Applied Mathematics, 2011, 2, 141-144
doi:10.4236/am.2011.21016 Published Online January 2011 (http://www.SciRP.org/journal/am)
Integral Mean Estimates for Polynomials Whose Zeros
are within a Circle
Yash Paul1, Wali Mohammad Shah2, Gulshan Singh1
1Bharathiar University Coimbatore, Tamil Nadu, India
2Department of Mat hem at i cs, K ash mi r Uni versity Srinagar , In di a
Received December 28, 2010, revised January 12, 2011, accepted January 18, 2011
Abstract
Let

Pz be a polynomial of degree n having all its zeros in 1zK

, then for each >0r, >1p,
>1q with 11
=1
pq
, Aziz and Ahemad (1996) proved that





11
1
22 2
00 0
1.
rpr
qr
rpr
qr
iii
nPedKed Ped
 




 
In this paper, we extend the above inequality to the class of polynomials

=
:= ,
n
nnj
nnj
jm
Pzaza z


1mn

,
having all its zeros in 1zK, and obtain a generalization as well as refinement of the above result.
Keywords: Derivative of a Polynomial, Integral Mean Estimates, Complex Domain Inequalities
1. Introduction and Statement of Results
Let

Pz be a polynomial of degree n and


Pz

be
its derivative. If

Pz has all its zeros in 1z

, then
it was shown by Turan [1] that
 
=1=1 .
2
zz
n
M
axPzMaxP z
 (1)
Inequality (1) is best possible with equality for

=n Pz z

 , where =

 . As an extension of (1)
Malik [2] proved that if

Pz has all its zeros in zK

,
where 1K, then
 
=1=1 .
1
zz
n
M
axPzMaxP z
K
 (2)
Malik [3] obtained a generalization of (1) in the sense
that the right-hand side of (1) is replaced by a factor
involving the integral mean of

Pz on =1z. In
fact he proved the following theorem.
Theorem A. If

Pz has all its zeros in 1z

, then
for each >0r




11
22
=1
00
1.
rr
rr
ii
z
nPed edMaxPz






(3)
The result is sharp and equality in (3) holds for
 
=1
n
Pz z

.
If we let r in (3), we get (1).
As a generalization of Theorem A, Aziz and Shah [4]
proved the following:
Theorem B. If

=
:= n
nnj
nnj
jm
Pzaza z

, 1mn


is a polynomial of degree n having all its zeros in the
disk zK

, 1
K

, then for each >0r,




1
2
0
1
2
=1
01.
rr
i
rr
mi z
nPed
K
ed MaxPz








(4)
Aziz and Ahemad [5] generalized (3) in the sense that


=1z
M
axP z
 on =1z on the right-hand side of (3)
is replaced by a factor involving the integral mean of


P
z

on =1z and proved the following:
Theorem C. If


Pz is a polynomial of degree n
having all its zeros in 1zK, then for >0r,
>1p, >1q with 11
=1
pq
,

Y. PAUL ET AL. 142





1
2
0
1
1
22
00
1.
rr
i
pr
qr
p
r
qr
ii
nPed
KedP ed









(5)
If we let r and p (so that 1q) in (5),
we get (2).
In this paper, we extend Theorem B to the class of
polynomi als

=
:= ,
n
nnj
nnj
jm
Pzazaz

1,mn


having all the zeros in 1zK

, and thereby obtain a
more general result by proving the following.
Theorem 1. If

=
:= n
nnj
nnj
jm
Pzaza z

, 1mn


is a polynomial of degree n having all its zeros in the
disk zK

, 1K

, then for each >0r, >1s, >1t
with 11
=1
s
t
,




1
1 1
21
22 2
1
00 0
1,
sr sr
mm
rtr
rtr
nnm iii
m
nnm
na KmaK
nPeded Ped
na Kma
 



















 
(6)
If we take m = 1 in Theorem 1, we get the fol-
lowing:
Corollary 1. If

=0
:= n
j
j
j
Pz az
 is a polynomial of
degree n having all its zeros in the disk zK, 1K

,
then for each >0r, >1s, >1t with 11 =1
s
t
,




1
1 1
2
22 2
1
00 0
1
1.
sr sr
rtr
rtr
nn
iii
nn
na Ka
nPeded Ped
na a
 














 

 
(7)
The next result immediately follows from Theorem 1,
if we let t so that 1s
Corollary 2. If

=
:= ,
n
nnj
nnj
jm
Pzaza z


1mn


is a polynomial of degree n having all its zeros in the
disk zK

, 1
K

, then for each >0r,



1
121
22
=1
1
00
1.
rr
mm
rrnnm
ii z
m
nnm
na KmaK
nPed edMaxPz
na Kma

 









 







 (8)
Also if we let r in the Theorem 1 and note that


1
2
=1
0
1=
lim 2
rr
i
rz
PedMax Pz



  

.
We get the following:
Corollary 3. If

=
:= n
nnj
nnj
jm
Pzaza z

, 1mn


is a polynomial of degree n having all its zeros in the
disk zK, 1
K
, then



=1
1
=1
211 .
1
z
m
nnm z
mm m
nnm
MaxP z
na Kma
M
axP z
na KKmaK







(9)
For K = 1, Corollary 3 reduces to Inequality (1) (the
result of Turan[1]) .
2. Lemmas
For the proof of this theorem, we need the following
lemmas.
The first lemma is due to Qazi [6].
Lemma 1. If

0=
:= n
j
j
jm
Pz aaz is a polynomial
of degree n having no zeros in the disk <zK, 1K.
Then

