General Mathematics Vol. 18, No. 2 (2010), 95–119
On a subclass of certain p-valent starlike functions
with negative coefficients 1
S. M. Khairnar, N. H. More
Abstract
We introduce the subclass TΩ (n, p, λ, α, β) of analytic functions with
negative coefficients defined by salagean operators Dn . In this paper
we give some properties of functions in the class TΩ (n, p, λ, α, β) and
obtain numerous sharp results including coefficient estimates, distortion
theorems, closure theorems and modified Hadamard products of several
functions belonging to the class TΩ (n, p, λ, α, β). We also obtain radii of
close to convexity, starlikeness and convexity for the functions belonging
to the class TΩ (n, p, λ, α, β) and consider integral operators associated
with functions belonging to the class TΩ (n, p, λ, α, β).
2000 Mathematics Subject Classification: 30C45.
Key words and phrases: Analytic, Salagean Operator, Modified
Hadamard product, Negative coefficients.
1
Received 25 November, 2008
Accepted for publication (in revised form) 16 February, 2009
95
96
S. M. Khairnar, N. H. More
1
Introduction
Let A denote the class functions f (z) of the form
f (z) = z +
∞
X
ak z k
k=2
which are analytic in the unit disc U = {z : |z| < 1}.
For a function f (z) in A, we define
D0 (z) = f (z),
D1 (z) = Df (z) = zf 0 (z)
and
Dn (z) = D(Dn−1 f (z)) n ∈ N = {1, 2, 3, · · · }.
Note that
n
D (z) = D(D
n−1
f (z)) = z +
∞
X
k n ak z k n ∈ N0 = {0} ^ N.
k=2
the differential operator Dn was introduced by Salagean [5].
Let T (n, p) denote the class of functions f (z) of the form:
(1)
f (z) = z p −
∞
X
ak z k+p
k=n
(ak+p ≥ 0; p ∈ N = {1, 2, 3, · · · }; n ∈ N ), which are analytic in the unit disc
U = {z : |z| < 1}.
A function f (z) belonging to T (n, p) is in the class T (n, p, λ, α) if it satisfies
½
(2)
Re
zf 0 (z) + λz 2 f (z)
(1 − λ)f (z) + λzf (z)
¾
>α
for some α (0 ≤ α < 1) and λ (0 ≤ α ≤ 1), and for all z ∈ U [2].
On a subclass of certain p-valent starlike functions...
97
We can write the following equalities for the functions f (z) belong to the
class T (n, p)
D0 (z) = f (z),
"
D1 (z) = Df (z) = zf 0 (z) = z pz p−1 −
∞
X
#
(k + p)ak+p z k+p−1
k=n
= pz p −
∞
X
(k + p)ak+p z k+p−1 ,
k=n
and
DΩ (z) = D(DΩ−1 f (z)) n ∈ N = {1, 2, 3, · · · }.
Note that
DΩ (z) = D(DΩ−1 f (z)) = pΩ z p −
∞
X
(k + p)Ω ak+p z k+p .
k=2
We define a new subclass as follows:
Definition 1 A function f (z) belonging to T (n, p) is in the class TΩ (n, p,λ,α,β)
if and only if
¾
½
(1 − λ)z(DΩ f (z))0 + λz(DΩ+1 f (z))0
Re
(1 − λ)DΩ f (z) + λDΩ+1 f (z)
¯
¯
¯ (1 − λ)z(DΩ f (z))0 + λz(DΩ f (z))0 + λz(DΩ+1 f (z))0
¯
¯
>β¯
− 1¯¯ + α
Ω
Ω+1
(1 − λ)D f (z) + λD
f (z)
for some β ≥ 0, α (0 ≤ α < 1) and λ (0 ≤ λ < 1) and for all z ∈ U .
We note that by specializing the parameters n, p, λ, α and β we obtain the
following subclasses studied by various authors:
(i) TΩ (n, p, λ, α, 0) = TΩ (n, p, λ, α), (Kamali and Orhan [11])
(ii) T0 (n, p, λ, α) = T (n, p, λ, α), (Altintas et. al. [2])
(iii) T0 (n, 1, 0, α) = T0 (n), (Srivastava et. al. [8])
98
S. M. Khairnar, N. H. More
(iv) T0 (1, 1, 0, α) = T ∗ (α) (Silverman [7])
(v) T0 (1, 1, 1, α) = C(α), (Silverman [7])
(vi) T0 (n, 1, λ, α) = P (n, λ, al) (Altintas [1])
(vii) T0 (n, 1, λ, α) = C(n, λ, α). (Kamali and Akbulut [4]).
2
Coefficient Inequalities
Theorem 1 A function f (z) ∈ T (n, p) is in the class TΩ (n, p, λ, α, β) if and
only if
∞
X
(3)
(k + p)Ω [(k+p−α)+β(k+p−1)](λk+λp−λ+1)ak+p ≤ pn (p−α)(1+λp−λ)
k=n
(0 ≤ α < 1; 0 ≤ λ ≤ 1; p ≤ pΩ (p−α)(1+λp−λ)(p 6= 1); p ∈ N ; n ∈ N ; Ω ∈ N0 ).
The result is sharp.
Proof. Assume that the inequality (3) holds true. Note the fact that
Re(w) > β|w − 1| + α ⇔ Re{w(1 + βeiθ ) − βeiθ } > α,
− Λ ≤ θ ≤ Λ.
or equivalently
½
Re
(1 − λ)z(DΩ f (z))0 + λz(DΩ+1 f (z))0
(1 + βeiθ ) − βeiθ
(1 − λ)DΩ f (z) + λDΩ+1 f (z)
¾
> α.
