Polynomials and Polynomial Inequalities (Graduate Texts in Mathematics)

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For Theresa, Sophie, Alexandra, Jennifer, and Erika

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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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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

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

1. Introdu tion and Basi Properties

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

1.1 Polynomials and Rational Fun tions

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

1. Introdu tion and Basi Properties

℄ 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.

1.1 Polynomials and Rational Fun tions

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

1. Introdu tion and Basi Properties

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 )

1.1 Polynomials and Rational Fun tions

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

1. Introdu tion and Basi Properties

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

1. Introdu tion and Basi Properties

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

1.2 The Fundamental Theorem of Algebra 13

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 := 1 + :

Proof. See E.6.

nX1 j =0

9

!q=p 1=q = aj p : ; an p

j j j j

ut

Comments, Exer ises, and Examples. The Fundamental Theorem of Algebra appears to have been given its name by Gauss, although the result was familiar long before; it resisted rigorous proof by d'Alembert (1740), Euler (1749), and Lagrange (1772). It was more

ommonly formulated as a real theorem, namely: every real polynomial fa tors ompletely into real linear or quadrati fa tors. (This is an essential result for the integration of rational fun tions.) Girard has a laim to priority of formulation. In his \Invention Nouvelle en L'Algebra" of 1629 he wrote \every equation of degree n has as many solutions as the exponent of the highest term." Gauss gave the rst satisfa tory proof in 1799 in his do toral dissertation, and he gave three more proofs during his lifetime. His rst proof, while titled \A new proof that every rational integral fun tion of one variable an be resolved into real fa tors of the rst or se ond degree," was in fa t the rst more-or-less satisfa tory proof. Gauss' rst proof is a geometri argument that the real and imaginary parts of a polynomial, u and v , have the property that the urves u = 0 and v = 0 interse t, and by modern standards has some topologi al problems. His third proof of 1816 amounts to showing that

14

1. Introdu tion and Basi Properties

p0 (z ) dz jzj r p(z )

Z

=

must vanish if p has no roots, whi h leads to a ontradi tion and is a genuinely analyti proof (see Boyer [68℄, Burton [85℄, and Gauss [1866℄). An almost purely algebrai proof using Galois theory, but based on ideas of Legendre, may be found in Stewart [73℄. The \geometry of polynomials" is extensively studied in Marden [66℄ and Walsh [50℄, where most of the results of the se tion and mu h more may be a

essed. See also Barbeau [89℄ and Polya and Szeg}o [76℄. Theorem 1.2.2 is due to Kakeya [12℄. It is a spe ial ase of Corollary 1.2.3, due to Enestrom [1893℄. The Enestrom-Kakeya theorem and related matters are studied thoroughly in Anderson, Sa, and Varga [79℄ and [81℄ and in Varga and Wu Wen-da [85℄, and a number of interesting properties are explored. For example, it is shown in the rst of the above papers that the zeros of all p satisfying the assumption of Corollary 1.2.3 are dense in the annulus fz 2 C : r  jz j  r g. 1

2

E.1 Basi Theorems in Complex Analysis. We olle t a few of the basi theorems of omplex analysis that we need. (Proofs may be found in any

omplex variables text su h as Ahlfors [53℄ or Ash [71℄.) a℄ Cau hy's Integral Formula. Let Dr := fz 2 C : jz j < rg. Suppose f is analyti on Dr and ontinuous on the losure D r of Dr : Let Dr denote the boundary of Dr : Then 0=

f (z ) = and

f

(

n) (z ) =

Z

Dr

f (t) dt ;

Z

1 f (t) dt ; 2i Dr t z

n! 2i

Z

Dr

f (t) (t z )n

+1

z 2 Dr ; dt ;

z 2 Dr :

Unless otherwise spe i ed, integration on a simple losed urve is taken anti lo kwise. (We may repla e Dr and Dr by any simple losed urve and its interior, respe tively, though for most of our appli ations ir les suÆ e.) b℄ Rou he's Theorem. Suppose f and g are analyti inside and on a

simple losed path (for most purposes we may use a ir le). If

jf (z )

g(z )j < jf (z )j

for every z 2 ; then f and g have the same number of zeros inside ( ounting multipli ities).

1.2 The Fundamental Theorem of Algebra 15

A fun tion analyti on C is alled entire.

℄ Liouville's Theorem. A bounded entire fun tion is onstant. d℄ Maximum Prin iple. An analyti fun tion on an open set U



C

assumes its maximum modulus on the boundary. Moreover, if f is analyti and takes at least two distin t values on an open onne ted set U  C ; then

jf (z )j < sup jf (z )j ; z2U

z2U:

e℄ Uni ity Theorem. Suppose f and g are analyti on an open onne ted

set U. Suppose f and g agree on S, where S is an in nite ompa t subset of U, then f and g agree everywhere on U: E.2 Division. a℄ Suppose p is a polynomial of degree n and p() = 0. Then there exists a polynomial q of degree n 1 su h that

p(x) = (x

)q(x) :

Hint: Consider the usual division algorithm for polynomials. b℄ A polynomial of degree n has at most n roots.

ut

This is the easier part of the Fundamental Theorem of Algebra. The remaining ontent is that every non onstant polynomial has at least one omplex root. The next exer ise develops the basi omplex analysis tools mostly for polynomials on ir les. The point of this exer ise is to note that the proofs in this ase are parti ularly straightforward. E.3 Polynomial Complex Analysis. a℄ Dedu e Cau hy's integral formula for polynomials on ir les. Hint: Integrate z n on Dr . ut Qn b℄ If p(z ) = an i (z i ); then the number of indi es i for whi h ji j < r is Z 1 p0 (z ) dz ; 2i Dr p(z ) provided no i lies on Dr : Hint: We have n p0 (z ) X 1 = p(z ) i z i =1

=1

and

Z

dz 1 = 2i Dr z i



1 if 0 if

ji j < r ji j > r :

ut

16

1. Introdu tion and Basi Properties

℄ 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

1.2 The Fundamental Theorem of Algebra 17

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 <

k=0

 jan jjz jn :1  jan jjz j

n

℄

jak jjz jk k=0

8 <

p !1=p a k a

nX1

n

!1=p

nX1

1

:

k=0

p ak a n

jz j jz j

nX1

kq nq

k=0

(jz jq

1

!1=q 9 = ; 9 =

: 1) =q ; 1

When is the last expression positive?

E.7 The Number of Positive Zeros of a Polynomial. Suppose

p(z ) :=

n X j =0

aj z j

has m positive real roots. Then

m

2

 2n log

ja j + jp a j +


+ jan j j a an j 0

1

!

:

0

This result is due to S hur though the proof more or less follows Erd}os and Turan [50℄. It requires using Muntz's theorem from Chapter 4. a℄ Suppose

p(z ) = an

n Y

k=1

(z

rk eik )

1. Introdu tion and Basi Properties

18

and

q(z ) := Note that for jz j = 1;

jz

n Y k=1

rei j jr j

 jz

2

Use this to dedu e that

eik ) :

(z

ei j : !2

a j +


+ jan j jq(z )j  jjpa(za)j j  ja j + jp j a an j n 2

0

2

1

0

0

whenever jz j = 1: b℄ Sin e p has m positive real roots q has m roots at 1: Use the hange of variables x := z + z applied to z n q (z )q (z ) to show that 1

1

kqkfjzj g  min k(z 1)m (z n m + bn m z n m +


+ b z + b )kfjzj fbk g  min kxm (xn m + n m xn m +


+ x + )k ; f k g = 4n min kxm (xn m + dn m xn m +


+ d x + d )k ; fd g 2

=1

1

1

1

1

1

1

1

k

1

0

1

2

0

g

=1

[0 4℄

0

[0 1℄

4n
; 2n + 1 n nm where the last inequality follows by E.2 ℄ of Se tion 4.2.

℄ Show that ! 4n
 m =n log p 2n + 1 n nm and nish the proof of the main result.

p

2

+

2

2

+

1.3 Zeros of the Derivative The most basi and important theorem linking the zeros of the derivative of a polynomial to the zeros of the polynomial is variously attributed to Gauss, Lu as, Gra e, and others, but is usually alled Lu as' theorem [1874℄. Theorem 1.3.1 (Lu as' Theorem). Let p 2 Pn . All the zeros of p0 are on-

tained in the losed onvex hull of the set of zeros of p.

The proof of this theorem follows immediately from the following lemma by onsidering the interse tion of the halfplanes ontaining the onvex hull of the zeros of p.

1.3 Zeros of the Derivative 19

Lemma 1.3.2. Let p 2 Pn . If p has all its zeros in a losed halfplane, then

p0n also has all its zeros in the same losed halfplane.

Proof. On onsideration of the ee t of the transformation z 7! z + ; by whi h any losed halfplane may be mapped to H l := fz : Re(z )  0g; it suÆ es to prove the lemma under the assumption that p has all its zeros in H l : If p has all its zeros in H l ; then n 1 p0 (z ) X = ; p(z ) k z k

k 2 H l :

=1

But if z 2 Hr := fz 2 C : Re(z ) > 0g; then

z

1

k

and it follows that

2 Hr n X k=1

In parti ular,

k 2 H l ;

for ea h

z

n X k=1

z

1

k

1

k

2 Hr : 6= 0 ;

whi h nishes the proof.

ut

There is a sharpening of Lu as' theorem for real polynomials formulated by Jensen. We need to introdu e the notion of Jensen ir les for a polynomial p 2 Pn . For p 2 Pn the nonreal roots of p ome in onjugate pairs. For ea h su h pair,  + i ,  i , form the ir le entered at  with radius j j. So this ir le entered on the x-axis at  has  + i and  i on the opposite ends of its perpendi ular diameter. The olle tion of all su h

ir les are alled the Jensen ir les for p. Theorem 1.3.3 (Jensen's Theorem). Let p

lies in or on some Jensen ir le for p.

2 Pn . Ea h nonreal zero of p0

The proof, whi h is similar to the proof of Lu as' theorem, is left for the reader as E.3. We state the following pretty generalization of Lu as' theorem due to Walsh [21℄. The proof is left as E.4. Proofs an also be found in Marden [66℄ and Polya and Szeg}o [76℄.

20

1. Introdu tion and Basi Properties

Theorem 1.3.4 (Walsh's Two-Cir le Theorem). Suppose p 2 Pn has all its

has n zeros in the disk D with enter and radius r . Suppose q 2 Pm all its m zeros in the disk D with enter and radius r . Then a℄ All the zeros of (pq )0 lie in D [ D [ D , where D is the disk with

enter and radius r given by n + m nr + mr

:= ; r := : n+m n+m b℄ Suppose (n 6= m). Then all the zeros of (p=q )0 lie in D [ D [ D ; where D is the disk with enter and radius r given by nr + mr n m ; r :=

:= n m jn mj : 1

1

1

2

2

1

3

2

2

3

3

3

2

1

2

3

1

3

1

3

3

2

3

3

2

2

1

1

3

3

Comments, Exer ises, and Examples. Lu as proved his theorem in 1874, although it is an easy and obvious onsequen e of an earlier result of Gauss. Jensen's theorem is formulated in Jensen [13℄ and proved in Walsh [20℄. Mu h more on erning the geometry of zeros of the derivative an be found in Marden [66℄. E.1 A Remark on Lu as' Theorem. Show that p0 2 Pn has a zero  on the boundary of the onvex hull of the zeros of p if and only if  is a multiple zero of p. E.2 Laguerre's Theorem. Suppose p 2 Pn has all its zeros in a disk D.

Let  2 C . Let w be any zero of q(z ) := np(z ) + ( z )p0 (z ) (q is alled the polar derivative of p with respe t to ). a℄ If  2= D; then w lies in D. Hint: Consider r(z ) := p(z )(z  ) n ; where p has all its zeros in D and  62 D. Then n r0 (z ) p0 (z ) = + r(z ) p(z )  z and if q (w) = 0 with w 2= D, then r0 (w) = 0. Now observe that r is of the form



1



; s 2 Pn ;  where s0 ((w  ) ) = 0. Note that  2= D implies that e := f(z  ) : z 2 Dg D e and so does s0 by Lu as' theorem. is a disk. Then s has all its zeros in D e , so s0 ((w  ) ) 6= 0, a ontradi However, w 62 D implies (w  ) 62 D r(z ) = s

z

1

1

1

1

tion. ut b℄ If p(w) 6= 0; then any ir le through w and  either passes through all the zeros of pn or separates them.

1.3 Zeros of the Derivative 21

E.3 Proof of Jensen's Theorem. Prove Theorem 1.3.3. Hint: Suppose p 2 Pn n Pn and denote the zeros of p by z ; z ; : : : ; zn : Then n 1 p0 (z ) X = : p(z ) k z zk 1

1

2

=1

If zk = k + i k with k ; k Im



x + iy

2 R; and z = x + iy with x; y 2 R; then

1

k

i k x + iy k + i k 2y ((x k ) + y k ) ; k ) + (y k ) )
((x k ) + (y + k ) ) 2

=

((x



1

+

2

2

2

2

2

2

and so outside all the Jensen ir les and o the x-axis, 



p0 (z ) sign Im pn (z )



= sign(y ) 6= 0 :

ut E.4 Proof of Walsh's Theorem. Prove Theorem 1.3.4. a℄ Prove Theorem 1.3.4 a℄. Hint: Let z be a zero of p0 q + q0 p outside D and D . Let 0

1

 := z 1

0

2

np(z ) and  := z p0 (z ) 0

2

0

0

mq(z ) q0 (z ) 0

0

(p0 (z ) 6= 0 and q 0 (z ) 6= 0 by Lu as' theorem). Observe that   2 D by E.2, and n + m : z = n+m 0

2

0

1

2D

1

and

2

2

1

0

ut

b℄ Prove Theorem 1.3.4 b℄. Hint: Pro eed as in the hint to part a℄, starting from a zero z of p0 q q0 p outside D and D . ut

℄ If in Theorem 1.3.4 a℄ D , D , and D are disjoint, then D ontains n 1 zeros, D ontains m 1 zeros, and D ontains 1 zero of (pq)0 : Hint: By a ontinuity argument we may redu e the general ase to the ase where p(z ) = (z )n and q (z ) = (z )m . ut d℄ If in Theorem 1.3.4 b℄ n = m and D and D are disjoint, then D [ D

ontains all the zeros of (p=q )0 . 0

1

2

1

2

3

2

1

3

1

2

1

2

1

2

22

1. Introdu tion and Basi Properties

E.5 Real Zeros and Poles. a℄ If all the zeros of p 2 Pn are real, then all the zeros of p0n are also real. b℄ Suppose all the zeros of both p 2 Pn and q 2 Pm are real, and all the zeros of pn are smaller than any of the zeros of qn . Show that all the zeros of (p=q )0 are real. Hint: Consider the graph of (p=q )0 p0 = (p=q ) p

q0 : q

ut

De ne W (p), the Wronskian of p, by

W (p)(z ) = p(z )p00 (z ) (p0 (z )) p(z ) p0 (z ) = 0 p (z ) p00 (z )   0 p (z ) 0 : = p (z ) p(z )

2

2

℄ Prove that if p 2 Pn has only distin t real zeros, then W (p) has no real zeros. In Craven, Csordas, and Smith [87℄ it is onje tured that, for p 2 Pn , the number of real zeros of W (p)=p does not ex eed the number of nonreal zeros of p (a question they attribute to Gauss). d℄ Let p 2 Pn . Show that any real zero of W (p) lies in or on a Jensen

ir le of p. Proof. See Dil her [91℄. ut e℄ Show that Lu as' theorem does not hold for rational fun tions. Hint: Consider r(x) = x=( x ). ut The next exer ise is a weak form of Des artes' rule of signs. 2

2

2

E.6 Positive Zeros of Muntz Polynomials. Suppose Æ < Æ < and 0

1




< Æn

f (x) := a xÆ0 + a xÆ1 +


+ an xÆn ; ak 2 R : 0

1

Show that either f = 0 or f has at most n zeros in (0; 1). Hint: Pro eed by indu tion on n.

ut

1.3 Zeros of the Derivative 23

E.7 Apolar Polynomials and Szeg}o's Theorem. Two polynomials

f (x) : = and

g(x) : = are alled apolar if

n X k=0

ak

n X

bk

k=0

 

n k x ; k

an 6= 0

 

n k x ; k

bn 6= 0

 

n X

n

( 1)k ak bn k = 0: k k =0

a℄ A Theorem of Gra e [02℄. Suppose that f and g are apolar polynomials. If f has all its zeros in a ( losed or open) disk D; then g has at least one zero in D. Hint: Let  ;  ; : : : ; n and ; ; : : : ; n denote the zeros of f and g, respe tively. Suppose that the zeros of g are all outside D. Let f (x) := nf (x) + ( x)f 0 (x) 1

2

1

2

1

1

and for k = 2; 3; : : : ; n, let

fk (x) := (n

k + 1)fk (x) + ( k 1

x)fk0 (x) :

Then, by E.2, ea h fk has all its zeros in D: Now ompute

fn ( n ) = 1

n! bn

 

n 0

a bn

 

0

n 1

a bn 1

1

+




+ ( 1)n

 

n a b n n

 0

= 0;

where the vanishing follows by apolarity. This is a ontradi tion. ut b℄ If f and g are apolar, then the losed onvex hull of the zeros of f interse ts the losed onvex hull of the zeros of g .

℄ A Theorem of Szeg}o [22℄. Suppose

f (x) := g(x) :=

n X k=0 n X k=0

ak bk

and h(x) :=

 

n k x ; k

an 6= 0 ;

n k x ; k

bn 6= 0 ;

 

n X k=0

a k bk

 

n k x : k

Suppose f has all its zeros in a losed disk D, and g has zeros ; : : : ; n : Then all the zeros of h are of the form i i with i 2 D: 1

1. Introdu tion and Basi Properties

24

Hint: Suppose Æ is a zero of h. Then n X k=0

ak bk

 

n k Æ = 0: k

So the polynomial

r(x) :=

n X k=0

( 1)k

 

n b Æ k xn k k

k

is apolar to f , and thus has a zero  in D. But then  = Æ= i for some i sin e r(x) = xn g ( Æ=x): ut E.8 Zeros of the Integral. Suppose p 2 Pn n Pn has all its zeros in D := fz 2 C : jz j  1g: R a℄ Show that the polynomial q de ned by q (x) := x p(t) dt has all its zeros in D := fz 2 C : jz j  2g: Hint: Apply E.7 ℄. Take 1

1

0

2

f (x) := p(x); g(x) := Then

h(x) =

1

x

Z x

q(x) :=

Z x Z tm 0

0

1

1

Z tm

2

0

has all its zeros in Drm;n := fz the zero of






k=0

xk

k k+1

:

p(t) dt :

0

Note that g (x) = (n + 1) x ((1 + x)n b℄ Show that 1

n   X n

+1

Z t1

1) has all its zeros in D : 2

p(t) dt dt

1

0




dtm

2

dtm

ut

1

2 C : jz j  rm;n g, where rm;n  m + 1 is

n X

k=0





m+n k x m+k

with the largest modulus. Note that q is the mth integral of p normalized so that the onstants of integration are all zero. Proof. See Borwein, Chen, and Dil her [95℄. ut

1.3 Zeros of the Derivative 25

E.9 Gra e's Complex Version of Rolle's Theorem. Suppose  and are

zeros of p 2 Pn n Pn . Then p0 has at least one zero in the disk 1

D( ; r) := fz 2 C : jz

where

+

:=

and

2

j  r g ;

j

r :=

j

2

ot

 : n

Hint: Assume, without loss of generality, that  = 1 and = 1: Let n

n

X X xk p0 (x) = ak xk ; that is , p(x) = + ak : k+1 k k 1

=0

1

+1

=0

Apply E.7 a℄. Note that 0=

p(1)

So

p( 1)

=a +

2

0

1)n

f (z ) := (z

and p0 are apolar.

a

2

3

+

a

n

 

n

+


:

(z + 1)n

E.10 Corollaries of Szeg}o's Theorem. Suppose  

4

5

ut

 

n a zn ; 0 1 n n       n n n g(z ) := b + b z +


+ b zn ; 0 1 n n

f (z ) :=

a + 0

0

and h(z ) :=

 

n 0

ab + 0

0

a z +


+ 1

1

 

n 1

a b z +


+ 1

 

1

n a b zn n nn

with an bn 6= 0. a℄ If f has all its zeros in a onvex set S ontaining 0 and g has all its zeros in [ 1; 0℄; then h has all its roots in S: b℄ If f and g have all their zeros in [ 1; 0℄; then so does h. E.11 Another Corollary of Szeg}o's Theorem. If

in D := fz 1

2 C : jz j  1g; then so does

has all its zeros in D :

n X k=0

n X ak z
k n . k=0

k

ak z k has all its zeros

In parti ular,

n X

z k


n k=0 k

1

The results of the next exer ise were rst proved by M. Riesz (see, for example, Mignotte [92℄) and were redis overed by Walker [93℄.

26

1. Introdu tion and Basi Properties

E.12 Conse utive Zeros of p0 for p 2 Pn with Real Zeros. For a polynomial n Y p(x) := (x i ) ;  <  <


< n ; n  2 1

i=1

2

with only real zeros, let


(p) := min (i in 1

1

i ) :

+1

By Rolle's theorem n

Y p0 (x) = n (x i ) ;  < <  < <


< n 1

1

i=1

1

2

2

1

< n :

a℄ Suppose n  3. Prove that
(p) <
(p0 ):

Outline. It is required to show that j 2  j  n be xed. Sin e

j

1

>
(p) for ea h j

 2. Let

f 0 ( j ) f 0 ( j ) = =0 f ( j ) f ( j ) 1

1

we have

n X i=1

( j

1

1

i )( j

i )

= 0:

Now let uj := j j ; vj := j j : Also for ea h i; let di := j j i ; ei := j i j : Then the above an be rewritten as 1

+

j 1 X i=1

n j

(di

X 1 1 1 + + uj )(di + vj ) ( uj vj ) i (ei + uj )(ei =1

vj )

= 0:

De ne

F (u; v) :=

n j

j 1 X i=1

(di

X uv uv + u)(di + v) i (ei + u)(ei =1

Note that F is in reasing in ea h variable (0 observe that F (uj ; vj ) = 1 :

v)

:

 u < d ; 0  v < e ) and 1

1

To prove the result, it suÆ es to show that if u and v are nonnegative numbers satisfying u + v =
(p); then F (u; v ) < 1.

1.3 Zeros of the Derivative 27

Now show that j 1 X

F (u; v) uv

(d i=1 i j 1 X

1 u)(di

+1

u)

+

n Xj i=1

(ei



!

1 v)(ei

+1

v)

nXj 

1 1 1 1 
uv(p) + di u di u ei v ei v i i     1 1 uv + 
uv(p) v1 + u1  1 <
(p) d u e v +1

=1

1

!

