17.2 Weight Functions and Orthogonal Polynomials 245 en, g 2 0 for all n Z. From the above convergence result we deduce, for all⟨ f ⟩ =1([0, 2π]), ∈ ∈ C N

f , g 2 lim cn(f )en, g 2 0. ⟨ ⟩ = N ⟨ ⟩ = →∞ n N !=− Since 1([0, 2π]) is known to be dense in L2([0, 2π], dx) it follows that g 0, by CorollaryC 17.2, hence by Theorem 17.2, this system is an orthonormal basis= of L2([0, 2π], dx). Therefore, every f L2([0, 2π], dx) has a Fourier expansion, which converges (in the sense of the L2-topology).∈ Thus, convergence of the Fourier series in the L2-topology is “natural,” from the point of view of having convergence of this series for the largest class of functions.

17.2 Weight Functions and Orthogonal Polynomials

Not only for the interval I [0, 2π] are the Hilbert spaces L2(I,dx) separable, but for any interval I [a, b=], afunction ρ : I R on the interval I which is assumed to have the following properties: → 1. On the interval I, the function ρ is strictly positive: ρ(x) > 0 for all x I. 2. If the interval I is not bounded, there are two positive constants α and∈ C such that ρ(x)eα x C for all x I. | | ≤ ∈ The strategy to prove that the Hilbert space L2(I,dx) is separable is quite simple. A n first step shows that the countable set of functions ρn(x) x ρ(x), n 0, 1, 2, ... is total in this Hilbert space. The Gram–Schmidt orthonormalization= then= produces easily an orthonormal basis.

Lemma 17.1 The system of functions ρn : n 0, 1, 2, ... is total in the Hilbert space L2(I, dx), for any interval I. { = } 2 Proof For the proof we have to show: If an element h L (I,dx) satisfies ρn, h 2 0 for all n, then h 0. ∈ ⟨ ⟩ = = In the case I R we consider h to be be extended by 0 to R I and thus get a 2 ̸= \ function h L (R,dx). On the strip Sα p u iv C : u, v R, v < α , introduce the∈ auxiliary function = { = + ∈ ∈ | | }

F (p) ρ(x)h(x)eipx dx. = "R The growth restriction on the weight function implies that F is a well-defined holo- morphic function on Sα (see Exercises). Differentiation of F generates the functions 246 17 Separable Hilbert Spaces

ρn in this integral: dnF F (n)(p) (p) in h(x)ρ(x)xneipx dx = dpn = !R (n) n for n 0, 1, 2, ..., and we deduce F (0) i ρn, h 2 0 for all n. Since F is = = ⟨ ⟩ = holomorphic in the strip Sα it follows that F (p) 0 for all p Sα (see Theorem = ∈ 9.5) and thus in particular F (p) 0 for all p R. But F (p) √2π (ρh)(p) where is the inverse Fourier= transform (see∈ Theorem 10.1),= and weL know L 2 f , g 2 f , g 2 for all f , g L (R,dx) (Theorem 10.7). It follows that ⟨L L ⟩ = ⟨ ⟩ ∈ 2 ρh, ρh 2 (ρh), (ρh) 2 0 and thus ρh 0 L (R,dx). Since ρ(x) > 0 for⟨ x ⟩I this= ⟨ impliesL Lh 0⟩ and= we conclude. = ∈ ✷ Technically∈ it is simpler= to do the orthonormalization of the system of functions 2 ρn : n N not in the Hilbert space L (I,dx) directly but in the Hilbert space L{ 2(I, ρd∈x),} which is defined as the space of all equivalence classes of measurable functions f : I K such that f (x) 2ρ(x)dx< equipped with the inner prod- → I | | ∞ uct f , g f (x)g(x)ρ(x)dx. Note that the relation f , g ρf , ρg ρ I " ρ √ √ 2 holds⟨ for⟩ all=f , g L2(I, ρdx). It implies that the Hilbert⟨ spaces⟩ =L⟨2(I, ρdx) and⟩ L2(I,dx) are (isometrically)" ∈ isomorphic under the map L2(I, ρdx) f √ρf L2(I,dx). ∋ (→ ∈ This is shown in the Exercises. Using this isomorphism, Lemma 17.1 can be restated as saying that the system of powers of x, xn : n 0, 1, 2, ... is total in the Hilbert space L2(I, ρdx). { = } We proceed by applying the Gram–Schmidt orthonormalization to the system of powers xn : n 0, 1, 2, ... in the Hilbert space L2(I, ρdx). This gives a sequence { = } of polynomials Pk of degree k such that Pk, Pm ρ δkm. These polynomials are ⟨ ⟩ =0 defined recursively in the following way: Q0(x) x 1, and when for k 1 the = = ≥ polynomials Q0, ..., Qk 1 are defined, we define the polynomial Qk by − k 1 k − Qn, x ρ Qk(x) xk ⟨ ⟩ Qn. = − Q , Q n 0 n n ρ #= ⟨ ⟩ Finally, the polynomials Qk are normalized and we arrive at an orthonormal system of polynomials Pk: 1 Pk Qk, k 0, 1, 2, ... . = Qk ρ = ∥ ∥ Note that according to this construction, Pk is a polynomial of degree k with positive coefficient for the power xk. Theorem 17.1 and Lemma 17.1 imply that the system of polynomials Pk : k 0, 1, 2, ... is an orthonormal basis of the Hilbert space L2(I, ρdx). If we{ now introduce= the} functions

