Polynomials and Polynomial Inequalities

For Theresa, Sophie, Alexandra, Jennifer, and Erika

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.

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

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 4 b b + 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

3

4

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

r

; = while

a

2

b+y;

4

by + (a + 4d)y

3

2

3a 4

2

r

R

2

2b 

2

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

1 X 2 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 ). tu

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

5). where  := (1 + 5) and  := (1 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 is isomorphi to the Eu lidean ve tor spa e C n , so these norms are also equivalent to ea h other. Let +1

+1

pn (x) :=

n X k=0

ak 2 R :

ak 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

)

1=p

j

p(t) p dt

kpkl1 := max fjak jg 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 n so that 1

kp0 k

i=0

3

 C kpk

1 1℄

;

;

jai j  C kpk

1 1℄

;

;

;

[

1 1℄

n X

2

1

[

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 6 p2Pn kpk sup

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;i= 6 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) p(x) = Hen e

kf

(

+1)

( ) ! (x) :

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

pk a;b [

1 fn (n + 1)! (

℄

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

 

x : k


k p(0)

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 jak 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

< jan jxn

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

jak 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

1

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

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  jz j  k max =: r : r := min ;::: ;n ak k ;::: ;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 ;

= 1. Then the poly-

an 6= 0

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

j j j j

aj p an p

!q=p 91=q = ;

:

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 Z

p0 (z ) dz jzj r p(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 ; z 2U

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

1 dz = 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 g0 (z ) 1 f 0(z ) 1 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

jxp

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

jak jp

!1=p

k=1

jbk jq

!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

k=0

8 <

 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 ja 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 jrj

 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 ja 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 g 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

k

n

1

1

1

 p2n +41

1

0

1

2

0

=1

[0 4℄

0

[0 1℄


; 2n n+m

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. 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 p0 (z ) X 1 = ; 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 n m nr + mr

:= ; 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

1

2

3

1

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 r0 (z ) p0 (z ) n = + 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

2

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

0

2

mq(z ) q0 (z ) 0

0

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

 

n k x ; k

an 6= 0

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

1

n! ( 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

ak 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 Æk 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

n   X n k=0

q(x) :=

Z x Z tm 0

0

1

1

Z tm 0

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






:

p(t) dt : +1

2

k k+1

0

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

xk

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  rg ;

j

r :=

j

2

ot

 : n

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

n

k+1

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

=0

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

4

5

(z + 1)n

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

 

a + 0

n

ut

 

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


+ b zn ; 0 1 n n

f (z ) :=

n

+


:

a z +


+

0

1

and h(z ) :=

 

n 0

ab + 0

0

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

p(x) :=

n Y

i=1

i ) ;  <  <


<  n ; n  2

(x

1

2

with only real zeros, let


(p) := min (i in 1

1

i ) :

+1

By Rolle's theorem

p0 (x) = n

nY1 i=1

i ) ;  < <  < <


< n

(x

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 ; i j : Then the above an be rewritten as

ei := j

1

+

n j

j 1 X i=1

(di

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

vj )

= 0:

De ne

F (u; v) :=

j 1 X i=1

n j

(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)

+

nXj i=1

(ei

+1



!

1 v)(ei

v)

nXj 

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

2 C so that  


n  )(  )


(n

Then p has at least one zero z

jz j 

 2

1

aa = 6 0: 0

1

0

2

(

0

ak 2 C ;

ak z k ;

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 ) = a

0

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 ) =  a z n n 0

1

1

+

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

n= bX 2

2 k

p

1)n + (x

x

2

( 1)k

=0

p

x 2 [ 1; 1℄ ;



x2C;

1)n ;

x

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

;

[

1 1℄

= ( 1)n j ;

0

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

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

[

= k2 n Tn k 1

;

1 1℄

[

;

1 1℄

=2 n; 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

h(x) := 2

q(x)k

[

1

nT

n (x)

;

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℄ ;

jyj < 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

k (x2

2

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 xk :

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 Tn k ; : 1

[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

x a

a : 2a 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

=

1 2

= n(1

s ( )) 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

0

+ a z +


+ an z n kD 1

 max ja j kn k 0

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

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

(2

j

i=(2n) + e

1)

(2

j

1)

i=(2n) ;

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

n Y

Fh (x) :=

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 sin n 1 ; 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

1


n+1

p

2 x e℄ Orthogonality. Show that 2



Z1

p

x

p


n+1

x

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

k (x2

2

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

i;jn i6 j

jzi

zj j :

1

=

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

+1

jzi

i;jn i6 j

(E )) n+1 :

zj j 
n (E ) +1

1

=

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

jzi

1

+1

=

jzk

in i6 k

+1

=

in i6 k zi j :

1

Y


n (E )

Y

zj j

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)

(

p 2 P : p(z ) =

n

(

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

n and

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 ))

=n

1

exist.

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 (E )) +1

=

1 2

= abs

1 1 . .. 1

: : : zn : : : zn

z z

1

2

.. .

..

2

.

p(z ) p(z )

1

1

1

1

.. .

.. .

: : : znn

zn

p(zn

1 +1

+1

2

+1

)

:

Expanding the above determinant with respe t to its last olumn, we obtain (
n (E ))

 (
n (E ))

=

1 2

+1

=

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

Then

2


1=n

dnn

+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

=

1 2

=

Z

jf (x)j

2

X

1=2

d(x)

is nite. The equivalen e lasses are de ned by the equivalen e relation f  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. 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. 1 2

2

2

2

2

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 ; 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 kkgk + 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

!1=2

jgj

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

!1=2

jf j

2

d(x)

Z b

a

!1=2

jgj

2

d

:

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 1 2

is an orthonormal olle tion of ve tors in V: Let f n

X

min

i 2C

i=1

i fi

2 V: Then

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 fi

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

0

(

max

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 n X

jq(x )j  0

2

i=0

However, if

j i j

i = so

!

n X

2

i=0

jpi (x )j 0

pi (x ) 0

P

n j =0

!

2

jpj (x )j

2

0

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

(

m) (x

0

)j 

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

1=p

p d

;

)

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

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

kf + gkLp   kf kLp  (

)

(

)

+ kg kLp  (

)

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

sign(f (x)) ;

1

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

1=p

X X

 2 =p 1

(jf j + jg j)p d

jf j

p d + Z

1

X

Z

X

jgj

1=p

p d

jf jp d

1=p

+

Z

X

1=p !

jgjp d

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

(

1 X

:= x = (xk )1 k 1 : xk 2 C ;

`

2

=

and

1 X

kxk` := 2

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

k

2

= rk  k 2

 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

g(x) := sign(f (x)) := we obtain

Z 



8 > < > :

f (x)

if f (x) 6= 0

0

if f (x) = 0 ;

jf (x)j

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

itx

f (x)e

itx d(x)

f (x)e

2 R; we have

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

)

0

=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 )k

(t i)k

1Z X k=0 Z

k!

jt

0

t

k!

R

0



it0 x xk

f (x)e



d(x)

jk jf (x)jjxjk d(x)

jf (x)jejt t jjxj d(x) 0

R

Z

t

0

jf (x)j

2

R

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

1 2

on [ 1; 1℄ :

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

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

x)

=

2 1 2

on [ 1; 1℄ :

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 ; rn

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

is an inner produ t on Pn . ut 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 sequen e (qn )1 is alled a sequen e of orthogonal polynomials asso iated 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

dn

dn

1

hpn (x); n

1 n

dn

+1

n

1

dn dn

1

xn i 1

+1

Hen e

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) 1 a = (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) p0n (x)pn (x) : +1

Proof. Let y ! x in Theorem 2.3.3.

+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

Let (Kn (x ; x))1 be the sequen e of asso iated kernel fun tions (as in 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! n

n + 1)

 

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

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

=0







+n = n m m   n+ = F n =0

2





n+ (x n m

1)n m (x + 1)m

+ +n+m m

1



x 1

m

2

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) Cn Pn; (x) ; (

)

(

(

)

)

1

+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 dx

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

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

Ln (x) := (

)

+

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

jan 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

jbn 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 0

(3) ) (1). If Z Z

x

[ n

2

R

n X1

2

1

jak j + max jbk j k n 0

1

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

n

2

Z 2

R

d(x) ;

whi h implies supp()  [ K; K ℄.

ut

E.10 A Condition for supp()  [0; 1). Let (pn )1 be the sequen e 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 P = nk=0 ak xk .

for every p 2 Pn of the form p(x) 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

ak 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

 p (x) := .

n

f℄ Show that



1

..

1



2

.. .

n n

+1

(pn q) = 0 ;

::: ..

n .. .

.

::: 

1

n

2

1

.. . n

x

:

x

q 2 Pn : 1

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 (n )1 that onverges pointwise to a nonde reasing real-valued fun tion 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 dnk (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

jaj jbj



 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 Pn : 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 xk ; 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 + 

;n ;

 = 2; 3; : : : ; n :

1

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 ..

.

:::



;

n = 0; 1; : : : :

n

2

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 .. .

.

::: 

1

n

2

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

1

a b a

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

jbn

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 lim a n!1 n

Then where F

 R n [b

=:

a 2

>0

and

lim b n!1 n

=: b 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

a

 w(t) dt  = + +


x t x x 0

1

2

=

B

1

x A

B

2

B

1

x

A

2

x A

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



p

max

x2[

;

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

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

2

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

n+1

Pn (x) : = x =x

+ n+1 +

n

2

2

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

n+1



2

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

+

n+1

2

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 of (Æn )1 . Sin e Æ = 2 + Æ n n ; we have 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 ) p  p p 2 (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

aj (1

2(

x)j (1 + x)d

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 1

p (z ) =

1



n Y

j =1 jarg(zj )j>

where arg(z ) is de ned so that arg(z ) negative oeÆ ients.

2[

1



z ; zj

; ): Then p (z ) has all non-

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



exp i x 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 x)j (1 + x)d j ;

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



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 ) 2n n = > : (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

y ) 2

(1

2





n

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

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



2

n

n 16d 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)

x ) + 4(x + 1)

2(1

2

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

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 0

0

0

that

p(xi ) = yi ;

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

℄ 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 A. 0

0

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 Sm O i and Æ 2 (0; Æ ); i 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

=0

;::: ;n+1

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 f (x) : : : fn(x) f (x) f (z ) f (z ) : : : fn (z ) p (x) = .. .. .. .. . . . . f (zn ) f (zn ) : : : fn (zn ) 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

0

1

1

0

1

1

1

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 ;

Æi = 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

1

xi1 : : : xim x : : : xm

x

0



1

=

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 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 and



D () := D =

2 1 2

2

0

0

1

1

1

ei0 t x

ei1 t : : : x :::

eim t xm

1

1

1

2

2

2

0

1

1

0

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

Hint: Multiply both sides above by

(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 1

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


> b. Show that 1

ut

2





1 1

;

1

x 

2

x

;:::



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


< a: Show that 1

2



1

x 

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

p q = aj fj

bj f j +

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): 0

0

Now observe that

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 ): P P b℄ Let  > 0. Let p(x) = nk 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

1

+1

=0

=0

0

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 := 


+2

n X i=1

ai fi

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 fk ;

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, Tn(i ) = ( 1)i ; i = 0; 1; : : : ; n : Note that Tn p has at least n zeros in [ ; n ℄; one in ea h [i ; i ℄; 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 p0 (b) > Tn0 (b): ut [

℄

0

1

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, 0

1

ut

whi h is impossible.

Theorem 3.3.3. Suppose (f ; f ; : : : ) is a Markov system on [a; b℄ with as0

1

so iated Chebyshev polynomials

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;Pb℄ 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  i for ea h i with stri t inequality for at least one

a = 0. Suppose i 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 0 (a)j > jT 0 (a)j : jTn; n;

 n; su h that

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

[

[

2 (b; 1) :

x

0

℄

0

℄

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

j m j  2

n





2n 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 1)m 2 n

dm;n = ( If j m j > dm;n kpk

[

;

1 1℄

n


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

=2

m;

n





2n : 2m

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. of omplex numbers, an element of Given a sequen e  := (i )1 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 t + k + 1 xt dt 1 ; 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 ;j 6 k k j ) 1

Qn

=0

=

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 n is the ontent of the main theorem of 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

(x)xm dx =

Z

nY 1 t + k + 1 2i jtj R k t k (t

dt n )(t + m + 1)

1

=

n

=0

Æm;n

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

1

0

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

j =0

xk ;j 6 k (k =

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

1 0 n (x)) = 2i

n L

Z

Qn

k=0 (t + k

On multiplying both sides by

xn

+

n

1 +1

2

Qn

k=0 (t xn +n 1

+ 1) (t + n k )

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

n L

(x

n (x))

0 = (xn

1 +1

Ln (x))0 : 1

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

a℄ xL0n (x) = n Ln (x) +

nX1 k=0

 b℄ xL0 n (x) = n Ln (x) +

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

q

n + n + 1

and

℄ xL00n (x) = (n

1)L0n (x) +

nX1 k=0

nX1 q

k + k + 1 Lk (x) ;

k=0

(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)L0 (1) + (k + k + 1)L0 (1) :

℄ L00n (1) = (n

n

1

k

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 1 X n  = 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

n X

Ln (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 =

1

Z

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

1+

℄ If 0 =  <  <


; then 0

1

pn (0) = ( 1)n and (

 f (x))0 :

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

1)n p

n

n Y j =1

j

1

is stri tly de reasing on [0; 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.

E.5 On the Maximum of Ln on [0; 1℄. If  = (i )1 is a sequen e of 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 i be as in (3.4.4), and let 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

1

0

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

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

1g ;

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 gnk 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;n ( )

=

Qn

i k;i

k;n x ); ( )

j =0 (k;i + j Qn j =0;j 6=k (k;i 1

+ 1) 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  2 (0; 1) and a subsequen e (k;ij )1 of (k;i )1 su h that the Muntzj 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

1

2

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

in

0

 min j j n 0

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;

sin(k + )y + sin y : 2 sin y 1

1 2

2

1 2

p

jg(x)j  p1 2 x :

Let f (x) := g ( nx

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


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

Qn

=0

2

=

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

2

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

1

; ::: ;

x 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 

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

os  a os  a

os  an

(3:5:6)

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

(3:5:7)

j k j < 1 ;

ak = ( k + k ) ; 1

1

2

that is, (3:5:8)

k := ak

Note that



q

ak + ak 2

p

q

ak



ak

1

j k j < 1 :

1;

2

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

!1=2

k )(z

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 ) : z n 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

1

z

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

1

that

(3:5:15)

Ten () = Tn ( os ) = =

e

2 Pn and p 2 Pn 2

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 (3:5:16)

Uen () = Un ( os ) sin  = =

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 fn = 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 (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

A ;n A +


+ n;n x a x an

;n +

1

0

1

and (3:5:20)

where

Un (x) = A

;n

0

Ak;n =



and Bk;n =

2

1

1

2

( 1)n ( 2

=

k

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

;n

0



1 Mn (z ) z nMn (z ) + n z! 2 z Mn (z ) Mn (z )

= lim Tn (x) = lim x!1

(

=

1

1

0

1)n

2

( 1

1

1 2






n

1

+ 1

2




n )

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 Y n

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

11=2

n

Y (t j )(t j ) 1  A Tn (x) = 2i j (1 j t)(1 j t)

t

t

2

=1

x

2tx + 1

dt

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

=1

p

k = ak

1

1

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 1

2




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

1

0




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

Rn (x)Rm (x) p Z

1

1

dx

1 x




n ) + n ( 1)n ( 1

2

dx

2

Rn (x)

2

1

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

x

2

1. Also 1

This implies that Z

1

x

1

= 0;

2

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

p

q n (z )

Mn (z ) := and

ut

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

(ei )) =

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



1 Mn (ei ) Mn (ei ) + 2 Mn (ei ) Mn (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,

Ten0 = Ben Uen ; and

℄ Show that

Uen0 = Ben Ten ;

(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

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

!1=2

(z

ak ) !1=2

(z

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) x!1 n

1

= 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) x!1 n

= 0:

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,

Tn0 = Bn Un ; and

℄ Show that

Un0 = Bn Tn ;

(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

jaj 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

;n +

0

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 jxi in 1

xi

+1

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 ) Sn (d) Sn ( )  2n : 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  n = 1; 2; : : : Mn  ; 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 n!1 n

: [a; b℄) = 0 ;

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

j j:[

lim E ( x n!1 n

1; 1℄) = 0 :

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,

z=1

1

jxj =

p

x = 2

p

1

2

(1

3

x) 2

1 1 1
3 (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 =1

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

gj) :

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 2 C [0; 1℄ is de ned by

f

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 1℄

=0

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 f (t + x) +

os kt dt : 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 k + t sin nt = 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 1 d℄ The set T := [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

 21 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 ( )

( )

0

+1



Mn (Æ) := max in 1

+1

Æ xi;n

xiÆ

( )



( ) 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)

;n

1

= 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 1 log n 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)

(0)

;n

2



;n

1

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 argu=0

ment; see Rudin [87℄. The se ond statement follows from the rst and the Weierstrass approximation theorem; see E.1. ut

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 )

:=

Bn (f )k

;

[0 1℄

 < : 2

n      X 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

xk (1 x)n

k

k

nX1 k=1

=

xk (1 x)n

1

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 ; : : : ) 0

1

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 se=1

quen e of distin t positive numbers. Then

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

i

2

+1

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

Theorem 4.2.2 (Full Muntz Theorem in L [0; 1℄). Suppose (i )1 is a 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 

1

0

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=0 i

i

2

1 X

= 1 if and only if

+1

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)

Ln (x) = an xm + where

jan j =

p

1 + 2m

nX1 i=0

ai xi ;

m + i + 1 m 

nY1 i=0

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

nY1 i=0

m i = m +  + 1

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  = (i )1 are allowed to be arbitrary distin t real numbers. This is the 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 quotients and produ ts of Muntz polynomials from a given Muntz spa e are dis ussed in Se tions 4.4 and 6.2, respe tively. Some of the literature on the multivariate versions of Muntz's theorem an be found in Ogawa and Kitahara [87℄, Bloom [90℄, and Kroo and Szabados [94℄. =0

=0

E.1 Another Proof of Some Cases of Muntz's Theorem. P a℄ Golits kek's Proof of Muntz's Theorem when 1 i 1=i = 1. Sup1 pose that (  ) is a sequen e of distin t, positive real numbers satisfying i i P1 1 = = 1 . Golits hek [83℄ gives the following simple argument to i i show that spanf1; x1 ; x2 ; : : : g is dense in C [0; 1℄: Proof. Assume that m 6= k ; k = 0; 1; : : : , and de ne the fun tions Qn indu tively: Q (x) := xm and =1

=1

=1

0

m)xn

Qn(x) := (n

Z

1

Qn (t)t 1

x

1

n dt ;

n = 1; 2; : : : :

Show, by indu tion on n, that ea h Qn is of the form

Qn

(x) = xm

Show also that

kQ k 0

so

;

[0 1℄

=1

n X i=0

and

kQn k

;

[0 1℄



n Y 1 i=0

an;i 2 R :

an;i xi ;

kQnk

;

[0 1℄





m !0 i

1



m
kQn n

1

k

;

[0 1℄

;

as n ! 1 :

ut

176 4. Denseness Questions

b℄ Another Proof of Muntz's Theorem when i ! > 0. Suppose that (i )1 is a sequen e of distin t positive real numbers that onverges to i

> 0. Show, without using the arguments given in the proof of Muntz's theorem, that spanf1; x1 ; x2 ; : : : g is dense in C [a; b℄, a > 0. Hint: Let k be a nonnegative integer. Use divided dieren es to approximate x logk x uniformly on [a; b℄. Finish the proof by using the Weierstrass approximation theorem (see E.1 of Se tion 4.1). ut =1

E.2 Another Proof of Muntz's Theorem in L [0; 1℄. 2

a℄ Gram's Lemma. Let (V; h
;
i) be an inner produ t spa e, and let g 2 V:

Suppose ff ; : : : ; fng is a basis for an n-dimensional subspa e P of V: Then the distan e dn from g to P is given by 1

dn := inf fhg

p; g

pi

=



1 2

G(f ; f ; : : : ; fn ; g) : p 2 Pg = G(f ; f ; : : : ; fn) 1

2

1

2

where G is the Gram determinant G(f ; f ; : : : ; fm ) := 1

2



hf ; f i 1

::: ...

1

.. . hfm ; f

1

1=2

;



hf ; fm i 1

.. : . : : : hfm ; fm i

i

Proof. As in Theorem 2.2.3, the best approximation to g from P is given

by

f =

n X i=1

i fi ;

where the i are uniquely determined by the orthogonality onditions

hf  Sin e

g; fk i = 0 ; dn = hg 2

k = 1; 2; : : : ; n : f ; g

f i ;

we are led to a system of n + 1 equations n X i=1

and

i hfi ; fk i = hg; fk i ; n X i=1

k = 1; 2; : : : ; n

i hfi ; gi + dn = hg; gi : 2

Solving this system by using Cramer's rule, we get the desired result.

ut

4.2 Muntz's Theorem 177

b℄ As in E.3 of Se tion 3.2, 1  + 1 1 .. . 1 n + 1

::: ..



Q

1 1 + n

.. .

.

:::

1

n + n



(j i<jmQ

=

i )( j

1

i;jn

1

i )

(i + j )

for arbitrary omplex numbers i and j with i + j 6= 0:

℄ Let ;  ; : : : ; n be distin t real numbers greater than 1=2: Then the L [0; 1℄ distan e dn from x to spanfx0 ; : : : ; xn g is given by 0

2

1 dn = p 2 + 1

n Y

i=0

+ i

+ 1

i

:

Hint: In L [0; 1℄; 2

hxa ; xb i =

Z

1

xa xb dx =

0

1

a; b 2 ( 1=2; 1) :

;

a+b+1

ut

Now apply parts a℄ and b℄. d℄ Complete the proof of Muntz's theorem in L [0; 1℄: 2

E.3 More on Muntz's Theorem in the Nondense Case. We assume throughout this exer ise that (i )1 i is a sequen e of nonnegative real numbers satisfying 1 1 X =0

<1

 i=1 i

and the gap ondition inffi

i

: i 2 Ng > 0

1

holds. Some of the results of this exer ise hold even if the above gap ondition is removed (see the later exer ises). a℄ Show that 1  +  Y m i 0< = exp( m m ) ; i=0 i6=m

i

m

where m ! 0 as m ! 1: Hint: First show that the above in nite produ t exists. Write the above produ t as

1 Y

i=0 i <m

i 

i



+ m

m

1 Y

i=0 i 2(m ;2m )

i 

i



+ m

m

and estimate the three fa tors above separately.

1 Y

i=0 i 2m

i 

i



+ m

m

ut

178 4. Denseness Questions

b℄ Dedu e that if i 6= m for ea h i; then



x m

p(x)

L2 [0;1℄

 p2

1

1 ( Y i (



+ ) (m + ) i + ) + (m + )

m + 1 i=0 = exp( m (m )) ; xn ; where m

1

1

2

2

1

1

2

2

for every p 2 spanfx0 ; : : : g ! 0 as m ! 1:

℄ Show that for every  > 0 there is a onstant  depending only on  and (i )1 i (but not on the number of terms in p) su h that =0

jai j  (1 + )i kpkL ; P for every p 2 spanfx ; x ; : : : g of the form p(x) = ni 2 [0 1℄

0

1

=0

Hint: Use part b℄.

ai xi :

ut

d℄ Bounded Bernstein-Type Inequality. Let  = 0 and   1. Show that for every  2 (0; 1) there is a onstant  depending only on  and (i )1 i (but not on the number of terms in p) su h that 0

=0

kp0 k

;

[0 1

and hen e

kp0k

;

[0 1

℄

  kpkL

℄

1

;

2 [0 1℄

  kpk

;

[0 1℄

for every p 2 spanf g: Hint: Use part ℄. ut The result of the next part is due to Clarkson and Erd}os [43℄. e℄ The Closure of a Nondense Muntz Spa e. Suppose f 2 C [0; 1℄ and there exist pn 2 spanfx0 ; x1 ; : : : g of the form

x0 ; x1 ; : : :

pn (x) = su h that lim kpn n!1

kn X i=0

ai;n xi ;

ai;n 2 R ;

n = 1; 2; : : :

f k ; = 0: Show that f is of the form 1 X f (x) = ai xi ; ai;n 2 R ; x 2 [0; 1) : [0 1℄

i=0

Show also that f an be extended analyti ally throughout the region

fz 2 C n( 1; 0℄ : jz j < 1g

and

lim a n!1 i;n

= ai ;

i = 0; 1; : : : :

If (i )1 is a sequen e of distin t nonnegative integers, then f an be i extended analyti ally throughout the open unit disk. Hint: Use part ℄. ut =0

4.2 Muntz's Theorem 179

If (i )1 is la unary (that is, inf fi =i : i 2 N g > 1), then the i uniform losure of spanfx0 ; x1 ; : : : g on [0; 1℄ is exa tly +1

=0

(

f

2 C [0; 1℄ : f (x) =

1 X i=0

)

ai

xi ;

x 2 [0; 1℄ :

If (i )1 i is not la unary, then this fails, namely, there exists a fun tion f of the form 1 =0

f (x) =

X

i=0

x 2 [0; 1)

ai xi ;

in the uniform losure of spanfx0 ; x1 ; : : : g on [0; 1℄ su h that the righthand side does not onverge at the endpoint 1; see Clarkson and Erd}os [43℄. f℄ Bounded Chebyshev-Type Inequality. Show that for every  2 (0; 1) there exists a onstant  depending only on  and (i )1 i (but not on the number of terms in p) su h that =0

kpk

;

[0 1℄

 kpk

[1

;1℄

for every p 2 spanfx0 ; xi ; : : : g : Outline. Using the s aling x ! x =1 , without loss of generality we may assume that  = 1: Suppose there exists a sequen e 1

1

(pm )1 m

=1

 spanfx ; x ; : : : ; g ;

su h that

0

1

0 < Am := kpm k ;

[0 1℄

while

kpmk

[1

;1℄

= 1;

Let qm := pm =Am : Note that kqm k ; as m ! 1: Then, by part d℄,

[0 1℄

kqm0 k

;

[0 1

m = 1; 2; : : :

!1

m = 1; 2; : : : : = 1 for ea h m; and kqm k

[1

Æ℄

;1℄

!0

 Æ

for every Æ 2 (0; 1): Hen e (qm )1 is a sequen e of uniformly bounded m and equi ontinuous fun tions on losed subintervals of [0; 1); and by the Arzela-As oli theorem (see, for example, Rudin [87℄) we may extra t a uniformly onvergent subsequen e on [0; 1 =2℄: This subsequen e, by part e℄, onverges uniformly to a fun tion F analyti on (0; 1 =2); but sin e kqm k ; ! 0; F must be identi ally zero. This is a ontradi tion sin e kqm k ; = 1 and kqm k ;  = kqm k ; for every suÆ iently large m: ut =0

[1

1℄

[0 1℄

[0 1

℄

[0 1℄

g℄ Suppose (qm )1  spanfx0 ; x1 ; : : : g and kqmk a;b  1 for ea h m m; where 0  a < b: Show that there is a subsequen e of (qm )1 that m

onverges uniformly on every losed subinterval of [0; b): Hint: Use parts f℄ and d℄ and the Arzela-As oli theorem. ut =1

[

℄

=1

180 4. Denseness Questions

E.4 Muntz's Theorem with Real Exponents on [a; b℄; a > 0. Suppose (i )1 i 1 is a set of distin t real numbers satisfying =

1 X

1

<1 1 ji j

i= i 6=0

with i < 0 for i < 0 and i  0 for i  0. Suppose that the gap ondition inf fi

i

1

: i 2 Zg > 0

holds. Asso iated with

p(x) := let

p (x) :=

1 X

i= n

n X i= n

ai

ai xi ;

xi

and

n = 0; 1; : : :

p (x) := +

n X i=0

ai xi :

Let 0 < a < b: a℄ Show that there exists a onstant depending only on a; b; and (i )1 i 1 (but not on the number of terms in p) su h that =

kp k a;b  kpk a;b for every p 2 spanfxi g1 i 1: +

[

℄

[

℄

and

kp k a;b  kpk a;b [

℄

[

℄

=

Outline. It is suÆ ient to prove only the rst inequality; the se ond inequality follows from the rst by the substitution y = x : If the rst inequality i 1 fails to hold, then there exists a sequen e (pn )1 n  spanfx gi 1 su h 1

=1

that

kpn k a;b +

[

℄

= 1;

n = 1; 2; : : : ;

and

=

lim kpn k a;b = 0 :

n!1

[

℄

Sin e p = p + p ; the above relations imply that +

kpn k a;b  K < 1 ; [

℄

n = 1; 2; : : : :

1 By E.3 g℄ and E.3 e℄, there exists a subsequen e (ni )1 i su h that (pni )i

onverges uniformly on every losed subinterval of [0; b) to a fun tion f analyti on Db := fz 2 C n( 1; 0℄ : jz j < bg +

=1

of the form

f (z ) =

1 X i=0

ai z i ;

z 2 Db ;

=1

4.2 Muntz's Theorem 181

while (pni )1 i onverges uniformly on every losed subinterval of (a; 1) to a fun tion g analyti on =1

Ea := fz 2 C n (

1; 0℄ : jz j > ag

of the form

g(z ) =

1 X

1

i=

z 2 Ea ;

ai z i ;

xlim !1 g(x) = 0 : x2R

Now lim kpni k a;b = 0 and pni = pni + pni imply that f + g = 0 on (a; b): i!1 Show that  z Rez < log b h(z ) := f (ge(e)z;) ; Rez > log a [

+

℄

is a well-de ned bounded entire fun tion, and hen e h = 0 on C by Liouville's theorem. From this, dedu e that

f = 0 on [0; b)

g = 0 on (a; 1) :

and

Hen e, for every y 2 (a; b), lim kpni k a;y = 0

i!1

and

+

[

℄

lim kpni k y;b = lim kpni i!1 +

i!1

[

℄

Therefore

pni k y;b = 0 : [

℄

lim kpni k a;b = 0 ; +

i!1

[

℄

whi h ontradi ts kpn k a;b = 1; n = 1; 2; : : : : ut b℄ The Closure of Muntz Polynomials. Let f 2 C [0; 1℄, and suppose there exist Muntz polynomials pn 2 spanfxi g1 i 1 of the form +

[

℄

=

pn (x) = su h that lim kpn n!1

kn X i= kn

n = 1; 2; : : :

f k a;b = 0: Show that f is of the form [

f (x) = where

ai;n xi ;

℄

1 X i=

1

ai xi ;

x 2 (a; b) ;

182 4. Denseness Questions

f (x) := +

f (x) :=

1 X i=1

1 X

i=

x 2 [0; b) ;

ai xi ;

1

x 2 (a; 1) ;

ai xi ;

lim f (x) = 0 ;

x!1

f an be extended analyti ally throughout the region fz 2 C n ( 1; 0℄ : a < jz j < bg ; and

lim a n!1 i;n

i 2 Z:

= ai ;

ut

Hint: Use part a℄ and E.3 e℄.

E.5 Removing the GapP Conditions. Assume throughout this exer ise that 0   <  <


and 1 i 1=i < 1: a℄ Bounded Chebyshev-Type Inequality. Show that for every  2 (0; 1) there is a onstant  depending only on  and (i )1 (but not on the i number of terms in p) su h that 0

1

=1

=0

kpk

;

[0 1℄

 kpk

;1℄

[1

g: (This is the inequality of E.3 f℄ without for every p 2 spanf the gap ondition inf fi i : i 2 N g > 0:) Hint: Assume, without loss of generality, that  = 0: Observe that lim  =i = 1. Choose m 2 N su h that i > 2i whenever i > m: Dei!1 i ne := ( i )1 i by  fi ; ig ; i = 0; 1; : : : ; m

i := min i + i ; i = m + 1; m + 2; : : : : Then 1 1 X 0 = < <


; < 1;

i  i ; i = 0; 1; : : : x0 ; x1 ; : : :

1

0

=1

1

2

0

i=1 i

1

and inf f i i : i 2 N g > 0: Now use E.3 g℄ of Se tion 3.3 with [a; b℄ = [1 ; 1℄ and E.3 f℄ of this se tion. ut b℄ Bounded Bernstein-Type Inequality. Suppose  = 0 and   1: Prove that for every  2 (0; 1) there is a onstant  depending only on  and (i )1 i (but not on the number of terms in p) su h that 1

0

=0

kp0k

;

[0 1

℄

  kpk

;

[0 1℄

1

for every p 2 spanf g: (This is the se ond inequality of E.3 d℄ without the gap ondition inf fi i : i 2 N g > 0:) Hint: De ne the sequen e as in the hint given to part a℄. Now use E.3 f℄ of Se tion 3.3 with [a; b℄; a 2 (0; 1 ℄; E.3 g℄ of this se tion, and part a℄ of this exer ise. ut

x0 ; x1 ; : : :

1

4.2 Muntz's Theorem 183

℄ Let 0  a < b: Show that spanfx0 ; x1 ; : : : g is not dense in C [a; b℄: Hint: Use part a℄ and Theorem 4.2.1 (full Muntz's theorem in C [0; 1℄). d℄ Let

ut

 k := k + 2 k ; k = 1; 2; : : : ; 2 k a k := 2 ; k = 1; 2; : : : : k
P1 Show that the fun tion f (x) := i a i x2i 1 +Pa i x2i is a welli de ned ontinuous fun tion on [0; 1℄: Show also that 1 i ai x does not

onverge for any x 2 (0; 1) (hen e the on lusions of E.3 e℄ are not valid  a

2

k

2

1

1

:= k ; 2 := 2k ; 2

2

2

2

2

2

=1

1

2

=1

without a gap ondition).

E.6 A Comparison Theorem. Let 0  k  n be xed integers. Assume

 <  <


< k < 0 < k < k <


< n ;

< <


< k < 0 < k < k <


< n ; 0

1

0

1

+1

+2

+1

and

+2

j i j  ji j ;

i = 0; 1; : : : ; n 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

and let 0 < a < b: Then min k1

pk a;b < min k1 pk a;b : p2Hn

p2Gn

[

℄

[

Hint: Let q 2 Hn be the best approximation to x r(x) = ( 1)x + 0

n X i=0

x i

0

℄

 1 on [a; b℄: Let

2 spanfx ; : : : ; x k ; x ; x k 0

0

+1

; : : : ; x n g

interpolate

q

1 2 spanfx0 ; : : : ; xk ; x ; xk+1 ; : : : ; xn g 0

at the n + 1 distin t zeros, x ; x ; : : : ; xn , of q  1 on [a; b℄ (see Theorem 3.1.6). Use Theorem 3.2.5 to show that 1

2

+1

jr(x)j  jq(x)j ; x 2 [a; b℄ with stri t inequality for x = 6 xi : Finally show that if p := r + 1; then min k1 pk a;b  k1 p k a;b < k1 q  k a;b = min k1 pk a;b : p2Gn p2Hn [

℄

[

℄

[

℄

[

℄

ut

184 4. Denseness Questions

E.7 Full Muntz Theorem on [a; b℄; a > 0. Let (i )1 be a sequen e of i distin t real numbers, and let 0 < a < b. Show that spanfx0 ; x1 ; : : : g is dense in C [a; b℄ if and only if =0

1 X

1

i=0 i 6=0

ji j = 1 :

Hint: Distinguish the following ases. Case 1: The sequen e (i )1 i has a luster point 0 6=  2 R: Use Theorem 4.2.1 to show that spanfx0 ; x1 ; : : : g is dense in C [a; b℄: Case 2: The point 0 is a luster point of (i )1 i : Use Case 1 to show rst that spanfx0 ; x1 ; : : : g is dense in C [a; b℄; and re all that a > 0: Case 3: The sequen e (i )1 i does not have any ( nite) luster points, and either 1 1 1 1 X X =1 or i ji j = 1 : i i =0

+1

=1

+1

=0

=0

=0

i >0

i <0  0 x ; x1 ; : : :

Use Theorem 4.2.1 to show that spanf Case 4: 1 1 X i=0 i 6=0

g is dense in C [a; b℄:

ji j < 1 :

Without loss of generality we may assume that 0 2= fi g1 i (why?). By a

hange of s aling, we may also assume that [a; b℄ = [1 ; 1℄. Let =0

1 fei g1 i 1 = fi gi

; where


< e < e < 0 < e < e <


: Show that there is a sequen e ( i )1 i 1 satisfying


< < < 0 < < <


; 1 1 X j i j < jei j ; i 2 Z ; j j < 1 ; i 1 i =

2

=0

1

0

1

=

2

1

0

1

=

and the gap ondition

inf f i

i

1

: i 2 Zg > 0 :

Use E.4 a℄, E.5 a℄, and E.2 ℄ to show that

1 2= spanfx i g1 i 1;

1 i where spanfx gi 1 denotes the uniform losure of the span on [a; b℄: Finally use E.6 to show that =

=

i 1 1 2= spanfxi g1 i 1 = spanfx gi : e

=

=0

ut

4.2 Muntz's Theorem 185

E.8 Further Results for Nonnegative Sequen es with No P Gap Condition. Assume throughout this exer ise that 0   <  <


; 1 i 1=i < 1, and 0  a < b: a℄ Show that for every  2 (0; b) there is a onstant  depending only on ; a; b; and (i )1 i (but not on the number of terms in p) su h that 0

1

=1

=0

kpk

;b ℄

[0

 

Z b

a

jp(x)j dx

for every p 2 spanfx0 ; x1 ; : : : g: Hint: Assume that b = 1; the general ase an be redu ed to this by s aling. Use parts a℄ and b℄ of E.5 with

pe(x) :=

Z x 0

p(t) dt 2 spanfx0 ; x1 ; : : : g : +1

+1

ut

b℄ Assume

(pn )1 n

=1

 spanfx ; x ; : : : g 0

1

onverges to an f 2 C [a; b℄ uniformly on [a; b℄: Show that f an be extended analyti ally throughout the region

Db := fz 2 C n (

1; 0℄ : 0 < jz j < bg

and the onvergen e is uniform on every losed subset of Db : Hint: This part of the exer ise is diÆ ult. A proof of a more general statement an be found in S hwartz [59, pp. 38{48℄. ut  1 

℄ Suppose (pn )n  spanfx 0 ; x 1 ; : : : g and kpn k a;b  1 for ea h n: Show that there is a subsequen e of (pn )1 that onverges uniformly on n every losed subinterval of [0; b): (So the on lusion of E.3 g℄ holds without the gap ondition inf fi i : i 2 N g > 0:) Hint: Use parts a℄ and b℄ of E.5 and the Arzela-As oli theorem. ut d℄ Let K be a losed subset of Db de ned in part b℄. Show that there is a

onstant K depending only on K; a; b; and (i )1 i (but not on the number of terms in p) su h that [

=1

℄

=1

1

=0

kpkK  K kpk a;b [

℄

for every p 2 spanfx0 ; x1 ; : : : g: Hint: Use parts ℄ and b℄. (If the gap ondition inf fi i : i 2 N g > 0 holds, then the simpler result of E.3 e℄ an be used instead of part b℄ of this exer ise.) ut 1

186 4. Denseness Questions

E.9 Full Muntz Theorem on [a; b℄; a > 0; in Lq Norm. S hwartz [59℄ gives the following results: Suppose (i )1 i 1 is a sequen e of distin t real numbers. For a nite set of integers and =

p(x) = let

p (x) :=

X

i2 i <0

X

i2

ai xi ;

ai 2 R ;

and

p (x) :=

ai xi

+

Let 0 < a < b and 1  q  1: Theorem 4.2.5. Suppose

1 X i= 1 i 6=0

X

i2 i 0

ai xi :

1

ji j < 1 :

Then there exists a onstant depending only on a; b; q; and (i )1 i 1 (but not on the number of terms in p) su h that kp kLq a;b  kpkLq a;b and kp kLq a;b  kpkLq a;b for every p 2 spanfxi g1 i 1: =

+

[

℄

[

℄

[

℄

[

℄

=

Theorem 4.2.6. Suppose that 0 < a < b: Then spanfx0 ; x1 ; : : : g is dense in Lq [a; b℄ if and only if

1 X

(4:2:4)

i=0 i 6=0

1

ji j = 1 :

a℄ Prove the two above results under the gap ondition inf fi i : i 2 Zg > 0 : 1

Hint to Theorem 4.2.5: When q = 1 see E.4 a℄. If 1  q < 1; then modify

the proof suggested in the hints to E.4 a℄, by using E.8 a℄. ut Hint to Theorem 4.2.6: If (4.2.4) holds, then the fa t that spanfx0 ; x1 ; : : : g is dense in Lq [a; b℄ follows from E.7 and the obvious inequality

kpkLq a;b  (b [

℄

a)

=q

1

kpk a;b : [

℄

Now suppose (4.2.4) does not hold. Use Theorem 4.2.5 and E.8 a℄ to show that for every  2 (0; (b a)) there exists a onstant  depending only on a; b; q; and (i )1 i 1 (but not on the number of terms in p) su h that 1

2

=

kpk a ;b    kpkLq a;b 1 for every p 2 spanf gi 1 : Now show that the above inequality implies that spanfx0 ; x1 ; : : : g is not dense in Lq [a; b℄: ut [ +

xi

=

℄

[

℄

4.2 Muntz's Theorem 187

E.10 Denseness of spanf os k g and spanfz k g. Throughout this exer ise the span is assumed to be over C : Let

DR := fz 2 C : jz j < Rg

CR := fz 2 C : jz j = Rg :

and

a℄ Show that ( os n)1 n is a omplete orthogonal system in L [0;  ℄: So no term, os n; an be removed if we wish to preserve denseness of the span in L [0;  ℄: b℄ Let A := fei :  2 [0; Æ ℄g. Suppose (k )1 is a sequen e of distin t k

omplex numbers satisfying 2

=0

2

=0

jk j  ke

Æ;

1

k = 1; 2; : : : :

If Æ 2 [1; 2 ℄; then spanfz 0 ; z 1 ; : : : g is dense in L (A): The proof of part b℄ is outlined in parts ℄, d℄, and e℄.

℄ Jensen's Formula. Suppose h is a nonnegative integer and 2

f (z ) =

1 X k=h

h 6= 0

k z k ;

is analyti on a disk of radius greater than R; and suppose that the zeros of f in DR n f0g are a ; a ; : : : ; an ; where ea h zero is listed as many times as its multipli ity. Then 1

2

log j h j + h log R +

n X k=1

R 1 log jak j = 2

Z 2 0

log jf (Rei )j d :

Proof. This is a simple onsequen e of Poisson's formula (see, for example, Ahlfors [53℄), whi h states that log jF (0)j =

1 2

Z 2 0

log jF (Rei )j d

whenever the fun tion F is analyti and zero-free in an open region ontaining the losed disk DR . Now, in the above notation, if we let

F (z ) := f (z )



R z

h n Y

k=1

R ak z R(z ak ) 2

and apply Poisson's formula to F; we get the required result by noting that jF (z )j = jf (z )j whenever jz j = R: (The ase where f has zeros on the boundary of D requires an additional limiting argument.) ut

188 4. Denseness Questions

d℄ Let 1  a < b  1: Suppose (fk )1 is a sequen e in L (a; b) k and spanff ; f ; : : : g is not dense in L (a; b). Then there exists a nonzero g 2 L (a; b) su h that 2

=0

0

1

2

2

Z b

a

fk (x)g(x) dx = 0 ;

k = 0; 1; : : : :

This is an immediate onsequen e of the Riesz representation theorem (see E.7 g℄ of Se tion 2.2) and the Hahn-Bana h theorem. The se ond theorem says that if spanff ; f ; : : : g is not dense in a Bana h spa e, then there exists a nonzero ontinuous linear fun tional vanishing on ff ; f ; : : : g: The rst theorem gives the form of the fun tional; see Rudin [73℄. e℄ Prove b℄ as follows: Suppose spanfz 0 ; z 1 ; : : : g is not dense in L (A). Then by d℄ there exists a g 2 L [0; Æ ℄ su h that 0

1

0

1

2

2

f (z ) :=

Z Æ

exp(i(z + 1))g () d

0

vanishes at z = k , k = 0; 1; : : : : Also observe that f is an entire fun tion, and there is an absolute onstant  > 0 su h that

jf (z )j   exp(Æjz j) : Use E.8 a℄ of Se tion 2.2 to show that f 6= 0: Let R > j j be an integer. Applying ℄ on DR and exponentiating, we obtain 0

j h j exp((Æ

1)R)

RR R!

+1

 j h jRh

R Y k=0 k 6=0

R

jk j   exp(ÆR) ;

where h is the rst nonvanishing oeÆ ient of the Taylor series expansion of f around 0. However,

RR = 1; R!1 R! eR +1

lim

ut

whi h is a ontradi tion and nishes the proof.

f℄ Let A := [0; Æ ℄. Suppose (k )1 k is a sequen e of distin t omplex numbers satisfying jk j  ke Æ ; k = 1; 2; : : : : =0

1

Show that if Æ 2 [1;  ℄; then spanf os  ; os  ; : : : g is dense in L (A). Hint: Pro eed as in the proof of part b℄. ut 0

1

2

4.2 Muntz's Theorem 189

g℄ Suppose (k )1 k is a sequen e of distin t real numbers satisfying =0

0  k

k;

k = 0; 1; : : : :

Then spanf os  ; os  ; : : : g is dense in L [0;  0

1

2

℄ for any  > 0:

ut

Proof. This is harder; see Boas [54, p. 235℄.

h℄ Suppose (k )1 is a sequen e of distin t omplex numbers satisfying k 0  jk j  k: Suppose f is an entire fun tion su h that kf kDR  eR for all R > 0 with an absolute onstant  > 0, and spanff ( z ) : 2 C g is dense in L (C ): Then spanff ( z ); f ( z ); : : : g is dense in L (C ): =0

2

1

0

1

2

1

E.11 On the Hardy Spa e H1 . We denote by H1 the lass of fun tions that are analyti and bounded on D := fz 2 C : jz j < 1g: We let

kf kH1 := kf kD = sup jf (u)j : u2D

a℄ If f

2 H1 ; then f (z ) =

where

1 X n=0

z 2 D;

an z n ;

jan j  kf kH1 :

Hint: By Cau hy's integral formula

Z



f (t) jdtj  R n kf kH jan j = n1! jf n (0)j  21 1 jtj R tn (

)

=

+1

holds for every R 2 (0; 1): b℄ If f 2 H1 ; then

jf 0 (z )j  and



jf n (z )j  n! (

℄

ut

1 

)

H1 is a Bana h algebra.

Hint: See Rudin [73℄.

1

1

2

jz j kf kH1 ;

1

n+1

jz j

kf kH1 ;

jz j < 1 jz j < 1 : ut

190 4. Denseness Questions

E.12 Blas hke Produ ts. A produ t of the form

B (z ) := z k

1 Y



i=1

i

z 1 i z



ji j ;

i 2 C

i

n f0g

with k 2 Z is alled a Blas hke produ t. Let D := fz 2 C : jz j < 1g: a℄ Let  z ' (z ) := ; 2C: 1 z Show that j' (z )j = 1 whenever jz j = 1 and

'0 (z ) =

jj

1

2

:

z )

(1

2

b℄ Show that if jj < 1; then ' (z ) maps the losed unit disk D one-to-one onto itself.

℄ A Minimization Property. Let ; ; : : : ; n be xed omplex numbers with j i j > 1; i = 1; : : : n: Show that 1

min

1 ai 2C

n X i=1

ai

z

2



i

n Y

=

i=1

D

j i j

1

and that the minimum is attained by the normalized nite Blas hke produ t 1

n X i=1

ai = z i

n Y i=1

!

1

i

n Y i=1

i

z 1 i z 1

1

!

:

Hint: Suppose that the statement is false. Then there are some ai 2 C su h that

1

n X i=1

z

ai

i

<

n Y i=1

!

j i j

1

=

1

n X i=1



ai z i

for all z 2 C with jz j = 1. Now Rou he's theorem implies that n X ai i=1

z

ai i

has n zeros in the disk D; whi h is a ontradi tion. d℄ Suppose ( n )1 n is a sequen e in D satisfying =1

= =


= k = 0 ; 1

2

n 6= 0 ;

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

ut

4.2 Muntz's Theorem 191

and

1 X

(1

n=1

Then

j n j) < 1 :

1  Y

B (z ) := z k

n

z 1 nz

n=k+1



j n j n

de nes a bounded analyti fun tion on D (that is, B 2 H1 ), whi h vanishes at z if and only if z = j for some j = 1; 2; : : : ; in whi h ase the multipli ity of z in B (z ) is the same as the multipli ity of j in ( n )1 n : 1 e℄ Suppose ( n )n is a sequen e in D satisfying 0

=1

=1

1 X

(1

n=1

j n j) = 1 :

Denote the multipli ity of j in ( n )1 by mj : Suppose f 2 H1 has a n zero at ea h j with multipli ity mj : Then f = 0: Hint: Suppose kf kD > 0: Without loss of generality we may assume that f (0) 6= 0: By Jensen's formula (see E.10 ℄), =1

1 X j nnj
log

R 1 j n j + log jf (0)j = 2

Z 2 0

log jf (Rei j d  log kf kD

for every R 2 (0; 1). Letting R tend to 1, we obtain

1 X n=1

log

1

j n j < log kf kD

Hen e

1 X

(1

n=1

log jf (0)j < 1 :

j n j) < 1 ;

whi h ontradi ts the assumption. ut Note that the on lusion of part e℄ holds for the larger Nevanlinna

lass N; whi h is de ned as the set of those analyti fun tions f on D for whi h Z  1 sup log jf (Rei )jd < 1 ; R2 ; 2 2

+

(0 1)

0

where log x := maxflog x; 0g: +

192 4. Denseness Questions

f℄ Let

B (z )

:= z k

n Y i=1



i

z 1 i z



ji j ;

i 2 C

i

n f0g

be a nite Blas hke produ t. Show that

jB 0 (z )j = k +

n X i=1

ji j jz i j 1

2 2

jz j = 1 :

;

Hint: Consider B 0 =B; where jB (z )j = 1 whenever jz j = 1: g℄ Suppose

1 X i=1

(1

ji j) < 1 ;

and

B (z ) :=

1 Y i=1

i

ut

i 2 (0; 1)

z 1 i z



ji j : i

Show that

k((1 z ) B (z ))0 kD 2



4+2

!

1 X i=1

(1

i ) 2

kB kD : ut

Hint: Use f℄.

E.13 Yet Another Proof of Muntz's Theorem when inf fi : i 2 N g > 0. As in E.10, this proof requires a onsequen e of the Hahn-Bana h theorem and the Riesz representation theorem whi h we state in a℄. For details, the reader is referred to Feinerman and Newman [76℄ and Rudin [87℄. We assume throughout the exer ise that  := 0 and that (k )1 k is a sequen e of distin t positive numbers satisfying inf fk : k 2 N g > 0: a℄ spanf1; x1 ; x2 ; : : : g is not dense if and only if there exists a nonzero nite Borel measure  on [0; 1℄ with 0

Z

1

tk d(t) = 0 ;

=1

k = 0; 1; 2 : : : :

0

1 2 b℄ Show that 1 k 1=k = 1 implies that spanf1; x ; x ; : : : g is dense in C [0; 1℄: Outline. Suppose there is a nonzero nite Borel measure  on [0; 1℄ su h that Z tk d(t) = 0 ; k = 0; 1; : : : : P

=1

1

0

4.2 Muntz's Theorem 193

Let

f (z ) =

Z

1

tz d(t) ;

Re(z ) > 0 :

0

Show that and



1+z g(z ) := f 1 z









2 H1

k 1  1 g k =0 with k = 1; 2; : : : :  + 1 < 1 ; k + 1 k P Note that 1 1= = 1 and inf f : k 2 N g > 0 imply

k

k=1

k

1 X k=1



k 



1 = 1: + 1 k

1

Hen e E.12 e℄ yields that g = 0 on the open unit disk. Therefore f (z ) = 0 whenever Re(z ) > 0; so

f (n) =

Z

1

tn d(t) = 0 ;

n = 1; 2; : : : :

0

Note that

Z

1

t d(t) = 0 0

0

also holds be ause of the hoi e of : Now the Weierstrass approximation theorem yields that Z 1

f (t) d(t) = 0

0

for every f 2 C [0; 1℄, whi h ontradi ts the fa t that the Borel measure  is nonzero. So part a℄ implies that spanf1; x1 ; x2 ; : : : g is dense in C [0; 1℄: P1

℄ Show that k 1=k < 1 implies that spanf dense in C [0; 1℄. Outline. Show under the above assumption that

1; x1 ; x2 : : :

=1

f (z ) =

Z

1

tz

0



1 2

Z

1 1

f ( 1 + is)e



is log t ds

dt ;

if f is de ned by

f (z ) :=

1 Y

z

(2 + z ) n 3

=1

n

z

2 + n + z

:

ut

g is not

Re(z ) > 1

194 4. Denseness Questions

Show that

1 1 f ( 1 + is)e is t ds dt d(t) = 2 1 de nes a nonzero nite Borel measure, , on [0; 1℄ su h that 



Z

log

Z

1

tk d(t) = 0 ;

k = 0; 1; 2; : : :

0

as is required by part a℄. For the above, show that

f (z ) =

1 2

Z

and use that

1 f (is 1) ds ; 1 is 1 z 1

1+z

=

is

Z

1

tz

Re(z ) > 1 is dt :

0

ut

E.14 Another Proof of Denseness of Muntz Spa es when i ! 0. Suppose  := (i )1 is a sequen e of distin t positive numbers with lim i = 0: i i!1 =1

Show that

M () := spanf1; x1 ; x2 ; : : : g P is dense in C [0; 1℄ if and only if 1 i i = 1 : P1 Hint: If i i = 1; then lim i = 0 implies that i!1 1   1  X = 1: 1 i  + 1 i i =1

=1

=1

So the outline of the proof of E.13 b℄ yields that M () is dense in C [0; 1℄: P If  := 1 i i < 1; then, by Theorem 6.1.1, the inequality =1

kxp0 (x)k

;

[0 1℄

 9 kpk

;

[0 1℄

holds for every p 2 M (). Use this inequality to show that M () fails to be dense in C [0; 1℄: ut E.15 Denseness of Muntz Spa es with Complex Exponents. Suppose  := (i )1 i is a sequen e of omplex numbers satisfying =1

Re(i ) > 0 ; Show that if

1 X n=1



1

i = 1; 2; : : : :

n 



1 = 1; n + 1

then spanf1; x1 ; x2 ; : : : g is dense in C [0; 1℄. (In this exer ise the span is taken over C , and C [0; 1℄ denotes the set of all omplex-valued ontinuous fun tions on [0; 1℄.)

4.2 Muntz's Theorem 195

E.16 Christoel Fun tions for Nondense Muntz Spa es. Let  = (i )1 i for ea h i: As be a sequen e of distin t omplex numbers with Re(i ) > in Se tion 3.4, let Lk := Lk f ; : : : ; k g denote the asso iated orthonormal Muntz-Legendre polynomials on [0; 1℄: a℄ Let Kn be de ned by =0

1

2

0

1 := inf Kn (y)

Z

1

jp(t)j

2

0

dt : p 2 spanf

x0 ; x1 ; : : :

Show that

Kn (y) =

n X k=0

jLk (y)j

; xn

g;



p(y) = 1 :

:

2

The fun tion 1=Kn is alled the nth Christoel fun tion asso iated with . Hint: Pro eed as in the hint to E.13 of Se tion 2.3. ut 1 In the rest of the exer ise we assume that (i )i is a sequen e of nonnegative integers. We use this assumption for treating (higher) derivatives, although some weaker assumptions would lead to the same on lusions. P b℄ Suppose 1 i 1=i < 1: Show that for every  2 (0; 1) and m 2 N ; there exists a onstant ;m depending only on ; , and m su h that =0

=1

kp m k (

)

;

[0 1

℄

 ;m kpkL

;

2 [0 1℄

for every p 2 spanfx0 ; x1 ; : : : g: Hint: Use E.3 ℄.

℄ Show that the following statements are equivalent: (1) spanfx0 ; x1 ; : : : g is not dense in C [0; 1℄ : P1 (2) i 1=i < 1 :

ut

=1

P1

 onverges uniformly on [0; 1 ℄ for all  2 (0; 1) : P  (4) There exists an x 2 [0; 1) so that 1 k (Lk (x)) < 1 : Outline. The equivalen e of (1) and (2) is the ontent of Muntz's theorem

(3)

k=0 (Lk )

2

2

=0

(Theorem 4.2.1). To see that (2) implies (3), rst observe that

1 X (L )(m) (y ) 2 k=0

k

= sup

n p(m) (y ) 2

: p 2 spanfx0 ; x1 ; : : : g;

o

kpkL

;

2 [0 1℄

=1 ;

whi h an be proved similarly to part a℄. Hen e by part b℄, for every  (0; 1), there exists a onstant  depending only on  su h that

2

196 4. Denseness Questions

1 X k=0

(Lk (x))

2

 

P

1 X

and


0

k=0

(L0 k (x))

2

  ;

x 2 [0; 1 ℄ :

P

n  Sin e = 2 nk Lk L0 k on [0; 1); applying the Cau hyk (Lk ) S hwarz inequality, we obtain that 2

=0

=0

Pn



k=0 (Lk )

2


0



(x)  2  ;

P

x 2 [0; 1 ℄ :

Therefore the fun tions nk (Lk ) ; n = 1; 2; : : : ; are uniformly bounded and equi ontinuous on [0; 1 ℄; whi h implies the uniform onvergen e of the fun tions Kn on [0; 1 ℄ by the Arzela-As oli theorem. Sin e (3) obviously implies (4), what remains to be proven is that (4) implies (1). This an be easily done by part a℄. ut P1 d℄ Let  2 (0; 1) and m 2 N be xed. Show that if i 1=i < 1; then P1  m k ((Lk ) ) onverges uniformly on [0; 1 ℄: Hint: Modify the argument given in the hints to part ℄. ut P1 e℄ Show that if i 1=i < 1; then 2

=0

=0

(

) 2

=0

=0

lim k(Lk ) m

k!1

(

)

k

;

[0 1

℄

=0

for every  2 (0; 1) and m 2 N : E.17 Chebyshev-Type Inequality with Expli it Bound via the PaleyWiener Theorem. The method outlined in this exer ise was suggested by Halasz. A fun tion f is alled entire if it is analyti on the omplex plane. An entire fun tion f is alled a fun tion of exponential type Æ if there exists a onstant depending only on f su h that

jf (z )j  exp(Æjz j) ;

z2C:

The olle tion of all su h entire fun tions of exponential type Æ is denoted be E Æ . The Paley-Wiener theorem hara terizes the fun tions F that an be written as the Fourier transform of some fun tion f 2 L [ Æ; Æ ℄. 2

Theorem (Paley-Wiener). Let Æ 2 (0; 1): Then F only if there exists an F 2 L [ Æ; Æ℄ su h that

2 E Æ \ L (R) 2

if and

2

F (z ) =

Z Æ

Æ

f (t)eitz dt :

For a proof see, for example, Rudin [87℄. In the rest ofPthe exer ise let  = (k )1 be an in reasing sequen e k with  = 0 and 1 k 1=k < 1: =0

0

=1

4.2 Muntz's Theorem 197

E.5 says that

C (; ) := sup



kpk kpk

: p 2 spanf

; ;1℄

x0 ; x1 ; : : :

[0 1℄

[1



g

<1

holds for every  2 (0; 1): In this exer ise we establish an expli it bound for C (; ): a℄ Show that

C (; ) = sup





jp(0)j kpk ; [1

: p 2 spanfx0 ; x1 ; : : : g

1℄

for every  2 (0; 1): Hint: Use  = 0, E.4 ℄ of Se tion 3.3, and the monotoni ity of the Chebyshev polynomial 0

Tnfx0 ; x1 ; : : : ; xn ; [1 ; 1℄g on [0; 1 ℄: b℄ Assume that (1) F 2 E Æ \ L (R); (2) F (ik ) = 0; k = 1; 2; : : : (i is the imaginary unit); and (3) F (0) = 1: Show that jP (1)j  kF kL2 R kP kL2 Æ;Æ 2

(

)

[

℄

for every P 2 spanfe 0 t ; e 1 t ; : : : g: Outline. By the Paley-Wiener theorem

F (z ) = for some f

2L [ 2

Z Æ

Æ

f (t)eitz dt

Æ; Æ℄: Now if P (t) = a + 0

n X k=1

ak e

k t ;

then Z Æ

Æ

Z Æ

f (t)P (t) dt = a

0

Æ

f (t) dt +

= a F (0) + 0

n X k=1

n X k=1

ak

Z Æ

Æ

f (t)e

k t dt

ak F (ik ) = a = P (1) : 0

ut

198 4. Denseness Questions

Hen e by the Cau hy-S hwarz inequality and the L inversion theorem of Fourier transforms, we obtain that 2

jP (1)j  kf kL

2[

Given Æ 2 (0; 1), let N

Æ;Æ℄

kP kL

 kF kL

2 (R)

kP kL

2[

Æ;Æ℄ :

ut

2 N be hosen so that 1 X

1

k=N +1

Let

Æ;Æ℄

2[

k := Ak

k

 3Æ : Æ : 3N

A :=

with

Let

F (z ) :=



sin(Æz=3) Æz=3

N Y

k=1



1



z sin(k z=k ) ik k z=k

1 Y



1

k=N +1



sin(z=k ) sin i

4 !

;

where i is the imaginary unit.

℄ Show that F 2 E Æ : d℄ Observe that F (0) = 1; F (ik ) = 0; k = 1; 2; : : : ; and

jF (t)j 



N sin(Æt=3) Y 1 2+ Æt=3 k k



t 2 R:

;

=1

e℄ Show that

jP (1)j  3Æ

N  Y k=1

2+

1



k

kP k

[

Æ;Æ℄

for every P 2 spanfe 0 t ; e 1 t ; : : : g with := kt sin tkL2 R : Hint: Use parts b℄, ℄, and d℄. ut  f℄ Let k := k ;  > 1: Show that there exists a onstant  depending only on  su h that 1

kpk

;

[0 1℄

 exp

 

=

1 (1

)


kpk

[1

(

;1℄

for every p 2 spanfx0 ; x1 ; : : : g and for every  2 (0; 1=2℄:

)

4.2 Muntz's Theorem 199

Proof. Let

1 log(1 ) : 2 Observe that N in part e℄ an be hosen so that

Æ :=

(4.2.5)

N :=

(4.2.6)

$

Æ (

1)

3

1=(1 ) %

+1:

Also, k in part d℄ is of the form k = Æk  (3N ) : Let M +1 be the smallest value of k 2 N for whi h 1 3N < 1; that is ,   1: 1

k

Note that

k Æ

$

M := If 0 < M < N; then N Y k=1



2+

1



=

k

 =



N Y



k=1 M Y

2+ 9N

3N

1= %

Æ

!

!

N Y k=M +1

 9e N M



Æ

 (3(2e) )M 3N

3





   9N M M M N M 3

Æ

e

M 3N M

M

Æ

:



Æk

Æk k=1   M 9e N 

3N

1 3N 2 Æ

1= ! M

3N M

 (3(2e) )N ;

M

and the theorem follows by (4.2.5), (4.2.6), and part e℄. If N  M; then N  Y

k=1

2+

1

k



=

 

N  Y

k=1 N Y  

k=1

2+ 9N

3N

Æk !

Æk

 9e N

Æ

9e



 N 







   9N N N N

Æ(

Æ (

Æ

3

e

1)

1=(1

 1) N

=



9e N

! ) (1 )N

 (3e ( 1))N ; 3 and the theorem follows by (4.2.5), (4.2.6), and part e℄. 

Æ

(1

Æ

) N

200 4. Denseness Questions

If M = 0; then N Y



2+

k=1

1



k



N Y k=1

3 = 3N ;

ut

and the theorem follows by (4.2.5), (4.2.6), and part e℄.

The next part of the exer ise shows that the result of part f℄ is lose to sharp. g℄ Let k := k  ;  > 1: Let  2 (0; 1=2℄: Show that there exists a onstant

 depending only on  > 0 so that sup



jp(0)j kpk ; [1

1℄



: p 2 spanfx0 ; x1 ; : : : g

 exp

 

=

1 (1

)
:

Proof. Let n 2 N be a xed. We de ne k := kn ; k = 0; 1; : : : . Let Tn (x) := ((x 1)=2)n and 1

Qn (x) := Tn



2xn (1 )n

1

1

1



)n )n

1 + (1 1 (1

1

!