+1

=1

1

whenever u and v are nonnegative numbers satisfying u + v =
(p):

ut

b℄ Suppose n  3 and 2 R. Show that
(p0 p) has only real zeros and
(p0 p) >
(p):

℄ What happens when p has only real zeros but they are not ne essarily distin t? E.13 Fejer's Theorem on the Zeros of Muntz Polynomials. The following pretty results of Fejer may also be found in Polya and Szeg}o [76℄: Suppose that (k )1 k is an in reasing sequen e of nonnegative integers with  = 0. a℄ Let =0

0

n X

p(z ) :=

k=0

ak 2 C ;

ak z  k ;

2 C so that  


n  )(  )


(n

aa = 6 0: 0

1

Then p has at least one zero z

0

jz j 



2

(

0

2

1

3

3



1

1

1=1 1=1 a0 a ) 1

:

Outline. We say that z 2 C is not less than z 2 C if jz j  jz j: Studying q(z ) := z n p(z ); we need to show that the largest zero of 1

2

2

1

1

q(z ) = a0 xn +

n X k=1

ak xn

k

is not less than 

(

2

 )(  )


(n  


n 1

3

2

1

3

) 1

1=1 1=1 a1 a 0

:

We prove this statement by indu tion on n: The statement is obviously true for n = 1: Now assume that the statement is true for n 1: It follows from Lu as' theorem that if q is a polynomial with omplex oeÆ ients, then the largest zero of q 0 is not greater than the largest zero of q:

1. Introdu tion and Basi Properties

28

By the above orollary of Lu as' theorem, it is suÆ ient to prove that the largest zero of

z n

n +1 q 0 (z ) = 

1

 na z n 1 + 0

nX1 k=1

(n

k )ak z n

1

k

is not less than 

(

2

 )(  )


(n  


n 1

3

2

1

3

) 1

1=1 1=1 a1 a 0

ut

However, this is true by the indu tive hypothesis. b℄ Suppose

f (z ) =

1 X

k=0

ak z k ;

:

ak 2 C

is an entire fun tion so that 1 k 1=k < 1; that is, the entire fun tion f satis es the Fejer gap ondition. Show that there is a z 2 C so that f (z ) = 0: Hint: Use part a℄. ut P

=1

0

0

This is page

29

Printer: Opaque this

2

Some Spe ial Polynomials

Overview Chebyshev polynomials are introdu ed and their entral role in problems in the uniform norm on [ 1; 1℄ is explored. Sequen es of orthogonal fun tions are then examined in some generality, although our primary interest is in orthogonal polynomials (and rational fun tions). The third se tion of this

hapter is on erned with orthogonal polynomials; it introdu es the most

lassi al of these. These polynomials satisfy many extremal properties, similar to those of the Chebyshev polynomials, but with respe t to (weighted) L norms. The nal se tion of the hapter deals with polynomials with positive oeÆ ients in various bases. 2

2.1 Chebyshev Polynomials The ubiquitous Chebyshev polynomials lie at the heart of many analyti problems, parti ularly problems in C [a; b℄, the spa e of real-valued ontinuous fun tions equipped with the uniform (supremum) norm, k
k a;b . Throughout this book, for any real- or omplex-valued fun tion f de ned on [a; b℄, kf k a;b := sup jf (x)j : [

[

℄

x2[a;b℄

℄

30

2. Some Spe ial Polynomials

The Chebyshev polynomials are de ned by

Tn (x) : = os(n ar

os x) ; (2.1.1)

= =



1 (x + 2

n

bX n= 2

2 k

p

x

=0

1)n + (x

2

( 1)k

p

x

x 2 [ 1; 1℄ ;



x2C;

1)n ;

2

(n k 1)! (2x)n k ; k!(n 2k)!

x2C:

2

These elementary equivalen es are left for the reader (see E.1). The nth Chebyshev polynomial has the following equios illation property on [ 1; 1℄. There exist n + 1 points i 2 [ 1; 1℄ with 1 = n < n <


<  = 1 so that 1

(2:1:2)

Tn(j ) = ( 1)n

j

kTnk

[

n j; ; = ( 1)

0

j = 0; 1; : : : ; n :

1 1℄

In other words Tn 2 Pn takes the values kTn k ; with alternating sign the maximum possible number of times on [ 1; 1℄: (These extreme points are just the points os(k=n), k = 0; 1; : : : ; n.) The Chebyshev polynomial Tn satis es the following extremal property: [

1 1℄

Theorem 2.1.1. We have min

p2Pn

1

kxn

p(x)k

[

k

1 n Tn ; = 2

1 1℄

k

[

1 n ; ; =2

1 1℄

where the minimum is uniquely attained by p(x) = xn

2 n Tn (x): 1

Proof. Observe that, while the minimum is taken over Pn ; we need only

onsider p 2 Pn ; sin e taking the real part of a p 2 Pn an only improve the estimate. From the above formulas for Tn we have 1

1

1

s 2 Pn

2 n Tn (x) = xn + s(x) ; 1

Now suppose there exists q 2 Pn

kxn

(2:1:3) Then

2 n Tn (x) 1

(xn

1

1

:

with

q(x)k

[

;

1 1℄

<2

1

n:

q(x)) = s(x) + q(x) 2 Pn

1

hanges sign between any two onse utive extrema of Tn ; hen e it has at least n zeros in ( 1; 1); and thus it must vanish identi ally. This ontradi ts (2.1.3), and we are done up to proving uniqueness (this is left as E.2). u t

2.1 Chebyshev Polynomials 31

Comments, Exer ises, and Examples. The Chebyshev polynomials Tn are named after the versatile Russian mathemati ian, P. L. Chebyshev (1821{1894). The T omes from the spelling T heby hef (or some su h variant; there are many in the literature). A wealth of information on these polynomials may be found in Rivlin [90℄. Throughout later se tions of this book the Chebyshev polynomials will keep re urring. The initial exer ises explore elementary properties of the Chebyshev polynomials. Erd}os [39℄ proved that for t 2 Tn with ktkR  1; the length of the graph of t on [0; 2 ℄ is the longest if and only if t is of the form t() = os(n + ) with some  2 R (see E.6). He onje tured that for any p 2 Pn with kpk ;  1; the maximum ar length is attained by the nth Chebyshev polynomial Tn . This is proved in Bojanov [82b℄. Kristiansen [79℄ also laims a proof. In E.9 the redu ibility of Tn is onsidered, and in E.11 the basi properties of the trans nite diameter are established. [

1 1℄

E.1 Basi Properties. a℄ Establish the equivalen e of the three representations of Tn given in equation (2:1:1): Hint: os n = [( os  + i sin )n + (( os  i sin )n ℄: To get the third representation, use E.3 b℄. ut b℄ The zeros of Tn are pre isely the points 1

2

xk = os

k

(2

2



1)

n

;

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

℄ The extrema of Tn (x) in [ 1; 1℄ are pre isely the points

k = os k n ;

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

d℄ Observe that the zeros of Tn and Tn

+1

interla e, as do the extrema.

E.2 Uniqueness of the Minimum in Theorem 2.1.1. Prove the uniqueness of the minimum in Theorem 2.1.1. Hint: Assume that q 2 Pn and 1

kxn Then

q(x)k

[

h(x) := 2

1

n T (x) n

;

1 1℄

2

1

Re(xn

n:

q(x))

de nes a polynomial from Pn on R having at least n zeros ( ounted a

ording to their multipli ities). Thus 1

2 n Tn (x) = Re(xn 1

q(x)) ;

x 2 R;

32

2. Some Spe ial Polynomials

whi h, together with the previous inequality, implies that q (x) is real whenever Tn (x) = 1: Now E.6 of Se tion 1.1 (Lagrange interpolation) yields that q has real oeÆ ients. Hen e 2 n Tn (x) = xn

x 2 R:

q(x) ;

1

ut

E.3 Further Properties of Tn . a℄ Composition. Show that Tnm (x) = Tn (Tm (x)): b℄ Three-Term Re ursion. Show that

Tn (x) = 2xTn (x) 1

Tn (x) ;

n = 2; 3; : : : :

2

℄ Verify that

T T T T T T

0 1 2 3

4 5

(x) = 1 (x) = x (x) = 2x (x) = 4x (x) = 8x (x) = 16x 2

3

4

1 3x 8x + 1 20x + 5x : 2

5

3

Note that Tn is even for n even and odd for n odd. d℄ Another Formula for Tn : Show that Tn (x) = osh(n osh (x)) for every x 2 R n [ 1; 1℄: e℄ Dierential Equation. Show that 1

x )Tn00 (x)

(1

2

f℄ An Identity. Show that

Tn (x) =

xTn0 (x) + n Tn(x) = 0 : 2

Tn0 (x) 2n + 2 +1

g℄ Orthogonality. Show that 2



Z

1

1

Tn0 (x) : 2n 2 1



Tn (x)Tm (x)dx 0; n 6= m p = Æn;m := 1; n = m > 0 : 1 x 2

h℄ Generating Fun tion. Show that

1 X 1 yx = Tn (x)yn ; 1 2yx + y n 2

x 2 [ 1; 1℄ ;

jy j < 1 :

=0

Hint: Set x = os  and sum.

ut

2.1 Chebyshev Polynomials 33

i℄ Another Representation of Tn : Show that

bX n=  2

Tn(x) =

k=0



n xn 2k

2

k (x2

1)k :

j℄ Another Identity. Show that


Tn (x + x ) = (xn + x n ) : 1

1

1

2

2

E.4 Approximation to xk on [0; 1℄. a℄ Let Tn (x) = Tn (2x 1) be the nth Chebyshev polynomial shifted to the interval [0; 1℄: Suppose

Tn (x) =

n X k=0

bk x k :

Show that for ea h k = 0; 1; : : : ; n; min

j 2R

k

xk

n X j =0 j 6=k

j xj k

; =

[0 1℄

kbk

1

Tnk

;

[0 1℄

:

Hint: Pro eed as in the proof of Theorem 2.1.1 and use E.6 of Se tion 1.3.

ut

b℄ Why does this not hold for Tn on [ 1; 1℄ ?

E.5 A Composition Chara terization. Suppose (pn )1 n is a sequen e of polynomials of degree n and for all positive integers n and m =1

pn Æ pm = pn
m : Then there exists a linear transformation w(x) = x + so that

w Æ pn Æ w or

w Æ pn Æ w

1

1

= xn ;

n = 1; 2; : : :

= Tn ;

n = 1; 2; : : : :

This result is due to Blo k and Thielman [51℄. The proof outlined in this exer ise follows Rivlin [90℄.

34

2. Some Spe ial Polynomials

a℄ Let

q(x) := a + a x + a x ; a = 6 0 ; and v(x) := 0

1

2

2

2

a : 2a

x a

1

2

2

Then

v (q(v(x))) = x + with := a a + (a =2) (a =4) : 1

2

0

2

1

2 1

b℄ Let q (x) = a + a x + a x ; a 6= 0: Then there is at most one polynomial pn of degree exa tly n so that 0

1

2

2

2

pn (q(x)) = q(pn (x)) : Hint: By a℄ we may assume q(x) = x + : Now suppose r, s 2 Pn nPn ; 2

1

r(x + ) = r (x) + 2

and

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

s 2 Pn

Then u := r

2

1

2

satis es

u(x + ) = u(x)(r(x) + s(x)) 2

from whi h we dedu e, by omparing degrees on both sides, that n = 0: (Note that the above onditions imply r and s moni .) ut

℄ Finish the proof of the initial statement of this exer ise. This is a spe ial ase of a more general theorem of Ritt [23℄ that lassi es all rational fun tions r and s that ommute in the sense that r Æ s = s Æ r: d℄ Another Composition Chara terization. Suppose p 2 Pn has the property that the losure of the set

Ip := fz 2 C : p k (z ) = 0 for some k = 1; 2; : : : g [ ℄

is the interval [ 1; 1℄; where p k is the k th iterate of p; that is, [ ℄

p := p and p k := p Æ p k [1℄

Then p(x) = Tn (x): e℄ Let

[ ℄

[

for k = 2; 3; : : : :

1℄

rn (x) = tan(n tan (x)) : Show that rn is a rational fun tion in Rn;n ; and observe that 1

rn Æ rm = rn
m :

2.1 Chebyshev Polynomials 35

E.6 Trigonometri Polynomials of Longest Ar Length. Theorem 5.1.3 (Bernstein-Szeg}o inequality) asserts that

t0 () + n t ()  n ktkR 2

2 2

2

2

for every t 2 Tn and  2 R: Use this to prove the following result of Erd}os. For t 2 Tn with ktkR  1; the length of the graph of t on [0; 2 ℄ is the longest if and only if it is of the form t() = os(n + ) for some  2 R: Hint: Suppose t 2 Tn with ktkR = 1: Let s() := os n: If 1 < t( ) = s( ) < 1 1

2

holds, then by the Bernstein-Szeg}o inequality (see also E.5 of Se tion 5.1)

jt0 ( )j  n(1 1

t ( )) 2

1

= = n(1

s ( ))

1 2

2

2

= =

1 2

js0 ( )j ; 2

and if equality holds for one pair of  ,  ; then it holds for all pairs, and t()  os(n + ) for some  2 R: Suppose tn () 6 os(n + ): Let  and  be monotone ar s of the graphs of y = t() and y = s(); respe tively, with endpoints of ea h having the same ordinates y and y : Let j j and j j be the length of  and ; respe tively, and let jx j and jx j be the length of the proje tion of  and ; respe tively, on the x-axis. Show that 1

2

1

2

j j < jj + (jx j jx j) by approximating  and  by a polygonal line orresponding to a subdivision of the interval with endpoints, y and y on the y -axis. ut 1

2

E.7 Moni Polynomials with Minimal Norm on an Interval. a℄ The unique moni polynomial p 2 Pn minimizing kpk a;b is given by [

p(x) = 2



b a 4

n

Tn



2x

℄



a b : b a

b℄ Let 0 < a < b. Find all moni polynomials p 2 Pn minimizing

kpk

[

b; a℄[[a;b℄ :

(For two intervals of dierent lengths this is a mu h harder problem. The problem was originally due to Zolotarev and is solved in terms of ellipti fun tions. See Todd [88℄, Fis her [92℄, and Peherstorfer [87℄.)

36

2. Some Spe ial Polynomials

E.8 Lower Bound for the Norm of Polynomials on the Unit Disk. Let D be the open unit disk of C : Show that

ka

+ a z +


+ an z n kD



max jak j

kn for arbitrary omplex numbers a ; a ; : : : ; an : Thus z n plays the role of the nth Chebyshev polynomial on the unit disk. Hint: If p(z ) := a + a z +


+ an z n; then 0

1

0

0

0

1

1

1 am = 2i

Z

p(z ) dz : zm +1

D

ut

The next exer ise supposes some familiarity with the rudiments of redu ibility over Q and basi properties, su h as irredu ibility of y lotomi polynomials over Q (see Clark [71℄). Details of the following observation of S hur's are in Rivlin [90℄. E.9 On the Redu ibility of Tn over Q . Let n 2 N be xed. a℄ The zeros of Tn (x=2) are all of the form

xj := e

j

(2

i=(2n) + e

1)

j

(2

i=(2n) ;

1)

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

b℄ If n  3 and  is a primitive nth root of unity, then  +  is of degree '(n)=2: (Here ' is the Euler ' fun tion.)

℄ Thus if Tn is irredu ible over Q ; then n must be a power of 2: d℄ For a positive integer h; let 1

Fh (x) :=

n Y j =1 g d(2j 1;2n)=h

(x

xj ) :

(Here g d(m; n) denotes the greatest ommon divisor of m and n:) Show that if h is odd, then Fh is irredu ible over Q : e℄ The Fa torization of Tn : 2Tn (x=2) =

Y

hjn ; h odd

Fh (x) :

So if n is odd, Tn has '(n) fa tors, while if n is even, then Tn has '(m) fa tors, where m is the largest odd divisor of n: f℄ Let n  3 be odd. Then Tn (x)=x is irredu ible over Q if and only if n is prime.

2.1 Chebyshev Polynomials 37

E.10 Chebyshev Polynomials of the Se ond Kind. Let the Chebyshev polynomials of the se ond kind be de ned by 1 sin n Un (x) := Tn0 (x) = ; n sin 

x = os  :

1

a℄ Un (x) = 2Tn (x) + Un (x) . b℄ Tn (x) = Un (x) xUn (x) . 2

n P

℄ Un (x) =

k=0

d℄ Un (x) =

1

xk Tn k (x) .

p

x+ x

2


n+1

1

p

2 x e℄ Orthogonality. Show that 2



Z1

p

x

p

x


n+1

1

2

1

2

Un (x)Um (x) 1 x dx = Æn;m :=



2

1

.

n 6= m n = m > 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

2. Some Spe ial Polynomials

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

2. Some Spe ial Polynomials

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

2.2 Orthogonal Fun tions 43

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

2. Some Spe ial Polynomials

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

2.2 Orthogonal Fun tions 45

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

2. Some Spe ial Polynomials

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 k
k := 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

2.2 Orthogonal Fun tions 47

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

2. Some Spe ial Polynomials

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℄:

2.2 Orthogonal Fun tions 49

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

2. Some Spe ial Polynomials

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 k
kLp 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

2.2 Orthogonal Fun tions 51

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 > <

f (x)

if f (x) 6= 0

0

if f (x) = 0 :

sign(f (x)) := jf (x)j > :

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

2. Some Spe ial Polynomials

(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

2.2 Orthogonal Fun tions 53

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

2. Some Spe ial Polynomials

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

<1

2

1

x = (xk )11 ;

;

xk 2 C :

1 Show that the fun tion I : L [ ;  ℄ ! ` de ned by I (f ) := fb := (fb(k))11 k=

2

)

2

is one-to-one and onto, and

kI (f )k` = kf kL 2

;℄ :

2[

Hint: Use part a℄ to show that I is one-to-one. Use the Riesz-Fis her theorem (E.7 e℄) to show that I is onto. The norm-preserving property is the

ontent of part b℄. ut Part d℄ shows that the stru ture of L [ ;  ℄ is the same as that of ` : Hen e L [ ; ℄ is a separable Hilbert spa e, that is, it has a ountable dense subset. So if  > 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

2.2 Orthogonal Fun tions 55

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 > <

f (x)

if f (x) 6= 0

0

if f (x) = 0 ;

g(x) := sign(f (x)) := jf (x)j > :

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

2. Some Spe ial Polynomials

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

2.3 Orthogonal Polynomials 59

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

2. Some Spe ial Polynomials

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.

2.3 Orthogonal Polynomials 61

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

2. Some Spe ial Polynomials

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 ;

2.3 Orthogonal Polynomials 63

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

2. Some Spe ial Polynomials

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; : : : :

2.3 Orthogonal Polynomials 65

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

2. Some Spe ial Polynomials

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

f℄ Generating Fun tion.

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 :

2.3 Orthogonal Polynomials 67

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

f℄ Generating Fun tion.

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

2. Some Spe ial Polynomials

℄ 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.

2.3 Orthogonal Polynomials 69

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

2. Some Spe ial Polynomials

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; : : : :

2.3 Orthogonal Polynomials 71

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

2. Some Spe ial Polynomials

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

2.3 Orthogonal Polynomials 73

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

2. Some Spe ial Polynomials

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

2.3 Orthogonal Polynomials 75

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

2. Some Spe ial Polynomials

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

:

2.3 Orthogonal Polynomials 77

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

2. Some Spe ial Polynomials

℄ 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

2. Some Spe ial Polynomials

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

2.4 Polynomials with Nonnegative CoeÆ ients 81

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

2
4

n

nY +1 j =2

(1 + 2
2 j )  2e4 n :

82

2. Some Spe ial Polynomials

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

2.4 Polynomials with Nonnegative CoeÆ ients 83

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

2. Some Spe ial Polynomials

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

2.4 Polynomials with Nonnegative CoeÆ ients 85

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

0  p0 (x) <

n p(x) : x

1 b℄ If (pn )1 n is a sequen e of polynomials with pn 2 P := [k Pk and 1 (pn )n onverges to f uniformly on [0; 1℄; then f is the restri tion to [0; 1℄ of a fun tion analyti in D := fz 2 C : jz j < 1g of the form +

+

=1

=0

=1

f (z ) =

1 X n=0

an  0 :

an z n ;

Hint: Sin e (pn (1))1 n onverges and ea h pn has nonnegative oeÆ ients, there is a onstant C su h that =1

kpn kD  pn (1)  C ;

n = 1; 2; : : : :

Now Montel's theorem (see, for example, Ash [71℄) implies that (pn )1 n has a lo ally uniformly onvergent subsequen e on D. Dedu e that this subsequen e onverges to an f with nonnegative oeÆ ients. ut =1

E.3 Nonnegative-Valued Polynomials and Sums of Squares. a℄ Suppose p 2 P n is nonnegative on R. Then there exist s, t 2 Pn su h that p(x) = s (x) + t (x) : 2

2

2

Hint: If p 2 P and p is nonnegative, then for some real numbers  and , 2

p(x) = (x

) + : 2

2

Now use the identity (a + b )( + d ) = (a + bd) + (ad 2

2

2

2

b ) :

2

2

ut

b℄ If p 2 P n is nonnegative for x  0; then there exist s, t, u, v 2 Pn su h that p(x) = s (x) + t (x) + xu (x) + xv (x) : 2

2

℄ Suppose t su h that

2

2

2

2 Tn is nonnegative on R. Show that there exists a q 2 Pn

t() = jq(ei )j ;  2 R: Show also that if, in addition, t 2 Tn is even, then there exists a q

su h that the above holds.

2

2 Pn

86

2. Some Spe ial Polynomials

d℄ If p 2 Pn is nonnegative on [ 1; 1℄; then there exist s, t 2 Pn su h that

p(x) = s (x) + (1 x )t (x) : 2

2

2

Hint: Write, by ℄, p( os ) = js( os ) + it( os ) sin j : 2

ut

The above exer ise follows Polya and Szeg}o [76℄; see also E.1 of Se tion 7.2 where this result is extended. The following two exer ises dis uss results proved in Erdelyi and Szabados [88℄, [89b℄, and Erdelyi [91 ℄. E.4 Lorentz Degree of Polynomials. Given a polynomial p 2 Pn ; let d = d(p) be the minimal nonnegative integer for whi h the polynomial p is of the form

p(x) = 

d X j =0

aj (1 x)j (x + 1)d j ; aj

If there is no su h d; then let d(p) := degree of the polynomial p: a℄ Let p 2 Pn n Pn be of the form

 0:

1: We all d := d(p) the Lorentz

1

p(x) =

n X j =0

bj 2 R :

bj (1 x)j (x + 1)n j ;

For m  n; let the numbers bj;m , j = 0; 1; : : : ; m; be de ned by

p(x) =

n X

x)j (x + 1)n j

bj (1

j =0 m X = bj;m(1 j =0

!

