Polynomials and Polynomial Inequalities (Graduate Texts in Mathematics)

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Prefa e

Polynomials pervade mathemati s, and mu h that is beautiful in mathemati s is related to polynomials. Virtually every bran h of mathemati s, from algebrai number theory and algebrai geometry to applied analysis, Fourier analysis, and omputer s ien e, has its orpus of theory arising from the study of polynomials. Histori ally, questions relating to polynomials, for example, the solution of polynomial equations, gave rise to some of the most important problems of the day. The subje t is now mu h too large to attempt an en y lopedi overage. The body of material we hoose to explore on erns primarily polynomials as they arise in analysis, and the te hniques of the book are primarily analyti . While the onne ting thread is the polynomial, this is an analysis book. The polynomials and rational fun tions we are on erned with are almost ex lusively of a single variable. We assume at most a senior undergraduate familiarity with real and

omplex analysis (indeed in most pla es mu h less is required). However, the material is often tersely presented, with mu h mathemati s explored in the exer ises, some of whi h are quite hard, many of whi h are supplied with opious hints, some with omplete proofs. Well over half the material in the book is presented in the exer ises. The reader is en ouraged to at least browse through these. We have been mu h in uen ed by Polya and Szeg}o's lassi \Problems and Theorems in Analysis" in our approa h to the exer ises. (Though unlike Polya and Szeg}o we hose to in orporate the hints with the exer ises.)

viii

Prefa e

The book is mostly self- ontained. The text, without the exer ises, provides an introdu tion to the material, but mu h of the ri hness is reserved for the exer ises. We have attempted to highlight the parts of the theory and the te hniques we nd most attra tive. So, for example, Muntz's lovely

hara terization of when the span of a set of monomials is dense is explored in some detail. This result epitomizes the best of the subje t: an attra tive and nontrivial result with several attra tive and nontrivial proofs. There are ex ellent books on orthogonal polynomials, Chebyshev polynomials, Chebyshev systems, and the geometry of polynomials, to name but a few of the topi s we over, and it is not our intent to rewrite any of these. Of ne essity and taste, some of this material is presented, and we have attempted to provide some a

ess to these bodies of mathemati s. Mu h of the material in the later hapters is re ent and annot be found in book form elsewhere. Students who wish to study from this book are en ouraged to sample widely from the exer ises. This is de nitely \hands on" material. There is too mu h material for a single semester graduate ourse, though su h a ourse may be based on Se tions 1.1 through 5.1, plus a sele tion from later se tions and appendi es. Most of the material after Se tion 5.1 may be read independently. Not all obje ts labeled with \E" are exer ises. Some are examples. Sometimes no question is asked be ause none is intended. O

asionally exer ises in lude a statement like, \for a proof see : : : "; this is usually an indi ation that the reader is not expe ted to provide a proof. Some of the exer ises are long be ause they present a body of material. Examples of this in lude E.11 of Se tion 2.1 on the trans nite diameter of a set and E.11 of Se tion 2.3 on the solvability of the moment problem. Some of the exer ises are quite te hni al. Some of the te hni al exer ises, like E.4 of Se tion 2.4, are in luded, in detail, be ause they present results that are hard to a

ess elsewhere.

A knowledgments We would like to thank Di k Askey, Weiyu Chen, Carl de Boor, Karl Dil her, Jens Happe, Andras Kroo, Doron Lubinsky, Gua-Hua Min, Paul Nevai, Allan Pinkus, Jozsef Szabados, Vilmos Totik, and Ri hard Varga. Their thoughtful and helpful omments made this a better book. We would also like to thank Judith Borwein, Maria Fe Elder, and Chiara Veronesi for their expert assistan e with the preparation of the manus ript.

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ix


Contents

PREFACE CHAPTER 1

vii Introdu tion and Basi Properties

1.1 Polynomials and Rational Fun tions 1.2 The Fundamental Theorem of Algebra 1.3 Zeros of the Derivative CHAPTER 2 2.1 2.2 2.3 2.4

1 11 18

Some Spe ial Polynomials

29

Chebyshev Polynomials Orthogonal Fun tions Orthogonal Polynomials Polynomials with Nonnegative CoeÆ ients

29 41 57 79

CHAPTER 3 3.1 3.2 3.3 3.4 3.5

1

Chebyshev and Des artes Systems

Chebyshev Systems Des artes Systems Chebyshev Polynomials in Chebyshev Spa es Muntz-Legendre Polynomials Chebyshev Polynomials in Rational Spa es

91 92 100 114 125 139

x

Contents

CHAPTER 4 4.1 4.2 4.3 4.4

Denseness Questions

154

Variations on the Weierstrass Theorem Muntz's Theorem Unbounded Bernstein Inequalities Muntz Rationals

154 171 206 218

CHAPTER 5

Basi Inequalities

5.1 Classi al Polynomial Inequalities 5.2 Markov's Inequality for Higher Derivatives 5.3 Inequalities for Norms of Fa tors CHAPTER 6

Inequalities in Muntz Spa es

6.1 Inequalities in Muntz Spa es 6.2 Nondense Muntz Spa es CHAPTER 7

Inequalities for Rational Fun tion Spa es

7.1 Inequalities for Rational Fun tion Spa es 7.2 Inequalities for Logarithmi Derivatives

227 227 248 260 275 275 303 320 320 344

APPENDIX A1

Algorithms and Computational Con erns

356

APPENDIX A2

Orthogonality and Irrationality

372

APPENDIX A3

An Interpolation Theorem

382

APPENDIX A4

Inequalities for Generalized Polynomials in Lp

392

APPENDIX A5

Inequalities for Polynomials with Constraints

417

BIBLIOGRAPHY

448

NOTATION

467

INDEX

473

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1


1

Introdu tion and Basi Properties

Overview The most basi and important theorem on erning polynomials is the Fundamental Theorem of Algebra. This theorem, whi h tells us that every polynomial fa tors ompletely over the omplex numbers, is the starting point for this book. Some of the intri ate relationships between the lo ation of the zeros of a polynomial and its oeÆ ients are explored in Se tion 2. The equally intri ate relationships between the zeros of a polynomial and the zeros of its derivative or integral are the subje t of Se tion 1.3. This

hapter serves as a general introdu tion to the body of theory known as the geometry of polynomials. Highlights of this hapter in lude the Fundamental Theorem of Algebra, the Enestrom-Kakeya theorem, Lu as' theorem, and Walsh's two- ir le theorem.

1.1 Polynomials and Rational Fun tions The fo us for this book is the polynomial of a single variable. This is an extended notion of the polynomial, as we will see later, but the most important examples are the algebrai and trigonometri polynomials, whi h we now de ne. The omplex (n + 1)-dimensional ve tor spa e of algebrai polynomials of degree at most n with omplex oeÆ ients is denoted by Pn .

2

1. Introdu tion and Basi Properties

If C denotes the set of omplex numbers, then (

P

n :=

(1:1:1)

p : p(z ) =

n X k=0

ak

zk;

ak 2 C

)

:

When we restri t our attention to polynomials with real oeÆ ients we will use the notation

Pn :=

(1:1:2)

(

p : p(z ) =

n X k=0

)

ak

zk;

ak 2 R ;

where R is the set of real numbers. Rational fun tions of type (m; n) with

omplex oeÆ ients are then de ned by

R

:= m;n

(1:1:3)

p

; q 2 P ; : p 2 Pm n q

while their real ousins are denoted by

Rm;n :=

(1:1:4)

p : p 2 Pm ; q 2 Pn : q

The distin tion between the real and omplex ases is parti ularly important for rational fun tions (see E.4). The set of trigonometri polynomials Tn is de ned by (1:1:5)

T

n :=

(

t : t() :=

n X k= n

ak

eik ;

ak 2 C

)

:

A real trigonometri polynomial of degree at most n is an element of Tn taking only real values on the real line. We denote by Tn the set of all real trigonometri polynomials of degree at most n. Other hara terizations of Tn are given in E.9. Note that if z := ei , then an arbitrary element of Tn is of the form (1:1:6)

z

n

2n X

k=0

bk z k ;

bk 2 C

and so many properties of trigonometri polynomials redu e to the study of algebrai polynomials of twi e the degree on the unit ir le in C : The most basi theorem of this book, and arguably the most basi nonelementary theorem of mathemati s, is the Fundamental Theorem of Algebra. It says that a polynomial of exa t degree n (that is, an element

1.1 Polynomials and Rational Fun tions

3

of Pn nPn ) has exa tly n omplex zeros ounted a

ording to their multipli ities. 1

Theorem 1.1.1 (Fundamental Theorem of Algebra). If

p(z ) :=

n X i=0

ai 2 C ; an 6= 0 ;

ai z i ;

then there exist ; ; : : : ; n 2 C su h that 1

2

p(z ) = an

n Y

i=1

(z

i ) :

Here the multipli ity of the zero at i is the number of times it is repeated. So, for example, (z 1) (z + i) is a polynomial of degree 5 with a zero of multipli ity 3 at 1 and with a zero of multipli ity 2 at i: The polynomial 3

p(z ) :=

n X i=0

ai z i ;

2

ai 2 C ; an 6= 0

is alled moni if its leading oeÆ ient an equals 1. There are many proofs of the Fundamental Theorem of Algebra based on elementary properties of omplex fun tions (see Theorem 1.2.1 and E.4 of Se tion 1.2). We will explore this theorem more substantially in the next se tion of this hapter. Comments, Exer ises, and Examples. The importan e of the solution of polynomial equations in the history of mathemati s is hard to overestimate. The Greeks of the lassi al period understood quadrati equations (at least when both roots were positive) but

ould not solve ubi s. The expli it solutions of the ubi and quarti equations in the sixteenth entury were due to Ni

olo Tartaglia ( a 1500{1557), Ludovi o Ferrari (1522{1565), and S ipione del Ferro ( a 1465{1526) and were popularized by the publi ation in 1545 of the \Ars Magna" of Girolamo Cardano (1501{1576). The exa t priorities are not entirely lear, but del Ferro probably has the strongest laim on the solution of the ubi . These dis overies gave western mathemati s an enormous boost in part be ause they represented one of the rst really major improvements on Greek mathemati s. The impossibility of nding the zeros of a polynomial of degree at least 5, in general, by a formula ontaining additions, subtra tions, multipli ations, divisions, and radi als would await Niels Henrik Abel (1802{1829) and his 1824 publi ation of \On the Algebrai Resolution of Equations." Indeed, so mu h algebra, in luding Galois theory, analysis, and parti ularly omplex analysis, is born out of these ideas that it is hard to imagine how the ow of mathemati s might have pro eeded without these issues being raised. For further history, see Boyer [68℄.

4


E.1 Expli it Solutions. a℄ Quadrati Equations. Verify that the quadrati polynomial x + bx + has zeros at p p b b + b 4 b 4 ; : 2 2 2

2

2

b℄ Cubi Equations. Verify that the ubi polynomial x + bx + has zeros at p p + + +i 3 ; i 3 ; + ; 2 2 2 2 3

where

s

= and

2 s

=

3

3

r

+

2

+

4

r

2

2

+

4

b

3

27

b

3

27

:

℄ Show that an arbitrary ubi polynomial, x + ax + bx + ; an be transformed into a ubi polynomial as in part b℄ by a transformation x 7! ex + f . d℄ Observe that if the polynomial x + bx + has three distin t real zeros, then and are ne essarily nonreal and hen e 4b + 27 is negative. So, in this simplest of ases one is for ed to deal with omplex numbers (whi h was a serious te hni al problem in the sixteenth entury). e℄ Quarti Equations. The quarti polynomial x + ax + bx + x + d has zeros at a R a R + ; + ; 4 2 2 4 2 2 where r 3

2

3

3

2

4

R= y is any root of the resolvent ubi y and

r

; = while

a

2

2

3a 4

2

r

R

2

2b

2

b+y;

4

by + (a + 4d)y

3

3

4ab

a d + 4bd ; 2

8 4R

2

a

3

;

R 6= 0 ;

p 3a 2b 2 y 4d ; R = 0: 4 These unwieldy equations are quite useful in onjun tion with any symboli manipulation pa kage.

; =

2

2


5

E.2 Newton's Identities. Write (x

) (x

)(x 1

n ) = xn

2

xn 1

1

+ xn

+(

2

2

1)n n :

The oeÆ ients k are, by de nition, the elementary symmetri fun tions in the variables ; : : : ; n : a℄ For positive integers k , let 1

sk := k + k + + kn : 1

2

Prove that

sk = ( 1)k k k + ( 1)k +1

and

sk = ( 1)k

+1

kX1 j =k n

kX1 j =1

kn

( 1)j k j sj ;

( 1)j k j sj ;

k > n:

Here, and in what follows, an empty sum is understood to be 0. A polynomial of n variables is a fun tion that is a polynomial in ea h of its variables. A symmetri polynomial of n variables is a polynomial of n variables that is invariant under any permutation of the variables. b℄ Show by indu tion that any symmetri polynomial in n variables (with integer oeÆ ients) may be written uniquely as a polynomial (with integer

oeÆ ients) in the elementary symmetri fun tions f ; f ; : : : ; fn . Hint: For a symmetri polynomial f in n variables, let 1

(f ) := ( ; ; : : : ; n ) ; 1

2

1

2

n 0 2

if

f (x ; x ; : : : ; xn ) = 1

2

2 1 X X 1 =0 2 =0

n X n =0

1 ;2 ;::: ;n x1 x2 xnn 1

2

and 1 ;2 ;::: ;n = 6 0. If

(f ) = ( ; ; : : : ; n ) and (g) = (e ; e ; : : : ; en ) ; 1

2

1

2

then let (f ) < (g ) if j ej for ea h j with a stri t inequality for at least one index. This gives a (partial) well ordering of symmetri polynomials in n variables, that is, every set of symmetri polynomials in n variables has a minimal element. Now use indu tion on (f ). ut

6


℄ Show that

p 1+ 5 k 2

!0

(mod 1) :

(By onvergen e to zero (mod 1) we mean that the quantity approa hes integral values.) Hint: Consider the integers

sk := k + k ; 1

p

2

p

where := (1 + 5) and := (1 5). d℄ Find another algebrai integer with the property that 1

1

1

2

2

2

k ! 0

ut

(mod 1) :

Su h numbers are alled Salem numbers (see Salem [63℄). It is an open problem whether any nonalgebrai numbers > 1 satisfy k ! 0 (mod 1). E.3 Norms on Pn . Pn is a ve tor spa e of dimension n + 1 over R: Hen e Pn equipped with any norm is isomorphi to the Eu lidean ve tor spa e Rn , and these norms are equivalent to ea h other. Similarly, Pn is a ve tor spa e of dimension n + 1 over C : Hen e Pn equipped with any norm +1

is isomorphi to the Eu lidean ve tor spa e C n , so these norms are also equivalent to ea h other. Let +1

pn (x) :=

n X k=0

ak 2 R :

a k xk ;

Some ommon norms on Pn and Pn are

kpkA := sup jp(x)j

supremum norm

x2A

:=kpkL1 A (

kpkLp A

Z

(

)

:=

A

j

L1 norm

)

j

p(t) p dt

1=p

fjak jg kpkl1 := max k X n

kpklp :=

k=0

jak jp

1=p

Lp norm; p 1 l1 norm lp norm; p 1 :

In the rst ase A must ontain n + 1 distin t points. In the se ond ase A must have positive measure.


7

a℄ Con lude that there exist onstants C , C , and C depending only on 1

n so that

kp0 k

1 1℄

;

;

jai j C kpk

1 1℄

;

;

1 1℄

1

[

2

i=0

3

C kpk

;

[

n X

2

kpk

;

[

1 1℄

[

C kpkL 3

2[

;

1 1℄

for every p 2 Pn , and, in parti ular, for every p 2 Pn . These inequalities will be revisited in detail in later hapters, where pre ise estimates are given in terms of n. b℄ Show that there exist extremal polynomials for ea h of the above inequalities. That is, for example,

kp0 k p2Pn kpk

sup

6

0=

; ;

[

1 1℄

[

1 1℄

is a hieved. E.4 On Rn;m . a℄ Rn;m is not a ve tor spa e be ause it is not losed under addition. b℄ Partial Fra tion De omposition. Let rn;m 2 R n;m be of the form Qm0

k=1

p(x) ; (x k )mk

p 2 Pn ; k distin t ; p(k ) 6= 0 :

Then there is a unique representation of the form 0

rn;m (x) = q(x) +

mk m X X k=1 j =1

(x

ak;j ; k )j

q 2 Pn

m;

ak;j

2C

(if m > n; then Pn m is meant to be f0g). Hint: Consider the type and dimension of expressions of the above form.

℄ Show that if

ut

rn;m 2 R n;m ;

then

Re(rn;m ()) 2 Rn m; m : +

2

This is an important observation be ause in some problems a rational fun tion in R n;n an behave more like an element of R n; n than Rn;n . 2

2


8

E.5 Horner's Rule. a℄ We have n X i=0

ai xi = ( ((an x + an )x + an )x + + a )x + a : 1

2

1

0

So every polynomial of degree n an be evaluated by using at most n additions and n multipli ations. (The onverse is learly not true; onsider n x .) b℄ Show that every rational fun tion of type (n 1; n) an be put in a form so that it an be evaluated by using n divisions and n additions. 2

E.6 Lagrange Interpolation. Let zi and yi be arbitrary omplex numbers ex ept that the zi must be distin t (zi 6= zj ; for i 6= j ). Let

lk (z ) :=

Qn

i=0;i6=k (z i=0;i6=k (zk

zi ) ; zi )

Qn

k = 0; 1; : : : ; n :

a℄ Show that there exists a unique p 2 Pn that takes n + 1 spe i ed values at n + 1 spe i ed points, that is,

p(zi ) = yi ;

i = 0; 1; : : : ; n :

This p 2 Pn is of the form

p(z ) =

n X k=0

yk lk (z )

and is alled the Lagrange interpolation polynomial. If all the zi and yi are real, then this unique interpolation polynomial is in Pn : b℄ Let

!(z ) :=

n Y

i=0

(z

zi ) :

Show that lk is of the form

lk (z ) = and

p(z ) =

(z

n X k=0

(z

!(z ) zk )!0 (zk ) yk !(z ) : zk )!0 (zk )


9

℄ An Error Estimate. Assume that the points zi 2 [a; b℄; i = 0; 1; : : : ; n; are distin t and f 2 C n [a; b℄ (that is, f is an n + 1 times ontinuously dierentiable real-valued fun tion on [a; b℄). Let p 2 Pn be the Lagrange interpolation polynomial satisfying +1

p(zi ) = f (zi ) ;

i = 0; 1; : : : ; n :

Show that for every x 2 [a; b℄ there is a point 2 (a; b) so that

f (x) Hen e

kf

p(x) =

pk a;b [

℄

1 fn (n + 1)! (

+1)

( ) ! (x) :

(n +1 1)! kf n k a;b k!k a;b : (

Hint: Choose so that ' := f

p

+1)

[

℄

[

℄

w vanishes at x; that is,

:= (f (x)

p(x))=!(x) :

Then repeated appli ations of Rolle's theorem yield that

'n (

+1)

=f n (

+1)

(n + 1)!

ut

has a zero in (a; b):

E.7 Hermite Interpolation. a℄ Let zi 2 C ; i = 1; 2; : : : k , be distin t. Let mi ; i = 1; 2; : : : ; k , be positive P integers with n + 1 := ki mi , and let =1

yi;j

2C;

i = 1; 2; : : : ; k ; j = 0; 1; : : : ; mi

1

be xed. Show that there is a unique p 2 Pn , alled the Hermite interpolation polynomial, so that

p j (zi ) = yi;j ; i = 1; 2; : : : ; k ; j = 0; 1; : : : ; mi ( )

1:

If all the zi and yi;j are real, then this unique interpolation polynomial is in Pn . Hint: Use indu tion on n. ut n b℄ Assume that the points zi 2 [a; b℄ are distin t and f 2 C [a; b℄. Let p 2 Pn be the Hermite interpolation polynomial satisfying 1

1

p(zi ) = f j (zi ) ; i = 1; 2; : : : ; k ; j = 0; 1; : : : ; mi ( )

Show that for every x 2 [a; b℄ there is a point 2 (a; b) so that

1:

10


f (x)

p(x) =

with

!(x) := Hen e

kf

pk a;b [

℄

k Y

1 n f ( ) !(x) n! (

)

xi )mi

(x

i=1

1

:

n1! kf n k a;b k!k a;b : (

)

[

℄

[

℄

ut

Hint: Follow the hint given for E.6 ℄.

Polynomial interpolation and related topi s are studied thoroughly in Davis [75℄; Lorentz, Jetter, and Riemens hneider [83℄; and Szabados and Vertesi [92℄. E.8 On the Zeros of a p 2 Pn . Show that if p 2 Pn , then the nonreal zeros of p form onjugate pairs (that is, if z is a zero of p; then so is z ). E.9 Fa torization of Trigonometri Polynomials. a℄ Show that t 2 Tn (or t 2 Tn ) if and only if t is of the form

t(z ) = a + 0

n X k=1

ak ; bk 2 R (or C ) :

(ak os kz + bk sin kz ) ;

b℄ Show that if t 2 Tn nTn ; then there are numbers z ; z ; : : : ; z n and 0 6= 2 C su h that 1

t(z ) =

1

2n Y

j =1

sin

z

2

zj

2

2

:

Show also that the nonreal zeros zj of t form onjugate pairs. E.10 Newton Interpolation and Integer-Valued Polynomials. Let k f (x) be de ned indu tively by

f (x) := f (x) ; f (x) = f (x) := f (x + 1) f (x) 0

and Let

1

k f (x) := (k f (x)) ; +1

x x(x := k

1) (x k!

k = 1; 2; : : : : k + 1)

:

a℄ Show that xk is a polynomial of degree k that takes integer values at all integers.

1.2 The Fundamental Theorem of Algebra 11

b℄ Let f be an m times dierentiable fun tion on [a; a + m℄. Show that there is a 2 (a; a + m) su h that

m f (a) = f

(

m) ( ) :

℄ Show that if p 2 Pn ; then

p(x) =

n X k=0

k p(0)

x : k

d℄ Suppose p 2 Pn is integer-valued at all integers. Show that

p(x) =

n X k=0

ak

x k

for some integers a ; a ; : : : ; an : Note that this hara terizes su h polynomials. e℄ Show that if p 2 Pn takes integer values at n + 1 onse utive integers, then p takes integer values at every integer. f℄ Suppose 2 R and n is an integer for every n 2 N : Use part b℄ to show that is a nonnegative integer. 0

1

1.2 The Fundamental Theorem of Algebra The following theorem is a quantitative version of the Fundamental Theorem of Algebra due to Cau hy [1829℄. We oer a proof that does not assume the Fundamental Theorem of Algebra, but does require some elementary

omplex analysis. Theorem 1.2.1. The polynomial

p(z ) := an z n + an z n 1

1

+ +a

0

2 Pn ;

an 6= 0

has exa tly n zeros. These all lie in the open disk of radius r entered at the origin, where j ak j : r := 1 + max kn jan j 0

1

Proof. We may suppose that a = 6 0, or we may rst divide by z k for some k. Now observe that 0


12

g(x) := ja j + ja jx + + jan jxn jan jxn satis es g (0) > 0 and lim g (x) = 1. So by the intermediate value thex!1 orem, g has a zero in (0; 1) (whi h is, on onsidering (g (x)=xn )0 , in fa t unique). Let s be this zero. Then for jz j > s; (1:2:1) jp(z ) an z nj ja j + ja z j + + jan z n j < jan z n j : This, by Rou he's theorem (see E.1), shows that p(z ) and an z n have exa tly the same number of zeros, namely, n, in any disk of radius greater than s. It remains to observe that if x r; then g (x) < 0 so s < r. Indeed, 0

1

0

g(x) jan j

xn

< j a n j xn

1

1+

1+

0 for

1

1

x 1+

1

1

1

max

k=0;::: ;n

max

k=0;::: ;n

max

k=0;::: ;n

jak j nX xk n jan j k jak j 1 jan j x 1

1

1

1

!

=0

j ak j jan j :

ut

The exa t relationship between the oeÆ ients of a polynomial and the lo ation of its zeros is very ompli ated. Of ourse, the more information we have about the oeÆ ients, the better the results we an hope for. The following pretty theorem emphasizes this: Theorem 1.2.2 (Enestrom-Kakeya). If

p(z ) := an z n + an z n 1

1

+ +a

0

with

a a an > 0 ; then all the zeros of p lie outside the open unit disk. 0

Proof. Consider (1 z )p(z ) = a + (a 0

1

a )z + + (an

1

an )z n

0

1

an z n

+1

:

Then

j(1

z )p(z )j a [(a a )jz j + + (an an )jz jn + an jz jn ℄ : Sin e ak ak 0, the right-hand expression above de reases as jz j in reases. Thus, for jz j < 1, j(1 z )p(z )j > a [(a a ) + + (an an ) + an ℄ = 0 ; 0

0

1

+1

1

+1

0

and the result follows.

0

1

1

ut


Corollary 1.2.3. Suppose

p(z ) := an z n + an z n 1

1

+ +a

0

with ak > 0 for ea h k. Then all the zeros of p lie in the annulus ak ak r := min jz j k max =: r : k ;::: ;n ak ;::: ;n ak 1

=0

1

=0

+1

2

1

+1

ut

Proof. Apply Theorem 1.2.2 to p(r z ) and z n p(r =z ). 1

2

This is a theme with many variations, some of whi h are explored in the exer ises. Theorem 1.2.4. Suppose p > 1; q > 1, and p

nomial h 2 P of the form

n

h(z ) = an z n + an z n 1

1

1

+q

1

++a ; 0

= 1. Then the poly-

an 6= 0

has all its zeros in the disk fz 2 C : jz j rg, where 8
r :

ut

16


℄ Dedu e Rou he's theorem from part b℄ for polynomials f and g given by their fa torizations, and for ir les . Hint: Let h := 1 + (g f )=f . So fh = g and Z Z Z 1 1 1 g0 (z ) f 0(z ) h0 (z ) dz = dz + dz : 2i Dr g (z ) 2i Dr f (z ) 2i Dr h(z )

Show that the last integral is zero by expanding h and applying b℄. u t d℄ Dedu e E.1 ℄ and E.1 d℄ from E.1 a℄. e℄ Observe that the uni ity theorem an be sharpened for polynomials as follows. If p; q 2 Pn and p(z ) = q (z ) for n + 1 distin t values of z 2 C ; then p and q are identi al, that is, p(z ) = q(z ) for every z 2 C : Equivalently, a polynomial p 2 Pn is either identi ally 0 or has at most n zeros. (This is trivial from the Fundamental Theorem of Algebra, but as in E.2, it does not require it.) 1

E.4 The Fundamental Theorem of Algebra. Every non onstant polyno-

mial has at least one omplex zero.

Prove this dire tly from Liouville's theorem. E.5 Pellet's Theorem. Suppose ap 6= 0; jap

+1

g(x) := ja j + ja jx + + jap 0

1

1

j xp

1

j + + jan j > 0,

jap jxp + jap jxp +1

+1

and

+ + jan jxn

has exa tly two positive zeros s < s . Then 1

2

f (z ) := an z n + an z n 1

1

++a

0

2 Pn

has exa tly p zeros in the disk fz 2 C : jz j s g and no zeros in the annulus fz 2 C : s < jz j < s g. 1

1

2

Proof. Let s < t < s : Then g(t) < 0; that is, 1

2

n X j =0 j 6=p

jaj jtj < jap jtp :

Now apply Rou he's theorem to the fun tions

F (z ) := ap z p

and

G(z ) :=

n X j =0

aj z j :

ut


E.6 Proof of Theorem 1.2.4. a℄ Holder's inequality (see E.7 of Se tion 2.2) asserts that n X k=1

where p

1

+q

1

n X

jak bk j

k=1

j ak j p

!1=p

k=1

j bk j q

!1=q

;

= 1 and p 1. So if

p(z ) := an z n + an z n

1

1

then

n X

nX1 k=1

jak jjz j k

nX1 k=0

jak j

p

++a

2 Pn ;

0

!1=p

nX1

jz j

kq

k=0

!1=q

:

b℄ Thus, for jz j > 1; nX1

jp(z )j jan jjz jn 8
0:

0; 1;

f℄ Three-Term Re ursion. Show that

U (x) = 1 ; U (x) = 2x ; 0

1

Un (x) = 2xUn (x)

Un (x) ; n = 2; 3; : : : :

1

2

(Note that this is the same re ursion as for Tn :) g℄ The CoeÆ ients of Un . Show that

bX n= 2

Un (x) =

k=0

( 1)k

n

k

k

(2x)n k : 2

h℄ Another Form of Un . Show that

bX n= 2

Un (x) =

k=0

( 1)k

n+1 n x 2k + 1

2

k (x2

1)k :

The on epts of trans nite diameter and apa ity play a entral role in potential theory, harmoni analysis, and other areas of mathemati s.

38

2. Some Spe ial Polynomials

E.11 Trans nite Diameter. Let E be a ompa t subset of C : Let

n (E ) :=

Y

max

z1 ;::: ;zn 2E

1

i;jn i6 j

jzi

zj j :

=

The points zi at whi h the above maximum are obtained are alled nth Fekete points for E: If the points zi are the nth Fekete points for E; then

the polynomial

n Y

qn (z ) :=

i=1

(z

zi )

is alled an nth (moni ) Fekete polynomial for E: The trans nite diameter or logarithmi apa ity of E is de ned by 1

ap(E ) := lim (n (E )) n(n n!1

1)

;

where the limit exists by part ℄ (below). a℄ Let z ; z ; : : : ; zn be nth Fekete points for E: Then 1

2

(n (E )) = = abs 1 2

1 1 . .. 1

z z

1 2

.. .

1 2 .. . zn 1

: : : zn : : : zn

1

1

..

.

zn : : :

:

n

Hint: See E.2 b℄ (Vandermonde determinant) of Se tion 3.2. Q b℄ Let qn (z ) := ni (z zi ) be an nth Fekete polynomial for E: Let mn := min jq0 (zi )j and Mn := kqn kE : =1

i=1;::: ;n n

Then

Mn

n (E ) n (E ) +1

Outline. We have

jqn (z )j n (E ) = 2

and

n (E ) = +1

n Y i=1

1=2

mn (n +1

jz

Y

zi j

2 1

i;jn i6 j

jzi

1

+1

(E )) n+1 :

zj j n (E ) +1

=

Y 1i;j n+1 i6=k;j 6=k;i6=j

jzi

1

1

in i6 k =

in i6 k zi j :

+1

=

Y

n (E )

Y

zj j

jzk +1

2

jzk

zi j

2

ut

2.1 Chebyshev Polynomials 39

From the rst line above,

Mn n (E ) n (E ) : 2

+1

From the se ond line above,

n (E ) n (E )mn 2

+1

1

(n (E )) n(n

+1

:

ut

is de reasing, so the limit exists in the de ni ℄ Show that tion of ap(E ): d℄ The Fekete points lie on the boundary of E: So ap(E ) = ap( (E )): Hint: Use the maximum prin iple (see E.1 d℄ of Se tion 1.2). ut e℄ If E F; where F C is also ompa t, then ap(E ) ap(F ): f℄ Chebyshev Constants. Let

Mn := and fn := M

Let and

1)

(

p2P

n:

p(z ) =

(

p 2 Pn : p(z ) =

n Y j =1 n Y j =1

(z

zj ) ; zj

2C

)

)

(z

zj ) ; zj 2 E :

n (E ) := inf fkpkE : p 2 Mng fn g : en (E ) := inf fkpkE : p 2 M

Show that the in mum in the de nition of n (E ) and en (E ) is a tually minimum. Show also that

and

n

+

m (E )

n (E )m (E )

en

+

m (E )

en (E )em (E )

for any two nonnegative integers n and m: Finally show that the above inequalities imply that

(E ) := nlim !1(n (E ))

1

exist.

=n

and e(E ) := lim (en (E )) =n n!1

1

40


The numbers (E ) and e(E ) are alled the Chebyshev onstant and modi ed Chebyshev onstant, respe tively, asso iated with E: Obviously (E ) e(E ): g℄ Trans nite Diameter and Chebyshev Constants Are the Same:

ap(E ) = (E ) = e(E ) :

Proof. Without loss of generality we may assume that E ontains in nitely many points. Part b℄ yields e(E ) ap(E ): Therefore, sin e (E ) e(E ); it is suÆ ient to prove that ap(E ) (E ): Note that if p 2 Mn and z ; z ; : : : ; zn 2 E are the (n + 1)th Fekete points for E; then 1

2

+1

(n+1 (E )) = = abs 1 2

1 1 . .. 1

: : : zn : : : zn

z z

1 2

.. .

zn

..

+1

2

.

p(z ) p(z )

1

1

1

1

.. .

.. .

: : : znn

p(zn

1 +1

2

+1

)

:

Expanding the above determinant with respe t to its last olumn, we obtain

(n+1 (E )) =

1 2

(n (E )) =

1 2

nX +1 j =1

jp(zj )j

(n + 1)(n (E )) = kpkE ; 1 2

so

(n (E )) =

1 2

+1

(n + 1)(n (E )) = n (E ) : 1 2

For the sake of brevity let

n := (n + 1) (n (E )) 2

2

Then

dnn

1=n

+1 +1

2

and dn := (n (E )) n(n

n dnn

1

1)

:

:

Sin e E ontains in nitely many points, n > 0 and dn > 0 hold for ea h n = 2; 3; : : : : Multiplying the above inequalities for n = 1; 2; : : : ; k; we obtain after simpli ation that (d d 2

3

dk

1

+1

) k 1 (dk ) k

k

+1

1

(d ) k

2

2

1

( 2

3

k ) k

1

1

:

Sin e lim dk = ap(E ) and lim k = ((E )) ; we on lude k!1

2

k!1

( ap(E ))

2

whi h nishes the proof.

((E ))

2

;

ut

2.2 Orthogonal Fun tions 41

h℄ Show that ap([a; b℄) = (b

a):

1

4

i℄ Show that ap(D ) = ; where

D := fz 2 C : jz j g : j℄ Show that ap(A ) = sin(=4); where A is an ar of the unit ir le C of length , 0 2: Hint: Without loss of generality we may assume that the ar A is symmetri with respe t to the x-axis and 1 2 A : Now use part h℄ and the transformation x = (z + z ): ut 1

1

2

2.2 Orthogonal Fun tions The most basi properties of orthogonal fun tions are explored in this se tion. The following se tion spe ializes the dis ussion to polynomials. In this se tion the fun tions are omplex-valued and the ve tor spa es are over the omplex numbers. All the results have obvious real analogs and in many later appli ations we will restri t to these orresponding real

ases. An inner produ t on a ve tor spa e V is a fun tion h; i from V V to C that satis es, for all f; g; h 2 V and ; 2 C ; (2:2:1) hf; f i > 0 unless f = 0 (positivity)

(2:2:2) hf; gi = hg; f i ( onjugate symmetry) (2:2:3) hf + g; hi = hf; hi + hg; hi (linearity). A ve tor spa e V equipped with an inner produ t is alled an inner produ t spa e. It is a normed linear spa e with the norm k k := h; i = : The anoni al example for us will be the spa e C [a; b℄ of all omplexvalued ontinuous fun tions on [a; b℄ with the inner produ t 1 2

hf; gi :=

(2.2.4)

Z b

a

f (x)g(x)w(x) dx ;

where w(x) is a nonnegative integrable fun tion on [a; b℄ that is positive ex ept possibly on a set of measure zero. It is a normed linear spa e with the norm (2.2.5)

kf kL

2(

w) :=

hf; f i

= =

1 2

Z b

a

!1=2

jf (x)j w(x) dx 2

:

42


More generally, if (X; ) is a measure spa e (with nonnegative), then

hf; gi :=

(2.2.6)

Z

X

f (x)g(x) d(x)

is an inner produ t on the spa e L () of square integrable fun tions. More pre isely, L () denotes the spa e of equivalen e lasses of measurable fun tions for whi h 2

2

kf kL

:=

2( )

hf; f i

= =

Z

jf (x)j

2

1 2

X

1=2

d(x)

is nite. The equivalen e lasses are de ned by the equivalen e relation g if f = g -almost everywhere on X: If V is a ve tor spa e equipped with an inner produ t h; i; then a metri an be de ned on V by (f; g ) := hf g; f g i = : The fa t that this is a metri on V is an immediate onsequen e of (2.2.1) and Theorem 2.2.1 b℄. If this metri spa e (V; ) is omplete (that is, if every Cau hy sequen e in (V; ) onverges to some x 2 V ), then V is alled a Hilbert spa e. It an be shown that L () is a Hilbert spa e for every measure spa e (X; ) (see Rudin [87℄), while C [a; b℄ equipped with the inner produ t (2.2.4), where w(x) 1; is not a Hilbert spa e (see E.1). When we write L [a; b℄ we always mean L () where is the Lebesgue measure on X = [a; b℄: The fa t that the inner produ t gives a norm is part of the next theorem.

f

1 2

2

2

2

Theorem 2.2.1. If (V; h; i) is an inner produ t spa e equipped with the norm k k := h; i = ; then for all f; g 2 V; a℄ jhf; gij kf k kgk Cau hy-S hwarz inequality b℄ kf + gk kf k + kgk triangle inequality

℄ kf + gk + kf gk = 2 kf k + 2 kgk parallelogram law. Proof. Let f; g 2 V be arbitrary. To prove the Cau hy-S hwarz inequality, without loss of generality we may assume hg; g i = 1 and we may assume hf; gi is real (why?). Let := hf; gi and note that by (2:2:1) and (2:2:3); 0 hf g; f g i = hf; f i 2 hf; g i + hg; g i = kf k hf; g i ; 1 2

2

2

2

2

2

2

2

whi h nishes the proof of part a℄. Using the Cau hy-S hwarz inequality, we obtain kf + gk = hf + g; f + gi = hf; f i + 2Re(hf; gi) + hg; gi kf k + 2 kf k kgk + kgk (kf k + kgk) ; whi h is the triangle inequality. 2

2

2

2


The parallelogram law follows from

kf + gk

2

+ kf g k = hf; f i + 2 Re(hf; g i) + hg; g i + hf; f i 2

2 Re(hf; g i) + hg; g i :

ut For the spa e L () of all square integrable fun tions, the Cau hyS hwarz inequality be omes 2

Z b fg d a

Z b

a

jf j

!1=2

2

d(x)

Z b

jg j

!1=2

2

a

:

d

Applying this with f and g repla ed by jf j and jg j; we obtain (2.2.7)

Z b

a

jfgj d

Z b

a

jf j

!1=2

2

d(x)

Z b

a

jg j

2

d

!1=2

:

A olle tion of ve tors ff : 2 Ag in an inner produ t spa e (V; h; i) is said to be orthogonal if (2.2.8)

hf ; f i = 0 ;

; 2 A ; 6= :

If hf ; f i = 0; then we write f ?f . The olle tion is alled orthonormal if, in addition to being orthogonal,

hf ; f i = 1 ;

(2.2.9)

2 A:

An orthogonal olle tion ff : 2 Ag of nonzero ve tors in an inner produ t spa e an always be orthonormalized as fkf k f : 2 Ag: The ve tor spa e over C generated by ff : 2 Ag is denoted by 1

spanff : 2 Ag : So spanff : 2 Ag is just the set of all nite linear ombinations ( n X

i=1

i fi : i 2 A ; i 2 C ; n 2 N

)

:

Any linearly independent olle tion of ve tors an be orthonormalized, as the next theorem shows.

44


Theorem 2.2.2 (Gram-S hmidt). Let (V; h; i) be an inner produ t spa e with norm k k := h; i = : Suppose ffi g1 i is a linearly independent olle 1 2

=1

tion of ve tors in V: Let

g := 1

f

1

kf k 1

and (indu tively) let nX1

un := fn

k=1

hfn ; gk igk

and

gn :=

un kun k :

Then fgn g1 n is an orthonormal olle tion, and for ea h n; =1

spanfg ; g ; : : : ; gn g = spanff ; f ; : : : ; fn g : 1

2

1

2

Proof. This an be proved easily by indu tion where the indu tive step is: for m < n;

hun ; gmi = hfn ; gmi = hfn ; gm i

nX1

hfn ; gk ihgk ; gm i

k=1 nX1 k=1

hfn ; gk iÆk;m = 0 : ut

The key approximation theoreti property orthonormal sets have is en apsulated in the following result: Theorem 2.2.3 (Best Approximation by Linear Combinations). Let (V; h; i) be an inner produ t spa e with norm k k := h; i = : Suppose ff ; : : : ; fng is an orthonormal olle tion of ve tors in V: Let f 2 V: Then 1 2

n

X

min

i 2C

i=1

i f i

1

f

is attained if and only if

i = hf; fi i ;

i = 1; 2; : : : ; n :

P

In other words, the sum ni hf; fi ifi is the best approximation to f from spanff ; : : : ; fn g in the norm h; i = : =1

1

1 2


Proof. Fix f

2 V; and let i be as above. Let g :=

n X i=1

i f i

and let h 2 spanff ; : : : ; fn g: Note that 1

f )?fi ;

(g

i = 1; 2; : : : ; n

sin e by orthonormality

hg

f; fi i =

Thus and so

n X j =1

(g

kh

f k = k(h = kh = kh kg 2

j hfj ; fi i

i = 0 :

f )?(h g)

g) + (g f )k gk + 2Re(hh g; g gk + kg f k fk 2

2

2

f i) + kg

fk

2

2

2

with stri t inequality unless h = g: This nishes the proof.

ut

Note that the above theorem gives the following orollary: Corollary 2.2.4. If ff ; : : : ; fn g is an orthonormal olle tion, then every 1

g 2 spanff ; : : : ; fng 1

an be written as g=

n X i=1

hg; fi ifi :

Comments, Exer ises, and Examples. The theory of orthogonal fun tions, and in parti ular orthogonal polynomials, is old and far-rea hing. As we will see in the next se tion, the names asso iated with the lassi al orthogonal polynomials in luding Chebyshev, Laguerre, Legendre, Hermite, Ja obi, and Stieltjes, are the \who's who" of nineteenth entury analysis. Various aspe ts of this beautiful body of theory are explored in the exer ises of this and the next se tion. Mu h of this material is available in G. Szeg}o's [75℄ lassi al treatise \Orthogonal Polynomials." Of ourse, orthogonal polynomials are intimately onne ted to Fourier series and parts of harmoni and fun tional analysis generally. The standard fun tional analysis in the following exer ises is available in many sour es. See, for example, Rudin [73, 87℄.

46


E.1 C [0; 1℄ Is Not a Hilbert Spa e. Constru t a sequen e of ontinuous fun tions (fn )1 n on [0; 1℄ for whi h =1

kfn

f kL2

!0

;

[0 1℄

with somef 2= C [0; 1℄ (in the sense that f annot be modi ed on a set of measure zero to be in C [0; 1℄). So C [0; 1℄ equipped with the inner produ t (2.2.4), where [a; b℄ = [0; 1℄ and w(x) is identi ally 1; is not a Hilbert spa e. It an be shown that there is no way of putting a norm on C [0; 1℄ that preserves the uniform topology and makes C [0; 1℄ into a Hilbert spa e, essentially be ause C [0; 1℄ is not re exive (see Rudin [73℄, Chapter 4). This, in fa t, shows that C [0; 1℄ is not isomorphi to Lp [0; 1℄ for any p 2 (1; 1): For the de nition of Lp [0; 1℄; see E.7. E.2 On L (w). Consider 2

hf; gi =

Z b

a

f (x)g(x)w(x) dx :

What onditions on w guarantee that hf; g i is an inner produ t on C [a; b℄ ? E.3 Cau hy-S hwarz Inequality for Sequen es. Show that 2 n X i i

i=1

n X i=1

ji j

!

n X

2

i=1

j i j

!

2

for all ; : : : ; n ; ; : : : ; n 2 C : Equality holds if and only if there exists a 2 C so that either i = i for ea h i or i = i for ea h i: Hint: C n is a Hilbert spa e with inner produ t 1

1

h( ; ; : : : ; n ); ( ; ; : : : ; n )i = 1

2

1

2

n X i=1

i i :

ut

E.4 Bessel's Inequality. Let (V; h; i) be an inner produ t spa e with norm kk := h; i = : Suppose ffi g1 i is a ountable olle tion of orthonormal ve tors in V: a℄ Show that 1 2

=1

1 X i=1

jhfi ; f ij kf k 2

2

:

Hint: With h := 0 in the last expression of the proof of Theorem 2.2.3

kf k

2

= kg k + kg 2

fk : 2

ut


b℄ Suppose

f=

1 X i=1

hfi ; f ifi

in the sense that the partial sums of the right-hand side onverge to f in the norm k k: Show that

kf k

2

=

X

i

jhfi ; f ij

2

:

E.5 The Kernel Fun tion. Let fpi gni be a olle tion of orthonormal fun tions in L [a; b℄ with respe t to the inner produ t de ned by (2.2.6), where X := [a; b℄: De ne the kernel fun tion by Kn (x ; x) := p (x )p (x) + p (x )p (x) + + pn (x )pn (x) : =0

2

0

0

0

0

1

0

1

0

a℄ Reprodu ing Property. If q 2 spanfp ; : : : ; pn g; then Z b

a

0

Kn (t; x)q(t) d(t) = q(x) :

Hint: Expand q in terms of p ; : : : ; pn as in Corollary 2.2.4. ut b℄ (Kn (x ; x )) = Kn (x ; x) solves the following maximization problem: 0

(

max

0

1 2

0

0

jq(x )j : q 2 spanfp ; p ; : : : ; pn g and 0

0

Z b

1

a

)

jq(x)j

2

d(x) = 1 :

P

Outline. Write q = ni i pi : Then, as in E.4 b℄, kqkL2 = j j + j j + + j n j = 1 : =0

2

(

0

)

2

1

2

2

The Cau hy-S hwarz inequality of E.3 yields that

jq(x )j 0

2

However, if

n X i=0

j i j

i = so

!

n X

2

i=0

jpi (x )j 0

pi (x ) 0

P

n j =0

jpj (x )j 0

2

!

2

1=2

= Kn (x ; x ) : 0

0

;

Kn (x ; x) ; (Kn (x ; x )) = then equality holds in the above inequality. ut

℄ Show, as in a℄, that if q 2 spanfp ; : : : ; pn g and p ; : : : ; pn are m times dierentiable at x ; then q(x) =

0

0

0

1 2

0

0

0

jq

(

j

m) (x ) 0

When does equality hold?

m X k=0

j

j

pkm) (x0 ) 2 (

!1=2

kqkL

2( )

:

48


E.6 Completeness. Let ff : 2 Ag be an orthonormal olle tion in a Hilbert spa e H: The olle tion ff : 2 Ag is alled a maximal orthonormal set in H if there is no f 6= 0 so that hf; f i = 0 for every 2 A: The following statements are equivalent: (1) The set of all nite linear ombinations of f ; 2 A; is dense in H: P (2) kf k = 2A jhf ; f ij for all f 2 H: 2

2

P

(3) hf; g i = 2A hf ; f ihf ; g i for all f; g 2 H: (4) ff : 2 Ag is a maximal orthonormal set in H: If any of the above holds, then the orthonormal olle tion is alled a omplete orthonormal system. (See, for example, Rudin [87℄.) a℄ Dedu e (1) ) (2) from Theorem 2.2.3. b℄ Dedu e (2) ) (3) from the simple identity 4hf; g i = kf + g k

2

kf

gk + ikf + igk 2

2

ikf

igk : 2

The above identity is alled polarization.

℄ Prove (3) ) (4). d℄ Prove (4) ) (1) by ontradi tion. Equality (3) is alled Parseval's identity. The remaining exer ises assumes some familiarity with measure theory. E.7 Basi Theory of Lp Spa es. Let (X; ) be a measure spa e ( is nonnegative) and p 2 (0; 1℄: The spa e Lp () is de ned as the olle tion of equivalen e lasses of measurable fun tions for whi h kf kLp < 1; where (

kf kLp (

)

:=

Z

X

jf j

p d

1=p

;

)

p 2 (0; 1)

and

kf kL1 (

)

:= supf 2 R : (fx 2 X : jf (x)j > g) > 0g < 1 :

In any of the ases the equivalen e lasses are de ned by the equivalen e relation f g if f = g -almost everywhere on X . When we write Lp [a; b℄ we always mean Lp (); where is the Lebesgue measure on X = [a; b℄: The notations Lp (a; b); Lp [a; b); and Lp (a; b℄ are also used analogously to Lp [a; b℄:


a℄ Holder's Inequality. Suppose 1 p < q 1 and p that Z

1

+q

1

= 1: Show

X

fg d kf kLp kgkLq (

)

(

)

for every f 2 Lp () and g 2 Lq (): If 1 < p; q < 1; then equality holds if and only if jf jp = jg jq almost everywhere on X for some ; 2 R with + > 0; and there is a 2 C with j j = 1 so that fg is nonnegative -almost everywhere on X: Holder's inequality was proved by Rogers [1888℄ before Holder [1889℄ proved it independently. Hint: If the right-hand side is 0; then the inequality is obvious. If it is dierent from 0; then let 2

F :=

jf j kf kLp (

and G := )

2

jg j kgkLq (

If x 2 X is su h that 0 < F (x) < real numbers s and t su h that

: )

1 and 0 < G(x) < 1; then there are

F (x) = es=p and G(x) = et=q : Use the onvexity of the exponential fun tion to show that

es=p

+

t=q

p

1

es + q et : 1

Apply this with the above hoi es of s and t; and integrate both sides on

X with respe t to :

b℄ Minkowski's Inequality for p 2 [1; 1℄: Let p 2 [1; 1℄: Show that

kf + gkLp kf kLp (

)

(

)

+ kg kLp (

ut

)

for every f; g 2 Lp (): If 1 < p; q < 1; then equality holds if and only if f = g -almost everywhere on X for some ; 2 R with + > 0: Hint: The ases p = 1 and p = 1 are straightforward. Let p 2 (1; 1): Then 2

jf + gjp jf j jf + gjp

1

2

+ jg j jf + g jp

1

and apply Holder's inequality (part a℄) to ea h term separately.

ut

50


By part b℄, Lp () is a ve tor spa e and k kLp is a norm on Lp () whenever p 2 [1; 1℄: If p 2 (0; 1); then k kLp is still alled a norm in the literature, however, for p 2 (0; 1) the subadditive property, in general, fails. In fa t, if p 2 (0; 1); then kkLp a;b is superadditive for Riemann integrable fun tions in Lp [a; b℄; see Polya and Szeg}o [76℄.

℄ Assume (X ) < 1. Show that Lq () Lp () for every 0 < p < q 1: If (X ) 1; then prove that (

(

[

)

)

℄

kf kLp kf kLq (

)

(

)

for every measurable fun tion f . d℄ Assume f 2 Lq () for some q > 0: Show that lim kf kLp = kf kL1 :

p!1

(

)

(

)

e℄ Riesz-Fis her Theorem. Show that if 1 p 1; then (Lp (); ) is a

omplete metri spa e, where

(f; g) := kf

gkLp : (

)

Hint: Use the monotone onvergen e theorem and Minkowski's inequality (part b℄); see Rudin [87℄ for details. ut If p 2 [1; 1℄; then q 2 [1; 1℄ de ned by p + q = 1 is alled

onjugate to p: f℄ Bounded Linear Fun tionals on Lp (). Let 1 p < 1 and g 2 Lq (); where q is the onjugate exponent to p: Show that 1

g (f ) :=

Z

X

1

fg d

is a bounded linear fun tional on Lp (): Hint: Use Holder's inequality (part a℄). ut g℄ Riesz Representation Theorem. Suppose 1 p < 1, is (-) nite and is a bounded linear fun tional on Lp(): Then there is a unique g 2 Lq (); where q is the onjugate exponent to p; so that

(f ) =

Z

X

fg d ;

f

2 Lp () :

Moreover, if and g are related as above, then

kk := maxf(f ) : f 2 Lp (); kf kLp (

)

= 1g = kg kLq :

Proof. See, for example, Rudin [87℄ or Royden [88℄.

(

)

ut


If X is a lo ally ompa t Hausdor spa e, then the hara terization of bounded linear fun tionals on the spa e C (X ) of ontinuous fun tions with ompa t support equipped with the uniform norm is also known as the Riesz representation theorem, and its proof may be found in, for example, Rudin [87℄. h℄ Orthogonality in Lp (). Suppose 1 p < 1; is ( -) nite, Y is a nite-dimensional subspa e of Lp (): The fun tion f 2 Lp () is said to be orthogonal to Y in Lp (); written f ? Y; if

kf kLp kf + hkLp for every h 2 Y: Show that an element f 2 Lp () is orthogonal to Y (

only if

Z

jf jp

X

for every h 2 Y; where

)

(

1

)

if and

sign(f )h d = 0

8 >
:

Outline. Suppose that the integral vanishes for every h 2 Y: Let q be the

onjugate exponent to p. de ned by p + q = 1: Observe that 1

g := jf jp and

Z

X

1

1

sign(f ) 2 Lq ()

jgjq d =

Z

X

jf jp d :

Without loss of generality we may assume that kf kLp = 1: Then for every h 2 Y; Holder's inequality yields that (

kf kLp (

)

=1=

Z

X

fg d =

Z

X

(f + h)g d

kf + hkLp kgkLq (

)

)

(

)

= kf + hkLp ; (

)

proving that f ? Y: (Observe that this argument is also valid for p = 1:) Suppose now f ? Y: Without loss of generality we may assume that f 62 Y: By a standard orollary to the Hahn-Bana h theorem (see, for example, Rudin [87℄), there exists a linear fun tional M on Lp () su h that M (f ) = 1, M (h) = 0 for every h 2 Y; and kM k = kf kLp : This M is then representable by some element g 2 Lq (); that is, 1

(

M (f ) =

Z

X

fg d;

kgkLp (

= kf kLp 1

)

(

)

)

52


(see part g℄). Therefore Z

fg d = kf kLp kgkLq (

X

)

(

) ;

and by the onditions for equality to hold in Holder's inequality,

g(x)f (x) 0 and

a:e: [℄ on X

jg(x)jq = jf (x)jp

a:e: [℄ on X

for a suitable onstant > 0: Hen e

g(x) = jf (x)jp

1

sign(f (x)) ;

and so M (h) = 0, h 2 Y; implies Z

jf jp

X

1

sign(f )h d = 0

for every h 2 Y: ut The statement of part h℄ remains valid for losed subspa es Y instead of nite-dimensional subspa es, see, for example, Shapiro [71℄. i℄ Minkowski's Inequality for p 2 (0; 1). Show that

kf + gkLp 2 =p (

1

)

kf kLp

1

(

)

+ kg kLp (

)

for every f; g 2 Lp () and p 2 (0; 1): Hint: Verify that

kf + gkLp (

Z

)

Z

X X

2 =p 1

(jf j + jg j)p d

jf j

p d + Z

1

X

Z

X

1=p

jg j

jf jp d

p d

1=p

1=p

+

Z

X

jgjp d

1=p !

whenever f; g 2 Lp () and p 2 (0; 1): Further properties of Lp () spa es may be found in Rudin [87℄.

ut


E.8 Fourier Series. a℄ Show that

in e

p

:n2Z 2 is a maximal orthonormal olle tion in L [ ; ℄: Hint: The orthonormality is obvious. In order to show the maximality, rst note that L [ ; ℄ is a Hilbert spa e by the Riesz-Fis her theorem (E.7 e℄). Hen e, by E.6, it is suÆ ient to show that the set T of all omplex trigonometri polynomials is dense in L [ ; ℄: By the Stone-Weierstrass theorem (E.2 of Se tion 4.1) T is dense in C [ ; ℄; where C [ ; ℄ is the spa e of all omplex-valued 2 -periodi ontinuous fun tions on R equipped with the uniform norm on R: Finally, it is a standard measure theoreti argument to show that C [ ; ℄ is dense in L [ ; ℄; see, for example, Rudin [87℄. ut 2

2

2

2

by

The k th Fourier oeÆ ient fb(k ) of a fun tion f

fb(k)

Z

1 := 2

f ()e

The (formal) Fourier series of a fun tion f

f

1 X k=

The fun tions

Sn () :=

1

2L [ 1

; ℄ is de ned

ik d :

2L [ 1

; ℄ is de ned by

fb(k)eik :

n X k= n

fb(k)eik

are alled the nth partial sums of the Fourier series of f . b℄ Show that if f 2 L [ ; ℄; then 2

1 X k=

1

jfb(k)j

2

= kf kL2 ; < 1 : 2

[

℄

ut

Hint: Use part a℄, E.6, and E.7 e℄.

℄ Show that if f 2 L [ ; ℄; then 2

lim kf

n!1

Sn kL2

[

;℄ = 0 ;

so f is the L [ ; ℄ limit of the partial sums of its formal Fourier series. Hint: Use part b℄. ut 2

54


Carleson, in 1966, solved Luzin's problem by showing that Sn ! f almost everywhere on [ ; ℄ for every f 2 L [ ; ℄. Earlier, Kolmogorov showed that there is a fun tion f 2 L [ ; ℄ so that Sn diverges everywhere on [ ; ℄: d℄ Isometry of L [ ; ℄ and ` . Let 2

1

2

2

(

` 2 :=

x = (xk )1

1 : xk 2 C ;

k=

and

1 X

kxk` := 2

1 X

jxk j

!1=2

2

k=

jxk j

0 is xed, then any olle tion ff : 2 Ag from L [ ; ℄ for whi h kf f kL2 ; ; ; 2 A ; 6= must be ountable. e℄ The Riemann-Lebesgue Lemma. If f 2 L [ ; ℄; then fb(k ) ! 0 as 2

2

2

2

[

℄

k ! 1: Hint: First prove it for step fun tions, then extend the result to every f 2 L [ ; ℄ by using the fa t that step fun tions form a dense set in L [ ; ℄: ut 1

1

1

f℄ Show that

1 X

n=1

1

n

2

=

2

6

;

1 X n=1

1

n

4

=

4

90

; and

1 X n=1

for every k = 3; 4; ; where rk is a rational number. Hint: Let f be the 2 periodi fun tion de ned by

f () := Apply part b℄.

