A CONVOLUTION APPROACH ON PARTIAL SUMS
OF CERTAIN ANALYTIC AND UNIVALENT
FUNCTIONS
Partial Sums
K. K. Dixit and Saurabh Porwal
K. K. DIXIT AND SAURABH PORWAL
vol. 10, iss. 4, art. 101, 2009
Department of Mathematics
Janta College, Bakewar, Etawah
(U.P.) India-206124
EMail: kk.dixit@rediffmail.com
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saurabh.840@rediffmail.com
Contents
Received:
03 May, 2009
Accepted:
28 September, 2009
Communicated by:
JJ
II
S.S. Dragomir
J
I
2000 AMS Sub. Class.:
30C45.
Page 1 of 17
Key words:
Analytic functions, Univalent functions, Convolution, Partial Sums.
Abstract:
Acknowledgements:
n
f (z)∗ψ(z)
fn (z)∗ψ(z)
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o
In this paper, we determine sharp lower bounds for Re
and
n
o
(z)∗ψ(z)
Re ffn(z)∗ψ(z)
. We extend the results of ([1] – [5]) and correct the conditions for the results of Frasin [2, Theorem 2.7], [1, Theorem 2], Rosy et al. [4,
Theorems 4.2 and 4.3], as well as Raina and Bansal [3, Theorem 6.2].
The authors are thankful to the referee for his valuable comments and suggestions.
The present investigation was supported by the University grant commission under grant No. F- 11-12/2006(SA-I).
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Contents
1
Introduction
3
2
Main Results
6
Partial Sums
K. K. Dixit and Saurabh Porwal
vol. 10, iss. 4, art. 101, 2009
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1.
Introduction
Let A denote the class of functions f of the form
(1.1)
f (z) = z +
∞
X
ak z k ,
k=2
which are analytic in the open unit disc U = {z : |z| < 1}. Further, by S we shall
denote the class of all functions in A which are univalent in U . A function f (z) in
S is said to be starlike of order α (0 ≤ α < 1), denoted by S ∗ (α), if it satisfies
0 zf (z)
Re
> α (z ∈ U ),
f (z)
and is said to be convex of order α (0 ≤ α < 1), denoted by K (α), if it satisfies
zf 00 (z)
> α (z ∈ U ) .
Re 1 + 0
f (z)
Let T ∗ (α) and C (α) be subclasses of S ∗ (α) and K (α), respectively, whose
functions are of the form
(1.2)
f (z) = z −
∞
X
ak z k ,
ak ≥ 0.
k=2
A sufficient condition for a function of the form (1.1) to be in S ∗ (α) is that
(1.3)
∞
X
k=2
(k − α) |ak | ≤ 1 − α
Partial Sums
K. K. Dixit and Saurabh Porwal
vol. 10, iss. 4, art. 101, 2009
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and to be in K (α) is that
(1.4)
∞
X
k (k − α) |ak | ≤ 1 − α.
k=2
For functions of the form (1.2), Silverman [6] proved that the above sufficient
conditions are also necessary.
Let φ(z) ∈ S be a fixed function of the form
Partial Sums
K. K. Dixit and Saurabh Porwal
(1.5)
φ(z) = z +
∞
X
ck z k ,
vol. 10, iss. 4, art. 101, 2009
(ck ≥ c2 > 0, k ≥ 2) .
k=2
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Very recently, Frasin [2] defined the class Hφ (ck , δ) consisting of functions f (z),
of the form (1.1) which satisfy the inequality
(1.6)
∞
X
ck |ak | ≤ δ,
k=2
where δ > 0.
He shows that for suitable choices of ck and δ, Hφ (ck , δ) reduces to various
known subclasses of S studied by various authors (for a detailed study, see [2] and
the references therein).
n
o
In the present paper, we determine sharp lower bounds for Re ffn(z)∗ψ(z)
and
(z)∗ψ(z)
n
o
(z)∗ψ(z)
Re ffn(z)∗ψ(z)
, where
n
X
fn (z) = z +
ak z k
k=2
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is a sequence of partial sums of a function
f (z) = z +
∞
X
ak z k
k=2
belonging to the class Hφ (ck , δ) and
ψ(z) = z +
∞
X
Partial Sums
λk z k ,
(λk ≥ 0)
K. K. Dixit and Saurabh Porwal
k=2
vol. 10, iss. 4, art. 101, 2009
is analytic in open unit disc U and the operator “*” stands for the Hadamard product
or convolution of two power series, which is defined for two functions f, g ∈ A,
where f (z) and g (z) are of the form
f (z) = z +
∞
X
ak z k
and
g(z) = z +
k=2
as
(f ∗ g) (z) = f (z) ∗ g (z) = z +
∞
X
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bk z k
k=2
∞
X
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ak b k z k .