Then for 0 < |z| = r < 1,

(4)
Re
[(1−λ)z(DΩ f (z))0 +λz(D Ω+1 f (z))0 ](1−βeiθ )−βeiθ (1−λ)DΩ f (z)+λDΩ+1 f (z)
(1−λ)Ω f (z)+λDΩ+1 f (z)
Let
A(z) = [(1 − λ)z(DΩ f (z))0 + λz(DΩ+1 f (z))0 ](1 + βeiθ )
− βeiθ [(1 − λ)DΩ f (z) + λDΩ+1 f (z)]
ff
>α
On a subclass of certain p-valent starlike functions...
and
B(z) = (1 − λ)DΩ f (z) + λDΩ+1 f (z).
Then (4) is equivalent to
|A + (1 − α)B| > |A − (1 + α)|, 0 ≤ α < 1.
For A(z) and B(z) as above, we have
|A + (1 − α)B| = |[p(1 + β) − (α + β − 1)](1 + λp − λ)pn z p
∞
X
−
(λk + λp + 1 − λ)(k + p)n [(k + p)(1 + β) − αβ]ak+p z k+p |
k=n
≥ (p − α)(1 + λp − λ)pn |z|p −
∞
X
(λk + λp + 1 − λ)(k + p)n ×
k=n
k+p
[(k + p)(1 + β) − αβ]ak+p |z|
≥ (p − α)(1 + λp − λ)pn |z|p −
∞
X
(λk + λp + 1 − λ)(k + p)n ×
k=n
k+p
[k + p − α + β(k + p − 1)|ak+p |z|
and similarly
|A − (1 − α)B| = |(p + βp − (β − 1 − α)(1 + λp − λ)pn z p
∞
X
(λk + λp + 1 − λ)(k + p)n [(k + p)(1 + β) − αβ]ak+p z k+p |
−
k=n
< (α − p)(1 + λp − λ)pn |z|p +
∞
X
(λk + λp + 1 − λ)(k + p)n ×
k=n
[(k + p − α + β(k + p − 1)ak+p |z|k+p .
99
100
S. M. Khairnar, N. H. More
Therefore,
|A + (1 − α)B| − |A − (1 + α)B|
≥ 2(p − α)(1 + λp − λ)pn |z|p − 2
∞
X
(λk + λp + 1 − λ)(k + p)n ×
k=n
[k + p − α + β(k + p − 1)]ak+p |z|k+p ≥ 0.
Letting r → 1, we obtain
∞
X
(λk +λp+1−λ)(k +p)n [k +p−α+β(k +p−1)]ak+p ≤ (p−α)(1+λp−λ)pn
k=n
which yields (3).
On the other hand we must have
¾
½
(1 − λ)z(DΩ f (z))0 + λz(DΩ+1 f (z))0
iθ
iθ
(1 + βe ) − βe
> α.
Re
(1 − λ)DΩ f (z) + λDΩ+1 f (z)
Upon choosing the values of z on the positive real axis where 0 < |z| = r < 1,
the above inequality reduces to
8
9
∞
P
Ω+1 −
>
(λk+λp+1−λ)(k + p)Ω+1 [k+p−α+β(k+p−1)]ak+p rk+p >
>
>
< (p−α)(1+λp−λ)p
=
Re
k=n
>
>
:
(1+λp−λ)pΩ −
∞
P
(λk+λp+1−λ)(k + p)Ω [(k+p−α+β(k+p−1)]ak+p rk+p
>
>
;
> 0.
k=n
Letting r → 1, we get the desired result.
Finally, we note that the assertion (3) is sharp, the extremal function being
(5)
f (z) = z p −
pn (p − α)(1 + λp − λ)
.
(k + p)Ω [(k + p − α) + β(k + p − 1)](λk + λp − λ + 1)
Corollary 1 Let the function f (z) defined by (1) be in the class TΩ (n, p, λ, α, β).
Then
(6) ak+p ≤
(p − α)(1 + λp − λ)pn
, (k ≥ j+1).
(λk + λp + 1 − λ)(k + p)n [k + p − α + β(k + p − 1)]
The equality in (6) is attained from the function f (z) given by (5).
On a subclass of certain p-valent starlike functions...
101
By taking β = 0 in above Theorem 1, we get
Corollary 2 Let the function f (z) be defined by (1) then f (z)
∈
TΩ (n, p, λ, α, 0) if and only if
(7)
∞
X
(λk + λp + 1 − λ)(k + p)n (k + p − α)ak+p ≤ (p − α)(1 + λp − λ)pn .
k=n
3
Distortion Theorems
Theorem 2 Let the function f (z) be defined by (1) be in the class
TΩ (n, p, λ, α, β). Then for |z| = r < 1,
(8)
|f (z)| ≥ |z|p −
(p−α)(1+λp−λ)pΩ
|z|p+n
(n + p)Ω (λn+λp+1−λ)[n+p−α+β(n+p−1)]
and
(9)
|f (z)| ≤ |z|p +
(p−α)(1+λp−λ)pΩ
|z|p+n
(n + p)Ω (λn+λp+1−λ)[n+p−α+β(n + p − 1)]
for z ∈ U . The inequalities in (8) and (9) are attained for the function f (z)
given by
(p − α)(1 + λp − λ)pΩ
z p+n .
(n + p)Ω (λn + λp + 1 − λ)[n + p − α + β(n + p − 1)]
(10) f (z) = z p −
Proof. Note that
(11)
(n + p)Ω (λn + λp + 1 − λ)[n + p − α + β(n + p − 1)]
∞
X
ak+p
k=n
≤
∞
X
(k + p)Ω (λk + λp + 1 − λ)[k + p − α + β(k + p − 1)]ak+p
k=n
≤ (p − α)(1 + λp − λ)pΩ ,
this last inequality following from Theorem 1. Thus
∞
∞
X
X
|f (z)| ≤ |z|p +
|ak+p ||z|k+p ≤ |z|p + |z|n+p
|ak+p |
k=n
k=n
(p − α)(1 + λp − λ)pΩ
|z|n+p .
