THIRD HANKEL DETERMINANT FOR A SUBCLASS OF
ALPHA CONVEX FUNCTIONS
Gagandeep Singh
Department of Mathematics, M.S.K. Girls College, Bharowal(Tarn-Taran), Punjab,
India
Gurcharanjit Singh
Department of Mathematics, Guru Nanak Dev University College,Chungh(Tarn-Taran), Punjab, India
Abstract
:
In this paper a sharp upper bound of third Hankel determinantH
3
1
for the functions belonging to a subclass of alpha convex functions is established. By giving the particular values to alpha, it is easy to obtain the upper bound of
1
3
H
for starlike and convex functions.Mathematics Subject classification
:
30C45Keywords
:
Analytic functions, Starlike functions, Convex functions, Alpha convex functions, Third Hankel determinant.1.
Introduction
Let A be the class of analytic functions of the form
2
k k k
z
a
z
z
f
(1.1)By S we denote the class of functions
f
z
A
and univalent in E.
M
denote the class of functionsf
z
A
and satisfying the condition
0
,
0
1
,
.
1
Re
2
E
z
z
f
z
z
f
z
f
z
z
f
z
(1.2)
The class
M
is the subclass of alpha-convex functions studied by Singh [11]. AlsoM
0
S
, the class of starlike functions andM
1
K
, the class of convex functions.For the complex sequence
a
n,
a
n1,
a
n2,...,
the Hankel matrix, named after Herman Hankel(1839-1873), is the infinite matrix whose
i
,
j
th entrya
ij is defined by
a
ij
a
nij2
i
,
j
,
n
N
.
The qth Hankel matrix (q ϵ N \ {1} ) is by definition, the following q x q square sub matrix:
.
...
...
...
...
...
...
...
...
...
...
2 2 1
1
1 1
q n q
n n
q n n
n
a
a
a
a
a
a
We observe that the Hankel matrix has constant positive slopping diagonals whose entries also satisfy:
a
ij
a
i1,j1
i
N
\
1
;
j
N
.
This also describes the Hankel matrix without reference to a particular sequence. The determinant of the qth Hankel matrix,
usually denoted by
,
...
...
...
...
...
...
...
...
...
...
2 2 1
1
1 1
q n q
n n
q n n
n
q
a
a
a
a
a
a
n
H
is called the qthHankel determinant. In the particular cases
H
2
1
a
3
a
22 andH
2
2
a
2a
4
a
32.
In this paper, we consider the Hankel determinant in the case q = 3 and n = 1,
1
.
5 4 3
4 3 2
3 2 1 3
a
a
a
a
a
a
a
a
a
H
2
2
H
andH
3
1
are respectively called second and third Hankel determinants.For
f
S
,
a
1
1
so that,
2
2 3 5 3 2 4 4 2 3 4 2 3
3
1
a
a
a
a
a
a
a
a
a
a
a
H
and by using the triangle inequality, we have
1
3 2 4 32 4 2 3 4 5 3 22.
3
a
a
a
a
a
a
a
a
a
a
a
H
(1.3)
Second Hankel determinant for various classes has been extensively studied by various authors including Singh[11,12],
Mehrok and Singh[8] and Janteng et al.[3,4,5]. But Third Hankel determinant has been studied by some of the Researchers
including Babalola [1] and Shanmugam et al.[10].
For our discussion in this paper, we consider the third Hankel determinant and obtain an upper bound to the functional
1
3
H
for the functions in the classM
. Results due to Babalola [1] follows as special cases.
2.
Preliminary Results
Let P be the family of all functions p analytic in E for which
Re
p
z
0
and
p
z
1
p
1z
p
2z
2
...
forz
E
.
Lemma 2.1.[9] If
p
P , thenp
k
2
k
1
,
2
,
3
,...
.Lemma 2.2.[6,7] If
p
P , then
2
4
12
,
21
2
p
p
x
p
for some x and z satisfying
x
1
,
z
1
andp
1
0
,
2
. Lemma 2.3.[2] Ifp
P , then
.
2
1
2
,
2
0
2
,
0
1
2
2
2 1 2
if
if
if
p
p
Lemma 2.4.[11] If
f
z
M
, then
.
3
1
1
1
2 3 4
2
a
a
a
3.
Main Results
Theorem 3.1 If
f
M
, then
,
1
2
2
a
,
2
1
3
3
a
1
3
4
4
a
and
.
