Volume 10 (2009), Issue 4, Article 101, 9 pp.
A CONVOLUTION APPROACH ON PARTIAL SUMS OF CERTAIN ANALYTIC
AND UNIVALENT FUNCTIONS
K. K. DIXIT AND SAURABH PORWAL
D EPARTMENT OF M ATHEMATICS
JANTA C OLLEGE , BAKEWAR , E TAWAH
(U.P.) I NDIA -206124
kk.dixit@rediffmail.com
saurabh.840@rediffmail.com
Received 03 May, 2009; accepted 28 September, 2009
Communicated by S.S. Dragomir
o
n
A BSTRACT. In this paper, we determine sharp lower bounds for Re ffn(z)∗ψ(z)
and
(z)∗ψ(z)
o
n
(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].
Key words and phrases: Analytic functions, Univalent functions, Convolution, Partial Sums.
2000 Mathematics Subject Classification. 30C45.
1. I NTRODUCTION
Let A denote the class of functions f of the form
∞
X
(1.1)
f (z) = z +
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)
> α (z ∈ U ),
Re
f (z)
and is said to be convex of order α (0 ≤ α < 1), denoted by K (α), if it satisfies
zf 00 (z)
Re 1 + 0
> α (z ∈ U ) .
f (z)
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)..
121-09
2
K. K. D IXIT AND S AURABH P ORWAL
Let T ∗ (α) and C (α) be subclasses of S ∗ (α) and K (α), respectively, whose functions are
of the form
∞
X
(1.2)
f (z) = z −
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
∞
X
(1.3)
(k − α) |ak | ≤ 1 − α
k=2
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
(1.5)
φ(z) = z +
∞
X
ck z k ,
(ck ≥ c2 > 0, k ≥ 2) .
k=2
Very recently, Frasin [2] defined the class Hφ (ck , δ) consisting of functions f (z), of the form
(1.1) which satisfy the inequality
∞
X
(1.6)
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] andnthe references
o therein).
n
o
(z)∗ψ(z)
f (z)∗ψ(z)
,
In the present paper, we determine sharp lower bounds for Re fn (z)∗ψ(z) and Re ffn(z)∗ψ(z)
where
n
X
fn (z) = z +
ak z k
k=2
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
λk z k ,
(λk ≥ 0)
k=2
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
∞
∞
X
X
k
f (z) = z +
ak z and g(z) = z +
bk z k
k=2
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 101, 9 pp.
k=2
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PARTIAL S UMS
as
(f ∗ g) (z) = f (z) ∗ g (z) = z +
3
∞
X
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.
2. M AIN R ESULTS
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
cn+1
fn (z) ∗ ψ (z)
(2.2)
Re
≥
(z ∈ U ) ,
f (z) ∗ ψ (z)
cn+1 + λn+1 δ
where
(
λk δ
if k = 2, 3, . . . , n,
ck ≥
λk cn+1
if k = n + 1, n + 2, . . .
λn+1
The results (2.1) and (2.2) are sharp with the function given by
δ n+1
(2.3)
f (z) = z +
z ,
cn+1
where 0 < δ ≤ λcn+1
.
n+1
Proof. Define the function ω (z) by
(2.4)
cn+1
f (z) ∗ ψ (z)
cn+1 − δλn+1
1 + ω (z)
−
=
1 − ω (z)
(λn+1 ) δ fn (z) ∗ ψ (z)
cn+1
Pn
cn+1 P∞
k−1
1 + k=2 λk ak z
+ (λn+1 )δ k=n+1 λk ak z k−1
P
.
=
1 + nk=2 λk ak z k−1
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
(λn+1 )δ
P
P∞
ω (z) =
.
n+1
k−1
2 + 2 nk=2 λk ak z k−1 + (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 |
.
Now |ω (z)| ≤ 1 if
∞
n
X
cn+1 X
2
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.
(λn+1 ) δ k=n+1
k=2
n
X
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 101, 9 pp.
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4
K. K. D IXIT AND S AURABH P ORWAL
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
λn+1 δ
k=2
k=n+1
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
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
cn+1 +λn+1 δ
λn+1 δ
Pn
k=2 λk |ak | −
P
∞
k=n+1
λk |ak |
cn+1 −λn+1 δ P∞
λn+1 δ
k=n+1 λk |ak |
≤ 1.
This last inequality is equivalent to
∞
cn+1 X
λk |ak | +
λk |ak | ≤ 1.
(λ
n+1 ) δ
k=2
k=n+1
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).
Taking ψ (z) =
z
1−z
in Theorem 2.1, we obtain the following result given by Frasin in [2].
Corollary 2.2. If f ∈ Hφ (ck , δ), then
f (z)
cn+1 − δ
(2.7)
Re
≥
fn (z)
cn+1
(z ∈ U )
and
Re
(2.8)
fn (z)
f (z)
≥
cn+1
cn+1 + δ
(z ∈ U ) ,
where
(
ck ≥
δ
if k = 2, 3, . . . , n,
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:
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 101, 9 pp.
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PARTIAL S UMS
5
Corollary 2.3. If f ∈ Hφ (ck , δ), then
(2.9)
Re
f 0 (z)
cn+1 − (n + 1) δ
≥
0
fn (z)
cn+1
(z ∈ U )
Re
cn+1
fn0 (z)
≥
0
f (z)
cn+1 + (n + 1) δ
(z ∈ U ) ,
and
(2.10)
where
(
(2.11)
ck ≥
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).