:

1

Then Qn 2 spanfx 0 ; : : : x n g; and by E.3 g℄ of Se tion 3.3 we obtain that sup



jp(0)j kpk ; [1

: p 2 spanf

x0 ; x1 ; : : :

1℄



g  kQjQkn (0)j n ; =

[1



1

(1

1℄

1

= jQn (0)j

)n

Now let n be the smallest integer satisfying n   : Sin e (1 bounded away from 0 on (0; 1=2℄; the result follows. 1

1

n 1

)

: =

1

is

ut

E.18 Completion of the Proof of Theorem 4.2.1. The ase when i  1 for ea h i has already been proved. The only real remaining diÆ ulty is part d℄. a℄ Prove Theorem 4.2.1 in the ase when inf fi : i 2 N g > 0: Hint: Use the s aling x ! x =Æ and the already proved ase. ut 1 b℄ Show that if (i )i  (0; 1) has a luster point  2 (0; 1); then spanf1; x1 ; x2 ; : : : g is dense in C [0; 1℄: Hint: Use part a℄. ut

℄ Suppose (i )1  (0; 1)Pand i ! 0: Then spanf1; x1 ; x2 ; : : : g is i dense in C [0; 1℄ if and only if 1 i i = 1: Hint: This is the ontent of E.14. ut 1

=1

=1

=1

4.2 Muntz's Theorem 201

d℄ Suppose

fi : i 2 N g = fi : i 2 N g [ f i : i 2 N g

with

lim i = 0 i!1 1; x1 ; x2 ; : : :

Show that spanf

1 X

(4:2:7)

i=1

lim i = 1 :

and

i!1

g is dense in C [0; 1℄ if and only if

i +

1 X i=1

1

i

= 1:

Outline. If (4.2.7) holds, then the denseness of spanf1; x1 ; x2 ; : : : g in C [0; 1℄ follows from parts a℄ and ℄. Now assume that (4.2.7) does not hold, so 1 1 1 X X i < 1 and < 1: i i i =1

=1

For notational onvenien e, let

Tn; := Tn f1; x1 ; : : : ; xn : [0; 1℄g ; Tn; := Tn f1; x 1 ; : : : ; x n : [0; 1℄g ; T n;; := T n f1; x1 ; : : : ; xn ; x 1 ; : : : ; x n : [0; 1℄g 2

2

(we use the notation introdu ed in Se tion 3.3). It follows from Theorem 6.1.1 (Newman's inequality) and the Mean Value Theorem that for every  > 0 there exists a k () 2 N depending only on (i )1 i and  (and not on n) su h that Tn; has at most k () zeros in [; 1) and at least n k () zeros in (0; ): Similarly, E.5 b℄ and the Mean Value Theorem imply that for every  > 0 there exists a k () 2 N depending only on ( i )1 i and  (and not on n) so that Tn; has at most k () zeros in (0; 1 ℄ and at least n k () zeros (1 ; 1): Now, on ounting the zeros of Tn; T n;; and Tn; T n;; , we

an dedu e that for every  > 0 there exists a k () 2 N depending only on (i )1 i and  (and not on n) so that T n;; has at most k () zeros in [; 1 ℄.
Let  := and k := k . Pi k k + 4 points 1

1

=1

1

2

=1

2

2

2

2

2

=1

1

1

4

4

1 4

and a fun tion f while

<  <  <


< k 0

1

+3

<

3 4

2 C [0; 1℄ su h that f (x) = 0 for all x 2 f (i ) := 2
( 1)i ;

i = 0; 1; : : : :



0;

1 4



[

3 4



;1 ;

202 4. Denseness Questions

Assume that there exists a p 2 spanf1; x1 ; x2 ; : : : g su h that

kf

pk

;

[0 1℄

< 1:

Then p T n;; has at least 2n + 1 zeros in (0; 1): However, for suÆ iently large n, p T n;; 2 spanf1; x1 ; : : : ; x2n g 2

2

so it an have at most 2n zeros in [0; 1): This ontradi tion shows that spanf1; x1 ; x2 ; : : : g is not dense in C [0; 1℄: ut e℄ Prove Theorem 4.2.1 in full generality. Outline. Combine parts a℄ to d℄. ut E.19 Proof of Theorem 4.2.3. Prove Theorem 4.2.3. Proof. Assume that spanfx0 ; x1 ; : : : g is dense in L [0; 1℄. Let m be a xed nonnegative integer. Let  > 0. Choose a p 2 spanfx0 ; x1 ; : : : g 1

su h that Now let

kxm q(x) :=

Z x 0

Then

p(x)kL1

;

[0 1℄

< :

p(t) dt 2 spanfx0 ; x1 ; : : : g : +1

m+1

x

m + 1

q(x)

;

+1

< :

[0 1℄

So the Weierstrass approximation theorem yields that spanf1; x0 ; x1 ; : : : g +1

+1

is dense in C [0; 1℄; and Theorem 2.1 implies that

1 X i=0

i + 1 = 1: (i + 1) + 1 2

Now assume that (4:2:8)

1 X i=0

i + 1 = 1: (i + 1) + 1 2

By the Hahn-Bana h theorem and the Riesz representation theorem

4.2 Muntz's Theorem 203

spanfx0 ; x1 ; : : : g is not dense in L [0; 1℄ if and only if there exists a 0 6= h 2 L1 [0; 1℄ satisfying 1

Z

1

ti h(t) dt = 0;

i = 0; 1; : : : :

0

Suppose there exists a 0 6= h 2 L1 [0; 1℄ su h that Z

1

ti h(t) dt = 0 ;

i = 0; 1; : : : :

0

Let

f (z ) :=

Z

1

tz h(t) dt ;

Re(z ) > 1 :

0

Then



1+z g(z ) := f 1 z



1

is a bounded analyti fun tion on the open unit disk that satis es

g



i



i + 2

=0

with

i  + 2 <

i

1;

i = 0; 1; : : : :

Note that (4.2.8) implies

1 X i=1

1

 i  + 2

i

= 1:

Hen e Blas hke's theorem (E.12 e℄) yields that g = 0 on the open unit disk. Therefore f (z ) = 0 whenever Re(z ) > 1; so

f (n) =

Z

1

tn h(t) dt = 0 ;

n = 0; 1; : : : :

0

Now the Weierstrass approximation theorem yields Z

1

u(t)h(t) dt = 0

0

for every u 2 C [0; 1℄; whi h ontradi ts the fa t that 0 6= h: So spanfx0 ; x1 ; : : : g is dense in L [0; 1℄: 1

ut

204 4. Denseness Questions

E.20 Proof of Theorem 4.2.4. a℄ Show that if

1 X

(4:2:9)

i=0

i + p

1



i + p 1

2

+1

= 1;

then spanfx0 ; x1 ; : : : g is dense in Lp [0; 1℄: Outline. By the Hahn-Bana h theorem and the Riesz representation theorem spanfx0 ; x1 ; : : : g is not dense in Lp [0; 1℄ if and only if there exists a 0 6= h 2 Lq [0; 1℄ satisfying Z

1

ti h(t) dt = 0 ;

i = 0; 1; : : : ;

0

where q is the onjugate exponent of p de ned by p + q Suppose there exists a 0 6= h 2 Lq [0; 1℄ su h that 1

Z

1

ti h(t) dt = 0 ;

1

= 1.

i = 0; 1; : : : :

0

Let

f (z ) :=

Z

1

tz h(t) dt ;

Re(z ) >

0

p: 1

Use Holder's inequality to show that

g(z ) := f



1+z 1 z

1



p

is a bounded analyti fun tion on the open unit disk that satis es !

i + p 1 g =0 i + p + 1 1 1

with

i i



+ p 1 < 1; + p + 1 1

1

i = 0; 1; : : : :

Note that (4.2.9) implies

1 X i=1

1

i i

!

+ p 1 = 1: + p + 1 1

1

Hen e Blas hke's theorem (E.12 e℄) yields that g = 0 on the open unit disk. Therefore f (z ) = 0 whenever Re(z ) > p ; so 1

4.2 Muntz's Theorem 205

f (n) =

Z

1

tn h(t) dt = 0 ;

n = 0; 1; : : : :

0

Now the Weierstrass approximation theorem yields Z

1

u(t)h(t) dt = 0

0

for every u 2 C [0; 1℄; whi h ontradi ts the fa t that 0 6= h: So spanfx0 ; x1 ; : : : g

is dense in Lp [0; 1℄: b℄ Show that if

1 X i=0

i + p

1



i + p 1

2

+1

ut

< 1;

then spanfx0 ; x1 ; : : : g is not dense in Lp [0; 1℄. Outline. This follows from E.7 of Se tion 4.3. ut 1

℄ Suppose (i )i is a sequen e of distin t positive numbers. Let p 2 [1; 1). Show that spanfe 0 t ; e t ; : : : g is dense in Lp [0; 1) if and only if =0

1 X i=0

i = 1: i + 1 2

ut

Outline. Use parts a℄ and b℄ and the substitution x = e t :

E.21 Muntz Theorem on [a; b℄ with a < 0 < b. Suppose  := (i )1 is i a sequen e of distin t nonnegative integers, and suppose a < 0 < b: Then spanf1; x1 ; x2 ; : : : g is dense in C [a; b℄ if and only if =1

1 X i=1

i is even

1

i

=1

and

1 X i=1

i is odd

1

i

= 1:

E.22 The Zeros of the Chebyshev Polynomials in Nondense Muntz Spa es. Let (P i )1 i be a sequen e of distin t nonnegative real numbers with  := 0 and 1 i 1=i < 1: Let Tn := Tn f ;  ; : : : ; n ; [0; 1℄g be the Chebyshev polynomials for spanfx0 ; : : : ; xn g on [0; 1℄: Let Z := fx 2 [0; 1℄ : Tn (x) = 0 for some n 2 N g : 0

=0

=1

0

1

Let Z 0 be the set of all limit points of Z and Z 00 be the set of all limit points of Z 0 . Show that Z 00 = f1g: Hint: Use the bounded Bernstein-type inequality of E.5 b℄ and the interla ing property of the zeros of the Chebyshev polynomials Tn . ut

206 4. Denseness Questions

4.3 Unbounded Bernstein Inequalities In Se tion 4.1 we hara terized the denseness of C Markov spa es by the behavior of the zeros of their asso iated Chebyshev polynomials. The prin ipal result of this se tion is a hara terization of denseness of Markov spa es by whether or not they have an unbounded Bernstein inequality. 1

De nition 4.3.1 (Unbounded Bernstein Inequality). Let A be a subset of C [a; b℄. We say that A has an everywhere unbounded Bernstein inequality if  0  k p k ; sup : 0 6= p 2 A = 1 1

[

kpk a;b = 6 : [

for every [; ℄  [a; b℄;

℄

℄

The subset

A := fx p(x) : p 2 Pn ;

n = 0; 1; : : : g

2

has an everywhere unbounded Bernstein inequality despite the fa t that

f 0 (0) = 0 for every f

2 A:

The next result shows that in most instan es the Chebyshev polynomial is lose to extremal for Bernstein-type inequalities. This is a theme that will be explored further in later hapters. Theorem 4.3.2 (A Bernstein-Type Inequality for Chebyshev Spa es). Let (1; f ; : : : ; fn ) be a Chebyshev system on [a; b℄ su h that ea h fi is dierentiable at x 2 [a; b℄: Let 1

0

Tn := Tn f1; f ; : : : ; fn ; [a; b℄g 1

be the asso iated Chebyshev polynomial. Then 2 jp0 (x )j  0 kpk a;b 1 jTn (x )j jTn (x )j 0

0

[

℄

0

for every 0 6= p 2 spanf1; f ; : : : ; fn g; provided jTn (x )j 6= 1: 1

0

Proof. Let a = y < y <


< yn = b denote the extreme points of Tn ,

that is,

0

1

Tn(yi ) = ( 1)n i ;

i = 0; 1; : : : ; n

(see the de nition and E.1 a℄ in Se tion 3.3). Let yk

 x  yk

0 6= p 2 Hn := spanf1; f ; : : : ; fn g : 1

0

+1

and

4.3 Unbounded Bernstein Inequalities 207

If p0 (x ) = 0; then there is nothing to prove. Assume that p0 (x ) 6= 0: Then we may normalize p so that 0

0

kpk a;b [

℄

=1

sign(p0 (x )) = sign(Tn (yk )

and

0

Tn (yk )) :

+1

Let Æ := jTn (x )j: Let  2 (0; 1) be xed. Then there exists a onstant  with j j  Æ + (1 Æ ) su h that 0 1

2

 + (1 Æ)(1 )p(x ) = Tn(x ) : 1

0

2

Now let

q(x) :=  + (1

Æ)(1 )p(x) :

1

2

Then

kqk a;b [

and

℄

< 1;

0

q(x ) = Tn (x ) ; 0

0

sign(q 0 (x )) = sign(Tn (yk ) 0

+1

Tn(yk )) :

If the desired inequality did not hold for p; then for a suÆ iently small

>0

jq0 (x )j > jTn0 (x )j ; 0

so

0

h(x) := q(x)

Tn(x)

would have at least three zeros in (yk ; yk ): But h has at least one zero in ea h of (yi ; yi ): Hen e h 2 Hn has at least n + 2 zeros in [a; b℄; whi h is a ontradi tion. ut +1

+1

We now state the main result. Theorem 4.3.3 (Chara terization of Denseness by Unbounded Bernstein Inequality). Suppose M := (f ; f ; : : : ) is an in nite Markov system on [a; b℄ with ea h fi 2 C [a; b℄, and suppose that (f =f )0 does not vanish on (a; b): Then span M is dense in C [a; b℄ if and only if span M has an 0

1

2

1

0

everywhere unbounded Bernstein inequality.

Proof. The only if part of this Theorem is obvious. A good uniform approximation on [a; b℄ to a fun tion with uniformly large derivative on a subinterval [; ℄  [a; b℄ must have large derivative at some points in [; ℄. In the other dire tion we use Theorems 4.3.2 and 4.1.1 in the following way. Without loss of generality we may assume that f = 1 (why?). If span M is not dense in C [a; b℄; then, by Theorem 4.1.1, there exists a subinterval [; ℄  [a; b℄; where all elements of a subsequen e of the sequen e of asso iated Chebyshev polynomials, 0

208 4. Denseness Questions

(Tn f1; f ; : : : ; fn ; [a; b℄g) ; 1

have no zeros. It remains to show that from this subsequen e we an pi k another subsequen e (Tni ) and a subinterval [ ; d℄  [; ℄ with

kTni k ;d

(4:3:1)

[

℄

<1 Æ

and

kTn0 i k ;d

(4:3:2)

[

<

℄

for some absolute onstants Æ > 0 and > 0: The result will now follow from Theorem 4.3.2. A proof that the above hoi e of (Tni ) is possible is outlined in E.1. ut Theorem 4.3.3 has the following interesting orollary. Corollary 4.3.4. Suppose (k )1 k

numbers. Then

=1



span 1 ;

 Rn[

1; 1℄ is a sequen e of distin t real

1

1

x  is dense in C [ 1; 1℄ if and only if

1

1q X k=1

(Here, unlike in Se tion 3.5, k 1: )

p

2

;

x



2

; :::



1 = 1:

k 2

k

1 denotes the prin ipal square root of

2

Proof. A ombination of Theorem 4.3.3 and Corollary 7.1.3 yields the only if part of the orollary. The Chebyshev polynomials Tn (of the rst kind) and Un (of the se ond

kind) for the Chebyshev spa e 

span 1 ;

1

x



1

; ::: ;

1



x n

on [ 1; 1℄ were introdu ed in Se tion 3.5. The properties of

Ten () := Tn ( os )

and

Uen () := Un ( os ) sin  ;

established in Se tion 3.5, in lude (4:3:3)

kTenkR = 1

and

kUen kR = 1 ;

4.3 Unbounded Bernstein Inequalities 209

(4:3:4)

Ten + Uen = 1 ; (Ten0 ) + (Uen0 ) = Ben ; Te0 = Ben Uen ; 2

(4:3:5)

2

(4:3:6) and (4:3:7) where

2

2

n

Uen0 = Ben Ten ; Ben ()

p

2

p

n X

:=

jk

k=1

k 2

1 ;

os j

2R

( k 1 denotes the prin ipal square root of k hold on the real line. Suppose 2

2

1q X

Then (4:3:8)

1 = 1:

k

k=1

lim min B () n!1 2[; ℄ n

1) and the identities

2

= 1;

0<  < < :

Assume that there is an interval [a; b℄  ( 1; 1) su h that sup kTn0 k a;b < 1 : n2N

[

n2N

[

℄

Let  := ar

os b and := ar

os a: Then sup kTen0 k ; < 1 : ℄

It follows from properties (4.3.6) and (4.3.8) that lim kUen k ; = 0 ;

n!1

[

and hen e, by property (4.3.4),

lim kTen

1k ; = 0 :

2

n!1

℄

[

℄

Thus, by properties (4.3.7) and (4.3.8),

lim min jUen0 ()j = 1 ;

n!1 2[; ℄

that is,

lim jUen ( )

n!1

Uen ()j = 1 ;

whi h ontradi ts property (4.3.3). Hen e

kTn0 k a;b n2N kTn k ; sup

[

[

℄

1 1℄

= sup kTn0 k a;b = 1 n2N

[

℄

for every [a; b℄  ( 1; 1), whi h, together with Theorem 4.3.3 shows the if part of the orollary. ut

210 4. Denseness Questions

Comments, Exer ises, and Examples. We followed Borwein and Erdelyi [95a℄ in this se tion. Corollary 4.3.4, to be found in A hiezer [56℄, is proved by using entirely dierent methods. E.1 The Cru ial Detail in the Proof of Theorem 4.3.3. Suppose that M := (1; f ; f ; : : : ) is an in nite Markov system of C fun tions on [a; b℄ and f 0 does not vanish on (a; b): Suppose that the sequen e of asso iated Chebyshev polynomials (Tn ) has a subsequen e (Tni ) with no zeros on some subinterval [; ℄ of [a; b℄: Show that there exists another subinterval [ ; d℄ and another in nite subsequen e (Tni ) su h that for some Æ > 0 and > 0, and for ea h ni ; 1

2

2

1

kTni k ;d [

℄

<1 Æ

kTn0 i k ;d

and

[

℄

< :

Outline. For both inequalities rst hoose a subinterval [ ; d ℄  [; ℄ and a subsequen e (ni; ) of (ni ) su h that ea h alternation point of ea h Tni;1 is outside [ ; d ℄: Then hoose a subsequen e (ni; ) of (ni; ) so that either ea h Tni;2 is in reasing or ea h Tni;2 is de reasing on [ ; d ℄. Study the rst

ase; the se ond is analogous. Let [ ; d ℄ be the middle third of [ ; d ℄: If the rst inequality fails to hold with [ ; d ℄ and (ni; ); then there is a subsequen e (ni; ) of (ni; ) su h that kTni;3 k 2 ;d2 ! 1 as ni; ! 1. Hen e, there is a subsequen e (ni; ) of (ni; ) su h that either 1

1

1

1

1

2

1

1

2

2

1

2

3

2

2

[

4

1

1

2

3

℄

3

max Tni;4 (x) ! 1

min Tni;4 (x) ! 1 :

or

2 xd2

2 xd2

On e again, study the rst ase; the se ond is analogous. Sin e ea h Tni;3 is in reasing on [ ; d ℄; 1

1

lim

ni;4 !1

k1

Tni;4 k d2 ;d1 = 0 : [

℄

Now hoose g := a + a f + a f so that g has two distin t zeros  and  in [d ; d ℄; kgk 1;2 < 1; and g is positive on ( ;  ): Let := max g(x) and ge := g + 1 : Show that Tni;4 ge has at least n + 1 1 x2 distin t zeros in [a; b℄ if ni; is large enough, whi h is a ontradi tion. 0

1

2

2

1

1

[

1

2

2

1

℄

2

4

For the se ond inequality, note that E.4 of Se tion 3.2 implies that (f 0 ; : : : ; fn0 ) is a weak Chebyshev system on [a; b℄; and so is 1



    0 0 ! T0 0 T0 0 T ; ; : : : ; n0 ; T0 T0 T 2

3

1

1

n = 2; 3; : : : :

1

From this dedu e that ea h (Tn0 i;2 =T 0 )0 has at most one sign hange in [ ; d ℄: Choose a subinterval [ ; d ℄  [ ; d ℄ and a subsequen e (ni; ) of 1

2

2

3

3

2

2

5

4.3 Unbounded Bernstein Inequalities 211

(ni; ) so that none of (Tn0 i;5 =T 0 )0 hanges sign in [ ; d ℄: Choose a subsequen e (ni; ) of (ni; ) so that either ea h Tn0 i;6 =T 0 is in reasing or ea h Tn0 i;6 =T 0 is de reasing on [ ; d ℄: Again, onsider the rst ase; the se ond is analogous. Let [ ; d ℄ be the middle third of [ ; d ℄: If the se ond inequality fails to hold with [ ; d ℄ and (ni; ); then there is a subsequen e (ni; ) of (ni; ) su h that either Tn0 i;7 (x) lim max =1 ni;7 !1 4 xd4 T 0 (x) or Tn0 i;7 (x) lim min = 1: ni;7 !1 4 xd4 T 0 (x) On e again we just treat the rst ase; the se ond is analogous. In this ase, for every K > 0 there is an N 2 N su h that for every ni;  N we have Tn0 i;7 (x) > K ; x 2 [d ; d ℄ : 2

3

1

6

5

1

3

1

4

3

4

3

4

7

3

4

3

6

6

1

1

7

4

Hen e

d )

K (d

3

4

Z d3

d4

3

Tn0 i;7 (x) dx = Tni;7 (d ) Tni;7 (d )  2 ; 4

3

ut

whi h is a ontradi tion.

E.2 On the Uniform Closure of Nondense Markov Spa es. Suppose that M = (f ; f ; : : : ) is an in nite Markov system on [a; b℄ with ea h fi 2 C [a; b℄; and suppose that (f =f )0 does not vanish on (a; b): Suppose that span M fails to be dense in C [a; b℄: Show that there exists a subinterval [; ℄  [a; b℄;  < , su h that every g 2 C [a; b℄ in the uniform losure of span M on [a; b℄ is dierentiable on [; ℄: Hint: By Theorem 4.3.3 there exists an interval [; ℄  [a; b℄ and a onstant  2 R so that 0

1

2

1

0

kh0 k ;  khk a;b for every h 2 span M. Suppose g 2 C [a; b℄ and lim kh gk a;b = 0: n!1 n Choose ni 2 N su h that kg hni k a;b  2 i ; i = 0; 1; : : : : [

℄

[

[

[

g = hn0 +

1 X i=1

(hni

hni 1 )0 k ; it follows that g is dierentiable on [; ℄.

k(hni

℄

℄

Then Sin e

℄

[

hni 1 ) : ℄

 2

1

i;

ut

212 4. Denseness Questions

E.3 An Analog of Theorem 4.3.3 with Appli ations. Suppose that M = (f ; f ; : : : ) is an extended omplete Chebyshev (ECT) system of C 1 fun tions on [a; b℄, as in E.3 of Se tion 3.1. Note that E.3 a℄ of Se tion 3.1 implies that f does not vanish on [a; b℄: a℄ Show that the dierential operator D de ned by 0

1

0

 f 0 ; D(f ) := f 

2C

f

0

1

[a; b℄

maps M to MD ; where 

MD :=

f f

1

0 

0

0

f ; f

2 0

!

; :::

and MD is on e again an ECT system of C 1 fun tions on [a; b℄: Hint: Use E.8 b℄ of Se tion 3.2. ut n n b℄ The dierential operators D (f ) are de ned for every f 2 C [a; b℄ as follows. Let (

Fi; := fi ;  F Fi;n := i F

)

i = 0; 1; 2; : : : ; 0 ;n ; i = 0; 1; 2; : : : ; n = 1; 2; : : : ; ;n  n 0 D (f ) n D (f ) := f; D (f ) := ; n = 1; 2; : : : : F ;n 0

+1

1

0

1

(

(0)

(

1)

)

0

1

Let

MD

(1)

:= MD

MD n

and

( )

:=

MD n (


1)

D;

n = 2; 3; : : : :

Show that if span MD(n) is dense in C [a; b℄; then so is span M:

℄ Suppose that span M fails to be dense in C [ ; d℄ for every subinterval [ ; d℄  [a; b℄; < d: Show that for ea h n 2 N , there exists an interval [n ; n ℄  [a; b℄; n < n ; su h that (

sup

kD n (f )k n ; n kf k a;b (

)

[

[

℄

℄

: 0 6= f

2 span M

Hint: Use Theorem 4.3.1 and indu tion on n:

)

< 1:

ut

d℄ Suppose that span M fails to be dense in C [ ; d℄ for every subinterval [ ; d℄  [a; b℄ < d: Show that for ea h n 2 N ; there exists an interval [n ; n ℄  [a; b℄; n < n , where every g 2 C [a; b℄ in the uniform losure of span M on [a; b℄ is n times ontinuously dierentiable.

4.3 Unbounded Bernstein Inequalities 213

Hint: Use part ℄. The argument is similar to the one given in the hint to E.2. ut e℄ Suppose that span M fails to be dense in C [ ; d℄ for every subinterval [ ; d℄  [a; b℄; < d: Show that every fun tion in the uniform losure of span M on [a; b℄ is C 1 on a dense subset of [a; b℄. (This is the ase for Muntz systems

M := (x ; x ; : : : ) ; 0

i 2 R ;

1

1 X i=1 i 6=0

1

ji j < 1

on [a; b℄; 0  a < b; see E.7 of Se tion 4.2.) E.4 Bounded Bernstein-Type Inequality for Nondense Muntz Spa es. Suppose (i )1 i is a sequen e of distin t positive numbers satisfying =1

1 X  i=1 i

i

2

+1

< 1:

Then for every  > 0, there is a onstant  su h that

jp0 (x)j  x kpk

;

[0 1℄

for every

x 2 (0; 1 ℄

;

p 2 spanf1; x1 ; x2 ; : : : g :

To prove this pro eed as follows. Let  := 0; and let 0

Tn := Tn f ;  ; : : : ; n ; [0; 1℄g 0

1

be the Chebyshev polynomial for spanf1; x1 ; : : : ; xn g on [0; 1℄: Let

M () := spanf1; x1 ; x2 ; : : : g: a℄ Observe that for every  > 0 there exists a k 2 N depending only on (i )1 i and  (and not on n) su h that Tn has at most k zeros in [; 1 ℄. Hint: This is proved in the outline of the proof of E.18 d℄ of Se tion 4.2. u t b℄ Show that every nonempty (a; b)  (0; 1) ontains a nonempty subinterval (; ) for whi h there are integers 0 < n < n <


su h that none of the Chebyshev polynomials Tni vanishes on (; ). Hint: Use part a℄. ut =1

1

2

214 4. Denseness Questions

℄ Show that every nonempty (a; b) interval (; ) su h that sup



kp0 k ; kpk ; [



(0; 1) ontains a nonempty sub

℄

[0 1℄

: 0 6= p 2 M () < 1 :

Hint: Modify the proof of Theorem 4.3.3. d℄ Finish the proof of the initial statement of the exer ise. Hint: Use part ℄ and a linear s aling.

ut ut

E.5 The Closure of Nondense Muntz Spa es in C [0; 1℄. Suppose (i )1 i is a sequen e of distin t positive numbers satisfying

=1

1 X i=1

Show that

i < 1: i + 1 2

spanf1; x1 ; x2 ; : : : g  C 1 (0; 1) ;

that is, if f is the uniform limit of a sequen e from spanf1; x1 ; x2 ; : : : g; then f is in nitely many times dierentiable on (0; 1): Hint: Use E.4 with the substitution x = e t : ut E.6 A Nondense Markov Spa e with Unbounded Bernstein Inequality on a Subinterval. One may in orre tly suspe t that nondense Markov spa es on [a; b℄ an be hara terized by an everywhere bounded Bernstein inequality on (a; b), at least under the assumptions of Theorem 4.3.3. The purpose of this exer ise is to show that this is far from true, and in a sense, Theorem 4.3.3 is the best possible result. The same onstru tion an be used to give a nondense Markov spa e on [a; b℄ su h that the set

Z := fx 2 [a; b℄ : Tn(x) = 0 for some n 2 N g is neither dense nor nowhere dense in [a; b℄. Here (Tn )1 n is the sequen e of asso iated Chebyshev polynomials on [a; b℄. We onstru t an in nite Markov system on ( 1; 1) as follows. Suppose  := (i )1 i is a sequen e of even integers satisfying =0

=0

0 =  <  <  <


; 0

Suppose m > 0: Let 'k

1

2

1 X i=1

1

i

< 1:

2 C ( 1; 1), k = 0; 1; : : : , be de ned by

4.3 Unbounded Bernstein Inequalities 215 

k m 0 ' k (x) := x x2k ;; xx  0 k 2 +

2

2

and

'

k

2 +1

(x) :=



x

k m+1 ; x2k+1 ;

2 +

k

2 +1

x0 x  0;

where ( i )1 is a sequen e of positive numbers asso iated with a xed i sequen e of integers 0 := n < n < n <


and a onstant Æ > 0, and it is hosen as follows. Let =0

0

mj := nj

1

nj

+1

2

1;

j = 0; 1; : : : :

Let

Tn (x) := os(n ar

os(2x

1)) =:

n X i=0

i;n xi

be the nth Chebyshev polynomial on [0; 2℄: Now hoose the onstants i > 0 su h that +1 1 1 njX X

j =0 i=nj

i Æi ji

nj ;mj

j  1:

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

1

X

p(x) =

then

n X

p(x) = while

p(x) =

i=0

n X i=0

i=0

ai 'i (x) ;

ai xi ;

( 1)i ai xi ;

1; 1):

ai 2 R ;

x 2 [0; 1) ; x2(

1; 0℄ :

Now apply Theorem 3.2.4 (Des artes' rule of signs). ut b℄ Show that spanf' ; ' ; : : : g is not dense in C [ Æ; 2℄:

℄ Show that there is a sequen e of integers 0 := n < n < n <


su h that  0  k p k ; sup : 0 6= p 2 spanf' ; ' ; : : : g = 1 0

1

0

kpk

[

[

sup

jp0 (x)j kpk Æ; [

for every x 2 ( Æ; 0):

2

℄

0

Æ;2℄

1

for every nonempty interval [; ℄  [0; 2℄, while 

1

2℄

:



0 6= p 2 spanf' ; ' ; : : : g < 1 0

1

216 4. Denseness Questions

d℄ Suppose the sequen e ('i )1 i is de ned asso iated with a xed sequen e of integers 0 := n < n < n <


and Æ := 2: Let =0

0

1

2

Tn := Tnf' ; ' ; : : : ; 'n ; [ 2; 2℄g 0

1

be the Chebyshev polynomial for spanf' ; ' ; : : : ; 'n g on [ 2; 2℄: Let 0

1

Z := fx 2 [ 2; 2℄ : Tn(x) = 0 for some n 2 N g and let Z 0 be the set of all limit points of Z , and let Z 00 be the set of all limit points of Z 0 . Show that the sequen e of integers 0 := n < n < n <


in the de nition of ('i )1 i an be hosen so that 0

1

2

=0

Z 00 \ ( 2; 0) = ;

[0; 2℄ [ f 2g  Z 00 :

and

E.7 Nikolskii-Type Inequalities for Nondense Muntz Spa es. Suppose that p 2 [1; 1℄: Suppose (i )1 is a sequen e of distin t real numbers i greater than 1=p satisfying =0

1 X i=0

i + p

1



i + p 1

2

+1

< 1:

Show that for every  > 0, there exists a onstant  > 0 depending only on  and p so that jq(x)j  x =p kqkLp ; 1

[0 1℄

for every q 2 spanfx0 ; x1 ; : : : g and for every x 2 [0; 1 ℄: In parti ular, for every  > 0; there exists a onstant  > 0 depending only on  so that

kqk ; [

1

℄

  kqkLp

;

[0 1℄

for every q 2 spanfx0 ; x1 ; : : : g: Thus, spanfx0 ; x1 ; : : : g is not dense in Lp [0; 1℄: Hint: Use Holder's inequality to show, as in Operstein [to appear℄, that the operator J : Lp [0; 1℄ 7! L1 [0; 1℄ de ned by

J (')(0) := 0 ;

J (')(x) := x

1

=p

1

Z x

'(t) dt ; x > 0

0

is a bounded linear operator. That is, there is a onstant  0 su h that

kJ (')kL1

;

[0 1℄

 k'kLp

;

[0 1℄

4.3 Unbounded Bernstein Inequalities 217

for every ' 2 Lp [0; 1℄: Now observe that q 2 spanfx0 ; x1 ; : : : g implies is a that J (q ) 2 spanfx0 ; x1 ; : : : g; where i := i + p , and so (i )1 i sequen e of positive numbers satisfying 1

=0

1 X i=0

i < 1: i + 1 2

Therefore, by E.4, for every  > 0, there exists a onstant  > 0 su h that

j(J (q))0 (x)j  x kJ (q)k

 x kqkLp

;

[0 1℄

;

[0 1℄

for every q 2 spanfx0 ; x1 ; : : : g and for every x 2 (0; 1 x 2 (0; 1); (J (q ))0 (x) = x

=p

1

where

Z x q ( t ) dt

=x

1

1

0

Therefore

x

1

=p

jq(x)j



q(x)

1



1

=p

1

1



p

x

=p

1

p

1



kqkLp

;

q(t) dt ;

0

jJ (q)(x)j  x

1

Z x

2

℄. Note that for

[0 1℄

=p

1

kqkLp

  kqkLp

for every q 2 spanfx0 ; x1 ; : : : g and for every x 2 (0; 1

;

[0 1℄

:

;

[0 1℄

℄:

ut

E.8 The Closure of Nondense Muntz Spa es 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 satisfying 1 X i + p =0

1

i=0



i + p 1

2

+1

< 1:

Show that if f is a fun tion in the Lp [0; 1℄ losure of spanfx0 ; x1 ; : : : g ; then f 2 C 1 (0; 1); that is, f is in nitely many times dierentiable on (0; 1). Hint: Combine E.5 and E.7. ut

218 4. Denseness Questions

4.4 Muntz Rationals A surprising and beautiful theorem, onje tured by Newman and proved by Somorjai [76℄, states that rational fun tions derived from any in nite Muntz system are always dense in C [a; b℄; a  0: More spe i ally, we have the following result. Theorem 4.4.1 (Denseness of Muntz Rationals). Let (i )1 i be any sequen e of distin t real numbers. Suppose a > 0. Then =0

 Pn ai xi Pin=0 bi xi

i=0

: ai ; bi 2 R ; n 2 N



is dense in C [a; b℄: The same result holds when a = 0, however, the proof in this ase requires a few more te hni al details; see E.1 b℄. The proof of this theorem, primarily due to Somorjai, rests on the next theorem. We introdu e the following notation. A fun tion Z de ned on (a; b) is alled an -zoomer ( > 0) at  2 (a; b) if

Z (x) > 0 ; x 2 [a; b℄ ; Z (x)   ; x <  ; Z (x)   ; x >  +  :

(4:4:1)

1

While (approximate) Æ -fun tions are the building blo ks for polynomial approximations, the existen e of -zoomers is all that is needed for rational approximations. More pre isely, we have the following result. Theorem 4.4.2 (Existen e of Zoomers and Denseness). Let S be a linear subspa e of C [a; b℄: Suppose that S ontains an -zoomer for every  > 0 at every  2 (a; b): Then

R(S ) :=



p : p; q 2 S q



is dense in C [a; b℄: Proof. It suÆ es to onsider the ase [a; b℄ = [0; 1℄: Let n 2 N and  > 1=n be xed. We onstru t a partition of unity indu tively as follows. Let Z be any positive fun tion in S and hoose fun tions Z ; Z ; : : : ; Zn 2 S positive on [0; 1℄ so that k Zk (x) < Zk (x) ; x<  n 0

1

2

1

and

Zk (x) >  Zk (x) ; 1

1

x>

k

n

1

+

4.4 Muntz Rationals 219

(whi h the existen e of -zoomers allows for). Let (4:4:2)

Zk (x) ; i Zi (x)


k (x) :=

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

Pn

=0

So

x 2 [0; 1℄


k (x) > 0 ;

and

n X k=0

Sin e for every x 2 [0; 1℄; n X k k=0 n >x

Zk (x) +


k (x) = 1 :

n X k k=0 n + < x

Zk (x) < 2

n X k=0

Zk (x) ;

we also know that n X

j Now let f

k=0 k n x >


k (x)  2 ;

x 2 [0; 1℄ :

j

2 C [0; 1℄; and onsider the approximation n X k=0

f

k
n
k (x)

2 R(S ) :

Then

n X
k f n
k (x) f (x) k=0 n n X X

k f (x) f n
k (x) + k=0 k=0 k k n x  n x >



j

j

 !f () + 2 !f (1) ;

j

j

whi h nishes the proof.

f (x)

f



k

n
k (x)

ut

We an now nish the proof of Theorem 4.4.1:

Proof of Theorem 4.4.1. We may suppose, on passing to a subsequen e if ne essary, that (i )1 i is a onvergent sequen e (possibly onverging to in nity). We may also assume that a := 1=b with b > 1: Sin e C [1=b; b℄ is =0

220 4. Denseness Questions

invariant under x ! 1=x, we may assume that (i )1 has a nonnegative i (possibly in nite) limit. Case 1: The sequen e (i )1 untz's i has a nite nonnegative limit. Then M theorem on [a; b℄, a > 0 (see E.7 of Se tion 4.2), yields that the Muntz polynomials themselves asso iated with (i )1 i are already dense in C [1=b; b℄. 1 Case 2: The sequen e (i )i tends to in nity. In this ase (x= )i is an -zoomer at  2 (a; b) for suÆ iently large i , and the result follows from Theorem 4.4.2. ut =0

=0

=0

=0

Comments, Exer ises, and Examples. A omparison between Muntz's theorem and the main result of this se tion shows the power of a single division in approximation. In what other

ontexts does allowing a division reate a spe ta ularly dierent result? Newman [78, p. 12℄ onje tures that if M is any in nite Markov system on [0; 1℄; then the set  

p : p; q 2 span M q of rational fun tions is dense in C [0; 1℄:

Newman alls this a \wild onje ture in sear h of a ounterexample." It does, however, hold for both

M = (x ; x ; : : : ) ; 0

(see E.1 b℄) and

M=



x

1



1

;

x

i  0 are distin t

1

1



2

; :::



;

i 2 R n [0; 1℄ are distin t

(see E.2). A ounterexample to the full generality of this onje ture is presented in E.6. However, the hara terization of the lass of Markov systems for whi h it holds remains as an interesting question. In parti ular, it is open if Newman's onje ture holds for Des artes systems. The reader is referred to Newman [78℄ for an extensive treatment of these matters; see also Zhou [92a℄. In [78, p. 50℄ Newman asks about the P P 2 2 denseness of the produ ts ( ai xi )( bi xi ) in C [0; 1℄ (see E.3). He spe ulates that this \extra" multipli ation of Muntz polynomials should not arry the utility of the \extra" division. This is proved in Se tion 6.2, where it is shown that produ ts pq of Muntz polynomials from nondense Muntz spa es never form a dense set in C [0; 1℄. E.1 Denseness of Muntz Rationals on [0; 1℄. A fun tion C de ned on [a; b℄ is alled an - rasher,  > 0, at  2 (a; b) if

C (x) > 0 ; C (x)   ; C (x)   ; 1

x 2 [a; b℄ ; x >  +; x   :

4.4 Muntz Rationals 221

a℄ Let S be a linear subspa e of C [a; b℄: Suppose that S ontains an  rasher for every  > 0 at every  2 (a; b): Show that

R(S ) :=



p : p; q 2 S q



is dense in C [a; b℄: Hint: The argument is a trivial modi ation of the proof of Theorem 4.4.2.

ut

Let (i )1 be a sequen e of distin t real numbers. Let R() be the i spa e of fun tions f 2 C [0; 1℄ of the form Pn a xi f (x) = Pin i i ; ai ; bi 2 R ; n 2 N ; x 2 (0; 1℄ : i bi x b℄ Show that if (i )1 i is a sequen e of distin t nonnegative real numbers with  := 0; then R() is dense in C [0; 1℄: Hint: As in the proof of Theorem 4.4.1, we may suppose, on passing to a subsequen e if ne essary, that (i )1 is a onvergent sequen e (possibly i =0

=0

=0

=0

0

onverging to 1). Distinguish the following two ases. Case 1: lim i = 1: Given  > 0 and  2 (0; 1); show that =1

i!1

Z (x) :=



2

+

 i

x 

is an -zoomer at  on [0; 1℄ if i is large enough. Use Theorem 4.4.2 to nish the proof. Case 2: (i )1 i is a sequen e with nite limit. Given  > 0 and  2 (0; 1); show that =1





+ ( 1)n Tn f ;  ; : : : ; n ; [; 1℄g(x) 2 4 is an - rasher at  on [0; 1℄ if n is large enough. This an be proved by using Muntz's theorem (E.7 of Se tion 4.2) on [ ; 1℄ with  2 (0;  ); E.3

℄ of Se tion 3.3, and the monotoni ity of C on [0;  ℄: Finish the proof by part a℄. ut The result of E.1 b℄ is due to Bak and Newman [78℄. Zhou [92b℄ extends this result to sequen es of arbitrary distin t real numbers.

C (x) :=

0

1

E.2 Another Markov-System with Dense Rationals. Let ( i )1 be any i sequen e of distin t numbers in R n [0; 1℄: Show that =1

( Pn

ai i=1 x i Pn bi i=1 x i

is dense in C [0; 1℄:

: ai ; bi 2 R ; n 2 N

)

222 4. Denseness Questions

E.3 Produ ts of Muntz Spa es. Asso iated with a sequen e  := (i )1 i of real numbers, let

=0

(

n X

M () := 2

i=0

!

ai

n X

xi

i=0

!

bi

xi

: ai ; bi 2 R ; n 2 N

)

:

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

=0

=0

2

=0

2

=0

2

2

2

=1

 Pn ai xi Pin=0 bi xÆi

i=0

: ai ; bi 2 R ; n 2 N



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

 k < Æk 2

+1

< k

+1

;

k = 0; 1; : : :

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


: Let a > 0: Show that 0

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

2 R;

n2N

1



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

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

Rn (f ) :=

(

inf

f (x)

Pn

Pin=0

i=0

ai xi

bi xi

;

: ai ; bi 2 R

[0 1℄

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

)

;

4.4 Muntz Rationals 223

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

Rn (f ) = C!f

1

n




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

2

n X

f (x)
1

k=0



k
n
k (x)

f

 6 !f

1

n




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

Rn (f ) = O !f

1

n





=0

?

Hint: Estimate the \degree" of the zoomers Zk de ned in Theorem 4.4.2.

ut

Use the zoomers de ned in Case 1 of the hint for E.1 b℄.

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




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

1

1

2

1

=0

2

=0

=0

=0

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

=0

0 =  <  <  <


; 0

Let 'k

2 C[

1

2

1 X i=1

1

i

< 1:

1; 1℄; k = 0; 1; : : : , be de ned by

' k (x) := x2k 2

and

'

k (x) :=

2 +1



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

224 4. Denseness Questions

1; 1):

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

1

n X

p(x) = then

n X

p(x) = while

p(x) =

i=0

i=0

ai 2 R ;

ai 'i (x) ;

x 2 [0; 1) ;

ai xi ;

n X i=0

x2(

( 1)i ai xi ;

1; 0℄ :

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

n

X

ai xi



i=0

L2 [0;1℄



n

X

ai xi



i=0

L2 [1=2;1℄

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

℄ The inequalities n

1X jai j 3i

2

Z



n X

1

i=0

0

=0

p

ai 2i

!2

+ 1 xi

dx 

ut n 5X jai j 3i

2

=0

hold for all hoi es of ai 2 R:

Proof. We have Z

1

0

n X i=0

p

!2

ai 2i

+ 1 xi =

n X i=0

dx

jai j

2

+2

n nXk X k=1 j =0

p

aj aj

k

+

p

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

+

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

p

+

+

+

2

2 +2

2 +

2 +2

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

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

4

k=1

7 k

i=0

jai j  23 2

n X i=0

jai j

2

;

ut

and the result follows. d℄ The inequality

n

X

ai xi



i=0

L2 [0;1℄



p

5

n

X

(



1)i ai xi

i=0

L2 [0;1℄

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

Pin=0

i=0

ut

ai 'i ; ai ; bi 2 R ; bi 'i

n2N

are not dense in C [ 1; 1℄:

Proof. Consider f

2 C[

1; 1℄ de ned by

f (x) :=

8 > < > :

1 0

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

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



Pn

i=0 ai 'i

Pn

b'

f

i=0 i i

[

;

<  < 1;

ai ; bi 2 R :

1 1℄

This implies (4:4:3)

Pn

i=0 ai xi

Pn

b xi

i=0 i

<

;

[0 1℄

and (4:4:4)

Pn

Pin=0 (

(

i=0

1)i ai xi 1)i bi xi



1

= ;

[1 2 1℄

< :

226 4. Denseness Questions

Without loss of generality we may assume that

n

X

 i bi x



(4:4:5)

i=0

L2 [0;1℄

= 1:

From (4.4.3) and (4.4.5) it follows that

n

X

 i ai x



(4:4:6)

i=0

L2 [0;1℄

< :

Part d℄, together with (4.4.3) and (4.4.5), implies that

n

X

(

(4:4:7)

i=0



p

1)i ai xi

L2 [0;1℄

< 5:

Part d℄, together with (4.4.5), also implies that

n

X

(

(4:4:8)

i=0



1)i bi xi

L2 [0;1℄

 p1

5

:

Combining part b℄ and (4.4.8), we obtain that

n

X

(

(4:4:9)

i=0



1)i bi xi

L2 [1=2;1℄

 p1

5

:

Now (4.4.7) and (4.4.9) yield the right-hand side of 1



Pn

i=0 ( Pn

k

1)i ai xi L2 = ; i  i ( 1) bi x i kL2 = ; =0

[1 2 1℄

 5 

[1 2 1℄

while (4.4.4) yields the left-hand side of it. This shows that  > 0 annot be arbitrarily small. ut

This is page

227

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5

Basi Inequalities

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

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

1 1℄

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

228 5. Basi Inequalities

Theorem 5.1.1 (Remez Inequality). The inequality

kpk

[

;

1 1℄

 Tn



2+s 2 s



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

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

k
k denote the

1; 1℄ : jp(x)j  1g)  2

sg :

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

kp k =

sup

p2Pn (s)

kpk :

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



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

2





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

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

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

q (x) := p 1





2

1

+

 +1 2

x



and q (x) := p 2



 +1 2

+



2

1



x

5.1 Classi al Polynomial Inequalities 229

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

2 Pn(s) is extremal, the

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

2

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

2

1

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

(5:1:2)

2

1 < x < 1;

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

jp (x)j  1 ;

(5:1:3)

1x1

s:

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

8 > <

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

2

1

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

1

2

2

1

3

3

2

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

2

2

1

3

p (x) := 1

m Y j =1

(x

xj ) ;

p := 2

p : p 1

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

2

2

 has the 3

230 5. Basi Inequalities

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

2

1

2

1

2

p (x) = Tn



2x + s  2

s

are the only extremal polynomials. In parti ular,

kpk = Tn





2+s : 2 s

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

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

s:

The inequality 

ktkK  Tn 22 + 



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



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

2 Tn

and s

2 os  +   ; 2



2

(0; 2 ) satisfying (5.1.4), and

 = 1 os(s=2) :

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

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

5.1 Classi al Polynomial Inequalities 231

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

ktkK  Tn 22 + 



;

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

e t()

is even and

1

2

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

2s :

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

e t

L1 (K )

where  := 1

 Tn



2+ 2 



;

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



p

Tn (x)  x + x we have 

2+ Tn 2 





p

 p  exp n 2 +  exp(4ns)

= et(0)

n

1 + 2 + =2 1 =2

for every s 2 (0; =2℄: Therefore

ktkK = t(0)

1

2



 Tn 22 + 



7 4

!n



x  1;

;





 exp



p

1 + 2 +  7

n

2

 exp

n s+ s 7

2





4

n s+ s 7 4

2





 exp(4ns)

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

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

n s+ s 7

2





4

ut

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

K:

232 5. Basi Inequalities

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

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

2 2

2

2K

2

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

(5:1:5)

2

2 2

2

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

Tn;() = t()

): It is easy to see that

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

and

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

2

2

2

0 ()j jt0 ()j > jTn;

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

and

Hen e E.4 yields that 0 6= t Tn; 2 Tn has at least 2n + 2 distin t zeros in K; whi h is a ontradi tion. To nd all the extremal polynomials, see the hint to E.5. ut As a orollary of Theorem 5.1.3 we have kt0 kK  nktkK for every t 2 Tn ; and by indu tion on m we obtain the following theorem: Theorem 5.1.4 (Bernstein's Inequality). The inequality

kt m kK  nm ktkK (

)

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

t 2 Tn :

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

)

(

)

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

)

(

)

(

)

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

5.1 Classi al Polynomial Inequalities 233

Corollary 5.1.6. The inequality

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

jp0 (z )j = j

i t0 ( )

j  nktkK = nkpkD ;

ie

z = ei :

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

ut

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

jp0 (x)j  p

1

n

x

2

kpk

[

;

1 1℄

;

1<x<1

holds for every p 2 Pn : The next theorem improves the previous result if x is lose to 1: Theorem 5.1.8 (Markov's Inequality). The inequality

kp0 k

;

[

1 1℄

 n kpk 2

[

;

1 1℄

holds for every p 2 Pn : A proof an be given as a simple ombination of Theorem 5.1.7 and the next theorem. Theorem 5.1.9 (S hur's Inequality). The inequality

kpk

;

[

p

1 1℄



 n p(x)

1

k

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

x

2

[

;

1 1℄

holds for every p 2 Pn : 1

Proof. Let

xk := os

(2

1)

n

2



;

so the numbers xk are the zeros of the Chebyshev polynomial

Tn (x) = os(n ar

os x) :

234 5. Basi Inequalities

Assume that p 2 Pn

1

p

and kp(x) 1

jp(y)j  (1

y)   = sin 2n 2

x =

2

k

 1: If jyj  xn ; then

;

[

1 1℄

 (1 xn )    2 2n

1 2

1

2

=

1 2 1

= n:

Now let jy j 2 (xn ; 1℄: Without loss of generality we may assume that y 2 (xn ; 1℄: The Lagrange interpolation polynomial of a p 2 Pn with nodes x ; x ; : : : ; xn is just p itself, hen e E.6 of Se tion 1.1 yields 1

1

2

jp(y)j = =

xk )

n X T (y ) p(xk ) 0 n T ( x n k )(y k=1 n q 1 X p(xk ) 1 x2k n

k=1

n 1X Tn(y)

n

k=1

y

xk



Tn (y) y xk

 n1 Tn0 (y)  n1 Tn0 (1) = n ;

where we use the fa ts that

Tn(y) > 0; y xk

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

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

ut

2

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

1

and 5.1.9 yield

kp0k

[

;

1 1℄

 n p0 (x)

and the theorem is proved.

p

1

x

2



[

;

1 1℄

and Theorems 5.1.4

 n kpk 2

[

;

1 1℄

;

ut

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

5.1 Classi al Polynomial Inequalities 235

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

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

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

1 1℄

[

1 1℄

[

1 1℄

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

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

[

1 1℄

1 1℄

1

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

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

[

[

℄

℄

1

1

2

1

2

2

236 5. Basi Inequalities

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

ontinuous fun tions on K := R (mod 2); dierentiable at a xed  2 K; and suppose f and g have the properties (1) kf kK = 1; kg kK < 1; (2) there are   x < x <


< x n <  so that f (xj ) = ( 1)j ; (3) f ()) = g (); jg 0 ()j > jf 0 ()j; and sign(g 0 () = sign(f 0 ()): Then f g has at least 2n + 2 distin t zeros on K: 1

2

2

E.5 Sharpness of Theorem 5.1.3. Let  2 K be xed. Suppose t 2 Tn and

t0 () + n t () = n ktkK : 2

2 2

2

2

Show that either jt()j = ktkK or t is of the form

t( ) = os(n

; 2 R :

) ;

Hint: Suppose ktkK = 1 and jt()j < 1: Choose  2 K su h that (5.1.6) is satis ed and show that t Tn; 2 Tn has at least 2n + 2 zeros on K ( ounting multipli ities). ut E.6 Sharpness of Theorem 5.1.4. Show that Theorem 5.1.4 is sharp and equality holds if and only if t is of the form

t( ) = os(n

; 2 R :

) ;

E.7 Sharpness of Corollary 5.1.6. Show that Corollary 5.1.5 is sharp and equality holds if and only if p is of the form p(z ) = z n ; 2 C : E.8 Sharpness of Theorem 5.1.7. Show that for a xed integer n  1; Theorem 5.1.7 is sharp if and only if x is a zero of the Chebyshev polynomial Tn ; that is, k = 1; 2; : : : ; n ; x = os k n  ; and p = Tn for some 2 R: (2

1)

2

E.9 Sharpness of Theorem 5.1.9. Show that Theorem 5.1.9 is sharp and equality holds if and only if p = Un for some 2 R; where Un is the Chebyshev polynomial of the se ond kind de ned in E.10 of Se tion 2.1. E.10 Sharpness of Theorem 5.1.8. Show that Theorem 5.1.8 is sharp and equality holds if and only if p = Tn for some 2 R: E.11 A Property of the Zeros of t 2 Tn . Let  2 K be xed. Show that every t 2 Tn has at most

M := enrjt()j

1

ktkK

zeros ( ounting multipli ities) in the interval [

r;  + r℄; r > 0:

5.1 Classi al Polynomial Inequalities 237

Proof. Assume that t 2 Tn has m > M zeros in [ r;  + r℄: Interpolate t at these m zeros by a Hermite interpolation polynomial of degree at most m 1 (see E.7 of Se tion 1.1). This gives the identi ally zero polynomial. The formula for the remainder term of the Hermite interpolation polynomial and Theorem 5.1.4 (Bernstein's inequality) yield that there exists a  2 ( r;  + r) su h that 

m

jt()j = m1 ! rm jt m ( )j < me rm nm ktkK  enr m ktkK  jt()jm ktkK m  jt()j ; < M (

)

1

whi h is impossible.

ut

E.12 A Property of the Zeros of a p 2 Pn . Show that every p 2 Pn has at most e p M := p n r jp(1)j kpk ; 2 zeros ( ounting multipli ities) in [1 r; 1℄; r > 0: Hint: Use the substitution x = os ; E.11, and the inequality 1

os r < 1

1 4

[

1 1℄

0 < r  2:

r ; 2

ut E.13 Riesz's Lemma. a℄ Suppose t 2 Tn and t() = ktkK = 1 for some  2 K: Then

t()  os(n(

 ; +   ; n 2n  ;  +   if and 2n 2n 

)) ; 

2 

2

and equality holds for a xed  2  only if t is of the form t(
) = os(n( )). In parti ular, t does not vanish in  n ;  + n : Hint: If this were false, then 2

2

q( ) := t( )

os(n(

))

would have more than 2n zeros on K ( ounting multipli ities). b℄ Suppose p 2 Pn and p(1) = kpk ; = 1: Then that [

ut

1 1℄



p(x)  Tn (x) ;

x 2 os





 ; 1 n

2

;

and equality holds for a xed x 2 os n ; 1 if and only if p = Tn, where Tn is the Chebyshev polynomial ofdegree n as de ned by (2.1.1). In parti ular, p does not vanish in os n ; 1 : 2

2

The next two exer ises follows Erdelyi [88℄ and Erdelyi and Szabados [89b℄.

238 5. Basi Inequalities

E.14 A Markov-Type Inequality for Trigonometri Polynomials on [ !; ! ℄. Show that there exists a onstant 0 <  16 su h that

ks0k



!;!℄

[



n+



!

!

n



2

ksk

!;!℄

[

for every s 2 Tn and ! 2 (0;  ℄:

Proof. If 

! > (2n) ; then 1

(3 tan (!=2) + 1) = < 8n ; 2

1 2

and E.19 ℄ gives the result. If  !  (2n) ; then Theorem 5.1.4 (Bernstein's inequality) ombined with the Mean Value Theorem yields 1

ksk

[

and hen e

;℄

ksk

[

 ksk

[

!;!℄

!)n ksk

!;!℄ + (

 (1

!)) ksk

n(

;℄ ;

[

[

;℄

for every s 2 Tn : Therefore, using Theorem 5.1.4 (Bernstein's inequality) and  !  (2n) ; we get 1

ks0 k

[

!;!℄

 ks0k ;  n ksk  1 n(n !) ksk [

℄

[

[

;℄ !;!℄

 (n + 2(

!)n )ksk 2

[

!;!℄

for every s 2 Tn :

ut

E.15 S hur-Type Inequality for Tn on [ !; ! ℄. Let ! 2 (0; 2 ℄. Show that

ksk

[

!;!℄

2n + 1  sin( s( ) !=2)

1 2


1=2

( os 

os ! )

[

!;!℄

for every s 2 Tn , and equality holds if and only if s is of the form h

= sin (2n + 1) ar

os != s( ) = ( os  os ! ) = sin(

2)

sin(

2)

1 2

i

;

2 R:

Note that the right-hand side of the above is an element of Tn : Hint: De ne 2n + 1 distin t points in ( !; !) by 

k := 2 ar sin sin

! 2

sin



k ; 2n + 1

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

Distinguish two ases in estimating jsn ()j for  2 [ !; ! ℄:

5.1 Classi al Polynomial Inequalities 239

Case 1: jj  jn j. Use the inequality 1 ( os  2

os ! ) >

sin (!=2) (2n + 1) 2

2

to get the desired result. Case 2: n < jj  !: Let

Mn(s) := max fjs(k )j(( os k jkjn

os ! )=2) = g: 1 2

Let n 2 N , ! 2 (0;  ℄, and  2 [ !; n ) [ (n ; ! ℄ be xed. Show that there is an sen 2 Tn su h that

jsen ()j = max js()j : Mn (sen ) s2Tn Mn(s) Show by a variational method that sen is of the form h

sen ( ) =

sin (2n + 1) ar

os ( os 

=2) !=2)

sin(

i

sin(

os ! ) =

1 2

;

2 R:

ut E.16 Another Proof of S hur's Inequality. a℄ Prove Theorem 5.1.9 (S hur's inequality) by using the method given in the hint to E.15. b℄ Prove the result of E.15 by using interpolation, as in the proof of Theorem 5.1.9. Proof. See Erdelyi and Szabados [89b℄. ut E.17 Growth of Polynomials in the Complex Plane. Let

D := fz 2 C : jz j < 1g

D% := fz 2 C : jz j < %g:

and

a℄ Show that

jp(z )j  jz jn kpkD for every p 2 Pn and z 2 C n D: Find all p 2 Pn for whi h equality holds.

Hint: Apply the maximum prin iple (see E.1 d℄ of Se tion 1.2) with D and q(z ) := z n p(z ) 2 Pn : 1

ut

240 5. Basi Inequalities

b℄ Show that the transformation z = M (w) := (w + w ) an be written as x = (% + % ) os  ; y = (% % ) sin  ; 1

1

2

1

1

1

2

1

2

where z = x + iy; x; y 2 R and w = %ei ; % > 0;  2 K:

℄ For % > 0; let E% be the image of the ir le D% under M: Show that E% is the ellipse

x y + = 1 with semiaxes a := (% + % ) and b := j% % a b 2

2

2

2

1

1

1

2

2

1

j:

Furthermore, E% = E% 1 ; and E is the interval [ 1; 1℄ overed twi e. d℄ Show that jp(z )j  %n kpk ; 1

for every p 2 P

n;

[

z 2 E% ; % > 1:

1 1℄

Proof. Applying the maximum prin iple (see E.1 d℄ of Se tion 1.2) with D = D and
Q(w) := wn p (w + w ) 2 P n ; 1

1

1

2

2

we obtain

jQ(w)j  kQkD

= kpk

1

;

[

1 1℄

w 2 D% 1 :

;

This, together with part ℄, yields

jp(z )j  %n kpk

[

;

1 1℄

z 2 E% :

;

ut e℄ Show that there exists an absolute onstant su h that

jp(z )j  kpk

;

1 1℄

p 2 Pn

;

whenever

z = x + iy ; x; y 2 R ; (j1

jxj  1 + n

2

;

jyj  j1 n x j 2

+

+

1

n

2

:= maxf1 x ; 0g). Hint: Use part d℄. ut f℄ Prove the following Markov-Bernstein inequality. There is a onstant

(m) depending only on m su h that

x

2

j

p

1

2

+

jp

(

m) (x)

j  (m)



min

for every p 2 Pn and x 2 [ 1; 1℄:



n ;p 2

1

n

m

x

2

kpk

[

;

1 1℄

5.1 Classi al Polynomial Inequalities 241

Hint: Use part e℄ and Cau hy's integral formula (see E.1 a℄ of Se tion 1.2).

ut

Bernstein established Theorem 5.1.4 in order to prove inverse theorems of approximation. Bernstein's method is presented in the proof of the next exer ise, whi h is one of the simplest ases. However, several other inverse theorems of approximation an be proved by straightforward modi ations of the proof of this exer ise. That is why Bernstein- and Markov-type inequalities play a signi ant role in approximation theory. Dire t and inverse theorems of approximation and related matters may be found in many books on approximation theory, in luding Cheney [66℄, Lorentz [86a℄, and DeVore and Lorentz [93℄. E.18 An Inverse Theorem of Approximation. Let Lip ,  2 (0; 1℄, denote the family of all real-valued fun tions g de ned on K satisfying sup For f



jg(x) g(y)j : x 6= y 2 K  < 1: jx yj

2 C (K ), let En (f ) := inf fkt f kK : t 2 Tn g:

An example for a dire t theorem of approximation is stated in part a℄. Part b℄ deals with its inverse result. a℄ Suppose f is m times dierentiable on K and f m 2 Lip for some  2 (0; 1℄: Then there is a onstant C depending only on f so that (

En (f )  Cn

m+) ;

(

)

n = 1; 2; : : : :

Proof. See, for example, Lorentz [86a℄. b℄ Suppose m is a nonnegative integer,  2 (0; 1), and f there is a onstant C > 0 depending only on f su h that En (f )  Cn

m+) ;

(

ut

2 C (K ). Suppose

n = 1; 2; : : : :

Then f is m times ontinuously dierentiable on K and f

m)

(

2 Lip :

Outline. We show only that f is m times ontinuously dierentiable on K: The rest an be proved similarly, but its proof requires more te hni al details. See, for example, Lorentz [86a℄. For ea h k 2 N ; let Q k 2 T k be hosen so that 2

kQ k 2

2

f kK  C 2

k(m+) :

242 5. Basi Inequalities

Then

kQ k

Q

2 +1

Now

f ( ) = Q 0 ( ) + 2

1 X k=1

k  2C 2

k K

2

(Q

k(m+) :

2K;

Q k )() ;

k

2 +1

2

and by Theorem 5.1.4 (Bernstein's inequality)

jQ j j

( ) 20

1 X + (Q2k+1

Q2k ) j

( )

k=1

 kQ kK + 1

 kQ kK + 1

()

1 X k=1

1 X k=1

 kQ kK + 2j 1

+1

2k

+1

2k

+1


j
j

kQ k

Q

2 +1

k

k K

2

2C 2 k m  (

+

)

1 X

(2j m  )k < 1

C

k=1

for every  2 K and j = 0; 1; : : : ; m; sin e  > 0: Now we an on lude that f j () exists and ( )

f j () = Q j () + ( )

( ) 1

1 X k=1

(Q

Q k ) j ()

k

( )

2 +1

2

for every  2 K and j = 0; 1; : : : ; m: The fa t that f m seen by the Weierstrass M -test. (

)

2 C (K ) an be ut

The next exer ise follows Videnskii [60℄. E.19 Videnskii's Inequalities. The main results of this exer ise are the Bernstein- (part b℄) and Markov-type (part ℄) inequalities for trigonometri polynomials on an interval shorter than the period. Let ! 2 (0;  ); 



sin(=2) tn () := Qn;! () = os 2n ar

os sin(!=2) and



un() := sin 2n ar

os a℄ Re all that tn 2 Tn by E.3 a℄.



sin(=2) sin(!=2)



:



;

5.1 Classi al Polynomial Inequalities 243

b℄ Show that

js0n ()j  jt0n () + iu0n()j ksn k 

os (!=2)

os (=2) 2

=n 1

[



!;!℄ =

1 2

ksn k

!;!℄

[

2

for every sn 2 Tn and  2 ( !; ! ): Equality holds if and only if sn = tn for some 2 R and sn () = 0: Hint: First show that for every n 2 N ; ! 2 (0; ℄; and  2 [ !; !℄; there exists an se 2 Tn su h that

jse0 ()j ksek !;! [

js0n ()j sn 2Tn ksn k !;!

:

= max ℄

[

℄

Use a variational method to show that either

ksen k

[

!;!℄

or there exist  2 ( ; ! ℄ and

2 R; where, as in Se tion 3.3,

= ksen k ; [

℄

2 [!; ) su h that sen = Tn for some

Tn := Tn f1; os ; sin ; : : : ; os n; sin n ; [; ℄g is the Chebyshev polynomial for Tn on [; ℄: In the rst ase use Theorem 5.1.4 (Bernstein's inequality). In the se ond ase observe that

Tn( ) = tn with

:= ( + )



sin(( )=2) sin(!e =2)

!e := (

and

1 2

1 2

℄ Show that if 2n > (3 tan (!=2) + 1) 2

ks0n k

[

!;!℄

 t0n (!)ksn k

[

!;!℄



=

1 2

) :

ut

; then

= 2n ot(!=2)ksn k !;! 2

[

℄

for every sn 2 Tn ; and equality holds if and only if sn = tn ; 2 R:

Outline. Let ! =  <  <


<  0

1

tn (j ) = ( 1)j ; and

un(j ) = 0 ;

2

n

= ! be the points where

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

244 5. Basi Inequalities

and let  <  <


<  n be the zeros of tn (whi h lie in ( !; ! )): Note that u0n (j ) = 0 ; j = 1; 2; : : : ; 2n : 1

2

2

Step 1. Show that 2n > (3 tan (!=2) + 1) = implies t00n () > 0 for every  2 [ n ; !℄; so t0n is in reasing on [ n ; !℄: Step 2. Use part b℄ to show that if  2 [ ;  n ℄; then 2

2

1 2

1

2

1

1

2

jt0n ()j  jt0n () + iu0n()j  jt0n ( n ) + iu0n ( n )j = jt0n ( n )j : 2

2

2

Step 3. Dedu e from Steps 1 and 2 that

jt0n ()j < jt0n (!)j ;

 2 ( !; !) :

Step 4. Show that there is an se 2 Tn su h that

ksen0 k ksenk

[

[

!;!℄ !;!℄

ks0 k s2Tn ksk

= max

[

[

!;!℄ !;!℄

:

For the rest of the proof let sen be normalized by ksek !;! = 1 and let  2 [ !; !℄ be hosen so that [

jsen0 ( )j = ksen0 ()k

[

℄

!;!℄ :

Let (a < a <


< am ) be an alternation sequen e of maximal length for sen 2 C [ !; !℄ on [ !; !℄: We would like to show that m = 2n + 1: Clearly m < 2n + 2: Step 5. Use a variational method to show that 2n  m: Step 6. Use a variational method to show that  = ! implies m = 2n +1; so m = 2n implies  2 ( !; ! ): Step 7. Show by a variational method that 1

2

js0n (!)j  jt0n (!)j ksnk

[

!;!℄

for every sn 2 Tn ; and equality holds if and only if sn = tn ; parti ular, if m = 2n; then

jsen0 (!)j < jt0n (!)j : Step 8. Use part b℄ and Step 3 to show that

jsen0 ()j  t0n ( n ) < t0n (!) ; 2

 2 [ ;  n ℄ :

Step 9. Use part b℄ to show that m = 2n implies

1

2

2 R:

In

5.1 Classi al Polynomial Inequalities 245

jsen0 (j )j < jt0n (j )j = (

1)j t0n (j ) ;

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

Step 10. Suppose m = 2n: Use Steps 6 and 8 to show that

 2 ( !;  ) [ ( n ; !) : 1

2

Step 11. Suppose m = 2n: Use the de ning property of sen and Steps 1 and 10 to show that

jsen0 ( )j = ksen0 k

[

!;!℄

 jt0n (!)j > jt0n ( )j :

Step 12. Suppose m = 2n and sn0 ( )  0: Use Steps 7 to 11 to show that t0n se0n has at least 2n + 1 distin t zeros in ( !; !); a ontradi tion. Step 13. Show that m = 2n + 1 and sen = tn . ut E.20 Inequalities for Entire Fun tions of Exponential Type. Entire fun tions of (exponential) type  are de ned in E.17 of Se tion 4.2. Denote by E the set of all entire fun tions of exponential type at most : This exer ise

olle ts some of the interesting inequalities known for E : Sin e a trigonometri polynomial of degree n belongs to En ; these results an be viewed as extensions of the orresponding inequalities for trigonometri polynomials. More on various inequalities for entire fun tions of exponential type may be found in Rahman and S hmeisser [83℄. a℄ Bernstein's Inequality. The inequality

kf 0kR   kf kR ;

x2R

holds for every f 2 E : Proof. See Bernstein [23℄ or Rahman and S hmeisser [83℄. b℄ Extension of the Bernstein-Szeg}o Inequality. The inequality (f 0 (x)) + (f (x)) 2

2

  kf kR ; 2

2

x2R

holds for every f 2 E taking real values on the real line. Proof. See DuÆn and S haeer [37℄.