≤ |z|p +
(n + p)Ω (λn + λp + 1 − λ)[n + p − α + β(n + p − 1)]
102
S. M. Khairnar, N. H. More
Similarly,
∞
X
p
|f (z)| ≥ |z| −
k+p
|ak+p ||z|
p
≥ |z| − |z|
k=n
≥ |z|p −
n+p
∞
X
|ak+p |
k=n
(p − α)(1 + λp − λ)pΩ
|z|n+p .
(n + p)Ω (λn + λp + 1 − λ)[n + p − α + β(n + p − 1)]
This completes the proof of Theorem 2.
Corollary 3 Let the function f (z) be defined by (1) then f (z) ∈ TΩ (n, p, λ, α).
Then for |z| = r1 ,
(12)
|f (z)| ≥ |z|p −
(p − α)(1 + λp − λ)pΩ
|z|p+n
(n + p)Ω (λn + λp + 1 − λ)(n + p − α)
|f (z)| ≤ |z|p +
(p − α)(1 + λp − λ)pΩ
|z|p+n
(n + p)Ω (λn + λp + 1 − λ)(n + p − α)
and
(13)
for z ∈ U . The inequalities in (8) and (9) are attained for the function f (z)
given by
|f (z)| = |z|p −
(14)
(p − α)(1 + λp − λ)pΩ
|z|p+n
(n + p)Ω−1 (λn + λp + 1 − λ)(n + p − α)
Proof. Taking β = 0 in Theorem 2, we immediately obtain (12) and (13).
Theorem 3 Let the function f (z) ∈ TΩ (n, p, λ, α, β) be defined by (1). Then
for |z| = r < 1,
(15)
|f 0 (z)| ≥ p|z|p−1
−
(p − α)(1 + λp − λ)pΩ
|z|p+n−1
(n + p)Ω−1 (λn + λp + 1 − λ)[n + p − α + β(n + p − 1)]
and
(16)
|f 0 (z)| ≤ p|z|p−1
+
(p − α)(1 + λp − λ)pΩ
|z|p+n−1 .
(n + p)Ω−1 (λn + λp + 1 − λ)[n + p − α + β(n + p − 1)]
On a subclass of certain p-valent starlike functions...
103
Proof. We have
|f 0 (z)| ≤ p|z|p−1 +
(17)
∞
X
(k + p)ak+p |z|k+p−1
k=n
∞
X
n+p−1
≤ p|z|p−1 + |z|
(k + p)ak+p .
k=n
In view of Theorem 1, we have
∞
X
(k+p)Ω [(k+p−α)+β(k+p−1)](λk+λp−λ+1)ak+p ≤ pΩ (p−α)(1+λp−λ)
k=n
and then
(n + p)Ω−1 [(n + p − α) + β(n + p − 1)](λn + λp − λ + 1)
∞
X
(k + p)ak+p
k=n
≤
∞
X
(k + p)Ω [(k + p − α) + β(k + p − 1)](λk + λp − λ + 1)ak+p
k=n
≤ pΩ (p − α)(1 + λp − λ)
or
(18)
∞
X
(k+p)ak+p ≤
k=n
pΩ (p−α)(1+λp−λ)
.
(n+p)Ω−1 [(n+p−α)+β(n+p−1)](λn+λp−λ+1)
A substitution of (18) into (17) yields
|f 0 (z)| ≤ p|z|p−1 +
(p−α)(1+λp−λ)pΩ
|z|p+n−1 .
(n+p)Ω−1 (λn+λp+1−λ)[n+p−α+β(n+p−1)]
On the other hand,
|f 0 (z)| ≥ p|z|p−1 −
∞
X
(k + p)ak+p |z|k+p−1
k=n
∞
X
n+p−1
≥ p|z|p−1 − |z|
(k + p)ak+p
k=n
≥ p|z|p−1 −
(p − α)(1 + λp − λ)pΩ
|z|p+n−1 .
(n + p)Ω−1 (λn + λp + 1 − λ[(n + p − α + β(n + p − 1)]
104
S. M. Khairnar, N. H. More
Corollary 4 Let the function f (z) ∈ TΩ (n, p, λ, α) be defined by (1). Then
for |z| = r < 1,
(19)
|f 0 (z)| ≥ p|z|p−1 −
(p − α)(1 + λp − λ)pΩ
|z|p+n−1
(n + p)Ω−1 (λn + λp + 1 − λ)(n + p − α)
|f 0 (z)| ≤ p|z|p−1 +
(p − α)(1 + λp − λ)pΩ
|z|p+n−1 .
(n + p)Ω−1 (λn + λp + 1 − λ)(n + p − α)
and
(20)
Proof. Taking β = 0 in Theorem 3, we obtain (19) and (20).
4
Closure Theorems and Extreme Points
Let the function fj (z) be defined, for j = 1, 2, · · · , m by
(21)
fj (z) = z p −
∞
X
ak+p,j z k+p (ak+p,j ≥ 0)
k=n
for z ∈ U .
We shall prove that the following results for the closure of functions in the
class TΩ (n, p, λ, α, β).
Theorem 4 Let the functions fj (z) defined by (21) be in the class TΩ (n,p,λ,α,β)
for every j = 1, 2, · · · , m. Then the functions h(z) defined by
h(z) =
m
X
cj fj (z), (cj ≥ 0)
j=1
is also in the class TΩ (n, p, λ, α, β), where
m
X
j=1
cj = 1.