2

k

;

1

n

2

2k k = rk

2 [0; 2) :

ut


E.9 Denseness of Polynomials in L () on R. a℄ Let be a nite Borel measure on [a; b℄ and f 2

Z b

a

f (x)eitx d(x) = 0 ;

t=

2 L (): Show that if 1

2k ; k = 0; 1; 2; : : : ; b a

then f (x) = 0 a.e. [℄ on [a; b℄. Outline. Use the fa t that the set T of all omplex trigonometri polynomials is dense in C [ ; ℄ (see the hint given for E.8 a℄) and standard measure theoreti arguments to show that the assumption of part a℄ implies Z b

f (x)g(x) d(x) = 0

a

for every bounded measurable fun tion g de ned on [a; b℄: Now, hoosing 8 >
:

we obtain

Z

jf (x)j d(x) = 0 ;

and the result follows. b℄ Let be a nite Borel measure on R and f Z R

f (x)eitx d(x) = 0 ;

2 L (): Show that if

ut

1

t 2 R;

then f (x) = 0 a.e. [℄ on R: Hint: Use part a℄ to show that the assumption of part b℄ implies Z R

f (x)g(x) d(x) = 0

for every bounded measurable fun tion de ned on R ( rst assume that g has ompa t support, then eliminate this assumption). Finish the proof as in part a℄. ut

℄ Let be a Borel measure on R satisfying Z R

erjxj d(x) < 1

with some r > 0: Show that the set P of all omplex algebrai polynomials is dense in L (): 2

56


Outline. First observe that the assumption on implies P L (): The fa t that L () is a Hilbert spa e (see E.7 e℄), Theorem 2.2.2 (GramS hmidt), and E.6 imply that it is suÆ ient to prove that if f 2 L () 2

2

2

and

Z R

f (x)xk d(x) = 0 ;

k = 0; 1; 2; : : : ;

then f (x) = 0 a.e. [℄ on R: Assume that f 2 L () satis es the above orthogonality relation. Use Theorem 2.2.1 a℄ (Cau hy-S hwarz inequality) to show that Z 2

F (t) :=

R

is well-de ned on R: For every t

0

f (x)e

itx d(x)

2 R; we have

itx = f (x)e it0 x e i(t t0 )x 1 X (t t0 )k = ( i)k f (x)e it0 x xk :

f (x)e

k!

k=0

Note that if jt t j r=2; then the integral of the right-hand side with respe t to (x) on R an be al ulated by integrating term by term sin e 0

1 Z X ( k=0

R

=

Therefore, if jt

(t i)k

1Z X k=0 Z

R

Z

0

R

jf (x)j

0

k!

t

k!

jf (x)jejt

R

t

jt

t )k

2

0

it0 x xk

f (x)e

d(x)

jk jf (x)jjxjk d(x)

t0 jjxj d(x)

d(x)

Z

e jt 2

R

1=2

t0 jjxj d(x)

< 1:

j r=2; then

F (t) =

1 X k=0

( i)k

(t

t )k

k!

Z

0

R

f (x)e

it0 x xk d(x) :

This means that F has a Taylor series expansion about every t 2 R with radius of onvergen e at least r=2: Also, with the hoi e t = 0; by the assumed orthogonality relations, we have F (t) = 0 whenever jtj r=2: We

an now dedu e that F (t) = 0 for every t 2 R: Hen e it follows from part b℄ that f (x) = 0 a.e. [℄ on R: ut 0

0

2.3 Orthogonal Polynomials 57

2.3 Orthogonal Polynomials

The lassi al orthogonal polynomials arise on orthogonalizing the sequen e (1; x; x ; : : : ) 2

with respe t to various parti ularly ni e weights, w(x); on an interval, whi h, after a linear transformation, may be taken to be one of [ 1; 1℄, [0; 1); or ( 1; 1). The main examples we onsider are the Ja obi polyno-

mials

(2.3.1) Pn; (x) ; where w(x) := (1 x) (1+ x) on [ 1; 1℄ ; ; > 1 : (

)

When = = 1=2 the Ja obi polynomials are the Chebyshev polynomials of the rst kind, (2.3.2)

Tn(x) ; where w(x) := (1

x) 2

= on [ 1; 1℄ :

1 2

When = = 1=2 they are the Chebyshev polynomials of the se ond kind, (2.3.3)

Un(x) ; where w(x) := (1

x)

= on [ 1; 1℄ :

2 1 2

Another spe ial ase of importan e is = = 0; whi h gives the Legendre polynomials, (2.3.4)

Pn (x) ; where w(x) = 1 on [ 1; 1℄ :

The Laguerre polynomials are (2.3.5)

Ln(x) ; where w(x) := e

x on [0;

1) :

The Hermite polynomials are (2.3.6)

Hn (x) ; where w(x) := e

x2 on (

1 ; 1) :

The above notation is traditionally used to denote orthogonal polynomials with a standard normalization; see the exer ises. It is not usually the ase that this normalization gives orthonormality. All of these mu h studied polynomials arise naturally, as do all the spe ial fun tions, in the study of dierential equations. We atalog some of the spe ial properties of these

lassi al orthogonal polynomials in the exer ises. In general, a nonde reasing bounded fun tion (typi ally the distribution fun tion of a nite measure) de ned on R is alled an m-distribution if it takes in nitely many distin t values, and its moments, that is, the improper Stieltjes integrals

58

2. Some Spe ial Polynomials Z1

1

xn d(x) =

Z!2

lim

!1 ! 1 !2 !+1 !1

xn d(x) ;

exist and are nite for n = 0; 1; : : : : Theorem 2.3.1 (Existen e and Uniqueness of Orthonormal Polynomials).

For every m-distribution there is a unique sequen e of polynomials (pn )1 n with the following properties: =0

(i) (ii)

pn (x) = n xn + rn (x) ;

n > 0 ; r n

1

Z R

pn (x)pm (x) d(x) = Æn;m =

1 0

1

2 Pn

1

;

for n = m for n 6= m :

Proof. The result follows from Theorem 2.2.2 (Gram-S hmidt). Note that the de ning property of an m-distribution ensures that

hp; qi :=

Z

R

pq d

ut is an inner produ t on Pn . The sequen e (pn )1 n de ned by Theorem 2.3.1 is alled the sequen e of orthonormal polynomials asso iated with an m-distribution . The seis alled a sequen e of orthogonal polynomials asso iated quen e (qn )1 n with an m-distribution if =0

=0

qn = n pn ; 0 6= n 2 C ;

n = 0; 1; : : : ;

where (pn )1 n is the sequen e of orthonormal polynomials asso iated with . The support supp() of an m-distribution is de ned as the losure of the set fx 2 R : is in reasing at xg : If is absolutely ontinuous on R; then =0

d(x) = w(x) dx with some 0 w 2 L (1; 1) 1

in whi h ase may be identi ed as a nonnegative weight fun tion w 2 L ( 1; 1) whose integral takes in nitely many distin t values. If (a; b) is an interval and w 2 L [a; b℄ has an integral that takes in nitely many distin t values, then the sequen e of orthogonal (orthonormal) polynomials asso iated with (a; b) we(x) = w(0x) ifif xx 2 62 (a; b) is said to be orthogonal (orthonormal) with respe t to the weight w. One thing distinguishing orthogonal polynomials from general orthogonal systems is the existen e of a three-term re ursion. 1

1


Theorem 2.3.2 (Three-Term Re ursion). Suppose (pn )1 n is a sequen e of =0

orthonormal polynomials with respe t to an m-distribution . Then xpn (x) = an pn (x) + bn pn (x) + an pn (x) ; +1

1

n = 0; 1; : : : ;

1

where p

1

:= 0 ;

a

1

= 0;

n > 0 ; bn 2 R ;

n

an =

n = 0; 1; : : :

+1

( n is the leading oeÆ ient of pn ). This theorem has a onverse due to Favard [35℄; see E.12.

Proof. Sin e xpn (x) 2 Pn ; we may write +1

xpn (x) =

(2.3.7)

nX +1 k=0

dk 2 R :

dk pk (x) ;

For notational onvenien e, let

hp; qi :=

Z R

p(x)q(x) d(x)

for any two polynomials p and q . Sin e hpn ; q i = 0 for every q have hxpn (x); q(x)i = hpn (x); xq(x)i = 0

2 Pn

1

; we

for every q 2 Pn . In parti ular, 2

hxpn (x); pk (x)i = 0 ;

k = 0; 1; : : : ; n

2:

On the other hand, using (2.3.7) and the orthonormality of (pn )1 n ; we obtain hxpn (x); pk (x)i = dk hpk ; pk i = dk : =0

Hen e dk = 0 for ea h k = 0; 1; : : : ; n (2:3:8)

2 and

xpn (x) = dn pn (x) + dn pn(x) + dn pn (x) : +1

+1

1

1

Here the lead oeÆ ient of the left-hand side polynomial is n ; while the lead oeÆ ient of the right-hand side polynomial is dn n ; so +1

an := dn In order to show that an := dn orthonormality of (pn )1 n imply 1

=0

+1

1

= n = n

+1

+1

:

= n = n ; note that (2.3.8) and the 1

60


0 = hpn ; pn i 1 = hxpn (x); pn (x)i d +1

1

n+1

= =

1

dn

+1

hpn (x); n

1 n

dn

dn

dn

1

+1

n

1

dn dn

1

Hen e

xn i 1

+1

dn dn

hpn ; pn i 1

+1

dn dn

1 +1

hpn

1

; pn

1

i

1 +1

:

an

1

:= dn

1

= n

n

1

:

ut

Theorem 2.3.3 (Christoel-Darboux Formula). With the notation of the

previous theorem, n X k=0

pk (x)pk (y) =

n

n

pn (x)pn (y) pn (x)pn (y) x y +1

+1

+1

for all x 6= y 2 C : Proof. Theorem 2.3.2 (three-term re ursion) yields that k : = pk (x)pk (y) pk (x)pk (y) a 1 = (x y )pk (x)pk (y ) + k (pk (x)pk (y ) ak ak +1

+1

1

1

pk (x)pk (y)) : 1

So

k k = pk (x)pk (y ) + ak ; x y x y and we sum the above from 0 to n to get the desired formula. ak

1

1

ut

Corollary 2.3.4. In the notation of Theorem 2.3.2 n X k=0

pk (x) = 2

n

n

+1

p0n (x)pn (x) +1

Proof. Let y ! x in Theorem 2.3.3.

p0n (x)pn (x) : +1

ut

We an dedu e quite easily from this that orthogonal polynomials asso iated with an m-distribution have real interla ing zeros lying in the interior of the smallest interval ontaining supp(); see E.1 and E.2.


Comments, Exer ises, and Examples. Askey, in omments following an outline of the history of orthogonal polynomials by Szeg}o [82, vol. III℄, writes: \The lassi al orthogonal polynomials are mostly attributed to someone other than the person who introdu ed them. Szeg}o refers to Abel and Lagrange and Ts hebys he in [75, hapter 5℄ for work on the Laguerre polynomials Ln (x). Abel's work was published posthumously in 1881. Probably the rst published work on these polynomials that uses their orthonormality was by Murphy (1833). Hermite polynomials were studied extensively by Lapla e in onne tion with work on probability theory. Hermite's real ontribution to these polynomials was to introdu e Hermite polynomials in several variables. Lagrange ame a ross the re urren e relation for Legendre polynomials." Perhaps this is not very surprising given the many diverse ways in whi h these polynomials an arise. There are many sour es for the basi properties of orthogonal polynomials. In parti ular, Askey and Ismail [84℄, Chihara [78℄, Erdelyi et al. [53℄, Freud [71℄, Nevai [79b℄, [86℄, Szeg}o [75℄, and, in tabular form, Abramowitz and Stegun [65℄ are su h sour es. Exer ises in lude a treatment of the elementary properties of the most familiar orthogonal polynomials. The onne tions linking orthogonal polynomials, the moment problem, and Favard's

onverse theorem to the three-term re ursion are also examined in the exer ises. 0

E.1 Simple Real Zeros. Let (pn )1 n be the sequen e of orthonormal polynomials asso iated with an m-distribution . Show that ea h pn has exa tly n simple real zeros lying in the interior of the smallest interval ontaining supp(): Hint: Suppose the statement is false. Then pn has at most n 1 sign hanges on [a; b℄; hen e there exists 0 6= q 2 Pn so that =0

1

pn (x)q(x) 0 ;

x 2 [a; b℄ :

Show that this ontradi ts the orthogonality relation 0=

Z R

pn (x)q(x) d(x) =

Z b

a

pn (x)q(x) d(x) :

ut E.2 Interla ing of Zeros. Let (pn )1 be the sequen e of orthonormal n polynomials asso iated with an m-distribution . Then the zeros of pn and pn stri tly interla e. That is, there is exa tly one zero of pn stri tly between any two onse utive zeros of pn . =0

+1

+1

62


Hint: From Corollary 2.3.4, p0n (x)pn (x)

p0n (x)pn (x) +1

+1

is positive on R: Sin e pn has n + 1 simple real zeros (see E.1), we see that if and Æ are two onse utive zeros of pn ; then +1

+1

sign(p0n ( )) = sign(p0n (Æ )) ; +1

and hen e

+1

sign(pn ( )) = sign(pn (Æ )) :

ut

1 be the sequen e of E.3 Orthogonality of (Kn (x ; x))1 n . Let (pn )n orthonormal polynomials asso iated with an m-distribution . Let 0

=0

=0

x < min supp() or x > max supp() : 0

0

be the sequen e of asso iated kernel fun tions (as in Let (Kn (x ; x))1 n E.5 of Se tion 2.2). Show that 0

=0

Z R

Kn (x ; x)Km (x ; x)jx 0

x j d(x) = 0 ;

0

0

for any two nonnegative integers n 6= m: E.4 Hypergeometri Fun tions. We introdu e the following standard notation: the rising fa torial (or Po hammer symbol) (a)n := a(a + 1) (a + n

1) ;

(a) := 1 0

for a 2 C and n = 1; 2; : : : ; the binomial oeÆ ient

a a(a 1) (a n + 1) ; := n! n

a 0

:= 1

for a 2 C and n = 1; 2; : : : ; and the Gaussian hypergeometri series 2

F (a; b ; ; z ) := F (a; b ; ; z ) := 1

1 X n=0

(a)n (b)n z n ( )n n!

for a; b; 2 C : a℄ For Re( ) > Re(b) > 0 ;

F (a; b ; ; z ) =

( ) (b) ( b)

Z

1

tb (1 t) 1

0

b

1

(1

tz )

a dt ;


is, as usual, the gamma fun tion de ned by

where

(z ) :=

1

Z

tz e t dt ;

Re(z ) > 0 :

1

0

ut

Proof. See, for example, Szeg}o [75℄.

b℄ Hypergeometri Dierential Equation. The fun tion y = F (a; b ; ; z ) satis es dy dy + [ (a + b + 1)z ℄ aby = 0 : z (1 z ) 2

dz

dz

2

ut

Proof. See, for example, Szeg}o [75℄.

In E.5, E.6, and E.7 we atalog some of the basi properties of some of the lassi al orthogonal polynomials. Proofs are available in Szeg}o [75℄, for example. E.5 Ja obi Polynomials. a℄ Rodrigues' Formula. Let

Pn; (x) := ( 1)n (

)

2 n dn (1 x) (1+x) n (1 n! dx

x) (1 + x) (1 x )n : 2

Then (Pn; )1 n is a sequen e of orthogonal polynomials on [ 1; 1℄ asso iated with the weight fun tion (

)

=0

1 < ; < 1 :

w(x) := (1 x) (1 + x) ; That is,

Pn; ) (

2 Pn

Z

and

1

Pn; Pm; (1 x) (1 + x) dx = 0 (

1

)

(

)

for any two nonnegative integers n 6= m.

In the rest of the exer ise, the polynomials Pn; are as in part a℄. b℄ Normalization. We have (

Pn; (1) = (

)

)

n+ ( + 1)n = n n!

and Z

1

(Pn; (x)) (1 (

1

)

2

=

x) (1 + x) dx

(n + + 1) (n + + 1) 2 : 2n + + + 1 (n + 1) (n + + + 1) +

+1


64

℄ Expli it Form.

Pn; (x) = (

)

n 1 X n+ 2n m m

=0

=

n+ (x n m

n X +n +

n m n+ = F n m=0

2

1

1)n m (x + 1)m

+n+m

m

x

2

1 m

n; n + + + 1 ; + 1 ;

1

2

x

:

d℄ Dierential Equation. The fun tion y = Pn; (x) satis es (

(1

x) 2

dy + [ dx 2

2

( + + 2)x℄

)

dy + n(n + + + 1)y = 0 : dx

e℄ Re urren e Relation. The sequen e (Pn; (x))1 n satis es (

)

=0

Dn Pn; (x) = (An + Bn x)Pn; (x) (

)

(

)

+1

Cn Pn; (x) ; (

)

1

where

P

(

; )

0

=1

and

P

(

; )

1

(x) = [ 1

2

+ ( + + 2)x℄

and

Dn = 2(n + 1)(n + + + 1)(2n + + ) An = (2n + + + 1)( ) Bn = (2n + + + 2)(2n + + + 1)(2n + + ) Cn = 2(n + )(n + )(2n + + + 2) : 2

2

f℄ Generating Fun tion.

1 X

p

n=0

Pn; (x)z n = (

)

2 ; z + R) (1 + z + R) +

R(1

where R = 1 2xz + z : There are various spe ial ases, some of whi h we have previously de ned. The Legendre polynomials Pn are de ned by 2

Pn := Pn ; ; (0 0)

n = 0; 1; : : : :


The Chebyshev polynomials Tn de ned in Se tion 2.1 satisfy 4n

= ;

n Pn n

Tn =

(

1 2

2

=

1 2)

;

n = 0; 1; : : : :

The ultraspheri al (or Gegenbauer) polynomials Cn are de ned by (

(2 + n) ( + ) = ; = P ; (2) ( + n + ) n 1

Cn := (

)

)

(

2

1 2

n = 0; 1; : : : :

1 2)

1

2

In terms of Cn ; the Chebyshev polynomials of the rst and se ond kind are given by (

)

n Tn = Cn 2

(0)

and Un = Cn ; (1)

n = 0; 1; : : : :

E.6 Hermite Polynomials. a℄ Rodrigues' Formula. Let

Hn (x) :=

( 1)n dn exp( x ) : exp( x ) dxn 2

2

Then (Hn )1 n is a sequen e of orthogonal polynomials on (

iated with the weight fun tion =0

1; 1) asso-

w(x) := exp( x ) : 2

That is,

Hn 2 Pn and

Z

Hn (x)Hm (x) exp( x ) dx = 0 2

R

for any two nonnegative integers n 6= m. In the rest of the exer ise, the polynomials Hn are as in part a℄. b℄ Normalization. We have Z

1

1 and

H

2

p

(Hn (x)) exp( x ) dx = 2n n! 2

2

H n (0) = ( 1)n

n+1 (0) = 0 ;

2

(2n)! : n!

℄ Expli it Form.

bX n= 2

Hn (x) = n!

m=0

( 1)m (2x)n m : m!(n 2m)! 2

66


d℄ Dierential Equation. The fun tion y = Hn (x) satis es

dy + 2ny = 0 : dx e℄ Re urren e Relation. The sequen e (Hn (x))1 n satis es dy dx 2

2x

2

=0

Hn (x) = 2xHn (x)

2nHn (x)

H (x) = 1

H (x) = 2x :

+1

with

1

and

0

1


1 X

n=0

Hn (x)

zn = exp(2xz n!

z ): 2

E.7 Laguerre Polynomials. Let 2 ( 1; 1). a℄ Rodrigues' Formula. Let

Ln (x) := (

)

dn (e x x n ) : n!e xx dxn 1

+

Then (Ln )1 n is a sequen e of orthogonal polynomials on [0; 1) asso iated with the weight fun tion (

)

=0

w(x) := x exp( x) : That is,

Ln (

)

2 Pn

1

Z

and

0

Ln (x)Lm (x)x exp( x) dx = 0 (

)

(

)

for any two nonnegative integers n 6= m.

In the rest of the exer ise, the polynomials Ln are de ned as in part a℄. b℄ Normalization. We have (

1

Z

0

)

( + n + 1) (Ln (x)) x e x dx = n! (

and

)

2

Ln (0) = n + n (

)

:

℄ Expli it Form.

Ln (x) = (

)

n X m=0

( 1)m m!

n+ m n m x :


d℄ Dierential Equation. The fun tion y = Ln (x) satis es (

x

)

dy dy + ( + 1 x) + ny = 0 : dx dx 2

2

e℄ Re urren e Relation. The sequen e (Ln (x))1 n satis es (

)

=0

(n + 1)Ln (x) = [(2n + + 1) (

with

L =1 (

(

(

)

)

1

L (x) = x + + 1 :

and

)

0

(n + )Ln (x)

x℄Ln (x)

)

+1

(

)

1


1 X n=0

Ln) (x)z n = exp

(

z

xz

1

z)

(1

1

:

E.8 Christoel Numbers and Gauss-Ja obi Quadrature. Let (pn )1 n be the sequen e of orthonormal polynomials asso iated with an m-distribution . Let x;n , = 1; 2; : : : ; n; denote the zeros of pn . Let =0

1

;n := 0 pn (x;n )

Z Rx

pn (x) d(x) ; x;n

= 1; 2; : : : ; n :

The numbers ;n are alled the Christoel or Cotes numbers. a℄ Show that, for any q 2 P n ; 2

Z R

1

q(x)d(x) =

n X =1

;n q(x;n ) :

Hint: First show the equality for every q 2 Pn by using the Lagrange interpolation formula (E.6 of Se tion 1.1). If q 2 P n ; then q = spn + r with some s; r 2 Pn ; where s is orthogonal to pn . ut b℄ Show that ;n > 0 for every = 1; 2; : : : ; n: Hint: Use part a℄ to show that 1

2

1

1

;n =

Z

1

(p0n (x;n ))

2

R

pn (x) x x;n

2

d(x) :

ut

68


℄ Suppose [a; b℄ is a nite interval ontaining supp(). Let f Show that Z b f (x) d(x) a

n X =1

;n f (x;n )

2((b+)

(a )) min p2P2n

1

kf

2 C [a; b℄.

pk a;b : [

℄

Hint: Use parts a℄ and b℄ together with the observation n X =1

;n =

Z b

a

d(x) = (b+) (a ) :

d℄ Suppose supp() [a; b℄; where a; b 2 R. Show that n X =1

;n f (x;n ) n!1 !

Z b

a

ut

f (x)d(x)

for every Riemann-Stieltjes integrable fun tion on [a; b℄ with respe t to . Hint: First show that f is Riemann-Stieltjes integrable on [a; b℄ with respe t to if and only if for every > 0 there are g ; g 2 C [a; b℄ so that 1

g (x) f (x) g (x) ; 1

and

2

Z b

a

(g (x) 2

2

x 2 [a; b℄

g (x)) d(x) < : 1

Finish the proof by part ℄ and the Weierstrass approximation theorem (see E.1 of Se tion 4.1). ut e℄ Suppose supp() is ompa t. Let

Z := fx;n : = 1; 2; : : : ; n; n = 1; 2; : : : g : Show that supp() Z; where Z denotes the losure of Z . Hint: Use part d℄. f℄ Show by an example that supp() 6= Z is possible.

ut

E.9 Chara terization of Compa t Support. Using the notation of Theorem 2.3.2 and E.8, show that the following statements are equivalent: (1) supp() is ompa t. (2) supfjan j + jbn jg < 1. n2N

(3) The set Z := fx;n : = 1; 2; : : : ; n; n = 1; 2; : : : g is bounded.


Outline. (1) ) (2). Note that the orthogonality of fpng1 n implies =0

an =

Z

R

xpn (x)pn (x) d(x)

bn =

and

1

Z

xpn (x) d(x) : 2

R

So supp() [ K; K ℄; the Cau hy-S hwarz inequality, and the orthonormality of (pn )1 n yield =0

j an j K

Z K

jpn

K Z K

K

K

and

1

(x)pn (x)j d(x) !1=2

Z K

pn (x) d(x) 2

1

j bn j K

Z K

K

K

! 1=2

pn(x) d(x) 2

K

pn (x)d(x) = K : 2

(2) ) (3). Use Theorem 2.3.2 (three-term re ursion) to show that

x;n

nX1 k=0

pk (x;n ) 2 2

nX1 k=0

ak pk (x;n )pk (x;n ) + +1

+1

nX1 k=0

bk pk (x;n ) : 2

Hen e

jx;n j

nX1 k=0

pk (x;n ) 2 max kn 2

0

(3) ) (1). If Z Z R

x

[ 2

n

1

jak j +

0

max

kn

1

jbk j

n 1 X

k=0

pk (x;n ) : 2

K; K ℄; then by E.8 a℄ 2

d(x) =

n X =1

n ;n x;n 2

2

K

2

n

Z 2

R

d(x) ;

whi h implies supp() [ K; K ℄.

ut

be the sequen e E.10 A Condition for supp() [0; 1). Let (pn )1 n of orthonormal polynomials asso iated with an m-distribution . Suppose supp() is ompa t and =0

pn (0)pn (0) < 0 ; +1

n = 0; 1; : : : :

Show that supp() [0; 1). Hint: Use the interla ing property of the zeros of pn (E.2) to show that

Z := fx;n : = 1; 2; : : : ; n; n = 1; 2; : : : g [0; 1): Now use E.8 e℄ to obtain supp() [0; 1).

ut

70


E.11 The Solvability of the Moment Problem. Let (n )1 n be a sequen e of real numbers. We would like to hara terize those sequen es (n )1 n for whi h there exists an m-distribution so that =0

=0

Z

xn d(x) = n ;

R

Let

(p) :=

n = 0; 1; : : : :

n X

ak k k=0 Pn k n of the form p(x) = k=0 ak x .

for every p 2 P A polynomial p is alled nonnegative if it takes nonnegative values on the real line. The sequen e (n )1 n is alled positive de nite if =0

(p) :=

n X k=0

ak k > 0 ;

n = 0; 1; : : :

holds for every nonnegative polynomial p 2 Pn of the form

p(x) =

n X k=0

a k xk :

The aim of this exer ise is to outline the proof of Hamburger's hara terization of the solvability of the moment problem by the positive de niteness of the sequen e of moments. See part o℄. a℄ Show that if there exists an m-distribution so that Z

xn d(x) = n ;

R

n = 0; 1; : : : ;

then (n )1 n is positive de nite. Hint: An m-distribution is in reasing at in nitely many points. ut 1 b℄ Show that (n )n is positive de nite if and only if (p ) > 0 holds for every 0 6= p 2 Pn ; n = 0; 1; : : : : Hint: Use E.3 of Se tion 2.4. ut

℄ Show that (n )1 n is positive de nite if and only if =0

2

=0

=0

0 1 . . .

1

2

.. .

n n

+1

+1 .. > .

: : : n : : : n ..

.

:::

2

n

0;

n = 0; 1; : : : :


Hint: Use part b℄ and the law of inertia of Sylvester. See, for example, van

der Waerden [50℄. ut d℄ Helly's Sele tion Theorem. Suppose the fun tions fn ; n = 1; 2; : : : ;

are nonde reasing on R; and

sup kfn kR < 1 :

n2N

Then there exists a subsequen e of (fn )1 n that onverges for every x 2 R. That is, we an sele t a pointwise onvergent subsequen e. Hint: See, for example, Freud [71℄. ut e℄ Helly's Convergen e Theorem. Let [a; b℄ be a nite interval. Suppose the fun tions n , n = 1; 2; : : : ; are nonde reasing on [a; b℄ and =1

sup kn k a;b < 1 :

n2N

[

℄

Suppose also that (n (x))1 n onverges to (x) for every x 2 [a; b℄: Then =1

lim

Z b

n!1 a

f (x) dn (x) =

Z b

a

f (x) d(x)

for every f 2 C [a; b℄: Hint: See, for example, Freud [71℄.

ut

In the rest of the exer ise (ex ept for the last part) we assume that (n )1 is positive de nite. Our goal is to prove the onverse of part a℄. n Let : : : n 1 =0

0

pn (x) := ..

f℄ Show that

1

.

1

2

.. .

n n

+1

(pn q) = 0 ;

::: ..

n

1

.. .

.

:::

2

n

q 2 Pn

1

1

.. . n

x

:

x

:

g℄ Show that ea h pn has n simple real zeros. Hint: Use part f℄. Let x ;n > x ;n > > xn;n be the zeros of pn . Let 1

2

pn (x) ; l;n (x) := 0 pn (x;n )(x x;n )

= 1; 2; : : : ; n ; n = 1; 2; : : :

(see E.6 of Se tion 1.1), and let

;n := (l;n ) :

ut

72


h℄ Show that

(q) = for every q 2 P n : Hint: Use part f℄. i℄ Show that 2

n X =1

;n q(x;n )

1

ut

;n = (l;n ) > 0 ;

= 1; 2; : : : ; n ; n = 1; 2; : : : :

2

ut

Hint: Use part h℄. For x 2 R; let n (x) :=

X

f

:

x;n xg

;n ;

n = 1; 2; : : : :

j℄ Show that 0 n (x) on R for ea h n; and there is a subsequen e of that onverges pointwise to a nonde reasing real-valued fun tion (n )1 n on R. Hint: Use parts h℄, i℄, and d℄. ut k℄ Show that for every nite interval [a; b℄, 0

=1

lim

Z b

k!1 a

xm d

nk (x) =

Z b

a

xm d(x) ;

m = 0; 1; 2; : : : ;

where is de ned in part j℄. Hint: Use part e℄. ut l℄ Let m be a xed nonnegative integer and let r := bm=2 + 1. Show that if nk r + 1; a 1 and b 1; then Z a

1

xm dnk (x) +

1

Z

b

xm dnk (x)

ja1j + j1bj

Z

x r dnk (x) =

2

R

1

+

1

j a j j bj

r: 2

ut

Hint: Use part h℄. m℄ Show that Z R

xm d(x) = m ;

where is de ned in part j℄. Hint: Use parts k℄ and l℄.

m = 0; 1; 2; : : :

ut


n℄ Show that de ned in part j℄ is an m-distribution. o℄ There exists an m-distribution so that

n =

Z R

xn d(x)

if and only if (n )1 n is positive de nite, that is, if and only if =0

0 1 . . .

1

2

.. .

n n

+1

+1 .. .

: : : n : : : n ..

.

:::

2

> 0;

n = 0; 1; : : : :

n

ut

Hint: Combine parts a℄, ℄, m℄, and n℄.

Ne essary and suÆ ient onditions for the uniqueness of the solution of the moment problem are given in Freud [71℄, for example. E.12 Favard's Theorem. Given (an )1 n

polynomials pn 2 Pn are de ned by

=0

(0; 1) and (bn )1 n R; =0

the

xpn (x) = an pn (x) + bnpn (x) + an pn (x) ; p = 0; p = > 0: 1

1

0

+1

+1

0

Then there exists an m-distribution su h that Z R

pn (x)pm (x) d(x) = 0

for any two distin t nonnegative integers n and m. In other words, the

onverse of Theorem 2.3.2 is true. In order to prove Favard's theorem, pro eed as follows: a℄ Show that the polynomials pn are of the form

pn (x) = n xn + r(x) ;

n > 0 ; r 2 P n

1

b℄ Let pen := n pn ; n = 0; 1; : : : . The sequen e (n )1 n follows. Let := 1 ; (q) := if q = ; 2 R :

:

1

=0

is de ned as

0

If ; ; : : : ; n have already been de ned, then let 0

1

(q) :=

n X k=0

k k whenever q(x) =

n X k=0

k x k ; k 2 R

74


and let

n

:= (xn

+1

+1

pen (x)) : +1

Show that

(pen pem ) = 0 ; and

m = 0; 1; : : : ; n

(pen ) > 0 ;

1;

n = 0; 1; : : :

n = 0; 1; : : : :

2

Hint: It is suÆ ient to prove that (pen (x)xm ) = 0 ; and

m = 0; 1; : : : ; n 1 ; n = 0; 1; : : :

(pen (x)xn ) > 0 ;

n = 0; 1; : : : : These an be obtained from the de nition of and from Theorem 2.3.2 (three-term re ursion) by indu tion on n. ut

℄ Show that every q 2 Pn is of the form q= and if q 6= 0; then

n X k=0

(q ) = 2

k pek ; n X k=0

k 2 R

k (pek ) > 0 : 2

2

d℄ Show that (n )1 n is positive de nite in the sense of E.11. Hint: Use the previous part and E.11 b℄. ut e℄ Prove Favard's theorem. Hint: Use part o℄ of E.11, parts d℄ and b℄ of this exer ise, and the de nition of . ut =0

E.13 Christoel Fun tion. Let be an m-distribution. For a xed n 2 N ; the fun tion

n (z ) = inf

Z

q (x) d(x) : q 2 Pn 2

R

1

; jq(z )j = 1 ;

z2C

is alled the nth Christoel fun tion asso iated with . a℄ Show that

n (z ) :=

nX1 k=0

jpk (z )j

!

2

1

;

where (pn )1 n is the sequen e of orthonormal polynomials asso iated with . =0


Show also that the in mum in the de nition of n (z ) is a tually a minimum, and it is attained if and only if

q(x) = Hint: Write q=

Pn

k=0 pk (z )pk (x) 1 2 k=0 pk (z ) 1

Pn

nX1 k=0

j

j

:

k 2 C

k pk ;

and observe that the orthonormality of (pn )1 n implies =0

Z

q (x) d(x) = 2

R

nX1 k=0

j k j

2

:

Now use the Cau hy-S hwarz inequality (E.3 of Se tion 2.2) to nd the maximum of jq (z )j for polynomials q 2 Pn satisfying 1

Z

q (x) d(x) 1 2

R

where z 2 C is xed. b℄ Let ;n ; = 1; 2; : : : ; n; be the Christoel numbers asso iated with an m-distribution ; that is, the oeÆ ients in the Gauss-Ja obi quadrature formula, as in E.8. Show that

;n = n (x;n ) ;

= 1; 2; : : : ; n ;

that is, the Christoel numbers are the values of the Christoel fun tion at the zeros of the nth orthonormal polynomial pn . Hint: Use parts E.8 a℄ and E.8 b℄. ut

℄ Let x 2 R be xed. Show that

1 X n=0

pn (x) < 1 2

if and only if x is a mass point of ; that is, (x ) < (x+); in whi h ase

1 X n=0

pn (x) = ((x+) 2

(x ))

1

:

Hint: Use part a℄ and the Weierstrass approximation theorem. See E.1 of Se tion 4.1. ut

76


E.14 The Markov-Stieltjes Inequality. Let be an m-distribution with asso iated orthonormal polynomials (pn )1 n . Let x ;n > x ;n > > xn;n denote the zeros of pn . Let x ;n := 1 and xn ;n := 1. As in E.8 let ;n , = 1; 2; : : : ; n; be the Christoel numbers asso iated with . Show that Z x 1;n ;n d(x) ; = 1; 2; : : : ; n 1

=0

0

2

+1

x +1;n

and

Z x 1;n

x;n

d(x) ;n +

1

;n ;

= 2; 3; : : : ; n :

Hint: Let 1 k

n be xed. Use E.7 of Se tion 1.1 (Hermite interpolation) to nd polynomials P 2 P n and Q 2 P n with the following properties: (1) P (xj;n ) = Q(xj;n ) = 1 ; j = 1; 2; : : : ; k 1 ; (2) P (xk;n ) = 0 ; Q(xk;n ) = 1 ; (3) P (xj;n ) = Q(xj;n ) = 0 ; j = k + 1; k + 2; : : : ; n ; (4) P (x) 1; xk;n (x) Q(x) ; x 2 R; where 1 if 1 < x xk;n 1; xk;n (x) := 0 if xk;n < x < 1 : 2

(

1

2

1

℄

(

℄

ut

Now apply E.8 (Gauss-Ja obi quadrature formula) to P and Q.

E.15 Orthonormal Polynomials as Determinants. Suppose is an mdistribution with moments

n = Let

n :=

0 1 . . .

Z R

xn d(x) ;

1

2

.. .

n n

+1

n = 0; 1; : : : : +1 .. .

: : : n : : : n ..

.

:::

2

;

n = 0; 1; : : : :

n

a℄ Show that n > 0; n = 0; 1; 2; : : : : b℄ Show that the orthonormal polynomials pn asso iated with are of the form : : : n 1 0

pn (x) = (n n ) 1

1 1=2 .. .

1

2

.. .

n n

+1

::: ..

n

.

:::

2

1

.. .

n

1

.. . n

x

x

:


1

℄ Let (an )1 n (0; 1) and (bn )n R be the oeÆ ients in the threeterm re ursion for the sequen e (pn )1 n of orthonormal polynomials asso iated with as in Theorem 2.3.2. Show that the moni orthogonal polynomials pen := n pn are of the form =0

=0

=0

1

pen (x) = det (xIn

Jn )

where Jn is the tridiagonal n by n Ja obi matrix 0

b Ba B

Jn := B B

0 1

a b a

1

1

a b

1 2

2

2

..

.

a

3

..

.

C C C .. C .A

an bn

and In is the n by n unit matrix.

1 E.16 The Support of . Let (an )1 n (0; 1) and (bn )n R be the

oeÆ ient sequen es in the three-term re ursion for the sequen e of (pn )1 n of orthonormal polynomials asso iated with an m-distribution as in Theorem 2.3.2. a℄ Show that if supp() [b a; b + a℄ with some a > 0 and b 2 R; then =0

=0

=0

an a and

j bn

bj a ;

n = 0; 1; : : : :

Hint: Use the orthonormality of (pn )1 n to show that =0

an =

Z b+ a

b a

and

bn

b=

(x

Z b+ a

b a

b)pn (x)pn (x) d(x) 1

(x

b)pn (x) d(x) : 2

Now apply the Cau hy-S hwarz inequality, and use the orthonormality of (pn )1 ut n again. b℄ Show that supp() [ K; K ℄ =0

where

K := 2 supfan : n 2 N g + supfjbnj : n 2 Ng

(the suprema are taken over all nonnegative integers). Hint: Suppose K < 1; otherwise there is nothing to prove. Combine E.9, the inequality in the hint to the dire tion (2) ) (3) of E.9, and E.8 e℄. u t

78


℄ Blumenthal's Theorem. Assume that

a

lim a =: n!1 n 2

Then where F

R n [b

>0

and

lim b =: b n!1 n

2 R:

supp() = [b a; b + a℄ [ F a; b + a℄ is a ountable bounded set for whi h

F n (b a ; b + a + )

is nite for every > 0. Proof. See Nevai [79b℄ or Mate, Nevai, and Van Ass he [91℄. ut d℄ Rakhmanov's Theorem. Suppose supp() [b a; b + a℄ with some a > 0 and b 2 R. Suppose also that 0 (x) > 0 a.e. in [b a; b + a℄. Then a lim a = and nlim n!1 n !1 bn = b : 2 Proof. See, for example, Mate, Nevai, and Totik [85℄, or Nevai [91℄.

ut

There is an analogous theory of orthogonal polynomials on the unit

ir le initiated by Geronimus, Shohat, and A hiezer. An important ontribution, alled Szeg}o theory, may be found in Freud [71℄. E.17 A Theorem of Stieltjes [14℄. Let w be a positive ontinuous weight fun tion on [ a; a℄. Denote the nth moment by Z a

n :=

a

xn w(x) dx :

Let pen 2 Pn denote the nth moni orthogonal polynomial on [ a; a℄ asso iated with the weight w. Then (pen )1 n satis es a three-term re ursion =0

pen (x) = (x

An )pen (x)

Bn pen (x)

1

2

with pe (x) = 1 and pe (x) := x A ; see Theorem 2.3.2. Suppose the sequen e of polynomials (qn )1 n satis es the same re ursion ommen ing with q (x) := 0 and q (x) := B . Stieltjes' theorem (see, for example, Cheney [66℄) states the following. 0

1

1

=0

0

1

1

Theorem. For any x 2= [ a; a℄; Z a

w(t) dt = + + x x a x t 0

1

2

=

x

A

B

1

B

2

1

x

A

2

x A

B 3

3

:

2.4 Polynomials with Nonnegative CoeÆ ients 79

Furthermore, the nth onvergent qn =pen satis es qn (x) = pen (x)

B

1

x

A

B

2

...

1

Bn : x An

E.18 Completeness of Orthogonal Polynomials. Let (pn )1 be the sen quen e of orthonormal polynomials asso iated with an m-distribution . If supp() [a; b℄; where [a; b℄ is a nite interval, then (pn )1 n is a maximal orthogonal olle tion in L [a; b℄. Hint: Use the Weierstrass approximation theorem (E.1 of Se tion 4.1) on [a; b℄. ut =0

=0

2

E.19 Bounds for Ja obi Polynomials. For all Ja obi weight fun tions w(x) = (1 x) (1 + x) with 1=2 and 1=2; the inequalities max

x2[

p (x) Pn n 2 2

;

1 1℄

k=0 pk (x)

and max

x2[

p

;

1 1℄

p

4 2+ + 2n + + + 2

1

x w(x)pn (x) 2

2

2e 2 +

2

2

p

+ 2

2

hold, where (pn )1 is the sequen e of orthonormal Ja obi polynomials n asso iated with the weight fun tion w. Proof. See Erdelyi, Magnus, and Nevai [94℄. ut =0

2.4 Polynomials with Nonnegative CoeÆ ients A quadrati polynomial x + x + with real oeÆ ients has both roots in the halfplane fz 2 C : Re(z ) 0g if and only if 0 and 0. This easy onsequen e of the quadrati formula gives the following lemma: 2

Lemma 2.4.1. If p 2 Pn has all its zeros in fz 2 C : Re(z ) 0g; then either

p or p has all nonnegative oeÆ ients.

The onverse of this is far from true. Indeed, the following result of Meissner holds (see Polya and Szeg}o [76℄). We denote by Pn the set of polynomials in Pn ; that have all nonnegative oeÆ ients. +

80


Theorem 2.4.2. If p

2 Pn and p(x) > 0 for x > 0; then p = s=t; where s

and t are both polynomials with all nonnegative oeÆ ients.

Sin e a polynomial p that is real-valued on the positive real axis has real oeÆ ients, and sin e p(x) > 0 for all x > 0 implies that the leading

oeÆ ient of p is positive, Theorem 2.4.2 will follow immediately from the next lemma. Lemma 2.4.3. Suppose ;

2

and suppose x x + has no nonnegative root. Then x x + = p(x)=q(x); where p; q 2 Pm both have all nonnegative oeÆ ients, and where 2

R

2

m 10 4

2

=

1 2

:

Proof. The quadrati polynomial x x + has no positive root if and only if < 4 . We set := = and note that < 4. Consider 2

2

2

(x

2

x + )(x + x + ) = x + (2 )x + = x + (2 )x + : 2

4

2

4

2

2

2

2

If 2 we have the desired fa torization. If > 2; onsider (x + (2

)x + )(x

4

2

2

4

(2 )x + ) = x + (2 (2 ) )x + : 2

8

If 2 Let

(2

2

2

2

4

4

) > 0 we are nished. In general, we pro eed as follows: 2

Pn (x) : = x =x

2

2

n+1

+ n+1 +

n

2

n 2

1

2

(2

(2 (2 1 n n

n x + ; 2

) )

(2

2 2

) )x n + n 2

2

2

where n has n nested terms. Let

Qn (x) := x Note that, sin e n

+1

2

n+1

2

n

1

n

n

n x + : 2

2

n

=2

2

n

n

( +1)+1

n+1

n

Pn (x)Qn (x) = x n x + 2 x n+2 n n+1 n+1 =x + n x + = Pn (x) : 2

2

2

2

+1

2

+1

2

2

2

2

n+1

2

+

2

n+1

2


Consider the smallest n (if it exists) su h that n is nonnegative. Then (x

x + )(x + x + ) = P (x)

2

2

and

PQQ

1

Qn

= Pn ; where Q Q Qn 2 P n+1 sin e ea h k < 0 for k < n; and where Pn 2 P n+1 sin e n 0. Thus, we have the desired representation 1

1

2

1

2

1

+

1

2

4

+

2

Pn (x) ; (x + x + )(Q Q Qn )(x) where n is the smallest integer su h that n > 0: Now suppose ; : : : ; n , n are all nonpositive. Then x

x + =

2

1

1

2

1

1

p

k = and = 2

2

k

2

+1

;

k = 1; 2; : : : ; n 1

imply

1

> 2 + (2 + (2 + (2 + 2

(2:4:1)

= ) = ) =

1 2 1 2 1 2

) =

1 2

=: Æn ;

where the above formula ontains n iterations. Sin e, by assumption, < 4; and sin e Æn ! 4 as n ! 1; it is lear that (2:4:1) is not satis ed for some n; and eventually some n is greater than zero. The estimate on the degree requires analyzing the rate of onvergen e p . Sin e Æ = 2 + of (Æn )1 Æ ; we have n n n 1

=0

4

p

Æn = 2

Æn

1

=

4 Æn p 2 + Æn

1 1

4

Æn

2

1

:

By repeated appli ations of the above,

Æ 1 = n : 2n 2 Now we an improve the above estimate as follows: We have 4

4

Æn =

4 Æn p 2 + Æn

Æn

=

1

4

0

1

p

Æn

4

2

p

(2 + Æn )(2 + Æn ) 4 Æ p p = p (2 + Æn )(2 + Æn ) (2 + Æ ) 2 p p p (2 + 4 2 n )(2 + 4 2 n ) (2 + 4 2 n (2 + 2 2 n )(2 + 2 22 n) (2 + 2 2n n ) 1

1

2

0

1

2

0

2

1

2 4

n

nY +1 j =2

3

(

+1)

n)

2

1 1 2 j

24

n

nY +1 j =2

(1 + 2 2 j ) 2e4 n :

82


So if

m := 2n+1

then

Æn 2e4

4

that is, Æn :

p

p2 2e ; 4 n

4

;

ut

We note that in the above proof a little additional eort yields a slightly better onstant than 10. Let k 2 N and 2 (0; ). It follows easily from Lemma 2.3.4 that if p 2 Pk has no zeros in the one fz 2 C : j arg(z )j < g ; then there are s; t 2 Pm with m k so that p = s=t; see E.1 d℄. The essential sharpness of this upper bound is shown by E.1 e℄. An easier proof of Theorem 2.4.2 that gives a weaker bound for the degree of the numerator and denominator in the representation is given by E.1 f℄. A similar sort of representation theorem due to Bernstein [15℄ is the following: 5

+

1

4

Theorem 2.4.4. If p

representation

2 Pn

and p(x) > 0 for x

p(x) =

d X j =0

2(

x)j (1 + x)d

aj (1

1; 1); then there is a j

with ea h aj 0. (The smallest d := d(p) for whi h su h a representation exists is alled the Lorentz degree of p.) It suÆ es to prove this result for quadrati polynomials; this is left as an exer ise; see E.1 f℄. The proof of the following interesting result of Barnard et al. [91℄ is surprisingly ompli ated, and we do not reprodu e it here. Theorem 2.4.5. Suppose that p 2 Pn has all nonnegative oeÆ ients. Suppose that the zeros of p are z ; z ; : : : ; zn 2 C : For 0; let n Y z ; 1 p (z ) = zj j j zj j> where arg(z ) is de ned so that arg(z ) 2 [ ; ): Then p (z ) has all nonnegative oeÆ ients. 1

1

=1

arg(

)

It follows from this result that if p 2 Pn has all nonnegative oeÆ ients and if q (x) = x + x + is a quadrati polynomial with zeros forming a pair of onjugate zeros of p that have least angular distan e from the positive x-axis, then p=q also has all nonnegative oeÆ ients. 2


Comments, Exer ises, and Examples. Polynomials with all nonnegative oeÆ ients have a number of distinguishing properties that are explored in the exer ises. For example, only analyti fun tions with all nonnegative oeÆ ients an be approximated uniformly on [0; 1℄ by su h polynomials; see E.2. So a Weierstrass-type theorem does not hold for these polynomials. This is quite dierent from approximation by polynomials of the form X

(2:4:2)

ai;j (x + 1)i (1

x)j ;

ai;j

0:

Sin e every polynomial that is stri tly positive on ( 1; 1) has su h a representation (E.1 b℄), it follows from the Weierstrass approximation theorem that all nonnegative fun tions from C [ 1; 1℄ are in the uniform losure. It follows from Theorem 2.4.2 and the Weierstrass approximation theorem (see E.1 of Se tion 4.1) that fra tions of polynomials with all nonnegative oeÆ ients form a dense set in the uniform norm on [0; 1℄ in the set of nonnegative ontinuous fun tions on a nite losed interval [0; 1℄. Hen e they have a mu h larger uniform losure on [0; 1℄ than that of the polynomials with all nonnegative oeÆ ients. Various inequalities for polynomials of the form (2:4:2) are onsidered in Appendix 5. E.1 Remarks on Theorem 2.4.2. a℄ Suppose ; 2 R; 2 (0; ); and suppose x + x + has no zeros in the one fz 2 C : j arg(z )j < g : 2

Show that there are p; q 2 Pm with m 5

+

1

2

x + x + = 2

su h that

p(x) : q(x)

Hint: This is a reformulation of Theorem 2.4.3 by introdu ing the angle between the positive x-axis and the zero of the quadrati polynomial. u t b℄ Let n 2 N : Show that if p 2 Pn ; then p has no zeros in the one +

fz 2 C : jarg(z )j < =(2n)g : This is sharp, as the example p(x) := xn + 1 shows.

℄ Show that the result of part a℄ is sharp up to the onstant . Hint: Let n 2 N . Consider 5

2

x + x + = x 2

x exp i n 2

2

exp

i : n

2 2

84


Show that if there are p; q 2 Pm so that x + x + = p(x)=q (x); then m n 1. ut d℄ Let 2 (0; ) and k 2 N : Suppose p 2 Pk has no zeros in the one fz 2 C : j arg(z )j < g : Show that there are s; t 2 Pm with m k so that p = s=t : e℄ Let 2 (0; ) and k 2 N . Let fk (x) := ((x z )(x z ))k ; where arg(z ) = : Assume that fk = s=t; where s; t 2 Pm : Show that m (log 2)k : +

2

5

+

1

4

0

0

+

0

1

Hint: First observe that s(y) s(y + yÆm ) eÆ s(y) for every s 2 Pm ; y 2 (0; 1); and Æ 2 (0; 1): Therefore fk (y + yÆm ) eÆ fk (y) for every y 2 (0; 1) and Æ 2 (0; 1): Now let y jz j be hosen so that jy z j = : Applying the above inequality with this y and Æ := m; we 1

+

1

0

0

obtain

fk (y + y) emfk (y) ; hen e 2k em ; that is, k log 2 m. ut f℄ Prove that if r 2 Pn and r(x) > 0 for all x > 0; then there is an integer d n su h that q(x) ; r(x) = (1 + x)d n where q 2 Pd : Hint: Let ; 2 R and < 4 : Consider +

2

(x

2

x + )(1 + x)d =

d+2 X j =0

j xj

and ompute j expli itly. g℄ If p 2 Pn and p(x) > 0 for all x 2 ( 1; 1); then it is of the form

p(x) = for some d n: Hint: Apply a℄ to

d X j =0

aj (1

r(u) := (1 + u)n p

x)j (1 + x)d j ;

1 u 1+u

;

aj

u :=

ut

0

1 x : 1+x

ut


E.2 Polynomials with Nonnegative CoeÆ ients. a℄ If p 2 Pn ; then for x > 0 +

0 p0 (x)
0 for every > 0: So the above limit relations imply that D() > 0 for every large enough ; hen e for every > 0. In parti ular,

D(1) = D sinhx i0 t sinhx i1 t :: :: :: sinhx im t > 0 ; m 0

1

ut

whi h nishes the proof. e℄ Suppose 0 < < < . Show that 0

1

( osh t; osh t; : : : ) 0

1

is a Des artes system on (0; 1): Hint: Pro eed as in the outline for part d℄.

ut

E.3 Rational Systems. a℄ Cau hy Determinants. Show that 1 + 1 1 .. . 1 m + 1

::: ..

1 1 + m

.. .

.

:::

1

m + m

Q

(j i )( j i ) i<j m Q : = (i + j ) i;jm 1

1

Q

(i + j ) and observe that both i;jm sides are polynomials of the same degree, m 1; in ea h variable i ; i . Also both sides vanish exa tly when i = j or i = j : So up to a onstant

Hint: Multiply both sides above by

1

both sides are the same. Now show that the onstant is 1. b℄ Let > > > b. Show that 1

ut

2

1 1

;

x

1

2

x

;:::

is a Des artes system on [a; b℄: Let < < < a: Show that 1

2

x

1

1

;

x

1

2

; :::

is a Des artes system on [a; b℄ (see also E.6 e℄). E.4 Proof of Theorem 3.2.5. Assume the notation of this Theorem 3.2.5. a℄ Let 0 Æ < Æ < < Æ n and a < x < x < < x < b: Show P that there exists a unique p = fÆ + i ai fÆi su h that 0

1

1

1

=0

2

3.2 Des artes Systems 107

(1) p(xi ) = 0 , i = 1; 2; : : : ; : Show also that the above p has the following properties: (2) p(x) hanges sign at ea h xi ; (3) p(x) 6= 0 if x 2= fx ; x ; : : : ; x g ; (4) ai ai < 0 ; i = 0; 1; : : : ; 1 ; a := 1 ; (5) p(x) > 0 ; x 2 (x ; b℄ : Hint: Sin e (fÆ0 ; : : : ; fÆ ) and (fÆ0 ; : : : ; fÆ 1 ) are Chebyshev systems, E.11 of Se tion 3.1 shows that there exists a p of the desired form satisfying (1). Sin e (fÆ0 ; : : : ; fÆ 1 ) is a Chebyshev system, this p is unique. Now E.11 of Se tion 3.1 yields that p satis es (2) and (3). By Theorem 3.2.4, p satis es (4). The fa t that (5) holds for p follows from expanding the determinant 1

2

+1

D fxÆ0 fxÆ1 :: :: :: fÆx 1 fxÆ by Cramer's rule. This determinant is just p(x) with some > 0 sin e it vanishes at ea h xi ; and the oeÆ ient of fÆ is positive; see De nition 3.2.3. Also, the above determinant is positive for all x 2 (x ; b℄; see De nition 1

2

ut

3.2.3 again. b℄ Prove Theorem 3.2.5.

Outline. For notational simpli ity assume that = n (hen e m = k); the general ase is analogous. Further, we may assume that there is an index j su h that

j < j and i = i whenever i 6= j

sin e the result follows from this by a nite number of pairwise omparisons. So we assume

p = fn + aj fj + and

q = fn + bj f j +

k X i=1 i6=j

k X i=1 i6=j

ai fi

bi fi ;

where 0 < < < k < n and 0 j < j < j for some 1 j k (of ourse, the inequality j < j holds only if j is de ned, that is, only if j 2). Then 1

2

1

1

bj f j +

p q = aj fj

1

k X i=1 i6=j

(ai

bi )fi

has exa tly k zeros on [a; b℄ at x ; x ; : : : ; xk be ause p dimensional Chebyshev subspa e. 1

2

q is in a (k + 1)-

108 3. Chebyshev and Des artes Systems

By property (5) in part a℄ applied to p and q; respe tively, p(x) > 0 and q(x) > 0 ; x 2 (xk ; b℄ : Now by property (4) in part a℄ applied to (p q ), where is hosen so that the lead oeÆ ient of (p q ) is 1, and by the fa t that p and p q have the same oeÆ ient for fj , the lead oeÆ ient of p q (ak bk provided k > j ) is negative. So property (5) in part a℄ implies that p(x) q(x) < 0; x 2 (xk ; b℄ : Hen e 0 < p(x) < q (x); x 2 (xk ; b℄: Now use property (3) in part a℄ and the fa t that all of p; q; and p q

hange sign only at xi to nish the proof. ut The following extension of part a℄ will be used later:

℄ Suppose 0 Æ < Æ < < Æ n ; a x x x b ; a < x ; x < b, and xiP< xi , i = 1; 2; : : : ; 2. Show that there exists a unique p = fÆ + i ai fÆi (with ai 2 R) su h that (1) p(xi ) = 0, i = 1; 2; : : : ; ; (2) p(x) hanges sign at xi if and only if xi 62 fa; b; xi ; xi g : Show also that (3) p(x) 6= 0 if x 2= fx ; x ; : : : ; x g ; (4) ai ai 0 ; i = 0; 1; : : : ; 1 ; a := 1 ; (5) p(x) > 0 ; x 2 (x ; b) ; (6) ( 1) p(x) > 0 ; x 2 (a; x ) ; (7) ( 1) i p(x) > 0 ; x 2 (xi ; xi ) ; i = 1; 2; : : : ; 1 : Hint: Use part a℄ and a limiting argument. The uniqueness follows from E.10 of Se tion 3.1. ut The next exer ise provides a solution to a problem of Lorentz, whi h is settled in Borosh, Chui, and Smith [77℄. 0

2

1

1

1

2

+2

1

=0

1

1

+1

2

+1

1

+1

E.5 A Problem of Lorentz on Best Approximation to x . Suppose that [a; b℄ [0; 1); n 2 N ; and p 2 (0; 1℄ are xed. Let be a nite Borel measure on [a; b℄. a℄ Suppose ; ; : : : ; n are arbitrary xed real numbers if a > 0, or xed real numbers greater than 1=p if a = 0. Let f 2 Lp () be xed. Show that 1

2

Ep; ( ; ; : : : ; n ; f ) := min

f (x) ai 2R 1

exists and is nite.

