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  • Lecture Notes

    Orthogonal Polynomials

    Summer Term 2012

    Peter Junghanns

    Remark: The present lecture notes contain only a framework of the contents of thelectures. The lectures themselves present detailed expositions, proofs and examples.

  • 2

  • Contents

    1 Introduction 7

    1.1 Orthogonality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

    1.2 Generating function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

    1.3 Birth and death process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

    1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

    2 Basic Theory of Orthogonal Polynomials 15

    2.1 Moment functionals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

    2.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

    2.3 Recursion formulas and the formula of Christoffel-Darboux . . . . . . . . . . . . 18

    2.4 Zeros and Gauss quadrature rule . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

    2.5 The Jacobi polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

    2.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

    3 Singular Integral Operators 27

    3.1 Cauchy singular integral operators . . . . . . . . . . . . . . . . . . . . . . . . . . 27

    3.2 Singular integro-differential operators . . . . . . . . . . . . . . . . . . . . . . . . . 28

    3.3 Weakly singular integral operators . . . . . . . . . . . . . . . . . . . . . . . . . . 29

    4 Continued Fractions and Orthogonal Polynomials 31

    4.1 Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

    4.2 Jacobi fractions and orthogonal polynomials . . . . . . . . . . . . . . . . . . . . . 33

    4.3 Chain sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

    3

  • 4 CONTENTS

    4.4 Chain sequences and orthogonal polynomials . . . . . . . . . . . . . . . . . . . . 37

    5 The representation Theorem 41

    5.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

    5.2 The Representation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

    5.3 On the Location of the Spectral Points of a Representation . . . . . . . . . . . . 43

    5.4 On the Determinacy of a Representation . . . . . . . . . . . . . . . . . . . . . . . 44

    5.5 Classical Moment Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

    6 Integral equations 47

    6.1 The Nystrom method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

    6.2 Collectively compact operator sequences . . . . . . . . . . . . . . . . . . . . . . . 48

    6.3 The case = = 0 and X = C[1, 1] . . . . . . . . . . . . . . . . . . . . . . . . 49

    6.4 The application of weighted spaces of continuous functions . . . . . . . . . . . . . 50

    7 Orthogonal Polynomials in the Complex Plane 53

    7.1 Definitions. Location of the Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

    7.2 Asymptotic Properties of the Zeros . . . . . . . . . . . . . . . . . . . . . . . . . . 54

  • Bibliography

    [1] D. Berthold, W. Hoppe, B. Silbermann, A fast algorithm for solving the generalized airfoilequation, J. Comp. Math., 43 (1992), 185-219.

    [2] T. S. Chihara, An Introduction to Orthogonal Polynomials, Gordon & Breach, New York,1978.

    [3] G. Freud, Orthogonale Polynome, Berlin, 1969.

    [4] P. Junghanns, EAGLE-GUIDE Orthogonale Polynome, Edition am Gutenbergplatz, Leipzig,2009.

    [5] P. Junghanns, Hilbertraummethoden, Skript zur Vorlesung, SS 2011,

    http://www-user.tu-chemnitz.de/ peju/

    [6] I. P. Natanson, Konstruktive Funktionentheorie, Berlin, 1955.

    [7] P. G. Nevai, Orthogonal Polynomials, Amer. Math. Soc., Rhode Island, 1979.

    [8] P. Nevai, Orthogonal Polynomials, Theory and Practice, NATO ASI Series C, Vol. 294, 1990.

    [9] N. Obreschkoff, Verteilung und Berechnung der Nullstellen reeller Polynome, Berlin, 1963.

    [10] R. Remmert, Funktionentheorie 2, Springer-Verlag, Berlin, . . ., 1992.

    [11] G. Szego, Orthogonal Polynomials, Amer. Math. Soc., Rhode Island, 1939.

    5

  • 6 BIBLIOGRAPHY

  • Chapter 1

    Introduction

    1.1 Orthogonality

    By K[x] we denote the linear space of all polynomials of the form

    p(x) = anxn + an1xn1 + . . .+ a1x+ a0, (1.1)

    where ak K are from a field (for example, C or R) and n N0 = {0, 1, 2, . . .} . hence, theelements of C[x] are polynomials with complex coefficients, which we also consider as mapsp : R C, x 7 p(x) from the set of real numbers into the set of complex numbers. Thus, theindependent variable x sei also eine beliebige reelle Zahl. If an in (1.1) is different from zero,then deg p = n is called degree of the polynomial p(x) and an the leading coefficient ofthis polynomial. The linear subspace of C[x] of all polynomials of degree less than n will bedenoted by Cn[x], n N = {1, 2, . . .} .

    For any n N , the system Mk(x) = xk, k = 0, 1, . . . , n 1, forms a basis in Cn[x], whichmeans that the representation of p(x) from Cn+1[x] in (1.1) is unique. This also means thateach finite subsystem of the system (Mn)

    n=0 is linearly independent, because of which we also

    call this infinite system linearly independent. As an example, let us define on C[x] a scalarproduct, also called inner product ,

    p, q = 11p(x)q(x) dx, p, q C[x]. (1.2)

    I we apply the well known Schmidts orthogonalization method to the system (Mn)n=0 , then

    we get a system (Pn)n=0 of polynomials Pn(x) with the properties degPn = n,

    span {P0, . . . , Pn1} = Cn[x], n N,and

    Pk, Pj = jk ={

    0 : j 6= k1 : j = k

    , j, k N0. (1.3)

    With span {P0, . . . , Pn1} we refer to the set of all linear combinations of the polynomials Pj ,j = 0, 1 . . . , n 1 . Since this set is equal to span {M0, . . . ,Mn1} , it is easily seen that, fork 6= j , the orthogonality relations (1.3) are equivalent to

    Mk, Pn = 0, k = 0, . . . , n 1, n N. (1.4)

    7

  • 8 CHAPTER 1. INTRODUCTION

    Consider the polynomials fn(x) = (1 x2)n und Qn(x) = f (n)n (x). Using partial integrationleads to

    Mk, Qn = 11xk

    dn

    dxn(1 x2)n dx = k

    11xk1

    dn1

    dxn1(1 x2)n dx ,

    where n N und 0 < k n . Repeating this yields

    Mk, Qn = (1)kk! 11

    dnk

    dxnk(1 x2)n dx =

    {0 : k = 0, . . . , n 1,

    (1)nn!n : k = n,

    with n =

    11

    (1 x2)n dx. Again by partial integration we get, for n N ,

    n = n1 11

    [(1 x2)n1x]x dx = n1 1

    2nn

    and consequently, since 0 = 2 ,1

    n =2n

    2n+ 1n1 = . . . =

    2nn!

    (2n+ 1)!!0 =

    2n+1n!

    (2n+ 1)!!.

    Now we normalize Qn(x) in order to satisfy the condition Pn, Pn = 1 . To this end, we makethe ansatz Pn(x) = nQn(x) and require (for definiteness), that the leading coefficient of Pn(x)

    is positive. The leading coefficient of Qn(x) =dn

    dxn[(1)nx2n + . . .+ 1] equals (1)n (2n)!

    n!.

    Because of

    2n

    11

    [Qn(x)]2 dx = 2n(1)n

    (2n)!

    n!

    11xnQn(x) dx =

    2n(2n)!n =

    2n

    2 (2nn!)2

    2n+ 1

    we get that, for

    n = cn(1)n2nn!

    mit cn =

    2n+ 1

    2,

    the condition Pn, Pn = 1 is fulfilled. Now, for the leading coefficient of the polynomial Pn(x) =nx

    n + . . . we get n =cn2n

    (2n

    n

    ). The poynomial

    Ln(x) =(1)n2nn!

    dn

    dxn(1 x2)n (1.5)

    is called nth Legendre polynomial. Up to the constant cn this polynomialis the nth poly-nomial in orthonormal system of polynomials (Pn)

    n=0 w.r.t. the inner product defined in

    (1.2). The formula (1.5) is due to Rodrigues.

    The sequence (Pn)n=0 of polynomials is a solution of the following moment problem:

    Given is a sequence (n)n=0 of numbers n =

    11xn dx =

    1 (1)n+1n+ 1

    . On the linear set C[x]

    of polynomials of the form (1.1) define a linear functional L by

    L[p] = ann + an1n1 + . . .+ a11 + a00.1(2n+ 1)!! = 1 3 . . . (2n+ 1)

  • 1.1. ORTHOGONALITY 9

    Find a system of polynomials Pn(x) with degPn = n, n N0 , such that the orthogonalityrelations

    L[PmPn] = 0 fur m 6= nare fulfilled. If the polynomials Pn(x) also satisfy

    L[P 2n ] 6= 0, n N0 ,

    then the sequence (Pn(x))n=0 is called orthogonal polynomial system (OPS) w.r.t. the

    moment functional L. In the following chapter we will answer the question under whichconditions on the moment sequence (n)

    n=0 such an OPS exists.

    Another OPS we can find in the following way:

    Using the trigonometric relation

    2 cosm cosn = cos(m+ n) + cos(m n) (1.6)

    one get

    pi0

    cosm cosn d =

    0 , m 6= n ,pi , m = n = 0 ,

    pi

    2, m = n N .

    (1.7)

    With x = cos and Tn(x) = cosn , n N0 , from (1.6) for m = 1 we obtain

    2xTn(x) = Tn+1(x) + Tn1(x) (1.8)

    or

    Tn+1(x) = 2xTn(x) Tn1(x) , n N (1.9)Since T0(x) = 1 and T1(x) = x the recursive relation (1.9) shows, that Tn(x) is a polynomial ofdegree n with leading coefficient 2n1 for n N . From (1.7) it follows

    11Tm(x)Tn(x)

    dx1 x2 =

    0 , m 6= n ,pi , m = n = 0 ,

    pi2 , m = n N .

    (1.10)

    Tn(x) is called Chebyshev polynomial of first kind and of degree n, the function (1x2) 12Chebyshec weight of first kind .

    Considering the polynomials xPn(x) = M1(x)Pn(x) and using the orthogonality relations(1.4) we can obtain a recursion formula for the polynomials Pn(x) analogous to (1.8). Indeed,we have the representation

    M1Pn =

    n+1k=0

    nkPk,

    where nk = M1Pn, Pk = Pn,M1Pk = 0 for k = 0, . . . , n 2 . Since x[Pn(x)]2 is an odd func-tion, we also get nn = 0. For the remaining tw coefficients we obtain n,n+1 = nMn+1, Pn+1 =n

    1n+1Pn+1, Pn+1 = n1n+1 and n,n1 = n1Pn,Mn = n11n . Hence,

    n+1Pn+1(x) = xPn(x) nPn1, n N0, (1.11)

  • 10 CHAPTER 1. INTRODUCTION

    with n = n11n is proved, where w set P1(x) 0 . We will see that such a three termrecurrence relation is a typical property of orthogonal polynomial systems. Moreover, such arelation enables us to compute interesting numerical parameters of orthogonal polynomials. Forexample, the zeros of orthogonal polynomials are eigenvalues of tridiagonal matrices. Indeed,writing down the relation (1.11) for n = 0, 1, . . . ,m 1 in matrix language gives

    0 1 0

    1 0 2

    . . .. . .

    . . .

    m2 0 m1

    0 m1 0

    P0(x)

    P1(x)

    ...

    Pm2(x)

    Pm1(x)

    = x

    P0(x)

    P1(x)

    ...

    Pm2(x)

    Pm1(x)

    0

    0

    ...

    0

    mPm(x)

    , (1.12)

    and it is easily seen that a zero of the mth orthogonal polynomial Pm(x) is an eigenvalue of thematrix in (1.12).

    Relation (1.8) written in matrix language gives

    0 2 0

    1 0 1

    . . .. . .

    . . .

    1 0 1

    0 1 0

    T0(x)

    T1(x)

    ...

