Some L
p
Inequalities for B-Operators
Nisar Ahmad Rather, Sajad Hussain Ahangar Department of mathematics, University of Kashmir, Harzarbal, Sringar, India.
Email: [email protected], [email protected]
Received November 8, 2012; revised December 8, 2012; accepted December 15, 2012
ABSTRACT
If P z
is a polynomial of degree at most n having all its zeros in z 1, then it was recently claimed by Shah and Liman ([1], estimates for the family of $B$-operators, Operators and Matrices, (2011), 79-87) that for every ,,
1
R
1
p
0, ,1 2
0
,1
n
p p
p
R
B P z P z
z
where B is a n-operator with parameters 0, ,1 2 in
the sense of Rahman [2],
z Rz and
3
2 8
n n
2
0 1 2 0 1 2
1
, , n
1
. Unfortunately the proof of this re-
sult is not correct. In this paper, we present certain more general sharp Lp-inequalities for -operators which not only
provide a correct proof of the above inequality as a special case but also extend them for as well.
n
0 p
Keywords: Lp-Inequalities; -Operators; Polynomials
n
1. Introduction and Statement of Results
Let n denote the space of all complex polynomials
0n j
j j
P z
a z of degree at most n. For Pn,define
2π
i0 0
1
: exp log e d ,
2π
P z P
2π
i 10 1
: e , 0
2π
p p
p
P z P p
1
: Max ,
z
P z P z
and denote for any complex function : the composite function of P and , defined by
, as
P
z :P
z
z
P.A famous result known as Bernstein’s inequality (for reference, see [3, p. 531], [4, p. 508] or [5] states that if
, then
n
P
,P z n P z (1.1)
whereas concerning the maximum modulus of P z
onthe circle z R 1, we have
n
, 1P Rz R P z R , (1.2)
(for reference, see [6, p. 442] or [3, Vol. 1, p. 137]).
Inequalities (1.1) and (1.2) can be obtained by letting in the inequalities
p
,p p
P z n P z p1 (1.3)
and
n
, 1, 0p p
P Rz R P z R p , (1.4)
respectively. Inequality (1.3) was found by Zygmund [7] whereas inequality (1.4) is a simple consequence of a result of Hardy [8] (see also [9, Th. 5.5]). Since in- equality (1.3) was deduced from M. Riesz’s interpolation formula [10] by means of Minkowski’s inequality, it was not clear, whether the restriction on p was indeed essen-
tial. This question was open for a long time. Finally Arestov [11] proved that (1.3) remains true for 0 p 1
as well.
If we restrict ourselves to the class of polynomials
n
P having no zero in z 1, then Inequalities (1.1)
and (1.2) can be respectively replaced by
,2
n
P z P z (1.5) and
1
1.2
n
R
P Rz P z R (1.6)
Ankey and Ravilin [13].
Both the Inequalities (1.5) and (1.6) can be obtain by letting p in the inequalities
,1
p p
p
P z
P z n p
z
0 (1.7)
and for R1,p0,
1
.1
n p
p p
p
R z
P Rz P z
z
(1.8)
Inequality (1.7) is due to De-Bruijn [14] for . Rahman and Schmeisser [15] extended it for
1
p
0 p 1
whereas the Inequality (1.8) was proved by Boas and Rahman [16] for and later it was extended for
by Rahman and Schmeisser [15].
1
p
0 p 1
Q. I. Rahman [2] (see also Rahman and Schmeisser [4, p. 538]) introduced a class n of operators that
carries a polynomial into
B
n
P
0 1
2 2
:
2 1!
,
2 2!
P z nz
B P z P z
P z
nz
(1.9)
where 0, 1 and 2 are such that all the zeros of
20 1 2
: ,1 , 2
U z C n zC n z (1.10)
where
!
, 0
! !
n
C n r r n
r n r
, lie in half plane
2 .
z z n
As a generalization of Inequality (1.1) and (1.5), Q. I. Rahman [2, inequality 5.2 and 5.3] proved that if
and then for
,
n
P Bn z 1,
n
0, ,1 2
,B P z P z (1.11)
and if Pn, P z
0 in z 1, then z 1,
0 1 2
0
1 , , ,
2 n
B P z P z (1.12) where
0 1 2
0 1 2 2 3
1
, , .