℄ The Growth of f 2 E . The inequality

jf (x + iy)j  e jyjkf kR ; holds for every f 2 E . Proof. See Rahman and S hmeisser [83℄.

ut

ut

x; y 2 R

ut

246 5. Basi Inequalities

2 E Taking Real Values on R. The inequality jf (x + iy)j  ( osh y) kf kR ; x; y 2 R holds for every f 2 E taking real values on the real line.

d℄ The Growth of f

Proof. See S haeer and DuÆn [38℄. e℄ Bernstein-Type Inequality in Lp . Let p 2 (0; 1): The inequality kf 0 kL R   kf kL R p(

p(

)

ut

)

holds for every f 2 E . Proof. See Rahman and S hmeisser [90℄.

ut

E.21 Markov-Type Inequality on Conne ted Subsets of the Complex Plane. Let E be a onne ted ompa t set of the omplex plane. Then

n kpk kp0 kE  2e ap( E) E 2

for every p 2 Pn : Proof. See Pommerenke [59 ℄. ut e Erd}os onje tured that the onstant in the above inequality an be repla ed by : This result would ontain Theorem 5.1.8 (Markov's inequality) as a spe ial ase. However, Rassias, Rassias, and Rassias [77℄ disproved the onje ture. Erd}os still spe ulates that e in Pommerenke's inequality may be repla ed by (1 + o(1)): The result of the next exer ise is formulated so that its proof is elementary at the expense of pre ision and generality. 2

1 2

2

1

2

E.22 The Interval where the Sup Norm of a Weighted Polynomial Lives. Suppose w = exp( Q); where (1) Q : R ! R is ontinuous and even, (2) Q0 is ontinuous and positive in (0; 1), (3) tQ0 (t) is in reasing in (0; 1), and (4) lim tQ0 (t) = 0 and lim tQ0 (t) = 1: t!0+

Let an > 0 be hosen so that

t!1

ann w(an ) = max jxn w(x)j : x2R a℄ Show that

n = an Q0 (an )

and

0 < a < a < a <


: 1

2

3

5.1 Classi al Polynomial Inequalities 247

b℄ Show that

kpwkR  kpwk

a n ;2a2n ℄

[

for every p 2 Pn :

2 2

Proof. Using Chebyshev's inequality (see E.2 of Se tion 5.1) and the expli it form (2.1.1) of the Chebyshev polynomial Tn , we an dedu e that

jp(x)j 

x a

n 1

r

+

 x 2

a

kpk

[

a;a℄





 2jxj n

a

kpk

[

a;a℄

for every p 2 Pn , x 2 R n [ a; a℄; and a > 0: Choosing a := an , and using the fa t that w is de reasing on [0; 1), we obtain 

j(pw)(x)j 

 2jxj n w(x) kpwk an;an an w(an ) [

℄

for every p 2 Pn and x 2 R n [ an ; an ℄. Now if x 2 R n [ 2a n ; 2a n ℄; then 2

jxjn w(x)  exp

2

!

Z jxj

d log(tn w(t)) dt n an w(an ) an dt ! Z jxj n tQ0 (t) dt = exp t an

 exp

Z 2a2n

n

a2n

t

2n



dt = 2

n;

where we used a n  an , tQ0 (t)  2n for t  a n , and tQ0 (t)  n for t  an : Combining the above inequality with the previous one, we obtain that 2

j(pw)(x)j  kpwk

2

[

x 2 R n [ 2a n; 2a n ℄

an ;an ℄ ;

2

for every p 2 Pn ; from whi h the result follows.

2

ut

℄ Let Q(x) := jxj ;  > 0. Show that Q satis es the assumptions of the exer ise, and  n  = an = : 1



The idea of in nite- nite range inequalities, of whi h E.22 b℄ is an example, goes ba k to Freud (an is the Freud number); see Nevai's survey paper [86℄. The sharp form of these is due to Mhaskar and Sa [85℄; see also Lubinsky and Sa [88℄ and Sa and Totik [to appear℄. They are not that diÆ ult to prove, but need the maximum prin iple for subharmoni fun tions.

248 5. Basi Inequalities

E.23 A Theorem of Markov. Let Tn be the Chebyshev polynomial of degree n de ned by (2.1.1). Then

Tn (x) =

n X k=0

k;n xk :

The Markov numbers Mk;n ; 0  k  n; are de ned by

Mk;n :=



j k;n j j k;n j 1

if k  n (mod 2) if k  n 1 (mod 2) :

These an be expli itly omputed from (2.1.1). a℄ The inequalities

jak;n j  Mk;n kpk

hold for every p 2 Pn of the form

p(x) =

n X k=0

ak;n xk ;

;

[

1 1℄

ak;n 2 R :

ut

Proof. See, for example, Natanson [64℄. b℄ Show that

jp0 (0)j  (2n

for every p 2 P n : Hint: Use part a℄.

℄ Show that

1) kpk

[

;

1 1℄

2

ut

jp0 (x)j  12n jx1j kpk for every p 2 P n and x 2 ( 1; 1):

[

;

1 1℄

2

ut

Hint: Use part b℄ and a linear transformation.

Of ourse, part ℄ gives a better result than Theorem 5.1.7 (Bernstein's Inequality) only if x is very lose to 0: This is exa tly the ase we need in our appli ation of part ℄ in E.4 ℄ of Se tion 6.1.

5.2 Markov's Inequality for Higher Derivatives From Theorem 5.1.8 (Markov's Inequality), by indu tion on m it follows that kp m k ;  (n(n 1)


(n m + 1)) kpk ; (

for every p 2 Pn :

)

[

1 1℄

2

[

1 1℄

5.2 Markov's Inequality for Higher Derivatives 249

However, this is not the best possible result. The main result of this se tion is the following inequality of DuÆn and S haeer [41℄, whi h gives a sharp improvement of the above. E.2 d℄ extends the following result to polynomials with omplex oeÆ ients: Theorem 5.2.1. If p 2 Pn satis es p




os j n

 1;

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

then for every m = 1; 2; : : : ; n;

jp m (x + iy)j  jTnm (1 + iy)j ; (

)

(

x 2 [ 1; 1℄ ; y 2 R ;

)

where Tn is the Chebyshev polynomial of degree n de ned by (2.1.1). Equality an o

ur only if p = Tn: To prove this inequality we need three lemmas, whi h are of some interest in their own right. Lemma 5.2.2. Suppose  ;  ; : : : ; n are distin t real numbers, 1

2

q(z ) :=

n Y

(z

j =1

0 6= 2 R ;

j ) ;

and p 2 Pn satis es jp0 (j )j  jq0 (j )j ;

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

Then, for every m 2 N ,

jp m (x)j  jq m (x)j

(5:2:1)

(

whenever x is a zero of q m (

1)

)

(

)

:

Proof. For m = 1, inequality (5.2.1) is simply a restatement of the assumption of the lemma, so onsider the ase m = 2: The Lagrange interpolation formula (see E.6 of Se tion 1.1) gives

(5:2:2)

n 0 n X p0 (z ) X p (j ) 1 Æj = = ; q(z ) j q0 (j ) z j j z j =1

=1

where, by the hypotheses of the lemma, jÆj j  1: There is a similar expression for q 0 (z )=q (z ) in whi h ea h Æj is equal to 1. On dierentiating (5.2.2), we obtain n X p00 (z )q(z ) p0 (z )q0 (z ) Æj = : q(z ) (z j ) j 2

2

=1

250 5. Basi Inequalities

Thus at the points x where q 0 (x) = 0 we have 00 p (x) q (x)

=

n X j =1

2

Æj (x j )



n X j =1

(x

1

j )

2

=

00 q (x) q (x)

;

and it follows that the lemma is true for m = 2. The proof for m > 2 is by indu tion. Let jp m (x)j  jq m (x)j at the zeros of q m (whi h are real and distin t). Applying the previous argument to p m and q m instead of p0 and q 0 ; we obtain (

(

)

(

)

1)

(

)

(

)

jp m (

+1)

(x)j  jq m (

+1)

(x)j

ut

at the zeros of q m : This ompletes the indu tion. (

)

Lemma 5.2.3. Let q 2 Pn have n distin t zeros in ( 1; b); and suppose that in a strip of the omplex plane it satis es the inequality

jq(x + iy)j  jq(b + iy)j ; x 2 [a; b℄ ; Suppose also that p0 2 Pn satis es (5:2:4) jp0 (x)j  jq0 (x)j (5:2:3)

y 2 R:

1

whenever x is a zero of q. Then the derivatives of p and q satisfy (5:2:5)

jp m (x + iy)j  jq m (b + iy)j ; (

)

(

x 2 [a; b℄ ; y 2 R :

)

Proof. First we show that at every point x + iy in the strip jp0 (x + iy )j  jq0 (b + iy )j : 0

0

Let

q(z ) :=

n Y

(z

0

0

0 6= 2 R ;

j ) ;

j =1

0

j

2 ( 1; b) :

Let h(z ) be another polynomial with the same leading oeÆ ient as q and whose zeros are obtained by re e ting about x those zeros of q that lie to the right of x . Thus 0

0

h(z ) := where (5:2:6)

j :=



2x

0

j

n Y

j =1

(z

j ) ;

j if j > x if j  x : 0

0

5.2 Markov's Inequality for Higher Derivatives 251

Then on the line z = x + iy; y 2 R; we have jz

j j = jz

0

jh(x

(5:2:7)

0

j j; so

+ iy )j = jq (x + iy )j : 0

We now show that

jp0 (x

(5:2:8)

0

+ iy )j  jh0 (x + iy )j : 0

0

0

Note that n h0 (z ) X 1 = ; h(z ) j z j

(5:2:9)

=1

and re alling (5.2.2), we have n p0 (z ) X Æj = q(z ) j z j

(5:2:10)

=1

with Æj 2 [ 1; 1℄ for ea h j . Comparing the right-hand sides of (5.2.9) and (5.2.10), respe tively, at z = x + iy , we obtain 0

n X j =1

Æj x + iy 0

0

j

=

 =

n X j =1 n X j =1 n X j =1

sin e by onstru tion jx

0

Æj (x j ) j ) + y 0

(x (x

2 0

2

0

x

0

j

i

j ) + y

2 0

2

0

i

x

1 j + iy0

0

n X j =1

j =1

j j = x

j : Therefore 0 h (x0 + iy0 ) h(x + iy )

0

0

0



0

0

n X

0 p (x0 + iy0 ) q (x + iy )

0

(x

0

(x

0

y0 Æj 2 2 j ) + y0 y0 2 2 j ) + y0

;

and (5.2.7) yields (5.2.8). Let  2 C ; jj < 1; be an arbitrary onstant and let

'(z ) := q(z ) h(z + x

0

b) :

Let be the simple losed urve onsisting of a segment of the line z = b + iy; y 2 R; and the portion of a ir le with enter at b and radius % that lies to the right of this line. Relations (5.2.3) and (5.2.7) show that on the line segment of ;

252 5. Basi Inequalities

jq(z )j > jh(z + x

b)j :

0

If % is suÆ iently large, the same inequality is true for the ir ular portion of sin e q and h have the same leading oeÆ ient. Thus Rou he's theorem gives that ' and h have the same number of zeros inside : We on lude that ' has no zeros on or to the right of the line z = b + iy; y 2 R: The last statement, together with Theorem 1.3.1, implies that '0 has no zeros on the line z = b + iy; y 2 R: Thus for jj < 1;

q0 (b + iy ) 0

h0 (x + iy ) 6= 0 ; 0

0

whi h, together with (5.2.8), yields

jp0 (x

0

+ iy )j  jh0 (x + iy )j  jq 0 (b + iy )j : 0

0

0

0

This proves the lemma for m = 1: We turn now to the ase m > 1: Applying the lemma with m = 1 when p = q; we have

jq0 (x + iy)j  jq0 (b + iy)j ;

x 2 [a; b℄ ; y 2 R :

Thus q 0 satis es all the requirements that are imposed on the interpolating polynomial q (with n repla ed by n 1) in the lemma, and by Lemma 5.2.2, jp00 (x)j  jq 00 (x)j at the n 1 zeros of q 0 : Applying the lemma to p0 (x) instead of p in the ase m = 1, for whi h it has already been proved, we have jp00 (x + iy)j  jq00 (b + iy)j ; x 2 [a; b℄ ; y 2 R ; whi h proves that inequality (5.2.5) is true for m = 2: Repetition of this argument ompletes the proof for larger values of m. ut Lemma 5.2.4. Suppose p 2 Pn and jp(j )j  1 for

j := os j n ;

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

Then

jp0 (xk )j  p

1

n

xk 2

for xk := os

k

(2



1)

n

2

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

For every xed k; equality holds if and only if p = Tn for some 2 C with j j = 1. Proof. Let f (x) = (1

n

Y
x )Tn0 (x) = a x os j n : 2

j =0

5.2 Markov's Inequality for Higher Derivatives 253

Then by the dierential equation for Tn (see E.3 e℄ of Se tion 2.1), we have f 0 (x) = xTn0 (x) n Tn(x) : Dierentiating the Lagrange interpolation formula for p (see E.6 of Se tion 1.1) gives n X p(j ) (x j )f 0 (x) f (x) : p0 (x) = f 0 (j ) (x j ) j 2

2

=0

For the zeros of Tn ; this redu es to

p0 (x) = Tn0 (x)

(5:2:11)

sin e at these points (x j )f 0 (x)

n X j =0

p(j ) (1 xj ) f 0 (j ) (x j ) 2

j )Tn0 (x) (1 x )Tn0 (x) = (1 xj )Tn0 (x) : In the same way, we obtain for the zeros of Tn ; n X Tn (j ) (1 xj ) Tn0 (x) = Tn0 (x) f 0 (j ) (x j ) j and sin e Tn (j ) and f 0 (j ) are of opposite sign (f 0 (j ) = n Tn (j )), this f (x) = x(x

2

2

=0

2

gives





n X 1 (1 xj ) Tn0 (x) = Tn0 (x) 0 : f (j ) (x j ) j

(5:2:12)

2

=0

Sin e jp(j )j  1 in (5.2.11), on omparing (5.2.11) and (5.2.12), we obtain for every zero x of Tn that

jp0 (x)j  jTn0 (x)j = p

n

:

1 x For a xed zero of Tn ; the equality o

urs if and only if j = 0; 1; : : : ; n p(j ) = Tn (j ) ; for some 2 C ; j j = 1; that is, if and only if p = Tn for some 2 C with j j = 1. ut 2

Proof of Theorem 5.2.1. Suppose p 6= Tn satis es the assumption of the theorem. Then by Lemma 5.2.4 there exists a onstant  > 1 su h that jp0 (x)j  jTn0 (x)j at the zeros of Tn: Applying Lemma 5.2.3 with p and q repla ed by p and Tn (assumption (5.2.3) is satis ed with [a; b℄ := [ 1; 1℄ by E.1 b℄), we obtain

jp m (x + iy)j   (

)

1

(m) T (1 + iy ) ;

n

x 2 [ 1; 1℄ ; y 2 R

for every m = 1; 2; : : : ; n: If p = Tn ; then we have the same inequality with  repla ed by 1: ut 1

254 5. Basi Inequalities

Comments, Exer ises, and Examples. The inequality

kp m k (

)

[

 Tnm (1)
kpk (

;

1 1℄

)

;

[

1 1℄

; p 2 Pn

was rst proved by V. A. Markov [16℄. He was the brother of the more famous A. A. Markov who proved the above inequality for m = 1 in [1889℄ (see Theorem 5.1.8). However, their ingenious proofs are rather ompli ated. Bernstein presented a shorter variational proof of V. A. Markov's inequality in 1938 (see Bernstein [58℄, whi h in ludes a omplete list of Bernstein's publi ations). Our dis ussion in this se tion follows DuÆn and S haeer [41℄. E.1 A Property of Chebyshev Polynomials. a℄ Let (i )i n be a sequen e of 2n nonnegative numbers, and let (0i ) be a rearrangement of this sequen e a

ording to magnitude, 2 =1

0

1

 0 


 0 n  0 : 2

2

Show that for every y  0, (  + y )(  + y )


( n  n + y )

(5:2:13)

1

2

3

4

2

1

2

is not greater than (0 0 + y )(0 0 + y )


(0 n 0 n + y ) : 1

2

3

4

2

1

2

Proof. If  and  are at least as large as any of the remaining numbers i ; then 1

3

(  + y )(  + y ) 1

3

2

4

(  + y )(  + y ) = y ( 1

2

3

4

 )(

1

4

3

 )  0: 2

This shows that the numbers i in (5.2.13) an be rearranged so that the two largest o

ur in the same fa tor without de reasing (5.2.13). Then the two largest of the remaining numbers i may be brought into the same fa tor without de reasing (5.2.13), and so on. ut b℄ Show that the Chebyshev polynomials Tn de ned by (2.1.1) satisfy the inequality

jTn (x + iy)j  jTn (1 + iy)j ;

x 2 [ 1; 1℄ ; y 2 R :

Proof. We have

jTn(x + iy)j

2

=

2

n Y j =1

((x

os j ) + y ) ; 2

2

5.2 Markov's Inequality for Higher Derivatives 255

where j = j n  and = 2n : With x = os  we write (2

1)

1

2

jTn (x + iy)j

2

=

2

n  Y 1 i e 4

j =1

ij 2 ei

e

2

eij + y

2



:

Geometri ally, eij ; j = 1; 2; : : : ; n; represent 2n points equally distributed on the unit ir le. Conne t these points by hords to the point ei : Then the lengths of these 2n hords are given by jei eij j: If  is in reased or de reased by any multiple of n ; we obtain a new set of hords, but the aggregate of their lengths is un hanged. Choose ' su h that 
n

'   mod

 n

;

2

'

 : n

2

If x = os '; then

Tn (x + iy) =

2

n  Y 1 i' e

j =1

e

4



2

ij 2 ei'

eij + y

2



;

where the numbers jei' eij j are simply a rearrangement of the numbers jei eij j : Use part a℄ to show that 2

2

jTn (x + iy)j  jTn (x + iy)j : Note that os   x  1; where os  is the right most zero of Tn : Hen e jTn (x + iy)j  jTn (1 + iy)j ; and the proof is nished. ut E.2 Markov's Inequality for Higher Derivatives for Pn . a℄ Show that if p and q satisfy the onditions of Lemma 5.2.3 with p 2 Pn repla ed by p 2 Pn ; then jp m (x)j  jq m (b)j ; x 2 [a; b℄ for every m 2 N : 2

2

1

1

2

(

)

(

2

)

Hint: After dierentiating (5.2.2) m 1 times, we obtain p m (x) = (

)

n X j =1

Æj

dm dxm

1 1





q(x) ; x j

where jÆj j = jp0 (j )=q 0 (j )j  1: It is evident that if x 2 (a; b) is xed, then jp m (x)j attains its maximum when Æj = 1 for ea h j; in whi h ase p0 has real oeÆ ients. Now use Lemma 5.2.3. ut (

)

256 5. Basi Inequalities

b℄ Show that if p satis es the assumption of Theorem 5.2.1 with p 2 Pn repla ed by p 2 Pn ; then

kp m k (

)

;

[

1 1℄

 jTnm (1)j (

)

for every m 2 N : The equality an hold only if p = Tn for some j j = 1. Hint: Modify the proof of Theorem 5.2.1 by using part a℄.

℄ Show that (

1)(n 2 )


(n (m 1
3
5


(2m 1)

n (n 2

Tnm (1) = )

2

2

2

1) )

2

2

2 C; ut

:

Hint: Dierentiating the se ond-order dierential equation for Tn (see E.3 e℄ of Se tion 2.1) m 1 times gives (1

x )Tnm 2

(

+2)

from whi h

(x)

(2m + 1)xTnm (

(2m + 1) Tnm (

+1)

+1)

m )Tnm (x) = 0

(x) + (n

2

(1) = (n

2

2

(

)

m ) Tnm (1) 2

(

)

ut

follows. Use indu tion and Tn (1) = 1 to nish the proof. d℄ The Main Inequality. Suppose p 2 Pn satis es p




os j n

 1;

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

Show that for m = 1; 2; : : : ; n;

kp m k (

[

;

1 1℄

1)(n 2 )


(n (m 1
3
5


(2m 1)

 n (n 2

)

2

2

2

2

1) ) 2

;

and the equality an o

ur only if p = Tn for some 2 C ; j j = 1: Hint: Combine parts ℄ and b℄. ut A slightly weaker version of Markov's inequality for higher derivatives is mu h easier to prove. E.3 A Weaker Version of Markov's Inequality. Show that

kp m k (

)

[

;

1 1℄

 2mn m kpk 2

[

;

1 1℄

for every p 2 Pn : Hint: First show by a variational method that the extremal problem

jp0 (1)j 6 p2Pn kpk ; max

0=

[

1 1℄

5.2 Markov's Inequality for Higher Derivatives 257

is solved by the Chebyshev polynomial Tn ; hen e

jp0 (1)j  n kpk 2

:

;

[

1 1℄

Then, by using a linear transformation, show that

jp0 (y)j  1 2 y n kpk y;  2n kpk 2

and

[

jp0 (y)j  1 +2 y n kpk 2

[

2

1℄

1

;y℄

;

[

1 1℄

 2n kpk 2

[

1y0

;

;

1 1℄

0  y  1:

;

Hen e the inequality of the exer ise is proved when m = 1: For larger values of m use indu tion. ut E.4 Weighted Bernstein and Markov Inequalities. Let w 2 C [ 1; 1℄ be stri tly positive on [ 1; 1℄: a℄ Show that for every  > 0 there exists an n depending on  and w su h that p

0

p (x)w(x) 1 x ;  n(1 + )kpwk ; 0

2

[

[

1 1℄

1 1℄

for every p 2 Pn ; n  n : 0

Proof. By the Weierstrass approximation theorem, for every  > 0 there is a q 2 Pk su h that w(x)  q(x)  (1 + )w(x) ;

x 2 [ 1; 1℄ :

Let m := minfw(x) : x 2 [ 1; 1℄g: Applying Theorem 5.1.7 (Bernstein's inequality) to pq 2 Pn k and then to q 2 Pk ; we obtain +

p p 0 p (x)w(x) 1 x2 p0 (x)q (x) 1 x2 p p (pq )0 (x) 1 x2 + p(x)q 0 (x) 1



  (n + k)kpqk ; + kpk  (n + k)(1 + )kpwk ;  n(1 + )kpwk ; [

1 1℄

[

[

[

1 1℄

;

1 1℄

+

kkqk

1

m

[

x

2



;

1 1℄

kpwk

[

;

1 1℄

k(1 + )kwk

[

;

1 1℄

1 1℄

for every p 2 Pn ; provided  > 0 is suÆ iently small and n  n : 0

ut

258 5. Basi Inequalities

b℄ Show that for every  > 0 there exists an n depending on  and w su h that kp0wk ;  n (1 + )kpwk ; 0

[

2

1 1℄

[

1 1℄

for every p 2 Pn ; n  n : Hint: Use the idea given in the previous proof. ut S haeer and DuÆn [38℄ prove an extension of Theorem 5.1.7 (Bernstein's inequality) to higher derivatives. They show that 0

m d dxm







dm exp(in ar

os x) ; m dx

p(x) 

x 2 ( 1; 1)

for every p 2 Pn : The following exer ise gives a slightly weaker version of this, whi h is mu h simpler to prove. Some of this follows La han e [84℄. E.5 Bernstein's Inequality for Higher Derivatives. a℄ Show that there exists a onstant (m) depending only on m su h that

jp

m) (x)

(



j  (m) p

1

n

m

x

kpk

[

2

;

1 1℄

x 2 ( 1; 1)

;

for every p 2 Pn : (That we an hoose (m)  2m is shown in parts ℄, d℄, and e℄.) Hint: For j = 1; 2; : : : ; m; let (m

aj := x

j )(1 + x) m

bj := x +

and

(m

j )(1 x) : m

Use Corollary 5.1.5 to show that there are onstants j (m) depending only on m su h that

kp j k aj ;bj  j (m) p ( )

[

℄

1

n

x

kp j k aj (

2

1)

[

1

;bj

1℄

for every p 2 Pn and j = 1; 2; : : : ; m: ut b℄ Show that there exists a onstant (m) > 0 depending only on m su h that jp m (x)j  (m) min n ; p n m sup 1 x 6 p2Pn kpk ; for every x 2 [ 1; 1℄ and m = 1; 2; : : : ; n: Hint: First show that (

)

2

0=

[

k

k

 (m)

(1

jxj), x +

Tnm) I (x)

where I (x) := [x

(

1 2

2

1 1℄





min n ; p 2

1 2

(1

1

n

m

x

2

;

jxj)℄; then use a shift and a s aling. ut

5.2 Markov's Inequality for Higher Derivatives 259

In the rest of the exer ise we show that (m) = 2m is a suitable hoi e for the onstant in part a℄.

℄ Let k be a positive integer. Then

kp(x)(1

x )k

=

2 (

1) 2

k

[

;

1 1℄

 n +k k kp(x)(1

x )k= 2

for every p 2 Pn : Hint: Let p 2 Pn be normalized so that kp(x)(1 x )k= Theorem 5.1.3 (Bernstein-Szeg}o inequality) with 2

t() := p( os ) sink  2 Tn

2

2

k

[

k

;

[

1 1℄

= 1: Apply

;

1 1℄

k:

+

If x = os  denotes a relative extreme point for p on ( 1; 1); then p0 ( os  ) = 0; and after simpli ation we obtain that 0

0

0

(p( os 

0

) sink

1

) 0

2



(n + k ) ((n + k ) k ) sin  + k 2

2

2

2

0

2





n+k k

2

ut

d℄ Let k be a positive integer. Then

kp0(x)(1

x )k

=

2 ( +1) 2

k

[

;

1 1℄

:

 2(n + k)kp(x)(1

x )k= 2

2

k

[

;

1 1℄

for every p 2 Pn :

Proof. Let p 2 Pn be normalized so that kp(x)(1

x )k= k ; = 1: Applying Theorem 5.1.4 (Bernstein's inequality) with m = 1 and t() := p( os ) sink  2 Tn

2

2

[

1 1℄

k;

+

we obtain

jp0 ( os ) sink

+1

 + p( os )k sink

1

 os j  n + k :

Now the triangle inequality and part ℄ yield

jp0 ( os ) sink

+1





j  (n + k) + k p( os ) sink   (n + k) + k n +k k  2(n + k) : 1

ut

260 5. Basi Inequalities

e℄ Show that

jp

(

m) (x)



j

p 2n 1 x

m

kpk

[

2

;

1 1℄

for every p 2 Pn and x 2 ( 1; 1): Hint: Use indu tion on m; Theorem 5.1.7, and part d℄.

ut

5.3 Inequalities for Norms of Fa tors A typi al result of this se tion is the following inequality due to Kneser [34℄.

and r 2 P . Then Theorem 5.3.1. Suppose p = qr; where q 2 Pm n m

kq k

[

where

;

1 1℄

kr k

[

;

1 1℄

Cn;m := 2m



Cn;m Cn;n

1 2

m  Y

m

kpk

[

;

;

1 1℄



1 + os k n  : (2

1)

2

k=1

Furthermore, for any n and m  n the inequality is sharp in the ase that p is the Chebyshev polynomial Tn of degree n de ned by (2.1.1), and the

is hosen so that q vanishes at the m zeros of p losest to fa tor q 2 Pm 1. Before proving the above theorem, we establish an asymptoti formula for Cn;m and formulate a orollary. If f 2 C [a; b℄; then by the midpoint rule of numeri al integration 2

Z b

a

f (x) dx = (b

a)f

1 2

(b




(a + b) +

a) 00 f ( ) 3

24

for some  2 [a; b℄: Let f (x) := log(2 + 2 os x): Then

f 00 (x) =



2

(1 + os x)

2

:

On applying the midpoint rule to the above f; we obtain Z m=n 0

log(2 + 2 os x) dx =

m 1X

nk



log 2 + 2 os k n  (2

1)

2

=1



+O m

1 1 n 24 n (n m) 4

3

4





5.3 Inequalities for Norms of Fa tors 261

for all integers 0  m  n: Thus (5:3:1)

m  Y

Cn;m = exp log

2 + 2 os

k=1 m X

nk





= exp O where

I (n; m) :=

n

log 2 + 2 os

=1



mn (n m) 2

4

Z m=n

1)



!



2



1

= exp

k

(2

k

(2



1)

n

!!n 

2

(exp(I (n; m)))n ;

log(2 + 2 os x) dx :

0

So (5:3:2)

Cn;bn=


1=n

 exp

2

!

Z 1=2

log(2 + 2 os x) dx = 1:7916 : : :

0

and (5:3:3)

Cn;b

n=3


1=n

 exp

2

!

Z 2=3

log(2 + 2 os x) dx = 1:9081 : : : :

0

We use the notation an  bn and an lim sup an =bn  1; respe tively.

. bn to mean nlim !1 an =bn = 1 and

n!1

On estimating Cn;m Cn;n m in Theorem 5.3.1 and using (5.3.2), we obtain the following: 1

2

Corollary 5.3.2. Let p

and r: Then

kqk

[

;

1 1℄

krk

[

2 Pn ;

1 1℄

and suppose p = qr for some polynomials q



1 2

bn= Y  2

 2n

1

1 + os k n  (2

1)

2

2

k=1

Cn;bn= kpk 2

[

2

kpk

[

;

1 1℄

;

1 1℄

and equality holds when p is the Chebyshev polynomial Tn of degree n, and

is hosen so that m := bn=2 and q vanishes at the m the fa tor q 2 Pm zeros of Tn losest to 1: Here Cn;=n bn=  3:20991 : : : ; hen e 2

2



kqk

[

;

1 1℄

kpk

[

krk

[

;

1 1℄

;

1 1℄

1=n

. 3:20991 : : : :

262 5. Basi Inequalities

The proof of Theorem 5.3.1 pro eeds through a number of lemmas. For the remainder of the proof we assume that 0 < m < n are xed. Now

krk ; : kqrk ; = 1 ; q 2 Pm ; r 2 Pn mg

and r 2 P : We pro eed to show that there is attained for some q 2 Pm n m

and r 2 P are extremal polynomials q 2 Pm n m su h that p := qr is the

(5:3:4)

supfkq k

[

;

1 1℄

[

1 1℄

[

1 1℄

Chebyshev polynomial Tn of degree of n; and that the fa tors q and r are as advertised, that is,

p(x) = (qr)(x) = Tn (x) =

n

1Y  2 x 2k



os k n  ; (2

1)

2

=1

and the extremal fa tors, q and r, are given by

and

m Y

p1

q(x) :=

2k

os k n  (2

1)



2

=1

n Y

p1

r(x) :=



2 x 



os k n  ;

2 x

(2

1)

2k m respe tively. Note that for the above q and r we have kqk ; = jq( 1)j = p1 Cn;m 2 and krk ; = jr(1)j = p1 Cn;n m : 2 =

[

[

2

+1

1 1℄

1 1℄

2 Pm

First we show that there exist extremal polynomials q r 2 Pn m su h that (5:3:5)

jq (

1)j = kq k

1 1℄

To see this, hoose , 2 [ 1; 1℄ su h that

jq()j = kqk

and

;

[

jr(1)j = krk

and

;

[

1 1℄

;

[

jr( )j = krk

[

;

1 1℄

1 1℄

and

:

;

where, onsidering q ( z ) and r( z ) if ne essary, we may assume that   . Note that  = annot happen, so  < . We have

kqk ; krk ;  kqk ; krk kqrk ; kqrk ; [

℄

[

[

℄

[

1 1℄

℄

[

[

;

1 1℄

1 1℄

sin e the numerators are equal and

kqrk ;  kqrk [

℄

[

;

1 1℄

:

Let qe 2 Pm be de ned by shifting q from [; ℄ to [ 1; 1℄ linearly so that  ! 1: Let re 2 Pn m be de ned by shifting r from [; ℄ to [ 1; 1℄ linearly

and re 2 P so that ! 1: Then qe 2 Pm n m are extremal polynomials for whi h (5.3.5) holds.

5.3 Inequalities for Norms of Fa tors 263

and r 2 Pn m are extremal polynomials

and for whi h (5.3.5) holds. Then there are extremal polynomials qe 2 Pm

re 2 Pn m having only real zeros for whi h (5.3.5) holds. Lemma 5.3.3. Suppose q

2 Pm

Proof. Let qe(z ) be de ned by repla ing every fa tor z  with nonreal  by z (j +1j 1) in the fa torization of q: Let re(z ) be de ned by repla ing every fa tor z  with nonreal  by z (1 j 1j) in the fa torization

and re 2 P of r: Now it is elementary geometry to show that qe 2 Pm n m are extremal polynomials for whi h (5.3.5) holds, and all the zeros of both qe and re are real. ut

2 Pm and r 2 Pn m are extremal polynomials having only real zeros for whi h (5.3.5) holds. Then there are extremal poly and re 2 P nomials qe 2 Pm n m having all their zeros in [ 1; 1℄ for whi h (5.3.5) holds.

Lemma 5.3.4. Suppose q

Proof. Let qe(z ) be de ned by repla ing every fa tor z  by z 1 if  > 1, and by 1 if  < 1; in the fa torization of q: Let re(z ) be de ned by repla ing every fa tor z  by z + 1 if  < 1; and by 1 if  > 1: Now it is again

and qe 2 P elementary geometry to show that qe 2 Pm n m are extremal polynomials having all their zeros in [ 1; 1℄ for whi h (5.3.5) holds. ut

and r 2 P So we now assume that q 2 Pm n m are extremal polynomials having all their zeros in [ 1; 1℄ for whi h (5.3.5) holds. We may also assume that deg(q ) = m and deg(r) = n m; otherwise we would study

qe(z ) := z m and

re(z ) := z n

m

q

q(z ) 2 Pm

r

r(z ) 2 Pn

deg( )

deg( )

m;

whi h are also extremal polynomials having all their zeros in [ 1; 1℄ for whi h (5.3.5) holds. It is now lear that if q and r are extremal polynomials with the above properties, then the smallest zero of q is not less than the largest zero of r: Indeed, if there were numbers 1   <  1 so that q () = r( ) = 0; then the polynomials

qe(z ) := q(z ) and

z z



2 Pm

z  2 Pn m z would ontradi t the extremality of q and r sin e re(z ) := r(z )

264 5. Basi Inequalities

kqek

[

1 1℄

;

 jqe(

krek

[

1 1℄

;

 jre(1)j > jr(1)j = krk

and

1)j > jq ( 1)j = kq k [

kqerek

;

[

1 1℄

= kqrk

;

[

1 1℄

So now we have extremal polynomials q form

p

m Y

q(z ) = a

(z

k )

k=1

;

[

1 1℄

;

1 1℄

;

:

2 Pm and r 2 Pn p

r (z ) = a

and

;

nYm k=1

(z

of the

m

k )

satisfying

jq(

1)j = kq k

[

jr(1)j = krk

and

;

1 1℄

[

;

1 1℄

;

where 1

  


 n

  


 m  1 and the onstant a > 0 is hosen so that for p := qr we have kpk 1

2

m

1

2

[

Now we are ready to prove Theorem 5.3.1.

;

1 1℄

= 1:

Proof of Theorem 5.3.1. We show three properties of p = qr: (1) jp( 1)j = 1 and kp(1)j = 1: (2) kp(x)k i ;i+1 = 1; i = 1; 2; : : : ; n m 1, kp(x)k i; i+1 = 1; i = 1; 2; : : : ; m 1. (3) kpk n m ; 1 = 1: These three fa ts show that p is indeed the Chebyshev polynomial Tn de ned by (2.1.1) sin e Tn are the only polynomials of degree at most n that equios illate n + 1 times on [ 1; 1℄ with uniform norm 1: To prove (1), assume to the ontrary that jp( 1)j < 1: Then there is a Æ < 1 su h that kpk Æ; = kpk ; : Sin e jq j is stri tly de reasing on [Æ; 1℄; [

℄

[

℄

[

℄

[

1℄

[

kqk Æ;  jq(Æ)j > jq( [

and, of ourse,

1)j = kq k

1℄

krk Æ;  krk [

1℄

1 1℄

[

[

;

1 1℄

:

;

1 1℄

and re 2 P Let qe 2 Pm n m be the polynomials q and r shifted linearly from [Æ; 1℄ to [ 1; 1℄ so that 1 7! 1: By the previous observations, these qe

5.3 Inequalities for Norms of Fa tors 265

and re ontradi t the extremality of q and r; hen e jp( 1)j = 1; and we are nished. A similar argument shows that jp(1)j = 1: To prove (2) let

s(z ) := (z

i )(z

i ) : +1

Given  > 0; we an nd 0 < Æ ; Æ <  su h that 1

t(z ) := (z satis es

2

(i

Æ ))(z

kt

t(1) = s(1) ;

and

jt(x)j < js(x)j ; Suppose kpk i ;i < 1: Let [

(i

1

+1

sk

;

[

x 2 [ 1; i

1 1℄

+ Æ )) 2

< ;

Æ ℄ [ [i 1

+ Æ ; 1) :

+1

2

+1 ℄

qb(z ) := q(z ) 2 Pm

and

rb(z ) := r(z )

t(z ) s(z )

2 Pn

m:

If  > 0 is suÆ iently small, then

kqbrbk

[

;

1 1℄

 kqrk

[

j(qbrb)(

and

;

1 1℄

1)j < 1 :

The se ond inequality guarantees that there exists a Æ < 1 su h that

kqbrbk Æ; [

1℄

= kqbrbk

[

;

1 1℄

1; 1℄; 1)j = jq (

:

Sin e jqbj is (stri tly) de reasing on (

kqbkÆ;  jqb(Æ)j > jqb(

1)j = kq k

[

1 1℄

;

:

 jrb(1)j = jr(1)j = krk

[

1 1℄

;

:

1℄

Also

krbk Æ;  krbk

and re 2 P Now let qe 2 Pm n [

1℄

[

;

1 1℄

m be the polynomials pb and qb shifted linearly from [Æ; 1℄ to [ 1; 1℄ so that 1 ! 1: By the previous observations, these qe and re ontradi t the extremality of q and r: Hen e kpk i;i+1 = 1; and the proof is nished. The proof of kpk i; i+1 = 1 is identi al. To prove (3) assume that kpk n m; 1 < 1: Let [

[

[

p

qe(z ) := a(z

℄

℄

( + )) 1

m Y k=2

(z

k )

℄

266 5. Basi Inequalities

and

p

re(z ) := a(z

(n m

))

nY m

1

(z

k=1

k ) :

If  > 0 is suÆ iently small, then

kqek

1 1℄

krek

1 1℄

[

[

;

 jqe(

;

 jre(1)j > jr(1)j = krk

and

1)j > jq ( 1)j = kq k

;

[

;

[

kqerek

;

[

1 1℄

 kqrk

[

;

1 1℄

1 1℄

1 1℄

;

;

;

whi h ontradi ts the extremality of q and r: Hen e kpk n deed. [

m; 1 ℄

= 1; in-

ut

is a moni fa tor of Theorem 5.3.5. Suppose p 2 Pn is moni and q 2 Pm

p: Then

jq(

)j  m n 2n

1

m  Y

1 + os k n  (2

1)



2

k=1

kpk

[

; ℄

for every > 0: Equality holds if p is the Chebyshev polynomial Tn; of

degree n on [ ; ℄ (normalized to be moni ), and the moni fa tor q 2 Pm is hosen so that q vanishes at the m zeros of Tn; losest to . Note that Tn; (x) = n Tn (x= ); where Tn is de ned by (2.1.1). The proof of Theorem 5.3.5 is outlined in E.1.

is a moni fa tor of Corollary 5.3.6. Suppose p 2 Pn is moni and q 2 Pm

p: Then

jq(

2)j  2m

1

m  Y k=1

1 + os k n  (2

1)

2



kpk

[

;

2 2℄

1 = Cn;m kpk 2

and the inequality is sharp for all m  n: Here, for all m  n; =n  C =n Cn;m n;b n=  1:9081 : : : ; 1

1

2

and hen e



jq( kpk

[

2)j ;

2 2℄

3

1=n

. 1:9081 : : : :

[

;

2 2℄

5.3 Inequalities for Norms of Fa tors 267

Proof. Take = 2 in Theorem 5.3.5. Note that 2m

m  Y k=1

1 + os k n  (2

1)



 2 n (

2

)

Y (n) 



1 + os k n  ; (2

1)

k=1

2

where (n)+1 := b n + is the smallest k 2 N for whi h os k n  So, for every m = 0; 1; : : : ; n; 2

3

3

2

(2

1)

2



1 2

:

Cn;m  Cn; n (

and by (5.3.3)

=n . C Cn;m n;b 1

)

. 1:9081 : : : :

n=3

2

ut

2 Pn is moni and has a moni fa tor of the

: Then form qr; where q 2 P and r 2 Pm Theorem 5.3.7. Suppose p

m1

jq(

)jjr( )j  m1



+

2

m2 n 2n

1

m1  Y

k=1

1 + os k n  (2

1)

m2  Y

2

k=1

1 + os k n  (2

1)



2

kpk

[

; ℄ :

Equality holds if p is the Chebyshev polynomial Tn; of degree n on [ ; ℄

and r 2 P are hosen normalized to be moni , and the fa tors q 2 Pm m2 1 so that q vanishes at the m zeros of Tn; losest to ; while r vanishes at the m zeros of Tn; losest to : 1

2

The proof of Theorem 5.3.7 is analogous to the proof of Theorem 5.3.1 and is left as an exer ise (see E.2).

2 Pn is moni and has a moni fa tor of the


qj ; where qi 2 Pm i and m := m + m +


+ mj  n. Then

Theorem 5.3.8. Suppose p

form q q 1

j Y i=1

kqi k

2

[

1

; ℄

 m 

n 2n

1

bm= Y  2

k=1

2

1 + os

(2

k

1)

n

2



 dm= Y2e 

k=1

1 + os k n  (2

1)

2



kpk

[

; ℄

for every > 0. Equality holds if p is the Chebyshev polynomial Tn; of degree n on [ ; ℄ normalized to be moni , j = 2; and the fa tors q 2 Pb m= and q 2 Pd m= e are hosen so that q vanishes at the bm=2 zeros of Tn; losest to ; while q vanishes at the dm=2e zeros of Tn;

losest to : 1

2

2

1

2

2

268 5. Basi Inequalities

Proof. For ea h qi , write

qi = ri si ;

where ri is the moni fa tor of qi omposed of the roots of qi with negative real part, while si is the moni fa tor of qi omposed of the roots of qi with nonnegative real part. So

kri k

[

Thus

; ℄ j Y i=1

= jri ( )j

kqi k

[

; ℄

ksi k

and j Y



i=1

[

j Y

ri ( )


i=1

; ℄

= jsi ( )j :

si ( ) :

We now apply Theorem 5.3.7 to the two fa tors nish the proof.

Qj

i=1 ri

and

Qj

i=1 si

to

ut

As before, let D := fz 2 C : jz j < 1g: We now derive inequalities on the disk from those on the interval. A ontinuous fun tion on D has the same uniform norm on both D and D and it is notationally onvenient to state

; and v 2 P the remaining theorems over D. Suppose t 2 Pn , s 2 Pm n m are moni , and t = sv: By the maximum prin iple, t, s; and v a hieve their maximum on D somewhere on D. Now onsider

p(x) := t(z )t(z ) ; 1

q(x) := s(z )s(z ) ; with

r(x) := v(z )v(z )

and

1

x := z + z

1

1

:

The ee t of this transformation on linear fa tors is (z

)(z

) = x + 1 +  ;

1

2

, r 2 P ; and p = qr: Also so p 2 Pn , q 2 Pm n m

kpk

[

;

2 2℄

 ktkD : 2

If t(0) 6= 0; then the modulus of the leading oeÆ ient of p is jt(0)j; while the modulus of the leading oeÆ ient of q is js(0)j; and the modulus of the leading oeÆ ient of r is jv (0)j: From these transformations and the interval inequalities we an dedu e the next three theorems.

5.3 Inequalities for Norms of Fa tors 269

Theorem 5.3.9. Let t

and v 2 P

n m:

Then

2 Pn

be moni and suppose t = sv; where s

jv(0)j = kskD  1 2

1 2

Cn;m


1=2

2 Pm

ktkD ;

where Cn;m is the same as in Theorem 5.3.1. This bound is attained when

vanishes at m adja ent zeros m  n are even, t(z ) = z n + 1; and s 2 Pm of t on the unit ir le. Proof. We may assume, by performing an initial rotation if ne essary, that

kskD = js(

1)j :

So from Corollary 5.3.6 we dedu e that (5:3:6)

kskD = js( 1)j = jq(  js(0)=t(0)j2m 2

2)j

2

m  Y

1 + os k n 

1

 js(0)=t(0)j2m

(2

k=1



1 + os k n  (2

1)

2

kpk

[

2

k=1 m  Y

1

1)



;

2 2℄

ktkD ; 2

ut

where s(0)=t(0) = 1=v (0): Theorem 5.3.10. Suppose t = sv, where s 2 Pm and r 2 Pn m . Then

kskD kvkD 

1 2

Cn;m Cn;n

m


1=2

ktkD ;

where Cn;m is the same as in Theorem 5.3.1 and (Cn;m Cn;n m ) = n 1 (2

)

 Cn;=nbn=  1:7916 : : : : 1

2

This bound is attained when m  n are even, t(z ) = z n + 1; and s 2 Pm vanishes at the m zeros of t losest to 1 and v 2 Pn m vanishes at the n m zeros of t losest to 1:

Proof. From Theorem 5.3.1 we an dedu e that if a; b 2 D; then

js(a)j jv(b)j 2

2

= js(a)s(a )jjv (b)v (b )j = jq (a + a )jjr(b + b )j

  

1 2 1 2 1 2

1

1

1

1

Cn;m Cn;n Cn;m Cn;n Cn;m Cn;n

kpk a a m kpk ; m ktkD ; m

[ + [

1

;b+b

1℄

2 2℄

2

where, without loss of generality, we may assume that a + a  b + b . The result now follows on hoosing a and b to be points on D where s and v, respe tively, a hieve their uniform norm on D: ut 1

In the multifa tor ase we have the following theorem:

1

270 5. Basi Inequalities

2 Pn is of the form t = vs s


sj ; where with m := m + m +


+ mj  n: Then

Theorem 5.3.11. Suppose t

si 2 Pmi and v 2 Pn

jv(0)j

=

1 2

j Y i=1

m

ksi kD  2 m

1) 2

bm= Y 



2

k=1

2

2

=

(

0

1

1

1 + os

(2

k



1)

n

 dm= Y2e 

2

k=1

1 + os

k

(2

n

2

11=2 

 A

1)

ktkD :

Equality holds if t(z ) = z n + 1, j = 2, m = m := m=2 and n are even, and the fa tors q 2 Pm= and q 2 Pm= are hosen so that q vanishes at the m=2 zeros of t losest to 1 and q vanishes at the m=2 zeros of t losest to 1: 1

1

2

2

2

2

1

2

Proof. This follows from Theorem 5.3.8 in exa tly the same way as Theorem 5.3.10 follows from Theorem 5.3.1. ut Comments, Exer ises, and Examples. The rst result of this se tion is due to Kneser [34℄ and in part to Aumann [33℄. The proof follows Borwein [94℄, as does most of the se tion. There are many variations and generalizations. See Boyd [92℄, [93a℄, [93b℄, [94a℄, and [94b℄; Beauzamy and En o [85℄; Beauzamy, Bombieri, En o, and Montgomery [90℄; Gel'fond [60℄; Glesser [90℄; Granville [90℄; Mahler [60℄, [62℄, and [64℄; and Mignotte [82℄. Some of these are presented in the exer ises. In parti ular, E.6 reprodu es a very pretty proof of Boyd [92℄ that

kqkD krkD  (1:7916 : : : )n kpkD ;

and r 2 P : (Note that we where p 2 Pn and p = qr with some q 2 Pm n m have not assumed real oeÆ ients unlike in Theorem 5.3.10, and we have  instead of . :)

E.1 Proof of Theorem 5.3.5.

Outline. Let m < n and > 0 be xed. The value 

sup

jq( kpk

)j [

; ℄

: q2P

m

and p 2 P are moni and q divides p

n



and p 2 P . We an now argue, exa tly is attained for some moni q 2 Pm n as in the proof of Lemma 5.3.3, that there are extremal polynomials p 2 Pn

su h that all the zeros of p are real and lie in [ ; 1): Arguing and q 2 Pm as in Lemma 5.3.4 gives that p has all its roots in [ ; ℄: Thus q must be

omposed of the m roots of p losest to :

5.3 Inequalities for Norms of Fa tors 271

The argument of the proof of Theorem 5.3.1 now applies essentially verbatim and proves that an extremal p 2 Pn an be hosen to be the Chebyshev polynomial on [ ; ℄ normalized to be moni . Thus on [ 1; 1℄

p(x) =

n  Y k=1

x

os k n  (2

1)



q(x) =

and

2

m  Y k=1

x os

k

(2

1)

n





2

from whi h the result follows (on onsidering n p(x= ) and m q (x= ) on [ ; ℄). ut E.2 Proof of Theorem 5.3.7. Hint: Pro eed as in the proof of Theorem 5.3.1 (or E.1).

ut

E.3 A Version of Theorem 5.3.10 for Complex Polynomials. Suppose

and v 2 P : Then t = sv; where s 2 Pm n m

js(

1)jjv (1)j 

1 2

Cn;m Cn;n

m


1=2

ktkD

and if t is moni

jt(0)j = kskD kvkD  1 2

1 2

(Cn;m Cn;n m ) = ktkD : 1 2

2

Hint: The rst inequality follows as in the proof of Theorem 5.3.10 with a := 1 and b := 1: The se ond part is immediate from Theorem 5.3.9. u t E.4 Mahler's Measure. Let F : C k F be de ned by

Mk (F ) := exp

Z

1

0






Z

1

! C , and let the Mahler measure of

log jF (e it1 ; : : : ; e itk )j dt 2

0

2

1




dtk



if the integral exist. a℄ Show that if

p(z ) = then

n Y i=1

(z

M (p) = j j 1

i ) ; n Y i=1

; i 2 C ;

maxf1; ji jg :

Hint: Use Jensen's formula (see E.10 ℄ of Se tion 4.2). b℄ Show that if F := F (z ; : : : ; zk ) and G := G(z ; : : : ; zk ); then 1

1

Mk (F G) = Mk (F )Mk (G) :

ut

272 5. Basi Inequalities

℄ Show that

M (ax + b) = max(jaj; jbj) ; M (1 + x + y) = M (maxf1; j1 + xjg) ; M (1 + x + y xy) = M (maxfj1 xj; j1 + xjg) : 1

2

1

2

1

d℄ One an numeri ally he k that

M (1 + x + y) = 1:381356 : : : 2

and

M (1 + x + y 2

xy) = 1:791622 : : : :

E.5 The Norm of a Fa tor of a p 2 Pn on the Unit Disk. Suppose p 2 Pn

. Then is moni and has a moni fa tor q 2 Pm

kqkD  n kpkD ;

where := M (1 + x + y ) = 1:3813 : : : : 2

Outline. Let p(x) :=

n Y i=1

(x

i )

and

q(x) :=

Suppose kq kD = jq (u)j; where u 2 D: Then m Y

kqkD = jq(u)j =

i=1

ju

i j 

n Y

i=1

m Y

(x

i=1

maxfju

i ) ;

i 2 C :

i j; 1g

= M (p(x + u))  M (maxf1; jx + ujn gkpkD ) ; 1

1

where the last equality holds by E.4 a℄, and the last inequality follows be ause

jp(z )j  maxf1; jz jng
kpkD holds for every p 2 Pn and z 2 C by E.18 a℄ of Se tion 5.1. Now using E.4 (5:3:7)

b℄ and

M (maxf1; jx + ujg) = M (maxf1; jx + 1jg) ; 1

1

we obtain

kqkD  M (maxf1; jx + ujng) kpkD = M ((maxf1; jx + ujg)n ) kpkD = M ((maxf1; jx + 1jg)n ) kpkD = (M (maxf1; jx + 1jg))n kpkD = (M (1 + x + y ))n kpkD = n kpkD : 1

1

1

1 2

ut

5.3 Inequalities for Norms of Fa tors 273

E.6 Another Inequality for the Fa tors of a p 2 Pn on the Unit Disk. Let

and r 2 P : Then p = qr; where q 2 Pm n m

kqkD krkD  Æn kpkD where Æ := M (1 + x + y

xy) = 1:7916 : : : .

2

Outline. Without loss of generality we may assume that q and r are moni . Let

q(x) :=

m Y

(x

i=1

i )

r(x) :=

and

n Y

(x

i 2 C :

i ) ;

i=m+1

Choose u 2 D and v 2 D su h that jq (u)j = kq kD and jr(v )j = krkD : Then, using E.4 b℄ and ℄, we obtain

kqkD krkD = jq(u)jjr(v)j =  =

m Y i=1 n Y i=1 n Y i=1

ju

i j

n Y

maxfju

ju

jv

i j

i j; jv

i jg

i=m+1

i j max



(x

1)n p

max jx

1j; x

=M

1

v 1; u





ux v x 1



i i



:

Now, by (5.3.7), (x

1)n p



xu v x 1







n



v on kpkD ; u

hen e

kqkD krkD  M ((maxfjx = (M (maxfjx  (M (maxfj1

1j; jx v=ujg)n )kpkD 1j; jx v=ujg))n kpkD xj; j1 + xjg))n kpkD = (M (1 + x + y xy ))n kpkD = (1:7916 : : : )n kpkD : 1

1 1 2

ut

274 5. Basi Inequalities

E.7 Bombieri's Norm. For Q(z ) := de ned by n X

[Q℄p :=

Pn

k=0 ak z

 1 p

n k

k=0

n for

jak j

p

k

the Bombieri p norm is

!1=p

:

Note that this is a norm on P every p 2 [1; 1); but it varies with varying n: The following remarkable inequality holds (see Beauzamy et al.

, and S 2 P ; then [90℄). If Q = RS with Q 2 Pn , R 2 Pm n m [R℄ [S ℄ 2

2



 1=2

n m

[Q℄

2

and this is sharp. One feature of this inequality is that it extends naturally to the multivariate ase. See Beauzamy, En o, and Wang [94℄ and Rezni k [93℄ for further dis ussion.

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6

Inequalities in Muntz Spa es

Overview Versions of Markov's inequality for Muntz spa es, both in C [a; b℄ and Lp [0; 1℄, are given in the rst se tion of this hapter. Bernstein- and Nikolskii-type inequalities are treated in the exer ises, as are various other inequalities for Muntz polynomials and exponential sums. The se ond se tion provides inequalities, in luding most signi antly a Remez-type inequality, for nondense Muntz spa es.

6.1 Inequalities in Muntz Spa es We rst present a simpli ed version of Newman's beautiful proof of an essentially sharp Markov-type inequality for Muntz polynomials. This simpli ation allows us to prove the Lp analogs of Newman's inequality. Then, using the results of Se tion 3.4 on orthonormal Muntz-Legendre polynomials, we prove an L version of Newman's inequality for Muntz polynomials with omplex exponents. Some Nikolskii-type inequalities for Muntz polynomials are studied. The exer ises treat a number of other inequalities for Muntz polynomials and exponential sums. Throughout this se tion we use the notation introdu ed in Se tion 3.4. Unless stated otherwise, the span always denotes the linear span over R: 2

276 6. Inequalities in Muntz Spa es

Theorem 6.1.1 (Newman's Inequality). Let  := (i )1 be a sequen e of i =0

distin t nonnegative real numbers. Then n 2X j 3j



kxp0 (x)k kpk ;

sup

6 p2Mn 

0=

=0

;

[0 1℄

(

[0 1℄

)

9

n X j =0

j

for every n 2 N ; where Mn() := spanfx0 ; x1 ; : : : ; xn g: Proof. It is equivalent to prove that n

2X j 3j

(6:1:1)



=0

sup

6 P 2En 

0=

(

)

n kP 0k ;1  9 X j ; kP k ;1 j [0

)

[0

)

=0

where En () := spanfe 0 t ; e 1 t ; : : : ; e n t g: Without loss of generality we may assume that  := 0: By a hange of s ale we may also assume that Pn j j = 1: We begin with the rst inequality. We de ne the Blas hke produ t 0

=0

B (z ) := and the fun tion (6:1:2)

T (t) :=

1 2i

Z

n Y

j =1

z j z + j

e zt dz ; B (z )

:= fz 2 C : jz

1j = 1g :

By the residue theorem

T (t) := and hen e T

n X j =1

(B 0 (j )) e j t ; 1

2 En (): We laim that jB (z )j  13 ;

(6:1:3)

Indeed, it is easy to see that 0  j z j z + 

j

 22 + j j

z2 :

 1 implies

=

1 j ; 1 + j 1

z2 :

2 1

2

So, for z 2 ;

jB (z )j 

n Y j =1

1 j 1 + j 1

2

1

2



1 1+

1 2

1 2

n P j =1 n P j =1

j j

=

1 1+

1 2 1 2

=

1 : 3

6.1 Inequalities in Muntz Spa es 277

Here the inequality 1 x 1 y 1 (x + y ) 2xy (x + y ) = + 1 + x 1 + y 1 + (x + y ) (1 + x)(1 + y )(1 + (x + y )) y)  11 + ((xx + ; x; y  0 + y) is used. From (6.1.2) and (6.1.3) we an dedu e that (6:1:4)

jT (t)j 

Also

1 2

Z zt e B (z )

1 T 0(t) = 2i

and (6:1:5)

1 2i

T 0 (0) =

Z

jdz j  21 3(2) = 3 ;

Z

t  0:

ze zt dz B (z ) Z

z 1 dz = B (z ) 2i jzj

=1

z dz : B (z )

Now, for jz j > max j n j ; we have the Laurent series expansion 1

n n 1   k ! Y Y X z 1 + j =z j 1+2 (6:1:6) =z =z B (z ) 1  =z z j j j k =1

0

n X

= z 1 + 2

j =1

= z + 2 + 2z

1

!

=1

j z

+


;

1

=1

+2

n X j =1

j

!2

z

2

1

+


A

whi h, together with (6.1.5), yields that T 0 (0) = 2: Hen e, by (6.1.4), n jT 0 (0)j  2 = 2 X kT k ;1 3 3 j j ; [0

)

=1

so the lower bound of the theorem is proved. To prove the upper bound in (6.1.1), rst we show that if 1 U (t) := 2i

Z

(1

e

zt

z )B (z )

then (6:1:7)

1

Z 0

dz ;

:= fz 2 C : jz

jU 00 (t)j dt  6 :

1j = 1g ;

278 6. Inequalities in Muntz Spa es

Indeed, observe that if z = 1 + ei ; then jz j = 2 + 2 os ; so (6.1.3) and Fubini's theorem yield that 2

1

Z 0

1 1 Z z e zt dt dz 2 (1 z )B (z ) Z Z 1 1  jz j je zt j 2 jB (z )j d dt Z 1Z  3 (2 + 2 os )e 2

Z

jU 00 (t)j dt =

0

2



0

2

0

2



(1+ os

0

Z 2

3 2

= Now we show that

1

Z

(6:1:8)

2

(2 + 2 os )

0

j t U 00 (t) dt

e

)t d dt

0

0

1 d = 6 : 1 + os 

= j

3:

To see this we write the left-hand side as

1

Z

e

0

j t U 00 (t) dt =

0

1Z z e

Z

1 2i Z 1 = 2i jzj

=

1

Z

2

j t

e

z j )t

( +

=2

Z

z e zt dz dt (1 z )B (z ) 2

dz dt =

Z

1 2i

z )B (z ) z z 1 dz ; z + j 1 z B (z ) (1

0

1 2i

z

2

(z + j )(1

z )B (z )

dz

where in the third equality Fubini's theorem is used again. Here, for jz j > 1; we have the Laurent series expansions

z = 1 j z z + j z = 1 z 1 z

1

+ j z 2

z

1

2

+


;




;

2

and, as in (6.1.6), 1

B (z )

= 1 + 2z

1

+ 2z

2

+


:

Now (6.1.8) follows from the residue theorem (see, for example, Ash [71℄). Let P 2 En () be of the form

P (t) =

n X j =0

j e

j t ;

j

2 R:

6.1 Inequalities in Muntz Spa es 279

Then

1

Z

P (t + a)U 00 (t) dt =

0

n 1X

Z

j =0

0

=

n X j =0

j a

j e

1

Z

j e e

j t U 00 (t) dt =

Z

1

n X j =0

0

= P 0 (a)

3P (a)

and so

jP 0 (a)j  3jP (a)j +

(6:1:9)

j a e j t U 00 (t) dt

0

3)e j a

j (j

jP (t + a) U 00 (t)j dt :

Combining this with (6.1.7) gives

kP 0 k ;1  3 kP k ;1 [0

)

[0

)

+ 6 kP k ;1 = 9 kP k ;1 ; [0

)

[0

)

ut

and the theorem is proved.

The next theorem establishes an Lp extension of Newman's inequality. Theorem 6.1.2 (Newman's Inequality in Lp ). Let p 2 [1; 1): If  := (i )1 i is a sequen e of distin t real numbers greater than 1=p; then

=0

0

kxP 0 (x)kLp

;

[0 1℄

  p1 + 12

n  X j =0

j +

1

1



p

A

kP kLp

;

[0 1℄

for every P 2 Mn () := spanfx0 ; x1 ; : : : ; xn g : If := ( i )1 i is a sequen e of distin t positive real numbers, then =0

kP 0 kLp ;1  12 [0

for every P

2 En (

)

n X j =0

!

j

kP kLp ;1 [0

)

) := spanfe 0 t ; e 1 t ; : : : ; e n t g:

Proof. First we show that the rst statement of the theorem follows from the se ond. Indeed, if (i )1 i is a sequen e of distin t real numbers greater than 1=p and i := i + p for ea h i; then ( i )1 i is a sequen e of distin t positive real numbers. Let Q 2 Mn (): Applying the se ond inequality with =0 1

=1

P (t) := Q(e t )e

t=p

2 En (

and using the substitution x = e t , we obtain

)

280 6. Inequalities in Muntz Spa es Z

1

0

 0 p x x1=p Q(x) x

1

dx

1=p

n X

 12



j +

j =0

1

!

p

kQkLp

;

[0 1℄

:

Now the produ t rule of dierentiation and Minkowski's inequality yield 0

kxQ0 (x)kLp

;

[0 1℄



1 

p

+ 12

n X



j +

j =0

1

1



kQkLp

A

p

;

[0 1℄

;

whi h is the rst statement of the theorem. We prove the se ond statement. Let P 2 En ( ) and p 2 [1; 1) be xed. As in the proof of Theorem P 6.1.1, by a hange of s ale, without loss of generality we may assume that nj j = 1: It follows from (6.1.9) and Holder's inequality (see E.7 a℄ of Se tion 2.2) that =0

jP 0 (a)jp  2p

 1

3p

j

j

P (a) p +

 6p jP (a)jp + 2p

1

Z

1

1

Z 0

j

j

P (t + a) p

0

p 

jP (t + a)jU 00 (t)j dt

1

1=p Z

jU 00 (t)j dt

0

1=q !p

jU 00 (t)j dt

for every a 2 [0; 1), where q 2 (1; 1℄ is the onjugate exponent to p de ned by p + q = 1: Combining the above inequality with (6.1.7), we obtain 1

1

jP 0 (a)jp  6pjP (a)jp + 2p

6p=q

1

1

Z 0

jP (t + a)jp jU 00 (t)j dt

for every a 2 [0; 1): Integrating with respe t to a; then using Fubini's theorem and (6.1.7), we on lude that

kP 0 kpLp ;1  6pkP kpLp ;1 [0

)

[0

 kP k 6p

)

+ 2p 6p=q

p p Lp [0;1) + 2

 6pkP kpLp ;1  6pkP kpLp ;1 = (6p + 2p

[0

)

[0

)

and the proof is nished.