1

2

x

+

x+1

m n

2

x)j (x + 1)m j :

Show that if d(p) is nite, then it is the smallest value of m for whi h ea h bj;m is nonnegative or ea h bj;m is nonpositive. b℄ Show that if p 2 P n P has no zeros in ( 1; 1); then d(p) = 1.

℄ Show that if p 2 P nP has no zeros in the open unit disk, then d(p) = 2. d℄ Show that if the zeros of a polynomial p 2 P n P lie on the ellipse 1

2

0

1

2



1

B := z = x + iy : y =  (1 x ); x 2 ( 1; 1)

with  2 (0; 1℄; then

2



2

2

 d(p) < 2

2

2

+ 1:



2.4 Polynomials with Nonnegative CoeÆ ients 87

e℄ Let  2 (0; 1) be su h that 

p (x) := x + 2 2

1

is an integer. Let

2

3 1 8 5 + 1 x+ : 1  1  2

4

2

2

2

Show that p has its zeros on B de ned in d℄, and d(p ) =  : f℄ Let  2 (0; 1) be su h that 2 is an integer. Let 1

2

1

2

p (x) := x 2

2

2

3  5 +  x+  )(2  ) (1  )(2

2

(1

2

8

2

6

2

4

2

8 + 4 : ) 2

2 2

Show that p has its zeros on B de ned in part d℄, and d(p ) = 2 : g℄ Show that d(pq )  d(p) + d(q ) for any two polynomials p and q: h℄ Let  2 (0; 1℄. Show that if p 2 Pn has no zeros in 2

2

2





D := z = x + iy : y <  (1 x ); x 2 ( 1; 1); y 2 R ; 2

then

2

d(p) < 2n

2

2

+ n < 3n

2

:

i℄ Let p be a polynomial. Show that d(p) < 1 if and only if p = 0 or p has no zeros in ( 1; 1): j℄ Show that   q p y + i (1 y2 )n 2 d



 2n p

y

n
d

4

for every p 2 Bd ( 1; 1), 1  n  d; and y 2 [0; 1) (i is the imaginary unit). Hint: Modify the proof of Lemma A.5.4. ut k℄ Let b 2 [0; 1℄. Show that

p0 (b)  d p(b) for every p 2 Bd ( 1; 1); positive in ( 1; 1): Hint: If qj;d(x) := (1 x)j (x + 1)d j ; then 

0 (b) = qj;d (b) d j qj;d 1+b for every b 2 [0; 1℄: l℄ Show that d(p) 

1 17

p(x) = ((x

n

2

1

j



b

 d qj;d (b) ;

j = 0; 1; : : : ; d

ut

whenever

z )(x z ))n ; z 0

where B is de ned in part d℄.

0

0

2 B ;

 2 (0; 1℄ ;

88

2. Some Spe ial Polynomials

Proof. Let z = y + i(1 y ) = , y 2 ( 1; 1): Without loss of generality it may be assumed that 0  y < 1: Distinguish two ases. 2 1 2

0

2

Case 1: 1

2

 y < 1. By part k℄,

p0 (1) 2n(1 y ) = p(1) (1 y) +  (1 y ) n 2n : > = (1 y ) +  (1 + y ) 2

d(p) 

2

2

2

2

Case 2: 0  y < 1

2

2 : Applying part k℄ with 2

b := y + (1 y )

=

2 1 2

dedu e that

d(p) 

(2:4:3)

2 [0; 1℄

p0 (b) 2n(b y ) n = = p(b) (b y) +  (1 y ) (1 y ) 2

2

2

=

2 1 2

:

Use part j℄ to obtain (2:4:4)

(1

(1

y ) 2

2





y )n n  2n (1 y ) + 16nd 4d 2

n

2

2

2

2

;

where d := d(p) and n = deg(p): If 1

2

y )

(1

2

y )n ; 8d

 (1

2

2

then there is nothing to prove. Therefore assume that (2:4:5)

y )

(1

2

y )n : 8d

 (1

2

2

Now (2.4.3) to (2.4.5) yield 

(1

and so

(1

Sin e, by (2.4.3), n d 2

y )n 8d 2

1 17

n : 2



n 16d

2

n

2

y ) +

(1

2

2



 2 (1 y ) + 16nd :  (1 y ) ; the above inequality implies 2

2

 2n 

y )n 8d

2

2

2

2

2

(1 and so d 

n

y )n 8d 2

2

 2 17 (1 16

y ) 2

2

ut

2.4 Polynomials with Nonnegative CoeÆ ients 89

m℄ Show that p 2 Pn n Pn and d(p) = n imply that the zeros z ; z ; : : : ; zn of p satisfy jz z


zn j  1. n℄ Show that d(pq ) < maxfd(p); d(q )g an happen. Hint: Let 1

1

2

1

p(x) := (1 x)

2

2(1 x ) + 4(x + 1)

2

2

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

2

1

2

x) :

Show that d(p) = 4, d(q ) = 1; and d(pq ) = 3: ut o℄ Show that if p 2 Pk n Pk has no zeros in ( 1; 1), z 2 C ; jz j > 1; and 1

z )(x z ))m q(x) ;

p(x) = ((x

then d(p) = deg(p) = k + 2m if m is suÆ iently large. This shows that polynomials p with the property d(p) = deg(p) an have arbitrary many pres ribed zeros in C n ( 1; 1): E.5 Lorentz Degree of Trigonometri Polynomials. Given ! 2 (0;  ℄ and a real trigonometri polynomial t 2 Tn ; let d = d! (t) be the minimal nonnegative integer for whi h t is of the form

t() = 

2d X

j =0

aj sinj

!



2

+! sin d j ; 2

aj

2

If there is no su h d; then let d! (t) := degree of t. a℄ Let t 2 Tn n Tn be of the form

 0:

1. We all d = d! (t) the Lorentz

1

t() =

2n X

j =0

bj sinj

!

2



+! sin n j : 2 2

For m  n let the numbers bj;m , j = 0; 1; : : : ; 2m; be de ned by

t() =





0 2n X  bj sinj

j =0



!

2



1

 + !A sin n j 2 2

1 !  +! +! !  + 2 os ! sin sin + sin sin 2 2 2 2 sin ! m X +! !  sin m j : = bj;m sinj 2 2 j

m n

2

2

2

2

2

=0

Show that if d! (t) is nite, then it is the smallest value of m for whi h ea h bj;m is nonnegative or ea h bj;m is nonpositive.

90

2. Some Spe ial Polynomials

Hint: The se ond fa tor in the representation of t is identi ally 1. We introdu e the notation

ut

G := fz = x + iy :   x < ; y 2 Rg and

G! := fz = x + iy : os ! osh y  os x;

  x < ; y 2 Rg :

b℄ Let t 2 T n T . Show that d! (t) = 1 if and only if t has its zeros in G! :

℄ Assume 0 < ! < =2, t 2 T ; and t(z ) = 0 for some 1

0

1

z := x + iy 2 G n (G! [ ( !; !)) : Show that

d! (t) < max

4 sin(!  x)(sin ! osh y  sin x)

os ! sinh y 2

1;

where the maximum is to be taken over both sets of signs. d℄ Suppose =2  !  ; and t 2 T , t(z ) = 0 for some z 2 G n G! . Show that d! (t) = 1: e℄ Show that d! (t t )  d! (t ) + d! (t ) for any two trigonometri polynomials t and t : f℄ Let 0 < ! < =2 and 0 <  < 1. Show that if t 2 Tn has no zeros in 1

1 2

1

1

2

2

E!; := fz = x + iy : y <  (! 2

then

d! (t)  n



2

x ) ; x 2 ( !; !) ; y 2 Rg ;

2

4 

os !

2



2

+ 2 tan ! + 1 :

g℄ Let p be a trigonometri polynomial and 0 < ! < =2. Show that d! (t) < 1 if and only if t = 0 or t has no zeros in ( !; !): (Note that part d℄ shows that this on lusion fails to hold when =2  !  :) h℄ Show that there is an absolute onstant > 0 (independent of n, ! and z ) so that d! (t)  n whenever 2

0



t() = sin



2

z

0

sin



2

z

0

n

with z 2 E!; n f !; ! g, 0 <  < 1; where E!; denotes the boundary of E!; de ned in part f℄. 0

This is page

91

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3

Chebyshev and Des artes Systems

Overview A Chebyshev spa e is a nite-dimensional subspa e of C (A) of dimension n + 1 that has the property that any element that vanishes at n + 1 points vanishes identi ally. Su h spa es, whose prototype is the spa e Pn of real algebrai polynomials of degree at most n; share with the polynomials many basi properties. The rst se tion is an introdu tion to these Chebyshev spa es. A basis for a Chebyshev spa e is alled a Chebyshev system. Two spe ial families of Chebyshev systems, namely, Markov systems and Des artes systems, are examined in the se ond se tion. The third se tion examines the Chebyshev \polynomials" asso iated with Chebyshev spa es. These asso iated Chebyshev polynomials, whi h equios illate like the usual Chebyshev polynomials, are extremal for various problems in the supremum norm. The fourth se tion studies parti ular Des artes systems


x0 ; x1 ; : : : ;

 <  < ::: 0

1

on (0; 1) in detail. These systems, whi h we all Muntz systems, an be very expli itly orthonormalized on [0; 1℄, and this orthogonalization is also examined. The nal se tion onstru ts Chebyshev \polynomials" asso iated with the Chebyshev spa es 

span 1 ;

x

1

a

1

; ::: ;

x

1



an

;

on [ 1; 1℄ and explores their various properties.

ai 2 R n [ 1; 1℄

92

3. Chebyshev and Des artes Systems

3.1 Chebyshev Systems

From an approximation theoreti point of view an essential property that polynomials of degree at most n have is that they an uniquely interpolate at n + 1 points. This is equivalent to the fa t that a polynomial of degree at most n that vanishes at n + 1 points vanishes identi ally. Any (n + 1)dimensional ve tor spa e of ontinuous fun tions with this property is alled a Chebyshev spa e or sometimes a Haar spa e. Many basi approximation properties extend to these spa es. The pre ise de nition is the following. De nition 3.1.1 (Chebyshev System). Let A be a Hausdor spa e. The sequen e (f ; : : : ; fn ) is alled a real (or omplex) Chebyshev system or Haar system of dimension n + 1 on A if f ; : : : ; fn are real- (or omplex-) valued ontinuous fun tions on A, spanff ; : : : ; fn g over R (or C ) is an (n +1)-dimensional subspa e of C (A), and any element of spanff ; : : : ; fn g that has n + 1 distin t zeros in A is identi ally zero. If (f ; : : : ; fn ) is a Chebyshev system on A; then spanff ; : : : ; fn g is

alled a Chebyshev spa e or Haar spa e on A. Chebyshev systems and spa es will be assumed to be real, unless we expli itly spe ify otherwise. If A  R; then the topology on A is always meant to be the usual metri topology. 0

0

0

0

0

0

Impli it in the de nition is that A ontains at least n + 1 points. Being a Chebyshev system is a property of the spa e spanned by the elements of the system, so every basis of a Chebyshev spa e is a Chebyshev system. A point x 2 (a; b) is alled a double zero of an f 2 C [a; b℄ if f (x ) = 0 and f (x )f (x + ) > 0 for all suÆ iently small  > 0 (in other words, if f vanishes without hanging sign at x ). It is easy to see that if (f ; : : : ; fn ) is a Chebyshev system on [a; b℄  R; then every 0 6= f 2 spanff ; : : : ; fn g has at most n zeros even if ea h double zero is ounted twi e; see E.10. Chebyshev spa es are de ned via zero ounting, and many of the theorems in the theory of Chebyshev spa es are proved by zero ounting arguments. So it is important to make the agreement that, unless it is stated expli itly otherwise, we ount the zeros of an element f from a Chebyshev spa e on [a; b℄ so that ea h double zero of f is ounted twi e. The following simple equivalen es hold: 0

0

0

0

0

0

0

Proposition 3.1.2 (Equivalen es). Let f ; : : : ; fn be real- (or omplex-) valued ontinuous fun tions on a Hausdor spa e A ( ontaining at least n + 1 0

points). Then the following are equivalent: a℄ Every 0 6= p 2 spanff ; : : : ; fn g has at most n distin t zeros in A: 0

3.1 Chebyshev Systems 93

b℄ If x ; : : : ; xn are distin t elements of A and y ; : : : ; yn are real (or

omplex) numbers, then there exists a unique p 2 spanff ; : : : ; fn g su h that p(xi ) = yi ; i = 1; 2; : : : ; n : 0

0

0

℄ If x ; : : : ; xn are distin t points of A; then 0

D(x ; : : : 0

f0 (x0 ) ; xn ) := ... f0 (xn )



: : : fn (x ) .. 6= 0 : ... . : : : fn (xn ) 0

Proof. These equivalen es are all elementary fa ts in linear algebra.

ut

On an interval there is a sign regularity to the determinants in ℄. Proposition 3.1.3. Suppose (f ; : : : ; fn ) is a (real) Chebyshev system on [a; b℄  R: Then there exists a Æ := 1 or Æ := 1 su h that 0

f0 (x0 ) Æ ... f0 (xn )



: : : fn (x ) .. > 0 ... . : : : fn (xn ) 0

for any a  x < x <


< xn  b: 0

1

Proof. This follows immediately from part ℄ of the previous proposition and ontinuity onsiderations. That is, if D(x ; : : : ; xn ) < 0 while D(y ; : : : ; yn ) > 0; then for some  2 (0; 1) 0

0

D(x + (1 )y ; : : : ; xn + (1 )yn ) = 0 ; 0

0

ut

whi h is impossible.

The intimate relationship between Chebyshev systems and best approximation in the uniform norm is indi ated by the next result. In order to state it we need to introdu e the notion of an alternation sequen e. De nition 3.1.4 (Alternation Sequen e). Let A  R and let

x < x <


< xn 0

1

be n + 1 points of A: Then (x ; x ; : : : ; xn ) is said to be an alternation sequen e of length n + 1 for a real valued f 2 C (A) if 0

1

jf (xi )j = kf kA ; and

sign(f (xi )) = sign(f (xi )) ; +1

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

1:

94

3. Chebyshev and Des artes Systems

De nition 3.1.5 (Best Approximation). Suppose that U is a ( nitedimensional) subspa e of a normed spa e (V; k
k). If g 2 V and p 2 U satisfy kg pk = inf kg hk ; h2U

then p is said to be a best approximation to g from U . As a result of the nite dimensionality of the subspa e U; at least one best approximation to any g 2 V from U exists. This is straightforward sin e T := fp 2 U : kpk  kgk + 1g is a ompa t subset of U; so any sequen e (pj ) of approximations to g from U satisfying kg pj k  j + inf kg hk 1

h2U

has a onvergent subsequen e with limit in U: This limit is then a best approximation to g from U . Theorem 3.1.6 (Alternation of Best Approximations). Suppose (f ; : : : ; fn ) is a Chebyshev system on [a; b℄  R: Let A be a losed subset of [a; b℄

ontaining at least n + 2 distin t points. Then p 2 Hn := spanff ; : : : ; fng is a best approximation to g 2 C (A) from Hn in the uniform norm on A if and only if there exists an alternation sequen e of length n + 2 for g p on 0

0

A.

Proof. The proof of the only if part of the theorem is mostly an example

of a standard type of perturbation argument that will re ur later. The perturbation argument goes as follows. Suppose p is a best approximation of required type and suppose a alternation sequen e of maximal length for g p is (x < x ; <


< xm ) 0

1

where xi 2 A and where m is stri tly less than n + 1. Suppose, without loss of generality, that g(x ) p(x ) > 0 0

0

(otherwise multiply by 1). Now let

Y := fx 2 A : jg(x)

p(x)j = kg

pkA g :

Note that Y is ompa t. Sin e (x < x < : : : < xm ) is an alternation sequen e of maximal length, we an divide Y into m + 1 disjoint ompa t subsets Y ; Y ; : : : ; Ym with 0

0

1

1

x

0

2Y

0

; x

1

2Y

1

; : : : ; xm 2 Ym

3.1 Chebyshev Systems 95

so that

p(x)) = sign(g(y) p(y)) 6= 0 ; Now hoose a p 2 spanff ; : : : ; fn g satisfying

x 2 Yi ; y 2 Yi

sign(g (x)

+1

:

0

signx2Yi (p (x)) = ( 1)i ; i = 0; 1; : : : ; m : This an be done by hoosing points zi with max Yi

1

< zi < min Yi ;

i = 1; 2; : : : ; m

and then applying E.11. We now laim that, for Æ > 0 suÆ iently small,

kg

(3.1.1)

(p + Æp )kA < kg

pkA ;

whi h ontradi ts the fa t that p is a best approximation, and so there must exist an alternation set of length n + 2 for g p on A: To verify (3:1:1) we pro eed as follows: For ea h i = 0; 1; : : : ; m hoose an open set Oi  [a; b℄ (in the usual metri topology relative to [a; b℄) ontaining Yi so that for every x 2 O i ;

p(x)) = sign(p (x))

sign(g (x)

(3.1.2) and

jg(x)

(3.1.3) Now pi k a Æ

1

p(x)j 

kg pkA : S > 0 su h that for every x 2 B := A n m i Oi and Æ 2 (0; Æ ); jg(x) (p(x) + Æp (x))j < kg pkA ; 1

2

=0

1

whi h an be done sin e B is ompa t and by onstru tion

kg

pkB < kg

pkA :

Note that (3:1:2) and (3.1.3) allow us to pi k a Æ > 0 su h that for x i Oi and Æ 2 (0; Æ );

Sm

2

=0

2

jg(x)

(p(x) + Æp (x))j < jg (x)

2

p(x)j :

This veri es (3.1.1) and nishes the dire t half of the theorem. The proof of the onverse is simple. Suppose there is an alternation sequen e of length n + 2 for g p on A, and suppose there exists a p with

kg

p kA < kg

pkA :  Then p p has at least n +1 zeros on [a; b℄, one between any two onse utive alternation points for g p on A, and hen e it vanishes identi ally. This

ontradi tion nishes the proof. E.5.

ut

In the setting of Theorem 3.1.6 the best approximation is unique; see

96

3. Chebyshev and Des artes Systems

Comments, Exer ises, and Examples. The terminology is not entirely standard in the literature with Chebyshev systems often referred to as Haar systems on intervals of R: There are various proofs of Theorem 3.1.6, of whi h ours is by no means the most elegant (see Cheney [66℄, for example). The point of this proof is that it easily modi es to deal with hara terizations of extremal fun tions for various other extremal problems. Many good books over this standard material. See, for example, Cheney [66℄, Lorentz [86a℄, or Pinkus [89℄; see also Appendix 3. An extensive treatment of Chebyshev systems is available in Karlin and Studden [66℄ or Nurnberger [89℄, where E.3 an be found. E.4 shows that real Chebyshev systems are intrinsi ally one-dimensional. E.1 Examples of Chebyshev systems. a℄ Suppose 0 =  <  <


< n : Show that 0

1

(x0 ; x1 ; : : : ; xn )

is a Chebyshev system on [0; 1). b℄ Suppose  <  <


< n : Show that 0

1

(x0 ; x1 ; : : : ; xn )

is a Chebyshev system on (0; 1):

℄ Suppose  <  <


< n . Show that 0

1

(x0 ; x0 log x ; x1 ; x1 log x ; : : : ; xn ; xn log x)

is a Chebyshev system on (0; 1). d℄ Suppose  <  <


< n . Show that 0

1



1

1



; ::: ; x  x n is a Chebyshev system on ( 1; 1) n f ;  ; : : : ; n g: e℄ Suppose  <  <


< n . Show that x



0

;

1

1

0

0

1

1

(e0 x ; e1 x ; : : : ; en x ) is a Chebyshev system on ( f℄ Show that

1; 1):

(1; os ; sin ; os 2; sin 2; : : : ; os n; sin n) is a Chebyshev system on [0; 2 ). g℄ Show that (1; os ; os 2; : : : ; os n) is a Chebyshev system on [0;  ):

3.1 Chebyshev Systems 97

E.2 More Examples. a℄ If (f ; : : : ; fn ) is a Chebyshev system on A; then it is also a Chebyshev system on any subset B of A ontaining at least n + 1 points. b℄ If (f ; : : : ; fn ) is a Chebyshev system on A and g 2 C (A) is stri tly positive on A; then (gf ; : : : ; gfn ) is also a Chebyshev system on A:

℄ If (f ; : : : ; fn ) is a Chebyshev system on [0; 1℄; then 0

0

0

0



1;

Z x

f (t) dt ; : : : ;

Z x

0

0

0



fn (t) dt

is also a Chebyshev system on [0; 1℄: See E.8 of Se tion 3.2, whi h treats the ee t of dierentiation on a Chebyshev system. E.3 Extended Complete Chebyshev Systems. Let (g ; : : : ; gn ) be a sequen e of fun tions in C n [a; b℄. De ne the Wronskian determinant 0

W (g ; : : : ; gm )(t) 0

:=

g (t) g0 (t)

g (t) g0 (t)

0

0

1

1

.. .

.. .

::: ::: ..