2

n X i=1

ai xi

Lp ()


Outline. Use a standard ompa tness argument. ut b℄ Suppose 1 < p < 1, the support of ontains at least n + 1 points, ; ; : : : ; n ; are arbitrary xed distin t real numbers if a > 0; or xed real numbers greater than 1=p if a = 0: Show that if (eai )ni R satis es 1

2

=1

n X

Ep; ( ; ; : : : ; n ; x ) = x 1

2

then

f (x) := x

n X i=1

i=1

e ai xi Lp () ;

e ai xi

has exa tly n sign hanges on (a; b): Hint: Sin e (x1 ; : : : ; xn ; x ) is a Chebyshev system, it is suÆ ient to prove that f has at least n sign hanges on (a; b). Suppose f has at most n 1 sign hanges on [a; b℄: Then, sin e (x1 ; : : : ; xn ) is a Chebyshev system, by E.11 of Se tion 3.1 there exists an element

h 2 spanfx1 ; : : : ; xn g su h that

jf (x)jp

1

sign(f (x))h(x) 0

on [a; b℄ with stri t inequality at all but n points (at every point where f does not vanish). Using that the support of ontains at least n +1 distin t points, this implies Z b

a

jf jp

1

sign(f )h d > 0 ;

whi h ontradi ts E.7 h℄ of Se tion 2.2. ut

℄ Suppose p = 1, supp() = [a; b℄, ; ; : : : ; n ; are arbitrary xed distin t real numbers if a > 0; or xed distin t nonnegative real numbers if a = 0: Show that if (eai )ni R satis es 1

2

=1

E1; ( ; ; : : : ; n ; x ) =

x 1

2

then

f (x) := x has exa tly n sign hanges on (a; b): Hint: Use Theorem 3.1.6.

n X i=1

n X i=1

e ai xi

L1 ()

;

e ai xi

ut


d℄ Best Approximation to x from Certain Classes of Muntz Polynomials. Let 1 < p 1: Suppose the support of ontains at least n + 1 distin t points if 1 < p < 1 or supp() = [a; b℄ if p = 1: Let > > > be arbitrary xed real numbers if a > 0; or xed nonnegative real numbers if a = 0: Suppose we wish to minimize 1

2

Ep; ( ; ; : : : ; n ; x ) 1

2

for all sets of n distin t real numbers > > > n satisfying 1

2

f ; ; : : : ; n g f ; ; : : : g : 1

2

1

2

Show that the minimum o

urs if and only if

f ; ; : : : ; n g = f ; ; : : : ; n g : 1

2

1

2

Hint: Let n X

Ep; ( ; ; : : : ; n ; x ) =

x 1

2

i=1

e ai xi

Lp ()

;

where f ; ; : : : ; n g is a set of n distin t real numbers for whi h the minimum is taken; see part a℄. By parts b℄ and ℄ 1

2

f (x) := x

n X i=1

e ai xi

has exa tly n sign hanges x ; x ; : : : ; xn on (a; b): Let 1

2

g 2 spanfx 1 ; x 2 ; : : : ; x n g interpolate x at the points x ; x ; : : : ; xn : Now use Theorem 3.2.5 to nish the proof. ut 1

2

E.6 Stri tly Totally Positive Kernels (Karlin [68℄). A ( ontinuous) fun tion K (s; t) is an STP kernel on [a; b℄ [ ; d℄ if K (s0 ; t0 ) .. . K (sn ; t0 )

: : : K (s ; tn ) ..

0

.

:::

.. .

K (sn ; tn )

>0

for all a s < < sn b, t < < tn d; and for all n > 0: 0

0


a℄ Observe that E.3 b℄ implies that 1 K (s; t) = is STP on [a; b℄ [a; b℄ ; s+t Observe also that E.2 b℄ implies that

a > 0:

1 ; 1) ( 1 ; 1) :

K (s; t) = est is STP on (

b℄ Suppose K is STP on [a; b℄ [ ; d℄; and (f ; : : : ; fn ) is a Chebyshev system on [a; b℄: Show that if 0

vi (x) =

Z b

a

K (t; x)fi (t) dt ;

i = 0; 1; : : : ; n ;

then (v ; : : : ; vn ) is a Chebyshev system on [ ; d℄:

℄ Variation Diminishing Property. Suppose K is STP on [a; b℄ [ ; d℄ and suppose f 2 C [a; b℄. Let 0

g(x) :=

Z b

a

K (t; x)f (t) dt :

Then g has no more sign hanges on [ ; d℄ than f has on [a; b℄: d℄ The Lapla e transform of a fun tion f 2 C [0; 1) \ L [0; 1)

L(f )(x) :=

1

Z

1

f (t)e

tx dt

0

has no more sign hanges on [0; 1) than f does.

Proof. This follows from parts a℄ and ℄. It may also be proved dire tly by indu tion as follows. Suppose f has exa tly n sign hanges on [0; 1); one at x . Then g (x) := (x x)f (x) has exa tly n 1 sign hanges on [0; 1): Now observe that 0

0

d x0 x (e L(f )(x)) = L(g )(x) ; dx so L(f ) has at most one more sign hange on [0; 1) than L(g ) does. e

x0 x

ut

e℄ Use part d℄ and E.2 a℄ to reprove that

1 1 ; ; ::: x+ x+ 1

2

;

a < < < 1

ut

satis es Des artes' rule of signs on [a; b℄: Proof. Observe that Z1

for every x 2 ( i ; 1):

0

e

i t e tx dt =

2

1

x + i

ut


E.7 Des artes' Rule of Signs for Polynomials. P a℄ Prove by indu tion that nk ak xk 2 Pn has no more zeros in (0; 1) (repeated zeros are ounted a

ording to their multipli ities) than the number of sign hanges in (a ; a ; : : : ; an ): =0

0

1

Pn

P

b℄ Let > 0. Let p(x) = k ak xk and q (x) := (x )p(x) = nk bk xk : Show that if the number of sign hanges in fa ; a ; : : : ; an g is m; then the number of sign hanges in fb ; b ; : : : ; bn g is at least m + 1:

℄ Give another proof of a℄ based on b℄. d℄ In part a℄ the number of sign hanges in (a ; a ; : : : ; an ) ex eeds the number of positive zeros by an even integer. Hint: See Polya and Szeg}o [76℄. ut Re nements of the above exer ise are presented in E.6 of Appendix 1, where Cau hy indi es are dis ussed. The rst part of the following exer ise is a version of a result from Zielke [79℄: =0

0

0

+1

=0

1

1

+1

0

1

E.8 The Ee t of Dierentiation on Weak Markov Systems. The system (f ; : : : ; fn ) is alled a weak Chebyshev system on [a; b℄ if fi 2 C [a; b℄ for ea h i and every f 2 spanff ; : : : ; fn g has at most n sign hanges on [a; b℄ (so the only dieren e between a Chebyshev system and a weak Chebyshev system is that in the de nition of the latter, zeros without sign hange are not ounted). Analogously, the system (g ; : : : ; gn ) is alled a weak Markov system on [a; b℄ if gi 2 C [a; b℄ for ea h i and (g ; : : : ; gm ) is a weak Chebyshev system on [a; b℄ for every m = 0; 1; : : : ; n (so the only dieren e between a Markov system and a weak Markov system is that in the de nition of the latter, zeros without sign hange are not ounted). a℄ Suppose (1; f ; : : : ; fn ) is a weak Markov system of C fun tions on [a; b℄. Show that (f 0 ; : : : ; fn0 ) is a weak Markov system on [a; b℄: Outline. Pro eed by indu tion on n. If n = 0; then the statement is obvious. Suppose that the statement is true for n 1. By the indu tive hypothesis (f 0 ; : : : ; fn0 ) is a weak Markov system on [a; b℄, hen e Rolle's theorem implies that (1; f ; : : : ; fn ) is a Markov system on [a; b℄: Suppose that g 2 spanff 0 ; : : : ; fn0 g is of the form 0

0

0

0

1

1

1

1

1

1

1

1

g :=

n X i=1

1

ai fi0 ;

ai 2 R

and g has at least n sign hanges on [a; b℄: Then there exist n + 2 distin t


points

a < x < x < < xn 1

2

and = 1 su h that

F :=

n X i=1

0 ;

i = 1; 2; : : : ; n + 1 :

Sin e (1; f ; : : : ; fn ) is a Markov system on [a; b℄; by Proposition 3.1.2, there exist fun tions 1

1

GÆ 2 spanf1; f ; : : : ; fn 1

1

g

su h that

GÆ (xi ) = F (xi ) + Æ( 1)i ;

i = 2; 3; : : : ; n + 1

for every Æ > 0: Then by the indu tive hypothesis, G0Æ has at most n sign hanges on [a; b℄: It follows that if Æ > 0 is suÆ iently small, then ( 1)(GÆ (x ) 2

and

( 1)n (GÆ (xn ) +1

+2

2

GÆ (x )) < 0 1

GÆ (xn )) < 0 ; +1

otherwise G0Æ would have at least n sign hanges on [a; b℄: Now show that for suÆ iently small Æ > 0

F

GÆ 2 spanf1; f ; : : : ; fng 1

has at least n + 1 sign hanges, whi h is a ontradi tion. ut b℄ Suppose that (1; f ; : : : ; fn ) is an ECT system on [a; b℄ and suppose that ea h fi 2 C n [a; b℄ (see E.3 of Se tion 3.1). Show that (f 0 ; : : : ; fn0 ) is also an ECT system on [a; b℄ with ea h fi0 2 C n [a; b℄: Hint: Use the de nition given in E.3 of Se tion 3.1. ut

℄ Suppose (1; f ; : : : ; fn ) is a weak Markov system on [a; b℄ with ea h fi 2 C [a; b℄: Show that (1; f ; : : : ; fn) is a Markov system on [a; b℄: Hint: Use Rolle's theorem and part a℄. ut 1

1

1

1

1

1


3.3 Chebyshev Polynomials in Chebyshev Spa es Suppose

Hn := spanff ; f ; : : : ; fn g 0

1

is a Chebyshev spa e on [a; b℄; and A is a ompa t subset of [a; b℄ with at least n + 1 points. We an de ne the generalized Chebyshev polynomial

Tn := Tn ff ; f ; : : : ; fn; Ag 0

1

for Hn on A by the following three properties:

Tn 2 spanff ; f ; : : : ; fng

(3:3:1)

0

1

there exists an alternation sequen e (x < x < < xn ) for Tn on A; that is, 0

(3:3:2)

sign(Tn (xi )) = sign(Tn (xi )) = kTnkA +1

for i = 0; 1; : : : ; n (3:3:3)

1

1; and

kTnkA = 1

with Tn (max A) > 0 :

Of ourse the existen e and uniqueness of su h a Tn has to be proved. Note that if together with spanff ; : : : ; fn g, spanff ; : : : ; fn g is also a Chebyshev spa e, then Theorem 3.1.6 implies that 0

0

nX1

Tn = fn where the numbers a ; a ; : : : ; an 0

1

fn

(3:3:4)

1

k=0

1

!

ak f k ;

2 R are hosen to minimize

nX1 k=0

ak fk

; A

satis es properties (3.3.1) and (3.3.2), and the normalization onstant 2 R

an be hosen so that Tn satis es property (3.3.3) as well. In E.1 we outline the proof of the existen e and uniqueness of a Tn satisfying properties (3.3.1) to (3.3.3) without assuming that spanff ; : : : ; fn g is a Chebyshev spa e. Note that if (f ; : : : ; fn ) is a Des artes system on [a; b℄; then the normalization onstant (that is, the lead oeÆ ient) in Tn is positive. This follows from E.4 of Se tion 3.2. 0

0

1

3.3 Chebyshev Polynomials in Chebyshev Spa es 115

On intervals, with fi (x) := xi ; the de nition (3.3.1) to (3.3.3) gives the usual Chebyshev polynomials; see E.7 of Se tion 2.1. The Chebyshev polynomials Tn for Hn on A en ode mu h of the information of how the spa e Hn behaves with respe t to the uniform norm on A. Many extremal problems are solved by the Chebyshev polynomials. When (f ; f ; : : : ) is a Markov system on [a; b℄ we an introdu e the sequen e (Tn )1 n of asso iated Chebyshev polynomials 0

1

=0

Tn := Tn ff ; f ; : : : ; fn; [a; b℄g for Hn on [a; b℄: Then (T ; T ; : : : ) is a Markov system on [a; b℄ again with 0

0

1

1

the same span. (One reason for not always hoosing this as a anoni al basis is that it is never a Des artes system.) The denseness of Markov spa es in C [a; b℄ is intimately tied to the lo ation of the zeros of the asso iated Chebyshev polynomials; see Se tion 4.1. An example of an extremal problem solved by the Chebyshev polynomials is the following: Theorem 3.3.1. Suppose Hn := spanff ; : : : ; fn g is a Chebyshev spa e on [a; b℄ with asso iated Chebyshev polynomial 0

Tn := Tn ff ; f ; : : : ; fn; [a; b℄g and ea h fi is dierentiable at b: Then maxfjp0 (b)j : p 2 Hn ; kpk a;b 1 ; p(b) = Tn (b)g is attained by Tn: 0

1

[

℄

Proof. Suppose p 2 Hn , kpk a;b 1; and p(b) = Tn (b): We need to show that jp0 (b)j jTn0 (b)j: Let a < < < n b be the points of alternation for Tn ; that is, [

℄

0

1

Tn(i ) = ( 1)i ; i = 0; 1; : : : ; n : p has at least n zeros in [ ; n ℄; one in ea h [i ; i ℄;

Note that Tn i = 1; 2; : : : ; n (we ount ea h internal zero without sign hange twi e, as in E.10 of Se tion 3.1). So if b 6= n ; then Tn p has n + 1 zeros on [a; b℄ in luding the zero at b, hen e p = Tn , and the proof is nished. We may thus assume that Tn (b) = 1. Assume that jp0 (b)j > jTn0 (b)j: Sin e Tn0 (b) 0, without loss of generality we may assume that p0 (b) > Tn0 (b); otherwise we study p: Then Tn p has two zeros on [n ; b℄, and hen e has n + 1 zeros in [a; b℄ (again, we ount ea h internal zero without sign hange twi e). Thus by E.10 of Se tion 3.1 we have p = Tn , whi h ontradi ts the assumption ut p0 (b) > Tn0 (b): 0

E.3.

1

An extension of the above theorem to interior points is onsidered in


Theorem 3.3.2. Suppose (f ; : : : ; fn ; g ) and (f ; : : : ; fn ; h) are both Chebyshev systems on [a; b℄ with asso iated Chebyshev polynomials 0

1

0

1

Tn := Tn ff ; f ; : : : ; fn ; g; [a; b℄g 0

and

1

1

Sn := Tn ff ; f ; : : : ; fn ; h; [a; b℄g ; 0

1

1

respe tively. Suppose (f ; f ; : : : ; fn ; g; h) is also a Chebyshev system. Then the zeros of Tn and Sn interla e (there is exa tly one zero of Sn between any two onse utive zeros of Tn ). 0

1

1

Proof. Sin e (f ; : : : ; fn ; g; h) is a Chebyshev system on [a; b℄; Tn Sn has at most n + 1 zeros. However, between any two onse utive alternation points of Tn ; of whi h there are n + 1; there is a zero of Tn Sn (whi h may be at an internal alternation point of Tn only if it is a zero without sign

hange, whi h is then ounted twi e). Likewise, there is a zero of Tn Sn between any two onse utive alternation points of Sn : Thus between any three su

essive alternation points of say Tn there an be at most three zeros of Tn Sn : However, if Sn had two zeros between two onse utive zeros of Tn ; then there would be three onse utive alternation points of either Tn or Sn with at least four zeros of either Tn + Sn or Tn Sn between them, whi h is impossible. ut 0

1

Theorem 3.3.3. Suppose (f ; f ; : : : ) is a Markov system on [a; b℄ with asso iated Chebyshev polynomials 0

1

Tn := Tnff ; f ; : : : ; fn ; [a; b℄g : 0

1

Then the zeros of Tn and Tn stri tly interla e (there is exa tly one zero of Tn stri tly between any two onse utive zeros of Tn). 1

1

ut

Proof. The proof is analogous to that of Theorem 3.3.2.

Theorem 3.3.4 (Lexi ographi Property). Let (f ; f ; : : : ) be a Des artes system on [a; b℄. Suppose < < < n and < < < n are nonnegative integers satisfying 0

0

i i ; Let and

1

1

0

i = 0; 1; : : : ; n :

Tn := Tn ff0 ; f1 ; : : : ; fn ; [a; b℄g Sn := Tn ff 0 ; f 0 ; : : : ; f n ; [a; b℄g

denote the asso iated Chebyshev polynomials.

1


Let

< < < n and < < < n 1

2

1

2

denote the zeros of Tn and Sn ; respe tively. Then i i ;

i = 1; 2; : : : ; n

with stri t inequality if i 6= i for at least one index i: (In other words, the zeros of Tn lie to the left of the zeros of Sn :) Proof. It is learly suÆ ient to prove the theorem in the ase that i = i for ea h i 6= m, and m < m < m for a xed index m; and then to pro eed by a sequen e of pairwise omparisons (if m = n; then m is meant to be repla ed by 1). So suppose +1

+1

Tn := Tnff0 ; : : : ; fm ; : : : ; fn ; [a; b℄g =

n X i=0

i fi

and

Sn := Tnff0 ; : : : ; f m ; : : : ; fn ; [a; b℄g = dm f m +

n X i=0 i6=m

di fi

with m < m < m . Then by Theorem 3.3.2 the zeros of Sn and Tn interla e and all that remains to prove is that the largest zero of Sn is larger than the largest zero of Tn : That is, we must show that n < n : For this we argue as follows. It follows from Theorem 3.2.5 that the lead oeÆ ient of Tn is less than the lead oeÆ ient of Sn ( n < dn provided m < n). Sin e both Tn and Sn have an alternation sequen e of length n + 1 on [a; b℄; and sin e kTnk a;b = kSn k a;b = 1 ; Tn (b) > 0 ; Sn (b) > 0 ; +1

[

℄

[

℄

it follows from E.1 b℄ that

Tn 2 spanff0 ; f1 : : : ; fn ; f m g

Sn

has n +1 zeros x x xn on [a; b℄: Therefore, it follows from E.4

℄ of Se tion 3.2 that (Sn Tn )(x) > 0 on (xn ; b) and (Sn Tn )(x) < 0 on (xn ; xn ): Hen e the assumption n n would imply that Sn Tn has at least n + 2 zeros on [a; b℄ ( ounting ea h internal zero without sign hange twi e), whi h is a ontradi tion. (Draw a pi ture and use the alternation

hara terization of the Chebyshev polynomials Tn and Sn to make the proof of the above statement transparent.) So n > n ; indeed, and the proof is nished. ut 1

2

+1

+1

+1


Comments, Exer ises, and Examples. If Hn := span(f ; : : : ; fn ) is a Chebyshev spa e on [a; b℄, A is a ompa t subset of [a; b℄, and p 2 (0; 1); then 0

nX1

Tn = fn with a ; a ; : : : ; an 0

1

1

k=0

!

ak fk

2 R minimizing Z fn

nX1

A

k=0

p

ak fk

is alled an Lp Chebyshev polynomial for Hn on A: When A = [a; b℄ and p 2 (1; 1℄; the properties of the zeros of these Lp Chebyshev polynomials are explored in Pinkus and Ziegler [79℄, where mu h of the material of this se tion may be found. For example, an Lp analog of Theorem 3.3.2 still holds. E.1 Existen e and Uniqueness of Chebyshev Polynomials. Let A [a; b℄ be a ompa t set ontaining at least n + 1 points. Let (f ; : : : ; fn ) be a Chebyshev system on [a; b℄: a℄ Existen e of Chebyshev Polynomials. Show that there exists a Tn satisfying properties (3.3.1) to (3.3.3). Hint: If A ontains exa tly n + 1 points, then the existen e of Tn is just the interpolation property of a Chebyshev spa e formulated in Proposition 3.1.2 b℄. So assume that A ontains at least n + 2 points. Then there is a Æ > 0 so that A \ [a; ℄ ontains at least n + 1 points for every 2 (b Æ; b): Show that for every 2 (b Æ; b); there is a g 2 spanff ; : : : ; fn g for whi h 0

0

supff (b) : f

2 spanff ; f ; : : : ; fn g ; kf k a; 0

1

[

℄

= 1g

is attained. Use a variational method to show that g satis es properties (3.3.1) to (3.3.3) with A repla ed by A \ [a; ℄. Now let ( k )1 k be a sequen e of numbers from (b Æ; b) that onverges to b. Let g k 2 spanff ; : : : ; fn g satisfy properties (3.3.1) to (3.3.3) with A repla ed by A \ [a; k ℄: Show that there is a subsequen e of (g k )1 that k

onverges to a g 2 spanff ; : : : ; fn g uniformly on [a; b℄: Show that Tn := g satis es properties (3.3.1) to (3.3.3). ut b℄ A Lemma for Part ℄. Suppose f; g 2 C [a; b℄ with kf kA = kg kA > 0 and there are alternation sequen es =1

0

=1

0

(x < x < < xn ) and (y < y < < yn ) 1

2

+1

for f and g , respe tively, on A:

1

2

+1


Suppose also that sign(f (x )) = sign(g (y )): 1

1

Show that f g has at least n + 1 zeros on [a; b℄:

℄ Uniqueness of Chebyshev Polynomials. Show that the Chebyshev polynomials Tnff ; f ; : : : ; fn ; Ag 0

1

satisfying properties (3.3.1) to (3.3.3) are unique. Hint: Use part a℄ and E.10 of Se tion 3.1.

ut

E.2 More on Chebyshev Polynomials. Let Hn := spanff ; : : : ; fn g be a Chebyshev spa e on [a; b℄ with asso iated Chebyshev polynomial denoted by Tn := Tn ff ; : : : ; fn ; [a; b℄g: Show the following statements. a℄ If 1 2 Hn ; then jTn (a)j = jTn (b)j = 1: b℄ If 1 2 Hn ; then Tn is monotone between two su

essive points of its alternation sequen e. Note that the on lusions of parts a℄ and b℄ do not ne essarily hold in general. P

℄ If Tn =: ni ai fi ; ai 2 R; then the oeÆ ient sequen e of Tn =am solves n 0

0

=0

min

fm +

bi 2R

X

i=0 i6=m

bi fi

[

a;b℄

uniquely, provided that ff ; : : : fm ; fm ; : : : fn g is also a Chebyshev system on [a; b℄: (So this applies to ordinary polynomials on [0; 1℄ but not on [ 1; 1℄:) d℄ Suppose (f ; : : : ; fn ) is a Des artes system on [a;P b℄ with asso iated Chebyshev polynomial Tn := Tn ff ; : : : ; fn ; [a; b℄g =: ni ai fi ; ai 2 R: Show that an > 0 and ai ai < 0 for ea h i = 0; 1; : : : ; n 1: Hint: Use E.4 a℄ of Se tion 3.2. ut 0

1

+1

0

0

=0

+1

E.3 Extension of Theorem 3.3.1. Let Hn := spanff ; : : : ; fn g be a Chebyshev spa e on [a; b℄ with asso iated Chebyshev polynomial 0

Tn := Tn ff ; f ; : : : ; fn; [a; b℄g 0

1

and suppose ea h fi is dierentiable at x 2 [a; b℄. a℄ If Tn0 (x ) > 0; then maxfp0 (x ) : p 2 Hn ; kpk a;b 1 ; p(x ) = Tn (x )g 0

0

0

is attained only by Tn :

[

℄

0

0


b℄ If Tn0 (x ) < 0; then 0

minfp0 (x ) : p 2 Hn ; 0

kpk a;b 1 ; [

p(x ) = Tn (x )g 0

℄

is attained and only by Tn : Hint: Consider the number of zeros of Tn

0

tu Let [a; b℄ [0; 1):

p:

E.4 More Lexi ographi Properties of Muntz Spa es. Suppose < < < n and < < < n 0

1

0

1

are arbitrary real numbers if a > 0; or arbitrary nonnegative numbers if a = 0. Suppose i i for ea h i with stri t inequality for at least one index i: Let

Hn := spanfx0 ; x1 ; : : : ; xn g and Gn := spanfx 0 ; x 1 ; : : : ; x n g : Denote the asso iated Chebyshev polynomials for Hn and Gn on [a; b℄ by

Tn; := Tn fx0 ; x1 ; : : : ; xn ; [a; b℄g and

Tn; := Tn fx 0 ; x 1 ; : : : ; x n ; [a; b℄g ;

respe tively. a℄ Show that n 0 implies Tn; (b) = 1: 0 has at least n + 1 distin t zeros in Hint: Tn; (b) 6= 1 would imply that Tn; [a; 1) if > 0 and at least n distin t zeros if = 0: ut b℄ Let x = a or x = b. If x = a = 0; then assume that = 0 and = 1: Show that 0

0

0

0

0

0

1

maxfjp0 (x )j : p 2 Hn ; 0

kpk a;b 1g [

℄

is attained uniquely by Tn; :

℄ Let x 2 [0; 1) n [a; b℄. If x = 0; then assume that = 0: Show that 0

0

maxfjp(x )j : p 2 Hn ; 0

0

kpk a;b 1g [

℄

is attained uniquely by Tn; : Hint for b℄ and ℄: First prove that an extremal p 2 Hn exists. Then show, by a variational method, that p equios illates n + 1 times between 1 on [a; b℄: ut d℄ Let n 0, n 0; and a > 0: Show that

0 (b)j < jT 0 (b)j : jTn; n;


Show also that if a > 0 and there exists an index k , 0 k

k = k = 0; then

n; su h that

0 (a)j > jT 0 (a)j : jTn; n;

e℄ Let n 0, n 0; and a > 0: Show that

jTn; (x )j < jTn; (x )j ; x 2 (b; 1) : Show also that if 0; 0; and a > 0; then jTn;(x )j > jTn; (x )j ; x 2 [0; a) 0

0

0

0

0

0

0

0

(when x = 0; we need the assumption = = 0). Hint for d℄ and e℄: Suppose to the ontrary that one of the inequalities of parts d℄ and e℄ fails. Assume, without loss of generality, that there is an index m su h that i = i whenever i 6= m; and m < m : Note that by a℄, n 0 and n 0 imply Tn; (b) = Tn; (b) = 1: Also, by E.1 a℄, k = k = 0 implies Tn; (a) = Tn; (a) = ( 1)n : Now use Theorem 3.3.4 to show that Tn; Tn; 2 spanfx0 ; x1 ; : : : ; xn ; x m g 0

0

0

has at least n + 2 zeros in (0; 1) (in the ases when n 0 and n 0 are assumed) or in [0; 1) (in the ases when k = k = 0 is assumed). This

ontradi tion nishes the proof. ut f℄ Let 0 < a < b: Show that if n 0 and n 0; then

jp0 (b)j p2Hn kpk a;b max

[

℄

jq0 (b)j : < max q2Gn kq k a;b [

℄

Show also that if there exists an index k , 0 k n, su h that k = k = 0; then jp0 (a)j > max jq0 (a)j : max

kpk a;b

p2Hn

[

℄

q2Gn

kqk a;b [

℄

Hint: Combine parts b℄ and d℄. g℄ Let 0 < a < b. Let n 0 and n 0: Show that max

p2Hn

jp(x )j kpk a;b

jq(x )j ; < max q2Gn kq k a;b

0

[

x

0

℄

[

0

℄

2 (b; 1) :

Show also that

jp(x )j p2Hn kpk a;b max

0

[

℄

jq(x )j ; > max q2Gn kq k a;b

x

0

[

℄

0

2 [0; a)

(when x = 0, we need the assumption = = 0). 0

0

0

ut


Hint: Combine parts ℄ and e℄. ut h℄ Extend the validity, with < and > repla ed by and , respe tively, of the inequalities of parts f℄ and g℄ to the ase when the interval [a; b℄ (0; 1) is repla ed by [0; b℄: i℄ Suppose there exists an index k , 0 k n, su h that k = k = 0: Let x 2 (0; a): Show that the se ond inequalities of parts e℄ and g℄ hold true. Hint: Modify the arguments given in the hints to parts d℄, e℄, f℄, and g℄. Note that 1 2 Hn \ Gn ensures, as in E.1 a℄, that 0

Tn; (b) = Tn; (b) = 1

Tn; (a) = Tn; (a) = ( 1)n :

and

ut E.5 Lexi ographi Properties of (sinh t; : : : ; sinh n t). Let 0

0 < < < < n and 0 < < < < n : 0

1

0

1

Suppose i i for ea h i: Let

Hn := spanfsinh t; sinh t; : : : ; sinh n tg 0

and

1

Gn := spanfsinh t; sinh t; : : : ; sinh n tg : 0

1

Denote the asso iated Chebyshev polynomials for Hn and Gn on [0; 1℄ by

Tn; := Tnfsinh t; sinh t; : : : ; sinh n t; [0; 1℄g 0

and

1

Tn; := Tn fsinh t; sinh t; : : : ; sinh n t; [0; 1℄g ; 0

respe tively. a℄ Let

1

< < : : : < n and < < < n 1

2

1

2

denote the zeros of Tn; and Tn; , respe tively. Show that

i i ;

i = 1; 2; : : : ; n

(in other words, the zeros of Tn; lie to the left of the zeros of Tn; ). Outline. By E.2 d℄ of Se tion 3.2, (sinh t; : : : ; sinh n t) is a Des artes system on (0; 1): Hen e, by Theorem 3.3.3, the zeros of 0

Tn;;Æ := Tn fsinh t; sinh t; : : : ; sinh n t; [Æ; 1℄g 0

on [Æ; 1℄ lie to the left of the zeros of

1


Tn; ;Æ := Tnfsinh t; sinh t; : : : ; sinh n t; [Æ; 1℄g 0

1

on [Æ; 1℄ for every Æ 2 (0; 1): Show that lim kTn;

Tn;;Æ k = lim kTn; Æ!

Æ !0

0

Tn; ;Æ k = 0 ;

hen e the desired result follows by a ontinuity argument. b℄ Show that maxfjp0 (0)j : p 2 Hn ; kpk ; 1g is attained uniquely by Tn; :

℄ Show that Tn;(1) = Tn; (1) = 1 :

ut

[0 1℄

Hint: Tn;(1) 6= 1 would imply that 0 2 spanf osh t; osh t; : : : ; osh n tg Tn; 0

1

has at least n + 1 distin t zeros in (0; 1): This is impossible, sin e by E.2 e℄ of Se tion 3.2, ( osh t; : : : ; osh n t) is a Des artes (hen e Chebyshev) system on (0; 1): ut d℄ Show that 0 (0)j jT 0 (0)j : jTn; n; 0

Hint: Suppose to the ontrary that the above inequality fails to hold. Assume, without loss of generality, that there is an index m su h that i = i whenever i 6= m; and m < m : Obviously Tn;(0) = Tn; (0) = 0 : Part ℄ implies that Tn; (1) = Tn; (1) = 1 : Now use part a℄ and the above observation to show that

Tn;

Tn; 2 spanfsinh t; sinh t; : : : ; sinh n t; sinh m tg 0

1

has at least n + 2 zeros in (0; 1): This ontradi ts E.2 d℄ of Se tion 3.2. e℄ Show that jp0 (0)j max jq0 (0)j : max

ut

Hint: Combine parts b℄ and d℄.

ut

6 p2Hn kpk

0=

;

[0 1℄

6 q2Gn kqk

0=

;

[0 1℄

The result of the following exer ise has been observed independently by Lubinsky and Ziegler [90℄ and Kroo and Szabados [94℄. Various oeÆ ient estimates for polynomials are dis ussed in Milovanovi , Mitrinovi , and Rassias [94℄. An estimate for the oeÆ ients of polynomials having a given number of terms is obtained in Baishanski and Bojani [80℄. Approximation by su h polynomials is studied in Baishanski [83℄.


E.6 CoeÆ ient Bounds for Polynomials in a Spe ial Basis. Show that

n 2n

j m j 2

2m

kpk

[

;

;

1 1℄

m = 0; 1; : : : ; n

for every polynomial p of the form

p(x) =

n X m=0

m (1 x)m (x + 1)n

m 2 R :

m;

Outline. By E.5 a℄ and b℄ of Se tion 2.3 Tn (x) =

n X m=0

dm;n (1 x)m (x + 1)n

where

dm;n = ( 1)m 2 n If j m j > dm;n kpk

[

;

1 1℄

n

m;

n 1=2 1=2 2n m n m n =2 : n 1=2 2m n

for some index m; then the polynomial

q(x) =

Tn (x) dm;n

p(x)

m

has at least n distin t zeros in ( 1; 1): However,

q(x) =

n X j =0 j 6=m

aj (1

an have at most n

x)j (x + 1)n

j = (1

x)n

n X j =0 j 6=m

aj

1+x n j 1 x

1 distin t zeros in ( 1; 1) sin e

(u ; u ; : : : un m ; un m ; un m ; : : : ; un ) 0

1

1

+1

+2

is a Chebyshev system on (0; 1) by E.1 a℄ of Se tion 3.1.

ut

E.7 On the Zeros of the Chebyshev Polynomials for Muntz Spa es. Let 0 =: < < < n , and let 0

1

Hn := spanfx0 ; x1 ; : : : ; xn g : Denote the asso iated Chebyshev polynomials for Hn on [0; 1℄ by

Tn := Tn x0 ; x1 ; : : : ; xn ; [0; 1℄ :

3.4 Muntz-Legendre Polynomials 125

a℄ Let 2 0; . Suppose < are two onse utive zeros of Tn lying in [; 1 ℄. Show that 1

2

n

2

:

Hint: It is lear that Qn (x) := x(1 x)Tn (x) has two onse utive zeros

< Æ in [; 1℄ su h that Æ : n

Show that if

2; then M () is not dense in C [0; 1℄: 1 b℄ Show that with (k )1 k = (k )k P the nondenseness of M () does not follow from Muntz's theorem sin e k2 1=k = 1; where is the set of natural numbers k of the form k = n + m with nonnegative integers n and m: It is shown in Se tion 6.2 that M () is not denseP in C [0; 1℄ whenever is a sequen e of nonnegative real numbers satisfying 1 k 1=k < 1; so the \extra multipli ation" is of no spe ta ular utility. It is not always possible to extend Theorem 4.4.2 to the ase when the numerators and the denominators are oming from dierent in nite Muntz spa es. Somorjai [76℄ shows that 2

=0

=0

2

=0

2

=0

2

2

2

=1

Pn ai xi Pin=0 bi xÆi

i=0

: a i ; bi 2 R ; n 2 N

is not dense in C [0; 1℄ when, for example,

k < Æk 2

+1

< k

+1

;

k = 0; 1; : : :

for some > 1: E.4 Nondense Ratios of Muntz Spa es. Suppose 0 < < : Let a > 0: Show that 0

Pn ai xi Pni=0 i : ai ; bi i=0 bi x P1

2 R;

n2N

1

is dense in C [a; b℄ if and only if i 1=i = 1: Hint: For one dire tion use Muntz's theorem. For the other dire tion use E.5 a℄ and b℄ and E.8 b℄ and ℄ of Se tion 4.2. ut =1

E.5 On the Rate of Approximation by Muntz Rationals. Let (i )1 i be a xed sequen e of nonnegative real numbers. We wish to estimate the error of the best uniform approximation to f 2 C [0; 1℄ on [0; 1℄ from spanfx0 ; : : : ; xn g: We let =0

Rn (f ) :=

(

inf

f (x)

Pn

Pin=0

i=0

ai xi

bi xi

;

: ai ; bi 2 R

[0 1℄

where the in mum is taken for all ai ; bi 2 R; i = 0; 1; : : : ; n:

)

;

4.4 Muntz Rationals 223

In the ase when n n 1; n = 0; 1; : : : , Newman [78℄ laims (without proof) that there exists a onstant C independent of n su h that +1

Rn (f ) = C!f

1

n

and that this is the best possible. He onje tures, in Newman [78℄, that of distin t nonnegative real this estimate holds for every sequen e (i )1 i numbers. a℄ Observe in the last line of the proof of Theorem 4.4.2, with = n ; we have =0

2

n X

f (x) 1

k=0

k n k (x)

f

6 !f

1

n

sin e !f (1) n!f n : b℄ What growth onditions on the sequen e (i )1 i guarantees that

Rn (f ) = O !f

1

n

=0

?

Hint: Estimate the \degree" of the zoomers Zk de ned in Theorem 4.4.2. Use the zoomers de ned in Case 1 of the hint for E.1 b℄. ut

℄ Let x sin(1=x) ; x 2 R n f0g f (x) := 0; x = 0: Show that

Rn (f ) n !f n for every sequen e (i )1 of nonnegative real numbers, where and i are positive onstants independent of (i )1 i : Pn Pn i i ut = i bi x has at most n zeros on [0; 1℄: Hint: i ai x 1

1

1

2

1

=0

2

=0

=0

=0

E.6 A Markov System with Nondense Rationals. This example outlines a onstru tion of Markov systems on [ 1; 1℄ whose rationals are not dense in C [ 1; 1℄: It follows and orre ts Borwein and Shekhtman [93℄. We onstru t an in nite Markov system on ( 1; 1) as follows. Supi pose that := (i )1 i , where 2i + 1 = 7 : Then is a sequen e of even integers satisfying 2

=0

0 = < < < ; 0

Let 'k

2 C[

1

2

1 X i=1

1

i

< 1:

1; 1℄; k = 0; 1; : : : , be de ned by

' k (x) := x2k 2

and

'

k

2 +1

(x) :=

x2k+1 ; x 0 x2k+1 ; x 0 :

224 4. Denseness Questions

1; 1):

a℄ Show that (' ; ' ; : : : ) is a Markov system on ( Hint: If 0

1

n X

p(x) = then

n X

p(x) = while

p(x) =

i=0

i=0

ai 2 R ;

ai 'i (x) ;

x 2 [0; 1) ;

ai xi ;

n X i=0

x2(

( 1)i ai xi ;

1; 0℄ :

Now apply Theorem 3.2.4 (Des artes' rule of signs). ut b℄ There exists an absolute onstant > 0 (independent of n) su h that

n

X

ai x i

i=0

L2 [0;1℄

n

X

ai x i

i=0

L2 [1=2;1℄

for all hoi es of ai 2 R: Hint: Use E.8 a℄ of Se tion 4.2 and Holder's inequality.

℄ The inequalities n 1X jai j 3i

2

Z

n X

1

i=0

0

=0

p

ai 2i

!2

+ 1 xi

dx

ut n 5X j ai j 3i

2

=0

hold for all hoi es of ai 2 R:

Proof. We have Z

1

0

n X i=0

p

!2

ai 2i

+ 1 xi =

n X i=0

dx

j ai j

2

+2

n nXk X k=1 j =0

p

aj aj

+

k

p

2j + 1 2j k + 1 : j + j k + 1 +

+

Here p

p 2j + 1 2j k + 1 7j 7j k 7j k =2 j < 2 = 27 k; j + j k + 1 7 +7 j k 7j k +

+

+

2

2 +2

2 +

2 +2

so on applying the Cau hy-S hwarz inequality n times, we get

4.4 Muntz Rationals 225 p p n n k X X 2j + 1 2j +k + 1 2 aj aj +k j + j+k + 1 k=1 j =0 ! n n X X

4

k=1

7 k

i=0

jai j 23 2

n X i=0

j ai j

2

;

ut

and the result follows. d℄ The inequality

n

X

ai xi

i=0

L2 [0;1℄

p

5

n

X

(

1)i ai xi

i=0

L2 [0;1℄

holds for all hoi es of ai 2 R: Hint: Use part ℄. e℄ The rational fun tions of the form Pn

Pin=0

i=0

ut

a i 'i ; a i ; bi 2 R ; bi 'i

n2N

are not dense in C [ 1; 1℄:

Proof. Consider f

2 C[

1; 1℄ de ned by

f (x) :=

8 > < > :

1 0

if x 2 [ 1; 1=2℄ if x 2 [0; 1℄ 2x if x 2 [ 1=2; 0℄ :

We show that f is not uniformly approximable on [ 1; 1℄ by the above rational fun tions. Suppose that

Pn

i=0 ai 'i

Pn

b'

f

i=0 i i

[

;

< < 1;

ai ; bi 2 R :

1 1℄

This implies (4:4:3)

Pn

i=0 ai xi

Pn

b xi

i=0 i

0 annot be arbitrarily small. ut

This is page

227


5

Basi Inequalities

Overview The lassi al inequalities for algebrai and trigonometri polynomials are treated in the rst se tion. These in lude the inequalities of Remez, Bernstein, Markov, and S hur. The se ond se tion deals with Markov's and Bernstein's inequalities for higher derivatives. The nal se tion is on erned with the size of fa tors of polynomials.

5.1 Classi al Polynomial Inequalities We start with the lassi al inequalities of Remez, Bernstein, Markov, and S hur. The most basi and general of these is probably due to Remez. How large an kpk ; be if p 2 Pn and [

1 1℄

m(fx 2 [ 1; 1℄ : jp(x)j 1g) 2 s holds? The inequality of Remez [36℄ answers this question. His inequality and its trigonometri analog an be extended to generalized nonnegative polynomials (dis ussed in Appendix 4) by a simple density argument. These extensions also play a entral role in the proof of various other Bernstein, Markov, Nikolskii, and S hur type inequalities for generalized nonnegative polynomials, where simple density arguments do not work.

228 5. Basi Inequalities

Theorem 5.1.1 (Remez Inequality). The inequality

kpk

[

;

1 1℄

Tn

2+s 2 s

holds for every p 2 Pn and s 2 (0; 2) satisfying m (fx 2 [ 1; 1℄ : jp(x)j 1g) 2 s : Here Tn is the Chebyshev polynomial of degree n de ned by (2:1:1). Equality holds if and only if 2x + s : p(x) = Tn 2 s Proof. The proof is essentially a perturbation argument that establishes that an extremal polynomial is of the required form. Let k k denote the uniform norm on [ 1; 1℄; and let (5:1:1)

Pn (s) := fp 2 Pn : m (fx 2 [

1; 1℄ : jp(x)j 1g) 2

sg :

The set Pn (s) is ompa t, say, in the uniform norm on [ 1; 1℄; by E.1, and the fun tion p ! kpk is ontinuous. Hen e there is a p 2 Pn (s) su h that

kp k =

sup

p2Pn (s)

kpk :

First assume that p (1) = kp k: We laim that all the zeros of p are real and lie in [ 1; 1): Indeed, if p vanishes at a nonreal z; then for suÆ iently small > 0 and > 0;

q(x) := (1 + )p (x) 1

(x 1) (x z )(x z ) 2

is in Pn (s) and ontradi ts the maximality of p : If p has a real zero z outside [ 1; 1℄; then, in similar fashion,

1 x q(x) := (1 + )p (x) 1 sign(z ) z x

ontradi ts the maximality of p : Now we show that jp ( )j = kp k annot o

ur with 2 ( 1; 1): To see this, assume, without loss of generality, that p ( ) = kp k. Then the polynomials

q (x) := p 1

2

1

+

+1 2

x

and q (x) := p 2

+1 2

+

2

1

x

5.1 Classi al Polynomial Inequalities 229

satisfy qj (1) = kqj k = kp k; j = 1; 2: Sin e p Lebesgue measure m( j ) of

2 Pn(s) is extremal, the

j := fx 2 [ 1; 1℄ : jqj (x)j > 1g is at least s; otherwise qj + with suÆ iently small > 0 ontradi ts the maximality of p : On the other hand, 1+ 1 m( ) + m( ) = m (fx 2 [ 1; 1℄ : p(x) > 1g) s : 2 2 1

2

Hen e m( ) = m( ) = s; whi h means that qj 2 Pn (s), j = 1; 2, are extremal polynomials attaining their uniform norm on [ 1; 1℄ at 1. It now follows from the rst part of the proof that q and q have all their zeros in [ 1; 1℄; whi h is impossible sin e the number of zeros of qj ; j = 1; 2; in [ 1; 1℄ is equal to the number of zeros of p in [ 1; ) and in (; 1℄; respe tively. This ontradi tion proves that jp ( )j < kp k for every 2 ( 1; 1). Hen e either p (1) = kpk or p ( 1) = kpk: Without loss of generality we may assume that p (1) = kp k; otherwise we onsider p ( x) 2 Pn (s): We now have 1

2

1

p (1) = kpk > jp (x)j ;

(5:1:2)

2

1 < x < 1;

and ea h zero of p lies in [ 1; 1): Next we prove that

jp (x)j 1 ;

(5:1:3)

1x1

s:

Assume to the ontrary that for some 1 < < < 1; 3

8 >
1 ; jp (x)j 1 ; > : jp (x)j > 1 ;

2

1

x 2 I := ( ; 1℄ x 2 I := [ ; ℄ x 2 I := ( ; ) : 1

1

2

2

1

3

3

2

Let x ; x ; : : : ; xm be the zeros of p in ( ; ): Sin e all zeros of p are in [ 1; 1); we have m 1; otherwise p0 would vanish at an x larger than the largest zero of p ; whi h is a ontradi tion. The remaining n m zeros of p lie in [ 1; ): We set 1

2

2

1

3

p (x) := 1

m Y j =1

(x

xj ) ;

p := 2

p : p 1

The polynomial q (x) := p (x + h)p (x) with 0 < h < following properties: 1

2

2

has the 3


(1) If jp (x)j 1 for some x 2 [ 1; ℄; then jq (x)j 1: (2) For ea h x 2 I we have jq (x h)j jp (x)j 1: (3) q (1) = p (1 + h)p (1) > p (1)p (1) = p (1) = kp k: These properties show that q 2 Pn (s) ontradi ts the maximality of p : By E.2, among all polynomials p 2 Pn with kpk 1, the Chebyshev polynomial Tn in reases fastest for x > 1: Hen e, by a linear transformation, we see that the four polynomials 3

2

1

2

1

2

p (x) = Tn

2x + s 2

s

are the only extremal polynomials. In parti ular,

kpk = Tn

2+s : 2 s

ut The next theorem establishes a Remez-type inequality for trigonometri polynomials. Throughout this se tion, as before, K := R (mod 2 ). Theorem 5.1.2. The inequality

ktkK exp(4ns) holds for every t 2 Tn and s 2 (0; =2℄ satisfying (5:1:4) m(f 2 [ ; ) : jt()j 1g) 2

s:

The inequality

ktkK Tn 22 +

also holds for every even t equality holds if and only if t() = Tn

; = 1 os(s=2) ; 0 < s < 2 ;

2 Tn

and s

2 os + ; 2

2

(0; 2 ) satisfying (5.1.4), and

= 1 os(s=2) :

Proof. We prove the se ond part rst. Suppose t 2 Tn is even and satis es jt()j 1 on K n with m( ) s: Then the polynomial p 2 Pn de ned by t() := p( os ) satis es jp(x)j 1 on [ 1; 1℄ n 0 , where

0 := fx 2 [ 1; 1℄ : x = os ; 2 \ [0; ℄g :


It is easy to see that m( 0 ) 1 os(s=2) =: ; where equality holds if and only if := [ s=2; s=2℄: Hen e, Theorem 5.1.1 implies that

ktkK Tn 22 +

;

where equality holds if and only if p is of the form given in the theorem. The rst part of the theorem an be easily obtained from the se ond part as follows. Let t 2 Tn satisfy (5:1:4). Without loss of generality we may assume that t(0) = ktkK : The polynomial e t()

is even and

:= (t() + t( )) 2 Tn 1

2

m(f 2 [ ; ) : jet()j 1g) 2

2s :

Hen e, the se ond part of the theorem yields that

e t

where := 1

L1 (K )

Tn

2+ 2

;

os s = 2 sin (s=2) 1 for every s 2 (0; =2℄: Sin e 2

p

Tn (x) x + x we have

2+ Tn 2

p

1 + 2 + =2 1 =2

p exp n 2 + exp(4ns)

for every s 2 (0; =2℄: Therefore

ktkK = t(0)

n

1

2

= et(0)

Tn 22 +

7 4

!n

x 1;

;

exp

p

1 + 2 + 7

n

2

exp

n s+ s 7

2

4

n s+ s 7

4

2

exp(4ns)

for every s 2 (0; =2℄; and the theorem is proved. Note that we have proved slightly more than we laimed in the statement of the theorem. That is,

ktkK exp for every t 2 Tn satisfying (5:1:4):

n s+ s 7

2

4

ut

We now prove the basi inequality that bounds the derivative of a trigonometri polynomial in terms of its maximum modulus on the period

K:


Theorem 5.1.3 (Bernstein-Szeg}o Inequality). The inequality

t0 () + n t () n ktkK ; 2

2 2

2

2K

2

holds for every t 2 Tn : Equality holds if and only if jt()j = ktkK or t is of the form t( ) = os(n ) with , 2 R: Proof. Assume that there are t 2 Tn and 2 K su h that ktkK < 1 and t0 () + n t () > n :

(5:1:5)

2

2 2

2

For the sake of brevity let Tn; ( ) := os(n there exists an 2 K su h that (5:1:6)

Tn;() = t()

): It is easy to see that

0 ()) = sign(t0 ()) : sign(Tn;

and

0 () + n Tn; () = n , (5:1:5) and (5:1:6) imply that Sin e Tn; 2

2

2

2

0 ()j jt0 ()j > jTn;

0 () = sign(t0 ()) : sign(Tn;

and

Hen e E.4 yields that 0 6= t Tn; 2 Tn has at least 2n + 2 distin t zeros in K; whi h is a ontradi tion. To nd all the extremal polynomials, see the

ut

hint to E.5.

As a orollary of Theorem 5.1.3 we have kt0 kK nktkK for every t 2 Tn ; and by indu tion on m we obtain the following theorem: Theorem 5.1.4 (Bernstein's Inequality). The inequality

kt m kK nm ktkK (

)

holds for every t 2 Tn : Corollary 5.1.5. The inequality of Theorem 5.1.4 remains true for all

t 2 Tn :

Proof. Choose 2 R su h that ei t m attains the value kt m kK ; say, at = . Now et() := Re(ei t()) 2 Tn and ketkK ktkK : On applying Theorem 5.1.4 to et 2 Tn , we obtain (

)

(

)

kt m kK = ei t m ( ) = et m ( ) nm ketkK nmktkK : (

)

(

)

(

)

ut The above orollary implies the following algebrai polynomial version of Bernstein's inequality on the unit disk.


Corollary 5.1.6. The inequality

kp0 kD n kpkD holds for every p 2 Pn ; where D := fz 2 C : jz j < 1g: Proof. If p 2 Pn then t() := p(ei ) 2 Tn and by Corollary 5.1.5 we have

jp0 (z )j = j

i t0 ( )

j nktkK = nkpkD ;

ie

z = ei :

The maximum prin iple (see E.1 d℄ of Se tion 1.2) nishes the proof.

ut

From Corollary 5.1.5, by the substitution x = os ; we get the algebrai polynomial ase of Bernstein's inequality on [ 1; 1℄: Theorem 5.1.7 (Bernstein's Inequality). The inequality

jp0 (x)j p

1

n

x

2

kpk

[

;

1 1℄

;

1<x 0; y xk

k = 1; 2; : : : ; n ;

Tn0 is in reasing on (xn ; 1), and Tn0 (1) = n :

ut

2

Proof of Theorem 5.1.8. Let p 2 Pn . Then p0 2 Pn

1

and 5.1.9 yield

kp0k

[

;

1 1℄

p

n p0 (x)

and the theorem is proved.

1

x

2

[

;

1 1℄

and Theorems 5.1.4

n kpk 2

[

;

1 1℄

;

ut

Comments, Exer ises, and Examples. Every result of this se tion was proved by the person it is named after; see Remez [36℄, Bernstein [12℄, Szeg}o [28℄ or [82℄, A. A. Markov [1889℄, S hur [19℄, and M. Riesz [14℄. However, earlier less omplete versions of these basi polynomial inequalities also appear in the literature. Some of the proofs were simpli ed later. For example, in the proof of Theorem 5.1.1 we followed the method given in Erdelyi [89b℄, while in the proof of Theorem 5.1.8 a method of Polya and Szeg}o [76℄ is used. Theorem 5.1.3 is also obtained in Corput and S haake [35℄, however, it follows from an earlier result of Szeg}o [28℄. Theorem 5.1.2 was established in Erdelyi [92a℄ in a slightly weaker form. Various extensions of the inequalities of this se tion are dis ussed in Se tions 5.2 to 7.2 and in the appendi es. Rahman and S hmeisser [83℄ also


oers a olle tion of Markov- and Bernstein-type inequalities. Some further

lassi al inequalities are in E.5 of Appendix 3. An interesting extension of Markov's inequality is due to Bojanov [82a℄. It states that

kp0 kLq ; kTn0 kLq ; kpk ; for every p 2 Pn and q 2 [1; 1℄; where Tn is the Chebyshev polynomial of [

1 1℄

[

1 1℄

[

1 1℄

degree n as in (2.1.1). The following result is due to Szeg}o [25℄. The inequality jp0 (0)j n kpkD holds for every p 2 Pn ; where is an absolute onstant and D := fz 2 C : jz j 1; j arg(z )j (1 )g ; 2 (0; 1℄ : 2

Throughout the exer ises Tn denotes the Chebyshev polynomial of degree n as de ned by (2.1.1). E.1 A Detail in the Proof of Theorem 5.1.1. Show that the sets Pn (s) de ned by (5:1:1) are ompa t in the uniform norm on [ 1; 1℄ for every xed n 2 N and s 2 (0; 2): E.2 Chebyshev's Inequality. Prove that jp(y)j jTn (y)j kpk ; ; y 2 R n [ 1; 1℄ for every p 2 Pn ; and equality holds if and only if p = Tn for some 2 R: Extend the above inequality to every p 2 Pn and nd all p 2 Pn for whi h equality holds. Hint: If p 2 Pn , kpk ; = 1, and jp(y)j > jTn(y)j for some y 2 R n [ 1; 1℄; then with := Tn (y )p(y ) 2 [ 1; 1℄; the polynomial p Tn 2 Pn has at least n + 1 zeros ( ounting multipli ities). ut [

[

1 1℄

1 1℄

1

E.3 Trigonometri Chebyshev Polynomials on Subintervals of K . a℄ For n 2 N and ! 2 (0; ); let sin(=2) Qn;! () := T n : sin(!=2) Show that Qn;! 2 Tn attains the values kQn;! k !;! = 1 with alternating sign 2n + 1 times on [ !; ! ℄: b℄ Prove that jt()j Qn;! ()ktk !;! ; 2 K n [ !; !℄ for every t 2 Tn ; and for every xed 2 K n [ !; ! ℄: Equality holds if and only if p = Qn;! for some 2 R:

℄ Show that there exist absolute onstants > 0 and > 0 su h that exp( n( ! )) kQn;! kK = Qn;! ( ) exp( n( ! )) : 2

[

[

℄

℄

1

1

2

1

2

2


E.4 A Zero Counting Lemma. Let 2 K be xed. Suppose f and g are

ontinuous fun tions on K := R (mod 2); dierentiable at a xed 2 K;

and suppose f and g have the properties (1) kf kK = 1; kg kK < 1; (2) there are x < x < < x n < so that f (xj ) = ( 1)j ; (3) f ()) = g (); jg 0 ()j > jf 0 ()j; and sign(g 0 () = sign(f 0 ()): Then f g has at least 2n + 2 distin t zeros on K: 1

2

2

E.5 Sharpness of Theorem 5.1.3. Let 2 K be xed. Suppose t 2 Tn and

t0 () + n t () = n ktkK : 2

2 2

2

2

Show that either jt()j = ktkK or t is of the form

t( ) = os(n

; 2 R :

) ;

Hint: Suppose ktkK = 1 and jt()j < 1: Choose 2 K su h that (5.1.6) is satis ed and show that t Tn; 2 Tn has at least 2n + 2 zeros on K ( ounting multipli ities). ut E.6 Sharpness of Theorem 5.1.4. Show that Theorem 5.1.4 is sharp and equality holds if and only if t is of the form

t( ) = os(n

; 2 R :

) ;

E.7 Sharpness of Corollary 5.1.6. Show that Corollary 5.1.5 is sharp and equality holds if and only if p is of the form p(z ) = z n ; 2 C : E.8 Sharpness of Theorem 5.1.7. Show that for a xed integer n 1; Theorem 5.1.7 is sharp if and only if x is a zero of the Chebyshev polynomial Tn ; that is, x = os k n ; k = 1; 2; : : : ; n ; and p = Tn for some 2 R: (2

1)

2

E.9 Sharpness of Theorem 5.1.9. Show that Theorem 5.1.9 is sharp and equality holds if and only if p = Un for some 2 R; where Un is the Chebyshev polynomial of the se ond kind de ned in E.10 of Se tion 2.1. E.10 Sharpness of Theorem 5.1.8. Show that Theorem 5.1.8 is sharp and equality holds if and only if p = Tn for some 2 R: E.11 A Property of the Zeros of t 2 Tn . Let 2 K be xed. Show that every t 2 Tn has at most

M := enrjt()j

1

ktkK

zeros ( ounting multipli ities) in the interval [

r; + r℄; r > 0:


Proof. Assume that t 2 Tn has m > M zeros in [ r; + r℄: Interpolate t at these m zeros by a Hermite interpolation polynomial of degree at most m 1 (see E.7 of Se tion 1.1). This gives the identi ally zero polynomial. The formula for the remainder term of the Hermite interpolation polynomial and Theorem 5.1.4 (Bernstein's inequality) yield that there exists a 2 ( r; + r) su h that

m

jt()j = m1 ! rm jt m ( )j < me rm nm ktkK enr m < ktkK jt()jm ktkK m jt()j ; M (

)

1

whi h is impossible.

ut

E.12 A Property of the Zeros of a p 2 Pn . Show that every p 2 Pn has at most e p M := p n r jp(1)j kpk ; 2 zeros ( ounting multipli ities) in [1 r; 1℄; r > 0: Hint: Use the substitution x = os ; E.11, and the inequality 1

os r < 1

1 4

[

1 1℄

0 < r 2:

r ; 2

ut E.13 Riesz's Lemma. a℄ Suppose t 2 Tn and t() = ktkK = 1 for some 2 K: Then

t() os(n(

)) ;

; + ; n 2n ; + if and 2n 2n

2

2

and equality holds for a xed 2 only if t is of the form t() = os(n( )). In parti ular, t does not vanish in n ; + n : Hint: If this were false, then 2

2

q( ) := t( )

os(n(

))

would have more than 2n zeros on K ( ounting multipli ities). b℄ Suppose p 2 Pn and p(1) = kpk ; = 1: Then that [

ut

1 1℄

p(x) Tn (x) ;

x 2 os

; 1 n

2

;

and equality holds for a xed x 2 os n ; 1 if and only if p = Tn, where Tn is the Chebyshev polynomial ofdegree n as de ned by (2.1.1). In parti ular, p does not vanish in os n ; 1 : 2

2

The next two exer ises follows Erdelyi [88℄ and Erdelyi and Szabados [89b℄.


E.14 A Markov-Type Inequality for Trigonometri Polynomials on [ !; ! ℄. Show that there exists a onstant 0 < 16 su h that

ks0k

!;!℄

[

n+

!

