Applied Mathematics, 2011, 2, 321-328
doi:10.4236/am.2011.23038 Published Online March 2011 (http://www.scirp.org/journal/am)
p
L
Inequalities for Polynomials
Abdul Aziz, Nisar A. Rather
Department of Mathematics, Kashmir University, Srinagar, India
Received July 9, 2010; revised January 14, 2011; accepted January 17, 2011
Abstract
In this paper we consider a problem of investigating the dependence of


p
PRz Prz   on


p
Pz
for every real or complex number

with 1 

, >1
R
r, >0p and present certain compact generali-
zations which, besides yielding some interesting results as corollaries, include some well-known results, in
particular, those of Zygmund, Bernstein, De-Bruijn, Erdös-Lax and Boas and Rahman as special cases.
Keywords:
p
L-Inequalities, Polynomials, Complex Domain
1. Introduction
Let

n
Pz denote the space of all complex polynomials

=0
=n
j
j
j
Pz az
 of degree at most n. For n
PP

,
define


1
2
0
1
:=,1 <
2
p
p
i
p
PzPep







and
 
=1
:= max.
z
Pz Pz

A famous result known as Bernstein’s inequality(for
reference,see[1] or [2]) states that if n PP,then
 
'
Pz nPz
 (1)
whereas concerning the maximum modulus of


Pz on
the circle =>1zR , we have
 
,
n
PRzR Pz

 (2)
(for reference, see [3]). Inequalities (1) and (2) can be
obtained by letting p in the inequalities
 
,1
'
p
p
PznPz p (3)
and



,>1,>0,
n
pp
PRzRPzRp (4)
respectively. Inequality (3) was found by Zygmund [4]
whereas inequality (4) is a simple consequence of a re-
sult of Hardy [5] (see also [6]). Since Inequality (3) was
deduced from M.Riesz's interpolation formula [7] by
means of Minkowski’s inequality,it was not clear, whe-
ther the restriction onpwas indeed essential. This
question was open for a long time. Finally Arestov [8]
proved that (3) remains true for 0< <1p as well. Both
the Inequalities (3) and (4) can be sharpened if we res-
trict ourselves to the class of polynomials having no zero
in <1z. In fact, if n
PP

and

0Pz in <1z,
then Inequalities (3) and (4) can be respectively replaced
by
 
,0
1
p
'
pp
Pz
Pz np
z


 (5)
and
 
1,>1,>0.
1
n
p
pp
p
Rz
PRzPz Rp
z

(6)
Inequality (5) is due to De-Bruijn [9] for 1p and
Rahman and Schmeisser [10] extended it for 0< <1p
whereas the Inequality (6) was proved by Boas and
Rahman [11] for 1p and later it was extended for
0< <1p by Rahman and Schmeisser[12]. For =p

,
the Inequality (5) was conjectured by Erdös and later
verified by Lax [13] whereas Inequality (6) was proved
by Ankeny and Rivlin [14].
Recently the Authors in [12] (see also [15]) investi-
gated the dep endence o f

 
on
p
p
PRz PzPz
for >1R, 1.p As a compact generalization of
Inequalities (3) and (4), they have shown that if n
PP

,
then for every >1R and 1,p

A. AZIZ ET AL. 322
 



1.
n
p
p
PRz PzRPz (7)
It is natural to seek the corresponding an alog of (7) for
polynomials n PP having no zero in <1z and
which is a compact generalization of Inequalities (5) and
(6). In the present paper we consider a more general pro-
blem of investigating the dependence of
 
on
p
p
PRz PrzPz  
for every real or complex number

with 1 

,
>1Rr, >0p and develop a unified method for arri-
ving at these results. We first present the following inter-
esting result and a compact generalization of Inequalities
(3) and (4), which also extends Inequality (7) for
0< <1p as well.
Theorem 1. If n PP, then for every real or complex
number

with 1  , >1Rr and >0p,
 
.
nn
p
p
PRzPrzRr Pz   (8)
The result is best possible and equality in (8) holds for

