• Non ci sono risultati.

= 0 for all n ∈ Z. From the above convergence result we deduce, for all f ∈ C

N/A
N/A
Protected

Academic year: 2021

Condividi "= 0 for all n ∈ Z. From the above convergence result we deduce, for all f ∈ C"

Copied!
8
0
0

Testo completo

(1)

⟨e

n

, g ⟩

2

= 0 for all n ∈ Z. From the above convergence result we deduce, for all f ∈ C

1

([0, 2π]),

⟨f , g⟩

2

= lim

N→∞

!

N n=−N

c

n

(f )e

n

, g ⟩

2

= 0.

Since C

1

([0, 2π]) is known to be dense in L

2

([0, 2π], dx) it follows that g = 0, by Corollary

17.2, hence by Theorem 17.2, this system is an orthonormal basis

of L

2

([0, 2π], dx). Therefore, every f ∈ L

2

([0, 2π], dx) has a Fourier expansion, which converges (in the sense of the L

2

-topology). Thus, convergence of the Fourier series in the L

2

-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 L

2

(I , dx) separable, but for any interval I = [a, b], −∞ ≤ a < b ≤ +∞, as the results of this section will show. Furthermore an orthonormal basis will be constructed explicitly and some interesting properties of the elements of such a basis will be investigated.

The starting point is a weight function 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) e

α|x|

≤ C for all x ∈ I.

The strategy to prove that the Hilbert space L

2

(I , dx) is separable is quite simple. A first step shows that the countable set of functions ρ

n

(x) = x

n

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 .

Proof

For the proof we have to show: If an element h ∈ L

2

(I , dx) satisfies ⟨ρ

n

, 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 function h ∈ L

2

( R, dx). On the strip S

α

= {p = u + iv ∈ C : u, v ∈ R, |v| < α}, introduce the auxiliary function

F (p) =

"

R

ρ(x)h(x) e

ipx

dx.

The growth restriction on the weight function implies that F is a well-defined holo-

morphic function on S

α

(see Exercises). Differentiation of F generates the functions

(2)

246 17 Separable Hilbert Spaces

ρn in this integral:

F(n)(p)= dnF

dpn(p)= in

!

Rh(x)ρ(x)xneipxdx

for n = 0, 1, 2, . . . , and we deduce F(n)(0) = in⟨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) = √

2πL(ρh)(p) where L is the inverse Fourier transform (see Theorem 10.1), and we know

⟨Lf , Lg⟩2 = ⟨f , g⟩2 for all f , g ∈ L2(R, dx) (Theorem 10.7). It follows that

⟨ρh, ρh⟩2 = ⟨L(ρh), L(ρh)⟩2 = 0 and thus ρh = 0 ∈ L2(R, dx). Since ρ(x) > 0

for x ∈ I this implies h = 0 and we conclude. ✷

Technically it is simpler to do the orthonormalization of the system of functions {ρn : n∈ N} not in the Hilbert space L2(I , dx) directly but in the Hilbert space L2(I , ρdx), which is defined as the space of all equivalence classes of measurable functions f : I→K such that"

I |f (x)|2ρ(x) dx < ∞ equipped with the inner prod- uct ⟨f , g⟩ρ = "

I f(x)g(x)ρ(x) dx. Note that the relation⟨f , g⟩ρ = ⟨√ρf , √ρg⟩2 holds for all f , g ∈ L2(I , ρdx). It implies that the Hilbert spaces L2(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, Lemma17.1can 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, Pmρ = δkm. These polynomials are defined recursively in the following way: Q0(x) = x0 = 1, and when for k ≥ 1 the polynomials Q0, . . ., Qk−1 are defined, we define the polynomial Qk by

Qk(x)= xk

k−1

#

n=0

⟨Qn, xkρ

⟨Qn, QnρQn.

Finally, the polynomials Qk are normalized and we arrive at an orthonormal system of polynomials Pk:

Pk = 1

∥QkρQk, k= 0, 1, 2, . . . .

Note that according to this construction, Pk is a polynomial of degree k with positive coefficient for the power xk. Theorem 17.1and Lemma17.1imply 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.

(3)

Theorem 17.3 For any interval I = (a, b), −∞ ≤ a < b ≤ +∞ the Hilbert space L2(I , dx) is separable, and the above system{ek : k = 0, 1, 2, . . .} is an orthonormal basis.

Proof Only the existence of a weight function for the interval I has to be shown.