!

n

2

ksk

!;!℄

[

for every s 2 Tn and ! 2 (0; ℄:

Proof. If

! > (2n) ; then 1

(3 tan (!=2) + 1) = < 8n ; 2

1 2

and E.19 ℄ gives the result. If ! (2n) ; then Theorem 5.1.4 (Bernstein's inequality) ombined with the Mean Value Theorem yields 1

ksk

[

and hen e

;℄

ksk

[

ksk !;!℄

!)n ksk

!;!℄ + (

[

(1

!)) ksk

n(

;℄ ;

[

[

;℄

for every s 2 Tn : Therefore, using Theorem 5.1.4 (Bernstein's inequality) and ! (2n) ; we get 1

ks0 k

[

!;!℄

ks0k ; n ksk 1 n(n !) ksk [

℄

[

[

;℄ !;!℄

(n + 2(

!)n )ksk 2

[

!;!℄

for every s 2 Tn :

ut

E.15 S hur-Type Inequality for Tn on [ !; ! ℄. Let ! 2 (0; 2 ℄. Show that

ksk

[

!;!℄

2n + 1 sin( s( ) !=2)

1 2

1=2

( os

os ! )

[

!;!℄

for every s 2 Tn , and equality holds if and only if s is of the form h

s( ) =

=2) sin(!=2)

os ! )1=2

sin (2n + 1) ar

os ( os

sin(

i

;

2 R:

Note that the right-hand side of the above is an element of Tn : Hint: De ne 2n + 1 distin t points in ( !; !) by

k := 2 ar sin sin

! 2

sin

k ; 2n + 1

k = 0; 1; : : : ; n :

Distinguish two ases in estimating jsn ()j for 2 [ !; ! ℄:


Case 1: jj jn j. Use the inequality 1 ( os 2

os ! ) >

sin (!=2) (2n + 1) 2

2

to get the desired result. Case 2: n < jj !: Let

Mn(s) := max fjs(k )j(( os k jkjn

os ! )=2) = g: 1 2

Let n 2 N , ! 2 (0; ℄, and 2 [ !; n ) [ (n ; ! ℄ be xed. Show that there is an sen 2 Tn su h that

jsen ()j = max js()j : Mn (sen ) s2Tn Mn(s) Show by a variational method that sen is of the form h

sen ( ) =

sin (2n + 1) ar

os ( os

=2) !=2)

sin( sin(

os ! ) =

i

1 2

;

2 R:

ut E.16 Another Proof of S hur's Inequality. a℄ Prove Theorem 5.1.9 (S hur's inequality) by using the method given in the hint to E.15. b℄ Prove the result of E.15 by using interpolation, as in the proof of Theorem 5.1.9. Proof. See Erdelyi and Szabados [89b℄. ut E.17 Growth of Polynomials in the Complex Plane. Let

D := fz 2 C : jz j < 1g

D% := fz 2 C : jz j < %g:

and

a℄ Show that

jp(z )j jz jn kpkD for every p 2 Pn and z 2 C n D: Find all p 2 Pn for whi h equality holds.

Hint: Apply the maximum prin iple (see E.1 d℄ of Se tion 1.2) with D and q(z ) := z n p(z ) 2 Pn : 1

ut


b℄ Show that the transformation z = M (w) := (w + w ) an be written as x = (% + % ) os ; y = (% % ) sin ; 1

1

2

1

1

1

2

1

2

where z = x + iy; x; y 2 R and w = %ei ; % > 0; 2 K:

℄ For % > 0; let E% be the image of the ir le D% under M: Show that E% is the ellipse

x y + = 1 with semiaxes a := (% + % ) and b := j% % a b 2

2

2

2

1

1

1

2

2

1

j:

Furthermore, E% = E% 1 ; and E is the interval [ 1; 1℄ overed twi e. d℄ Show that jp(z )j %n kpk ; 1

for every p 2 P

n;

[

z 2 E% ; % > 1:

1 1℄

Proof. Applying the maximum prin iple (see E.1 d℄ of Se tion 1.2) with D = D and Q(w) := wn p (w + w ) 2 P n ; 1

1

1

2

2

we obtain

jQ(w)j kQkD

1

= kpk

;

[

1 1℄

w 2 D% 1 :

;

This, together with part ℄, yields

jp(z )j %n kpk

[

;

1 1℄

z 2 E% :

;

ut e℄ Show that there exists an absolute onstant su h that

jp(z )j kpk

;

1 1℄

p 2 Pn

;

whenever

z = x + iy ; x; y 2 R ; (j1

j xj 1 + n

2

;

jy j j1 n x j 2

+

+

1

n

2

:= maxf1 x ; 0g). Hint: Use part d℄. ut f℄ Prove the following Markov-Bernstein inequality. There is a onstant

(m) depending only on m su h that

x

2

j

p

1

2

+

jp m (x)j (m) (

)

min n ; p

for every p 2 Pn and x 2 [ 1; 1℄:

2

1

n

m

x

2

kpk

[

;

1 1℄


Hint: Use part e℄ and Cau hy's integral formula (see E.1 a℄ of Se tion 1.2).

ut

Bernstein established Theorem 5.1.4 in order to prove inverse theorems of approximation. Bernstein's method is presented in the proof of the next exer ise, whi h is one of the simplest ases. However, several other inverse theorems of approximation an be proved by straightforward modi ations of the proof of this exer ise. That is why Bernstein- and Markov-type inequalities play a signi ant role in approximation theory. Dire t and inverse theorems of approximation and related matters may be found in many books on approximation theory, in luding Cheney [66℄, Lorentz [86a℄, and DeVore and Lorentz [93℄. E.18 An Inverse Theorem of Approximation. Let Lip , 2 (0; 1℄, denote the family of all real-valued fun tions g de ned on K satisfying sup For f

jg(x) g(y)j : x 6= y 2 K < 1: jx yj

2 C (K ), let En (f ) := inf fkt f kK : t 2 Tn g:

An example for a dire t theorem of approximation is stated in part a℄. Part b℄ deals with its inverse result. a℄ Suppose f is m times dierentiable on K and f m 2 Lip for some 2 (0; 1℄: Then there is a onstant C depending only on f so that (

En (f ) Cn

(

m+) ;

)

n = 1; 2; : : : :

Proof. See, for example, Lorentz [86a℄. b℄ Suppose m is a nonnegative integer, 2 (0; 1), and f there is a onstant C > 0 depending only on f su h that En (f ) Cn

(

m+) ;

ut

2 C (K ). Suppose

n = 1; 2; : : : :

Then f is m times ontinuously dierentiable on K and f

(

m)

2 Lip :

Outline. We show only that f is m times ontinuously dierentiable on K: The rest an be proved similarly, but its proof requires more te hni al details. See, for example, Lorentz [86a℄. For ea h k 2 N ; let Q k 2 T k be hosen so that 2

kQ k 2

2

f kK C 2

k(m+) :


Then

kQ k

Q

2 +1

Now

f () = Q () + 20

1 X k=1

k 2C 2

k K

2

(Q

k(m+) :

2K;

Q k )() ;

k

2 +1

2

and by Theorem 5.1.4 (Bernstein's inequality)

jQ j j

( )

20

1 X + (Q2k+1

Q2k ) j () ( )

k=1

kQ kK + 1

kQ kK + 1

1 X k=1

1 X

k=1

kQ kK + 2j 1

+1

2k

+1

2k

+1

j j

kQ k

Q

2 +1

2

2C 2 k m (

+

k

k K

)

1 X

(2j m )k < 1

C

k=1

for every 2 K and j = 0; 1; : : : ; m; sin e > 0: Now we an on lude that f j () exists and ( )

f j () = Q j () + ( )

( )

1

1 X k=1

(Q

Q k ) j ()

k

( )

2 +1

2

for every 2 K and j = 0; 1; : : : ; m: The fa t that f m seen by the Weierstrass M -test. (

)

2 C (K ) an be ut

The next exer ise follows Videnskii [60℄. E.19 Videnskii's Inequalities. The main results of this exer ise are the Bernstein- (part b℄) and Markov-type (part ℄) inequalities for trigonometri polynomials on an interval shorter than the period. Let ! 2 (0; );

sin(=2) tn () := Qn;! () = os 2n ar

os sin(!=2) and

un() := sin 2n ar

os a℄ Re all that tn 2 Tn by E.3 a℄.

sin(=2) sin(!=2)

:

;


b℄ Show that

js0n ()j jt0n () + iu0n()j ksn k

os (!=2)

os (=2) 2

=n 1

[

!;!℄ =

1 2

ksn k

!;!℄

[

2

for every sn 2 Tn and 2 ( !; ! ): Equality holds if and only if sn = tn for some 2 R and sn () = 0: Hint: First show that for every n 2 N ; ! 2 (0; ℄; and 2 [ !; !℄; there exists an se 2 Tn su h that

jse0 ()j ksek !;! [

js0n ()j sn 2Tn ksn k !;!

:

= max ℄

[

℄

Use a variational method to show that either

ksen k

[

k k

!;!℄ = sen

or there exist 2 ( ; ! ℄ and

2 R; where, as in Se tion 3.3,

[

;℄

2 [!; ) su h that sen = Tn for some

Tn := Tn f1; os ; sin ; : : : ; os n; sin n ; [; ℄g is the Chebyshev polynomial for Tn on [; ℄: In the rst ase use Theorem 5.1.4 (Bernstein's inequality). In the se ond ase observe that

Tn( ) = tn with

:= ( + )

sin(( )=2) sin(!e =2)

!e := (

and

1

2

ks0n k

[

!;!℄

1 2

t0n (!)ksn k

[

) :

1

2

℄ Show that if 2n > (3 tan (!=2) + 1) = 2

ut

; then

ksnk

2 !;!℄ = 2n ot(!=2)

[

!;!℄

for every sn 2 Tn ; and equality holds if and only if sn = tn ; 2 R:

Outline. Let ! = < < < n = ! be the points where 0

1

tn (j ) = ( 1)j ; and

un(j ) = 0 ;

2

j = 0; 1; : : : ; 2n j = 1; 2; : : : ; 2n 1 ;


and let < < < n be the zeros of tn (whi h lie in ( !; ! )): Note that u0n (j ) = 0 ; j = 1; 2; : : : ; 2n : 1

2

2

Step 1. Show that 2n > (3 tan (!=2) + 1) = implies t00n () > 0 for every 2 [ n ; !℄; so t0n is in reasing on [ n ; !℄: Step 2. Use part b℄ to show that if 2 [ ; n ℄; then 2

2

1

1 2

2

1

1

2

jt0n ()j jt0n () + iu0n()j jt0n ( n ) + iu0n ( n )j = jt0n ( n )j : 2

2

2

Step 3. Dedu e from Steps 1 and 2 that

jt0n ()j < jt0n (!)j ;

2 ( !; !) :

Step 4. Show that there is an se 2 Tn su h that

ksen0 k ksenk

[

[

!;!℄ = max s2Tn !;!℄

ks0 k ksk

[

[

!;!℄ !;!℄

:

For the rest of the proof let sen be normalized by ksek !;! = 1 and let 2 [ !; !℄ be hosen so that [

jsen0 ( )j = ksen0 ()k

[

℄

!;!℄ :

Let (a < a < < am ) be an alternation sequen e of maximal length for sen 2 C [ !; !℄ on [ !; !℄: We would like to show that m = 2n + 1: Clearly m < 2n + 2: Step 5. Use a variational method to show that 2n m: Step 6. Use a variational method to show that = ! implies m = 2n +1; so m = 2n implies 2 ( !; ! ): Step 7. Show by a variational method that 1

2

js0n (!)j jt0n (!)j ksnk

[

!;!℄

for every sn 2 Tn ; and equality holds if and only if sn = tn ; parti ular, if m = 2n; then

jsen0 (!)j < jt0n (!)j : Step 8. Use part b℄ and Step 3 to show that

jsen0 ()j t0n ( n ) < t0n (!) ; 2

2 [ ; n ℄ :

Step 9. Use part b℄ to show that m = 2n implies

1

2

2 R:

In


jsen0 (j )j < jt0n (j )j = (

1)j t0n (j ) ;

j = 1; 2; : : : ; 2n :

Step 10. Suppose m = 2n: Use Steps 6 and 8 to show that

2 ( !; ) [ ( n ; !) : 1

2

Step 11. Suppose m = 2n: Use the de ning property of sen and Steps 1 and 10 to show that

jsen0 ( )j = ksen0 k

[

!;!℄

jt0n (!)j > jt0n ( )j :

Step 12. Suppose m = 2n and sn0 ( ) 0: Use Steps 7 to 11 to show that t0n se0n has at least 2n + 1 distin t zeros in ( !; !); a ontradi tion. Step 13. Show that m = 2n + 1 and sen = tn . ut E.20 Inequalities for Entire Fun tions of Exponential Type. Entire fun tions of (exponential) type are de ned in E.17 of Se tion 4.2. Denote by E the set of all entire fun tions of exponential type at most : This exer ise

olle ts some of the interesting inequalities known for E : Sin e a trigonometri polynomial of degree n belongs to En ; these results an be viewed as extensions of the orresponding inequalities for trigonometri polynomials. More on various inequalities for entire fun tions of exponential type may be found in Rahman and S hmeisser [83℄. a℄ Bernstein's Inequality. The inequality

kf 0kR kf kR ;

x2R

holds for every f 2 E : Proof. See Bernstein [23℄ or Rahman and S hmeisser [83℄. b℄ Extension of the Bernstein-Szeg}o Inequality. The inequality (f 0 (x)) + (f (x)) 2

2

kf kR ; 2

2

x2R

holds for every f 2 E taking real values on the real line. Proof. See DuÆn and S haeer [37℄.

℄ The Growth of f 2 E . The inequality

jf (x + iy)j e jyjkf kR ; holds for every f 2 E . Proof. See Rahman and S hmeisser [83℄.

ut

ut

x; y 2 R

ut


2 E Taking Real Values on R. The inequality jf (x + iy)j ( osh y) kf kR ; x; y 2 R holds for every f 2 E taking real values on the real line. d℄ The Growth of f

Proof. See S haeer and DuÆn [38℄. e℄ Bernstein-Type Inequality in Lp . Let p 2 (0; 1): The inequality kf 0 kL R kf kL R p(

p(

)

ut

)

holds for every f 2 E . Proof. See Rahman and S hmeisser [90℄.

ut

E.21 Markov-Type Inequality on Conne ted Subsets of the Complex Plane. Let E be a onne ted ompa t set of the omplex plane. Then

n kpk kp0 kE 2e ap( E) E 2

for every p 2 Pn : Proof. See Pommerenke [59 ℄. ut e Erd}os onje tured that the onstant in the above inequality an be repla ed by : This result would ontain Theorem 5.1.8 (Markov's inequality) as a spe ial ase. However, Rassias, Rassias, and Rassias [77℄ disproved the onje ture. Erd}os still spe ulates that e in Pommerenke's inequality may be repla ed by (1 + o(1)): The result of the next exer ise is formulated so that its proof is elementary at the expense of pre ision and generality. 2

1

2

2

1

2

E.22 The Interval where the Sup Norm of a Weighted Polynomial Lives. Suppose w = exp( Q); where (1) Q : R ! R is ontinuous and even, (2) Q0 is ontinuous and positive in (0; 1), (3) tQ0 (t) is in reasing in (0; 1), and (4) lim tQ0 (t) = 0 and lim tQ0 (t) = 1: t!0+

Let an > 0 be hosen so that

t!1

ann w(an ) = max jxn w(x)j : x2R a℄ Show that

n = an Q0 (an )

and

0 < a < a < a < : 1

2

3


b℄ Show that

kpwkR kpwk

for every p 2 Pn :

a n ;2a2n ℄

[

2 2

Proof. Using Chebyshev's inequality (see E.2 of Se tion 5.1) and the expli it form (2.1.1) of the Chebyshev polynomial Tn , we an dedu e that

jp(x)j

x a

r

+

n 1

x 2

a

kpk

[

a;a℄

2jxj n

a

kpk

[

a;a℄

for every p 2 Pn , x 2 R n [ a; a℄; and a > 0: Choosing a := an , and using the fa t that w is de reasing on [0; 1), we obtain

j(pw)(x)j

2jxj n w(x) kpwk an;an an w(an ) [

℄

for every p 2 Pn and x 2 R n [ an ; an ℄. Now if x 2 R n [ 2a n ; 2a n ℄; then 2

jxjn w(x) exp

2

!

Z jxj

d log(tn w(t)) dt n an w(an ) an dt ! Z jxj n tQ0 (t) = exp dt t an

exp

Z 2a2n

n

a2n

t

2n

dt = 2

n;

where we used a n an , tQ0 (t) 2n for t a n , and tQ0 (t) n for t an : Combining the above inequality with the previous one, we obtain that 2

2

j(pw)(x)j kpwk

[

x 2 R n [ 2a n; 2a n ℄

an ;an ℄ ;

2

for every p 2 Pn ; from whi h the result follows.

2

ut

℄ Let Q(x) := jxj ; > 0. Show that Q satis es the assumptions of the exer ise, and n = an = : 1

The idea of in nite- nite range inequalities, of whi h E.22 b℄ is an example, goes ba k to Freud (an is the Freud number); see Nevai's survey paper [86℄. The sharp form of these is due to Mhaskar and Sa [85℄; see also Lubinsky and Sa [88℄ and Sa and Totik [to appear℄. They are not that diÆ ult to prove, but need the maximum prin iple for subharmoni fun tions.


E.23 A Theorem of Markov. Let Tn be the Chebyshev polynomial of degree n de ned by (2.1.1). Then

Tn (x) =

n X k=0

k;n xk :

The Markov numbers Mk;n ; 0 k n; are de ned by

Mk;n :=

j k;n j j k;n j 1

if k n (mod 2) if k n 1 (mod 2) :

These an be expli itly omputed from (2.1.1). a℄ The inequalities

jak;n j Mk;n kpk

hold for every p 2 Pn of the form

p(x) =

n X k=0

ak;n xk ;

;

[

1 1℄

ak;n 2 R :

ut

Proof. See, for example, Natanson [64℄. b℄ Show that

jp0 (0)j (2n

for every p 2 P n : Hint: Use part a℄.

℄ Show that

1) kpk

[

;

1 1℄

2

ut

jp0 (x)j 12n jx1j kpk for every p 2 P n and x 2 ( 1; 1):

[

;

1 1℄

2

ut

Hint: Use part b℄ and a linear transformation.

Of ourse, part ℄ gives a better result than Theorem 5.1.7 (Bernstein's Inequality) only if x is very lose to 0: This is exa tly the ase we need in our appli ation of part ℄ in E.4 ℄ of Se tion 6.1.

5.2 Markov's Inequality for Higher Derivatives From Theorem 5.1.8 (Markov's Inequality), by indu tion on m it follows that kp m k ; (n(n 1) (n m + 1)) kpk ; (

for every p 2 Pn :

)

[

1 1℄

2

[

1 1℄

5.2 Markov's Inequality for Higher Derivatives 249

However, this is not the best possible result. The main result of this se tion is the following inequality of DuÆn and S haeer [41℄, whi h gives a sharp improvement of the above. E.2 d℄ extends the following result to polynomials with omplex oeÆ ients: Theorem 5.2.1. If p 2 Pn satis es p

os j n

1;

j = 0; 1; : : : ; n ;

then for every m = 1; 2; : : : ; n;

jp m (x + iy)j jTnm (1 + iy)j ; (

)

(

x 2 [ 1; 1℄ ; y 2 R ;

)

where Tn is the Chebyshev polynomial of degree n de ned by (2.1.1). Equality an o

ur only if p = Tn: To prove this inequality we need three lemmas, whi h are of some interest in their own right. Lemma 5.2.2. Suppose ; ; : : : ; n are distin t real numbers, 1

2

q(z ) :=

n Y j =1

(z

0 6= 2 R ;

j ) ;

and p 2 Pn satis es jp0 (j )j jq0 (j )j ;

j = 1; 2; : : : ; n :

Then, for every m 2 N ,

jp m (x)j jq m (x)j

(5:2:1)

(

whenever x is a zero of q m (

1)

)

(

)

:

Proof. For m = 1, inequality (5.2.1) is simply a restatement of the assumption of the lemma, so onsider the ase m = 2: The Lagrange interpolation formula (see E.6 of Se tion 1.1) gives (5:2:2)

n 0 n X p (j ) 1 Æj p0 (z ) X = = ; q(z ) j q0 (j ) z j j z j =1

=1

where, by the hypotheses of the lemma, jÆj j 1: There is a similar expression for q 0 (z )=q (z ) in whi h ea h Æj is equal to 1. On dierentiating (5.2.2), we obtain n X p00 (z )q(z ) p0 (z )q0 (z ) Æj = : q(z ) (z j ) j 2

2

=1


Thus at the points x where q 0 (x) = 0 we have 00 p (x) q (x)

=

n X j =1

Æj j )2

(x

n X j =1

(x

1

j )

2

=

00 q (x) q (x)

;

and it follows that the lemma is true for m = 2. The proof for m > 2 is by indu tion. Let jp m (x)j jq m (x)j at the zeros of q m (whi h are real and distin t). Applying the previous argument to p m and q m instead of p0 and q 0 ; we obtain (

(

)

(

)

1)

(

)

(

)

jp m (

+1)

(x)j jq m (

+1)

(x)j

ut

at the zeros of q m : This ompletes the indu tion. (

)

2 Pn have n distin t zeros in ( 1; b); and suppose that in a strip of the omplex plane it satis es the inequality

Lemma 5.2.3. Let q

jq(x + iy)j jq(b + iy)j ; x 2 [a; b℄ ; Suppose also that p0 2 Pn satis es (5:2:4) jp0 (x)j jq0 (x)j

(5:2:3)

y 2 R:

1

whenever x is a zero of q. Then the derivatives of p and q satisfy (5:2:5)

jp m (x + iy)j jq m (b + iy)j ; (

)

(

x 2 [a; b℄ ; y 2 R :

)

Proof. First we show that at every point x + iy in the strip 0

jp0 (x Let

q(z ) :=

n Y j =1

(z

0

0

+ iy )j jq 0 (b + iy )j : 0

0

0 6= 2 R ;

j ) ;

j

2 ( 1 ; b) :

Let h(z ) be another polynomial with the same leading oeÆ ient as q and whose zeros are obtained by re e ting about x those zeros of q that lie to the right of x . Thus 0

0

h(z ) := where (5:2:6)

j :=

2x

j

0

n Y

j =1

(z

j ) ;

j if j > x if j x : 0

0


Then on the line z = x + iy; y 2 R; we have jz

j j = jz

0

jh(x

(5:2:7)

0

j j; so

+ iy )j = jq (x + iy )j : 0

We now show that

jp0 (x

(5:2:8)

0

+ iy )j jh0 (x + iy )j : 0

0

0

Note that n h0 (z ) X 1 = ; h(z ) j z j

(5:2:9)

=1

and re alling (5.2.2), we have n p0 (z ) X Æj = q(z ) j z j

(5:2:10)

=1

with Æj 2 [ 1; 1℄ for ea h j . Comparing the right-hand sides of (5.2.9) and (5.2.10), respe tively, at z = x + iy , we obtain 0

n X j =1

Æj x + iy 0

0

j

=

=

n X j =1 n X j =1 n X j =1

sin e by onstru tion jx

0

Æj (x j ) j ) + y

(x (x

x

0

x

0

0

0 p (x0 + iy0 ) q (x + iy ) 0

0

j

i i

j ) + y

2 0

2

0

j j = x

0

2 0

2

0

1 j + iy0 0

n X j =1

(x

0

n X j =1

(x

0

y0 Æj 2 2 j ) + y0 y0 2 2 j ) + y0

j : Therefore 0 h (x0 + iy0 ) h(x + iy ) 0

0

;

and (5.2.7) yields (5.2.8). Let 2 C ; jj < 1; be an arbitrary onstant and let

'(z ) := q(z ) h(z + x

0

Let

b) :

be the simple losed urve onsisting of a segment of the line

z = b + iy; y 2 R; and the portion of a ir le with enter at b and radius % that lies to the right of this line. Relations (5.2.3) and (5.2.7) show that on the line segment of ;


jq(z )j > jh(z + x

b)j :

0

If % is suÆ iently large, the same inequality is true for the ir ular portion of sin e q and h have the same leading oeÆ ient. Thus Rou he's theorem gives that ' and h have the same number of zeros inside : We on lude that ' has no zeros on or to the right of the line z = b + iy; y 2 R: The last statement, together with Theorem 1.3.1, implies that '0 has no zeros on the line z = b + iy; y 2 R: Thus for jj < 1;

q0 (b + iy ) 0

h0 (x + iy ) 6= 0 ; 0

0

whi h, together with (5.2.8), yields

jp0 (x

0

+ iy )j jh0 (x + iy )j jq 0 (b + iy )j : 0

0

0

0

This proves the lemma for m = 1: We turn now to the ase m > 1: Applying the lemma with m = 1 when p = q; we have

jq0 (x + iy)j jq0 (b + iy)j ;

x 2 [a; b℄ ; y 2 R :

Thus q 0 satis es all the requirements that are imposed on the interpolating polynomial q (with n repla ed by n 1) in the lemma, and by Lemma 5.2.2, jp00 (x)j jq 00 (x)j at the n 1 zeros of q 0 : Applying the lemma to p0 (x) instead of p in the ase m = 1, for whi h it has already been proved, we have jp00 (x + iy)j jq00 (b + iy)j ; x 2 [a; b℄ ; y 2 R ; whi h proves that inequality (5.2.5) is true for m = 2: Repetition of this argument ompletes the proof for larger values of m. ut Lemma 5.2.4. Suppose p 2 Pn and jp(j )j 1 for

j := os j n ;

j = 0; 1; : : : ; n :

Then

jp0 (xk )j p

1

n

xk 2

for xk := os

k

(2

2

1)

n

; k = 0; 1; : : : ; n :

For every xed k; equality holds if and only if p = Tn for some 2 C with j j = 1. Proof. Let f (x) = (1

n

Y x x )Tn0 (x) = a 2

j =0

os j n :


Then by the dierential equation for Tn (see E.3 e℄ of Se tion 2.1), we have f 0 (x) = xTn0 (x) n Tn(x) : Dierentiating the Lagrange interpolation formula for p (see E.6 of Se tion 1.1) gives n X p(j ) (x j )f 0 (x) f (x) : p0 (x) = f 0 (j ) (x j ) j 2

2

=0

For the zeros of Tn ; this redu es to

n X p(j ) (1 xj ) p0 (x) = Tn0 (x) f 0 (j ) (x j ) j

(5:2:11)

2

=0

sin e at these points (x j )f 0 (x)

j )Tn0 (x) (1 x )Tn0 (x) = (1 xj )Tn0 (x) : In the same way, we obtain for the zeros of Tn ; n X Tn (j ) (1 xj ) Tn0 (x) = Tn0 (x) f 0 (j ) (x j ) j and sin e Tn (j ) and f 0 (j ) are of opposite sign (f 0 (j ) = n Tn (j )), this f (x) = x(x

2

2

=0

2

gives

n X 1 (1 xj ) : Tn0 (x) = Tn0 (x) 0 f ( j ) (x j ) j

(5:2:12)

2

=0

Sin e jp(j )j 1 in (5.2.11), on omparing (5.2.11) and (5.2.12), we obtain for every zero x of Tn that

jp0 (x)j jTn0 (x)j = p

n

:

1 x For a xed zero of Tn ; the equality o

urs if and only if p(j ) = Tn (j ) ; j = 0; 1; : : : ; n for some 2 C ; j j = 1; that is, if and only if p = Tn for some 2 C with j j = 1. ut 2

Proof of Theorem 5.2.1. Suppose p = 6 Tn satis es the assumption of the theorem. Then by Lemma 5.2.4 there exists a onstant > 1 su h that jp0 (x)j jTn0 (x)j at the zeros of Tn: Applying Lemma 5.2.3 with p and q repla ed by p and Tn (assumption (5.2.3) is satis ed with [a; b℄ := [ 1; 1℄ by E.1 b℄), we obtain

jp m (x + iy)j (

)

1

(m) T (1 + iy ) ;

n

x 2 [ 1; 1℄ ; y 2 R

for every m = 1; 2; : : : ; n: If p = Tn ; then we have the same inequality with repla ed by 1: ut

1


Comments, Exer ises, and Examples. The inequality

kp m k (

)

[

Tnm (1) kpk (

;

1 1℄

)

;

[

1 1℄

; p 2 Pn

was rst proved by V. A. Markov [16℄. He was the brother of the more famous A. A. Markov who proved the above inequality for m = 1 in [1889℄ (see Theorem 5.1.8). However, their ingenious proofs are rather ompli ated. Bernstein presented a shorter variational proof of V. A. Markov's inequality in 1938 (see Bernstein [58℄, whi h in ludes a omplete list of Bernstein's publi ations). Our dis ussion in this se tion follows DuÆn and S haeer [41℄. E.1 A Property of Chebyshev Polynomials. a℄ Let (i )i n be a sequen e of 2n nonnegative numbers, and let (0i ) be a rearrangement of this sequen e a

ording to magnitude, 2 =1

0

1

0 0 n 0 : 2

2

Show that for every y 0, ( + y )( + y ) ( n n + y )

(5:2:13)

1

2

3

4

2

1

2

is not greater than (0 0 + y )(0 0 + y ) (0 n 0 n + y ) : 1

2

3

4

2

1

2

Proof. If and are at least as large as any of the remaining numbers i ; then 1

3

( + y )( + y ) 1

3

2

4

( + y )( + y ) = y ( 1

2

3

4

)(

1

4

3

) 0: 2

This shows that the numbers i in (5.2.13) an be rearranged so that the two largest o

ur in the same fa tor without de reasing (5.2.13). Then the two largest of the remaining numbers i may be brought into the same fa tor without de reasing (5.2.13), and so on. ut b℄ Show that the Chebyshev polynomials Tn de ned by (2.1.1) satisfy the inequality

jTn (x + iy)j jTn (1 + iy)j ;

x 2 [ 1; 1℄ ; y 2 R :

Proof. We have

jTn(x + iy)j

2

=

2

n Y j =1

((x

os j ) + y ) ; 2

2


where j = j n and = 2n : With x = os we write (2

1)

1

2

jTn (x + iy)j

2

=

2

n Y 1 i e 4

j =1

2

ij 2 ei

e

eij + y

2

:

Geometri ally, eij ; j = 1; 2; : : : ; n; represent 2n points equally distributed on the unit ir le. Conne t these points by hords to the point ei : Then the lengths of these 2n hords are given by jei eij j: If is in reased or de reased by any multiple of n ; we obtain a new set of hords, but the aggregate of their lengths is un hanged. Choose ' su h that n

' mod

n

;

2

'

n

2

:

If x = os '; then

Tn (x + iy) =

2

n Y 1 i' e

j =1

e

4

2

ij 2 ei'

eij + y

2

;

where the numbers jei' eij j are simply a rearrangement of the numbers jei eij j : Use part a℄ to show that 2

2

jTn (x + iy)j jTn (x + iy)j : Note that os x 1; where os is the right most zero of Tn : Hen e jTn (x + iy)j jTn (1 + iy)j ; and the proof is nished. ut E.2 Markov's Inequality for Higher Derivatives for Pn . a℄ Show that if p and q satisfy the onditions of Lemma 5.2.3 with p 2 Pn repla ed by p 2 Pn ; then 2

2

1

1

2

2

jp m (x)j jq m (b)j ; (

)

(

for every m 2 N : Hint: After dierentiating (5.2.2) m

p m (x) = (

)

n X j =1

Æj

x 2 [a; b℄

)

1 times, we obtain

dm dxm

1 1

q(x) ; x j

where jÆj j = jp0 (j )=q 0 (j )j 1: It is evident that if x 2 (a; b) is xed, then jp m (x)j attains its maximum when Æj = 1 for ea h j; in whi h ase p0 has real oeÆ ients. Now use Lemma 5.2.3. ut (

)


b℄ Show that if p satis es the assumption of Theorem 5.2.1 with p 2 Pn repla ed by p 2 Pn ; then

kp m k (

2 N:

for every m

j j = 1.

)

;

[

1 1℄

jTnm (1)j (

)

The equality an hold only if p = Tn for some

2 C; ut

Hint: Modify the proof of Theorem 5.2.1 by using part a℄.

℄

Show that (

1)(n 2 ) (n (m 1 3 5 (2m 1)

n (n 2

Tnm (1) = )

2

2

2

1) )

2

2

:

Hint: Dierentiating the se ond-order dierential equation for Tn (see E.3 e℄ of Se tion 2.1) m 1 times gives (1

x )Tnm 2

(

+2)

from whi h

(x)

(2m + 1)xTnm (

(2m + 1) Tnm (

+1)

+1)

(x) + (n

2

(1) = (n

2

m )Tnm (x) = 0 2

(

)

m ) Tnm (1) 2

(

)

ut

follows. Use indu tion and Tn (1) = 1 to nish the proof. d℄ The Main Inequality. Suppose p 2 Pn satis es p

os j n

1;

j = 1; 2; : : : ; n :

Show that for m = 1; 2; : : : ; n;

kp m k (

)

[

2

;

1 1℄

1)(n 2 ) (n (m 1 3 5 (2m 1)

n (n

2

2

2

2

1) ) 2

;

and the equality an o

ur only if p = Tn for some 2 C ; j j = 1: Hint: Combine parts ℄ and b℄. ut A slightly weaker version of Markov's inequality for higher derivatives is mu h easier to prove. E.3 A Weaker Version of Markov's Inequality. Show that

kp m k (

)

;

[

1 1℄

2mn m kpk 2

[

;

1 1℄

for every p 2 Pn : Hint: First show by a variational method that the extremal problem

6

jp0 (1)j p2Pn kpk ;

max

0=

[

1 1℄


is solved by the Chebyshev polynomial Tn ; hen e

jp0 (1)j n kpk 2

:

;

[

1 1℄

Then, by using a linear transformation, show that

jp0 (y)j 1 2 y n kpk y; 2n kpk 2

and

[

jp0 (y)j 1 +2 y n kpk 2

[

2

1℄

1

;y℄

;

[

2n kpk

1 1℄

2

[

1y0

;

;

1 1℄

0 y 1:

;

Hen e the inequality of the exer ise is proved when m = 1: For larger values of m use indu tion. ut E.4 Weighted Bernstein and Markov Inequalities. Let w 2 C [ 1; 1℄ be stri tly positive on [ 1; 1℄: a℄ Show that for every > 0 there exists an n depending on and w su h that p

0

p (x)w(x) 1 x ; n(1 + )kpwk ; 0

2

[

[

1 1℄

1 1℄

for every p 2 Pn ; n n : 0

Proof. By the Weierstrass approximation theorem, for every > 0 there is a q 2 Pk su h that w(x) q(x) (1 + )w(x) ;

x 2 [ 1; 1℄ :

Let m := minfw(x) : x 2 [ 1; 1℄g: Applying Theorem 5.1.7 (Bernstein's inequality) to pq 2 Pn k and then to q 2 Pk ; we obtain +

p p 0 p (x)w(x) 1 x2 p0 (x)q (x) 1 x2 p p (pq )0 (x) 1 x2 + p(x)q 0 (x) 1

(n + k)kpqk ; + kpk (n + k)(1 + )kpwk ; n(1 + )kpwk ; [

1 1℄

[

[

[

1 1℄

;

1 1℄

+

kkqk

1

m

[

x

2

;

1 1℄

kpwk

[

;

1 1℄

k(1 + )kwk

[

;

1 1℄

1 1℄

for every p 2 Pn ; provided > 0 is suÆ iently small and n n : 0

ut


b℄ Show that for every > 0 there exists an n depending on and w su h that kp0wk ; n (1 + )kpwk ; 0

[

2

1 1℄

[

1 1℄

for every p 2 Pn ; n n : Hint: Use the idea given in the previous proof. ut S haeer and DuÆn [38℄ prove an extension of Theorem 5.1.7 (Bernstein's inequality) to higher derivatives. They show that 0

m d dxm

dm exp(in ar

os x) ; m dx

p(x)

x 2 ( 1; 1)

for every p 2 Pn : The following exer ise gives a slightly weaker version of this, whi h is mu h simpler to prove. Some of this follows La han e [84℄. E.5 Bernstein's Inequality for Higher Derivatives. a℄ Show that there exists a onstant (m) depending only on m su h that

jp

(

j (m) p

m) (x)

1

n

m

x

kpk

[

2

;

1 1℄

x 2 ( 1; 1)

;

for every p 2 Pn : (That we an hoose (m) 2m is shown in parts ℄, d℄, and e℄.) Hint: For j = 1; 2; : : : ; m; let (m

aj := x

j )(1 + x) m

bj := x +

and

(m

j )(1 x) : m

Use Corollary 5.1.5 to show that there are onstants j (m) depending only on m su h that

kp j k aj ;bj j (m) p ( )

[

℄

1

n

x

kp j k aj (

2

1)

[

1

;bj

1℄

for every p 2 Pn and j = 1; 2; : : : ; m: ut b℄ Show that there exists a onstant (m) > 0 depending only on m su h that jp m (x)j (m) min n ; p n m sup 1 x 6 p2Pn kpk ; for every x 2 [ 1; 1℄ and m = 1; 2; : : : ; n: Hint: First show that (

)

2

0=

[

k

k

(m)

(1

jxj), x +

Tnm) I (x)

where I (x) := [x

(

1 2

2

1 1℄

min n ; p 2

1 2

(1

1

n

m

x

2

;

jxj)℄; then use a shift and a s aling. ut


In the rest of the exer ise we show that (m) = 2m is a suitable hoi e for the onstant in part a℄.

℄ Let k be a positive integer. Then

kp(x)(1

x )k

=

2 (

1) 2

k

[

;

1 1℄

n +k k kp(x)(1

x )k= 2

for every p 2 Pn : Hint: Let p 2 Pn be normalized so that kp(x)(1 x )k= Theorem 5.1.3 (Bernstein-Szeg}o inequality) with 2

t() := p( os ) sink 2 Tn

+

2

2

k

k

;

[

1 1℄

; = 1: Apply

[

1 1℄

k:

If x = os denotes a relative extreme point for p on ( 1; 1); then p0 ( os ) = 0; and after simpli ation we obtain that 0

0

0

(p( os

0

) sink

1

) 0

2

(n + k ) ((n + k ) k ) sin + k 2

2

2

2

0

2

n+k k

2

ut

d℄ Let k be a positive integer. Then

kp0(x)(1

x )k

=

2 ( +1) 2

k

[

;

1 1℄

:

2(n + k)kp(x)(1

x )k= 2

2

k

[

;

1 1℄

for every p 2 Pn :

Proof. Let p 2 Pn be normalized so that kp(x)(1

x )k= k ; = 1: Applying Theorem 5.1.4 (Bernstein's inequality) with m = 1 and t() := p( os ) sink 2 Tn

+

2

2

[

1 1℄

k;

we obtain

jp0 ( os ) sink

+1

+ p( os )k sink

1

os j n + k :

Now the triangle inequality and part ℄ yield

jp0 ( os ) sink

+1

j (n + k) + k p( os ) sink (n + k) + k n +k k 2(n + k) : 1

ut


e℄ Show that

jp

(

j

m) (x)

p 2n 1 x

m

kpk

[

2

;

1 1℄

for every p 2 Pn and x 2 ( 1; 1): Hint: Use indu tion on m; Theorem 5.1.7, and part d℄.

ut

5.3 Inequalities for Norms of Fa tors A typi al result of this se tion is the following inequality due to Kneser [34℄.

and r 2 P . Then Theorem 5.3.1. Suppose p = qr; where q 2 Pm n m

kqk

[

where

;

1 1℄

krk

[

;

1 1℄

Cn;m := 2m

1 2

Cn;m Cn;n

m Y

m

kpk

[

;

1 1℄

;

1 + os k n : (2

1)

2

k=1

Furthermore, for any n and m n the inequality is sharp in the ase that p is the Chebyshev polynomial Tn of degree n de ned by (2.1.1), and the

is hosen so that q vanishes at the m zeros of p losest to fa tor q 2 Pm 1. Before proving the above theorem, we establish an asymptoti formula for Cn;m and formulate a orollary. If f 2 C [a; b℄; then by the midpoint rule of numeri al integration 2

Z b

a

f (x) dx = (b

a)f

1 2

(a + b) +

(b

a) 00 f ( ) 3

24

for some 2 [a; b℄: Let f (x) := log(2 + 2 os x): Then

f 00 (x) =

2

(1 + os x)

2

:

On applying the midpoint rule to the above f; we obtain Z m=n 0

log(2 + 2 os x) dx =

m

1X

nk

log 2 + 2 os k n (2

1)

2

=1

+O m

n 1 1 24 n (n m) 4

3

4

5.3 Inequalities for Norms of Fa tors 261

for all integers 0 m n: Thus m Y

Cn;m = exp log

(5:3:1)

2 + 2 os

k=1 m X

nk

= exp O where

I (n; m) :=

mn (n m) 2

4

1)

n

log 2 + 2 os

=1

Z m=n

2

1

= exp

k

(2

!

k

(2

2

1)

n

!!n

(exp(I (n; m)))n ;

log(2 + 2 os x) dx :

0

So (5:3:2)

Cn;bn=

1=n

exp

2

!

Z 1=2

log(2 + 2 os x) dx = 1:7916 : : :

0

and (5:3:3)

Cn;b

2

n=3

1=n

!

Z 2=3

exp

log(2 + 2 os x) dx = 1:9081 : : : :

0

We use the notation an bn and an lim sup an =bn 1; respe tively.

. bn to mean nlim !1 an =bn = 1 and

n!1

On estimating Cn;m Cn;n m in Theorem 5.3.1 and using (5.3.2), we obtain the following: 1

2

Corollary 5.3.2. Let p

and r: Then

kqk

[

;

1 1℄

krk

[

2 Pn ;

1 1℄

and suppose p = qr for some polynomials q

1 2

bn= Y 2

2n

1

1 + os k n (2

1)

2

2

k=1

Cn;bn= kpk 2

2

[

kpk

[

;

1 1℄

;

1 1℄

and equality holds when p is the Chebyshev polynomial Tn of degree n, and

is hosen so that m := bn=2 and q vanishes at the m the fa tor q 2 Pm zeros of Tn losest to 1: Here Cn;=n bn= 3:20991 : : : ; hen e 2

2

kqk

[

;

1 1℄

kpk

[

krk

[

;

1 1℄

;

1 1℄

1=n

. 3:20991 : : : :


The proof of Theorem 5.3.1 pro eeds through a number of lemmas. For the remainder of the proof we assume that 0 < m < n are xed. Now

krk ; : kqrk ; = 1 ; q 2 Pm ; r 2 Pn mg

and r 2 P : We pro eed to show that there is attained for some q 2 Pm n m

and r 2 P are extremal polynomials q 2 Pm n m su h that p := qr is the (5:3:4)

supfkq k

[

;

1 1℄

[

1 1℄

[

1 1℄

Chebyshev polynomial Tn of degree of n; and that the fa tors q and r are as advertised, that is,

p(x) = (qr)(x) = Tn (x) =

n

1Y 2 x 2k

os k n ; (2

1)

2

=1

and the extremal fa tors, q and r, are given by

and

m Y

p1

q(x) :=

2k

os k n (2

1)

2

=1

n Y

p1

r(x) :=

2 x

os k n ;

2 x

(2

1)

2k m respe tively. Note that for the above q and r we have kqk ; = jq( 1)j = p1 Cn;m 2 and krk ; = jr(1)j = p1 Cn;n m : 2 =

[

[

2

+1

1 1℄

1 1℄

2 Pm

First we show that there exist extremal polynomials q r 2 Pn m su h that (5:3:5)

jq (

1)j = kq k

1 1℄

To see this, hoose , 2 [ 1; 1℄ su h that

jq()j = kqk

and

;

[

jr(1)j = krk

and

;

[

1 1℄

;

[

jr( )j = krk

[

;

1 1℄

1 1℄

and

:

;

where, onsidering q ( z ) and r( z ) if ne essary, we may assume that . Note that = annot happen, so < . We have kqk ; krk ; kqk ; krk ; kqrk ; kqrk ; [

℄

[

[

℄

[

1 1℄

℄

[

[

1 1℄

1 1℄

sin e the numerators are equal and

kqrk ; kqrk [

℄

[

;

1 1℄

:

Let qe 2 Pm be de ned by shifting q from [; ℄ to [ 1; 1℄ linearly so that ! 1: Let re 2 Pn m be de ned by shifting r from [; ℄ to [ 1; 1℄ linearly

and re 2 P so that ! 1: Then qe 2 Pm n m are extremal polynomials for whi h (5.3.5) holds.


and r 2 Pn m are extremal polynomials

and for whi h (5.3.5) holds. Then there are extremal polynomials qe 2 Pm

re 2 Pn m having only real zeros for whi h (5.3.5) holds. Lemma 5.3.3. Suppose q

2 Pm

Proof. Let qe(z ) be de ned by repla ing every fa tor z with nonreal by z (j +1j 1) in the fa torization of q: Let re(z ) be de ned by repla ing every fa tor z with nonreal by z (1 j 1j) in the fa torization

and re 2 P of r: Now it is elementary geometry to show that qe 2 Pm n m are extremal polynomials for whi h (5.3.5) holds, and all the zeros of both qe and re are real. ut

and r 2 P Lemma 5.3.4. Suppose q 2 Pm n m are extremal polynomials having only real zeros for whi h (5.3.5) holds. Then there are extremal poly and re 2 P nomials qe 2 Pm n m having all their zeros in [ 1; 1℄ for whi h (5.3.5) holds.

Proof. Let qe(z ) be de ned by repla ing every fa tor z by z 1 if > 1, and by 1 if < 1; in the fa torization of q: Let re(z ) be de ned by repla ing every fa tor z by z + 1 if < 1; and by 1 if > 1: Now it is again

and qe 2 P elementary geometry to show that qe 2 Pm n m are extremal polynomials having all their zeros in [ 1; 1℄ for whi h (5.3.5) holds. ut

and r 2 P So we now assume that q 2 Pm n m are extremal polynomials having all their zeros in [ 1; 1℄ for whi h (5.3.5) holds. We may also assume that deg(q ) = m and deg(r) = n m; otherwise we would study

qe(z ) := z m and

re(z ) := z n

m

q q (z )

2 Pm

r r(z )

2 Pn

deg( )

deg( )

m;

whi h are also extremal polynomials having all their zeros in [ 1; 1℄ for whi h (5.3.5) holds. It is now lear that if q and r are extremal polynomials with the above properties, then the smallest zero of q is not less than the largest zero of r: Indeed, if there were numbers 1 < 1 so that q () = r( ) = 0; then the polynomials

qe(z ) := q(z ) and

z z

2 Pm

z 2 Pn m z would ontradi t the extremality of q and r sin e re(z ) := r(z )


kqek krek

;

jqe(

;

jre(1)j > jr(1)j = krk

[

1 1℄

[

1 1℄

and

1)j > jq ( 1)j = kq k

kqerek

; =

[

1 1℄

kqrk

;

[

1 1℄

So now we have extremal polynomials q form

p

m Y

q(z ) = a

(z

k )

k=1

;

[

[

1 1℄

;

1 1℄

;

:

2 Pm and r 2 Pn p

r(z ) = a

and

;

nYm k=1

(z

m of the

k )

satisfying

jq (

1)j = kq k

[

jr(1)j = krk

and

;

1 1℄

[

;

1 1℄

;

where 1

n

m 1 and the onstant a > 0 is hosen so that for p := qr we have kpk 1

2

m

1

2

[

Now we are ready to prove Theorem 5.3.1.

; = 1:

1 1℄

Proof of Theorem 5.3.1. We show three properties of p = qr: (1) jp( 1)j = 1 and kp(1)j = 1: (2) kp(x)k i ;i+1 = 1; i = 1; 2; : : : ; n m 1, kp(x)k i; i+1 = 1; i = 1; 2; : : : ; m 1. (3) kpk n m ; 1 = 1: These three fa ts show that p is indeed the Chebyshev polynomial Tn de ned by (2.1.1) sin e Tn are the only polynomials of degree at most n that equios illate n + 1 times on [ 1; 1℄ with uniform norm 1: To prove (1), assume to the ontrary that jp( 1)j < 1: Then there is a Æ < 1 su h that kpk Æ; = kpk ; : Sin e jq j is stri tly de reasing on [Æ; 1℄; [

℄

[

℄

[

℄

[

1℄

[

kqk Æ; jq(Æ)j > jq( [

and, of ourse,

1)j = kq k

1℄

krk Æ; krk [

1℄

1 1℄

[

[

;

1 1℄

:

;

1 1℄

and re 2 P Let qe 2 Pm n m be the polynomials q and r shifted linearly from [Æ; 1℄ to [ 1; 1℄ so that 1 7! 1: By the previous observations, these qe


and re ontradi t the extremality of q and r; hen e jp( 1)j = 1; and we are nished. A similar argument shows that jp(1)j = 1: To prove (2) let

s(z ) := (z

i )(z

i ) : +1

Given > 0; we an nd 0 < Æ ; Æ < su h that 1

t(z ) := (z satis es

2

(i

Æ ))(z

kt

t(1) = s(1) ;

and

jt(x)j < js(x)j ; Suppose kpk i ;i < 1: Let [

(i

1

sk

;

[

x 2 [ 1; i

+1

1 1℄

+ Æ )) 2

< ;

Æ ℄ [ [i 1

+ Æ ; 1) :

+1

2

+1 ℄

qb(z ) := q(z ) 2 Pm

and

rb(z ) := r(z )

t(z ) s(z )

2 Pn

m:

If > 0 is suÆ iently small, then

kqbrbk

[

;

1 1℄

kqrk

[

j(qbrb)(

and

;

1 1℄

1)j < 1 :

The se ond inequality guarantees that there exists a Æ < 1 su h that

kqbrbk Æ; [

1℄

= kqbrbk

[

;

1 1℄

1; 1℄; 1)j = jq (

:

Sin e jqbj is (stri tly) de reasing on (

kqbkÆ; jqb(Æ)j > jqb( 1℄

Also

krbk Æ; krbk

and re 2 P Now let qe 2 Pm n [

1℄

[

;

1 1℄

1)j = kq k

jrb(1)j = jr(1)j = krk

[

1 1℄

;

:

[

1 1℄

;

:

m be the polynomials pb and qb shifted linearly from [Æ; 1℄ to [ 1; 1℄ so that 1 ! 1: By the previous observations, these qe and re ontradi t the extremality of q and r: Hen e kpk i;i+1 = 1; and the proof is nished. The proof of kpk i; i+1 = 1 is identi al. To prove (3) assume that kpk n m; 1 < 1: Let [

[

[

p

qe(z ) := a(z

℄

℄

( + )) 1

m Y k=2

(z

k )

℄


and

p

re(z ) := a(z

(n m

))

nY m

1

k=1

(z

k ) :

If > 0 is suÆ iently small, then

kqek

[

1 1℄

;

jqe(

krek

[

1 1℄

;

jre(1)j > jr(1)j = krk

and

1)j > jq ( 1)j = kq k

;

[

kqerek

;

[

1 1℄

kqrk

[

;

1 1℄

[

1 1℄

;

1 1℄

;

;

;

whi h ontradi ts the extremality of q and r: Hen e kpk n deed. [

m; 1 ℄

= 1; in-

ut

is a moni fa tor of Theorem 5.3.5. Suppose p 2 Pn is moni and q 2 Pm

p: Then

jq (

)j m n 2n

1

m Y

1 + os k n (2

1)

2

k=1

kpk

[

; ℄

for every > 0: Equality holds if p is the Chebyshev polynomial Tn; of

degree n on [ ; ℄ (normalized to be moni ), and the moni fa tor q 2 Pm is hosen so that q vanishes at the m zeros of Tn; losest to . Note that Tn; (x) = n Tn (x= ); where Tn is de ned by (2.1.1). The proof of Theorem 5.3.5 is outlined in E.1.

is a moni fa tor of Corollary 5.3.6. Suppose p 2 Pn is moni and q 2 Pm

p: Then

jq (

2)j 2m

1

m Y k=1

1 + os k n (2

1)

2

kpk

[

; =

2 2℄

1 C kpk 2 n;m

and the inequality is sharp for all m n: Here, for all m n; =n C =n Cn;m n;b n= 1:9081 : : : ; 1

1

2

and hen e

jq ( kpk

2)j [

;

2 2℄

3

1=n

. 1:9081 : : : :

[

;

2 2℄


Proof. Take = 2 in Theorem 5.3.5. Note that 2m

m Y k=1

1 + os k n (2

1)

2 n (

2

)

Y (n )

1 + os k n ; (2

1)

k=1

2

where (n)+1 := b n + is the smallest k 2 N for whi h os k n So, for every m = 0; 1; : : : ; n; 2

3

3

2

(2

Cn;m Cn; n (

and by (5.3.3)

1)

2

=n . C Cn;m n;b 1

2

2

:

)

. 1:9081 : : : :

n=3

1

ut

2 Pn is moni and has a moni fa tor of the

: Then form qr; where q 2 P and r 2 Pm Theorem 5.3.7. Suppose p

m1

jq (

)jjr( )j m1

+

2

m2 n 2n

1

m1 Y

k=1

1 + os k n (2

1)

m2 Y

2

k=1

1 + os k n (2

1)

2

kpk

[

; ℄ :

Equality holds if p is the Chebyshev polynomial Tn; of degree n on [ ; ℄

and r 2 P are hosen normalized to be moni , and the fa tors q 2 Pm m2 1 so that q vanishes at the m zeros of Tn; losest to ; while r vanishes at the m zeros of Tn; losest to : 1

2

The proof of Theorem 5.3.7 is analogous to the proof of Theorem 5.3.1 and is left as an exer ise (see E.2).

2 Pn is moni and has a moni fa tor of the qj ; where qi 2 Pm i and m := m + m + + mj n. Then

Theorem 5.3.8. Suppose p

form q q 1

j Y i=1

kqi k

[

2

1

; ℄

m

n 2n

1

bm= Y 2

k=1

2

1 + os

k

(2

2

1)

n

dm= Y2e

k=1

1 + os k n (2

1)

2

kpk

[

; ℄

for every > 0. Equality holds if p is the Chebyshev polynomial Tn; of degree n on [ ; ℄ normalized to be moni , j = 2; and the fa tors q 2 Pb m= and q 2 Pd m= e are hosen so that q vanishes at the bm=2 zeros of Tn; losest to ; while q vanishes at the dm=2e zeros of Tn;

losest to : 1

2

2

1

2

2


Proof. For ea h qi , write

qi = ri si ;

where ri is the moni fa tor of qi omposed of the roots of qi with negative real part, while si is the moni fa tor of qi omposed of the roots of qi with nonnegative real part. So

kri k

[

Thus

; ℄ = j Y i=1

jri ( )j

kqi k

[

; ℄

and j Y

i=1

ksi k

[

j Y

ri ( )

i=1

; ℄ =

jsi (

)j :

si ( ) :

We now apply Theorem 5.3.7 to the two fa tors nish the proof.

Qj

i=1 ri and

Qj

i=1 si to

ut

As before, let D := fz 2 C : jz j < 1g: We now derive inequalities on the disk from those on the interval. A ontinuous fun tion on D has the same uniform norm on both D and D and it is notationally onvenient to state

; and v 2 P the remaining theorems over D. Suppose t 2 Pn , s 2 Pm n m are moni , and t = sv: By the maximum prin iple, t, s; and v a hieve their maximum on D somewhere on D. Now onsider

p(x) := t(z )t(z ) ; 1

q(x) := s(z )s(z ) ; with

r(x) := v(z )v(z )

and

1

x := z + z

1

1

:

The ee t of this transformation on linear fa tors is (z

)(z

) = x + 1 + ;

1

2

, r 2 P ; and p = qr: Also so p 2 Pn , q 2 Pm n m

kpk

[

;

2 2℄

ktkD : 2

If t(0) 6= 0; then the modulus of the leading oeÆ ient of p is jt(0)j; while the modulus of the leading oeÆ ient of q is js(0)j; and the modulus of the leading oeÆ ient of r is jv (0)j: From these transformations and the interval inequalities we an dedu e the next three theorems.


Theorem 5.3.9. Let t

and v 2 P

n m:

Then

2 Pn

be moni and suppose t = sv; where s

jv(0)j = kskD 1 2

1 2

Cn;m

1=2

2 Pm

ktkD ;

where Cn;m is the same as in Theorem 5.3.1. This bound is attained when

vanishes at m adja ent zeros m n are even, t(z ) = z n + 1; and s 2 Pm of t on the unit ir le. Proof. We may assume, by performing an initial rotation if ne essary, that

kskD = js(

1)j :

So from Corollary 5.3.6 we dedu e that (5:3:6)

kskD = js( 1)j = jq( js(0)=t(0)j2m 2

2)j

2

m Y

1

js(0)=t(0)j2m

1 + os k n

k=1

1)

2

k=1 m Y

1

(2

1 + os k n (2

1)

2

kpk

[

;

2 2℄

ktkD ; 2

ut

where s(0)=t(0) = 1=v (0): Theorem 5.3.10. Suppose t = sv, where s 2 Pm and r 2 Pn m . Then

kskD kvkD

1 2

Cn;m Cn;n

m

1=2

ktkD ;

where Cn;m is the same as in Theorem 5.3.1 and (Cn;m Cn;n m ) = n 1 (2

)

Cn;=nbn= 1:7916 : : : : 1

2

This bound is attained when m n are even, t(z ) = z n + 1; and s 2 Pm vanishes at the m zeros of t losest to 1 and v 2 Pn m vanishes at the n m zeros of t losest to 1:

Proof. From Theorem 5.3.1 we an dedu e that if a; b 2 D; then

js(a)j jv(b)j 2

2

= js(a)s(a )jjv (b)v (b )j = jq (a + a )jjr(b + b )j

1 2 1 2 1 2

1

1

1

1

Cn;m Cn;n Cn;m Cn;n Cn;m Cn;n

kpk a a m kpk ; m ktkD ; m

[ + [

1

;b+b

1℄

2 2℄

2

where, without loss of generality, we may assume that a + a b + b . The result now follows on hoosing a and b to be points on D where s and v, respe tively, a hieve their uniform norm on D: ut 1

In the multifa tor ase we have the following theorem:

1


2 Pn is of the form t = vs s sj ; where with m := m + m + + mj n: Then

Theorem 5.3.11. Suppose t

si 2 Pmi and v 2 Pn

j

j

v(0) =

1 2

j Y i=1

m

ksi kD 2 m (

0

1

1

=

1) 2

bm= Y

2

k=1

2

2

1 + os

k

(2

2

1)

n

11=2

dm= Y2e

k=1

1 + os

k

(2

2

n

A

1)

ktkD :

Equality holds if t(z ) = z n + 1, j = 2, m = m := m=2 and n are even, and the fa tors q 2 Pm= and q 2 Pm= are hosen so that q vanishes at the m=2 zeros of t losest to 1 and q vanishes at the m=2 zeros of t losest to 1: 1

1

2

2

2

2

1

2

Proof. This follows from Theorem 5.3.8 in exa tly the same way as Theorem

ut

5.3.10 follows from Theorem 5.3.1.