=,0.
n
Pzaz a
Remark 1. For =0

, Theorem 1 reduces to Inequality
(4) and for =1

, =1r, it validates Inequality (7) for
each >0p.
If we set =1

in Inequality (8), we immediately get
the following generalization of Inequality (7).
Corollary 1. If n PP, then for >1Rr and
>0p
 


.
nn
p
p
PRz PrzR r Pz (9)
The result is best possible and equality in (9) holds for

=,0.
n
Pzaz a
If we divide the two sides of Inequality (9) by

Rr and let Rr, we get:
Corollary 2. If n PP, then for 1r and >0p,
 
1. 'n
p
p
PrznrPz

 (10)
Remark 2. For =1r, Corollary 2 reduces to
Zygmund’s Inequality (3) for each >0p.
The following result which is a compact generalization
of Inequalities of (1) and (2) follows from Theorem 1 by
letting p in Inequality (8).
Corollary 3. If n PP, then for every real or
complex number

with 1 

and >1Rr,
 
=1
maxfor=1.
nn
z
PRzPrz RrPzz  
(11)
The result is best possible and equality in (11) ho lds
for

=,0.
n
Pzaz a
Remark 3. For =0

, Corollary 3 reduces to
Inequality (2) and for =1

, if we divide the two sides
of (11) by Rr

and let Rr, it follows that if
n
PP

, then for 1r,


1
=1
maxfor =1.
'n
z
Prz nrPzz

 (12)
Inequality (12) reduces to Bernstein’s Inequality (1)
for =1r.
For polynomials n PP

having no zero in <1z,
we next prove the following interesting improvement of
(8) which among other things include De-Bruijn’s theo-
rem (Inequality (5)) and a result of Boas and Rahman
(Inequality (6)) as special cases.
Theorem 2. If n PP

and

Pz does not va nish in
<1z, then for every real or complex number

with
1 

, >1Rr and >0p
 



1.
1
nn
p
p
p
p
Rrz
PRz PrzPz
z



 
(13)
The result is best possible and equality in (13) ho lds
for


=,==1.
n
Pzaz bab
For =0

, Theorem 2 reduces to Inequality (6). A
variety of interesting results can be easily deduced from
Theorem 2. Here we mention a few of these. The
following corollary immediately follows from Theorem
2 by taking =1

.
Corollary 4. If n PP

and

Pz does not vanish
in <1z, then for >1Rr and >0p,
 


.
1
nn
p
p
p
Rr
PRz PrzPz
z


 (14)
The result is sharp and equality in (14) holds for
()=, ==1.
n
Pzaz bab
Remark 4. For =1r, if we divide the two sides of
(14) by 1R

and let 1R,we immediately get
De-Bruijn’s theorem (Inequality (5)) for each >0p.
Next we mention the following compact generaliza-
tion of a theorem of Erdös and Lax (Inequality (5) for
p

) and a result of Ankeny and Rivlin (Inequality (5)
for p

) which immediately follows from Theorem 2
by letting p in (13).
Corollary 5. If n PP

and

Pz does not vanish
in <1z, then for every real or complex number

with 1


and >1Rr,
 
=1
1max
2
for=1.
nn
z
Rr
PRz PrzPz
z



 (15)
The result is best possible and equality in (15) ho lds
for


=,==1.
n
Pzaz ba b

A. AZIZ ET AL. 323
Remark 5. For =1

, if we divide the two sides of
(15) by Rr and let Rr,we get
 
1
=1
maxfor=1.
2
'n
z
n
Prz rPzz

 (16)
For =1r, Inequality (16) was conjectured by Erdös
and later verified by Lax[10]. If we take =0

in
(15),we immediately get
 
1,>1.
2
n
R
PRzPz R


 (17)
Inequality (17) is due to Ankeny and Rivlin [1].
A polynomial n
PP is said to be self-inversive if
 
=PzuQz for allzCwhere =1uand


=Qz
(1 )
n
zP z. It is known[16, 17] that if n PP is self-
inversive polynomial, then for every 1p,
 
