orthpoly engl
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OrthPoly EnglTRANSCRIPT
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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.
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2
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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
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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
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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.
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6 BIBLIOGRAPHY
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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)
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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)
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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)
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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.
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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 .
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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
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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)
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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.
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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 .
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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 ,
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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 .
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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} .
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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 .
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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
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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) .
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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)
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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) .
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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) ,
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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) .
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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
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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