On a subclass of certain p-valent starlike functions...
105
Proof. According to the definition h(z), we can write
"
#
m
∞
X
X
cj z p −
h(z) =
ak+p,j z k+p
j=1

= 
m
X

cj  z p −
j=1
= zp −
k=n
∞
X
k=n


∞
X
k=n
m
X


m
X

cj ak+p,j  z k+p
j=1

cj ak+p,j  z k+p .
j=1
Further, since fj (z) are in TΩ (n, p, λ, α, β) for every j = 1, 2, · · · , m we get
∞
X
(k+p)Ω (λk+λp+1−λ)[k+p−α+β(k+p−1)]ak+p,j ≤ (p−α)(1+λp−λ)pΩ
k=n
for every j = 1, 2, · · · , m. Hence we can see that
∞
X

(k + p)Ω (λk + λp + 1 − λ)[k + p − α + β(k + p − 1)] 
m
X
cj ak+p,j 
j=1
k=n
=

∞
X
(k + p)Ω (λk + λp + 1 − λ)[k + p − α + β(k + p − 1)]
k=n
×(c1 ak+p,1 + c2 ak+p,2 + · · · + cm ak+p,m )
∞
X
(k + p)Ω (λk + λp + 1 − λ)[k + p − α + β(k + p − 1)]ak+p,1
= c1
k=n
+c2
∞
X
(k + p)Ω (λk + λp + 1 − λ)[k + p − α + β(k + p − 1)]ak+p,2
k=n
+ · · · cm
∞
X
(k + p)Ω (λk + λp + 1 − λ)[k + p − α + β(k + p − 1)]ak+p,m
k=n
≤ c1 (p − α)(1 + λp − λ)pΩ + c2 (p − α)(1 + λp − λ)pΩ + · · ·


m
X
+cm (p − α)(1 + λp − λ)pΩ = 
cj  (p − α)(1 + λp − λ)pΩ
j=1
= (p − α)(1 + λp − λ)pΩ
which implies that h(z) ∈ TΩ (n, p, λ, α, β). Thus we have the Theorem.
106
S. M. Khairnar, N. H. More
Theorem 5 TΩ (n, p, λ, α, β) is a convex set.
Proof. Let the function
(22)
fv (z) = z −
∞
X
ak+p,v z k (ak+p,v ≥ 0; v = 1, 2)
k=n
be in the class TΩ (n, p, λ, α, β). It is sufficient to show that the function h(z)
defined by
(23)
H(z) = µf1 (z) + (1 − µ)f2 (z) (0 ≤ µ ≤ 1)
is also in the class TΩ (n, p, λ, α, β). Since, for 0 ≤ µ ≤ 1,
(24)
h(z) = z −
∞
X
[µak+p,1 + (1 − µ)ak+p,2 ]z k
k=n
with the aid of Theorem 1, we have
(25)
∞
X
(k+p)Ω (λk+λp+1−λ)[k+p−α+β(k+p−1)][µak+p,1 +(1−µ)ak+p,2 ]
k=n
=µ
∞
X
(k + p)Ω (λk + λp + 1 − λ)[k + p − α + β(k + p − 1)]ak+p,1
k=n
+(1 − µ)
∞
X
(k + p)Ω (λk + λp + 1 − λ)[k + p − α + β(k + p − 1)]ak+p,2
k=n
≤ µ(p − α)(1 + λp − λ)pΩ + (1 − µ)(p − α)(1 + λp − λ)pΩ
= (p − α)(1 + λp − λ)pΩ
which implies that h(z) ∈ TΩ (n, p, α, β). Hence, TΩ (n, p, λ, α, β) is a convex
set.
As a consequence of Theorem 6 there exists the extreme points of the
TΩ (n, p, λ, α, β).
On a subclass of certain p-valent starlike functions...
107
Theorem 6 Let
fn−1 (z) = z p
(26)
and
fk (z) = z p −
(27)
pΩ (p−α)(1+λp−λ)
zk ,
(n+p)Ω−1 [(n+p−α)+β(n+p−1)](λn+λp−λ+1)
(k ≥ n), for 0 ≤ α < 1 and 0 ≤ λ ≤ 1 and n ∈ N . Then f (z) is in the class
TΩ (n, p, λ, α, β) if and only if it can be expressed in the form
(28)
f (z) =
∞
X
ηk fk (z),
k=n−1
where
(29)
ηk ≥ 0 (k ≥ n − 1) and
∞
X
ηk = 1.
k=n−1
Proof. Assume that
(30)
f (z) =
∞
X
ηk fk (z).
k=n−1
Then
∞
X
f (z) =
ηk fk (z) = ηn−1 fn−1 (z) +
k=n−1
∞
X
p
Ã
=
!
ηk
zp −
(n+p)Ω−1 [(n+p−α)+β(n+p−1)](λn+λp−λ+1)
∞
X
ηk
k=n
k=n−1
pΩ (p−α)(1+λp−λ)
p
k=n
∞
X
ηk fk (z)
k=n
·
ηk z −
= ηn−1 z +
∞
X
pΩ (p−α)(1+λp−λ)
z k+p .
(n+p)Ω−1 [(n+p−α)+β(n+p−1)](λn+λp−λ+1)
Thus
∞
X
ηk
k=n
pΩ (p − α)(1 + λp − λ)
(n + p)Ω [(n + p − α) + β(n + p − 1)](λn + λp − λ + 1)
(n + p)Ω [(n + p − α) + β(n + p − 1)](λn + λp − λ + 1)
pΩ (p − α)(1 + λp − λ)
∞
∞
X
X
=
ηk =
ηk − ηk−1 = 1 − ηn−1 ≤ 1,
×
k=n
¸
z k+p
k=n−1
108
S. M. Khairnar, N. H. More
so, by Theorem 1, f (z) ∈ TΩ (n, p, λ, α, β).