Comments, Exer ises, and Examples. The rst result of this se tion is due to Kneser [34℄ and in part to Aumann [33℄. The proof follows Borwein [94℄, as does most of the se tion. There are many variations and generalizations. See Boyd [92℄, [93a℄, [93b℄, [94a℄, and [94b℄; Beauzamy and En o [85℄; Beauzamy, Bombieri, En o, and Montgomery [90℄; Gel'fond [60℄; Glesser [90℄; Granville [90℄; Mahler [60℄, [62℄, and [64℄; and Mignotte [82℄. Some of these are presented in the exer ises. In parti ular, E.6 reprodu es a very pretty proof of Boyd [92℄ that

kqkD krkD (1:7916 : : : )n kpkD ;

and r 2 P : (Note that we where p 2 Pn and p = qr with some q 2 Pm n m have not assumed real oeÆ ients unlike in Theorem 5.3.10, and we have instead of . :)

E.1 Proof of Theorem 5.3.5.

Outline. Let m < n and > 0 be xed. The value

sup

jq ( kpk

[

)j ; ℄

: q2P

m and

p2P

n are moni and

q divides p

and p 2 P . We an now argue, exa tly is attained for some moni q 2 Pm n as in the proof of Lemma 5.3.3, that there are extremal polynomials p 2 Pn

su h that all the zeros of p are real and lie in [ ; 1): Arguing and q 2 Pm as in Lemma 5.3.4 gives that p has all its roots in [ ; ℄: Thus q must be

omposed of the m roots of p losest to :


The argument of the proof of Theorem 5.3.1 now applies essentially verbatim and proves that an extremal p 2 Pn an be hosen to be the Chebyshev polynomial on [ ; ℄ normalized to be moni . Thus on [ 1; 1℄

p(x) =

n Y k=1

x

os k n (2

1)

2

m Y

q(x) =

and

k=1

os k n

x

(2

1)

2

from whi h the result follows (on onsidering n p(x= ) and m q (x= ) on [ ; ℄). ut E.2 Proof of Theorem 5.3.7. Hint: Pro eed as in the proof of Theorem 5.3.1 (or E.1).

ut

E.3 A Version of Theorem 5.3.10 for Complex Polynomials. Suppose

and v 2 P : Then t = sv; where s 2 Pm n m

js(

1)jjv (1)j

1 2

Cn;m Cn;n

m

1=2

ktkD

and if t is moni

jt(0)j = kskD kvkD 1 2

1 2

(Cn;m Cn;n m ) = ktkD : 1 2

2

Hint: The rst inequality follows as in the proof of Theorem 5.3.10 with a := 1 and b := 1: The se ond part is immediate from Theorem 5.3.9. u t E.4 Mahler's Measure. Let F : C k F be de ned by

Mk (F ) := exp

Z

1

0

Z

1

! C , and let the Mahler measure of

log jF (e it1 ; : : : ; e itk )j dt 2

2

0

1

dtk

if the integral exist. a℄ Show that if

p(z ) = then

n Y i=1

(z

M (p) = j j 1

i ) ; n Y i=1

; i 2 C ;

maxf1; ji jg :

Hint: Use Jensen's formula (see E.10 ℄ of Se tion 4.2). b℄ Show that if F := F (z ; : : : ; zk ) and G := G(z ; : : : ; zk ); then 1

1

Mk (F G) = Mk (F )Mk (G) :

ut


℄ Show that

M (ax + b) = max(jaj; jbj) ; M (1 + x + y) = M (maxf1; j1 + xjg) ; M (1 + x + y xy) = M (maxfj1 xj; j1 + xjg) : 1

2

1

2

1

d℄ One an numeri ally he k that

M (1 + x + y) = 1:381356 : : : 2

and

M (1 + x + y 2

xy) = 1:791622 : : : :

E.5 The Norm of a Fa tor of a p 2 Pn on the Unit Disk. Suppose p 2 Pn

. Then is moni and has a moni fa tor q 2 Pm

kqkD n kpkD ;

where := M (1 + x + y ) = 1:3813 : : : : 2

Outline. Let p(x) :=

n Y i=1

(x

i )

and

q(x) :=

Suppose kq kD = jq (u)j; where u 2 D: Then m Y

kqkD = jq(u)j =

i=1

ju

i j

n Y

i=1

m Y

(x

i=1

maxfju

i ) ;

i 2 C :

i j; 1g

= M (p(x + u)) M (maxf1; jx + ujn gkpkD ) ; 1

1

where the last equality holds by E.4 a℄, and the last inequality follows be ause

jp(z )j maxf1; jz jng kpkD holds for every p 2 Pn and z 2 C by E.18 a℄ of Se tion 5.1. Now using E.4 (5:3:7)

b℄ and

M (maxf1; jx + ujg) = M (maxf1; jx + 1jg) ; 1

1

we obtain

kqkD M (maxf1; jx + ujng) kpkD = M ((maxf1; jx + ujg)n ) kpkD = M ((maxf1; jx + 1jg)n ) kpkD = (M (maxf1; jx + 1jg))n kpkD = (M (1 + x + y ))n kpkD = n kpkD : 1

1

1

1

2

ut


E.6 Another Inequality for the Fa tors of a p 2 Pn on the Unit Disk. Let

and r 2 P : Then p = qr; where q 2 Pm n m

kqkD krkD Æn kpkD where Æ := M (1 + x + y

xy) = 1:7916 : : : .

2

Outline. Without loss of generality we may assume that q and r are moni . Let

q(x) :=

m Y

(x

i=1

i )

r(x) :=

and

n Y

(x

i 2 C :

i ) ;

i=m+1

Choose u 2 D and v 2 D su h that jq (u)j = kq kD and jr(v )j = krkD : Then, using E.4 b℄ and ℄, we obtain

kqkD krkD = jq(u)jjr(v)j = =

m Y i=1 n Y i=1 n Y i=1

ju

i j

n Y

maxfju

ju

jv

i j

i j; jv

i jg

i=m+1

i j max (x

1)n p

max jx

1j; x

=M

1

v 1; u

ux v x 1

i i

:

Now, by (5.3.7), (x

1)n p

xu v x 1

n

v on kpkD ; u

hen e

kqkD krkD M ((maxfjx = (M (maxfjx (M (maxfj1

1j; jx v=ujg)n )kpkD 1j; jx v=ujg))n kpkD xj; j1 + xjg))n kpkD = (M (1 + x + y xy ))n kpkD = (1:7916 : : : )n kpkD : 1

1

1

2

ut


E.7 Bombieri's Norm. For Q(z ) := de ned by n X

[Q℄p :=

Pn

k k=0 ak z the

1 p

n k

jak jp

!1=p

k=0 Note that this is a norm on n for every

Bombieri p norm is

:

p 2 [1; 1); but it varies with P varying n: The following remarkable inequality holds (see Beauzamy et al.

, and S 2 P ; then [90℄). If Q = RS with Q 2 Pn , R 2 Pm n m [R ℄ [S ℄ 2

2

1=2

n m

[Q℄

2

and this is sharp. One feature of this inequality is that it extends naturally to the multivariate ase. See Beauzamy, En o, and Wang [94℄ and Rezni k [93℄ for further dis ussion.

This is page

275


6

Inequalities in Muntz Spa es

Overview Versions of Markov's inequality for Muntz spa es, both in C [a; b℄ and Lp [0; 1℄, are given in the rst se tion of this hapter. Bernstein- and

Nikolskii-type inequalities are treated in the exer ises, as are various other inequalities for Muntz polynomials and exponential sums. The se ond se tion provides inequalities, in luding most signi antly a Remez-type inequality, for nondense Muntz spa es.

6.1 Inequalities in Muntz Spa es We rst present a simpli ed version of Newman's beautiful proof of an essentially sharp Markov-type inequality for Muntz polynomials. This simpli ation allows us to prove the Lp analogs of Newman's inequality. Then, using the results of Se tion 3.4 on orthonormal Muntz-Legendre polynomials, we prove an L version of Newman's inequality for Muntz polynomials with omplex exponents. Some Nikolskii-type inequalities for Muntz polynomials are studied. The exer ises treat a number of other inequalities for Muntz polynomials and exponential sums. Throughout this se tion we use the notation introdu ed in Se tion 3.4. Unless stated otherwise, the span always denotes the linear span over R: 2

276 6. Inequalities in Muntz Spa es

Theorem 6.1.1 (Newman's Inequality). Let := (i )1 be a sequen e of i =0

distin t nonnegative real numbers. Then n 2X j 3j

kxp0 (x)k kpk ;

sup

6 p2Mn

0=

=0

;

[0 1℄

(

[0 1℄

)

9

n X j =0

j

for every n 2 N ; where Mn() := spanfx0 ; x1 ; : : : ; xn g: Proof. It is equivalent to prove that n

2X j 3j

(6:1:1)

=0

sup

;1) [0;1) [0

6 P 2En

0=

kP 0k kP k

(

)

9

n X j =0

j ;

where En () := spanfe 0 t ; e 1 t ; : : : ; e n t g: Without loss of generality we may assume that := 0: By a hange of s ale we may also assume that P n j j = 1: We begin with the rst inequality. We de ne the Blas hke produ t 0

=0

B (z ) := and the fun tion (6:1:2)

T (t) :=

1 2i

Z

n Y

j =1

z j z + j

e zt dz ; B (z )

:= fz 2 C : jz

1j = 1g :

By the residue theorem

T (t) := and hen e T

n X j =1

(B 0 (j )) e j t ; 1

2 En (): We laim that jB (z )j 13 ;

(6:1:3)

Indeed, it is easy to see that 0 j z j z +

j

22 + j j

z2 :

1 implies

=

j 1 ; 1 + j 1

z2 :

2

1

2

So, for z 2 ;

jB (z )j

n Y j =1

1 j 1 + j 1

2

1

2

1 1+

1 2

1 2

n P j =1 n P j =1

j j

=

1 1+

1 2 1 2

=

1 : 3

6.1 Inequalities in Muntz Spa es 277

Here the inequality 2xy (x + y ) 1 x 1 y 1 (x + y ) = + 1 + x 1 + y 1 + (x + y ) (1 + x)(1 + y )(1 + (x + y )) 11 + ((xx ++ yy)) ; x; y 0 is used. From (6.1.2) and (6.1.3) we an dedu e that (6:1:4)

jT (t)j

Also

1 2

Z zt e B (z )

1 T 0(t) = 2i

and (6:1:5)

1 2i

T 0 (0) =

Z

jdz j 21 3(2) = 3 ;

Z

t 0:

ze zt dz B (z ) Z

z 1 dz = B (z ) 2i jzj

=1

z dz : B (z )

Now, for jz j > max j n j ; we have the Laurent series expansion 1

n n 1 k ! Y Y X 1 + j =z z j =z =z 1+2 (6:1:6) B (z ) 1 =z z j j j k =1

0

n X

= z 1 + 2

j =1

= z + 2 + 2z

1

!

=1

j z

+ ;

1

=1

+2

n X j =1

j

!2

z

2

1

+ A

whi h, together with (6.1.5), yields that T 0 (0) = 2: Hen e, by (6.1.4), n jT 0 (0)j 2 = 2 X kT k ;1 3 3 j j ; [0

)

=1

so the lower bound of the theorem is proved. To prove the upper bound in (6.1.1), rst we show that if 1 U (t) := 2i

Z

(1

e

zt

z )B (z )

then (6:1:7)

Z 0

1

dz ;

:= fz 2 C : jz

jU 00 (t)j dt 6 :

1j = 1g ;


Indeed, observe that if z = 1 + ei ; then jz j = 2 + 2 os ; so (6.1.3) and Fubini's theorem yield that 2

Z 0

1

1 1 Z z e zt dt dz 2 (1 z )B (z ) Z Z 1 1 jz j je zt j 2 jB (z )j d dt Z 1Z 3 (2 + 2 os )e 2

Z

jU 00 (t)j dt =

0

2

0

2

0

2

(1+ os

0

Z 2

3 2

= Now we show that

1

Z

(6:1:8)

2

(2 + 2 os )

0

1 d = 6 : 1 + os

j t U 00 (t) dt = j

e

0

)t d dt

0

3:

To see this we write the left-hand side as Z

1

e

0

j t U 00 (t) dt = Z

1

0

1Z z e

1 2i Z 1 = 2i jzj

=

Z

0

=2

2

j t 1

e

Z

z e zt dz dt (1 z )B (z ) 2

2i

z j )t

( +

dz dt =

z )B (z ) 1 z z dz ; z + j 1 z B (z ) (1

Z

1 2i

z

2

(z + j )(1

z )B (z )

dz

where in the third equality Fubini's theorem is used again. Here, for jz j > 1; we have the Laurent series expansions

z = 1 j z z + j z = 1 z 1 z

1

+ j z 2

z

1

2

+ ;

;

2

and, as in (6.1.6), 1

B (z )

= 1 + 2z

1

+ 2z

2

+ :

Now (6.1.8) follows from the residue theorem (see, for example, Ash [71℄). Let P 2 En () be of the form

P (t) =

n X j =0

j e

j t ;

j

2 R:


Then Z

1

P (t + a)U 00 (t) dt =

0

n 1X

Z

j =0

0

=

n X j =0

j a

j e

Z

1

e

j t U 00 (t) dt =

Z

1

n X j =0

0

= P 0 (a)

3P (a)

and so

jP 0 (a)j 3jP (a)j +

(6:1:9)

j a e j t U 00 (t) dt

j e

0

3)e j a

j (j

jP (t + a) U 00 (t)j dt :

Combining this with (6.1.7) gives

kP 0 k

;1)

[0

3 kP k

;1) + 6

[0

kP k

;1) = 9

[0

kP k

;1) ;

[0

ut

and the theorem is proved.

The next theorem establishes an Lp extension of Newman's inequality. Theorem 6.1.2 (Newman's Inequality in Lp ). Let p 2 [1; 1): If := (i )1 i is a sequen e of distin t real numbers greater than 1=p; then

=0

0

kxP 0 (x)kLp

;

[0 1℄

p1 + 12

n X j =0

j +

1

1 A

p

kP kLp

;

[0 1℄

for every P 2 Mn () := spanfx0 ; x1 ; : : : ; xn g : If := ( i )1 i is a sequen e of distin t positive real numbers, then =0

kP 0 kL

p[0;1)

for every P

2 En (

12

n X j =0

!

j

kP kLp

;1)

[0

) := spanfe 0 t ; e 1 t ; : : : ; e n t g:

Proof. First we show that the rst statement of the theorem follows from the se ond. Indeed, if (i )1 i is a sequen e of distin t real numbers greater than 1=p and i := i + p for ea h i; then ( i )1 i is a sequen e of distin t positive real numbers. Let Q 2 Mn (): Applying the se ond inequality with =0 1

=1

P (t) := Q(e t )e

t=p

2 En (

and using the substitution x = e t , we obtain

)

280 6. Inequalities in Muntz Spa es Z

1

0

0 p x x1=p Q(x) x

1

dx

1=p

n X

12

j =0

j +

1

!

p

kQkLp

;

[0 1℄

:

Now the produ t rule of dierentiation and Minkowski's inequality yield 0

kxQ0 (x)kLp

;

[0 1℄

1

p

+ 12

n X j =0

j +

1

1

kQkLp

A

p

;

[0 1℄

;

whi h is the rst statement of the theorem. We prove the se ond statement. Let P 2 En ( ) and p 2 [1; 1) be xed. As in the proof of Theorem P 6.1.1, by a hange of s ale, without loss of generality we may assume that nj j = 1: It follows from (6.1.9) and Holder's inequality (see E.7 a℄ of Se tion 2.2) that =0

jP 0 (a)jp 2p

1

3p

j

j

P (a) p +

6p jP (a)jp + 2p

1

Z

1

j

0

1

Z 0

j

P (t + a) p

p

jP (t + a)jU 00 (t)j dt

1

1=p Z

jU 00 (t)j dt

0

1=q !p

jU 00 (t)j dt

for every a 2 [0; 1), where q 2 (1; 1℄ is the onjugate exponent to p de ned by p + q = 1: Combining the above inequality with (6.1.7), we obtain 1

1

jP 0 (a)jp 6pjP (a)jp + 2p

1

6p=q

1

Z 0

jP (t + a)jp jU 00 (t)j dt

for every a 2 [0; 1): Integrating with respe t to a; then using Fubini's theorem and (6.1.7), we on lude that

kP 0 kpLp

;1)

[0

6pkP kpLp kP k 6p

p 6p=q ;1 + 2

[0

)

p p Lp [0;1) + 2

6pkP kpLp 6pkP kpLp

Z

1

1

6p=q

1Z 1

Z0 1 Z0 1 0

0

jP (t + a)jp jU 00 (t)j dt da jP (t + a)jp jU 00 (t)j da dt

p 6p=q kP kp ;1 + 2 Lp ;1

Z

1

[0

)

[0

)

k k k k

0

1

jU 00 (t)j dt

p 1 6p=q+1 P p ;1) + 2 Lp [0;1) p p p 1 p p = (6 + 2 6 ) P Lp[0;1) 12 P pLp [0;1) [0

and the proof is nished.

k k

ut

The following Nikolskii-type inequality follows from Theorem 6.1.1 quite simply:


Theorem 6.1.3 (Nikolskii-Type Inequality). Suppose 0 < q < p 1: If := (i )1 i is a sequen e of distin t real numbers greater than 1=q; then =0

ky

1

=q

1

kLp

=p P (y )

;

[0 1℄

18 2q

n X j =0

1

j +

!1=q

1

=p

q

kP kLq

;

[0 1℄

for every P 2 Mn () := spanfx0 ; x1 ; : : : ; xn g: If := ( i )1 i is a sequen e of distin t positive real numbers, then =0

kP kLp

(6:1:10)

2 En (

for every P

;1)

[0

18 2q

n X j =0

!1=q

1

=p

j

kP kLq

;1)

[0

) := spanfe 0 t ; e 1 t ; : : : ; e n t g:

Proof. First we show that the rst statement of the theorem follows from is a sequen e of distin t real numbers greater than the se ond. If (i )1 i is a sequen e of distin t 1=q and i := i + 1=q for ea h i; then ( i )1 i positive real numbers. Let Q 2 Mn (): Applying (6.1.10) with =0

=1

P (t) := Q(e t )e

t=q

2 En (

)

and using the substitution x = e t ; we obtain

ky =q 1

1

kLp

=p Q(y )

;

[0 1℄

(18 2q )1=q

1

=p

n X j =0

j +

1

!1=q

q

1

=p

kQkLq

;

[0 1℄

;

whi h is the rst statement of the theorem. It is suÆ ient to prove (6.1.10) when p = 1; and then a simple argument gives the desired result for arbitrary 0 < q < p < 1: To see this, assume that there is a onstant C so that

kP k

C =q kP kLq ) and 0 < q < 1: Then ;1)

1

[0

for every P

2 En (

kP k

Z

1

;1)

[0

jP (t)jp q q dt kP kp ;1q kP kqLq C p=q kP kpLq q ;1 kP kqLq ;1

p Lp[0;1) =

+

[0

0

)

1

[0

and therefore for every f

2 En (

kP kLp

℄

[0

C =q =p kP kLq ) and 0 < q < p 1. ;1)

[0

1

1

)

;1)

[0

;1)

[0


When p = 1, (6:1:10) an be proven as follows. Let P 2 En ( ); and let y 2 [0; 1) be hosen so that jP (y )j = kP k ;1 : From Theorem 6.1.1 and the Mean Value Theorem, we an dedu e that jP (t)j > kP k ;1 for every [0

)

1

[0

2

t 2 I := y; y + (18 )

1

; where :=

Thus

kP k

q Lq [0;1℄

Z

j I

j

P (t) q dt

18

n X j =0

!

1

j

n X j =0

)

j :

2 q kP kq ;1 ; [0

)

ut


Theorem 6.1.3 immediately implies the following result, whi h is a spe ial ase of Theorem 4.2.4:

2 [1; 1): Let (i )1 i be a sequen e of distin t real numbers greater than 1=p satisfying 1 X 1 j + < 1: p j

Theorem 6.1.4 (Muntz-Type Theorem in Lp ). Let p

=0

=0

Then

f

span x0 ; x1 ; : : :

g is not dense in Lp [0; 1℄:

The next theorem oers an L analog of Theorem 6.1.1 even for omplex exponents. It also improves the multipli ative onstant 12 in the L inequality of Theorem 6.1.2 and shows that the L inequality of Theorem 6.1.2 is essentially sharp. 2

2

2

Theorem 6.1.5. If := (i )1 is a sequen e of distin t omplex numbers i with Re(i ) > 1=2 for ea h i; then kxp0 (x)kL2 ; sup =0

kpkL

6 p2Mn

0=

(

[0 1℄

;

2 [0 1℄

)

n X

j =0

jj j

2

+

n X

n X

j =0

k=j +1

(1 + 2 Re(j ))

!1=2

(1 + 2 Re(k ))

for every n 2 N; where Mn() denotes the linear span of fx0 ; x1 ; : : : ; xn g over C : If := (i )1 i is a sequen e of distin t nonnegative real numbers, then n n X kxp0 (x)kL2 ; p1 X p1 j sup (1 + 2j ) 2 30 j 2j 6 p2Mn kpkL2 ; =0

[0 1℄

=0

0=

(

)

[0 1℄

=0

for every n 2 N, where Mn () denotes the linear span of fx0 ; x1 ; : : : ; xn g over R:


Proof. Let p 2 Mn () with kpkL2 p(x) =

n X k=0

and

; = 1: Then

[0 1℄

ak Lk (x)

n X

with

k=0

jak j

2

=1

n

X xp0 (x) = ak xL0 k (x) ;

k=0

where Lk

2 Mk () denotes the kth orthonormal Muntz-Legendre polynomials on [0; 1℄: Using the re urren e formula of Corollary 3.4.5 b℄ for the terms xL0 k (x) in the above sum, we obtain xp0 (x) =

n X j =0

q

aj j + 1 + j + j

!

q

n X k=j +1

ak 1 + k + k Lj (x) :

Hen e

kxp0 (x)kL2

; =

2

[0 1℄

n X aj j j =0

+

q

1 + j + j

n X k=j +1

q

2

ak 1 + k + k :

If we apply theP Cau hy-S hwarz inequality to ea h term in the rst sum and re all that nk jak j = 1; we see that 2

=0

kxp0 (x)k

2

L2 [0;1℄

n X

j =0

jj j

2

n X

+ (1 + j + j ) !2

n 1 X (1 + 2jj j) 2 j

k=j +1

(1 + k + k )

;

=0

whi h proves the rst part and the upper bound in the se ond part of the theorem. Now we prove the lower estimate in the se ond part of the theorem. With the sequen e := (i )1 of distin t nonnegative real numbers, we i asso iate ! =0

n p X

q(x) :=

k=0

k

k X j =0

j Lk (x) 2 Mn () :

Sin e the system (Lk )1 k is orthonormal on [0,1℄, we have =0

(6:1:11)

kqkL 2

; =

2 [0 1℄

n X k=0

k

k X j =0

!2

j

n X j =0

!3

j

:


Furthermore n

k X

k=0

j =0

Xp xq0 (x) = k

!

j xL0 k (x) =

n X m=0

bm Lm(x) ;

where, by the re urren e formula of Corollary 3.4.5 b℄, m X

p

bm := m m

p

m

n X

j =0

k =m

k

p

j + 1 + 2m k X j =0

n p X k=m+1

k (1 + 2k )

L2 [0;1℄ = 2

=

n X m=0

bm

n X

2

X

j =0

m=0

m

X

mn jk mk;k0 n j 0 k0 0

n X k =m

k

k X j =0

!2

j

m k j k0 j0

0

0

X

0

j

j :

Hen e

kxq0 (x)k

k X

mjj0 kk0 n

m k j k0 j0

n

1 X j 5! j

!5

:

=0

This, together with (6.1.11), yields the lower bound in the se ond part of the theorem. ut Comments, Exer ises, and Examples. Theorem 6.1.1 is due to Newman [76℄. We presented a modi ed version of Newman's original proof of Theorem 6.1.1. He worked with T instead of U , and instead of (6.1.9) he established a more ompli ated identity involving the se ond derivative of P . Therefore, he needed an appli ation of Kolmogorov's inequality (see E.1) to nish his proof. It an be proven that if the exponents j are distin t nonnegative integers, then kxp0 (x)k ; in Theorem 6.1.1 an be repla ed by kp0 k ; (see E.3). Theorems 6.1.2 to 6.1.4 were proved by Borwein and Erdelyi [to appear 6℄, while Theorem 6.1.5 is due to Borwein, Erdelyi, and J. Zhang [94b℄. It is shown in E.8 that Theorem 6.1.2 is essentially sharp for every with a gap ondition, and for every p 2 [2; 1). The interval [0; 1℄ plays a spe ial role in this se tion, analogs of the results on [a; b℄; a > 0; annot be obtained by a linear transformation. E.10 deals with the nontrivial extension of Newman's inequality to intervals [a; b℄; a > 0: A onje ture of Lorentz about the \right" Bernstein-type inequality for exponential sums with n terms is settled in E.4. [0 1℄

[0 1℄


E.1 Kolmogorov's Inequality. Show that

kf 0 k

;1)

2 [0

4kf k

;1)

[0

kf 00 k

;1)

[0

for every f 2 C [0; 1): Hint: By Taylor's theorem 2

f (x + h) = f (x) + f 0 (x)h + f 00 ( )h ; 1

h>0

2

2

with some 2 (x; x + h): Hen e

kf 0k

;1)

[0

2h kf k 1

kf 00 k

;1) + (h=2)

[0

;1) :

[0

Now minimize the right-hand side by taking

h := 2

kf k

;1)

[0

kf 00 k

1

;1)

1=2

[0

:

ut

that

The onstant 4 in E.1 is not the best possible. Kolmogorov [62℄ proved

kf k kR K (n; k)kf kR k=nkf n kk=n R 1

( )

(

)

for every f 2 C n (R) and 0 < k < n and found the best possible onstants K (n; k); see also DeVore and Lorentz [93℄. This generalizes a result of Landau, who proved p the above inequality for n = 2, k = 1, and showed that K (2; 1) = 2: Various multivariate extensions of Kolmogorov's inequality have also been established; see, for example, Ditzian [89℄. E.2 Nikolskii-Type Inequalities. a℄ Suppose (V; kk) is an (n +1)-dimensional real or omplex Hilbert spa e, (pk )nk V is an orthonormal system, and ' 6= 0 is a linear fun tional on V . Then ! =0

j'(p)j

n X

k=0

j'(pk )j

2

=

1 2

kpk

for every p 2 V: Equality holds if and only if

p=

n X j =0

'(pk )pk ;

2 R or 2 C :

Hint: Write p as a linear ombination of the orthonormal elements pk ; use the linearity of '; then apply the Cau hy-S hwarz inequality. ut


b℄ Suppose := (i )1 is a set of distin t omplex numbers satisfying i Re(i ) > 1=2 for ea h i; and y 2 [0; 1) is xed. Show that =0

jp

(

n X

j

m) (y )

k=0

jLk m (y)j (

)

!1=2 2

kpkL

;

2 [0 1℄

for every p 2 Mn (); where Mn () denotes the linear span of

fx ; x ; : : : ; xn g 1

0

over C ; and Lk 2 Mk () is the k th orthonormal Muntz-Legendre polynomial on [0; 1℄: Show that if there exists a q 2 Mn () with q m (y ) 6= 0; then equality holds if and only if (

p=

n X k=0

Lk m (y)Lk ; (

)

2C:

)

℄ Under the assumptions of part b℄ show that

jy = p(y)j kpkL ; 1 2

2 [0 1℄

n X j =0

!1=2

(1 + 2 Re(j ))

and

p0 (y)j

jy kpkL =

3 2

;

2 [0 1℄

0 n X (1 + 2 Re(k )) k

k=0

+

kX1 j =0

11=2 2 (1 + 2 Re(j )) A

for every 0 6= p 2 Mn () and y 2 [0; 1℄: Hint: When y = 1, use part b℄ and substitute the expli it values of Lk (1) and L0 k (1) (see Corollary 3.4.6 and formula (3.4.8)). If 0 < y < 1; then the s aling x ! yx redu es the inequality to the ase y = 1: ut d℄ Show that if n 1 and p 6= 0; then equality holds in the inequalities of part b℄ if and only if y = 1 and

p=

n X k=0

Lk (1)Lk

respe tively, with some 0 6= 2 C :

or

p=

n X k=0

L0 k (1)Lk ;


e℄ Show that

6

kpk k p2Pn pkL

sup

0=

;

[

=

1 1℄

;

2[

n+1

p

1 1℄

2

:

Hint: Show that there is an extremal polynomial pe for the above extremal problem for whi h kpek ; is a hieved at 1: Now use parts ℄ and d℄ to [

show that

1 1℄

n X jpe(1)j = 21 (1 + 2k ) k

!Z

1

1 2

2

1

=0

Z

pe (t) dt = (n + 1)

2

1

2

pe (t) dt 2

1

ut

and the result follows. E.3 An Improvement of Newman's Inequality. a℄ Suppose := (k )1 k is a sequen e with = 0 and k ea h k: Show that 0

=0

kp0k

;

[0 1℄

n X

18

j =1

!

j

kpk

+1

k 1 for

;

[0 1℄

for every p 2 Mn () := spanfx0 ; x1 ; : : : ; xn g: Hint: Let y 2 [0; 1℄: To estimate jp0 (y)j distinguish two ases. If y 1; use Theorem 6.1.1 (Newman's inequality), and if 0 y ; use E.3 f℄ of Se tion 3.3 and Theorem 5.1.8 (Markov's inequality) transformed to [y; 1℄ to show that 1

1

2

2

jp0 (y)j 2n

2

1

y

kpk y; 8 [

1℄

n X j =1

!

j

kpk

;

[0 1℄

for every p 2 Mn (): ut The next exer ise is based on an example given by Bos. b℄ Show that for every Æ 2 (0; 1) there exists a sequen e := (k )1 k with = 0, 1; and =0

0

1

k

+1

k Æ ;

i = 0; 1; 2; : : :

su h that lim

sup

n!1 06=p2Mn ()

P

jp0 (0) j kpk

n j =0 j

;

= 1:

[0 1℄

Outline. Let Qn be the Chebyshev polynomial Tn transformed linearly from [ 1; 1℄ to [0; 1℄, that is, Qn (x) = os(n ar

os(2x 1)) ;

x 2 [0; 1℄ :


Choose natural numbers u and v so that Æ < u=v < 1: Let := (k )1 k be de ned by := 0, := 1; and =0

0

1

k := 1 + Let

pn (x) := x

1

u

ku ; v

k = 1; 2; : : : :

Qn (xu=v )

Then

( 1)n

jp0n (0)j =


2n

2

v

qn (x) := Qn (xu=v )

v

2 Mnv

v () :

:

( 1)n :

Use Theorem 5.1.8 (Markov's inequality) and the Mean Value Theorem to show that

kqn k y; kqn k [

with

1

1℄

y := 2n

Thus, if n is odd, then

kpnk

;

[0 1℄

kpn k y; y y u kqn k [

1

P

nv j =0

jp0n (0) j v j kpn k

1

;

=

[0 1℄

v=u

:

kqn k y;

u(

1

1℄

2

and

2

;

[0 1℄

2

;

v

[0 1℄

P nv v

j =0 v=u 2

[

1℄

= 2n

)v

2

(u

v=u

(2n )v 1 + j uv 2n

1)

2n (1 + nu)nv

2

2

2n

2

v=u

uv

(u 1

v=u

1)

! 1:

n!1

ut

In his book Nonlinear Approximation Theory, Braess [86℄ writes the following: \The rational fun tions and exponential sums belong to those

on rete families of fun tions whi h are the most frequently used in nonlinear approximation theory. The starting point of onsideration of exponential sums is an approximation problem often en ountered for the analysis of de ay pro esses in natural s ien es. A given empiri al fun tion on a real interval is to be approximated by sums of the form n X j =1

aj ej t ;

where the parameters aj and j are to be determined, while n is xed." The next exer ise treats inequalities for exponential sums of n + 1 terms.


E.4 Nikolskii- and Bernstein-Type Inequalities for Exponential Sums. Let := (i )1 i =1

be a sequen e of distin t nonzero real numbers. Let (

En () := f : f (t) = a + 0

and

En :=

[

n X j =1

(

En () = f : f (t) = a + 0

aj

n X j =1

aj 2 R

ej t ;

aj

ej t ;

)

aj ; j

2R

)

;

that is, En is the olle tion of all (n + 1)-term exponential sums with onstant rst term. S hmidt [70℄ proved that there is a onstant (n) depending only on n su h that

kf 0k a Æ;b Æ (n)Æ kf k a;b for every p 2 En and Æ 2 0; (b a) . Lorentz [89℄ improved S hmidt's [ +

1

℄

[

℄

1

2

result by showing that for every > ; there is a onstant () depending only on su h that (n) in the above inequality an be repla ed by

()n n , and he spe ulated that there may be an absolute onstant su h that S hmidt's inequality holds with (n) repla ed by n. Part d℄ of this exer ise shows that S hmidt's inequality holds with (n) = 2n 1: A weaker version of this showing that S hmidt's inequality holds with (n) = 8(n+1) is obtained in part b℄ and uses a Nikolskii-type inequality for exponential sums established in part a℄. Part e℄ shows that the result of part d℄ is sharp up to a multipli ative absolute onstant. a℄ Let p 2 (0; 2℄: Show that 1

2

log

2

kf k a

[ +

for every f

Æ;b Æ℄

2 En and Æ 2

2 2 =p

0; (b 1

2

2

n+1 Æ

1=p

kf kLp a;b [

℄

a) :

Proof. Take the orthonormal sequen e (Lk )nk on ; , that is, t t t (1) Lk 2 spanf1; e 1 ; e 2 ; : : : ; e k g ; k = 0; 1; : : : ; n ; =0

and (2)

Z 1=2

=

1 2

Li Lj = Æi;j ;

1

1

2

2

0 i j n;

where Æi;j is the Krone ker symbol. On writing f 2 En () as a linear

ombination of L ; L ; : : : ; Ln ; and using the Cau hy-S hwarz inequality 0

1


and the orthonormality of (Lk )nk fashion that

on

=0

max

jf (t )j

6 f 2En kf kL2

0=

(

)

Sin e

= ; =

[

Z 1=2 Pn

0

max

1 2

;

1

(

Observe that if f

)

, we obtain in a standard

2

Lk (t )

!1=2

;

2

0

t

0

2 R:

su h that n X

=

0

[

1

2

jf (t )j

6 p2En kf kL2

0=

2

;

k=0 Lk (x) dx = n + 1 ;

=

1 2

2

there exists a t

k=0

1 2 1 2℄

1 2

n X

=

0

= ; =

k=0

1 2 1 2℄

!1=2

Lk (t ) 2

0

p

n + 1:

2 En (); then g(t) := f (t t ) 2 En (); so jf (0)j pn + 1 : max 6 f 2En kf kL ; 0

0=

Let

(

C :=

)

2[

1 1℄

jf (0)j : max 6 f 2En kf kLp ;

0=

(

)

[

2 2℄

Then max

1

6 f 2En

0=

jf (y)j C 2 =p 2 =p C ; kf kLp ; 2 jy j

(

)

y 2 [ 1; 1℄ :

1

[

Therefore, for every f

2 2℄

2 En ();

p jf (0)j n + 1 kf kL ; = p n + 1 kf kpLp ; kf k p; p n + 1 kf kpLp ; 2 =p C p kf kLp p p n + 1 2 =p C p= kf kLp ; p = 2 =p = n + 1 C p= kf kLp ; : 2[

1 1℄

1 2

2

[

[

1 1℄

1 1℄

2

1

[

1

Hen e

1 1℄

1

1

2

[

2 2℄

2

[

1

1 2

2

2 2℄ [

2 2℄

jf (0)j 2 =p = pn + 1 C 6 f 2En kf kLp ; and we on lude that C 2 =p =p (n + 1) =p : So C=

0=

max

1

(

)

[

2

2

1 2

2 2℄

1

;

1=2

1

1

p=2


jf (0)j 2 =p

kf kLp ; for every f 2 En (): Now let f 2 En () and t 2 [a + Æ; b 2

2

1

=p (n + 1)1=p

[

2 2℄

0

the above inequality to

g(t) := f

1 2

Æt + t

0

Æ℄: If we apply

2 En () ;

we obtain

kf k a

[ +

Æ;b Æ℄

2 2 =p 2

1

=p (n + 1)1=p

1=p

2

Æ

kf kLp a;b [

℄

ut


The following Bernstein-type inequality an be obtained as a simple

orollary of part a℄: b℄ Show that kf 0 k a Æ;b Æ 8(n + 1) Æ kf k a;b [ +

2

℄

1

[

℄

2 En and Æ 2 (0; (b a)): Note that f 2 En () implies f 0 2 En (): Applying part a℄ to f 0

for every f

1

2

Proof. with p = 1; we obtain

jf 0 (0)j 2(n + 1)kf 0kL ; = 2(n + 1)Var ; (f ) 4(n + 1) kf k ; for every f 2 En (): If f 2 En () and t 2 [a + Æ; b Æ ℄; then on applying 1[

2 2℄

[

2

2 2℄

[

2 2℄

0

the above inequality to

g(t) := f

1 2

Æt + t

0

2 En () ; ut

we obtain the desired result.

℄ Lorentz's Conje ture. Show that

6

0=

where

jf 0 (0)j f 2Ee n kf k ;

(

Ee n := f : f (t) = a + 2

= 2n

sup

0

2

n X j =1

[

1;

1 1℄

aj ej t + bj e j t ;

Proof. First we prove that

jf 0 (0)j (2n

1) kf k

[

;

1 1℄

aj ; bj ; j

2R

)

:


for every f

2 Ee n . So let f 2 spanf1; e t ; e t ; : : : ; en t g 2

1

2

with some nonzero real numbers ; ; : : : ; n ; where, without loss of generality, we may assume that 1

2

0 < < < < n : 1

Let

2

g(t) := (f (t)

f ( t)) :

1

2

Observe that

g 2 spanfsinh t ; sinh t ; : : : ; sinh n tg : 1

2

It is also straightforward that

g0 (0) = f 0 (0)

kgk

and

;

[0 1℄

kf k

;

[

1 1℄

:

For a given > 0; let

Hn; := spanfsinh t ; sinh 2t ; : : : ; sinh ntg and

Kn; := sup jh0 (0)j : h 2 Hn; ; khk

:

; =1

[0 1℄

The inequality of E.5 e℄ in Se tion 3.3 is the key to the proof. It shows that it is suÆ ient to prove that inf fKn; : > 0g 2n

1:

Observe that every h 2 Hn; is of the form

h(t) = e

nt P (et ) ;

P

2P n: 2

Therefore, using E.23 ℄ of Se tion 5.1, we obtain for every h 2 Hn; that

jh0 (0)j = jP 0 (1) nP (1)j 1(2n e 1) kP k e ;e It follows that

[

℄

+ n kP k e ;e [

(2n 1) + n en khk 1 e

[

Kn;

So inf fKn; : > 0g 2n

;

1 1℄

℄

:

(2n 1) + n en : 1 e

1; and the upper bound follows.


Now we prove that

6

jf 0 (0)j 2n f 2Ee n kf k ;

1:

sup

0=

[

2

1 1℄

Let > 0 be xed. We de ne

Q where T n by 2

2

nt T

n; (t) := e

2

n

et

e

1

e

1

1

1

;

denotes the Chebyshev polynomial of degree 2n

1

T

2

n

1

1 de ned

1) ar

os x); x 2 [ 1; 1℄ :

(x) = os((2n

It is simple to he k that Q n; 2 Ee n ; 2

2

kQ n;k 2

and

[

;

1 1℄

jQ0 n;(0)j 2n 2

ent n :

1

ut

Now the result follows by letting de rease to 0: d℄ Show that

kf 0k a Æ;b Æ (2n 1)Æ kf k a;b for every f 2 En and Æ 2 0; (b a) . Proof. Observe that En Ee n . Hen e the result follows from part ℄ by a linear substitution. ut e℄ Let a < b and y 2 (a; b). Suppose that n 2 N is odd. Let Tn be the [ +

1

℄

[

℄

1

2

2

Chebyshev polynomial of degree n de ned by (2.1.1). Let

Qn (t) := Qn;y (t) := Tn and

Rn (t) := Rn;y (t) := Tn Show that Qn ; Rn 2 En and

jQ0n (y)j kQnk a;b [

=

℄

for every y 2 (a; b) :

1

n

e 1b y

e

e 1

e

e

exp

t a exp 1 a y

and

t b b y

e 1

1

e 1

jRn0 (y)j kRn k a;b [

1

℄

=

1

e 1y

:

n

a


f℄ Let a < b and y 2 (a; b): Con lude that

n 1 e 1 minfy a; b yg 1

2n 1 jf 0 (y)j minfy a; b f 2En kf k a;b

sup

6

0=

[

℄

yg

:

In the rest of the exer ise let (

l

X En := f : f (t) = Pkj (t)ej t ; j

2 R;

j =1

Pkj

2 Pkj ;

l X j =1

)

(kj + 1) = n :

g℄ Extend the inequalities of parts a℄, ℄, and f℄ to En : h℄ Let [a; b℄ be a nite interval. Let g 2 C [a; b℄: Show that the value inf

kg

f k a;b : f [

℄

2 En

is attained by an fe 2 En : Hint: Use S hmidt's inequality (or its improved form given by part ℄). For the nontrivial details, see Braess [86℄. ut i℄ Let [a; b℄ be a nite interval. Let p 2 [1; 1) and g 2 Lp [a; b℄: Show that the value inf kg f kLp a;b : f 2 En [

℄

is attained by an fe 2 En :

Hint: Use part a℄ with p = 1 and Holder's inequality. On e again, for the details, see Braess [86℄. The following result is from Borwein and Erdelyi [95 ℄:

ut

E.5 Upper Bound for the Derivative of Exponential Sums with Nonnegative Exponents. The equality sup p

jp0 (a)j kpk a;b [

℄

=

2n

2

b a

holds for every a < b; where the supremum is taken over all exponential sums 0 6= p 2 En with nonnegative exponents. The equality sup p

jp0 (a)j kpk a;b [

℄

=

2n

2

a(log b log a)

also holds for every 0 < a < b; where the supremum is taken over all Muntz polynomials 0 6= p of the form

p(x) = a + 0

n X j =1

aj xj ;

aj

2 R;

j > 0 :


Outline. It is suÆ ient to prove only the se ond statement, the rst an be obtained from it by the hange of variable x = et : For > 0; de ne := (j)1 j : Let =0

Tn; := Tnf1; x; x ; : : : ; xn ; [a; b℄g 2

be the Chebyshev polynomial for Mn ( ) on [a; b℄: Use E.3 b℄ and E.3 f℄ of Se tion 3.3 to show that

0 (a)j jp0 (a)j lim jTn; kpk a;b ! kTn;k a;b [

0+

℄

[

0 (a)j = lim jTn; !0+

℄

for every p of the form

p(x) = a + 0

n X j =1

aj xj ;

aj

2 R;

j > 0 :

From the de nition and uniqueness of Tn; it follows that

Tn;(x) = Tn

b

2

a

b + a b a

x

where Tn (y ) = os(n ar

os y ); y 2 [ 1; 1℄: Therefore 2 a 1)j b a 2n = (b 1) (a

0 (a)j = jT 0 ( jTn; n

1

2

1

1

1)

a

1

! !

0+

2n

2

a(log b log a)

ut

and the proof is nished. The next exer ise follows Turan [84℄. E.6 Turan's Inequalities for Exponential Sums. a℄ Let

g( ) := Suppose

n X j =1

bj zj ;

jzj j 1 ;

bj ; zj 2 C :

j = 1; 2; : : : ; n :

Then max

=m+1;::: ;m+n

jg( )j

for every nonnegative integer m:

n 2e(m + n)

2

jb

1

+ b + + bn j 2


Proof. Let (6:1:12)

f (z ) :=

n Y

j =1

n X z =: z : zj

1

=0

Sin e f (z ) has all its zeros outside the open unit disk, g := 1=f is of the form (6:1:13)

g(z ) =

1 X =0

jz j < 1 :

z ;

Let (6:1:14)

gm (z ) :=

m X =0

z

and

h := 1 fgm 2 Pn

(6:1:15)

+

m:

Note that (6:1:16)

1 X

h(z ) = 1 f (z ) g(z ) = f (z )

1 X =m+1

=m

!

z

z

so h is of the form (6:1:17)

h(z ) =

m +n X =m+1

z :

Observe that f (zj ) = 0 and (6.1.13) imply h(zj ) = 1; that is, m +n X =m+1

zj = 1 ;

j = 1; 2; : : : ; n :

Multipli ation with bj and summation over j yield the fundamental identity (6:1:18) This immediately gives

m +n X =m+1

g( ) =

n X j =1

bj :


(6:1:19)

max

=m+1;::: ;m+n

!

n X

jg( )j

=m+1

j j

n X bj

j =1

:

It follows from (6.1.15), (6.1.17), (6.1.12), and (6.1.14) that (6:1:20)

m +n X =m+1

n X

j j

=0

j j

!

m X =0

!

j j

:

Sin e ea h zj is of modulus at least 1, (6.1.12) yields that n X

(6:1:21)

=0

j j 2n :

Also, (6.1.13) implies that

=

X

1

i1 ++in =

:

in i1 i2 z z zn 2

1

Again using that ea h zj is of modulus at least 1; we obtain

j j

i1 ++in =

Hen e (6:1:22)

m X =0

X

j j

+n 1 1= : n 1

m+n n

e

m+n n

n

:

By (6.1.20) to (6.1.22) we on lude that m +n X =m+1

j j

2e

m+n n

n

;

ut

whi h, together with (6.1.19), nishes the proof. b℄ Let

f (t) := Suppose Show that

n X j =1

bj ej t ;

Re(j ) 0 ;

jf (0)j

for every a > 0 and d > 0:

bj ; j

2C:

j = 1; 2; : : : ; n :

2e(a + d) n

d

kf k a;a [

d

+ ℄


Hint: First observe that the result of part a℄ an be formulated as max

m m+ 2N

jg( )j n

n 2e(m + n)

n

jb

+ b + + bn j ;

1

2

where m is an arbitrary positive (not ne essarily integer) number. Now apply the above inequality with

dj an ; zj := exp d n

m :=

; j = 1; 2; : : : ; n :

ut

℄ Let

n X

p(z ) := Show that

j =1

bj z j ;

max jp(z )j

jzj

bj

2C;

4e n

Æ

=1

j

2 R;

max

jzj

=1

arg(z)+Æ

z = ei :

jp(z )j

for every 0 < + Æ 2: Hint: Use part b℄. ut The inequalities of the above exer ise and their variants play a entral role in the book of Turan [83℄, where many appli ations are also presented. The main point in these inequalities is that the exponent on the right-hand side is only the number of terms n; and so it is independent of the numbers j : An inequality, say in part ℄, of type max jp(z )j (Æ )n

jzj

=1

jp(z )j ;

max

jzj=1 arg(z)+Æ

where 0 < < < n are integers and (Æ ) depends only on Æ;

ould be obtained by a simple dire t argument, but it is mu h less useful than the inequality of E.6 ℄. 1

2

E.7 Nikolskii-Type Inequality for Muntz Polynomials. Suppose that i 1 for ea h i. Show that

:= (i )1 is a sequen e with := 0 and i i 0

=0

0

kP kLp for every P

;

[0 1℄

18 2q

n X j =0

11=q

j A

1

+1

=p

kP kLq

;

[0 1℄

2 Mn() := spanfx ; x ; : : : ; xn g and 0 < q < p 1: 0

1


Proof. It is suÆ ient to study the ase when p = 1 (see the omment in the proof of Theorem 6.1.3). Let P 2 Mn () and let x 2 [0; 1℄ be hosen so that jP (x )j = kP k ; . Combining E.3 a℄ and the Mean Value Theorem, we obtain jP (x)j 21 kP k ; ; x 2 I ; where 0

0

[0 1℄

[0 1℄

h

I := x

(36)

0

1

; x + (36)

1

i

with :=

0

n X j =0

j :

So Z

1

0

jP (x)jq dx

1=q

Z

I

jP (x)jq dx

1=q

(18)

1

1 kP k ; 2

q 1=q

[0 1℄

ut


E.8 Sharpness of Theorem 6.1.2. Suppose := (i )1 i is a sequen e with := 0 and i i 1 for ea h i: Show that there exists an absolute

onstant > 0 (independent of and p) su h that =0

0

+1

sup

P 2Mn ()

kxP 0 (x)kLp kP kLp ;

;

[0 1℄

[0 1℄

n X k=0

k

for every p 2 [2; 1), where Mn () := spanfx0 ; x1 ; : : : ; xn g:

Proof. Let Lk 2 Mk () be the kth orthonormal Muntz-Legendre polynomial on [0; 1℄: Let p 2 [2; 1) and n X

P :=

k=0


:=

L0 k (1)Lk :

n X j =0

j :

Using Theorem 3.4.3 (orthogonality), (3.4.8), and Corollary 3.4.6, we obtain (6:1:23)

n

X P 0 (1) = (L0 (1))

k=0

2

k

k X j =0

2

!2

j

n X

k X

k=0

j =0

(2k + 1)

1 8

3

!2

j


and (6:1:24)

kP kL

n X

; =

2 [0 1℄

k=0

0

!1=2

(L0 k (1))

2

!2 11=2

k X

n X

(2k + 1) k p = 32 :

(2j + 1)

j =0

=0

A

3 2

A ombination of E.7 and (6.1.24) yields (6:1:25)

kP kLp

(72 ) = =p kP kL ; p 72 = =p 32 =p 48 1 2

;

[0 1℄

1 2

1

2 [0 1℄

1

2

1

2

1

=p :

From E.3 a℄, E.7, and (6.1.24) we an dedu e that (6:1:26)

kP 0k

;

[0 1℄

18 kP k ; 18 (72 ) = kP kL ; p 18 (72 ) = 32 = 18 48 : [0 1℄

1 2

2 [0 1℄

1 2

3 2

3

Applying E.3 a℄ with P 0 , we get (6.1.27)

kP 00 k

;

[0 1℄

18 kP 0k

;

[0 1℄

18 48 2

4

:

Now (6.1.23), (6.1.27), and the Mean Value Theorem give

jP 0 (x)j where

I := 1

1 16

x2I;

;

18

2

3

48 16

1

;1 :

So (6:1:28)

kxP 0 (x)kLp

;

[0 1℄

Z

I

0 xP (x) p dx

1=p

48 16 =p 18 48 16 32 p 128 9 3 =p : 18

2

1

1

32

1

2

1

3

=

3

1 2

p 1=p

3

1

=p

1

Combining (6.1.25) p and (6.1.28), we obtain the required result with a onut stant = (128 9 3) : 1


E.9 On the Interval Where the Sup Norm of a Muntz Polynomial Lives. Let := (j )1 be an in reasing sequen e of nonnegative real numbers. j Let 0 6= p 2 Mn () := spanfx0 ; x1 ; : : : xn g; and let q (x) := xk p(x); where > 0 and k is a nonnegative integer. Let 2 [0; 1℄ be hosen so that jq( )j = kqk ; : a℄ Suppose = 1 and ea h j is an integer. Then =0

[0 1℄

k

2

k+n

0

rA := supfx 2 [0; 1) : m(A \ (x; 1)) > 0g ; then every fun tion f 2 C (A) from the uniform losure of M () on A is of the form 1 X x 2 A \ [0; rA ): aj xj ; f (x) = j =0

If the above gap ondition does not hold, then every fun tion f 2 C (A) from the uniform losure of M () on A an still be extended analyti ally throughout the region

fz 2 C n ( 1; 0℄ : jz j < rA g :

The proof of the above theorem rests on the following bounded Remeztype inequality:


Theorem 6.2.2 untz Spa es). Let P (Remez-Type Inequality for Nondense M := 0 and 1 1 = < 1 : Then there exists a

onstant

depending only i i on and s (and not on %, A, or the number of terms in p) su h that 0

=0

kpk

;%℄

[0

kpkA

for every p 2 spanfx0 ; x1 ; : : : g and for every ompa t set A Lebesgue measure at least s > 0:

[%; 1℄ of

To prove Theorem 6.2.2 we need a few lemmas. We use the notation introdu ed in Se tion 3.3. However, for notational onvenien e, we let

Tn f ; ; : : : ; n ; Ag := Tn fx0 ; x1 ; : : : ; xn ; Ag 0

1

for ompa t sets A [0; 1): By the unique interpolation property of Chebyshev spa es (see Proposition 3.1.2), asso iated with 0 < x < x < < xn ; 0

1

we an de ne

`k fx ; x ; : : : ; xn g 2 Mn () ; 0

k = 0; 1; : : : ; n

1

su h that

k `k fx ; x ; : : : ; xn g(xj ) = Æj;k := 10 ifif jj = 6= k : 0

1

Lemma 6.2.3. Let 0 < x < x < < xn 0

1

and

0 < xe < xe < < xen : 0

1

Suppose 0 k n and xj xej if j = 0; 1; : : : ; k 1 ; xj = xej if j = k ; xj xej if j = k + 1; k + 2; : : : ; n : For notational onvenien e, let `k := `k fx ; x ; : : : ; xn g 0

Then

1

and

èk := `k fxe ; xe ; : : : ; xen g :

j`k (0)j jèk (0)j :

0

1

6.2 Nondense Muntz Spa es 305

Proof. It is suÆ ient to prove the lemma in the ase where there is an index m su h that 1 m n, m 6= k, and xj = xej if j = 0; 1; : : : ; n; j 6= m ; xm < xem if m < k ; xm > xem if m > k : The general ase of the lemma then follows from repeated appli ations of the above spe ial ases. Note that in the above spe ial ases

`k

èk 2 Mn ()

has a zero at ea h of the points

xj ; j = 0; 1; : : : ; n ; j 6= m ; hen e it hanges sign at ea h of these points, and it has no other zero in [0; 1) (see E.10 of Se tion 3.1). It is also obvious that

x 2 (0; x ) ;

sign(`k (x)) = sign(èk (x)) ;

0

whi h, together with the previous observation and the inequality x yields that j`k (0)j jèk (0)j ;

0

xe ; 0

ut


By a simple s aling we an extend Lemma 6.2.3 as follows. We use the notation introdu ed in Lemma 6.2.3. Lemma 6.2.4. Let 0 < x < x < < xn 0

1

and

0 < xe < xe < < xen : 0

1

Suppose 0 k n, 0; and xj xej xj = xej xj xej Then

if j = 0; 1; : : : ; k 1 ;

if j = k ;

if j = k + 1; k + 2; : : : ; n :

j`k (0)j jèk (0)j :


Proof. If = 0; then Lemma 6.2.3 yields the lemma. So we may assume that > 0: Let x xe := k = k ; xek xek xj := xej ; j = 0; 1; : : : ; n ; and

`k := `k fx ; x ; : : : ; xn g ; 0

Obviously and

k = 0; 1; : : : ; n :

1

èk ( x) = `k (x) ;

x 2 [0; 1)

xj xj if j = 0; 1; : : : ; k 1 ; xj = xj if j = k ; xj xj if j = k + 1; k + 2; : : : ; n :

Hen e Lemma 6.2.3 implies that

j`k (0)j j`k (0)j = jèk (0)j ut


Lemma 6.2.5. Let A be a losed subset of [0; 1℄ with Lebesgue measure at least s 2 (0; 1): Then

jp(0)j jTnf ; ; : : : ; n ; [1 0

1

s; 1℄g(0)j kpkA

for every p 2 M (): Proof. If 0 2 A; then the statement is trivial. So assume that 0 2= A: Let xe < xe < < xen 0

1

denote the extreme points of

Tn := Tnf ; ; : : : ; n ; [1 s; 1℄g 0

in [1

1

s; 1℄; that is, Tn (xej ) = ( 1)n j ;

Let xj

2 A;

j = 0; 1; : : : ; n :

j = 0; 1; : : : ; n; be de ned so that m([xj ; 1℄ \ A) = m([xej ; xen ℄) = xen

xej :

Sin e A is a losed subset of [0; 1℄ with m(A) s; su h points xj Let p 2 Mn (): Then, by Lemma 6.2.4, we an dedu e that

2 A exist.


jp(0)j =

n X p(xk )`k (0) k=0 ! n X

k=0 n X

k=0 n X

=

k=0 n X

=

j`k (0)j kpkA !

j

j kpkA

èk (0)

!

( 1)n k èk (0) !

Tn (xek )èk (0)

k=0 = Tn (0)

j

kpkA

kpkA

j kpkA

ut

whi h proves the lemma.

Lemma 6.2.6. Suppose := 0: Let A be a losed subset of [0; 1℄ with Lebesgue measure at least s 2 (0; 1): Then 0

jp(y)j jTn f ; ; : : : ; n ; [1 for every p 2 Mn () and y 2 [0; inf A℄: 0

1

s; 1℄g(0)j kpkA

Proof. For notational onvenien e, let

Tn;A := Tn f ; ; : : : ; n ; Ag : 0

1

Note that = 0 implies that jTn;A j is de reasing on [0; inf A℄; otherwise 0

0 2 spanfx1 ; x2 ; : : : ; xn Tn;A 1

1

1

g

would have at least n + 1 zeros in (0; 1℄; whi h is impossible. Hen e, it follows from E.3 and Lemma 6.2.5 that jp(y)j jTn;A (y)j = jT (y)j jT (0)j n;A n;A

kpkA

kTn;AkA jTn f ; ; : : : ; n ; [1 0

1

for every 0 6= p 2 Mn (): This nishes the proof.

s; 1℄g(0)j

ut

Proof of Theorem 6.2.2. Lemma 6.2.6 and E.5 a℄ of Se tion 4.2 yield the theorem.

ut

Proof of Theorem 6.2.1. The theorem follows from E.5 of this se tion and E.3 e℄ and E.8 b℄ of Se tion 4.2. ut Our next theorem is an interesting hara terization of la unary sequen es.


