anl p inequality for polynomials

7/30/2019 AnL p inequality for polynomials

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J. Math. Anal. Appl. 336 (2007) 14561465

www.elsevier.com/locate/jmaa

An Lp inequality for polynomials

M.A. Qazi

Department of Mathematics, Tuskegee University, Tuskegee, AL 36088, USA

Received 26 November 2006

Available online 28 March 2007

Submitted by Richard M. Aron

Abstract

LetPn be the class of all polynomials of degree at most n, and let Mp(g; ) denote the Lp mean ofg onthe circle of radius centered at the origin. We specify a number (0, 1), depending on n and k, suchthat for any f Pn, the ratio Mp(f(k); )/Mp(f; 1) is maximized by f(z) := zn for all [, )and p 1. The interest of the result lies in the fact that is strictly less than 1. 2007 Elsevier Inc. All rights reserved.

Keywords: Polynomials; Bernsteins inequality; Zygmunds inequality

1. Introduction and statement of results

For any entire function , let

Mp(; r) :=

1

2

reip d

1/p

(0 < p < , r > 0). (1)

By a classical result of G.H. Hardy [4], Mp(; r) is a strictly increasing function ofr except inthe case where is a constant. It is well known (see, for example, [5, p. 143]) that for any givenr > 0,

Mp(; r) max|z|=r

(z)

as p .

E-mail address: [email protected].

0022-247X/$ see front matter 2007 Elsevier Inc. All rights reserved.doi:10.1016/j.jmaa.2007.03.065


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M.A. Qazi / J. Math. Anal. Appl. 336 (2007) 14561465 1457

Hence,

M(; r) := max|z|=r

(z)

(r > 0)

may be seen as M(; r). We add that, by the maximum modulus principle, M(; r) is astrictly increasing function ofr unless f is a constant.Let Pn be the class of all polynomials of degree at most n. It is clear that iff belongs to Pn

then so does f(z) := znf (1/z), and so for R > 1 and any p > 0, we have

Mp(f; R) = RnMp

f;

1

R

RnMp

f; 1

= RnMp(f; 1),

where equality holds in the inequality only iff is a constant. Hence,

Mp(f; R ) < RnMp(f; 1) (0 < p ; 1 < R < ), (2)

unless f is a constant multiple ofzn.It is also known that for any polynomial f of degree at most n, we have

Mp(f; 1) nMp(f; 1) (0 < p ). (3)

The inequality is sharp. It becomes an equality for f (z) := czn.The case p = of (3) is known as S. Bernsteins inequality (see [8, Chapter 14]). It was

proved by A. Zygmund [10] for p [1, ). The fact that (3) is also true for 0 < p < 1 is a resultof V.V. Arestov [1].

The following result is an immediate consequence of(2) and (3).

Theorem A. Letf be a polynomial of degree at most n. Then

Mp(f; ) nn1Mp(f; 1) (0 < p ; 1 < ). (4)

The inequality is sharp. It becomes an equality for f(z) := czn.

We show that the restriction on in (4) can be relaxed. In order to state our result we need tointroduce some notation.

Let

Am() :=

1 m1m

1mm1

0m1

m1 2

mm21

mm1

... ... . . . ... ...1

mm12

mm2 1 m1

m

0 1mm1

m1m

1

(m+1)(m+1)

, (5)

and denote by detAm() the determinant of this matrix. It may be noted that detAm() is a poly-nomial in 1/ and so a continuous function of for 0 < < .

Notation 1. Let m = m,m+1 be the largest root of detAm() = 0 in (0, 1).

It is clear that detAm

() is a polynomial of degree m2 1 in 1/. For large m, it may be hardto compute m explicitly. We would have liked to give an asymptotic estimate for it as m tendsto infinity, but at this time we do not have one.

We shall see that detAm() is strictly positive in (m, ). Now, we are ready to state ourextension of Theorem A.


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1458 M.A. Qazi / J. Math. Anal. Appl. 336 (2007) 14561465

Theorem 1. LetM(; r) := max|z|=r |(z)|, and for 0 < p < , letMp(; r) be as in (1).Furthermore, as indicated in Notation 1, let m be the largest root of detAm() = 0 in (0, 1).Then, for any polynomial f of degree at most n, and1 p , we have

Mp

f(k); n!