ek(x) Pk(x) ρ(x), x I = ∈ we obtain an orthonormal basis of the Hilbert$ space L2(I,dx). This shows Theorem 17.3. 17.2 Weight Functions and Orthogonal Polynomials 247

Theorem 17.3 For any interval I (a, b), a 0. ✷ ∈ Naturally, the orthonormal polynomials Pk depend on the interval and the weight function. After some general properties of these polynomials have been studied we will determine the orthonormal polynomials for some intervals and weight functions explicitly.

Lemma 17.2 If Qm is a polynomial of degree m, then Qm, Pk ρ 0 for all k>m. ⟨ ⟩ = 2 Proof Since Pk : k 0, 1, 2, ... is an ONB of the Hilbert space L (I, ρdx) the { = } polynomial Qm has a Fourier expansion with respect to this ONB: Qm n∞ 0 cnPn, k = = cn Pn, Qm ρ . Since the powers x , k 0, 1, 2, ... are linearly independent = ⟨ ⟩ = ! functions on the interval I and since the degree of Qm is m and that of Pn is n, the m coefficients cn in this expansion must vanish for n>m, i.e., Qm n 0 cnPn and = = thus Pk, Qm ρ 0 for all k>m. ✷ ⟨ ⟩ = ! Since, the orthonormal system Pk : k 0, 1, 2, ... is obtained by the Gram– Schmidt orthonormalization from the{ system= of powers}xk for k 0, 1, 2, ... with = respect to the inner product , ρ , the polynomial Pn 1 is generated by multiplying ⟨· ·⟩ + the polynomial Pn with x and adding some lower order polynomial as correction. Indeed one has Proposition 17.1 Let ρ be a weight for the interval I (a, b) and denote the complete system of orthonormal polynomials for this weight= and this interval by Pk : k 0, 1, 2, ... . Then, for every n 1, there are constants An, Bn, Cn such that{ = } ≥ Pn 1(x) (Anx Bn)Pn(x) CnPn 1(x) x I. + = + + − ∀ ∈ k Proof We know Pk(x) akx Qk 1(x) with some constant ak > 0 and some = + − polynomial Qk 1 of degree smaller than or equal to k 1. Thus, if we define An an 1 − − = + , it follows that Pn 1 AnxPn is a polynomial of degree smaller than or equal an + − to n, hence there are constants cn,k such that

n

Pn 1 AnxPn cn,kPk. + − = k 0 "= Now calculate the inner product with Pj , j n: ≤ n

Pj , Pn 1 AnxPn ρ cn,k Pj , Pk ρ cn,j . ⟨ + − ⟩ = ⟨ ⟩ = k 0 "= 248 17 Separable Hilbert Spaces

Since the polynomial Pk is orthogonal to all polynomials Qj of degree j k 1 ≤ − we deduce that cn,j 0 for all j

Proof Per construction the orthonormal polynomials Pk have real coefficients, have the degree k, and the coefficient ck is positive. The fundamental theorem of algebra (Theorem 9.4) implies: The polynomial Pk has a certain number m k of simple ≤ real roots x1, ..., xm and the roots which are not real occur in pairs of complex conjugate numbers, (zj , zj ), j m 1, ..., M with the same multiplicity nj , m M = + + 2 j m 1 nj k. Therefore the polynomial Pk can be written as = + = ! m M nj nj Pk(x) ck (x xj ) (x zj ) (x zj ) . = − − − j 1 j m 1 "= ="+ m Consider the polynomial Qm(x) ck j 1 (x xj ). It has the degree m and ex- = = − M 2nj actly m real simple roots. Since Pk(x) Qm(x) j m 1 x zj , it follows that =# = + | − | Pk(x)Qm(x) 0 for all x I and PkQm 0, hence Pk, Qm ρ > 0. If the degree ≥ ∈ ̸= # ⟨ ⟩ m of the polynomial Qm would be smaller than k, we would arrive at a contradiction to the result of the previous lemma, hence m k and the pairs of complex conjugate roots cannot occur. Thus we conclude. = ✷ In the Exercises, with the same argument, we prove the following extension of this proposition.