    Tm2(x)

    Tm1(x)

    = 2x

    T0(x)

    T1(x)

    ...

    Tm2(x)

    Tm1(x)

    0

    0

    ...

    0

    Tm(x)

    .

    hence, the eigenvalues of the m m matrix on the left-hand side of this equation are just thenumbers

    2 cos(2k 1)pi

    2m, k = 1, . . . ,m.

    If we normalize the polynomials Tn(x) by Tn(x) =

    2piTn(x) , n N und T0(x) =

    1pi , relation

    (1.9) can be written as

    1

    2Tn+1(x) = xTn(x) 1

    2Tn1(x) , n = 2, 3, . . . ,

    1

    2T2(x) = xT1(x) 1

    2T0(x) ,

    12T1(x) = xT0(x) .

    Consequently, the numbers cos 2k12m , k = 1, . . . ,m are the eigenvalues of the symmetric tridiag-onal matrix

    0 12

    0

    12

    0 12

    . . .. . .

    . . .

    12 0

    12

    0 12 0

    associated to these last recursion formulas.

  • 1.2. GENERATING FUNCTION 11

    1.2 Generating function

    Consider the following function of two variables (a 6= 0)

    G(x,w) = eaw(1 + w)x =m=0

    (a)mwmm!

    n=0

    (x

    n

    )wn . (1.13)

    The Cauchy product of the two series in (1.13) is equal to

    G(x,w) =

    n=0

    Pn(x)wn (1.14)

    with

    Pn(x) =nk=0

    (x

    k

    )(a)nk(n k)! . (1.15)

    Due to

    (x

    k

    )=x(x 1) (x k + 1)

    k!the function Pn(x) is a polynomial of degree n . The func-

    tion G(x,w) is called generating function of the polynomials Pn(x) , the so-called Charlierpolynomials. Relation (1.13) implies

    axG(x, v)G(x,w) = ea(v+w)[a(1 + v)(1 + w)]x

    and, consequently,

    k=0

    akG(k, v)G(k,w)

    k!= ea(v+w)

    k=0

    [a(1 + v)(1 + w)]k

    k!

    (1.16)

    = ea(v+w)ea(1+v)(1+w) = eaeavw =n=0

    eaan(vw)n

    n!.

    Using (1.14) we get

    k=0

    akG(k, v)G(k,w)

    k!=

    k=0

    ak

    k!

    m,n=0

    Pm(k)Pn(k)vmwn

    (1.17)

    =

    m,n=0

    ( k=0

    Pm(k)Pn(k)ak

    k!

    )vmwn .

    A comparison of the coefficients in (1.16) and (1.17) leads to

    k=0

    Pm(k)Pn(k)ak

    k!=

    0 , m 6= n ,

    eaan

    n!, m = n .

    (1.18)

    If we define

    n = L[Mn] =k=0

    knak

    k!,

    then, due to (1.18), it follows

    L[PmPn] = eaan

    n!mn , mn :=

    {0 , m 6= n ,1 , m = n .

  • 12 CHAPTER 1. INTRODUCTION

    1.3 Birth and death process

    We model a birth and death process (a special Markov process the state space of which is theset N0 of the nonnegative integers). By pmn(t), m, n N0, t 0, we denote the so calledtransition probabilities. This means that pmn(t) is the probability that the system (for example,the size of a population) changes from the state m to the state n during the time period t . Thgematrix P (t) = [ pmn(t) ]

    m,n=0 is called transition matrix. Here, pmn(t) really depends only on

    m,n, t and does not depend on how the system reached the stae m (stationary process). Thisis equivalent to the relation

    P (s+ t) = P (s)P (t). (1.19)

    For t +0 we assume that the transition probabilies satisfy 2

    pmn(t) =

    mt+ o(t) : n = m+ 1,

    1 mt mt+ o(t) : n = m,mt+ o(t) : n = m 1

    (1.20)

    andm2n=0

    pmn(t) +

    n=m+2

    pmn(t) = o(t). (1.21)

    The coefficients m and m are called birth and death rate at state m, respectively, and areassumed to have the properties

    m > 0, m+1 > 0, m N0, 0 0. (1.22)

    Now, let t > 0. The conditon (1.19) leads to P (t+ t) = P (t)P (t), i.e., in view of (1.20),

    pmn(t+ t) =

    k=0

    pmk(t)pkn(t)

    = pm,n1(t)n1t+ pm,n+1(t)n+1t

    +pmn(t)[1 (n + n)t] + o(t).

    (Quatities with negative indizees are asumed to be zero.) It follows

    pmn(t+ t) pmn(t)t

    = n1pm,n1(t) + n+1pm,n+1(t)

    (n + n)pmn(t) + o(t)t

    ,

    and, for t +0 ,

    pmn(t) = n1pm,n1(t) + n+1pm,n+1(t) (n + n)pmn(t) .

    Analogously, using the relation P (t+ t) = P (t)P (t) leads to

    pmn(t) = mpm+1,n(t) + mpm1,n(t) (m + m)pmn(t).2By o(t) we denote quatities satisfying limt+0 o(t)/t = 0

  • 1.4. EXERCISES 13

    the last two equations ar called backward and forward Chapman-Kolmogorov equations,respectively. Taking the separarion ansatz pmn(t) = f(t)QmFn , from the backward Chapman-Kolmogorov equations we get

    f (t)f(t)

    =n1Fn1 + n+1Fn+1 (n + n)Fn

    Fn=: x.

    hence, up to a multiplicative constant f(t) = ex t. Obviously, the Fns depend on x and wewrite Fn(x) . With the agreement F1(x) 0 we get

    n+1Fn+1(x) = (n + n x)Fn(x) n1Fn1(x), (1.23)n N0. The function F0(x) can be chosen arbitrarily. Hence, the functions Fn(x) are uniquelydetermined by, for example, the initial conditions F1(x) 0, F0(x) 1 . Analogously, thefunctions Qn(x) are determined uniquely by the initial consditions Q1(x) 0, Q0(x) 1 andby the recursive relation

    nQn+1(x) = (n + n x)Qn(x) nQn1(x),n N0 . The polynomial system (nQn) n=0 with

    0 = 1 und n =0 n11 n , n N,

    satisfies the same initial conditions and recurrence relation as the polynomial system (Fn)n=0 .

    Thus, our separation ansatz leads to

    pmn(t) =1

    mex tFm(x)Fn(x),

    where x 0 in order to guarantee that pmn(t) remains bounded for t + . Remark that weagain arrive at a system of polynomials which fulfills a recurrence relation analogous to (1.11) (cf.(1.23)). We will see that such a recursion formula implies the existence of a moment sequence(n)

    n=0 , w.r.t. which the system (Fn)

    n=0 is an OPS, and that a respective distribution function

    (x) with the property

    n =

    0

    xn d(x), n N0exists.

    1.4 Exercises

    1. Define3

    Tn(x) =(1)n1 x2

    (2n 1)!! Dn(1 x2)n 12 , n N , D = d

    dx, T0(x)01 .

    (a) Prove thatTn(x) = Tn(x) , n N0 ,

    holds true.

    (b) Use this representation of Tn(x) to prove the respective orthogonality relations (1.10).

    3(2n 1)!! = 1 3 (2n 1)

  • 14 CHAPTER 1. INTRODUCTION

    2. The Chebyshev polynomials Un(x) of second kind can be defined by

    Un(x) =sin[(n+ 1)]

    sin, x = cos, n N0 .

    (a) Show that Un(x) is a polynomial of degree n and that 11Um(x)Un(x)

    1 x2 dx = pi

    2mn . (1.24)

    (b) Prove the formula

    Un(x) =(1)n(n+ 1)

    (2n+ 1)!!

    1 x2 Dn(1 x2)n+ 12 , n N0 ,

    and use this formula to prove the orthogonality relations (1.24).

    3. For x = cos , define

    Rn(x) =cos[(n+ 12)]

    cos 2, n N0 .

    Prove

    (a) Rn(x) =Tn(x) + Tn+1(x)

    1 + x, n N0 ,

    (b) R0(x) = 1 , R1(x) = 2x 1 ,Rn+1(x) = 2xRn(x)Rn1(x) , n N ,

    (c)

    11

    Rm(x)Rn(x)

    1 + x

    1 x dx = pimn .

    4. Let F (x,w) = e(xw)2 . Show that

    (a) Hn(x) = ex2

    n

    wnF (x, 0) = ex

    2(1)nDnex2

    is a polynomial of degree n ,

    (b) G(x,w) := e2xww2 =n=0

    Hn(x)wn

    n!,

    (c)

    +

    G(x, v)G(x,w)ex2dx =

    pi e2vw ,

    (Hint:

    +

    ex2dx =

    pi ,)

    (d)

    +

    Hm(x)Hn(x)ex2dx =

    pi 2nn! mn .right)

    Hn(x) is the so-called Hermite polynomial of degree n.

  • Chapter 2

    Basic Theory of OrthogonalPolynomials

    2.1 Moment functionals.Existence of orthogonal systems of polynomials

    Definition 2.1 For a given sequence (n)n=0 C of numbers we define the respective moment

    functional L on the linear space C[x] of all (algebraic) polynomials in x byL[Mn] = n , n N0 , Mn(x) = xn ,

    andL[1pi1 + 2pi2] = 1L[pi1] + 2L[pi2] , j C , pij C[x] .

    The number n is called moment of nth-order.

    The coefficients of the polynomials are in general complex numbers, but the independent variablewill be considered here as real.

    Definition 2.2 A sequence (Pn)n=0 C[x] is called orthogonal polynomial system (OPS)

    w.r.t. the moment functional L , if m,n N0 the following conditions are satisfied:

    1. degPn = n ,

    2. L[PmPn] = knmn , kn 6= 0 .

    In case kn = 1 , n N0 , the system (Pn) n=0 is a orthonormal polynomial system (ONPS).

    It is easy to see that an OPS does not exist for all sequences (n)n=0 C , for example if 0 = 0 .

    But, for example, also in case 0 = 1 = 2 = 1 there does not exist an OPS, since otherwisefor P0(x) = a and P1(x) = bx+ c we have

    0 = L[P0P1] = a(b+ c) , d.h. c = b ,which implies

    L[P 21 ] = L[b2(x 1)2] = b2(2 21 + 0) = 0 .

    15

  • 16 CHAPTER 2. BASIC THEORY OF ORTHOGONAL POLYNOMIALS

    Proposition 2.3 Let L be a moment functional and (Pn) n=0 C[x] be a sequence of polyno-mials with deg Pn = n . Then the following statements are equivalent:

    (a) (Pn)n=0 is an OPS w.r.t. L .

    (b) For all n N0 ,L[piPn] = 0 pi Cn[x]

    andL[piPn] 6= 0 pi C[x] with deg pi = n .

    (c) For all n N0 ,L[MmPn] = knmn , m = 0, 1, . . . , n , kn 6= 0 .

    Remark 2.4 Let (Pn)n=0 be an OPS w.r.t. the moment functional L .

    (a) If pi =

    nk=0

    kPk C[x] then k = L[piPk]L[P 2k ], k = 0, 1, . . . , n .

    (b) If (Qn)n=0 is a further OPS w.r.t. L , then, due to (a) and Prop. 2.3,(b),

    Qn = nPn , n 6= 0 , n N0 .

    (c) If all Pn(x) are monic, i.e. Pn(x) = xn + , then (Pn) n=0 is called monic OPS.

    (d) The polynomial system (pn)n=0 with

    pn(x) =(L[P 2n ]) 12 Pn(x) , n N0 ,

    is an ONPS w.r.t. L , where by (L[P 2n ]) 12 we refer to one solution of z2 = L[P 2n ] .For a given moment sequence (n)

    n=0 , we define

    n = det[j+k

    ] nj,k=0

    =

    0 1 n1 2 n+1...