k=2
In this paper, we extend the results of Silverman [5], Frasin ([1], [2]) Rosy et al. [4]
as well as Raina and Bansal [3] and we point out that some conditions on the results
of Frasin ([2, Theorem 2.7], [1, Theorem 2]), Rosy et al. ([4, Theorem 4.2, 4.3]),
Raina and Bansal ([3, Theorem 6.2]) are incorrect and we correct them. It is seen
that this study not only gives a particular case of the results ([1] – [5]) but also gives
rise to several new results.
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2.
Main Results
P
k
Theorem 2.1. If f ∈ Hφ (ck , δ) and ψ (z) = z + ∞
k=2 λk z , λk ≥ 0, then
f (z) ∗ ψ (z)
cn+1 − λn+1 δ
(2.1)
Re
≥
(z ∈ U )
fn (z) ∗ ψ (z)
cn+1
and
Partial Sums
(2.2)
Re
fn (z) ∗ ψ (z)
f (z) ∗ ψ (z)
≥
cn+1
cn+1 + λn+1 δ
K. K. Dixit and Saurabh Porwal
(z ∈ U ) ,
where
(
ck ≥
λk δ
if k = 2, 3, . . . , n,
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λk cn+1
λn+1
if k = n + 1, n + 2, . . .
Contents
The results (2.1) and (2.2) are sharp with the function given by
f (z) = z +
(2.3)
where 0 < δ ≤
vol. 10, iss. 4, art. 101, 2009
δ
cn+1
z
n+1
,
cn+1
.
λn+1
Proof. Define the function ω (z) by
1 + ω (z)
cn+1
f (z) ∗ ψ (z)
cn+1 − δλn+1
(2.4)
=
−
1 − ω (z)
(λn+1 ) δ fn (z) ∗ ψ (z)
cn+1
Pn
P
∞
c
n+1
1 + k=2 λk ak z k−1 + (λn+1 )δ k=n+1 λk ak z k−1
P
.
=
1 + nk=2 λk ak z k−1
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It suffices to show that |ω (z)| ≤ 1. Now, from (2.4) we can write
cn+1 P∞
k−1
k=n+1 λk ak z
Pn (λn+1 )δk−1
P
.
ω (z) =
∞
n+1
k−1
2 + 2 k=2 λk ak z
+ (λcn+1
k=n+1 λk ak z
)δ
Hence we obtain
|ω (z)| ≤
cn+1
(λn+1 )δ
2−2
P∞
k=n+1 λk |ak |
cn+1 P∞
k=2 λk |ak | − (λn+1 )δ
k=n+1
Pn
λk |ak |
.
Partial Sums
K. K. Dixit and Saurabh Porwal
vol. 10, iss. 4, art. 101, 2009
Now |ω (z)| ≤ 1 if
2
∞
n
X
cn+1 X
λk |ak | ≤ 2 − 2
λk |ak |
(λn+1 ) δ k=n+1
k=2
or, equivalently,
(2.5)
∞
cn+1 X
λk |ak | +
λk |ak | ≤ 1.
(λ
n+1 ) δ
k=2
k=n+1
n
X
P
ck
It suffices to show that the L.H.S. of (2.5) is bounded above by ∞
k=2 δ |ak |, which
is equivalent to
n ∞ X
X
ck − δλk
λn+1 ck − cn+1 λk
(2.6)
|ak | +
|ak | ≥ 0.
δ
λ
n+1 δ
k=2
k=n+1
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To see that the function given by (2.3) gives a sharp result we observe that for
z = reiπ/n
f (z) ∗ ψ (z)
δ
δ
=1+
λn+1 z n → 1 −
λn+1
fn (z) ∗ ψ (z)
cn+1
cn+1
cn+1 − δλn+1
=
cn+1
Partial Sums
−
when r → 1 .
To prove the second part of this theorem, we write
1 + ω (z)
cn+1 + λn+1 δ fn (z) ∗ ψ (z)
cn+1
=
−
1 − ω (z)
λn+1 δ
f (z) ∗ ψ (z)
cn+1 + λn+1 δ
Pn
cn+1 P∞
k−1
1 + k=2 λk ak z
− λn+1 δ k=n+1 λk ak z k−1
P
,
=
k−1
1+ ∞
k=2 λk ak z
where
|ω (z)| ≤
2−2
Pn
cn+1 +λn+1 δ
λn+1 δ
k=2
λk |ak | −
P
∞
k=n+1
λk |ak |
cn+1 −λn+1 δ P∞
λn+1 δ
k=n+1
K. K. Dixit and Saurabh Porwal
vol. 10, iss. 4, art. 101, 2009
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Page 8 of 17
λk |ak |
≤ 1.