1

6p )

1Z 1

Z

1

1

6p=q

Z0 1 Z0 1 0

0

jP (t + a)jp jU 00 (t)j dt da jP (t + a)jp jU 00 (t)j da dt

+ 2p 6p=q kP kpLp ;1

1

Z

1

[0

0

jU 00 (t)j dt

kP kpLp ;1 p p ;1  12 kP kLp ;1

+ 2p 6p=q 1

kP k

)

p Lp[0

)

+1

[0

)

[0

)

ut

The following Nikolskii-type inequality follows from Theorem 6.1.1 quite simply:

6.1 Inequalities in Muntz Spa es 281

Theorem 6.1.3 (Nikolskii-Type Inequality). Suppose 0 < q < p  1: If  := (i )1 i is a sequen e of distin t real numbers greater than 1=q; then =0

ky

=q

=p P (y )

1

1

kLp

;

[0 1℄



18


2q

n X



j =0

j +

1

!1=q

=p

1

q

kP kLq

;

[0 1℄

for every P 2 Mn () := spanfx0 ; x1 ; : : : ; xn g: If := ( i )1 i is a sequen e of distin t positive real numbers, then =0

kP kLp ;1 

(6:1:10)

[0

2 En (

for every P

18


)

2q

n X j =0

!1=q

=p

1

j

kP kLq ;1 [0

)

) := spanfe 0 t ; e 1 t ; : : : ; e n t g:

Proof. First we show that the rst statement of the theorem follows from the se ond. If (i )1 is a sequen e of distin t real numbers greater than i 1=q and i := i + 1=q for ea h i; then ( i )1 is a sequen e of distin t i positive real numbers. Let Q 2 Mn (): Applying (6.1.10) with =0

=1

P (t) := Q(e t )e

t=q

2 En (

)

and using the substitution x = e t ; we obtain

ky =q

=p Q(y )

1

1

kLp

;

[0 1℄

 (18


2q )1=q

=p

1

n  X

j +

j =0

1

!1=q

q

=p

1

kQkLq

;

[0 1℄

;

whi h is the rst statement of the theorem. It is suÆ ient to prove (6.1.10) when p = 1; and then a simple argument gives the desired result for arbitrary 0 < q < p < 1: To see this, assume that there is a onstant C so that

kP k ;1  C =q kP kLq ;1 ) and 0 < q < 1: Then [0

for every P

2 En (

kP k

p Lp[0;1)

1

)

1

Z

[0

)

jP (t)jp q q dt  kP kp ;1q kP kqLq ;1  C p=q kP kpLq q ;1 kP kqLq ;1

=

+

[0

0

)

1

[0

and therefore

2 En (

[0

)

kP kLp ;1  C =q =p kP kLq ;1 ) and 0 < q < p  1. 1

[0

for every f

℄

)

1

[0

)

[0

)

282 6. Inequalities in Muntz Spa es

When p = 1, (6:1:10) an be proven as follows. Let P 2 En ( ); and let y 2 [0; 1) be hosen so that jP (y )j = kP k ;1 : From Theorem 6.1.1 and the Mean Value Theorem, we an dedu e that jP (t)j > kP k ;1 for every [0

)

1

[0

2



t 2 I := y; y + (18 )

1



; where :=

Thus

kP k

q Lq [0;1℄

Z

 j

j

P (t) q dt

I



18

n X j =0

!

1

j

n X j =0

)

j :

2 q kP kq ;1 ; [0

)

ut

and the result follows.

Theorem 6.1.3 immediately implies the following result, whi h is a spe ial ase of Theorem 4.2.4: Theorem 6.1.4 (Muntz-Type Theorem in Lp ). Let p

2 [1; 1): Let (i )1 i

=0

be a sequen e of distin t real numbers greater than 1=p satisfying  1 X 1 < 1: j + p j =0

Then spanf

x0 ; x1 ; : : :

g is not dense in Lp [0; 1℄:

The next theorem oers an L analog of Theorem 6.1.1 even for omplex exponents. It also improves the multipli ative onstant 12 in the L inequality of Theorem 6.1.2 and shows that the L inequality of Theorem 6.1.2 is essentially sharp. 2

2

2

Theorem 6.1.5. If  := (i )1 is a sequen e of distin t omplex numbers i with Re(i ) > 1=2 for ea h i; then kxp0 (x)kL2 ; sup =0

6 p2Mn 

0=

(

[0 1℄

kpkL

;

2 [0 1℄

)

n X



j =0

jj j

2

+

n X

n X

j =0

k=j +1

(1 + 2 Re(j ))

!1=2

(1 + 2 Re(k ))

for every n 2 N; where Mn() denotes the linear span of fx0 ; x1 ; : : : ; xn g over C : If  := (i )1 i is a sequen e of distin t nonnegative real numbers, then n n X kxp0 (x)kL2 ;  p1 X p1 j  sup (1 + 2j ) 2 30 j 2j 6 p2Mn  kpkL2 ; =0

[0 1℄

=0

0=

(

)

[0 1℄

=0

for every n 2 N, where Mn () denotes the linear span of fx0 ; x1 ; : : : ; xn g over R:

6.1 Inequalities in Muntz Spa es 283

Proof. Let p 2 Mn () with kpkL2 p(x) =

n X k=0

and

;

[0 1℄

= 1: Then

ak Lk (x)

n X

with n X

xp0 (x) =

k=0

where L

k=0

jak j

2

=1

ak xL0 k (x) ;

untz-Legendre polynok 2 Mk () denotes the k th orthonormal M mials on [0; 1℄: Using the re urren e formula of Corollary 3.4.5 b℄ for the terms xL0 k (x) in the above sum, we obtain

xp0 (x) =

n X j =0

q

aj j + 1 + j + j

!

q

n X k=j +1

ak 1 + k + k Lj (x) :

Hen e

kxp0 (x)k

2

L2 [0;1℄

=

n X aj j j =0

+

q

1 + j + j

n X k=j +1

q

2

ak 1 + k + k :

If we apply theP Cau hy-S hwarz inequality to ea h term in the rst sum and re all that nk jak j = 1; we see that 2

=0

kxp0 (x)k

2

L2 [0;1℄

n X



j =0

jj j

2

+ (1 + j + j ) !2

n 1 X (1 + 2jj j) 2 j



n
X k=j +1

(1 + k + k )

;

=0

whi h proves the rst part and the upper bound in the se ond part of the theorem. Now we prove the lower estimate in the se ond part of the theorem. With the sequen e  := (i )1 of distin t nonnegative real numbers, we i asso iate ! =0

n p X

q(x) :=

k=0

k

k X j =0

j Lk (x) 2 Mn () :

Sin e the system (Lk )1 k is orthonormal on [0,1℄, we have =0

(6:1:11)

kqkL 2

;

2 [0 1℄

=

n X k=0

k

k X j =0

!2

j



n X j =0

!3

j

:

284 6. Inequalities in Muntz Spa es

Furthermore n

k X

k=0

j =0

Xp k xq0 (x) =

!

j xL0 k (x) =

n X m=0

bm Lm(x) ;

where, by the re urren e formula of Corollary 3.4.5 b℄, m X

p

bm := m m



p

m

n X

j =0

k=m

k

p

j + 1 + 2m k X j =0

n p X k=m+1

k (1 + 2k )

2

L2 [0;1℄

= =

n X m=0

bm 

n X

2

X

j =0

m=0

m

X

mn jk mk;k0 n j 0 k0 0

n X k=m

k

k X j =0

!2

j

m k j k0 j0

0

0

X



j

j :

Hen e

kxq0 (x)k

k X

mjj0 kk0 n

m k j k0 j0



n

1 X j 5! j

!5

:

=0

0

This, together with (6.1.11), yields the lower bound in the se ond part of the theorem. ut Comments, Exer ises, and Examples. Theorem 6.1.1 is due to Newman [76℄. We presented a modi ed version of Newman's original proof of Theorem 6.1.1. He worked with T instead of U , and instead of (6.1.9) he established a more ompli ated identity involving the se ond derivative of P . Therefore, he needed an appli ation of Kolmogorov's inequality (see E.1) to nish his proof. It an be proven that if the exponents j are distin t nonnegative integers, then kxp0 (x)k ; in Theorem 6.1.1 an be repla ed by kp0 k ; (see E.3). Theorems 6.1.2 to 6.1.4 were proved by Borwein and Erdelyi [to appear 6℄, while Theorem 6.1.5 is due to Borwein, Erdelyi, and J. Zhang [94b℄. It is shown in E.8 that Theorem 6.1.2 is essentially sharp for every  with a gap ondition, and for every p 2 [2; 1). The interval [0; 1℄ plays a spe ial role in this se tion, analogs of the results on [a; b℄; a > 0; annot be obtained by a linear transformation. E.10 deals with the nontrivial extension of Newman's inequality to intervals [a; b℄; a > 0: A onje ture of Lorentz about the \right" Bernstein-type inequality for exponential sums with n terms is settled in E.4. [0 1℄

[0 1℄

6.1 Inequalities in Muntz Spa es 285

E.1 Kolmogorov's Inequality. Show that

kf 0 k ;1  4kf k ;1 kf 00 k ;1 2

[0

[0

)

)

[0

)

for every f 2 C [0; 1): Hint: By Taylor's theorem 2

f (x + h) = f (x) + f 0 (x)h + f 00 ( )h ; 1

h>0

2

2

with some  2 (x; x + h): Hen e

kf 0k ;1  2h kf k ;1 1

[0

)

[0

)

+ (h=2)kf 00 k ;1 : [0

)

Now minimize the right-hand side by taking

h := 2



1=2

kf k ;1 kf 00 k ;1 1

[0

)

[0

)

:

ut

that

The onstant 4 in E.1 is not the best possible. Kolmogorov [62℄ proved

kf k kR  K (n; k)kf kR k=nkf n kk=n R 1

( )

(

)

for every f 2 C n (R) and 0 < k < n and found the best possible onstants K (n; k); see also DeVore and Lorentz [93℄. This generalizes a result of Landau, who proved p the above inequality for n = 2, k = 1, and showed that K (2; 1) = 2: Various multivariate extensions of Kolmogorov's inequality have also been established; see, for example, Ditzian [89℄. E.2 Nikolskii-Type Inequalities. a℄ Suppose (V; k
k) is an (n +1)-dimensional real or omplex Hilbert spa e, (pk )nk  V is an orthonormal system, and ' 6= 0 is a linear fun tional on V . Then ! =0

j'(p)j 

n X k=0

j'(pk )j

2

=

1 2

kpk

for every p 2 V: Equality holds if and only if

p=

n X j =0

'(pk )pk ;

2 R or 2 C :

Hint: Write p as a linear ombination of the orthonormal elements pk ; use the linearity of '; then apply the Cau hy-S hwarz inequality. ut

286 6. Inequalities in Muntz Spa es

b℄ Suppose  := (i )1 is a set of distin t omplex numbers satisfying i Re(i ) > 1=2 for ea h i; and y 2 [0; 1) is xed. Show that =0

n X

jp m (y)j  (

)

k=0

!1=2

jL m (y)j (

k

)

2

kpkL

;

2 [0 1℄

for every p 2 Mn (); where Mn () denotes the linear span of

fx ; x ; : : : ; xn g 0

1

over C ; and Lk 2 Mk () is the k th orthonormal Muntz-Legendre polynomial on [0; 1℄: Show that if there exists a q 2 Mn () with q m (y ) 6= 0; then equality holds if and only if (

p=

n X k=0

Lk m (y)Lk ; (

)

2C:

)

℄ Under the assumptions of part b℄ show that

jy = p(y)j  kpkL ; 1 2

2 [0 1℄

!1=2

n X j =0

(1 + 2 Re(j ))

and

p0 (y)j

jy kpkL =

3 2

;

2 [0 1℄

0

n X



k=0

(1 + 2 Re(k )) k

+

kX1 j =0

11=2 2 (1 + 2 Re(j )) A

for every 0 6= p 2 Mn () and y 2 [0; 1℄: Hint: When y = 1, use part b℄ and substitute the expli it values of Lk (1) and L0 k (1) (see Corollary 3.4.6 and formula (3.4.8)). If 0 < y < 1; then the s aling x ! yx redu es the inequality to the ase y = 1: ut d℄ Show that if n  1 and p 6= 0; then equality holds in the inequalities of part b℄ if and only if y = 1 and

p=

n X k=0

Lk (1)Lk

respe tively, with some 0 6= 2 C :

or

p=

n X k=0

 L0 k (1)Lk ;

6.1 Inequalities in Muntz Spa es 287

e℄ Show that

kpk k 6 p2Pn pkL sup

;

[

0=

=

1 1℄

;

2[

n+1

p

2

1 1℄

:

Hint: Show that there is an extremal polynomial pe for the above extremal problem for whi h kpek ; is a hieved at 1: Now use parts ℄ and d℄ to [

show that

1 1℄

!Z

n

X jpe(1)j = 12 (1 + 2k ) k

1

1 2

2

1

=0

Z

pe (t) dt = (n + 1)

2

1

2

pe (t) dt 2

1

ut

and the result follows. E.3 An Improvement of Newman's Inequality. a℄ Suppose  := (k )1 k is a sequen e with  = 0 and k ea h k: Show that 0

=0

kp0k

;

[0 1℄

n X

 18

j =1

!

j

kpk

+1

k  1 for

;

[0 1℄

for every p 2 Mn () := spanfx0 ; x1 ; : : : ; xn g: Hint: Let y 2 [0; 1℄: To estimate jp0 (y)j distinguish two ases. If  y  1; use Theorem 6.1.1 (Newman's inequality), and if 0  y  ; use E.3 f℄ of Se tion 3.3 and Theorem 5.1.8 (Markov's inequality) transformed to [y; 1℄ to show that 1

1

2

2

jp0 (y)j  2n

2

1

y

kpk y;  8 [

1℄

n X j =1

!

j

kpk

;

[0 1℄

for every p 2 Mn (): ut The next exer ise is based on an example given by Bos. b℄ Show that for every Æ 2 (0; 1) there exists a sequen e  := (k )1 k with  = 0,   1; and =0

0

1

k

+1

k  Æ ;

i = 0; 1; 2; : : :

su h that lim

sup

n!1 06=p2Mn ()

P

jp0 (0) j  j kpk

n j =0

;

= 1:

[0 1℄

Outline. Let Qn be the Chebyshev polynomial Tn transformed linearly from [ 1; 1℄ to [0; 1℄, that is, Qn (x) = os(n ar

os(2x 1)) ;

x 2 [0; 1℄ :

288 6. Inequalities in Muntz Spa es

Choose natural numbers u and v so that Æ < u=v < 1: Let  := (k )1 k be de ned by  := 0,  := 1; and =0

0

1

k := 1 + Let

pn (x) := x

1

u

ku ; v

k = 1; 2; : : : :

Qn (xu=v )

Then

( 1)n

jp0n (0)j =

For the sake of brevity let

2n

2


v

qn (x) := Qn (xu=v )


v

2 Mnv

v () :

:

( 1)n :

Use Theorem 5.1.8 (Markov's inequality) and the Mean Value Theorem to show that

kqn k y;  kqn k [

with

1

1℄

y := 2n

2

Thus, if n is odd, then

kpnk

;

[0 1℄

 kpn k y;  y  y u kqn k [

1

2

and P

nv j =0

jp0n (0) j  v j kpn k

;

=

[0 1℄




v=u

:

kqn k y;

u(

1

1℄

1

;

[0 1℄

2

;


v

[0 1℄

P nv v

j =0
v=u 2

[

1℄

)v

= 2n

2


(u

v=u

1)

(2n )v
 1 + j uv 2n

2n (1 + nu)nv

2

2



2n

2


v=u

uv


(u 1

1)

v=u

! 1:

n!1

ut

In his book Nonlinear Approximation Theory, Braess [86℄ writes the following: \The rational fun tions and exponential sums belong to those

on rete families of fun tions whi h are the most frequently used in nonlinear approximation theory. The starting point of onsideration of exponential sums is an approximation problem often en ountered for the analysis of de ay pro esses in natural s ien es. A given empiri al fun tion on a real interval is to be approximated by sums of the form n X j =1

aj ej t ;

where the parameters aj and j are to be determined, while n is xed." The next exer ise treats inequalities for exponential sums of n + 1 terms.

6.1 Inequalities in Muntz Spa es 289

E.4 Nikolskii- and Bernstein-Type Inequalities for Exponential Sums. Let  := (i )1 i =1

be a sequen e of distin t nonzero real numbers. Let (

En () := f : f (t) = a + 0

and

En :=

[



n X j =1

(

En () = f : f (t) = a + 0

)

aj

n X j =1

aj 2 R

ej t ;

)

aj

ej t ;

aj ; j

2R

;

that is, En is the olle tion of all (n + 1)-term exponential sums with onstant rst term. S hmidt [70℄ proved that there is a onstant (n) depending only on n su h that

kf 0k a Æ;b Æ  (n)Æ kf k a;b
for every p 2 En and Æ 2 0; (b a) . Lorentz [89℄ improved S hmidt's [ +

1

℄

[

℄

1

2

result by showing that for every  > ; there is a onstant () depending only on  su h that (n) in the above inequality an be repla ed by

()n n , and he spe ulated that there may be an absolute onstant su h that S hmidt's inequality holds with (n) repla ed by n. Part d℄ of this exer ise shows that S hmidt's inequality holds with (n) = 2n 1: A weaker version of this showing that S hmidt's inequality holds with (n) = 8(n+1) is obtained in part b℄ and uses a Nikolskii-type inequality for exponential sums established in part a℄. Part e℄ shows that the result of part d℄ is sharp up to a multipli ative absolute onstant. a℄ Let p 2 (0; 2℄: Show that 1

2

log

2

kf k a

Æ;b Æ℄

[ +

for every f

2 En and Æ 2

2

0; (b 1

2

=p2



2

n+1 Æ

1=p

kf kLp a;b [

℄




a) : 



Proof. Take the orthonormal sequen e (Lk )nk on ; , that is,  t  t  t (1) Lk 2 spanf1; e 1 ; e 2 ; : : : ; e k g ; k = 0; 1; : : : ; n ; =0

and (2)

Z 1=2

=

1 2

Li Lj = Æi;j ;

1

1

2

2

0  i  j  n;

where Æi;j is the Krone ker symbol. On writing f 2 En () as a linear

ombination of L ; L ; : : : ; Ln ; and using the Cau hy-S hwarz inequality 0

1

290 6. Inequalities in Muntz Spa es

and the orthonormality of (Lk )nk fashion that

on

=0

max

jf (t )j

6 f 2En  kf kL2

0=

(

)

Sin e

= ; =

[

0

;

1 2

1

max

(

Observe that if f

)

0

;

t

0

2 R:

su h that n X

=

0

[

!1=2 2

2

jf (t )j

6 p2En  kf kL2

0=



2

1

2

k=0 Lk (x) dx = n + 1 ;

=

1 2



2

there exists a t



; , we obtain in a standard

Lk (t )

k=0

1 2 1 2℄

Z 1=2 Pn

1 2

n X

=

0



= ; =

k=0

1 2 1 2℄

!1=2

Lk (t ) 2

0



p

n + 1:

2 En (); then g(t) := f (t t ) 2 En (); so jf (0)j  pn + 1 : max 6 f 2En  kf kL ; 0

0=

Let

(

C :=

)

2[

1 1℄

jf (0)j : max 6 f 2En  kf kLp ;

0=

(

)

[

2 2℄

Then max

1

(

y 2 [ 1; 1℄ :

1

6 f 2En 

0=

jf (y)j  C  2  =p  2 =p C ; kf kLp ; 2 jy j

)

[

Therefore, for every f

2 2℄

2 En ();

p jf (0)j  n + 1 kf kL ;   = p  n + 1 kf kpLp ; kf k p;  p
 n + 1 kf kpLp ; 2 =p C p kf kLp p p
 n + 1 2 =p C p= kf kLp ; p = 2 =p = n + 1 C p= kf kLp ; : 2[

1 1℄

1 2

2

[

1 1℄

[

1 1℄

2

1

[

1

Hen e

1 1℄

1

1

2

[

2 2℄

2

[

1 2

1

2

2 2℄ [

2 2℄

jf (0)j  2 =p = pn + 1 C 6 f 2En  kf kLp ; and we on lude that C  2 =p =p (n + 1) =p : So C=

0=

max

1

(

)

[

2

2

1 2

2 2℄

1

;

1=2

1

1

p=2

6.1 Inequalities in Muntz Spa es 291

jf (0)j  2 =p

kf kLp ; for every f 2 En (): Now let f 2 En () and t 2 [a + Æ; b =p (n + 1)1=p

2

2

1

[

2 2℄

0

the above inequality to

g(t) := f

1 2




Æt + t

0

Æ℄: If we apply

2 En () ;

we obtain

kf k a

Æ;b Æ℄

[ +

2

=p2

=p (n + 1)1=p

2

 1=p

2

1

Æ

kf kLp a;b [

℄

ut

and the result follows.

The following Bernstein-type inequality an be obtained as a simple

orollary of part a℄: b℄ Show that kf 0 k a Æ;b Æ  8(n + 1) Æ kf k a;b [ +

2

℄

1

[

℄

2 En and Æ 2 (0; (b a)): Note that f 2 En () implies f 0 2 En (): Applying part a℄ to f 0

for every f

Proof.

1

2

with p = 1; we obtain

jf 0 (0)j  2(n + 1)kf 0kL ; = 2(n + 1)Var ; (f )  4(n + 1) kf k ; for every f 2 En (): If f 2 En () and t 2 [a + Æ; b Æ ℄; then on applying 1[

2 2℄

[

2

2 2℄

[

2 2℄

0

the above inequality to

g(t) := f

1 2




Æt + t

0

2 En () ; ut

we obtain the desired result.

℄ Lorentz's Conje ture. Show that

jf 0 (0)j 6 f 2Ee n kf k ;

= 2n

sup

0=

where

(

Ee n := f : f (t) = a + 2

0

2

n X j =1

[

1;

1 1℄


aj ej t + bj e j t ;

Proof. First we prove that

jf 0 (0)j  (2n

1) kf k

[

;

1 1℄

aj ; bj ; j

2R

)

:

292 6. Inequalities in Muntz Spa es

for every f

2 Ee n . So let f 2 spanf1; e t ; e t ; : : : ; en t g 2

1

2

with some nonzero real numbers  ;  ; : : : ; n ; where, without loss of generality, we may assume that 1

2

0 <  <  <


< n : 1

Let

2

g(t) := (f (t)

f ( t)) :

1

2

Observe that

g 2 spanfsinh  t ; sinh  t ; : : : ; sinh n tg : 1

2

It is also straightforward that

g0 (0) = f 0 (0)

kgk

and

;

[0 1℄

 kf k

;

[

1 1℄

:

For a given  > 0; let

Hn; := spanfsinh t ; sinh 2t ; : : : ; sinh ntg and



Kn; := sup jh0 (0)j : h 2 Hn; ; khk



=1 :

;

[0 1℄

The inequality of E.5 e℄ in Se tion 3.3 is the key to the proof. It shows that it is suÆ ient to prove that inf fKn; :  > 0g  2n

1:

Observe that every h 2 Hn; is of the form

h(t) = e

nt P (et ) ;

P

2P n: 2

Therefore, using E.23 ℄ of Se tion 5.1, we obtain for every h 2 Hn; that

jh0 (0)j = jP 0 (1) nP (1)j  1(2n e 1)  kP k e  ;e  It follows that



[

℄

+ n kP k e  ;e [

(2n 1) + n en khk 1 e 

[

Kn; 

So inf fKn; :  > 0g  2n





;

1 1℄

℄

:



(2n 1) + n en : 1 e 

1; and the upper bound follows.

6.1 Inequalities in Muntz Spa es 293

Now we prove that

jf 0 (0)j  2n 6 f 2Ee n kf k ;

1:

sup

0=

[

2

1 1℄

Let  > 0 be xed. We de ne

Q where T n by 2



nt T

n; (t) := e

2

2

n

et

e

1

e

1



1

1

;

denotes the Chebyshev polynomial of degree 2n

1

T

2

n

1

1 de ned

1) ar

os x); x 2 [ 1; 1℄ :

(x) = os((2n

It is simple to he k that Q n; 2 Ee n ; 2

2

kQ n;k 2

and

[

;

1 1℄

jQ0 n;(0)j  2n 2

 ent n :

1

ut

Now the result follows by letting  de rease to 0: d℄ Show that

kf 0k a Æ;b Æ  (2n 1)Æ kf k a;b
for every f 2 En and Æ 2 0; (b a) . Proof. Observe that En  Ee n . Hen e the result follows from part ℄ by a linear substitution. ut e℄ Let a < b and y 2 (a; b). Suppose that n 2 N is odd. Let Tn be the [ +

1

℄

[

℄

1

2

2

Chebyshev polynomial of degree n de ned by (2.1.1). Let

Qn (t) := Qn;y (t) := Tn and

Rn (t) := Rn;y (t) := Tn Show that Qn ; Rn 2 En and

jQ0n (y)j kQnk a;b [

= ℄

for every y 2 (a; b) :

1

n

e 1b y



e

e 1



e

e

exp





t a exp 1 a y

and



t b b y



e 1



1



e 1

jRn0 (y)j kRn k a;b [

1

= ℄

1

e 1y

:

n

a

294 6. Inequalities in Muntz Spa es

f℄ Let a < b and y 2 (a; b): Con lude that

n 1 e 1 minfy a; b yg 1



jf 0 (y)j  2n 1 6 f 2En kf k a;b minfy a; b sup

0=

[

℄

yg

:

In the rest of the exer ise let (

l

X En := f : f (t) = Pkj (t)ej t ; j

2 R;

j =1

Pkj

2 Pkj ;

l X j =1

)

(kj + 1) = n :

g℄ Extend the inequalities of parts a℄, ℄, and f℄ to En : h℄ Let [a; b℄ be a nite interval. Let g 2 C [a; b℄: Show that the value inf



kg

f k a;b : f [

℄

2 En



is attained by an fe 2 En : Hint: Use S hmidt's inequality (or its improved form given by part ℄). For the nontrivial details, see Braess [86℄. ut i℄ Let [a; b℄ be a nite interval. Let p 2 [1; 1) and g 2 Lp [a; b℄: Show that the value  inf kg f kLp a;b : f 2 En [

℄

is attained by an fe 2 En :

Hint: Use part a℄ with p = 1 and Holder's inequality. On e again, for the details, see Braess [86℄. The following result is from Borwein and Erdelyi [95 ℄:

ut

E.5 Upper Bound for the Derivative of Exponential Sums with Nonnegative Exponents. The equality sup p

jp0 (a)j kpk a;b [

℄

=

2n

2

b a

holds for every a < b; where the supremum is taken over all exponential sums 0 6= p 2 En with nonnegative exponents. The equality sup p

jp0 (a)j kpk a;b [

℄

=

2n

2

a(log b log a)

also holds for every 0 < a < b; where the supremum is taken over all Muntz polynomials 0 6= p of the form

p(x) = a + 0

n X j =1

aj xj ;

aj

2 R;

j > 0 :

6.1 Inequalities in Muntz Spa es 295

Outline. It is suÆ ient to prove only the se ond statement, the rst an be obtained from it by the hange of variable x = et : For  > 0; de ne  := (j)1 j : Let =0

Tn; := Tnf1; x; x  ; : : : ; xn ; [a; b℄g 2

be the Chebyshev polynomial for Mn ( ) on [a; b℄: Use E.3 b℄ and E.3 f℄ of Se tion 3.3 to show that

0 (a)j jp0 (a)j  lim jTn; kpk a;b ! kTn;k a;b [

0+

℄

[

0 (a)j = lim jTn; !0+

℄

for every p of the form

p(x) = a + 0

n X j =1

aj xj ;

aj

2 R;

j > 0 :

From the de nition and uniqueness of Tn; it follows that

Tn;(x) = Tn



2

b

a

b + a b a

x



where Tn (y ) = os(n ar

os y ); y 2 [ 1; 1℄: Therefore

0 (a)j = jT 0 ( jTn; n =

1)j  b

2

a

a

2n 1)  (a

1

2



1

(b

1

1)

a

1

! !

0+

2n

2

a(log b log a)

ut

and the proof is nished. The next exer ise follows Turan [84℄. E.6 Turan's Inequalities for Exponential Sums. a℄ Let

g( ) := Suppose

n X j =1

bj zj ;

jzj j  1 ;

bj ; zj 2 C :

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

Then max

 =m+1;::: ;m+n

jg( )j 



for every nonnegative integer m:

n 2e(m + n)

2

jb

1

+ b +


+ bn j 2

296 6. Inequalities in Muntz Spa es

Proof. Let (6:1:12)

f (z ) :=

n Y



z zj

1

j =1



n X

=:

 z  :

 =0

Sin e f (z ) has all its zeros outside the open unit disk, g := 1=f is of the form (6:1:13)

g(z ) =

1 X  =0

jz j < 1 :

 z  ;

Let (6:1:14)

gm (z ) :=

m X  =0

 z 

and

h := 1 fgm 2 Pn

(6:1:15)

m:

+

Note that (6:1:16)

1 X

h(z ) = 1 f (z ) g(z ) = f (z )

1 X  =m+1

 =m

!



z

 z 

so h is of the form (6:1:17)

h(z ) =

m +n X  =m+1

 z  :

Observe that f (zj ) = 0 and (6.1.13) imply h(zj ) = 1; that is, m +n X  =m+1

 zj = 1 ;

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

Multipli ation with bj and summation over j yield the fundamental identity (6:1:18) This immediately gives

m +n X  =m+1

 g( ) =

n X j =1

bj :

6.1 Inequalities in Muntz Spa es 297

(6:1:19)

max

 =m+1;::: ;m+n

!

n X

jg( )j

 =m+1

j  j 

n X bj

j =1

:

It follows from (6.1.15), (6.1.17), (6.1.12), and (6.1.14) that (6:1:20)

m +n X  =m+1

n X

j  j 

 =0

!

m X

j j

 =0

!

j  j

:

Sin e ea h zj is of modulus at least 1, (6.1.12) yields that n X

(6:1:21)

 =0

j j  2n :

Also, (6.1.13) implies that

 =

X

1

:

i1 i2 in  z z


zn

i1 +


+in =

1

2

Again using that ea h zj is of modulus at least 1; we obtain

j  j  m X  =0



 +n 1 1= : n 1 

i1 +


+in =

Hen e (6:1:22)



X

j  j 





m+n n





e

m+n n

n

:

By (6.1.20) to (6.1.22) we on lude that m +n X  =m+1

j  j 



2e

m+n n

n

;

ut

whi h, together with (6.1.19), nishes the proof. b℄ Let

f (t) := Suppose Show that

n X j =1

bj ej t ;

Re(j )  0 ;

jf (0)j 

for every a > 0 and d > 0:



bj ; j

2C:

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

 2e(a + d) n

d

kf k a;a [

d

+ ℄

298 6. Inequalities in Muntz Spa es

Hint: First observe that the result of part a℄ an be formulated as

jg( )j  m m n  2N max



+

n 2e(m + n)

n

jb

+ b +


+ bn j ;

1

2

where m is an arbitrary positive (not ne essarily integer) number. Now apply the above inequality with 

an dj ; zj := exp d n

m :=



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

ut

℄ Let

n X

p(z ) := Show that

j =1

bj z j ;

max jp(z )j 

jz j

bj 

2C;

 4e n

=1

Æ

j

2 R;

max

jzj=1 arg(z)+Æ

z = ei :

jp(z )j

for every 0   <  + Æ  2: Hint: Use part b℄. ut The inequalities of the above exer ise and their variants play a entral role in the book of Turan [83℄, where many appli ations are also presented. The main point in these inequalities is that the exponent on the right-hand side is only the number of terms n; and so it is independent of the numbers j : An inequality, say in part ℄, of type max jp(z )j  (Æ )n

jzj

=1

jp(z )j ;

max

jzj=1 arg(z)+Æ

where 0   <  <


< n are integers and (Æ ) depends only on Æ;

ould be obtained by a simple dire t argument, but it is mu h less useful than the inequality of E.6 ℄. 1

2

E.7 Nikolskii-Type Inequality for Muntz Polynomials. Suppose that i  1 for ea h i. Show that

 := (i )1 is a sequen e with  := 0 and i i 0

=0

0

kP kLp for every P

;

[0 1℄

 18


2q

+1

n X j =0

11=q

j A

=p

1

kP kLq

;

[0 1℄

2 Mn() := spanfx ; x ; : : : ; xn g and 0 < q < p  1: 0

1

6.1 Inequalities in Muntz Spa es 299

Proof. It is suÆ ient to study the ase when p = 1 (see the omment in the proof of Theorem 6.1.3). Let P 2 Mn () and let x 2 [0; 1℄ be hosen so that jP (x )j = kP k ; . Combining E.3 a℄ and the Mean Value Theorem, 0

we obtain

0

[0 1℄

jP (x)j  21 kP k

where h

I := x

(36)

0

1

x2I;

;

;

[0 1℄

; x + (36)

1

i

with  :=

0

n X j =0

j :

So Z

1

0

1=q

jP (x)jq dx



Z

1=q

I

jP (x)jq dx





(18)

 1

1 kP k ; 2

q 1=q

[0 1℄

ut

and the result follows.

E.8 Sharpness of Theorem 6.1.2. Suppose  := (i )1 i is a sequen e with  := 0 and i i  1 for ea h i: Show that there exists an absolute

onstant > 0 (independent of  and p) su h that =0

0

+1

sup

P 2Mn ()

kxP 0 (x)kLp kP kLp ;

;

[0 1℄

[0 1℄



n X k=0

k

for every p 2 [2; 1), where Mn () := spanfx0 ; x1 ; : : : ; xn g:

Proof. Let Lk 2 Mk () be the kth orthonormal Muntz-Legendre polynomial on [0; 1℄: Let p 2 [2; 1) and n X

P :=

k=0

For the sake of brevity let

 :=

 L0 k (1)Lk :

n X j =0

j :

Using Theorem 3.4.3 (orthogonality), (3.4.8), and Corollary 3.4.6, we obtain (6:1:23)

n

X P 0 (1) = (L0 (1))

k=0

 2

2

k

k X j =0



!2

j

n X

k X

k=0

j =0



1 8

(2k + 1)



3

!2

j

300 6. Inequalities in Muntz Spa es

and (6:1:24)

kP kL

;

2 [0 1℄

n X

=

 

k=0

(L0 k (1))

0 n X  (2k

p

32 

2

!2 11=2

k X

+ 1)

k=0

=

!1=2

j =0

(2j + 1)

A

:

3 2

A ombination of E.7 and (6.1.24) yields (6:1:25)

kP kLp

;

[0 1℄

 (72 ) = =p kP kL ; p  72 = =p 32  =p  48  1 2

1 2

1

2 [0 1℄

2

1

1

=p :

2

1

From E.3 a℄, E.7, and (6.1.24) we an dedu e that (6:1:26)

kP 0 k

;

[0 1℄

 18  kP k ;  18 (72 ) = kP kL ; p  18 (72 ) = 32  =  18
48  : [0 1℄

1 2

2 [0 1℄

1 2

3 2

3

Applying E.3 a℄ with P 0 , we get (6.1.27)

kP 00 k

;

[0 1℄

 18  kP 0k

;

[0 1℄

 18
48  2

4

:

Now (6.1.23), (6.1.27), and the Mean Value Theorem give

jP 0 (x)j  where



I := 1

1 16

18

x2I;

 ;

2

3





48
16 

1



;1 :

So (6:1:28)

kxP 0 (x)kLp

;

[0 1℄

   

Z 

I

1=p 0 xP (x) p dx



48
16 
=p 18
48
16 32 p
128
9 3  =p : 18

2

1

1

32

1

2

1

3



3

=

1 2


p 1=p



3

1

=p

1

Combining (6.1.25) p and (6.1.28), we obtain the required result with a onstant = (128
9 3) : ut 1

6.1 Inequalities in Muntz Spa es 301

E.9 On the Interval Where the Sup Norm of a Muntz Polynomial Lives. Let  := (j )1 be an in reasing sequen e of nonnegative real numbers. j Let 0 6= p 2 Mn () := spanfx0 ; x1 ; : : : xn g; and let q (x) := xk p(x); where  > 0 and k is a nonnegative integer. Let  2 [0; 1℄ be hosen so that jq( )j = kqk ; : a℄ Suppose  = 1 and ea h j is an integer. Then =0

[0 1℄



k

2

k+n

<:

ut

Proof. See Sa and Varga [81℄.

The above result is sharp in a ertain limiting sense, whi h is des ribed in detail in Sa and Varga [78℄. b℄ Suppose j = j for ea h j: Use part a℄ to show that 

k

2=

k+n

<:

℄ Suppose j = j for ea h j: Use E.11 of Se tion 5.1 to show, without using part a℄, that there exists an absolute onstant > 0 su h that 

k

2=

k+n

<:

d℄ Suppose j  j for ea h j: Use part b℄ and E.3 g℄ of Se tion 3.3 to show that the on lusion of part b℄ remains valid. e℄ Suppose j  j for ea h j: Use part ℄ and E.3 g℄ of Se tion 3.3 to show that the on lusion of part ℄ remains valid. The following extension of Newman's inequality is in Borwein and Erdelyi [to appear 3℄. E.10 Newman's Inequality on [a; b℄  (0; 1). a℄ Let  := (j )1 j be an in reasing sequen e of nonnegative real numbers. Assume that there exists an  > 0 su h that j  j for ea h j: Suppose that [a; b℄  (0; 1): Show that there exists a onstant (a; b; ) depending only on a, b, and  su h that =0

kp0k a;b [

℄

 (a; b; )

n X j =0

!

j

for every p 2 Mn () := spanfx0 ; x1 ; : : : xn g:

kpk a;b [

℄

302 6. Inequalities in Muntz Spa es

Proof. We base the proof on E.9 d℄, however; it may also be based on E.9 e℄. Let p 2 Mn (): We want to estimate p0 (y ) for every y 2 [a; b℄: First let y 2 (a + b); b : We de ne q (x) := xmn p(x); where m is the smallest 1

2

positive integer satisfying

a

a+b



2

m m+1

2=

:

S aling Newman's inequality from [0; 1℄ to [0; y ℄; then using E.9 d℄, we obtain n X

jq0 (y)j  y9 =

j =0 n X

9

yj

=0

(j + mn)kq k ;y

℄

(j + mn)kq khy(

m 2= m+1 ) ;y

[0

!

n X

 (a; b; ) 1

kqk a;y

j

j =0

[

i

℄

with a onstant (a; b; ) depending only on a, b, and : Hen e 1



mn +

jp0 (y)j  q0 (y)y y

mn

1

mn jp(y)j y n X

(a; b; ) n X

 (a; b; ) 2

j =0 n X

 (a; b; ) 2

j =0

j =0

!

j

!

kqk a;y

j

[

kpk a;y [

!

j

℄

+

mn kpk a;y y [

℄

℄

kpk a;b [

℄

with a onstant (a; b; ) depending only on a, b; and :   Now let y 2 a; (a + b) : Then, by E.3 b℄ and f℄ of Se tion 3.3, we

an dedu e that 2

1

2





jp0 (y)j  Tn0 fx ; x ; x  ; : : : ; xn ; [y; b℄g(y) kpk y;b 2y  =  n kpk y;b  (a; b; )n kpk y;b b y 0

2

[

℄

1

2

[

n X

 (a; b; ) 4

j =0

℄

!

j

2

3

[

kpk y;b [

℄

℄

with onstants (a; b; ) and (a; b; ) depending only on a, b; and : This nishes the proof. ut 3

4

6.2 Nondense Muntz Spa es 303

b℄ Show that if the gap ondition j  j in part a℄ is dropped, then the on lusion of part a℄ fails to hold with (a; b; ) repla ed by a onstant

(a; b) depending only on a and b: Hint: Study    Tn0 x ; x ; x  ; : : : ; xn ; ; 1 and let  ! 0 + : ut 0

2

1

2

6.2 Nondense Muntz Spa es Throughout this se tion we assume that  := (i )1 is a sequen e i of distin t nonnegative real numbers, Mn () denotes the linear span of fx0 ; x1 ; : : : ; xn g over R; and =0

M () :=

1 [

n=0

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

If A  [0; 1℄ is ompa t, then a ombination of Tietze's and Muntz's theoP rems yields that M () is dense in C (A) whenever 1 i 1=i = 1: (Re all that Tietze's theorem guarantees that if A  [0; 1℄ is ompa t, then for every f 2 C (A) there exists an fe 2 C [0; 1℄ su h that fe(x) = f (x) for all x 2 A:) If the Lebesgue measure m(A) of A is positive, then the onverse is also true. More pre isely, we have the following. =1

Theorem 6.2.1 (Muntz P Theorem on Compa t Sets of Positive Measure). Suppose  := 0 and 1 i 1=i < 1: Let A  [0; 1℄ be a ompa t set with positive Lebesgue measure. Then M () is not dense in C (A): Moreover, if 0

the gap ondition holds and

=1

inf fi

i

1

: i 2 Ng > 0

rA := supfx 2 [0; 1) : m(A \ (x; 1)) > 0g ; then every fun tion f 2 C (A) from the uniform losure of M () on A is of the form 1 X f (x) = aj xj ; x 2 A \ [0; rA ): j =0

If the above gap ondition does not hold, then every fun tion f 2 C (A) from the uniform losure of M () on A an still be extended analyti ally throughout the region

fz 2 C n ( 1; 0℄ : jz j < rA g :

The proof of the above theorem rests on the following bounded Remeztype inequality:

304 6. Inequalities in Muntz Spa es

Theorem 6.2.2 untz Spa es). Let P (Remez-Type Inequality for Nondense M  := 0 and 1 1 = < 1 : Then there exists a

onstant

depending only i i on  and s (and not on %, A, or the number of terms in p) su h that 0

=0

kpk

[0

;%℄

 kpkA

for every p 2 spanfx0 ; x1 ; : : : g and for every ompa t set A Lebesgue measure at least s > 0:

 [%; 1℄ of

To prove Theorem 6.2.2 we need a few lemmas. We use the notation introdu ed in Se tion 3.3. However, for notational onvenien e, we let

Tn f ;  ; : : : ; n ; Ag := Tn fx0 ; x1 ; : : : ; xn ; Ag 0

1

for ompa t sets A  [0; 1): By the unique interpolation property of Chebyshev spa es (see Proposition 3.1.2), asso iated with 0 < x < x <


< xn ; 0

1

we an de ne

`k fx ; x ; : : : ; xn g 2 Mn () ; 0

k = 0; 1; : : : ; n

1

su h that 

k `k fx ; x ; : : : ; xn g(xj ) = Æj;k := 10 ifif jj = 6= k : 0

1

Lemma 6.2.3. Let 0 < x < x <


< xn 0

1

and

0 < xe < xe <


< xen : 0

1

Suppose 0  k  n and xj  xej if j = 0; 1; : : : ; k 1 ; xj = xej if j = k ; xj  xej if j = k + 1; k + 2; : : : ; n : For notational onvenien e, let `k := `k fx ; x ; : : : ; xn g 0

Then

1

and

`ek := `k fxe ; xe ; : : : ; xen g :

j`k (0)j  j`ek (0)j :

0

1

6.2 Nondense Muntz Spa es 305

Proof. It is suÆ ient to prove the lemma in the ase where there is an index m su h that 1  m  n, m 6= k, and xj = xej if j = 0; 1; : : : ; n; j 6= m ; xm < xem if m < k ; xm > xem if m > k : The general ase of the lemma then follows from repeated appli ations of the above spe ial ases. Note that in the above spe ial ases

`k

`ek 2 Mn ()

has a zero at ea h of the points

xj ; j = 0; 1; : : : ; n ; j 6= m ; hen e it hanges sign at ea h of these points, and it has no other zero in [0; 1) (see E.10 of Se tion 3.1). It is also obvious that

x 2 (0; x ) ;

sign(`k (x)) = sign(`ek (x)) ;

0

whi h, together with the previous observation and the inequality x yields that j`k (0)j  j`ek (0)j ;

0

 xe ; 0

ut

whi h nishes the proof.

By a simple s aling we an extend Lemma 6.2.3 as follows. We use the notation introdu ed in Lemma 6.2.3. Lemma 6.2.4. Let 0 < x < x <


< xn 0

1

and

0 < xe < xe <


< xen : 0

1

Suppose 0  k  n,  0; and xj  xej xj = xej xj  xej Then

if j = 0; 1; : : : ; k 1 ;

if j = k ;

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

j`k (0)j  j`ek (0)j :

306 6. Inequalities in Muntz Spa es

Proof. If = 0; then Lemma 6.2.3 yields the lemma. So we may assume that > 0: Let x xe ; := k = k xek xek xj := xej ; j = 0; 1; : : : ; n ; and

`k := `k fx ; x ; : : : ; xn g ; 0

Obviously and

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

1

`ek ( x) = `k (x) ;

x 2 [0; 1)

xj  xj if j = 0; 1; : : : ; k 1 ; xj = xj if j = k ; xj  xj if j = k + 1; k + 2; : : : ; n :

Hen e Lemma 6.2.3 implies that

j`k (0)j  j`k (0)j = j`ek (0)j ut

whi h nishes the proof.

Lemma 6.2.5. Let A be a losed subset of [0; 1℄ with Lebesgue measure at least s 2 (0; 1): Then

jp(0)j  jTnf ;  ; : : : ; n ; [1 0

1

s; 1℄g(0)j
kpkA

for every p 2 M (): Proof. If 0 2 A; then the statement is trivial. So assume that 0 2= A: Let xe < xe <


< xen 0

1

denote the extreme points of

Tn := Tnf ;  ; : : : ; n ; [1 s; 1℄g 0

in [1

1

s; 1℄; that is, Tn (xej ) = ( 1)n j ;

Let xj

2 A;

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

j = 0; 1; : : : ; n; be de ned so that m([xj ; 1℄ \ A) = m([xej ; xen ℄) = xen

xej :

Sin e A is a losed subset of [0; 1℄ with m(A)  s; su h points xj Let p 2 Mn (): Then, by Lemma 6.2.4, we an dedu e that

2 A exist.

6.2 Nondense Muntz Spa es 307

jp(0)j = 

n X p(xk )`k (0) k=0 ! n X

j`k (0)j kpkA



k=0 n X

j

j kpkA

(

1)n k `ek (0)

k=0 n X

=

!

k=0 n X

=

!

`ek (0)

!

Tn (xek )`ek (0)

k=0 Tn (0)

=j

kpkA

kpkA

j
kpkA

ut

whi h proves the lemma.

Lemma 6.2.6. Suppose  := 0: Let A be a losed subset of [0; 1℄ with Lebesgue measure at least s 2 (0; 1): Then 0

jp(y)j  jTn f ;  ; : : : ; n ; [1 for every p 2 Mn () and y 2 [0; inf A℄: 0

1

s; 1℄g(0)j
kpkA

Proof. For notational onvenien e, let

Tn;A := Tn f ;  ; : : : ; n ; Ag : 0

1

Note that  = 0 implies that jTn;A j is de reasing on [0; inf A℄; otherwise 0

0 2 spanfx1 ; x2 ; : : : ; xn Tn;A 1

1

1

g

would have at least n + 1 zeros in (0; 1℄; whi h is impossible. Hen e, it follows from E.3 and Lemma 6.2.5 that jp(y)j  jTn;A (y)j = jT (y)j  jT (0)j n;A n;A

kpkA

kTn;AkA  jTn f ;  ; : : : ; n ; [1 0

1

for every 0 6= p 2 Mn (): This nishes the proof.

s; 1℄g(0)j

ut

Proof of Theorem 6.2.2. Lemma 6.2.6 and E.5 a℄ of Se tion 4.2 yield the theorem.

ut

Proof of Theorem 6.2.1. The theorem follows from E.5 of this se tion and E.3 e℄ and E.8 b℄ of Se tion 4.2.

ut

Our next theorem is an interesting hara terization of la unary sequen es.

308 6. Inequalities in Muntz Spa es

Theorem 6.2.7 (Chara terization of La unary Muntz Spa es). Suppose  := (i )1 with 0   <  <


. There exists a onstant depending i =0

only on  su h that

0

1

jaj;n j  kpk

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

;

;

[0 1℄

for every p 2 M () of the form p(x) =

n X j =0

aj;n xj

if and only if  is la unary, that is, if and only if the elements i of  satisfy inf fi =i : i 2 N g > 1 : +1

To prove Theorem 6.2.7 we need the following result of Hardy and Littlewood [26℄ whose proof we do not reprodu e: Theorem 6.2.8. Suppose 0 = < <


is a la unary sequen e, that is, 0

1

inf f i = i : i 2 N g > 1 : +1

Suppose the fun tion f is of the form 1 X f (x) = ai x i ; ai 2 R ; i=0

and A := xlim ! f (x) exists and is nite. Then 1

x 2 [0; 1)

P1

i=0 ai

= A:

Proof of Theorem 6.2.7. Suppose  is la unary and suppose there exists a  M () su h that if Pk is of the form

sequen e (Pk )1 k

=1

Pk (x) =

nk X j =0

aj;nk xj ;

aj;nk

2 R;

then (6:2:1)

kPk k

;

[0 1℄

=1

and

max jaj;nk j  k ; 2

jnk

0

k = 1; 2; : : : :

We may assume, without loss of generality, that  = 0: Choose a sequen e (k )1 k of positive integers su h that 0

=1

 = 1; 1

 k 1

+1

> 2k nk ;

Now let the fun tion f be de ned by

k = 1; 2; : : : :

6.2 Nondense Muntz Spa es 309

1 X

k Pk (xk ) 

1 X

0  x  1:

f (x) =

Note that f let

2 C [0; 1℄ by the Weierstrass M-test. For notational onvenien e,

k=1

2

m := 0;

mk :=

0

k X i=1

k=1

k

< 1;

(6:2:2)

2

ni ;

k = 1; 2; : : : :

Further, let

:= 0 and

mk

a :=

and

0

j

1+

0

1 X k=1

k a 2

0

;nk

=

:= k j ; j = 1; 2; : : : ; nk ;

1 X k=1

k Pk (0) 2

k = 1; 2; : : : :

2 R is well-de ned sin e jPk (0)j  1 for ea h k 2 N : Also inf f i = i : i 2 N g  minf2; inf fi =i : i 2 N gg > 1 : := ( i )1 i . Then f 2 C [0; 1℄ de ned by (6.2.2) is in the uniform

Observe that a

0

+1

+1

Let

losure of M ( ) on [0; 1℄; hen e, by the Clarkson-Erd}os theorem (see E.3 e℄ of Se tion 4.2), f is of the form =0

f (x) =

1 X i=0

ai x i ; x 2 [0; 1) :

Sin e f 2 C [0; 1℄; Theorem 6.2.8 implies that A := 1 i ai exists and is nite. Re alling (6.2.2) and the hoi e of k ; and using E.3 e℄ of Se tion 4.2, we an dedu e that ea h P

=0

k aj;nk ;

j = 1; 2; : : : ; nk ; k = 1; 2; : : :

2

is equal to one of the oeÆ ients a ; a : : : : Sin e ja ;nk j = jPk (0)j  1 for ea h k 2 N ; from (6.2.1) and (6.2.2) we see that P jai j  1 holds for in nitely many i 2 N ; whi h ontradi ts the fa t that 1 i ai onverges. This nishes the if part of the theorem. Now assume that  is not la unary. Then for every  > 0 there is an n 2 N su h that n =n > 1 : Observe that Pn (x) := xn xn 1 a hieves its maximum modulus on [0; 1℄ at 1

2

0

=0

1

x= and hen e



n n

1

1=(n n

1)

310 6. Inequalities in Muntz Spa es

kPn k

;

[0 1℄





n n

1

n

= n n

1 (

1)



n n

1

 1

n < ; n

1

1

whi h shows that the lead oeÆ ient an;n of

Tn f ;  ; : : : ; n ; [0; 1℄g 0

1

is at least 1=; otherwise an;n Tn Pn 2 Mn () would have at least n zeros on (0; 1), whi h is a ontradi tion. The proof of the only if part of the theorem is now nished. ut 1

1

From the above proof it also follows that under the assumptions of Theorem 6.2.7, the Chebyshev polynomials

Tn f ;  ; : : : ; n ; [0; 1℄g 0

1

have uniformly bounded oeÆ ients if and only if  is la unary. As an appli ation of Theorem 6.2.7 we derive the following Bernsteintype inequality.

 1; and

Theorem 6.2.9 (Bernstein-Type Inequality). Suppose  := 0, 

suppose  := (i )1 i is la unary, that is,

0

1

=0

inf fi =i : i 2 N g > 1 : +1

Then there exists a onstant depending only on  (and not on y or the number of terms in p) su h that jp0 (y)j  1 y kpk ; [0 1℄

for every p 2 M () = spanfx0 ; x1 ; : : : g and for every y 2 [0; 1): Proof. Let p 2 M () be of the form p(x) = a

0

;n +

n X j =1

kpk

aj;n xj ;

;

[0 1℄

= 1:

Theorem 6.2.7 and the assumptions on  yield n

X jp0 (y)j  aj;n j yj

j =1 n X



1

j =1

j yj

1

1





n X j =1



1 X 2

j =0

jaj;n j j yj

yj =

1

2

y

1

;

where and depend only on ; and the theorem is proved. 1

2

ut

6.2 Nondense Muntz Spa es 311

Comments, Exer ises, and Examples. The results of this se tion have been obtained by Borwein and Erdelyi [91, 93, 95b, to appear 1℄. In E.1 we present some of the several important

onsequen es of our entral result, Theorem 6.2.2. In E.6 we oer another proof of the rst part of Theorem 6.2.1 when  is la unary, while E.9 shows that the Bernstein-type inequality of Theorem 6.2.9 \almost" hara terizes the la unary Muntz spa es. Note that if A  [0; 1℄ ontains an interval, then the rst part of Theorem 6.2.1 follows immediately from E.3 of Se tion 4.2. A typi al ase that does not follow from that exer ise is when A  [0; 1℄ is a \fat" Cantor-type set of positive measure. E.1 Some Consequen es of Theorem 6.2.2. Let A  [0; 1) be a set of positive Lebesgue measure, and let rA be the essential supremum of A as de ned in Theorem 6.2.1. Suppose q 2 (0; 1) and suppose w is a R nonnegative-valued, integrable weight fun tion on A with A w > 0: Let Lq (w) := Lq (); where d = w dt; and where Lq () is de ned in E.7 of Se tion 2.2. Let  := (i )1 be a sequen e of distin t nonnegative real i numbers with i 6= 0 for ea h i = 1; 2; : : : : P a℄ Suppose 1 i 1=i < 1: Then M () is not dense in Lq (w): Moreover, if the gap ondition inf fi i : i 2 N g > 0 holds, then every fun tion f 2 Lq (w) belonging to the Lq (w) losure of M () an be represented as =0

=1

1

f (x) = where

1 X i=0

ai xi ;

(

ai 2 R ;

rw := sup y 2 [0; 1) :

x 2 A \ [0; rw ) ; )

Z

A\(y;1)

w(x) dx > 0 :

If the above gap ondition does not hold, then every fun tion f 2 Lq (w) belonging to the Lq (w) losure of M () an still be represented as an analyti fun tion on

fz 2 C n ( 1; 0℄ : jz j < rw g restri ted to A:

Proof. Suppose f su h that

2 Lq (w) and suppose there is a sequen e (pi )1 i  M () =1

lim kf

i!1

pi kLq

(

w)

= 0:

Minkowski's inequality (see E.7 b℄ and E.7 i℄ of Se tion 2.2.) yields that (pi )1 i is a Cau hy sequen e in Lq (w): The assumptions on w imply that for every Æ 2 (0; rw ) there exists an  > 0 su h that =1

312 6. Inequalities in Muntz Spa es

B := fx 2 A \ (Æ; 1) : w(x) > g is of positive Lebesgue measure. Let s := m(B ) > 0: Observe that if kpkLq w < "; then (

)

(

x 2 B : jp(x)j 

m so

(

x 2 B : jp(x)j <

m



2"

1=q )!

s 

2"

1=q )!

s

 2s ; s > : 2

Hen e, by Theorem 6.2.2, (pi )1 i is uniformly Cau hy on [0; Æ ℄: The proof

an now be nished as that of Theorem 6.2.1. ut =1

b℄ PMuntz-Type Theorem in Lq (w). M () is dense in Lq (w) if and only if 1 i 1=i = 1: =1

P1

i 1=i = 1: Let f 2 Lq (w): It is standard measure theory to show that for every " > 0 there exists a g 2 C [0; 1℄ su h that g(0) = 0 and

Proof. Suppose

=1

" gkLq w < :

kf

2 Now Theorem 4.2.1 (full Muntz theorem in C [0; 1℄) implies that there exists a p 2 M () su h that

kg

pkLq w (

)

 kg

pkA

(

Z

A

w

)

1=q

 kg

pk

Z

;

[0 1℄

A

w

1=q

" < : 2

Therefore M () is dense in Lq (w): P Suppose now that 1 i 1=i < 1: Then part a℄ yields that M () is not dense in Lq (w): ut =1

℄ Convergen e in M (). Suppose

P1

pi (x) ! f (x) ;

i=1 1=i

< 1, (pi )1 i

=1

 M (); and

x 2 A:

Then (pi )1 i onverges uniformly on every losed subinterval of [0; rA ): =1

Proof. Let Æ 2 (0; rA ) be xed. Egoro's theorem (see, for example, Royden [88℄) and the de nition of rA imply the existen e of a set B  A \ (Æ; 1) of positive Lebesgue measure so that (pi )1 i onverges uniformly on B and hen e is uniformly Cau hy on B: Now Theorem 6.2.2 yields that (pi )1 i is uniformly Cau hy on [0; Æ ℄; and the result follows. ut =1

=1

6.2 Nondense Muntz Spa es 313

1 d℄ Suppose 1 i 1=i = 1: Show that there is a sequen e (pi )i  M () that onverges pointwise on [0; 1) but does not onverge uniformly on A \ [0; a℄ for some a 2 (0; rA ): Hint: Use Theorem 4.2.1 (Muntz's theorem). ut P1 e℄ Suppose i 1=i < 1 and inf fi i : i 2 N g > 0 : Let P () denote the olle tion of all real-valued fun tions f de ned on [0; 1) by a power series P

=1

=1

=1

1

f (x) =

1 X i=0

ai 2 R ;

ai xi ;

x 2 [0; 1) :

Suppose that A  [0; 1℄ with rA = 1: Show that if (fi )1 i fi (x) ! f (x) ; x 2 A; then fi (x) ! fe(x) ; x 2 [0; 1) ; e where f 2 P (). Hint: Use part ℄ and E.3 e℄ of Se tion 4.2.

=1

 P () and

ut

E.2 On the Smallest Zero of Chebyshev Polynomials in Nondense Muntz P Spa es. Suppose  := 0 and 1 1 = i < 1: Show that there exists a i

onstant > 0 depending only on  := (i )1 (and not on n) su h that i the smallest positive zero of Tn f0;  ;  ; : : : ; n ; [0; 1℄g ; n = 1; 2; : : : is greater than : Hint: If   1; then use the Mean Value Theorem, E.1 a℄ of Se tion 3.3, and E.5 b℄ of Se tion 4.2. If 0 <  < 1; then the s aling x ! x =1 redu es the problem to the ase  = 1: ut 0

=1

=0

1

2

1

1

1

1

E.3 Extremal Fun tions for the Remez-Type Inequality of Theorem 6.2.2. Suppose 0   <  <


< n ; 0 < %; A  [%; 1) is a ompa t set

ontaining at least n + 1 points, and y 2 (0; %) is xed. Let 0

1

Mn () := spanfx0 ; x1 ; : : : ; xn g : a℄ Show that there is a 0 6= p 2 Mn () su h that jp (y)j = sup jp(y)j : kp kA 6 p2Mn  kpkA 0=

(

)

Hint: Use a ompa tness argument. b℄ Show that p = Tn f ;  ; : : : ; n ; Ag for some 2 R: Hint: Use a perturbation argument. 0

1

ut ut

314 6. Inequalities in Muntz Spa es

E.4 A Lexi ographi Property of Chebyshev Polynomials in Dierent Muntz Spa es. Let 0 :=  <  <


< n ; 0 := < <


< n ; and j  j ; j = 0; 1; : : : ; n : 0

1

0

1

Let % > 0 and let A  [%; 1) be ompa t ontaining at least n + 1 points, and let

Tn; := Tn f ;  ; : : : ; n ; Ag and Tn; := Tnf ; ; : : : ; n ; Ag : 0

1

0

1

a℄ Show that jTn; (y )j  jTn; (y )j for every y 2 [0; %): Hint: Suppose, without loss of generality, that there is an index m, 1  m  n; su h that m < m and j = j if j 6= m: We hoose an Rn; 2 Mn () that interpolates Tn; at the n zeros of Tn; and is normalized so that Rn; (0) = Tn; (0): Use Theorem 3.2.5 to show that jRn; (x)j  jTn; (x)j for every x 2 [0; 1); in parti ular for every x 2 A: Now use E.3 to show that jTn; (0)j = jRn; (0)j  jTn; (0)j; whi h gives the desired result for y = 0: Using this, we an dedu e that jTn; (y )j  jTn; (y )j for every y 2 [0; %); otherwise

Tn;

Tn;

2 spanfx ; x ; : : : ; xn ; x m g 0

1

would have at least (n + 2) zeros in (0; 1); whi h is a ontradi tion. b℄ Show that max

p2Mn (

)

ut

jp(y)j  max jp(y)j kpkA p2Mn  kpkA (

)

for every y 2 [0; %); where

Mn ( ) := spanfx 0 ; x 1 ; : : : ; x n g and

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

ut

Hint: Combine part a℄ and E.4.

E.5 Theorem 6.2.1 Follows from Theorem 6.2.2. Under the assumptions of Theorem 6.2.1 show that if (pj )1  M () is uniformly Cau hy in j C (A); then it is uniformly Cau hy in C [0; y℄ for every y 2 (0; rA ), where rA is de ned as in Theorem 6.2.1. Hint: Use Theorem 6.2.2. ut =1

6.2 Nondense Muntz Spa es 315

E.6 Some Corollaries of Theorem 6.2.9 in the La unary Case. Suppose  := 0,   1; and  := (i )1 i is la unary. a℄ Show that the Chebyshev polynomials Tn := Tn f ;  ; : : : ; n : [0; 1℄g ; n = 1; 2; : : : have the following property: There is a onstant 2 (0; 1) depending only on  (and not on n) su h that if y 2 [0; 1) and jTn (y )j = 1; then jTn (x)j  for every x 2 [y; y + (1 y )℄: Hint: Use the Mean Value Theorem and the Bernstein-type inequality of Theorem 6.2.9. ut b℄ Show that there is a onstant 2 (0; 1) depending only on  (and not on n) so that if a < b are two onse utive zeros of Tn ; then 1 b < (1 a):

℄ Let  2 (0; 1): Show that there is an n 2 N depending only on  (and not on n) so that every Tn has at most n zeros in [0; 1 ℄: d℄ Give a new proof of the rst part of Theorem 6.2.1 based on parts a℄ and ℄. 0

1

=0

0

1

1

1

2

1

2

2

0

0

Outline. By Lebesgue's density theorem (see Royden [88℄), it may be sup-

posed, without loss of generality, that the left-hand side Lebesgue density of A at 1 is 1: Choose  2 (0; 1) so that A \ [0; 1 ) ontains in nitely many points and m(A \ [y; 1℄) (6:2:5) > 1 y for every y 2 [1 ; 1℄; where 2 (0; 1) is the same as in part a℄. For this ; hoose n a

ording to part ℄. Now de ne g 2 C (A) so that g alternates n + 3 times in A \ [0; 1 ) between 2 and 2 and is identi ally zero on [1 ; 1℄: Assume that there exists a p 2 Mn () su h that kp g kA  : Use part a℄ and (6.2.5) to show that p Tn 2 Mn () has more than n distin t zeros in [0; 1℄; whi h is a ontradi tion. ut 1

1

0

0

1

4

The following simple appli ation of Theorem 6.2.9 was pointed out by Woj ieszyk: e℄ Suppose A  [0; 1℄ is a measurable set and the left-hand side Lebesgue density of A at 1 is 1: Show that there is a onstant > 0 depending only on  and A so that

kpk

 kpkA g:

[0;1℄   x 0; x 1; : : :

for every p 2 M () = spanf Hint: Use the Mean Value Theorem, the Bernstein-type inequality of Theorem 6.2.9, and the Chebyshev-type inequality of E.3 f℄ of Se tion 4.2. u t f℄ Use part e℄ to give another proof of Theorem 6.2.1. The following exer ise onstru ts quasi-Chebyshev polynomials Pn for Mn () if the la unarity onstant of  is large:

316 6. Inequalities in Muntz Spa es

E.7 Quasi-Chebyshev Polynomials in Very La unary Muntz Spa es. Let  = 0,  = 2; and i =i  16 for i = 1; 2; : : : : Let 0

1

+1

n X

Pn (x) := 1 + 2

j =1

( 1)j xj ;

n = 1; 2; : : : :

Let yi := (4i ) . Prove the following statements: a℄ kPn k ; = 1 and Pn (1) = ( 1)n : b℄ Pn has exa tly n zeros, x ;n < x ;n <


< xn;n , in (0; 1):

℄ jPn0 ( )j  2n for every  2 [xn;n ; 1℄. d℄ We have 1

[0 1℄

1

Pn (x) 

1 if 1 5

y

2

2

k

y

x1

k

2

2

and 1  2k  n

and

Pn (x) 

1 if 1 5

y

k

2 +1

x1

y

and 1  2k + 1  n :

k

2 +1

2

Hint: Part a℄ is obvious. Prove the rest together, by indu tion on n: The next exer ise follows Borwein and Erdelyi [95b℄.

ut

E.8 Produ ts of Muntz Spa es. Asso iated with  := (j )1 j ; let =0

(

M k () :=

p=

)

k Y j =1

pj : pj

2 M ()

;

k = 1; 2; : : : :

Is M () dense in C [0; 1℄ for  := (j )1 j ? 2

2

=0

Note that M k (); k  2 is not the linear span of monomials, and Muntz's theorem does not give the answer. This exer ise establishes Remez-, Bernstein-, and Nikolskii-typePinequalities for M k (): From any of these it follows immediately that if 1 j 1=j < 1 and A  [0; 1℄ is a set of positive Lebesgue measure, then M k () is not dense in C (A): P1 Throughout parts a℄ to d℄ of the exer ise we assume 0 =  <  <


, j 1=j < 1; and s 2 (0; 1): =1

0

1

=1

a℄ Remez-Type Inequality for M k (). There exists a onstant depending only on ; s; and k (and not on % or A) su h that

kpk

[0

;%℄

 kpkA

for every p 2 M k () and for every ompa t set A measure at least s > 0:



[%; 1℄ of Lebesgue

6.2 Nondense Muntz Spa es 317

Proof. Theorem 6.2.2 implies that there exists a onstant  > 0 depending only on , s; and k su h that s m(fx 2 [y; 1℄ : jp(x)j >  jp(y)jg)  1 y 2k 1

for every p 2 M () and y 2 [0; 1

p= Then, for every y 2 [0; 1

k Y j =1

s℄: Now let p 2 M k (), that is,

pj ;

pj

2 M () :

s℄;

m(fx 2 [y; 1℄ : jp(x)j > 

m 1

k \

j =1

y

k

jp(y)jg)

!

fx 2 [y; 1℄ : jpj (x)j >  jpj (y)jg 1

k

s =1 y 2k

s

2

:

Hen e y 2 [0; inf A℄ and m(A)  s imply that

m(fx 2 A : jp(x)j > 

k

jp(y)jg)  2s > 0 ut

and the inequality follows with = k .

b℄ Solution to Newman's Problem. Let A  [0; 1℄ be a set of positive Lebesgue measure. Then M k () is not dense in C (A):

ut

Proof. This follows from part a℄.