Conversely, assume that the function f (z) defined by (1) belongs to the
class TΩ (n, p, λ, α, β). Then
(31)
ak+p ≤
pΩ (p − α)(1+λp−λ)
, (k ≥ n).
(n + p)Ω [(n+p−α)+β(n+p−1)](λn+λp−λ+1)
Setting
(32)
ηk =
(n + p)Ω [(n + p − α) + β(n + p − 1)](λn + λp − λ + 1)
pΩ (p − α)(1 + λp − λ)
and
(33)
ηn−1 = 1 −
∞
X
ηk .
k=n
Then
p
f (z) = z −
∞
X
ak+p z k+p
k=n
= zp −
∞
X
pΩ (p−α)(1+λp−λ)
ηk z k+p
(k + p)Ω [(k+p−α)+β(k+p−1)](λk+λp−λ + 1)
k=n
∞
X
= zp −
ηk [z p − fk (z)] = z p −
ηk z p +
k=n
k=n
= ηn−1 z p +
∞
X
∞
X
ηk fk (z) = ηn−1 fn−1 (z) +
k=n
∞
X
ηk fk (z)
k=n
∞
X
k=n
ηk fk (z) =
∞
X
ηk fk (z).
k=n−1
This completes the proof.
Theorem 7 Let the function f (z) defined by (1) be in the class TΩ (n, p, λ, α, β).
Then f (z) is close-to-convex of order ρ (0 ≤ ρ < 1) in |z| < r, where
(34)
r = r(n, p, λ, α, β, ρ)
"µ
¶
µ
¶·
¸
(k + p − α) + β(k + p − 1)
k + p Ω−1 p − ρ
= inf
k
p
p−α
(k + p − ρ) + β(k + p − 1)
µ
¶¸ 1
λk + λp − λ + 1 k
, (k ≥ j + 1)
λp − λ + 1
On a subclass of certain p-valent starlike functions...
109
Proof. We must show that
¯ 0
¯
¯ zf (z)
¯
¯
¯
¯ f (z) + 1 − p¯ ≤ p − ρ,
(0 ≤ ρ < 1) for |z| < r. We have
¯ ¯ 0
¯ 0
¯
¯ zf (z)
¯ ¯ zf (z) + (1 − p)f (z) ¯
¯
¯
¯
¯
¯ f (z) + 1 − p¯ = ¯
¯
f (z)
¯
∞
P
¯
¯ p(p − 1)z p−1 −
(k + p)(k + p − 1)ak+p z k+p−1
¯
k=n
= ¯¯
∞
P
¯
(k + p)ak+p z k+p−1
pz p−1 −
¯
k=n
p(p −
1)z p−1
+
−
pz p−1 −
∞
P
¯
(p − 1)(k + p)ak+p
¯
z k+p−1 ¯
k=n
∞
P
(k + p)ak+p z k+p−1
k=n
¯
¯
¯
¯
¯
¯
¯
∞
∞
P
P
¯
¯
¯ −
(k + p)kak+p z k+p−1 ¯
k(k + p)ak+p |z|k
¯
¯
k=n
k=n
¯≤
= ¯¯
.
∞
∞
¯
¯ pz p−1 − P (k + p)ak+p z k+p−1 ¯ p − P (k + p)ak+p z k
¯
¯
k=n
Thus
k=n
¯ 0
¯
¯ zf (z)
¯
¯
¯ ≤ p − ρ,
+
1
−
p
¯ f (z)
¯
if
∞
X
(k + p − ρ)(k + p)
ak+p |z|k ≤ 1.
p(p − ρ)
(35)
k=n
But, by Theorem 1 confirms that
∞
X
(k + pΩ [(k + p − α) + β(k + p − 1)](λk + λp − λ + 1)
ak+p ≤ 1.
pn (p − α)(1 + λp − λ)
k=n
Hence (34) will be true if
(k+p−ρ)(k+p) k (k + p)Ω [(k + p − α)+β(k + p − 1)](λk+λp−λ+1)
|z| ≤
p(p−ρ)
pn (p−α)(1+λp−λ)
110
S. M. Khairnar, N. H. More
that is, if
(36)
·
(k+p)Ω [(k+p−α)+β(k + p − 1)](λk+λp−λ+1)p(p−ρ)
|z| ≤
pn (p−α)(1+λp−λ)(k+p−ρ)(k + p)
¸ k1
, (k ≥ n).
Theorem 7 follows easily from (36).
Theorem 8 Let the function f (z) defined by (1) be in the class Tj (n, m, λ, α, β).
Then f (z) starlike of order ρ (0 ≤ ρ < 1) in |z| < r2 , where
(37) r2 = r2 (n, m, λ, α, β, ρ)
·
¸ 1
(1 − ρ)k n−1 [(k − α) + β(k − 1)](1 + (k m − 1)λ) k−1
= inf
, (k ≥ j + 1).
k
1−α
The result is sharp, with the extremal function f (z) given by (7).
Proof. It is sufficient to show that
¯ 0
¯
¯ zf (z)
¯
¯
¯
¯ f (z) − 1¯ ≤ 1 − ρ
for |z| < r2 (n, m, λ, α, β, ρ), where r2 (n, m, λ, α, β, ρ) is given by (37). Indeed
we find again from the Definition 1, that
¯ 0
¯
¯ zf (z)
¯
¯
¯≤
−
1
¯ f (z)
¯
∞
P
(k − 1)ak |z|k−1
k=j+1
1−
∞
P
k=j+1
Thus
.
ak |z|k−1
¯
¯ 0
¯ zf (z)
¯
¯≤1−ρ
¯
−
1
¯ f (z)
¯
if
(38)
¶
∞ µ
X
k−ρ
ak |z|k−1 ≤ 1.