    ......

    ...

    n n+1 2n

    .

    Proposition 2.5 Let L be a moment functional with the moment sequence (n) n=0 . Then,there exists an OPS w.r.t. L if and only if

    n 6= 0 n N0 .

    Proof. Let (Pn)n=0 be an OPS w.r.t. L , Pn(x) =

    nk=0

    nkxk . Prop. 2.3,(c) leads to

    L[MmPn] =nk=0

    nkk+m = knmn , m n , kn 6= 0 ,

  • 2.1. MOMENT FUNCTIONALS 17

    i.e.

    0 1 n1 2 n+1...

    ......

    ...

    n1 n 2n1n n+1 2n

    n0

    n1

    ...

    n,n1

    nn

    =

    0

    0

    ...

    0

    kn

    . (2.1)

    Remark 2.4,(a),(b) implies, that Pn(x) is uniquely determined if kn is given, which yields theunique solvability of the linear system (2.1). If, vice versa, n 6= 0 , then (2.1) is uniquelysolvable, which means that Pn(x) exists, where

    nn =knn1

    n6= 0 , n N0 , (2.2)

    (1 := 1).

    Let (Pn)n=0 be an OPS w.r.t. L and pin C[x] be a polynomial of degree n with the leading

    coefficient n , i.e. pin(x) = nxn + . Then, due to (2.2),

    L[pinPn] = nkn = nnnn1

    , (2.3)

    where n denotes the leading coefficient of Pn(x) .

    Definition 2.6 A moment functional L is called positive definite, if L[pi] > 0 for all poyno-mials pi C[x] with pi(x) 0 , x R , and pi(x) 6 0 .

    If L is positive definite, then 2n = L[M2n] > 0 , n N0 , and, since

    0 < L[(x+ 1)2n] =2nk=0

    (2n

    k

    )k ,

    also 2n1 R , n N . Moreover, by p, q := L[pq] there is defined a (positive definite) innerproduct on C[x] . Schmidts orthogonalization procedure applied to the system {M0,M1,M2, . . .}yields an ONPS (pn)

    n=0 with pn R[x] .

    Lemma 2.7 Let pi C[x] . Then pi(x) 0 , x R , if and only if there exist polynomialsp, q R[x] such that pi(x) = [p(x)]2 + [q(x)]2 .

    Proposition 2.8 A moment functional L is positive definite, if and only if all moments arereal and n > 0 for all n N0 .

    Corollary 2.9 The proof of Prop. 2.8 shows, that a moment functional L is positive definite ifan OPS (Pn)

    n=0 R[x] exists with L[P 2n ] > 0 , n N0 .

    Definition 2.10 A moment functional L is called quasi-definite, if n 6= 0 , n N0 .

  • 18 CHAPTER 2. BASIC THEORY OF ORTHOGONAL POLYNOMIALS

    2.2 Exercises

    1. Assume that L[Mn] = an , n 0 , where a C \ {0} . Show that there exists no OPSw.r.t. L .

    2. Show that there is no moment functional w.r.t. which (Mn)n=0 is an OPS.

    3. Let L be a moment functional, for which an OPS exists and let {cn} be a sequence ofnumbers different from 0 . Show that each of the following conditions defines uniquely anOPS (Pn(x))

    n=0 w.r.t. L :

    (a) L[MnPn] = cn , n N0 ,(b) lim

    x0xnPn(1/x) = cn , n N0 .

    4. Use known OPSs to determine polynomial systems (Pn(x))n=0 satisfying the following

    conditions:

    (a)

    10Pm(x)Pn(x) (1 x) 12x 12dx = knmn , k0 = pi , kn = pi

    2, n N0 ,

    (b)

    +

    Pm(x)Pn(x) ex2/2dx =

    2pi n! mn ,

    (c)

    10Pm(x)Pn(x) dx =

    1

    2n+ 1mn .

    5. Let L be a quasi-definite moment functional with the moment sequence {n} . Prove that{(n1)1Dn(x)} with

    Dn(x) = det

    0 1 . . . n

    ......

    ......

    n1 n . . . 2n1

    1 x . . . xn

    is the monic OPS w.r.t. L . Determine an ONPS w.r.t. L .

    6. Let L be a quasi-definite moment functional. Prove: If deg pin = n , n N0 , andL[pimpin] = 0 for m 6= n , then (pin) n=0 is an OPS w.r.t. L .

    7. Let L be a quasi-definite moment functional. Show that pi C[x] and L[piMn] = 0 ,n N0 , implies pi(x) 0 .

    8. Let L be positive definite and (Pn) n=0 the monic OPS w.r.t. L . Prove that the inequalityL[P 2n ] < L[|pi|2] holds for all monic polynomials pi(x) 6 Pn(x) with deg pi(x) = n .

    2.3 Recursion formulas and the formula of Christoffel-Darboux

    Proposition 2.11 If the moment functional L is quasi-definite and (Pn) n=0 is the respectivemonic OPS, then there exists numbers n, n C with n 6= 0 and

    Pn+1(x) = (x n)Pn(x) nPn1(x) , n N0 , (2.4)where P1(x) 0 , P0(x) 1 and 0 is arbitrary. In case of a positive definite momentfunctional L we have n R and n > 0 .

  • 2.3. RECURSION FORMULAS AND THE FORMULA OF CHRISTOFFEL-DARBOUX 19

    Corollary 2.12 From (2.4) and from the proof of Prop. 2.11 we get the following relations:

    1. It holds

    n =n2n

    2n1=L[P 2n ]L[P 2n1]

    , n N .

    2. If we agree that 0 = 0 = 0 , then

    L[P 2n ] = 01 n , n N0 .

    3. We have

    n =L[M1P 2n ]L[P 2n ]

    , n N0 .

    4. The nth monic orthogonal polynomial admits the representation

    Pn(x) = xn (0 + 1 + . . .+ n1)xn1 + . . . ,

    since, for Pn+1(x) = xn+1 + dnx

    n + . . . , we get dn = dn1 n from (2.4).

    Example 2.13 The monic OPS(Tn(x)

    ) n=0

    w.r.t. the Chebyshev weight of first kind is given

    by

    T0(x) = T0(x) und Tn(x) = 21nTn(x) , n N .

    Using (1.9) we get

    T1(x) = x T0(x) , T2(x) = x T1(x) 12T0(x)

    and

    Tn+1(x) = x Tn(x) 14Tn1(x) , n = 2, 3, . . . ,

    so that in this case n = 0 for n 0 and 1 = 12 as well as n = 14 for n 2 .

    Proposition 2.14 (Favard/Shohat/Natanson) For arbitrary sequences (n)n=0 , (n)

    n=0

    of complex numbers and a polynomial system {Pn}n=1 with P1(x) 0 and P0(x) 1 , satis-fying (2.4), there exists exactly one moment functional L with the properties

    L[P0] = 0 und L[PmPn] = 0 , m 6= n .

    This moment functional L is quasi-definite if and only if n 6= 0 for all n N0 and positivedefinite if and only if (n)

    n=0 R and n > 0 for all n N0 .

    Proposition 2.15 (Christoffel/Darboux) If the polynomial system {Pn}n=1 satisfies therecursion formula (2.4) with n 6= 0 , n N0 , then

    nk=0

    Pk(x)Pk(t)

    0 k =1

    0 nPn+1(x)Pn(t) Pn(x)Pn+1(t)

    x t . (2.5)

  • 20 CHAPTER 2. BASIC THEORY OF ORTHOGONAL POLYNOMIALS

    From a monic OPS (Pn(x))n=0 w.r.t. a positive definite moment functional we get a respective

    ONPS (pn(x))n=0 via the formula

    pn(x) = knPn(x) , kn = (0 n)12 .

    From (2.4) it followsn+1pn+1(x) = (x n)pn(x)

    npn1(x) , n N0 (2.6)

    and by (2.5)nk=0

    pk(x)pk(t) =knkn+1

    pn+1(x)pn(t) pn(x)pn+1(t)x t . (2.7)

    Moreover, from (2.5) we get for t xnk=0

    [Pk(x)]2

    0 k =P n+1(x)Pn(x) P n(x)Pn+1(x)

    0 n . (2.8)

    Consequently, in case of a positive definite moment functional,

    P n+1(x)Pn(x) P n(x)Pn+1(x) > 0 . (2.9)

    2.4 Zeros and Gauss quadrature rule

    Definition 2.16 We say that E R is a supporting set of L , if P (x) 0 on E and P (x) 6 0on E implies L[P ] > 0 . In this case L is called positive definite on E .

    In what follows L is assumed to be positive definite and (Pn) n=0 to be the respective monicOPS.

    Proposition 2.17 Let (a, b) be a supporting set of L and n N . Then all zeros of Pn(x) arereal, simple and in (a, b) .

    Provided that the assumptions of Prop. 2.17 are fulfilled, we denote the zeros of Pn(x) by xnk ,where xnn < xn,n1 < < xn1 . It follows sgnPn(x) = 1 for x > xn1 and sgnPn(x) = (1)nfor x < xnn . The polynomial P

    n(x) possesses exactly one zero in the interval (xnk, xn,k1) ,

    k = 2, . . . , n , which impliessgnP n(xnk) = (1)k1 . (2.10)

    Proposition 2.18 We have xn+1,k+1 < xnk < xn+1,k , k = 1, . . . , n .

    Corollary 2.19 For every k 1 , the sequence {xnk} n=k is monotone increasing and the se-quence {xn,nk+1} n=k is monotone decreasing. Consequently, the limits

    k := limnxnk und k := limnxn,nk+1

    exist (k = + or/and k = is possible) .

  • 2.4. ZEROS AND GAUSS QUADRATURE RULE 21

    Definition 2.20 The interval (1, 1) is named support of the moment functional L .

    By `nk(x) we denote the kth Lagrange fundamental polynomial

    `nk(x) =n

    j=1,j 6=k

    x xnjxnk xnj =

    Pn(x)

    (x xnk)P n(xnk), k = 1, . . . , n .

    Obviously, `nk(xnj) = jk . For a continuous function f : (1, 1) C , i.e. f C(1, 1) , wedefine the Lagrange interpolation polynomial

    (Lnf)(x) =

    nk=1

    f(xnk)`nk(x)

    and the functional, the so-called Gaussian quadrature rule ,

    Ln[f ] = L[Lnf ] =nk=1

    Ankf(xnk) with Ank = L[`nk] .

    Proposition 2.21 We have

    Ln[P ] = L[P ] P C2n[x] .

    Corollary 2.22 It holds

    nk=1

    Ank = 0 und Ank > 0 , k = 1, . . . , n .

    Corollary 2.23 By Corollary 2.22 and Lnp = p for p C[x] and for all n n0(p) we get that(1, 1) is a supporting set of L .

    Now, in case < 1 < 1

  • 22 CHAPTER 2. BASIC THEORY OF ORTHOGONAL POLYNOMIALS

    Proposition 2.26 The moment functional Lz is quasi-definite and (P zn) n=0 is the respectivemonic OPS. The moment functional Lz is positive definite if and only if z 1 .

    In view of (2.11) it holds

    Kn(z, x) :=1

    01 n Pn(z)Pzn(x) =

    nk=0

    pk(x)pk(z) , x, z R .

    For w, z C , we defineKn(z, w) =

    nk=0

    pk(w)pk(z) .

    Proposition 2.27 For all z0 C and n N0 ,1

    Kn(z0, z0)= min

    {L[|pi|2] : pi Cn+1[x], pi(z0) = 1}and

    Ank =1

    Kn1(xnk, xnk)=

    1

    Kn(xnk, xnk).