This last inequality is equivalent to
∞
cn+1 X
λk |ak | +
λk |ak | ≤ 1.
(λn+1 ) δ k=n+1
k=2
n
X
Making use of (1.6), we get (2.6). Finally, equality holds in (2.2) for the function
f (z) given by (2.3).
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z
1−z
Taking ψ (z) =
Frasin in [2].
in Theorem 2.1, we obtain the following result given by
Corollary 2.2. If f ∈ Hφ (ck , δ), then
f (z)
cn+1 − δ
(z ∈ U )
(2.7)
Re
≥
fn (z)
cn+1
and
fn (z)
cn+1
(2.8)
Re
≥
(z ∈ U ) ,
f (z)
cn+1 + δ
(
where
δ
if k = 2, 3, . . . , n,
ck ≥
cn+1 if k = n + 1, n + 2, . . .
The results (2.7) and (2.8) are sharp with the function given by (2.3).
If we put ψ (z) =
z
(1−z)2
in Theorem 2.1, we obtain:
Corollary 2.3. If f ∈ Hφ (ck , δ), then
(2.9)
Re
cn+1 − (n + 1) δ
f 0 (z)
≥
0
fn (z)
cn+1
(z ∈ U )
Re
cn+1
fn0 (z)
≥
0
f (z)
cn+1 + (n + 1) δ
(z ∈ U ) ,
where
(
(2.11)
ck ≥
K. K. Dixit and Saurabh Porwal
vol. 10, iss. 4, art. 101, 2009
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and
(2.10)
Partial Sums
kδ
if k = 2, 3, . . . , n,
kcn+1
n+1
if k = n + 1, n + 2, . . .
The results (2.9) and (2.10) are sharp with the function given by (2.3).
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Remark 1. Frasin has shown in Theorem 2.7 of [2] that for f ∈ Hφ (ck , δ), inequalities (2.9) and (2.10) hold with the condition
(
kδ
if k = 2, 3, . . . , n,
(2.12)
ck ≥
n+1
if k = n + 1, n + 2, . . .
kδ 1 + cn+1
However, it can be easily seen that the condition (2.12) for k = n + 1 gives
cn+1
cn+1 ≥ (n + 1) δ 1 +
(n + 1) δ
Partial Sums
K. K. Dixit and Saurabh Porwal
vol. 10, iss. 4, art. 101, 2009
or, equivalently δ ≤ 0, which contradicts the initial assumption δ > 0. So Theorem
2.7 of [2] does not seem suitable with the condition (2.12), but our condition (2.11)
remedies this problem.
k+λ−1
z
, where λ ≥ 0, β ≥ 0, −1 ≤
Taking ψ (z) = 1−z
, ck = [(1+β)k−(α+β)]
1−α
k
α < 1 and δ = 1 in Theorem 2.1, we obtain the following result given by Rosy et al.
in [4].
P
Corollary 2.4. If f is of the form (1.1) and satisfies the condition ∞
k=2 ck |ak | ≤ 1,
[(1+β)k−(α+β)] k+λ−1
where ck =
, λ ≥ 0, β ≥ 0, −1 ≤ α < 1, then
1−α
k
f (z)
cn+1 − 1
(2.13)
Re
≥
(z ∈ U )
fn (z)
cn+1
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and
(2.14)
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Re
fn (z)
f (z)
≥
cn+1
cn+1 + 1
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(z ∈ U ) .
The results (2.13) and (2.14) are sharp with the function given by
1 n+1
(2.15)
f (z) = z +
z .
cn+1
Taking
[(1 + β) k − (α + β)]
ck =
1−α
z
ψ (z) =
,
(1 − z)2
k+λ−1
k
,
where λ ≥ 0, β ≥ 0, −1 ≤ α < 1 and δ = 1 in Theorem 2.1, we obtain
Corollary 2.5. If f is of the form (1.1) and satisfies the condition
Partial Sums
∞
X
K. K. Dixit and Saurabh Porwal
ck |ak | ≤ 1,
vol. 10, iss. 4, art. 101, 2009
k=2
where
[(1 + β) k − (α + β)]
ck =
1−α
k+λ−1
k
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,
(λ ≥ 0, β ≥ 0, − 1 ≤ α < 1),
then
(2.16)
Re
f 0 (z)
fn0 (z)
≥
cn+1 − (n + 1)
cn+1
(z ∈ U )
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Re
fn0 (z)
f 0 (z)
≥
cn+1
cn+1 + (n + 1)
(z ∈ U ) ,
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where
(
(2.18)
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Page 11 of 17
and
(2.17)
Contents
ck ≥
k
if k = 2, 3, . . . , n,
kcn+1
n+1
if k = n + 1, n + 2, . . .