,
1
p
'
pp
Pz
Pzn z  (18)
Finally, we present the following result which include
some well-known results for self-inversive polynomials
as special cases.
Theorem 3. If n PP is self-inversive polynomial,
then for every real or complex number

with 1


,
>1Rr and >0,p
 


1.
1
nn
p
p
p
p
Rrz
PRz PrzPz
z



 
(19)
The result is best possible and equality in (19) ho lds
for

=1
n
Pz z.
Remark 6. Taking =0

in Theorem 3, it follows
that if n
PP is self-inversive polynomial, then for
>1R and >0,p



1.
1
n
p
p
p
p
Rz
PRz Pz
z

 (20)
The result is sharp.
Many interesting results can be deduced from Th eor em
3 in exactly the same way as we have deduced from The-
orem 2.
2. Lemmas
For the proofs of these theorems, we need the following
lemmas.
Lemma 1. If n PP and

Pz has all its zeros in
zk where 1k, then for every 1Rr and
=1z,
 
.
n
Rk
PRz Prz
rk




 (21)
Proof of Lemma 1. Since all the zeros of


Pz lie
in zk

, we write


=1
=ni
j
j
j
PzCz re



where j rk

. Now for 0<2

 , 1Rr , we
have


12
22
22
2
=2
i
ij
jjjj
i
ijjj j
j
RereRrRr Cos
rr rrCos
rer e













 




,=1,2,, .
j
j
Rr Rk jn
rr rk


 







Hence



=1
=
i
ii
j
nj
i
ii
j
jj
PRe Rer e
Pre rer e







=1
=
n
n
j
Rk Rk
rk rk






for 0<2



. This implies for =1z and >1Rr,
 
,
n
Rk
P
RzP rz
rk





which completes the proof of Lemma 1.
Lemma 2. If n PP

and

Pz does not vanish in
<1z, then for every real or complex number

with
1 

,1Rr, and =1z,




P
Rz PrzQRz Prz   (22)
where


=(1)
n
QzzP z .The result is sharp and
equality in (22) holds for

=1
n
Pz z.
Proof of Lemma 2. For the case =Rr, the result
follows by observing that




Pz Qz for 1z.
Henceforth, we assume that >Rr. Since the polynomi-
al


Pz has all its zeros in 1z, therefore, for every
real or complex number

with >1  , the polynomi-
al






=
f
zPz Qz  , where

=(1)
n
QzzP z, has
all its zeros in 1z

. Applying Lemma 1 to the
polynomial


f
z with =1k, we obtain for every
>1Rr and 0<2



,
 
1.
1
n
ii
R
fRe fre
r





 (23)
Since


0
i
fRe


for every >1,0 <2Rr

 
and 1> 1Rr

, it follows from (23) that

A. AZIZ ET AL. 324
  
1
>1
n
iii
r
f
Ref Refre
R


 



for every >1Rr and 0<2

 . This gives
 
<
f
rzfRz ,
for 1z and >1Rr.
Using Rouche’s theorem and noting that all the zeros
of

f
Rz lie in 1<1, zR  we conclude that the
polynomial
 
 

 

=
=
Tz fRzfrz
PRzPrzQ RzQ rz

 

  (24)
has all its zeros in <1z for every real or complex
number ,

 with 1, >1   and >1Rr. This
implies
 
P
RzPrzQ RzQ rz   (25)
for 1z and >1Rr. If Inequality (25) is not true,
then exist a point =
z
w with 1w such that
 
>.PRwP rwQ RwQ rw  
But all the zeros of

Qz lie in 1z, therefore, it
follows (as in case of

f
z) that all the zeros of
 
QRz Qrz   lie in <1z. Hence
 
0QRw Qrw   with 1w. We take
 
 
=,
PRw Prw
QRwQrw




then

is a well defined real or complex number with
>1  and with this choice of

, from (24) we obtain

=0Tw where 1w. This contradicts the fact that
all the zeros of ()Tz lie in <1z. Thus
 
P
RzPrzQ RzQ rz  
for 1z and >1Rr. This proves Lemma 2.
Next we describe a result of Arestov.
For

01
=,,,
n

   and

=0
=nj
j
n
j
Pzaz P

,
we define

=0
=.
n
j
jj
j
Pz az

 
The operator

 is said to be admissible if it pre-
serves one of the following properties:
1)