1−ρ
k=j+1
On a subclass of certain p-valent starlike functions...
111
But, by Theorem 1, (38) will be true if
µ
¶
k−ρ
k n [(k − α) + β(k − 1)](1 + (k m − 1)λ)
|z|k−1 ≤
1−ρ
1−α
that is, if
(39)
·
(1 − ρ)k n−1 [(k − α) + β(k − 1)](1 + (k m − 1)λ)
|z| ≤
(k − ρ)(1 − α)
1
¸ k−1
, (k ≥ j + 1).
Theorem 8 follows easily from (39).
Corollary 5 Let the function f (z) defined by (1) be in the class Tj (n, m, λ, α, β).
Then f (z) is convex of order ρ (0 ≤ ρ < 1) in |z| < r3 , where
(40) r3 = r3 (n, m, λ, α, β, ρ)
¸ 1
·
(1 − ρ)k n−1 [(k − α) + β(k − 1)](1 + (k m − 1)λ) k−1
= inf
, (k ≥ j + 1).
k
(k − ρ)(1 − α)
The result is sharp, with the extremal function f (z) given by (5).
5
Modified Hadmard Products
Let the function f (z) be defined by (1) and function g(z) be defined by
g(z) = z p −
∞
X
bk+p z k+p (bk+p ≥ 0; p ∈ N, n ∈ N )
k=n
be in the same class TΩ (n, p, α, λ, β). We define the modified Hadamard product of the functions f (z) and g(z) is defined by
(41)
p
f ∗ g(z) = z −
∞
X
ak+p bk+p z k+p .
k=j+1
Theorem 9 Let each of the functions f (z) and g(z) be in the class TΩ (n,p,α,λ,β).
Then
f ∗ g(z) ∈ TΩ (n, p, δ, λ, β)
112
S. M. Khairnar, N. H. More
where
(42)
δ≤
p(k+p)Ω [(k+p−α)+β(k+p−1)]2 (λk+λp−λ+1)−pΩ (p−α)(1+λp−λ)[k+p+β(k+p−1)]
(k + p)Ω [(k+p−α)+β(k+p−1)]2 (λk+λp−λ+1)−pΩ (p−α)2 (1+λp−λ)
The result is sharp.
Proof. Employing the technique used earlier by Schild and Silverman, we
need to find the largest δ such that
(43)
∞
X
(k + p)Ω [(k + p − δ) + β(k + p − 1)](λk + λp − λ + 1)
ak+p bk+p ≤ 1.
pΩ (p − δ)(1 + λp − λ)
k=n
Since
(44)
∞
X
(k + p)Ω [(k + p − α) + β(k + p − 1)](λk + λp − λ + 1)
ak+p ≤ 1
pΩ (p − α)(1 + λp − λ)
k=n
and
(45)
∞
X
(k + p)Ω [(k + p − α) + β(k + p − 1)](λk + λp − λ + 1)
bk+p ≤ 1
pΩ (p − α)(1 + λp − λ)
k=n
by the Cauchy-Schwarz inequality, we have
(46)
∞
X
(k+p)Ω [(k+p−α)+β(k+p−1)](λk+λp−λ+1) p
ak+p bk+p ≤ 1.
pΩ (p−α)(1+λp−λ)
k=n
Thus it is sufficient to show that
(k + p)Ω [(kp − α) + β(k + p − 1)](λk + λp − λ + 1)
ak+p bk+p
pΩ (p − α)(1 + λp − λ)
(k + p)Ω [(kp − α) + β(k + p − 1)](λk + λp − λ + 1) p
≤
ak+p bk+p .
pΩ (p − α)(1 + λp − λ)
That is that
(47)
p
ak+p bk+p ≤
[(k + p − α) + β(k + p − 1)](p − δ)
, (k ≥ n).
[(k + p − δ) + β(k + p − 1)](p − α)
On a subclass of certain p-valent starlike functions...
113
Note that
p
(48)
ak+p bk+p ≤
pΩ (p−α)(1+λp−λ)
, (k ≥ n).
(k+p)Ω [(k+p−α)+β(k+p−1)](λk+λp−λ+1)
Consequently, we need only to prove that
pΩ (p − α)(1 − λp − λ)
(k +
+ p − α) + β(k + p − 1)](λk + λp − 1)
[(k + p − α) + β(k + p − 1)](p − δ)
≤
(k ≥ n),
[(k + p − δ) + β(k + p − 1)](p − α)
p)Ω [(k
or, equivalently, that
(49)
δ≤
p(k+p)Ω [(k+p−α)+β(k+p−1)]2 (λk+λp−λ+1)−pΩ (p−α)(1+λp−λ)[k+p+β(k+p−1)]
.
(k+p)Ω [(k+p−α)+β(k+p−1)]2 (λk+λp−λ+1)−pΩ (p−α)2 (1+λp−λ)
Since
(50)
Φ(k) =
p(k+p)Ω [(k+p−α)+β(k+p−1)]2 (λk+λp−λ+1)−pΩ (p−α)(1+λp−λ)[k+p+β(k+p−1)]
(k+p)Ω [(k+p−α)+β(k+p−1)]2 (λk+λp−λ+1)−pΩ (p−α)2 (1+λp−λ)
is an increasing function of k (k ≥ n), letting k = n in (50) we obtain
(51)
δ≤
p(n+p)Ω [(n+p−α)+β(n+p−1)]2 (λn+λp−λ+1)−pΩ (p−α)(1+λp−λ)[nv +p+β(n+p−1)]
(n+p)Ω [(n+p−α)+β(n+p−1)]2 (λn+λp−λ+1)−pΩ (p−α)2 (1+λp−λ)
which proves the pain assertion of Theorem.