(n k)!nkMp(f; 1) (k = 1, . . . , n ) (6)

ifk1

j=0 nj < .

Remark. Let f (z) := zn + zn1. Then M(f; 1) = 2 and

M

f(k);

=n!

(n k)!nk +

(n 1)!

(n k 1)!nk1 (k = 1, . . . , n 1).

Hence,

M

f(k);

> n!(n k)!

nkM(f; 1)

for 0 < 1 k/n. This shows thatk1

j=0 nj cannot be smaller than 1 k/ n.

In the case where p = 2 we can easily prove the following result which leaves little to bedesired.

Theorem 2. For any polynomial f(z) :=n

=0 a z of degree at most n and k = 1, . . . , n, we

have

M2

f(k);

2 +

n!

(n k)!

22n2k

k1=0

|a |2

n!

(n k)!

22n2k

M2(f; 1)

2 (7)

if 1 k/ n. The inequality may not hold for any < 1 k/ n. Besides, the coefficient ofk1=0 |a |

2 cannot be replaced by a larger number.

Inequality (7) says in particular that

M2(f; ) nn1

M2(f; 1)

2 |a0|2

1

1

n

.

2. Some auxiliary results

We start with a standard notation.For a function g that is defined and bounded on the unit circle we shall sometimes write g

to denote sup{|g(z)|: |z| = 1}.Next, we recall the notion of Hadamard product of two holomorphic functions defined in

concentric disks.Let A(z) =

=0 z

and B(z) =

=0 z be holomorphic in the disks |z| < R1 and

|z| < R2, respectively. Then, =0 z

defines a function holomorphic in the disk |z| 0 for arbitrary values of the variablesx1, . . . , xn, not all equal to zero.

Here is then [8, Theorem 12.2.8] a necessary and sufficient condition for a polynomial Q tobelong to B0

n

.

Lemma 1. The polynomial Q(z) := 1 +n

=1 c z belongs to the class B0n introduced in Defini-

tion 1, if and only if the Hermitian form

x := (x0, . . . , xn) ,

cxx (c0 = 1, ck = ck, 1 k n)

is positive semidefinite.

Let us recall that a principal minor of a matrix A is a minor whose diagonal is part of the

diagonal ofA.The definiteness of a Hermitian form can be decided in terms of the principal minors ofits matrix. The precise relationship is summarized in Lemma 2 ([3, p. 337], see Theorems 19and 20). An alternative reference could be [6, Chapter XIII]. However, as a cautionary note, itmay be mentioned that in [6] the definition of semidefinite is slightly different.

Lemma 2.

(i) A Hermitian form

H(x,x) :=

n,=1

h xx

is positive definite if and only if its leading principal minors are all positive, that is

Dk :=

h11 h1k

hk1 hkk

> 0 (k = 1, . . . , n ) .

(ii) A Hermitian form is positive semidefinite if and only if all its principal minors are non-negative.

The next lemma, which is a result of E. Fischer [2], simplifies certain things for us in asignificant way. For its proof we refer the reader to [6, p. 420].


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Lemma 3. Let (cj k) be a positive definite Hermitian (n + 1) (n + 1) matrix. Then, for =1, . . . , n, we have

c1,1 c1,

c,1 c,

c+1,+1 c+1,n+1

cn+1,+1 cn+1,n+1

c1,1 c1,n+1

cn+1,1 cn+1,n+1

.

For the proof of Theorem 1, we shall also need the following auxiliary result [7, Lemma 3].

Lemma 4. LetPn be the linear space of all polynomials of degree at mostn, normed by f :=

max|z|=1 |f (z)|. Also let t0, . . . , t n be an arbitrary set of n + 1 numbers in C, and denote by Lthe linear functional defined on Pn, by

f t0a0 + + tnan

f (z) :=

n=0

a z

.

Furthermore, letN := L. Then, for any non-decreasing convex function on [0, ), we have

|n

=0 t a ei|

N

d

n

=0

a ei

d .

3. Proofs

Proof of Theorem 1. By Theorem A, inequality (6) holds for all p (0, ] and any 1. So,we only need to prove that it also holds for

k1j=0 nj < 1 if p [1, ].

