Gen. Math. Notes, Vol. 9, No. 2, April 2012, pp.56-68
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2012
www.i-csrs.org
Available free online at http://www.geman.in
Differential Subordinations for New Subclass
of Multivalent Functions Defined by Generalized
Derivative Operator
Ahmed Sallal Joudah
Department of Mathematics, College of Computer Science and Mathematics,
University of Al-Qadisiya, Diwiniya- Iraq
E-mail: ahmedhiq@yahoo.com
(Received: 31-3-12/ Accepted: 1-4-12)
Abstract
In this paper, we introduce new subclass of multivalent functions by making
use of the principle of subordination between these functions and generalized
derivative operator. We study several properties like, coefficient estimates,
distortion theorem, partial sums and integral means inequalities’.
Keywords: Multivalent function, Differential subordination, Fractional
calculus, Integral means.
1
Introduction
Let Σp,δ denote the class of functions of the form:
f (z) = z p+δ +
∞
X
ak z k+p+δ ,
(0 ≤ δ < 1)
(1)
k=2
which are analytic in the unit disk U = {z ∈ C : |z| < 1}.
Let Ap denote the subclass of Σp,δ consisting of functions of the form:
f (z) = z p+δ −
∞
X
k=2
ak z k+p+δ ,
(ak ≥ 0, 0 ≤ δ < 1)
(2)
57
Differential Subordinations for New Subclass...
which are analytic in the unit disk U .For the Hadamard product or (convolution) of two power series f defined in (2) and the function g where:
g(z) = z p+δ −
∞
X
bk z k+p+δ ,
(bk ≥ 0, 0 ≤ δ < 1, z ∈ U )
k=2
Is:
(f ∗ g)(z) = z p+δ −
∞
X
ak bk z k+p+δ = (g ∗ f )(z),
(z ∈ U )
(3)
k=2
Note that the authors defined and studied some classes of analytic functions take form (1) in [3]
We consider the following definitions of fractional integrals and fractional derivatives are due to Owa [5] and Srivastava and Owa [8]
Definition 1. The fractional integral of order λ is defined for a function
f by:
Dz−λ f (z) =
1 Zz
f (t)
dt
Γ(λ) 0 (z − t)1−λ
(4)
where λ > 0, f is an analytic function in a simply connected region of the
z-plane containing the origin and the multiplicity of (z − t)λ−1 is removed by
requiring log(z − t) to be real when (z − t) > 0.
Definition 2. The fractional derivative of order λ is defined for a function
f by:
Dzλ f (z) =
1
d Z z f (t)
dt
Γ(1 − λ) dz 0 (z − t)λ
(5)
where 0 ≤ λ < 1, f is an analytic function in a simply connected region of the
z-plane containing the origin and the multiplicity of (z − t)−λ is removed as in
Definition 1 above.
Definition 3. Under the hypothesis of Definition 2 , the fractional derivative of order n + λ is defined , for a function f , by
Dzn+λ f (z) =
dn λ
D f (z),
dz n z
(0 ≤ λ < 1; n ∈ N0 = N ∪ {0}).
It readily follows from Definition 2 that
Dzλ z k+p+δ =
Γ(k + p + δ + 1) k+p+δ−λ
z
, (0 ≤ λ < 1).
Γ(k + p + δ − λ + 1)
(6)
58
Ahmed Sallal Joudah
Motivated with the definition of salagean operator [1], we need to introduce
a generalized derivative operator such that class can be defined by means of
this derivative operator. For a function f ∈ Ap given by (2) , we define the
derivative operator (Dzλ,p,δ )n f (z) by:
(Dzλ,p,δ )0 f (z) =
Γp+δ−λ+1 λ λ
z Dz f (z)
Γp+δ+1
(Dzλ,p,δ )1 f (z) =
z
p+δ
0
Γp+δ−λ+1 λ λ
z
D
f
(z)
z
Γp+δ+1
(Dzλ,p,δ )n f (z) =
(Dzλ,p,δ )(Dzλ,p,δ )n−1 f (z)
= z p+δ −
∞
X
(φ(k, p, δ, λ))n ak z k+p+δ ,
(7)
k=2
Γ(k+p+δ+1)Γ(p+δ−λ+1) k+p+δ
where φ(k, p, δ, λ) = Γ(p+δ+1)Γ(k+p+δ−λ+1)
.