6
Integral Operators
Theorem 10 Let the function f (z) defined by (1) be in the class TΩ (n,p,λ,α,β),
and let c be a real number such that c > −p. Then the function F (z) defined
by
(52)
c+p
F (z) =
zc
Z
z
tc−1 f (t)dt, (c > −p)
0
also belongs to the class TΩ (n, p, λ, α, β).
114
S. M. Khairnar, N. H. More
Proof. From the representation (52) of F (z), it follows that
p
F (z) = z −
∞
X
bk+p z k ,
k=n
where
µ
bk =
c+p
c+p+k
¶
ak+p .
Therefore, we have
∞
X
k=n
∞
X
(k + p)Ω [(k + p − α) + β(k + p − 1)](λk + λp − λ + 1)bk+p
µ
Ω
(k+p) [(k+p−α)+β(k+p−1)](λk+λp−λ+1)
k=n
∞
X
≤
c+p
c+p+k
¶
ak+p
(k + p)Ω [(k + p − α) + β(k + p − 1)](λk + λp − λ + 1)ak+p
k=n
Ω
≤ p (p − α)(1 + λp − λ),
since f (z) ∈ TΩ (n, p, λ, α, β). Hence, by Theorem 1, F (z) ∈ TΩ (n, p, λ, α, β),
since F (z) ∈ TΩ (n, p, λ, αβ.
Hence by Theorem 1, F (z) ∈ TΩ (n, p, λ, α, β).
Theorem 11 Let the function f (z) be in the class TΩ (n, p, λ, α, β) and let c
be a real number such that c > −p. Then the function f (z) given by (52) is
univalent in |z| < Rp∗ , where
(µ
(53)
¶
c+p
= inf
k
c+p+k
·
¸µ
¶¾ 1
k
(k + p − α) + β(k + p − 1)
λk + λp − λ + 1
, (k ≥ n).
p−α
λp − λ + 1
Rp∗
The result is sharp.
k+p
p
¶Ω−1 µ
On a subclass of certain p-valent starlike functions...
115
Proof. From (52) we have
z 1−c (z c F (z))0
c+1
¶
∞ µ
X
c+p+k
= z−
ak+p z k+p , (c > −p)
c+p
f (z) =
k=n
in order to obtain the required result, it suffices to show that
¯ 0
¯
¯ f (z)
¯
¯
¯
¯ z p−1 − p¯ ≤ p
whenever |z| < Rp∗ , where Rp∗ is given by (53). Now
¯
¯
³
´
∞
P
¯
¯
c+p+k
k+p−1
¯ ¯ pz p−1 −
¯
¯ 0
(k + p) c+p ak+p z
¯ ¯
¯
¯ f (z)
k=n
¯ ¯
¯
− p¯¯
¯ z p−1 − p¯ = ¯
p−1
z
¯
¯
¯
¯
µ
µ
¶
¶
∞
∞
X
X
c+p+k
c+p+k
(k + p)
(k + p)
−
ak+p z k ≤
ak+p |z|k .
c+p
c+p
k=n
k=n
Thus
¯ 0
¯
¯ f (z)
¯
¯
¯
¯ z p−1 − p¯ ≤ p
if
¶µ
¶
∞ µ
X
k+p
c+p+k
(54)
k=n
p
c+p
ak+p |z|k ≤ 1.
But Theorem 1 confirms that
(55)
∞
X
(k + p)Ω [(k + p − α) + β(k + p − 1)](λk + λp − λ + 1)
ak+p ≤ 1.
pΩ (p − α)(1 + λp − λ)
k=n
Hence, (54) will be satisfied if
µ
¶µ
¶
k+p
c+p+k
|z|k
p
c+p
µ
¶
·
¸µ
¶
k + p Ω−1 (k + p − α) + β(k + p − 1)
λk + λp − λ + 1
≤
(k ≥ n)
p
p−α
∂p − λ + 1
116
S. M. Khairnar, N. H. More
that is if
"µ
(56)
¶
c+p
|z| <
c+p+k
·
¸µ
¶¸ 1
(k + p − α) + β(k + p − 1)
λk + λp − λ + 1 k
, (k ≥ n).
p−α
λp − λ + 1
k+p
p
¶Ω−1 µ
The required result follows from (56). Sharpness of the result follows if we
take
(57)
pΩ (p − α)(1 + λp − λ)
f (z) = z p −
·
(k + p)Ω [(k + p − α + β(k + p − 1)](λk + λp − λ + 1)
µ
¶
c+p+k
z k+p (k ≥ n).
c+p
Theorem 12 Let the function f (z) defined by (1) be in the class TΩ (n,p,λ,α,β).
Then f (z) is close-to-convex of order ρ (0 ≤ ρ < 1) in |z| < r, where
r = r(n, p, λ, α, β, ρ)
"µ
¶
µ
¶·
¸
(k + p − α) + β(k + p − 1)
k + p Ω−1 p − ρ
= inf
k
p
p−α
(k + p − ρ) + β(k + p − 1)
¶¸ 1
µ
λk + λp − λ + 1 k
, (k ≥ +1).