.

gm (t) 0 (t) gm .. .

g m (t) g m (t) : : : gmm (t) (

)

(

0

)

(

)



:

1

We say that (g ; : : : ; gn ) is an extended omplete Chebyshev system (ECT system) on [a; b℄ if 0

W (g ; : : : ; gm)(t) > 0 ; m = 0; 1; : : : ; n ; t 2 [a; b℄ : 0

a℄ Let spanfg ; : : : ; gn g be an (n + 1)-dimensional subspa e of C n [a; b℄: Show that the following statements are equivalent: (i) For every m = 0; 1; : : : ; n, 0 6= f 2 spanfg ; : : : ; gm g has at most m zeros in [a; b℄ ounting multipli ities (x 2 [a; b℄ is a zero of f with multipli ity k if f (x ) = f 0 (x ) =


f k (x ) = 0 and f k (x ) 6= 0). (ii) For ea h i = 0; 1; : : : ; n, there exists a hoi e of Æi := 1 or Æi := 1 su h that (Æ g ; Æ g ; : : : ; Æn gn ) 0

0

0

0

(

0

0

0

1

1)

0

( )

0

1

is an ECT system on [a; b℄: In parti ular, every ECT system on [a; b℄ is a Chebyshev system on [a; b℄. Proof. For details see Karlin and Studden [66℄. ut

98

3. Chebyshev and Des artes Systems

b℄ Chara terization Theorem. The following statements are equivalent: (i) (g ; : : : ; gn ) is an ECT system on [a; b℄: (ii) There exist wi 2 C n i [a; b℄, i = 0; 1; : : : ; n, with wi stri tly positive 0

su h that

g (t) = w (t) ; 0

0

g (t) = w (t) 1

0

.. . gn (t) = w (t) 0

Z t

w (t ) dt ; 1

a Z t

1

w (t ) 1

a

1

1

Z t1

a

w (t )


2

Z tn

2

1

wn (tn ) dtn


dt dt : 2

a

1

Proof. This is proved by indu tion on n. See Karlin and Studden [66℄. u t   0 n

℄ Suppose  <  <


< n : Show that (x ; : : : ; x ) is an ECT system on [a; b℄ provided a > 0: 0

1

E.4 Railway Tra k Theorem. Real Chebyshev systems exist only on very spe ial subsets of Rm . Indeed, real Chebyshev systems intrinsi ally live on one-dimensional subsets. a℄ Suppose A  Rm ontains three distin t ar s that join at a point x : Then, for n  2; there exists no real Chebyshev system (f ; : : : ; fn ) on A: 0

0

Proof. Suppose there exists a real Chebyshev system (f ; : : : ; fn ) on su h a set A: Let V (x; y) := D(x; y; x ; x ; : : : ; xn ) 0

2

3

(D is de ned in Proposition 3.1.2) whi h is never zero for distin t points x; y; x ; x ; : : : ; xn . Choose distin t points x ; x ; : : : ; xn on one of the three distin t ar s so that x is adja ent to x . Pi k the points z 6= x and z 6= x so that z , z , and x are on dierent ar s. Now onsider inter hanging x := x and y := x by moving x from x to z , y from x to z , x from z to x , and y from z to x . Sin e x; y; x ; x ; : : : ; xn remain distin t, V (x; y ) does not vanish in this pro ess. This ontradi ts the fa t that V (x; y ) is

ontinuous and V (x ; x ) = V (x ; x ): ut 2

3

0

0

1

0

1

2

1

1

1

0

2

0

1

1

0

2

0

0

2

1

1

1

1

2

1

3

0

The following more general result of Mairhuber [56℄ also holds: b℄ Mairhuber's Theorem. If (f ; : : : ; fn ) is a real Chebyshev spa e on A; 0

then A is homeomorphi to a subset of the unit ir le.

E.5 Uniqueness of Best Approximations. Prove that a best approximation from a Chebyshev spa e satisfying the onditions of Theorem 3.1.6 is unique.

3.1 Chebyshev Systems 99

Hint: Suppose f has two best approximations p 2 Hn and p 2 Hn . Then, by the alternation hara terization, p p 2 Hn has at least n + 1 zeros on [a; b℄ (we ount ea h internal zero without sign hange twi e). Now E.10 implies that p p = 0: ut 1

1

1

2

2

2

E.6 De la Vallee Poussin Theorem. Suppose Hn is a Chebyshev spa e of dimension (n + 1) on [a; b℄. If p 2 Hn and there exist n + 2 points

a  x < x <


< xn 0

1

+1

b

so that sign(f (xi )

then

p(xi )) = sign(f (xi )

p(xi )) ; i = 0; 1; : : : ; n ;

+1

inf

p2Hn

kf

pk a;b [

℄

i

+1

min

;::: ;n+1

=0

jf (xi )

p(xi )j :

E.7 Haar's Chara terization of Chebyshev Spa es. The following pretty theorem is due to Haar (for a proof, see E.3 of Appendix 3):

2 C (A) where A is a ompa t Hausdor spa e

ontaining at least n + 1 points. Then (f ; : : : ; fn ) is a Chebyshev system on A if and only if every g 2 C (A) has a unique best approximation from spanff ; : : : ; fn g in the uniform norm on A: Theorem. Let f ; : : : ; fn 0

0

0

E.8 Best Approximation to xn . Reprove Theorem 2.1.1 by using the alternation hara terization of best approximations. E.9 Best Rational Approximations. Let f 2 C [a; b℄. Then p=q 2 Rn;m is a best approximation to f from Rn;m in C [a; b℄ if and only if f p=q has an alternation set of length at least 2 + maxfn + deg(q ); m + deg(p)g

[a; b℄: (Here we must assume p=q is written in a redu ed form.) The proof of this is a fairly ompli ated variant of the proof of Theorem 3.1.6 (see, for example, Cheney [66℄). E.10 Zeros of Fun tions in Chebyshev Spa es. As before, we all the point x 2 (a; b) a double zero of f 2 C [a; b℄ if f (x ) = 0 and 0

0

f (x

0

)f (x + ) > 0 0

for all suÆ iently small  > 0 (in other words, if f vanishes without hanging sign at x ). Let (f ; : : : ; fn ) be a Chebyshev system on [a; b℄  R. Show that every 0 6= f 2 spanff ; : : : ; fn g has at most n zeros even if ea h double zero is ounted twi e. Hint: Use Proposition 3.1.2 b℄. ut 0

0

0

100 3. Chebyshev and Des artes Systems

E.11 Fun tions in a Chebyshev Spa e with Pres ribed Sign Changes. Let (f ; : : : ; fn ) be a Chebyshev system on [a; b℄; and let 0

a < z < z <


< zm < b ; 0  m  n: Show that there is a fun tion p 2 spanff ; : : : ; fn g su h that (i) p (x) = 0 if and only if x = zi for some i = 1; 2; : : : ; m; (ii) p (x) hanges sign at ea h zi ; i = 1; 2; : : : ; m. 1

2

0

Hint: If m = n; then use Proposition 3.1.3 and a ontinuity argument to show that

p (x) =

f0 (x) f0 (z1 ) .. .

f (x) : : : fn(x) f (z ) : : : fn (z ) 1

1

1

1

.. .. .. . . . f (zn ) f (zn ) : : : fn (zn ) 0

1

satis es the requirements. If m < n; then use the already proved ase, a limiting argument, and E.10 to show that there are pj 2 spanff ; : : : ; fn g; j = 1; 2, su h that (1) pj (x) hanges sign at x if and only if x = zi , i = 1; 2; : : : ; m, (2) p (x) 6= 0 for every x 2 [a; zm ℄ n fz ; z ; : : : ; zm g, (3) p (x) 6= 0 for every x 2 [z ; b℄ n fz ; z ; : : : ; zm g. Now show that either p := p + p or p := p p satis es the requirements. ut 0

1

1

2

1

1

1

2

2

2

1

2

E.12 The Dimension of a Chebyshev Spa e on a Cir le. Let (f ; : : : ; fn ) be a Chebyshev system on a ir le C . Show that n must be even. Observe that su h Chebyshev systems exist. Hint: Show that for every set of n distin t points x ; x ; : : : ; xn on the

ir le there is a p 2 spanff ; : : : ; fn g su h that p(x) = 0 if and only if x 2 fx ; x ; : : : ; xn g and p(x) hanges sign at ea h xi . ut 0

1

2

0

1

2

3.2 Des artes Systems Chebyshev systems apture some of the essential properties of polynomials. There are two additional types of systems that apture some additional properties. De nition 3.2.1 (Markov System). We say that (f ; : : : ; fn ) is a Markov system on a Hausdor spa e A if ea h fi 2 C (A), and ff ; : : : ; fmg is a Chebyshev system for ea h m = 0; 1; : : : ; n. (We allow n to tend to +1; in whi h ase we all the system an in nite Markov system on A.) 0

0

3.2 Des artes Systems 101

A Markov system is just a Chebyshev system with ea h initial segment also a Chebyshev system. Being a Chebyshev system is a property of the spa e not of the basis. However, the Markov system depends on the basis. For example,  <  <


(x0 ; x1 ; : : : ) ; 0

1

is a Markov system on any A  (0; 1) ontaining in nitely many points, but not every basis of spanfx0 ; x1 ; : : : g is a Markov system on A (see E.1). Proposition 3.2.2. (f ; : : : ; fn ) is a Markov system on an interval [a; b℄ if and only if for ea h i = 0; 1; : : : ; n, there exists a hoi e of Æi := 1 or Æi := 1 su h that with gi := Æi fi ; 0



D xg

0 0

g x

: : : gm : : : xm

1 1



:=

g0 (x0 ) .. . g0 (xm )



::: ...

gm(x ) .. > 0 . : : : gm(xm ) 0

for every a  x < x <


< xm  b and m = 0; 1; : : : ; n: 0

1

Proof. This is an easy onsequen e of Proposition 3.1.3 by indu tion on n. ut A stronger property that a system on an interval an have is the following: De nition 3.2.3 (Des artes System). The system (f ; : : : ; fn ) is said to be a Des artes system (or order omplete Chebyshev system) on an interval I if ea h fi 2 C (I ) and 0



D fxi0 fxi1 :: :: :: fxim m 0



1

>0

for any 0  i < i <


< im  n and x < x <


< xm from I . (On e again we allow n to tend to 1.) 0

1

0

1

This again is a property of the basis. It implies that any nitedimensional subspa e generated by some basis elements is a Chebyshev spa e on I . The anoni al example of a Des artes system on [a; b℄, a > 0; is (x0 ; x1 ; : : : ) ;

 <  <


0

1

(see E.2). A Des artes system on I is obviously a Des artes system on any subinterval of I .

102 3. Chebyshev and Des artes Systems

The following version of Des artes' rule of signs holds for Des artes systems. Theorem 3.2.4 (Des artes' Rule of Signs). If (f ; : : : ; fn ) is a Des artes system on [a; b℄; then the number of distin t zeros of any 0

0 6= f =

n X i=0

ai 2 R

ai fi ;

is not greater than the number of sign hanges in (a ; : : : ; an ): A sign hange o

urs between ai and ai k exa tly when ai ai ai = ai =


= ai k = 0: 0

+

+1

+2

+

+

k

< 0 and

1

Proof. Suppose (a ; : : : ; an ) has p sign hanges. Then we an partition 0

fa ; : : : ; an g into exa tly p + 1 blo ks so that ea h blo k is of the form 0

anm ; anm ; : : : ; anm+1 ; +1

m = 0; 1; : : : ; p

+2

(n := 1; np := n), where all of the oeÆ ients in ea h of the blo ks are of the same sign and not all the oeÆ ients in a blo k vanish. Now let 0

+1

gm :=

nX m+1 i=nm +1

jai jfi ;

m = 0; 1; : : : ; p :

Then, for a  x < x <


< xp  b, 0



D xg

0 0

g x

1 1

=

1

: : : gp : : : xp n1 X

i0 =n0 +1








nX p+1 ip =np +1

jai j


jaip jD 0



fi0 fi1 : : : fip x x : : : xp 0

1



>0

sin e ea h of the determinants in the sum is positive. Thus fg ; : : : :gp g is a (p + 1)-dimensional Chebyshev system on [a; b℄; and hen e 0

f :=

p X i=0

Æi gi ;

Æ i = 1

has at most p zeros. This nishes the proof.

ut

A re ned version of Des artes' rule of signs for ordinary polynomials is presented in the exer ises. The following omparison theorem due to Pinkus and, independently, Smith [78℄ will be of use later.

3.2 Des artes Systems 103

Theorem 3.2.5. Suppose (f ; : : : ; fn ) is a Des artes system on [a; b℄: Sup0

pose

p = f +

k X i=1

ai fi ;

and

q = f +

k X i=1

bi f i ;

where 0   <  <


< k  n; 0  < <


< k  n, 1

2

1

0  i  i <  ;

and

 < i  i  n ;

2

i = 1; 2; : : : ; m ; i = m + 1; m + 2; : : : ; k

with stri t inequality for at least one index i = 1; 2; : : : ; k. If p(xi ) = q(xi ) = 0 ;

i = 1; 2; : : : ; k ;

where xi 2 [a; b℄ are distin t, then

jp(x)j  jq(x)j for all x 2 [a; b℄ with stri t inequality for x 6= xi : The proof is left as a guided exer ise (see E.4) with some interesting

onsequen es presented in E.5. Comments, Exer ises, and Examples. Theorem 3.2.4 hara terizes Des artes systems; see Karlin and Studden [66, p. 25℄. Some aution must be exer ised sin e, as in the previous se tion, definitions are not entirely standard. We will explore two parti ular Des artes systems in greater detail later; see E.2 and E.3. For further material, the reader is referred to Karlin and Studden [66℄, Karlin [68℄, and Nurnberger [89℄. E.1 Distin tions. a℄ Given  <  <


and A  (0; 1) with in nitely many points, show that (x0 ; x1 ; : : : ) is a Markov system on A; but there is a basis for spanfx0 ; x1 ; : : : g; whi h is not a Markov system on A. b℄ Find a Markov system that is not a Des artes system. 0

1

E.2 Examples of Des artes Systems. a℄ Suppose  <  <


: Show that the Muntz system 0

1

(x0 ; x1 ; : : : ) is a Des artes system on (0; 1):

104 3. Chebyshev and Des artes Systems

Hint: For every 0  i < i <


< im; the determinant 0

D

  x i0

x

1

xi1 x

0

 xim

::: :::

1

xm

=

i x 0 0 . ..  i0

xm



: : : xim ..

0

.

:::

.. .

im x

m

is nonzero for any 0 < x < x <


< xm < 1 by Proposition 3.1.2 and E.1 a℄ of Se tion 3.1. It only remains to prove that it is positive whenever 0 < x < x <


< xm < 1. Observe that the exponents ij an be varied

ontinuously (for xed xi ) without hanging the sign of the determinant provided no two ever be ome equal. Now perturb (i0 ; i1 ; : : : ; im ) into (0; 1; : : : ; m) and observe that the determinant be omes a Vandermonde determinant, as in part b℄, whi h in this ase is positive. ut b℄ Vandermonde Determinant. Show that 0

0

1

1

1 1 . . .

x x

0 1

.. .

: : : xm : : : xm 0

..

m 1

.

1 xm : : :

.. .

xm

Y

= 0

i<jm

(xj

xi ) :

Hint: The determinant is a polynomial in x ; x ; : : : ; xm of degree m in ea h variable that vanishes whenever xi = xj : ut

℄ Suppose  <  <


. Show that the exponential system 0

0

1

1

(e0 t ; e1 t ; : : : ) is a Des artes system on ( 1; 1): Hint: Use part a℄. d℄ Suppose 0 <  <  <


. Show that 0

ut

1

(sinh  t; sinh  t; : : : ) 0

1

is a Des artes system on (0; 1): Outline. Let 0  i < i <


< im be xed integers. First we show that 0

1

(sinh i0 t; sinh i1 t; : : : ; sinh im t) is a Chebyshev system on (0; 1): Indeed, let 0 6= f Then

2 spanfsinh i t; sinh i t; : : : ; sinh im tg :

0 6= f

0

1

2 spanfei t ; ei t ; : : : ; eim t g 0

1

and by E.1 e℄ of Se tion 3.1, f has at most 2m zeros in ( is odd, it has at most m zeros in (0; 1):

1; 1). Sin e f

3.2 Des artes Systems 105

Sin e for every 0  i < i <


< im , (sinh i0 t; : : : ; sinh im t) is a Chebyshev system on (0; 1); the determinant 0

1



D sinh i0 t sinh i1 t : : : x

x

0

=

:::

1



sinh im t

xm

sinh i0 x sinh i1 x : : : sinh im x sinh i0 x sinh i1 x : : : sinh im x .. .. .. .. . . . . sinh i0 xm sinh i1 xm : : : sinh im xm 0

0

0

1

1

1

is nonzero for any 0 < x < x <


< xm < 1 Proposition 3.1.2. So it only remains to prove that it is positive whenever 0 < x < x <


< xm < 1. Now let 0

1

0



D() := D sinh i0 t sinh i1 t : : : x

x

0

sinh im t

:::

1

1



xm

sinh i1 x : : : sinh im x sinh i0 x sinh i1 x : : : sinh im x sinh i0 x = .. .. .. .. . . . . sinh i0 xm sinh i1 xm : : : sinh im xm

and



D () := D =

1

2

2

0

0

1

1

ei0 t x

ei1 t : : : x :::

eim t xm

1

1

1

2

2

2

0

2 1

1

0

1

ei0 x0 ei0 x1 .. .

ei0 xm

1

1 2 1 2

1 2

ei1 x0 ei1 x1 .. .

::: :::

ei1 xm

..

.

:::



2 .. . 1 im xm e 1

2 1

eim x0 eim x1

2

where 0 < x < x <


< xm < 1 are xed. Sin e 0

1

(sinh i0 t; sinh i1 t; : : : ; sinh im t) and

(ei0 t ; ei1 t ; : : : ; eim t )

are Chebyshev systems on (0; 1), D() and D () are ontinuous nonvanishing fun tions of  on (0; 1). Now, observe that lim jD()j = lim jD ()j = 1

!1

By part ℄,

!1

and

(ei0 t ; ei1 t ; : : : ; eim t )

D() = 1: !1 D () lim

106 3. Chebyshev and Des artes Systems

is a Des artes system on ( 1; 1); hen e D () > 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 ()

3.2 Des artes Systems 109

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

110 3. Chebyshev and Des artes Systems

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

3.2 Des artes Systems 111

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

112 3. Chebyshev and Des artes Systems

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

3.2 Des artes Systems 113

points

a < x < x <


< xn 1

2

and  = 1 su h that

F := 

n X i=1


+2

ai f i

satis es ( 1)i (F (xi ) +1

F (xi )) > 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

114 3. Chebyshev and Des artes Systems

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

116 3. Chebyshev and Des artes Systems

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

3.3 Chebyshev Polynomials in Chebyshev Spa es 117

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

118 3. Chebyshev and Des artes Systems

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

3.3 Chebyshev Polynomials in Chebyshev Spa es 119

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

120 3. Chebyshev and Des artes Systems

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;

3.3 Chebyshev Polynomials in Chebyshev Spa es 121

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

122 3. Chebyshev and Des artes Systems

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

3.3 Chebyshev Polynomials in Chebyshev Spa es 123

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℄.

124 3. Chebyshev and Des artes Systems

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



<

then

 ; n 2




Rn (x) := Tn(x) Tn x  2 Hn has at least n + 1 zeros on [0; = ℄; whi h is a ontradi tion. b℄ Denote the zeros of Tn in (0; 1) by x < x <


< xn : Show that 1

log xk

+1

log xk

 log xk

log xk

ut

2

1

;

k = 2; 3; : : : ; n 1 :

Hint: Use a zero ounting argument, as in the hint to part a℄.

ut

3.4 Muntz-Legendre Polynomials We examine in some detail the system (x0 ; x1 ; : : : ) on [0; 1℄ whi h we all a Muntz system. In parti ular, we expli itly onstru t orthogonal \polynomials" for this system. This allows us to derive various extremal properties of these systems and leads to a very simple proof of the

lassi al Muntz-Szasz theorem in Se tion 4.2. We adopt the following de nition for x : (3:4:1)

x = e

log

x;

x 2 (0; 1) ;  2 C

with value at 0 de ned to be the limit of x as x ! 0 from (0; 1) whenever the limit exists. Given a sequen e  := (i )1 of omplex numbers, an element of i spanfx0 ; x1 ; : : : ; xn g is alled a Muntz polynomial or a -polynomial. We denote the set of all su h polynomials by Mn (); that is, =0

(3:4:2)

Mn () := spanfx0 ; x1 ; : : : ; xn g ;

126 3. Chebyshev and Des artes Systems

where the linear span is over R or C a

ording to ontext. Let (3:4:3)

M () :=

1 [

n=0

Mn () = spanfx0 ; x1 ; : : : g :

For the L [0; 1℄ theory of Muntz systems, we onsider 2

 = (i )1 i ; Re(i ) > 1=2 ; and i 6= j ; i 6= j ;

(3:4:4)

=0

where Re() denotes the real part of : This ensures that the -polynomials

p(x) =

n X k=0

ak xk ;

ak 2 C

are in L [0; 1℄. We an then de ne the orthogonal -polynomials with respe t to Lebesgue measure. We all these Muntz-Legendre polynomials. Although we often assume (3:4:4), the following de nition requires neither the distin tness of the numbers i nor the assumption Re(i ) > 1=2: 2

De nition 3.4.1 (Muntz-Legendre Polynomials). Let  := (i )1 i be a sequen e of omplex numbers. We de ne the nth Muntz-Legendre polynomial on (0; 1) by =0

(3:4:5)

Ln (x) := Ln f ; : : : ; n g(x) Z nY 1 t + k + 1 xt dt := ; n = 0; 1; : : : ; 2i t k t n k 0

1

=0

where the positively oriented, simple losed ontour surrounds the zeros of the denominator in the integrand, and k denotes the onjugate of k : The orthogonality of the above fun tions with respe t to the Lebesgue measure is proved in Theorem 3.4.3. However, rst we give a simple expli it representation of Ln in the ase that the numbers i are distin t. This is dedu ed immediately from evaluating the above integral by the residue theorem. Proposition 3.4.2. Let  := (i )1 i be a sequen e of distin t omplex num=0

bers. Then

Lnf ; : : : ; n g(x) =

(3:4:6)

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

where Ln f ; : : : ; n g(x) is de ned by (3.4.5). 0

x 2 (0; 1)

3.4 Muntz-Legendre Polynomials 127

So Ln f ; : : : ; n g is indeed a  polynomial provided the numbers i are distin t. Its value at x = 0 is de ned if for all i either Re(i ) > 0 or i = 0: From either De nition 3.4.1 or the above proposition it is obvious that the order of  ; : : : ; n in Ln f ; : : : ; n g does not make any dieren e, as long as n is kept last. For example, 0

0

1

0

L f ;  ;  g = L f ;  ; 

g while both, in general, are dierent from L f ;  ;  g. For a xed sequen e 2

0

1

2

2

1

2

0

0

2

2

1

, we let Ln(), or simply Ln , denote the nth Muntz-Legendre polynomial Ln f ; : : : ; n g, whenever there is no ambiguity. 0

An analog of Proposition 3.4.2 an be established even if the numbers

i are not distin t, however, in the nondistin t ase, Ln () does not belong to the spa e Mn (); see E.7 b℄. In the very spe ial ase that all the indi es

are the same we re over the Laguerre polynomials; see E.1. The orthogonality of fLn g1 is the ontent of the main theorem of n this se tion. =0

Theorem 3.4.3 (Orthogonality). Let  = (i )1 i be a sequen e of omplex numbers with Re(i ) > 1=2 for i = 0; 1; : : : . The fun tions Ln de ned by =0

(3.4.5) satisfy

Z

(3:4:7)

1

0

Ln (x)Lm (x) dx =

Æn;m 1 + n + n

for all nonnegative integers n and m: (Here Æn;m is the Krone ker symbol.) Proof. We may assume that the numbers i are distin t. Note that Lnf ;  ; : : : ; n g(x) 0

1

is uniformly ontinuous in  ; : : : ; n for x in losed subintervals of (0; 1℄; and the nondistin t i ase an be handled by a limiting argument. We may further suppose that m  n: Sin e Re(i ) > 1=2; we an pi k a simple

losed ontour su h that lies ompletely to the right of the verti al line Re(t) = 1=2 and surrounds all zeros of the denominator of the integrand in (3.4.5). When t 2 ; we have Re(t + m ) > 1; and 0

Z

1

xt

+

m dx =

0

1

t + m + 1

for every m = 0; 1; : : : : Hen e Fubini's theorem yields Z

1

0

Ln(x)xm dx =

1 2i

Z nY1

k=0

t + k + 1 dt : t k (t n )(t + m + 1)

128 3. Chebyshev and Des artes Systems

Noti e that for m < n; the new fa tor, t + m + 1; in the denominator

an be an elled, and for m = n the new pole (n + 1) is outside sin e Re( n 1) < 1=2: Changing the ontour from to jtj = R with R > maxf j + 1; : : : ; jn j + 1g; gives 0

Z

0

1

Ln

1 (x)xm dx =

Z

nY1

2i jtj R k =

n

=0

Æm;n

t + k + 1 dt t k (t n )(t + m + 1) nY1

n

1

n + k : n 1 k

k=0

On letting R ! 1; we see that the integral on the right-hand side is a tually 0; whi h gives Z

1

0

Ln(x)xm dx =

nY Æn;m n k : n + n + 1 k n + k + 1 1

=0

Therefore Proposition 3.4.2 and m  n yield Z

0

1

Ln (x)Lm (x) dx =

Z

0

1

m X

Ln (x)

= m;m

Z

1

0

k=0

k;m xk dx

Ln (x)xm dx =

and the proof is nished.