(p ∈ N ; 0 ≤ δ < 1)
p+δ
λ,p,0 n
Note that ,when δ = 0 the operator (Dz ) was studied recently by S.Porwal
,P.Dixit and V.Kumar [6] .
Clearly, (7) yields :
f ∈ Ap ⇒ (Dzλ,p,δ )n f ∈ Ap .
For two functions f and g analytic in U , we say that the function f is
subordinate to g in U ,and write f ≺ g (z ∈ U ) if there exists a Schwarz
function w(z) ,which is analytic in U with
w(0) = 0
, |w(z)| < 1
(z ∈ U )
(8)
Such that
f (z) = g(w(z))
(z ∈ U ).
(9)
Indeed ,it is know that
f (z) ≺ g(z)
, (z ∈ U ) ⇒ f (0) = g(0),
f (U ) ⊂ g(U ).
(10)
Furthermore ,if the function g is univalent in U ,then we have the following
equivalence :
f (z) ≺ g(z)
, (z ∈ U ) ⇔ f (0) = g(0) , f (U ) ⊂ g(U )
(11)
By making use of the operator (Dzλ,p,δ )n and the above mentioned principle
of subordination between analytic functions , we introduce and investigate the
following new subclass of the class Ap of (p + δ)-valent analytic functions.
59
Differential Subordinations for New Subclass...
Definition 4. A function f ∈ Ap be given by (2) is said to be in the class
if it satisfies the following subordination condition:
Aαp,δ (k, λ, A, B)


λ,p,δ n
p+δ
 (1 − α) (Dz ) f (z) + αz

(p + δ) 
z (Dzλ,p,δ )n f (z)
0

≺
1 + Bz
,
1 + Az
(12)
where 0 ≤ α ≤ 1, p ∈ N, 0 ≤ δ < 1, n ∈ N0 , 0 ≤ λ < 1, −1 ≤ A < B < 1 and
(Dzλ,p,δ )n f given by (7).
2
Coefficient Estimates
Theorem 1. Let the function f ∈ Ap be defined by (2). Then f is in the class
Aαp,δ (k, λ, A, B) if and only if
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − B))
≤ 1.
(p + δ)(B + A)
k=2
∞
X
(13)
where 0 ≤ α ≤ 1, p ∈ N, 0 ≤ δ < 1, n ∈ N0 , 0 ≤ λ < 1, −1 ≤ A < B < 1.
Proof. Let f ∈ Aαp,δ (k, λ, A, B) , then by the definition of subordination ,
we can write (12) as


p+δ
λ,p,δ n

 (1 − α) (Dz ) f (z) + αz
(p + δ) 
z
0
(Dzλ,p,δ )n f (z)

1 + Bw(z)
1 + Aw(z)
=
(w(z) ∈ U ),
which gives
(p + δ)
(1−α)((Dzλ,p,δ )n f (z))+αz p+δ
z ((Dzλ,p,δ )n f (z))

0


=
B − A(p + δ) 
!
−1
(1 − α)
z
(Dzλ,p,δ )n f (z)
(Dzλ,p,δ )n f (z)
From (14) , we obtain
P∞
(p+δ)z p+δ − k=2 (φ(k,p,δ,λ))n ((p+δ)(1−α)ak z k+p+δ
P∞
(p+δ)z p+δ − k=2 (φ(k,p,δ,λ))n (k+p+δ)ak z k+p+δ
+ αz
0
−1
p+δ


 w(z).
