removed as in definition 2.6
Definition 2.8. Under the hypothesis of the definition 2.7, the fractional
derivative of order n + δ is defined, for a function f, by
dn
{Dδ f (z)}
(0 ≤ δ < 1,
dz n z
By the virtue of Definitions 3, 4, and 5, we have
Dzn+δ f (z) =
Dz−δ z k =
and
Dzδ z k =
Γ(k + 1) k+δ
z
Γ(k + δ + 1)
Γ(k + 1) k−δ
z
Γ(k − δ + 1)
n ∈ N0 ).
(k ∈ N ; δ > 0)
(k ∈ N ; 0 ≤ δ < 1).
Theorem 2.9. Let f (z) defined by equation (13) be in the class
Ẽm,n (Φ, Ψ; A, B, α, β, γ). Then,
1+δ
−δ
Dz f (z) ≥ |z|
Γ(2 + δ)




2(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
1−

 (2 + δ{(1 − B)(1 + β)[(γ + 1)m λk (γ) − (γ + 1)n µk (γ)]
+(A − B)(1 − α)(γ + 1)n µk (γ)})|z|
(δ > 0;





z ∈ U)
and
1+δ
−δ
Dz f (z) ≤ |z|
Γ(2 + δ)



2(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
1+

 (2 + δ{(1 − B)(1 + β)[(γ + 1)m λk (γ) − (γ + 1)n µk (γ)]
+(A − B)(1 − α)(γ + 1)n µk (γ)})|z|