Lemma 17.3 The polynomial Qk(x, λ) Pk(x) λPk 1(x) has k simple real roots, = + − for any λ R. ∈ Lemma 17.4 There are no points x0 I and no integer k 0 such that Pk(x0) ∈ ≥ = Pk 1(x0) 0. − = Proof Suppose that for some k 0 the orthonormal polynomials Pk and Pk 1 have a ≥ − common root x0 I: Pk(x0) Pk 1(x0) 0. Since we know that these orthonormal ∈ = − = polynomials have simple real roots, we know in particular Pk′ 1(x0) 0 and thus − ̸= Pk′ (x0) we can take the real number λ0 − to form the polynomial Qk(x, λ0) Pk′ 1(x0) = − = Pk(x) λ0Pk 1(x). It follows that Q(x0, λ0) 0 and Qk′ (x0) 0, i.e., x0 is a root + − = = of Qk( , λ) with multiplicity at least two. But this contradicts the previous lemma. · Hence there is no common root of the polynomials Pk and Pk 1. ✷ − Theorem 17.4 (Knotensatz) Let Pk : k 0, 1, 2, ... be the orthonormal basis for { = } some interval I and some weight function ρ. Then the roots of Pk 1 separate the − roots of Pk, i.e., between two successive roots of Pk there is exactly one root of Pk 1. − Proof Suppose that α < β are two successive roots of the polynomial Pk so that Pk(x) 0 for all x (α, β). Assume furthermore that Pk 1 has no root in the open ̸= ∈ − interval (α, β). The previous lemma implies that Pk 1 does not vanish in the closed − 17.3 Examples of Complete Orthonormal Systems for L2(I, ρdx) 249 interval [α, β]. Since the polynomials Pk 1 and Pk 1 have the same system of roots, − − − we can assume that Pk 1 is positive in [α, β] and Pk is negative in (α, β). Define the Pk (x−) function f (x) − . It is continuous on [α, β] and satisfies f (α) f (β) 0 Pk 1(x) = − = = and f (x) > 0 for all x (α, β). It follows that λ0 sup f (x):x [α, β] f (x0) ∈ = { ∈ } = for some x0 (α, β). Now consider the family of polynomials Qk(x, λ) Pk(x) ∈ = + λPk 1(x) Pk 1(x)(λ f (x)). Therefore, for all λ λ0, the polynomials Qk( , λ) − = − − ≥ · are nonnegative on [α, β], in particular Qk(x, λ0) 0 for all x [α, β]. Since ≥ ∈ λ0 f (x0), it follows that Qk(x0, λ0) 0, thus Qk( , λ0) has a root x0 (α, β). = = · ∈ Since f has a maximum at x0,weknow0 f ′(x0). The derivative of f is easily calculated: = Pk′(x)Pk 1(x) Pk(x)Pk′ 1(x) f (x) − − − . ′ 2 = − Pk 1(x) − Thus f ′(x0) 0 implies Pk′(x0)Pk 1(x0) Pk(x0)Pk′ 1(x0) 0, and therefore = − − − = Qk′ (x0) Pk′(x0) f (x0)Pk′ 1(x0) 0. Hence the polynomial Qk( , λ0) has a root of = + − = · multiplicity 2 at x0. This contradicts Lemma 17.3 and therefore the polynomial Pk 1 − has at least one root in the interval (α, β). Since Pk 1 has exactly k 1 simple real − − roots according to Proposition 17.2, we conclude that Pk 1 has exactly one simple root in (α, β) which proves the theorem. − ✷ Remark 17.1 Consider the function

n 2 k F (Q) Q(x) ρ(x)dx, Q(x) akx . = = I k 0 ! "= Since we can expand Q in terms of the orthonormal basis Pk : k 0, 1, 2, ... , n { = } Q k 0 ckPk, ck Pk, Q ρ the value of the function F can be expressed in = = = ⟨ ⟩ n 2 terms of the coefficients ck as F (Q) k 0 ck and it follows that the orthonormal # = = polynomials Pk minimize the function Q F (Q) under obvious constraints (see Exercises). # '→

17.3 Examples of Complete Orthonormal Systems for L2(I, ρdx)

For the intervals I R, I R+ [0, ), and I [ 1, 1] we are going to construct explicitly an= orthonormal= basis= by∞ choosing a suitable= − weight function and applying the construction explained above. Certainly, the above general results apply to these concrete examples, in particular the “Knotensatz.”

x2 17.3.1 I R, ρ(x) e− : Hermite Polynomials = =

x2 Evidently, the function ρ(x) e− is a weight function for the real line. Therefore, = 2 n x by Lemma 17.1, the system of functions ρn(x) x e 2 generates the Hilbert space = − 250 17 Separable Hilbert Spaces

L2(R,dx). Finally the Gram–Schmidt orthonormalization produces an orthonormal basis hn : n 0, 1, 2, ... . The elements of this basis have the form (Rodrigues’ formula){ = }

2 n 2 n x d x2 x hn(x) ( 1) cne 2 e− cnHn(x)e− 2 (17.1) = − dx = ! " # $ with normalization constants

n 1/2 cn (2 n √π)− n 0, 1, 2, ... . = ! =

Here the functions Hn are polynomials of degree n, called Hermite polynomials and the functions hn are the Hermite functions of order n.

