Applied Mathematics E-Notes, 15(2015), 73-79 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Order Of Schlictness Of Certain Linear Sums
Kunle Oladeji Babalolay, Mashood Sidiqz, A…s Saliux, Catherine Ngozi Ejieji{
Received 27 March 2014
Abstract
In this paper, we …nd the largest real number such that the real part of
certain geometric quantity greater than
implies schlictness of certain linear
sums.
1
Introduction
Let H denote the class of functions:
f (z) = z + a2 z 2 + :::
(1)
which are holomorphic in the unit disk E = fz 2 C : jzj < 1g. Let f; g 2 H, we say that
f (z) is subordinate to g(z) (written as f
g) if there exists a holomorphic function
w(z) (not necessarily schlict) in E, satisfying w(0) = 0 and jw(z)j < 1 such that
f (z) = g(w(z)), z 2 E. For example, it is well known that any Caratheodory function
p(z) = 1 + c1 z +
is subordinate to the M•
obius function L0 (z) = (1 + z)=(1 z).
In [2], Babalola proved that holomorphic functions satisfying
Re (1
)
Dn+1 f (z)
Dn f (z)
+
nz
Dn f (z)
also satisfy
Re
for all
such that
Dn f (z)
> 0;
nz
< 1, and that if
Re
1
z 2 E;
(2)
z 2 E;
= 21 , then for all
Dn f (z)
1
> ;
nz
2
Dn is de…ned as Dn f (z) = D(Dn
implied by both results.
> ;
0<
f (z)) = z(Dn
1
0,
1:
f (z))0 and for n
1, schlictness is
Mathematics Subject Classi…cations: 35C45, 35C50.
Department of Physical Sciences, Al-Hikmah University, Ilorin. Permanent: Department
of Mathematics, University of Ilorin, Ilorin, Nigeria, kobabalola@gmail.com
z Department of Mathematics, University of Ilorin, Ilorin, Nigeria, mashoodsidiq@yahoo.com
x Department of Mathematics, Gombe State Univeristy, Gombe, Nigeria, a…s.saliu66@gmail.com
{ Department of Mathematics, University of Ilorin, Ilorin, Nigeria, ejieji.cn@unilorin.edu.ng
y Current:
73
74
Order of Schlictness of Linear Sums
DEFINITION. We say f 2 Tn; ( ) if and only if f (z) satis…es the inequality (2).
In this paper, we shall …nd the largest real number
Re
Dn f (z)
> ;
nz
such that
z2E
given that f 2 Tn; ( ) for 0
< 1. To do this we shall employ the technique
of Briot-Bouquet di¤erential subordination. A holomorphic function p(z) is said to
satisfy Briot-Bouquet di¤erential subordination if
p(z) +
zp0 (z)
p(z) +
h(z);
z2E
(3)
for complex constants and , and a complex function h(z) with h(0) = 1 such that
Re [ h(z) + ] > 0 in E. If the di¤erential equation
q(z) +
zq 0 (z)
q(z) +
= h(z);
q(0) = 1
(4)
has the schlict solution q(z) in E, then
p(z)
q(z)
h(z)
and q(z) is the best dominant.
A schlict function q~(z) is said to be the best dominant of the di¤erential subordination (3) if q(z) and q~(z) are dominants of (3) and q~(z) q(z) for all the dominants
q(z) of (3). For more on the technique of di¤erential subordination, see [3,4,5,7].
We state and prove our main result in Section 3.
2
Preliminary Lemmas
In this section, we present Lemmas 1–4 as follows.
LEMMA 1 ([1]). Let f 2 H and
> 0 be real, if
Dn+1 f (z)
Dn f (z)
is independent of n for z 2 E, then
Dn+1 f (z)
=
Dn f (z)
Dn+1 f (z)
:
Dn f (z)
LEMMA 2 ([3]). Let and be complex constants and h(z) be a convex schlict
function in E satisfying h(0) = 1 and Re [ h(z) + ] > 0. Suppose p 2 P satis…es the
di¤erential subordination (3). If the di¤erential equation (4) has schlict solution q(z)
Babalola et al.