    2.5 The Jacobi polynomials

    Let > 1 , > 1 . For n N0 , the Jacobi polynomials can be defined by Rodriguesformula (cf. also Section 1.4, Exercise 1)

    P,n (x) =(1 x)(1 + x)

    (2)n n!dn

    dxn

    [(1 x)n+(1 + x)n+

    ]. (2.12)

    Lemma 2.28 It holds

    nk=0

    (n+

    n k)(

    n+

    k

    )=

    (2n+ +

    n

    ), n N0

    Proof. Using

    (1 + z) =

    j=0

    (

    j

    )zj

    we get

    j=0

    (2n+ +

    j

    )zj = (1 + z)n+(1 + z)n+ =

    j=0

    jk=0

    (n+

    j k)(

    n+

    k

    )zj .

    Comparing the coefficients at zn proves the lemma.

    From (2.12) and

    dn

    dxn

    [(1 x)n+(1 + x)n+

    ]

  • 2.5. THE JACOBI POLYNOMIALS 23

    =nk=0

    (n

    k

    )[dnk

    dxnk(1 x)n+

    ] [dk

    dxk(1 + x)n+

    ]

    = (1 x)(1 + x)nk=0

    n!

    k!(n k)! (1)nk(n+ ) (k + + 1)(1 x)k

    (n+ ) (n+ k + 1)(1 + x)nk

    = (1)n(1 x)(1 + x)n!nk=0

    (n+

    n k)(

    n+

    k

    )(x 1)k(x+ 1)nk

    it follows

    P,n (x) =nk=0

    (n+

    n k)(

    n+

    k

    )(x 1

    2

    )k (x+ 12

    )nk. (2.13)

    In view of Lemma 2.28, P,n (x) has the leading coefficient

    k,n = 2n

    nk=0

    (n+

    n k)(

    n+

    k

    )= 2n

    (2n+ +

    n

    ). (2.14)

    The monic Jacobi polynomials we denote by P,n (x) , i.e.

    P,n (x) =1

    k,nP,n (x) .

    With the help of partial integration one can prove the following Lemma.

    Lemma 2.29 For n N0 , 11xkP,n (x)(1 x)(1 + x) dx

    {= 0 , k = 0, . . . , n 1 ,> 0 , k = n .

    Hence,(P,n (x)

    ) n=0

    is an OPS. Let us look for formulas of the coefficients n and n in the

    recursion formula

    P,n+1(x) = (x n)P,n (x) nP,n1(x) , n = 0, 1, . . . (2.15)of the monic Jacobi polynomials. Define

    ()n :=

    1 , n = 0 ,

    nk=1

    (k + ) , n N

    Relation (2.13) implies

    P,n (1) =

    (n+

    n

    )and P,n (1) = (1)n

    (n+

    n

    ),

    so that

    P,n (1) =

    2n(n+

    n

    )(

    2n+ +

    n

    ) = 2n()n(+ )n(+ )2n

  • 24 CHAPTER 2. BASIC THEORY OF ORTHOGONAL POLYNOMIALS

    and

    P,n (1) =(2)n

    (n+

    n

    )(

    2n+ +

    n

    ) = (2)n()n(+ )n(+ )2n

    .

    Consequently,

    1 0 = P1(1) = 2(1 + )2 + +

    , d.h. 0 =

    2 + + .

    For n N , we solve the system

    P,n+1(1) = (1 n)P,n (1) nP,n1(1) ,

    P,n+1(1) = (1 + n)P,n (1) nP,n1(1) ,

    which is a consequence of (2.15) for x = 1 and which can be written in the form

    4()n+1(+ )n+1(+ )2n+2

    = (1 n) 2()n(+ )n(+ )2n

    n ()n1(+ )n1(+ )2n2

    , (2.16)

    4()n+1(+ )n+1(+ )2n+2

    = (1 + n)2()n(+ )n

    (+ )2n n ()n1(+ )n1

    (+ )2n2. (2.17)

    Multiply equation (2.16) by ()n1 , equation (2.17) by ()n1 , and substract the first from thesecond one to obtain

    2()n1()n1(+ )n+1(2n+ + + 1)( )(+ )2n+2

    =()n1()n1(+ )n[ + n(2n+ + )]

    (+ )2n,

    where we took into account

    (n+ )(n+ 1 + ) (n+ )(n+ 1 + ) = (2n+ + + 1)( ) .

    Thus,

    n =2 2

    (2n+ + )(2n+ + + 2), n N

    Now, we multiply (2.16) by ()n , (2.17) by ()n and consider the sum of both equations. Itfollows

    4()n()n(+ )n+1(+ )2n+1

    =4()n()n(+ )n

    (+ )2n n ()n1()n1(+ )n1(2n+ + )

    (+ )2n2,

    so that

    n =

    4(1 + )(1 + )

    (2 + + )2(3 + + ), n = 1

    4n(n+ )(n+ )(n+ + )

    (2n+ + 1)(2n+ + )2(2n+ + + 1) , n = 2, 3, . . .

  • 2.6. EXERCISES 25

    2.6 Exercises

    In what follows by (Pn)n=0 we denote the monic OPS w.r.t. the quasi-definite moment functional

    L with the moment sequence {n} and the respective recursion formula (2.4).

    1. Prove that the following assertions are equivalent:

    (a) L is symmetric, i.e. 2n+1 = 0 n N0 .(b) Pn(x) = (1)nPn(x) x R , n N0 .(c) In the recursion formula (2.4) there holds n = 0 n N0 .

    2. Define Qn(x) = anPn(ax+ b) (a 6= 0) and show that

    (a) Qn+1(x) =

    (x n b

    a

    )Qn(x) n

    a2Qn1(x) ,

    (b) (Qn)n=0 is OPS w.r.t. the moment sequence {n} with

    n = an

    nk=0

    (n

    k

    )(b)nkk .

    3. Show that the polynomials Pn(x) in formula (1.15) satisfy the recursion formula

    Qn+1(x) = (x n a)Qn(x) anQn1(x), n = 0, 1, . . . , Qn(x) := n!Pn(x) .

    4. Let n = 0 and n < 0 , n N0 . Then (Pn) n=0 is OPS w.r.t. a quasi-definit momentfunctional L. Define L[Mn] := inL[Mn] and prove that L is positive definite. Determinethe respective monic OPS.

    5. Let n = 0 , n < 0 , n N , and 0 R \ {0} . Define Rn(x) = Re [inPn(ix)] andIn(x) = Im [i

    nPn(ix)] . Show that (Rn) n=0 and(10 In+1

    ) n=0

    are monic OPS w.r.t. somepositive definite moment functionals.

    6. Prove that

    (a)1 xw

    1 2xw + w2 =n=0

    Tn(x)wn ,

    (b)1

    1 2xw + w2 =n=0

    Un(x)wn .

    7. Show that a monic OPS (Pn)n=0 fulfils a condition of the form

    Pn1(x)Pn(x) + Pn1(x)Pn(x) = an 6= 0 , n N ,if and only if n 6= 0 , n > 0 , and n = 0 , n 1 , 0 6= 0 in the recursion formula (2.4).Moreover, show that the respective moment functional is positive definite if and only if(1)na1an < 0 , n 1 , and 0 > 0 .

    8. Let (Pn)n=0 be an OPS andM a moment functional withM[P0] 6= 0 andM[Pn] = 0, n

    N . Prove that {Pn} is an OPS w.r.t. M .9. Show that the weights in the Gaussian rule satisfy An,nk+1 = Ank if the moment func-

    tional is symmetric.

  • 26 CHAPTER 2. BASIC THEORY OF ORTHOGONAL POLYNOMIALS

    10. Let L be positive definit and Kn(z, x) be defined as in Section 2.4. Show that pi(t) =L[piKn(., t)] for each polynomial pi C[x] and n deg pi(x) .

    11. Show, that the normalized Jacobi polynomials p,n (x) are given by

    p,n (x) =[h,n

    ]1P,n (x) (2.18)

    with

    h,n =

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

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

    (cf. [4, Equation (3.1)]).

    12. Prove, that, for n N , the formulas

    dP,n (x)

    dx=

    1

    2(n+ + + 1)P+1,+1n1 (x) , (2.19)

    d p,n (x)

    dx= ,n p

    +1,+1n1 (x) (2.20)

    with ,n =n(n+ + + 1) and

    (1 x)(1 + x)p,n (x) = 1

    ,n

    d

    dx

    [(1 x)+1(1 + x)+1p+1,+1n1 (x)

    ](2.21)

    (cf. [4, Equation (3.4) and Exercise 3.1]) are valid.

  • Chapter 3

    Singular Integral Operators

    3.1 Cauchy singular integral operators

    Define

    (Su)(x) = 1pi

    11

    u(y) dy

    y x := lim+0( x1

    +

    1x+

    )u(y) dy

    y x , 1 < x < 1 , (3.1)

    where we assume that this Cauchy principal value integral exists.

    Lemma 3.1 For 0 < < 1 and 0 < t 1 , + = 1 , and 1 < x < 1 ,

    1

    pi

    11

    v,(y) dy

    y x = cot(pi)v,(x) , (3.3)

    where v,(x) = (1 x)(1 + x) .

    For , > 1 , define

    (S,u) (x) = cos(pi)v,(x)u(x) + sin(pi)pi

    11

    v,(y)u(y)

    y x dy , 1 < x < 1 .

    Proposition 3.3 For , > 1 , + = 1 , and 1 < x < 1 ,(S,P,n

    )(x) = P,n1 (x) , n N0 . (3.4)

    27

  • 28 CHAPTER 3. SINGULAR INTEGRAL OPERATORS

    Remark 3.4 Proposition 3.3 can be generalized in the following sense. Let a, b R , a ib =eipi0 , 0 (0, 1) , , Z with := + 0 (1, 1) and := 0 (1, 1) . Then, for

    (Au)(x) = av,(x)u(x) + bpi

    11

    v,(y)u(y)

    y x dy , (3.5)

    we have the relations(AP,n

    )(x) = (1)P,n (x) , n N0 , 1 < x < 1 ,

    and (Ap,n

    )(x) = (1)p,n (x) , n N0 , 1 < x < 1 , (3.6)

    where = (+ ) = (+ ) (cf. [4, Corollary 3.7, Exercises 3.8, 3.9]).

    Corollary 3.5 Let L2, denote the Hilbert space of w.r.t. the Jacobi weight v,(x) square

    integrable functions u : (1, 1) C equipped with the inner product

    u, v, = 11u(x)v(x)v,(x) dx .

    Then the operator A defined by (3.5) on all polynomials can be uniquely extended to a boundedlinear operator A : L2, L2, . This operator is invertible if = 0 , left-sided invertible if = 1 , and right-sided invertible if = 1 .

    For s 0 , define the weighted Sobolev spaces L2,s, as

    L2,s, =

    {u L2,s, :

    n=0

    (1 + n)2su, p,n ,2 0 , the operator A : L2, L2, defined in Remark 3.4 is alsobounded from L2,s, into L

    2,s, .

    Remark 3.7 For s 0 and > 0 , the space L2,s+, is compactly embedded into the spaceL2,s, .

    3.2 Singular integro-differential operators

    Remark 3.8 ([1], Section 2) If r N0 then u L2,r, if and only if u(k)k L2, for allk = 0, . . . , r , where (x) =

    1 x2 . Moreover, the norms u,,s and

    rk=0

    u(k)k,

    are

    equivalent.

  • 3.3. WEAKLY SINGULAR INTEGRAL OPERATORS 29

    Lemma 3.9 For s 0 and , > 1 , the operator D of generalized differentiation is a con-tinuous isomorphism from L2,s+1,0, onto L

    2,s+1,+1 , where L

    2,s,0, =

    {u L2,s, : u, 1, = 0

    }.

    With the help of the operator A given in Remark 3.4 we define B = DA .

    Proposition 3.10 We have

    Bp,n = (1)

    (n )(n+ 1) p1,1n1 , n N0 .

