Applied Mathematics E-Notes, 13(2013), 92-99 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
On Coe¢ cient Determinants With Fekete-Szegö
Parameter
Kunle Oladeji Babalolay
Received 17 April 2013
Abstract
In this paper we introduce Hankel determinants involving the Fekete-Szegö
parameter and other similar determinants for the coe¢ cients of analytic functions
in the open unit disk. We investigate bounds on such determinants for the class
of functions of bounded turning.
1
Introduction
The Fekete-Szegö functional ja3
a22 j for normalized univalent functions
f (z) = z + a2 z 2 +
is well known for its rich history in the theory of geometric functions. Its origin was in
the disproof by Fekete and Szegö of the 1932 conjecture of Littlewood and Parley that
the coe¢ cients of odd univalent functions are bounded by unity. The functional has
since received great attention, particularly in many subclasses of the family of univalent
functions.
For integers n
1 and q
Hq (n) =
1, the q-th Hankel determinant, de…ned as
an
an+1
an+1
..
.
..
.
an+q
1
..
.
an+q
..
.
..
.
an+2(q
1
;
1)
which include the Fekete-Szegö functional as a special case ( = 1), has also received
the attention of many researchers for wide range of subclasses of functions. One
natural consequence of the continuous investigations is that, for function classes de…ned by other function classes (for example the classes of close-to-star, close-to-convex,
quasi-convex, -starlike, -convex, -close-to-star, -close-to-convex whose de…nitions
involve other function classes), coe¢ cient functionals of the form ja2 a3
a4 j and
Mathematics Subject Classi…cations: 30C45, 30C50.
Department of Physical Sciences, Al-Hikmah University, Ilorin, Nigeria; Permanent:
Department of Mathematics, University of Ilorin, Ilorin, Nigeria
y Current:
92
K. O. Babalola
93
ja2 a4
a33 j (and possibly more) for the de…ning function classes have frequently appeared to be resolved in the investigations of Hankel determinants for the desired classes
of functions.
In this work, therefore, we are motivated by such emerging functionals to de…ne
what we call the Hankel determinants with Fekete-Szegö parameter as follows:
DEFINITION 1. Let be a nonnegative real number. Then for integers n 1 and
q 1, we de…ne the q-th Hankel determinants with Fekete-Szegö parameter , that is
Hq (n), as
Hq (n) =
an
an+1
an+1
..
.
..
.
an+q
q
an+q
..
.
..
.
..
.
an+2(q
1
1
:
1)
DEFINITION 2. Let be a nonnegative real number. Then for integers n
1, we de…ne the Bq (n) determinants as
Bq (n) =
an
an+q
an+2q
..
.
an+q(q
an+1
an+q+1
an+2q+1
..
.
..
.
an+q
an+2q
an+3q
..
.
an+q2
1)
1 and
1
1
1
:
1
The investigation of the determinants Hq (n) and Bq (n) is of interest for many
classes of functions. We investigate in this paper, the determinants
H2 (2) =
a2
a3
a3
= ja2 a4
a4
a23 j
B2 (1) =
1
a3
a2
= ja2 a3
a4
a4 j
and
for the class of functions whose derivatives have positive real parts, known as functions
of bounded turning. Those are functions satisfying Re f 0 (z) > 0 in the open unit disk,
and are denoted by R.
In the next section we state the lemmas we shall use to establish the desired bounds
in Section 3.
94
On Coe¢ cient Determinants with Fekete-Szegö Parameter
2
Preliminary Lemmas
Let P denote the class of functions p(z) = 1 + c1 z + c2 z 2 +
which are regular in E
and satisfy Re p(z) > 0, z 2 E. To prove the main results in the next section we shall
require the following two lemmas.
LEMMA 1 ([2]). Let p 2 P . Then jck j 2, k = 1; 2; :::, and the inequality is sharp.
Equality is realized by the Möbius function L0 (z) = (1 + z)=(1 z).