The results (2.16) and (2.17) are sharp with the function given by (2.15).
Remark 2. Rosy et al. has obtained inequalities (2.16) & (2.17) in Theorem 4.2 &
4.3 of [4] without any restriction on ck . However, when we critically observe the
proof of Theorem 4.2 we find that inequality (4.16) of [4, Theorem 4.2]
n
∞ X
X
cn+1 k
(ck − k) |ak | +
ck −
|ak | ≥ 0
n
+
1
k=2
k=n+1
cannot hold if condition (2.18) does not occur. So Theorems 4.2 & 4.3 of [4] are not
proper and proper results are mentioned in Corollary 2.5.
z
Taking ψ (z) = 1−z
, ck = λk − αµk , δ = 1 − α, where 0 ≤ α < 1, λk ≥ 0,
µk ≥ 0, and λk ≥ µk (k ≥ 2) in Theorem 2.1, we obtain the following result given
by Frasin in [1].
Partial Sums
K. K. Dixit and Saurabh Porwal
vol. 10, iss. 4, art. 101, 2009
Title Page
Corollary 2.6. If f is of the form (1.1) and satisfies the condition
∞
X
(λk − αµk ) |ak | ≤ 1 − α,
k=2
then
Re
f (z)
fn (z)
λn+1 − αµn+1 − 1 + α
≥
λn+1 − αµn+1
(z ∈ U )
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and
(2.20)
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(2.19)
Contents
Re
fn (z)
f (z)
where
≥
(
λk − αµk ≥
λn+1 − αµn+1
λn+1 − αµn+1 + 1 − α
1−α
Close
(z ∈ U ) ,
if k = 2, 3, . . . , n,
λn+1 − αµn+1 if k = n + 1, n + 2, . . .
The results (2.19) and (2.20) are sharp with the function given by
f (z) = z +
(2.21)
1−α
z n+1 .
λn+1 − αµn+1
z
Taking ψ (z) = (1−z)
2 , ck = λk − αµk , δ = 1 − α where 0 ≤ α < 1, λk ≥ 0,
µk ≥ 0, and λk ≥ µk (k ≥ 2) in Theorem 2.1, we obtain:
Partial Sums
Corollary 2.7. If f is of the form (1.1) and satisfies the condition
∞
X
K. K. Dixit and Saurabh Porwal
vol. 10, iss. 4, art. 101, 2009
(λk − αµk ) |ak | ≤ 1 − α,
k=2
Title Page
then
(2.22)
Re
f 0 (z)
fn0 (z)
Contents
λn+1 − αµn+1 − (n + 1) (1 − α)
≥
λn+1 − αµn+1
(z ∈ U )
and
(2.23)
Re
fn0 (z)
f 0 (z)
≥
λn+1 − αµn+1
λn+1 − αµn+1 + (n + 1) (1 − α)
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(z ∈ U ) ,
where
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(
(2.24)
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λk − αµk ≥
k (1 − α)
if k = 2, 3, . . . , n,
k(λn+1 −αµn+1 )
n+1
if k = n + 1, n + 2, . . .
The results (2.22) and (2.23) are sharp with the function given by (2.21).
Close
Remark 3. Frasin has obtained inequalities (2.22) & (2.23) in Theorem 2 of [1] under
the condition
(
k (1 − α)
if k = 2, 3, . . . , n,
(2.25) λk+1 − αµk+1 ≥
−αµn+1 )
k (1 − α) + k(λn+1n+1
if k = n + 1, n + 2, . . .
However, when we critically observe the proof of Theorem 2 of [1], we find that
the last inequality of this theorem
(2.26)
n X
k=2
λk − αµk
− k |ak |
1−α
∞ X
λn+1 − αµn+1
λk − αµk
+
− 1+
k |ak | ≥ 0
1−α
(n + 1) (1 − α)
k=n+1
cannot hold for the function given by (2.21) for supporting the sharpness of the
results (2.22) & (2.23). So condition 2.25 of Theorem 2 in [1] is incorrect and the
corrected results are mentioned in Corollary 2.7.