    Hence, in case = = 12 , i.e. a = 0 , b = 1 , = 0 , and = 1 , this leads to

    ddx

    1

    pi

    11

    1 y2 Un(y)y x dy = (n+ 1)Un(x) , n N0 , 1 < x < 1 .

    Consequently, the operator DS : L2,s+112, 12

    L2,s12, 12

    is an isometrical isomorphism.

    3.3 Weakly singular integral operators

    We use the same notations as in Remark 3.4 and define

    (Wu)(x) = a x1v,(y)u(y) dy b

    pi

    11v,(y) ln |y x|u(y) dy , 1 < x < 1 .

    In case + = 1 we have, due to (2.21), x1v,(y)p,n (y) dy =

    1

    nv+1,+1(x)p+1,+1n1 (x) , n N ,

    and, by partial integration, 11v,(y) ln |y x| p,n (y) dy =

    1

    n

    11

    v+1,+1(y)p+1,+1n1 (y)y x dy , 1 < x < 1 , n N .

    Hence,

    Wp,n = 1

    np1,1n , n N .

    Remark 3.11 (cf. [5], Example 3.27) In case = = 12 it yields, for 1 < x < 1 ,

    1pi

    11

    ln |y x| p12, 1

    2n (y)

    dy1 y2 =

    ln 2 p

    12, 1

    20 (x) : n = 0 ,

    1

    np 1

    2, 1

    2n (x) : n N .

    Further considerations in this direction one find in [4, Section 3.4].

  • 30 CHAPTER 3. SINGULAR INTEGRAL OPERATORS

  • Chapter 4

    Continued Fractions and OrthogonalPolynomials

    4.1 Basics

    By an (infinite) continued fraction we refer to a tripel(

    (an)n=1 , (bn)

    n=0 , (cn)

    n=0

    )of number

    sequences, where

    c0 = b0

    c1 = b0 +a1b1

    c2 = b0 +a1

    b1 +a2b2

    ...

    cn = b0 +a1

    b1 +a2

    b2 +.. .

    +anbn

    The number cn is called nth approximant of the infinite continued fraction

    b0 +a1

    b1 +a2

    b2 +.. .

    +an

    bn +.. .

    (4.1)

    In what follows, cn is written shortly as

    cn = b0 +a1||b1 +

    a2||b2 + +

    an||bn

    31

  • 32 CHAPTER 4. CONTINUED FRACTIONS AND ORTHOGONAL POLYNOMIALS

    and (4.1) as

    b0 +a1||b1 +

    a2||b2 + +

    an||bn +

    In case of ak = dk , we write dk||bk instead of +dk||bk .

    Definition 4.1 We say that the continued fraction (4.1) converges to K , if at most finitelymany approximants cn are not defined and if

    limn cn = K .

    In this case, we also write

    b0 +a1||b1 +

    a2||b2 + +

    an||bn + = K .

    We write cn in the form

    cn =AnBn

    , n = 0, 1, 2, . . . ,

    where, for example,

    A0 = b0 , B0 = 1 ,

    A1 = b0b1 + a1 B1 = b1 ,

    A2 = b0b1b2 + b0a2 + a1b2 , B2 = b1b2 + a2 .

    In general, it is possible to define the sequences (An)n=0 and (Bn)

    n=0 can be defined in such a

    way that

    An = bnAn1 + anAn2 , n = 1, 2, . . . , A1 = 1 , A0 = b0 , (4.2)

    and

    Bn = bnBn1 + anBn2 , n = 1, 2, . . . , B1 = 0 , B0 = 1 . (4.3)

    An und Bn are called the nth partial numerator and denominator of the continued fraction(4.1), respectively. We have

    AnBn1 BnAn1 = (1)n+1a1a2 an , n = 1, 2, . . . , (4.4)

    andAnBn

    = b0 +

    nk=1

    (1)k+1a1a2 akBk1Bk

    (4.5)

    provided that Bk 6= 0 , k = 1, . . . , n

    Lemma 4.2 If m0 = 0 , then the nth partial denominator of the continued fraction

    1 1||1 (1m0)m1|

    |1 (1m1)m2|

    |1

    equals

    Bn = (1m1) (1mn1) , n = 1, 2, . . . (4.6)

  • 4.2. JACOBI FRACTIONS AND ORTHOGONAL POLYNOMIALS 33

    Lemma 4.3 Let an = (1mn1)mn , m0 = 0 and 0 < mn < 1 , n = 1, 2, . . . Then

    1 a1||1 a2||1

    a3||1 =

    1

    1 + L,

    where

    L =

    n=1

    m1m2 mn(1m1)(1m2) (1mn) .

    Proposition 4.4 Let bn = (1 gn1)gn , 0 g0 < 1 and 0 < gn < 1 , n = 1, 2, . . . Then

    1 b1||1 b2||1

    b3||1 = g0 +

    1 g01 +G

    , (4.7)

    wobei

    G =

    n=1

    g1g2 gn(1 g1)(1 g2) (1 gn) .

    Example 4.5 If

    b0 +a1||b1 +

    a2||b2 +

    a3||b3 + = K 6= 0

    then

    b1 +a0||b0 +

    a2||b2 +

    a3||b3 + = b1 +

    a0K.

    Example 4.6 Assuming that the continued fraction

    1 +1||1 +

    1||1 +

    1||1 +

    converges we show that its value is equal to1 +

    5

    2.

    4.2 Jacobi fractions and orthogonal polynomials

    Let n and n be given numbers with n 6= 0 . Let us denote the nth partial denominator of theso-called Jacobi fraction

    0||x 0

    1||x 1

    2||x 2 (4.8)

    by Pn(x) . Formula (4.3) gives

    Pn+1(x) = (x n)Pn(x) nPn1(x) , n = 0, 1, 2, . . . , P0(x) = 1 , P1(x) = 0 . (4.9)

    The nth partial numerator An(x) satisfies the recursion formula

    An+1(x) = (x n)An(x) nAn1(x) n = 1, 2, . . . , A1(x) = 0 , A0(x) = 0 ,

  • 34 CHAPTER 4. CONTINUED FRACTIONS AND ORTHOGONAL POLYNOMIALS

    where 10 An(x) is a monic polynomial of degree n 1 , which is independent of 0 . For thisreason we write Qn(x) =

    10 An+1(x) , n = 1, 0, 1, . . . Then

    Qn+1(x) = (x n+1)Qn(x) n+1Qn1 , n = 0, 1, 2, . . . , (4.10)Q0(x) = 1 , Q1(x) = 0 . The polynomials Qn(x) are called monic numerator polynomialsw.r.t. the polynomial system (Pn(x))

    n=0 . From (4.4) we deduce

    Pn+1(x)Qn1(x) Pn(x)Qn(x) = 12 n , n = 1, 2, . . . (4.11)

    Proposition 4.7 If n R and n > 0 , then the zeros xnk and ynk of the polynomials Pn(x)and Qn(x) , respectively, fulfil the relations

    xn+1,k+1 < ynk < xn+1,k .

    Corollary 4.8 Let (1, 1) and (11,

    11) be the supports of the OPS

    (Pn(x))n=0 und (Qn(x))

    n=0 ,

    respectively. Then (11, 11) (1, 1) . Moreover, if, for example, 11 < 1 , then, for all sufficiently

    large n , there lies exactly one zero of Pn(x) in the interval (11 , 1) .

    Proposition 4.9 If the numbers n are real and the numbers n are positive, then

    0Qn1(x)Pn(x)

    =nk=1

    Ankx xnk .

    where the Ank-s denote the weights in the Gaussian quadrature rule w.r.t. the OPS (Pn(x))n=0 .

    Example 4.10 The monic Chebyshev polynomials 2nUn(x) , n N0 are the numerator poly-nomials for 21nTn(x) , n N and T0(x) .

    4.3 Chain sequences

    Definition 4.11 A sequence (an)n=1 of the form

    an = (1 gn1)gn mit 0 g0 < 1 and 0 < gn < 1 , n = 1, 2, . . . ,is called chain sequence. The sequence (gn)

    n=0 is named a parameter sequence and g0 a

    initial parameter of the chain sequence (an)n=1 .

    Example 4.12 The constant sequence

    (1

    4

    ) n=1

    is a chain sequence, where

    (n

    2(n+ 1)

    ) n=0

    as

    well as the constant sequence

    (1

    2

    ) n=0

    are parameter sequences. The equations

    a =

    (1 1

    1 4a2

    )11 4a

    2=

    (1 1 +

    1 4a2

    )1 +

    1 4a2

    show that each constant sequence (a) n=1 with 0 < a 1

    4is a chain sequence.

  • 4.3. CHAIN SEQUENCES 35

    Lemma 4.13 Let (gn)n=0 and (hn)

    n=0 be parameter sequences of the chain sequence (an)

    n=1 .

    Then gn < hn , n = 1, 2, . . . , if and only if g0 < h0 .

    Lemma 4.14 If a chain sequence (an)n=1 possesses the parameter sequence (gn)

    n=0 with g0 >

    0 , then, for each h0 [0, g0] , there exists a parameter sequence (hn) n=0 of (an) n=1 .

    Corollary 4.15 Each chain sequence possesses a parameter sequence (mn)n=0 with m0 = 0 .

    It holds mn < gn , n = 0, 1, 2, . . . , for each other parameter sequence (gn)n=0 of this chain

    sequence. The sequence (mn)n=1 is called minimal parameter sequence of the respective

    chain sequence. A parameter sequence (Mn)n=0 , for which Mn gn , n = 0, 1, 2, . . . , holds for

    any parameter sequence (gn)n=0 , is named maximal parameter sequence of the respective

    chain sequence.

    Lemma 4.16 To each chain sequence there exists a maximal parameter sequence.

    In what follows (mn)n=0 denotes the minimal parameter sequence and (Mn)

    n=0 the maximal

    parameter sequence of the chain sequence (an)n=1 .

    Proposition 4.17 If (bn)n=1 is a chain sequence with the parameter sequence (hn)

    n=0 and if

    an bn , n = 1, 2, . . . , thenmn hn Mn , n = 0, 1, 2, . . .

    Lemma 4.18 Let the chain sequence (an)n=1 be monotonously non-decreasing. Then the min-

    imal parameter sequence (mn)n=0 is monotonously increasing and the maximal parameter se-

    quence (Mn)n=0 monotonously non-increasing.

    Corollary 4.19 If (an)n=1 =

    (1

    4

    ) n=1

    then (Mn)n=0 =

    (1

    2

    ) n=0

    .

    Corollary 4.20 Let (an)n=1 be a chain sequence. If there exists an index N N with an

    1

    4,

    n = N,N + 1, . . . , then limn an =

    1

    4. Consequently, if bn b > 1

    4, n = N,N + 1, . . . , then

    (bn)n=1 is not a chain sequence.

    Proposition 4.21 (comparison test) If (an)n=1 is a chain sequence and if 0 < cn an ,

    n = 1, 2, . . . , then (cn)n=1 is a chain sequence, too.

    Lemma 4.22 If (an)n=1 is a chain sequence, then

    nk=1

    (ak 1

    2

    ) 0 .

    Now, consider the recursion formulas

    Pn+1(x) = (x n)Pn(x) nPn1(x) , P1 0, P0 1 ,Qn+1(x) = (x n)Qn(x) nQn1(x) , Q1 0, Q0 1 ,Sn+1(x) = xSn(x) nQn1(x) , S1 0, S0 1 ,

    0 = L[1] , 0 = L0[1] , 0 =M[1] . Then,

    Qm+1(x) = (x 2m+1 2m+2)Qm(x) 2m2m+1Qm1(x) , m N0 ,Pm+1(x) = (x 2m 2m+1)Pm(x) 2m12mPm1(x) , m N ,P1(x) = (x 1)P0(x) .

    Lemma 4.31 The moment functionals L , L0 and M are positive definite if and only if n > 0n N0 .