(14)
60
Ahmed Sallal Joudah
n
k+p+δ
(p + δ)z p+δ − ∞
k=2 (φ(k, p, δ, λ)) (p + δ)(1 − α)ak z
= B − A(p + δ)
P
n
k+p+δ
(p + δ)z p+δ − ∞
k=2 (φ(k, p, δ, λ)) (k + p + δ)ak z
"
P
!#
w(z)
which yields
−
P∞
=
n
k
k=2 (φ(k, p, δ, λ)) ((p + δ)(1 − α) − (k + p + δ))ak z
P
n
k
(p + δ) − ∞
k=2 (φ(k, p, δ, λ)) (k + p + δ)ak z
"
P∞
(p + δ)(B − A) − k=2 (φ(k, p, δ, λ))n (B(k + p + δ) − A(p +
P
n
k
(p + δ) − ∞
k=2 (φ(k, p, δ, λ)) (k + p + δ)ak z
δ)(1 − α)ak z k
#
w(z).
Since |w(z)| ≤ 1,
∞
X
n
k
−
(φ(k,
p,
δ,
λ))
((p
+
δ)(1
−
α)
−
(k
+
p
+
δ))a
z
k k=2
∞
X
≤ (p + δ)(B − A) −
(φ(k, p, δ, λ))n (B(k + p + δ) − A(p + δ)(1 − α)ak z k k=2
−
Letting |z| → 1 , through real values ,we have
∞
X
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − β))ak ≤ (p + δ)(B − A)
k=2
and , therefore ,
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − B))
≤ 1.
(p + δ)(B + A)
k=2
∞
X
Conversely ,assume that (13) be true .From (14) we see that since |w(z)| ≤ 1,
0 λ,p,δ n
(p + δ) (1 − α) (Dzλ,p,δ )n f (z) + αz p+δ
−
z
(D
)
f
(z)
z
0
λ,p,δ n
λ,p,δ n
Bz (Dz ) f (z) − A(p + δ) (1 − α) (Dz ) f (z) + αz p+δ P
n
k
− ∞
k=2 (φ(k, p, δ, λ)) ((p + δ)(1 − α) − (k + p + δ))ak z
= P∞
n
k
(p + δ)(B − A) −
(φ(k, p, δ, λ)) (B(k + p + δ) − A(p + δ)(1 − α)ak z ≤ 1. (15)
k=2
We need to prove that (15) is true .
By applying the hypothesis (13) and letting |z| = 1, we find that
P
n
k
− ∞
k=2 (φ(k, p, δ, λ)) ((p + δ)(1 − α) − (k + p + δ))ak z
P∞
n
k
(p + δ)(B − A) −
k=2 (φ(k, p, δ, λ)) (B(k + p + δ) − A(p + δ)(1 − α)ak z
P∞
n
k
≤
(p + δ)(B
(p + δ)(B
≤
(p + δ)(B
k=2 (φ(k, p, δ, λ)) ((p + δ)(1 − α) − (k + p + δ))ak z
P
n
− A) − ∞
k=2 (φ(k, p, δ, λ)) (B(k + p + δ) − A(p + δ)(1
P∞
− A) − k=2 (φ(k, p, δ, λ))n (B(k + p + δ) − A(p + δ)(1
P
n
− A) − ∞
k=2 (φ(k, p, δ, λ)) (B(k + p + δ) − A(p + δ)(1
Hence we find that (15) is true . Therefore f ∈ Aαp, (k, λ, A, B) .
This completes the proof of the theorem.
− α)ak z k
− α)ak z k
≤ 1.
− α)ak z k
Differential Subordinations for New Subclass...
3
61
Distortion Theorem
Theorem 2. If f ∈ Aαp,δ (k, λ, A, B) , then
(
)
(p + δ)(B − A)
r
−r
((p + δ)(1 − α)(1 − A) − (p + δ + 2)(1 − B)) (φ(2, p, δ, λ))n
(
)
(p + δ)(B − A)
p+δ
p+δ+2
≤ |f (z)| ≤ r
+r
((p + δ)(1 − α)(1 − A) − (p + δ + 2)(1 − B)) (φ(2, p, δ, λ))n
p+δ
p+δ+2
(|z| = r < 1).