Theorem 6.2.7 (Chara terization of La unary Muntz Spa es). Suppose := (i )1 with 0 < < . There exists a onstant depending i =0

only on su h that

0

1

jaj;n j kpk

j = 0; 1; : : : ; n ; n 2 N

;

;

[0 1℄

for every p 2 M () of the form p(x) =

n X j =0

aj;n xj

if and only if is la unary, that is, if and only if the elements i of satisfy inf fi =i : i 2 N g > 1 : +1

To prove Theorem 6.2.7 we need the following result of Hardy and Littlewood [26℄ whose proof we do not reprodu e: Theorem 6.2.8. Suppose 0 = < < is a la unary sequen e, that is, 0

1

inf f i = i : i 2 N g > 1 : +1

Suppose the fun tion f is of the form 1 X ai 2 R ; f (x) = ai x i ; i=0

and A := xlim ! f (x) exists and is nite. Then 1

x 2 [0; 1)

P1

i=0 ai = A:

Proof of Theorem 6.2.7. Suppose is la unary and suppose there exists a M () su h that if Pk is of the form

sequen e (Pk )1 k

=1

Pk (x) =

nk X j =0

aj;nk xj ;

aj;nk

2 R;

then (6:2:1)

kPk k

; =1

[0 1℄

and

0

max jaj;nk j k ; 2

jnk

k = 1; 2; : : : :

We may assume, without loss of generality, that = 0: Choose a sequen e (k )1 k of positive integers su h that 0

=1

= 1; 1

k 1

+1

> 2k nk ;

Now let the fun tion f be de ned by

k = 1; 2; : : : :


1 X

1 X

0 x 1:

Note that f let

2 C [0; 1℄ by the Weierstrass M-test. For notational onvenien e, m := 0;

mk :=

0

k X i=1

k=1

k

< 1;

f (x) =

k=1

k Pk

(xk )

(6:2:2)

2

2

ni ;

k = 1; 2; : : : :

Further, let

:= 0 and

m k

a :=

and

0

1+

0

j := k j ;

1 X k=1

k a 2

0

;nk =

1 X k=1

k Pk (0) 2

j = 1; 2; : : : ; nk ; k = 1; 2; : : : :

2 R is well-de ned sin e jPk (0)j 1 for ea h k 2 N : Also inf f i = i : i 2 N g minf2; inf fi =i : i 2 N gg > 1 : := ( i )1 i . Then f 2 C [0; 1℄ de ned by (6.2.2) is in the uniform

Observe that a

0

+1

+1

Let

losure of M ( ) on [0; 1℄; hen e, by the Clarkson-Erd}os theorem (see E.3 e℄ of Se tion 4.2), f is of the form =0

f (x) =

1 X i=0

ai x i ; x 2 [0; 1) :

Sin e f 2 C [0; 1℄; Theorem 6.2.8 implies that A := 1 i ai exists and is nite. Re alling (6.2.2) and the hoi e of k ; and using E.3 e℄ of Se tion 4.2, we an dedu e that ea h P

=0

k aj;nk ;

j = 1; 2; : : : ; nk ; k = 1; 2; : : :

2

is equal to one of the oeÆ ients a ; a : : : : Sin e ja ;nk j = jPk (0)j 1 for ea h k 2 N ; from (6.2.1) and (6.2.2) we see that P jai j 1 holds for in nitely many i 2 N ; whi h ontradi ts the fa t that 1 i ai onverges. This nishes the if part of the theorem. Now assume that is not la unary. Then for every > 0 there is an n 2 N su h that n =n > 1 : Observe that Pn (x) := xn xn 1 a hieves its maximum modulus on [0; 1℄ at 1

2

0

=0

1

x= and hen e

n n

1

1=(n n

1)


kPn k

;

[0 1℄

n n

1

n

= n n

1 (

1)

n n

1

1

n < ; n

1

1

whi h shows that the lead oeÆ ient an;n of

Tn f ; ; : : : ; n ; [0; 1℄g 0

1

is at least 1=; otherwise an;n Tn Pn 2 Mn () would have at least n zeros on (0; 1), whi h is a ontradi tion. The proof of the only if part of the theorem is now nished. ut 1

1

From the above proof it also follows that under the assumptions of Theorem 6.2.7, the Chebyshev polynomials

Tn f ; ; : : : ; n ; [0; 1℄g 0

1

have uniformly bounded oeÆ ients if and only if is la unary. As an appli ation of Theorem 6.2.7 we derive the following Bernsteintype inequality. Theorem 6.2.9 (Bernstein-Type Inequality). Suppose := 0, suppose := (i )1 i is la unary, that is, 0

1

1; and

=0

inf fi =i : i 2 N g > 1 : +1

Then there exists a onstant depending only on (and not on y or the number of terms in p) su h that jp0 (y)j 1 y kpk ; [0 1℄

for every p 2 M () = spanfx0 ; x1 ; : : : g and for every y 2 [0; 1): Proof. Let p 2 M () be of the form p(x) = a

0

;n +

n X j =1

kpk

aj;n xj ;

; = 1:

[0 1℄

Theorem 6.2.7 and the assumptions on yield n

X jp0 (y)j aj;n j yj

j =1 n X

1

j =1

j yj

1

1

n X j =1

1 X 2

j =0

jaj;n j j yj

yj =

1

2

y

1

;

where and depend only on ; and the theorem is proved. 1

2

ut


Comments, Exer ises, and Examples. The results of this se tion have been obtained by Borwein and Erdelyi [91, 93, 95b, to appear 1℄. In E.1 we present some of the several important

onsequen es of our entral result, Theorem 6.2.2. In E.6 we oer another proof of the rst part of Theorem 6.2.1 when is la unary, while E.9 shows that the Bernstein-type inequality of Theorem 6.2.9 \almost" hara terizes the la unary Muntz spa es. Note that if A [0; 1℄ ontains an interval, then the rst part of Theorem 6.2.1 follows immediately from E.3 of Se tion 4.2. A typi al ase that does not follow from that exer ise is when A [0; 1℄ is a \fat" Cantor-type set of positive measure. E.1 Some Consequen es of Theorem 6.2.2. Let A [0; 1) be a set of positive Lebesgue measure, and let rA be the essential supremum of A as de ned in Theorem 6.2.1. Suppose q 2 (0; 1) and suppose w is a R nonnegative-valued, integrable weight fun tion on A with A w > 0: Let Lq (w) := Lq (); where d = w dt; and where Lq () is de ned in E.7 of Se tion 2.2. Let := (i )1 be a sequen e of distin t nonnegative real i numbers with i 6= 0 for ea h i = 1; 2; : : : : P a℄ Suppose 1 i 1=i < 1: Then M () is not dense in Lq (w): Moreover, if the gap ondition inf fi i : i 2 N g > 0 holds, then every fun tion f 2 Lq (w) belonging to the Lq (w) losure of M () an be represented as =0

=1

1

f (x) = where

1 X i=0

ai xi ;

(

ai 2 R ;

rw := sup y 2 [0; 1) :

x 2 A \ [0; rw ) ; )

Z

A\(y;1)

w(x) dx > 0 :

If the above gap ondition does not hold, then every fun tion f 2 Lq (w) belonging to the Lq (w) losure of M () an still be represented as an analyti fun tion on

fz 2 C n ( 1; 0℄ : jz j < rw g restri ted to A:

Proof. Suppose f

su h that

2 Lq (w) and suppose there is a sequen e (pi )1 i M () =1

lim kf

i!1

pi kLq

(

w) = 0 :

Minkowski's inequality (see E.7 b℄ and E.7 i℄ of Se tion 2.2.) yields that (pi )1 i is a Cau hy sequen e in Lq (w): The assumptions on w imply that for every Æ 2 (0; rw ) there exists an > 0 su h that =1


B := fx 2 A \ (Æ; 1) : w(x) > g is of positive Lebesgue measure. Let s := m(B ) > 0: Observe that if kpkLq w < "; then (

)

(

m so

(

m

x 2 B : jp(x)j x 2 B : jp(x)j
: 2

Hen e, by Theorem 6.2.2, (pi )1 i is uniformly Cau hy on [0; Æ ℄: The proof

an now be nished as that of Theorem 6.2.1. ut =1

b℄ PMuntz-Type Theorem in Lq (w). M () is dense in Lq (w) if and only if 1 i 1=i = 1: =1

P1

i 1=i = 1: Let f 2 Lq (w): It is standard measure theory to show that for every " > 0 there exists a g 2 C [0; 1℄ su h that g(0) = 0 and

Proof. Suppose

=1

" gkLq w < :

kf

2 Now Theorem 4.2.1 (full Muntz theorem in C [0; 1℄) implies that there exists a p 2 M () su h that

kg

pkLq w (

)

kg

pkA

(

Z

A

w

)

1=q

kg

pk

Z

;

[0 1℄

A

w

1=q

" < : 2

Therefore M () is dense in Lq (w): P Suppose now that 1 i 1=i < 1: Then part a℄ yields that M () is not dense in Lq (w): ut =1

℄ Convergen e in M (). Suppose

P1

pi (x) ! f (x) ;

i=1 1=i

< 1, (pi )1 i

=1

M (); and

x 2 A:

Then (pi )1 i onverges uniformly on every losed subinterval of [0; rA ): =1

Proof. Let Æ 2 (0; rA ) be xed. Egoro's theorem (see, for example, Royden [88℄) and the de nition of rA imply the existen e of a set B A \ (Æ; 1) of positive Lebesgue measure so that (pi )1 i onverges uniformly on B and hen e is uniformly Cau hy on B: Now Theorem 6.2.2 yields that (pi )1 i is uniformly Cau hy on [0; Æ ℄; and the result follows. ut =1

=1


1 d℄ Suppose 1 i 1=i = 1: Show that there is a sequen e (pi )i M () that onverges pointwise on [0; 1) but does not onverge uniformly on A \ [0; a℄ for some a 2 (0; rA ): Hint: Use Theorem 4.2.1 (Muntz's theorem). ut P1 e℄ Suppose i 1=i < 1 and inf fi i : i 2 N g > 0 : Let P () denote the olle tion of all real-valued fun tions f de ned on [0; 1) by a power series P

=1

=1

=1

1

f (x) =

1 X i=0

ai 2 R ;

ai xi ;

x 2 [0; 1) :

Suppose that A [0; 1℄ with rA = 1: Show that if (fi )1 i fi (x) ! f (x) ; x 2 A; then fi (x) ! fe(x) ; x 2 [0; 1) ; e where f 2 P (). Hint: Use part ℄ and E.3 e℄ of Se tion 4.2.

=1

P () and

ut

E.2 On the Smallest Zero of Chebyshev Polynomials in Nondense Muntz P Spa es. Suppose := 0 and 1 1 = i < 1: Show that there exists a i

onstant > 0 depending only on := (i )1 (and not on n) su h that i the smallest positive zero of Tn f0; ; ; : : : ; n ; [0; 1℄g ; n = 1; 2; : : : is greater than : Hint: If 1; then use the Mean Value Theorem, E.1 a℄ of Se tion 3.3, and E.5 b℄ of Se tion 4.2. If 0 < < 1; then the s aling x ! x =1 redu es the problem to the ase = 1: ut 0

=1

=0

1

2

1

1

1

1

E.3 Extremal Fun tions for the Remez-Type Inequality of Theorem 6.2.2. Suppose 0 < < < n ; 0 < %; A [%; 1) is a ompa t set

ontaining at least n + 1 points, and y 2 (0; %) is xed. Let 0

1

Mn () := spanfx0 ; x1 ; : : : ; xn g : a℄ Show that there is a 0 6= p 2 Mn () su h that jp (y)j = sup jp(y)j : kp kA 6 p2Mn kpkA 0=

(

)

Hint: Use a ompa tness argument. b℄ Show that p = Tn f ; ; : : : ; n ; Ag for some 2 R: Hint: Use a perturbation argument. 0

1

ut ut


E.4 A Lexi ographi Property of Chebyshev Polynomials in Dierent Muntz Spa es. Let 0 := < < < n ; 0 := < < < n ; and j j ; j = 0; 1; : : : ; n : 0

1

0

1

Let % > 0 and let A [%; 1) be ompa t ontaining at least n + 1 points, and let

Tn; := Tn f ; ; : : : ; n ; Ag and Tn; := Tnf ; ; : : : ; n ; Ag : 0

1

0

1

a℄ Show that jTn; (y )j jTn; (y )j for every y 2 [0; %): Hint: Suppose, without loss of generality, that there is an index m, 1 m n; su h that m < m and j = j if j 6= m: We hoose an Rn; 2 Mn () that interpolates Tn; at the n zeros of Tn; and is normalized so that Rn; (0) = Tn; (0): Use Theorem 3.2.5 to show that jRn; (x)j jTn; (x)j for every x 2 [0; 1); in parti ular for every x 2 A: Now use E.3 to show that jTn; (0)j = jRn; (0)j jTn; (0)j; whi h gives the desired result for y = 0: Using this, we an dedu e that jTn; (y )j jTn; (y )j for every y 2 [0; %); otherwise

Tn;

Tn;

2 spanfx ; x ; : : : ; xn ; x m g 0

1

would have at least (n + 2) zeros in (0; 1); whi h is a ontradi tion. b℄ Show that max

p2Mn (

)

ut

jp(y)j max jp(y)j kpkA p2Mn kpkA (

)

for every y 2 [0; %); where

Mn ( ) := spanfx 0 ; x 1 ; : : : ; x n g and

Mn() := spanfx0 ; x1 ; : : : ; xn g :

ut

Hint: Combine part a℄ and E.4.

E.5 Theorem 6.2.1 Follows from Theorem 6.2.2. Under the assumptions of Theorem 6.2.1 show that if (pj )1 M () is uniformly Cau hy in j C (A); then it is uniformly Cau hy in C [0; y℄ for every y 2 (0; rA ), where rA is de ned as in Theorem 6.2.1. Hint: Use Theorem 6.2.2. ut =1


E.6 Some Corollaries of Theorem 6.2.9 in the La unary Case. Suppose := 0, 1; and := (i )1 i is la unary. a℄ Show that the Chebyshev polynomials Tn := Tn f ; ; : : : ; n : [0; 1℄g ; n = 1; 2; : : : have the following property: There is a onstant 2 (0; 1) depending only on (and not on n) su h that if y 2 [0; 1) and jTn (y )j = 1; then jTn (x)j for every x 2 [y; y + (1 y )℄: Hint: Use the Mean Value Theorem and the Bernstein-type inequality of Theorem 6.2.9. ut b℄ Show that there is a onstant 2 (0; 1) depending only on (and not on n) so that if a < b are two onse utive zeros of Tn ; then 1 b < (1 a):

℄ Let 2 (0; 1): Show that there is an n 2 N depending only on (and not on n) so that every Tn has at most n zeros in [0; 1 ℄: d℄ Give a new proof of the rst part of Theorem 6.2.1 based on parts a℄ and ℄. 0

1

=0

0

1

1

1

2

1

2

2

0

0

Outline. By Lebesgue's density theorem (see Royden [88℄), it may be supposed, without loss of generality, that the left-hand side Lebesgue density of A at 1 is 1: Choose 2 (0; 1) so that A \ [0; 1 ) ontains in nitely many points and m(A \ [y; 1℄) (6:2:5) > 1 y for every y 2 [1 ; 1℄; where 2 (0; 1) is the same as in part a℄. For this ; hoose n a

ording to part ℄. Now de ne g 2 C (A) so that g alternates n + 3 times in A \ [0; 1 ) between 2 and 2 and is identi ally zero on [1 ; 1℄: Assume that there exists a p 2 Mn () su h that kp g kA : Use part a℄ and (6.2.5) to show that p Tn 2 Mn () has more than n distin t zeros in [0; 1℄; whi h is a ontradi tion. ut 1

1

0

0

1

4

The following simple appli ation of Theorem 6.2.9 was pointed out by Woj ieszyk: e℄ Suppose A [0; 1℄ is a measurable set and the left-hand side Lebesgue density of A at 1 is 1: Show that there is a onstant > 0 depending only on and A so that

kpk

kpkA g:

[0;1℄ M () = span x 0 ; x 1 ; : : :

for every p 2 f Hint: Use the Mean Value Theorem, the Bernstein-type inequality of Theorem 6.2.9, and the Chebyshev-type inequality of E.3 f℄ of Se tion 4.2. u t f℄ Use part e℄ to give another proof of Theorem 6.2.1. The following exer ise onstru ts quasi-Chebyshev polynomials Pn for Mn () if the la unarity onstant of is large:


E.7 Quasi-Chebyshev Polynomials in Very La unary Muntz Spa es. Let = 0, = 2; and i =i 16 for i = 1; 2; : : : : Let 0

1

+1

n X

Pn (x) := 1 + 2

j =1

( 1)j xj ;

n = 1; 2; : : : :

Let yi := (4i ) . Prove the following statements: a℄ kPn k ; = 1 and Pn (1) = ( 1)n : b℄ Pn has exa tly n zeros, x ;n < x ;n < < xn;n , in (0; 1):

℄ jPn0 ( )j 2n for every 2 [xn;n ; 1℄. d℄ We have 1

[0 1℄

1

Pn (x)

1 if 1 5

2

y

2

k

y

x1

2

k

2

and 1 2k n

and

Pn (x)

1 if 1 5

y

k

2 +1

x1

y

and 1 2k + 1 n :

k

2 +1

2

Hint: Part a℄ is obvious. Prove the rest together, by indu tion on n: The next exer ise follows Borwein and Erdelyi [95b℄.

ut

E.8 Produ ts of Muntz Spa es. Asso iated with := (j )1 j ; let =0

(

M k () :=

p=

)

k Y j =1

pj : pj

2 M ()

;

k = 1; 2; : : : :

Is M () dense in C [0; 1℄ for := (j )1 j ? 2

2

=0

Note that M k (); k 2 is not the linear span of monomials, and Muntz's theorem does not give the answer. This exer ise establishes Remez-, Bernstein-, and Nikolskii-typePinequalities for M k (): From any of these it follows immediately that if 1 j 1=j < 1 and A [0; 1℄ is a set of positive Lebesgue measure, then M k () is not dense in C (A): P1 Throughout parts a℄ to d℄ of the exer ise we assume 0 = < < , j 1=j < 1; and s 2 (0; 1): =1

0

1

=1

a℄ Remez-Type Inequality for M k (). There exists a onstant depending only on ; s; and k (and not on % or A) su h that

kpk

;%℄

[0

kpkA

for every p 2 M k () and for every ompa t set A measure at least s > 0:

[%; 1℄ of Lebesgue


Proof. Theorem 6.2.2 implies that there exists a onstant > 0 depending only on , s; and k su h that s m(fx 2 [y; 1℄ : jp(x)j > jp(y)jg) 1 y 2k 1

for every p 2 M () and y 2 [0; 1

p= Then, for every y 2 [0; 1

k Y j =1

s℄: Now let p 2 M k (), that is,

pj ;

2 M () :

pj

s℄;

m(fx 2 [y; 1℄ : jp(x)j >

m 1

k \

j =1

y

j

jg

k p(y ) )

fx 2 [y; 1℄ : jpj (x)j > jpj (y)jg

!

1

k

s =1 y 2k

s

2

:

Hen e y 2 [0; inf A℄ and m(A) s imply that

m(fx 2 A : jp(x)j >

j

jg 2s > 0

k p(y ) )

ut

and the inequality follows with = k .

b℄ Solution to Newman's Problem. Let A [0; 1℄ be a set of positive Lebesgue measure. Then M k () is not dense in C (A):

ut

Proof. This follows from part a℄.

℄ Bernstein-Type Inequality for M k (). Suppose 1: There exists a

onstant depending only on , s; and k (and not on % and A) su h that 1

kp0k

;%℄

[0

kpkA

for every p 2 M k () and for every ompa t set A [%; 1℄ of Lebesgue measure at least s > 0: Hint: Use the produ t rule of dierentiation, and estimate ea h term separately. Pro eed as in the proof of part a℄. Use Theorem 6.2.2 and E.5 a℄ and b℄ of Se tion 4.2. ut k d℄ Nikolskii-Type Inequality for M (). There exists a onstant depending only on , s, k , q , and w (and not on % and A) su h that Z

kpkq ;% jp(x)jq w(x) dx [0

℄

A


for every p 2 M k (); for every ompa t set A [%; 1℄ of Lebesgue measure at least s > 0; for every fun tion w measurable and positive a.e. on [0; 1℄; and for every q 2 (0; 1): Hint: Use part a℄. ut e℄ Asso iated with

j = (i;j )1 i ;

j = 1; 2; : : : ; k ;

=0

let

8
1 depending only on su h that n n :

Outline. Let

Tn := Tn f ; ; : : : ; n ; [0; 1℄g and denote its zeros in (0; 1) by x ;n > x ;n > > xn;n : Use the Mean 0

1

1

2

Value Theorem and the assumed Bernstein-type inequality to show that there is a onstant 2 (0; 1) depending only on su h that 1

xj;n (1 xj

j = 1; 2; : : : ; n 1 ; n 2 N ;

;n ) ;

+1

hen e 1 x ;n n : On the other hand, use the Mean Value Theorem and Theorem 6.1.1 (Newman's inequality) to show that 1

1

x

1

;n

1+9

n X j =1

!

j

1

(9(n + 1)n )

Finally, ombine the lower and upper bounds for 1

n

n : 9(n + 1)

x

1

1

:

;n to on lude that

ut


E.10 Polynomials in xn . Given n 2 N and n 2 R; let

Pn (n ) := fpn (xn ) : pn 2 Pn g (as in E.6 of Se tion 4.1). Suppose n 1 for all n 2 N : Let Æ 2 R be de ned by log n 1 1 lim sup = log : n 2 Æ n Suppose Æ > 0: a℄ Bounded Remez-Type Inequality. Suppose 0 < Æe < Æ: Show that there exists a onstant depending only on Æe (and not on n, y; or A) su h that

jp(y)j kpkA for every p 2 [1 n Pn (n ); for every A [0; 1℄ of Lebesgue measure at least e and for every y 2 [0; inf A℄: 1 Æ; Hint: Use Lemma 6.2.6 and E.6 a℄ of Se tion 4.1. ut =1

b℄ Muntz-Type Theorem. If 0 Æe < Æ and A [0; 1℄ is a set of Lebesgue e then [1 Pn (n ) is not dense in C (A). measure at least 1 Æ; n Hint: Use part a℄. ut =1

This is page

320


7

Inequalities for Rational Fun tion Spa es

Overview Pre ise Markov- and Bernstein-type inequalities are given for various lasses of rational fun tions in the rst se tion of this hapter. Extensions of the inequalities of Lax, S hur, and Russak are also presented, as are inequalities for self-re ipro al polynomials. The se ond se tion of the hapter is

on erned with metri inequalities for polynomials and rational fun tions.

7.1 Inequalities for Rational Fun tion Spa es Sharp extensions of most of the polynomial inequalities of Se tion 5.1 are established for rational fun tion spa es on K := R (mod 2 ), on the interval [ 1; 1℄, on the unit ir le of C , and on the real line. The lassi al inequalities of Se tion 5.1 are then re overed as limiting ases. A sharp extension of Lax's inequality is also given. Essentially sharp Markov- and Bernsteintype inequalities for self-re ipro al and antiself-re ipro al polynomials are presented in the exer ises. Let D := fz 2 C : jz j < 1g and D := fz 2 C : jz j = 1g; as before. We study the rational fun tion spa es:

Tn (a ; a ; : : : ; a n ; K ) := 1

2

2

(

Q2n

k=1

t() : t 2 Tn j sin(( ak )=2)j

)

7.1 Inequalities for Rational Fun tion Spa es 321

and (

T

n (a1 ; a2 ; : : :

; a n ; K ) :=

on K with a ; a ; : : : ; a n 2 C 1

2

2

1; 1℄) :=

Pn (a ; a ; : : : ; an ; [

1; 1℄) :=

2

k=1

)

n R;

Pn (a ; a ; : : : ; an ; [ 1

Q2n

2

t() : t 2 Tn sin(( ak )=2)

Qn

p(x) jx ak j : p 2 Pn

Qn

p(x) : p 2 Pn (x ak )

k=1

and 1

2

on [ 1; 1℄ with a ; a ; : : : ; an 2 C 1

2

Pn (a ; a ; : : : ; an ; D) := 1

k=1

n[

1

2

p(z ) : p 2 Pn (z ak )


Qn

Qn

p(z ) : p 2 Pn (z ak )

Qn

k=1

n D; and

Pn (a ; a ; : : : ; an ; R) := 1

1; 1℄;

2

on D with a ; a ; : : : ; an 2 C

2

k=1

and

P

n (a1 ; a2 ; : : :

; an ; R) :=

on R with a ; a ; : : : ; an 2 C 1

2

k=1

n R:

The Chebyshev polynomials Ten , Uen , and

2K

V := ( os )Ten + (sin )Uen ;

for the rational fun tion spa e Tn (a ; a ; : : : ; a n ; K ) are de ned in E.3 of Se tion 3.5, and they play a entral role in this se tion. 1

2

2

322 7. Inequalities for Rational Fun tion Spa es

Theorem 7.1.1 (Bernstein-Szeg}o-Type Inequality on K ). Given (ak )kn 2

=1

C nR;

let Ben () :=

Im(ak ) > 0 ;

n

1 X 1 jeiak j : 2 k jeiak ei j 2

2

2

=1

Then

f 0 () + Ben ()f () Ben ()kf kK ; 2K for every f 2 Tn (a ; a ; : : : ; a n ; K ): Equality holds if and only if either is a maximum point of jf j (that is, f () = kf kK ) or f is a linear ombination of Ten and Uen (with real

oeÆ ients) as de ned in E.3 of Se tion 3.5. 2

2

1

2

2

2

2

2

Corollary 7.1.2 (Bernstein-Type Inequality on K , Real Case). Given (ak )kn 2

=1

C nR;

Im(ak ) > 0 ;

let the Bernstein fa tor Ben be de ned as in Theorem 7.1.1. Then jf 0 ()j Ben ()kf kK ; 2 K

for every f 2 Tn (a ; a ; : : : ; a n K ): Equality holds if and only if f is a linear ombination of Ten and Uen (with real oeÆ ients) as de ned in E.3 of Se tion 3.5, and f () = 0: 1

2

2

;

Theorem 7.1.1 and Corollary 7.1.2 an be easily obtained from the extension of Theorem 3.5.3 given by E.3 of Se tion 3.5, whi h gives expli it formulas for the Chebyshev polynomials for these lasses Tn (a ; a ; : : : ; a n ; K ): The arguments are outlined in E.1. The following two results an be obtained from Theorem 7.1.1 and Corollary 7.1.2 by the substitution x = os ; see E.2. 1

2

2

Corollary 7.1.3 (Bernstein-Szeg}o-Type Inequality on [ 1; 1℄). Asso iated C n [ 1; 1℄; let the Bernstein fa tor Bn be de ned by

with (ak )nk

=1

Bn (x) := p

n X k=1

p

Re

ak

ak

1

2

x

!

;

p

ak 1 < 1: Then where the hoi e of ak 1 is determined by ak x 2 [ 1; 1℄ (1 x )f 0 (x) + Bn (x)f (x) Bn (x)kf k ; ; 2

2

2

2

2

2

2 [ 1 1℄

2

for every f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄): Equality holds if and only if either x is a maximum point of jf j (that is, f (x) = kf k ; ) or f = Tn with 2 R; where Tn is de ned as in Se tion 3.5. 1

[

1 1℄

2


Corollary 7.1.4 (Bernstein-Type Inequality on [ 1; 1℄, Real Case). Given (ak )nk C n [ 1; 1℄; let the Bernstein fa tor Bn be de ned as in Corollary =1

7.1.3. Then

jf 0 (x)j pBn (x) kf k 1

x

2

[

;

1 1℄

x 2 ( 1; 1)

;

for every f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄): Equality holds if and only if f = Tn with 2 R; where Tn is de ned as in Se tion 3.5, and f (x) = 0: (Note that Bn (x) > 0 for every x 2 ( 1; 1).) 1

2

Our next result follows from Theorem 7.1.1; see the hints to E.3. Corollary 7.1.5 (Bernstein-Szeg}o-Type Inequality on R). Given (ak )nk

=1

C nR;

Im(ak ) > 0 ;

let the Bernstein fa tor Bn be de ned by Bn (x) := Then

n X k=1

Im(ak )

jx

ak j

2

:

f 0 (x) + Bn (x)f (x) Bn (x)kf kR ; 2

2

2

2

2

x2R

for every f 2 Pn (a ; a ; : : : ; an ; R): Equality holds if and only if either x is a maximum point of jf j (that is, jf (x)j = kf kR) or f is a linear ombination of Tn and Un (with real

oeÆ ients) de ned in E.5 of Se tion 3.5. 1

2

Corollary 7.1.6 (Bernstein-Type Inequality on R, Real Case). Asso iated with (ak )nk C n R ; let the Bernstein fa tor Bn be de ned as in Corollary =1

7.1.5. Then

jf 0 (x)j Bn (x)kf kR ;

x2R

for every f 2 Pn (a ; a ; : : : ; an ; R): Equality holds if and only if f is a linear ombination of Tn and Un (with real oeÆ ients) de ned in E.5 of Se tion 3.5, and f (x) = 0: 1

2

To formulate our next theorem we introdu e some notation. For a polynomial

q(z ) := we de ne

n Y

k=1

(z

ak ) ;

ak 2 C ;

324 7. Inequalities for Rational Fun tion Spa es n

Y q (z ) := (1 ak z ) =: z n q n (z ) : 1

k=1

Then

jq(z )j = jq (z )j ;

(7:1:1) The fun tion

Sn (z ) :=

z 2 D :

n q (z ) Y 1 ak z = q(z ) k z ak =1

is the Blas hke produ t asso iated with (ak )nk : =1

Theorem 7.1.7 (Bernstein-Type Inequality on D, Complex Case). Given (ak )nk C n D ; let the Bernstein fa tor Bn be de ned by =1

Bn (z ) := maxfBn (z ); Bn (z )g +

with Bn (z ) := +

n X

2

jakk=1 j>1

Then

j ak j 1 j ak z j

and

2

Bn (z ) :=

jf 0 (z )j Bn (z )kf kD ;

n X

jakk=1 j 1: If the se ond sum is not less than the rst sum for a xed z 2 D; then equality holds for f = Sn with

2 C ; where Sn is the Blas hke produ t asso iated with those ak for whi h jak j < 1: 1

2

+

+

Proof. For reasons of symmetry it is suÆ ient to prove the theorem only for z = 1: Without loss of generality we may assume that (7:1:2)

Re

n X k=1

1

1

!

ak

6= n2 ;

the remaining ases follow from this by a limiting argument. Let Q := D (equipped with the usual metri topology), V := Pn (a ; a ; : : : ; an ; D) ; and L(f ) := f 0 (1) for f 2 V: We show in this situation that n + 1 r in Theorem A.3.3 (interpolation of linear fun tionals). Suppose to the ontrary that r n: By Theorem A.3.3, there are distin t points x ; x ; : : : ; xr on D, and there are onstants ; ; : : : ; r 2 C su h that 1

2

1

1

2

2


(7:1:3)

r p(x ) p0 (1)q(1) q0 (1)p(1) X =

k k ; q(1) q(xk ) k

p 2 Pn ;

2

=1

where (7:1:4)

q(z ) :=

n Y

(z

k=1

ak ) :

We laim that xk 6= 1 for ea h k = 1; 2; : : : ; r: Indeed, if there is index k su h that xk = 1; then Theorem A.3.3 implies that

p(z ) := (z + 1)n

r

r Y k=1

xk ) 2 Pn

(z

has a zero at 1 with multipli ity at least two, whi h is a ontradi tion. Applying (7.1.3) with the above p; we obtain

p0 (1)q(1)

q0 (1)p(1) = 0

and sin e p(1) 6= 0 and q (1) 6= 0, this is equivalent to

q0 (1) p0 (1) = ; q(1) p(1) that is, in terms of the zeros of pn and qn ; n X

(7:1:5) Sin e xk (7:1:6)

k=1

1

1

=

ak

n r 2

+

r X k=1

1

1

xk

:

2 D and xk 6= 1; we have Re

1

1

xk

=

1 ; 2

k = 1; 2; : : : ; r :

It follows from (7.1.5) and (7.1.6) that Re

n X k=1

1

1

!

ak

=

n 2

;

whi h ontradi ts assumption (7.1.2). So n + 1 r; indeed. that

A ompa tness argument shows that there is a fun tion fe 2 V su h

kfekD = 1

and

jL(f )j : L(fe) = kLk := max 6 f 2V kf kD 0=


Theorem A.3.3 implies jfe(xk )j = 1 for every k = 1; 2; : : : ; r: Hen e, if

pe fe = ; q

pe 2 Pn ;

q(z ) =

n Y k=1

(z

ak ) ;

then

h(z ) := jpe(z )j

(7:1:7)

jq(z )j 0 ;

2

z 2 D

2

and (7:1:8)

h(xk ) = 0 ;

k = 1; 2; : : : ; r :

Note that t() := h(ei ) 2 Tn vanishes at ea h k , where the numbers k 2 [ ; ) are de ned by xk = eik ; k = 1; 2; : : : ; r: Be ause of (7.1.7), ea h of these zeros is of even multipli ity. Hen e, n + 1 r implies that t 2 Tn has at least 2n + 2 zeros and therefore t = 0: From this we an dedu e that h(z ) = 0 for every z 2 D; so

jpe(z )j = jq(z )j ;

(7:1:9)

z 2 D :

We now have

z npe(z )pe (z ) = jpe(z )j = jq(z )j = z nq(z )q (z ) ; 2

2

z 2 D ;

so by the uni ity theorem for analyti fun tions (see E.1 e℄ of Se tion 1.2)

pe pe = qq : From this, it follows that there exists a onstant 0 6= 2 C su h that

fe(z ) =

m Y z k pe(z ) = ; q(z ) z k k 1

z2C;

q(z ) 6= 0

=1

with some m n and

k := ajk ;

k = 1; 2; : : : ; m ;

1 j < j < < jm n : 1

2

A straightforward al ulation gives that

m e0 X 1 jfe0 (1)j = fe(1) = 1 k f (1) k m X =

k=1


1

1

=1

1 1 2

1

k

jk j jk j maxfBn (z ); Bn (z )g ; 2

+

ut


Corollary 7.1.8 (Bernstein-Type Inequality on K , Complex Case). Given (ak )kn C n R; let the Bernstein fa tor Bn be de ned by 2

=1

Ben () := maxfBen (); Ben ()g +

with Ben () :=

jeiak j

2n X

+

k=1 Im(ak )0

jeiak

jeiak j

2

ei j

2

:

2K

for every f 2 Tn (a ; a ; : : : ; a n ; K ): If the rst sum is not less than the se ond sum for a xed 2 K; then equality holds for f () = S n (ei ) with 2 C ; where S n is the Blas hke produ t asso iated with those eiak for whi h Im(ak ) < 0: If the se ond sum is not less than the rst sum for a xed 2 K; then equality holds for f () = S n (ei ) with 2 C ; where S n is the Blas hke produ t asso iated with those eiak for whi h Im(ak ) > 0: Note that S (ei ) 2 T (a ; a ; : : : ; a n ; K ) : 1

2

2

+

+

2

2

2

2

2

n

n

1

2

2

Proof. Observe that if h() :=

2n Y

j =1

aj )=2) 2 Tn

sin((

and tn 2 Tn ; then there are p 2 P n and q 2 P n su h that 2

2

t() p(ei )e = h() q(ei )e where q is of the form

q(z ) =

2n Y

in p(ei ) = in q(ei )

;

eiaj )

(z

j =1

with some 2 C : So the orollary follows from Theorem 7.1.7.

ut

Corollary 7.1.9 (Bernstein-Type Inequality on [ 1; 1℄, Complex Case). C n [ 1; 1℄, let the Bernstein fa tor Bn (x) be de ned

Given fak gnk by

=1

Bn (x) = max

( n X

k=1

j k j j k z j 1

2 2

;

n X k=1

j k j j k

1

2

1

zj

2

)

;


where k and z are determined by

j k j < 1 ;

ak := ( k + k ) ; x := (z + z ) ; 1

1

2

1

Im(z ) > 0 :

1

2

Then

jf 0 (x)j pBn (x) kf k x

1

for every f

2

2 Pn (a ; a ; : : : ; an ; [ 1

n X k=1

2

j k j j k z j 1

2

p

where the hoi e of

ak

;

1 1℄

x 2 ( 1; 1)

;

1; 1℄): Note that

n X

= Re

2

[

p

k=1

ak 1 ak x 2

!

;

x 2 [ 1; 1℄ ;

p

1 is determined by ak

2

ak 2

1 < 1:

Proof. The orollary follows from Theorem 7.1.7 by the substitution ut x = (z + z ): 1

1

2

Bernstein's lassi al polynomial inequalities dis ussed in Se tion 5.1 are ontained in Theorem 7.1.7 and Corollaries 7.1.8 and 7.1.9 as limiting

ases. In Theorem 7.1.7 and Corollary 7.1.9 we take (a m ; a m ; : : : ; anm ) C (

)

(

1

so that

)

(

lim jakm j = 1 ; (

m!1

)

2

nD

k = 1; 2; : : : ; n :

)

In Corollary 7.1.8 we take (a m ; a m ; : : : ; a m n )C (

1

)

(

(

)

)

2

2

nR

so that

anm k = akm (

)

(

and

)

+

lim jIm(akm )j = 1 ; (

m!1

)

k = 1; 2; : : : ; n :

To formulate our next result we introdu e the Blas hke produ t

Qn (z ) := asso iated with (a ; a ; : : : ; an )

z 2 R:

1

2

n Y k=1

z z

ak ak

C n R: Obviously jQn (z )j = 1 for every


Corollary 7.1.10 (Bernstein-Type Inequality on R, Complex Case). Given (ak )nk C n R; let the Bernstein fa tor Bn (x) be de ned by =1

Bn (x) := maxfBn (x); Bn (x)g +

with Bn (x) := +

n X

2 jIm(ak )j

k=1 ak )>0

jx

ak j

2

n X

and Bn (x) :=

Im(

k=1 ak ) 0g : +

If the se ond sum is not less than the rst sum for a xed x 2 R; then equality holds for f = Qn with 2 C ; where Qn is the Blas hke produ t asso iated with the poles ak lying in the open lower half-plane H := fz 2 C : Im(z ) < 0g : Corollary 7.1.10 follows from Theorem 7.1.7; see E.4. The next theorem improves the Bernstein-type inequality of Theorem 7.1.7 in the ase when fak gnk C n D and f has all its zeros in C n D: It extends Lax [44℄. =1

Theorem 7.1.11 (Lax-Type Inequality). Given (ak )nk

=1

Bernstein fa tor Bn be, as in Theorem 7.1.7, de ned by Bn (z ) := Then

jh0 (z )j

1 2

n X k=1

C n D,

let the

jak j 1 : jak z j 2

Bn (z )khkD ;

2

z 2 D

for every h 2 Pn (a ; a ; : : : ; an ; D) having all its zeros in C n D: Equality holds for h = (Sn + 1) with 2 C ; where Sn is the Blas hke produ t asso iated with (ak )nk . 1

2

=1


Note that Bn (z ) = jSn0 (z )j: Note also that

6= 0

h := (Sn + 1) 2 Pn (a ; a ; : : : ; an ; D) ; 1

2

has all its zeros on D: Proof. First assume that ea h zero of h is on D; the general ase an be redu ed to this (see E.5). Thus let h := p=q; where p 2 Pn has all its zeros on D and where q(z ) := Let

n Y k=1

(z

q (z ) :=

We study

jak j > 1 :

ak ) ; n Y k=1

ak z ) :

(1

p(e i )e in jq(e i )j =

p(e i ) p q(e i q (e i ) p for 2 R; where the square roots are taken so that q is analyti in a p neighborhood of the losed unit disk, and q is analyti in a neighborhood of the omplement of the open unit disk. Sin e p 2 Pn has all its zeros on D; there exists a 0 6= 2 C su h that u() :=

2

2

p

2

2

t() := p(e i )e

)

2

in

2

is a real trigonometri polynomial of degree at most n (see E.5 a℄). Also

j

j

q(e i) = 2

n

Y jq (e i )j = j1 ak e

=

2

2

2n Y

k=1

k=1

j sin((

i

j

k )=2)j ;

where > 0;

ei k = ak

=

1 2

;

Im( k ) > 0 ;

k = 1; 2; : : : ; n

and

ei k = ak

=

1 2

;

Im( k ) > 0 ;

k = n + 1; n + 2; : : : ; 2n :

Applying Theorem 7.1.1 to

u 2 Tn ( ; ; : : : ; n ; K ) ; 1

1

2

2


we obtain

ju0 ()

(7:1:10)

iBen ()u()j Ben ()kukK ;

2K;

where n

1X Ben () = 2k

(7:1:11)

n X

k=1

jak = j ak j

1

jak

Observe that

p(e i ) u() = q(e i ) 2

(7:1:12)

2

e

1

2

p

2

2

i

p

1

ei j

1 2

=1

=

jak j

1

=

j

2

n X k=1

j

ak

jak j

jak

ei j

=

1 2

2

1

2

e

2

i

!

1

j

2

:

q(e i ) p(e i ) fn (ei ) ; = q (e i ) q(e i ) 2

2

2

2

where

p

(7:1:13)

+

jak j

1

fn (z ) :=

p

q(z ) : q (z ) 2

2

A simple al ulation (see E.4. of Se tion 3.5) shows that

Ben () = ei

(7:1:14)

fn0 (ei ) ; fn (ei )

2K:

Also, sin e jfn (ei )j = 1 for every 2 K; we have

p

K =

q

kuk

(7:1:15)

D

= khkD :

Now (7.1.10) to (7.1.15) yield d p(e2i ) i ) f ( e d q (e2i ) n

iei

fn0 (ei ) p(e i ) fn (ei ) Ben ()khkD : i i fn (e ) q(e ) 2

2

So

j2ie i h0 (e i )fn (e i ) + iei fn0 (ei )h(e i ) iei fn0 (ei )h(e i )j Ben ()khkD : 2

2

2

2

2

Thus

2 jh0 (e i )j Ben ()khkD ; 2

whi h, together with (7.1.11), nishes the proof.

ut


Comments, Exer ises, and Examples. Most of the results in this se tion have been proved in Borwein and Erdelyi [to appear 4℄ and in Borwein, Erdelyi, and Zhang [94a℄. A weaker version of Corollary 7.1.9 has been obtained by Russak (see Petrushev and Popov [87℄). Theorem 7.1.11 ontains, as a limiting ase, an inequality of Lax [44℄

onje tured by Erd}os. Lax's inequality establishes the sharp Bernstein-type inequality on the unit disk for polynomials p 2 Pn having no zeros in the open unit disk. That is,

kp0 kD n2 kpkD

for su h polynomials. Various extensions of this inequality are given by Ankeny and Rivlin [55℄, Govil [73℄, Malik [69℄, and others. We dis uss some of these in E.16 of Appendix 5. E.1 Proof of Theorem 7.1.1 and Corollary 7.1.2. Given (ak )kn let Tn;a := Tn (a ; a ; : : : ; a n ; K ) : 2

=1

1

2

C n R;

2

a℄ Show that Tn;a is a Hermite interpolation spa e. That is, if the points x ;x P ; : : : ; xk 2 K are distin t, and m ; m ; : : : ; mk are positive integers with ki mi 2n + 1; then for any hoi e of real numbers yi;j , there is a fun tion f 2 Tn;a su h that 1

2

1

2

=1

f j (xi ) = yi;j ;

i = 1; 2; : : : ; k ; j = 0; 1; : : : ; mi

( )

1:

ut

Hint: See the hint to E.7 of Se tion 1.1. b℄ Show that for every xed 2 K; the value f 0 () + Bn ()f () 6 f 2Tn;a kf kK max

0=

2

2

2

2

is attained by an fe 2 Tn;a : Hint: Use a ompa tness argument. ut

℄ Show that fe = V; where 2 R and V is one of the Chebyshev polynomials for Tn;a de ned in Theorem 3.5.3 and E.3 of Se tion 3.5. Hint: Use a variational method with the help of part a℄. ut d℄ Prove Theorem 7.1.1. Hint: Use part ℄ and E.4 of Se tion 3.5. ut e℄ Prove Corollary 7.1.2. E.2 Proof of Corollaries 7.1.3 and 7.1.4. a℄ Prove the inequality of Corollary 7.1.3.


Hint: Use Theorem 7.1.1 with the substitution x = os : Note that f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄) implies 1

2

g() := f ( os ) = Tn ( ; ; ; ; : : : ; n ; n ; K ) ; 1

where the numbers k

ei k = ak

q

2C

ak

2

2

are de ned by

ak = (ei k + e

1;

2

1

1 2

i k );

Im( k ) > 0 :

Verify that if Ben is the Bernstein fa tor given in Theorem 7.1.1 asso iated with ( ; ; ; ; : : : ; n ; n ) ; 1

then

Ben () = p

n X k=1

1

2

2

p

Re

ak

ak 2

1

os

!

2K;

;

p

where the hoi e of ak 1 is determined by ak ak 1 < 1: ut b℄ Given x 2 [ 1; 1℄; prove that equality holds in the inequality of Corollary 7.1.3 if and only if either x is a maximum point of jf j (that is, f (x) = kf k ; ) or f = Tn with 2 R; where Tn is de ned in Se tion 3.5. Hint: Observe that V = ( os )Ten + (sin )Uen [

2

2

1 1℄

is even if and only if V = Ten and use Theorem 7.1.1.

℄ Prove Corollary 7.1.4.

ut

E.3 Proof of Corollaries 7.1.5 and 7.1.6. a℄ Prove Corollary 7.1.5.

ei + 1 Hint: Use Theorem 7.1.1 and the substitution x = i i ; whi h maps K e 1 onto R [ f1g: ut b℄ Prove Corollary 7.1.6.

E.4 Proof of Corollary 7.1.10. Prove Corollary 7.1.10. z+1 Hint: Use Theorem 7.1.7 and the substitution x = i ; whi h maps D z 1 onto R [ f1g : ut E.5 Completion of the Proof of Theorem 7.1.11. a℄ Show that if p 2 Pn has all its zeros on the unit ir le, then there is a 0 6= 2 C su h that g () := e in p(e i ) is a real trigonometri polynomial of degree at most n: 2


For f = p=q with

q(z ) := and

p(z ) :=

m Y k=1

(z

n Y k=1

ak ) ;

ak 2 C

bk 2 C ;

2C;

(z

bk ) ;

m n;

we de ne f := p =q; where m

Y p (z ) := (1 zbk ) :

k=1

Let D := fz 2 C : jz j < 1g: b℄ Show that jf (z )j = jf (z )j for every z 2 D: In ea h of the remaining parts of the exer ise suppose that jak j > 1 for ea h k:

℄ Show that if 2 D and jbk j 1 for ea h k; then f + f has all its zeros on the unit ir le. d℄ Show that if jbk j 1 for ea h k; then

jf 0 (z )j jf 0 (z )j ;

z 2 D :

Hint: First observe that it is suÆ ient to study the ase z = 1: We have 0 Re f (1) f (1)

=

=

Re Re

n 2

=n = Re = Re =

Re

n X k=1 n X k=1

Re Re n X

1 1

1 1

!

bk

1

k=1 n X k=1

k=1 1 n X k=1 1

Re

1

1 1

f 0 (1) f (1)

Re !

1

n

Re

! ak

2

n X

Re

k=1

n X k=1

n X

k=1 n X k=1

1 1

k=1

n X

Re

Re

bk

1

1

bk

!

!

1

bk

bk

+

1

1

k =1 n n + 2 2 !

!

bk

n X

n X

1 1

1

ak

!

ak

1

1

ak

!

1 1

!

1

ak

! ak !


and 0 Im f (1) f (1)

! ! n n X X 1 1 = Im Im 1 b 1 a k k k =1 k =1 ! ! n n X 1 X 1 = Im n Im 1 b 1 a k k k =1 k =1 ! ! n n X X 1 1 = Im + Im 1 1 ak bk k =1 1 k =1 0 f (1) = Im : f (1)

The result now follows from the ombination of part b℄ and the above two inequalities. ut e℄ Prove that if jbk j 1 for ea h k; then 2jf 0 (z )j jf 0 (z )j + jf 0 (z )j Bn (z )kf kD ;

z 2 D ;

where Bn (z ) is the Bernstein fa tor de ned in Theorem 7.1.11. Hint: Use parts ℄ and d℄ and the already proved part of Theorem 7.1.11 (when f has all its zeros on the unit ir le). ut f℄ Show that if jbk j 1 for ea h k; then

jf 0 (z )j 12 Bn (z ) zmax jf (z )j 2D

min jf (z )j ;

z2D

z 2 D ;

where Bn (z ) is the Bernstein fa tor de ned in Theorem 7.1.7. This extends a result of Aziz and Dawood [88℄. Hint: Assume that kf kD = 1: Let m := minz2D jf (z )j: Let be a onstant of modulus less than 1: Let g (z ) := f (z ) m: Observe that the argument of an be hosen so that

jg0 (z )j = jf 0 (z )j jjmBn (z ) : By Rou he's theorem, g has no zeros in D: So parts d℄ and e℄ imply that 2 jf 0 (z )j

2 jj mBn (z ) = 2 jg 0 (z )j jg 0 (z )j + jg 0 (z )j = jf 0 (z )j + jf 0 (z )j jjmBn (z ) Bn (z ) jjmBn (z ) :

Sin e jj an be hosen arbitrarily lose to 1, the result follows.

ut


E.6 Extensions of Russak's Inequalities. a℄ Given (ak )kn C n R, show that 2

=1

kf 0kL

1(

2n kf kK

K)

2 Tn (a ; a ; : : : ; a n ; K ) and kf 0kL K 4n kf kK for every f 2 Tn (a ; a ; : : : ; a n ; K ): b℄ Given (ak )nk C n R, show that kf 0 kL R n kf kR for every f 2 Pn (a ; a ; : : : ; an ; R; ) and kf 0kL R 2n kf kR for every f 2 Pn (a ; a ; : : : ; an ; R). for every f

1

2

2

1(

1

2

)

2

=1

1(

1

)

2

1(

1

)

2

Hint: Use Corollaries 7.1.2, 7.1.8, 7.1.6, and 7.1.10. Write the Bernstein

fa tors in a form so that the integral (of ea h term in the maximum if the Bernstein fa tor is de ned by a maximum) an be evaluated by the residue theorem (in part a℄) and by nding the antiderivative (in part b℄). ut

℄ Are any of the inequalities of parts a℄ and b℄ sharp? If so, in whi h

ases? E.7 Markov-Type Inequality. Given (ak )nk

=1

kf 0k for every f by

[

;

1 1℄

n

n

n X

1 k

2 Pn (a ; a ; : : : ; an ; [ 1

k := ak

q

2

ak 2

1;

=1

1 + j k j 1 j k j

R n[ !2

kf k

[

1; 1℄; show that ;

1 1℄

1; 1℄); where the numbers k are de ned

ak = ( k + k ) ; 1

1

2

j k j < 1 :

Pro eed as follows: a℄ Given (ak )nk R n [ 1; 1℄; let =1

8 >
k : + 1 y 1 y and let k (y ) be de ned by

if

0y1

if

1y0


j k (y)j < 1 :

ak (y) =: ( k (y) + k (y) ) ; 1

1

2

Show that

n X 2 1 + k (y ) 1 + jy j k 1 k (y ) for every f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄). Hint: Show by a variational method that

jf 0 (y)j

!2

kf k

[

;

1 1℄

=1

1

2

max f

jf 0 (1)j kf k ; [

= jTn0 (1)j ;

1 1℄

where the maximum is taken for all 0 6= f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄); and Tn is the Chebyshev polynomial for Pn (a ; a ; : : : ; an ; [ 1; 1℄) de ned in Se tion 3.5. Now the result follows from E.1 ℄ of Se tion 3.5 by a linear shift from [ 1; 1℄ to [ 1; y ℄ if 0 y 1; or to [y; 1℄ if 1 y 0: ut b℄ Given (ak )nk R n [ 1; 1℄; show that 1

1

2

2

=1

jf 0 (y)j 1 njyj kf k

;

[

1 1℄

y 2 ( 1; 1)

;

for every f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄): Hint: When y = 0; this follows from Corollary 7.1.3. When y 2 ( 1; 1) is arbitrary, use a linear shift from [ 1; 1℄ to [2y 1; 1℄ if 0 y 1; or to [ 1; 2y + 1℄ if 1 < y 0: ut

℄ Prove the Markov-type inequality of the exer ise. Hint: Combine parts a℄ and b℄. Note that 1

jak (y)j jak j holds for every y 2 [

2

and

j k (y)j j k j < 1 ;

ut

1; 1℄:

E.8 S hur-Type Inequality. Given fak gnk

=1

kf k

[

;

1 1℄

k = 1; 2; : : : ; n

maxfjUn (1)j; jUn (

R n[

p

1)jg f (x) 1

1; 1℄; show that

x

2

[

;

1 1℄

for every f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄); where Un is the Chebyshev polynomial (of the se ond kind) for Pn (a ; a ; : : : ; an ; [ 1; 1℄) de ned in Se tion 3.5, and n p X ak 1 jUn (1)j = a 1 k k 1

2

1

2

2

p

=1

with the hoi e of ak 1 determined by ak equality holds if and only if f = Un ; 2 R: 2

p

ak 2

1 < 1: Show that


Hint: First show that if fe 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄) is extremal for jf (1)j p ; max

f f (x) 1 x ; 1

2

2

[

1 1℄

where the maximum is taken for all 0 6= f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄); then f = Un with some 2 R: Observe that Un (1) an be evaluated by L'Hospital's rule sin e 1

2

Tn (x) ; 1 x hen e jUn (1)j = jTn0 (1)j = jBn (1)j jUn (1)j with the notation of Se tion 3.5. Thus jUn (1)j = jBn (1)j : If y 2 [ 1; 1℄ is arbitrary, then use a linear shift from [ 1; 1℄ to [ y; y ℄ Un (x) = 2

1

2

2

2

ut

(some aution must be exer ised about the hange of poles).

E.9 Extension of Lax's Inequality on the Half-Plane. Asso iated with (ak )nk H := fz 2 C : Im(z ) > 0g; let the Bernstein fa tor Bn be, as in Corollary 7.1.10, de ned by +

=1

Bn (x) := Show that for every h 2 P

jh0 (x)j

n (a1 ; a2 ; : : :

1 2

n X k=1

2 Im(ak )

jx

ak j

Bn (x)khkR ;

2

: x2R

; an; R) having all its zeros in H : Equality holds for h = (Sen + 1) with 2 C ; where Sen is the Blas hke produ t asso iated with (ak )nk : Note that Bn (z ) = jSen0 (z )j: Note also that +

=1

h=

(Se + 1)

2 Pn (a ; a ; : : : ; an ; R) ; 1

2

has all its zeros on R:

Hint: Use Theorem 7.1.11 with the substitution x = i

6= 0

z+1 : z 1

ut

E.10 Remarks on Theorem 7.1.7 and Corollary 7.1.10. a℄ Given (ak )nk D and z 2 D; show that equality holds in the inequality of Theorem 7.1.7 if and only if f = Sn with 2 C ; where Sn is the Blas hke produ t asso iated with (ak )nk : Hint: Analyze the proof of Theorem 7.1.7. ut b℄ Given (ak )nk H = fz 2 C : Im(z ) > 0g and x 2 R; show that equality holds in the inequality of Corollary 7.1.10 if and only if f = Qn with 2 C ; where Qn is the Blas hke produ t asso iated with (ak )nk : Note that the only if parts of E.10 a℄ and E.10 b℄ above are not laimed in the general ase of Theorem 7.1.7 and Corollary 7.1.10 (why?). =1

=1

+

=1

=1


E.11 Markov-Bernstein-Type Inequality for SR n and ASR n . Let SR n denote the set of all self-re ipro al polynomials p 2 Pn satisfying

p(z ) z np(z ) : 1

Let SRn denote the set of all real self-re ipro al polynomials of degree at most n; that is, SRn := SR n \ Pn : For a polynomial p 2 Pn of the form (7:1:16)

p(z ) =

n X j =0

j z j ;

j

2C ;

p 2 SR n if and only if

j = n j ;

j = 0; 1; : : : ; n :

Let ASR n denote the set of all antiself-re ipro al polynomials p 2 Pn satisfying p(z ) z np(z ) : 1

Let ASRn denote the set of all real antiself-re ipro al polynomials of degree at most n; that is, ASRn := ASR n \ Pn : Let ASRn := ASR n \ Pn . For a polynomial p 2 Pn of the form (7.1.16) p 2 ASR n if and only if

j = n j ;

j = 0; 1; : : : ; n :

a℄ There exists an absolute onstant su h that

jp0 (x)j n min

(1 + log n); log

e

1

x

2

kpk

[

;

1 1℄

holds for every x 2 [ 1; 1℄ and for every p 2 Pn satisfying (7:1:17)

jp(x)j (1 + jxjn )kpk

[

;

1 1℄

x2R;

;

in parti ular, for every p 2 SR n and for every p 2 ASR n : The inequality

jp0 (x)j n(1 + log n) kpk

[

;

1 1℄

for all p 2 SRn and for all p 2 ASRn was rst obtained by Kroo and Szabados [94a℄. They also showed that up to the onstant > 0 the above inequality is sharp for both SRn and ASRn : Here we present a distin t proof. The sharpness is studied in E.12 f℄.