Æm;n ; n +  n + 1

ut

An alternative proof of orthogonality is suggested in E.3. If we let

Ln := (1 + n + n )

(3:4:8)

=

1 2

Ln ;

then we get an orthonormal system, that is, Z

1

0

Ln(x)Lm (x) dx = Æm;n ;

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

We all these Ln the orthonormal Muntz-Legendre polynomials. There is also a Rodrigues-type formula for the Muntz-Legendre polynomials (see E.2). Let

pn (x) =

n X k=0

Qn

xk ;j 6 k (k

j =0

=

j )

:

Then

Ln (x) = (D0 D1


Dn 1 )(pn )(x) ; where the dierential operators D are de ned by d D (f )(x) := x  (x  f (x)) : dx 1+

The following is a dierential re urren e formula for (Ln )1 n : =0

3.4 Muntz-Legendre Polynomials 129

Theorem 3.4.4. For a xed sequen e  := (i )1 of omplex numbers, let i =0

Ln be de ned by (3.4.5). The identity (3:4:9) xL0n (x) xL0n (x) = n Ln (x) + (1 + n )Ln (x) holds for every x 2 (0; 1) and n = 1; 2; : : : : 1

1

Proof. From (3:4:5) we get (x

n L (x))0 = 1 n

Z

Qn

k=0 (t + k + 1) (t +  n 1 k=0 (t k ) 2

Qn

2i

On multiplying both sides by xn n +

xn

+

n

1 +1

1 +1

1

1

+ 1)xt n

1

dt :

; we obtain

(x n Ln (x))0 Z Qn 1 k (t + k + 1) (t +  = Qn n 2i k (t k ) 2

=0

1

1

+ 1)xt n 1 dt ; +

=0

and again by the de nition of Ln ; 1

xn +n

1 +1

(x

n L (x))0 = (xn n

1 +1

Ln (x))0 : 1

We nish the proof by simplifying by the produ t rule and dividing both ut sides by xn 1 :

Corollary 3.4.5. For a xed sequen e  := (i )1 of omplex numbers, let i Ln and Ln be de ned by (3.4.5) and (3.4.8), respe tively. Then for every x 2 (0; 1) and for every n = 0; 1; : : : , =0

nX1

a℄ xL0n (x) = n Ln (x) +

k=0

 b℄ xL0 n (x) = n Ln (x) +

q

(k + k + 1)Lk (x) ;

n +  n + 1

and

℄ xL00n (x) = (n

1)L0n (x) +

nX1 k=0

nX1 q k=0

k + k + 1 Lk (x) ;

(k + k + 1)L0k (x) :

Proof. The rst identity follows from Theorem 3.4.4 on expressing xL0n (x) xL0 (x) 0

as a teles oping sum. From this and the relation

Lk = (k + k + 1)

=

1 2

Lk

we get part b℄. Dierentiating the identity of part a℄ gives part ℄.

ut

The values and derivative values of the Muntz-Legendre polynomials at 1 an now all be al ulated.

130 3. Chebyshev and Des artes Systems

Corollary 3.4.6. For a xed sequen e  := (i )1 of omplex numbers, let i =0

Ln be de ned by (3.4.5). Then a℄ Ln (1) = 1 ; b℄ L0n (1) = n +

and

nX1 k=0

(k + k + 1) ; n

X 1)L0n (1) + (k + k + 1)L0k (1) :

℄ L00n (1) = (n

1

k=0

Proof. It suÆ es to show that Ln (1) = 1; the rest follows from Corollary 3.4.5. Noti e that

Ln (1) =

1 2i

Z nY1

k=0

t + k + 1 dt : t k t n

Now, sin e surrounds all zeros of the denominator, and the degree of the denominator is one higher than that of the numerator, we an evaluate the integral on ir les of radius R ! 1 to get the result. ut Comments, Exer ises, and Examples. P Muntz polynomials are just exponential polynomials ak e k t under t the hange of variables x = e and have re eived onsiderable s rutiny (S hwartz [59℄ is a monograph on this topi ). The orthogonalizations of Muntz systems exist in the Russian literature (see, for example, Badalyan [55℄ and Taslakyan [84℄; it has been further explored in M Carthy, Sayre, and Shawyer [93℄). Borwein, Erdelyi, and Zhang [94b℄ ontains most of the

ontent of this se tion. Various properties of the Muntz-Legendre polynomials are examined in the exer ises. Note that if  <  <  <


; then the Muntz system (x0 ; x1 ; : : : ) is a Des artes system on [a; b℄; a > 0; and so we an apply Theorem 3.3.4 to the asso iated Chebyshev polynomials on [a; b℄ to dedu e how the zeros shift when the exponents are varied lexi ographi ally. Similar results are given for the Muntz-Legendre polynomials in E.7. 0

1

2

E.1 Laguerre Polynomials. a℄ Let Ln f ; : : : ; n g(x) be de ned by (3.4.5). If  =


= n = ; then 0

0

Ln f ;  ; : : : ; n g(x) = x Ln ( (1 +  + ) log x) ; 0

1

where Ln is the nth Laguerre polynomial orthonormal with respe t to the weight e x on [0; 1) with Ln (0) = 1 (see E.7 of Se tion 2.3, where Ln is denoted by Ln as is standard).

3.4 Muntz-Legendre Polynomials 131

Proof. Sin e k = , (3.4.5) yields 1 Ln f ;  ; : : : ; n g(x) = 2i 0

1

xt (t +  + 1)n dt ; (t )n +1

an be taken to be any ir le entered at : By the

where the ontour residue theorem,

Ln f ;  ; : : : ; n g(x) = 0

Z

1

 





1 dn t (x (t +  + 1)n n! dtn t 

n n  1 X x (log x)k [n(n = n! k k =0

= x

n X

k=0

=

1)


(k + 1)℄( +  + 1)k

 

1 n (1 +  + )k logk x : k! k

ut

See also part b℄. b℄ Let

Ln (x) :=

n X k=0

 

1 n ( x)k : k! k

Then (Ln )1 is an orthonormal sequen e of polynomials on [0; 1) with n =0

respe t to the inner produ t

hf; gi =

Z

1

f (x)g(x)e

x dx :

0

Dedu e the orthonormality from a℄ and Theorem 3.4.3 by substituting

y = (1 +  + ) log x : E.2 Rodrigues-Type Formula. Let  = (i )1 i be a sequen e of distin t

omplex numbers. Let Ln be de ned by (3.4.5). a℄ Let =0

pn (x) := Show that where

pn(x) =

n X

k=0

xk ;j 6 k (k

Qn

1 2i

j =0

=

Z Qn

j )

:

xt dt ; (t j )

j =0

is any ontour surrounding  ;  ; : : : ; n : Use this to show that 0

pnk (1) = 0 ; ( )

1

k = 0; 1; : : : ; n

1:

132 3. Chebyshev and Des artes Systems

b℄ Show that

Ln (x) = (D0 D1


Dn 1 )(pn )(x) ;

where

D (f )(x) := x  (x

℄ If 0 =  <  <


; then 0

 f (x))0 :

1+

1

pn (0) = ( 1)n

n Y

j =1 n and ( 1) pn is stri tly de reasing on [0; 1℄:

j

1

E.3 Another Proof of Orthogonality. Dedu e the orthogonality of the sequen e (Ln )1 on [0; 1℄ from Theorem 3.4.4 by using integration by n parts and indu tion. =0

E.4 Integral Re ursion. For a given sequen e  := (i )1 of omplex i numbers satisfying (3.4.4), let Ln be de ned by (3.4.5). Show that =0

Ln (x) = Ln (x) 1

(n + n

1

+ 1)xn

Z

1

x

t

n

1

Ln (t) dt ; x 2 (0; 1℄ : 1

ut

Hint: Use Theorem 3.4.4.

is a sequen e of E.5 On the Maximum of Ln on [0; 1℄. If  = (i )1 i nonnegative numbers satisfying =0

(3:4:10)

n 

nX1 k=0

(1 + 2k ) ;

and Ln is de ned by (3.4.5), then

jLn (x)j < Ln (1) = 1 ;

n = 1; 2; : : :

x 2 [0; 1) ; n = 2; 3; : : : :

Hint: Use Theorem 3.4.4. ut p k If k =  ; then (3:4:10) holds if and only if   2 + 3: E.6 The Reprodu ing Kernel. Let  = (i )1 be as in (3.4.4), and let i Ln and Ln be de ned by (3.4.5) and (3.4.8), respe tively. Then for every p 2 Mn (); we have =0

p(x) =

Z

0

1

Kn (x; t)p(t) dt ;

where

Kn (x; t) :=

n X k=0

Lk (x)Lk (t)

is the nth reprodu ing kernel (see also E.5 of Se tion 2.2).

3.4 Muntz-Legendre Polynomials 133

E.7 On the Zeros of Muntz-Legendre Polynomials. Assume that ( ;  ; : : : ; n )  0

1

1

2




;1 :

a℄ For a fun tion f 2 C (0; 1) let S (f ) and Z (f ) denote the number of sign hanges and the number of zeros, respe tively, of f in (0; 1) (we ount ea h zero without sign hange twi e). Let  and 2 C (0; 1). Show that if

n  S ( + )  Z ( + )  n + 1 for every real  and with  + > 0; then the zeros of  and stri tly interla e. Proof: This result is due to Pinkus and Ziegler [79℄. ut b℄ Assume that 2

2

f ;  ; : : : ; n g = fe ; e ; : : : ; em g ; 0

1

0

1

where the numbers e ; e ; : : : ; em are distin t, and let mj ; j = 0; 1; : : : ; m; be the number of indi es i = 0; 1; : : : ; n for whi h i = ej : Show that Ln f ; : : : ; n g is in the spa e 0

1

0

1g ;

Hn := spanfxj (log x)i : j = 0; 1; : : : ; m; i = 0; 1; : : : ; mj

whi h is a Chebyshev spa e on (0; 1): Hint: Use the de nition and the residue theorem. ut n

℄ Show that fLk f ; : : : ; k ggk is a basis for the Chebyshev spa e Hn de ned in part b℄. Hint: Use Theorem 3.4.3 (orthogonality). ut d℄ Show that Ln := Ln f ; : : : ; n g has exa tly n distin t zeros in (0; 1) and Ln hanges sign at ea h of these zeros. Hint: Assume to the ontrary that the number of sign hanges of Ln is less than n. Use part ℄ to nd a fun tion p 2 spanfLk gkn that hanges sign R exa tly at those points in (0; 1) where Ln hanges sign. Then Ln p 6= 0; whi h ontradi ts Theorem 3.4.3. ut  e℄ Suppose n < n : Show that the zeros of 0

=0

0

1

=0

1

0

 := Ln f ;  ; : : : ; n ; n g 0

and

1

1

:= Lnf ;  ; : : : ; n ; n g 0

in (0; 1) stri tly interla e.

1

1

134 3. Chebyshev and Des artes Systems

Hint: Note that Theorem 3.4.3 (orthogonality) implies Z

1

( + )p = 0

0

for every p 2 Hn , where Hn is de ned by part b℄ with respe t to the sequen e ( ;  ; : : : ; n ): Use the hint given to part d℄ to show that  + has at least n sign hanges in (0; 1) whenever  and are real with  + > 0: Use part b℄ to obtain that  + annot have more than n + 1 zeros in (0; 1) whenever  and are real with  + > 0: Finish the proof by part a℄. ut f℄ Let  ; : : : ; k ; k ; : : : ; n be xed distin t numbers. Suppose 1

0

2

1

1

1

2

2

0

1

2

+1

(k;i )1 i

=1

(

1=2; 1)

is a sequen e with lim k;i = 1: Show that the largest zero of i!1

Ln;k;i := Ln f ; : : : ; k ; k;i ; k ; : : : ; n g 0

1

+1

in (0; 1) tends to 1: Outline. Assume, without loss, that k;i is greater than ea h of the numbers j ; j = 0; 1; : : : ; n; j 6= k. Let

gi (x) := k;i (Ln;k;i (x) where

i k;i

k;n x ); ( )

Qn 1 (k;i + j + 1) i

k;n = Qjn=0 j =0;j 6=k (k;i j ) ( )

is the oeÆ ients of xk;i in Ln;k;i : Use (3.4.6) to show that the fun tions gi onverge uniformly on [Æ; 1℄, Æ 2 (0; 1), to a fun tion 0 6= g 2 Hn

1

:= spanfx0 ; : : : ; xk 1 ; xk+1 ; : : : ; xn g :

i Use Ln;k;i (1) = 1 (see Corollary 3.4.6) and the expli it formula for k;n to show that g (1)  0 and that the fun tions ( )

Ln;k;i (x) = (Ln;n;i (x)

i k;i i k;i

k;n x ) + k;n x ( )

( )

onverge to g (x); as i ! 1; for every x 2 (0; 1): Now assume that the statement of part f℄ is false. Then there is an of (k;i )1 su h that the Muntz 2 (0; 1) and a subsequen e (k;ij )1 j i Legendre polynomials Ln;k;ij have no zeros in [1 ; 1℄: Dedu e from this and =1

=1

3.4 Muntz-Legendre Polynomials 135

L0n;k;ij (1) > 0, n;ij > 0 (see Corollary 3.4.6 a℄), that gij is nonde reasing on [1 ; 1℄ whenever n;ij > 0: Therefore g is nonde reasing on [1 ; 1℄; whi h, together with 0 6= g 2 Hn and g(1)  0; implies that g(1 ) < 0: Hen e Ln;k;i (1 ) < 0 if i is large enough. Sin e Ln;k;i (1) = 1 (see Corollary 3.4.6), ea h Ln;k;i has a zero in (1 ; 1); provided i is large enough, whi h ontradi ts our 1

ut

assumption. g℄ Let  and be as in part e℄. Let

x < x <


< xn and x < x <


< xn 1

2

2

1

be the zeros of  and ; respe tively, in (0; 1): Show that n < n implies that xj < xj ; j = 1; 2; : : : ; n :

ut

Hint: Combine parts e℄ and f℄. h℄ Let k 6= n : Show that the zeros of  := Lnf ; : : : ; k ; k ; k 0

and

1

+1

:= Ln f ; : : : ; k ; n ; k 0

1

+1

: : : ; n ; n g 1

: : : ; n ; k g 1

in (0; 1) stri tly interla e. Hint: Use part a℄ and arguments similar to those given in the hints to part e℄. Note that

(1) = (1) = 1

and

0 (1) 0 (1) = n

k 6= 0

(see Corollary 3.4.6 a℄) imply that  + is not identi ally 0 whenever  and are real with  + > 0: ut i℄ Let  and be as in part h℄. Let 2

2

x < x <


< xn and x < x <


< xn 1

2

1

2

be the zeros of  and ; respe tively, in (0; 1). Show that k < n implies that xj < xj ; j = 1; 2; : : : ; n :

Hint: By part h℄ it is suÆ ient to prove that xn



xn : Let Hn be the

Chebyshev spa e de ned in part b℄. Corollary 3.4.6 implies that

(1) = (1) = 1

and

0 (1)

0 (1) = n

k > 0 :

Dedu e from this and part i℄ that xn < xn would imply that 0 6=  2 Hn has at least n + 1 distin t zeros in (0; 1℄; whi h is a ontradi tion. ut

136 3. Chebyshev and Des artes Systems

j℄ Lexi ographi Property. Suppose max i

0

in



0

min j

jn

and i < j for some indi es i and j: Let

x < x <


< xn and x < x <


< xn 1

2

1

2

be the zeros of

Ln f ;  ; : : : ; n g and Ln f ;  ; : : : ; n g ; 0

1

0

1

respe tively, in (0; 1): Show that

xj < xj ;

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

ut

Hint: Repeated appli ations of parts g℄ and i℄ give the desired result. k℄ Let  < n : Let 0

x < x <


< xn and x < x <


< xn 1

2

1

2

be the zeros of

Ln f ;  ; : : : ; n g and Ln fn ; n ; : : : ;  g ; 0

1

1

respe tively, in (0; 1). Show that (xj )nj

=1

xj < xj ;

and (xj )nj

=1

0

stri tly interla e and

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

ut

Hint: Use parts h℄ and i℄ and the omment given after Proposition 3.4.2. l℄ Show that the zeros of

 := Ln

1

f ; : : : ; n g 0

1

and := Ln f ; : : : ; n g 0

in (0; 1) stri tly interla e. Hint: Use part a℄ and arguments similar to those given in the hints for part e℄. ut E.8 A Global Estimate for the Zeros. Let (i )ni  ( 1=2; 1). Assume that x < x <


< xn are the zeros of Ln f ; : : : ; n g in (0; 1): Then =1

1

exp



2

4n + 2 1 + 2

0



< x < x <


< xn < exp 1

2



j

2 1

(1 + 2 )(4n + 2)



;

3.4 Muntz-Legendre Polynomials 137

where  := minf ; : : : ; n g,  := maxf ; : : : ; n g and j > 3=4 is the smallest positive zero of the Bessel fun tion 0

0

J (z ) := 0

1 X k=0

1

( z )k : (k ! 2k ) 2

2

Proof. Let Ln be the nth Laguerre polynomial with respe t to the weight e x on [0; 1); and let the zeros of Ln be z < z <


< zn : Then by 1

[Szeg}o [75℄, pp. 127{131℄)

2

j < z < z <


< zn < 4n + 2 ; 4n + 2 2 1

1

2

where the upper estimate is asymptoti ally sharp, and the lower estimate is sharp up to a multipli ative onstant (not ex eeding 4 =(9 )). Now use E.1 and E.7 j℄. ut 4

2

E.9 The Order of the Zero at 1 of Certain Polynomials. This exer ise, due in part to Kos, gives pre ise estimates on the maximum order of the zero at 1 of a polynomial whose oeÆ ients are bounded in modulus by the leading

oeÆ ient. a℄ Suppose a ; a ; : : : ; an are omplex numbers with modulus at most 1; and suppose an = 1: Then the multipli ity of the zero of 0

1

p

1

p(x) = a + a x + a x +


+ an xn 0

1

2

2

at 1 is at most 5 n:

Proof. If p has a zero at 1 of multipli ity m, then for every polynomial f of degree less than m, we have (3.4.11)

a f (0) + a f (1) +


+ an f (n) = 0 : 0

1

p

We onstru t a polynomial f of degree at most 5 n; for whi h

f (n) > jf (0)j + jf (1)j +


+ jf (n 1)j : Equality (3.4.11) annot hold with this f , so the multipli ity of the zero of p at 1 is at most the degree of f: Let T be the  th Chebyshev polynomial de ned by (2.1.1). Let k 2 N ; and let g := T + T +


+ Tk 2 Pk : 0

1

Note that g (1) = k + 1: Also, for 0 < y  ;

138 3. Chebyshev and Des artes Systems

g( os y) = 1 + os y + os 2y +


+ os ky = Hen e, for 1  x < 1;

1

1

2

2

1

2

p p jg(x)j  1 2 x :

Let f (x) := g ( nx

sin(k + )y + sin y : 2 sin y

1): Then f (n) = g (1) = (k + 1) and n 1 1 n X X 4 jf (0)j + jf (1)j +


+ jf (n 1)j  < n = :
j j 6 j j 4

2

4

4

2

2

2

2

n

2

2

pn ; then f (n) > jf (0)j + jf (1)j +


+ jf (n p In this ase the degree of f is 4k  5 n: =1

If k := b( =6) = 2

=1

1 4

1)j :

ut

The result of part a℄ is essentially sharp. b℄ For every n 2 N ; there exists a polynomial 2 pn (x) = a + a x +


+ a n2 x n su h that a n2 = 1; ja j; ja j; : : : ; ja n2 j are real numbers with modulus at most 1; and pn has a zero at 1 with multipli ity at least n: 0

0

2

1

1

Proof. De ne

2

2

2

1

Z

xt dt (n!) Qn Ln (x) := ; n = 0; 1; : : : ; 2i k (t k ) where the simple losed ontour surrounds the zeros of the denominator of the integrand. Then Ln is a polynomial of degree n with a zero of multipli ity at least n at 1: (This an easily be seen by repeated dierentiation and then evaluation of the above ontour integral by expanding the ontour to in nity.) Also, by the residue theorem, 2

2

=0

2

Ln (x) = 1 + where

k;n = It follows that Hen e,

n X

k=1

k;n xk

2

( 1)k 2(n!) ( 1)n (n!) = : j ) (n k)!(n + k)! j ;j 6 k (k =0

2

2

Qn

2

=

j k;n j  2 ; Pn (x) :=

2

k = 1; 2; : : : ; n : Ln (x) + Ln (x ) 2

2 is a polynomial of degree 2n with a zero of order n at 1: Also Pn has

onstant oeÆ ient 1 and ea h of its other2 oeÆ ients is a real number of modulus less than 1: Now let pn (x) := x n Pn (1=x): ut 2

2

3.5 Chebyshev Polynomials in Rational Spa es 139

℄ For every n 2 N ; there exists a polynomial

pn (x) = a + a x +


+ an xn 0

1

su h that an = 1; a ; a ; : : : ; an are real numbers pof modulus less than 1; and p has a zero at 1 with multipli ity at least b n=2 : 0

1

1

3.5 Chebyshev Polynomials in Rational Spa es There are very few situations where Chebyshev polynomials an be expli itly omputed. Indeed, only the lassi al ase of Se tion 2.1 is well known. However, the expli it formulas for the Chebyshev polynomials for the trigonometri rational system (3:5:1)



1;

1  sin  1  sin  1  sin  ; ; ::: ;

os  a os  a

os  an 1



2

;  2 [0; 2)

and therefore also for the rational system 

(3:5:2)

1;

x

1

a

1

;

x

1

a

2

; ::: ;

x

1



an

; x 2 [ 1; 1℄

with distin t real poles outside [ 1; 1℄ are impli itly ontained in A hiezer [56℄. The ase (3:5:1) does not perfe tly t our dis ussion of Se tion 3.3 be ause of the periodi ity or be ause [0; 2 ) is not a ompa t subset of R. This leads to nonuniqueness of the Chebyshev polynomials. Note that ordinary polynomials arise as a limiting ase of the span of system (3:5:2) on letting all the poles tend to 1: We are primarily interested in the linear span of (3:5:2) and its trigonometri ounterpart obtained with the substitution x = os : Let

Pn (a ; a ; : : : ; an) :=

(3:5:3)

1



2

Qn

p(x) jx ak j : p 2 Pn



k=1

and

Tn (a ; a ; : : : ; an ) :=

(3:5:4)

1

where (ak )nk

=1

 C n[

2

 Qn

k=1

t() j os 



ak j

1; 1℄ is a xed sequen e of poles.