λp − λ + 1
Proof. We must show that
¯
¯ 0
¯
¯ zf (z)
¯
¯
¯ f (z) + 1 − p¯ ≤ p − ρ
(0 ≤ ρ < 1) for |z| < r. We have
¯ ¯ 0
¯
¯ 0
¯ zf (z)
¯ ¯ zf (z) + (1 − p)f 0 (z) ¯
¯ ¯
¯
¯
¯ f (z) + 1 − p¯ = ¯
¯
f (z)
¯
∞
P
¯
¯ p(p − 1)z p−1 −
(k + p)(k + p − 1)ak+p z k+p−1
¯
k=n
= ¯¯
∞
P
¯
p−1 −
(k + p)ak+p z k+p−1
pz
¯
k=n
On a subclass of certain p-valent starlike functions...
p(p −
1)z p−1
+
pz p−1 −
∞
P
k=n
∞
P
k=n
117
¯
(p − 1)(k + p)ak+p
¯
z k+p−1 ¯
(k + p)ak+p z k+p−1
¯
¯
¯
¯
¯
¯
¯
∞
∞
P
P
¯
¯
¯ −
(k + p)kak+p z k+p−1 ¯
k(k + p)ak+p |z|k
¯
¯
k=n
¯ ≤ k=n
= ¯¯
.
∞
∞
¯
P
P
¯ pz p−1 −
¯
k+p−1
k
(k + p)ak+p z
(k + p)ak+p z
¯
¯ p−
k=n
Thus
k=n
¯ 0
¯
¯ zf (z)
¯
¯
¯
¯ f (z) + 1 − p¯ ≤ p − ρ,
if
∞
X
(k + p − ρ)(k + p)
ak+p |z|k ≤ 1.
p(p − ρ)
(58)
k=n
But, by Theorem 1 confirms that
∞
X
(k + p)Ω [(k + p − α) + β(k + p − 1)](λk + λp − λ + 1)
ak+p ≤ 1.
pn (p − α)(1 + λp − λ)
k=n
Hence (58) will be true if
(k+p−ρ)(k+p) k (k+p)Ω [(k+p−α)+β(k+p−1)](λk+λp−λ+1)
|z| ≤
p(p−ρ)
pn (p−α(1+λp−λ)
that is, if
(59)
·
(k+p)Ω [(k+p−α)+β(k+p−1)](λk+λp−λ+1)p(p−ρ)
|z| ≤
pn (p−α)(1+λp−λ)(k+p−ρ)(k+p)
Acknowledgements:
¸ k1
, (k ≥ n).
The first author is thankful and acknowledges the
support from the research projects funded from the Department of Science
and Technology, (Ref. No. SR/S4/MS:544/08), Government of India, National Board of Higher Mathematics, Department of Atomic Energy, (Ref.
No.NBHM/DAE/R.P.2/09), Government of India and BCUD, University of
Pune (UOP), Pune, (Ref No.BCUD, Engg.2009), and UGC New Delhi.
118
S. M. Khairnar, N. H. More
References
[1] S. A. Halim, A. Janteng, M. Darus,Coefficient properties for classes with
negative coefficients and starlike with respect to other points, Proceeding
of 13-th Mathematical Sciences National Symposium, UUM, 2, 2005, 658663.
[2] Aini Janteng, M. Darus, Classes with negative coefficients and starlike
with respect to other points , International Mathematical Forum 2, 2007,
No. 46, 2261-2268.
[3] H. Silverman, Univalent functions with negative coefficients, Proc. Amer.
Math. Soc., 51(1), 1975, 109-116.
[4] J. Stankiewicz, Some remarks on functions starlike with respect to symmetric points, Ann. Univ. Marie Curie Sklodowska, 19(7), 1965, 53-59.
[5] E. K. Sakaguchi, On certain univalent mapping, J. Math. Soc. Japan, 11,
1959, 72-75.
[6] K. Al Shaqsi, M. Darus, On subclass of harmonic starlike functions with
respect to K-symmetric points, International Mathematical Forum, 2,
2007, No. 57, 2799-2805.
[7] K. Al Shaqsi, M. Darus, On subclass of close-to-convex functions, Int.
Jou. Contemp. Math. Sci., 2, 2007, No. 15, 745-757.
[8] M.I.S. Robertson, Applications of the subordination principle to univalent
functions, Pacific J. of Math., 11, 1961, 315-324.
[9] V. Ravichandran, Starlike and convex functions with respect to conjugate
points, Acta Math. Acad. Paedagog, Nyíregyházi, 20, 2004, 31-37.
On a subclass of certain p-valent starlike functions...
119
[10] R. M. El. Ashwah, D. K. Thomas, Some subclasses of close-to-convex
functions, J. Ramanujan Math. Soc., 2, 1987, 86-100.
[11] P. L. Duren, Univalent Functions, Grundlehren Math. Wiss., Vol. 259,
Springer, New York, 1983.
[12] S. Owa, Z. Wu, F. Ren, A note on certain subclass of Sakaguchi functions,
Bull. de la Royale de liege, 57 (3), 1988, 143-150.
[13] T. V. Sudharsan, P. Balasubrahmanayan and K. G. Subramanian, On
functions starlike with respect to symmetric and conjugate points, Taiwanese Journal of Mathematics, 2(1), 1998, 57-68.
[14] Z. Wu, On classes of Sakaguchi functions and Hadamard products, Sci.
Sinica Ser. A, 30, 1987, 128-135. O. Atlantas, Math. Japon., 36, 1991,
489-95.
S. M. Khairnar
Department of Mathematics
Maharashtra Academy of Engineering
Alandi, Pune - 412105, Maharashtra, India
e-mail: smkhairnar2007@gmail.com
N. H. More
Department of Applied Sciences
Rajiv Gandhi Institute of Technology
Andheri (west), Mumbai - 53, Maharashtra, India
e-mail: nhmore@rediffmail.com