75
in E, then p(z) q(z) h(z) and q(z) is the best dominant of (3). Furthermore, the
formal solution of (4) is given as
q(z) =
zF 0 (z)
=
F (z)
where
+
+
z
Z
H(z) = z exp
Z
F (z) =
H(z)
F (z)
(5)
z
t
1
H(t) dt
0
and
z
0
h(t)
t
1
dt :
LEMMA 3 ([6]). Let v be a positive measure on [0; 1] and let h be a complex valued
function de…ned on E [0; 1] such that h(:; t) is holomorphic in E for each t 2 [0; 1] for
all z 2 E. Suppose that Re[h(z; t)] > 0, h( r; t) is real and
1
for jzj
h( r; t)
1
h(z; t)
Re
If
h(z) =
Z
r < 1 and t 2 [0; 1]:
1
h(z; t)dv(t);
0
then
Re
1
:
h( r)
1
h(z)
For real or complex numbers a, b, c with c 6= 0; 1; 2; : : : , the hypergeometric
function is de…ned by
2 F1 (a; b; c; z)
=1+
a:b z
a(a + 1):b(b + 1) z 2
: +
: +
c 1!
c(c + 1)
2!
:
(6)
We note that the series converges absolutely for z 2 E and hence represents an holomorphic function in E. The following identities associated with the hypergeometric
series are well known.
LEMMA 4 ([6]). For real numbers a; b; c (c 6= 0; 1; 2; : : : ), we have
Z
0
1
tb
t)c b 1
dt =
(1 tz)a
1 (1
(b) (c b)
2 F1 (a; b; c; z);
(c)
2 F1 (a; b; c; z)
=
(c > b > 0);
2 F1 (b; a; c; z)
and
2 F1 (a; b; c; z)
= (1
z)
a
2 F1
a; c
b; c;
z
z
1
:
76
3
Order of Schlictness of Linear Sums
Main Result
In this section, we present our main result.
THEOREM 1. Let f 2 Tn; ( ). If
Dn f (z)
nz
0, then
where
q(z) =
(1
)
R1
0
1 + (1 2 )z
1 z
q(z)
(1
z)
2 (
1)
1 (1
s
sz)
2 (
1)
ds
= 1 + q1 z +
is the best dominant. Furthermore
Re
Dn f (z)
>
nz
where
=
2 F1
1;
2 (1
)
;
+ 1;
1
2
1
:
PROOF. Since f 2 Tn; ( ), we see that
(1
)
Dn f (z)
Dn+1 f (z)
+
= Bn (f ; )
nz
Dn f (z)
1 + (1 2 )z
:
1 z
(7)
Let
Dn f (z)
= p(z):
(8)
nz
Then p(z) is holomorphic in E with p(0) = 1. So taking the logarithmic di¤erentiations
in both sides of (8), we have
Dn+1 f (z)
=
Dn f (z)
+
zp0 (z)
p(z)
Using Lemma 1 and then simplify, we have
Dn+1 f (z)
zp0 (z)
=1+
n
D f (z)
p(z)
With (8) and (9) in (7), we have
"
+ (1
) p(z) +
If
+ (1
"
#
zp0 (z)
= Bn (f ; )
1
p(z)
) q(z) +
(9)
1 + (1 2 )z
:
1 z
#
1 + (1 2 )z
zq 0 (z)
=
;
1
1 z
q(z)
Babalola et al.
77
then
1
zq 0 (z)
=
1
)q(z)
(
q(z) +
+ (1 +
(1
)(1
2 )z
= h(z):
z)
(10)
Then h(0) = 1 and we take
(1
=
)
and
= 0:
It is easily veri…ed that h(z)+ has positive real part given that 0
by Lemma 2, p(z) satis…es the di¤erential subordination (3) and hence
Dn f (z)
nz
q(z)
. Therefore,
h(z)
where q(z) is the solution of di¤erential equation (10) obtained as follows:
2(
H(z) = z(1
1)
)
z) (1
:
Also, we have
F (z) =
(1
)
Z
z
1
t
(1
t)
2 (
1)
(1
)
:
dt
0
Now from (5), we have
q(z) =
where
Z
Q(z) =
H
=
F
(1
)Q(z)
2 (
1
1
s
0
1)
1 sz
1 z
ds
Next, we show that
inf fRe(q(z))g = q( 1); z 2 E;
(11)
jzj<1
To prove (11), we show that
Re
1
Q(z)
1
Q( 1)
and by Lemma 4 with some simpli…cation, we get
Q(z) =
(b)
z
2 F1 1; a; c;
(c)
z 1
where
a=
2 (1
)
;
b=
and c =
+ 1:
78
Order of Schlictness of Linear Sums
Hence, we see that
Q(z) =
Z
1
h(z; s)dv(s)
0
where
h(z; s) =
and
1 z
;0
(1 s)z
1
(b)
sa
(a) (c a)
dv(s) =
1
s
1
s)c
(1
a 1
ds
which is a positive measure on [0; 1].