LEMMA 2 ([4,5]). Let p 2 P . Then
2c2 = c21 + x(4
c21 )
(1)
and
4c3 = c31 + 2xc1 (4
for some x, z such that jxj
3
c21 )
x2 c1 (4
1 and jzj
1.
c21 ) + 2z(1
jxj2 )(4
c21 )
(2)
Main Result
Our …rst result is the following.
THEOREM 1. Let f 2 R. Then
8
1
>
<2
729 1152
a2 a4
a23
144(9
>
:4
+512
8 )
2
9
if
= 0;
if 0 <
27
32 ;
27
32 :
if
The inequalities are sharp. For each , equality is attained by f (z) given by
8
>
z + z 2 + 23 z 3 + 12 z 4 +
>
q
>
>
>
<z + 1 27 32 z 2 + 2 z 3 + p 81 64
2
9 8
3
8 (9 8 )(27
q
q
f (z) =
>
1
27
32
2 3
32
27
27
2
>
z
+
z
+
z
+
>
2
9
8
3
18
9
>
>
:
2 3
2 5
z + 3z + 5z +
if
4
32 )
32
8
z +
z4 +
= 0;
3
4;
if 0 <
if
if
3
4
27
32 ;
<
27
32 :
PROOF. Let f 2 R. Then there exists a p 2 P such that f 0 (z) = p(z). Equating
coe¢ cients of f 0 (z) and p(z), we …nd that 2a2 = c1 , 3a3 = c2 and 4a4 = c3 . Thus we
have
c22
c1 c3
:
(3)
a2 a4
a23 =
8
9
The trivial case
= 0 is a consequence of Lemma 1. In this case the extremal is
achieved by choosing c1 = c2 = c3 = 2.
K. O. Babalola
95
Now substituting for c2 and c3 using Lemma 2, we obtain
a23 =
a2 a4
By Lemma 1, jc1 j
c 2 [0; 2].
8 )c41
(9 8 )c21 (4
+
288
144
2
2 2
c1 (4 c1 )x
c1 (4 c21 )(1
+
32
16
(9
c21 )x
(4
jxj2 )z
c21 )2 x2
36
:
(4)
2. Then letting c1 = c, we may assume without restriction that
Now suppose 9 8 is nonnegative. Then applying the triangle inequality on (4),
with = jxj, we obtain
8 )c4
c2 (4 c2 ) (9 8 )c2 (4
+
+
288
16
144
2
[(9 8 )c
18c + 32 ](4 c2 ) 2
+
288
F ( ):
c2 )
(9
a23
a2 a4
=
Now we have
F 0( ) =
Observe that F 0 ( )
F 0( )
8 )c2 (4
144
(9
c2 )
+
[(9
8 )c2
18c + 32 ](4
144
c2 )
:
F 0 (1) > 0 since the …rst term is nonnegative, so that
(4
c2 )[2(9
8 )c2
144
18c + 32 ]
Thus F ( ) is increasing on [0; 1] so that F ( )
F( )
4
(27 32 )c2
+
9
72
(4
c2 )(2
8
c)
> 0:
F (1). That is
(9
8 )c4
= G(c):
144
It is easy to see that, if 27 32 is negative, that is
27=32, then G(c) is decreasing
on [0; 2] so that G(c) G(0) = 4 =9. Otherwise (that is
27=32), we have
8 )c3
:
36
p
Then the maximum of G(c) on [0; 2] occurs at c = (27 32 )=(9 8 ) and is given
by
!
r
27 32
729 1152 + 512 2
G
=
:
9 8
144(9 8 )
p
By setting c1 = c = (27 32 )=(9 8 ) and selecting x = p
1 in (1) we …nd that
c2 = 2. Then the admissible c3 is given by c3 = (81 64 )=[2 (9 8 )(27 32 )]
if 0 <
3=4 and for 3=4
27=32 we …nd the admissible c3 = [(32
G0 (c) =
(27
32 )c
36
(9
96
On Coe¢ cient Determinants with Fekete-Szegö Parameter
p
27) (27 32 )=(9 8 )]=18. Thus equality in these cases is attained by the function de…ned in the theorem.