Outline. Suppose p 2 Pn satis es (7.1.17). Then p(x) (7:1:18) f (x) := 1+x n 2

satis es

kf kR 2 kpk

(7:1:19)

;

[

1 1℄

:

Let

ak := ei

(7:1:20)

k

(2

=(2n) ;

k = 1; 2; : : : ; 2n

1)

be the zeros of the equation z n + 1 = 0: Now Corollary 7.1.10, together with (7.1.18) to (7.1.20), yields that 2

(7:1:21)

n X

jf 0 (x)j 2

k=1 n X

4

k=1

!

Im(ak )

jx

ak j

2

!

Im(ak )

jx

ak j

2

kf kR kpk

[

;

1 1℄

x 2 R:

;

Show that if n 2 N and x 2 [ 1; 1℄; then (7:1:22)

n X

4

k=1

Im(ak )

jx

ak j

n min

2

(1 + log n); log

e

1

x

2

;

where here the symbol means that there are absolute onstants > 0 and > 0 (independent of n 2 N and x 2 [ 1; 1℄) su h that the left-hand side is between times the right-hand side and times the right-hand side for every n 2 N and x 2 [ 1; 1℄: Combining (7.1.18), (7.1.21), and (7.1.22), we on lude that there is an absolute onstant su h that 1

2

1

0 p (x) 1 + x2 n

2

p(x) 2nx 1 + x2n 1 + x2n 2

n

n min 2

So if x 2 [ 1; 1℄; then

2

1

(1 + log n); log

1

jp0 (x)j 2 n min (1 + log n); log 2

e

x

e

1

kpk

;

[

2

x

2

1 1℄

+ 2n

b℄ There exists an absolute onstant > 0 su h that

for every p 2 P

: 2n

R

kpk

[

;

1 1℄

ut


0

p (x)

1 + x2 n

x 2 R:

;

n(1 + log n)

1 p+(xx) n

2

R


Proof. Asso iated with p 2 P n ; let 2

p(x) : 1+x n

f (x) := Let

ak := ei

k

2

=(2n) ;

k = 1; 2; : : : ; 2n be the zeros of the equation z n + 1 = 0: Using Corollary 7.1.10, we an dedu e that there are absolute onstants and su h that (2

1)

2

1

n X

kf 0kR

k=1

2 jIm(ak )j

! 1

2

kf kR

1

n(1 + log n)kf kR :

n X k

1

!

n

k=1

kf kR

2

Therefore

0 p (x) 1 + x2 n

p(x) p(x) 2nx n

n(1 + log n)

n n 1+x 1+x 1 + x n R 2

1

2

2

2

2

for every x 2 R; whi h implies

0

p (x)

1 + x2n

R

p(x)

( 2 n(1 + log n) + 2n) 1 + x2 n

R

;

ut


℄ For every m 2 N ; there exists a onstant (m) depending only on m so that kp m k ; (m)(n(1 + log n))m kpk ; (

)

[

for every p 2 P

1 1℄

[

1 1℄

n satisfying

jp(x)j (1 + jxjn )kpk

(7:1:23)

[

;

1 1℄

:

Proof. Using part b℄ and indu tion on m, we see that there exists a onstant

(m) depending only on m su h that 1

(m)

p (x)

1 + x 2n

R

1

p(x)

1 + x n R

(m)(n(1 + log n))m

2

for every p 2 P n : Note that if p 2 Pn satis es (7.1.23), then 2

p(x)

1 + x2 n

R


p(x)

2 1+ x n

jj

R

2kpk

[

;

1 1℄

;

ut


E.12 Quasi-Chebyshev Polynomials for SR n and ASR n . Let 2

ak := ei

k

=(2n) ;

(2

2

k = 1; 2; : : : n

1)

be the zeros of the equation z n + 1 = 0 in the upper half-plane. Let 2

M n (z ) := 2

n Y k=1

(z

2

P n (x) :=Re(M n (x)) ;

x 2 R;

Q n(x) :=Im(M n (x)) ;

x 2 R:

2

and

z2C;

ak ) ;

2

2

2

a℄ Show that if n is even, then P n 2 SR n and Q n is odd, then Q n 2 SR n and P n 2 ASR n : b℄ Show that jM n (x)j = 1 + x n ; x 2 R 2

2

2

2

2

2

2

2

x 2 R;

P n (x) + Q n (x) = (1 + x n ) ; 2

2 ASR n ; while if n

2

2

and

2

2

2

2

2

in parti ular

kP n k 2

[

;

1 1℄

2

kQ n k

and

2

[

;

1 1℄

2:

Proof. Note that n M n (x) Y x = 1+x n k x

(7:1:24)

ak ; ak

2

2

=1

whi h implies the rst equality. The rest is straightforward from the de nitions. ut

℄ Show that there are extended real numbers

1=z su h that

0

> z > > z n = 1

M n (zj ) = ( 1)j ; 1 + zj n 2

2

2

1

j = 0; 1; : : : ; 2n

(the value of the left-hand side at 1 is de ned by taking the limit when x ! 1). Hint: Use (7.1.24) and the argument prin iple (see, for example, Ash [71℄).

ut


d℄ Let n 2 N be even. Show that there exist 1 = x > y > x > y > > xn 0

1

1

2

1

> y n > xn = 1

su h that

P n (xj ) = ( 1)j 1 + xj n 2

+

2

n=2 ;

Q n (xj ) = 0 ; j = 0; 1; : : : ; n 2

and

Q n (yj ) = ( 1)j 1 + yj n 2

n=2 ;

+1+

2

P n (yj ) = 0 ; j = 1; 2; : : : n : 2

Formulate the analog statement when n 2 N is odd. e℄ Show that there exists an absolute onstant > 0 su h that

jP 0n (x)j + jQ0 n (x)j n min 2

2

(1 + log n); log

1

e

x

2

for every n 2 N and x 2 [ 1; 1℄:

Proof. If x 2 [ 1; 1℄; then

n

1 + x jP 0n (x)j + jQ0 n (x)j jM 0 n (x)j = jM 0 n (x)j M n (x) n 0 n X X Im(ak ) 2 n (x) = M 2 M n(x) k x ak j x ak j k n min (1 + log n); log 1 e x 2

2

2

2

2

2

2

2

2

=1

=1

2

with an absolute onstant > 0; where the last inequality follows from (7.1.22). ut The next part shows the sharpness of the inequality of E.11 a℄. f℄ Let > 0 be the same absolute onstant as in part e℄. For the sake of brevity let

Æn (x) := n min (1 + log n); log

1

e

x

Show that for every interval

In;x := [x; x + 8Æn (x)℄ [0; 1℄ ;

2

1

:


2 In;x and y 2 In;x su h that

there exist y

1

2

jP 0n (y )j

1

1

2

2

Æn (x)

Proof. Suppose that

jQ0 n (y )j

and

1

jP 0n (y)j

1

2

2

for every y 2 In;x : Therefore

kQ nk 2

[

;

1 1℄

1 2

R

In;x

jQ0 n (y)j dy > 2

1 2

8Æn (x) Æn (x) 1

1

2

= 2;

whi h ontradi ts the last inequality of part b℄. This nishes the proof of the rst inequality. The se ond inequality an be proven in the same way.

ut

g℄ Show that

p0 (1) = n p(1) 1

2

for every p 2 By using the quasi Chebyshev polynomials for SR n and ASR n , it an be shown that the inequality of E.11 ℄ is essentially sharp for the lasses SRn and ASR n for every m. This has been pointed out to us by Szabados. The argument requires some more te hni al details than the proof in the m = 1 ase dis ussed in the above exer ise. SR n :

2

2

2

7.2 Inequalities for Logarithmi Derivatives We derive a series of metri inequalities of the form

m

x2R:

r0 (x) r(x)

n ;

> 0;

where r is a rational fun tion of type (n; n) and is a onstant independent of n. Here m is the Lebesgue measure, although, sin e the sets in question are usually just nite unions of intervals, this is mostly a notational onvenien e. One of the interesting features of these inequalities is their easy extension from the polynomial ase to the rational ase. The basi inequality is due to Loomis [46℄. Note the invarian e of the measure of the set in this ase.

7.2 Inequalities for Logarithmi Derivatives 345

Theorem 7.2.1. If p 2 Pn has n real roots, then p0 (x) n m x2R: = ; p(x)

> 0:

Proof. By onsidering p instead of p; it is suÆ ient to prove the theorem for = 1: We rst onsider the ase where p has distin t roots, whi h are denoted by < < < n : Then n 1 p0 (x) X = : p(x) k x i Let < < < n be the roots of p p0 ; whi h must all be real. Note that these are the points where p0 (x)=p(x) = 1: It is now easy to see from 1

1

2

=1

1

2

the graph that

p0 (x) x2R: p(x)

and

m

1

x2R:

= [ ; ℄ [ [ ; ℄ [ [ [n ; n ℄ 1

p0 (x) p(x)

However, if p(x) := xn + an xn n X i=1

while

n X i=1

1 1

1

1

2

=

2

n X

n X

i

i=1

+ + a ; then

i=1

i :

0

i = an

i = (an

1

1

; n)

is 1 times the se ond oeÆ ient of p p0 : This gives the result for distin t roots. The ase when some of the roots of p are repeated an be handled by an easy limiting argument. ut Corollary 7.2.2. If i

then

(

m

2

x2R:

R;

i > 0; i = 1; 2; : : : ; n; and

n X i=1

x

i

)!

i

=

1

;

Pn

i=1 i = 1;

> 0:

Proof. For i rational this follows immediately from Theorem 7.2.1 on learing the denominators of i by multiplying by an integer. The real ase is an obvious limiting argument.

ut

In order to extend Theorem 7.2.1 to arbitrary polynomials we need the following generalization of E.3 of Se tion 2.4 due to Videnskii [51℄. The proof is indi ated in the exer ises.


Lemma 7.2.3. If p 2 Pn is positive on [a; b℄; then there exists q; s nonnegative on [a; b℄ with all real roots (in [a; b℄) so that

2 Pn

p(x) = q(x) + s(x) : We now prove the unrestri ted ase of Theorem 7.2.1. Theorem 7.2.4. Let p 2 Pn : Then

m

p0 (x) x2R: p(x)

2n ;

> 0:

Proof. Let > 0 and let pn 2 Pn : Choose a and b su h that p0 (x) [a; b℄ : x2R: p(x) By Lemma 7.2.3 we an nd polynomials q 2 P n and s 2 P 2

2

n su h that

p (x) = q(x) + s(x) 2

where, for x 2 [a; b℄;

0 q (x) p (x)

0 s(x) p (x)

and

2

2

and both q and s have only real roots. Now

p0 (x) x2R: p(x)

Also,

0 = x 2 R : (pp )(x(x) ) 2

2

2

:

(p )0 (x) 2 p (x) 2

holds exa tly when

2

q0 (x) + s0 (x) 2(q(x) + s(x)) :

By Theorem 7.2.1

s0 (x) q0 (x) 2 =m x2R : m x2R: q(x) s(x) Sin e q and s are nonnegative on [a; b℄, it follows that

2

=

2n m(fx 2 [a; b℄ : q0 (x) + s0 (x) 2(q(x) + s(x))g) ;


n :

ut

It an be shown that this inequality is asymptoti ally sharp to the extent that the onstant 2 annot be repla ed by any smaller onstant for large n; see Kristiansen [82℄. Theorem 7.2.4 extends easily to rational fun tions.


Theorem 7.2.5. If r = p=q, where p; q 2 Pn ; then

m

r0 (x) r(x)

x2R:

Proof. We have

r0 p0 = r p

8n ;

> 0:

q0 q

and for > 0;

x2R:

r0 (x) r(x)

p0 (x) p(x)

x2R:

2 [

x2R:

q0 (x) q(x)

2

:

By Theorem 7.2.4

m

p0 (x) x2R: p(x)

2

4n ;

and with s(x) := q ( x);

m

q0 (x) x2R: q(x)

It follows that

m

2

=m

r0 (x) x2R: r(x)

s0 (x) x2R: s(x)

8n

for every > 0:

2

4n :

ut

This inequality probably does not have the exa t onstant. It an be shown (see Borwein, Rakhmanov, and Sa [to appear℄) that the onstant 8

annot be repla ed by any onstant less than or equal to 2: Comments, Exer ises, and Examples. Many variants on the inequalities of this se tion are presented in E.2, E.3, E.4, and E.5. Some of these are in Borwein [82℄. E.5 explores some metri properties of the lemnis ate

E (p) := fz 2 C : jp(z )j 1g of a moni polynomial p 2 Pn of the form

p(z ) =

n Y i=1

(z

zi ) ;

zi 2 C :


The reader is referred to Erd}os, Herzog, and Piranian [58℄ for many results and open problems on erning the lemnis ate of moni polynomials. One in parti ular, whi h is still open, onje tures that for moni polynomials p 2 Pn ; the length of the boundary of E (p) is maximal for p(z ) := z n 1 and so is O(n): Pommerenke [61℄ has shown that this length is O(n ): Borwein [95℄ improves this to O(n); see E.7. E.9 solves another problem of Erd}os, namely, the diameter of E (p) for a moni polynomial p 2 Pn is always at least 2. Erd}os [76℄ ontains several other related open problems. 2

E.1 Polynomials as Sums of Polynomials with Real Roots. a℄ Suppose p 2 P n n P n ; and suppose that p > 0 on [a; b℄: Then 2

2

1

p(x) = (x

a)(b x)u (x) + v (x) 2

2

for some u 2 Pn and v 2 Pn ; whi h have all their zeros in [a; b℄: b℄ Suppose p 2 P n n P n and suppose that p > 0 on [a; b℄: Then 1

2

+1

2

p(x) = (b x)u (x) + (x 2

a)v (x) 2

for some u; v 2 Pn ; whi h have all their zeros in [a; b℄: Hint: Let p be a polynomial of degree 2n that is stri tly positive on [a; b℄: Let Tn be the Chebyshev polynomial for the Chebyshev system (

1 p ; p(x)

x p ; ::: ; p(x)

xn p p(x)

)

:

p

Then Tn (x) = v (x)= p(x) with some v 2 Pn : Show that, for part a℄, v and u de ned by (x a)(b x)u (x) = p(x) v (x) 2

2

are the required polynomials. Use a similar onstru tion for part b℄.

ut

E.2 Various Spe ializations. For the next exer ises we use the notation Pn to denote the polynomial of degree at most n with nonnegative oeÆ ients, and Pn" to denote those elements of Pn that are nonde reasing on [0; 1): a℄ If r = p=q; where p; q 2 Pn and both p and q have only real roots, then +

m

x2R:

r0 (x) r(x)

4n ;

> 0:


b℄ Let r(x) := xn =(4n

x)n : Then r0 (x) 1 = 4n : m x2R: r(x)

℄ If r = p=q; where p 2 Pn and q 2 Pn" ; then r0 (x) 2n m x0: ; > 0: r(x) d℄ If r = p=q; where p 2 Pn and q 2 Pn" ; then r0 (x) n m x0: ; > 0: r(x)

+

e℄ Let r(x) := xn . Then

m

x0:

r0 (x) r(x)

=

n ;

> 0:

E.3 Another Metri Inequality. If p 2 Pn has n real roots lying in the interval (a; b); then 0 p (x) 2n m x2R: ; > 0: = p(x) j(x a)(b x)j

Outline. Prove that m and

x2R:0

(x

(x

a)(b x)p0 (x) p(x) a)(b x)p0 (x) p(x)

n

=

: n First onsider the ase when p has distin t zeros. Let y < y < < yn denote the n + 1 roots of (x a)(b x)p0 (x), and let x < x < < xn denote the n roots of p(x): Then y < x y < xn < yn < xn := 1 : m

x2R :0

=

0

1

0

0

Sin e

0

1

1

1

1

p0 (x) = 1; x!xi p(x) we an dedu e that for ea h interval (yi ; xi ) there exists a point Æi 2 (yi ; xi ) lim (x

a)(b x)

su h that

(Æi a)(b Æi )p0 (Æi ) = p(Æi ) : Sin e the above equation an have at most n + 1 solutions, we have

m

x2R:0

(x

a)(b x)p0 (x) p(x)

Now pro eed as in the proof of Theorem 7.2.1.

=

n X i=0

(Æi

yi ) :

ut


E.4 Extensions to Pn . a℄ If p 2 Pn ; then

m

x2

0 p (x) R: p(x)

p

8 2n

;

> 0:

Hint: Write p as the sum of its real and imaginary parts. b℄ If r = p=q with p; q 2 Pn ; then m

x2

0 r (x) R: r(x)

p

32 2 n

;

ut

> 0:

E.5 On the Lemnis ate E (p). Let

p(z ) := and let

n Y i=1

(z

zi ) ;

zi 2 C

E (p) := fz 2 C : jp(z )j 1g :

a℄ Show that

m(E (p) \ R)) 4 2

1

=n

with equality only for the Chebyshev polynomial of degree n normalized to have lead oeÆ ient 1; see Polya [28℄. Hint: Analogously to the proof of the Remez inequality of Se tion 5.1, show that the Chebyshev polynomial transformed to an interval of length 4 is extremal for this problem. ut b℄ Let m () denote the planar Lebesgue measure. Show that 2

m (E (p)) 4 : 2

(In fa t, m (E (p)) ; whi h is exa t for z n ; this is due to Polya [28℄.) 2

E.6 Cartan's Lemma. Let

p(z ) :=

n Y j =1

(z

zj ) ;

zj

2C

and > 0 be xed. Then there exist at most n open disks, the sum of whose radii is at most 2 ; so that if z 2 C is outside the union of these open disks then n jp(z )j > e :


Prove Cartan's Lemma as follows: a℄ Let ; ; : : : ; be xed omplex numbers. Let > 0: Show that there exists a positive integer less than or equal to for whi h there exists an open disk with radius =n ontaining j for exa tly distin t values of j = 1; 2; : : : ; : Hint: Suppose to the ontrary that there is no su h positive integer . Show that this would imply the existen e of an open disk with radius =n

ontaining j for at least + 1 distin t values of j = 1; 2; : : : ; ; whi h is a

ontradi tion. ut b℄ Show that there exist open disks D ; D ; : : : ; Dk and positive integers m ; m ; : : : ; mk with the following properties: Pk (1) j mj = n ; m m mk ; 1

2

1

1

2

2

1

=1

2

m

(2) Dj has radius j ; n (3) Dj \ Ej ontains exa tly mj zeros of p, where

Ej := C n (D

1

[ D [ : : : [ Dj 2

1

);

(4) for every integer m > mj ; no open disk of radius m n ontains exa tly m zeros of p in Ej : Hint: Use part a℄. ut

℄ Let D ; D ; : : : ; Dk be the open disks spe i ed in part b℄. For j = 1; 2; : : : ; k; let Dj be the disk with the same enter as Dj and with twi e its radius. Show that for every 1

2

z 2 E := C n (D [ D [ [ Dk ) 2

1

there is a permutation ze ; ze ; : : : ; zen of the zeros z ; z ; : : : ; zn su h that 1

jz

2

zej j

1

j ; n

2

j = 1; 2; : : : ; n :

Hint: Let z 2 E be xed. Show, by indu tion on i; that for every i = 0; 1; : : : ; n 1; there are at least i + 1 zeros of p outside the open disk with

enter z and radius n ni : Distinguish the ases (1) n i > m ; (2) mj n i > mj ; j = 1; 2; : : : ; k 1 ; (3) mk n i > 0 : ut (

)

1

+1

d℄ Finish the proof of Cartan's lemma.


E.7 The Length of the Boundary of E (p). Let

p(z ) :=

n Y j =1

(z

zj ) ;

zj

2C

and let

E := E (p) := fz 2 C : jp(z )j 1g : Show that the boundary E of E is of length at most 4en: (In fa t, with E.10 a℄, the estimate an be improved to (5:2) n:) Outline. Pro eed as follows: a℄ Let L be an arbitrary line in the omplex plane. Show that the set E \ L ontains at most 2n distin t points. Hint: By performing a translation and a rotation, if ne essary, we may assume that L = R: Now observe that there is a polynomial P (x; y ) of degree at most 2n; in two real variables x and y; with omplex oeÆ ients, su h that

E = fz 2 C : jp(z )j = 1g = fz 2 C : p(z )p(z ) = 1g = fz = x + iy : x; y 2 R ; P (x; y ) = 1g : 2

Hen e

\

f 2

g

E R = x R : P (x; 0) = 1

; the result follows. 2n

and sin e P (x; 0) 2 P b℄ For a 2 C and r > 0; let Q be the square

Q := fz 2 C : jRe(z

a)j < r ; jIm(z

ut a)j < rg :

Show that Q \ E is of length at most 8rn: Hint: Divide Q \ E into sub urves C ; C ; : : : ; Cm so that every verti al and horizontal line ontains at most one point of ea h Cj . Let lxj and lyj denote the length of the interval obtained by proje ting Cj to the x axes and y axes, respe tively. Let 1

2

( )

lx :=

m X j =1

lxj

( )

Use part a℄ to show that lx length of Q \ En ; then

and

ly :=

m X j =1

( )

lyj : ( )

4nr and ly 4rn. Hen e, if l denotes the l lx + ly 8rn :

ut


℄ Show that the boundary E of E is of length at most 16en: Hint: Combine Cartan's lemma (see E.7) and part b℄. ut The rest of the exer ise is about improving 16en to 4en: d℄ Let B be the open disk with enter a 2 C and radius r > 0. Show that B \ E is of length at most 2rn: Outline. Let Q be the square

Q := fz 2 C : jRe(z )j < r ; jIm(z )j < rg : For 2 [0; 2 ) let Q be the square Q := fa + ei (z a) : z 2 Qg : Let l() denote the length of Q \ E: Let lx () and ly () denote the total length of the intervals obtained by proje ting Q \ E to the lines z = rei : r 2 R and z = rei = : r 2 R ; respe tively ( ounting multipli ities). The pre ise de nition of lx () and ly () an be formulated in the same way as in the hint to part b℄, whi h is (

left to the reader. Use part a℄ to show that

lx() + ly () 8rn ;

+

2)

2 [0; 2) :

Hen e

Z 1 l(0) = l(0)(j sin j + j os j) d 2 Z 1 = (lx () + ly ()) d 8rn ; 2 and l(0) 2rn follows. e℄ Prove the initial statement of the exer ise. Hint: Combine Cartan's lemma (see E.6) and part d℄.

4

2

0

2

0

ut ut

E.8 On the Length of Another Lemnis ate. Suppose p 2 Pn : Show that the length of the lemnis ate 0 p (z ) =n F = F (p) := z 2 C : p(z ) is at most 16n(1 + log n) (a tually at most 4n(1 + log n)). Hint: The arguments are very similar to those given in the outline to E.7. First show that if L is an arbitrary line in the omplex plane, then F \ L

ontains at most 2n distin t points. Next prove that if Dr is an open disk of radius r in the omplex plane, then Dr \ F is of length at most 8rn (a tually at most 2rn). Now use E.6 ℄ with = 1 + log n: ut The proof of the following exer ise requires some familiarity with harmoni fun tions.


E.9 On the Diameter of E (p). The diameter diam(A) of a nonempty set

A C is de ned by

diam(A) := supfjz

1

z j : z ;z 2

1

2

2 Ag :

Let p 2 Pn be an arbitrary moni polynomial of degree n; that is,

p(z ) :=

n Y j =1

(z

zj ) ;

zj

2C:

Then the diameter of the set

E (p) := fz 2 C : jp(z )j 1g is at least 2. Pro eed as follows: a℄ Let Cb := C [ f1g

:= fz 2 Cb : jz j > 1g :

and

Let p 2 Pn be moni . Let E := o(E (p)); that is, the onvex hull of E (p): Use the Riemann mapping theorem (see Ahlfors [53℄) to show that there exists a fun tion g of the form

g(z ) = bz +

1 X j =0

bj z

j;

z 2 ; b; bj 2 C

su h that g is analyti and one-to-one on ; and

g() = Cb n E : Hint: Note that Cb n E is simply onne ted. b℄ Let b be the same as in part a℄. Show that jbj 1:

ut

Outline. Be ause of the de nition of E; for every " > 0 there exists a Æ > 0 su h that " log jz np(g(z ))j ; jz j = 1 + Æ :

Sin e G(z ) := log jz n p(g (z ))j is harmoni on ; we have

G(1) =

1 2 (1 + Æ )

Z 2 0

G((1 + Æ)ei ) d

"

2 (1 + Æ )

for every " > 0; so G(1) 0: On the other hand, sin e p 2 Pn is a moni polynomial of degree n;


G(1) = log jbjn = n log jbj ; from whi h jbj 1 follows.

ut

℄ Show that diam(E ) 2 :

Outline. Assume to the ontrary that diam(E ) < 2. Then there exists a Æ > 0 su h that

jz

1

g( z ))j < 2 ;

(g (z )

jz j = 1 + Æ :

Sin e

F (z ) := z (g(z )

g( z )) = 2b + 2b z

1

2

1

+ 2b z 3

4

+

is analyti on , the maximum prin iple (E.1 d℄ of Se tion 1.2) yields 2jbj = jF (1)j max jF (z )j = max jz (g (z )

jzj

Æ

=1+

jzj

1

Æ

=1+

that is, jbj < 1; whi h ontradi ts part b℄.

g( z ))j < 2 ;

ut

d℄ Note that diam(A) = diam( o(A)) for every nonempty A C ; in parti ular, diam(E (p)) = diam(E ): The more general result that diam(A) 2 ap(A) for every nonempty A C is observed in Pommerenke [75℄. E.10 More on E (p). Suppose p is a moni polynomial with omplex oeÆ ients. As before, let

E := E (p) := fz 2 C : jp(z )j 1g : a℄ It follows from E.6 (Cartan's lemma) that the set E (p) an be overed by disks the sum of whose radii is at most 2e: It is onje tured in Erd}os, Herzog, and Piranian [58℄ that the orre t value in this problem is 2. (The

urrent best onstant is less than 2.6.) b℄ IfP E (p) is onne ted, then it is ontained in a disk with radius 2 entered at n nk zk , where z ; z ; : : : ; zn are the zeros of p. Proof. This is onje tured in Erd}os, Herzog, and Piranian [58℄ and proved in Pommerenke [59b℄. ut

℄ If E (p) is onne ted, then its ir umferen e is at least 2: Proof. This is also onje tured in Erd}os, Herzog, and Piranian [58℄, and proved in Pommerenke [59b℄. ut 1

=1

1

2

This is page

356


A1

Algorithms and Computational Con erns

Overview Appendix 1 presents some of the basi algorithms for omputing with polynomials and rational fun tions and dis usses some of the omplexity issues. In luded is a dis ussion of root nding methods. It requires very little ba kground and an essentially be read independently.

Algorithms and Computational Con erns Polynomials lend themselves to omputation perhaps more than any other obje t of analysis. Algorithms that involve spe ial fun tions, dierential equations, series, and the like usually must redu e at some point to a nite polynomial or rational approximation or trun ation. This often allows analyti problems to be redu ed to algebrai ones. This appendix will present, as a series of exer ises, some of the prin ipal algorithmi on erns. The reader is en ouraged to experiment with the algorithms. With urrent te hnology this is most omfortably done in any of the large symboli manipulation pa kages available. Code, a tual or s hemati , is not presented. Indeed, methods rather than algorithms are presented. Current \pra ti al" best methods date qui kly in this rapidly evolving area. It is also the authors' belief that today's theoreti al uriosities may be vital for tomorrow's

Algorithms and Computational Con erns 357

algorithms as larger instan es are al ulated on faster ma hines. Histori ally we have already seen this happen repeatedly with algorithms su h as the fast Fourier transform algorithm. One of the most interesting lessons to be learned from the last few de ades of revitalized interest in omputational mathemati s is that many of the most familiar mathemati al algorithms, su h as how to multiply large numbers, were very poorly understood, and indeed, still are in ompletely analyzed. Very many of the familiar pro esses of mathemati s, su h as multipli ation of large numbers or omputation of determinants, an be omputed far more expeditiously than allowed by the usual \s hool" algorithms. See, for example, Aho, Hop roft, and Ullman [74℄; Bini and Pan [92℄; Borodin and Munro [75℄; Borwein and Borwein [87℄; Brent [74℄; Knuth [81℄; Pan [92℄; Smale [85℄; and Wilf [86℄ for the omplexity side of the following exer ises. E.1 Complexity and Re ursion. We are on erned with measuring the size of an algorithm given an input of a ertain length. Unfortunately, there are many dierent ways of measuring this. (One ould, for example, use the length of the tapes of some well-de ned instantiation of a Turing ma hine.) We will settle for less. The measure of input size will usually be hosen to be a natural one, so for polynomials of degree n; the measure will often be n: The omplexity measure then depends a bit on the problem. For example, it might ount the number of additions of oeÆ ients (we do not distinguish subtra tion from addition) and multipli ations of oeÆ ients required to evaluate the polynomial at a point. (So, for example, by Horner's rule O(n) operations suÆ e.) Care is already required to ount naturally. Note that we have not spe i ed the size of the oeÆ ients (this may or may not be reasonable) and so we ould heat on addition of oeÆ ients by doing two additions as one addition of twi e the length. (Sin e a + b and + d an be de oded from (10m a + ) + (10m b + d); where m is larger than the number of digits in any of a, b, ; or d.) It is more reasonable in this ontext to x a pre ision (or to think of working to in nite pre ision or over some polynomial ring). Our ases are fairly simple, and the measures should be

lear in ontext. We adopt the following notations:

f (n) f (n) = O(g(n)) means lim sup 0; then Q P f 2 GAPN with N kj rj nj : Similarly, if f = kj jQj jrj with ea h P Qj 2 Tnj and rj > 0; then f 2 GTPN with N kj rj nj : =1

=1

=1

=1

Q

rj = b℄ Show that if f 2 GAPN is of the form (A.4.1), then f = m j jQj j with ea h Qj 2 P Qand 0 Qj on R: Similarly, if f 2 GTPN is of the form rj = with ea h Q 2 T and 0 Q on R: (A.4.3), then f = m j j j jQj j 2

=1

2

2

=1

1

408 A4. Inequalities for Generalized Polynomials in Lp

E.2 The Derivative of an f 2 GTPN with Real Zeros. Show that if f 2 GTPN is of the form (A.4.3) with ea h rj 1 and zj 2 R; then jf 0 j 2 GTPN is of the form (A.4.3) with ea h rj > 0 and zj 2 R; and at least one of any two adja ent (in K ) zeros of jf 0 j has multipli ity exa tly 1: E.3 Generalized Nonnegative Polynomials with Rational Exponents. Show that if f 2 GAPN is of the form (A.4.1) with rational exponents rj = qj =q , where qj ; q 2 N ; then f q 2 P qN ; while if f 2 GTPN is of the form (A.4.3) with rational exponents rj of the above form then f q 2 T qN : 2

2

2

2

E.4 Proof of (A.4.6). Prove (A.4.6) from the expli it formula (2.1.1) for the Chebyshev polynomial Tn : E.5 Sharpness of the Remez-Type Inequality for GAPN . Let

fn(x) := Tn

2x + s 2 s

N=n

2 GAPN ;

n = 1; 2; : : : :

Show that

m(fx 2 [ 1; 1℄ : fn (x) 1g) = m([ 1; 1 s℄) = 2 s and

p p !N p 2 + ps :

lim f (1) = n!1 n

s

2

The upper bound in Theorem A.4.1 is a tually not a hieved by an element of GAPN ; see Erdelyi, Li, and Sa [94℄. E.6 Proof of Theorem A.4.2. Hint: First assume that f 2 GTPN is of the form (A.4.3) with rational exponents rj = qj =q; where qj ; q 2 N : Then p := f q 2 T qN ; and the desired inequality an be obtained from Theorem 5.1.2 as in the proof of Theorem A.4.1. ut 2

2

E.7 Corollaries of Theorems A.4.1 and A.4.2. a℄ The inequality

p

m x 2 [ 1; 1℄ : f (x) exp( N s)kf k

[

;

1 1℄

2 GAPN and 0 < s < 2. In parti ular, m x 2 [ 1; 1℄ : f (x) e kf k ; (8N holds for every f 2 GAPN .

8s

holds for every f

1

[

1 1℄

2

+ 4)

1

Inequalities for Generalized Polynomials in Lp 409

b℄ The inequality

m (f 2 [ ; ) : f () exp( Ns)kf kK g) holds for every f

2 GTPN

and 0 < s < 2. In parti ular,

m 2 [ ; ) : f () e holds for every f

s

4

1

kf kK (4(N + 1))

1

2 GTPN :

E.8 Nonnegative Trigonometri Polynomials. Part a℄ restates E.3 ℄ of Se tion 2.4. Parts b℄ and ℄ dis uss simple related observations. a℄ Let t 2 Tn be nonnegative on R: Show that there is a p 2 Pn su h that t() = jp(ei )j for every 2 R: b℄ Let p 2 Pn and t() := jp(ei )j for every 2 R: Then t 2 Tn and t is nonnegative on R:

℄ Show that if t 2 Tn ; then there is a p 2 P n su h that jt()j = jp(ei )j for every 2 R: 2

2

2

E.9 A Cru ial Inequality in the Proof of Theorem A.4.5. Show that

jP (z )j

2 n jP (rz )j ; 1+r

z 2 D ; r 2 [0; 1℄

for every P 2 Pn having all its zeros outside the open unit disk D: Q Hint: Let P (z ) = m j (z zj ); where 2 C ; zj 2 C n D; and m Show that =1

jz

zj j

2 jrz 1+r

zj j ;

n:

j = 1; 2; : : : ; m ; r 2 [0; 1℄ :

ut E.10 f 2 GAPN Implies f ( os ) 2 GTPN . a℄ Show that if f 2 GAPN ; then g () = f ( os ) 2 GTPN : b℄ Show that if f 2 GAPN is of the form (8.1.1) with ea h rj 1; then g() = f ( os ) 2 GTPN is of the form (A.4.3) with ea h rj 1: E.11 A Property of Qn;! . Let Qn;! be de ned by (A.4.9) and (A.4.10). Show that there is an absolute onstant > 1 su h that Qn;! ( ) : Hint: Use the expli it formula (2.1.1) for the Chebyshev polynomial Tn . u t 6

6


E.12 A Property of the Zeros of a g 2 GTPN . Assume that g 2 GTPN is of the form (A.4.3) and let

M := Show that M

3Nkgk

[

1

;

1 1℄

m X j =1 zj 2[ ;+℄

rj :

:

Hint: First assume that ea h rj is rational with ommon denominator q 2 N ; and apply E.12 of Se tion 5.1 to p := g q 2 T qN : ut 2

2

E.13 Extremal Fun tions for the Bernstein-Type Inequality. a℄ Let rj 1; j = 1; 2; : : : ; m; be xed real numbers. Show that there exists an fe 2 GTPN of the form

fe(z )

(A:4:26)

=

m Y

j =1

su h that

j sin((z

ej )=2)jrj ;

ej

2C

jfe0 ()j = sup jf 0 ()j ; kfekK f kf kK

where the supremum in the right-hand side is taken for all f the form

f (z ) = j!j

(A:4:27)

m Y

j =1

j sin((z

zj )=2)jrj ;

zj

2C;

2 GTPN

of

0 6= ! 2 C :

Hint: Write ea h f of the form (A.4.27) for whi h the supremum is taken as

f (z ) = j!

0

j

m Y j =1

j!j sin((z

zj )=2) sin((z

zj )=2)jrj = ; 2

where the numbers !j > 0 are de ned by

k!j sin((z

zj )=2) sin((z

z j )=2)kK = 1 ;

j = 1; 2; : : : ; m ;

and then use a ompa tness argument for ea h fa tor separately. ut b℄ Let fe be as in part a℄. Show that ea h zero of fe is real, that is, ej 2 R for ea h j in (A.4.26). Hint: Assume that there is an index 1 j m su h that j 2 C n R: Then

fe(z )

:=

fe(z )

1

sin ((z )=2) sin((z j )=2) sin((z j )=2) 2

!rj

2 GTPN

e with suÆ iently small > 0 ontradi ts the maximality of f:

ut


E.14 A Corollary of the Remez-Type Inequality for GAPN . Suppose that := 1 (1 + N ) : Show that there is an absolute onstant > 0 su h 2

that

9

kf k

[

kf k

;

9

1 1℄

;℄

[

for every f 2 GAPN : Hint: This is a orollary of Theorem A.4.1.

ut

E.15 Proof of Theorem A.4.10.

2 GAPN , let

Outline. For f

p

I (f ) := x 2 [ 1; 1℄ : ((f (x)))p exp( 5pN 2s)k(f )kp [

;

1 1℄

:

From Theorem 5.1.1, 0 < s 1=2; and the assumptions pres ribed for it follows that m(I (f )) 2s: Let I := A \ I (f ): Sin e m([ 1; 1℄ n A) s; m(I ) s: Therefore Z [

;

nA

1 1℄

((f (x)))p dx

Z [

;

nA

1 1℄

exp 5pN exp 5pN

k(f )kp p Z

;

[

2s

p

2s

ZI

1 1℄

dx

((f (x)))p dx

A

((f (x)))p dx :

ut

E.16 Proof of Theorem A.4.11. Hint: For f 2 GTPN ; let

I (f ) := 2 [ ; ℄ : ((f ()))p exp( 8pNs)k(f )kpK : From Theorem 5.2.2, 0 < s =4; and the assumptions for ; it follows that m(I (f )) 2s: Now nish the proof as in the hint for E.1. ut E.17 Sharpness of Theorem A.4.10. a℄ Show that there exists a sequen e of polynomials Qn absolute onstant > 0 su h that Z

1

1

Z jQn (x)jp dx s 1 + exp pnps

1

1

s

2 Pn

and an

jQn (x)jp dx

for every n 2 N ; s 2 (0; 1℄; and p 2 (0; 1): Hint: Study the Chebyshev polynomials Tn transformed linearly from [ 1; 1℄ to [ 1; 1 s℄: ut


b℄ Show that there exist a sequen e of trigonometri polynomials tn 2 Tn and an absolute onstant > 0 su h that Z

jtn j

() p d

s(1 + exp( pns))

Z s=2

+s=2

for every p 2 N ; s 2 (0; ℄; and p 2 (0; 1): Hint: Study Qn;! de ned by (A.4.9) with ! :=

jtn ()jp d ut

s=2:

E.18 Sharpness of Corollary A.4.13 and Theorem A.4.14. a℄ Find a sequen e of trigonometri polynomials tn 2 Tn that shows the sharpness of Corollary A.4.13 up to the onstant > 0 for every p 2 (0; 1) simultaneously. Hint: Take tn () := os n: ut b℄ For every p 2 (0; 1); nd a sequen e of polynomials Qn;p 2 Pn whi h shows the sharpness of Theorem A.4.14 up to the onstant > 0:

Outline. Let Lk 2 Pk be the kth orthonormal Legendre polynomial on [ 1; 1℄ (see E.5 of Se tion 2.3), and let Gm (x) := where is hosen so that

Z

(A:4:28)

1

m X k=0

L0k (1)Lk (x) ;

Gm (x) dx = 1 : 2

1

Show that there exist absolute onstants > 0 and > 0 (independent of m) su h that 1

2

jGm (1)j m and jG0m (1)j m : For a xed n 2 N ; let u := b2=p + 1 and m := bn=u : If m 1; then let Qn;p := Gum 2 Pn ; otherwise let Qn;p(x) := x 2 Pn : If m = 0; then the

al ulation is trivial, so let m 1: Using (A.4.29) and the Nikolskii-type (A:4:29)

1

3

2

inequality of Theorem A.4.4, show that there exists an absolute onstant

> 0 su h that 3

Z

(A:4:30)

1

jQ0n;p (x)jp dx p

+1

3

1

mu (

p (1 + pn)

+2)

2

:

Use the inequality Z

1

1

jQn;p (x)jp dx =

Z

1

1

jGm (x)jup dx kGm kup ; [

Z

1

2

1 1℄

Gm (x) dx ; 2

1


the Nikolskii-type inequality of Theorem A.4.4, and (A.4.28) to show that there is an absolute onstant > 0 su h that 4

Z

(A:4:31)

1

jQn;p (x)jp dx p

+1

4

1

mup

2

:

Finally, ombine (A.4.30) and (A.4.31). Note that if p > 2; then m = bn ; while if p 2; then p(n 1) m 2pn: Note also that pp is between two positive bounds for p 2 (0; 2℄: ut 1

4

E.19 Sharpness of Theorems A.4.3 and A.4.4. a℄ Let q 2 (0; 1) be xed. Show that there exists a sequen e of polynomials Qn;q 2 Pn and an absolute onstant > 0 su h that

kQn;q k

;

[

1 1℄

=q (1 + qn)2=q

1+1

kQn;q kLq

;

[

1 1℄

for every n 2 N : Hint: Study Qn;p with p = q in the hint to the previous exer ise. ut b℄ Let Qn;q 2 Pn be the same as in part a℄. Show that there exist absolute

onstants > 0 and > 0 su h that 1

kQn;q kLp

2

[

;

1 1℄

=q 1=p

1+1 1

2

(1 + qn) =q (1 + pn) =p kQn;q kLp 2

2

;

[

1 1℄

for every n 2 N and 0 < q < p 1: Hint: Combine part a℄ and the Nikolskii-type inequality of Theorem A.4.4.

ut

℄ Let q 2 (0; 1) be xed. Show that there exists a sequen e of trigonometri polynomials tn;q 2 Tn and an absolute onstant > 0 su h that

ktn;q kK

=q (1 + qn)1=q

1+1

ktn;q kLq K (

)

for every n 2 N : Hint: Let tn;q () := Qn;q ( os ); where Qn;q are the same as in part a℄. Use part a℄ and the S hur-type inequality of Theorem A.4.5. ut d℄ Let tn;q 2 Tn be the same as in part ℄. Show that there exist absolute

onstants > 0 and > 0 su h that 1

2

ktn;q kLp K (

)

=q 1=p

1+1 1

2

(1 + qn) =q (1 + pn) =p ktn;q kLq K 1

1

(

)

for every n 2 N and 0 < q < p 1: Hint: Combine part a℄ and the Nikolskii-type inequality of Theorem A.4.3.

ut


E.20 Jensen's Inequality. Let be a real-valued onvex fun tion de ned on [0; 1); and let f and w be nonnegative Riemann integrable fun tions R on the interval [a; b℄; where ab w = 1: Show that

Z b

a

!

f (x)w(x) dx

Z b

a

(f (x))w(x) dx :

Hint: First assume that w is a step fun tion. Use the fa t that the onvexity of implies its ontinuity, and hen e the fun tions (f )w and fw are

ut

Riemann integrable.

E.21 A Pointwise Remez-Type Inequality for GAPN . a℄ Show that there exists an absolute onstant > 0 su h that (

jp(y)j exp

n min

p

)!

s

y

1

2

p ; s

y 2 [ 1; 1℄

;

for every p 2 Pn and s 2 (0; 1℄ satisfying

m(fx 2 [ 1; 1℄ : jp(x)j 1g) 2 s ; that is, with the notation of Theorem 5.1.1, for every p

s 2 (0; 1℄:

2 Pn (s)

with

Proof. Assume, without loss of generality, that y 2 [0; 1℄: Let a := y + (1 y );

:= ar

os a ;

1

2

:= 2 ar

os y

ar

os a ;

Qn;! () := T

2

n

! := sin(=2) sin(!=2)

1

2

(

) ;

;

where T n is the Chebyshev polynomial of degree 2n de ned by (2.1.1), and let Qn;; () := Qn;! ( (2 ( + ))) : 2

1

2

Asso iated with p 2 Pn (s); we introdu e

g() := p( os )Qn;; () 2 T n : 2

Obviously and

jQn;; ()j 1 ; kQn;; k

; ℄ = Qn;;

[0 2

2 [0; 2) n (; ) 1 2

( + ) = Qn;! ( ) :


The de nition of ! implies that there exist absolute onstants > 0 and 1

> 0 su h that 2

p 1

y

1

2

!

p

2

y :

1

2

This, together with E.3 ℄, yields that there are absolute onstants > 0 and > 0 su h that 3

4

p

Qn;! () exp n 1 y 3

whenever y 2 [0; 1

2

exp 5nps

s℄. It follows now from Theorem 5.1.1 that 4

jg()j exp 5nps Qn;! ()

for every 2 [0; 2 ) n (; ) and y 2 [0; 1

s℄: Furthermore 4

jg()j Qn;! () for every 2 (; ) for whi h jp( os )j 1: Note that

j os j 1 and hen e p 2 Pn (s) with s

onstant > 0 su h that

1 2

(1

y) ;

2 (; )

2 (0; 1℄ implies that there exists an absolute

5

m(f 2 (; ) : jp( os )j 1g) Therefore

f := (Qn;! ()) g 2 T 1

satis es

2

p

s 5

2

:

n

m(f 2 [0; 2) : jf ()j 1g) 2

with

y

1

s

se

; y 2 [0; 1 s℄ : y Applying Theorem 5.1.2 with f and se; we on lude that se :=

jp(y)j = p =f

os 1 2

p

1 2

5

1

4

2

( + )

( + )

= (Qn;! ( )) g 1

exp(4nse) exp

p

1 2

( + )

s

!

5

1

y

2

whenever se 2 (0; =2℄ and y 2 [0; 1 s℄: If s 2 (0; 1℄; but se > =2 or y 2 [1 s; 1℄; then Theorem 5.1.1 yields the required inequality. ut 4

4


b℄ Show that the inequality of part a℄ is sharp up to the absolute onstant

> 0: Hint: Assume, without loss, that y 2 [0; 1℄: Show that there exists an absolute onstant > 0 su h that the polynomials 1

Wn;y;s (x) := Tn

2

2x

x

+

2

s

2 Pn (s) ;

s

where Tn is the Chebyshev polynomial of degree n de ned by (2.1.1), satisfy

jWn;y;s (y)j exp( nps) s=2; 1℄; and s 2 (0; 1℄: 1

for every n 2 N ; y 2 [1 If y 2 [0; 1 s=2℄; then let

s a := y + ;

:= ar

os a ;

4

:= 2 ar

os y

ar

os a ;

! :=

2

sin(( + ( + )=2)=2) Qn;; () := T n sin(!=2) Rn;; () := (Qn;; () + Qn;; ( )) ;

2

; ;

1

2

and we de ne Wn;y;s for every n 2 N ; y 2 [0; 1

s=2); and s 2 (0; 1℄ by

Wn;y;s 2 Pn :

Rn;; () = Wn;y;s ( os ) ;

Show that Wn;y;s 2 Pn (s) and that there exists an absolute onstant > 0 su h that !

jWn;y;s (y)j exp

n p

1

s

y

2

for every n 2 N ; y 2 [0; 1 s=2); and s 2 (0; 1℄: ut

℄ Extend the validity of part a℄ to the lass GAPN ; that is, prove that there exists an absolute onstant > 0 su h that

jf (y)j exp for every f

(

N min

p

s 1

)!

y

2

p ; s

y 2 [ 1; 1℄

;

2 GAPN and s 2 (0; 1℄ satisfying m(fx 2 [ 1; 1℄ : f (x) 1g) 2

s:

This is page

417


A5

Inequalities for Polynomials with Constraints

Overview This appendix deals with inequalities for onstrained polynomials. Typi ally the onstraints are on the lo ation of the zeros, though various oef ient onstraints are also onsidered.

Inequalities for Polynomials with Constraints For integers 0 k n; let

Pn;k := fp 2 Pn : p has at most k zeros in Dg where, as before, D := fz 2 C : jz j < 1g: For a < b; let Bn (a; b) :=

n

p 2 Pn : p(x) =

For integers 0 k n; let

n X j =0

j (b x)j (x

Pen;k (a; b) := fp = hq : h 2 Bn

k (a; b) ;

a)n j ; j

q 2 Pk g :

Two useful relations, given in E.1, are

Pn; Bn ( 0

1; 1)

and

Pn;k Pen;k (

1; 1) :

0

o

:

418 A5. Inequalities for Polynomials with Constraints

Theorem A.5.1 (Markov-Type Inequality for Pen;k ). The inequality

kp m k (

)

[

;

1 1℄

(18n(k + 2m + 1))m kpk

2 Pen;k (

holds for every p repla ed by 9:)

[

;

1 1℄

1; 1): (When m = 1; the onstant 18 an be

Proof. First we study the ase m = 1. Applying Theorem 6.1.1 (Newman's inequality) with

( ; ; : : : ; k ) = (n 0

k; n k + 1; : : : ; n)

1

and using a linear shift from [0; 1℄ to [ 1; 1℄; we obtain that

jp0 (1)j

(A:5:1)

9 2

n(k + 1)kpk

[

;

1 1℄

for every p 2 Pen;k ( 1; 1) of the form

p(x) = (x + 1)n k q(x) ;

(A:5:2)

q 2 Pk :

Now let p 2 Pen;k ( 1; 1) be of the form p = hq with q 2 Pk and

h(x) =

nXk j =0

j (1 x)j (x + 1)n

k j

with ea h j

0:

Without loss of generality, we may assume that n k 1; otherwise Theorem 5.1.8 (Markov's inequality) gives the theorem. Using (A.5.1) and the fa t that ea h j 0; we get (A:5:3) jp0 (1)j

nXk = j (1 j =0 1 X = (j (1

j =0

9 2

n(k + 2) (x + 1)n

nXk

9 n ( k + 2) j (1

2

j =0

k

1

( (x + 1) + (1 0

x))q(x)

1

x)j (x + 1)n k j q(x)

[

[

;

1 1℄

;

1 1℄

n(k + 2)kpk ; : 2 [ 1; 1℄ be arbitrary. To estimate jp0 (y)j we distinguish two 9

2

Now let y

ases.

x)j (x + 1)n k j q(x) (1) x)j (x + 1)n k j q(x))0 (1) 0

[

1 1℄

Inequalities for Polynomials with Constraints 419

If y 2 [0; 1℄; then by a linear shift from [ 1; 1℄ to [ 1; y ℄; from (A.5.3) that (A:5:4) jp0 (y)j y +9 1 n(k + 2)kpk ;y 9n(k + 2)kpk for every p 2 Pen;k ( 1; y ): It follows from E.1 d℄ that Pen;k ( 1; 1) Pen;k ( 1; y) : So (A.5.4) holds for every p 2 Pen;k ( 1; 1): If y 2 [ 1; 0℄; then by a linear shift from [ 1; 1℄ to [y; 1℄; from (A.5.3) that (A:5:5) jp0 (y)j 1 9 y n(k + 2)kpk y; 9n(k + 2)kpk for every p 2 Pen;k (y; 1): By E.1 d℄ again, Pen;k ( 1; 1) Pen;k (y; 1) : [

1

[

℄

[

1℄

[

we obtain ;

1 1℄

we obtain ;

1 1℄

So (A.5.5) holds for every p 2 Pen;k ( 1; 1); whi h nishes the ase when m = 1: Now we turn to the ase when m 2: Note that an indu tion on m does not work dire tly for an arbitrary p 2 Pen;k ( 1; 1): However, it follows by indu tion on m that (A:5:6) kp m k ; (9n(k + m + 1))m kpk ; (

)

[

1 1℄

[

1 1℄

for every p 2 Pen;k ( 1; 1) of the form (A.5.2). Now let p of the form p = hq , where q 2 Pk and

h(x) =

nXk j =0

j (1

x)j (x + 1)n

j

For notational onvenien e let s := minfn fa t that ea h j 0, we get (A:5:7) jp m (1)j (

2 Pen;k (

with ea h j

1; 1) be

0:

k; mg. Using (A.5.6) and the

)

= =

nXk (j (1 j =0 s X (j (1

x)j (1 + x)n k j q(x))(m) (1)

x)j (1 + x)n k j q(x))(m) (1) j =0

s

X

(9n(k + s + m + 1))m j (1 x)j (x + 1)n k j q (x)

j =0 [

nXk

(9n(k + 2m + 1))m

j (1 x)j (x + 1)n k j q(x)

= (9n(k + 2m + 1))m

j =0

kpk

[

;

1 1℄

:

[

;

;

1 1℄

1 1℄


From (A.5.7), in a similar fashion to the ase when m = 1; we on lude that kp m k ; (18n(k + 2m + 1))m kpk ; (

)

[

1 1℄

[

1 1℄

for every p 2 Pen;k ( 1; 1); whi h nishes the proof.

ut

Theorem A.5.1 is essentially sharp as is shown in E.2 and E.4 f℄. However, a better upper bound an be given for jp m (y )j when y 2 ( 1; 1) is away from the endpoints 1 and 1: (

)

Theorem A.5.2 (Markov-Bernstein Type Inequality for Pn;k ). There exists a onstant (m) depending only on m su h that (

p

jp j (m) min n(k + 1) ; pn(k + 1) 1 y for every p 2 Pn;k and y 2 [ 1; 1℄: (

m) (y )

)!m

2

kpk

[

;

1 1℄

Theorem A.5.2 has been proved in Borwein and Erdelyi [94℄. Its proof is long, and we p do not reprodu e it here. However, the proof of a less sharp version, where 1 y is repla ed by 1 y ; is outlined in E.4 and E.5. p The fa tor n(k + 1) in Theorem A.5.2 is essentially sharp in the ase that m = 1; see E.7. 2

2

Theorem A.5.3 (Markov-Bernstein Type Inequality for Bn ( 1; 1)). There exists a onstant (m) depending only on m su h that

jp

(

j (m)

m) (y )

(

min n ;

p

pn

1

)!m

y

2

kpk

[

;

1 1℄

for every p 2 Bn ( 1; 1) (hen e for every p 2 Pn; ) and y 2 [ 1; 1℄: 0

Note that the uniform (Markov-type) upper bound of the above theorem is a spe ial ase (k = 0) of Theorem A.5.1. Our proof of Theorem A.5.3 oers another way to prove the ase of Theorem A.5.1 when k = 0: First we need a lemma. Lemma A.5.4. For n 2 N and y 2 R; let 1 n;y := 4

p

pny ) + n1

(1

2 +

!

;

where x := maxfx; 0g: Then +

p jp(y + i n;y )j 2e p y n for every p 2 Bn ( 1; 1); y 2 1 n ; 1 + n ; and

2 [0;1℄; where i is the imaginary unit. The + sign is taken if y 2 1 n ; 0 ; and the sign is taken if y 2 0; 1 + n : 1

2

1

8

1

8

1

8

1

8


Proof. It is suÆ ient to prove the lemma for the polynomials n2N;

pn;j (x) := (1 x)j (x + 1)n j ;

j = 0; 1; : : : ; n :

The general ase when

p=

n X j =0

j pn;j

0

with ea h j

0

or ea h j

an then be obtained by taking linear

ombinations. Without loss of gen erality, we may assume that y 2 0; 1 + n : Then 1

n;y 2

8

1 1 (1 y ) + 8 n n 2 +

2

(1 4ny)

+

1 8n

+

2

from whi h it follows that

jpn;j (y + i n;y )j jpn;j (y + in;y )j = (1

(1

= 1

2

2

2

y

1

2

2

n n

(n j )=2

1

1

2

j

1+ y

2

2n + 1 2n 1 and the lemma is proved.

j=2

y) + n;y (1 + y ) + n;y j y) + n (1 + y) + n n j

pn;j y

1 2

n

1 2

n

n j

p

1+y+ n 1+y n 1

n j

2

1

2

2epn;j y

1 2

n

;

ut

2 N and y 2 [ 1; 1℄; let Bn;y denote the

ir le of the omplex plane with enter y and radius n;y . Using E.4, Lemma A.5.4, and the maximum prin iple, we obtain Proof of Theorem A.5.3. For n

1

4

p jp(z )j 2e kpk ; 1; 1); z 2 Bn;y ; n 2 N ; and y 2 [

1 1℄

for every p 2 Bn ( Cau hy's integral formula,

jp

(

[ 1; 1℄: Hen e, by

Z

p( ) d j 2i m Bn;y ( y ) Z ! p( ) m j d j 2 Bn;y ( y )m p m ! m 2 n;y n;y 2ekpk 2 ( pn )!m kpk (m) min n ; p 1 y

m) (y ) = m!

+1

+1

1

1

4

4

(

+1)

2

[

;

[

1 1℄

;

1 1℄

for every p 2 Bn ( 1; 1) and y 2 [ 1; 1℄; whi h proves the theorem.

ut


For r 2 (0; 1℄; let

Dr := fz 2 C : jz (1 r)j < rg : be the set of all For n 2 N ; k = 0; 1; : : : ; n; and r 2 (0; 1℄; let Pn;k;r p 2 Pn that have at most k zeros ( ounting multipli ities) in Dr , and let Pn;k;r := Pn;k;r \Pn;k;r : The following result is proved in Erdelyi [89a℄. The +

proof in a spe ial ase is outlined in E.9.