2 8
n
n n
n
(1.13)
As a corresponding generalization of Inequalities (1.2) and (1.4), Rahman and Schmeisser [4, p. 538] proved that if Pn, then z 1,
0, ,1 2
.n n
B P z R P z (1.14) and if Pn, P z
0 in z 1, then as a special case of Corollary 14.5.6 in [4, p. 539], we have
0 1 2 0
1
, , ,
2
n n
B P z
R P
z
(1.15)
where
z :Rz R, 1 and n
0, ,1 2
is definedby (1.13).
Inequality (1.15) also follows by combining the Inequalities (5.2) and (5.3) due to Rahman [2].
As an extension of Inequality (1.14) to Lp-norm, re-
cently Shah and Liman [1, Theorem 1] proved:
Theorem A. If Pn, then for every and
,
1
R
1
p
1, ,2 3
,n
n p
p
B P z R P z (1.16)
where Bn,
z Rz and n
0, ,1 2
is de- fined by (1.13).While seeking the analogous result of (1.15) in Lp
norm, they [1, Theorem 2] have made an incomplete attempt by claiming to have proved the following result:
Theorem B. If Pn, and does not vanish
for
P z1,
z then for each p1, R1,
1, ,2 3
0
, 1
n n
p
R
B P z P z
z
(1.17)
p
where Bn,
z Rz and n
1, ,2 3
is de- fined by (1.13).Further, it has been claimed in [1] to have proved the Inequality (1.17) for self-inversive polynomials as well.
Unfortunately the proof of Inequality (1.17) and other related results including the key lemma [1, Lemma 4] given by Shah and Liman is not correct. The reason being that the authors in [1] deduce:
1) line 10 from line 7 on page 84,
2) line 19 on page 85 from Lemma 3 [1] and, 3) line 16 from line 14 on page 86,
by using the argument that if
: n
1P z z P z , then
for
z Rz, R1 and z 1,
,B P z B P z
which is not true, in general, for every R1 and
1
z . To see this, let
1 0n k
n k
P z a z a z a za
be an arbitrary polynomial of degree n, then
1
0 1
: 1
= .
n
n n n k
k n
P z z P z
a z a z a z a
Now with 1:1n 2 and 2:2n2 8, we have
0 1 2
0
1 ,
n
n k n k k
k
B P z
n k n k n k a z R
and in particular for z 1, we get
0 1 2
0
1 ,
n n
k n
k k
B P z R z
z
n k n k n k a
R
whence
0 1 2
0
1
k n
n
k k
B P z
z
R n k n k n k a
R
.
But
0 1 2
0
1 ,
k n
n
k k
B P z
z
R k k k a
R
so the asserted identity does not hold in general for every and
1
R z 1 as e.g. the immediate counterexample of
: n demonstrates in view ofz
P z P
z 1,
0B P z and
2
3
0 1 2 2 1 8
B P z n n n
for z 1.
Authors [1] have also claimed that Inequality (1.17) and its analogue for self-inversive polynomials are sharp has remained to be verified. In fact, this claim is also wrong.
The main aim of this paper is to establish Lp-mean
extensions of the inequalities (1.14) and (1.15) for and present correct proofs of the results mentioned in [1]. In this direction, we first present the following result which is a compact generalization of the Inequalities (1.1), (1.2), (1.14) and (1.16) and also extend
Inequality (1.17) for as well.
0 p
0 p
1
Theorem 1. If P n then for with 1,
and
0 p R r 1,
0, ,1 2
,R r p
n n
n p
B P z B P z
R r P z
(1.18)
where Bn, t
z tz and n
0, ,1 2
is givenby (1.13). The result is best possible and equality holds in (1.18) for
n.P z z
0
If we choose in (1.18), we get the following result which extends Theorem A to 0 p 1,
Corollary 1. If Pn then for 0 p and
1,
R
0, ,1 2
,n
n p
p
B P z R P z (1.19)
where Bn,
z Rz and n
0, ,1 2
is givenby (1.13).
Remark 1. Taking 0 0 2 in (1.19) and noting that in this case all the zeros of U(z) defined in (1.10) lie
in z z n2 , we get for R1 and 0 p
n 1
p p
P Rz nR P z ,
which includes (1.4) as a special case. Next if we choose
1 0 2
in (1.19), we get inequality (1.4). Inequality (1.11) also follows from Theorem 1 by letting
in (1.18).
p
Theorem 1 can be sharpened if we restrict ourselves to the class of polynomials P z
which does not vanish in1
z In this direction, we next present the following interesting compact generalization of Theorem B which yields Lp mean extension of the inequality (1.12) for
0 p which among other things includes a correct
proof of inequality (1.17) for 1 as a special
case.