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
kδ 1 + cn+1
if k = n + 1, n + 2, . . .
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
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
Taking ψ (z) = 1−z
, ck = [(1+β)k−(α+β)]
, where λ ≥ 0, β ≥ 0, −1 ≤ α < 1 and
1−α
k
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, where
[(1+β)k−(α+β)] k+λ−1
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
and
(2.14)
Re
fn (z)
f (z)
≥
cn+1
cn+1 + 1
(z ∈ U ) .
The results (2.13) and (2.14) are sharp with the function given by
f (z) = z +
(2.15)
1
cn+1
z n+1 .
Taking
z
,
ψ (z) =
(1 − z)2
[(1 + β) k − (α + β)]
ck =
1−α
k+λ−1
k
,
where λ ≥ 0, β ≥ 0, −1 ≤ α < 1 and δ = 1 in Theorem 2.1, we obtain
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 101, 9 pp.
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6
K. K. D IXIT AND S AURABH P ORWAL
Corollary 2.5. If f is of the form (1.1) and satisfies the condition
∞
X
ck |ak | ≤ 1,
k=2
where
[(1 + β) k − (α + β)]
ck =
1−α
f 0 (z)
fn0 (z)
k+λ−1
k
,
(λ ≥ 0, β ≥ 0, −1 ≤ α < 1),
then
Re
(2.16)
≥
cn+1 − (n + 1)
cn+1
(z ∈ U )
≥
cn+1
cn+1 + (n + 1)
(z ∈ U ) ,
and
Re
(2.17)
fn0 (z)
f 0 (z)
where
(
ck ≥
(2.18)
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
|ak | ≥ 0
(ck − k) |ak | +
ck −
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].
Corollary 2.6. If f is of the form (1.1) and satisfies the condition
∞
X
(λk − αµk ) |ak | ≤ 1 − α,
k=2
then
(2.19)
Re
f (z)
fn (z)
≥
λn+1 − αµn+1 − 1 + α
λn+1 − αµn+1
(z ∈ U )
≥
λn+1 − αµn+1
λn+1 − αµn+1 + 1 − α
(z ∈ U ) ,
and
(2.20)
Re
fn (z)
f (z)
where
(
λk − αµk ≥
1−α
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
1−α
(2.21)
f (z) = z +
z n+1 .
λn+1 − αµn+1
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 101, 9 pp.
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PARTIAL S UMS
7
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:
Corollary 2.7. If f is of the form (1.1) and satisfies the condition
∞
X
(λk − αµk ) |ak | ≤ 1 − α,
k=2
then
(2.22)
Re
f 0 (z)
fn0 (z)
≥
λn+1 − αµn+1 − (n + 1) (1 − α)
λn+1 − αµn+1
(z ∈ U )
≥
λn+1 − αµn+1
λn+1 − αµn+1 + (n + 1) (1 − α)
(z ∈ U ) ,
and
(2.23)
Re
fn0 (z)
f 0 (z)
where
(
(2.24)
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).
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
n X
λk − αµk
− k |ak |
(2.26)
1−α
k=2
∞ X
λk − αµk
λn+1 − αµn+1
+
− 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.
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
ψ (z) =
{(1 + β) k − (α + β)} µk
,
1−α
and hµk i∞
k=2 is a nondecreasing sequence such that
1−α
1−α
µ2 ≥
0<
< 1, −1 ≤ α < 1, β ≥ 0 ,
2+β−α
2+β−α
ck =
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 101, 9 pp.
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8
K. K. D IXIT AND S AURABH P ORWAL
then
Re
(2.27)
f (z)
fn (z)
≥
cn+1 − 1
cn+1
(z ∈ U )
≥
cn+1
cn+1 + 1
(z ∈ U ) .
and
Re
(2.28)
fn (z)
f (z)
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, β ≥ 0, µk ≥ 0
2 , ck =
1−α
(∀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 |ak | ≤ 1,
k=2
where
{(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, −1 ≤ α < 1, β ≥ 0 .
2+β−α
2+β−α
ck =
Then
(2.30)
Re
f 0 (z)
fn0 (z)
≥
cn+1 − (n + 1)
cn+1
(z ∈ U )
≥
cn+1
cn+1 + (n + 1)
(z ∈ U ) ,
and
(2.31)
Re
fn0 (z)
f 0 (z)
where
(
(2.32)
ck ≥
k
if k = 2, 3, . . . , n,
kcn+1
n+1
if k = n + 1, n + 2, . . .
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].
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 101, 9 pp.
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PARTIAL S UMS
9
R EFERENCES
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[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
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4(4) (2003), Art. 64. [ONLINE: http://jipam.vu.edu.au/article.php?sid=305]
[5] H. SILVERMAN, Partial sums of starlike and convex functions, J. Math. Anal. Appl., 209 (1997),
221–227.
[6] H. SILVERMAN, Univalent functions with negative coefficients, Proc. Amer. Math. Soc., 51 (1975),
109–116.
J. Inequal. Pure and Appl. Math., 10(4) (2009), Art. 101, 9 pp.
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