℄ Bernstein-Type Inequality for M k (). Suppose   1: There exists a

onstant depending only on , s; and k (and not on % and A) su h that 1

kp0 k

;%℄

[0

 kpkA

for every p 2 M k () and for every ompa t set A  [%; 1℄ of Lebesgue measure at least s > 0: Hint: Use the produ t rule of dierentiation, and estimate ea h term separately. Pro eed as in the proof of part a℄. Use Theorem 6.2.2 and E.5 a℄ and b℄ of Se tion 4.2. ut k d℄ Nikolskii-Type Inequality for M (). There exists a onstant depending only on , s, k , q , and w (and not on % and A) su h that Z

kpkq ;%  jp(x)jq w(x) dx [0

℄

A

318 6. Inequalities in Muntz Spa es

for every p 2 M k (); for every ompa t set A  [%; 1℄ of Lebesgue measure at least s > 0; for every fun tion w measurable and positive a.e. on [0; 1℄; and for every q 2 (0; 1): Hint: Use part a℄. ut e℄ Asso iated with

j = (i;j )1 i ;

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

=0

let

8 <

M ( ;  ; : : : ; k ) := p = 1

2

:

k Y j =1

pj : pj

9 =

2 M (j ); :

Formulate and prove the analogs of parts a℄ to d℄ for M ( ;  ; : : : ; k ): 1

2

E.9 A Weak Converse of Theorem 6.2.9. Suppose  := (i )1 is a i (stri tly) in reasing sequen e of nonnegative real numbers with  := 0 and   1: Suppose also that there exists a onstant depending only on  (and not on y or the number of terms in p) su h that =0 0

1

jp0 (y)j  1 y kpk

;

[0 1℄

for every p 2 M () = spanfx0 ; x1 ; : : : g and for every y 2 [0; 1): Show that there is a onstant  > 1 depending only on  su h that n  n :

Outline. Let

Tn := Tn f ;  ; : : : ; n ; [0; 1℄g and denote its zeros in (0; 1) by x ;n > x ;n >


> xn;n : Use the Mean 0

1

1

2

Value Theorem and the assumed Bernstein-type inequality to show that there is a onstant 2 (0; 1) depending only on  su h that 1

xj;n  (1 xj

+1

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

;n ) ;

hen e 1 x ;n  n : On the other hand, use the Mean Value Theorem and Theorem 6.1.1 (Newman's inequality) to show that 1

1

x

;n

1



1+9

n X j =1

!

j

1

 (9(n + 1)n )

Finally, ombine the lower and upper bounds for 1

n 

n : 9(n + 1)

x

;n

1

1

:

to on lude that

ut

6.2 Nondense Muntz Spa es 319

E.10 Polynomials in xn . Given n 2 N and n 2 R; let

Pn (n ) := fpn (xn ) : pn 2 Pn g (as in E.6 of Se tion 4.1). Suppose n  1 for all n 2 N : Let Æ 2 R be de ned by log n 1 1 = log : lim sup n 2 Æ n Suppose Æ > 0: a℄ Bounded Remez-Type Inequality. Suppose 0 < Æe < Æ: Show that there exists a onstant depending only on Æe (and not on n, y; or A) su h that

jp(y)j  kpkA for every p 2 [1 n Pn (n ); for every A  [0; 1℄ of Lebesgue measure at least e and for every y 2 [0; inf A℄: 1 Æ; Hint: Use Lemma 6.2.6 and E.6 a℄ of Se tion 4.1. ut =1

b℄ Muntz-Type Theorem. If 0  Æe < Æ and A  [0; 1℄ is a set of Lebesgue e then [1 Pn (n ) is not dense in C (A). measure at least 1 Æ; n Hint: Use part a℄. ut =1

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7

Inequalities for Rational Fun tion Spa es

Overview Pre ise Markov- and Bernstein-type inequalities are given for various lasses of rational fun tions in the rst se tion of this hapter. Extensions of the inequalities of Lax, S hur, and Russak are also presented, as are inequalities for self-re ipro al polynomials. The se ond se tion of the hapter is

on erned with metri inequalities for polynomials and rational fun tions.

7.1 Inequalities for Rational Fun tion Spa es Sharp extensions of most of the polynomial inequalities of Se tion 5.1 are established for rational fun tion spa es on K := R (mod 2 ), on the interval [ 1; 1℄, on the unit ir le of C , and on the real line. The lassi al inequalities of Se tion 5.1 are then re overed as limiting ases. A sharp extension of Lax's inequality is also given. Essentially sharp Markov- and Bernsteintype inequalities for self-re ipro al and antiself-re ipro al polynomials are presented in the exer ises. Let D := fz 2 C : jz j < 1g and D := fz 2 C : jz j = 1g; as before. We study the rational fun tion spa es:

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

2

2

(

Q2n

k=1

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

)

7.1 Inequalities for Rational Fun tion Spa es 321

and (

T

n (a1 ; a2 ; : : :

; a n ; K ) :=

on K with a ; a ; : : : ; a n 2 C 1

2

2

1; 1℄) :=

Pn (a ; a ; : : : ; an ; [

1; 1℄) :=

2

k=1

)

n R;

Pn (a ; a ; : : : ; an ; [ 1

Q2n

2

t() : t 2 Tn sin(( ak )=2)





Qn

p(x) jx ak j : p 2 Pn

Qn

p(x) : p 2 Pn (x ak )

k=1

and 1

2

on [ 1; 1℄ with a ; a ; : : : ; an 2 C 1

2

Pn (a ; a ; : : : ; an ; D) := 1



k=1

n[

1

2

p(z ) : p 2 Pn (z ak )



p(x) jx ak j : p 2 Pn



Qn



Qn

p(z ) : p 2 Pn (z ak )

Qn

k=1

n D; and

Pn (a ; a ; : : : ; an ; R) := 1

1; 1℄;



2

on D with a ; a ; : : : ; an 2 C





2

k=1

and

P

n (a1 ; a2 ; : : :

; an ; R) :=

on R with a ; a ; : : : ; an 2 C 1

2



k=1

n R:

The Chebyshev polynomials Ten , Uen , and

2K

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

for the rational fun tion spa e Tn (a ; a ; : : : ; a n ; K ) are de ned in E.3 of Se tion 3.5, and they play a entral role in this se tion. 1

2

2

322 7. Inequalities for Rational Fun tion Spa es

Theorem 7.1.1 (Bernstein-Szeg}o-Type Inequality on K ). Given (ak )kn 2

=1

 C nR;

let Ben () :=

Im(ak ) > 0 ;

n

1 X 1 jeiak j : 2 k jeiak ei j 2

2

2

=1

Then

f 0 () + Ben ()f ()  Ben ()kf kK ; 2K for every f 2 Tn (a ; a ; : : : ; a n ; K ): Equality holds if and only if either  is a maximum point of jf j (that is, f () = kf kK ) or f is a linear ombination of Ten and Uen (with real

oeÆ ients) as de ned in E.3 of Se tion 3.5. 2

2

1

2

2

2

2

2

Corollary 7.1.2 (Bernstein-Type Inequality on K , Real Case). Given (ak )kn 2

=1

 C nR;

Im(ak ) > 0 ;

let the Bernstein fa tor Ben be de ned as in Theorem 7.1.1. Then jf 0 ()j  Ben ()kf kK ;  2 K

for every f 2 Tn (a ; a ; : : : ; a n K ): Equality holds if and only if f is a linear ombination of Ten and Uen (with real oeÆ ients) as de ned in E.3 of Se tion 3.5, and f () = 0: 1

2

2

;

Theorem 7.1.1 and Corollary 7.1.2 an be easily obtained from the extension of Theorem 3.5.3 given by E.3 of Se tion 3.5, whi h gives expli it formulas for the Chebyshev polynomials for these lasses Tn (a ; a ; : : : ; a n ; K ): The arguments are outlined in E.1. The following two results an be obtained from Theorem 7.1.1 and Corollary 7.1.2 by the substitution x = os ; see E.2. 1

2

2

Corollary 7.1.3 (Bernstein-Szeg}o-Type Inequality on [ 1; 1℄). Asso iated  C n [ 1; 1℄; let the Bernstein fa tor Bn be de ned by

with (ak )nk

=1

Bn (x) := p

n X k=1

p

Re

ak

ak

1

2

x

!

;



p



where the hoi e of ak 1 is determined by ak ak 1 < 1: Then (1 x )f 0 (x) + Bn (x)f (x)  Bn (x)kf k ; ; x 2 [ 1; 1℄ 2

2

2

2

2

2

2 [ 1 1℄

2

for every f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄): Equality holds if and only if either x is a maximum point of jf j (that is, f (x) = kf k ; ) or f = Tn with 2 R; where Tn is de ned as in Se tion 3.5. 1

[

1 1℄

2

7.1 Inequalities for Rational Fun tion Spa es 323

Corollary 7.1.4 (Bernstein-Type Inequality on [ 1; 1℄, Real Case). Given (ak )nk  C n [ 1; 1℄; let the Bernstein fa tor Bn be de ned as in Corollary =1

7.1.3. Then

jf 0 (x)j  pBn (x) kf k 1

x

[

2

;

1 1℄

x 2 ( 1; 1)

;

for every f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄): Equality holds if and only if f = Tn with 2 R; where Tn is de ned as in Se tion 3.5, and f (x) = 0: (Note that Bn (x) > 0 for every x 2 ( 1; 1).) 1

2

Our next result follows from Theorem 7.1.1; see the hints to E.3. Corollary 7.1.5 (Bernstein-Szeg}o-Type Inequality on R). Given (ak )nk

=1

 C nR;

Im(ak ) > 0 ;

let the Bernstein fa tor Bn be de ned by Bn (x) := Then

n X k=1

Im(ak )

jx

ak j

2

:

f 0 (x) + Bn (x)f (x)  Bn (x)kf kR ; 2

2

2

2

2

x2R

for every f 2 Pn (a ; a ; : : : ; an ; R): Equality holds if and only if either x is a maximum point of jf j (that is, jf (x)j = kf kR) or f is a linear ombination of Tn and Un (with real

oeÆ ients) de ned in E.5 of Se tion 3.5. 1

2

Corollary 7.1.6 (Bernstein-Type Inequality on R, Real Case). Asso iated

with (ak )nk 7.1.5. Then

=1

 C n R ; let the Bernstein fa tor Bn be de ned as in Corollary jf 0 (x)j  Bn (x)kf kR ;

x2R

for every f 2 Pn (a ; a ; : : : ; an ; R): Equality holds if and only if f is a linear ombination of Tn and Un (with real oeÆ ients) de ned in E.5 of Se tion 3.5, and f (x) = 0: 1

2

To formulate our next theorem we introdu e some notation. For a polynomial

q(z ) := we de ne

n Y

k=1

(z

ak ) ;

ak 2 C ;

324 7. Inequalities for Rational Fun tion Spa es

q (z ) :=

n Y k=1

ak z ) =: z n qn (z ) :

(1

1

Then

jq(z )j = jq (z )j ;

(7:1:1) The fun tion

Sn (z ) :=

z 2 D :

n q  (z ) Y 1 ak z = q(z ) k z ak =1

is the Blas hke produ t asso iated with (ak )nk : =1

Theorem 7.1.7 (Bernstein-Type Inequality on D, Complex Case). Given (ak )nk  C n D ; let the Bernstein fa tor Bn be de ned by =1

Bn (z ) := maxfBn (z ); Bn (z )g +

with Bn (z ) := +

n X

2

jakk=1 j>1

Then

jak j 1 jak z j

and

2

Bn (z ) :=

jf 0 (z )j  Bn (z )kf kD ;

n X

jakk=1 j<1

jak j jak z j 1

2 2

:

z 2 D

for every f 2 Pn (a ; a ; : : : ; an ; D): If the rst sum is not less than the se ond sum for a xed z 2 D; then equality holds for f = Sn with 2 C , where Sn is the Blas hke produ t asso iated with those ak for whi h jak j > 1: If the se ond sum is not less than the rst sum for a xed z 2 D; then equality holds for f = Sn with

2 C ; where Sn is the Blas hke produ t asso iated with those ak for whi h jak j < 1: 1

2

+

+

Proof. For reasons of symmetry it is suÆ ient to prove the theorem only for z = 1: Without loss of generality we may assume that (7:1:2)

Re

n X k=1

1

1

!

ak

6= n2 ;

the remaining ases follow from this by a limiting argument. Let Q := D (equipped with the usual metri topology), V := Pn (a ; a ; : : : ; an ; D) ; and L(f ) := f 0 (1) for f 2 V: We show in this situation that n + 1  r in Theorem A.3.3 (interpolation of linear fun tionals). Suppose to the ontrary that r  n: By Theorem A.3.3, there are distin t points x ; x ; : : : ; xr on D, and there are onstants ; ; : : : ; r 2 C su h that 1

2

1

1

2

2

7.1 Inequalities for Rational Fun tion Spa es 325

(7:1:3)

r p(x ) p0 (1)q(1) q0 (1)p(1) X =

k k ; q(1) q(xk ) k

p 2 Pn ;

2

=1

where (7:1:4)

q(z ) :=

n Y

(z

k=1

ak ) :

We laim that xk 6= 1 for ea h k = 1; 2; : : : ; r: Indeed, if there is index k su h that xk = 1; then Theorem A.3.3 implies that

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

r

r Y k=1

xk ) 2 Pn

(z

has a zero at 1 with multipli ity at least two, whi h is a ontradi tion. Applying (7.1.3) with the above p; we obtain

p0 (1)q(1)

q0 (1)p(1) = 0

and sin e p(1) 6= 0 and q (1) 6= 0, this is equivalent to

q0 (1) p0 (1) = ; q(1) p(1) that is, in terms of the zeros of pn and qn ; n X

(7:1:5) Sin e xk (7:1:6)

k=1

1

1

=

ak

n r 2

+

r X k=1

1

1

xk

:

2 D and xk 6= 1; we have Re



1

1



xk

=

1 ; 2

k = 1; 2; : : : ; r :

It follows from (7.1.5) and (7.1.6) that Re

n X k=1

1

1

!

ak

=

n 2

;

whi h ontradi ts assumption (7.1.2). So n + 1  r; indeed. that

A ompa tness argument shows that there is a fun tion fe 2 V su h

kfekD = 1

and

jL(f )j : L(fe) = kLk := max 6 f 2V kf kD 0=

326 7. Inequalities for Rational Fun tion Spa es

Theorem A.3.3 implies jfe(xk )j = 1 for every k = 1; 2; : : : ; r: Hen e, if

pe fe = ; q

pe 2 Pn ;

q(z ) =

n Y k=1

(z

ak ) ;

then

h(z ) := jpe(z )j

(7:1:7)

jq ( z )j  0 ;

2

z 2 D

2

and (7:1:8)

h(xk ) = 0 ;

k = 1; 2; : : : ; r :

Note that t() := h(ei ) 2 Tn vanishes at ea h k , where the numbers k 2 [ ; ) are de ned by xk = eik ; k = 1; 2; : : : ; r: Be ause of (7.1.7), ea h of these zeros is of even multipli ity. Hen e, n + 1  r implies that t 2 Tn has at least 2n + 2 zeros and therefore t = 0: From this we an dedu e that h(z ) = 0 for every z 2 D; so

jpe(z )j = jq(z )j ;

(7:1:9)

z 2 D :

We now have

z npe(z )pe (z ) = jpe(z )j = jq(z )j = z nq(z )q (z ) ; 2

2

z 2 D ;

so by the uni ity theorem for analyti fun tions (see E.1 e℄ of Se tion 1.2)

pe pe = qq : From this, it follows that there exists a onstant 0 6= 2 C su h that

fe(z ) =

m Y z k pe(z ) = ; q(z ) z k k 1

z2C;

q(z ) 6= 0

=1

with some m  n and

k := ajk ;

1  j < j <


< jm  n :

k = 1; 2; : : : ; m ;

1

2

A straightforward al ulation gives that







m e0 X 1 jfe0 (1)j = fe(1) = 1 k f (1) k m X =

k=1

whi h nishes the proof.

1

1

=1

1 1 2

1

 k

jk j jk j  maxfBn (z ); Bn (z )g ; 2

+

ut

7.1 Inequalities for Rational Fun tion Spa es 327

Corollary 7.1.8 (Bernstein-Type Inequality on K , Complex Case). Given (ak )kn  C n R; let the Bernstein fa tor Bn be de ned by 2

=1

Ben () := maxfBen (); Ben ()g +

with Ben () := +

jeiak j

2n X

1

2

k=1 Im(ak )<0

Then

jeiak

and Ben () :=

ei j

2

jf 0 ()j  Bn ()kf kK ;

2n X

1

k=1 Im(ak )>0

jeiak

jeiak j

2

ei j

2

:

2K

for every f 2 Tn (a ; a ; : : : ; a n ; K ): If the rst sum is not less than the se ond sum for a xed  2 K; then equality holds for f () = S n (ei ) with 2 C ; where S n is the Blas hke produ t asso iated with those eiak for whi h Im(ak ) < 0: If the se ond sum is not less than the rst sum for a xed  2 K; then equality holds for f () = S n (ei ) with 2 C ; where S n is the Blas hke produ t asso iated with those eiak for whi h Im(ak ) > 0: Note that S  (ei ) 2 T (a ; a ; : : : ; a n ; K ) : 1

2

2

+

+

2

2

2

2

2

n

n

1

2

2

Proof. Observe that if h() :=

2n Y

j =1

aj )=2) 2 Tn

sin((

and tn 2 Tn ; then there are p 2 P n and q 2 P n su h that 2

2

t() p(ei )e = h() q(ei )e where q is of the form

q (z ) =

2n Y

j =1

in in

p(ei ) ; q(ei )

=

eiaj )

(z

with some 2 C : So the orollary follows from Theorem 7.1.7.

ut

Corollary 7.1.9 (Bernstein-Type Inequality on [ 1; 1℄, Complex Case). Given fak gnk  C n [ 1; 1℄, let the Bernstein fa tor Bn (x) be de ned

by

=1

Bn (x) = max

( n X

k=1

j k j j k z j 1

2 2

;

n X k=1

j k j j k

1

2

1

zj

2

)

;

328 7. Inequalities for Rational Fun tion Spa es

where k and z are determined by

j k j < 1 ;

ak := ( k + k ) ; x := (z + z ) ; 1

1 2 1

Im(z ) > 0 :

1

2

Then

jf 0 (x)j  pBn (x) kf k x

1

for every f

2 Pn (a ; a ; : : : ; an ; [ 1

n X k=1

2

j k j j k z j 1

2

p

where the hoi e of

ak

;

1 1℄

x 2 ( 1; 1)

;

1; 1℄): Note that

n X

= Re

2

[

2

p

k=1

ak 1 ak x 2

!

;

x 2 [ 1; 1℄ ;



p

1 is determined by ak

2

ak 2



1 < 1:

Proof. The orollary follows from Theorem 7.1.7 by the substitution x = (z + z ): ut 1

1

2

Bernstein's lassi al polynomial inequalities dis ussed in Se tion 5.1 are ontained in Theorem 7.1.7 and Corollaries 7.1.8 and 7.1.9 as limiting

ases. In Theorem 7.1.7 and Corollary 7.1.9 we take (a m ; a m ; : : : ; anm )  C (

)

(

1

so that

)

(

lim jakm j = 1 ; (

m!1

)

2

nD

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

)

In Corollary 7.1.8 we take (a m ; a m ; : : : ; a m n )C (

1

)

(

)

(

2

)

2

nR

so that

anm k = akm (

)

(

and

)

+

lim jIm(akm )j = 1 ; (

m!1

)

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

To formulate our next result we introdu e the Blas hke produ t

Qn (z ) := asso iated with (a ; a ; : : : ; an )

z 2 R:

1

2

n Y k=1

z z

ak ak

 C n R: Obviously jQn (z )j = 1 for every

7.1 Inequalities for Rational Fun tion Spa es 329

Corollary 7.1.10 (Bernstein-Type Inequality on R, Complex Case). Given (ak )nk  C n R; let the Bernstein fa tor Bn (x) be de ned by =1

Bn (x) := maxfBn (x); Bn (x)g +

with Bn (x) := +

n X

2 jIm(ak )j

k=1 ak )>0

jx

ak j

2

n X

and Bn (x) :=

k=1 ak )<0

2 jIm(ak )j

jx

ak j

2

Im(

Im(

for every x 2 R: Then

jf 0 (x)j  Bn (x)kf kR ;

x2R

for every f 2 Pn (a ; a ; : : : ; an ; R): If the rst sum is not less than the se ond sum for a xed x 2 R; then equality holds for f = Qn with 2 C ; where Qn is the Blas hke produ t asso iated with the poles ak lying in the open upper half-plane 1

2

+

+

H := fz 2 C : Im(z ) > 0g : +

If the se ond sum is not less than the rst sum for a xed x 2 R; then equality holds for f = Qn with 2 C ; where Qn is the Blas hke produ t asso iated with the poles ak lying in the open lower half-plane H := fz 2 C : Im(z ) < 0g : Corollary 7.1.10 follows from Theorem 7.1.7; see E.4. The next theorem improves the Bernstein-type inequality of Theorem 7.1.7 in the ase when fak gnk  C n D and f has all its zeros in C n D: It extends Lax [44℄. =1

Theorem 7.1.11 (Lax-Type Inequality). Given (ak )nk

=1

Bernstein fa tor Bn be, as in Theorem 7.1.7, de ned by Bn (z ) := Then

jh0 (z )j 

1 2

n X k=1

 C n D,

let the

jak j 1 : jak z j 2

Bn (z )khkD ;

2

z 2 D

for every h 2 Pn (a ; a ; : : : ; an ; D) having all its zeros in C n D: Equality holds for h = (Sn + 1) with 2 C ; where Sn is the Blas hke produ t asso iated with (ak )nk . 1

2

=1

330 7. Inequalities for Rational Fun tion Spa es

Note that Bn (z ) = jSn0 (z )j: Note also that h := (Sn + 1) 2 Pn (a ; a ; : : : ; an ; D) ; 1

6= 0

2

has all its zeros on D: Proof. First assume that ea h zero of h is on D; the general ase an be redu ed to this (see E.5). Thus let h := p=q; where p 2 Pn has all its zeros on D and where q(z ) := Let

n Y k=1

(z

q (z ) :=

We study

jak j > 1 :

ak ) ; n Y k=1

ak z ) :

(1

p(e i )e in jq(e i )j =

p(e i ) p q(e i q (e i ) p for  2 R; where the square roots are taken so that q is analyti in a p neighborhood of the losed unit disk, and q  is analyti in a neighborhood of the omplement of the open unit disk. Sin e p 2 Pn has all its zeros on D; there exists a 0 6= 2 C su h that u() :=

2

2

2

p

2

t() := p(e i )e

)

2

in

2

is a real trigonometri polynomial of degree at most n (see E.5 a℄). Also

jq(e

i)

2

n

Y j = jq (e i )j = j1 ak e 2

=

2n Y

k=1

i

2

k=1

j sin((

j

k )=2)j ;

where > 0;

ei k = ak

=

1 2

;

Im( k ) > 0 ;

k = 1; 2; : : : ; n

and

ei k = ak

=

1 2

;

Im( k ) > 0 ; k = n + 1; n + 2; : : : ; 2n :

Applying Theorem 7.1.1 to

u 2 Tn ( ; ; : : : ; n ; K ) ; 1

1

2

2

7.1 Inequalities for Rational Fun tion Spa es 331

we obtain

ju0 ()

(7:1:10)

iBen ()u()j  Ben ()kukK ;

2K;

where n

1X Ben () = 2k

(7:1:11)

n X k=1

jak = jak j

1

jak

Observe that

p(e i ) u() = q(e i ) 2

(7:1:12)

2

e

1

2

i

2

p

1

ei j

1 2

=1

=

jak j

1

j

=

2

+

n X k=1

j

ak

jak j

jak

ei j

=

1 2

2

1

2

i

e

2

!

1

j

2

:

q(e i ) p(e i ) fn (ei ) ; = q (e i ) q(e i )

p

2

2

2

2

where

p

(7:1:13)

2

jak j

1

fn (z ) :=

p

q (z ) : q (z ) 2

2

A simple al ulation (see E.4. of Se tion 3.5) shows that

Ben () = ei

(7:1:14)

fn0 (ei ) ; fn (ei )

2K:

Also, sin e jfn (ei )j = 1 for every  2 K; we have

kuk

(7:1:15)



p

K =

q

D

= khkD :

Now (7.1.10) to (7.1.15) yield   d p(e2i ) i d q (e2i ) fn (e )

f 0 (ei ) iei n i fn (e )



p(e i ) i )  B en ()khkD : f ( e n q(e i ) 2

2

So

j2ie i h0 (e i )fn (e i ) + iei fn0 (ei )h(e i ) iei fn0 (ei )h(e i )j  Ben ()khkD : 2

2

2

2 2

Thus

2 jh0 (e i )j  Ben ()khkD ; 2

whi h, together with (7.1.11), nishes the proof.

ut

332 7. Inequalities for Rational Fun tion Spa es

Comments, Exer ises, and Examples. Most of the results in this se tion have been proved in Borwein and Erdelyi [to appear 4℄ and in Borwein, Erdelyi, and Zhang [94a℄. A weaker version of Corollary 7.1.9 has been obtained by Russak (see Petrushev and Popov [87℄). Theorem 7.1.11 ontains, as a limiting ase, an inequality of Lax [44℄

onje tured by Erd}os. Lax's inequality establishes the sharp Bernstein-type inequality on the unit disk for polynomials p 2 Pn having no zeros in the open unit disk. That is,

kp0 kD  n2 kpkD

for su h polynomials. Various extensions of this inequality are given by Ankeny and Rivlin [55℄, Govil [73℄, Malik [69℄, and others. We dis uss some of these in E.16 of Appendix 5. E.1 Proof of Theorem 7.1.1 and Corollary 7.1.2. Given (ak )kn let Tn;a := Tn (a ; a ; : : : ; a n ; K ) : 2

=1

1

2

 C n R;

2

a℄ Show that Tn;a is a Hermite interpolation spa e. That is, if the points x ;x P ; : : : ; xk 2 K are distin t, and m ; m ; : : : ; mk are positive integers with ki mi  2n + 1; then for any hoi e of real numbers yi;j , there is a fun tion f 2 Tn;a su h that 1

2

1

2

=1

f j (xi ) = yi;j ;

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

( )

1:

ut

Hint: See the hint to E.7 of Se tion 1.1. b℄ Show that for every xed  2 K; the value f 0 () + Bn ()f () 6 f 2Tn;a kf kK max

0=

2

2

2

2

is attained by an fe 2 Tn;a : Hint: Use a ompa tness argument. ut

℄ Show that fe = V; where 2 R and V is one of the Chebyshev polynomials for Tn;a de ned in Theorem 3.5.3 and E.3 of Se tion 3.5. Hint: Use a variational method with the help of part a℄. ut d℄ Prove Theorem 7.1.1. Hint: Use part ℄ and E.4 of Se tion 3.5. ut e℄ Prove Corollary 7.1.2. E.2 Proof of Corollaries 7.1.3 and 7.1.4. a℄ Prove the inequality of Corollary 7.1.3.

7.1 Inequalities for Rational Fun tion Spa es 333

Hint: Use Theorem 7.1.1 with the substitution x = os : Note that f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄) implies 1

2

g() := f ( os ) = Tn ( ; ; ; ; : : : ; n ; n ; K ) ; 1

where the numbers k

ei k = ak

2C

q

ak

2

2

are de ned by

ak = (ei k + e

1;

2

1

1

2

i k );

Im( k ) > 0 :

Verify that if Ben is the Bernstein fa tor given in Theorem 7.1.1 asso iated with ( ; ; ; ; : : : ; n ; n ) ; 1

then

Ben () = p

n X k=1

1

2

2

p

Re

ak

ak 2

1

os 

!

 2K;

;

p



where the hoi e of ak 1 is determined by ak ak 1 < 1: ut b℄ Given x 2 [ 1; 1℄; prove that equality holds in the inequality of Corollary 7.1.3 if and only if either x is a maximum point of jf j (that is, f (x) = kf k ; ) or f = Tn with 2 R; where Tn is de ned in Se tion 3.5. Hint: Observe that V = ( os )Ten + (sin )Uen [

2

2

1 1℄

is even if and only if V = Ten and use Theorem 7.1.1.

℄ Prove Corollary 7.1.4.

ut

E.3 Proof of Corollaries 7.1.5 and 7.1.6. a℄ Prove Corollary 7.1.5.

ei + 1 Hint: Use Theorem 7.1.1 and the substitution x = i i ; whi h maps K e 1 onto R [ f1g: ut b℄ Prove Corollary 7.1.6.

E.4 Proof of Corollary 7.1.10. Prove Corollary 7.1.10. z+1 Hint: Use Theorem 7.1.7 and the substitution x = i ; whi h maps D z 1 onto R [ f1g : ut E.5 Completion of the Proof of Theorem 7.1.11. a℄ Show that if p 2 Pn has all its zeros on the unit ir le, then there is a 0 6= 2 C su h that g () := e in p(e i ) is a real trigonometri polynomial of degree at most n: 2

334 7. Inequalities for Rational Fun tion Spa es

For f = p=q with n Y

q(z ) := and

p(z ) :=

m Y k=1

(z

k=1

ak ) ;

ak 2 C

bk 2 C ;

2C;

(z

bk ) ;

m  n;

we de ne f  := p =q; where

p (z ) :=

m Y k=1

zbk ) :

(1

Let D := fz 2 C : jz j < 1g: b℄ Show that jf  (z )j = jf (z )j for every z 2 D: In ea h of the remaining parts of the exer ise suppose that jak j > 1 for ea h k:

℄ Show that if  2 D and jbk j  1 for ea h k; then f + f  has all its zeros on the unit ir le. d℄ Show that if jbk j  1 for ea h k; then

jf 0 (z )j  jf 0 (z )j ;

z 2 D :

Hint: First observe that it is suÆ ient to study the ase z = 1: We have  0  Re f (1) f (1)

=

 =

Re Re

n 2

=n = Re = Re

n X k=1 n X k=1

Re Re n X

1 1

1 1

!

bk

1

k=1 n X k=1

k=1 1 n X

1

1 1

bk

bk

+

!

!

Re !

1 1

Re

! ak

2

n X

Re

n X

k=1 n X k=1

1 1

1 1

1

1 a k=1 ! k

n X k=1

k=1

n X

Re

Re

bk

1

n

1

1

k =1 n n + 2 2 !

!

1

1 bk 0 (1)  f = Re f  (1) k=1 

Re

bk

n X

n X

1 1

1

ak

1

ak

!

ak

!

! ak !

7.1 Inequalities for Rational Fun tion Spa es 335

and  0  Im f (1) f (1)

! ! n n X X 1 1 = Im Im 1 b 1 a k k k =1 k =1 ! ! n n X X 1 1 Im = Im n 1 b 1 a k k k =1 k =1 ! ! n n X X 1 1 = Im + Im 1 1 a k 1 b k =1 k =1 k  0  f (1) = Im : f  (1)

The result now follows from the ombination of part b℄ and the above two inequalities. ut e℄ Prove that if jbk j  1 for ea h k; then 2jf 0 (z )j  jf 0 (z )j + jf 0 (z )j  Bn (z )kf kD ;

z 2 D ;

where Bn (z ) is the Bernstein fa tor de ned in Theorem 7.1.11. Hint: Use parts ℄ and d℄ and the already proved part of Theorem 7.1.11 (when f has all its zeros on the unit ir le). ut f℄ Show that if jbk j  1 for ea h k; then 

jf 0 (z )j  21 Bn (z ) zmax jf (z )j 2D



min jf (z )j ;

z2D

z 2 D ;

where Bn (z ) is the Bernstein fa tor de ned in Theorem 7.1.7. This extends a result of Aziz and Dawood [88℄. Hint: Assume that kf kD = 1: Let m := minz2D jf (z )j: Let  be a onstant of modulus less than 1: Let g (z ) := f (z ) m: Observe that the argument of  an be hosen so that

jg0 (z )j = jf 0 (z )j jjmBn (z ) : By Rou he's theorem, g has no zeros in D: So parts d℄ and e℄ imply that 2 jf 0 (z )j

2 jj mBn (z ) = 2 jg 0 (z )j  jg 0 (z )j + jg 0 (z )j = jf 0 (z )j + jf 0 (z )j jjmBn (z )  Bn (z ) jjmBn (z ) :

Sin e jj an be hosen arbitrarily lose to 1, the result follows.

ut

336 7. Inequalities for Rational Fun tion Spa es

E.6 Extensions of Russak's Inequalities. a℄ Given (ak )kn  C n R, show that 2

=1

kf 0kL

1(

 2n kf kK

K)

2 Tn (a ; a ; : : : ; a n ; K ) and kf 0kL K  4n kf kK for every f 2 Tn (a ; a ; : : : ; a n ; K ): b℄ Given (ak )nk  C n R, show that kf 0 kL R  n kf kR for every f 2 Pn (a ; a ; : : : ; an ; R; ) and kf 0kL R  2n kf kR for every f 2 Pn (a ; a ; : : : ; an ; R). for every f

1

2

2

1(

1

2

)

2

=1

1(

1

)

2

1(

1

)

2

Hint: Use Corollaries 7.1.2, 7.1.8, 7.1.6, and 7.1.10. Write the Bernstein

fa tors in a form so that the integral (of ea h term in the maximum if the Bernstein fa tor is de ned by a maximum) an be evaluated by the residue theorem (in part a℄) and by nding the antiderivative (in part b℄). ut

℄ Are any of the inequalities of parts a℄ and b℄ sharp? If so, in whi h

ases? E.7 Markov-Type Inequality. Given (ak )nk

=1

kf 0 k for every f by

[

;

1 1℄

n

n

n X

1 k

2 Pn (a ; a ; : : : ; an ; [ 1

k := ak

q

2

ak 2

1;

=1

1 + j k j 1 j k j

R n[ !2

kf k

[

1; 1℄; show that ;

1 1℄

1; 1℄); where the numbers k are de ned

ak = ( k + k ) ; 1

1 2

j k j < 1 :

Pro eed as follows: a℄ Given (ak )nk  R n [ 1; 1℄; let =1

8 > <

2ak 1 y + 1+y 1+y ak (y) := > : 2ak + 1 + y 1 y 1 y and let k (y ) be de ned by

if

0y1

if

1y0

7.1 Inequalities for Rational Fun tion Spa es 337

j k (y)j < 1 :

ak (y) =: ( k (y) + k (y) ) ; 1

1

2

Show that

n

X 1 + k (y ) 2 1 + jy j k 1 k (y ) for every f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄). Hint: Show by a variational method that

jf 0 (y)j 

!2

kf k

[

;

1 1℄

=1

1

2

max f

jf 0 (1)j kf k ; [

= jTn0 (1)j ;

1 1℄

where the maximum is taken for all 0 6= f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄); and Tn is the Chebyshev polynomial for Pn (a ; a ; : : : ; an ; [ 1; 1℄) de ned in Se tion 3.5. Now the result follows from E.1 ℄ of Se tion 3.5 by a linear shift from [ 1; 1℄ to [ 1; y ℄ if 0  y  1; or to [y; 1℄ if 1  y  0: ut b℄ Given (ak )nk  R n [ 1; 1℄; show that 1

1

2

2

=1

jf 0 (y)j  1 njyj kf k

;

[

1 1℄

y 2 ( 1; 1)

;

for every f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄): Hint: When y = 0; this follows from Corollary 7.1.3. When y 2 ( 1; 1) is arbitrary, use a linear shift from [ 1; 1℄ to [2y 1; 1℄ if 0  y  1; or to [ 1; 2y + 1℄ if 1 < y  0: ut

℄ Prove the Markov-type inequality of the exer ise. Hint: Combine parts a℄ and b℄. Note that 1

jak (y)j  jak j holds for every y 2 [

2

and

j k (y)j  j k j < 1 ;

ut

1; 1℄:

E.8 S hur-Type Inequality. Given fak gnk

=1

kf k

[

;

1 1℄

k = 1; 2; : : : ; n

 maxfjUn (1)j; jUn (

R n[

p

1)jg
f (x) 1

1; 1℄; show that

x

2



[

;

1 1℄

for every f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄); where Un is the Chebyshev polynomial (of the se ond kind) for Pn (a ; a ; : : : ; an ; [ 1; 1℄) de ned in Se tion 3.5, and n p X ak 1 jUn (1)j = a  1 k k 1

2

1

2

2

p

=1



with the hoi e of ak 1 determined by ak equality holds if and only if f = Un ; 2 R: 2

p

ak 2



1 < 1: Show that

338 7. Inequalities for Rational Fun tion Spa es

Hint: First show that if fe 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄) is extremal for jf (1)j p max

; f f (x) 1 x ; 1

2

2

[

1 1℄

where the maximum is taken for all 0 6= f 2 Pn (a ; a ; : : : ; an ; [ 1; 1℄); then f = Un with some 2 R: Observe that Un (1) an be evaluated by L'Hospital's rule sin e 1

2

Tn (x) ; 1 x hen e jUn (1)j = jTn0 (1)j = jBn (1)j
jUn (1)j with the notation of Se tion 3.5. Thus jUn (1)j = jBn (1)j : If y 2 [ 1; 1℄ is arbitrary, then use a linear shift from [ 1; 1℄ to [ y; y ℄ Un (x) = 2

1

2

2

2

ut

(some aution must be exer ised about the hange of poles).

E.9 Extension of Lax's Inequality on the Half-Plane. Asso iated with (ak )nk  H := fz 2 C : Im(z ) > 0g; let the Bernstein fa tor Bn be, as in Corollary 7.1.10, de ned by +

=1

Bn (x) := Show that

jh0 (x)j 

for every h 2 P

n (a1 ; a2 ; : : :

1 2

n X k=1

2 Im(ak )

jx

ak j

2

Bn (x)khkR ;

: x2R

; an; R) having all its zeros in H : Equality holds for h = (Sen + 1) with 2 C ; where Sen is the Blas hke produ t asso iated with (ak )nk : Note that Bn (z ) = jSen0 (z )j: Note also that +

=1

h=

(Se + 1)

2 Pn (a ; a ; : : : ; an ; R) ; 1

2

has all its zeros on R:

Hint: Use Theorem 7.1.11 with the substitution x = i

6= 0

z+1 : z 1

ut

E.10 Remarks on Theorem 7.1.7 and Corollary 7.1.10. a℄ Given (ak )nk  D and z 2 D; show that equality holds in the inequality of Theorem 7.1.7 if and only if f = Sn with 2 C ; where Sn is the Blas hke produ t asso iated with (ak )nk : Hint: Analyze the proof of Theorem 7.1.7. ut b℄ Given (ak )nk  H = fz 2 C : Im(z ) > 0g and x 2 R; show that equality holds in the inequality of Corollary 7.1.10 if and only if f = Qn with 2 C ; where Qn is the Blas hke produ t asso iated with (ak )nk : Note that the only if parts of E.10 a℄ and E.10 b℄ above are not laimed in the general ase of Theorem 7.1.7 and Corollary 7.1.10 (why?). =1

=1

+

=1

=1

7.1 Inequalities for Rational Fun tion Spa es 339

E.11 Markov-Bernstein-Type Inequality for SR n and ASR n . Let SR n denote the set of all self-re ipro al polynomials p 2 Pn satisfying

p(z )  z np(z ) : 1

Let SRn denote the set of all real self-re ipro al polynomials of degree at most n; that is, SRn := SR n \ Pn : For a polynomial p 2 Pn of the form (7:1:16)

p(z ) =

n X j =0

j z j ;

j

2C ;

p 2 SR n if and only if

j = n j ;

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

Let ASR n denote the set of all antiself-re ipro al polynomials p 2 Pn satisfying p(z )  z np(z ) : 1

Let ASRn denote the set of all real antiself-re ipro al polynomials of degree at most n; that is, ASRn := ASR n \ Pn : Let ASRn := ASR n \ Pn . For a polynomial p 2 Pn of the form (7.1.16) p 2 ASR n if and only if

j = n j ;

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

a℄ There exists an absolute onstant su h that

jp0 (x)j  n min



(1 + log n); log



e

1



x

2

kpk

[

;

1 1℄

holds for every x 2 [ 1; 1℄ and for every p 2 Pn satisfying (7:1:17)

jp(x)j  (1 + jxjn )kpk

[

;

1 1℄

x2R;

;

in parti ular, for every p 2 SR n and for every p 2 ASR n : The inequality

jp0 (x)j  n(1 + log n) kpk

[

;

1 1℄

for all p 2 SRn and for all p 2 ASRn was rst obtained by Kroo and Szabados [94a℄. They also showed that up to the onstant > 0 the above inequality is sharp for both SRn and ASRn : Here we present a distin t proof. The sharpness is studied in E.12 f℄.

340 7. Inequalities for Rational Fun tion Spa es

Outline. Suppose p 2 Pn satis es (7.1.17). Then p(x) (7:1:18) f (x) := 1+x n 2

satis es

kf kR  2 kpk

(7:1:19)

;

[

1 1℄

:

Let

ak := ei

(7:1:20)

(2

k

=(2n) ;

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

1)

be the zeros of the equation z n + 1 = 0: Now Corollary 7.1.10, together with (7.1.18) to (7.1.20), yields that 2

(7:1:21)

n X

jf 0 (x)j  2

jx

k=1 n X

4

!

Im(ak )

ak j

2

!

Im(ak )

jx

k=1

ak j

kf kR kpk

;

[

2

1 1℄

x 2 R:

;

Show that if n 2 N and x 2 [ 1; 1℄; then (7:1:22)

n X

4

Im(ak )

jx

k=1

 n min

ak j

2



(1 + log n); log



e

1



x

2

;

where here the  symbol means that there are absolute onstants > 0 and > 0 (independent of n 2 N and x 2 [ 1; 1℄) su h that the left-hand side is between times the right-hand side and times the right-hand side for every n 2 N and x 2 [ 1; 1℄: Combining (7.1.18), (7.1.21), and (7.1.22), we on lude that there is an absolute onstant su h that 1

2

1

0 p (x) 1 + x2n

2

2nx p(x) 1 + x2n 1 + x2n 

n

2

 n min 2

So if x 2 [ 1; 1℄; then 

2

1

(1 + log n); log 



1

jp0 (x)j  2 n min (1 + log n); log 2



e

x



1

e

kpk

;

[

2



x

2

1 1℄



+ 2n

b℄ There exists an absolute onstant > 0 su h that

for every p 2 P

: 2n



R

kpk

[

;

1 1℄

ut

and the proof is nished.

0

p (x)

1 + x2n

x 2 R:

;



 n(1 + log n)

1 p+(xx) n

2

R

7.1 Inequalities for Rational Fun tion Spa es 341

Proof. Asso iated with p 2 P n ; let 2

p(x) : 1+x n

f (x) := Let

ak := ei

k

2

=(2n) ;

k = 1; 2; : : : ; 2n be the zeros of the equation z + 1 = 0: Using Corollary 7.1.10, we an dedu e that there are absolute onstants and su h that (2

1)

2

n

1

n X

kf 0kR 

k=1

2 jIm(ak )j

! 1

2

n X

kf kR 

1

 n(1 + log n)kf kR :

 

k n

k=1

1

!

kf kR

2

Therefore

0 p (x) 1 + x2n





p(x) 2nx n p(x)



n(1 + log n)

n n 1+x 1+x 1 + x n R 2

1

2

2

2

2

for every x 2 R; whi h implies

0

p (x)

1 + x2n

R





p(x)

( 2 n(1 + log n) + 2n) 1 + x2n

R

;

ut

and the result follows.

℄ For every m 2 N ; there exists a onstant (m) depending only on m so that kp m k ;  (m)(n(1 + log n))m kpk ; (

)

[

1 1℄

[

1 1℄

for every p 2 P satisfying

n

jp(x)j  (1 + jxjn )kpk

(7:1:23)

[

;

1 1℄

:

Proof. Using part b℄ and indu tion on m, we see that there exists a onstant

(m) depending only on m su h that 1

(m)

p (x)

1 + x2n

R



1

p(x)

1 + x n R

(m)(n(1 + log n))m

2

for every p 2 P n : Note that if p 2 Pn satis es (7.1.23), then 2



p(x)

1 + x2n

and the result follows.

R





p(x)

2 1+ x n

jj

R

 2kpk

[

;

1 1℄

;

ut

342 7. Inequalities for Rational Fun tion Spa es

E.12 Quasi-Chebyshev Polynomials for SR n and ASR n . Let 2

ak := ei

(2

k

1)

=(2n) ;

2

k = 1; 2; : : : n

be the zeros of the equation z n + 1 = 0 in the upper half-plane. Let 2

M n (z ) := 2

n Y k=1

(z

2

P n (x) :=Re(M n (x)) ;

x 2 R;

Q n(x) :=Im(M n (x)) ;

x 2 R:

2

and

z2C;

ak ) ;

2

2

2

a℄ Show that if n is even, then P n 2 SR n and Q n is odd, then Q n 2 SR n and P n 2 ASR n : b℄ Show that jM n (x)j = 1 + x n ; x 2 R 2

2

2

2

2

2

2

2

x 2 R;

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

2 ASR n ; while if n

2

2

and

2

2

2

2

2

in parti ular

kP n k 2

[

;

1 1℄

2

and

kQ n k 2

[

;

1 1℄

 2:

Proof. Note that n M n (x) Y x ak = ; 1 + x n k x ak

(7:1:24)

2

2

=1

whi h implies the rst equality. The rest is straightforward from the de nitions. ut

℄ Show that there are extended real numbers

1=z

0

su h that

> z >


> z n = 1

M n (zj ) = ( 1)j ; 1 + zj n 2

2

2

1

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

(the value of the left-hand side at 1 is de ned by taking the limit when x ! 1). Hint: Use (7.1.24) and the argument prin iple (see, for example, Ash [71℄).

ut

7.1 Inequalities for Rational Fun tion Spa es 343

d℄ Let n 2 N be even. Show that there exist 1 = x > y > x > y >


> xn 0

1

1

2

1

> yn > xn = 1

su h that

P n (xj ) = ( 1)j 1 + xj n 2

n=2 ;

+

2

Q n (xj ) = 0 ; j = 0; 1; : : : ; n 2

and

Q n (yj ) = ( 1)j 1 + yj n 2

n=2 ;

+1+

2

P n (yj ) = 0 ; j = 1; 2; : : : n : 2

Formulate the analog statement when n 2 N is odd. e℄ Show that there exists an absolute onstant > 0 su h that

jP 0n (x)j + jQ0 n (x)j  n min 2

2



(1 + log n); log



1

e



x

2

for every n 2 N and x 2 [ 1; 1℄:

Proof. If x 2 [ 1; 1℄; then



1 + x n jP 0n (x)j + jQ0 n (x)j  jM 0 n (x)j = jM 0 n (x)j M n (x) n 0 n X M n (x) X 2  M (x) = x a  2 jxIm(aak )j n k k k k     n min (1 + log n); log 1 e x 2

2

2

2

2

2

2

2

2

=1

=1

2

with an absolute onstant > 0; where the last inequality follows from (7.1.22). ut The next part shows the sharpness of the inequality of E.11 a℄. f℄ Let > 0 be the same absolute onstant as in part e℄. For the sake of brevity let 



Æn (x) := n min (1 + log n); log



1

e



x

Show that for every interval

In;x := [x; x + 8Æn (x)℄  [0; 1℄ ;

2

1

:

344 7. Inequalities for Rational Fun tion Spa es

2 In;x and y 2 In;x su h that

there exist y

1

2

jP 0n (y )j 

1

1

2

2

Æn (x)

Proof. Suppose that

jQ0 n (y )j 

and

1

jP 0n (y)j <

1

jQ0 n (y)j >

1

2

2

2

2

Æn (x)

1

Æn (x)

1

1 2

Æn (x)

1

:

for every y 2 In;x : Then, from part e℄ we an dedu e that 2

2

for every y 2 In;x : Therefore

kQ n k 2

[

;

1 1℄



1 2

R

In;x

jQ0 n (y)j dy > 2

1 2

8Æn (x) Æn (x) 1

1

2

= 2;

whi h ontradi ts the last inequality of part b℄. This nishes the proof of the rst inequality. The se ond inequality an be proven in the same way.

ut

g℄ Show that

p0 (1) = n p(1) 1

2

for every p 2 SR n : By using the quasi Chebyshev polynomials for SR n and ASR n , it an be shown that the inequality of E.11 ℄ is essentially sharp for the lasses SRn and ASR n for every m. This has been pointed out to us by Szabados. The argument requires some more te hni al details than the proof in the m = 1 ase dis ussed in the above exer ise. 2

2

2

7.2 Inequalities for Logarithmi Derivatives We derive a series of metri inequalities of the form

m



x2R:

r0 (x) r(x)





 n ; 

 > 0;

where r is a rational fun tion of type (n; n) and is a onstant independent of n. Here m is the Lebesgue measure, although, sin e the sets in question are usually just nite unions of intervals, this is mostly a notational onvenien e. One of the interesting features of these inequalities is their easy extension from the polynomial ase to the rational ase. The basi inequality is due to Loomis [46℄. Note the invarian e of the measure of the set in this ase.

7.2 Inequalities for Logarithmi Derivatives 345

Theorem 7.2.1. If p 2 Pn has n real roots, then   p0 (x) n m x2R:   = ; p(x) 

 > 0:

Proof. By onsidering  p instead of p; it is suÆ ient to prove the theorem for  = 1: We rst onsider the ase where p has distin t roots, whi h are denoted by  <  <


< n : Then 1

1

2

n 1 p0 (x) X = : p(x) k x i Let < <


< n be the roots of p p0 ; whi h must all be real. Note that these are the points where p0 (x)=p(x) = 1: It is now easy to see from =1

1

2

the graph that 

p0 (x) x2R: p(x)

and

m



1

x2R:



= [ ; ℄ [ [ ; ℄ [


[ [n ; n ℄ 1

p0 (x) p(x)

However, if p(x) := xn + an xn n X i=1

while

n X i=1

2



1 1

1

1

=

2

n X

n X

i

i=1

+


+ a ; then

i=1

i :

0

i = an

i = (an

1

1

; n)

is 1 times the se ond oeÆ ient of p p0 : This gives the result for distin t roots. The ase when some of the roots of p are repeated an be handled by an easy limiting argument. ut Corollary 7.2.2. If i

then

(

m

2

x2R:

R;

i > 0; i = 1; 2; : : : ; n; and

n X i=1

x

i

i



)!

=

1



;

Pn

i=1 i

= 1;

 > 0:

Proof. For i rational this follows immediately from Theorem 7.2.1 on learing the denominators of i by multiplying by an integer. The real ase is an obvious limiting argument. ut In order to extend Theorem 7.2.1 to arbitrary polynomials we need the following generalization of E.3 of Se tion 2.4 due to Videnskii [51℄. The proof is indi ated in the exer ises.

346 7. Inequalities for Rational Fun tion Spa es

Lemma 7.2.3. If p 2 Pn is positive on [a; b℄; then there exists q; s nonnegative on [a; b℄ with all real roots (in [a; b℄) so that

2 Pn

p(x) = q(x) + s(x) : We now prove the unrestri ted ase of Theorem 7.2.1. Theorem 7.2.4. Let p 2 Pn : Then

m



p0 (x) x2R: p(x)





 2n ;

 > 0:

Proof. Let  > 0 and let pn 2 Pn : Choose a and b su h that   p0 (x) x2R:    [a; b℄ : p(x) By Lemma 7.2.3 we an nd polynomials q 2 P n and s 2 P 2

n

2

su h that

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

where, for x 2 [a; b℄;

0  q (x)  p (x)

0  s(x)  p (x)

and

2

2

and both q and s have only real roots. Now 

p0 (x) x2R: p(x)

Also,





0   = x 2 R : (pp )(x(x) ) 2

2

 2



:

(p )0 (x)  2 p (x) 2

holds exa tly when

2

q0 (x) + s0 (x)  2(q(x) + s(x)) :

By Theorem 7.2.1 

  q0 (x) s0 (x) m x2R:  2 =m x2R : q(x) s(x) Sin e q and s are nonnegative on [a; b℄, it follows that

 2



=

2n m(fx 2 [a; b℄ : q0 (x) + s0 (x)  2(q(x) + s(x))g)  ; 

and the proof is nished.

n : 

ut

It an be shown that this inequality is asymptoti ally sharp to the extent that the onstant 2 annot be repla ed by any smaller onstant for large n; see Kristiansen [82℄. Theorem 7.2.4 extends easily to rational fun tions.

7.2 Inequalities for Logarithmi Derivatives 347

Theorem 7.2.5. If r = p=q, where p; q 2 Pn ; then

m



r0 (x) x2R: r(x)

Proof. We have





r0 p0 = r p

 8n ;

 > 0:

q0 q

and for  > 0; 

x2R:

r0 (x) r(x)



 



p0 (x) p(x)

x2R:



 2 [



x2R:

q0 (x) q(x)



 2



:

By Theorem 7.2.4

m



p0 (x) x2R: p(x)



2



 4n ;

and with s(x) := q ( x);

m



q0 (x) x2R: q(x)

It follows that

m







2

=m





r0 (x) x2R: r(x)

s0 (x) x2R: s(x)





 8n

for every  > 0:



 2



 4n :

ut

This inequality probably does not have the exa t onstant. It an be shown (see Borwein, Rakhmanov, and Sa [to appear℄) that the onstant 8

annot be repla ed by any onstant less than or equal to 2: Comments, Exer ises, and Examples. Many variants on the inequalities of this se tion are presented in E.2, E.3, E.4, and E.5. Some of these are in Borwein [82℄. E.5 explores some metri properties of the lemnis ate

E (p) := fz 2 C : jp(z )j  1g of a moni polynomial p 2 Pn of the form

p(z ) =

n Y i=1

(z

zi ) ;

zi 2 C :

348 7. Inequalities for Rational Fun tion Spa es

The reader is referred to Erd}os, Herzog, and Piranian [58℄ for many results and open problems on erning the lemnis ate of moni polynomials. One in parti ular, whi h is still open, onje tures that for moni polynomials p 2 Pn ; the length of the boundary of E (p) is maximal for p(z ) := z n 1 and so is O(n): Pommerenke [61℄ has shown that this length is O(n ): Borwein [95℄ improves this to O(n); see E.7. E.9 solves another problem of Erd}os, namely, the diameter of E (p) for a moni polynomial p 2 Pn is always at least 2. Erd}os [76℄ ontains several other related open problems. 2

E.1 Polynomials as Sums of Polynomials with Real Roots. a℄ Suppose p 2 P n n P n ; and suppose that p > 0 on [a; b℄: Then 2

2

1

p(x) = (x

a)(b x)u (x) + v (x) 2

2

for some u 2 Pn and v 2 Pn ; whi h have all their zeros in [a; b℄: b℄ Suppose p 2 P n n P n and suppose that p > 0 on [a; b℄: Then 1

2

+1

2

p(x) = (b x)u (x) + (x a)v (x) 2

2

for some u; v 2 Pn ; whi h have all their zeros in [a; b℄: Hint: Let p be a polynomial of degree 2n that is stri tly positive on [a; b℄: Let Tn be the Chebyshev polynomial for the Chebyshev system (

1 p ; p(x)

x p ; ::: ; p(x)

xn p p(x)

)

:

p

Then Tn (x) = v (x)= p(x) with some v 2 Pn : Show that, for part a℄, v and u de ned by (x a)(b x)u (x) = p(x) v (x) 2

2

are the required polynomials. Use a similar onstru tion for part b℄.

ut

E.2 Various Spe ializations. For the next exer ises we use the notation Pn to denote the polynomial of degree at most n with nonnegative oeÆ ients, and Pn" to denote those elements of Pn that are nonde reasing on [0; 1): a℄ If r = p=q; where p; q 2 Pn and both p and q have only real roots, then +

m



x2R:

r0 (x) r(x)





 4n ;

 > 0:

7.2 Inequalities for Logarithmi Derivatives 349

b℄ Let r(x) := xn =(4n

x)n : Then  r0 (x) m x2R:  1 = 4n : r(x)

℄ If r = p=q; where p 2 Pn and q 2 Pn" ; then   2n r0 (x)    ;  > 0: m x0: r(x)  d℄ If r = p=q; where p 2 Pn and q 2 Pn" ; then   r0 (x) n m x0:    ;  > 0: r(x)  

+

e℄ Let r(x) := xn . Then

m



x0:

r0 (x) r(x)





=

n ; 

 > 0:

E.3 Another Metri Inequality. If p 2 Pn has n real roots lying in the interval (a; b); then 0   p (x)  2n  = ;  > 0: m x2R: p(x) j(x a)(b x)j 

Outline. Prove that m

and



x2R:0

(x

a)(b x)p0 (x) p(x)





 n

=



 : n First onsider the ase when p has distin t zeros. Let y < y <


< yn denote the n + 1 roots of (x a)(b x)p0 (x), and let x < x <


< xn denote the n roots of p(x): Then y < x  y <


xn < yn < xn := 1 : m

(x

a)(b x)p0 (x) p(x)

x2R :0





=

0

1

0

0

Sin e

0

1

1

1

1

p0 (x) = 1; x!xi p(x) we an dedu e that for ea h interval (yi ; xi ) there exists a point Æi 2 (yi ; xi ) lim (x

a)(b x)

su h that

(Æi a)(b Æi )p0 (Æi ) = p(Æi ) : Sin e the above equation an have at most n + 1 solutions, we have

m



x2R:0

(x

a)(b x)p0 (x) p(x)



Now pro eed as in the proof of Theorem 7.2.1.





=

n X i=0

(Æi

yi ) :

ut

350 7. Inequalities for Rational Fun tion Spa es

E.4 Extensions to Pn . a℄ If p 2 Pn ; then

m



x2

0 p (x) R: p(x)







p

8 2n



;

 > 0:

Hint: Write p as the sum of its real and imaginary parts. b℄ If r = p=q with p; q 2 Pn ; then m



x2R:

0 r (x) r(x)







p

32 2 n



;

ut

 > 0:

E.5 On the Lemnis ate E (p). Let

p(z ) := and let

n Y i=1

(z

zi ) ;

zi 2 C

E (p) := fz 2 C : jp(z )j  1g :

a℄ Show that

m(E (p) \ R))  4
2

=n

1

with equality only for the Chebyshev polynomial of degree n normalized to have lead oeÆ ient 1; see Polya [28℄. Hint: Analogously to the proof of the Remez inequality of Se tion 5.1, show that the Chebyshev polynomial transformed to an interval of length 4 is extremal for this problem. ut b℄ Let m (
) denote the planar Lebesgue measure. Show that 2

m (E (p))  4 : 2

(In fa t, m (E (p))  ; whi h is exa t for z n ; this is due to Polya [28℄.) 2

E.6 Cartan's Lemma. Let

p(z ) :=

n Y j =1

(z

zj ) ;

zj

2C

and > 0 be xed. Then there exist at most n open disks, the sum of whose radii is at most 2 ; so that if z 2 C is outside the union of these open disks then  n jp(z )j > e :

7.2 Inequalities for Logarithmi Derivatives 351

Prove Cartan's Lemma as follows: a℄ Let  ;  ; : : : ;  be xed omplex numbers. Let > 0: Show that there exists a positive integer  less than or equal to  for whi h there exists an open disk with radius  =n ontaining j for exa tly  distin t values of j = 1; 2; : : : ; : Hint: Suppose to the ontrary that there is no su h positive integer . Show that this would imply the existen e of an open disk with radius  =n

ontaining j for at least  + 1 distin t values of j = 1; 2; : : : ; ; whi h is a

ontradi tion. ut b℄ Show that there exist open disks D ; D ; : : : ; Dk and positive integers m ; m ; : : : ; mk with the following properties: Pk (1) j mj = n ; m  m 


 mk ; 1

2

1

1

2

2

1

=1

2

m

(2) Dj has radius j ; n (3) Dj \ Ej ontains exa tly mj zeros of p, where

Ej := C n (D

1

[ D [ : : : [ Dj 2

1

);

(4) for every integer m > mj ; no open disk of radius m n ontains exa tly m zeros of p in Ej : Hint: Use part a℄. ut

℄ Let D ; D ; : : : ; Dk be the open disks spe i ed in part b℄. For j = 1; 2; : : : ; k; let Dj be the disk with the same enter as Dj and with twi e its radius. Show that for every 1

2

z 2 E  := C n (D [ D [


[ Dk ) 1

2

there is a permutation ze ; ze ; : : : ; zen of the zeros z ; z ; : : : ; zn su h that 1

jz

2

zej j 

1

j ; n

2

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

Hint: Let z 2 E  be xed. Show, by indu tion on i; that for every i = 0; 1; : : : ; n 1; there are at least i + 1 zeros of p outside the open disk with

enter z and radius n ni : Distinguish the ases (1) n i > m ; (2) mj  n i > mj ; j = 1; 2; : : : ; k 1 ; (3) mk  n i > 0 : ut (

)

1

+1

d℄ Finish the proof of Cartan's lemma.

352 7. Inequalities for Rational Fun tion Spa es

E.7 The Length of the Boundary of E (p). Let

p(z ) :=

n Y j =1

(z

zj ) ;

zj

2C

and let

E := E (p) := fz 2 C : jp(z )j  1g : Show that the boundary E of E is of length at most 4en: (In fa t, with E.10 a℄, the estimate an be improved to (5:2) n:) Outline. Pro eed as follows: a℄ Let L be an arbitrary line in the omplex plane. Show that the set E \ L ontains at most 2n distin t points. Hint: By performing a translation and a rotation, if ne essary, we may assume that L = R: Now observe that there is a polynomial P (x; y ) of degree at most 2n; in two real variables x and y; with omplex oeÆ ients, su h that

E = fz 2 C : jp(z )j = 1g = fz 2 C : p(z )p(z ) = 1g = fz = x + iy : x; y 2 R ; P (x; y ) = 1g : 2

Hen e

E \ R = fx 2 R : P (x; 0) = 1g

and sin e P (x; 0) 2 P the result follows. b℄ For a 2 C and r > 0; let Q be the square

ut

; 2n

Q := fz 2 C : jRe(z

a)j < r ; jIm(z

a)j < rg :

Show that Q \ E is of length at most 8rn: Hint: Divide Q \ E into sub urves C ; C ; : : : ; Cm so that every verti al and horizontal line ontains at most one point of ea h Cj . Let lxj and lyj denote the length of the interval obtained by proje ting Cj to the x axes and y axes, respe tively. Let 1

2

( )

lx :=

m X j =1

lxj

( )

Use part a℄ to show that lx length of Q \ En ; then

and

ly :=

m X j =1

( )

lyj : ( )

 4nr and ly  4rn. Hen e, if l denotes the l  lx + ly  8rn :

ut

7.2 Inequalities for Logarithmi Derivatives 353

℄ Show that the boundary E of E is of length at most 16en: Hint: Combine Cartan's lemma (see E.7) and part b℄. ut The rest of the exer ise is about improving 16en to 4en: d℄ Let B be the open disk with enter a 2 C and radius r > 0. Show that B \ E is of length at most 2rn: Outline. Let Q be the square

Q := fz 2 C : jRe(z )j < r ; jIm(z )j < rg : For  2 [0; 2 ) let Q be the square Q := fa + ei (z a) : z 2 Qg : Let l() denote the length of Q \ E: Let lx () and ly () denote the total length of the intervals obtained by proje ting Q \ E to the lines   z = rei : r 2 R and z = rei  = : r 2 R ; respe tively ( ounting multipli ities). The pre ise de nition of lx () and ly () an be formulated in the same way as in the hint to part b℄, whi h is (

left to the reader. Use part a℄ to show that

lx() + ly ()  8rn ;

+

2)

 2 [0; 2) :

Hen e

Z  1 l(0) = l(0)(j sin j + j os j) d  2 Z  1 = (lx () + ly ()) d  8rn ; 2 and l(0)  2rn follows. e℄ Prove the initial statement of the exer ise. Hint: Combine Cartan's lemma (see E.6) and part d℄.

4

2

0

2

0

ut ut

E.8 On the Length of Another Lemnis ate. Suppose p 2 Pn : Show that the length of the lemnis ate 0   p (z ) F = F (p) := z 2 C : =n p(z ) is at most 16n(1 + log n) (a tually at most 4n(1 + log n)). Hint: The arguments are very similar to those given in the outline to E.7. First show that if L is an arbitrary line in the omplex plane, then F \ L

ontains at most 2n distin t points. Next prove that if Dr is an open disk of radius r in the omplex plane, then Dr \ F is of length at most 8rn (a tually at most 2rn). Now use E.6 ℄ with = 1 + log n: ut The proof of the following exer ise requires some familiarity with harmoni fun tions.

354 7. Inequalities for Rational Fun tion Spa es

E.9 On the Diameter of E (p). The diameter diam(A) of a nonempty set A  C is de ned by diam(A) := supfjz

1

z j : z ;z 2

1

2

2 Ag :

Let p 2 Pn be an arbitrary moni polynomial of degree n; that is,

p(z ) :=

n Y j =1

(z

zj ) ;

zj

2C:

Then the diameter of the set

E (p) := fz 2 C : jp(z )j  1g is at least 2. Pro eed as follows: a℄ Let Cb := C [ f1g


:= fz 2 Cb : jz j > 1g :

and

Let p 2 Pn be moni . Let E := o(E (p)); that is, the onvex hull of E (p): Use the Riemann mapping theorem (see Ahlfors [53℄) to show that there exists a fun tion g of the form

g(z ) = bz +

1 X j =0

bj z

j;

z 2
; b; bj 2 C

su h that g is analyti and one-to-one on
; and

g(
) = Cb n E : Hint: Note that Cb n E is simply onne ted. b℄ Let b be the same as in part a℄. Show that jbj  1:

ut

Outline. Be ause of the de nition of E; for every " > 0 there exists a Æ > 0 su h that

"  log jz np(g(z ))j ;

jz j = 1 + Æ :

Sin e G(z ) := log jz n p(g (z ))j is harmoni on
; we have

G(1) =

1 2 (1 + Æ )

Z 2 0

G((1 + Æ)ei ) d 

"

2 (1 + Æ )

for every " > 0; so G(1)  0: On the other hand, sin e p 2 Pn is a moni polynomial of degree n;

7.2 Inequalities for Logarithmi Derivatives 355

G(1) = log jbjn = n log jbj ; from whi h jbj  1 follows.

ut

℄ Show that diam(E )  2 :

Outline. Assume to the ontrary that diam(E ) < 2. Then there exists a Æ > 0 su h that

jz

1

g( z ))j < 2 ;

(g (z )

jz j = 1 + Æ :

Sin e

F (z ) := z (g(z )

g( z )) = 2b + 2b z

1

2

1

+ 2b z 3

4

+




is analyti on
, the maximum prin iple (E.1 d℄ of Se tion 1.2) yields 2jbj = jF (1)j  max jF (z )j = max jz (g (z )

jzj

Æ

=1+

1

jzj

Æ

=1+

that is, jbj < 1; whi h ontradi ts part b℄.

g( z ))j < 2 ;

ut

d℄ Note that diam(A) = diam( o(A)) for every nonempty A  C ; in parti ular, diam(E (p)) = diam(E ): The more general result that diam(A)  2 ap(A) for every nonempty A  C is observed in Pommerenke [75℄. E.10 More on E (p). Suppose p is a moni polynomial with omplex oeÆ ients. As before, let

E := E (p) := fz 2 C : jp(z )j  1g : a℄ It follows from E.6 (Cartan's lemma) that the set E (p) an be overed by disks the sum of whose radii is at most 2e: It is onje tured in Erd}os, Herzog, and Piranian [58℄ that the orre t value in this problem is 2. (The

urrent best onstant is less than 2.6.) b℄ IfP E (p) is onne ted, then it is ontained in a disk with radius 2 entered at n nk zk , where z ; z ; : : : ; zn are the zeros of p. Proof. This is onje tured in Erd}os, Herzog, and Piranian [58℄ and proved in Pommerenke [59b℄. ut

℄ If E (p) is onne ted, then its ir umferen e is at least 2: Proof. This is also onje tured in Erd}os, Herzog, and Piranian [58℄, and proved in Pommerenke [59b℄. ut 1

=1

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2

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A1

Algorithms and Computational Con erns

Overview Appendix 1 presents some of the basi algorithms for omputing with polynomials and rational fun tions and dis usses some of the omplexity issues. In luded is a dis ussion of root nding methods. It requires very little ba kground and an essentially be read independently.