It will be noted that Re fh(z; s)g > 0, h( r; s) is real for 0
1
h(z; s)
Re
for jzj
1
= Re
(1 s)z
1 z
r < 1, and
1 + (1 s)r
1
=
1+r
h( r; s)
r < 1 and s 2 [0; 1]. So using Lemma 3, we have
Re
1
Q(z)
1
;
Q( r)
Re
1
Q(z)
1
:
Q( 1)
and letting r ! 1 , we obtain
Hence we have
Dn f (z)
nz
Re
> q( 1);
that is,
Dn f (z)
nz
Re
>
where
=
(1
)Q( 1)
=
a
(1
)
2
Z
1
1
b 1
s
a
(1 + s)
ds
0
By Lemma 4, we see that
=
(1
1
(b) (c b)
1
2 F1 1; a; c;
(c)
2
)
can be simplify as
=
The bound
2 F1 1;
2 (1
is the best possible.
)
;
(1
)
+ 1;
1
2
1
:
:
Babalola et al.
4
79
Remark
We …rst remark that for n
1, our result shows that holomorphic function f (z);
satisfying the geometric condition (2) de…ned by linear sum, is schlict of order in the
unit disk. Then by (6), we can write in series form:
1
=1+
k
1
X
)+j
1 Y 2 (1
;
k
2 j=1 (1
) + (j + 1)
k=1
from which it follows that
1
=1+
= 0 yields
k
1
X
1 Y
2k j=1
k=1
Thus, if
(1
1
(1
(1
=
for all . The series can be rewritten as
) + (j + 1)
(1
)
+
) + (j + 1)
(1
) + (j + 1)
:
); we have
>1+
1
k
X
1 Y (1
2k j=1 (1
k=1
1
X 1
) + (j + 1)
>1+
> 2:
) + (j + 1)
2k
k=1
It follows that > 1=2 whenever 0 <
which guarantees that
(1
)
1
given that
.
This deduction shows some improvement on the earlier result of Babalola [2] for
in the real interval [0; ].
Acknowledgment. The authors are grateful to the referee for his helpful comments.
References
[1] K. O. Babalola, On some n-starlike integral operators, Kragujevac Journal of Mathematics, 34(2010), 61–71.
[2] K. O. Babalola, On a linear combination of some geometric expressions, Indian
Journal of Mathematics, Special Memorial Volume, 51(1)(2009), 1–8.
[3] P. Eenigenburg, S. S. Miller, P. T. Mocanu and M. O. Reade, On a Briot-Bouquet
di¤erential surbordination, Rev. Roumaine Math. Pures Appl., 29(1984), 567–573.
[4] S.S Miller and P.T. Mocanu, Schlict solution of Briot-Bouquet di¤erential equations,
Lecture Notes in Mathematics, Springer Berlin/Heidelberg, 1013(1983), 292–310.
[5] J. Patel and S. Rout, An application of di¤erential subordinations, Rendiconti di
Matematica, Serie VII, Roma 14(1994), 367–384.
[6] J. Patel, On certain subclass of p-valently Bazilevic functions, J. Ineq. Pure and
Applied Math., 6(1)(2005), Art. 16.
[7] H. M. Srivastava and A. Y. Lashin, Some applications of the Briot-Bouquet di¤erential subordination, J. Ineq. Pure and Appl. Math., 6(2)(2005) Art. 41.