p
Theorem 2. If Pn and does not vanish
for
P z1
z then for with 1, 0 p and R r 1,
0, ,1 2 1 0 1
R r p
n n
n
p p p
B P z B P z
R r z
P z z
(1.20)
where Bn, t
z tz and n
0, ,1 2
is de-fined by (1.13). The result is best possible and equality holds in (1.18) for P z
aznb, a b 1.If we take 0 in (1.20), we get the following result which is the generalization of Theorem B for
but also extends it for
1
p 0 p
Corollary 2. If Pn and does not vanish
for
P z1
z then for 0 p and R1,
0, ,1 2
0
, 1
n
n p
p p
p
R z
B P z P z
z
(1.21)
,
n
B
z Rz and n
0, ,1 2
is defined by (1.13).By triangle inequality, the following result is an immediately follows from Corollary 2.
Corollary 3. If Pn and does not vanish
for
P z1
z then for 0 p and R1,
0, ,1 2
0
1
n n p
p
R
B P z P z
z
(1.22)
p
,
n
B
z Rz and n
0, ,1 2
is defined by (1.13).Remark 2. Corollary 3 establishes a correct proof of a result due to Shah and Liman [1, Theorem 3] for and also extends it for
1
p
Remark 3. If we choose 0 0 2
in (1.21), we
get for R1 and 0 p ,
1
1
n
p p
p
nR
P Rz P z
z
which, in particular, yields Inequality (1.7). Next if we take 1 0 2
in (1.21), we get Inequality (1.8). Ine-quality (1.12) can be obtained from corollary 2 by letting
in (1.20). p
By using triangle inequality, the following result im- mediately follows from Theorem 2.
Corollary 4. If and does not vanish
for
n
P P z
1,
z then for with 1 0 p
and R r 1,
0, ,1 2 1 0 1
R r p
n n
n
p p
B P z B P z
R r
P z z
(1.23)
,
n
B t
t tz and n
0, ,1 2
is defined by (1.13).A polynomial is said be self-inversive if
where n
P
P z P z 1 and is the con-
jugate polynomial of , that is,
P z
P z P z :z Pn
1z .Finally in this paper, we establish the following result for self-inversive polynomials , which includes a correct proof of an another result of Shah and Liman [1, Theo- rem 2] as a special case.
Theorem 3. If and is a self-inversive
polynomial, then for
n
P P z
with 1 0 p
and R r 1,
0, ,1 2 1 0
, 1
R r p
n n
n
p
p p
B P z B P z
R r z
P z z
(1.24) where Bn, t
t tz and n
0, ,1 2
is given by (1.13). The result is sharp and an extremal polynomialis
n
c az
0,
P z a , ac0.
For we get the following result.
Corollary 5. If and is a self-inversive
polynomial, then for and
n
P
0 p
P z
R1,
0, ,1 2 0
, 1
p
n
n p
p p
B P z
R z
P z z
(1.25)
where Bn,
z Rz and n
0, ,1 2
is givenby (1.13).
The following result is an immediate consequence of
Corollary 5.
Corollary 6 If Pn
0 p
and is a self-inversive polynomial, then for
P z and R1,
0, ,1 2 0
, 1
p
n n
p p
B P z
R
P z z
(1.26)
where Bn,
z Rz and n
0, ,1 2
is givenby (1.13).
Remark 4. Corollary 6 establishes a correct proof of a result due to Shah and Liman [1, Theorem 3] for and also extends it for
1
p
0 p 1 as well.
Remark 5. A variety of interesting results can be eas-ily deduced from Theorem 3 in the same way as we have deduced from Theorem 2. Here we mention a few of these. Taking 0 0 2
p
= in (1.25), we g r
1
R
et fo and 0 ,
1
1
n
p p
p
nR
P Rz P z
z
,
which, in particular, yields a result due to Dewan and Govil [17] and A. Aziz [18] for polynomials n .
Next if we choose
P
1 0 2
in (1.25), we get for
1
R ; 0 p
1
1
n p
p p
p
R z
P Rz P z
z
.
The above inequality is a special case of a result proved by Aziz and Rather [19].
Lastly letting p in (1.25), it follows that if
P z , is a self-inversive polynomial then
0 1 2 0
1, , ,
2
n n
B P z
R P
z
(1.27)
where Bn,
z Rz and n
0, ,1 2
isde-fined by (1.13). The result is sharp.