Algorithms and Computational Con erns Polynomials lend themselves to omputation perhaps more than any other obje t of analysis. Algorithms that involve spe ial fun tions, dierential equations, series, and the like usually must redu e at some point to a nite polynomial or rational approximation or trun ation. This often allows analyti problems to be redu ed to algebrai ones. This appendix will present, as a series of exer ises, some of the prin ipal algorithmi on erns. The reader is en ouraged to experiment with the algorithms. With urrent te hnology this is most omfortably done in any of the large symboli manipulation pa kages available. Code, a tual or s hemati , is not presented. Indeed, methods rather than algorithms are presented. Current \pra ti al" best methods date qui kly in this rapidly evolving area. It is also the authors' belief that today's theoreti al uriosities may be vital for tomorrow's

Algorithms and Computational Con erns 357

algorithms as larger instan es are al ulated on faster ma hines. Histori ally we have already seen this happen repeatedly with algorithms su h as the fast Fourier transform algorithm. One of the most interesting lessons to be learned from the last few de ades of revitalized interest in omputational mathemati s is that many of the most familiar mathemati al algorithms, su h as how to multiply large numbers, were very poorly understood, and indeed, still are in ompletely analyzed. Very many of the familiar pro esses of mathemati s, su h as multipli ation of large numbers or omputation of determinants, an be omputed far more expeditiously than allowed by the usual \s hool" algorithms. See, for example, Aho, Hop roft, and Ullman [74℄; Bini and Pan [92℄; Borodin and Munro [75℄; Borwein and Borwein [87℄; Brent [74℄; Knuth [81℄; Pan [92℄; Smale [85℄; and Wilf [86℄ for the omplexity side of the following exer ises. E.1 Complexity and Re ursion. We are on erned with measuring the size of an algorithm given an input of a ertain length. Unfortunately, there are many dierent ways of measuring this. (One ould, for example, use the length of the tapes of some well-de ned instantiation of a Turing ma hine.) We will settle for less. The measure of input size will usually be hosen to be a natural one, so for polynomials of degree n; the measure will often be n: The omplexity measure then depends a bit on the problem. For example, it might ount the number of additions of oeÆ ients (we do not distinguish subtra tion from addition) and multipli ations of oeÆ ients required to evaluate the polynomial at a point. (So, for example, by Horner's rule O(n) operations suÆ e.) Care is already required to ount naturally. Note that we have not spe i ed the size of the oeÆ ients (this may or may not be reasonable) and so we ould heat on addition of oeÆ ients by doing two additions as one addition of twi e the length. (Sin e a + b and + d an be de oded from (10m a + ) + (10m b + d); where m is larger than the number of digits in any of a, b, ; or d.) It is more reasonable in this ontext to x a pre ision (or to think of working to in nite pre ision or over some polynomial ring). Our ases are fairly simple, and the measures should be

lear in ontext. We adopt the following notations:

f (n) f (n) = O(g(n)) means lim sup <1 n!1 g (n) and

g(n) < 1: f (n) = (g(n)) means lim sup f n!1 (n)

Parts of these exer ises are reprinted from Borwein and Borwein [87℄, with permission from Wiley.

358 A1. Algorithms and Computational Con erns

So the rst measure gives an upper bound, while the se ond gives a lower bound. Many algorithms are analyzed re ursively. For example, addition of two polynomials of degree at most 2n redu es to two additions of polynomials of degree at most n, plus perhaps an \overhead" for reassembling the pie es. In other words, for the omplexity of addition, we have

A(2n)  2A(n) + ; from whi h one an dedu e that

A(n) = O(n) : This general re ursive strategy of breaking a problem in half is often alled \divide and onquer." We introdu e the following fun tions. Here n is the maximum degree of the polynomials p and q: Additions, multipli ations, and so on are performed in the underlying eld of oeÆ ients (in our ase C or R) and are all performed to some predetermined xed pre ision (possibly in nite).

A(n) := the maximum number of  ;  ;  ; to ompute p  q ; M (n) := the maximum number of  ;  ;  ; to ompute pq ; e(n) := the maximum number of  ;  ;  ; to evaluate p() for an arbitrary xed  2 C ; E (n) := the maximum number of  ;  ;  ; to evaluate p( ); : : : ; p(n ) for arbitrary xed  ; : : : ; n 2 C : 1

1

These are the omplexity fun tions for polynomial addition, polynomial multipli ation, polynomial evaluation at a single point, and polynomial evaluation at n points, respe tively. The input for the omputation is the

oeÆ ients (and the evaluation points for e(n) and E (n)). So the input may be onsidered to be in C n (or more generally an (n + 1)-dimensional ve tor spa e over an in nite eld). In the rst two ases the output is the sequen e of oeÆ ients. In the last two ases, respe tively, the output is the evaluation and the sequen e of evaluations. a℄ Show that usual algorithms give +1

A(n)  2n + 2 = O(n) ; M (n) = O(n ) ; e(n) = O(n) (Horner's rule) ; E (n) = O(n ) : 2

2

E.2 and E.3 of this appendix provide better upper bounds for the last three fun tions above.

Algorithms and Computational Con erns 359

b℄ Show that

A(n)  n + 1 ; M (n)  n + 1 ; e(n)  n + 1 ; E (n)  n + 1 : Hint: In all ases this is a uniqueness argument. At least one operation must be performed for ea h oeÆ ient, otherwise the algorithm will not distinguish various dierent sequen es of input. ut

℄ Some Re ursive Bounds. Let a; b > 0 and ; d > 1: Suppose that f is monotone on (0; 1): If f (n)  af (n= ) + bn; then

f (n) = O(n) if a < ; f (n) = O(n log n) if a = ; f (n) = O(n a ) if a > : log

If f (n)  df (n=d) + bn(log n) ; then 1

f (n) = O(n(log n) ) : Hint: Analyze the equality ase. Then establish the general prin iple that

ut

the equality solution is the maximal solution.

E.2 The (Finite) Fast Fourier Transform (FFT). This is undoubtedly one of the most widely used algorithms. It has, in its various forms, tremendous pra ti al and theoreti al appli ations. Let w be a primitive (n + 1)th root of unity in either C or a nite eld F m ; that is, wn = 1 and wk 6= 1 for k = 1; 2; : : : ; n: In the omplex ase we may take w := e i= n : Consider the following two problems. Interpolation Problem. Given n + 1 numbers,  ;  ; : : : ; n ; nd the oeÆ ients of the unique polynomial +1

2

(

+1)

0

1

p(z ) := a + a z +


+ an z n 0

1

of degree n that satis es

p(wk ) = k ;

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

Evaluation Problem. Given the oeÆ ients of a polynomial pn of degree at most n; al ulate the n + 1 values

p(wk ) ;

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

360 A1. Algorithms and Computational Con erns

These are the two dire tions of the ( nite) Fourier transform. The

lassi al approa hes to either part of the Fourier transform problem have

omplexity at least n : This is the omplexity, for example, of evaluating pn at n + 1 points using Horner's rule. Both dire tions an, however, be solved with omplexity O(n log n). a℄ If n + 1 = 2m ; then both the interpolation and the evaluation problem have omplexity O(n log n): (Here we are ounting the number of additions and multipli ations in the underlying oeÆ ient eld, whi h for most of our purposes is C :) 2

Outline. We treat the evaluation rst. Suppose p(x) := a + a x +


+ an xn : 0

Let and

1

q(x ) := a + a x + a x +


+ an xn 2

0

2

2

4

4

1

1

x r(x ) := x(a + a x +


+ an xn ) : 2

Then, with y := x ;

1

2

3

1

2

p(x) = x r(y) + q(y) ;

where r and q are both polynomials of degree at most 2m 1: The observation that makes the proof work is that for w an (n + 1)th root of unity,
(w k ) = w n = k : 1

2

(

2

+1) 2 +

Hen e, evaluating p(x) at the n + 1 roots of unity redu es to evaluating r and q ea h at the (n + 1) points (w ) ; (w ) ; : : : ; (w ) n = and amalgamating the results. Observe that w is a primitive (2m )th root of unity, so we an iterate this pro ess. Let F (2m ) be the number of additions and multipli ations required to evaluate a polynomial of degree at most 2m 1 at the 2m points wk ; k = 1; 2; : : : ; 2m ; where w is a primitive (2m )th root of unity. Then 1

2 1

2

F (1) = 1

and

2 2

2

2 (

+1) 2

1

F (2m) = 2F (2m ) + 2
2m ; 1

m = 1; 2; : : : :

The se ond term omes from the single addition and multipli ation required to al ulate ea h p(wk ) from r(w k ) and q (w k ). This re ursion solves as 2

2

F (2m ) = 2m m ; +1

and the bound for the evaluation problem is established. The interpolation problem is equivalent to evaluation. This an be seen as follows. Let w be a primitive (n + 1)th root of unity, and let

Algorithms and Computational Con erns 361 0

::: 1 1 : : : wn C C ::: w n C C:

1 1 1 B1 w w B 1 w w W := B B. .. ..  .. . . 1 wn w n 2

2

4

2

..

.

: : : wn

2

Then

.. A .2

0

1

1 1 1 ::: 1 n C B1 w w : : : w C 1 B B1 w w ::: w nC W = B .. .. .. C n + 1  .. .. . . . . . 2A n n 1 w w ::: w n and w is also a primitive (n+1)th root of unity. The interpolation problem

an be formulated as follows. Find (a ; : : : ; an ) so that 1

1

2

2

4

2

2

1

0

W (a ; a ; : : : ; an ) = ( ;  ; : : : ; n ): 0

1

0

1

However, this an be solved by

W ( ;  ; : : : ; n ) = (a ; a ; : : : ; an ) ; 1

0

1

0

1

whi h is exa tly the evaluation problem.

ut

See Borodin and Munro [75℄ and Borwein and Borwein [87℄. Versions of FFT exist in a plethora of shapes and sizes. We have just exposed the tip of the i eberg. As an appli ation we onstru t a fast polynomial multipli ation. b℄ Fast Polynomial Multipli ation. Suppose the polynomials p; q of degree at most n 1 are given. Compute the oeÆ ients of the produ t pq as follows: b1℄ Use an FFT to evaluate p and q at the primitive (2n)th roots of unity

w ; w ; : : : ; w n: 1

2

b2℄ Form

2

p(wk )q(wk ) ;

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

b3℄ Find the oeÆ ients of the produ t pq by solving the interpolation problem on e again by using the FFT. Show that this algorithm requires

O(n log n) additions, multipli ations, and divisions (of omplex numbers) and therefore M (n) = O(n log n): This is the best-known upper bound on the serial omplexity of polynomial multipli ation. The only known lower bound is the trivial one O(n): In parallel (on a PRAM) polynomial multipli ation an be done in O(log n) time on O(N ) pro essors; see Pan [92℄. The same bounds apply for the FFT in a℄.

362 A1. Algorithms and Computational Con erns

E.3 Other Elementary Operations. a℄ Fast Polynomial Division. For polynomials p of degree n and q of degree m  n; it is possible to nd polynomials u and r with deg(r) < deg(q) su h that p(x) = u(x)q(x) + r(x) in O(n log n) additions and multipli ations (of omplex numbers).

Outline. Simplify by observing that it suÆ es to al ulate u sin e r may then be omputed by E.2 b℄. If we repla e x by 1=x then p(x ) r(x ) = u(x ) + q (x ) q(x ) 1

1

1

1

1

and so, for some h  1;

p (x) r (x) = u (x) + xn m h  ; where v (x) := x v v(x ) :  q (x) q (x) To al ulate u (and hen e to al ulate u) it suÆ es to al ulate the rst n m (= deg(u)) Taylor oeÆ ients of 1=q: This an be done by Newton's method (see the next exer ise) as follows: Suppose deg si = j 1 and +

deg( )

1

1

si (x) = O(xj ) :

q (x) Establish that

 1 si (x)q (x) =  [1 si (x)q (x)℄ = O(x j ) : q (x) Note that we may assume q  (0) 6= 0: Now the omputation of

1

q (x)



2si (x)

2

2

si

+1

:= 2si

2

si q 2

an be performed by using an FFT-based polynomial multipli ation and it needs only be performed by using the rst 2j 1 oeÆ ients of q  and si : By starting with an appropriate rst estimate of s (say, s (x) := 1=q  (0)), and pro eeding indu tively as above (doubling the number of oeÆ ients used at ea h stage) we an show that the required number of terms of the expansion an be al ulated in O(n log n) additions and multipli ations (of

omplex numbers). 0

b℄ Fast Reversion of Power Series. Let

f (x) =

1 X k=0

ak xk ;

a = 6 0 0

0

Algorithms and Computational Con erns 363

be a formal power series with given oeÆ ients. Show that the rst n oeÆ ients of the formal Taylor expansion of 1=f an be omputed in O(n log n) additions, multipli ations, and divisions (of omplex numbers).

℄ Fast Polynomial Expansion. Given  ;  ; : : : ; n show that the oQ eÆ ients of ni (x i ) an all be al ulated in O(n(log n) ) additions, multipli ations, and divisions (of omplex numbers). Hint: Pro eed re ursively by dividing the problem into two parts of roughly half the size. Re ombine the pie es using part b℄ of the previous exer ise. 1

2

2

=1

ut

d℄ Fast Polynomial Expansion at Arbitrary Points. Given a polynomial

p of degree at most n; and n + 1 distin t points x ; x ; : : : ; xn ; show that p(x ); p(x ); : : : ; p(xn ) an all be evaluated in O(n(log n) ) multipli ations 0

0

1

2

1

and additions. Hint: Let

bn=Y 2

q (x) := 1

i=0

1

(x

xi )

and let r be the remainder on dividing p by q : Note that r (xi ) = p(xi ) for ea h i < n=2: Similarly, use 1

1

q (x) := 2

n Y

bn=

(x

1

xi ):

2

Thus two divisions redu e the problem to two problems of half the size. u t e℄ Extend d℄ to rational fun tions. f℄ Evaluation of xn . The S -and-X binary method for al ulating xn is the following algorithm. Suppose n has binary representation Æ Æ Æ : : : Æk with Æ = 1: Given symbols S and X; de ne 0

1

2

0

Si := and onstru t the rule



SX if Æi = 1 S if Æi = 0

SS 1

2




Sk :

Now let S be the operation of squaring, and let X be the operation of multiplying by x. Let S S


Sk operate from left to right beginning with x: For example, for n = 27, 1

2

Æ Æ Æ Æ Æ = 11011 0

and

1

2

3

4

S S S S = (SX )(S )(SX )(SX ) : 1

2

3

4

364 A1. Algorithms and Computational Con erns

The sequen e of al ulations of x

27

x!x

2

is then

!x !x !x !x !x !x 3

6

12

13

26

27

:

f1℄ Prove that the above method omputes xn and observe that it only requires storing x; n, and one partial produ t. f2℄ Show that the number of multipli ations in the S -and-X method is less than 2blog n : f3℄ Show that the S -and-X method is optimal for omputation of x m ( onsidering only multipli ations). f4℄ Show that the S -and-X method is not optimal for omputing x : An extended dis ussion of this interesting and old problem is presented in Knuth [81℄. Mu h further material on the omplexity of polynomial operations and omplexity generally may be found in Aho, Hop roft, and Ulman [74℄; Borodin and Munro [75℄; Pan [92℄; and Wilf [86℄. 2

2

15

E.4 Newton's Method. One of the very useful algorithms, both in theory and pra ti e, for zero nding is Newton's method. a℄ Suppose f is analyti in a ( omplex) neighborhood of z ; and suppose f (z ) = 0 and f 0 (z ) 6= 0: Show that the iteration 0

0

0

xn

+1

:= xn

f (xn ) f 0 (xn )

onverges lo ally uniformly quadrati ally, that is, with a onstant independent of n;

jxn

z j < jxn

+1

0

z

0

j

2

for initial values x in some neighborhood of z : As before, we all this lo ally quadrati onvergen e. Hint: Note that 0

0

f (xn ) = f (z ) + (xn 0

z )f 0 (z ) + O((xn 0

0

z ) ); 0

2

whi h implies

xn

+1

z = (xn 0



 f 0 (xn ) f 0 (z ) z) + O((xn f 0 (xn ) 0

0

z ) ): 0

2

ut

Algorithms and Computational Con erns 365

b℄ Show that the iteration

xn

+1

f (x ) := xn + 0 n f (xn )

onverges lo ally quadrati ally to the simple poles of a meromorphi fun tion f: Note that this is Newton's method with the sign hanged.

℄ The iteration

xn

+1

:= xn

f (xn )f 0 (xn ) f 0 (xn ) f (xn )f 00 (xn ) 2

onverges lo ally quadrati ally to a zero of an analyti f independent of its multipli ity; see Henri i [74℄. d℄ Let g := f

1

: The iteration xn

g n (xn ) (xn ) (

+1

:= xn + (n + 1) n g (

)

+1)

onverges lo ally uniformly to a zero of an analyti fun tion f with (n + 2)th order. Newton's method is n := 0; Halley's method is n := 1 (see Householder [70℄). Newton's method and its variants work tremendously well provided that a good starting value an be found. This is a problem. On a real interval a bise tion method an be used initially to lo alize the zeros. In the plane, life is more ompli ated as is seen in the next exer ise. Another drawba k to Newton's method is the need to ompute the derivative. Of

ourse, this is not a problem for polynomials, but in a general setting it is usually repla ed by an approximation su h as

f (xn ) f (xn ) xn xn 1

1

(whi h yields the so- alled se ant method). p e℄ Consider Newton's method for omputing x starting with x := 1: This gives   1 x xn := xn + : 2 xn 0

+1

Show that rn (x) := xn is a rational fun tion in x with numerator px hasofa degree 2n and denominator of degree 2n 1: Show that rn (x) zeropof order 2n at 1: So rn is in fa t the (2n ; 2n 1) Pade approximant to x. (This implies, though not obviously, that rn has all real negative roots and poles.) +1

+1

366 A1. Algorithms and Computational Con erns

An attra tive feature of Newton's method is that it is \self- orre ting." So, for example, to ompute a root to a large pre ision, one an start at a small pre ision and double the pre ision at ea h iteration. This is a substantial savings both pra ti ally and theoreti ally. The same feature applies to Newton's method solutions over formal power series, as in E.3 a℄. This allows for doubling the number of terms used at ea h stage. Mu h additional material on Newton's method is available in Borwein and Borwein [87℄, Henri i [74℄, Householder [70℄, and Traub [82℄. E.5 Newton's Method in Many Variables. a℄ Let f : C n ! C n ; and suppose f has Ja obian

J :=

fi z

j

where f := (f ; f ; : : : ; fn ) with fi : C n 1

2

;

! C : Let

x := (x ; x ; : : : ; xn ) 2 C n : 1

2

The fun tion J (x) is the Ja obian evaluated at x. Then Newton's method be omes xk = xk sk +1

where sk solves J (xk )sk = f (xk ). This iteration onverges lo ally uniformly quadrati ally to a zero z0 of f; that is, with a onstant independent of n;

jxn

+1

z j < jxn

z

0

j

0

2

for initial values x0 in some neighborhood of z0 ; provided in a neighborhood of z ; f is ontinuously dierentiable, J exists and is bounded in norm, and J satis es a Lips hitz ondition. We all this lo ally quadrati

onvergen e. For a polynomial f; we require only that J exists in a neighborhood of the zero z : (For examples and detail, see Dennis and S hnabel [83℄.) b℄ Finding All Zeros of a Polynomial. Let 1

0

1

0

p(z ) := a + a z +


+ an z n ; 0

an := 1 :

1

Let fi (x ; : : : ; xn ) be the ith oeÆ ient of 1

g(x) := and let

n Y i=1

(x

xi )

n X i=0

ai xi

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

2

Algorithms and Computational Con erns 367

Then the iteration of part a℄ applied to f onverges lo ally quadrati ally to z := (z ; z ; : : : ; zn ); where z ; z ; : : : ; zn are the zeros of p; provided the zeros of p are distin t.

℄ Another Approa h to Finding All Zeros. Let xj (k ) denote the k th approximation to the j th root of p; where p is a polynomial of degree n with n distin t roots. Let p(xj (k)) ; j = 1; 2; : : : ; n : xj (k + 1) = xj (k) Qk ( i6 j xj (k ) xi (x)) 0

1

2

1

2

=

Show that for suÆ iently good hoi e of (x (1); : : : ; x (k )); the above iteration onverges lo ally quadrati ally to a sequen e of the n distin t zeros of p: Hint: This is really just the single variable version of Newton's method for ea h root, where the derivative is approximated by the derivative of the k th estimate. ut n d℄ Observe that the iteration of part ℄ fails to onverge for p(x) := x 1; n > 2; if the starting values are all taken to be real. In pra ti e, the iteration of part ℄ works rather well for reasonably

hosen starting values. One an use the te hniques of the next exer ise to lo alize the zeros rst. With are, an algorithm an be given that omputes a zero of a polynomial with an error < 2 b in O(n log b log n) time and all zeros in O(n log b log n) time; see Pan [92℄. On a parallel ma hine (PRAM) an algorithm requiring O(log (nb)) time and O(nb)O pro essors an be given for omputing all zeros. e℄ Modify the iteration of E.4 ℄ as given in the previous exer ise to get a method that omputes all roots even in the presen e of repeated roots. For further dis ussion, see Aberth [73℄, Durand [60℄, Kerner [66℄, and Werner [82℄. 0

0

2

3

(1)

E.6 Lo alizing Zeros. a℄ Cau hy Indi es. Let r be a real rational fun tion with a real pole . The Cau hy index of r at  is 8 if limx! r(x) = 1 and limx! r(x) = 1 > <1 1 if limx! r(x) = 1 and limx! r(x) = 1 > : 0 otherwise : +

+

The Cau hy index of r on an interval [a; b℄ is the sum of the Cau hy indi es of the poles of r in (a; b). (We demand that neither a nor b be poles of r.) We denote this by Iab (r): b℄ The Eu lidean Algorithm. Let p and p be nonzero polynomials. De ne polynomials p ; p ; : : : ; pm and q ; q ; : : : ; qm (by the usual division algorithm) so that 0

0

1

1

1

2

368 A1. Algorithms and Computational Con erns

p (z ) = p (z )q (z ) p (z ) = p (z )q (z ) 0

1

1

1

2

2

p (z ) ; p (z ) ;

deg(p ) < deg(p ) ; deg(p ) < deg(p ) ;

2 3

.. . pm (z ) = pm (z )qm (z )

2

1

3

2

0;

1

the algorithm stops the rst time that the remainder is zero. This is alled the Eu lidean algorithm. Show that pm is the greatest ommon fa tor of p and p .

℄ Let p ; p ; : : : ; pm be the polynomials generated by the Eu lidean algorithm in part b℄. Let Pi := pi =pm ; i = 0; 1; : : : ; m: Suppose p (a)p (b) 6= 0: Show that if Pk ( ) = 0 for some k and a   ; then 0

1

0

1

0

0

P ( ) 6= 0 if k = 0 1

and

Pk ( )Pk ( ) < 0 if 1  k  m 1

+1

1:

d℄ Show that for real polynomials p and p with p (a)p (b) 6= 0; 0

Iab (p =p ) = v(a) 1

0

1

0

v(b) ;

0

a < b;

where v () is the number of sign hanges in the sequen e (p (); p (); : : : ; pm ()) 0

1

and where the polynomials pi are generated by the Eu lidean algorithm as in part b℄. (As before, by a sign hange we mean that pi ()pi k () < 0 and pi () = pi () =


= pi k () = 0:) Hint: Without loss of generality, we may assume that pm = 1 (why?), so Pk = pk for ea h k: First, note that v(x) may hange magnitude only if pi (x) = 0 for some i: By ontinuity of the polynomials pi ; and by part ℄, v(x) is onstant on any subinterval of [a; b℄ that does not ontain a zero of +

+1

+2

+

1

p: 0

We are now redu ed to onsidering the behavior of v (x) at the zeros of p : Consider the various possibilities for the behavior of p =p at the zeros of p by onsidering the four possible hanges of signs of p and p at the zeros of p and the ee t this has on the Cau hy index and the in rease and de rease of v (x): (Note that v (x) de reases by 1 if the Cau hy index is 1 while v (x) in reases by 1 if the Cau hy index is 1:) ut e℄ Zeros on an Interval. Suppose p is a real polynomial and p(a)p(b) 6= 0: Then Iab (p0 =p) equals the number of distin t zeros of p in [a; b℄: This also equals v (a) v (b); where v (x) is as in d℄ with p := p and p := p0 : 0

1

0

0

0

1

0

0

1

Algorithms and Computational Con erns 369

Hint: Use

p0 (z ) X mk = p(z ) z k and note that ea h distin t pole in (a; b) ontributes +1 to the Cau hy

index. ut f℄ Show that the number of zeros of a real polynomial on the real line equals lim (v ( a) v (a)) ; a!1

where v (x) is omputed as in part e℄. (That is, v (x) is the number of sign

hanges in the sequen e (p ; p ; : : : ; pm ) generated by the Eu lidean algorithm with p := p and p := p0 :) g℄ The Number of Zeros in H := fz 2 C : Im(z ) > 0g. Suppose that the moni polynomial p 2 Pn has (exa tly) k real zeros ounting multipli ities. Write p(x) = r(x) + i s(x) ; r; s 2 Pn : 0

0

1

1

Then the number of zeros of p in H is 1 n 2

k



I 11 (s=r) ;

and this an be omputed as in part d℄ by using  I 11 (s=r) = lim !1 I  (s=r) :

Hint: First onsider the ase where k = 0: Consider p on a ounter lo kwise semi ir ular ontour with base [ ; ℄ and radius : Consider the argument of p as the ontour is traversed. On the half- ir le, for large , the argument in reases by n asymptoti ally; while on the axis the hange is I  (s=r); from whi h the result follows. ut h℄ Budan-Fourier Theorem. Let p 2 Pn . Let V (x) be the number of sign

hanges in the sequen e


p(x); p0 (x); : : : ; p n (x) : (

)

Then the number of zeros of p in the interval [a; b℄, ounting multipli ities, is V (a) V (b) 2m for some nonnegative integer m: We have followed Henri i [74℄ in this dis ussion. E.7 Zeros in a Disk.

z+i a℄ The transform maps the unit ir le to the real axis and maps 1 + iz the open unit disk D to the upper half-plane fz 2 C : Im(z ) > 0g: So the algorithms of the previous exer ise apply after transformation.

370 A1. Algorithms and Computational Con erns

b℄ Zero Counting by Winding Number. Let !m := e i=m , where m is a positive integer. If p 2 Pn then 2

zp :=

Z m 0 k X 1 p0 (z ) k p (!m ) dz  !m k) 2i D p(z ) p(!m k =1

ounts the number of zeros of p in the open unit disk D (assuming no zeros on the boundary D). More pre isely, we have

zp = mlim !1

m X k=1

k !m

k) p0 (!m k) : p(!m

(Note that this lends itself to rapid evaluation by FFT methods.)

℄ Show that Z

m k X 1 1 !m = + m () ; k 2i D z  k !m  =1

where

m () 

jjm 1 jj 1

jj < 1

if

and

m () 

jj

1

Hint: Write z

1



m

jj =

1

if

1

jj > 1 :

1 zk 1X

k

=0

k

and use the fa t that 0=

Z

D

p(z ) dz =

m X k=1

k p(! k ) !m m

: for every p 2 Pm d℄ If p 2 Pn has no zeros in the annulus   z is greater than $ % log n  + 1; log  1

ut

 1= and if m in part b℄

1

then the error in estimating zp by the sum is less than 1: So this provides an algorithm.

Algorithms and Computational Con erns 371

E.8 Computing General Chebyshev Polynomials. Given a Chebyshev system M := (f ; : : : ; fn ) of C fun tions on, say, [0; 1℄ how does one

ompute the asso iated Chebyshev polynomial Tn ? That is, how does one P nd the unique equios illating form T := ni ai fi of Se tion 3.3? a℄ The Remez Algorithm Step 1. Choose 1

0

=0

x := (0 =: x < x and nd P 2 span M su h that 0

(0)

(0)

0

1

<


< xn := 1) (0)

0

P (xi ) = ( 1)i ;

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

(0)

0

Step 2. Indu tively set xm := (0 =: x m < x m <


< xnm := 1); where (

)

(

0

)

(

)

1

Pm0 (xim ) = 0 ; i = 1; 2; : : : ; n 1 : (That is, nd the extrema of Pm .) Step 3. Find Pm 2 span M with (

)

1

1

Pm (xim ) = ( 1)i ; i = 0; 1; : : : ; n : (

)

Then, provided the initial estimate x is suÆ iently good, Pm ! Tn quadrati ally (see Veidinger [60℄). This is reasonably easy to ode. It involves solving an interpolation problem in Steps 1 and 3. The zero nding at Step 2 an be done quite easily sin e one an nd very good starting values for Newton's method, namely (xi m )ni . b℄ This algorithm modi es to solve the best approximation problem min k!p f k a;b 0

(

1)

=0

p2span M

[

℄

for !; f 2 C [a; b℄; where ! is positive on [a; b℄: One uses the Remez agorithm to nd an equios illating form

P (x) = f (x)

!(x)

n X i=0

i fi (x)

at n + 2 points. To do this, one solves the system

f (xk )

!(xk )

n X i=0

i fi (xk ) + ( 1)k  ;

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

for both the i and : (This works reasonably well, provided that at ea h stage kP k a;b o

urs at one of the xk : If not, an extra point must be inserted where the maximum norm o

urs and one of the original points must be dropped. This is ee ted in su h a way as to maintain the alternations in sign of the error.) For details see Cody, Fraser, and Hart [68℄ and Veidinger [60℄. [

℄

This is page

372

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A2

Orthogonality and Irrationality

Overview This appendix is an appli ation of orthogonalization of parti ular Muntz systems to the proof of the irrationality of  (3) and some other familiar numbers. It reprodu es Apery's remarkable proof of the irrationality of  (3) in the ontext of orthogonal systems.

Orthogonality and Irrationality Apery's wonderful proof of the irrationality of  (3) amounts to showing that 0 < jdn an  (3) bn j ! 0 ; 3

where bn is an integer,

an := and

n X k=0



n+k k

2  2

n k

;

dn := l mf1; 2; : : : ; ng :

Here l m denotes the least ommon multiple; see van der Poorten [79℄ and Beukers [79℄. In [81℄ Beukers re ast the proof using Pade approximations.

Orthogonality and Irrationality 373

Many, maybe most, irrationality proofs may be based on approximation by Pade approximants and related orthogonal polynomials; see, for example, Borwein [91a℄, [92℄ or Chudnovsky and Chudnovsky [84℄. It is the intention of this appendix to try to put the proof of the irrationality of  (3) into the framework of orthogonality. From the general orthogonalization of the system (x0 ; x1 ; x2 ; : : : ) on [0; 1℄; where the numbers j are nonnegative and distin t, spe ializing to the ase when

 j =j



and

2

j

2 +1

= j +;

j = 0; 1; : : :

where  de reases to 0; is a very natural thing to do. This is how one should interpret orthogonalizing the system (x ; x ; x ; x ; : : : ): This leads to orthogonal fun tions that generalize the Legendre polynomials and are of the form pn (x) log x + qn (x) with polynomials pn ; qn 2 Pn of degree n: Legendre polynomials are losely tied to irrationality questions on erning log (see Borwein and Borwein [87℄, Chapter 11), and higher-order P analogs prove
to be the basis of dealing n with the irrationality of the trilog 1 n x =n for some values of x: We think that the proof of the irrationality of  (3) ows quite naturally from this point of view. Although in the end (Lemma A.2.3) we get ba k to Beukers' integral approa h to the irrationality of  (3) (as indeed we must). What follows, up to one appli ation of the prime number theorem, is a self- ontained proof of the irrationality of  (3): The orthogonalization in question is the ontent of the rst theorem. 0

0

1

1

3

=1

Theorem A.2.1. Let 1 Gn (x) := 2i

where

Z

Qn

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

2

(t + n + 1)xt dt ;

is any simple ontour ontaining t = 0; 1; : : : ; n: Then Gn (x) = pn (x) log x + qn (x) ;

where pn (x) =

 n  2  X n n+k 2

k=0

k

k

(n + k + 1) xk :

Furthermore, we have the orthogonality relations Z

Z 0

0 1

1

Gn (x)xk dx = 0 ;

Gn (x)(log x)xk dx = 0 ;

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

374 A2. Orthogonality and Irrationality

and Z

1

Gn (x) dx = 2

0

1 : 2n + 1

Proof. As in Se tion 3.4, the representation of Gn is just the evaluation of the integral at the poles, t = 0; 1; : : : ; n; by the residue theorem. The proof of the orthogonality onditions is a straightforward exer ise on evaluating Z

1

0

xk (log x)j Gn (x) dx

by inter hanging the order of integration as in the proof of Theorem 3.4.3.

ut

We need to modify these forms marginally to give a zero at 1; as in the next result. Theorem A.2.2. Let Gn and

Fn (x) := =

Z x

1

xn+1 1 2i

Z

0

be de ned as in Theorem A.2.1. Then

yn Gn (y) dy

Qn

(t + k + 1) t x dt = An (x) log x + Bn (x) ; k (t k ) 1

2

k=0 Q n

2

=0

where An (x) := and

   n  X n+k 2 n 2

k

k=0

n X

Bn (x) := with



n+k

k := k

k=0

2  2 (n 1 X n

k

i=0

k

k xk n X

2

k+i+1

Furthermore, Z 0

Z

1

0

and

1

2Æn;m Fn (x)Fm (x) dx = ; (2n + 1) 3

Fn (x)xk (log x)j dx = 0 ; Fn (1) = 0 ;

i=0 i6=k

Æn;m :=

k = 0; 1; : : : ; n

Fn0 (1) = 1 ;

xk

k



1;

0 1

2

)

i

:

if n 6= m if n = m ; j = 0; 1 ;

Fn00 (1) = 2n + 2n 1 : 2

Orthogonality and Irrationality 375

Proof. This follows mu h as in Theorem A.2.1. The fa t that Fn (1) = 0 is just the statement that Gn is orthogonal to xn : ut The fa t that Fn (1) = 0 is riti al in what follows. It immediately gives that Bn (x) has a zero at 1, so Bn (x)=(1 x) is a polynomial, as in part ℄ of the following orollary. (This is the only part of the orollary we need, but the orollary is of some interest in its own right.) Corollary A.2.3. Let Fn be de ned as in Theorem A.2.2. Then a℄ Fn has 2n + 1 zeros on (0; 1℄. b℄ An (x) has all real negative zeros.

℄ Bn (x)=(1 x) is a polynomial with all real negative zeros that interla e the zeros of An (x):

Proof. The orthogonality onditions give 2n zeros of Fn on (0; 1) in a standard fashion, and there is one zero at 1. The real negative zeros of An (x) and Bn (x) and their interla ing follow from the fa t that B (x) log x + n An (x) has 2n + 1 zeros on (0; 1℄; and known results on interpolating Stieltjes transforms by rational fun tions (see, for example, Baker and Graves Morris [81℄ or Borwein [83℄). ut One an, from the integral representation, dedu e the next orollary, whi h is also not a tually needed in the proof of the irrationality of  (3): Corollary A.2.4. Both Fn and An satisfy

x (yn00

yn00 )

2

2nx(yn0 + yn0 ) + x(yn0

1

1

yn0 ) + n (yn 2

1

yn ) = 0 : 1

From Theorem A.2.2 we obtain (see E.2 b℄) the following: Corollary A.2.5. Let dn := l mf1; 2; : : : ; ng: Then the polynomial dn Bn (x) has integer oeÆ ients. We get an approximation to  (3) by integration over the unit square. Theorem A.2.6. 1 2

Z

1

0

Z

1

0

Fn (xy) dx dy = An (1) (3) + Rn ; 1 xy

where 2dn Rn is an integer. 3

376 A2. Orthogonality and Irrationality

Proof. Re all that

sin e

1 2

Z

1

Z

0

1

Z

1

0

Z

1

0

log(xy ) =  (3) 1 xy

(xy )n log(xy ) dx dy =

0

2 : (n + 1) 3

We have

Fn (xy) An (xy) An (1) A (1) B (xy) = log(xy ) + n log(xy ) + n : 1 xy 1 xy 1 xy 1 xy The rst term of the last expression is a polynomial multiple of log(xy ); and the last term is a polynomial. Both have degree n 1 in xy: Here we use part ℄ of Corollary A.2.3 in an essential way. Integrating the above equation with respe t to x on [0; 1℄ and with respe t to y on [0; 1℄; we get the identity of the theorem. The fa t that dn Rn is an integer an be seen as follows. One dn arises from ea h of the two integrations of a polynomial of degree n 1 with integer oeÆ ients and one dn omes from Corollary A.2.5. ut 3

Theorem A.2.7. The number  (3) is irrational.

Proof. It now suÆ es to show that there is an  > 0 for whi h Z

0 <

1

Z

0

1

0









Fn (xy) 1 1 dx dy = O =o 1 xy dn e n



3

(3+ )

sin e by the prime number theorem, l mf1; 2; : : : ; ng = O(e  n ) for every  > 0; see Borwein and Borwein [87, p. 377℄. (We use the notation bn = o(an ) if bn = n an with limn!1 n = 0:) This an be proven in a number of ways, we hose to onne t this proof via Pade approximants to the integral estimate due to Beukers. This is the ontent of the following results and, in parti ular, Lemma A.2.10. From Lemma A.2.10, the above estimates are easy sin e the integrand in the right-hand side in Lemma A.2.10 satis es (1+ )

0<

xyv(1

x)(1

y)(1 v) 1 (1 xy )v

on the open unit ube 0 < x; y; v < 1:

p

(

2

1)

4

ut

The following lemma gives standard representations for the Pade approximants to log and an be he ked by expanding the integrals; see E.3.

Orthogonality and Irrationality 377

Lemma A.2.8 (Pade Approximants). We have Z

(n!) 2i

2

Z

Qn

xt dt vn (1 v)n dv = (x 1) n (1 (1 x)v )n (t k ) = pn (x) log x + qn (x) = O((x 1) 2

2

k=0

1

+1

+1

0

n+1 ) ;

2

where is a simple ontour ontaining t = 0; 1; : : : ; n; and pn and qn are polynomials of degree n: (So pn =qn is the (n; n) Pade approximant to log x at 1:) Lemma A.2.9 (Rodrigues-Type Formula). With Z dn dn xn yn

Fn (xy) =

dyn dxn 2i Z 1 dn dn = (n!) dy n dxn 2

(xy )t dt k (t k ) (xy )n v n (1 v )n (xy 1) n (1 (1 xy )v )n

Qn

2

=0

1

as in Theorem A.2.1,

2

+1

+1

0

dv :

Proof. This follows from dierentiating the two representations derived from Lemma A.2.8 oupled with Theorem A.2.2. ut Lemma A.2.10. We have Z

1

Z

0

1

0

Fn (xy) dx dy 1 xy

Z

=

1

Z

0

1

Z

0

1

[xyv (1

y)(1 v)℄n dx dy dv : (1 (1 xy )v )n x)(1

+1

0

Proof. For k, n nonnegative integers Z

1

Z

1

(xy )n k (1

x)n (1 y)n dx dy

+

0

0

1 (n!)

=

Z 2

1

Z

0

1

0

(Both sides equal

dn dn (xy )n k (xy 1 xy dy n dxn 1



+

n!(n + k)! (2n + k + 1)!

2

1) n 2

+1

dx dy :

;

though this is not ompletely transparent; see E.4.) So 1 (n!)

2

Z

1

0

Z

1

0

dn dn n n x y (xy 1 xy dy n dxn 1

=

Z

1

0

Z

0

1

[xy (1

1) n 2

k dx dy

+1+

x)(1

y)℄n (xy

1)k dx dy :

378 A2. Orthogonality and Irrationality

Hen e, on expanding (1 1 (n!)

2

Z

1

Z

0

n

xy)v)

(1

1

in the following integrands,

dn dn xn yn (xy 1) n dx dy 1 xy dy n dxn (1 (1 xy )v )n Z Z [xy (1 x)(1 y )℄n = dx dy ; (1 (1 xy )v )n 1

1

2

+1

+1

0

1

0

1

+1

0

ut

and with Lemma A.2.9 we are done.

Comments, Exer ises, and Examples. The approa h of this appendix follows Borwein, Dykshoorn, Erdelyi, and Zhang [to appear℄. Beukers' [79℄ very elegant re asting of Apery's proofs of the irrationality of  (2) and  (3) is also presented in Borwein and Borwein [87℄. E.5 re asts the irrationality of  (2) =  into a form similar to the proof of the irrationality of  (3): E.6 treats the irrationality of log 2: Mahler [31℄ asts trans enden e results for exp and log in terms of general systems of Pade approximants. 1

2

6

E.1 Proof of Corollaries A.2.4 and A.2.5. a℄ Prove Corollary A.2.4 from the expli it representations of Theorem A.2.2. b℄ Prove Corollary A.2.5 from the formula for k in Theorem A.2.2. Hint: Observe that if
p is a prime and n < p  n + k for some integer ; then p divides n k k . This is fairly straightforward from Euler's formula for the largest power of a prime dividing a fa torial. ut +

E.2 Formulas for  (n). a℄ Show that Z

1

0






Z

1

Z

0

1 1 X dx dx


dxn =  (n) := : 1 x x


xn kn k

1

1

1

0

b℄ Show that

2

Z

1

0

2

=1

log x dx = 1 x



2

6

:

℄ Find a losed formula for Z

1

0






Z

1

0

Z

1

0

log(x x 1 xx 1

1

2




xn ) dx


xn

1

2

dx

2




dxn

in terms of  (n + 1): E.3 Proof of Lemma A.2.8. Give a proof of Lemma A.2.8.

Orthogonality and Irrationality 379

E.4 The Identity in the Proof of Lemma A.2.10. Prove the identity Z

1

Z

1

(xy )n k (1

x)n (1 y)n dx dy

+

0

0

Z

Z

1 1 dn dn (xy )n k (xy (n!) 1 xy dy n dxn   n!(n + k)! = ; (2n + k + 1)! =

1

1

1) n

+

2

0

2

+1

dx dy

0

2

where k and n are nonnegative integers. Follow parts a℄ to ℄. Let k and n be nonnegative integers. a℄ Show that

An;k :=

Z

1

Z

1

(xy )n k (1 +

0

x)n (1

y)n dx dy =



0

n!(n + k)! (2n + k + 1)!

2

:

b℄ Let

Bn;k :=

Z

1 (n!)

2

1

Z

0

1

0

dn dn n k x (xy (1 xy ) dy n dxn 1

1) n

+

2

+1

dx dy :

Show that

n Bn;k =[(n + k) + (2n + 1)(2n + 2k + 1) + (2n + 1)(2n)℄Bn [2(n + k ) + (2n + 1)(2n + 2k + 1)℄Bn ;k + (n + k ) Bn ;k : 2

2

2

1

2

;k+2

1

+1

1

℄ Show that A ;k = B ;k : Show that the values 0

0



n!(n + k)! An;k = (2n + k + 1)!

2

satisfy the re urren e relation established by part b℄ for the values Bn;k : E.5 The Irrationality of  . Consider 2

n! Fn (x) := (2i) where

Z

Qn (t + k ) Qnk=1

k=0 (t

k)

2

xt dt ;

is any simple ontour ontaining the poles at 0; 1; : : : ; n:

380 A2. Orthogonality and Irrationality

a℄ Show that Z

1

0



n X Fn (x) n+k dx = 1 x k k

 2

n  (2) + Rn ; k

=0

where dn Rn is an integer (dn := l mf1; 2; : : : ; ng as before). Hint: Show that Fn (x) = An (x) log x + Bn (x) ; 2

where

An (x) :=

n X k=0



n+k k

and

Bn (x) := with



n+k

k := k

n X k=0

 2 ( n X n

k

2  2

n k x k

k xk

1 k+i

i=1

n X i=0 i6=k

k

2

)

i

:

Write

A (1) B (x) Fn (x) An (x) An (1) = log x + n + n : 1 x 1 x 1 x 1 x Show that Fn (1) = 0; hen e Bn (1) = 0: Re all that Z

1

0

1

log x

x dx

=  (2) :

Now the on lusions follow, as in the proof of Theorem A.2.6. b℄ Show that Z

1

0

Fn (x) dx = 1 x

Z

1

0

Z

1

xn yn(1

x)n (1 y)n dx dy : (1 xy )n +1

0

Hint: Use Lemma A.2.8.

℄ Show that there exists a onstant independent of n su h that 0<

Z

1

0



Fn (x) dx  1 x

p

5 1 2

!5n

and dedu e that  (2) =  irrational. Hen e  is irrational. 1

6

2

ut

ut

Orthogonality and Irrationality 381

E.6 The Irrationality of log 2. Let 1 dn n [x (1 pn (x) := n! dxn

x)n ℄ =

  n  X n+k n

k=0

k

k

( 1)k xk

be the nth Legendre polynomial on [0; 1℄: a℄ Show that Z

1

0

Z

pn( 1) dx + 1+x = pn ( 1) log 2 + Rn ;

pn (x) dx = 1+x

1

pn (x)

Z

0

1

0

pn ( 1) dx 1+x

where dn Rn is an integer (dn := l mf1; 2; : : : ; ng as before). b℄ Show that Z Z pn (x) xn (1 x)n dx = dx : 1+x (1 + x)n 1

0

1

0

+1

℄ Use parts a℄ and b℄ to show that log 2 is irrational.

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A3

An Interpolation Theorem

Overview Appendix 3 presents an interpolation theorem for linear fun tions that is used in Se tion 7.1. From this Haar's hara terization of Chebyshev spa es follows, as do alternate proofs of many of the basi inequalities.

An Interpolation Theorem The main result of this se tion, Theorem A.3.3, is an interpolation theorem that plays an important role in Se tion 7.1. Further appli ations are given in the exer ises. Throughout this appendix we use the following notation. Let Q be a

ompa t Hausdor spa e. Let C (Q) be the spa e of all real- or omplexvalued ontinuous fun tions de ned on Q. Let P be a (usually nitedimensional) linear subspa e of C (Q) over R if C (Q) is real or over C if C (Q) is omplex. The fun tion f 2 C (Q) is said to be orthogonal to P , written as f ? P , if

kf kQ  kf + pkQ

for all p 2 P :

This is an L1 analog of the more usual L notion of orthogonality, as in Se tion 2.2. The following two lemmas give ne essary and suÆ ient onditions for the relation f ? P . 2

An Interpolation Theorem 383

Lemma A.3.1. Let 0 6= f 2 C (Q). The fun tion f is orthogonal to P if and only if there exists no p 2 P su h that


(A:3:1)

Re f (x)p(x) > 0

holds on E := E (f ) := fx 2 Q : jf (x)j = kf kQg :

(A:3:2)

Proof. Assuming there exists p (A.3.2), we show that

2P

kf

satisfying (A.3.1) on E de ned by

pkQ < kf kQ




for some  > 0: Sin e the set E de ned by (A.3.2) is ompa t, Re f (x)p(x) attains its positive minimum, say, 2Æ > 0, on E , and there exists an open set G ontaining E su h that


x 2 G:

Re f (x)p(x) > Æ > 0 ;

Sin e G := Q n G is also ompa t, there exists an  > 0 su h that

jf (x)j < (1

)kf kQ ;

x 2 G :

Thus, with a suÆ iently small  > 0; 2Æ +  kpkQ < kf kQ ;

jf (x)

p(x)j

jf (x)

p(x)j  (1 )kf kQ + kpkQ < kf kQ ;

2

 kf kQ 2

2

2

2

x 2 G;

while

Therefore

kf

x 2 G :

pkQ < kf kQ

if  > 0 is small enough. Conversely, if f is not orthogonal to P; then there exists p that kf pkQ < kf kQ ; so 2

2




2 Re f (x)p(x) > kpkQ  0 ; and the proof is nished.

2

2P

su h

x2E;

ut

384 A3. An Interpolation Theorem

Lemma A.3.2. Let 0 6= f 2 C (Q) and let P be an n-dimensional linear subspa e of C (Q) over R if C (Q) is real or over C if C (Q) is omplex. Then the

fun tion f is orthogonal to P if and only if there exist points x ; x ; : : : ; xr in E (f ) de ned by (A.3.2) and positive real numbers ; ; : : : ; r ; where 1  r  n +1 when C (Q) is real, and 1  r  2n +1 when C (Q) is omplex, su h that 1

1

r X

(A:3:3)

i=1

2

2

p2P:

i f (xi )p(xi ) = 0 ;

Proof. P Suppose (A.3.3) holds with some positive real ; ; : : : ; r satisfying ri i = 1: As jf (xi )j = kf kQ ; we have 1

2

=1

kf kQ = 2

r X i=1

i f (xi )f (xi ) =

 kf kQ

r X i=1

r X i=1

i f (xi )(f (xi ) p(xi ))

i max jf (xj ) jr 1

p(xj )j  kf kQkf

pkQ

for every p 2 P; so f ? P: (Note that r  n + 1 or r  2n + 1; respe tively, was not needed for this part of the proof, so the suÆ ien y of (A.3.3) is valid with no hypothesis about r:) Conversely, suppose f ? P: Let f' ; ' ; : : : ; 'n g be a basis for P over R (or C ; respe tively), and onsider the map T : Q ! Rn (or C n ; respe tively) de ned by 1

2

T (x) = f (x)(' (x); ' (x); : : : ; 'n (x)) : 1

2

Observe that the origin is in the onvex hull of

T (E ) := fT (x) : x 2 E g ; otherwise by the prin iple of separating hyperplanes (a orollary of the Hahn-Bana h theorem; see Rudin [73℄), there would exist omplex numbers a ; a ; : : : ; an su h that 1

2

Re Hen e, with p :=

Pn

n X i=1

!

ai f (x)'i (x) > 0 ;

i=1 ai 'i

x2 E:

2 P;


Re f (x)p(x) > 0 ;

x2E

and f is not orthogonal to P by Lemma A.3.1, whi h ontradi ts our assumption.

An Interpolation Theorem 385

Now it follows from Caratheodory's lemma (see E.1) that there exist points x ; x ; : : : ; xr in E and positive real numbers ; ; : : : ; r ; where 1  r  n + 1 when C (Q) is real and 1  r  2n + 1 when C (Q) is omplex, su h that r 1

2

1

X

i=1

i f (xi )'k (xi ) = 0;

Therefore

r X i=1

2

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

i f (xi )p(xi ) = 0;

p2P;

ut

and the lemma is proved.

Theorem A.3.3 (Interpolation of Linear Fun tionals). Let C (Q) be the set

of real- ( omplex-) valued ontinuous fun tions on the ompa t Hausdor spa e Q. Let P be an n-dimensional linear subspa e of C (Q) over R (C ). Let L 6= 0 be a real- ( omplex-) valued linear fun tional on P . Then there exist points x ; x ; : : : ; xr in Q, and nonzero real ( omplex) numbers a ; a ; : : : ; ar , where 1  r  n in the real ase and 1  r  2n 1 in the omplex ase, su h that 1

1

2

2

(A:3:4)

L(p) =

r X i=1

ai p(xi ) ;

p2P

and

kLk =

(A:3:5)

where

r X i=1

jai j ;

kLk := supfjL(p)j : p 2 P ; kpkQ  1g :

Proof. Be ause of the nite dimensionality of P; there exists an element p 2 P ( alled an extremal element for L) su h that kp kQ = 1 and L(P  ) = kLk: Let P denote the null-spa e of L, so 0

P := fp 2 P : L(p) = 0g : 0

Now p is orthogonal to P be ause if 0

kp + p kQ < kp kQ = 1 2 P; then g := p + p satis es kgkQ < 1 and L(g) = kLk; 0

for some p whi h is impossible. Note that the dimension of P over R is n 1 in the real ase and 2n 2 in the omplex ase. So by Lemma A.3.2 there exist points x ; x ; : : : ; xr in 0

0

0

1

2

386 A3. An Interpolation Theorem

E := fx 2 Q : jp (x)j = 1g and positive real numbers ; ; : : : ; r ; where 1  r and 1  r  2n 1 in the omplex ase, su h that 1

r X i=1

Sin e L(p)p

L(p)

2

i p (xi )p (xi ) = 0 ; 0

p

0

2P

 n in the real ase

0

:

L(p )p 2 P for all p 2 P; we have 0

r X i=1

r

X

i jp (xi )j = L(p ) i p(xi )p(xi ) ; 2

i=1

p2P:

Sin e L(p ) = kLk and jp (xi )j = 1 for ea h i; we obtain (A.3.4) by taking

ai =

i p (xi ) kLk : j j

Pr

=1

Using the fa t that jp (xi )j = 1 for ea h i in the above formula for ai ; we get (A.3.5). ut Comments, Exer ises, and Examples. We have followed Shapiro [71℄, whi h gives a long dis ussion of questions related to the best uniform approximation of a fun tion f 2 C (Q) from a (usually nite-dimensional) linear subspa e P  C (Q): Some of these, together with other appli ations, are dis ussed in the exer ises. E.1 Caratheodory's Lemma. If A  Rn ; then every point from the onvex hull o(A) of A an be written as a onvex linear ombination of at most n + 1 points of A:

Proof. Let x 2 A: After a translation if ne essary, we may assume that x = 0: Suppose (A:3:6)

0=

r X i=1

xi 2 A ; i > 0 ; r > n + 1 :

i xi ;

Sin e r > n + 1; the elements x ; x ; : : : ; xr are linearly dependent, so there exist real numbers i ; i = 2; 3; : : : ; r; not all zero, su h that 2

3

r X i=2

i xi = 0 :

Let := 0: For all  2 R; we have 1

An Interpolation Theorem 387

0=

r X i=1

i xi + 

r X i=1

i xi =

r X i=1

(i +  i )xi :

When  = 0; ea h term in the last sum is positive. We now de ne

:= min ji = i j; where the minimum is taken for all indi es i for whi h i 6= 0: If the index j is hosen so that jj = j j = ; and if  := j = j ; then at least one of the numbers i +  i is zero, and all are nonnegative. Also  +  =  > 0: We have thus obtained a representation of the same form as (A.3.6), but with s terms, where 1  s  r 1: If s > n + 1; then the pro ess an be repeated, and after a nite number of steps we obtain the desired representation. ut 1

1

1

E.2 Reformation in Terms of Integrals. Lemma A.3.2 an be reformulated as follows. Under the assumptions of Lemma A.3.2, f 2 C (Q) is orthogonal to P if and only if there exists a nonzero nonnegative Borel measure  on Q whose support onsists of r points of E (f ) de ned by (A.3.2), where 1  r  n + 1 in the real ase and 1  r  2n + 1 in the

omplex ase, su h that f (x) d(x) annihilates P , that is, Z

(A:3:7)

Q

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

p2P:

This reformation is not only a notational onvenien e, but it is essential in generalizations where P is no longer nite-dimensional. Moreover, (A.3.7) with any nonzero nonnegative Borel measure (not ne essarily dis rete) is suÆ ient for f ? P: This is often useful, even when P is nite-dimensional; see Shapiro [71℄. E.3 Haar's Chara terization of Chebyshev Spa es. a℄ Let f ; f ; : : : ; fn be real- or omplex-valued ontinuous fun tions de ned on a (not ne essarily ompa t) Hausdor spa e Q: Show that P := spanff ; : : : ; fn g; where the span is taken over R (or C ), is a Chebyshev spa e if and only if there exists no real (or omplex) measure on Q annihilating P whose support onsists of less than n + 1 points. Hint: Use Proposition 3.1.2. ut b℄ Let P be an n-dimensional linear subspa e of C (Q), the spa e of real(or omplex-) valued ontinuous fun tions de ned on a ompa t Hausdor spa e Q ontaining at least n points. The spa e P is a Chebyshev spa e if and only if for ea h f 2 C (Q); there is a unique best uniform approximation to f from P: 0

1

0

Proof. First suppose P is a Chebyshev spa e of dimension n, and p and p are best uniform approximations to some f 2 C (Q) from P . Then p := (p + p ) is also a best uniform approximation to f from P . As 1

2

3

1

2

1

2

388 A3. An Interpolation Theorem

f p ? P; Lemma A.3.2 yields that jf (x) p (x)j attains its maximum on Q at r points, x ; x ; : : : ; xr ; that support an annihilating measure for P; where r  n + 1 by part a℄. Note that 3

3

1

p (xj )

2

f (xj ) = p (xj )

1

2

f (xj ) ;

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

and hen e, sin e P is a Chebyshev spa e, p p = 0: Conversely, if P is not a Chebyshev spa e, then there exist n distin t points x ; x ; : : : ; xn in Q su h that the system of homogeneous linear equations 0 1 g (x ) : : : g (xn ) 0 a 1 B . .. .. C ..  . . . A  .. A ; an gn (x ) : : : gn (xn ) 1

1

2

2

1

1

1

1

1

where fg ; : : : ; gn g is a basis for P , has a nontrivial solution. Then also the homogeneous system formed with the transposed matrix has a nontrivial solution, so there exist onstants bi , not all zero, so that 1

n X i=1

Thus, with g :=

bi gi (xj ) = 0 ;

Pn

i=1 bi gi ;

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

we have

g(xj ) = 0 ;

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

We may assume, without loss, that kg kQ = 1: Some of the onstants a ; a ; : : : ; an may be zero; however, the set of indi es j for whi h aj 6= 0 is not empty. By Tietze's theorem there exists an f 2 C (Q) su h that kf kQ = 1 and a j2 : f (xj ) = j ; 1

2

jaj j

jg(x)j); we have

Setting h(x) := f (x)(1

h(xj ) = We laim that kh some p 2 P; then

jf (xj )

pkQ

j2 :

 1 for every p 2 P: Indeed, if kh

p(xj )j = jf (xj )j 2

aj

jaj j ;

2




pkQ < 1 for

2 Re p(xj )f (xj ) + jp(xj )j < 1

for every j 2 ; hen e Re(aj p(xj )) > 0 ;

j2 :

2

An Interpolation Theorem 389

Sin e aj = 0 if j 62 ; we have Re(aj p(xj )) = 0 for ea h j 62 : Thus 0

n X

Re  However, if p := n X j =1

Pn

i=1 i gi ;

aj p(xj ) =

n X j =1

j =1

1

aj p(xj )A > 0 :

then

aj

n X i=1

i gi (xj ) =

n X i=1

bi

n X j =1

ai gi (xj ) = 0 ;

whi h ontradi ts the previous inequality and shows that kh pkQ  1 for every p 2 P . Finally, for all  2 [0; 1℄, g is a best uniform approximation to h from P be ause

jh(x)

g(x)j  jf (x)j(1 jg(x)j) + jg(x)j  1 + ( 1)jg(x)j  1

for all x 2 Q; so the best uniform approximation to h not unique.

2 C (Q) from P

is

ut

E.4 Uni ity of the Extremal Fun tion. Assume the notation of Theorem A.3.3. Show that if P  C (Q) is an n-dimensional real Chebyshev spa e and r = n; then the extremal element p 2 P satisfying kp kQ = 1 and L(p ) = kLk is unique. The interesting relations of Theorem A.3.3 to the Riesz representation theorem, the Krein-Milman theorem, and the Hahn-Bana h theorem are dis ussed in Shapiro [71℄. E.5 Appli ations of the Interpolation Theorem. As before, let

D := fz 2 C : jz j < 1g

and

K := R (mod2) :

Prove the following statements. Ea h of them may be proven by hara terizing the extremal polynomial for the given inequality with the help of Theorem A.3.3. A detailed hint is given only to part a℄. a℄ Bernstein's Inequality.

jt0 ()j  nktkK ;

t 2 Tn ;  2 R :

Hint: Let  2 R be xed, and study the linear fun tional L(t) := t0 ( ), t 2 Tn . Observe that an extremal p in Lemma A.3.3 must satisfy jp(xi )j = 1 for ea h i = 1; 2; : : : ; r and r must equal 2n: Note that r  2n holds by 0

0

390 A3. An Interpolation Theorem

Theorem A.3.3, while the argument for r  2n is similar to the orresponding step in the proof of Theorem 7.1.7. Finally, show that the extremal element t satisfying L(t) = kLk is of the form

t() = os(n

for some  2 K: b℄ Markov's Inequality.

jp0 (1)j  n kpk 2

[

;

1 1℄

)

ut

p 2 Pn :

;

℄ Chebyshev's Inequality.

jp(x)j  jTn(x)j kpk

p 2 Pn ; x 2 R n [ 1; 1℄ ; where Tn is the Chebyshev polynomial of degree n de ned by (2.1.1). ;

[

1 1℄

;

d℄ Bernstein's Inequality.

jp(z )j  jz jn kpkD ;

p 2 Pn ; z 2 C n D :

e℄ Bernstein's Inequality.

jp0 (z )j  njz jn kpkD ; 1

p 2 Pn ; z 2 C n D :

Hint: Use Theorem 1.3.1 (Lu as' theorem). f℄ Riesz' Identity. There are real numbers ai with that

t0 () =

where

2n X

i=1

i :=

ai t( + i ) ; i

2

2

n

1

;

P2n

i=1

ut

jai j = n su h

t 2 Tn ;  2 R ; i = 1; 2; : : : ; 2n :

(This is, apart from the expli it determination of the number ai ; an identity dis overed by M. Riesz [14℄.) g℄ Show that in part f℄, 1 ai = ( 1)i ; i = 1; 2; : : : ; 2n : 4n sin (i =2) +1

2

h℄ Bernstein's Inequality in Lp . Z 2 0

jt0 ()jp d  np

Z 2 0

jt()jp d ;

t 2 Tn ; p  1 :

Hint: Use part f℄ and Jensen's inequality (see E.20 of Appendix 4).

ut

Arestov [81℄ shows that the inequality of part h℄ is valid for all p > 0: Golits hek and Lorentz [89℄ gives a simpler proof of this. i℄ Find all extremal polynomials in parts a℄ to e℄ and h℄.

An Interpolation Theorem 391

E.6 An Inequality of Szeg}o. If p 2 Pn and z ; z ; : : : ; z n are any equally spa ed points on the unit ir le D; then 1

2

2

kp0 kD  n max jp(zk )j : k n 1

2

Proof. See Frappier, Rahman, and Rus heweyh [85℄.

ut

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A4

Inequalities for Generalized Polynomials in Lp

Overview Many inequalities for generalized polynomials are given in this appendix. Of parti ular interest are the extensions of virtually all the basi inequalities to Lp spa es. The prin ipal tool is a generalized version of Remez's inequality.

Inequalities for Generalized Polynomials in Lp Generalized (nonnegative) polynomials are de ned by (A.4.1) and (A.4.3). The basi inequalities of Chapter 5 are extended to these fun tions by repla ing the degree with the generalized degree. The ru ial observation is that Remez's inequality extends naturally to this setting. This Remez inequality then plays a entral role in the extensions of the other inequalities. These generalizations allow for a simple general treatment of Lp inequalities, whi h is one main feature of this appendix. The fun tion (A:4:1)

f (z ) = j!j

m Y j =1

jz

zj jrj

with 0 < rj 2 R; zj 2 C ; and 0 6= ! 2 C is alled a generalized nonnegative (algebrai ) polynomial of (generalized) degree

Inequalities for Generalized Polynomials in Lp 393

(A:4:2)

N :=

m X j =1

rj :

The set of all generalized nonnegative algebrai polynomials of degree at most N is denoted by GAPN . The fun tion (A:4:3)

f (z ) = j!j

m Y

j =1

j sin((z

zj )=2)jrj

with 0 < rj 2 R; zj 2 C ; and 0 6= ! 2 C is alled a generalized nonnegative trigonometri polynomial of degree (A:4:4)

N :=

m

1X r : 2j j =1

The set of all generalized nonnegative trigonometri polynomials of degree at most N is denoted by GTPN : Throughout this se tion we will study generalized nonnegative polynomials restri ted to the real line. If the exponents rj in (A.4.1) or (A.4.3) are even integers, then f is a nonnegative algebrai or trigonometri polynomial, respe tively. Note that the lasses GAPN and GTPN are not linear spa es. Note also that if f 2 GAPN or f 2 GTPN is of the form (A.4.1) or (A.4.3), respe tively, with all rj  1; then the one-sided derivatives of f exist at every x 2 R with the same modulus, hen e jf 0 (x)j is well-de ned for every x 2 R: We use the notation jf 0 (x)j for f 2 GAPN or f 2 GTPN and x 2 R throughout this se tion with this understanding. If f 2 GAPN is of the form (A.4.1) or f 2 GTPN is of the form (A.4.3), where the zeros zj 2 C ; j = 1; 2; : : : ; m; are distin t, then rj is alled the multipli ity of zj in f: Our intention in this se tion is to extend most of the lassi al inequalities of Se tion 5.1 to generalized nonnegative polynomials. In addition, we prove Nikolskii-type inequalities for GAPN and GTPN : Theorem A.4.1 (Remez-Type Inequality for GAPN ). The inequality

kf k

[

holds for every f

;

1 1℄

p p !N  p2 + ps 2

s

2 GAPN and s 2 (0; 2) satisfying m(fx 2 [ 1; 1℄ : f (x)  1g)  2

s:

E.5 shows that this inequality is sharp. Note that if 0 < s  1; then

p p !N p2 + ps  exp(5N ps): 2

s

Throughout this se tion, as before, K := R (mod 2 ).