    Lemma 4.32 Let

    Rn+1(x) = (x n)Rn(x) nRn1(x) , n N0 , R1 0 , R0 1 ,

    with 0 > 0 and n 6= 0 , n N , as well as the respective moment functional N be given. Then,

  • 38 CHAPTER 4. CONTINUED FRACTIONS AND ORTHOGONAL POLYNOMIALS

    (a) the moment functional N is positive definite on (0,) if and only if there exists a numbersequence (n)

    n=0 with the properties

    0 0, n > 0, n N, n = 2n + 2n+1, n N0, n = 2n12n, n N .

    (b) In this case, we have 0 > 0 if and only if there exists a moment functional L with N = L0 .

    Lemma 4.33 Under the assumptions of Lemma 4.32, there holds:

    (a) The moment functional N is positive definit on (0,)if and only if n > 0 , n N0 andif there exists a number sequence (gn)

    n=1 with the properties

    0 g0 < 1 , 0 < gn < 1 , n N and n

    n1n= (1 gn1)gn , n N .

    (b) We have N = L0 with L positive definite on (0,) if and only if g0 > 0 .

    Corollary 4.34 It holds 1 t if and only if n < t , n = 0, 1, 2, . . . , and if (n(t)) n=1 is achain sequence.

    Corollary 4.35 We have 1 < n < 1 , n = 0, 1, 2, . . .

    Proposition 4.36 The support (1, 1) is bounded if and only if (n)n=0 and (n)

    n=0 are

    bounded sequences.

    Proposition 4.37 We have (1, 1) = (,) if and only if (n(x)) n=1 is not a chain sequencefor all x R .

    Consequently, each of the following conditions guarantees (1, 1) = (,):

    (a) inf {n : n = 0, 1, 2, . . .} = and sup {n : n = 0, 1, 2, . . .} = + (vgl. Folg. 4.35)

    (b) (n)n=0 is bounded and (n)

    n=0 is unbounded, since in that case, for all x R , the

    sequence (n(x))n=1 is unbounded

    (c) limnn = and lim infn

    nn1n

    >1

    4

    Consider (c): Since

    n(n1 x)(n x) =

    nn1 x

    n1n1 x

    nn x

    and due to Corollary 4.20, for all x R , the sequence (n(x)) n=1 cannot be a chain sequence.

  • 4.4. CHAIN SEQUENCES AND ORTHOGONAL POLYNOMIALS 39

    Proposition 4.38 Let x 6 (1, 1) , so that (n(x)) n=1 is not a chain sequence. Then therespective minimal parameter sequence (mn(x))

    n=0 is given by

    mn(x) = 1 Pn+1(x)(x n)Pn(x) , n = 0, 1, 2, . . .

    By (pn(x))n=0 we denote the ONPS of L , i.e.

    pn(x) = (0 n)12 Pn(x) , n = 0, 1, 2, . . .

    Proposition 4.39 If (1, 1) is bounded, then

    limn |pn(x)| = x 6 [1, 1] .

  • 40 CHAPTER 4. CONTINUED FRACTIONS AND ORTHOGONAL POLYNOMIALS

  • Chapter 5

    Distribution Functions and theRepresentation Theorem

    5.1 Preliminaries

    Definition 5.1 A bounded, non-decreasing function : R R is called a distributionfunction if all moments

    n :=

    xn d(x) , n = 0, 1, 2, . . . , (5.1)

    are finite. The spectrum of is defined by

    S() := {x R : (x+ ) (x ) > 0 > 0} .

    A point x S() is called a spectral point of .

    The spectrum of a distribution function is closed. If it consists of infinitely many points, then(n)

    n=0 (cf. (5.1)) defines a positive definite moment functional.

    Lemma 5.2 Let E be a countable set and let fn : E R be a sequence of functions such thatthe number sequence (fn(x))

    n=1 is bounded for all x E . Then there exists a subsequence of

    (fn)n=1 converging everywhere on E .

    Lemma 5.3 If (n)n=1 is a sequence of non-decreasing and uniformly bounded functions n :

    R R , then there exists a subsequence {nj}j=1 such that the limit (x) = limj nj (x)exists for all x R . The function : R R is bounded and non-decreasing.

    Lemma 5.4 Let the functions n : [a, b] R , n = 1, 2, . . . , be non-decreasing and uniformlybounded, where < a < b

  • 42 CHAPTER 5. THE REPRESENTATION THEOREM

    The examples

    n(x) =

    {0 : x < n

    1 : x n

    } (x) = 0 , x R ,

    and

    n(x) =

    0 : x < 0 ,

    x

    n: 0 x n ,

    1 : n < x ,

    (x) = 0 , x R ,show, that the assumption on the boundedness of [a, b] in Lemma 5.4 is essential. In the secondexample the supports of all n are infinite.

    5.2 The Representation Theorem

    Let L be positive-definite and let {Pn(x)} be a respective orthogonal polynomial system (OPS).Then (cf. Section 2.4)

    k = L[Mk] =nj=1

    Anjxknj , k = 0, 1, 2, . . . , 2n 1 . (5.2)

    Define

    n(x) =

    0 , x < xnn ,

    Ann + +Anp , xnp x < xn,p1 , n p > 1 ,0 , x xn1 .

    (5.3)

    Then S(n) = {xnj} nj=1 , n(xnj + 0) n(xnj 0) = Anj , and

    xk dn(x) =

    nj=1

    Anjxknj = k , k = 0, 1, 2, . . . , 2n 1 . (5.4)

    If [a, b] is a bounded supporting interval of L , then, by Lemma 5.3 and Lemma 5.4, we have theexistence of a subsequence

    {nj} j=1

    with nj (x) (x) , x R , and

    xk d(x) = k = L[Mk] , k = 0, 1, 2, . . . (5.5)

    The example

    n(x) =

    {0 , x < n ,

    1 , x n ,n(x) (x) = 0 , x R ,

    shows that the assertion of Lemma 5.4 may fail to be true in case of unbounded intervals.

    Proposition 5.5 Let L be positive-definite and n defined (5.3). Then there exists a subse-quence

    {nj} j=1

    with nj (x) (x) , x R , such that : R R is a distribution functionwith infinite spectrum satisfying (5.5).

    A distribution function : R R having infinite spectrum and satisfying (5.5) is called arepresentation of L .

  • 5.3. ON THE LOCATION OF THE SPECTRAL POINTS OF A REPRESENTATION 43

    Proposition 5.6 (representation theorem) Every positive-definite moment functional pos-sesses a representation (x) with S() [1, 1] . If (x) is a representation of L with S() (a, b) , then (1, 1) (a, b) (comp. Cor. 2.19).

    5.3 On the Location of the Spectral Points of a Representation

    In what follows let L be positive definite and let (Pn) n=0 be the respective monic OPS.

    Proposition 5.7 If is a representation of L , then

    S() (xn,j+1, xnj) 6= j = 1, . . . , n 1 , n = 2, 3, . . .

    A representation , which is the limit of a subsequence of (n)n=1 (comp. (5.3)), is called a

    natural representation of L .

    Proposition 5.8 Let be a natural representation of L , G R be an open subset of R , andassume that there is an index n0 such that Pn(x) 6= 0 for all x G and for all n > n0 . ThenS() G = .

    Define the setsX := {xnj : 1 j n , n = 1, 2, 3, . . .}

    andZ = {x R : Pn(x) = 0 for infinitely many n} ,

    X denotes the set of accumulation points of the set X .

    Corollary 5.9 Let be a natural representation of L and s S() . Then, for each > 0 andany index n N , there exists an N > n and a k {1, . . . , N} such that s < xNk < s + .Hence,

    S() X Z . (5.6)

    Recall the definition of the numbers j und j ,

    j = limnxnj , j = limnxn,nj+1 , j = 1, 2, . . .

    Obviously, := 0 1 2 2 1 0 := ,

    where := lim

    jj und := lim

    jj .

    We say that + belongs to X if for every A > 0 there exists an x X (A,+) . Analogously, X is defined.

  • 44 CHAPTER 5. THE REPRESENTATION THEOREM

    Proposition 5.10 Let be a representation of L .

    (a) If k < k+1 then S() (k, k+1] 6= .(b) If k = k+1 then k S() .(c) S() .

    Proposition 5.11 Assume that 1 > and let be a natural representation of L . Thenj S() j = 1, 2, . . . and S() (, ) = {j : j 1, j < } .

    Proposition 5.12 If p = p+1 for some index p , then p = .

    Consequently, only the following three situations are possible:

    1. = 1 = 2 = = ,2. < 1 < < p = p+1 = = ,3. < 1 < 2 < < .

    Thus, if is a natural representation of the positive-definite moment functional L , then thespectrum of can be represented in the form

    S() = S1 with

    =

    { , if = ,{j : j 1 , j < } , otherwise ,

    with the analogously defined set , and with a set

    S1

    [, ] : < < + ,(, ] : = < < + ,[,+) : < < = + ,

    (,+) : = < = + .

    5.4 On the Determinacy of a Representation

    The aim of our further considerations is to prove that, for any positive-definite moment functionalL with < 1 < 1 < , the difference of two representations is a constant at all commonpoints of continuity.

    Again, by (Pn)n=1 we denote the monic OPS. Define, for x0 6 (1, 1) ,

    Qn(x) = Pn+1(x) + aPn(x) with a = Pn+1(x0)Pn(x0)

    6= 0 .

    Due to Qn(xn+1,k) = aPn(xn+1,k) , the values Qn(xn+1,k) change the sign (comp. the proof ofProp. 2.18), so that Qn(x) has exactly n + 1 real zeros x

    nk (xn+1,k+1, xn+1,k) , k = 1, . . . , n ,

    and xn0 = x0 . Let be a representation of L .

  • 5.4. ON THE DETERMINACY OF A REPRESENTATION 45

    There exist numbers Ank such that the quadrature rule

    pi(x) d(x) =nk=0

    Ankpi(xnk)

    is true for all polynomials pi C2n+1[x] . Analogously to Proposition 2.27 one can prove that

    An0 =

    (nk=0

    [p(x0)]2

    )1, (5.7)

    where (pn(x))n=0 denotes the orthonormal polynomial system (ONPS) w.r.t. L . Moreover, the

    inequalitiesAn0 (x0) () , x0 1 , (5.8)

    andAn0 (+) (x0) , x0 1 , (5.9)

    are fulfilled. Taking into account (5.7), (5.8), (5.9), and Proposition 4.39 one can prove

    Corollary 5.13 If < 1 < 1 < + , then S() [1, 1] for any representation of apositive-definite moment functional L .

    Lemma 5.14 Let j : [a, b] R , j = 1, 2 , be two functions of bounded variation on thecompact interval [a, b] with the property b

    axn d1(x) =

    baxn d2(x) , n = 0, 1, 2, . . .

    Then there exists a constant c R such that 1(x) 2(x) = c for all common points ofcontinuity x [a, b] of 1 and 2 .

    A moment functional L is called determined if its representations (x) are unique in the senseof Lemma 5.14. In this case, (x) is called the essentially unique representation of L .

    Proposition 5.15 If [1, 1] is compact, then L is determinate.

    Proposition 5.16 For a determinate functional L with the essentially unique representation(x) , the spectrum S() is a subset of any closed supporting set of L .

    Example 5.17 (Stieltjes) The moment sequence

    n =pie

    (n+1)2

    4 , n N0defines a positive definite moment functional, which however possesses the essentially differentrepresentations

    (x) =

    0 : x 0 x

    0e ln

    2 t[1 + sin(2pi ln t)] dt : x > 0

    , 1 < < 1 .