4
1
5
5
a
Proof. Since
f
z
M
, then there existsp
P
such that
.
1
2
z
p
z
f
z
z
f
z
f
z
z
f
z
(3.1)
Equating coefficients in (3.1) yields
,
1
1 2
p
,
2
1
2
2
1
2
2 1 23
p
p
a
(3.3)
,
3
1
6
3
1
2
3
1
3
3 1 2 1 34
p
p
p
p
a
(3.4)
4
1
4
4
1
4
8
1
4
3
1
4
24
1
4
4 1 3 1 2 2 2 2 1 45
p
p
p
p
p
p
p
a
(3.5)and the results follow by triangle inequality and using Lemma 2.1.
Theorem 3.2 If
f
M
, then
.
.
1
0
3
1
3
2
0
2
4 3 2
if
if
a
a
a
Proof. From equations (3.2),(3.3) and (3.4), we obtain
1
1
2
1
3
3
1
3
.
6
2
6
2
3
1
2
1
1
2
2
3 31 2 2 1 2 4 3
2
a
p
p
p
p
a
a
(3.6)
Substituting for
p
2 andp
3 from Lemma 2.2 and lettingp
1
p
, we get
.
3
1
6
1
4
3
1
12
4
3
1
2
1
1
6
4
5
3
1
3
1
2
1
1
4
4
3
1
2 3 2 2 2 2 2 24 3
2
a
p
px
p
px
p
p
x
z
a
a
Since
p
p
1
2
by using Lemma 2.1, we may assume thatp
0
,
2
. Then using triangle inequality andz
1
withx
, we obtain
3
1
12
4
2
3
1
6
4
3
1
2
1
1
6
4
5
3
1
3
1
2
1
1
4
4
3
1
2 3 2 2 2 2 24 3 2
a
p
p
p
p
p
p
a
a
F
.
Then
.
3
1
6
4
2
3
1
2
1
1
6
4
5
3
1
2 2 2
p
p
p
p
Note that
F
F
1
0
.
Then there existsp
0
,
2
such thatF
0
forp
(
p
,
2
]
andF
0
otherwise. Then for
p
(
p
,
2
]
,F
F
1
.
But
3
1
2
1
1
2
3
1
2
1
1
4
3
1
1
3 2 2
p
p
F
G
p
.
If
0
,
we haveG
p
p
2
.
Otherwise
G
p
is maximum at 2 26
4
3
1
p
and is given by
.
6
4
3
1
3
1
2
1
1
3
4
3
1
2
2 2 2
p
G
If
p
[
0
,
p
]
, thenF
F
0
,
that is
3
1
6
4
3
1
2
1
1
4
4
3
1
2 3 2
p
p
F
G
p
.
Now
G
p
turns at p = 0 or
2
4
3
1
9
2
1
1
4
p
with its maximum at p = 0.Hence
.
3
1
3
2
4 3
2
a
a
a
For
0
, Theorem 3.2 agree with the following result due to Babalola [1].Corollary 3.2.1 If
f
z
S
, then
a
2a
3
a
4
2
.
For
1
, Theorem 3.4 gives the following result due to Babalola [1].
.
6
1
4 3
2
a
a
a
Theorem 3.3 If
f
M
, then
.
2
1
1
2 2
3
a
a
Proof. Since
f
z
M
, then using equations (3.2) and (3.3) we obtain
22 2
1 2
2 2 3
1
2
1
2
2
1
2
1
2
a
p
p
a
1
2
.
2
1
2
2
1
2
1
122 2 2
p
p
Using Lemma 2.3, with
1
2
,
2
1
2
0
2 2
we have
.
2
1
1
2 2
3
a
a
Theorem 3.4 If
f
M
, then
.
.
1
0
4
1
3
1
2
1
1
3
199
445
224
32
0
16
1
2 3 2
3
if
if
H
Proof. Using Lemma 2.4 and Theorems 3.1, 3.2 and 3.3 in inequality (1.3), the above result can be easily obtained.
For
0
, Theorem 3.4 agree with the following result due to Babalola [1].Corollary 3.4.1 If
f
z
S
, then
H
3
1
16
.
For
1
, Theorem 3.4 gives the following result due to Babalola [1].
.
24
15
1
3
H
4.
References
[1]
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1
3
H
Hankel determinant for some classes of univalent functions, Inequality
Theory and Applications, 6(2010), 1-7.
[2]
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[7]
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[8]
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[9]
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[10]