118
E.A. Oyekan et al.
(δ > 0;
z ∈ U ).
Each of these results is sharp.
Proof. Let
F (z) = Γ(2 + δ)z −δ Dzδ f (z) = z −
∞
X
Γ(k + 1)Γ(2 + δ)
k=2
=z−
∞
X
Γ(k + 1 + δ)
ak (γ)z k
Ω(k)ak (γ)z k
k=2
Γ(k + 1)Γ(2 + δ)
(k ∈ N {1}).
Γ(k + 1 + δ)
since Ω(k) is a decreasing function of k, we can write
where Ω(k) =
0 < Ω(k) ≤ Ω(2) =
2
.
2+δ
(21)
Furthermore, in view of Theorem 2.2, we have
(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
X
∞
ak (γ)
+ (A − B)(1 − α)(γ + 1) µ2 (γ)
n
k=2
∞
X
≤
{(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
k=2
+ (A − B)(1 − α)(γ + k − 1)n µk (γ)}ak (γ)
≤ (A − B)(1 − α)(γ)m − (1 − B)(1 + β)(γ m − γ n ),
which evidently yields
∞
X
k=2
ak (γ) ≤
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − γ + 1)n µ2 (γ)]
+(A − B)(1 − α)(γ + 1)n µ2 (γ)
Therefore, by using equation (21)and (22) we can see that
2
|F (z)| ≥ |z| − |Ω(2)| · |z|
∞
X
ak (γ)
k=2
≥ |z| −
2(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
|z|2
(2 + δ){(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
+(A − B)(1 − α)(γ + 1)n µ2 (γ)}
(22)
119
On Some Subclasses of Analytic Functions With...
and
2
|F (z)| ≤ |z| + |Ω(2)| · |z|
∞
X
ak (γ)
k=2
≤ |z| +
2(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
|z|2
(2 + δ){(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
+(A − B)(1 − α)(γ + 1)n µ2 (γ)}
which prove Theorem 2.9
Finally, since the equalities are attained for the function f (z) defined by
Dz−δ f (z) =
1−
z 1+δ
Γ(2 + δ)
2(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
z
(2 + δ){(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
+(A − B)(1 − α)(γ + 1)n µ2 (γ)}
or equivalently, by
f (z) = z −
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n ) 2
z ,
(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
+(A − B)(1 − α)(γ + 1)n µ2 (γ)
Our proof of Theorem 2.9 is completed.
Corollary 2.10. Under the hypothesis of Theorem 2.9, Dz−δ f (z) is included
in a disk with its center at the origin and radius r1 given by
1
2(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
r1 =
1−
.
(2 + δ){(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
Γ(2 + δ)
+(A − B)(1 − α)(γ + 1)n µ2 (γ)}
Theorem 2.11. Let the function f (z) defined by equation (13) be in the
class Ẽm,n (Φ, Ψ; A, B, α, β, γ). Then,
1−δ
δ
Dz f (z) ≥ |z|
Γ(2 − δ)
!
2(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
1−
|z|
(2 + δ{(1 − B)(1 + β)[(γ + 1)m λk (γ) − (γ + 1)n µk (γ)]
+(A − B)(1 − α)(γ + 1)n µk (γ)})
120
E.A. Oyekan et al.
(0 ≤ δ < 0;
z ∈ U)
and
1−δ
δ
Dz f (z) ≤ |z|
Γ(2 − δ)
1+
2(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
|z|
(2 + δ{(1 − B)(1 + β)[(γ + 1)m λk (γ) − (γ + 1)n µk (γ)]
+(A − B)(1 − α)(γ + 1)n µk (γ)})
(0 ≤ δ < 0;
z ∈ U ).
Each of these results is sharp.
Proof. Let
G(z) = Γ(2−δ)z
δ
Dzδ f (z)
∞
∞
X
X
Γ(k)Γ(2 − δ)
k
kak (γ)z = z−
Λ(k)kak (γ)z k
= z−
Γ(k
+
1
−
δ)
k=2
k=2
Γ(k)Γ(2 − δ)
(k = {2, 3, · · · }).
Γ(k + 1 − δ)
Since Λ(k) is a decreasing function of k, we can write
where Λ(k) =
0 < Λ(k) ≤ Λ(2) =
1
.
2−δ
(23)
Furthermore, in view of Theorem 2.2, we have
(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
+ (A − B)(1 − α)(γ + 1)
n−1
X
∞
µ2 (γ)
kak (γ)
k=2
≤
∞
X
{(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
k=2
+ (A − B)(1 − α)(γ + k − 1)n µk (γ)}ak (γ)
≤ (A − B)(1 − α)(γ)m − (1 − B)(1 + β)(γ m − γ n ),
which evidently yields
∞
X
k=2
kak (γ) ≤
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
.
(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
+(A − B)(1 − α)(γ + 1)n−1 µ2 (γ)
(24)
121
On Some Subclasses of Analytic Functions With...
Therefore, by using equation (23) and (24) we can see that
2
|G(z)| ≥ |z| + |Λ(2)| · |z|
∞
X
kak (γ)
k=2
≥ |z| −
2(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
|z|2
(2 + δ){(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
+(A − B)(1 − α)(γ + 1)n µ2 (γ)}
and
2
|G(z)| ≤ |z| + |Λ(2)| · |z|
∞
X
kak (γ)
k=2
≤ |z| +
2(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
|z|2
(2 + δ){(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
+(A − B)(1 − α)(γ + 1)n µ2 (γ)}
which together yields the inequalities asserted by Theorem 2.11. Equalities
are attained for the function f (z) defined by
Dzδ f (z) =
1−
z 1−δ
Γ(2 − δ)
2(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
z
(2 − δ){(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
+(A − B)(1 − α)(γ + 1)n µ2 (γ)}
Corollary 2.12. Under the hypothesis of Theorem 2.11, Dzδ f (z) is included
in a disk with its center at the origin and radius r2 given by
1