First we shall treat the case where p = and k = 1. Thus, with m as in Notation 1, we wishto prove that if

q(z) = qn, (z) := 0 + z + 2z2 + + 1z + + nn1zn ( > 0),

then, for any polynomial f of degree at most n, we have

q f := max|z|=1(q f )(z) n

n1f (n < 1). (8)

Iff belongs to Pn then so does f(z) := znf (1/z). Besides, f = f. Hence, (8) holdsif and only ifq f

nn1f (n < 1; f Pn). (9)

Since q f = (q f) and (q f) = q f inequality (9) is equivalent toq f nn1f (n < 1; f Pn).

Thus, setting

Q (z) :=

1

nn1 q

(z) = 1 +

n=1

(n )n1

nn1 z

we have to prove that

Q f 1 (n < 1; f Pn, f 1).


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Since Q (0) = 1, this means that Q belongs to the class B0n introduced in Definition 1. Hence, if

c0 = c0() := 1, and c () = c () :=(n )n1

nn1for = 1, . . . , n ,

then, in view of Lemma 1, it would suffice to show that the Hermitian form

(x0, . . . , xn) =,

cxx

is positive definite for n < 1. This is exactly what we shall do.We shall show that the leading principal minors of detAn() are all strictly positive for n 0 for n, < 1 and detAn,(n,) = 0.

We claim that

n,n+1 n, ( = 1, . . . , n ) . (10)

For this we first note that ifAn() is written as an (n + 1) (n + 1) matrix (cj k), then

c+1,+1 c+1,n+1

cn+1,+1 cn+1,n+1

=An,n+1 ( ) ( = 1, . . . , n ) .

For a proof (of (10)) by contradiction let us suppose that (10) is not true, i.e. there exists aninteger {1, . . . , n} such that

n, = max1n

n, > n,n+1, (11)

and so, by part (i) of Lemma 2, the matrix An() is positive definite for n, < 1. Then,Lemma 3 in conjunction with (11) implies that

detAn, () detAn,n+1 () detAn,n+1( ) (n, < 1). (12)

As n, the left-hand side of (12) tends to 0 whereas its right-hand side tends todetAn,n+1(n, )a strictly positive quantity if (11) was really true. This is a contradictionwhich proves (10). Hence, the matrix An() is positive definite for n < 1 and, by part (ii)of Lemma 2, all the principal minors ofAn() are non-negative for these values of . By con-tinuity, they remain non-negative for = n. Once again, part (ii) of Lemma 2 implies that thematrix An() is positive semidefinite for n 1. Applying Lemma 1 we conclude that

M(f; ) nn1M(f; 1) (13)

for n 1, which completes the proof of(6) in the case where p = and k = 1.Now, let us turn to the case where p [1, ). For any given [n, 1] let

t := 1 ( = 0, 1, . . . , n ) .

Then

f t0a0 + + tnan

f (z) :=

n=0

a z


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defines a linear functional L on the linear space Pn of all polynomials f of degree at most n.By (13), the norm L of this functional is nn1. Lemma 4 with N = nn1 shows that (6)holds also for p [1, ) ifk = 1.

Now we shall show how to prove (6) for k = 2, etc.Let 1 p < . We already know that1

2

fneip d1/p

nn1n

1

2

feip d1/p

.

So, applying the case k = 1 of(6) to the polynomial f(nz), which is a polynomial of degree atmost n 1, we see that for n1 < we have

n 12

fneip d

1/p

(n 1)n2 12

fneip d

1/p

n(n 1)n1n n2

1

2

feip d1/p

.

Thus, for any p [1, ) the inequality1

2

f

ei

pd

1/p n(n 1)n2

1

2

fei

pd

1/p(14)

holds if nn1 < . This proves (6) in the case where k = 2 and 1 p < . Lettingp in (14), we see that (6) holds also when p = . Repeated application of this argumentgives (6) for any k and any p [1, ]. 2

Proof of Theorem 2. Since

f(k)(z) =

nk=0

(n )!

(n k )!anz

nk,

we have

M2

f(k);

2=

nk=0

(n )!

(n k )!

2

|an|22(nk)

=

n!