Pz has all its zeros in

:1,zCz
2)

Pz has all its zeros in


:1,zCz
The result of Arestov may now be stated as follows.
Lemma 3. [8] Let
  
=
x
logx  where

is a
convex nondecreasing function on .R Then for all
n
PP

and each admissible operator

,








22
00
,
ii
Pe dCnPe d




  

where




0
,=ax, .
n
Cn


In particular, Lemma 3 applies with :p
x
x   for
every


0,p

. Therefore, we have



 


11
22
00
,.
p
p
pp
ii
PedCnPed



 





(26)
We use (26) to prove the following interesting result.
Lemma 4. If n PP

and

Pz does not vanish in
<1z, then for every real or complex number

with
1, >1Rr 

, >0p and

real,






 




2
0
2
0
1.
ii
p
inini
pp
nnii
PRe Pre
eRPeRrPe rd
Rre Ped

 








  


(27)
Proof of Lemma 4. Let
 
=1
n
QzzPz . Since


Pz does no t vanish in <1z, by Lemma 2, for every
real or complex number

with 1,>1Rr   and
=1z,we have




  
=nn
PRz Prz
QRzQrz RPzRrPzr




Now(as in the proof of Lemma 2) , the polynomial
  
==11
nn nn
HzQRzQrzRzPRzrzP rz  
has all its zeros in <1z for every real or complex
number

with 1 

and >Rr, it follows that the
polynomi al
  
1=
nn n
zH zRPzRrPzr  
has all its zeros in >1z. Hence the function



 
=nn
PRz Prz
fz RPzR rPzr




is analytic in 1z

and

1fz  for =1z. Since


f
z is not a constant, it follows by the Maximum
Modulus Principle that


<1 for <1,fzz
or equivalently,
  
< for <1.
nn
PRzPrz RPzRrPzrz  
(28)
A direct application of Rouche’s theorem shows that

A. AZIZ ET AL. 325
 

 
 

 


 
0
=
=1
(1
in n
nnin
n
in n
Pz PRzPrz
e RPzRrPzr
Rre az
eR ra











   
does not vanish in <1z for every

with 1, 

>1Rr and

real. Therefore,

 is admissibe
operator. Applying (26) of Lemma 3, the desired result
follows immediately for each >0p. This completes the
proof of Lemma 4.
From lemma 4, we deduce the following more general
lemma which is a result of independent interest with
variety of application.
Lemma 5. If n PP, then for every real or complex
number

with 1, >1Rr  , >0p and

real,
 

  



2
0
2
0
1.
ii
p
inini
pp
nnii
PRe Pre
eRPeRrPer d
Rre Ped

 








  


(29)
The result is sharp and equality in (29) holds for

=,0
n
Pz z  
Proof of Lemma 5. Since

Pz is a polynomial of
degree at most n, we can write


12 =1= 1
== ,1
kn
jj
jjk
PzPzPzzzzz k




where all the zeros of

1
Pz lie in 1z and all the
zeros of

2
Pz lie in <1z. First we suppose that

1
Pz has no zero on =1z so that all the zeros of

1
Pz lie in >1z. Let
 
22
=1
nk
Qzz Pz
, then all
the zeros of

2
Qz lie in >1z and



22
=Qz Pz
for =1z. Now consider the polynomial



12 =1= 1
== 1,
kn
jj
jjk
g
zPzQz zzzz



then all the zeros of


g
zlie in >1z and for =1z,
  


12 12
===.
g
zPzQzPzPzPz (30)
By the Maximum Modulus Principle, it follows that
 
for 1.Pz gzz (31)
We claim that the polynomial
 


=hzPzgz  
does not vanish in 1z for every

with >1  . If
this is not true, then

0=0hz for some 0
z with
01z. This gives


00
=.Pz gz 
Since


00gz

and >1  , it follows that




00 0
>with 1,Pzgzz 
which clearly contradicts (31). Thus

hz does not
vanish in 1z

for every

with >1  , so that all
the zeros of


hz lie in z

 for some >1

and
hence all the zeros of


hz

lie in 1z. Applying
(28) to the polynomial


hz

, we get
  
<
for<1,>1.
nn
hRzhrzRhzRrhzr
zRr
  

Taking =,0<2
i
ze


 , then

=1 <1z  as
>1

and we get






<,
iinini
hRehreRheRrher
 
 
0<2



, >1Rr and 1  . This implies






< for =1.
nn
hRzhrz RhzRrhzrz  
An application of Rouche’s theorem shows that the
polynomi al