2(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
r2 =
1−
(2 + δ){(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
Γ(2 − δ)
+(A − B)(1 − α)(γ + 1)n−1 µ2 (γ)}
2.3
!
.
Extreme Points for the Functions in Class
Ẽm,n (Φ, Ψ; A, B, α, β, γ)
Theorem 2.13. Let f1 (z) = z and
fk (z) = z −
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n ) k
z
(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
+(A − B)(1 − α)(γ + k − 1)n µ2 (γ)
(25)
122
E.A. Oyekan et al.
(k = {2, 3, · · · }).
Then, f (z) ∈ Ẽm,n (Φ, Ψ; A, B, α, β, γ) if and only if it can be expressed in the
form
∞
X
f (z) =
ηk fk (z),
k=1
where ηk ≥ 0
P∞
and
k=1
ηk = 1.
Proof. Suppose that
f (z) =
∞
X
ηk fk (z)
k=1
=z−
∞
X
k=2
ηk
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n ) k
z .
(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
+(A − B)(1 − α)(γ + k − 1)n µ2 (γ)
Then, from Theorem 2.2, we have
∞
X
{(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
k=2
+ (A − B)(1 − α)(γ + k − 1)n µk (γ)}
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
(1 − B)(1 + β)[(γ + 1)m λ2 (γ) − (γ + 1)n µ2 (γ)]
+(A − B)(1 − α)(γ + k − 1)n µ2 (γ)
!
= [(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )]
ηk
∞
X
ηk
k=2
m
m
n
= [(A − B)(1 − α)γ − (1 − B)(1 + β)(γ − γ )](1 − η1 )
≤ (A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n ).
Thus in view of Theorem 2.2, we find that
f (z) ∈ Ẽm,n (Φ, Ψ; A, B, α, β, γ).
Conversely, let us suppose that
Ẽm,n (Φ, Ψ; A, B, α, β, γ).
Then, since
ak (γ) ≤
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
+(A − B)(1 − α)(γ + k − 1)n µk (γ)
On Some Subclasses of Analytic Functions With...
123
(k = {2, 3, · · · }),
we may set
(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
+(A − B)(1 − α)(γ + k − 1)n µk (γ)
ηk =
ak (γ)
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
(k = {2, 3, · · · }),
P
and η1 = 1 − ∞
k=2 ηk .
Thus, clearly, we have
f (z) = z −
∞
X
ηk fk (z).
k=2
This completes the proof of Theorem 2.13
Corollary 2.14. The extreme points of the functions Ẽm,n (Φ, Ψ; A, B, α, β, γ)
are given by
f1 (z) = z
and
fk (z) = z −
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
zk
(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
+(A − B)(1 − α)(γ + k − 1)n µk (γ)
(k = {2, 3, · · · }).
2.4
Integral Means Inequalities for the Function Class
Ẽm,n (Φ, Ψ; A, B, α, β, γ)
In the year 1925, Littlewood [8] proved the following:
Lemma 2.15. If the function f (z) and g(z) are analytic in U with g(z) ≺
f (z), (z ∈ U ), then for p > 0 and z = reiθ , (0 < r < 1), we have
Z
2π
p
Z
|f (z)| dθ ≤
0
2π
|g(z)|p dθ.
0
We now make use of Lemma 2.15 to prove Theorem 2.16 below:
124
E.A. Oyekan et al.
Theorem 2.16. Let f (z) ∈ Ẽm,n (Φ, Ψ; A, B, α, β, γ). Suppose also that
fk (z) is defined by equation (25). If ∃ an analytic function ω(z) given by
(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
+(A − B)(1 − α)(γ + k − 1)n µk (γ)
=
ak (γ)
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
[ω(z)]k−1
×
∞
X
ak (γ)z k−1 ,
k=2
then, for z = riθ
and
Z
0<r<1,
2π
iθ
p
Z
|f (r )| dθ ≤
0
Z
2π
|fk (riθ )|p dθ
(p > 0).
0
Proof. We need to show that
p
∞
2π X
k−1 1 −
a
(γ)z
k
dθ ≤
0
k=2
2π
Z
0
p
m
m
n
(A
−
B)(1
−
α)γ
−
(1
−
B)(1
+
β)(γ
−
γ
)
k−1 1 −
z
dθ.
(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
+(A − B)(1 − α)(γ + k − 1)n µk (γ)
By applying Lemma 2.15 above, it would suffice to show that
1−
∞
X
ak (γ)z k−1
k=2
<1−
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
z k−1
(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
+(A − B)(1 − α)(γ + k − 1)n µk (γ)
(z ∈ U )
By setting
1−
∞
X
ak (γ)z k−1
k=2
=1−
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
[ω(z)]k−1 ,
(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)]
+(A − B)(1 − α)(γ + k − 1)n µk (γ)
On Some Subclasses of Analytic Functions With...
125
we find that
[ω(z)]k−1
∞
(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)] X
n
+(A − B)(1 − α)(γ + k − 1) µk (γ)
=
ak (γ)z k−1 ,
k=2
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
which readily yields ω(0) = 0.
Furthermore by using equation (17), we obtain
|[ω(z)]|k−1
(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)] ∞
X
k−1
n
+(A − B)(1 − α)(γ + k − 1) µk (γ)
ak (γ)z ≤ k=2
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
∞
(1 − B)(1 + β)[(γ + k − 1)m λk (γ) − (γ + k − 1)n µk (γ)] X
+(A − B)(1 − α)(γ + k − 1)n µk (γ)
≤
ak (γ)z k−1
k=2
(A − B)(1 − α)γ m − (1 − B)(1 + β)(γ m − γ n )
≤ |z|k−1 < 1.
This completes the proof of Theorem 2.16
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