Theorem A.5.5 (Markov-Type Inequality for Pn;k;r ). There exists a on-

stant (m) depending only on m su h that kp m k ; (m) n(kp+r 1) for every p 2 Pn;k;r ; m 2 N ; and r 2 (0; 1℄:

2

(

)

[

m

kpk

1 1℄

[

;

1 1℄

We state, without proof, the Lq analogs of Theorem A.5.2 for m = 1, established in Borwein and Erdelyi [to appear 2℄. Theorem A.5.6 (Lq Markov-Type Inequality for Pn;k ). There exists an absolute onstant su h that Z Z jp0 (x)jq dx q (n(k + 1))q jp(x)jq dx 1

1

+1

1

1

for every p 2 Pn;k and q 2 (0; 1):

Theorem A.5.7 (Lq Bernstein-Type Inequality for Pn;k ). There exists an

absolute onstant su h that Z jp0 ( os t) sin tjq dt q (n(k + 1))q= +1

2

for every p 2 Pn;k and q 2 (0; 1):

Z

jp( os t)jq dt

Both of the above inequalities are sharp up to the absolute onstant > 0: A Markov-type inequality for polynomials having at most k zeros in the disk Dr := fz 2 C : jz j < rg is given by the following theorem; see Erdelyi [90a℄. Theorem A.5.8 (Markov-Type Inequality for Polynomials with At Most k Zeros in Dr ). Let k 2 N and r 2 (0; 1℄: Then there exist onstants (k ) > 0 and (k) > 0 depending only on k so that 1

2

(k) n + (1 r)n 1

2

0 sup kkppkk

[

p

[

; 1;1℄

1 1℄

where the supremum is taken for all p 2 Dr :

(k) n + (1 2

r)n ; 2

Pn that have at most k zeros in


Comments, Exer ises, and Examples. Erd}os [40℄ proved that

kp0 k

;

[

1 1℄

n2

n

n

n

1

1

kpk

e < n kpk

;

[

2

1 1℄

[

;

1 1℄

for every p 2 Pn; n 2; having only real zeros. In this result, the assumption that p has only real zeros an be dropped. In E.11 we outline the proof of the above inequality for every p 2 Pn; ; n 2: The polynomials 0

0

pn (x) := (x + 1)n (1 x) ;

n = 1; 2; : : : ;

1

show that Erd}os's result is the best possible for Pn; : Erd}os [40℄ also showed that there exists an absolute onstant su h that 0

jp0 (y)j

p

n (1 y )

2 2

kpk

;

[

1 1℄

for every p 2 Pn; having only real zeros. Markov- and Bernstein-type inequalities for Bn ( 1; 1) were rst established by Lorentz [63℄, who proved Theorem A.5.3. Lorentz's approa h is outlined for m = 1 in E.8. The proof presented in the text follows Erdelyi [91 ℄. Up to the onstant (m) > 0 Theorem A.5.3 is sharp; see E.12. S hei k [72℄ found the best possible asymptoti onstant in Lorentz's Markov-type inequality for m = 1 and m = 2: He proved the inequalities 0

kp0 k

[

;

1 1℄

2e n kpk

[

and

;

1 1℄

kp00 k

;

[

1 1℄

2e n(n

for every p 2 Bn ( 1; 1) (and hen e for every p pn (x) := (x + 1)n (1 x); 1

kp0n k nkpn k

;

[

1 1℄ [

;

1 1℄

! 2e

and

kp00nk

[

;

1 1℄

n(n 1)kpk

[

;

1 1℄

1)kpk

[

;

1 1℄

2 Pn; ). Note that with 0

! 2e

as n ! 1 :

A slightly weaker version of Theorem A.5.1 was onje tured by Szabados [81℄, who gave polynomials pn:k 2 Pn;k with only real zeros so that

jp0n;k (1)j

1 3

n(k + 1)kpn;k k

[

;

1 1℄

for all integers 0 k n: After some results of Szabados and Varma [80℄ and Mate [81℄, Szabados's onje ture has been settled in Borwein [85℄, where it is shown that

kp0 k

[

;

1 1℄

9n(k + 1)kpk

[

;

1 1℄


for every p 2 Pn;k having n k zeros in R n ( 1; 1). The ru ial part of this proof is outlined in E.4 d℄ and e℄. The above inequality is extended to all p 2 Pen;k (and hen e to all p 2 Pn;k ) and to higher derivatives in Erdelyi [91b℄, whose approa h is followed in our proof of Theorem A.5.1. After partial results of Erdelyi and Szabados [88, 89b℄ and Erdelyi [91b℄, the \right" Markov-Bernstein-type analogs of Theorem A.5.2 for the lasses Pn;k has been proved in Borwein and Erdelyi [94℄. Note that Pn;n = Pn ; and hen e, up to the onstant (m); Theorem A.5.2 ontains the lassi al inequalities of Markov and Bernstein, and of ourse the ase k = 0 gives ba k Lorentz's inequalities for the lasses Pn; Bn ( 1; 1): The \right" Markov- and Bernstein-type inequalities of Theorems A.5.8 and A.5.9 for all lasses Pn;k in Lq , 0 < q < 1, are established in Borwein and Erdelyi [to appear 2℄. The Markov-type inequality for the lasses Pn;k;r given by Theorem A.5.5 is proved in Erdelyi [89a℄. A weaker version, when k = 0 and the fa tor r = is repla ed by a onstant (r) depending on r, is established in Rahman and Labelle [68℄. When k = 0; Theorem A.5.5 is sharp up to the

onstant (m) depending only on m; see E.10. 0

1 2

E.1 Relation Between Classes of Constrained Polynomials. a℄ Show that Pn; ( 1; 1) B ( 1; 1): (This is an observation of G. G. Lorentz.) Hint: Use the identities (x )(x ) = j1 + j (1 x) + (jj 1)(1 x ) + j1 j (1 + x) and ( + 1)(1 x) : x = (1 )(x + 1) 0

1

2

2

4

1

2

2

2

1

2

2

4

1

1

2

2

b℄ Show that Pn;k P 1; 1):

℄ Show that Bn (a; b) Bn ( ; d) whenever [ ; d℄ [a; b℄: Hint: First show that x a = (x ) + (d x) and b x = (x ) + (d with some nonnegative oeÆ ients. d℄ Show that Pen;k (a; b) Pen;k ( ; d) whenever [ ; d℄ [a; b℄:

ut

en;k (

1

2

3

4

x)

ut

E.2 Sharpness of Theorem A.5.1. Show that there exist polynomials

pn;k 2 Pn;k of the form pn;k (x) = (x + 1)n k qn;k (x) ; su h that

jp0n;k (1)j

1 6

for every n 2 N ; k = 0; 1; : : : ; n:

n(k + 1)kpn;k k

qn;k 2 Pk [

;

1 1℄


Hint: Use the lower bound in Theorem 6.1.1 (Newman's inequality) with k; n k + 1; : : : ; ng :

( ; ; : : : ; k ) = (n 0

1

ut E.3 A Te hni al Detail in the Proof of Lemma A.5.4. For n

y 2 [ 1; 1℄; let

Fn := z = a + ib : a 2 and

1

n ; 1 + 8n 1

1 8

2N

and

; b 2 ( n;a ; n;a ) ;

Bn;y := z 2 C : jz yj = n;y : Show that Bn;y Fn for every y 2 [ 1; 1℄: 1

4

E.4 Bernstein-Type Inequality for Pn;k . Prove that there exists an absolute onstant su h that p

jp0 (y)j 1n(ky+ 1) kpk

[

2

;

1 1℄

for every p 2 Pn;k ; and y 2 ( 1; 1): Pro eed as follows: a℄ Show that for every n 2 N and k = 0; 1; : : : ; n; there exists a polynomial Q 2 Pn;k su h that

jQ0 (0)j kQk ; [

jp0 (0)j p2Pn;k kpk ;

:

= sup

1 1℄

[

1 1℄

Hint: Use a ompa tness argument. Use Rou he's theorem to show that the uniform limit of a sequen e of polynomials from Pn;k on [ 1; 1℄ is also in Pn;k : ut b℄ Show that Q has only real zeros, and at most k + 1 of them ( ounting multipli ities) are dierent from 1: Hint: Use a variational method. For example, if z 2 C n R is a zero of Q; 0

then the polynomial

Q(x) := Q(x) 1

x

2

(x

z )(x 0

z ) 0

with suÆ iently small > 0 is in Pn;k and ontradi ts the maximality of

ut

Q:

℄ Let Æ := (36n(k + 3)) : Show that 1

kpk

[

Æ;1℄

2 kpk

;

[0 1℄

for every polynomial Pn;k having all its zeros in [0; 1) with at most k of them ( ounting multipli ities) in (0; 1):


Hint: Use the Mean Value Theorem and the result of Theorem A.5.1 transformed linearly from [ 1; 1℄ to [0; 1℄: ut = d℄ Let z := i (36n(k +3)) ; where i is the imaginary unit and 2 [0; 1℄: 1 2

0

Show that

jp(z )j

p

0

2 kpk

[

;

1 1℄

for every polynomial p 2 Pn having only real zeros with at most k of them ( ounting multipli ities) in ( 1; 1):

Outline. Let p(x) =

Qs

j =1 (x

uj ) with some s n: Applying part ℄ to s Y

q(x) :=

j =1

(x

uj ) ; 2

we obtain

jq (

(36n(k + 3)) )j 2kq k ; = 2kq (x )k ; = 2kp(x)p( x)k 2kpk ; : 1

2

[0 1℄

[0 1℄

[

;

1 1℄

2 [ 1 1℄

Observe that

jp(z )j jp(i(36n(k + 3)) 0

2

j

= )2=

1 2

= jq ( (36n(k + 3)) )j ;

s Y

(uj + (36n(k + 3)) ) 1

2

j =1

1

whi h, together with the previous inequality, yields the desired result.

ut

e℄ Let Q 2 Pn;k be the extremal polynomial of part a℄, and let n

Fn;k := z = a + ib : jaj 1 ; jbj (36n(k + 3)) Show that

jQ(z )j

p

2 kQk

[

= (1

1 2

o

j aj )

:

;

1 1℄

for every z 2 Fn;k : Hint: Use parts b℄ and d℄ and a linear shift from the interval [ 1; 1℄ to [2 Re(z ) 1; 1℄ if Re(z ) 0; or to [ 1; 2 Re(z ) + 1℄ if Re(z ) < 0: ut


f℄ Show that there exists an absolute onstant > 0 su h that

jp0 (0)j

p

n(k + 1) kpk

[

;

1 1℄

for every p 2 Pn;k : Hint: By part a℄ it is suÆ ient to prove the inequality when p = Q; in whi h

ase use part e℄ and Cau hy's integral formula. ut g℄ Prove the main result stated in the beginning of the exer ise. Hint: In order to estimate jp0 (y)j when, for example, y 2 [0; 1℄ use a linear shift from [ 1; 1℄ to [2y 1; 1℄ and apply part f℄. ut E.5 Bernstein-Type Inequality for Pn;k for Higher Derivatives. Prove that there exists a onstant (m) depending only on m so that

jp

(

p

n(k + 1) 1 y

j (m)

m) (y )

!m

kpk

2

[

;

1 1℄

for every p 2 Pn;k and y 2 ( 1; 1): Hint: First show that for every n; m 2 N, k = 0; 1; : : : ; n, and Æ there exists a polynomial QÆ 2 Pn;k su h that

jQÆm (0)j kQÆ k ; n Æ;Æ (

[

)

1 1℄ [

jp m (0)j p2Pn;k kpk ; n Æ;Æ (

= sup ℄

[

)

1 1℄ [

2 (0; 1℄;

: ℄

Show that QÆ has at most k +m zeros dierent from 1: Show that there is a polynomial Q 2 Pn;k having at most k + m zeros (by ounting multipli ities) dierent from 1 su h that

jQ m (0)j = sup jp m (0)j : kQk ; p2Pn;k kpk ; (

(

)

[

1 1℄

)

[

1 1℄

If y = 0; then use E.5 and indu tion on m to prove the inequality of the exer ise for all p 2 Pn;k having at most k + m zeros dierent from 1: For an arbitrary y 2 ( 1; 1) use a linear shift as is suggested in the hint to E.5 g℄. ut E.6 Sharpness of Theorem A.5.2. Show that there exist polynomials

pn;k 2 Pn;k and an absolute onstant > 0 su h that

jp0n;k (0)j

p

n(k + 1)kpn;k k

for every n 2 N and k = 0; 1; : : : ; n:

[

;

1 1℄

428 A5. Inequalities for Polynomials with Constraints 1

Hint: If k = 0; then let m :=

3

(n

1) and

+1+

1

If 1 k n; then let m := 1

3

3

pn;k (x) := (x

bpm :

1)m (x + 1)m

pn;k (x) := (x

n ; s :=

1

1)m T s

2

(k

1) ; and

r

m x : 2s + 1

3

2 +1

1

If n < k n; then let m := 1

3

3

ut

n and pn;k := pn;m:

E.7 Some Inequalities of Lorentz. (See Lorentz [63℄.) a℄ Show that

jp0 (x)j

n

n

n

n p x n

1

1

4n p x n 1

for every p 2 Bn ( 1; 1) (hen e for every p 2 Pn; ); n 2; and x 2 [ 1; 1℄; where in x n the + sign is taken if x 2 [ 1; 0); while the sign is taken if x 2 [0; 1℄: Hint: Observe that it is suÆ ient to prove the inequality only for 0

1

pn;j (x) := (1 x)j (x + 1)n j ;

n2N;

j = 0; 1; : : : ; n :

ut

b℄ Let

r

Æn (x) :=

1

n

x

2

n2N;

;

x 2 [ 1; 1℄ :

Show that there exists an absolute onstant > 0 su h that

jp0 (x)j (Æn (x))

1

max jp(x)j; p x Æn (x) 1

2

for every p 2 Bn ( 1; 1) (hen e for every p 2 Pn; ) and x 2 1 + n ; 1 Hint: Note that it is suÆ ient to prove the inequality only for 0

pn;j (x) := (1 x)j (x + 1)n j ;

n2N;

n :

j = 0; 1; : : : ; n :

ut

℄ Show that Z

1

1

jp0 (x)jq dx 2 4q nq

Z

1

1

jp(x)jq dx

for every p 2 Bn ( 1; 1) (hen e for every p 2 Pn; ) and q 2 (0; 1): 0


ut

Hint: Use part a℄. d℄ Show that there is an absolute onstant > 0 su h that Z

p 0 p (x) 1

1

1

Z

q dx

1

5 q nq= jp(x)jq dx 1; 1) (hen e for every p 2 Pn; ) and q 2 (0; 1): x

2

2

1

for every p 2 Bn ( Hint: Use parts b℄ and ℄. ut Analogs of parts a℄ and b℄ for higher derivatives are established by Lorentz [63℄. From these, analogs of parts ℄ and d℄ for higher derivatives

an be proven. 0

E.8 Theorem A.5.5 in a Spe ial Case. Show that there is an absolute

onstant su h that jp0 (1)j pn kpk ; 1

1

r

[

for every p 2 Pn having all its zeros in (

Proof. First assume that (A:5:8)

jp(1)j = kpk

1 1℄

1; 1 ;

[

1 1℄

2r℄; r 2 (0; 1℄:

:

Without loss of generality, we may assume that p 2 Pn n Pn : Denote the zeros of p by ( 1 0 depending only on m su h that

j

m

j (m) min pnr ; n

m) (1) pn;m;r (

kpn;m;r k

2

;

[

1 1℄

for every n 2 N ; r 2 (0; 1℄; and m = 1; 2; : : : ; n: Outline. If 0 < r m ; then with 4

xj := 1 and

4 mr os

2

j 2

n

1

zj := xj + ir ;

let

pn;m;r (x) :=

n Y j =1

;

j = 1; 2; : : : ; n

j = 1; 2; : : : ; n

(x

z j ) 2 Sn;r :

zj )(x

By E.1 of Se tion 5.2, jpn;m;r (1)j = kpn;m;r k

;

[

m (1)j jpn;m;r kpn;m;r k ; (

)

[

=

1 1℄

m (1)j jpn;m;r jpn;m;r (1)j (

1 1℄

: Prove that

)

m

2 1 + 4m p X n e2 1 j =m

1

m (1)j p1 jjqqn;m;r (1) 2 n;m;r j ! (

)

m

xj

m

(m) min pnr ; n

2

Q

;

where qn;m;r (x) := nj (x xj ) and (m) > 0 depends only on m: If < r 1; then let p ut n;m;r := pn;m;re; where re := m : m b℄ Con lude from Theorem A.5.5 and part a℄ that there exist two onstants

(m) > 0 and (m) > 0 depending only on m su h that =1

4

4

1

2

m

n

(m) min p ; n r 1

2

sup kpkpk k (

(m) 2

m)

;

[

[

min

1 1℄

;

1 1℄

m

pnr ; n

2

;

where the supremum is taken either for all p 2 Sn;r or for all p 2 Pn; ;r : 0


E.10 An Inequality of Erd}os. (See Erd}os [40℄.) Prove that

kp0 k

;

[

1 1℄

n

2

n

n

n

1

1

kpk

;

[

1 1℄

for every p 2 Pn; ; n 2: This extends a result of Erd}os [40℄ where the above inequality is proven under the additional assumption that p has only real zeros. a℄ Suppose p 2 Pn; ; all the zeros of p are real, p(1) = p( 1) = 0; and kpk ; = 1: Show that 0

0

[

1 1℄

p(x)

n

n

1

n 1

1+x 1+x

0

for every x 2 [ 1; 1℄; where x is the only point in ( 1; 1) with p0 (x ) = 0: 0

0

Proof. Without loss of generality, we may assume that deg(p) = n and 1 < x < x : Let d := x x: Let x := 1; x ; : : : ; xn denote the zeros of p: Then n 1+x Y d p(x) = 1 : p(x) = p(x ) 1 + x j x xj 0

0

1

0

2

0

0

=2

Sin e the geometri mean of n 1 nonnegative numbers is not greater than their arithmeti mean, we have n Y j =2

1

x

d 0

xj

n

1

n 1

1

1

n X

n 1

j =2

n 1+

x

d 0

!!n

xj

n

d

1

1

x +1 0

n

n

1

n 1

:

ut b℄ Under the assumptions of part a℄ show that

kp0 k

;

n2

X

1

[

1 1℄

n

n

1

n 1

kpk

;

[

1 1℄

:

Proof. Note that X

x xj j 1

1

x

0

=

xj

1

x

0

k

min 1 x ; xn +k1 xj 0

0

n2 ;

where k denotes the number of zeros of p in [1; 1): Hen e, by part a℄,

432 A5. Inequalities for Polynomials with Constraints n

X p0 (x) = p(x)

j =1

1

x

n

n

xj

1+x X 1 n 1 1 + x x x xj j n X 1 n n 1 11++xx 11++xx x xj xj

1

0

1

0

0

n2

n

n

1

0

1

n 1

for every x 2 [ 1; 1℄: Similarly

n p0 (x)

n

2

for every x 2 [ 1; 1℄:

n

n

1

1

ut

℄ Suppose p 2 Pn; has only real zeros, p0 does not vanish in [ 1; 1℄; p( 1) = 0; and p(1) = 1: Show that 0

kp0 k

;

[

Hint: Use the relation Pn;

1 1℄

n2 kpk

Bn (

0

p(x)

[

;

1 1℄

:

1; 1) to show that

x+1

n

2

; x 2 [ 1; 1℄ :

Denote the zeros of p by x = 1, x ; x ; : : : ; xn : Then 1

n

X 0 p0 (x) = p(x)

j =1

x

2

1

xj

3

x+1 2

n

n x+1

n2 = n2 kpk

[

;

1 1℄

for every x 2 [ 1; 1℄: ut d℄ Prove Erd}os's inequality for every p 2 Pn; having only real zeros. Hint: Redu e the general ase to either part b℄ or part ℄ by a linear transformation. ut e℄ Prove Erd}os's inequality for every p 2 Pn; : Hint: Show that for every n 2 N and y 2 [ 1; 1℄; there exists a polynomial Q 2 Pn; su h that 0

0

0

jQ0 (y)j kQk ; [

1 1℄

= sup

p2Pn;0

Show by a variational method that if y zeros. Now part d℄ nishes the proof.

jp0 (y)j : kpk ; 2 ( 1; 1); then Q has only real ut [

1 1℄


f℄ Show that

kp0 k

; =

n

n

n

1

kpk

; 2 n 1 holds for a p 2 Pn; having only real zeros if and only if either [

1 1℄

[

1 1℄

0

p(x) = (x + 1)n (1 x)

p(x) = (1 x)n (x + 1)

or

1

1

with some 0 6= 2 R: E.11 Sharpness of Theorem A.5.3. For every n 2 N ; m = 1; 2; : : : ; n; and 2 Bn ( 1; 1) having zeros only in

y 2 [ 1; 1℄; there are polynomials pn;m;y R n ( 1; 1) su h that m (y )j (m) jpn;m;y (

)

(

min n ;

p

pn

1

)!m

y

kpn;m;y k

[

2

;

1 1℄

with a onstant (m) > 0 depending only on m:

Outline. If

y 2 [ 1; 1℄ n

then let

1 + nm ; 1 2

m n

2

;

pn;m;y (x) := (x + 1)n :

In what follows, assume that

y 2 [ 1+

m n ;1

2

m n ℄:

2

x)j (x + 1)n j ; n 2 N ; j = 0; 1; : : : ; n: Show that

Let qn;j (x) := (1

m qn;j (x) Qn;j;m (x) ; = qn;j (x) (x 1)m (

)

m j n m;

2

where Qn;j;m is a polynomial of degree m with only real zeros and with leading oeÆ ient n nm : Let !

(

)!

(

n;y := max

1

n

p

;

1

pn

y

2

)

;

n2N;

y 2 [ 1; 1℄ :

Use the Mean Value Theorem and Theorem A.5.3 to show that there exists an absolute onstant 2 (0; 1) su h that

qn;j (x)

1 2

kqn;j k

[

;

1 1℄

;

mjn m

for every

x 2 Iy := [y

n;y ; y + n;y ℄ \ [ 1; 1℄ ;

y=1

j n

2

:


Let m su h that

jn j

j n be xed. Choose a point

m and y = 1

2

2 Iy

i j

n;y ; i = 1; 2; : : : ; m ; 2(m + 1) where the numbers i are the zeros of Qn;j;m : Now show that there exists a onstant (m) > 0 depending only on m su h that 1

j

(A:5:9)

(

j (m)

m) qn;j ( ) (

min n; p

1

Next show that if

y2

then there exists a point

2 y

1 2

1

)!m

y

kqn;j k

[

2

;

1 1℄

:

m n ;

1 + nm ; 1

2

2

jy j) ; y +

(1

pn

1 2

(1

jy j)

and a value of j , m j n m, su h that (A.5.9) holds. Polynomials pn;m;y with the desired properties an now be easily de ned by using linear

ut

transformations.

E.12 An Inequality of Turan. (See Turan [39℄.) Show that

kp0k kpk

[

[

for every p 2 Pn n Pn

1

; 1;1℄

1 1℄

>

1p n 6

having all its zeros in [ 1; 1℄:

Outline. Assume that p 2 Pn has all its zeros in [ 1; 1℄; and kpk ; = 1: Choose an a 2 [ 1; 1℄ su h that jp(a)j = 1. Without loss of generality, assume that p(a) = 1: pn: a℄ Show that if a = 1; then jp0 (1)j n > If a 2 ( 1; 1); then p0 (a) = 0. Without loss of generality, assume that a 2 [ 1; 0℄: Let I := a; a + 2n = [ 1; 1℄ : If n 3; then the result follows by the Mean Value Theorem; let n 4: pn on I; then b℄ Use the Mean Value Theorem to show that if jp0 (x)j p(x) on I: pn:

℄ Show that if jp00 (x)j > n on I; then jp0 (a + 2n = )j > [

1

1

2

6

1 1℄

1 2

1

6

2

3

1

1

1 2

12

6

d℄ The proof of Turan's inequality an now be nished as follows. Suppose p(x) > on I , and there exists a 2 I su h that p00 ( ) n: Note that 2

1

3

12

p0 (x)

2

p(x)p00 (x) = p(x)

2

n X k=1

(x

1

xk )

2

;


where x ; xP; : : : ; xn denote the zeros of p: Sin e ea h xk lies in [ 1; 1℄, the inequality nk (x xk ) n holds for every x 2 I: Sin e p(x) on I; this implies that 1

2

2

=1

2

p0 ( )

2

2

3

n p(x)p00 (x) ;

p0 (x) and hen e

1

4

x2I

9

n9 jp( )jjp00 ( )j > n9

n

12

=

n

36

:

ut

An extension of Turan's inequality to Lp norms is given by Zhou [92b℄. E.13 An Inequality of Erd}os and Turan. Let p 2 Pn be of the form

p(x) =

n Y

(x

1x

xj ) ;

j =1

1

x xn 1 : 2

Suppose p is onvex on [xk ; xk ℄ for some index k: Then 16 1

xk

xk

1

pn :

Pro eed as follows: Let a be the only point in [xk ; xk ℄ for whi h

p0 (a) = 0:

1

2 R su h that a 2n = < a < a + 2n = ; jp0 ( )j pn jp(a)j ; and jp0 ( )j pn jp(a)j :

a℄ Show that there exist ; 1

2

1 2

1

1 2

2

1

1

1

2

6

6

ut

Hint: Modify the outline of the proof of E.12. b℄ Show that

1

xk

1

p6n

xk

and

2

p6n :

Outline. To prove, say, the rst inequality, we may assume that xk < ; otherwise the inequality is trivial. Using the onvexity of p on [xk ; xk ℄; 1

1

1

we get

jp0 (x)j jp0 ( )j ;

x 2 [ xk ; ℄ :

1

1

Hen e

jp(a)j jp( )j = jp( ) 1

(

1

p(xk )j =

1

xk

1

1

)jp0 ( )j (


1

1

1

Z 1 0 p (x) dx xk 1

xk ) 1

1 6

pn jp(a)j ;

=

Z 1

xk

1

jp0 (x)j dx ut


℄ Con lude that

a xk

(a

1

) + ( 1

xk )

1

1

p2n + p6n = p8n ;

and similarly

a (

xk

)

a) + (xk

2

2

p2n + p6n = p8n :

ut

d℄ Er}od [39℄ establishes the sharp inequalities

xk

xk

1

xk

xk

1

p2n2

p2n2

3

if n is even,

p

n 2n 3 n 1

if n 3 is odd.

2

E.14 S hur-Type Inequality for Bn ( 1; 1). Let be an arbitrary positive real number. Show that

kpk

sup

6 p2Bn

0=

(

;

1 1)

kp(x)(1

[

;

=

1 1℄

x ) k 2

[

;

1 1℄

(n + 2)n (4) (n + )n +2

+

e (n + 2) : 4 The supremum is attained if and only if p(x) = (1 x)n ; 0 6= 2 R: Hint: Let x := n n : If jyj x ; then
0 su h that

kp0(x)e x k

;1)

[0

pn kp(x)e x k

;1)

[0

holds for every p 2 Pn having no zeros in the disk with diameter [0; 2℄: Proof. See Erdelyi [89 ℄, where this result is extended to various other general lasses of weight fun tions and onstrained polynomials. ut E.20 Inequalities for Generalized Polynomials with Constraints. For

s 2 (0; 2); let

Pen;k (s) :=

n

o

p 2 Pen;k : m(fx 2 [ 1; 1℄ : jp(x)j 1g) 2 s :

The polynomials Tn;k

2 Pn;k are de ned by

Tn;k (x) := Tnfn k; n k + 1; : : : ; n; [0; 1℄g

1 2

(x + 1) ;

where 0 k n are integers. (Note that the notation introdu ed in Se tion 3.3 de nes Tn;k only on [ 1; 1℄; and, to be pre ise, Tn;k denotes the polynomial de ned above on [ 1; 1℄:) a℄ Remez-Type Inequality for Pen;k . The inequality

kpk

[

2+s ; Tn;k 2 s

1 1℄

holds for every p 2 Pen;k (s) and s 2 (0; 2): Equality holds if and only if

p(x) Tn;k

2x

2

s

+

2

s

s

:

Proof. The proof is quite similar to that of Theorem 5.1.1 (Remez Inequality); see Borwein and Erdelyi [92℄. ut b℄ A Numeri al Version of Part a℄. Show that Tn;k

2+s 2 s

exp

for every s 2 (0; 1℄: Proof. See Borwein and Erdelyi [92℄.

p

5

nks + ns

ut


Let GAPN;K , 0 (A.4.1) for whi h

K N , be the set of all f 2 GAPN n X j =1 jzj j1

of the form

K:

rj

Note that if 0 K N are integers and p 2 PN;K ; then jpj 2 GAPN;K : For 0 K N and 0 < s < 2, let GAPN;K (s) denote the olle tion of all f 2 GAPN;K for whi h

m(fx 2 [ 1; 1℄ : f (x) 1g) 2 s :

The next part of the exer ise extends a numeri al version of part a℄ to the

lasses GAPN;K (s):

℄ Remez-Type Inequality for GAPN;K . There exists a onstant 5 su h that p kf k ; exp NKs + Ns 1

[

1

1 1℄

for every f 2 GAPN;K (s) and s 2 (0; 1℄: Hint: Let q 2 N be the ommon denominator of the numbers rj . Apply part a℄ with p := f q 2 Pe qN; qK (s), and then use part b℄. ut d℄ Nikolskii-Type Inequality for GAPN;K . Let be a nonnegative nonde reasing fun tion de ned on [0; 1) su h that (x)=x is nonin reasing on [0; 1): Then there exists an absolute onstant 25e su h that 2

2

2

2

2

k(f )kLp ; ( maxf1 ; q NK ; qN g) =q for every f 2 GAPN;K and 0 < q < p 1: [

1 1℄

2

2

1

1

=p

k(f )kLq

;

[

1 1℄

The ase K = N (when there is no restri tion on the zeros) of part d℄ is the ontent of Theorem A.4.4. If qK p 1; then the Nikolskii fa tor in the unrestri ted ase (K = N )pis like ( qN ) =q =p ; while in our restri ted p

ases it improves to ( q NK ) =q =p : 2

2

2

2

2

2

1, and then a similar argument, as in the proof of Theorems A.4.3 and A.4.4, gives the result for arbitrary 0 < q < p 1: Thus, in the sequel, let 0 < q < p = 1: Using the inequality of part ℄ with Proof. It is suÆ ient to prove the inequality when p =

s := minf1 ; ( q NK )

; ( qN ) g and re alling the onditions pres ribed for ; we on lude that m x 2 [ 1; 1℄ : ((f (x)))q e k(f )kq ; 2 1

2

1

1

1

2

m x2[ m x2[ s

[

1 1℄

1; 1℄ : f (x) e =q kf k ; p NKs + Ns kf k 1; 1℄ : f (x) exp 2

[

1

1 1℄

[

;

1 1℄


for every f where

2 GAPN;K : Now integrating only on the subset E e k(f )kq

((f (x)))q

2

[

;

1 1℄

of [ 1; 1℄

;

we obtain

k(f )k

q [

2

;

1 1℄

Z

me(E ) ((f (x)))q dx E maxf1 ; q NK ; qN gk(f )kqLq 2

2

for every f 2 GAPN;K , where := e (assuming we have 25e . 2 1

2

2

2

2

1

[

;

1 1℄

1). Sin e 5; ut 1

e℄ Remez-Type Inequality for Bn ( 1; 1). Show that

kpk

[

;

1 1℄

2

2

n

s

for every p 2 Bn ( 1; 1) satisfying

m(fx 2 [ 1; 1℄ : jp(x)j 1g) 2 s : Equality holds if and only if p is of the form

p(x) =

1x n : 2 s

ut

Proof. See Erdelyi [90b℄.

f℄ Markov-Type Inequality for GAPN;K . There exists an absolute onstant > 0 su h that

kf 0k

[

;

1 1℄

N (K + 1)kf k

[

;

1 1℄

for every f 2 GAPN;K of the form (A.4.1) with ea h rj 1: Re all that jf 0 (x)j is well-de ned for every f 2 GAPN;K and x 2 R, as the modulus of the one-sided derivative of f at x. The ondition that rj 1 for ea h j in (A.4.1) is needed to ensure that jf 0 (zj )j < 1 if zj 2 R. The above result generalizes the orresponding polynomial inequality for the lasses Pn;k . Proof. See Borwein and Erdelyi [92℄. ut


E.21 The Ilye-Sendov Conje ture. The following problem is known as the Ilye-Sendov onje ture. As before, let D := fz 2 C : jz j < 1g: Suppose P 2 Pn has all its zeros in the losed unit disk D. Then ea h

losed disk entered at a zero of P ontains a zero of of P 0 . Although the problem in this generality is still unanswered, several spe ial ases have been settled. The ase outlined in part a℄ was rst proved in Rubinstein [68℄. Vâjâitu and Zahares u [93℄ ontains stronger results; see also Miller [90℄. Milovanovi , Mitrinovi , and Rassias [94℄ has a dis ussion about the re ent status of the onje ture. a℄ Suppose

P (z ) =

n Y k=1

(z

jzk j 1 ; jz j = 1 :

zk ) ;

1

Show that P 0 has at least one zero in the losed disk entered at z : Proof. Without loss of generality, we may assume that z := 1: Then P is of the form P (z ) = (z 1)Q(z ); where 1

1

Q(z ) :=

n Y k=2

(z

jzk j 1 :

zk ) ;

Suppose the statement is false. Then R(z ) := P 0 (z + 1) has no zero in the

losed unit disk D: Hen e 0 R (0) R(0)

< n 1:

Observe that

R(0) = Q(1) Hen e

However,

0 Q (1) Q(1)

R0 (0) = 2Q0 (1) :

and

=

1 R0 (0) n 1 < : 2 R(0) 2

Q0 (1) = Re Re Q(1)

n X k=2

1

1

!

zk

n2 1:

This ontradi tion nishes the proof. b℄ The polynomial P (z ) := z n 1 shows the sharpness of part a℄.

ut


It follows from Theorem A.5.3 that there exists an absolute onstant

> 0 su h that

kp0k

[

n kpk

;

1 1℄

[

;

1 1℄

for every p 2 Pn with no zeros in the open unit disk. The polynomials pn (x) := (x+1)n show that up to the absolute onstant > 0 this inequality

is the best possible. The next exer ise shows that the \right" Markov fa tor on [ 1; 1℄ for polynomials of degree at most n with omplex oeÆ ients and with no zeros in the open unit disk is n(1 + log n) rather than n. This is an observation of Halasz.

E.22 Markov Inequality for Pn with No Zeros in the Unit Disk. Show that there exists an absolute onstant > 0 so that kp0 k ; n(1 + log n)kpk ;

for every p 2 Pn with no zeros in the open unit disk. Show also that there exist polynomials pn 2 Pn with no zeros in the open unit disk su h that kp0nk ; n(1 + log n)kpnk ; ; n = 1; 2; : : : with an absolute onstant > 0: Pro eed as follows: a℄ Show that if z 2 C is outside the open unit disk, then 1

[

[

1

1 1℄

[

2

1 1℄

[

1 1℄

1 1℄

2

jx jx

zj zj

1

j xj

for every x 2 R n [ 1; 1℄: b℄ Suppose p 2 Pn has no zeros in the open unit disk. Let x 2 R n [ 1; 1℄. Show that jp(x)j jxjn jp(x )j : 1

℄ Prove the upper bound of the exer ise. Hint: Use part b℄ and E.11 a℄. d℄ Let zk := e ik= n ; k = 1; 2; : : : ; n be the (2n + 1)th roots of unity in the open upper half-plane. Let 2

p

2

+1

2

+1

2

jp n 2

Show also that

+1)

n (z ) := p n (z ) := (z + 1)

Show that jp n (x)j = jx 2

(2

+2

( 1)j = kp n

0 p2n+1 ( p ( 2

n+1

n Y

(z

2

+1

k

2

[

; = 2:

1 1℄

1) n(1 + log n) 1)

with an absolute onstant > 0:

zk ) : 2

k=1 1 for every x R: Note that this implies

j

n+1

+1

ut

This is page

448


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This is page

467


Notation

De nitions of the more ommonly used spa es are given. The equation numbers here are the same as the equation numbers in the text. Throughout the book, unless otherwise stated, spans should be assumed to be real. Likewise, in fun tion spa es, unless otherwise stated, the fun tions should be assumed to be real valued. The Basi Spa es. (1:1:1)

Pn :=

(1:1:2)

Pn :=

(1:1:3)

R

(1:1:4)

(

p : p(z ) = (

p : p(z ) =

:= m;n

n X k=0 n X k=0

ak z k ; ak 2 C

)

: )

ak

zk;

ak 2 R :

p

; q 2 P : : p 2 Pm n q

Rm;n := pq : p 2 Pm; q 2 Pn :

468

Notation

(1:1:5)

T

(

n :=

t : t() := n

n X k= n

= t : t(z ) = a + 0

Tn : =

n

t : t(z ) = a + 0

ak

ak 2 C

eik ;

)

n X


k=1

n X


k=1

o

ak ; bk 2 C : o

ak ; bk 2 R :

Muntz Spa es. (3:4:2)

(3:4:3)

Mn () := spanfx0 ; x1 ; : : : ; xn g :

M () :=

1 [ n=0

Mn () = spanfx0 ; x1 ; : : : g :

Asso iated with a sequen e (i )1 i with Re(i ) > 1=2 for ea h i; the nth Muntz-Legendre polynomial is: =0

(3:4:5)

Ln (x) := Ln f ; : : : ; n g(x) Z nY 1 t + k + 1 xt dt := : 2i t k t n k 0

1

=0

For distin t i ; (3:4:6)

Ln f ; : : : ; n g(x) = 0

with

k;n :=

Qn

n X k=0

k;n xk ;

j =0 (k + j + 1) : j =0;j 6=k (k j ) 1

Qn

Also, (3:4:8)

Ln := (1 + n + n )

=

1 2

Ln

is the nth orthonormal Muntz-Legendre polynomial.

x 2 (0; 1)

Notation 469

Rational Spa es.

D := fz 2 C : jz j < 1g ;

K := R (mod 2) ; (3.5.3)

Pn (a ; a ; : : : ; an ) :=

(3.5.4)

Tn (a ; a ; : : : ; an ) :=

Also

1

1

p(x) jx ak j : p 2 Pn :

Qn

2

D := fz 2 C : jz j = 1g :

k=1

Qn

2

k=1

t() j os

ak j

: t 2 Tn :

Tn (a ; a ; : : : ; a n ; K ) := Tn (a ; a ; : : : ; a n ) 1

2

2

(

:= and

T

1

2

2

2

1

2

n R;

1; 1℄) :=

and

n (a1 ; a2 ; : : :

; an ; [ 1; 1℄) :=

on [ 1; 1℄ with a ; a ; : : : ; an 2 C 1

2

n (a1 ; a2 ; : : :

Q2n

Q2n

t() : t 2 Tn sin(( ak )=2)

on D with a ; a ; : : : ; an 2 C 1

2

n (a1 ; a2 ; : : :

; an ; R) :=

on R with a ; a ; : : : ; an 2 C 1

2

n R:

Qn


Qn

p(x) : p 2 Pn (x ak )

k=1

k=1

1; 1℄;

p(z ) : p 2 Pn (z ak )


Qn

p(z ) : p 2 Pn (z ak )

Qn

Qn

k=1

2

and

P

n D; and

Pn (a ; a ; : : : ; an ; R) := 1

k=1

n[

; an ; D) :=

2

t() : t 2 Tn j sin(( ak )=2)j

k=1

; a n ; K ) :=

Pn (a ; a ; : : : ; an ; [

P

2

(

n (a1 ; a2 ; : : :

on K with a ; a ; : : : ; a n 2 C

P

1

k=1

k=1

)

)

470

Notation

Generalized Polynomials. The fun tion m Y

f (z ) = j!j

(A:4:1)

j =1

jz

zj jrj

with 0 < rj 2 R; zj 2 C ; and 0 6= ! 2 C is alled a generalized nonnegative (algebrai ) polynomial of (generalized) degree (A:4:2)

N :=

m X j =1

rj :

The set of all generalized nonnegative algebrai polynomials of degree at most N is denoted by GAPN : The fun tion

f (z ) = j!j

(A:4:3)

m Y j =1

j sin((z

zj )=2)jrj

with 0 < rj 2 R; zj 2 C ; and 0 6= ! 2 C is alled a generalized nonnegative trigonometri polynomial of degree m

1X N := r : 2j j

(A:4:4)

=1

The set of all generalized nonnegative trigonometri polynomials of degree at most N is denoted by GTPN : Constrained Polynomials. The following lasses of polynomials with onstraints appear in Appendix 5:

Pn;k := fp 2 Pn : p has at most k zeros in Dg ; Bn (a; b) :=

n

p 2 Pn : p(x) =

n X j =0

Pen;k (a; b) := fp = hq : h 2 Bn

j (b x)j (x a)n k (a; b) ;

j;

q 2 Pk g ;

0 k n;

j

o

0

; a < b;

0 k n:

Notation 471

Fun tion Spa es. The uniform or supremum norm of a omplex-valued fun tion f de ned on a set A is de ned by kf kA := sup jf (x)j : x2A

The spa e of all real-valued ontinuous fun tions on a topologi al spa e

A equipped with the uniform norm is denoted by C (A): If A := [a; b℄ is equipped with the usual metri topology, then the notation C [a; b℄ := C (A)

is used. Let (X; ) be a measure spa e ( nonnegative) and p 2 (0; 1℄: The spa e Lp () is de ned as the olle tion of equivalen e lasses of real-valued measurable fun tions for whi h kf kLp < 1; where (

kf kLp (

)

:=

Z

X

jf j

p d

)

1=p

;

p 2 (0; 1)

and

kf kL1 (

)

:= supf 2 R : (fx 2 X : jf (x)j > g) > 0g < 1 :

The equivalen e lasses are de ned by the equivalen e relation f g if f = g -almost everywhere on X: When we write Lp [a; b℄ we always mean Lp (); where is the Lebesgue measure on X = [a; b℄: The notations Lp (a; b); Lp [a; b); and Lp (a; b℄ are also used analogously to Lp [a; b℄: Again, it is always our understanding that the spa e Lp () is equipped with the Lp () norm. Sometimes C (A) denotes the spa e of all omplex-valued ontinuous fun tions de ned on A equipped with the uniform norm. Similarly, Lp () may denote the spa e all omplex-valued ontinuous fun tions for whi h kf kLp < 1: This should always be lear from the ontext; many times, but not always, the reader is reminded if C [a; b℄ or Lp () is omplex. (

)

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472


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473


Index

Abel, Niels Henrik, 3, 61 Algorithms, 356{371 evaluation of xn , 363 fast Fourier transform, 359 fast polynomial division, 362 fast polynomial expansion, 363 fast polynomial multipli ation, 361 for Chebyshev polynomials, 371 for ounting zeros, 364{370 for polynomial evaluations, 363 for reversion of power series, 362 interpolation, 360 Newton's method, 362, 364{367 Remez, 371 zero nding for polynomials, 366{370 Alternation set, 93 Alternation theorem, 94 Apolar polynomials, 23, 24, 25 Ar length of algebrai polynomials, 31 Ar length of trigonometri polynomials, 35 Berman's formula, 166 Bernstein fa tor, 145, 150, 152,

322{328 Bernstein polynomials, 163{164 Bernstein-Szeg}o inequality, 231, 245, 259, 321{323 for entire fun tions of exponential type, 245 for rational fun tions on [ 1; 1℄, 322 for rational fun tions on K , 322 for rational fun tions on R, 323 for trigonometri polynomials, 232 Bernstein-type inequality bounded, 178, 182, 213{214 for Chebyshev spa es, 206 for onstrained polynomials, 420{447 for entire fun tions of exponential type, 245 for exponential sums, 289 for generalized polynomials, 392{416 for generalized polynomials in Lp , 401{417 for higher derivatives, 258 for nondense Muntz spa es, 213, 310 for polynomials, 232{233, 390

474 Index

for polynomials in Lp , 235, 390, 401{417 for produ ts of Muntz spa es, 317 for rational fun tions on [ 1; 1℄, 323, 327 for rational fun tions on D, 324 for rational fun tions on K , 322, 327 for rational fun tions on R, 323, 329 for self-re ipro al polynomials, 339 for trigonometri polynomials, 232 unbounded, 154, 206{217 weighted, 257 Bessel's inequality, 46 Best approximation, 94 by rationals, 99 to x , a problem of Lorentz, 108 to xn , 99 uniqueness, 98 Binomial oeÆ ient, 62 Blas hke produ t, 190, 324 Blumenthal's theorem, 78 Bombieri's norm, 274 Bounded Bernstein-type inequality; see Bernstein-type inequality Bounded Chebyshev-type inequality; see Chebyshev-type inequality Bounded linear fun tionals, 50 Budan-Fourier theorem, 369

Cardano, Girolamo, 3 Cartan's lemma, 350 Cau hy determinant, 106 Cau hy indi es, 367 Cau hy's integral formula, 14 Cau hy-S hwarz inequality, 42 for sequen es, 46 Cesaro means, 165 Chebyshev onstants, 39 Chebyshev, P.L., 31 Chebyshev polynomials algorithms for, 371 best approximation to xn , 30

omposition

hara terization, 33 expli it formulas, 30, 32

in Chebyshev spa es, 114{125 in rational spa es; see Chebyshev rationals orthogonality, 32 redu ibility, 36 se ond kind, 37 three-term re ursion, 32 trigonometri on subintervals, 235 uniqueness, 118 Chebyshev rationals, 139{153 and orthogonality, 147 derivative formulas, 146 in algebrai rational spa es, 142 in trigonometri rational spa es, 143 of the rst and se ond kind, 141 on the real line, 151 partial fra tion representation, 144 Chebyshev spa e, 92 dimension on the ir le, 100 fun tions with pres ribed sign

hanges, 100 Chebyshev system, 91{100 extended omplete, 97 Chebyshev's inequality, 235, 390 Chebyshev-type inequality bounded, 179, 182 expli it bounds via Paley-Weiner theorem, 196 for entire fun tions of exponential type, 245 Christoel fun tion, 74 for Muntz spa es, 195 Christoel numbers, 67 Christoel-Darboux formula, 60 CoeÆ ient bounds for polynomials in spe ial bases, 124 in nondense rational spa es, 153 of Markov, 248 Comparison theorem, 103, 120, 122, 183 Completeness, 48, 79 Complexity on erns, 356{371 Conse utive zeros of p0 , 26 Constrained polynomials, 417{447 Bernstein-type inequality, 420, 425, 427

Index 475 Lp inequalities, 422 Markov-type inequality, 417{447 Nikolskii-type, 444 Remez-type, 443{445 S hur-type, 436{437 Cotes numbers, 67 Cubi equations, 4

Fekete point, 38 Fekete polynomial, 38 Ferrari, Ludovi o, 3 Ferro, S ipione del, 3 Fourier oeÆ ient, 53 Fourier series, 53 Fundamental theorem of algebra, 3

d'Alembert, Jean le Rond, 13

Gamma fun tion, 63

de la Vallee Poussin theorem, 99 Denseness, 154{226 of Markov spa es, 206{217 of Muntz polynomials, 171{205 of Muntz rationals, 218{226 of polynomials, 154{170 Derivatives of Markov systems, 112 Des artes' rule of signs, 22, 102 Des artes system, 100{113 examples, 103 lexi ographi properties, 103 Divide and onquer, 358 Division of polynomials, 15, 362

Elementary symmetri fun tion, 5 Enestrom-Kakeya theorem, 12 Erd}os inequality, 431 Erd}os-Turan inequality, 435 Euler, Leonhard, 13 Evaluation of xn , 363 Exponential sums; see Muntz polynomials a problem of Lorentz, 291 Bernstein-type inequality, 291 Markov-type inequality, 294 Nikolskii-type inequality, 289 Turan's inequality, 295 with nonnegative exponents, 294 Exponential type, 196, 245 Extended omplete Chebyshev system, 97 Fa tor inequalities, 260{274

via Mahler's measure, 271{273 Fa torization, 10, 36 Fast Fourier transform, 359 Favards theorem, 73 Fejer gap, 27{28 Fejer operators, 164 Fejer's theorem, 165

Gauss, Carl Friedri h, 13 Gaussian hypergeometri series, 62 Gauss-Ja obi quadrature, 67, 75 Gauss-Lu as theorem, 18 Gegenbauer polynomials, 65 Generalized polynomials Bernstein-type inequality, 399, 407 Lp inequalities, 401{407 Markov-type inequality, 399, 407 Nikolskii-type inequality, 394{395 Remez-type inequality, 393{394, 414 S hur-type inequality, 395 weighted inequalites, 407 Girard, Albert, 13 Gra e's omplex version of Rolle's theorem, 25 Gra e's theorem, 18 Gram's lemma, 176 Gram-S hmidt, 44

Haar spa e, 92 Haar system, 92 Halley's method, 365 Hardy spa e, 189 Helly's onvergen e theorem, 71 Helly's sele tion theorem, 71 Hermite interpolation, 9 Hermite polynomials, 57 expli it formulas, 65 Hilbert spa e, 42 Holder's inequality, 17, 49 Horner's rule, 8 Hypergeometri dierential equation, 63 Hypergeometri fun tions, 62 Identity theorem, 15 Inequalities

476 Index

Bernstein-Szeg}o inequality; see Bernstein-Szeg}o inequality Bernstein-type; see Bernstein-type inequality Bessel's; see Bessel's inequality bounded Bernstein-type; see bounded Bernstein-type inequality bounded Chebyshev-type; see bounded Chebyshev-type inequality Cau hy-S hwarz; see Cau hyS hwarz inequality Chebyshev-type; see Chebyshevtype inequality for fa tors; see fa tor inequalities for Muntz polynomials; see Muntz polynomials Holder's; see Holder's inequality Lax-type; see Lax Markov-type; see Markov-type inequality metri ; see metri inequalities Minkowski's; see Minkowski's inequality Nikolskii-type; see Nikolskii-type inequality Remez-type; see Remez-type inequality Russak's; see Russak S hur-type; see S hur-type inequality Triangle; see Triangle inequality unbounded Bernstein-type; see unbounded Bernstein-type inequality Inner produ t, 41 Inner produ t spa e, 41 Integer-valued polynomials, 10 Interpolation Hermite; see Hermite interpolation Lagrange; see Lagrange interpolation Newton; see Newton interpolation

Ja obi polynomials, 57, 63 expli it formulas, 63 Jensen ir les, 19 Jensen's formula, 187

Jensen's inequality, 414 Jensen's theorem, 19

Kernel fun tion, 47, 132 Kolmogorov's inequality, 285 Korovkin's theorems, 163 La unary spa es, 308

quasi-Chebyshev polynomials, 316 Lagrange interpolation, 8 Laguerre polynomials, 57, 66, 130 expli it formulas, 66 Laguerre's theorem, 20 Lax-type inequality, 438 for rationals, 329 Malik's extension, 438 on a half-plane, 338 Legendre polynomials, 57 Lemnis ates of onstant modulus, 352{353 Lexi ographi properties, 116 for Muntz polynomials, 120, 314 for Muntz-Legendre polynomials, 136 for sinh systems, 122 Liouville's theorem, 15 Logarithmi apa ity, 38 Lorentz degree, 82 for polynomials, 86 for trigonometri polynomials, 89 Lorentz's problem, 108, 291 Lp norm, 6, 48, 471 lp norm, 6, 471 Lu as' theorem, 18

Mahler's measure, 271

Mairhuber theorem, 98 Markov system, 100

losure of nondense, 211 derivative of, 112 Markov-Stieltjes inequality, 76 Markov-type inequality for Pn , 233 for Pn , 255 for onstrained polynomials, 417{447 for onstrained polynomials in Lp , 422, 428{429

Index 477 for exponential sums, 276{280, 294{295 for generalized polynomials, 399{407, 445 for generalized polynomials in Lp , 401{407 for higher derivatives, 248{260 for monotone polynomials, 439{ 441 for Muntz polynomials, 276{279, 287{288 for Muntz polynomials in Lp , 279{280 for nonnegative polynomials, 420, 439 for rational fun tions, 336 for self-re ipro al polynomials, 339 for trigonometri polynomials on subintervals, 242{245 in the omplex plane, 235 weighted, 442{443 Maximum prin ipal, 15 m-distribution, 57 Mergelyan's theorem, 170 Mesh of zeros, 155 Metri inequalities, 344{355 for polynomials, 345{346 for rational fun tions, 347{349 Minkowski's inequality in Lp ; p 1, 49 in Lp ; p 1, 52 Moment, 57 Moment problem, 70 Monotone operator theorem, 163 Multipli ation of polynomials, 361 Muntz polynomials bounded Bernstein-type inequality, 178, 182, 213, 310, 317 bound for smallest zero, 313 lexi ographi properties of zeros, 116, 120 Newman's inequality, 276, 301 Newman's inequality in Lp , 279 Newman's problem, 317 Nikolskii-type inequality, 281, 298, 317 positive zeros, 22

Remez-type inequality, 304, 307, 316 where the sup norm lives, 301 Muntz rationals, 218{226 denseness, 218 Muntz spa e, 125{126 lexi ographi properties of zeros, 120 produ ts of, 222, 316 quotients of, 218{226 Muntz system, 125{136

losure, 171{205 nondense, 303{319 Muntz-Legendre polynomials, 125{138 de nition, 126 dierential re ursion, 129 global estimate of zeros, 136 integral re ursion, 132 lexi ographi properties of zeros, 133{136 orthogonality, 127, 132 orthonormality, 128 Rodrigues-type formula, 128, 131 zeros of, 133 Muntz's theorem, 171{205 another proof, 176, 192

losure of span in, 178, 181, 185 in C [0; 1℄, 171 in C [a; b℄, 180, 184 in L2 [0; 1℄, 171 in Lp [0; 1℄, 172 in Lp [a; b℄, 186 in Lp (w), 311 on sets of positive measure, 303

Newman's onje ture

on denseness of produ ts, 316 on denseness of quotients, 220, 223 Newman's inequality, 275{279 an improvement, 287 for Muntz polynomials, 276 for Muntz polynomials in Lp , 279 on positive intervals, 301 Newton's identities, 5 Newton Interpolation, 10 Newton's method, 362, 364{367 for x1=2 , 365 in many variables, 366

478 Index

Nikolskii-type inequality for onstrained polynomials, 444 for exponential sums, 289 for generalized polynomials, 394, 395 for Muntz polynomials, 281, 298, 317 for produ ts of Muntz spa es, 317 Nondense Muntz spa es, 303{319 Nonnegative polynomials, 70, 85, 417, 420 Nonnegative trigonometri polynomials, 85, 409 Norms, 471 Lp ; see Lp norm lp ; see lp norm supremum; see supremum norm

Orthogonal olle tion, 43 Orthogonal fun tions, 41{56 Orthogonal polynomials, 57{79 as ontinued fra tions, 79 as determinants, 76

hara terization of ompa t support, 77 Gegenbauer; see Gegenbauer polynomials Hermite; see Hermite polynomials interla ing of zeros, 61 Ja obi; see Ja obi polynomials Laguerre; see Laguerre polynomials Legendre; see Legendre polynomials Muntz-Legendre; see MuntzLegendre polynomials simple real zeros, 61 ultraspheri al; see Ultraspheri al polynomials Orthogonal rational fun tions, 147 Orthonormal set, 43 Paley-Weiner theorem, 196 Parallelogram law, 42 Parseval's identity, 48 Partial fra tion de omposition, 7, 144 Pellet's theorem, 16 Polar derivative, 20 Polynomials

as sums of squares, 85, 348 Bernstein; see Bernstein polynomials Chebyshev; see Chebyshev polynomials Gegenbauer; see Gegenbauer polynomials generalized; see generalized polynomials growth in the omplex plane, 239 Hermite; see Hermite polynomials in xn , 167 integer valued; see Integer valued polynomials Ja obi; see Ja obi polynomials Laguerre; see Laguerre polynomials Legendre; see Legendre polynomials Muntz; see Muntz polynomial Muntz-Legendre; see MuntzLegendre polynomials number of real roots, 17, 137 symmetri , 5 trigonometri ; see Trigonometri polynomial Ultraspheri al; see Ultraspheri al polynomials with integer oeÆ ients, 169 with nonnegative oeÆ ients, 79{90, 417 with real roots, 345, 347{348 Produ ts of Muntz spa es, 222, 316

Quadrati equations, 4

Quarti equations, 4 Quasi-Chebyshev polynomials, 316, 342

Railway tra k theorem, 98 Rakhmanov's theorem, 78 Rational fun tions algebrai , 139 Chebyshev polynomials of, 139{ 153

oeÆ ient bounds, 153 inequalities; see inequalities trigonometri , 139 Rational spa es

Index 479 of algebrai rational fun tions, 139{153, 320{321 of trigonometri rational fun tions, 139{153, 320{ 321 Re ursive bounds, 359 Remez's algorithm, 371 Remez-type inequality for algebrai polynomials, 228 for onstrained polynomials, 443, 445 for generalized polynomials, 393, 394 for generalized polynomials in Lp , 401{402 for Muntz spa es, 307 for nondense Muntz spa es, 304 for produ ts of Muntz spa es, 316 pointwise, 414 for trigonometri polynomials, 230 Reprodu ing kernel, 47, 132 Reversion of power series, 362 Riemann-Lebesgue lemma, 54 Riesz representation theorem, 50 Riesz's identity, 390 Riesz's lemma, 237 Riesz-Fis her theorem, 50 Rising fa torial, 62 Rolle's theorem, 25 Rou he's theorem, 14, 16 Russak's inequalities, 336

Salem numbers, 6 S hur's theorem, 17 S hur-type inequality for algebrai polynomials, 233 for onstrained polynomials, 436{437 for generalized polynomials, 395 for rational fun tions, 337 for trigonometri polynomials, 238 Self-re ipro al polynomials, 339 quasi-Chebyshev polynomials, 342 Somorjai's theorem, 218 Spa e Chebyshev; see Chebyshev spa e Des artes; see Des artes spa e Haar; see Haar spa e Muntz; see Muntz spa e

rational; see rational spa e Stieltjes' theorem, 78 Stone-Weierstrass theorem, 161 Sums of squares of polynomials, 85, 348 Supremum norm, 6, 29, 471 Symmetri fun tion, 5 Symmetri polynomial, 5 Szeg}o's inequality, 391 Szeg}o's theorem, 23, 235

Tartaglia, Ni

olo, 3

T heby hev; see Chebyshev Three-term re ursion, 59 Totally positive kernels, 110 Trans nite diameter, 38 Triangle inequality, 42 Trigonometri polynomial, 2 Trigonometri polynomials of longest ar length, 35 Turan's inequality, 434 Turan's inequality for exponential sums, 295

Ultraspheri al polynomials, 65 Unbounded Bernstein-type inequality, 206{217

hara terization of denseness, 207 Uni ity theorem, 15 Variation diminishing property, 111 Vandermonde determinant, 38, 103 Videnskii's inequalities, 242{245 Walsh's two ir le theorem, 20

Weierstrass' theorem, 154{170 for Markov systems, 155 for polynomials, 159 for polynomials in x , 167 for polynomials with integer

oeÆ ients, 169 for trigonometri polynomials, 165 in Lp , 169 on ar s, 170 Stone-Weierstrass theorem, 161 Wronskian, 22

Zeros, 11{18

algorithms for nding, 364{367

480 Index

omplexity of, 367

ounting by winding number, 370 in a disk, 369 in an interval, 368 in Chebyshev spa es, 99 lo alizing, 367 maximum number at one, 137 maximum number of positive, 17 of Chebyshev polynomials, 34, 116, 120, 122

of derivatives of polynomials, 18{28 of integrals of polynomials, 24 of Muntz polynomials, 120 of Muntz-Legendre polynomials, 133{136 of orthogonal polynomials, 61 Zolotarev, 35 Zoomers, 218

Polynomials and Polynomial Inequalities (Graduate Texts in Mathematics)

Polynomials and Polynomial Inequalities (Graduate Texts in Mathematics 161)

Polynomials and Polynomial Inequalities (Graduate Texts in Mathematics 161)