Theorem 17.5 The system of Hermite functions hn : n 0, 1, 2, ... is an or- { = } thonormal basis of the Hilbert space L2(R, dx). The statements of Theorem 17.4 apply to the Hermite polynomials. Using Eq. (17.1) one deduces in the Exercises that the Hermite polynomials satisfy the recursion relation

Hn 1(x) 2xHn(x) 2nHn 1(x) 0 (17.2) + − + − = and the differential equation (y Hn(x)) =

y′′ 2xy′ 2ny 0. (17.3) − + =

These relations show that the Hermite functions hn are the eigenfunctions of the 1 2 2 quantum harmonic oscillator with the Hamiltonian H 2 (P Q ) for the eigen- 1 1 = + value n 2 , Hhn (n 2 )hn, n 0, 1, 2..... For more details we refer to [2–4]. In these+ references= one also+ finds other= methods to prove that the Hermite functions form an orthonormal basis. Note also that the Hermite functions belong to the Schwartz test function space (R). S

x 17.3.2 I R+, ρ(x) e− : Laguerre Polynomials = =

x On the positive real line the exponential function ρ(x) e− certainly is a weight function. Hence our general results apply here and we obtain=

Theorem 17.6 The system of Laguerre functions ℓn : n 0, 1, 2, ... which { n x = } is constructed by orthonormalization of the system x e− 2 : n 0, 1, 2, ... in 2 { } = L (R+, dx) is an orthonormal basis. These Laguerre functions have the following form (Rodrigues’ formula):

n 1 x x d n x ℓn(x) Ln(x)e− 2 , Ln(x) e (x e− , n 0, 1, 2, ... . (17.4) = n = dx = ! ! " For the system Ln : n 0, 1, 2, ... of Laguerre polynomials Theorem 17.4 applies. { = } 17.3 Examples of Complete Orthonormal Systems for L2(I, ρdx) 251

In the Exercises we show that the Laguerre polynomials of different order are related according to the identity

(n 1)Ln 1(x) (x 2n 1)Ln(x) nLn 1(x) 0, (17.5) + + + − − + − = and are solutions of the second order differential equation (y Ln(x)) =

xy′′ (1 x)y′ ny 0. (17.6) + − + = In quantum mechanics this differential equation is related to the radial Schrödinger equation for the hydrogen atom.

17.3.3 I [ 1, 1], ρ(x) 1: Legendre Polynomials = − + = For any finite interval I [a, b],

n 1 d 2 n Pn(x) (x 1) , x [ 1, 1], n 0, 1, 2, ... (17.7) = 2nn dx − ∈ − = ! ! " is an orthogonal basis of the Hilbert space L2 ([ 1, 1], dx). The Legendre polynomials are normalized according to the relation − 2 Pn, Pm 2 δnm. ⟨ ⟩ = 2n 1 + Again one can show that these polynomials satisfy a recursion relation and a second order differential equation (see Exercises):

(n 1)Pn 1(x) (2n 1)xPn(x) nPn 1(x) 0, (17.8) + + − + + − = 2 (1 x )y′′ 2xy′ n(n 1)y 0, (17.9) − − + + = where y Pn(x). = 252 17 Separable Hilbert Spaces

Fig. 17.1 Legendre polynomials P3, P4, P5

Without further details we mention the weight functions for some other systems of orthogonal polynomials on the interval [ 1, 1]: − ν,µ µ Jacobi Pn ρ(x) (1 x) , ν, µ> 1, = − − λ 2 λ 1 Gegenbauer C ρ(x) (1 x ) 2 , λ > 1/2, n = − − − Tschebyschew 1st kind ρ(x) (1 x) 1/2, = − − Tschebyschew 2nd kind ρ(x) (1 x2)1/2. = − We conclude this section by an illustration of the Knotensatz for some Legendre polynomials of low order. This graph clearly shows that the zeros of the polynomial Pk are separated by the zeros of the polynomial Pk 1, k 3, 4. In addition the orthonormal polynomials are listed explicitly up to order+ n = 6. =