12
0
1
0
mm
m
m
m
na KmaKPz Qz
nama K









for =1z, 1mn

,
where

1
=n
QzzP z



and
0
1
m
m
a mK
na .
Lemma 2. If

=
:= n
nnj
nnj
jm
Pzaza z

 is a poly-
nomial of degree n having all its zeros in the disk
1zK

s then
 
21
1
mm
nnm
m
nnm
na KmaK
Qz Pz
na Kma











for =1z, 1.mn


Proof of Lemma 2

Y. PAUL ET AL. 143
Since all the zeros of

Pz lie in 1zK, there-
fore all the zeros of

1
=n QzzP z



lie in 11z
K
.
Hence applying lemma 1 to the polynomial

=
:= n
j
nnj
jm
Qzaa z

, we get
 
12
1
11
1
nnm
mm
nnm
m
na ma
KK
Qz Pz
na maK













.
Or, equivalently
 
21
1
mm
nnm
m
nnm
na KmaK
Qz Pz
na Kma






 



.
This proves lemma 2.
Remark 1: Lemma 3 of Govil and Mc Tume [7] is a
special case of this lemma when m = 1.
Proof of Theorem 1
Since

1
=n QzzP z



, therefore, we have

1
=n
PzzQ z



. This gives

12
11
=.
nn
Pznz QzQ
zz

 


 
 
(10)
Equivalently

1
11
=nn zPznz QzQ
zz






(11)
this implies
 


= =1.PznQz zQzforz

 (12)
Now by hypothesis,

Pz has all its zeros i n
1zK, therefore, by Lemma 2, we have for =1z
 
21
1, 1.
mm
nnm
m
nnm
na KmaK
QzPzm n
na Kma








(13)
Using (12) in (13), we get
 
21
1
=1, 1.
mm
nnm
m
nnm
na KmaK
QznQz zQz
na Kma
for zmn






 





(14)
Since


Pz has all its zeros in 1zK , by
Gauss-Lucas theorem all the zeros of

Pz
 also lie in
1z

, therefore, it follows that the polynomial
 
11= n
zPnQzzQz
z





 (15)
has all its zeros in 11zk  and hence, we conclude
that the function
 
  
1
21
=m
nnm
mm
nnm
na KmazQ z
Wz na KmaKnQ zzQz




 


 


(16)
is analytic for <1z, W(0) = 0 and by (14)


1Wz

for =1z. Thus the function

1
21
1m
nnm
mm
nnm
na KmaWz
na KmaK











is subordinate to the function
1
21
1m
nnm
mm
nnm
na Kmaz
na KmaK















for 1z

. Hence by a well known property of subor-
dination [8], we have for eac h >0r and 0<2

 ,

21
2
1
0
21
2
1
0
1
1.
r
mm
nnm i
m
nnm
r
mm
nnm i
m
nnm
na KmaKWe d
naKma
na KmaKed
na Kma



























(17)
Also from (16), we have
 
 
21
1
1=.
mm
nnm
m
nnm
na KmaKnQ z
Wz nQ zzQz
na Kma








 


Therefore,
 
21
1
=1 .
mm
nnm
m
nnm
na KmaK
nQ zWznQ zzQz
na Kma











(18)
Using (12) and the fact that
 
=QzPz for =1z, we get from (18)
  
21
1
=1for =1.
mm
nnm
m
nnm
na KmaK
nP zWzPzz
na Kma









 (19)

Y. PAUL ET AL. 144
Combining (17) and (19), we get
 
21
22
1
00
1 for >0.
r
mm
r r
nnm ri ii
m
nnm
na KmaK
n PedePedr
na Kma














 (20)
Now applying Hölder’s inequality for 1>s, 1>t, with 1=
11
ts to (20), we get
 

11
21
22 2
1
00 0
1 for >0.
sr s
mm
rtr t
nnm rii i
m
nnm
na KmaK
nPededPedr
na Kma
 



















 (21)
This is equivalent to




1
1 1
21
22 2
1
00 0
1 for >0.
sr sr
mm
rtr
rtr
nnm iii
m
nnm
na KmaK
nPededPedr
na Kma
 



















 (22)
which proves the desired resu lt.
3. Acknowledgements
The authors are grateful to the referee for useful sug-
gestions.
4. References
[1] P. Turan, “Über die Ableitung von Polynomen,” Compo-
sitio Mathematica, Vol. 7, 1939, pp. 89-95.
[2] M. A. Malik, “On the Derivative of a Polynomial,”
Journal of the London Mathematical Society, Vol. 2, No.
1, 1969, pp. 57-60. doi:10.1112/jlms/s2-1.1.57
[3] M. A. Malik, “An Integral Mean Estimate for Polyno-
mials,” Proceedings of the American Mathematical So-
ciety, Vol. 91, No. 2, 1984, pp. 281-284.
[4] A. Aziz and W. M. Shah, “An Integral Mean Estimate for
Polynomials,” Indian Journal of Pure and Applied Ma-
thematics, Vol. 28, No. 10, 1997, pp. 1413-1419.
[5] A. Aziz and N. Ahemad, “Integral Mean Estimates for
Polynomials Whose Zeros Are within a Circle,” Glasnik
Matematicki Series, III, Vol. 31, No. 2, 1996, pp.
229-237.
[6] M. A. Qazi, “On the Maximum Modulus of Polyno-
mials,” Proceedings of the American Mathematical So-
ciety, Vol. 115, 1990, pp. 337-343. doi:10.1090/S0002-
9939-1992-1113648-1
[7] N. K. Govil and G. N. McTume, “Some Generalizations
Involving the Polar Derivative for an Inequality of Paul
Turan,” Acta Mathematica Thunder, Vol. 104, No. 1-2,
2004, pp. 115-126. doi:10.1023/B:AMHU.0000034366.
66170.01
[8] E. Hille, “Analytic Function Theory, Vol. II,” Introduc-
tion to Higher Mathematics, Ginn and Company, New
York, 1962.