: t 2 Tn ;

140 3. Chebyshev and Des artes Systems

When the poles a ; a ; : : : ; an are distin t and real, (3:5:3) and (3:5:4) are simply the real spans of the systems 1



(3:5:5)

1;

2

1

x a

1

;

x

1

; ::: ;

a

2

x

1



on [ 1; 1℄

an

and 

(3:5:6)

1  sin  1  sin  1  sin  1; ; ; ::: ; ;

os  a os  a

os  an 1



on [0; 2 ) ;

2

respe tively. We an onstru t Chebyshev polynomials of the rst and se ond kinds, whi h are analogous to Tn and Un of Se tion 2.1, for the spa es Pn (a ; a ; : : : ; an ) and Tn (a ; a ; : : : ; an ) as follows. Given a sequen e (ak )nk  C n[ 1; 1℄; we de ne the sequen e ( k )nk by 1

2

1

2

=1

=1

j k j < 1 ;

ak = ( k + k ) ;

(3:5:7)

1

1

2

that is, (3:5:8)

k := ak

Note that



q

ak + ak 2

p

q

ak 2



1

j k j < 1 :

1;

ak

q



ak

1 = 1:

2

In what follows, ak 1 is always de ned by (3:5:8) (this spe i es the

hoi e of root). Let D := fz 2 C : jz j < 1g; let (3:5:9)

2

Mn (z ) :=

n Y k=1

(z

k )(z

!1=2

k )

;

where the square root is de ned so that Mn (z ) := z n Mn (z ) is an analyti fun tion in a neighborhood of the losed unit disk D; and let 1

(3:5:10)

fn (z ) :=

Mn (z ) n z Mn(z 1 )

:

Note that fn is a tually a nite Blas hke produ t (see E.12 of Se tion 4.2). Also, fn (z ) = fn (z ) whenever jz j = 1: 2

1

1

3.5 Chebyshev Polynomials in Rational Spa es 141

The Chebyshev polynomials of the rst kind for the spa es Pn (a ; a ; : : : ; an ) and Tn (a ; a ; : : : ; an ) are now de ned by (3:5:11) Tn (x) := (fn (z ) + fn (z ) ) ; where x := (z + z ) ; jz j = 1 and (3:5:12) Ten () := Tn ( os ) ;  2 R; respe tively. The Chebyshev polynomials of the se ond kind for these two spa es are de ned by f (z ) fn (z ) ; where x := (z + z ) ; jz j = 1 (3:5:13) Un (x) := n 1

2

1

1

2

1

1

2

1

2

1

z

z

1

1

2

1

and (3:5:14) Uen () := Un ( os ) sin  ;  2 R; respe tively. As we will see, these Chebyshev polynomials preserve many of the elementary properties of the lassi al trigonometri and algebrai Chebyshev polynomials. This is the ontent of the next three results. Theorem 3.5.1 (Chebyshev Polynomials of the First and Se ond Kinds in Trigonometri Rational Spa es). Given (ak )nk  C n[ 1; 1℄; let Ten and =1

Uen be de ned by (3.5.12) and (3.5.14), respe tively. Then the following statements hold: a℄ Ten 2 Tn (a ; a ; : : : ; an ) and Uen 2 Tn (a ; a ; : : : ; an ) : b℄ kTen kR = 1 and kUen kR = 1 :

℄ There exist 0 =  <  <


< n =  su h that Ten (j ) = Ten( j ) = ( 1)j ; j = 0; 1; : : : ; n : 1

2

1

0

2

1

d℄ There exist 0 <  <  <


< n <  su h that Uen(j ) = Uen( j ) = ( 1)j ; j = 1; 2; : : : ; n : 1

2

1

e℄ For every  2 R;

Ten () + Uen () = 1 : 2

2

Proof. Observe that there are polynomials p that

(3:5:15)

Ten () = Tn ( os ) = =

e

1

2 Pn

and p

2

2 Pn

in M 2 (ei ) + ein M 2 (e i ) n n 2Mn (ei )Mn (e i )

Qn

p ( os ) j os  ak j 1

k=1

1

su h

142 3. Chebyshev and Des artes Systems

and

Uen () = Un ( os ) sin  =

(3:5:16)

=

e

in M 2 (ei ) ein M 2 (e i ) n n 2iMn (ei )Mn (e i )

p ( os ) sin  : k j os  ak j 2

Qn

=1

Thus a℄ is proved. Sin e j k j < 1 and fn is a nite Blas hke produ t, we have 2

jfn (z )j = 1

(3:5:17)

whenever

jz j = 1 :

Now b℄ follows immediately from (3:5:10) to (3:5:14). Note that Ten () is the real part, and Uen () is the imaginary part of fn (ei ), that is,

 2 R;

fn (ei ) = Ten() + iUen() ; whi h together with (3:5:17) implies e℄.

To prove parts ℄ and d℄, we note that Ten () = 1 if and only if 1; and Uen () = 1 if and only if fn (ei ) = i: Sin e j k j < 1 for k = 1; 2; : : : ; n; fn has exa tly 2n zeros in the open unit disk D: Sin e fn is analyti in a region ontaining the losed unit disk D; ℄ and d℄ follow by the argument prin iple (see, for example, Ash [71℄). ut

fn (ei ) =

2

2

With the transformation x = os  = (z + z ) and z = ei ; Theorem 3.5.1 an be reformulated as follows: 1

1

2

Theorem 3.5.2 (Chebyshev Polynomials in Algebrai Rational Spa es). Given (ak )nk  C n [ 1; 1℄; let Tn and Un be de ned by (3.5.11) and =1

(3.5.13), respe tively. Then a℄ Tn 2 Pn (a ; a ; : : : ; an ) and Un 2 Pn (a ; a ; : : : ; an ) : p b℄ kTn k ; = 1 and k 1 x Un (x)k ; = 1 :

℄ There exist 1 = x > x >


> xn = 1 su h that 1

[

2

1

2

1 1℄

0

[

2

1 1℄

1

Tn (xj ) = ( 1)j ; j = 0; 1; 2; : : : ; n : d℄ There exist 1 > y > y >


> yn > 1 su h that 1

q

1

2

yj Un(yj ) = ( 1)j 2

e℄ For every x 2 [ 1; 1℄;

p

(Tn (x)) + ( 1 2

1

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

x Un (x)) = 1 : 2

2

3.5 Chebyshev Polynomials in Rational Spa es 143

Parts ℄ and d℄ of Theorems 3.5.1 and 3.5.2 establish the equios illation property of the Chebyshev polynomials, whi h also extends to ertain linear ombinations of Chebyshev polynomials. In the trigonometri polynomial ase this is the fa t that os  os n +sin  sin n = os(n ) equios illates 2n times on the unit ir le [0; 2 ℄: Our next theorem hara terizes the Chebyshev polynomials for Tn (a ; a ; : : : ; an ) and re ords a monotoni ity property that we require later. 1

2

Theorem 3.5.3 (Chebyshev Polynomials in Trigonometri Rational Spa es). Let (ak )nk  C n [ 1; 1℄: Then (i) and (ii) below are equivalent: =1

(i) There is an  2 R su h that

V = ( os ) Ten + (sin ) Uen ; where Ten and Uen are de ned by (3.5.12) and (3.5.14). (ii) V 2 Tn (a ; a ; : : : ; an ) has uniform norm 1 on R; and it equios illates 2n times on R (mod 2). That is, there exist 1

2

0   <  <


<  n 0

so that

1

2

1

< 2

V (j ) = ( 1)j ; j = 0; 1; : : : ; 2n 1 :

Furthermore, if V is of the form in (i) (or hara terized by (ii)), then V 0 = ( os ) Ten0 + (sin ) Uen0 does not vanish between any two onse utive alternation points of V (that is, between j and j for j = 1; 2; : : : ; 2n 1 and between  n and 2 +  ). 1

2

1

0

Proof. (i) ) (ii). By Theorem 3.5.1 e℄ and Cau hy's inequality, we have

j( os ) Ten + (sin ) Uen j  ( os

(3:5:18)

2

2

 + sin )(Ten + Uen ) = 1 2

2

2

on the real line. From Theorem 3.5.1 ℄, d℄, and e℄, we obtain that Ten =Uen os illates between +1 and 1 exa tly 2n times on R (mod 2 ), and hen e it takes the value ot  exa tly 2n times. At ea h su h point, (3:5:18) be omes an equality, namely, ( os ) Ten + (sin ) Uen = 1 with dierent signs for every two onse utive su h points. (ii) ) (i). Let V be as spe i ed in part (ii) of the theorem. Let  be a point where V a hieves its maximum on R, so V ( ) = 1: We want to show that V is equal to p := Ten ( )Ten + Uen ( )Uen : Sin e V ( ) = p( ) = 1 and V 0 ( ) = p0 ( ) = 0, V p has a zero at  with multipli ity at least 2: There are at least 2n 1 more zeros (we ount multipli ities) of V p 0

0

0

0

0

0

0

0

0

144 3. Chebyshev and Des artes Systems

in R (mod 2 ), with one between any two onse utive alternation points of p if the rst zero of p to the right of  is greater than the rst zero of V to the right of  : If the rst zero of V to the right of  is greater than or equal to the rst zero of p to the right of  ; then there is one zero of p V between any two onse utive alternation points of V: In any ase V p has at least 2n + 1 zeros in R (mod 2 ). Hen e V p is identi ally 0: To prove the nal part of the theorem let V 2 Tn (a ; a ; : : : ; an ) be su h that kV kR = 1 and V equios illates 2n times on R (mod 2 ) between 1: Assume there is a  2 [0; 2) su h that jV ( )j < 1 and V 0 ( ) = 0: Then V ( ) 6= 0; otherwise the numerator of V would have at least 2n + 1 zeros in R (mod 2 ), whi h is a ontradi tion. Observe that there is a trigonometri polynomial t 2 T n su h that 0

0

0

0

1

0

2

0

0

0

2

V ()

V ( ) =

2

2

0

Qn

k=1

t() ( os  ak )( os 

ak )

:

This t has at least 4n + 1 zeros in R (mod 2 ), whi h is a ontradi tion again. Therefore V 0 () 6= 0 if jV ()j < 1; whi h means that V is stri tly monotone between any two of its onse utive alternation points. ut Under some assumptions on (ak )nk it is easy to write down the expli it partial fra tion de ompositions for Tn and Un . =1

 C n [ 1; 1℄ be a sequen e of distin t numbers su h that its nonreal elements are paired by omplex onjugation. Let Tn and Un be the Chebyshev polynomials of the rst and se ond kinds de ned by (3.5.11) and (3.5.13), respe tively. Then

Theorem 3.5.4. Let (ak )nk

=1

(3:5:19)

Tn (x) = A

0

A ;n A +


+ n;n x a x an

;n +

1

1

and (3:5:20)

where

Un (x) = A

0

Ak;n =

k

and Bk;n =

2

1

1

2

( 1)n ( 2

;n =



B ;n B B ;n + +


+ n;n ; x a x a x an 1

1

2

k

k

1

2 n Y

1 2

2

1

n Y j =1 j 6=k

1

k j ;

k j

1

j =1 j 6=k

k




n

+ 1

2




n ) ;

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

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

k j

1

3.5 Chebyshev Polynomials in Rational Spa es 145

Proof. It follows from Theorems 3.5.1 a℄ and 3.5.2 a℄ that Tn and Un an be written as the partial fra tion forms above. Now it is quite easy to al ulate the oeÆ ients Ak;n and Bk;n . For example, A

0



z nMn (z ) Mn (z )

1 Mn (z ) T (x) = lim + ;n = xlim n n !1 z!0 2 z Mn (z 1 ) ( 1)n 1 1 =

n 1 + 1 2 n ) ( 1 2 2






1








and for k = 1; 2; : : : ; n ;

Ak;n = xlim !a (x

ak )Tn (x)

k

1 (z z! k 4

k )(1

= lim =



k

2

k

1

2 n Y

j =1 j 6=k

k z ) 1



1

Mn (z ) z n Mn(z ) + n z Mn (z ) Mn(z ) 1



1

k j ;

k j

1

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

The oeÆ ients Bk;n an be al ulated in the same fashion.

ut

Comments, Exer ises, and Examples. The expli it formulas of this se tion are tremendously useful. They allow, for example, derivation of sharp Bernstein-type inequalities for rational fun tions; see Se tion 7.1. Various further properties of these Chebyshev polynomials for rational fun tion spa es are explored in the exer ises, whi h follow, Borwein, Erdelyi, and Zhang [94b℄. In parti ular, the orthogonalization of su h rational systems on [ 1; 1℄ with respe t to the weight w(x) = (1 x ) = an be made expli it in terms of the Chebyshev polynomials. Various other aspe ts of these orthogonalizations may be found in A hiezer [56℄, Bultheel et al. [91℄, and Van Ass he and Vanherwegen [92℄. 2

1 2

E.1 Further Properties of Ten and Tn . Given (ak )nk ( k )nk be de ned by

=1

 C n[

=1

k := ak

q

ak

j k j < 1 ;

1;

2

as before. We introdu e the Bernstein fa tors

Bn (x) :=

n X k=1

p

Re

and

Ben () := Bn ( os ) ==

n X k=1

ak

ak

2

1

x

p

Re

!

ak

ak 2

1

os 

!

;

1; 1℄, let

146 3. Chebyshev and Des artes Systems p

where the hoi e of ak 1 is determined by the restri tion j k j < 1: Note that for x 2 [ 1; 1℄; we have p

Re

ak

ak

2

1

!

x

2

= Re k

k

k x

1 1

1 2

!

j k j )  (1 jj1 k j )(1 2 k xj 2

2

2

> 0:

The following result generalizes the trigonometri identities ( os nt)0 = n sin nt ; and

(sin nt)0 = n os nt ;

(( os nt)0 ) + ((sin nt)0 ) = n ; 2

2

2

whi h are limiting ases (if n 2 N and t 2 R are xed, then lim Ben (t) = n as all ak ! 1): a℄ Show that, on the real line,

Ten0 = Ben Uen ; and

Uen0 = Ben Ten ;

(Ten0 ) + (Uen0 ) = Ben : 2

2

2

Hint: For example, Te0 () n

b℄ If V



1 0 i f (e ) = 2 n ei fn0 (ei ) = fn (ei )

:= ( os ) Ten



fn0 (ei ) iei fn (ei ) fn (ei ) fn (ei ) 2i 2

+ (sin ) Uen

1

= Ben ()Uen () :

ut

for some  2 R; then

(V 0 ) + Ben V = Ben 2

2

2

2

holds on the real line.

℄ The Derivative of Tn at 1. Let Tn be de ned by (3:5:11). Then

T 0 (1) = n

n X k=1

and

Tn0 ( 1) = ( 1)n



1 + k Re 1 k n X k=1



!2

1 k Re 1 + k

!2

:

3.5 Chebyshev Polynomials in Rational Spa es 147

d℄ Contour Integral for Tn . Show that 0

Z

n Y

1  2i j

Tn (x) =

=1

11=2

(t j )(t j ) A (1 j t)(1 j t)

t

t x dt 2tx + 1

2

for every x 2 [ 1; 1℄; where is a ir le entered at the origin with radius 1 < r < minfj j j : 1  j  ng; and the square root in the integrand is an analyti fun tion of t in a neighborhood of : Hint: Cau hy's integral formula and the map x = (z + z ) give 1

1

1

2





1 Mn (z ) z n Mn (z ) + 2 z n Mn (z ) Mn(z )  Z 1 1 Mn (t) 1 1 = + n 2i 2 t Mn (t ) t z t z Z 1 t x Mn (t) = dt ; 2i tn Mn (t ) t 2tx + 1

Tn (x) =

1

1



1

1

1

dt

2

ut

where Mn is de ned by (3.5.9). E.2 Orthogonality. Given (ak )1 k

=1

ak = ( k + k ) ;

=1

p

k = ak

1

1

2

1; 1℄; let ( k )1 k be de ned by

 Rn[

ak

k 2 ( 1; 1)

1;

2

and let (Tn )1 n be de ned by (3.5.11). a℄ Show that =0

Z

1

1

Tn(x)Tm (x) p

dx

1

 = ( 1)n m (1 + 2 x +

2

2 1




m ) m


n 2

+1

for all integers 0  m  n: (The empty produ t is understood to be 1.) b℄ Given a 2 R n [ 1; 1℄; let 2 ( 1; 1) be de ned by

a = ( + ) ; 1

1

2

=a

p

a

2

1;

2 ( 1; 1) :

Show that Z

1

1

Tn (x)

x

1

dx p a 1

x

2

=

n Y

2



1

j =1



j : 1

j

148 3. Chebyshev and Des artes Systems

℄ Show that

Z

1

1

and

Z

1

1

Tn (x)

Tn (x) p

x

1

=1

0

1




n

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

1; 1℄; we de ne Rn := Tn + n Tn

1

Rn :=

p ;

2

 Rn[

0

R :=

1

2

Given a sequen e (ak )1 k R := 1 ; and

= ( 1)n 

2

p dx = 0 ; ak 1 x

1

x

dx

s

2

(1 n ) 2

(Tn + n Tn ) : 1

The following part of this exer ise indi ates that these simple linear ombinations of Tn and Tn give the orthogonalization of the rational system   1 1 ; ; ::: 1; 1

x

a

x

1

a

2

whenever (ak )1 k  R n [ 1; 1℄ is a sequen e of distin t real numbers. d℄ Show that, for all nonnegative integers n and m; =1

Z

1

1

 (x) p dx Rn (x)Rm

where Æm;n is the Krone ker symbol.

Proof. Let m  n. By part ℄, Z

holds for k = 1; 2; : : : ; n Z

1

1

Rn (x) p

dx

x

1

2

=

1

Z

1

Rn (x)

1

2

= Æm;n ;

p dx = 0 ak 1 x

(Tn (x) + n Tn (x)) p

1

= ( 1)n ( 1 2

Rn (x)Rm (x) p Z

1

2

1

dx

2

Rn (x)

= 0;

2

1

dx

1 x


n ) + n ( 1)n ( 1

1 x Finally, it follows from part a℄ that 1

x

2

1. Also 1

This implies that Z

1

x

1

p dx 1

1

2




n

1

) = 0:

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

x

2

= 1:

ut

e℄ Assume (ak )1 k  R n [ 1; 1℄: Then Tn and Rn have exa tly n zeros in [ 1; 1℄; and the zeros of Tn and Tn stri tly interla e. =1

1

3.5 Chebyshev Polynomials in Rational Spa es 149

E.3 Extension of Theorems 3.5.1 and 3.5.3. Given (ak )kn 2

=1

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

2

(

)

t() : t 2 Tn : jsin(( ak )=2)j

Q2n

2

 C n R; let

k=1

Without loss of generality we may assume that Im(ak ) > 0 ;

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

a℄ Show that there is a polynomial q n 2 P n of the form 2

q n (z ) = 2

2n Y

k=1

su h that

jq n (ei )j = 2

(z

2n Y

k=1

2

j k j < 1 ;

k ) ;

jsin((

2C

ak )=2)j ;

 2 R:

Hint: Use the fa t that jz j = j1 zj whenever jz j = 1 and 2 C : Asso iated with q n 2 P n de ned in part a℄, let 2

2

Mn (z ) := and

ut

p

q n (z ) 2

!1=2

n

Y Mn(z ) := (1 z ) 2

k=1

;

where the square roots are de ned so that Mn is analyti in a neighborhood of the losed unit disk, and Mn is analyti in a neighborhood of the

omplement of the open unit disk. Let

fn (z ) :=

Mn (z ) : Mn (z )

For  2 R; we de ne

Ten () and

:= Re(fn

 Mn (ei ) Mn(ei ) + 2 Mn (ei ) Mn (ei )

1 (ei )) =

Uen () := Im(fn (ei )) =





1 Mn (ei ) 2i Mn (ei )



Mn (ei ) : Mn (ei )

Using the new (extended) de nitions, show the following:

150 3. Chebyshev and Des artes Systems

b℄ Ten 2 Tn (a ; a ; : : : ; a n ) and Uen 2 Tn (a ; a ; : : : ; a n ) :

℄ kTen kR = 1 and kUen kR = 1 : d℄ There are numbers  <  <


<  n in [ ;  ) su h that 1

2

2

1

1

2

2

2

2

Te(j ) = ( 1)j ;

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

e℄ There are numbers  <  <


<  n in [ ;  ) su h that 1

2

2

Ue (j ) = ( 1)j ;

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

f℄ Te() + Ue () = 1 for every  2 R : 2

2

g℄ Both Ten and Uen have exa tly 2n simple zeros in the period [ ;  ), and the zeros of Ten and Uen stri tly interla e. h℄ The statements of Theorem 3.5.3 remain valid. E.4 Extension of the Bernstein Fa tor Ben . Let (ak )kn 2

=1

 C nR;

Im(ak ) > 0 :

With the notation of the previous exer ise we de ne

Ben () :=

ei fn0 (ei ) ; fn(ei )

 2 R:

a℄ Show that for every  2 R;

Ben () =

2n X

k=1

j k j

1

2

j k

ei

j

2

=

2n X

k=1

j

1

eiak

jeiak j

2

ei

j

2

b℄ Show that, on the real line,

Uen0 = Ben Ten ;

Ten0 = Ben Uen ; and

℄ Show that

(Ten0 ) + (Uen0 ) = Ben : 2

2

2

(V 0 ) + Ben V = Ben 2

2

2

2

holds on the real line for every V of the form

V = ( os ) Ten + (sin ) Uen ;

 2 R:

:

3.5 Chebyshev Polynomials in Rational Spa es 151

E.5 Chebyshev Polynomials for Pn (a ; a ; : : : ; an ) on R. Let 1

(ak )nk

=1

C nR

Im(ak ) > 0 ;

with

Let

Mn (z ) := and

M (z ) := n

2

n Y k=1 n Y

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

!1=2

(z

ak ) !1=2

(z

k=1

;

ak )

where the square roots are de ned so that Mn is analyti in a neighborhood of the losed upper half-plane, and Mn is analyti in a neighborhood of the

losed lower half-plane. Let

fn (z ) :=

Mn (z ) : Mn (z )

For x 2 R; we de ne

Tn (x) := Re(fn (x)) = and



1 Mn (x) Mn (x) + 2 Mn (x) Mn (x) 





Mn (x) : Mn (x)

1 Mn (x) Un (x) := Im(fn (x)) = 2i Mn (x)

Show the following: a℄ Tn 2 Pn (a ; a ; : : : ; an ) and Un 2 Pn (a ; a ; : : : ; an ) : b℄ kTn kR = 1 and kUn kR = 1 :

℄ There are real numbers x > x >


> xn su h that 1

2

1

1

Tn (xj ) = ( 1)j ;

2

2

lim T (x) = 1 ; x!1 n

1

and

d℄ There are real numbers y > y >


> yn 1

Un(yj ) = ( 1)j

2

+1

and

lim Tn (x) = ( 1)n :

x! 1

1

su h that

lim U (x) = 0 : x!1 n

e℄ Tn (x) + Un (x) = 1 for every x 2 R : f℄ Both Tn and Un have exa tly n simple zeros on R; and the zeros of Tn and Un stri tly interla e. 2

2

152 3. Chebyshev and Des artes Systems

g℄ The following statements are equivalent: (i) There exists an  2 R su h that

V = ( os ) Tn + (sin ) Un : (ii) V 2 Pn (a ; a ;


; an ) has uniform norm 1 on R; and it equios illates n times on the extended real line. That is, there are extended real numbers 1  z > z >


> zn > 1 su h that 1

1

2

2

V (zj ) = ( 1)j ; where

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

V (1) := xlim !1 V (x) :

h℄ With the notation of part g℄, V is stri tly monotone on ea h of the intervals (z ; 1); (z ; z ); : : : ; (zn ; zn ); ( 1; zn ) : 1

2

1

1

E.6 Bernstein Fa tor on R. Let (ak )nk

=1

Im(ak ) > 0 ;

 C n R with

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

With the notation of E.5 let

Bn (x) := a℄ Show that

Bn (x) =

n X k=1

fn0 (x) ; fn (x)

x 2 R:

2Im(ak )

jx

ak j

2

x 2 R:

;

b℄ Show that, on the real line,

Un0 = Bn Tn ;

Tn0 = Bn Un ; and

℄ Show that

(Tn0 ) + (Un0 ) = Bn : 2

2

2

(V 0 ) + Bn V = Bn 2

2

2

2

holds on the real line for every V of the form

V = ( os ) Tn + (sin ) Un :

3.5 Chebyshev Polynomials in Rational Spa es 153

E.7 CoeÆ ient Bounds in Nondense Rational Fun tion Spa es. Suppose (ak )1 k  R n [ 1; 1℄ is a sequen e of distin t numbers satisfying =1

1q X j =1

1

j aj j

2

< 1:

Show that there are numbers Kj > 0 su h that

jDj;n j  Kj kpk

[

;

1 1℄

;

j = 0; 1; : : : ; n ; n 2 N

for every p 2 Pn (a ; a ; : : : ; an ) of the form 1

p(x) = D

2

0

;n +

D ;n D +


+ n;n ; x a x an 1

1

Hint: Use E.2 ℄ of Se tion 3.3 and Theorem 3.5.4.

Dj;n 2 R :

ut

This is page

154

Printer: Opaque this

4

Denseness Questions

Overview We give an extended treatment of when various Markov spa es are dense. In parti ular, we show that denseness, in many situations, is equivalent to denseness of the zeros of the asso iated Chebyshev polynomials. This is the prin ipal theorem of the rst se tion. Various versions of Weierstrass'

lassi al approximation theorem are then onsidered. The most important is in Se tion 4.2 where Muntz's theorem on erning the denseness of spanf1; x1 ; x2 ; : : : g is analyzed in detail. The third se tion on erns the equivalen e of denseness of Markov spa es and the existen e of unbounded Bernstein inequalities. In the nal se tion we onsider when rational fun tions derived from Markov systems are dense. In luded is the surprising result that rational fun tions from a xed in nite Muntz system are always dense.

4.1 Variations on the Weierstrass Theorem Mu h of the utility of polynomials stems from the fa t that all ontinuous fun tions on a nite losed interval are uniform limits of them. This is the well-known Weierstrass approximation theorem. There are numerous proofs of this; several are presented in the exer ises. Another proof follows from the main theorem of this se tion.

4.1 Variations on the Weierstrass Theorem 155

Asso iated with a Markov system M := (f ; f ; : : : ) on [a; b℄ we de ne, as in Se tion 3.3, the Chebyshev polynomials 0

1

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

1

Denote the zeros of Tn by (a )x < x <


< xn ( b). Let x := a and xn := b. The mesh of Tn is de ned by 1

2

0

+1

(4:1:1)

Mn := Mn(Tn : [a; b℄) := max in 1

+1

jxi

xi

1

j:

This is a measure of the maximal gap between two onse utive zeros of Tn with respe t to the interval [a; b℄: For a sequen e (Tn )1 of Chebyshev polynomials asso iated with a n xed Markov system on [a; b℄; we have =0

lim Mn = 0

lim inf Mn = 0 :

if and only if

n!1

n!1

This follows from the fa t that if m < n; then Tm annot have more than one zero between any two onse utive zeros of Tn : Our main result shows the strong onne tion between the denseness of the real span of an in nite Markov system M of C fun tions on [a; b℄ in C [a; b℄ and the density of the zeros of the asso iated Chebyshev polynomials. 1

Theorem 4.1.1. Suppose M := (1; f ; f ; : : : ) is an in nite Markov system on [a; b℄ with ea h fi 2 C [a; b℄: Then span M is dense in C [a; b℄ if and only if lim M = 0 ; n!1 n where Mn is the mesh of the asso iated Chebyshev polynomials. 1

2

1

Proof. The only if part of this result is the easier part and we oer the following proof. Suppose span M is dense in C [a; b℄; while lim inf Mn > 0: n!1 Then there exists an interval [ ; d℄  [a; b℄ that ontains no zero of Tn for in nitely many n, say, for n < n <


: Consider the pie ewise linear fun tion F de ned as follows. Let < y < y < y < y < d; and let 1

2

1

F (x) :=

8 < :

2

3

4

0 ; x 2 fa; ; d; bg 2 ; x 2 fy ; y g 2 ; x 2 fy ; y g 1

3

2

4

and be linear elsewhere. Sin e span M is dense in C [a; b℄; there exists a

k 2 N and a p 2 spanf1; f ; : : : , fnk g with 1

(4:1:2)

kp

F k a;b < 1 : [

℄

156 4. Denseness Questions

Now p Tnk has at least nk 2 zeros on [a; ℄ [ [d; b℄ be ause Tnk has at least nk extrema on these intervals. The four extrema of F on ( ; d) together with (4.1.2) guarantee at least three more zeros of p Tnk on ( ; d): Hen e p Tnk has at least nk + 1 zeros and vanishes identi ally. This ontradi ts (4:1:2). The if part of the theorem follows from the next theorem and E.8 a℄ of Se tion 3.2. This exer ise shows that (f 0 ; f 0 ; : : : ) is a weak Markov system on [a; b℄: ut 1

2

The phenomenon formulated in Theorem 4.1.1 is quite general, and we prove a rather more general result than is needed for the pre eding theorem. The modulus of ontinuity !f of a fun tion f : [a; b℄ 7! R is de ned by

!f (Æ) := sup jf (x) jx yj<Æ x;y2 a;b

(4:1:3)

[

f (y)j :

℄

Theorem 4.1.2. Suppose that

Hn := spanf1; g ; g ; : : : ; gng 1

2

is a Chebyshev spa e on [a; b℄ with asso iated Chebyshev polynomial Tn : Suppose ea h gi 2 C [a; b℄ and (g0 ; : : : ; gn0 ) is a weak Chebyshev system on [a; b℄ (weak Chebyshev systems are de ned in E.8 of Se tion 3.2). Let Hn0 := spanfg0 ; : : : ; gn0 g: If f 2 C [a; b℄; then there exists an hn 2 Hn su h that p
khn f k a;b  C!f Æn ; 1

1

1

[

where

℄

Æn := Mn(Tn : [a; b℄) :

Here C is a onstant depending only on a and b: Proof. Suppose a < < d < b and Sn 2 Hn is the best uniform approximation from Hn to F on [a; ℄ [ [d; b℄; where F (x) :=



0; x 2 [a; ℄ 1; x 2 [d; b℄ :

We laim the following: (4:1:4)

Sn is monotone on [ ; d℄

and (4:1:5)

kSn

F k a; [ d;b [

℄

[

℄

 (d5Æn ) :

4.1 Variations on the Weierstrass Theorem 157

Let  := n + 1 be the dimension of the Chebyshev spa e Hn : Sin e Sn is a best approximation to F on [a; ℄ [ [d; b℄; there exist  + 1 points in this set where the maximum error

n := kF

(4:1:6)

Sn k a; [ d;b [

℄

[

℄

o

urs with alternating sign (see Theorem 3.1.6). Suppose m + 1 of these points y <


< ym lie in [a; ℄, and  m of these points ym <


< y lie in [d; b℄: Then Sn0 has at least m 1 sign hanges in (a; ) (one at ea h alternation point in [a; ℄ ex ept possibly at the endpoints a and ). Likewise, Sn0 has at least  m 2 sign hanges in (d; b): So Sn0 has at least  3 sign hanges in (a; ) [ (d; b). Note that this ount ex ludes ym and ym . Thus Sn0 has at most one more sign hange in (a; b) unless Sn0 vanishes identi ally (whi h is not possible for   2). Now suppose Sn0 has a sign hange on ( ; d). Then, sin e there is at most one sign hange of Sn0 in ( ; d); it annot be the ase that both ym = and ym = d and Sn0

hanges sign at neither nor d; otherwise 0

+1

+1

+1

sign(Sn ( )

f ( )) = sign(Sn (d)

f (d))

as a onsideration of the two ases shows. But if ym 6= or ym 6= d or Sn0 hanges sign at either or d; then we have a

ounted for all the sign

hanges of Sn0 by a

ounting for the (possible) one additional sign hange (either Sn0 vanishes with sign hange at or d or one of ym or ym is an interior alternation point of Sn where Sn0 vanishes). Thus Sn0 has no zeros with sign hange in ( ; d) and (4.1.4) is proved. To prove (4.1.5) we pro eed as follows. With n de ned by (4.1.6), +1

+1

Dn := n Tn

Sn

has at least m zeros on [a; ℄ and

Dn := Dn + 1 = 1 + n Tn

Sn

has at least  m 1 zeros on [d; b℄ ( ounting ea h internal zero without sign hange twi e). Thus Dn0 has at least  3 sign hanges on [a; ℄ [ [d; b℄: Suppose Tn has at least four alternation points on an interval [ ; Æ ℄  ( ; d); and suppose that Sn (Æ) Sn ( ) < 2n : Then, be ause of (4.1.4) and the os illation of Tn on [ ; Æ ℄;

Dn +

Sn ( ) + Sn (Æ) 2

= n Tn



has at least three zeros on [ ; Æ ℄ and hen e

Sn

Sn ( ) + Sn (Æ) 2



158 4. Denseness Questions  Sn ( ) + Sn (Æ) 0



Dn0 = Dn +

2 has at least two sign hanges on [ ; Æ ℄: This, however, gives that Dn0 2 Hn0 has a total of at least  1 = n sign hanges, whi h is impossible. In parti ular, Sn (Æ) Sn ( )  2n on any interval [ ; Æ ℄  ( ; d) where Tn has at least 4 alternation points. Thus, (d ) 2n : Sn (d) Sn ( )  5Æn However, sin e Sn is a best approximation to F on [a; ℄ [ [d; b℄;

Sn ( )  1 + 2n

Sn (d)

and we an dedu e (4.1.5) on omparing these last two inequalities and noting that n  : The proof is now a routine argument, whi h for simpli ity, is presented on the interval [a; b℄ := [0; 1℄: Let 1

2

V (x) := f (0) +

m X1 i=0

f

i+1
m

i

m Sn;i (x) ;

f

where, for i = 0; 1; : : : ; m 1, Sn;i 2 Hn is the best uniform approximation to
  0 ; x 2 0; im   Fn;i (x) := 1 ; x 2 im ; 1     on 0; mi [ im ; 1 : Let +1

+1

+1

fe(x) := f (0) +

m X1 i=0

f

i+1
m

f

i

m Fn;i (x) : 



Then repeated appli ations of (4.1.5) with the intervals [a; ℄ := 0; mi and  i [d; b℄ := m ; 1 yield for every x 2 [0; 1℄ that +1

jV (x)

f (x)j  jV (x)



m X1 i=0

f

fe(x)j + jfe(x) i+1
m

f

f (x)j

i

m (Sn;i (x)


Fn;i (x)) + !f

 (m 1)(5Ænm) !f m + 2 !f m Hen e, with m := bÆn = ; p
kV f k ;  C !f Æn : 1

1




1

m







+ !f m : 1

1 2

[0 1℄

ut

An immediate orollary to Theorem 4.1.1 is the Weierstrass theorem.

4.1 Variations on the Weierstrass Theorem 159

Corollary 4.1.3. The polynomials are dense in C [ 1; 1℄:

Proof. M = (1; x; x ; : : : ) is an in nite Markov system of C fun tions on [ 1; 1℄: The asso iated Chebyshev polynomials are just the usual Chebyshev polynomials Tn (see Se tion 2.1) and  Mn  ; n = 1; 2; : : : n 2

1

ut

is obvious from E.1 of Se tion 2.1.

Also from the last part of the proof of Theorem 4.1.2 we have the following orollary. Corollary 4.1.4. Suppose M := (1; f ; f ; : : : ) is an in nite Markov system on [a; b℄ with ea h fi 2 C [a; b℄. Then for ea h n 2 N ; there exists a 1

2

1

pn 2 spanf1; f ; f ; : : : ; fn g 1

su h that

kpn

2




f k a;b  C (1 + m Mn ) !f m for every m 2 N ; where C is a onstant depending only on a and b: [

℄

2

1

Comments, Exer ises, and Examples. The Weierstrass approximation theorem of 1885 (see Weierstrass [15℄) is one of the very basi theorems of approximation theory. It, of ourse, requires that lear distin tions be made about the nature of onvergen e (pointwise versus uniform) and the region of onvergen e (intervals versus omplex domains). Weierstrass, the preeminent analyst of the last third of the nineteenth entury, was prin ipal in insisting that su h distin tions be learly made. His famous and profoundly surprising example of a nowhere dierentiable ontinuous fun tion dates from 1872. A number of proofs of his approximation theorem and its many generalizations are explored in the exer ises. Theorem 4.1.1 was proved by Borwein [90℄. The only if part of this theorem an be found in Kroo and Peherstorfer [92℄. Appli ations of the methods and results of this se tion an be found in Borwein [91b℄, Borwein and Sa [92℄, and Lorentz, Golits hek, and Makovoz [92℄. The last two papers give an appli ation to weighted in omplete polynomials, where the zeros of the Chebyshev polynomials are often dense in a subinterval (see also Mhaskar and Sa [85℄). E.1 The Weierstrass Approximation Theorem. Every real-valued ontinuous fun tion on a nite losed interval [a; b℄ an be uniformly approximated

by polynomials with real oeÆ ients. Every omplex-valued ontinuous fun tion on a nite losed interval [a; b℄ an be uniformly approximated by polynomials with omplex oeÆ ients.

160 4. Denseness Questions

More pre isely, in the real ase, let

En := En (f : [a; b℄) := inf kf p2Pn

pk a;b : [

℄

The Weierstrass approximation theorem asserts that lim E (f : [a; b℄) = 0 ; n!1 n

f

2 C [a; b℄ :

The following steps outline an elementary proof basi ally due to Lebesgue [1898℄. Parts a℄ to d℄ deal with the real version ( rst statement) of the theorem. The omplex version (se ond statement) of the theorem an easily be redu ed to the real version; see part e℄. a℄ Every ontinuous fun tion on [a; b℄ an be uniformly approximated by pie ewise linear fun tions. Hint: Consider the pie ewise linear fun tion that interpolates f at n equally spa ed points and use the uniform ontinuity of f: ut b℄ It suÆ es to prove that jxj an be uniformly approximated by polynomials on [ 1; 1℄: Hint: Use part a℄. ut

℄ Approximation to jxj: Show that

jj

lim E ( x : [ 1; 1℄) = 0 : n!1 n

p

Hint: The Taylor series expansion of f (z ) := 1 z yields

p

1 1 1
3 z+ z z +


2 2
4 2
4
6 and the onvergen e is uniform for 0  z  1: (By Abel's theorem, a power series onverges uniformly on every losed subinterval of the set of points in R where it onverges; see, for example, Stromberg [81℄). Thus,

j xj =

p

z=1

1

x =

p

2

3

1 (1 x ) 1 1 1
3 =1 (1 x ) + (1 x ) (1 2 2
4 2
4
6 and the onvergen e is uniform for 1  x  1: d℄ An Alternative to ℄. Let 2

2

2

Q (x) := 1 0

and

2 2

1 2

Qn (x) := (1 +1

x ) +


2 3

ut

x + Qn (x)) : 2

2

Show that 0  Qn (x)  Qn (x)  1; +1

and Qn (x) ! 1

jxj uniformly on [

n = 0; 1; : : : ; x 2 [ 1; 1℄

1; 1℄ as n ! 1.

4.1 Variations on the Weierstrass Theorem 161

Hint: First show the pointwise onvergen e and then use Dini's theorem (see, for example, Royden [88℄). ut e℄ Complex Version of the Weierstrass Approximation Theorem. Every

omplex-valued ontinuous fun tion on a nite losed interval [a; b℄ an be

uniformly approximated by polynomials with omplex oeÆ ients. It an be shown that

En (jxj : [ 1; 1℄) 

; n

where 0:280168 < < 0:280174: Bernstein [13℄ established the above asymptoti with weaker bounds on ; namely, 0:278 < < 0:286; and observed that  = = 0:282 is roughly the average of these bounds. The stronger bounds on , due to Varga and Carpenter, show that 6=  = , but it is open whether or not is some familiar onstant; see Varga [90℄. 1

1 2

2

1

1 2

2

E.2 The Stone-Weierstrass Theorem. If X is a ompa t Hausdor spa e, then a subalgebra A of C (X ); whi h ontains f = 1 and separates points, is dense in C (X ): A subalgebra A of C (X ) is a ve tor spa e of fun tions that is losed under multipli ation (here, addition and multipli ation are pointwise). Separating points means that for any two distin t x, y 2 X; there exists an f 2 A su h that f (x) 6= f (y): a℄ Observe that the set P := [1 n Pn of all polynomials with real oeÆ ients is a subalgebra of C [a; b℄ that separates points, and hen e the StoneWeierstrass theorem implies the Weierstrass approximation theorem. b℄ Observe that the real polynomials in x form a subalgebra of C [ 1; 1℄ that does not separate points. We outline a standard proof of the Stone-Weierstrass theorem. Let A denote the losure of a subalgebra A  C (X ) in the uniform norm.

℄ If f 2 A; then jf j 2 A: =0

2

Proof. If f 2 A; then p(f ) 2 A for any polynomial p: Now hoose pn su h that pn (x) ! jxj on the interval [ kf k; kf k℄: ut d℄ Let (f _ g )(x) := maxff (x); g (x)g and (f ^ g )(x) := minff (x); g (x)g : Show that if f , g 2 A; then so are f _ g and f ^ g:

162 4. Denseness Questions

Hint: 1 2

f _ g = (f + g + jf

1 2

gj) ; and f ^ g = (f + g

e℄ If p; q 2 X are distin t and ;  2 R; then there exists f

f (p) = 

and

jf

g j) :

2 A with

ut

f (q) =  :

Hint: Let g 2 A be su h that g(p) 6= g(q) and onsider f :=

  g(p) g(q)
g+
1: g(p) g(q) g(p) g(q)

ut

f℄ Completion of Proof. Let f 2 C (X ): For ea h p; q 2 X; let fpq be an element of A with fpq (p) = f (p) and fpq (q ) = f (q ): Fix  > 0 and de ne open sets Vpq := fx 2 X : fpq (x) < f (x) + g :

Now fVpq : p 2 X g is an open over of the ompa t Hausdor spa e X; so for ea h q 2 X we an pi k a nite sub over

fVp q ; Vp q ; : : : ; Vpn q g 1

of X: We let Observe that fq

2

fq := minffp1 q ; fp2 q ; : : : ; fpnq g :

2 A by part e℄, and

fq (x) < f (x) +  ;

x2X:

g℄ Continued. Let

Vq := fx 2 X : fq (x) > f (x)

g ;

where fq is de ned in part f℄ for every q 2 X . Then fVq : q 2 X g is an open

over of the ompa t Hausdor spa e X , so we an extra t a nite sub over

fVq ; Vq ; : : : ; Vqm g 1

of X: Now let

2

g := maxffq1 ; fq2 ; : : : ; fqm g :

Note that g 2 A by part e℄, and

f (x) =  < g(x) < f (x) +  ; whi h nishes the proof.

x2X;

ut

4.1 Variations on the Weierstrass Theorem 163

The next exer ise presents pretty theorems due to Bohman [52℄ and Korovkin [53℄ on the onvergen e of sequen es of positive linear operators. The exer ise after that gives some appli ations that in lude dierent proofs of the Weierstrass theorem via onvergen e of spe ial polynomials, su h as the Bernstein polynomials. An operator L on C (X ) is alled monotone if

f (here f

g

implies L(f )  L(g )

 g means f (x)  g(x) for all x 2 X ).