394 A4. Inequalities for Generalized Polynomials in Lp

Theorem A.4.2 (Remez-Type Inequality for GTPN ). The inequality



kf kK  exp N s + s  exp(4Ns) holds for every f 2 GTPN and s 2 (0; =2℄ satisfying (A:4:5) m(fx 2 [ ; ) : f (x)  1g)  2 s : 7

2

4

The inequality

p p !N kf kK  p2 + p ;

holds for every even f

 = 1 os(s=2)



2

2 GTPN

and s 2 (0; 2) satisfying (A.4.5).

We do not dis uss what happens when s 2 (=2; 2 ) in the general ase be ause the ase when s 2 (0; =2℄ is satisfa tory for our needs.

Proof of Theorem A.4.1. First assume that f 2 GAPN is of the form (A.4.1) with rational exponents rj = qj =q; where qj ; q 2 N : Let k 2 N be an integer. Then (restri ted to R) p := f kq 2 P kqN and 2

2

m (fx 2 [ 1; 1℄ : jp(x)j  1g)  2 s : Hen e Theorem 5.1.1 yields

kf k

[

;

1 1℄

= kpk

= kq) [ 1;1℄ 1 (2





T



kqN

2

2+s 2 s

Sin e by E.4, (A:4:6)





2+s lim T kqN k!1 2 s

1=(2kq)

2

=

1=(2kq)

:

p p !N p2 + ps ; s

2

the theorem is proved. The ase when the exponents rj > 0 are arbitrary real numbers an be easily redu ed to the already proved rational ase by a straightforward density argument. ut Theorem A.4.2 follows from Theorem 5.1.2 in exa tly the same way that Theorem A.4.1 follows from Theorem 5.1.1; see E.6. Theorem A.4.3 (Nikolskii-Type Inequality for GTPN ). Let  be a nonnegative nonde reasing fun tion de ned on [0; 1) su h that (x)=x is nonin reasing on [0; 1): Then there is an absolute onstant > 0 su h that 1

k(f )kLp K  ( (1 + qN )) =q =p k(f )kLq K for every f 2 GTPN and 0 < q < p  1: If (x) = x; then  e(4) (

)

1

1

1

(

)

1

1

:

Inequalities for Generalized Polynomials in Lp 395

Theorem A.4.4 (Nikolskii-Type Inequality for GAPN ). Let  be a nonnegative nonde reasing fun tion de ned on [0; 1) su h that (x)=x is nonin reasing on [0; 1): Then there is an absolute onstant > 0 su h that 2

k(f )kLp for every f

2 GAPN

;

[

1 1℄

 ( (2 + qN )) =q 2

2

2

=p

k(f )kLq

[

;

1 1℄

and 0 < q < p  1: If (x) = x; then

2

 e (2) 2

1

:

In the proof of the se ond part of Theorem A.4.4 we will need the following S hur-type inequality, whi h is interesting in its own right. Theorem A.4.5 (S hur-Type Inequality for GAPN ). The inequality

kf kq [

holds for every f

;

1 1℄

p

 e(1 + qN )

2 GAPN

1

x f q (x) 2

;

[

1 1℄

and q > 0:

A

ording to Theorem 5.1.9 (S hur's inequality), if N 2 N , f 2 PN ; and q 2 N; then the onstant e in the above inequality an be repla ed by 1: It is suÆ ient to prove Theorems A.4.3 and A.4.4 when p = 1, and then a simple argument gives the required results for arbitrary exponents 0 < q < p < 1: To see this, say, in the trigonometri ase, assume that there is a onstant CN su h that

k(f )kK  CN=q k(f )kLq K for every f 2 GTPN and 0 < q < 1: Then k(f )kpLp K = k((f ))p q q kL K  k(f )kpK q k(f )kqLq K  CNp=q k(f )kpLq qK k(f )kqLq K 1

(

)

+

(

1(

)

)

(

)

1

(

and therefore

)

(

)

;

k(f )kLp K  CN=q =p k(f )kLq K for every f 2 GTPN and 0 < q < p  1: Proof of Theorem A.4.3 (when p = 1). Sin e  is nonnegative and nonde reasing and ((x)=x)q is nonin reasing on [0; 1); we have ((f ()))q  exp( qNs)k(f )kqK 1

(

whenever

1

)

f ()  exp( Ns)kf kK :

Hen e, by E.7 b℄, we an dedu e that

(

)

396 A4. Inequalities for Generalized Polynomials in Lp

m (f 2 [ ; ) : ((f ()))q for every f

 exp(

qNs)k(f )kqK g) 

2 GTPN and s 2 (0; 2): Choosing s := (1 + qN )



m  2 [ ; ) : ((f ()))q

1

s

4

; we get




 e k(f )kqK  (4(1 + qN )) 1

1

:

Hen e, integrating only on the subset I of K where

 e k(f )kqK ;

((f ()))q

1

we on lude that Z

k(f )k  4e(1 + qN ) ((f ()))q d I  4e(1 + qN )k(f )kqLq K ; q K

(

)

and the rst part of the theorem is proved. Now we turn to the se ond statement. Let

D := fz 2 C : jz j < 1g

D := fz 2 C : jz j = 1g :

and

If h is analyti in the open unit disk D and ontinuous on the losed unit disk D; then by Cau hy's integral formula we have

jrz j )h(rz ) = 21i

(1

Z

2

D

whenever z 2 D and r 2 [0; 1): Note that u j1 rzuj = ju rz j for all r 2 [0; 1): Hen e, if P (1

r )jP  (rz )jq  2

rzu du u rz

1

h(u)

2 D and z 2 D imply 2 Pn and 0 < q < 1; then

Z

1 jP  (u)jq jduj 2 D

whenever z 2 D and r 2 [0; 1℄; where P  is obtained from the fa torization of P by repla ing ea h fa tor (z ) of P with jj < 1 by (1 z ): Sin e 1 2

(1 + r)jz

j  jrz

j ;

jj > 1 ;


q deg(P )

jP (z )jq  21

z 2 D ; r 2 [0; 1℄ ;

we have (1

r) 2

1 2

(1 + r)

Z

D

jP  (u)jq jduj

whenever z 2 D and r 2 [0; 1℄: Maximizing the left-hand side for r 2 [0; 1℄ and using the fa t that jP  (z )j = jP (z )j for z 2 D; we on lude that

Inequalities for Generalized Polynomials in Lp 397

P ))e jP (z )jq  (2 + q deg( 8

Z



jP (ei )jq d ;

Hen e, by E.8,

kRk  q K

(1 + qn)e 4

Z



z 2 D :

jR()jq d

for every R 2 Tn : If f 2 GTPN is of the form (A.4.3) with rational exponents rj = j =; where j ;  2 N ; then on applying the above inequality to R := f  2 T N with q repla ed by q=(2); we on lude that 2

2

kf kqK  1 +4qN

Z 



f ()q d

and the se ond statement of the theorem is proved. The ase when the exponents rj > 0; j = 1; 2; : : : ; m; are arbitrary real numbers an be redu ed to the already proved rational ase by a straightforward density argument. ut

2 Pn and

Proof of Theorem A.4.5. Let P M

p := 1

x jP (x)jq 2

[

;

1 1℄

:

By E.8 ℄, there exists an R 2 P n su h that jR(ei )j = jP ( os )j;  2 R: We de ne R 2 P n from the fa torization of R by repla ing all the fa tors (z ) of R with jj < 1 by (1 z ): Note that j1 e i j = 2j sin j and jR(ei )j = jR (ei )j for all  2 R: Hen e the maximum prin iple yields that 2

2

j1

2

(rz ) jjR (rz )jq 2

 zmax j1 z jjR (z )jq 2D = max 2 j sin jjP ( os )jq = 2M : 2R 2

By E.9 we have

jR (z )j 



 2n

2 1+r

jR (rz )j ;

z 2 D ; r 2 [0; 1℄ :

Hen e nq

jR (z )jq  (1 +2 r) nq 1 2

2

1

r

2

z 2 D ; r 2 [0; 1℄ ;

2M ;

where the minimum on [0; 1℄ of the right-hand side is taken at

r :=

qn

1 + qn

:

398 A4. Inequalities for Generalized Polynomials in Lp

Estimating the right-hand side at this value of r; we get

kP kq [

If f

;

1 1℄

2 GAPN

= max jR (z )jq z2D

p

 e(1 + qn)

x jP (x)jq

1

2

;

[

1 1℄

:

is of the form (A.4.1) with rational exponents rj = j =; 2P N with q

j ;  2 N ; then applying the above inequality to P = f repla ed by q=(2); we get

kf kq [

p

;

1 1℄

 e(1 + qN )

2

x f q (x)

1

2

2

;

;

[

1 1℄

and the theorem is proved. The ase when the exponents rj > 0 are arbitrary real numbers an be easily redu ed to the already proved rational

ase on e again by a standard approximation. ut

Proof of Theorem A.4.4 (when p = 1). Sin e  is nonnegative and nonde reasing and ((x)=x)q is nonin reasing on [0; 1); we have ((f (x)))q whenever

 exp(

p

qN s)k(f )kq [

p

f (x)  exp( N s)kf k

[

So by E.7 a℄ we an dedu e that

m

n

x 2 [ 1; 1℄ : ((f (x)))q

;

1 1℄

;

1 1℄

:

p

 exp(

qN s)k(f )kq [

o

;

1 1℄

 8s

for all s 2 (0; 2): Choosing s := (1 + qN ) ; we obtain 2



m x 2 [ 1; 1℄ : ((f (x)))q

 e k(f )k 1

;

[

1 1℄

[

1 1℄




 8(1 +1qN )

2

:

Integrating on the subset I of [ 1; 1℄ where ((f (x)))q

 e k(f )k 1

we on lude that

k(f )kq [

;

;

Z

;

1 1℄

 8e(1 + qN ) ((f (x)))q dx I  8e(1 + qN ) k(f )kqLq ; : 2

2

[

1 1℄

Thus the rst part of the theorem is proved. To show that the given onstant works in the ase that (x) = x we use another method. Let h 2 GAPM : Then by E.10, g () = h( os ) 2 GTPM : On using the substitution x = os ; from Theorem A.4.3 we get

Inequalities for Generalized Polynomials in Lp 399

khk

[

;

1 1℄

 e(2)

1

Z

(1 + qM )

1

1

jh(x)jq (1

=

x) 2

1 2

dx :

If f 2 GAPN ; then h(x) = f (x)(1 x ) = q 2 GAPM with M = N + q ; so an appli ation of the above inequality yields 2

(A:4:7)

p

1

x

2

f q (x)

[

;

1 1℄

1 (2 )

 e(2)

1

1

(2 + qN )

Z

1

f q (x) dx :

1

(Note that the weaker assumption h 2 GAPM instead of f 2 GAPN already implies (A.4.7).) Now a ombination of Theorem A.4.5 and inequality (A.4.7) gives that if (x) = x; then the inequality of Theorem A.4.4 holds with := e (2 ) : 2

2

1

ut

Now we prove extensions (up to multipli ative absolute onstants) of Markov's and Bernstein's inequalities for generalized nonnegative polynomials. Theorem A.4.6 (Bernstein-Type Inequality for GTPN ). There exists an absolute onstant > 0 su h that 3

kf 0kK  N kf kK 3

for every f

2 GTPN

of the form (A.4.3) with ea h rj

 1.

Theorem A.4.7 (Bernstein-Type Inequality for GAPN ). The inequality

jf 0 (x)j  p N kf k

x 2 ( 1; 1) ; ; ; x holds for every f 2 GAPN of the form (A.4.1) with ea h rj is as in Theorem A.4.6. 3

1

2

[

1 1℄

 1; where

3

Theorem A.4.8 (Markov-Type Inequality for GAPN ). There exists an ab-

solute onstant > 0 su h that kf 0k ; 4

[

for every f

2 GAPN

1 1℄

 N kf k 4

2

[

;

1 1℄

of the form (A.4.1) with ea h rj

 1:

To prove Theorem A.4.6 we need the following lemma. Lemma A.4.9. Suppose g 2 GTPN is of the form (A.4.3) with ea h zj

2 R; and suppose at least one of any two adja ent (in K) zeros has multipli ity at least 1. Then there exists an absolute onstant > 0 su h that for every su h g there is an interval I  K of length at least N for whi h 5

5

min g ()  e 2I

1

kgkK :

1

400 A4. Inequalities for Generalized Polynomials in Lp

Proof. Take a g 2 GTPN satisfying the hypothesis of the lemma. Be ause of the periodi ity of g we may assume that g() = kgkK :

(A:4:8) De ne (A:4:9)



Qn;! () := T

2

n

sin(=2) sin(!=2)



with

n := bN

(A:4:10)

and

! := 

(3N )

1

;

where T n is the Chebyshev polynomial of degree 2n de ned by (2.1.1). By E.11 there exists an absolute onstant > 1 su h that Qn;! ( )  : Introdu e the set 2

6



A :=  2 [

6



(3N ) ;  + (3N ) ℄ : g ()  e g ( ) : 1

We study h := g jQn;! j sumption (A.4.8) yield

1

1

2 GTP N : The inequality Qn;! ()  2

6

h()  g()  Qn;! ()g() = 1

6

1

6

and as-

khkK

for all  2 [ !; ! ℄ = [  + (3N ) ;  (3N ) ℄: Further, the de nition of the set A; the fa t that kQn;! kK = Qn;! ( ); and (A.4.8) imply that 1

1

h()  e g()Qn;! () = e 1

1

khkK

for all  2 [ (3N ) ;  + (3N ) ℄ n A: From the last two inequalities we

on lude that 1

h() 

1

1

khk

;

[

7

1 1℄

for all  2 [ ;  ℄ n A ;

where := minf ; eg > 1 is an absolute onstant. Therefore, by E.7 b℄ 7

6

m(A)  N 8

1

with := 8

1 17

log > 0 : 7

Sin e g 2 GTPN is of the form (A.4.3) with ea h zj 2 R; and at least one of any two adja ent zeros of g has multipli ity at least 1, E.12 and assumption (8.1.8) imply that g annot have two or more distin t zeros in [ (3N ) ;  + (3N ) ℄: Hen e A is the union of at most two intervals. Therefore there exists an interval I  A su h that m(I )  (2N ) ; and the lemma is proved. ut 1

1

8

1

Inequalities for Generalized Polynomials in Lp 401

Proof of Theorem A.4.6. Let f 2 GTPN be of the form (A.4.3) with ea h rj  1: Without loss of generality we may assume that jf 0()j = kf 0kK ;

and it is suÆ ient to prove only that

jf 0 ()j  N kf kK : 3

By E.13 we may assume that ea h zj is real in (A.4.3). Hen e, by E.2, g := jf 0 j satis es the assumption of Lemma A.4.9. Denote the endpoints of the interval I oming from Lemma A.4.9 by a < b: We an now dedu e that

kf 0kK = jf 0 ()j  b e a 

eN

5

Z b

a

Z b

a

jf 0 ()j d

jf 0()j d = eN jf (b)

f (a)j  N kf kK 3

5

ut

with := e , and the proof is nished. 1

3

5

Proof of Theorem A.4.7. The theorem follows from Theorem A.4.6 by using the substitution x = os  and E.10 b℄. ut Proof of Theorem A.4.8. Let  := 1 (1 + N ) . Using Theorem A.4.7 and 2

then E.14, we obtain

kf 0k

[

;℄

 N (N + 1)kf k ;  N (N + 1) kf k ;  N kf k 3 3

[

9

1 1℄ [

4

℄

2

[

;℄ ;

and then the theorem follows by a linear transformation.

ut

Now we establish Remez-, Bernstein-, and Markov-type inequalities for generalized nonnegative polynomials in Lp : In the proofs we use the inequalities proved in this appendix so far, and the methods illustrate how one an ombine the \basi " inequalities in the proofs of various other inequalities for generalized nonnegative polynomials. First we state the main results. Theorem A.4.10 (Lp Remez-Type Inequality for GAPN ). Let  be a nonnegative nonde reasing fun tion de ned on [0; 1) su h that p (x)=x is nonin reasing on [0; 1): There exists an absolute onstant  5 2 su h that Z

1

1

p





((f (x)))p dx  1 + exp pN s

Z

A

((f (x)))p dx

for every f 2 GAPN , A  [ 1; 1℄ with m([ 1; 1℄ n A)  s  1=2; and for every p 2 (0; 1):

402 A4. Inequalities for Generalized Polynomials in Lp

Theorem A.4.11 (Lp Remez-Type Inequality for GTPN ). Let  be a nonnegative nonde reasing fun tion de ned on [0; 1) su h that (x)=x is nonin reasing on [0; 1): There exists an absolute onstant  8 su h that Z 



((f ()))p d  (1 + exp( pNs))

Z

A

((f ()))p d

for every f 2 GTPN , A  [ ; ℄ with m([ ; ℄ n A)  s  =2; and for every p 2 (0; 1): Theorem A.4.12 (Lq Bernstein-Type Inequality for GTPN ). Let  be a nonnegative, nonde reasing, onvex fun tion de ned on [0; 1). There exists

an absolute onstant su h that Z 



for every f q 2 (0; 1℄:

N

2 GTPN

q

jf 0 ()jq




d 

Z 



( f ()q ) d

of the form (A.4.3) with ea h rj

 1, and for every

Corollary A.4.13 (Lp Bernstein-Type Inequality for GTPN ). The inequal-

ity

Z 

Z 

jf 0 ()jp d  p N p +1





jf ()jp d

holds for every f 2 GTPN of the form (A.4.3) with ea h rj every p 2 (0; 1); where is as in Theorem A.4.12.

 1; and for

Theorem A.4.14 (Lp Markov-Type Inequality for GAPN ). There exists an

absolute onstant su h that Z

1

1

for every f p 2 (0; 1):

2 GAPN

jf 0 (x)jp dx  p

+1

N

2

p

Z

1

1

jf (x)jp dx

of the form (A.4.1) with ea h rj

 1; and for every

Theorems A.4.1 and A.4.2 an be easily obtained from their L1 analogs, Theorems A.4.1 and A.4.2, respe tively; see E.15 and E.16, where hints are given.

Proof of Theorem A.4.12. For n := bN let Dn () :=

n X

j= n



eij

be the modulus of the nth Diri hlet kernel. Choose q 2 (0; 1℄; and set m := 2q  2: Let g 2 GTPN be of the form (A.4.3) with ea h rj  1: On applying the Nikolskii-type inequality of Theorem A.4.3 to 1

Inequalities for Generalized Polynomials in Lp 403

G := gDnm 2 GTPN

nq

;

1

+2

we obtain (A:4:11)





kgDnm kqK  1 + q N + 2nq kgDnm kqLq K Z   (1 + 3N ) (g()Dnm ())q d 1

1

(

1

)



Z 

= (1 + 3N ) 1



gq ()Dn () d : 2

If g 2 GTPN is of the form (A.4.3) with ea h rj  1; then m  2 implies that G 2 GTPN is of the form (A.4.3) with ea h rj  1 as well. If we apply the Bernstein-type inequality of Theorem A.4.6 to G and use (A.4.11), we

an dedu e that 0 g ()Dm () + mDm

q

()Dn0 ()g () K

q

N + 2nq kgDnm kqK

n

1

n

  q N q

1

3

1 + 2q

3

1


q

(1 + 3N )

Z 

1



gq ()Dn () d 2

for every  2 K (we take one-sided derivatives everywhere). By putting  = 0; and noti ing that

Dn0 (0) = 0 and Dn(0)m = (2n + 1)

2

=q

N

=q ;

2

we get

jg0 (0)jq  N q

(A:4:12)

Z 



gq ()(2) (2n + 1) Dn () d 1

1

2

with an absolute onstant : Now let f 2 GTPn be of the form (A.4.3) with ea h rj  1: Let  2 K be xed. On applying (A.4.12) to g () := f ( +  ); we on lude that

jf 0 ( )jq  N q

Z 

Hen e (A:4:13) Sin e (A:4:14)

N

q

jf 0 ( )jq  Z 



f q ()(2) (2n + 1) Dn ( 1



Z 



1

f q ()(2) (2n + 1) Dn ( 1

(2 ) (2n + 1) Dn ( 1

 ) d :

2

1

2

1

 ) d = 1 ;

2

 ) d :

404 A4. Inequalities for Generalized Polynomials in Lp

Jensen's inequality (see E.7) and (A.4.3) imply that

N

q




jf 0 ( )jq 

Z 



( f q ())(2) (2n + 1) Dn ( 1

1

 ) d :

2

If we integrate both sides with respe t to ; Fubini's theorem and (A.4.14) (on inter hanging the role of  and  ) yield the inequality of the theorem.

ut

Proof of Corollary A.4.13. If 0 < p  1; then Theorem A.4.12 yields the

orollary with q = p and (x) = x: If 1  p < 1; then the orollary follows from Theorem A.4.12 again with q = 1 and (x) = xp . ut Proof of Theorem A.4.14. We distinguish two ases. Case 1: p  1: Let f 2 GAPN be of the form (A.4.1) with ea h rj  1: Then by E.10 b℄, g () := f ( os ) 2 GTPN is of the form (A.4.3) with ea h rj  1: With the substitution x = os ; Corollary A.4.13 and Theorem A.4.11 imply that (A:4:15)

Z

1

1

jf 0 (x)jp (1

x )p

=

2 (

 p N p

Z

1

1

1) 2

dx

f p (x)(1

2

1

 p N p exp( pNN

1

2

1

=

x) )

1 2

Z Æ

Æ

dx

f p (x)(1

x) 2

=

1 2

dx ;

where Æ := maxf1 N ; os(=16)g and and are appropriate absolute

onstants. Sin e p 1  0; it follows from (A.4.15) that 2

(A:4:16)

Z Æ

Æ

1

2

jf 0 (x)jp dx Z Æ

jf 0 (x)jp (1

 (1

Æ)

p)=2

 (1

Æ)

p)=2 p N p exp(

2 (1

2 (1

 p N p 3

= p N p 2

3

Æ

1

2

1

N pN

Z Æ

Æ

Z Æ

Æ

p)(1

x )p 2 (

Æ) 2

=

1) 2

=

1 2

dx Z Æ

Æ

f p (x) dx

f p (x) dx

f p (x) dx ;

where is also an absolute onstant. Sin e (A.4.16) is valid for every f 2 GAPN of the form (A.4.1) with ea h rj  1; the theorem follows by a linear shift from [ Æ; Æ ℄ to [ 1; 1℄: 3

Inequalities for Generalized Polynomials in Lp 405

Case 2: 0 < p  1: Let f 2 GAPN be of the form (A.4.1) with ea h rj  1: Using the inequality ja + bjp  jajp + jbjp for p 2 (0; 1℄; we an dedu e that Z

(A:4:17)

jf 0 ( os )jj sin j =p 1

A

Z AZ



+

+1


p

d p =p
0 d

f ( os )j sin j

1

f ( os )p

A

1




j sin j =p j os j p d 1

1

for every measurable subset A of [ ;  ): Applying Theorem A.4.12 (with (x) = x) to g() := f ( os )j sin j =p 2 GTPN =p ; 1

+1

then using (A.4.17) with A := [ Æ; Æ ℄, Æ := 1 the substitution x = os , that Z Æ

(A:4:18)

Æ

jf 0 (x)jp (1

 (N

2

x )p= dx 2

+ 1=p)p

1

(N + 1) ; we on lude, by

Z

2

1

f p (x) dx + p p

Z Æ

1

Æ

f p (x)(1

x) 2

p=2 dx ;

where is an absolute onstant. Note that Theorem A.4.10, 0 < p and the hoi e of Æ imply that 1

Z

(A:4:19)

1

Z Æ

f p (x) dx 

2

1

Æ

 1,

f p(x) dx

with an absolute onstant . A ombination of (A.4.18) and (A.4.19) yields 2

(A:4:20)

Z Æ

Æ

jf 0 (x)jp dx Z Æ

 (1

Æ)

p=2

 (1

Æ)

p=2 (

2

2

Æ 1

jf 0 (x)jp (1

(N 2

x )p= dx 2

+ 1=p)p + p p (1

 2p= (N + 1)p ( (N + 1=p)p + p 2

 N 3

1

p

2

Z Æ

Æ

2

2

Æ) 2

p 2p=2 (N

p=2 )

Z Æ

+ 1)p )

Æ

f p (x) dx

Z Æ

Æ

f p (x) dx

f p (x) dx ;

where is an absolute onstant. Sin e (A.4.20) is valid for every f 2 GAPN of the form (A.4.1) with ea h rj  1; the theorem follows by a linear shift from [ Æ; Æ ℄ to [ 1; 1℄: ut 3

406 A4. Inequalities for Generalized Polynomials in Lp

Comments, Exer ises, and Examples. Most of the results of this se tion an be found in Erdelyi [91a℄ and [92a℄; Erdelyi, Mate, and Nevai [92℄; and Erdelyi, Li, and Sa [94℄; however, the proofs are somewhat simpli ed here. For polynomials f 2 Pn and for arbitrary q 2 (0; 1); Theorem A.4.5 was also proved by Kemperman and Lorentz [79℄. An early version of Markov's inequality in Lp for ordinary polynomials is proven in Hille, Szeg}o, and Tamarkin [37℄. Weighted Markov- and Bernstein-type analogs of Theorems A.4.6 to A.4.8 are obtained in Erdelyi [92b℄. Appli ations of the inequalities of this se tion are given in Erdelyi, Magnus, and Nevai [94℄ and in Erdelyi and Nevai [92℄, where bounds are established for orthonormal polynomials and related fun tions asso iated with (generalized) Ja obi weight fun tions. Further appli ations in the theory of orthogonal polynomials may be found in Freud [71℄ and Erdelyi [91d℄. Lp extensions of Theorem 5.1.4 (Bernstein's inequality) and Theorem 5.1.8 (Markov's inequality) have been studied by a number of authors. The sharp Lp version of Bernstein's inequality for trigonometri polynomials was rst established by Zygmund [77℄ for p  1: Using an interpolation formula of M. Riesz, he proved that Z 

(A:4:21)



jt0 ()jp d  np

Z 



jt()jp d

for every t 2 Tn (see E.5 h℄ of Appendix 4). For 0 < p < 1; rst Klein [51℄ and later Osval'd [76℄ proved (A.4.21) with a multipli ative onstant (p): Nevai [79a℄ showed that (p)  8p : Subsequently, Mate and Nevai [80℄ showed the validity of (A.4.21) with a multipli ative absolute onstant, and then Arestov [81℄ proved (A.4.21) (with the best possible onstant 1) for every 0 < p < 1: Golits hek and Lorentz [89℄ gave a very elegant proof of Arestov's theorem. The Lp analog of Markov's inequality states that 1

(A:4:22)

Z

1

1

jQ0 (x)jp dx  p

+1

n

2

p

Z

1

1

jQ(x)jp dx

for every Q 2 Pn and 0 < p < 1; where is an absolute onstant. This

an be obtained from Arestov's theorem similarly to the way that Theorem A.4.14 is proven from Corollary A.4.13. To nd the best possible onstant in (A.4.22) is still an open problem even for p = 2 or p = 1: The magnitude of (A:4:23)

kf 0wk kfwk

[

[

; 1;1℄

1 1℄

;

Inequalities for Generalized Polynomials in Lp 407

jf 0 (y)w(y)j ; kfwk ;

(A:4:24)

[

1  y  1;

1 1℄

2 Pn

and their orresponding Lp analogs for f weight fun tions (A:4:25)

w (z ) =

k Y j =1

jz

zj jrj ;

zj

and generalized Ja obi

2C;

rj

2(

1; 1)

have been examined by several people. See, for example, Ditzian and Totik [88℄, Khalilova [74℄, Konjagin [78℄, Lubinsky and Nevai [87℄, Nevai [79a℄ and [79b℄, and Protapov [60℄, but a multipli ative onstant depending on the weight fun tion appears in these estimates. The magnitude of (A.4.23), (A.4.24), and their Lp analogs are examined in Erdelyi [92b℄ and [93℄, when both f and w are generalized nonnegative polynomials. In these inequalities only the degree of f , the degree of w, and a multipli ative absolute onstant appear. The most general results are the following: Theorem A.4.15. There exists an absolute onstant su h that Z 



and

jf 0 ()jp w() d  p kf 0 wk

[

;℄

+1

(N + M )p (Mp

1

+ 1)p

Z 



 (N + M )(M + 1)kfwk

[

jf ()jp w() d

;℄

for every f 2 GTPN of the form (A.4.3) with ea h rj w 2 GTPM ; and for every p 2 (0; 1):



1; for every

Theorem A.4.16. There exists an absolute onstant su h that Z

1

1

jf 0 (x)jp w(x) dx  p

and

kf 0wk

[

+1

(N + M ) p (Mp 2

1

+ 1) p

Z

1

2

1

;

1 1℄

 (N + M ) kfwk 2

[

;

1 1℄

for every f 2 GAPN of the form (A.4.1) with ea h rj w 2 GAPM ; and for every p 2 (0; 1):

jf (x)jp w(x) dx 

1; for every

E.1 Another Representation of Generalized Nonnegative Polynomials. Q a℄ Show that if f = kj jQj jrj with ea h Qj 2 Pnj and rj > 0; then P Q f 2 GAPN with N  kj rj nj : Similarly, if f = kj jQj jrj with ea h P Qj 2 Tnj and rj > 0; then f 2 GTPN with N  kj rj nj : =1

=1

=1

=1

Q

rj = b℄ Show that if f 2 GAPN is of the form (A.4.1), then f = m j jQj j with ea h Qj 2 P Qand 0  Qj on R: Similarly, if f 2 GTPN is of the form rj = with ea h Q 2 T and 0  Q on R: (A.4.3), then f = m j j j jQj j 2

=1

2

2

=1

1

408 A4. Inequalities for Generalized Polynomials in Lp

E.2 The Derivative of an f 2 GTPN with Real Zeros. Show that if f 2 GTPN is of the form (A.4.3) with ea h rj  1 and zj 2 R; then jf 0 j 2 GTPN is of the form (A.4.3) with ea h rj > 0 and zj 2 R; and at least one of any two adja ent (in K ) zeros of jf 0 j has multipli ity exa tly 1: E.3 Generalized Nonnegative Polynomials with Rational Exponents. Show that if f 2 GAPN is of the form (A.4.1) with rational exponents rj = qj =q , where qj ; q 2 N ; then f q 2 P qN ; while if f 2 GTPN is of the form (A.4.3) with rational exponents rj of the above form then f q 2 T qN : 2

2

2

2

E.4 Proof of (A.4.6). Prove (A.4.6) from the expli it formula (2.1.1) for the Chebyshev polynomial Tn : E.5 Sharpness of the Remez-Type Inequality for GAPN . Let

fn(x) := Tn



2x + s 2 s

 N=n

2 GAPN ;

n = 1; 2; : : : :

Show that

m(fx 2 [ 1; 1℄ : fn (x)  1g) = m([ 1; 1 s℄) = 2 s and lim f (1) n!1 n

p p !N p2 + ps :

=

s

2

The upper bound in Theorem A.4.1 is a tually not a hieved by an element of GAPN ; see Erdelyi, Li, and Sa [94℄. E.6 Proof of Theorem A.4.2. Hint: First assume that f 2 GTPN is of the form (A.4.3) with rational exponents rj = qj =q; where qj ; q 2 N : Then p := f q 2 T qN ; and the desired inequality an be obtained from Theorem 5.1.2 as in the proof of Theorem A.4.1. ut 2

2

E.7 Corollaries of Theorems A.4.1 and A.4.2. a℄ The inequality

p



m x 2 [ 1; 1℄ : f (x)  exp( N s)kf k

[

;




1 1℄

2 GAPN and 0 < s < 2. In parti ular, 
m x 2 [ 1; 1℄ : f (x)  e kf k ;  (8N holds for every f 2 GAPN .

 8s

holds for every f

1

[

1 1℄

2

+ 4)

1

Inequalities for Generalized Polynomials in Lp 409

b℄ The inequality

m (f 2 [ ; ) : f ()  exp( Ns)kf kK g)  holds for every f

2 GTPN

and 0 < s < 2. In parti ular,



m  2 [ ; ) : f ()  e holds for every f

s

4

1




kf kK  (4(N + 1))

1

2 GTPN :

E.8 Nonnegative Trigonometri Polynomials. Part a℄ restates E.3 ℄ of Se tion 2.4. Parts b℄ and ℄ dis uss simple related observations. a℄ Let t 2 Tn be nonnegative on R: Show that there is a p 2 Pn su h that t() = jp(ei )j for every  2 R: b℄ Let p 2 Pn and t() := jp(ei )j for every  2 R: Then t 2 Tn and t is nonnegative on R:

℄ Show that if t 2 Tn ; then there is a p 2 P n su h that jt()j = jp(ei )j for every  2 R: 2

2

2

E.9 A Cru ial Inequality in the Proof of Theorem A.4.5. Show that

jP (z )j 



 2 n jP (rz )j ; 1+r

z 2 D ; r 2 [0; 1℄

for every P 2 Pn having all its zeros outside the open unit disk D: Q Hint: Let P (z ) = m j (z zj ); where 2 C ; zj 2 C n D; and m Show that =1

jz

zj j 

2 jrz 1+r

zj j ;

 n:

j = 1; 2; : : : ; m ; r 2 [0; 1℄ :

ut E.10 f 2 GAPN Implies f ( os ) 2 GTPN . a℄ Show that if f 2 GAPN ; then g () = f ( os ) 2 GTPN : b℄ Show that if f 2 GAPN is of the form (8.1.1) with ea h rj  1; then g() = f ( os ) 2 GTPN is of the form (A.4.3) with ea h rj  1: E.11 A Property of Qn;! . Let Qn;! be de ned by (A.4.9) and (A.4.10). Show that there is an absolute onstant > 1 su h that Qn;! ( )  : Hint: Use the expli it formula (2.1.1) for the Chebyshev polynomial Tn . u t 6

6

410 A4. Inequalities for Generalized Polynomials in Lp

E.12 A Property of the Zeros of a g 2 GTPN . Assume that g 2 GTPN is of the form (A.4.3) and let

M :=

m X j =1 zj 2[ ;+℄

rj :

Show that M  3Nkg k ; : Hint: First assume that ea h rj is rational with ommon denominator q 2 N ; and apply E.12 of Se tion 5.1 to p := g q 2 T qN : ut 1

[

1 1℄

2

2

E.13 Extremal Fun tions for the Bernstein-Type Inequality. a℄ Let rj  1; j = 1; 2; : : : ; m; be xed real numbers. Show that there exists an fe 2 GTPN of the form (A:4:26)

fe(z ) =

m Y

j =1

su h that

j sin((z

ej )=2)jrj ;

ej

2C

jfe0 ()j = sup jf 0 ()j ; kfekK f kf kK

where the supremum in the right-hand side is taken for all f the form

f (z ) = j!j

(A:4:27)

m Y

j =1

j sin((z

zj )=2)jrj ;

zj

2C;

2 GTPN

of

0 6= ! 2 C :

Hint: Write ea h f of the form (A.4.27) for whi h the supremum is taken

as

f (z ) = j!

0

j

m Y j =1

j!j sin((z

zj )=2) sin((z

zj )=2)jrj = ; 2

where the numbers !j > 0 are de ned by

k!j sin((z

zj )=2) sin((z

z j )=2)kK = 1 ;

j = 1; 2; : : : ; m ;

and then use a ompa tness argument for ea h fa tor separately. ut b℄ Let fe be as in part a℄. Show that ea h zero of fe is real, that is, ej 2 R for ea h j in (A.4.26). Hint: Assume that there is an index 1  j  m su h that j 2 C n R: Then

fe(z )

:=

fe(z )

1

 sin ((z )=2) sin((z j )=2) sin((z  j )=2) 2

!rj

2 GTPN

e with suÆ iently small  > 0 ontradi ts the maximality of f:

ut

Inequalities for Generalized Polynomials in Lp 411

E.14 A Corollary of the Remez-Type Inequality for GAPN . Suppose that  := 1 (1 + N ) : Show that there is an absolute onstant > 0 su h 2

that

9

kf k

[

;

1 1℄

 kf k 9

;℄

[

for every f 2 GAPN : Hint: This is a orollary of Theorem A.4.1.

ut

E.15 Proof of Theorem A.4.10.

2 GAPN , let

Outline. For f

p



I (f ) := x 2 [ 1; 1℄ : ((f (x)))p  exp( 5pN 2s)k(f )kp [

;



1 1℄

:

From Theorem 5.1.1, 0 < s  1=2; and the assumptions pres ribed for  it follows that m(I (f ))  2s: Let I := A \ I (f ): Sin e m([ 1; 1℄ n A)  s; m(I )  s: Therefore Z [

nA

;

1 1℄

((f (x)))p dx



Z [

;

nA

1 1℄

 exp 5pN  exp 5pN

k(f )kp p
Z

;

[

2s

p




2s

ZI

1 1℄

dx

((f (x)))p dx

A

((f (x)))p dx :

ut

E.16 Proof of Theorem A.4.11. Hint: For f 2 GTPN ; let 



I (f ) :=  2 [ ; ℄ : ((f ()))p  exp( 8pNs)k(f )kpK : From Theorem 5.2.2, 0 < s  =4; and the assumptions for ; it follows that m(I (f ))  2s: Now nish the proof as in the hint for E.1. ut E.17 Sharpness of Theorem A.4.10. a℄ Show that there exists a sequen e of polynomials Qn absolute onstant > 0 su h that Z

1

1

jQn (x)jp dx  s 1 + exp pnps





Z

1

1

s

2 Pn

and an

jQn (x)jp dx

for every n 2 N ; s 2 (0; 1℄; and p 2 (0; 1): Hint: Study the Chebyshev polynomials Tn transformed linearly from [ 1; 1℄ to [ 1; 1 s℄: ut

412 A4. Inequalities for Generalized Polynomials in Lp

b℄ Show that there exist a sequen e of trigonometri polynomials tn 2 Tn and an absolute onstant > 0 su h that Z 



jtn j

() p d

 s(1 + exp( pns))

Z  s=2

+s=2

for every p 2 N ; s 2 (0;  ℄; and p 2 (0; 1): Hint: Study Qn;! de ned by (A.4.9) with ! := 

jtn ()jp d ut

s=2:

E.18 Sharpness of Corollary A.4.13 and Theorem A.4.14. a℄ Find a sequen e of trigonometri polynomials tn 2 Tn that shows the sharpness of Corollary A.4.13 up to the onstant > 0 for every p 2 (0; 1) simultaneously. Hint: Take tn () := os n: ut b℄ For every p 2 (0; 1); nd a sequen e of polynomials Qn;p 2 Pn whi h shows the sharpness of Theorem A.4.14 up to the onstant > 0:

Outline. Let Lk 2 Pk be the kth orthonormal Legendre polynomial on [ 1; 1℄ (see E.5 of Se tion 2.3), and let Gm (x) := where is hosen so that

Z

(A:4:28)

1

m X k=0

L0k (1)Lk (x) ;

Gm (x) dx = 1 : 2

1

Show that there exist absolute onstants > 0 and > 0 (independent of m) su h that 1

2

jGm (1)j  m and jG0m (1)j  m : For a xed n 2 N ; let u := b2=p + 1 and m := bn=u : If m  1; then let Qn;p := Gum 2 Pn ; otherwise let Qn;p(x) := x 2 Pn : If m = 0; then the

al ulation is trivial, so let m  1: Using (A.4.29) and the Nikolskii-type (A:4:29)

1

3

2

inequality of Theorem A.4.4, show that there exists an absolute onstant

> 0 su h that 3

Z

(A:4:30)

1

jQ0n;p (x)jp dx  p 3

1

+1

mu (

p (1 + pn)

+2)

2

:

Use the inequality Z

1

1

jQn;p (x)jp dx =

Z

1

1

jGm (x)jup dx  kGm kup ; [

Z

1

2

1 1℄

Gm (x) dx ; 2

1

Inequalities for Generalized Polynomials in Lp 413

the Nikolskii-type inequality of Theorem A.4.4, and (A.4.28) to show that there is an absolute onstant > 0 su h that 4

Z

(A:4:31)

1

jQn;p (x)jp dx  p

+1

4

1

mup

2

:

Finally, ombine (A.4.30) and (A.4.31). Note that if p > 2; then m = bn ; while if p  2; then p(n 1)  m  2pn: Note also that pp is between two positive bounds for p 2 (0; 2℄: ut 1

4

E.19 Sharpness of Theorems A.4.3 and A.4.4. a℄ Let q 2 (0; 1) be xed. Show that there exists a sequen e of polynomials Qn;q 2 Pn and an absolute onstant > 0 su h that

kQn;q k

;

[

1 1℄



=q (1 + qn)2=q

1+1

kQn;q kLq

;

[

1 1℄

for every n 2 N : Hint: Study Qn;p with p = q in the hint to the previous exer ise. ut b℄ Let Qn;q 2 Pn be the same as in part a℄. Show that there exist absolute

onstants > 0 and > 0 su h that 1

kQn;q kLp

2

[



1+1

;

1 1℄

=q 1=p

1

2

(1 + qn) =q (1 + pn) =p kQn;q kLp 2

2

;

[

1 1℄

for every n 2 N and 0 < q < p  1: Hint: Combine part a℄ and the Nikolskii-type inequality of Theorem A.4.4.

ut

℄ Let q 2 (0; 1) be xed. Show that there exists a sequen e of trigonometri polynomials tn;q 2 Tn and an absolute onstant > 0 su h that

ktn;q kK 

=q (1 + qn)1=q

1+1

ktn;q kLq K (

)

for every n 2 N : Hint: Let tn;q () := Qn;q ( os ); where Qn;q are the same as in part a℄. Use part a℄ and the S hur-type inequality of Theorem A.4.5. ut d℄ Let tn;q 2 Tn be the same as in part ℄. Show that there exist absolute

onstants > 0 and > 0 su h that 1

2

ktn;q kLp K 

=q 1=p

1+1

(

)

1

2

(1 + qn) =q (1 + pn) =p ktn;q kLq K 1

1

(

)

for every n 2 N and 0 < q < p  1: Hint: Combine part a℄ and the Nikolskii-type inequality of Theorem A.4.3.

ut

414 A4. Inequalities for Generalized Polynomials in Lp

E.20 Jensen's Inequality. Let  be a real-valued onvex fun tion de ned on [0; 1); and let f and w be nonnegative Riemann integrable fun tions R on the interval [a; b℄; where ab w = 1: Show that



Z b

a

!



f (x)w(x) dx

Z b

a

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

Hint: First assume that w is a step fun tion. Use the fa t that the onvexity of  implies its ontinuity, and hen e the fun tions (f )w and fw are

ut

Riemann integrable.

E.21 A Pointwise Remez-Type Inequality for GAPN . a℄ Show that there exists an absolute onstant > 0 su h that (

jp(y)j  exp

n min

p

)!

s

y

1

2

p ; s

y 2 [ 1; 1℄

;

for every p 2 Pn and s 2 (0; 1℄ satisfying

m(fx 2 [ 1; 1℄ : jp(x)j  1g)  2 s ; that is, with the notation of Theorem 5.1.1, for every p s 2 (0; 1℄:

2 Pn (s)

with

Proof. Assume, without loss of generality, that y 2 [0; 1℄: Let a := y + (1 y );

 := ar

os a ;

1 2

:= 2 ar

os y

ar

os a ;

Qn;! () := T



n

2

! :=  sin(=2) sin(!=2)

1



2

(

) ;

;

where T n is the Chebyshev polynomial of degree 2n de ned by (2.1.1), and let
Qn;; () := Qn;! ( (2 ( + ))) : 2

1

2

Asso iated with p 2 Pn (s); we introdu e

g() := p( os )Qn;; () 2 T n : 2

Obviously and

jQn;; ()j  1 ; kQn;; k

; ℄

[0 2

= Qn;;

 2 [0; 2) n (; ) 1 2




( + ) = Qn;! ( ) :

Inequalities for Generalized Polynomials in Lp 415

The de nition of ! implies that there exist absolute onstants > 0 and

> 0 su h that 1

2

p

1

y

1

2



!

p

y :

1

2

2

This, together with E.3 ℄, yields that there are absolute onstants > 0 and > 0 su h that 3

4

p

Qn;! ()  exp n 1 y 3

whenever y 2 [0; 1

2




 exp 5nps




s℄. It follows now from Theorem 5.1.1 that 4

jg()j  exp 5nps  Qn;! ()


for every  2 [0; 2 ) n (; ) and y 2 [0; 1

s℄: Furthermore 4

jg()j  Qn;! () for every  2 (; ) for whi h jp( os )j  1: Note that

j os j  1 and hen e p 2 Pn (s) with s

onstant > 0 su h that

1 2

(1

y) ;

 2 (; )

2 (0; 1℄ implies that there exists an absolute

5

m(f 2 (; ) : jp( os )j  1g)  Therefore

f := (Qn;! ()) g 2 T 1

satis es

2

s 5

p

2

:

n

m(f 2 [0; 2) : jf ()j  1g)  2

with

y

1

s

se

; y 2 [0; 1 s℄ : y Applying Theorem 5.1.2 with f and se; we on lude that se :=



jp(y)j = p =f

os 1 2

5

p

1

1 2

4

2

( + )

( + )








= (Qn;! ( )) g 1

 exp(4nse)  exp

1 2

( + )

s




!

p

5

1

y

2

whenever se 2 (0; =2℄ and y 2 [0; 1 s℄: If s 2 (0; 1℄; but se > =2 or y 2 [1 s; 1℄; then Theorem 5.1.1 yields the required inequality. ut 4

4

416 A4. Inequalities for Generalized Polynomials in Lp

b℄ Show that the inequality of part a℄ is sharp up to the absolute onstant

> 0: Hint: Assume, without loss, that y 2 [0; 1℄: Show that there exists an absolute onstant > 0 su h that the polynomials 1

Wn;y;s (x) := Tn



2

2x

x

+

2



s

2 Pn (s) ;

s

where Tn is the Chebyshev polynomial of degree n de ned by (2.1.1), satisfy

jWn;y;s (y)j  exp( nps) s=2; 1℄; and s 2 (0; 1℄: 1

for every n 2 N ; y 2 [1 If y 2 [0; 1 s=2℄; then let

s a := y + ;

 := ar

os a ;

4

:= 2 ar

os y

ar

os a ;

! := 





2



sin(( +  ( + )=2)=2) Qn;; () := T n sin(!=2) Rn;; () := (Qn;; () + Qn;; ( )) ;



2

; ;

1

2

and we de ne Wn;y;s for every n 2 N ; y 2 [0; 1

s=2); and s 2 (0; 1℄ by

Wn;y;s 2 Pn :

Rn;; () = Wn;y;s ( os ) ;

Show that Wn;y;s 2 Pn (s) and that there exists an absolute onstant > 0 su h that !

jWn;y;s (y)j  exp

n p

1

s

y

2

for every n 2 N ; y 2 [0; 1 s=2); and s 2 (0; 1℄: ut

℄ Extend the validity of part a℄ to the lass GAPN ; that is, prove that there exists an absolute onstant > 0 su h that

jf (y)j  exp for every f

(

N min

p

s 1

)!

y

2

p ; s

y 2 [ 1; 1℄

;

2 GAPN and s 2 (0; 1℄ satisfying m(fx 2 [ 1; 1℄ : f (x)  1g)  2

s:

This is page

417

Printer: Opaque this

A5

Inequalities for Polynomials with Constraints

Overview This appendix deals with inequalities for onstrained polynomials. Typi ally the onstraints are on the lo ation of the zeros, though various oef ient onstraints are also onsidered.

Inequalities for Polynomials with Constraints For integers 0  k  n; let

Pn;k := fp 2 Pn : p has at most k zeros in Dg where, as before, D := fz 2 C : jz j < 1g: For a < b; let Bn (a; b) :=

n

p 2 Pn : p(x) = 

For integers 0  k  n; let

n X j =0

x)j (x

j (b

Pen;k (a; b) := fp = hq : h 2 Bn

k (a; b) ;

a)n j ; j

q 2 Pk g :

Two useful relations, given in E.1, are

Pn;  Bn ( 0

1; 1)

and

Pn;k  Pen;k (

1; 1) :

o

0

:

418 A5. Inequalities for Polynomials with Constraints

Theorem A.5.1 (Markov-Type Inequality for Pen;k ). The inequality

kp m k (

holds for every p repla ed by 9:)

)

[

;

1 1℄

 (18n(k + 2m + 1))m kpk

[

2 Pen;k (

;

1 1℄

1; 1): (When m = 1; the onstant 18 an be

Proof. First we study the ase m = 1. Applying Theorem 6.1.1 (Newman's inequality) with

( ;  ; : : : ; k ) = (n 0

1

k; n k + 1; : : : ; n)

and using a linear shift from [0; 1℄ to [ 1; 1℄; we obtain that

jp0 (1)j 

(A:5:1)

9 2

n(k + 1)kpk

[

;

1 1℄

for every p 2 Pen;k ( 1; 1) of the form

p(x) = (x + 1)n k q(x) ;

(A:5:2)

q 2 Pk :

Now let p 2 Pen;k ( 1; 1) be of the form p = hq with q 2 Pk and

h(x) =

nXk j =0

j (1 x)j (x + 1)n

k j

with ea h j

 0:

Without loss of generality, we may assume that n k  1; otherwise Theorem 5.1.8 (Markov's inequality) gives the theorem. Using (A.5.1) and the fa t that ea h j  0; we get (A:5:3) jp0 (1)j

nXk
0 = j (1 x)j (x + 1)n k j q(x) (1) j =0 1 X = (j (1 x)j (x + 1)n k j q (x))0 (1) j =0

9 n(k + 2) (x + 1)n k 1 (0 (x + 1) + 1 (1 2

nXk

9 j (x + 1)n k j q (x) n ( k + 2)  (1 x )

j 2







j =0

[

[

;

1 1℄

;

1 1℄

 n(k + 2)kpk ; : 2 [ 1; 1℄ be arbitrary. To estimate jp0 (y)j we distinguish two 9 2

Now let y

ases.

x))q(x)

[

1 1℄

Inequalities for Polynomials with Constraints 419

If y 2 [0; 1℄; then by a linear shift from [ 1; 1℄ to [ 1; y ℄; we obtain from (A.5.3) that (A:5:4) jp0 (y)j  y +9 1 n(k + 2)kpk ;y  9n(k + 2)kpk ; for every p 2 Pen;k ( 1; y ): It follows from E.1 d℄ that Pen;k ( 1; 1)  Pen;k ( 1; y) : So (A.5.4) holds for every p 2 Pen;k ( 1; 1): If y 2 [ 1; 0℄; then by a linear shift from [ 1; 1℄ to [y; 1℄; we obtain from (A.5.3) that (A:5:5) jp0 (y)j  1 9 y n(k + 2)kpk y;  9n(k + 2)kpk ; for every p 2 Pen;k (y; 1): By E.1 d℄ again, Pen;k ( 1; 1)  Pen;k (y; 1) : So (A.5.5) holds for every p 2 Pen;k ( 1; 1); whi h nishes the ase when m = 1: Now we turn to the ase when m  2: Note that an indu tion on m does not work dire tly for an arbitrary p 2 Pen;k ( 1; 1): However, it follows by indu tion on m that (A:5:6) kp m k ;  (9n(k + m + 1))m kpk ; [

1

[

(

)

[

1 1℄

℄

[

1℄

[

[

1 1℄

for every p 2 Pen;k ( 1; 1) of the form (A.5.2). Now let p of the form p = hq , where q 2 Pk and

h(x) =

nXk j =0

j (1

x)j (x + 1)n

j

For notational onvenien e let s := minfn fa t that ea h j  0, we get (A:5:7) jp m (1)j (

1 1℄

1 1℄

2 Pen;k (

with ea h j

1; 1) be

 0:

k; mg. Using (A.5.6) and the

)

= =

 

nXk (j (1 j =0 s X (j (1



x)j (1 + x)n k j q(x))(m) (1)

x)j (1 + x)n k j q(x))(m) (1) j =0

s

X

(9n(k + s + m + 1))m j (1 x)j (x + 1)n k j q (x)

j =0 [

nXk

(9n(k + 2m + 1))m

j (1 x)j (x + 1)n k j q (x)



= (9n(k + 2m + 1))m

j =0

kpk

[

;

1 1℄

:

[

;

;

1 1℄

1 1℄

420 A5. Inequalities for Polynomials with Constraints

From (A.5.7), in a similar fashion to the ase when m = 1; we on lude that kp m k ;  (18n(k + 2m + 1))m kpk ; (

)

[

1 1℄

[

1 1℄

for every p 2 Pen;k ( 1; 1); whi h nishes the proof.

ut

Theorem A.5.1 is essentially sharp as is shown in E.2 and E.4 f℄. However, a better upper bound an be given for jp m (y )j when y 2 ( 1; 1) is away from the endpoints 1 and 1: (

)

Theorem A.5.2 (Markov-Bernstein Type Inequality for Pn;k ). There exists

a onstant (m) depending only on m su h that (

p

jp j  (m) min n(k + 1) ; pn(k + 1) 1 y for every p 2 Pn;k and y 2 [ 1; 1℄: m) (y )

)!m

(

kpk

[

2

;

1 1℄

Theorem A.5.2 has been proved in Borwein and Erdelyi [94℄. Its proof is long, and we p do not reprodu e it here. However, the proof of a less sharp version, where 1 y is repla ed by 1 y ; is outlined in E.4 and E.5. p The fa tor n(k + 1) in Theorem A.5.2 is essentially sharp in the ase that m = 1; see E.7. 2

2

Theorem A.5.3 (Markov-Bernstein Type Inequality for Bn ( 1; 1)). There exists a onstant (m) depending only on m su h that

jp

j  (m)

m) (y )

(

pn

(

min n ;

p

1

)!m

y

kpk

[

2

;

1 1℄

for every p 2 Bn ( 1; 1) (hen e for every p 2 Pn; ) and y 2 [ 1; 1℄: 0

Note that the uniform (Markov-type) upper bound of the above theorem is a spe ial ase (k = 0) of Theorem A.5.1. Our proof of Theorem A.5.3 oers another way to prove the ase of Theorem A.5.1 when k = 0: First we need a lemma. Lemma A.5.4. For n 2 N and y 2 R; let 1
n;y := 4

p

pny ) + n1

(1

2 +

!

;

where x := maxfx; 0g: Then +

p
jp(y + i
n;y )j  2e p y  n   for every p 2 Bn ( 1; 1); y 2 1 n ; 1 + n ; and

2 [0;
1℄; where i is  the imaginary unit. The + sign is taken if y 2 1 n ; 0 ; and the  sign is taken if y 2 0; 1 + n : 1

2

1

8

1

8

1

8

1

8

Inequalities for Polynomials with Constraints 421

Proof. It is suÆ ient to prove the lemma for the polynomials n2N;

pn;j (x) := (1 x)j (x + 1)n j ;

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

The general ase when

p=

n X

j pn;j

j =0

0

with ea h j

or ea h j

0

an then be obtained by taking linear

ombinations. Without loss of gen  erality, we may assume that y 2 0; 1 + n : Then 1

8



1 (1 y ) 1
n;y  + 8 n n from whi h it follows that 2 +



2

2

jpn;j (y + i
n;y )j  jpn;j (y + i
n;y )j

+


j=2

1 8n

+

2

y) +
n;y (1 + y ) +
n;y
j
n j (1 y ) + n (1 + y ) + n

= (1



2

2

2

1

y



1

2

n n



j

1+ y

2

2n + 1 2n 1 and the lemma is proved.

2


(n j )=2

1

2

= 1



 (1 4ny)

pn;j y

1

n

1

n



n j








p

1

n j

2

2

2

1+y+ n 1+y n 1

2

2epn;j y

1

n




2

;

ut

2 N and y 2 [ 1; 1℄; let Bn;y denote the

ir le of the omplex plane with enter y and radius
n;y . Using E.4, Lemma A.5.4, and the maximum prin iple, we obtain Proof of Theorem A.5.3. For n

1 4

p jp(z )j  2e kpk ; 1; 1); z 2 Bn;y ; n 2 N ; and y 2 [

for every p 2 Bn ( Cau hy's integral formula,

jp

j

m) (y )

(

Z m! = 2i Bn;y ( Z m! 2 Bn;y (



p( ) y)m p( ) y )m

+1

! m 2 2

1

 (m)

min n ;

4


n;y

(

+1

1 4


n;y p




pn

1

1 1℄

[ 1; 1℄: Hen e, by

d

j d j m+1)

(

p

2ekpk

)!m

y

2

;

[

kpk

[

1 1℄

;

1 1℄

for every p 2 Bn ( 1; 1) and y 2 [ 1; 1℄; whi h proves the theorem.

ut

422 A5. Inequalities for Polynomials with Constraints

For r 2 (0; 1℄; let

Dr := fz 2 C : jz  (1 r)j < rg :  be the set of all For n 2 N ; k = 0; 1; : : : ; n; and r 2 (0; 1℄; let Pn;k;r p 2 Pn that have at most k zeros ( ounting multipli ities) in Dr , and let Pn;k;r := Pn;k;r \Pn;k;r : The following result is proved in Erdelyi [89a℄. The +

proof in a spe ial ase is outlined in E.9.