  • 46 CHAPTER 5. THE REPRESENTATION THEOREM

    5.5 Classical Moment Problems

    1. (The Stieltjes moment problem) In 1894 T. J. Stieltjes formulated the followingmoment problem: Let the sequence (n)

    n=0 of real numbers be given. Look for necessary

    and sufficient conditions for the existence of a distribution function (x) with infinitespectrum in [0,) , such that

    0xn d(x) = n , n = 0, 1, 2, . . .

    2. (The Hamburger moment problem) About 1920/21 H. Hamburger stated the prob-lem analogous to the Stieltjes moment problem, where the interval [0,) is replaced by(,) .

    Define (comp. Section 2.1)

    n = det

    0 1 n1 2 n+1...

    ......

    ...

    n n+1 2n

    and (1)n = det

    1 2 n+12 3 n+2...

    ......

    ...

    n+1 n+2 2n+1

    ,n = 0, 1, 2, . . .

    Proposition 5.18 The Hamburger moment problem has a solution if and only if, for all n =0, 1, 2, . . . , n > 0 .

    Proposition 5.19 The Stieltjes moment problem has a solution if and only if the momentfunctional L w.r.t. the sequence (n) n=0 is positive-definite on [0,) .

    Proposition 5.20 The Stieltjes moment problem has a solution if and only if n > 0 and

    (1)n > 0 for all n = 0, 1, 2, . . .

    Proposition 5.21 (First general representation theorem) Let (n)n=0 R be an arbi-

    trary number sequence. Then, there exists a function : R R of bounded variation suchthat

    xn d(x) = n , n = 0, 1, 2, . . .

    Proposition 5.22 (Second general representation theorem) For arbitrarily chosen num-ber sequences (n)

    n=0 , (n)

    n=0 and the polynomial system {Pn(x)} , defined by the three-term

    recurrence relation

    P1 0 , P0 1 , Pn+1(x) = (x n)Pn(x) nPn1(x) , n = 1, 2, . . .there exists a function : R C of bounded variation such that

    Pm(x)Pn(x) d(x) = 01 nmn , m, n = 0, 1, 2, . . .

    The function (x) can be chosen as a real valued one if and only if the number sequences (n)n=0

    and (n)n=0 are real. It can be chosen as a distribution function if and only if n R and n > 0

    for all n = 0, 1, 2, . . .

  • Chapter 6

    On the numerical solution of integralequations

    6.1 The Nystrom method

    Consider an integral equation of the form (Fredholm integral equation of second kind)

    f(x) 11K(x, y)v,(y)f(y) dy = g(x) , 1 < x < 1 , (6.1)

    where : g(1, 1) R and K : (1, 1)2 R are given continuous functions and where thecontinuous function f : (1, 1) R is looked for. By v,(x) = (1 x)(1 + x) we denote aJacobi weight function, where 1 < , is assumed (cf. Section 2.5).

    We denote by x,nk , k = 1, . . . , n , x,nn < . . . < x

    ,n1 , the zeros of the nth Jacobi polynomial

    P,n (x) and by ,nk the respective weights in the Gaussian rule 1

    1u(x)v,(x) dx

    nk=1

    ,nk u(x,nk ) ,

    which are also callec Christoffel numbers. A first idea to get an approximation for the solutionof (6.1), is to replace the integral in (6.1) by the gaussian rule,

    fn(x)nk=1

    ,nk K(x, x,nk )fn(x

    ,nk ) = g(x) , 1 < x < 1 . (6.2)

    However, this equation is not completely discretized, i.e., we do not get a finite number ofeuqtions with a finite number of unknowns.

    If we assume that (6.2) has a solution fn : (1, 1) R then the vector[fn(x

    ,nk )

    ] nk=1

    is

    a solution of the linear sysrem of equations

    fn(x,nk )

    nk=1

    ,nk K(x,nj , x

    ,nk )fn(x

    ,nk ) = g(x

    ,nj ) , j = 1, . . . , n . (6.3)

    47

  • 48 CHAPTER 6. INTEGRAL EQUATIONS

    If (6.2) is uniquely solvable, then this is also true for (6.3), and with the help of the solution of(6.3), by

    fn(x) = g(x) +

    nk=1

    ,nk K(x, x,nk )fn(x

    ,nk )

    one gets the solution of (6.2), the so called Nystrom interpolant.

    Now, let us assume that we have found an appropriate Banach space X of continuous func-tions on (1, 1) to investigate the equation (6.1). Then, we write (6.1) in the form

    (I K)f = gwith the identity operator I : X X and the integral operator

    K : X X , f 7 11K( . , y)v,(y)f(y) dy .

    Equation (6.2) can be written as(I Kn)fn = g

    with the sequence (Kn) n=1 of operators

    Kn : X X , f 7nk=1

    ,nk K( . , x,nk )f(x

    ,nk ) .

    6.2 Collectively compact operator sequences

    A subset A E of a metric space (E, d) is called compact, if every covering of A by opensubsets of E contains a covering by finitely many of these sets. The set A is referred to berelatively compact, if the closure A is compact. This is equivalent to the fact that everysequence (xn)

    n=1 of points xn A has a convergent subsequence.

    Let (E, d) be a compact metric space. By C(E) we denote the banach space of all continuousfunctions f : E R , where the norm is given by

    f = f,E = max {|f(x)| : x E} .A subset F C(E) is called uniformly bounded, if F is bounded in (C(E), .) , i.e., ifthere is a constant M (0,) such that

    |f(x)| M x E , f F .We say that the set F is equicontinuous, if for any > there is a > 0 such that

    |f(x1) f(x2)| < , x1, x2 E : d(x1, x2) < , f F .We remember the Theorem of Arzela-Ascoli: A subset A C(E) is relatively compact in(C(E), .) if and only if it is uniformly bounded an equicontinuous.

    Example 6.1 Let the set {fn : n N} C(E) be uniformly bounded and equicontinuous. More-over, assume that there is a function f C(E) such that lim

    n fn(x) = f(x) for all x E . Then,limn fn f = 0 .

  • 6.3. THE CASE = = 0 AND X = C[1, 1] 49

    Now. let (X, .) be a Banach space and Kn : X X , n N be a sequence of linear oper-ators. This sequence is called collectively compact if the set {Knf : f X, f 1, n N}is relatively compact in X . This immediately implies that the sequence (Kn)

    n=1 is a sequence

    of uniformly bounded and compact operators.

    Proposition 6.2 Let X be a Banach space and K : X X as well as Kn : X X , n Nbe linear operators, for which (Kn)

    n=1 is relatively compact and limn Knf Kf = 0 for all

    f X . For given g X , consider the equations(I K)f = g (6.4)

    and(I Kn)fn = g . (6.5)

    Then:

    (a) limn (Kn K)Kn = 0(b) If the null space N(I K) is trivial, i.e., in case g = 0 , the equation (6.4) has in X only

    the trivial solution f = 0 , then there exists an n0 N , such that, for all n n0 , theequation (6.5) has a unique solution fn X . Moreover,

    fn f c Knf Kfwith an constant c (0,) independet of n n0 and g X and with the unique solutionf X of equation (6.4).

    6.3 The case = = 0 and X = C[1, 1]

    Let = = 0 . Consider equation (6.1) in the space C[1, 1] of all continuous functionsf : [1, 1] R and assume that K : [1, 1]2 R is continuous. Obviously,

    K max{ 11|K(x, y)| dy : 1 x 1

    }.

    We investigate the respective sequence Kn ,

    (Knf)(x) =nk=1

    nkK(x, xnk)f(xnk) ,

    where nk = 0,0nk and xnk = x

    0,0nk . The set

    A = {Knf : f C[1, 1], f 1, n N}is uniformly bounded and equicontinuous:

    We have|(Knf)(x)| K

    nk=1

    nk f = 2 K f , (6.6)

    where K = max{|K(x, y)| : (x, y) [1, 1]2} .

  • 50 CHAPTER 6. INTEGRAL EQUATIONS

    For any > 0 there exists a > 0 , such that K(x1, y) K(x2, y) < x1, x2y [1, 1]with x1 x2| < . This implies

    |(Knf)(x1) (Knf)(x2)| 2 f , x1, x2 [1] : |x 1 x 2| < . (6.7)

    Consequently, the sequence (Kn) n=1 is collectively compact in C[1, 1] . The relations (6.6) and(6.7) also show that, for any f C[1, 1] the set Af = {Knf : n N} is relatively compact inC[1, 1] . Furthermore, proposition 2.24 gives

    limn(Knf)(x) = Kf(x) , x [1, 1] ,

    so that, due to example 6.1, limn Knf Kf = 0 .

    Hence, Propositions 6.2 says: If the equation

    f(x) 11K(x, y)f(y)dy = 0 , 1 < x < 1 ,

    has only the trivial solution f 0 in C[1, 1] then there is an n0 N such that, for all n n0 ,the equation

    fn(x)nk=1

    nkK(x, xnk)fn(xnk) , 1 < x < 1 ,

    has a unique solution fn C[1, 1] . Moreover,

    fn f c Knf Kfwith a constant c (0,) , c 6= c(n, g) , and with the unique solution f C[1, 1] of theequation

    f(x) 11K(x, y)f(y)dy = g(x) , 1 < x < 1 .

    6.4 The application of weighted spaces of continuous functions

    By Cu we denote the Banach space of all continuous functions f : (1, 1) R , for whichuf : [1, 1] R is continuous (more precisely: possesses a continuous continuation into 1and +1), equipped with the norm

    f,u := uf = sup {|u(x)f(x)| : 1 < x < 1} .

    Here, u(x) = v,(x) = (1 x)(1 + x) with , [0, 1] .

    With respect to K(x, y) we assume the following:

    (A) K : [1, 1]2 R is continuous, where K(x, y) = v,(x)K(x, y)v1,1(y) .

    (B) c0 :=

    11

    v,(x) dx

    v,(x)v1,1(x) + 1 and + 1 > + 1 .

  • 6.4. THE APPLICATION OF WEIGHTED SPACES OF CONTINUOUS FUNCTIONS 51

    Lemma 6.3 Let +1 > 1 and +1 > 1 and let j N be fixed. Then, there is a constantc1 (0,) such that

    nk=1

    ,nk

    q(x,nk ) v1,1(x,nk ) c1 11 |Q(x)|v,(x)v1,1(x) dxfor all polynomials q R[x] with deg q jn , c1 6= c1(n, q) .

    By Cu we refer to the closed subspace of Cu of all f Cu withlim

    x10u(x)f(x) = 0 , if > 0 , und lim

    x1+0u(x)f(x) = 0 , if > 0 .

    Lemma 6.4 The set R[x] is dense in Cu .

    Lemma 6.5 If u = v, , 0 < + 1 , 0 < + 1 and f Cu then

    limn

    nk=1

    ,nk f(x,nk ) =

    11f(x)v,(x) dx .

    Lemma 6.6 Let the conditions (A) and (B) be fulfilled and let

    Kn : Cu Cu , f 7nk=1

    ,nk K( . , x,nk )f(x

    ,nk ) .

    Then, the sequence (Kn) n=1 is collectively compact and

    limn Knf Kf,u = 0 f Cu .

    Summarizing we can formulate the following: If the equation

    f(x) 11K(x, y)v,(y)f(y) dy = 0 , 1 < x < 1 ,

    has only the trivial solution in Cu , if the function K(x, y) = v,(x)K(x, y)v1,1(y) is continuous

    on [1, 1]2 and if 0 , 1 , 1 < , , + 1 < + 1 , + 1 < + 1 , then there exist ann0 N and a constant c 6= c(n, g) , such that the equation

    fn(x)nk=1

    ,nk K(x, x,nk )fn(x

    ,nk ) = g(x) , 1 < x < 1 ,

    possesses for all n n0 and for all g Cu a unique solution fn Cu and

    fn f,u c sup1

  • 52 CHAPTER 6. INTEGRAL EQUATIONS

    In order to estimate the rate of the convergence of the right hand side in (6.8), we consider the

    error abschatzen zu konnen, betrachten wir den Fehler R,n (f) of the Gaussian rule 11f(x)v,(x) dx =

    nk=1

    ,nk f(x,nk ) +R

    ,n (f) .