Inequality (1.27) is a special case of a result due to Rahman and Schmeisser [4, Cor. 14.5.6].
2. Lemma
For the proof of above theorems we need the following Lemmas:
The following lemma follows from Corollary 18.3 of [20, p. 86].
Lemma 1. If Pn and P z
has all zeros in1,
z then all the zeros of B P
z also lie in1.
z
Lemma 2. If Pn and P z
have all its zeros in1
1
. 1n
R
P Rz P rz
r
Proof. Since all the zeros of P z
lie in z 1, wewrite
i
1
e j ,
n j j
P z C z r
where rj1. Now for 0< 2π, R r 1, we have
1 2
i 2 2
i
i 2
i
2 cos
e e
2 cos
e e
1
,for 1, 2, , . 1
j j
j j j
j
j j j
j
j
j
R r Rr
R r
r r rr r r
R r R
j n
r r r
Hence
i
i i
i
i i
1
1
R e R e e
e e e
1 1 ,
1 1
j j
n
j
j j
n n
j
P r
P r r r
R R
r r
for 0 2π. This implies for z 1 and R r 1,
1
,1
n
R
P Rz P rz
r
which completes the proof of Lemma 2.
Lemma 3. If Pn and P z
has no zero in1,
z then for every with 1, and
> 1
R r
1
z ,
,R r
R r
B P z B P z
B P z B P z
(2.1)
where
: n
1P z z P z and t
z tz.
P z
Proof. Since the polynomial has all its zeros in
1,
z therefore, for every real or complex number
with 1, the polynomial f z
P z
P
z ,where
: n
1P z z P z has all zeros in z 1. App-
lying Lemma 2 to the polynomial f z
, we obtain for every R>r1 and 0 2π,
ei 1
ei 1n
R
f R f r
r
.
0
(2.2)
Since
ei for everyf R R r 1, 0 2π
and R 1 r 1, it follows from (2.2) that
ei 1
ei
1
n
R
f R f R f r
r
i e , for every R r 1 and 0 2π. This gives
<
for 1, and 1.f rz f Rz z Rr
Using Rouche’s theorem and noting that all the zeros of f Rz
lie in z 1 R1, we conclude that the polynomial
z f rz
P Rz P rz P Rz P rz
T z f R
1
z
has all its zeros in for every real or complex with 1 and R r 1.
Applying Lemma polynomial T z
and notitha
1 to ng
t B is a linear operator, it follows that all the zeros of polynomial
R r
R r
R r
f z B f z
B P z B P z
B P z B P z
B T z B
where t
z tz. 1z This implies
lie in
R r
B P P z
R r
z B
B P z B P z
(2.3)
for z 1 and If Inequality (2.3) is not true, there
1.
R r z
then exits a point z0 with z0 1 such that
0 0
R r
B P z P
0 0
R r
B z
B P z B P z
(2.4)
But all the zeros of P
Rz lie in z 1R1,therefore, it follows (as in case of f z
) thatzeros of
all the
P Rz P rz lie in z 1 Hence, by
Lemma 1, we have
z0 B P r
z0
0.R
B P
We take
00
0 0,
R r
R r
B P z B P z
B P z B P z
then is well defined real or complex number with
1
and with this choice of , we obtain
0 0B T z where z0 1. This contradicts the fact the zeros of
that all B T
z lie in z 1. Thus (2.3) holds true for 1 and r 1.Next we describe a result ov [1 For
R
of Arest 1].
,
n1
0, ,1 n and
n j
j 0 j nP z
a z , we define
n0
.
j j j j
P z a z
serves one of the following properties:
1) P z
has all its zeros in
z: z 1 .
2) P z
has all its zeros in
z: z 1 .
The result of Arestov [11] may now be stated as follows.
4] Let
Lemma 4. [11, Theorem
x
logx
where is a convex non decreasing function on . Then for all Pn and each admissible operator ,
2π i
0
i
e d
, e d ,
P
C n P
2π
0
where C
,n max
0 ,n
.In particular, Lemma 4 applies with : p
x x
for
every p
0,
. Therefore, we have
1
2π i
0 e d
1 2π i
0
, e d .
p p
p p
C n P
(2.5)
We use (2.5) to prove the following intere
Lemma 5. If and
P
sting result.