 



=in n TzhRzhrzeRhzRrhzr

 
does not vanish in 1z

for every real or complex
number

with 1, 

>1Rr and

real.
Replacing


hz by


Pz hz  , it follows that the
polynomi al
  




=in n TzPRzPrz e RPzRrPzr

 
 


 



inn
g
RzgrzeRgz Rrgz r

  
(32)
does not vanish in 1z

for every ,

 with 1 

and >1
 . This implies
 

  


    

in n
inn
PRzPrze RPzRrPzr
g
RzgrzeRgz Rrgz r




 
 
(33)
for 1z

, 1 

, >1Rr and

real. If Ine-
quality (33) is not true, then there is a point 0 =zz with
01z

such that
 


 

000 0
in n
PRzPrzeRPz RrPzr

  
  


000 0
>( ).
in n
g
RzgrzeRg zRrg zr

 
Since all the zeros of polynomials

g
z lie in >1z,
it follows (as before) that all the zeros of polynomial






 


inn
g
RzgrzeRg zRrgzr

  also li-

A. AZIZ ET AL. 326
e in >1z for every real or complex number

with
1  , >1Rr and

real. Hence
 

 


000 0
0
0
with 1.
in n
gRzgrzeRgzRrgz r
z

 

We take
 

 

 

 

0000
000 0
=
in n
in n
PRzPrzeRPz RrPzr
g
RzgrzeRg zRrg zr






 
so that

is a well-defined real or complex number
with >1  and with this choice of

, from (32) we
get

0=0Tz with 01z This clearly is a con-
tradiction to the fact that


Tz does not vanish in
1z. Thus for every

with 1  , >1Rr and

real,
 

 

    

in n
inn
PRzPrzeRPzRrPzr
g
Rzg rzeRg zRrg zr




 
 
for ||1z, which in particular gives for each >0p
and 0<2

 ,
 

 

  
 

2
0
2
0
ii
p
inini
ii
p
ini ni
PRe Pre
eRPeRrPer d
gRegre
eRgeRrge rd

 

 












Using lemma 4 and (30), it follows that for every

with 1  , >Rr, >0p and

real,
  
 







2
0
2
0
2
0
1
=1 .
ii
p
inini
pp
nnii
pp
nnii
PRe Pre
eRPeRrPer d
Rre ged
Rre Ped














  




(34)
Now if


1
Pz has a zero on =1z, then applying
(34) to the polynomial


*12
=Pz PtzPz where
<1t, we get for every

with 1  , >1Rr,
>0p and

real,
 

  
  
2**
0
**
2*
0
1.
ii
p
in in i
pp
nnii
PRe Pre
eRPeRrPer d
Rre Ped










  


(35)
Letting 1t in (35) and using continuity, the
desired result follows immediately and this proves
Lemma 5.
3. Proofs of the Theor ems
Proof of Theorem 1. Since

Pz is a polynomial of
degree at most n, we can write
 

12 =1= 1
== ,1
kn
jj
jjk
PzPzP zzzzzk



where all the zeros of


1
Pz lie in 1z and all the
zeros of


2
Pz lie in >1z. First we suppose that all
the zeros of


1
Pz lie in <1z. Let
 
22
=1
nk
Qzz Pz
,
then all the zeros of

2
Qz lie in <1z and




22
=Qz Pz for =1z. Now consider the poly-
nomial
 


12 =1= 1
== 1,
kn
jj
jjk
F
zPzQz zzzz



then all the zeros of


F
z lie in <1z and for =1z,





 