Partial Sums
K. K. Dixit and Saurabh Porwal
vol. 10, iss. 4, art. 101, 2009
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Taking
z
{(1 + β) k − (α + β)} µk
,
ck =
1−z
1−α
and δ = 1, where −1 ≤ α < 1, β ≥ 0, µk ≥ 0 (∀k ∈ N \ {1}) in Theorem 2.1, we
obtain the following result given by Raina and Bansal in [3].
P
Corollary 2.8. If f is of the form (1.2) and satisfies the condition ∞
k=2 ck |ak | ≤ 1,
where
{(1 + β) k − (α + β)} µk
,
ck =
1−α
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ψ (z) =
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and hµk i∞
k=2 is a nondecreasing sequence such that
1−α
1−α
µ2 ≥
< 1, −1 ≤ α < 1, β ≥ 0 ,
0<
2+β−α
2+β−α
then
(2.27)
Re
f (z)
fn (z)
≥
cn+1 − 1
cn+1
(z ∈ U )
K. K. Dixit and Saurabh Porwal
and
vol. 10, iss. 4, art. 101, 2009
(2.28)
Partial Sums
Re
fn (z)
f (z)
≥
cn+1
cn+1 + 1
(z ∈ U ) .
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The results (2.27) and (2.28) are sharp with the function given by
f (z) = z −
(2.29)
1
cn+1
z n+1 .
{(1+β)k−(α+β)}µk
z
Taking ψ (z) = (1−z)
and δ = 1, where −1 ≤ α < 1,
2 , ck =
1−α
β ≥ 0, µk ≥ 0 (∀k ∈ N \ {1}) in Theorem 2.1, we obtain the following result given
by Raina and Bansal in [3].
Corollary 2.9. If f is of the form (1.2) and satisfies the condition
∞
X
ck =
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ck |ak | ≤ 1,
k=2
where
Contents
{(1 + β) k − (α + β)} µk
,
1−α
and hµk i∞
k=2 is a nondecreasing sequence such that
2 (1 − α)
1−α
µ2 ≥
0<
< 1,
2+β−α
2+β−α
−1 ≤ α < 1, β ≥ 0 .
Then
(2.30)
Re
f 0 (z)
fn0 (z)
≥
cn+1 − (n + 1)
cn+1
(z ∈ U )
K. K. Dixit and Saurabh Porwal
vol. 10, iss. 4, art. 101, 2009
and
(2.31)
Partial Sums
Re
fn0 (z)
f 0 (z)
≥
cn+1
cn+1 + (n + 1)
(z ∈ U ) ,
Title Page
Contents
where
(
(2.32)
ck ≥
k
if k = 2, 3, . . . , n,
JJ
II
kcn+1
n+1
if k = n + 1, n + 2, . . .
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The results (2.30) and (2.31) are sharp with the function given by (2.29).
Remark 4. Raina and Bansal [3] have obtained inequalities (2.30) & (2.31) in Theorem 6.2 of [3] without any restriction on ck . However, we easily see that condition
(2.32) is must.
z
Remark 5. Taking ψ (z) = 1−z
, ck = (k − α), ck = k (k − α), δ = 1−α, 0 ≤ α < 1
in Theorem 2.1, we obtain Theorems 1-3 given by Silverman in [5].
z
Remark 6. Taking ψ (z) = (1−z)
2 , ck = (k − α), ck = k (k − α) , δ = 1 − α,
0 ≤ α < 1 in Theorem 2.1, we obtain Theorems 4-5 given by Silverman in [5].
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References
[1] B.A. FRASIN, Partial sums of certain analytic and univalent functions, Acta
Math. Acad. Paed. Nyir., 21 (2005), 135–145.
[2] B.A. FRASIN, Generalization of partial sums of certain analytic and univalent
functions, Appl. Math. Lett., 21(7) (2008), 735–741.
[3] R.K. RAINA AND D. BANSAL, Some properties of a new class of analytic
functions defined in terms of a Hadamard product, J. Inequal. Pure Appl. Math.,
9(1) (2008), Art. 22. [ONLINE: http://jipam.vu.edu.au/article.
php?sid=957]
[4] T. ROSY, K.G. SUBRAMANIAN AND G. MURUGUSUNDARAMOORTHY,
Neighbourhoods and partial sums of starlike functions based on Ruscheweyh
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Partial Sums
K. K. Dixit and Saurabh Porwal
vol. 10, iss. 4, art. 101, 2009
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