n
P P z
does not vanish in1,
z then for every p0, R r 1 and for real, 0 2π,
2π i i
0 B P R B P r e
i
i i
i
0 1 2 0
2π i
0
e e
e e d
, , e 1
e d ,
R r
p
n n
p
B P B P
R r
P
(2.6)
where z
n
B , t
z :t ,
:
t t
B P z B P z and
0, ,1 2
is defined by (1.13).f. Sinc and
Proo e Pn
:
1 nP z z P z , by
Lemma 3, we have for z 1,
,R
B P z B P r z
z P z
Also, since
R r
B P B
(2.7)
1
1
n n n n
P Rz P rz
R z P Rz r z P rz
0
1 1 2
1
1 1 2
2
2 2
1 3 2 4
2 1
1 1
1 1
2
1 1
1 1
2! 2
2 1 1 1
1 1 2 1
R r
n n n n
n n n n
n n n n
n n
n n n n
n n n n
B P z B P z
R z P Rz r z P rz
nz
nR z P Rz R z P Rz
nr z P rz r z P rz
nz
n n R z P Rz
n R z P Rz R z P Rz
n n r z P rz n r z
3
2 4
1 1
n n
P rz
r z P rz
and therefore,
3 2
0 1 2
2
1 1
1 2
2
2 2 2 2
2
1
2 8
1
2 4
. 8
R r
R
B P
r
n n
n n
n n
B P z B P z
z B P z
n n
n
R P z R r P z r
n n
n
R zP z R r zP z r
n
R z P z R r z P z r
(2.8) Also, for z 1,
= B .
R r
R r
B P z B P z
P z B P z
Using this in (2.7), we get for z 1,
.R r
B P z B P z
P z
As in the proof of Lemma 3, the polynomial
R r
B P z B
,R r
P z P z
,
has all its zeros in z 1
and by Lemma 1, R
z B P r
z ,
B P
also has all its zero in z 1, therefore,
r
z has all its zerR
B P z B P os in
1.
z Hence by the maximum modulus principle, for
1,
z
.B P z B P z
(2.9)
R r
R r
B P z B P
A direct application of Rouche’s theorem shows that
with
0,n n
P z a z a
i
e R
B P z B P z B P z
3 2
i
0 1 2 0
3 2
i
0 1 2 0
0
1
e 1
2 8
1
e 1
2 8
,
R r
r
n n
n n
n n
P z
B P z
n n n
R r a z
n n n
R r a
has all its zeros in
1
z for every real ,
0 2π. Therefore,
ma 4,
0
is an admissible ope .5) of Lem the desired result foll
h
ce the ra
result.
rator. ows Applying (2
immediately for eac p .
From Lemma 5, we dedu following more gene l
Lemma 6. If Pn, then for every p0,
1
R r and real 0 2π,
2π i i i
B
0
i e
R r
p
R
i
i
0 1 2 0
2π i
0
e e e
e d
, , e 1
e d ,
r
p
n n
p
P B P
B P B P
R r
P
(2.10)
Proof. Let and let be the zeros
of . If
n
P z z1, , ,2 zn
P z zj 1 for , then the
resu llows b a 5. assume that
zero i
all 1, 2, , Henceforth, we
n
j n
lt fo
P z ha
y Lemm
s at least one z 1 so that we can
write
1 2
,
k n
j j
P z P z P z
a z z
1 1
0 1, 0,
j j k
z z
k n a
where the zeros z z1, , ,2 zk of P z1
lie in z 1 and the zeros zk1,zk2, , zn of P z2
lie in z 1.First we suppose that P z1
has no zero on z 1 so that all the zeros of P z1
lie in z 1. Since all the lie in zeros of
nk
th degree polynomial P z2
1
z , all the ze jugate po omial
roes of its con lyn
2 n k 2 1
P z z P z lie in z 1 and
2 2
P z P z for z 1. Now consid e poly-
nomial
er th
1 2
f z P z P z
1 1
1 j ,
j j k
a z zz
k j
n
then all the zeroes of z
f z lie in z 1, and for
1,
z
1 2
1 2 .