12 12
===.
F
zPzQzPzPzPz (29)
By the Maximum Modulus Principle, it follows that




for 1.PzFzz


Since


0Fz

for 1z and >1  , a direct
application of Rouche’s theorem shows that the poly-
nomial




=
H
zPz Fz   has all its zeros in
<1z for every

with >1  . Applying lemma 1 to
the polynomial


H
z, we deduce (as before)


<
H
rzH Rz
for =1z and >1Rr.
Since all the zeros of


H
Rz lie in 1 <1,zR

we
conclude that for every ,

 with ||1

 and ||>1

,
all the zeros of polynomial






 

 

=
=
Gz HRzHrz
P RzP rzFRzFrz



 
lie in <1z. This implies (as in the case of Lemma 2)






for 1
and>1,
PRzPrzFRzFrzz
Rr


 

which in particular gives for >Rr
and >0,p
 
 
2
0
2
0
p
ii
p
ii
PRePre d
F
ReFred









 (30)

A. AZIZ ET AL. 327
Again, since all the zeros of

F
z lie in 1z

, as
before,
 
F
RzF rz   has all its zeros in 1z

for
every real or complex number

with 1  . There-
fore, the operator

 defined by
 




0
=
=1
nnn
n
Fz FRzFrz
Rrbz b




 
is admissible. Hence by (26) of Lemma (3), for each
>0p, we have
 

2
0
2
0.
p
ii
p
p
nn i
F
ReF red
Rr Fed










(31)
Combining Inequalities (37) and (38) and noting that

=
ii
F
ePe

, we obtain for >1Rr and >0p
 



1
2
0
1
2
0.
p
p
ii
p
p
nn i
PRePre d
Rr Ped








(32)
In case

1
Pz has a zero on =1z, the Inequality
(39) follows by using similar argument as in the case of
Lemma 5. This completes the proof of Theorem 1.
Proof of Theorem 2. By hypothesis n PP

and

Pz does not vanish in 1z

, therefore, b y Lemma 2
for every real or complex number

with 1 

,
0<2

  and >1Rr,



 
ii
ni ni
PRe Pre
RPe RrPe r





 (33)
Also, by Lemma 5,
 



2
0
2
0
1
p
i
pp
nnii
FeGd
Rre Ped







  

 (34)
where

 
=ii
F
PRe Pre


  and

 
=.
ni ni
GRPeRrPer

 
Integrating both sides of (41) with respect to

from
0 to 2

, we get for each >0p, >1Rr and

real,
 
22
00
p
i
F
eG dd
 

 






22
00
1pp
nnii
RredPed



 

(35)
Now for every real

, 1t and >0p, we have
22
00
1.
pp
ii
te de d



 

If


0F 

, we take

=tG F

 , then by (40)
1t and we get
 
 

 

 


2
0
2
0
2
0
2
0
2
0
=1
=
=
1.
p
i
p
pi
p
pi
p
pi
p
pi
FeGd
G
F
ed
F
G
F
ed
F
G
F
ed
F
Fed





























For


=0F  , this inequality is trivially true. Using
this in(42), we conclude that for every real or complex
number

with 1 

, >1Rr and

real,


 







22
00
22
00
1
1.
p
p
iii
pp
nnii
edPRePred
RredPed











(43)
Since
 








2
0
2
0
2
0
2
0
2
0
1
=1
=1
=1
=1,
p
nni
p
nni
p
nni
p
nni
p
nni
Rre d
Rre d
Rre d
Rre d
Rre d





















(44)
the desired result follows immediately by combining (43)
and (44). This completes the proof of Theorem 2.
Proof of Theorem 3. Since


Pz is a self-inversive
polynomial, we have


=PzuQz for all zC

where =1u and


=1
n
QzzPz . Therefore, for
every real or complex number

and >1Rr,






= for all PRzPrzQ RzQ rzzC 


so that
 


 
==1.
ii
ni ni
PRe Pre
GF RPeRrPer



 



A. AZIZ ET AL. 328
Using this in (41) with 1 

and proceeding simi-
larly as in the proof of Theorem 2, we get the desired
result. This completes the proof of Theorem 3.
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