Equality is attained for
(
f (z) = z
p+δ
−z
p+δ+2
)
(p + δ)(B − A)
.
((p + δ)(1 − α)(1 − A) − (p + δ + 2)(1 − B)) (φ(2, p, δ, λ))n
Proof. From Theorem 1, we have that
∞
X
ak ≤
k=2
(p + δ)(B − A)
.
((p + δ)(1 − α)(1 − A) − (p + δ + 2)(1 − B)) (φ(2, p, δ, λ))n
(16)
From (2) and (16) , it follows that
|f (z)| =
p+δ
z
≤ |z|p+δ +
−
∞
X
ak z
k+p+δ k=2
∞
X
ak |z|k+p+δ
k=2
∞
X
≤ rp+δ + rp+δ+2
ak
k=2
(
≤ r
p+δ
+r
p+δ+2
)
(p + δ)(B − A)
.
((p + δ)(1 − α)(1 − A) − (p + δ + 2)(1 − B)) (φ(2, p, δ, λ))n
Similarly ,
|f (z)| ≥ |z|p+δ −
∞
X
ak |z|k+p+δ
k=2
≥ rp+δ − rp+δ+2
∞
X
ak
k=2
(
≥ r
p+δ
−r
p+δ+2
This completes the proof.
)
(p + δ)(B − A)
.
((p + δ)(1 − α)(1 − A) − (p + δ + 2)(1 − B)) (φ(2, p, δ, λ))n
62
Ahmed Sallal Joudah
Partial Sums of the Function Class f ∈ Aαp,δ (k, λ, A, B)
4
Following the earlier work by Silverman [7] and recently Atshan and Joudah
[2],we investigate the ratio of real parts of functions involving (2) and its sequence of partial sums defined by:
(
Lm (z) =
z p+δ
, (m = 1)
Pm
k+p+δ
p+δ
, (m = 2, 3, ...; k ≥ 2; p ∈ N )
z
− k=2 ak z
n
o
n
(17)
o
(z)
and determine sharp lower bounds for R Lfm(z)
, R Lfm(z)
.
(z)
Theorem 3. Let f ∈ Ap and Lm (z) be given (2) and (17), respectively.
Suppose also that
∞
X
ϕk ak ≤ 1,
k=2
where
((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − B))(φ(k, p, δ, λ))n
. (18)
ϕk =
(p + δ)(B − A)
!
Then for, we have
(
f (z)
R
Lm (z)
)
>1−
1
,
(19)
ϕm+1
.
1 + ϕm+1
(20)
ϕm+1
and
(
Lm (z)
R
f (z)
)
>
The results are sharp for every m with the extremely functions given by
f (z) = z p+δ −
1
ϕm+1
z m+1
(21)
Proof. Under the hypothesis of the theorem , we can see from (18) that
ϕk+1 > ϕk > 1
(k ≥ 2).
Therefore , we have
m
X
k=2
ak + ϕm+1
∞
X
k=m+1
ak ≤
∞
X
k=2
ϕk ak ≤ 1,
(22)
63
Differential Subordinations for New Subclass...
by using hypothesis (18) again
Upon setting
"
f (z)
1
h1 (z) = ϕm+1
− 1−
Lm (z)
ϕm+1
P∞
k
ϕm+1 k=m+1 ak z
= 1−
.
P
k
1− m
k=2 ak z
!#
(23)
By applying (22) and (23) ,we find that
h (z) − 1 1
h1 (z) + 1 =
≤
P
k
−ϕm+1 ∞
k=m+1 ak z
Pm
P
∞
k −ϕ
k
2 − 2
a
z
a
z
m+1
k=2 k
k=m+1 k
P∞
k
2−2
(24)
ϕm+1 k=m+1 ak z
≤ 1 (z ∈ U ; k ≥ 2),
P∞
k
k
k=2 ak z − ϕm+1
k=m+1 ak z
Pm
which shows that R {h1 (z)} > 0 (z ∈ U ). From (23), we immediately obtain
the inequality (19).