E.3 Monotone Operator Theorems. Korovkin's First Theorem. Let (Ln )1 n be a sequen e of monotone linear operators on C (K ) (the set of ontinuous, 2 periodi , real-valued fun tions =1

on R). Let

f (x) := 1 ; f (x) := sin x ; f (x) := os x : 0

1

Then

2

lim kLn (f )

n!1

for all f

f kK = 0

2 C (K ) if and only if lim kLn (fi )

n!1

fi kK = 0 ;

i = 0; 1; 2 :

Korovkin's Se ond Theorem. Let (Ln )1 n be a sequen e of monotone linear operators on C [a; b℄: Let =1

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

Then

1

lim kLn (f )

f k a;b = 0

n!1

for all f

2

2

[

℄

2 C [a; b℄ if and only if lim kLn (fi )

n!1

fi k a;b = 0 ; [

℄

i = 0; 1; 2 :

Korovkin's theorem in a more general setting an be found in Lorentz [86a℄. a℄ Proof of Korovkin's Se ond Theorem. The only if part of the theorem is trivial. For the if part, observe that the pointwise onvergen e of (Ln )1 n

an be easily proved sin e, for any preassigned  > 0 at any xed x ; one an =1

0

164 4. Denseness Questions

nd parabolas y = p (x) := a x + b x + and y = p (x) := a x + b x + su h that p (x) < f (x) < p (x) ; x 2 [a; b℄ 1

1

2

1

1

with

jf (x )

2

2

2

2

2

2

p (x )j < 

0

1

1

jf (x )

and

0

p (x )j <  :

0

2

0

Now use the ontinuity of f and the ompa tness of [a; b℄ to make the above argument uniform on the interval [a; b℄: ut b℄ Proof of Korovkin's First Theorem. Hint: Modify the proof of Korovkin's se ond theorem. ut E.4 Bernstein Polynomials. The nth Bernstein polynomial for a fun tion f 2 C [0; 1℄ is de ned by

Bn (f )(x) := a℄ Let

n X k=0

f

  

k n

n k x (1 x)n k

k;

n = 1; 2; : : : :

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

1

2

2

Show that

Bn (f ) = f ; Bn (f ) = f ; Bn (f ) = 0

0

1

1

2

n

n

1

f + nf 1

2

1

for every n = 0; 1; 2; : : : : b℄ Use Korovkin's se ond theorem and part a℄ to show that lim kBn (f )

n!1

fk

; =0

[0 1℄

for every f 2 C [0; 1℄: For more on Bernstein polynomials, see Lorentz [86b℄. E.5 The Fourier and Fejer Operators. For f 1 Sn (f )(x) := 2

Z 

and

Fn (f )(x) :=

1

2n

2 C (K ); let


!

sin n + t f (t + x) 2 sin t  1

2

1

dt ;

n = 0; 1; : : :

2

Z 



sin nt f (t + x) sin t  1

2

1

2

dt ;

n = 0; 1; : : : :

2

The operator Sn is alled the Fourier operator, while the operator Fn is

alled the Fejer operator.

4.1 Variations on the Weierstrass Theorem 165

a℄ Show that Sn (f ) is the nth partial sum of the Fourier series of f; that is,

Sn (f )(x) = where

ak = and

bk = Hint:

a

0

2

1

k=1

Z 

 1

n X

+

f (t) os kt dt

 Z 



(ak os kx + bk sin kx) ;

f (t) sin kt dt :




n sin n + t 1 X

os kt = + 2 k 2 sin t 1

2

1

2

and

1

Sn (f )(x) =

=1

Z 

!

n

1 X +

os kt dt : f (t + x) 2 k 



=1

ut

b℄ Fn (f ) is the Cesaro mean of S ; S ; : : : Sn , that is, 0

Fn (f ) = Hint:

nX1 k=0

1

1

S (f ) + S (f ) +


+ Sn (f ) : n 0

1

1






sin nt sin k + t = sin t sin t 1

1

2

2

2

1

1

2

2

:

ut

℄ Fejer's Theorem. For every f 2 C (K ); Fn (f ) ! f uniformly on R: Hint: Ea h Fn is obviously a monotone operator on C (K ); so it suÆ es to prove the uniform onvergen e of (Fn )1 n on R only for fi , i = 0; 1; 2, as de ned in Korovkin's rst theorem. However, this is obvious, sin e =1

Fn (f ) = f ; Fn (f ) = 0

0

1

n

n

1

f ; Fn (f ) = 1

2

n

n

1

f

2

for every n = 1; 2; : : : : ut d℄ The set T := [1 n Tn of all real trigonometri polynomials is dense in C (K ); the set of all ontinuous, 2 periodi , real-valued fun tions. The set

T := [1 n Tn of all omplex trigonometri polynomials is dense in C (K ); the set of all ontinuous, 2 periodi , omplex-valued fun tions. =0

=0

166 4. Denseness Questions

Hint: This follows from Fejer's theorem. This is also a orollary of the Stone-

Weierstrass theorem (see E.2). ut The remaining parts of the exer ise follow Lorentz [86a℄. Suppose that Ln : C (K ) 7! Tn is a linear operator. We say that Ln preserves the elements of Tn if Ln (t) = t for every t 2 Tn : A anoni al example for su h a linear operator Ln is the Fourier operator Sn : The purpose of the remaining part of the exer ise is to show that the Fourier operator Sn is extremal among linear operators preserving the elements of Tn in the sense that it has the smallest norm. This leads to the result of Faber, Nikolaev, and Lozinskii (see part g℄) that for arbitrary linear operators Ln preserving the elements of Tn ; n = 1; 2; : : : , the sequen e (Ln (f ))1

annot onverge for every n f 2 C (K ): e℄ Berman's Generalization of a Formula of Faber and Mar inkiewi z. Let fa denote the a-translation of a fun tion f 2 C (K ); that is, fa (x) := f (x + a): Suppose Ln is a linear operator preserving Tn : Show that =1

1 2 for every f Hint: Let

Z 



Ln (ft )(x

t) dt = Sn (f )(x)

2 C (K ) and x 2 K: Z

1  An (x) := L (f )(x t) dt : 2  n t Show that An (f ) = Sn (f ) for every f 2 Tn : Prove that An (f ) = Sn (f ) for every f of the form f (x) = os mx or f (x) = sin mx; where m is an integer greater than n. Con lude that An (f ) = Sn (f ) for every f 2 T := [1 n Tn : Note that T is dense in C (K ): This means that to omplete the proof, it is suÆ ient to show that An : C (K ) ! Tn and Sn : C (K ) ! Tn are

ontinuous. Observe that kAn k  kLn k and kSn k  log n for some > 0; see also part f℄. ut f℄ The Norm of the Fourier Operator Sn . Show that =0

kSn k := sup



kSn (f )kK kf kK



1 : f 2 C (K ) = 2


Z  1 sin n + 2 t 1  2 sin 2 t

dt :

Use this to prove that there exist two onstants > 0 and > 0 independent of n su h that 1

log n  kSnk  log n ;

n = 2; 3; : : : :

A tually, it an be proved that kSnk = 4 log n + O(1) ;

n = 2; 3; : : : :

1

2



2

See, for example, Lorentz [86a℄.

2

4.1 Variations on the Weierstrass Theorem 167

g℄ The Norm of Operators that Preserve Trigonometri Polynomials. Let Ln : C (K ) ! Tn be a linear operator preserving the elements of Tn . Show that kLnk  kSn k  log n ; 1

where > 0 is a onstant independent of n: Hint: Use parts e℄ and f℄. 1

ut

E.6 Polynomials in xn . Given n 2 N and n 2 R; let

Pn (n ) := fpn(xn ) : pn 2 Pn g:  1 for all n 2 N : Then [1 n Pn (n ) is dense in

Suppose Æ 2 (0; 1) and n C [Æ; 1℄ if and only if

=1

lim sup n!1

log n

n

 12 log 1Æ :

To prove the above statement, pro eed as follows (see also Borwein [91b℄). Denote the Chebyshev polynomial for Pn (n ) on [Æ; 1℄ by Tn;Æ : Denote the zeros of Tn;Æ in [Æ; 1℄ by Æ : x Æ;n < x Æ;n <


< xn;n ( )

( )

1

2

( )

Let x Æ;n := Æ and xnÆ ;n := 1: Let ( )

( )

+1

0



Mn (Æ) := max in 1

+1

Æ xi;n

xi Æ

( )



( ) 1

;n

:

a℄ Show that

Tn;Æ (x) = Tn



2

Æn

1

1 + Æ n 1 Æ n

xn



x 2 [Æ; 1℄ ;

;

where Tn is the Chebyshev polynomial of degree n as de ned by (2.1.1). b℄ Let Æ := lim inf x ;n : Show that if n!1 (0) 1

lim inf n!1

 2

max

in



+1

xi;n

(0)

xi

(0) 1



;n

= 0;

then lim inf Mn (Æ ) = 0: n!1

Hint: Count the zeros of Tn;Æ

Tn;

0

2 Pn (n ) in [Æ; 1℄:

ut

168 4. Denseness Questions

℄ Let Æ := lim inf x ;n ; as in part b℄. Show that n!1 1 1 log n log = lim sup 2 Æ n!1 n whenever the right-hand side is nite. Hint: Use the expli it formula for Tn; given in part a℄. d℄ Let Æ be de ned by 1 log n 1 log = lim sup : 2 Æ n!1 n Suppose Æ > 0: Show that lim inf (x ;n x ;n ) = 0 : (0) 1

ut

0

n!1

(0)

(0)

2

1

Hint: Use parts a℄ and ℄. e℄ Let Æ := lim inf x ;n ; as in parts b℄ and ℄. Suppose Æ > 0: Show that n!1

ut

(0) 1

xi;n

(0)

xi

(0) 1



2 ;n  Æ

x

x

(0) 2

(0)

;n

1



;n

for every suÆ iently large n 2 N and for every i = 2; 3; : : : ; n + 1: Hint: Count the zeros of Tn; (x) Tn; (i;n x) 2 Pn (n ) ; where x 2 i;n := i ;n < 0

0

(0)

1

x

Æ

(0) 1

;n

for every suÆ iently large n 2 N and for every i = 2; 3; : : : ; n + 1: f℄ Let Æ be de ned by 1 1 log n log = lim sup : 2 Æ n n Suppose Æ > 0: Show that [1 n Pn (n ) is dense in C [Æ; 1℄: Hint: By parts a℄ to e℄, lim inf Mn (Æ) = 0: Now apply Theorem 4.1.2. =1

n!1

ut

ut

g℄ Let 0  y < Æ , where Æ is as in part f℄. Show that [1 n Pn (n ) is not dense in C [y; 1℄: Hint: Show that there exists a onstant depending only on Æ (and not on n or y) su h that jp(y)j  kpk Æ; 1 for every p 2 [n Pn (n ) and y 2 [0; Æ ℄: Now use E.4 ℄ of Se tion 3.3 and part a℄. ut E.10 of Se tion 6.2 extends part g℄ of the above exer ise. Namely, if 0  Æe < Æ; where Æ is the same as in part f℄ and A  [0; 1℄ is a set of e then [1 Pn (n ) is not dense in C (A): Lebesgue measure at least 1 Æ; n =1

[

=1

=1

1℄

4.1 Variations on the Weierstrass Theorem 169

E.7 The Weierstrass Theorem in Lp . Let [a; b℄ be a nite interval and p 2 (0; 1). Show that both C [a; b℄ \ Lp [a; b℄ and the set P := [1 n Pn of all real algebrai polynomials are dense in Lp [a; b℄: Hint: The proof of the rst statement is a routine measure theoreti argument; see Rudin [87℄. The se ond statement follows from the rst and the Weierstrass approximation theorem; see E.1. ut =0

E.8 Density of Polynomials with Integer CoeÆ ients. a℄ Suppose f 2 C [0; 1℄ and f (0) and f (1) are integers. Show that for every  > 0 there is a polynomial p with integer oeÆ ients su h that

kf

pk

< :

;

[0 1℄

Outline. By E.4, there is an integer n > 2= so that

kf Let

Ben (f )

:=

n X

Bn (f )k

;

[0 1℄

 < : 2

    

k n

f

k=0

n k

xk (1 x)n

k:

Show that if x 2 [0; 1℄; then

0  Bn (f )(x)

Ben (f )(x) 

n   X n

 n1

k=0

k

nX1 k=1

xk (1 x)n

k= 1

xk (1 x)n

n

k

 < : 2

Note that Ben (f ) is a polynomial with integer oeÆ ients, and

kf

Ben (f )k

;

[0 1℄

 kf <



2

Bn (f )k  + = :

; +

[0 1℄

2

kBn (f )

Be (f )k

;

[0 1℄

ut

b℄ Suppose the interval [a; b℄ does not ontain an integer. Show that polynomials with integer oeÆ ients form a dense set in C [a; b℄: Proof 1. This is an immediate onsequen e of part a℄. ut Proof 2. Assume, without loss of generality, that [a; b℄  (0; 1): By the Weierstrass approximation theorem, it is suÆ ient to prove that for every  > 0 there is a polynomial p with integer oeÆ ients su h that

170 4. Denseness Questions

1

p a;b <  ;

2

[

℄

sin e then all real numbers, and hen e all p 2 Pn , an be approximated by polynomials with integer oeÆ ients. The existen e of su h a polynomial p follows from the identity 1 = 2 1

(1

x

1

x))

2(1

=

1 X

k=0

x)(1

(1

x))k ;

2(1

where the in nite sum onverges uniformly on [a; b℄  (0; 1):

ut

E.9 Weierstrass Theorem on Ar s. Let D denote the unit ir le of the

omplex plane.

a℄ Show that the set P := [1 n Pn of all polynomials with omplex oef ients is not dense in C (D): Hint: Use the orthonormality of the system ((2) = ein )1 n 1 on [ ;  ℄ to show that if k is a positive integer and p 2 P ; then =0

1 2

=

2 kz k

p(z )kC D (

)

 ke

ik

p(ei )kL2

[

;℄

 2 :

So none of the fun tions z ; z ; : : : is in the uniform losure of P on D. ut

b℄ Let A  D be an ar of length less than 2: Then the set P of all polynomials with omplex oeÆ ients is dense in C (A). This is a spe ial ase of Mergelyan's theorem (see, for example, Rudin [87℄). 1

2

Proof. Without loss of generality, we may assume that A is symmetri with respe t to the real line. By E.5 d℄, it is suÆ ient to prove that f (z ) := z is in the uniform losure P of P on D (this already implies that ea h z k , k 2 Z, is in P ). By E.11 j℄ of Se tion 2.1, ap(A) < 1: By E.11 g℄ of Se tion 2.1, (A) = ap(A); where (A) denotes the Chebyshev onstant of A: Hen e 0  (A) <  < 1 with some . Re alling the de nition of (A), we an dedu e that there are moni polynomials pn 2 Pn su h that 1

kpn kA  n ;

n = 1; 2; : : : :

For n = 1; 2; : : : , let

qn (z ) := z n pn (1=z ) = z 1

where rn

1

2P kz

Hen e f (z ) = z

n

1

1

1

+ rn (z ) ; 1

: Sin e A is symmetri with respe t to the real line,

+ rn (z )kA = kqn kA = kpn kA  n

1

1

is in P ; whi h nishes the proof.

!

n!1

0:

ut

4.2 Muntz's Theorem 171

4.2 Muntz's Theorem

A very attra tive variant of the Weierstrass theorem hara terizes exa tly when the linear span of a system of monomials

M := (x ; x ; : : : ) 1

0

is dense in C [0; 1℄ or L [0; 1℄: 2

Theorem 4.2.1 (Full Muntz Theorem in C [0; 1℄). Suppose (i )1 i is a sequen e of distin t positive numbers. Then =1

spanf1; x1 ; x2 ; : : : g

is dense in C [0; 1℄ if and only if 1  X i = 1: +  i 1 i 2

=1

Note that when inf i i > 0;

1 X

i

 +1 i=1 i 2

1 X

= 1 if and only if

1

 i=1 i

= 1:

Muntz studied only this ase, and his theorem is usually given in terms of the se ond ondition. When i  1 for ea h i = 1; 2; : : : ; the above theorem follows by a simple tri k from the L version of Muntz's theorem. The proof of the general ase is left as a guided exer ise. The diÆ ult ase to deal with is the one where 0 and 1 are both luster points of the sequen e (i )1 i ; see E.18. 2

=0

is a Theorem 4.2.2 (Full Muntz Theorem in L [0; 1℄). Suppose (i )1 i sequen e of distin t real numbers greater than 1=2. Then 2

=0

spanfx0 ; x1 ; : : : g

is dense in L [0; 1℄ if and only if 1 X 2i + 1 = 1: (2  i + 1) + 1 i 2

2

=0

The proof of the following full L version of Muntz's theorem is presented as E.19. 1

172 4. Denseness Questions

Theorem 4.2.3 (Full Muntz Theorem in L [0; 1℄). Suppose (i )1 is a i sequen e of distin t real numbers greater than 1: Then 1

=0

spanfx0 ; x1 ; : : : g

is dense in L [0; 1℄ if and only if 1 X i + 1 = 1: (  i + 1) + 1 i 1

2

=0

Now we formulate a general Muntz-type theorem in Lp [0; 1℄; that ontains the above C [0; 1℄, L [0; 1℄, and L [0; 1℄ results as spe ial ases. The proof of this theorem is outlined in E.20. 2

1

Theorem 4.2.4 (Full Muntz Theorem in Lp [0; 1℄). Let p 2 [1; 1): Suppose (i )1 i is a sequen e of distin t real numbers greater than 1=p: Then =0

spanfx0 ; x1 ; : : : g

is dense in Lp [0; 1℄ if and only if

1 X

i + p 1



i + p

i=0

1

2

+1

= 1:

The full version of Muntz's theorem for arbitrary distin t real exponents on an interval [a; b℄, 0 < a < b; is given in E.7 and E.9.

Proof of Theorem 4.2.1 assuming Theorem 4.2.2 and ea h i  1. We need the following two inequalities:

(4:2:1)

m x

n X i=0

ai



Z x





0

Z

1

xi =

 

0

mtm

m mt

0 Z 

1

0

1

1

m mt

!

n X

ti

ai i i=0 n X ai i ti i=0

1

n X i=0

1

1



dt

dt

2  1 i ai  i t

11=2

dtA

for every x 2 [0; 1℄ and m = 1; 2; : : : , and (4:2:2)

0 Z  0

1

m t

n X i=0

2 ai ti

11=2

dtA





m

x

n X i=0

ai xi

;

[0 1℄

4.2 Muntz's Theorem 173

for every m = 0; 1; 2; : : : . The assumption that i that

1 X i=0

i

= 1 if and only if

i + 1 2

1 X i=0

 1 for ea h i implies

2(i 1) + 1 =1 (2(i 1) + 1) + 1 2

and

1 X

i

1 X

= 1 if and only if

 +1 i=0 i 2

i=0

2i + 1 = 1: (2i + 1) + 1 2

If 1 i i =(i + 1) = 1; then (4.2.1), together with Theorem 4.2.2 and the Weierstrass approximation theorem (see E.1 of Se tion 4.1), shows that spanf1; x1 ; x2 ; : : : g P

2

=0

is dense in C [0; 1℄: If the above span is dense in C [0; 1℄; then (4.2.2), together with E.7 of Se tion P 4.1, shows that it is also dense in L [0; 1℄: Hen e Theorem 4.2.2 implies 1 ut i i =(i + 1) = 1: 2

2

=1

Proof of Theorem 4.2.2. We onsider the approximation to xm by elements of spanfx0 ; : : : ; xn 1 g in L [0; 1℄; and we assume m > and m 6= i for any i: In the notation of Se tion 3.4 we de ne 1

2

2

 := ( ;  ; : : : ; n ; m) 0

1

1

and study Ln , the nth orthonormal Muntz-Legendre polynomial asso iated with : By (3.4.8) and (3.4.6) we have (with n := m) n

X Ln (x) = an xm + ai xi ; 1

i=0

where

j an j =

p

1 + 2m

nY1 i=0

m + i + 1 m 

i

:

It follows from kLn kL2 ; = 1 and orthogonality that Ln =an is the error term in the best L [0; 1℄ approximation to xm from spanfx0 ; : : : ; xn 1 g (why?). Therefore [0 1℄

2

min

xm bi 2C

nX1 i=0

bi xi





nY 1 m i : =p = j a j m +  + 1 1 + 2m i n i L2 ;

1

[0 1℄

1

=0

174 4. Denseness Questions

So, for a nonnegative integer m dierent from any of the exponents i ,

xm 2 spanfx0 ; x1 ; : : : g

(4:2:3)

(where span denotes the L [0; 1℄ losure of the span) if and only if 2

lim

n!1

m i = m +  + 1

nY1 i=0

i

0:

That is, (4.2.3) holds if and only if lim

n!1

nY1 i=0 i >m

1



2m + 1 m + i + 1

nY1 i=0 = <i m



1

2i + 1 = 0: m + i + 1

1 2

Hen e (4.2.3) holds if and only if either

1 X i=0 i >m

1 = 1 or 2i + 1

1 X i=0 1=2<i m

(2i + 1) = 1 ;

whi h is the ase if and only if

1 X i=0

2i + 1 = 1; (2i + 1) + 1 2

and the proof an be nished by the Weierstrass approximation theorem (see E.1 of Se tion 4.1). ut Comments, Exer ises, and Examples. Theorem 4.2.1 (in the ase when inf fi : i 2 N g > 0) and Theorem 4.2.2 were proved independently by Muntz [14℄ and Szasz [16℄. Szasz [16℄ proved the full version of Theorem 4.2.2. Theorem 4.2.1 is to be found in Borwein and Erdelyi [to appear 5℄. Mu h of Theorem 4.2.4 is stated in S hwartz [59℄ without proof and may be dedu ed by his methods. Indeed, S hwartz's method appears to give Theorem 4.2.4 for p 2 [1; 2℄. Johnson (private ommuni ation) and Operstein [to appear℄ show how to derive the full Theorem 4.2.4 from Theorem 4.2.1 as does E.20; see also E.7 of the next se tion. Less omplete versions of the results presented in this se tion are often

alled the Muntz-Szasz Theorems. A 1912 version due to Bernstein an be found in his olle ted works. A variant on our proof of Muntz's theorem is presented in E.2. A distin t proof based on possible zero sets of analyti fun tions may be found in Feinerman and Newman [76℄; see also E.10, where this method is explored for denseness questions for f os k g:

4.2 Muntz's Theorem 175

Extensions of Muntz's theorem abound. For example, generalizations to omplex exponents are onsidered in Luxemburg and Korevaar [71℄, to angular regions in Anderson [72℄ and with an exponential weight on [0; 1) in Fu hs [46℄. It is a nontrivial problem to establish a Muntz-type theorem on an interval [a; b℄, a > 0, in whi h ase the elements of the sequen e are allowed to be arbitrary distin t real numbers. This is the  = (i )1 i

ontent of E.7; it is due to Clarkson and Erd}os [43℄ (in the ase when ea h i is a nonnegative integer) and S hwartz [59℄ (in the general ase). It is shown in Se tion 6.2 that if  = (i )1 is an in reasing sequen e i of nonnegative real numbers, then the interval [0; 1℄ in Muntz's theorem (Theorem 4.2.1) an be repla ed by an arbitrary ompa t set A  [0; 1) of positive Lebesgue measure. The exer ises also explore in detail the losure of Muntz spa es in the nondense ases. This study was initiated by Clarkson and Erd}os [43℄, who treated the ase when the exponents are nonnegative integers. The

onsiderably harder general ase is due to S hwartz [59℄. Denseness questions about