Next we consider that 9
a2 a4
a23
=
8 is negative. Then we write (4) as
9)c41
(8
9)c21 (4 c21 )x
(4 c21 )2 x2
+
+
288
144
36
c21 (4 c21 )x2
c1 (4 c21 )(1 jxj2 )z
+
:
32
16
(8
Following the same argument as above, we have
a2 a4
a23
9)c4
c2 (4 c2 ) (8
9)c2 (4
+
+
288
16
144
[32
18c (8
9)c2 ](4 c2 ) 2
+
288
= F ( ):
(8
c2 )
Furthermore,
F( )
4
9
F (1) =
and G(c) is decreasing on [0; 2] so that G(c)
c2
= G(c)
8
G(0) = 4 =9.
In this case we set c1 = 0 and selecting x = 1 in (1) and (2), we …nd that c2 = 2
and c3 = 0 so that equality is attained by f (z) de…ned in the theorem and the proof is
complete.
COROLLARY 1 ([3]). Let f 2 R. Then
a2 a4
a23
4
:
9
The inequality is sharp. Equality is attained by
Z z
2
2
1 + t2
f (z) =
dt = z + z 3 + z 5 +
2
1
t
3
5
0
:
Next we have
THEOREM 2. Let f 2 R. Then
ja2 a3
a4 j
84
>
>
<
3
6
9
>
>
:9
4
18
4
54
q
2
3;
4
3;
if 0
9
q 3(3
2(9
4
2)
if
4)
if
2
3
<
4
3:
K. O. Babalola
97
The inequalities are sharp. For each , equality is attained by f (z) given by
8
z + z 2 + 32 z 3 + 12 z 4 +
>
>
q
q
<
1
9
4 2
9
4 4
9
8 3
1
f (z) = z + 2 9 6 z + 18 z
4
9
6z +
q
q
>
>
:
2(9
4)
2(9
4) 4
z + 12
z 2 + 9 18 8 z 3 14
z +
9
9
2
3;
4
3;
if 0
2
3
if
<
4
3:
if
PROOF. As is in the proof of Theorem 1, if f 2 R, then 2a2 = c1 , 3a3 = c2 and
4a4 = c3 . Thus we have
c3
c1 c2
:
(5)
ja2 a3
a4 j =
6
4
Substituting for c2 and c3 using Lemma 2, we obtain
ja2 a3
a4 j =
By Lemma 1, jc1 j
c 2 [0; 2].
3 )c31
(4 6 )c1 (4
+
48
48
(4 c21 )(1 jxj2 )z
:
8
(4
c21 )x
+
c21 )x2
c1 (4
16
(6)
2. Then letting c1 = c, we may assume without restriction that
Now suppose 4 6 is nonnegative. Then applying the triangle inequality on (6),
with = jxj, we obtain
ja2 a3
a4 j
3 )c3
(4 c2 ) (4
+
+
48
8
= F ( ):
(4
The extreme values of F ( ) are at
F 0( ) =
(4
= 0,
c2 )
6 )c(4
48
Now let
G1 (c) = F (0) =
(4
6 )c(4
48
= 1 and
+
(c
c2 )
+
(c
c2 )
2)(4
16
2
such that
2)(4
8
c2 )
= 0:
3 )c3
(4 c2 )
+
;
48
8
G2 (c) = F (1) =
(4
3 )c
12
and
G3 (c) = F
(2 3 )c
3 (2 c)
=
(4
3 )c3
(4 c2 ) c2 (c + 2)(2
+
+
48
8
144
3 )2
:
By elementary calculus, we …nd that G1 (c) G1 (0) = =2, G2 (c) G2 (2) = (4 3 )=6
and G3 (c) G3 (0) = =2 for all admissible c. Hence G(c) G2 (2) = (4 3 )=6.
98
On Coe¢ cient Determinants with Fekete-Szegö Parameter
By selecting c1 = c = 2 and selecting x = 1 in (1) and (2) we …nd that c2 = c3 = 2.