P z P P z
f z P z P z
z
(2.11)
Therefore, it follows by Rouche’s Theorem that the polynomial g z
P z
f z
has all its zeros in1
z for every , with 1 so of
that all the zeros
T z g z lie in z 1 for some 1. Apply- ing (2.9) and (2.8) to the polynomial T z
, we get for1
and
R z 1,
3
0 1 2
2
1 2
1 1
2
2 2 2 2
2
1
2 8
1
2 4
, 8
n n
n n
n n
z
n n
n
R T z R r T z r
n n
n
R zT z R r zT z r
n
R z T z R r z T z r
that is,
2 <
R r
R r
B T z B T z
B T z B T
3 2 0
1 2
2
1 2
1 1
2
2 2 2 2 2 2
2
1 <
2 8
1
2 4
, 8
R r
n n
n n
n n
B T z B T z
n n
n
R g z R r g z r
n n
n
R zg z R r zg z r
n
R z g z R r z g z r
(2.12)
for If ei ,0 < 2π
z , then z
1 11.
z
as 1 and we get
ei
ei
R r
R
B g z B g r
z has all zeros in
3 2
0 1 2
i i
2
1 2
1 i i 1 i i
2
2 i i 2 i i
2
1 <
2 8
e e
1
2 4
e e e e
e e e e ,
8
n n
n n
n n
n n
n
R g R r g r
n n
n
R g R r g r
n
R g R r g r
Equivalently, for
1
z and thus B g R
z B g r
z does not vanish in z 1.An application of Rouche’s theorem shows that the polynomial
i e
R r
R r
L z B g z B g z
B g z B g z
(2.13)
has all zeros in z 1. Writing in
:=
g z P z f z and noting that B is a linear
operator, it follows that the polynomial
1,
z
i
i
e
e ,
R r
R r
R R
R r
L z B g z B g z
B g z B g z
B f z B f z
B f z B f z
(2.14)
,R r
R
B g
r
B g z B g z
z B g z
where Since
.
t z tz
g z has all its zeros in z 1, it follows that g
z has its zeros in z 1 and hence (pro- rly as in proof of Lemma 3) the poly- nomceeding simila
ial gR
z gr
z also has all its zeros has all its zeros in z 1 for every with 1.in z 1. By Lemma 1, We claim
i
i e
e ,
R r R r
R R R r
B P z B P z B P z B P z
B f z B f z B f z B f z
(2.15)
for z 1. If Inequality (2.15) is not true, then there exists a point zz0 with z0 1 such that
i
0 0 0
i
0 0 0
e
e ,
R r R r
R R R r
B P z B P z B P z B P z
B f z B f z B f z B f z
0
0
Since f z
has all its zeros in z 1, proceeding similarly as in the proof of (2.13), it follows that
ei
0R R R r
B f z B f z B f z B f z for z 1 We take
i
0 0
i
0 0
e e
R r
B P z B P z
B f z B f z
0 0
0 0
R r
R R R r
B P z B P z
B f z B f z
so that is a well-defined real or complex number with 1 and with this choice of , from (2.14), we get . This clearly is a contradiction to the fact
that
0 0L z
L z has all its zeros in z 1. Thus (2.15) holds, which in particular gives for each p0 and real,
2π i i i i
e e e e d
p
R r r
B P B P B P B
i 0
2π i i i i i
0
e
e e e e e d .
R
p
R R R r
P
B f B f B f B f
,
f gives for each p0, Lemma 4 and (2.7) applied to
2π i i i i i
0
2π
2π
i i
0 1 2 0 0
e e e e e d
, ,
, , e 1 e d .
p
R r R r
p p
p p
n n
B P B P B P B P
R r P
i i
0 1 2 e 1 0 0 e d
n n
R r f
(2.16)
Now if
, then applying (2.16) to the polynomial P z
P1
z P z2
1
P z has a zero on z 1 where
0 1, we get for each p0, R r 1 and real,
2π i i i i i
0
2π
0
e e e e e d
d
p
R r R r
B P B P B P B P
i i
0, ,1 2 e 1 0 e .
p p
n n
R r P
(2.17)
tting
Le 1 in (2.17) and using continuity, the desired result follows immediately and this proves Lemma 6. Lemma 7. If Pn, then for every p0, R r 1 and 0 2π,
2π 2π i
0 0
2π i 2π i
0 1 2 0
0 0
e d
, , e 1 d e d ,
p
R r R r
p p
n n
B P z B P z B P z B P z
R r P
d
(2.18)
where Bn, t
z tz and n
0, ,1 2
0.is de by
tremal polynomial for any b
fined (1.13). The result is best possible and is an ex
Proof. By Lemma 6, for each
nP z bz
0
p , 0 2π and R r 1, the Inequality (2.6) holds. Since
R r
B P z B P z is the conjugate polynomial of B P R
z B P r
z ,
ei
eR r R r
B P z B P z B P B P i , and therefore for each p0, R r 1 and 0 2π , we have
2π i i i i
0 B P e B
i
2π i i i i i
0
2π i i i i
0
e e e e d
e e e e e d
e e e e d .
p
R r R