(n k)!

2 nk=0

(n )!

(n k )!

(n k)!

n!

2|an|

22(nk).

Now, note that

(n )!

(n k )!

(n k)!

n!

= 1=0 (n k )1

=0 (n )

=

1

=0

1 k

n 1

k

n

.

Hence,(n )!

(n k )!

(n k)!

n!

22(nk) 2(nk)

n k

n <

,


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which implies that if 1 kn

, then

M2f(k); 2 n!(n k)!

2

2(nk)nk

=0

|an|2.

From this (7) readily follows.We claim that (7) may not hold for 0 < < 1 k

n. In fact,

(n 1)!

(n k 1)!

22n2k2 >

n!

(n k)!

22n2k

0 < < 1

k

n

,

and so for any polynomial of the form f (z) := anzn + an1zn1 +k1

=0 a z with an1 = 0

and k n 1, we have

M2

f(k);

2+

n!

(n k)!

2

2n2kk1=0

|a |2

=

n!

(n k)!

2|an|

22n2k +

(n 1)!

(n k 1)!

2|an1|

22n2k2

+

n!

(n k)!

22n2k

k1=0

|a |2

> n!

(n k)!2

2n2k

M2(f; 1)2

0 < < 1

k

n

.

In order to see that the coefficient {(n!)/(n k)!}22n2k ofk1

=0 |a |2 in (7) cannot be

replaced by a larger quantity, we may consider any polynomial of the form f (z) := anzn +k1=0 a z

withk1

=0 |a | > 0. 2

4. The restriction on p in Theorem 1

It is known (see [5, p. 139]) that ifMp(; r) is as in (1), then for any given r > 0,

Mp(; r) exp

1

2

logreid

as r 0.

This is the motivation behind the definition

M0(; r) := exp

1

2

logreid

(0 < r < ).

Using Jensens theorem [9, p. 124] it can be shown that if f is a polynomial of the form

f(z) := amm

=1(z z), then

exp

1

2

logfeid

= |am|

m=1

max

|z|, 1

,


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and so

M0(f; 1) = |am|m

=1

max|z|, 1. (15)Now, let f(z) := (z 1)n. Then, we may apply (15) to conclude that M0(f; 1) = 1. The

same formula shows that

M0(f; ) = nn1

n1=1

max

1, 1

= n (0 < < 1),

that is M0(f; ) = nM0(f; 1) for 0 < < 1. So, there is no chance of (6) being true for allp [0, 1).

Note added in proof

Within the last four weeks I have learned that the case p = of Theorem 1 appeared firstin a paper entitled On a class of extremal problems for polynomials in the unit circle by H.S.Shapiro, published in Portugaliae Mathematica 20 (1961) 6793 (see Theorem 9 on page 91).However, our proof is totally diferent; it has nothing in common with his.

References

[1] V.V. Arestov, On integral inequalities for trigonometric polynomials and their derivatives, Math. USSR Izv. 18

(1982) 117; Izv. Akad. Nauk SSSR, Ser. Mat. 45 (1981) 322.[2] E. Fischer, ber den Hadamardschen Determinantensatz, Arch. Math. Phys. (3) 13 (1908) 3240.[3] F.R. Gantmacher, The Theory of Matrices, vol. I, Chelsea, New York, 1959.[4] G.H. Hardy, On the mean value of the modulus of an analytic function, Proc. London Math. Soc. (2) 14 (1915)

269277.[5] G.H. Hardy, J.E. Littlewood, G. Plya, Inequalities, Cambridge Univ. Press, 1934.[6] L. Mirsky, An Introduction to Linear Algebra, Dover, New York, 1990.[7] Q.I. Rahman, Functions of exponential type, Trans. Amer. Math. Soc. 135 (1969) 295309.[8] Q.I. Rahman, G. Schmeisser, Analytic Theory of Polynomials, London Math. Soc. Monogr. (N.S.), vol. 26, Claren-

don Press, Oxford, 2002.[9] E.C. Titchmarsh, The Theory of Functions, second ed., Oxford Univ. Press, 1939.

[10] A. Zygmund, A remark on conjugate series, Proc. London Math. Soc. (2) 34 (1932) 392400.

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