To see that the function f given by (21) gives the sharp result ,we observe
for z → 1− that
1 m−(p+δ)+1
1
f (z)
=1−
z
→1−
,
Lm (z)
ϕm+1
ϕm+1
which shows that the bound in (19) is the best possible .
Similarly , if we put
"
Lm (z)
ϕm+1
h2 (z) = (1 + ϕm+1 )
−
f (z)
1 − ϕm+1
P∞
(1 + ϕm+1 ) k=m+1 ak z k
= 1+
,
P
k
1− m
k=2 ak z
#
(25)
and make use of (22) , we can deduce that
h (z) − 1 2
h2 (z) + 1 =
≤
P
k
(1 + ϕm+1 ) ∞
k=m+1 ak z
Pm
P
∞
k + (ϕ
k
2 − 2
a
z
−
1)
a
z
m+1
k=2 k
k=m+1 k
P∞
k
2−2
(26)
(1 + ϕm+1 ) k=m+1 ak z
≤ 1 (z ∈ U ; k ≥ 2),
P∞
k
k
k=2 ak z + (ϕm+1 − 1)
k=m+1 ak z
Pm
which leads us immediately to assertion (20) of the theorem .
The bound in (20) is sharp with the extremal function given by (21).
The proof of theorem is thus completed.
64
Ahmed Sallal Joudah
5
Weighted Mean and Arithmetic Mean
For f and g belong to Ap , the weighted mean hj of f and g is given by:
hj (z) =
1
[(1 − j)f (z) + (1 + j)g(z)] .
2
(27)
In the theorem below we will show the weighted mean for this class.
Theorem 4. If f and g are in the class Aαp,δ (k, λ, A, B), then the weighted
mean of f and g is also in Aαp,δ (k, λ, A, B).
Proof. We have for hj be given (27),
∞
∞
X
X
1
bk z k+p+δ
ak z k+p+δ + (1 + j) z p+δ −
hj (z) =
(1 − j) z p+δ −
2
k=2
k=2
!
"
= z p+δ −
!#
.
∞
X
1
[(1 − j)ak + (1 + j)bk ] z k+p+δ .
2
k=2
Since f and g are in the class Aαp,δ (k, λ, A, B), so by Theorem1 we must prove
that
∞
X
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − β)) 1
[(1 − j)ak + (1 + j)bk ]
(p + δ)(B − A)
2
k=2
=
∞
X
1
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − β))
(1 − j)
2
(p + δ)(B − A)
k=2
∞
X
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − β))
1
(1 + j)
2
(p + δ)(B − A)
k=2
1
1
≤
(1 − j) + (1 + j) = 1.
2
2
+
The proof is complete.
Theorem 5. Let f1 , f2 , . . . , f` defined by:
fi (z) = z p+δ −
∞
X
ai,k z k+p+δ ,
(ai,k ≥ 0; i = 1, 2, . . . , `; k ≥ 2)
(28)
k=2
be in the class Aαp,δ (k, λ, A, B), then the arithmetic mean of fi (i = 1, 2, . . . , `)
defined by :
h(z) =
`
1X
fi (z),
` i=1
(29)
65
Differential Subordinations for New Subclass...
is also in the class Aαp,δ (k, λ, A, B).
Proof. By (28),(29) we can write
`
∞
X
1X
p+δ
h(z) =
z
−
ai,k z k+p+δ
` i=1
k=2
!
=z
p+δ
−
∞
X
k=2
`
1X
ai,k z k+p+δ .
` i=1
!
Since fi ∈ Aαp,δ (k, λ, A, B) for every i = 1, 2, . . . , `, so by using Theorem 1,
We prove that
∞
X
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − β))
(p + δ)(B − A)
k=2
`
1X
ai,k
` i=1
!
`
∞
X
1X
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − β))
=
ai,k
` i=1 k=2
(p + δ)(B − A)
≤
!