Theorem A.5.5 (Markov-Type Inequality for Pn;k;r ). There exists a on-

stant (m) depending only on m su h that  m kp k ;  (m) n(kp+r 1) for every p 2 Pn;k;r ; m 2 N ; and r 2 (0; 1℄:

2

(

)

[

m

kpk

1 1℄

[

;

1 1℄

We state, without proof, the Lq analogs of Theorem A.5.2 for m = 1, established in Borwein and Erdelyi [to appear 2℄. Theorem A.5.6 (Lq Markov-Type Inequality for Pn;k ). There exists an absolute onstant su h that Z Z jp0 (x)jq dx  q (n(k + 1))q jp(x)jq dx 1

1

+1

1

1

for every p 2 Pn;k and q 2 (0; 1):

Theorem A.5.7 (Lq Bernstein-Type Inequality for Pn;k ). There exists an

absolute onstant su h that Z  jp0 ( os t) sin tjq dt  q (n(k + 1))q= +1

2



for every p 2 Pn;k and q 2 (0; 1):

Z 



jp( os t)jq dt

Both of the above inequalities are sharp up to the absolute onstant > 0: A Markov-type inequality for polynomials having at most k zeros in the disk Dr := fz 2 C : jz j < rg is given by the following theorem; see Erdelyi [90a℄. Theorem A.5.8 (Markov-Type Inequality for Polynomials with At Most k Zeros in Dr ). Let k 2 N and r 2 (0; 1℄: Then there exist onstants (k ) > 0 and (k) > 0 depending only on k so that 1

2

(k) n + (1 r)n 1

2




0  sup kkppkk p

[

[

; 1;1℄

1 1℄

where the supremum is taken for all p 2 Dr :

 (k) n + (1 2




r)n ; 2

Pn that have at most k zeros in

Inequalities for Polynomials with Constraints 423

Comments, Exer ises, and Examples. Erd}os [40℄ proved that

kp0 k

;

[

1 1℄

 n2



n

n

n

1

1

kpk

e < n kpk

;

[

2

1 1℄

[

;

1 1℄

for every p 2 Pn; n  2; having only real zeros. In this result, the assumption that p has only real zeros an be dropped. In E.11 we outline the proof of the above inequality for every p 2 Pn; ; n  2: The polynomials 0

0

pn (x) := (x + 1)n (1 x) ;

n = 1; 2; : : : ;

1

show that Erd}os's result is the best possible for Pn; : Erd}os [40℄ also showed that there exists an absolute onstant su h that 0

jp0 (y)j 

p

n (1 y )

2 2

kpk

;

[

1 1℄

for every p 2 Pn; having only real zeros. Markov- and Bernstein-type inequalities for Bn ( 1; 1) were rst established by Lorentz [63℄, who proved Theorem A.5.3. Lorentz's approa h is outlined for m = 1 in E.8. The proof presented in the text follows Erdelyi [91 ℄. Up to the onstant (m) > 0 Theorem A.5.3 is sharp; see E.12. S hei k [72℄ found the best possible asymptoti onstant in Lorentz's Markov-type inequality for m = 1 and m = 2: He proved the inequalities 0

kp0 k

[

;

1 1℄

 2e n kpk

[

and

;

1 1℄

kp00 k

;

[

1 1℄

 2e n(n

for every p 2 Bn ( 1; 1) (and hen e for every p pn (x) := (x + 1)n (1 x); 1

kp0n k nkpn k

;

[

1 1℄ [

;

1 1℄

! 2e

and

kp00nk

[

;

1 1℄

n(n 1)kpk

[

;

1 1℄

1)kpk

[

;

1 1℄

2 Pn; ). Note that with 0

! 2e

as n ! 1 :

A slightly weaker version of Theorem A.5.1 was onje tured by Szabados [81℄, who gave polynomials pn:k 2 Pn;k with only real zeros so that

jp0n;k (1)j 

1 3

n(k + 1)kpn;k k

[

;

1 1℄

for all integers 0  k  n: After some results of Szabados and Varma [80℄ and Mate [81℄, Szabados's onje ture has been settled in Borwein [85℄, where it is shown that

kp0 k

[

;

1 1℄

 9n(k + 1)kpk

[

;

1 1℄

424 A5. Inequalities for Polynomials with Constraints

for every p 2 Pn;k having n k zeros in R n ( 1; 1). The ru ial part of this proof is outlined in E.4 d℄ and e℄. The above inequality is extended to all p 2 Pen;k (and hen e to all p 2 Pn;k ) and to higher derivatives in Erdelyi [91b℄, whose approa h is followed in our proof of Theorem A.5.1. After partial results of Erdelyi and Szabados [88, 89b℄ and Erdelyi [91b℄, the \right" Markov-Bernstein-type analogs of Theorem A.5.2 for the lasses Pn;k has been proved in Borwein and Erdelyi [94℄. Note that Pn;n = Pn ; and hen e, up to the onstant (m); Theorem A.5.2 ontains the lassi al inequalities of Markov and Bernstein, and of ourse the ase k = 0 gives ba k Lorentz's inequalities for the lasses Pn;  Bn ( 1; 1): The \right" Markov- and Bernstein-type inequalities of Theorems A.5.8 and A.5.9 for all lasses Pn;k in Lq , 0 < q < 1, are established in Borwein and Erdelyi [to appear 2℄. The Markov-type inequality for the lasses Pn;k;r given by Theorem A.5.5 is proved in Erdelyi [89a℄. A weaker version, when k = 0 and the fa tor r = is repla ed by a onstant (r) depending on r, is established in Rahman and Labelle [68℄. When k = 0; Theorem A.5.5 is sharp up to the

onstant (m) depending only on m; see E.10. 0

1 2

E.1 Relation Between Classes of Constrained Polynomials. a℄ Show that Pn; ( 1; 1)  B ( 1; 1): (This is an observation of G. G. Lorentz.) Hint: Use the identities (x )(x ) = j1 + j (1 x) + (jj 1)(1 x ) + j1 j (1 + x) and x  = (1 )(x + 1) ( + 1)(1 x) : 0

1

2

2

4

1

2

2

2

1

2

2

4

1

1

2

2

b℄ Show that Pn;k  P 1; 1):

℄ Show that Bn (a; b)  Bn ( ; d) whenever [ ; d℄  [a; b℄: Hint: First show that x a =  (x ) +  (d x) and b x =  (x ) +  (d with some nonnegative oeÆ ients. d℄ Show that Pen;k (a; b)  Pen;k ( ; d) whenever [ ; d℄  [a; b℄:

ut

en;k (

1

2

3

4

x)

ut

E.2 Sharpness of Theorem A.5.1. Show that there exist polynomials pn;k 2 Pn;k of the form pn;k (x) = (x + 1)n k qn;k (x) ; qn;k 2 Pk su h that jp0n;k (1)j  n(k + 1)kpn;k k ; for every n 2 N ; k = 0; 1; : : : ; n: 1

6

[

1 1℄

Inequalities for Polynomials with Constraints 425

Hint: Use the lower bound in Theorem 6.1.1 (Newman's inequality) with k; n k + 1; : : : ; ng :

( ;  ; : : : ; k ) = (n 0

1

ut E.3 A Te hni al Detail in the Proof of Lemma A.5.4. For n y 2 [ 1; 1℄; let 

Fn := z = a + ib : a 2 and



1

n ; 1 + 8n 1

1 8





2N

and



; b 2 (
n;a ;
n;a ) ;

Bn;y := z 2 C : jz yj =
n;y : Show that Bn;y  Fn for every y 2 [ 1; 1℄: 1 4

E.4 Bernstein-Type Inequality for Pn;k . Prove that there exists an absolute onstant su h that p

jp0 (y)j  1n(ky+ 1) kpk

[

2

;

1 1℄

for every p 2 Pn;k ; and y 2 ( 1; 1): Pro eed as follows: a℄ Show that for every n 2 N and k = 0; 1; : : : ; n; there exists a polynomial Q 2 Pn;k su h that

jQ0 (0)j kQk ; [

jp0 (0)j p2Pn;k kpk ;

:

= sup

1 1℄

[

1 1℄

Hint: Use a ompa tness argument. Use Rou he's theorem to show that the uniform limit of a sequen e of polynomials from Pn;k on [ 1; 1℄ is also in Pn;k : ut b℄ Show that Q has only real zeros, and at most k + 1 of them ( ounting multipli ities) are dierent from 1: Hint: Use a variational method. For example, if z 2 C n R is a zero of Q; 0

then the polynomial



Q(x) := Q(x) 1

x



2

(x

z )(x 0

z ) 0

with suÆ iently small  > 0 is in Pn;k and ontradi ts the maximality of

ut

Q:

℄ Let Æ := (36n(k + 3)) : Show that 1

kpk

[

Æ;1℄

 2 kpk

;

[0 1℄

for every polynomial Pn;k having all its zeros in [0; 1) with at most k of them ( ounting multipli ities) in (0; 1):

426 A5. Inequalities for Polynomials with Constraints

Hint: Use the Mean Value Theorem and the result of Theorem A.5.1 transformed linearly from [ 1; 1℄ to [0; 1℄: ut = d℄ Let z := i (36n(k +3)) ; where i is the imaginary unit and 2 [0; 1℄: 1 2

0

Show that

jp(z )j 

p

2 kpk

0

[

;

1 1℄

for every polynomial p 2 Pn having only real zeros with at most k of them ( ounting multipli ities) in ( 1; 1):

Outline. Let p(x) =

Qs

j =1 (x

uj ) with some s  n: Applying part ℄ to s Y

q(x) :=

j =1

(x

uj ) ; 2

we obtain

jq(

(36n(k + 3)) )j  2kq k ; = 2kq (x )k ; = 2kp(x)p( x)k  2kpk ; : 1

2

[0 1℄

[0 1℄

[

;

1 1℄

2 [

1 1℄

Observe that

jp(z )j  jp(i(36n(k + 3)) 0

2

=

1 2

)j = 2

= jq ( (36n(k + 3)) )j ;

s Y

(uj + (36n(k + 3)) ) 2

j =1

1

1

whi h, together with the previous inequality, yields the desired result.

ut

e℄ Let Q 2 Pn;k be the extremal polynomial of part a℄, and let n

Fn;k := z = a + ib : jaj  1 ; jbj  (36n(k + 3)) Show that

jQ(z )j 

p

2 kQk

[

=

1 2

(1

jaj)

o

:

;

1 1℄

for every z 2 Fn;k : Hint: Use parts b℄ and d℄ and a linear shift from the interval [ 1; 1℄ to [2 Re(z ) 1; 1℄ if Re(z )  0; or to [ 1; 2 Re(z ) + 1℄ if Re(z ) < 0: ut

Inequalities for Polynomials with Constraints 427

f℄ Show that there exists an absolute onstant > 0 su h that

jp0 (0)j 

p

n(k + 1) kpk

[

;

1 1℄

for every p 2 Pn;k : Hint: By part a℄ it is suÆ ient to prove the inequality when p = Q; in whi h

ase use part e℄ and Cau hy's integral formula. ut g℄ Prove the main result stated in the beginning of the exer ise. Hint: In order to estimate jp0 (y)j when, for example, y 2 [0; 1℄ use a linear shift from [ 1; 1℄ to [2y 1; 1℄ and apply part f℄. ut E.5 Bernstein-Type Inequality for Pn;k for Higher Derivatives. Prove that there exists a onstant (m) depending only on m so that

jp

(

p

n(k + 1) 1 y

j  (m)

m) (y )

!m

kpk

[

2

;

1 1℄

for every p 2 Pn;k and y 2 ( 1; 1): Hint: First show that for every n; m 2 N, k = 0; 1; : : : ; n, and Æ there exists a polynomial QÆ 2 Pn;k su h that

jQÆm (0)j kQÆ k ; n Æ;Æ (

[

)

1 1℄ [

jp m (0)j p2Pn;k kpk ; n Æ;Æ (

= sup ℄

[

)

1 1℄ [

2 (0; 1℄;

: ℄

Show that QÆ has at most k +m zeros dierent from 1: Show that there is a polynomial Q 2 Pn;k having at most k + m zeros (by ounting multipli ities) dierent from 1 su h that

jQ m (0)j = sup jp m (0)j : kQk ; p2Pn;k kpk ; (

)

[

(

1 1℄

)

[

1 1℄

If y = 0; then use E.5 and indu tion on m to prove the inequality of the exer ise for all p 2 Pn;k having at most k + m zeros dierent from 1: For an arbitrary y 2 ( 1; 1) use a linear shift as is suggested in the hint to E.5 g℄. ut E.6 Sharpness of Theorem A.5.2. Show that there exist polynomials pn;k 2 Pn;k and an absolute onstant > 0 su h that

jp0n;k (0)j 

p

n(k + 1)kpn;k k

for every n 2 N and k = 0; 1; : : : ; n:

[

;

1 1℄

428 A5. Inequalities for Polynomials with Constraints 1

Hint: If k = 0; then let m :=

3



(n

1) and

1)m (x + 1)m

pn;k (x) := (x

1 

If 1  k  n; then let m := 1

3

3

n ; s := 1)m T

pn;k (x) := (x

2

1

bpm :

+1+



(k

1) ; and

r

m x : 2s + 1

3

s

2 +1



1 

If n < k  n; then let m := 1 3

3

ut

n and pn;k := pn;m:

E.7 Some Inequalities of Lorentz. (See Lorentz [63℄.) a℄ Show that

jp0 (x)j 



n

n

n



n p x  n

1

1






 4n p x  n 1




for every p 2 Bn ( 1; 1) (hen e for every p 2 Pn; ); n  2; and x 2 [ 1; 1℄; where in x  n the + sign is taken if x 2 [ 1; 0); while the sign is taken if x 2 [0; 1℄: Hint: Observe that it is suÆ ient to prove the inequality only for 0

1

n2N;

pn;j (x) := (1 x)j (x + 1)n j ;

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

ut

b℄ Let

r

Æn (x) :=

1

x

2

n

n2N;

;

x 2 [ 1; 1℄ :

Show that there exists an absolute onstant > 0 su h that

jp0 (x)j  (Æn (x))

1








max jp(x)j; p x  Æn (x) 1 2



for every p 2 Bn ( 1; 1) (hen e for every p 2 Pn; ) and x 2 1 + n ; 1 Hint: Note that it is suÆ ient to prove the inequality only for 0

pn;j (x) := (1 x)j (x + 1)n j ;

n2N;

 n :

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

ut

℄ Show that Z

1

1

jp0 (x)jq dx  2
4q nq

Z

1

1

jp(x)jq dx

for every p 2 Bn ( 1; 1) (hen e for every p 2 Pn; ) and q 2 (0; 1): 0

Inequalities for Polynomials with Constraints 429

ut

Hint: Use part a℄. d℄ Show that there is an absolute onstant > 0 su h that Z

p 0 p (x) 1

1

1

Z

q dx

1

 5 q nq= jp(x)jq dx 1; 1) (hen e for every p 2 Pn; ) and q 2 (0; 1): x

2

2

1

for every p 2 Bn ( Hint: Use parts b℄ and ℄. ut Analogs of parts a℄ and b℄ for higher derivatives are established by Lorentz [63℄. From these, analogs of parts ℄ and d℄ for higher derivatives

an be proven. 0

E.8 Theorem A.5.5 in a Spe ial Case. Show that there is an absolute

onstant su h that jp0 (1)j  pn kpk ; 1

1

r

[

for every p 2 Pn having all its zeros in (

Proof. First assume that (A:5:8)

1 1℄

1; 1

jp(1)j = kpk

[

;

1 1℄

2r℄; r 2 (0; 1℄:

:

Without loss of generality, we may assume that p 2 Pn n Pn : Denote the zeros of p by ( 1 <)x  x 


 xn ( 1 2r): Let I := (1 2( + 1) r; 1 2 r℄ ;  = 1; 2; : : : : Using E.12 of Se tion 5.1, we obtain n 1 X 1 X jp0 (1)j = X 1 jp(1)j j 1 xj =  x 2I 1 xj 1

1

2

4

4

=1

=1



1 e X v=1

 2e

p

1 X

p

2

v=1

j



n 2( + 1) r 4

( + 1)

4

pnr  pnr

2



1 2 r 1

4

with an absolute onstant : Now we an drop assumption (A.5.8) as follows: Sin e p has all its zeros in ( 1; 1 2r℄; jpj and jp0 j are in reasing on [1 2r; 1): Pi k the unique y 2 [1; 1) satisfying jp(y )j = kpk ; = kpk ;y : Using a linear transformation, from the already proved ase we easily obtain 1

[



jp0 (1)j  jp0 (y)j 

1

 pnr kpk 1

[

;

1 1℄

2n 2r + y 1 y+1 y+1

1 1℄

[



=

1 2

1

℄

kpk

[

;y℄

1

:

ut

430 A5. Inequalities for Polynomials with Constraints

E.9 Sharpness of Theorem A.5.5. For n 2 N and r 2 (0; 1); let Sn;r be the family of all p 2 Pn that have no zeros in the strip

fz 2 C : jIm(z )j < rg : a℄ Show that there exist polynomials pn;m;r 2 Sn;r and a onstant (m) > 0 depending only on m su h that

j

m) (1) pn;m;r (



m



j  (m) min pnr ; n

kpn;m;r k

2

;

[

1 1℄

for every n 2 N ; r 2 (0; 1℄; and m = 1; 2; : : : ; n: Outline. If 0 < r  m ; then with 4

xj := 1 and

 mr 4







os j n  ; 2

j = 1; 2; : : : ; n

1

2

zj := xj + ir ;

let

pn;m;r (x) :=

n Y j =1

j = 1; 2; : : : ; n zj )(x z j ) 2 Sn;r :

(x

By E.1 of Se tion 5.2, jpn;m;r (1)j = kpn;m;r k

;

[

m jpn;m;r (1)j kpn;m;r k ; (

)

[

=

1 1℄

m jpn;m;r (1)j jpn;m;r (1)j (

1 1℄

: Prove that

)



j =m



m j p1 jjqqn;m;r (1) 2 n;m;r (1)j !

 m

 2 1 + 4m p X n  e2 1

1



(

)

m

xj

m

 (m) min pnr ; n

2

Q

;

where qn;m;r (x) := nj (x xj ) and (m) > 0 depends only on m: If  < r  1; then let p  ut n;m;r := pn;m;re; where re := m : m b℄ Con lude from Theorem A.5.5 and part a℄ that there exist two onstants

(m) > 0 and (m) > 0 depending only on m su h that =1

4

4

1

2





n

(m) min p ; n r 1

2

m

 sup kpkpk k m)

(

 (m) 2



;

[

[

min

1 1℄

;

1 1℄



m

pnr ; n

2

;

where the supremum is taken either for all p 2 Sn;r or for all p 2 Pn; ;r : 0

Inequalities for Polynomials with Constraints 431

E.10 An Inequality of Erd}os. (See Erd}os [40℄.) Prove that

kp0 k

;

[

1 1℄



n

2

n

n

n

1

1

kpk

;

[

1 1℄

for every p 2 Pn; ; n  2: This extends a result of Erd}os [40℄ where the above inequality is proven under the additional assumption that p has only real zeros. a℄ Suppose p 2 Pn; ; all the zeros of p are real, p(1) = p( 1) = 0; and kpk ; = 1: Show that 0

0

[

1 1℄

p(x) 



n

n

1

n 1

1+x 1+x

0

for every x 2 [ 1; 1℄; where x is the only point in ( 1; 1) with p0 (x ) = 0: 0

0

Proof. Without loss of generality, we may assume that deg(p) = n and 1 < x < x : Let d := x x: Let x := 1; x ; : : : ; xn denote the zeros of p: Then  n  1+x Y d p(x) = 1 : p(x) = p(x ) 1 + x j x xj 0

0

1

0

2

0

0

=2

Sin e the geometri mean of n 1 nonnegative numbers is not greater than their arithmeti mean, we have n  Y j =2

1

x

d 0





xj

n 



1

n 1

1 

1

n X

n 1

j =2

n 1+

d

x

0

!!n

xj

n

d

1

1

x +1 0





n

n

1

n 1

:

ut b℄ Under the assumptions of part a℄ show that

kp0 k

;

 n2

X

1

[

1 1℄



n

n

1

n 1

kpk

;

[

1 1℄

:

Proof. Note that X

x xj  j 1

1

x

0

=

xj 

1

x

0



k

 min 1 x ; xn +k1 xj 0

0



 n2 ;

where k denotes the number of zeros of p in [1; 1): Hen e, by part a℄,

432 A5. Inequalities for Polynomials with Constraints n X

p0 (x) = p(x) 

j =1

1

x

n

1+x X 1 n 1 1 + x x  x xj j   n n 1+x 1+x X 1  n 1 1+x 1+x x xj x



n

xj

1

0

1

0

0

 n2



n

n

j

1

0

1

n 1

for every x 2 [ 1; 1℄: Similarly



n p0 (x) 

n

2

for every x 2 [ 1; 1℄:

n

n

1

1

ut

℄ Suppose p 2 Pn; has only real zeros, p0 does not vanish in [ 1; 1℄; p( 1) = 0; and p(1) = 1: Show that 0

kp0 k

;

[

Hint: Use the relation Pn;

1 1℄

 n2 kpk

[

 Bn (

0



p(x) 

;

1 1℄

:

1; 1) to show that

x+1

n

2

; x 2 [ 1; 1℄ :

Denote the zeros of p by x = 1, x ; x ; : : : ; xn : Then 1

n

X 0  p0 (x) = p(x)

j =1

x

2

1

xj





3

x+1 2

n

n x+1

 n2 = n2 kpk

[

;

1 1℄

for every x 2 [ 1; 1℄: ut d℄ Prove Erd}os's inequality for every p 2 Pn; having only real zeros. Hint: Redu e the general ase to either part b℄ or part ℄ by a linear transformation. ut e℄ Prove Erd}os's inequality for every p 2 Pn; : Hint: Show that for every n 2 N and y 2 [ 1; 1℄; there exists a polynomial Q 2 Pn; su h that 0

0

0

jQ0 (y)j kQk ;

jp0 (y)j : p2Pn; kpk ; Show by a variational method that if y 2 ( 1; 1); then Q has only real zeros. Now part d℄ nishes the proof. ut [

1 1℄

= sup

0

[

1 1℄

Inequalities for Polynomials with Constraints 433

f℄ Show that

kp0 k

=



n

n

n

1

kpk

; 2 n 1 holds for a p 2 Pn; having only real zeros if and only if either ;

[

1 1℄

[

1 1℄

0

p(x) = (x + 1)n (1 x)

p(x) = (1 x)n (x + 1)

or

1

1

with some 0 6= 2 R: E.11 Sharpness of Theorem A.5.3. For every n 2 N ; m = 1; 2; : : : ; n; and y 2 [ 1; 1℄; there are polynomials pn;m;y 2 Bn ( 1; 1) having zeros only in R n ( 1; 1) su h that m (y )j  (m) jpn;m;y (

)

pn

(

min n ;

p

1

)!m

y

kpn;m;y k

[

2

;

1 1℄

with a onstant (m) > 0 depending only on m:

Outline. If



y 2 [ 1; 1℄ n

then let

1 + nm ; 1 2

m n

2

;

pn;m;y (x) := (x + 1)n :

In what follows, assume that

y 2 [ 1+

m n

2

;1

m n ℄:

2

x)j (x + 1)n j ; n 2 N ; j = 0; 1; : : : ; n: Show that

Let qn;j (x) := (1

m qn;j (x) Qn;j;m (x) = ; qn;j (x) (x 1)m (

)

m  j  n m;

2

where Qn;j;m is a polynomial of degree m with only real zeros and with leading oeÆ ient n nm : Let !

(

)!

(


n;y := max

1

n

p

;

1

pn

y

2

)

;

n2N;

y 2 [ 1; 1℄ :

Use the Mean Value Theorem and Theorem A.5.3 to show that there exists an absolute onstant 2 (0; 1) su h that

qn;j (x) 

1 2

kqn;j k

[

;

1 1℄

;

mjn m

for every

x 2 Iy := [y


n;y ; y +
n;y ℄ \ [ 1; 1℄ ;

y=1

j n

2

:

434 A5. Inequalities for Polynomials with Constraints

Let m su h that

jn j

j n

m and y = 1

be xed. Choose a point 

2

2 Iy




i j 

n;y ; i = 1; 2; : : : ; m ; 2(m + 1) where the numbers i are the zeros of Qn;j;m : Now show that there exists a onstant (m) > 0 depending only on m su h that 1

j

m) qn;j ( )

(A:5:9)

(

(

j  (m)

min n; p

1

Next show that if

y2



then there exists a point 

2 y

1 2

1

)!m

y

kqn;j k

[

2

;

1 1℄

:

m n ;

1 + nm ; 1 2

2

jyj) ; y +

(1

pn

1 2

(1

jyj)



and a value of j , m  j  n m, su h that (A.5.9) holds. Polynomials pn;m;y with the desired properties an now be easily de ned by using linear

ut

transformations.

E.12 An Inequality of Turan. (See Turan [39℄.) Show that

kp0k kpk

[

[

for every p 2 Pn n Pn

1

; 1;1℄

1 1℄

>

1p n 6

having all its zeros in [ 1; 1℄:

Outline. Assume that p 2 Pn has all its zeros in [ 1; 1℄; and kpk ; = 1: Choose an a 2 [ 1; 1℄ su h that jp(a)j = 1. Without loss of generality, assume that p(a) = 1: pn: a℄ Show that if a = 1; then jp0 (1)j  n > If a 2 ( 1; 1); then p0 (a) = 0. Without loss of generality, assume that a 2 [ 1; 0℄: Let   I := a; a + 2n =  [ 1; 1℄ : If n  3; then the result follows by the Mean Value Theorem; let n  4: pn on I; then b℄ Use the Mean Value Theorem to show that if jp0 (x)j  p(x)  on I: pn:

℄ Show that if jp00 (x)j > n on I; then jp0 (a + 2n = )j > [

1

1

2

6

1 1℄

1 2

1

6

2

3

1

1

1 2

12

6

d℄ The proof of Turan's inequality an now be nished as follows. Suppose on I , and there exists a  2 I su h that p00 ( )  n: Note that

p(x) >

2

1

3

12

p0 (x)

2

p(x)p00 (x) = p(x)

2

n X k=1

(x

1

xk )

2

;

Inequalities for Polynomials with Constraints 435

where x ; xP; : : : ; xn denote the zeros of p: Sin e ea h xk lies in [ 1; 1℄, the inequality nk (x xk )  n holds for every x 2 I: Sin e p(x)  on I; this implies that 1

2

2

=1

p0 (x) p0 ( )

2

2 3

p(x)p00 (x) 

2

and hen e

1

4

n 9

x2I

;

 n9 jp( )jjp00 ( )j > n9

n

12

=

n

36

:

ut

An extension of Turan's inequality to Lp norms is given by Zhou [92b℄. E.13 An Inequality of Erd}os and Turan. Let p 2 Pn be of the form

p(x) = 

n Y

j =1

(x

1x

xj ) ;

1

 x 


 xn  1 : 2

Suppose p is onvex on [xk ; xk ℄ for some index k: Then 16 1

xk

xk

1

 pn :

Pro eed as follows: Let a be the only point in [xk ; xk ℄ for whi h

p0 (a) = 0:

1

2 R su h that a 2n =   < a <   a + 2n = ; jp0 ( )j  pn jp(a)j ; and jp0 ( )j  pn jp(a)j :

a℄ Show that there exist  ;  1

2

1 2

1

1 2

2

1

1

1

2

6

6

ut

Hint: Modify the outline of the proof of E.12. b℄ Show that



1

xk

1

 p6n

xk

and



2

 p6n :

Outline. To prove, say, the rst inequality, we may assume that xk <  ; otherwise the inequality is trivial. Using the onvexity of p on [xk ; xk ℄; we get jp0 (x)j  jp0 ( )j ; x 2 [xk ;  ℄ : Hen e 1

1

1

1

jp(a)j  jp( )j = jp( ) 1

 (

1

p(xk )j =

1

xk

1

1

1

)jp0 ( )j  (

and the result follows.

1

1

1

Z 1 0 p (x) dx xk 1

xk ) 1

1 6

pn jp(a)j ;

=

Z 1

xk

1

jp0 (x)j dx ut

436 A5. Inequalities for Polynomials with Constraints

℄ Con lude that

a xk

 (a

1

 ) + (

xk ) 

a) + (xk

 )

1

1

1

p2n + p6n = p8n ;

and similarly

a  (

xk

2

2

p2n + p6n = p8n :

ut

d℄ Er}od [39℄ establishes the sharp inequalities

xk

xk

1

xk

xk

1

 p2n2

 p2n2

3

if n is even,

p

n 2n 3 n 1

if n  3 is odd.

2

E.14 S hur-Type Inequality for Bn ( 1; 1). Let  be an arbitrary positive real number. Show that

kpk

sup

6 p2Bn

0=

;

(

1 1)

kp(x)(1

[

;

=

1 1℄

x ) k 2

[

;

1 1℄

(n + 2)n  (4) (n + )n  +2

+





e (n + 2) : 4 The supremum is attained if and only if p(x) = (1  x)n ; 0 6= 2 R: Hint: Let x := n n  : If jyj  x ; then <

1

kp(x)(1

1

+2

jp(y)j x

2

)

k

[

;

1 1℄

 (1

1

)

 (1

1

x ) (n + 2)n  (n + 2)  = < :   (4) (n + ) (4) (n + )n  y

2

2 1

+2

2

+

If x < jy j  1; say x < y  1; then 1

(1

1

y)j (y + 1)n

j

 k(1

x)j (x + 1)n

j

k

[

;

1 1℄

j j (n

j )n j (n + 2)n = (1 j n  n (n + )n j  j  n

j )n

j

x )j (x + 1)n

j

=

2n j j (n

nn

1

1

j (n + ) n + 2 (1 x )j (x + 1)n (n j ) n+   n + 2 n  n +  (1 x )j (x + 1)n j



1

1

whenever 0  j  nn : +2

1

1

j

Inequalities for Polynomials with Constraints 437

x)j (x + 1)n

On the other hand, sin e the fun tion (1  de reasing in 1 nj ; 1 ; the inequality 2

(1

y)j (y + 1)n

j

x )j (x + 1)n

< (1

1

1

j

is monotone

j

follows whenever nn < j  n: Finally, use the representation +2

p(x) = with ea h aj

n X

aj (1 x)j (x + 1)n

j =0

j

 0 or ea h aj  0 to show that n+2

jp(y)j  (4(n)+(n2+) )n



+

p(x1 )(1



x ) 2 1

for every p 2 Bn ( 1; 1):

ut

E.15 S hur-Type Inequality for Tn ( !; ! ). For n 2 N and ! 2 (0;  ℄, let

Tn (

!; !) := ft 2 Tn : d! (t)  ng ;

where the Lorentz degree d! (t) is de ned in E.5 of Se tion 2.4. Let  be an arbitrary positive real number. Show that sup

6 t2Tn

0=

(

!;!) t()

ktk

!;!℄

[

1 2

( os 








os ! )

[

!;!℄

p

(n( 2! ) + !n) ; 0 < !  =2 (2!  + n = )  ; =2  !   ; 1 2

and the supremum is attained if and only if

t() = sin

2

n

! 2

;

0 6= 2 R :

Here the  symbol means that the ratio of the two sides is between two positive onstants depending only on  (and independent of n 2 N and ! 2 (0; ℄). E.16 Extensions and Variations of Lax's Inequality. Theorem 7.1.11 ontains, as a limiting ase, an inequality of Lax [44℄ onje tured by Erd}os; see part a℄. Various extensions of this inequality are given by Ankeny and Rivlin [55℄, Govil [73℄, Malik [69℄, and others. Parts b℄ to e℄ dis uss some of these. As before, let

D := fz 2 C : jz j < 1g

and

D := fz 2 C : jz j = 1g :

438 A5. Inequalities for Polynomials with Constraints

a℄ Lax's Inequality. The inequality

kp0 kD  n2 kpkD holds for all p 2 Pn that have no zeros in the open unit disk. Proof. This follows from Theorem 7.1.11 as a limiting ase. b℄ Malik's Extension. Asso iated with

p(z ) = let

Then

n Y j =1

(z

zj 2 C ; 0 6= 2 C ;

zj ) ;

p(z ) :=

n Y j =1

(1

ut

zzj ) = z n p(1=z) :


max jp0 (z )j + jp0 (z )j = n

z2D

for every 0 6= p 2 Pn : Proof. See Malik [69℄. ut

℄ An Observation of Kroo. Suppose p 2 Pn satis es that if p(z ) = 0 for some z 2 D; then p(1=z ) = 0 (there is no restri tion for the zeros of p outside D). Then

kp0kD  n2 kpkD :

Hint: Show that if p 2 Pn satis es the assumption of the lemma, then jp0 (z )j  jp0 (z )j for every z 2 D. Use part b℄ to nish the proof. ut d℄ An Inequality of Ankeny and Rivlin. Let r  1. The inequality max jp(z )j 

jzj

r

=

rn + 1 2

max jp(z )j

jzj

=1

holds for all p 2 Pn that have no zeros in the open unit disk. Proof. See Ankeny and Rivlin [55℄. e℄ An Inequality of Govil. Let r  1. The inequality

ut

kp0kD  1 +n r kpkD holds for all p 2 Pn that have no zeros in the disk fz 2 C : jz j < rg: Proof. See Govil [73℄.

ut

Inequalities for Polynomials with Constraints 439

f℄ Let

p(z ) := Show that

n Y k=1

(z

n X

kp0kD k 

0 6= 2 C :

zk ) ;

k=1

1 1 + jzk j

!

kpkD :

Proof. If z 2 D; then 





n X zp0 (z ) z Re = Re p(z ) z zk k





=1

n X k=1

1 ; 1 + jzk j

ut

and the result follows. g℄ Let r 2 (0; 1℄: The inequality

kp0kD  1 +n r kpkD holds for all p 2 Pn that have all their zeros in the disk fz 2 C : jz j  rg: Proof. Use part f℄. ut h℄ Another Inequality of Govil. Let r > 1. The inequality

kp0 kD  1 +n rn kpkD holds for all p 2 Pn whi h have no zeros in the disk fz 2 C : jz j < rg: Proof. See Govil [73℄.

ut

E.17 Markov-Type Inequality for Nonnegative Polynomials. Show that

kp0 k

 n2 kpk 2

[

;

1 1℄

[

;

1 1℄

for every p 2 Pn positive on [ 1; 1℄: Proof. Suppose p 2 Pn is positive on [ 1; 1℄ and kpk Theorem 5.1.8 (Markov's inequality) to q := p 1:

[

;

1 1℄

= 2: Apply

ut

E.18 Markov's Inequality for Monotone Polynomials. It has been observed by Bernstein that Markov's inequality for monotone polynomials is not essentially better than for arbitrary polynomials. He proved that if n is odd, then kp0k ; =  n + 1  ; sup 2 6 p kpk ; 2

0=

[

1 1℄

[

1 1℄

where the supremum is taken for all p 2 Pn that are monotone on [ 1; 1℄:

440 A5. Inequalities for Polynomials with Constraints

For even n, the inequality

kp0 k sup 6 p kpk 0=



; 1;1℄

[

1 1℄

[



n+1

2

2

still holds. Parts a℄ and b℄ of this exer ise outline a proof. a℄ Show that for every odd n; there is a p 2 Pn monotone on R for whi h

jp0 (1)j kpk ; [

Proof. Let m := (n q 2 Pm for whi h 1

2

m X

q (1) = 2

k=0



=

n+1 2

1 1℄

(1 + 2k )

!Z

1

q (t) dt = (m + 1) 2

Z x

1

q (t) dt : 2

0

Z

1 2 Obviously p 2 Pn ; p is monotone on R; and

q (t) dt

1

2

0

1 1℄

Z

2

0

p(x) :=

[

:

1): Use E.2 e℄ of Se tion 6.1 to show that there is a

Now let

jp0 (1)j kpk ;

2

q (t) dt : 2

0



q (1) n+1 = 2(m + 1) = 2 = R 2 q (t) dt 2

2

1

1 2

2

2

:

0

Now the proof an be nished by a linear transformation mapping [ 1; 1℄ to [0; 1℄: ut b℄ Let n be odd. Show that

kp0k 6 p kpk

sup

[

0=

[

; 1;1℄

1 1℄





n+1 2

2

;

where the supremum is taken for all p 2 Pn that are monotone on [ 1; 1℄: Hint: It is suÆ ient to prove that if n := 2m is even, then

kpk  21



n+2 2

2 Z

1

p(t) dt

1

for every p 2 Pn nonnegative on [ 1; 1℄: Show that there is an extremal polynomial pe for the above inequality for whi h kpek ; is a hieved at 1: Show, by a variational method, that this pe must have at least 2m zeros ( ounting multipli ities) in [ 1; 1): Sin e pe is nonnegative, it is of the form pe = q with a q 2 Pm : Now use E.2 e℄ of Se tion 6.1 to show that [

2

1 1℄

Inequalities for Polynomials with Constraints 441

q (1)  2

1 (m + 1) 2

Z



1

2

1 n+2 q (t) dt = 2 2

2 Z

1

q (t) dt ;

2

1

2

1

and the result follows. ut

℄ Let r = p=q; where p; q 2 Pn and q is positive and monotone nonde reasing on an interval [a; b℄: Show that

kr0 k a

[ +

 4n krk a;b 2

;b℄

[

℄

for every  2 (0; b a): Proof. The proof follows Borwein [80℄. Let 2 [a + ; b℄ be a point where jr0 ( )j = kr0 k a ;b : Let d 2 [a; ℄ be a point jp(d)j = kpk a; . Then [ +

℄

[

r0 ( ) =

p0 ( ) q( )

℄

q0 ( ) r( ) : q( )

Theorem 5.1.8 (Markov's inequality) and the monotoni ity of q imply that

jp0 ( )j  2n kpk a; = 2n jp(d)j  2n jp(d)j  2n jr( )j jq( )j ( a)jq( )j ( a)jq( )j a jq(d)j  2

and

[

2

℄

2

2

jq0 ( )j jr( )j  2n kqk a; jr( )j  2n jr( )j : jq( )j jq( )j  2

Thus

kr0 k a

;b℄

[ +

[

2

℄

= jr0 ( )j 

4n

2



krk a;b : [

℄

tu Let n be odd. By part a℄, there exists a p 2 Pn

d℄ Sharpness of Part ℄. su h that p is monotone in reasing on R; p(a) = 1; p(a + ) = 1, and 

jp0 (a + )j = 21 n +2 1

2

:

Let q := p + 2 and r := 1=q: Show that

krk a;1 [

℄

=1

and

jr0 (a + )j = 19
21



n+1 2

2

:

442 A5. Inequalities for Polynomials with Constraints

Weighted polynomial inequalities and their appli ations are beyond the s ope of this book. A thorough dis ussion requires serious potential theoreti ba kground, and proofs are usually quite long. Some of the main results in this area in lude von Golits hek, Lorentz, and Makovoz [92℄; He [91℄; Ivanov and Totik [90℄; Levin and Lubinsky [87a℄ and [87b℄; Lubinsky [89℄, Lubinsky and Sa [88a℄, [88b℄, and [89℄; Mhaskar and Sa [85℄; Nevai and Totik [86℄ and [87℄; Sa and Totik [to appear℄; and Totik [94℄. The following exer ise treats a weighted Markov-type inequality with respe t to the Laguerre weight on [0; 1): The method presented below works in more general ases; however, the te hni al details oming from the \right" polynomial approximation of the weight fun tion is typi ally far more ompli ated. E.19 A Weighted Markov-Type Inequality of Szeg}o. Part a℄ presents a simple weighted polynomial inequality of Szeg}o [64℄. a℄ Show that

kp0(x)e x k ;1  (8n + 2)kp(x)e xk ;1 [0

)

[0

)

for every p 2 Pn : Proof. Let p 2 Pn . First prove the inequality

jp0 (0)j  (8n + 2)kp(x)e x k ;1 [0

)

as follows. Apply Theorem 5.1.8 (Markov's inequality) on [0; n=2℄ (by a linear transformation) to 

x n n

q(x) := p(x) 1 and note that 

1

x n e n

x

and



1

x n

1

 2;

x 2 [0; n=2℄ :

The general ase an be easily redu ed to the ase dis ussed above. If

y 2 [0; 1) is xed, then let q(x) := p(x y) 2 Pn : On applying the already proved inequality with q; we obtain jp0 (y)e y j = jq0 (0)e y j

 e y (8n + 2) x2max jq(x)e x j ;1  (8n + 2) x2max;1 jp(x + y)e x y j  (8n + 2) x2max;1 jp(x)e x j ; [0

)

(

whi h nishes the proof.

[0

)

[0

)

+ )

ut

Inequalities for Polynomials with Constraints 443

b℄ There is an absolute onstant > 0 su h that

kp0(x)e x k ;1  pn kp(x)e x k ;1 [0

[0

)

)

holds for every p 2 Pn having no zeros in the disk with diameter [0; 2℄: Proof. See Erdelyi [89 ℄, where this result is extended to various other general lasses of weight fun tions and onstrained polynomials. ut E.20 Inequalities for Generalized Polynomials with Constraints. For s 2 (0; 2); let

Pen;k (s) :=

n

o

p 2 Pen;k : m(fx 2 [ 1; 1℄ : jp(x)j  1g)  2 s :

The polynomials Tn;k

2 Pn;k are de ned by

Tn;k (x) := Tnfn k; n k + 1; : : : ; n; [0; 1℄g

1 2




(x + 1) ;

where 0  k  n are integers. (Note that the notation introdu ed in Se tion 3.3 de nes Tn;k only on [ 1; 1℄; and, to be pre ise, Tn;k denotes the polynomial de ned above on [ 1; 1℄:)

a℄ Remez-Type Inequality for Pen;k . The inequality

kpk



[

2+s ;  Tn;k 2 s



1 1℄

holds for every p 2 Pen;k (s) and s 2 (0; 2): Equality holds if and only if

p(x)  Tn;k



2x

2

s

+

2

s



s

:

Proof. The proof is quite similar to that of Theorem 5.1.1 (Remez Inequality); see Borwein and Erdelyi [92℄. ut b℄ A Numeri al Version of Part a℄. Show that Tn;k



2+s 2 s



 exp

for every s 2 (0; 1℄: Proof. See Borwein and Erdelyi [92℄.

p

 

5



nks + ns

ut

444 A5. Inequalities for Polynomials with Constraints

Let GAPN;K , 0 (A.4.1) for whi h

 K  N , be the set of all f 2 GAPN n X j =1 jzj j1

of the form

K:

rj

Note that if 0  K  N are integers and p 2 PN;K ; then jpj 2 GAPN;K : For 0  K  N and 0 < s < 2, let GAPN;K (s) denote the olle tion of all f 2 GAPN;K for whi h

m(fx 2 [ 1; 1℄ : f (x)  1g)  2 s :

The next part of the exer ise extends a numeri al version of part a℄ to the

lasses GAPN;K (s):

℄ Remez-Type Inequality for GAPN;K . There exists a onstant  5 su h that  p  kf k ;  exp NKs + Ns 1

[

1

1 1℄

for every f 2 GAPN;K (s) and s 2 (0; 1℄: Hint: Let q 2 N be the ommon denominator of the numbers rj . Apply part a℄ with p := f q 2 Pe qN; qK (s), and then use part b℄. ut d℄ Nikolskii-Type Inequality for GAPN;K . Let  be a nonnegative nonde reasing fun tion de ned on [0; 1) su h that (x)=x is nonin reasing on [0; 1): Then there exists an absolute onstant  25e su h that 2

2

2

2

2

k(f )kLp ;  ( maxf1 ; q NK ; qN g) =q for every f 2 GAPN;K and 0 < q < p  1: [

1 1℄

2

2

1

=p

1

k(f )kLq

;

[

1 1℄

The ase K = N (when there is no restri tion on the zeros) of part d℄ is the ontent of Theorem A.4.4. If qK p  1; then the Nikolskii fa tor in the unrestri ted ase (K = N )pis like ( qN ) =q =p ; while in our restri ted p

ases it improves to ( q NK ) =q =p : 2

2

2

2

2

2

1, and then a similar argument, as in the proof of Theorems A.4.3 and A.4.4, gives the result for arbitrary 0 < q < p  1: Thus, in the sequel, let 0 < q < p = 1: Using the inequality of part ℄ with

Proof. It is suÆ ient to prove the inequality when p =

s := minf1 ; ( q NK )

; ( qN ) g and re alling the onditions pres ribed for ; we on lude that 
m x 2 [ 1; 1℄ : ((f (x)))q  e k(f )kq ; 2 1

2

1

1

1

2



m x2[  m x2[ s

[

1 1℄




1; 1℄ : f (x)  e =q kf k ; p

1; 1℄ : f (x)  exp NKs + Ns kf k 2

[

1

1 1℄

[

;

1 1℄




Inequalities for Polynomials with Constraints 445

for every f where

2 GAPN;K : Now integrating only on the subset E  e k(f )kq

((f (x)))q

2

[

;

1 1℄

of [ 1; 1℄

;

we obtain

k(f )k

q [

2

;

1 1℄

Z

 me(E ) ((f (x)))q dx E  maxf1 ; q NK ; qN gk(f )kqLq 2

2

for every f 2 GAPN;K , where := e (assuming we have  25e . 2 1

2

2

2

2

1

[

;

1 1℄

 1). Sin e  5; ut 1

e℄ Remez-Type Inequality for Bn ( 1; 1). Show that

kpk

[

;

1 1℄





2

2

n

s

for every p 2 Bn ( 1; 1) satisfying

m(fx 2 [ 1; 1℄ : jp(x)j  1g)  2 s : Equality holds if and only if p is of the form

p(x) = 



 1x n : 2 s

ut

Proof. See Erdelyi [90b℄.

f℄ Markov-Type Inequality for GAPN;K . There exists an absolute onstant > 0 su h that

kf 0k

[

;

1 1℄

 N (K + 1)kf k

[

;

1 1℄

for every f 2 GAPN;K of the form (A.4.1) with ea h rj  1: Re all that jf 0 (x)j is well-de ned for every f 2 GAPN;K and x 2 R, as the modulus of the one-sided derivative of f at x. The ondition that rj  1 for ea h j in (A.4.1) is needed to ensure that jf 0 (zj )j < 1 if zj 2 R. The above result generalizes the orresponding polynomial inequality for the lasses Pn;k . Proof. See Borwein and Erdelyi [92℄. ut

446 A5. Inequalities for Polynomials with Constraints

E.21 The Ilye-Sendov Conje ture. The following problem is known as the Ilye-Sendov onje ture. As before, let D := fz 2 C : jz j < 1g: Suppose P 2 Pn has all its zeros in the losed unit disk D. Then ea h

losed disk entered at a zero of P ontains a zero of of P 0 . Although the problem in this generality is still unanswered, several spe ial ases have been settled. The ase outlined in part a℄ was rst proved in Rubinstein [68℄. V^aj^aitu and Zahares u [93℄ ontains stronger results; see also Miller [90℄. Milovanovi , Mitrinovi , and Rassias [94℄ has a dis ussion about the re ent status of the onje ture. a℄ Suppose

P (z ) =

n Y k=1

(z

jzk j  1 ; jz j = 1 :

zk ) ;

1

Show that P 0 has at least one zero in the losed disk entered at z : Proof. Without loss of generality, we may assume that z := 1: Then P is of the form P (z ) = (z 1)Q(z ); where 1

1

Q(z ) :=

n Y k=2

(z

jzk j  1 :

zk ) ;

Suppose the statement is false. Then R(z ) := P 0 (z + 1) has no zero in the

losed unit disk D: Hen e 0 R (0) R(0)

< n 1:

Observe that

R(0) = Q(1) Hen e

However,

0 Q (1) Q(1) 



R0 (0) = 2Q0 (1) :

and

=



1 R0 (0) n 1 < : 2 R(0) 2

Q0 (1) Re = Re Q(1)

n X k=2

1

1

!

zk

 n2 1:

This ontradi tion nishes the proof. b℄ The polynomial P (z ) := z n 1 shows the sharpness of part a℄.

ut

Inequalities for Polynomials with Constraints 447

It follows from Theorem A.5.3 that there exists an absolute onstant

> 0 su h that

kp0k

[

;

1 1℄

 n kpk

[

;

1 1℄

for every p 2 Pn with no zeros in the open unit disk. The polynomials pn (x) := (x+1)n show that up to the absolute onstant > 0 this inequality is the best possible. The next exer ise shows that the \right" Markov fa tor on [ 1; 1℄ for polynomials of degree at most n with omplex oeÆ ients and with no zeros in the open unit disk is n(1 + log n) rather than n. This is an observation of Halasz. E.22 Markov Inequality for Pn with No Zeros in the Unit Disk. Show that there exists an absolute onstant > 0 so that kp0 k ;  n(1 + log n)kpk ;

for every p 2 Pn with no zeros in the open unit disk. Show also that there exist polynomials pn 2 Pn with no zeros in the open unit disk su h that kp0nk ;  n(1 + log n)kpnk ; ; n = 1; 2; : : : with an absolute onstant > 0: Pro eed as follows: a℄ Show that if z 2 C is outside the open unit disk, then 1

[

[

1

1 1℄

[

2

1 1℄

[

1 1℄

1 1℄

2

jx jx

zj zj

1

 jxj

for every x 2 R n [ 1; 1℄: b℄ Suppose p 2 Pn has no zeros in the open unit disk. Let x 2 R n [ 1; 1℄. Show that jp(x)j  jxjn jp(x )j : 1

℄ Prove the upper bound of the exer ise. Hint: Use part b℄ and E.11 a℄. d℄ Let zk := e ik= n ; k = 1; 2; : : : ; n be the (2n + 1)th roots of unity in the open upper half-plane. Let 2

p

(2

+1

2

Show that jp n (x)j = jx 2

+1

+2

n Y

(z

k=1 x R:

zk ) : 2

1j for every 2 Note that this implies jp n ( 1)j = kp n k ; = 2 : 2

2

Show also that

+1)

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

2

ut

n+1

+1

0 p2n+1 ( p (

n+1

2

2



+1

[

1 1℄

1)  n(1 + log n) 1)

with an absolute onstant > 0:

This is page

448

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Pommerenke, Ch., On some metri properties of omplex polynomials, Mi higan Math. J. 8 (1961), 97{115. Pommerenke, Ch., On some metri properties of polynomials with real zeros, I and II, Mi higan Math. J. 6 (1959), 25{27; and 8 (1961), 49{54. Pommerenke, Ch., Univalent Fun tions, Vandenhoe k & Rupre ht, Gottingen, 1975. Protapov, M.K., Some inequalities for polynomials and their derivatives, Vestnik Moskov. Univ. Ser. I Mat. Mekh. 2 (1960), 10{19 (in Russian). Rahman, Q.I., On extremal properties of the derivatives of polynomials and rational fun tions, J. London Math. So . 35 (1960), 334{336. Rahman, Q.I., On extremal properties of the derivatives of polynomials and rational fun tions, Amer. J. Math. 113 (1991), 169{177. Rahman, Q.I., & G. Labelle, L'in uen e des zeros sur la borne dans le theoreme de Marko pour les polyn^omes sur l'intervalle unite, J. London Math. So . 43 (1968), 183{185. Rahman, Q.I., & G. S hmeisser, Some inequalities for polynomials with a pres ribed zero, Trans. Amer. Math. So . 216 (1976), 91{103. Rahman, Q.I., & G. S hmeisser, Les Inegalites de Marko et de Bernstein, Les Presses de L'Universite de Montreal, 1983. Rahman, Q.I., & G. S hmeisser, Lp inequalities for entire fun tions of exponential type, Trans. Amer. Math. So . 320 (1990), 91{103. Rainville, E.D., Spe ial Fun tions, Chelsea, New York, NY, 1971. Rassias, G.M., J.M. Rassias, & Th.M. Rassias, A ounter-example to a onje ture of P. Erd}os, Pro . Japan A ad. S i. 53 (1977), 119{121. Remez, E.J., Sur une propriete des poly^omes de T hebys he, Comm. Inst. S i. Kharkow 13 (1936), 93{95.

Bibliography 463 Rezni k, B., An inequality for produ ts of polynomials, Pro . Amer. Math. So . (1993).

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Notation

De nitions of the more ommonly used spa es are given. The equation numbers here are the same as the equation numbers in the text. Throughout the book, unless otherwise stated, spans should be assumed to be real. Likewise, in fun tion spa es, unless otherwise stated, the fun tions should be assumed to be real valued. The Basi Spa es. (1:1:1)

P

(

n

:= p : p(z ) =

(1:1:2)

Pn :=

(1:1:3)

R

(1:1:4)

(

p : p(z ) =

m;n

n X k=0 n X k=0

ak

zk;

ak

zk;

ak 2 C

)

: )

ak 2 R :









p

; q 2 P : := : p 2 Pm n q

Rm;n := pq : p 2 Pm; q 2 Pn :

468

Notation

(1:1:5)

T

(

n

: = t : t() := n

n X

= t : t(z ) = a + 0

Tn : =

n

ak

k= n

t : t(z ) = a + 0

ak 2 C

eik ;

)

n X

o

k=1

(ak os kz + bk sin kz ) ; ak ; bk 2 C :

n X k=1

o

(ak os kz + bk sin kz ) ; ak ; bk 2 R :

Muntz Spa es. (3:4:2)

(3:4:3)

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

M () :=

1 [ n=0

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

Asso iated with a sequen e (i )1 i with Re(i ) > 1=2 for ea h i; the nth Muntz-Legendre polynomial is: =0

(3:4:5)

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

1

=0

For distin t i ; (3:4:6)

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

with

k;n :=

Qn

n X k=0

k;n xk ;

j =0 (k

+ j + 1) : j ;j 6 k (k j ) 1

Qn

=0

=

Also, (3:4:8)

Ln := (1 + n + n )

=

1 2

Ln

is the nth orthonormal Muntz-Legendre polynomial.

x 2 (0; 1)

Notation 469

Rational Spa es.

D := fz 2 C : jz j < 1g ;

K := R (mod 2) ; (3.5.3)

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

(3.5.4)

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

Also

1

1





p(x) jx ak j : p 2 Pn :

Qn

2

D := fz 2 C : jz j = 1g :

k=1

 Qn

2

k=1

t() j os 



ak j

: t 2 Tn :

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

2

2

(

:= and

T

2

Q2n

Q2n

t() : t 2 Tn sin(( ak )=2)

k=1

; a n ; K ) := 2

1

2

2

Pn (a ; a ; : : : ; an ; [ 1

2

n R;

1; 1℄) :=

and

n (a1 ; a2 ; : : :

; an ; [ 1; 1℄) :=

on [ 1; 1℄ with a ; a ; : : : ; an 2 C 1

2

P

n (a1 ; a2 ; : : :

on D with a ; a ; : : : ; an 2 C 1

2

n (a1 ; a2 ; : : :

; an ; R) :=

on R with a ; a ; : : : ; an 2 C 1

2

n R:



Qn

p(x) jx ak j : p 2 Pn

Qn

p(x) : p 2 Pn (x ak )

k=1







k=1

1; 1℄;

p(z ) : p 2 Pn (z ak )



p(x) jx ak j : p 2 Pn



Qn

p(z ) : p 2 Pn (z ak )



Qn

Qn

k=1



2

and

P



n D; and

Pn (a ; a ; : : : ; an ; R) := 1

k=1

n[

; an ; D) :=

2

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

(

n (a1 ; a2 ; : : :

on K with a ; a ; : : : ; a n 2 C

P

1

k=1



k=1

)

)

470

Notation

Generalized Polynomials. The fun tion m Y

f (z ) = j!j

(A:4:1)

j =1

jz

zj jrj

with 0 < rj 2 R; zj 2 C ; and 0 6= ! 2 C is alled a generalized nonnegative (algebrai ) polynomial of (generalized) degree (A:4:2)

N :=

m X j =1

rj :

The set of all generalized nonnegative algebrai polynomials of degree at most N is denoted by GAPN : The fun tion

f (z ) = j!j

(A:4:3)

m Y j =1

j sin((z

zj )=2)jrj

with 0 < rj 2 R; zj 2 C ; and 0 6= ! 2 C is alled a generalized nonnegative trigonometri polynomial of degree m 1X r : N := 2j j

(A:4:4)

=1

The set of all generalized nonnegative trigonometri polynomials of degree at most N is denoted by GTPN : Constrained Polynomials. The following lasses of polynomials with onstraints appear in Appendix 5:

Pn;k := fp 2 Pn : p has at most k zeros in Dg ; Bn (a; b) :=

n

p 2 Pn : p(x) = 

n X j =0

Pen;k (a; b) := fp = hq : h 2 Bn

0  k  n;

j (b x)j (x a)n j ; j k (a; b) ;

q 2 Pk g ;

o

0

; a < b;

0  k  n:

Notation 471

Fun tion Spa es. The uniform or supremum norm of a omplex-valued fun tion f de ned on a set A is de ned by kf kA := sup jf (x)j : x2A

The spa e of all real-valued ontinuous fun tions on a topologi al spa e A equipped with the uniform norm is denoted by C (A): If A := [a; b℄ is equipped with the usual metri topology, then the notation C [a; b℄ := C (A) is used. Let (X; ) be a measure spa e ( nonnegative) and p 2 (0; 1℄: The spa e Lp () is de ned as the olle tion of equivalen e lasses of real-valued measurable fun tions for whi h kf kLp  < 1; where (

kf kLp  (

)

:=

Z

X

jf j

)

1=p

p d

;

p 2 (0; 1)

and

kf kL1  (

)

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

The equivalen e lasses are de ned by the equivalen e relation f  g if f = g -almost everywhere on X: When we write Lp [a; b℄ we always mean Lp (); where  is the Lebesgue measure on X = [a; b℄: The notations Lp (a; b); Lp [a; b); and Lp (a; b℄ are also used analogously to Lp [a; b℄: Again, it is always our understanding that the spa e Lp () is equipped with the Lp () norm. Sometimes C (A) denotes the spa e of all omplex-valued ontinuous fun tions de ned on A equipped with the uniform norm. Similarly, Lp () may denote the spa e all omplex-valued ontinuous fun tions for whi h kf kLp  < 1: This should always be lear from the ontext; many times, but not always, the reader is reminded if C [a; b℄ or Lp () is omplex. (

)

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473

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Index

Abel, Niels Henrik, 3, 61 Algorithms, 356{371 evaluation of xn , 363 fast Fourier transform, 359 fast polynomial division, 362 fast polynomial expansion, 363 fast polynomial multipli ation, 361 for Chebyshev polynomials, 371 for ounting zeros, 364{370 for polynomial evaluations, 363 for reversion of power series, 362 interpolation, 360 Newton's method, 362, 364{367 Remez, 371 zero nding for polynomials, 366{370 Alternation set, 93 Alternation theorem, 94 Apolar polynomials, 23, 24, 25 Ar length of algebrai polynomials, 31 Ar length of trigonometri polynomials, 35 Berman's formula, 166 Bernstein fa tor, 145, 150, 152,

322{328 Bernstein polynomials, 163{164 Bernstein-Szeg}o inequality, 231, 245, 259, 321{323 for entire fun tions of exponential type, 245 for rational fun tions on [ 1; 1℄, 322 for rational fun tions on K , 322 for rational fun tions on R, 323 for trigonometri polynomials, 232 Bernstein-type inequality bounded, 178, 182, 213{214 for Chebyshev spa es, 206 for onstrained polynomials, 420{447 for entire fun tions of exponential type, 245 for exponential sums, 289 for generalized polynomials, 392{416 for generalized polynomials in Lp , 401{417 for higher derivatives, 258 for nondense Muntz spa es, 213, 310 for polynomials, 232{233, 390

474 Index

for polynomials in Lp , 235, 390, 401{417 for produ ts of Muntz spa es, 317 for rational fun tions on [ 1; 1℄, 323, 327 for rational fun tions on D, 324 for rational fun tions on K , 322, 327 for rational fun tions on R, 323, 329 for self-re ipro al polynomials, 339 for trigonometri polynomials, 232 unbounded, 154, 206{217 weighted, 257 Bessel's inequality, 46 Best approximation, 94 by rationals, 99 to x , a problem of Lorentz, 108 to xn , 99 uniqueness, 98 Binomial oeÆ ient, 62 Blas hke produ t, 190, 324 Blumenthal's theorem, 78 Bombieri's norm, 274 Bounded Bernstein-type inequality; see Bernstein-type inequality Bounded Chebyshev-type inequality; see Chebyshev-type inequality Bounded linear fun tionals, 50 Budan-Fourier theorem, 369

Cardano, Girolamo, 3 Cartan's lemma, 350 Cau hy determinant, 106 Cau hy indi es, 367 Cau hy's integral formula, 14 Cau hy-S hwarz inequality, 42 for sequen es, 46 Cesaro means, 165 Chebyshev onstants, 39 Chebyshev, P.L., 31 Chebyshev polynomials algorithms for, 371 best approximation to xn , 30

omposition

hara terization, 33 expli it formulas, 30, 32

in Chebyshev spa es, 114{125 in rational spa es; see Chebyshev rationals orthogonality, 32 redu ibility, 36 se ond kind, 37 three-term re ursion, 32 trigonometri on subintervals, 235 uniqueness, 118 Chebyshev rationals, 139{153 and orthogonality, 147 derivative formulas, 146 in algebrai rational spa es, 142 in trigonometri rational spa es, 143 of the rst and se ond kind, 141 on the real line, 151 partial fra tion representation, 144 Chebyshev spa e, 92 dimension on the ir le, 100 fun tions with pres ribed sign

hanges, 100 Chebyshev system, 91{100 extended omplete, 97 Chebyshev's inequality, 235, 390 Chebyshev-type inequality bounded, 179, 182 expli it bounds via Paley-Weiner theorem, 196 for entire fun tions of exponential type, 245 Christoel fun tion, 74 for Muntz spa es, 195 Christoel numbers, 67 Christoel-Darboux formula, 60 CoeÆ ient bounds for polynomials in spe ial bases, 124 in nondense rational spa es, 153 of Markov, 248 Comparison theorem, 103, 120, 122, 183 Completeness, 48, 79 Complexity on erns, 356{371 Conse utive zeros of p0 , 26 Constrained polynomials, 417{447 Bernstein-type inequality, 420, 425, 427

Index 475 p inequalities, 422 Markov-type inequality, 417{447 Nikolskii-type, 444 Remez-type, 443{445 S hur-type, 436{437 Cotes numbers, 67 Cubi equations, 4

Fekete point, 38 Fekete polynomial, 38 Ferrari, Ludovi o, 3 Ferro, S ipione del, 3 Fourier oeÆ ient, 53 Fourier series, 53 Fundamental theorem of algebra, 3

d'Alembert, Jean le Rond, 13

Gamma fun tion, 63 Gauss, Carl Friedri h, 13 Gaussian hypergeometri series, 62 Gauss-Ja obi quadrature, 67, 75 Gauss-Lu as theorem, 18 Gegenbauer polynomials, 65 Generalized polynomials Bernstein-type inequality, 399, 407 Lp inequalities, 401{407 Markov-type inequality, 399, 407 Nikolskii-type inequality, 394{395 Remez-type inequality, 393{394, 414 S hur-type inequality, 395 weighted inequalites, 407 Girard, Albert, 13 Gra e's omplex version of Rolle's theorem, 25 Gra e's theorem, 18 Gram's lemma, 176 Gram-S hmidt, 44

L

de la Vallee Poussin theorem, 99 Denseness, 154{226 of Markov spa es, 206{217 of Muntz polynomials, 171{205 of Muntz rationals, 218{226 of polynomials, 154{170 Derivatives of Markov systems, 112 Des artes' rule of signs, 22, 102 Des artes system, 100{113 examples, 103 lexi ographi properties, 103 Divide and onquer, 358 Division of polynomials, 15, 362

Elementary symmetri fun tion, 5 Enestrom-Kakeya theorem, 12 Erd}os inequality, 431 Erd}os-Turan inequality, 435 Euler, Leonhard, 13 Evaluation of xn , 363 Exponential sums; see Muntz polynomials a problem of Lorentz, 291 Bernstein-type inequality, 291 Markov-type inequality, 294 Nikolskii-type inequality, 289 Turan's inequality, 295 with nonnegative exponents, 294 Exponential type, 196, 245 Extended omplete Chebyshev system, 97 Fa tor inequalities, 260{274

via Mahler's measure, 271{273 Fa torization, 10, 36 Fast Fourier transform, 359 Favards theorem, 73 Fejer gap, 27{28 Fejer operators, 164 Fejer's theorem, 165

Haar spa e, 92 Haar system, 92 Halley's method, 365 Hardy spa e, 189 Helly's onvergen e theorem, 71 Helly's sele tion theorem, 71 Hermite interpolation, 9 Hermite polynomials, 57 expli it formulas, 65 Hilbert spa e, 42 Holder's inequality, 17, 49 Horner's rule, 8 Hypergeometri dierential equation, 63 Hypergeometri fun tions, 62 Identity theorem, 15 Inequalities

476 Index

Bernstein-Szeg}o inequality; see Bernstein-Szeg}o inequality Bernstein-type; see Bernstein-type inequality Bessel's; see Bessel's inequality bounded Bernstein-type; see bounded Bernstein-type inequality bounded Chebyshev-type; see bounded Chebyshev-type inequality Cau hy-S hwarz; see Cau hyS hwarz inequality Chebyshev-type; see Chebyshevtype inequality for fa tors; see fa tor inequalities for Muntz polynomials; see Muntz polynomials Holder's; see Holder's inequality Lax-type; see Lax Markov-type; see Markov-type inequality metri ; see metri inequalities Minkowski's; see Minkowski's inequality Nikolskii-type; see Nikolskii-type inequality Remez-type; see Remez-type inequality Russak's; see Russak S hur-type; see S hur-type inequality Triangle; see Triangle inequality unbounded Bernstein-type; see unbounded Bernstein-type inequality Inner produ t, 41 Inner produ t spa e, 41 Integer-valued polynomials, 10 Interpolation Hermite; see Hermite interpolation Lagrange; see Lagrange interpolation Newton; see Newton interpolation

Ja obi polynomials, 57, 63 expli it formulas, 63 Jensen ir les, 19 Jensen's formula, 187

Jensen's inequality, 414 Jensen's theorem, 19

Kernel fun tion, 47, 132

Kolmogorov's inequality, 285 Korovkin's theorems, 163

La unary spa es, 308

quasi-Chebyshev polynomials, 316 Lagrange interpolation, 8 Laguerre polynomials, 57, 66, 130 expli it formulas, 66 Laguerre's theorem, 20 Lax-type inequality, 438 for rationals, 329 Malik's extension, 438 on a half-plane, 338 Legendre polynomials, 57 Lemnis ates of onstant modulus, 352{353 Lexi ographi properties, 116 for Muntz polynomials, 120, 314 for Muntz-Legendre polynomials, 136 for sinh systems, 122 Liouville's theorem, 15 Logarithmi apa ity, 38 Lorentz degree, 82 for polynomials, 86 for trigonometri polynomials, 89 Lorentz's problem, 108, 291 Lp norm, 6, 48, 471 lp norm, 6, 471 Lu as' theorem, 18

Mahler's measure, 271

Mairhuber theorem, 98 Markov system, 100

losure of nondense, 211 derivative of, 112 Markov-Stieltjes inequality, 76 Markov-type inequality for Pn , 233 for Pn , 255 for onstrained polynomials, 417{447 for onstrained polynomials in Lp , 422, 428{429

Index 477 for exponential sums, 276{280, 294{295 for generalized polynomials, 399{407, 445 for generalized polynomials in Lp , 401{407 for higher derivatives, 248{260 for monotone polynomials, 439{ 441 for Muntz polynomials, 276{279, 287{288 for Muntz polynomials in Lp , 279{280 for nonnegative polynomials, 420, 439 for rational fun tions, 336 for self-re ipro al polynomials, 339 for trigonometri polynomials on subintervals, 242{245 in the omplex plane, 235 weighted, 442{443 Maximum prin ipal, 15 m-distribution, 57 Mergelyan's theorem, 170 Mesh of zeros, 155 Metri inequalities, 344{355 for polynomials, 345{346 for rational fun tions, 347{349 Minkowski's inequality in Lp ; p  1, 49 in Lp ; p  1, 52 Moment, 57 Moment problem, 70 Monotone operator theorem, 163 Multipli ation of polynomials, 361 Muntz polynomials bounded Bernstein-type inequality, 178, 182, 213, 310, 317 bound for smallest zero, 313 lexi ographi properties of zeros, 116, 120 Newman's inequality, 276, 301 Newman's inequality in Lp , 279 Newman's problem, 317 Nikolskii-type inequality, 281, 298, 317 positive zeros, 22

Remez-type inequality, 304, 307, 316 where the sup norm lives, 301 Muntz rationals, 218{226 denseness, 218 Muntz spa e, 125{126 lexi ographi properties of zeros, 120 produ ts of, 222, 316 quotients of, 218{226 Muntz system, 125{136

losure, 171{205 nondense, 303{319 Muntz-Legendre polynomials, 125{138 de nition, 126 dierential re ursion, 129 global estimate of zeros, 136 integral re ursion, 132 lexi ographi properties of zeros, 133{136 orthogonality, 127, 132 orthonormality, 128 Rodrigues-type formula, 128, 131 zeros of, 133 Muntz's theorem, 171{205 another proof, 176, 192

losure of span in, 178, 181, 185 in C [0; 1℄, 171 in C [a; b℄, 180, 184 in L2 [0; 1℄, 171 in Lp [0; 1℄, 172 in Lp [a; b℄, 186 in Lp (w), 311 on sets of positive measure, 303

Newman's onje ture

on denseness of produ ts, 316 on denseness of quotients, 220, 223 Newman's inequality, 275{279 an improvement, 287 for Muntz polynomials, 276 for Muntz polynomials in Lp , 279 on positive intervals, 301 Newton's identities, 5 Newton Interpolation, 10 Newton's method, 362, 364{367 for x1=2 , 365 in many variables, 366

478 Index

Nikolskii-type inequality for onstrained polynomials, 444 for exponential sums, 289 for generalized polynomials, 394, 395 for Muntz polynomials, 281, 298, 317 for produ ts of Muntz spa es, 317 Nondense Muntz spa es, 303{319 Nonnegative polynomials, 70, 85, 417, 420 Nonnegative trigonometri polynomials, 85, 409 Norms, 471 Lp ; see Lp norm lp ; see lp norm supremum; see supremum norm

Orthogonal olle tion, 43

Orthogonal fun tions, 41{56 Orthogonal polynomials, 57{79 as ontinued fra tions, 79 as determinants, 76

hara terization of ompa t support, 77 Gegenbauer; see Gegenbauer polynomials Hermite; see Hermite polynomials interla ing of zeros, 61 Ja obi; see Ja obi polynomials Laguerre; see Laguerre polynomials Legendre; see Legendre polynomials Muntz-Legendre; see MuntzLegendre polynomials simple real zeros, 61 ultraspheri al; see Ultraspheri al polynomials Orthogonal rational fun tions, 147 Orthonormal set, 43

Paley-Weiner theorem, 196

Parallelogram law, 42 Parseval's identity, 48 Partial fra tion de omposition, 7, 144 Pellet's theorem, 16 Polar derivative, 20 Polynomials

as sums of squares, 85, 348 Bernstein; see Bernstein polynomials Chebyshev; see Chebyshev polynomials Gegenbauer; see Gegenbauer polynomials generalized; see generalized polynomials growth in the omplex plane, 239 Hermite; see Hermite polynomials in xn , 167 integer valued; see Integer valued polynomials Ja obi; see Ja obi polynomials Laguerre; see Laguerre polynomials Legendre; see Legendre polynomials Muntz; see Muntz polynomial Muntz-Legendre; see MuntzLegendre polynomials number of real roots, 17, 137 symmetri , 5 trigonometri ; see Trigonometri polynomial Ultraspheri al; see Ultraspheri al polynomials with integer oeÆ ients, 169 with nonnegative oeÆ ients, 79{90, 417 with real roots, 345, 347{348 Produ ts of Muntz spa es, 222, 316

Quadrati equations, 4

Quarti equations, 4 Quasi-Chebyshev polynomials, 316, 342

Railway tra k theorem, 98

Rakhmanov's theorem, 78 Rational fun tions algebrai , 139 Chebyshev polynomials of, 139{ 153

oeÆ ient bounds, 153 inequalities; see inequalities trigonometri , 139 Rational spa es

Index 479 of algebrai rational fun tions, 139{153, 320{321 of trigonometri rational fun tions, 139{153, 320{ 321 Re ursive bounds, 359 Remez's algorithm, 371 Remez-type inequality for algebrai polynomials, 228 for onstrained polynomials, 443, 445 for generalized polynomials, 393, 394 for generalized polynomials in Lp , 401{402 for Muntz spa es, 307 for nondense Muntz spa es, 304 for produ ts of Muntz spa es, 316 pointwise, 414 for trigonometri polynomials, 230 Reprodu ing kernel, 47, 132 Reversion of power series, 362 Riemann-Lebesgue lemma, 54 Riesz representation theorem, 50 Riesz's identity, 390 Riesz's lemma, 237 Riesz-Fis her theorem, 50 Rising fa torial, 62 Rolle's theorem, 25 Rou he's theorem, 14, 16 Russak's inequalities, 336

Salem numbers, 6 S hur's theorem, 17 S hur-type inequality for algebrai polynomials, 233 for onstrained polynomials, 436{437 for generalized polynomials, 395 for rational fun tions, 337 for trigonometri polynomials, 238 Self-re ipro al polynomials, 339 quasi-Chebyshev polynomials, 342 Somorjai's theorem, 218 Spa e Chebyshev; see Chebyshev spa e Des artes; see Des artes spa e Haar; see Haar spa e Muntz; see Muntz spa e

rational; see rational spa e Stieltjes' theorem, 78 Stone-Weierstrass theorem, 161 Sums of squares of polynomials, 85, 348 Supremum norm, 6, 29, 471 Symmetri fun tion, 5 Symmetri polynomial, 5 Szeg}o's inequality, 391 Szeg}o's theorem, 23, 235

Tartaglia, Ni

olo, 3

T heby hev; see Chebyshev Three-term re ursion, 59 Totally positive kernels, 110 Trans nite diameter, 38 Triangle inequality, 42 Trigonometri polynomial, 2 Trigonometri polynomials of longest ar length, 35 Turan's inequality, 434 Turan's inequality for exponential sums, 295

Ultraspheri al polynomials, 65

Unbounded Bernstein-type inequality, 206{217

hara terization of denseness, 207 Uni ity theorem, 15

Variation diminishing property, 111 Vandermonde determinant, 38, 103 Videnskii's inequalities, 242{245 Walsh's two ir le theorem, 20

Weierstrass' theorem, 154{170 for Markov systems, 155 for polynomials, 159 for polynomials in x , 167 for polynomials with integer

oeÆ ients, 169 for trigonometri polynomials, 165 in Lp , 169 on ar s, 170 Stone-Weierstrass theorem, 161 Wronskian, 22

Zeros, 11{18

algorithms for nding, 364{367

480 Index

omplexity of, 367

ounting by winding number, 370 in a disk, 369 in an interval, 368 in Chebyshev spa es, 99 lo alizing, 367 maximum number at one, 137 maximum number of positive, 17 of Chebyshev polynomials, 34, 116, 120, 122

of derivatives of polynomials, 18{28 of integrals of polynomials, 24 of Muntz polynomials, 120 of Muntz-Legendre polynomials, 133{136 of orthogonal polynomials, 61 Zolotarev, 35 Zoomers, 218