    By Em(f),u we denote the error of the (with the help of u(x)) weighted best uniformapproximation of f by polynomials of degree < m , i.e.

    Em(f),u = inf{f p,u : p Rm[x]

    }.

    In case of u 1 we write simply Em(f) instead of Em(f),1 .

    Lemma 6.7 For f C[1, 1] , we have

    |R,n (f)| 2c,0 E2n(f) ,

    where c,0 =

    11v,(x) dx .

    Lemma 6.8 If 0 < + 1 , 0 < + 1 und f Cv, , then

    |R,n (f)| c,, E2n(f),v, ,

    where the constant c,, does not depend on n and f .

    Taking into account Lemma 6.7 and Lemma 6.8, in view of (6.8) one can conlcude (6.8) thatf+n f,u c sup {E2n(K(x, . )f),v : 1 < x < 1} ,if the function K( . , y)f(y) , 1 < y < 1 , has the respective properties w.r.t. an appropriateweight function v(y) . The function classes

    Wr,u ={g Cu :(r) r Cu

    }, r N ,

    where (x) =

    1 x2 , are of special interest. Indeed, for g Wr,u , one has the estimate

    En(g),u c1n

    g,u , c1 6= c1(n, g) ,which, for g Wr,u , can be iterated to get

    En(g),u crnr

    g(r),ru

    , cr 6= c1(n, g) .

  • Chapter 7

    Orthogonal Polynomials in theComplex Plane

    7.1 Definitions. Location of the Zeros

    Let be a positive measure on the Borel sets of the complex plane. We assume thatsupport S = S() = supp() of is compact and contains infinitely many points.

    Define the inner product

    f, g =f(z)g(z) d(z) , f, g C[z] .

    Examples: Let E C be an open and bounded set, f, g = E f(z)g(z)w(z) d(x, y) with z =x+ iy , x, y R , w(z) 0 on E , 0 < E w(z) d(x, y)

  • 54 CHAPTER 7. ORTHOGONAL POLYNOMIALS IN THE COMPLEX PLANE

    Proposition 7.2 All zeros of Pn(z;) lie in the convex hull conv(S) of the support of .

    Proposition 7.3 If conv(S) is not a segment, then all zeros of the polynomial Pn(z;) arelocated in the interior int conv (S) of the convex hull of the support S of .

    It can happen that all zeros of Pn(z) for all n N are not in S , for example, if S T :={z C : |z| = 1} .

    Set C := C {P} . By D(S) we refer to that connectivity component of C \ S , whichcontains the infinite point P (i.e. D(S) is open, connected and unbounded). The setPconv(S) := C \D(S) is called the polynomial convex hull of S .

    If S = T , then Pconv(S) = {z C : |z| 1} . But, if S = {z C : |z| = 1, Im z 0} , thenPconv(S) = S .

    Any function holomorphic on Pconv(S) can be uniformly approximated on S by polyno-mials (see, for example, [10, Chapter 12]).

    Lemma 7.4 Let E C be compact and E Pconv(S) = (i.e. E D(S)). Then, thereexist numbers m Z+ and (0, 1) , such that, for m arbitrary numbers z1, . . . , zm E , thereare numbers w1, . . . , wm C satisfying

    mk=1

    z wkz zk z S .

    Proposition 7.5 Let E C be a closed set with EPconv(S) = . Then, there exists a naturalnumber m0 such that

    |{z E : Pn(z;) = 0}| m0 n = 1, 2, . . . ,

    i.e. the number of zeros of Pn(z;) located in E is uniformly bounded.

    7.2 Asymptotic Properties of the Zeros

    Define

    vS = sup (|v(z)| : z S()) n=1and

    tn(S) := min {zn + S : zn + Cn+1[z]} .The (uniquely determined) polynomial Tn(z) = Tn(z;) = z

    n+ Cn+1[z] with tn(S) = TnSis called Chebyshev polynomial (w.r.t. ) of degree n . Due to Prop. 7.1 we have

    1

    n=

    (|Pn(z)|2 d

    ) 12

    (|Tn(z)|2 d

    ) 12

    [(S)] 12 tn(S) .

  • 7.2. ASYMPTOTIC PROPERTIES OF THE ZEROS 55

    The limit cheb(S) := limn

    ntn(S) (which exists) is called Chebyshev constant for S . Because

    of n 1tn(S)

    1

    [(S)]12

    we have

    lim infn

    nn 1

    cheb(S). (7.1)

    The measure is called completely regular if

    limn pn(z;)

    1nS = 1 . (7.2)

    Proposition 7.6 If the measure is completely regular, then

    limn

    nn =

    1

    cheb(S).

    In what follows we describe an alternative definition of the Chebyshev constant. Let E Cbe compact, let M(E) be the set of all Borel positive measures satisfying S() E and(E) = 1 . For M(E) , define the logarithmic potential by

    U(z) :=

    log |z t|1 d(t)

    and the energy of this potential by

    I[] :=

    U d =

    log |z t|1 d(t) d(z) .

    Set V (E) := inf {I[] : M(E)} . The number cap(E) := eV (E) is called the logarithmiccapacity of E .

    Proposition 7.7 (electrostatic problem) If cap(E) > 0 , then there exists a uniquely deter-mined measure E M(E) such that I[E ] = V (E) .

    This measure is said to be the equilibrium distribution for E . We have

    (a) S(E) E , where E = D(E) denotes the outer boundary of E ,(b) UE (z) V (E) z C ,(c) cap(E) = cheb(E) .

    Additionally, in (a) we have cap(E \S(E)) = 0 , and in (b) equality holds for all z E exeptof a possible set of capacity 0 .

    In what follows we assume that the outer boundary E consists of finitely many analyticarcs. Denote by gE(z) the Green function for D(E) with singularity at infinity definied by thefollowing conditions:

    gE(z) is harmonic in D(E) \ {P} , gE(z) 0 if z E, z D(E) ,

  • 56 CHAPTER 7. ORTHOGONAL POLYNOMIALS IN THE COMPLEX PLANE

    V C :(gE(z) log |z|

    ) V if |z| .

    By Greens formulae we get

    V gE(z) = 12pi

    E

    log |z t|1 n

    gE(t) |dt| =E

    log |z t|1 d ,

    where n denotes the outer normal for E and

    =1

    2pi

    ngE(t) |dt| .

    Proposition 7.8 We have V = V (E) , = E , and

    UE (z) = V (E) gE(z) = log 1cap(E)

    gE(z) .

    Example 7.9 If E = {z C : |z| = R} then

    gE(z) = log zR

    ,V = logR and cap(E) = R = cheb(E) . On E = E there holds |dt| = ds and

    ng(t) =

    1

    R,

    i.e. dE =ds

    2piR.

    Example 7.10 In case of E = [1, 1] one obtains V = log 12 , i.e. cap(E) = cheb(E) = 12 ,and

    dE =1

    pi

    dx1 x2 .

    For a polynomial Q(z) with the zeros z1, ..., zk ,

    Q(z) = a0(z z1)m1 ...(z zk)mk , n = m1 + ...+mk ,

    and for a set A C , defineQ(A) =

    1

    n

    zjA

    mj .

    Proposition 7.11 Let E C be compact with positive capacity. Suppose that the monic poly-nomials Qn(z) = z

    n + Cn+1[z] satisfy the conditions

    (a) lim supn Qn1nE cap(E) ,

    (b) limn Qn(A) = 0 for all closed sets A int Pconv(E) .

  • 7.2. ASYMPTOTIC PROPERTIES OF THE ZEROS 57

    Then

    limn

    f dQn =

    f dE (7.3)

    for all continuous functions f : C C with compact support, where E denotes the equilibriumdistribution for E .

    Condition (a) means that the polynomials Qn are asymptotic minimal w.r.t. the-norm (comp.the definition of cheb(S) and assertion (c) after Theorem 7.7). Condition (b) says that thenumber of zeros of Qn(z) , located in a closed set A int Pconv(S) , multiplied by 1n , tends tozero if n .

    Proposition 7.12 Let be a completely regular measure and assume that S() has positivecapacity. Moreover, suppose that int Pconv(S()) = . Then, Pn(.,) S() in the sense of(7.3).

    Proposition 7.13 Assume that S() T = {z C : |z| = 1} and lim supn

    |Pn(0;)|1n = 1 .

    Moreover, let N N be an infinite subset of N with limn,nN

    |Pn(0;)|1n = . Then, in the

    sense of (7.3)

    (a) if 0 < % < 1 , we have the convergence

    PnnN ds

    2pi%

    (b) if % = 1 we have the convergence

    PnnN ds

    2pi,

    if limn

    1

    n

    nk=0

    |Pk(0;)| = 0 .

    Hence, the limit measure is the equilibrium distribution for E = {z C : |z| = } .

    For the determination of in case of S T , one can use, for example, the following assertion:If d(eis) =

    D(eis)2 ds a.e. on [0, 2pi] , where D(z) is holomorphic in {z C : |z| < 1} , then is the smallest number such that

    1

    D(z)is holomorphic in

    {z C : |z| < 1

    }.

    Example 7.14 For

    d(eis) =sin s

    2

    4 = 116

    (1 eis)22 , 0 s 2pi ,we obtain = 1 and for

    d(eis) =

    (5

    4 cos s

    )ds =

    1 12 eis2 ds , 0 s 2pi ,

    = 12 .

  • Index

    D(S), 54Em(f), 52Em(f),u, 52I[], 55Mk(x), 7

    R,n (f), 52S(), 53Tn(x), 9C(E), 48Cu, 50jk, 7, 11`nk(x), 21k, 20C[x], 7, 15K[x], 7., ., 7k, 20Pconv(S), 54cap(E), 55cheb(S), 55

    approximant, 31Arzela-Ascoli, Theorem of, 48

    chain sequence, 34Chapman-Kolmogorov, equations of, 13Charlier polynomials, 11Chebyshev constant, 55Chebyshev polynomial, 54Chebyshev polynomial of first kind, 9Chebyshev polynomials of second kind, 14Chebyshev weight of first kind, 9Christoffel numbers, 47Christoffel-Darboux, formula of, 19collectively compact operators, 49compact set, 48comparison test, 35completely regular measure, 55continued fraction, 31convergent continued fraction, 32

    degree of a polynomial, 7determined moment functional, 45

    energy, 55equicontinuous functions, 48equilibrium distribution, 55

    Gaussian quadrature rule, 21generating function, 11

    Hermite polynomials, 14

    initial parameter, 34inner product, 7interpolation polynomial, 21

    Jacobi fraction, 33Jacobi polynomials, 22

    Lagrange fundamental polynomial, 21leading coefficient, 7Legendre, 8Legendre polynomials, 8logarithmic capacity, 55logarithmic potential, 55

    maximal parameter sequence, 35minimal parameter sequence, 35moment, 15moment functional, 9, 15moment problem, 8monic numerator polynomials, 34monic OPS, 16monic polynomial, 16

    numerator polynomials, 34Nystrom interpolant, 48

    ONPS, 15ONPS, orthonormal system of polynomials, 8OPS, 15OPS, orthogonal polynomial system, 9orthogonal polynomial, 15orthogonal polynomial system, 15orthonormal polynomial, 15orthonormal polynomial system, 15outer boundary, 55

    58

  • INDEX 59

    parameter sequence, 34partial denominator, 32partial numerator, 32polynomial convex hull, 54positive definite moment functional, 20positive definite momenten functional, 17

    quasi-definite moment functional, 17

    recursion formula, 18relatively compact set, 48Rodrigues, 8Rodrigues formula, 8, 22

    scalar product, 7support, 21supporting set, 20

    uniformly bounded functions, 48