Thus the extremal function for this case 0
2=3 is the function given in the
theorem.
Next we consider the case: 4 6 is negative while 4
2=3 <
4=3. In this case we write (6) as
ja2 a3
a4 j =
3 )c31
48
(4 c21 )(1
8
(4
(6
4)c1 (4
48
a4 j
+
c21 )x2
c1 (4
16
2
jxj )z
:
(7)
so that as in the …rst part with c1 = c 2 [0; 2] and
ja2 a3
c21 )x
3 is nonnegative, that is
3 )c3
(4 c2 ) (6
+
+
48
8
= F ( ):
(4
= jxj we have
4)c(4
48
c2 )
+
(c
2)(4
16
c2 )
2
Using the same extreme value technique, we …nd that the extreme value of F ( ) yielding
the best possible bound for functional is
G(c) = F (1) =
and that
(2
3 )c3
(9
4)c
+
12
12
(9
4)
3 )c2
+
:
4
12
p
Thus the maximum of G(c) on [0; 2] occurs at c = (9
4)=(9
6) and is given by
s
s
!
9
4
9
4
9
4
=
:
G
3(3
2)
18
3(3
2)
p
By setting c1 = c = (9
4)=(9
6) and selecting x = 1 in (2) we …nd that c3 = c1 .
The choice c3 = c1 is also appropriate as the context so admits. Then the admissible
c2 is given by c2 = (9
8)=6. Thus for 2=3 <
4=3, the given function in the
theorem attains the equality.
G0 (c) =
Finally we suppose 4
ja2 a3
(2
3 is negative, that is
(6
4)c1 (4
4)c31
+
48
48
2
2
(4 c1 )(1 jxj )z
+
:
8
a4 j =
(3
As we have shown above, with c1 = c 2 [0; 2] and
ja2 a3
a4 j
4=3. Then we write (6) as
4)c3
(4 c2 ) (6
+
+
48
8
= F ( ):
(3
c21 )x
c21 )x2
c1 (4
16
(8)
= jxj, we have
4)c(4
48
c2 )
+
(c
2)(4
16
c2 )
2
K. O. Babalola
99
Using the same extreme value technique, we …nd that
G(c) = F (1) =
and
(9
c3
8
4)c
12
3 c2
:
12
8
p
Thus the maximum of G(c) on [0; 2] occurs at c = 2(9
!
r
r
9
4 2(9
2(9
4)
=
G
9
54
G0 (c) =
(9
4)
4)=9 and is given by
4)
:
p
Finally setting c1 = c = 2(9
4)=(9 ) and selecting x = 1 in (2) we …nd that
c3 = c1 . Again the choice c3 = c1 is also appropriate as the context so admits. Thus
the admissible c2 is given by c2 = (9
8)=6. Thus for 2=3 <
4=3, the given
function in the theorem is admitted in the equality and the proof is now complete.
COROLLARY 2 ([1]). Let f 2 R. Then
ja2 a3
a4 j
5
18
The inequality is sharp. Equality is attained by
r
1 5 2
1
f (z) = z +
z + z3
2 3
18
r
5
:
3
1
4
r
5 4
z +
3
:
Acknowledgment. The author is thankful to the referees for their helpful comments and suggestions.
References
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[2] P. L. Duren, Univalent Functions, Springer Verlag, New York Inc., 1983.
[3] A. Janteng, S. Abdulhalim and M. Darus, Coe¢ cient inequality for a function whose
derivative has positive real part, J. Ineq. Pure and Applied Math., 7(2)(2006), 1–5.
[4] R. J. Libera and E. J. Zlotkiewicz, Early coe¢ cient of the inverse of a regular
convex function, Proc. Amer. Math. Soc., 85(2)(1982), 225–230.
[5] R. J. Libera and E. J. Zlotkiewicz, Coe¢ cient bounds for the inverse of a function
with derivative in P , Proc. Amer. Math., Soc., 87(2)(1983), 251–257.