`
1X
=1
` i=1
The proof is complete.
6
Integral Means Inequalities for the Class Aαp,δ (k, λ, A, B)
In 1925 ,Littlewood [4] proved the following subordination theorem .
Theorem 6. (Littlewood) If f and g are analytic functions in U with
f ≺ g , then for µ > 0 and z = reiϑ (0 < r < 1),
Z 2π
µ
|f (z)| dϑ ≤
0
Z 2π
|g(z)|µ dϑ,
(µ > 0; 0 < r < 1),
0
we will make use of the following theorem to prove
Theorem 7. Let f ∈ Aαp,δ (k, λ, A, B) and suppose that fn is defined by:
fn (z) = z p+δ −
(φ(k, p, δ, λ))n ((p
(p + δ)(B − A)
z k+p+δ .k ≥ 2
+ δ)(1 − α)(1 − A) − (k + p + δ)(1 − B))
If there exists an analytic function w give by
{w(z)}k =
∞
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − B)) X
ak z k ,
(p + δ)(B − A)
k=2
66
Ahmed Sallal Joudah
Then for z = reiϑ and (0 < r < 1),
Z 2π µ
f (reiϑ ) dϑ
≤
0
Z 2π µ
fn (reiϑ ) dϑ,
(µ > 0; 0 < r < 1),
0
(30)
Proof. We must show that
µ
Z 2π ∞
X
k
1 −
ak z dϑ
0
k=2
≤
Z 2π 1 −
0
µ
(p + δ)(B − A)
k
dϑ.
z
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − B)) By applying Littlewoods subordination theorem ,it would suffice to show that
1−
∞
X
ak z k ≺ 1 −
k=2
(φ(k, p, δ, λ))n ((p
(p + δ)(B − A)
zk .
+ δ)(1 − α)(1 − A) − (k + p + δ)(1 − B))
(φ(k, p, δ, λ))n ((p
(p + δ)(B − A)
{w(z)}k .
+ δ)(1 − α)(1 − A) − (k + p + δ)(1 − B))
By setting
1−
∞
X
ak z k = 1 −
k=2
We find that
{w(z)}k =
∞
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − B)) X
ak z k
(p + δ)(B − A)
k=2
which readily yields w(0)=0 .
Furthermore, using (13) , we obtain
k
|{w(z)}|
=
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k
(p + δ)(B − A)
∞
+ p + δ)(1 − B)) X
k=2
∞
X
ak z
k
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − B))
ak |z|k
(p + δ)(B − A)
k=2
≤ |z| ≤ 1.
≤
This completes the proof of the theorem.
Theorem 8. Let µ > 0 . If f ∈ Aαp,δ (k, λ, A, B) and
fn (z) = z p+δ −
(φ(k, p, δ, λ))n ((p
(p + δ)(B − A)
z k+p+δ .k ≥ 2
+ δ)(1 − α)(1 − A) − (k + p + δ)(1 − B))
then for z = reiϑ and (0 < r < 1),
Z 2π µ
0
f (reiϑ ) dϑ
0
≤
Z 2π µ
0
fn (reiϑ ) dϑ,
0
(31)
Differential Subordinations for New Subclass...
67
Proof. It is sufficient to show that
1−
∞
X
!
k+p+δ
ak z k
p+δ
k=2
(p + δ)(B − A)
≺ 1−
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − B))
!
k+p+δ k
z . (32)
p+δ
This follows because
k
|{w(z)}|
=
∞
X
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − B))
ak z k (33)
(p + δ)(B − A)
k=2
∞
X
(φ(k, p, δ, λ))n ((p + δ)(1 − α)(1 − A) − (k + p + δ)(1 − B))
≤ |z|
ak
(p + δ)(B − A)
k=2
k
≤ |z|k ≤ |z| .
The proof is complete.
Acknowledgements
The author, Ahmed Sallal, is thankful of his wife Lubna Hameed for her support of him in his work.
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