Then by the preceding discussion we conclude. A simple choice of a weight function for any of these intervals is for instance the exponential function ρ(x) = e−αx2,

x ∈ R, for some α > 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 Qmis a polynomial of degree m, then⟨Qm, Pkρ = 0 for all k > m.

Proof Since {Pk : k = 0, 1, 2, . . .} is an ONB of the Hilbert space L2(I , ρdx) the polynomial Qmhas a Fourier expansion with respect to this ONB: Qm =!

n=0cnPn, cn = ⟨Pn, Qmρ. Since the powers xk, 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 coefficients cn in this expansion must vanish for n > m, i.e., Qm =!m

n=0cnPn and

thus⟨Pk, Qmρ = 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.

Proof We know Pk(x) = akxk + Qk−1(x) with some constant ak > 0 and some polynomial Qk−1of degree smaller than or equal to k− 1. Thus, if we define An =

an+1

an , it follows that Pn+1− AnxPn is a polynomial of degree smaller than or equal to n, hence there are constants cn,k such that

Pn+1− AnxPn =

"n k=0

cn,kPk. Now calculate the inner product with Pj, j ≤ n:

⟨Pj, Pn+1− AnxPnρ =

"n k=0

cn,k⟨Pj, Pkρ = cn,j.

(4)

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 < n − 1, cn,n−1 = −An⟨xPn−1, Pnρ, and cn,n =

−An⟨xPn, Pnρ. The statement follows by choosing Bn = cn,nand Cn = cn,n−1. ✷ Proposition 17.2 For any weight function ρ on the interval I , the kth orthonormal polynomial Pk has exactly k simple real zeroes.

Proof Per construction the orthonormal polynomials Pkhave 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+ 2!M

j=m+1nj = k. Therefore the polynomial Pk can be written as Pk(x) = ck

"m j=1

(x− xj)

"M j=m+1

(x− zj)nj(x− zj)nj.

Consider the polynomial Qm(x) = ck#m

j=1(x − xj). It has the degree m and ex- actly m real simple roots. Since Pk(x) = Qm(x)#M

j=m+1|x − zj|2nj, it follows that Pk(x)Qm(x) ≥ 0 for all x ∈ I and PkQm ̸= 0, hence ⟨Pk, Qmρ > 0. If the degree mof the polynomial Qmwould 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 Pkand Pk−1have 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 we can take the real number λ0 = P−P k(x0)

k−1(x0) to form the polynomial Qk(x, λ0) = 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

(5)

interval [α, β]. Since the polynomials Pk−1and−Pk−1have the same system of roots, we can assume that Pk−1is positive in [α, β] and Pk is negative in (α, β). Define the function f (x) = P−Pk−1k(x)(x). It is continuous on [α, β] and satisfies f (α) = f (β) = 0 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, we know 0 = f(x0). The derivative of f is easily calculated:

f(x) = −Pk(x)Pk−1(x)− Pk(x)Pk−1(x) Pk−1(x)2 .

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 Lemma17.3and therefore the polynomial Pk−1 has at least one root in the interval (α, β). Since Pk−1has 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

F(Q)=

!

I

Q(x)2ρ(x) dx, Q(x) =

"n k=0

akxk.

Since we can expand Q in terms of the orthonormal basis {Pk : k = 0, 1, 2, . . . }, Q = #n

k=0ckPk, ck = ⟨Pk, Q⟩ρ the value of the function F can be expressed in terms of the coefficients ck as F (Q)=#n

k=0ck2and 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 L

2

(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.”

17.3.1 I = R, ρ(x) = e

−x2

: Hermite Polynomials

Evidently, the function ρ(x)= e−x2 is a weight function for the real line. Therefore, by Lemma17.1, the system of functions ρn(x)= xnex22 generates the Hilbert space

(6)

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)

hn(x) = (−1)ncnex22 ! d dx

"n# e−x2$

= cnHn(x) ex22 (17.1) with normalization constants

cn = (2nn!√

π)−1/2 n= 0, 1, 2, . . . .

Here the functions Hnare polynomials of degree n, called Hermite polynomials and the functions hnare 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 quantum harmonic oscillator with the Hamiltonian H = 12(P2+ Q2) for the eigen- value n+ 12, H hn = (n + 12)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 S(R).

17.3.2 I = R

+

, ρ(x) = e

−x

: Laguerre Polynomials

On the positive real line the exponential function ρ(x) = e−x 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 is constructed by orthonormalization of the system {xnex2} : n = 0, 1, 2, . . . in L2(R+, dx) is an orthonormal basis. These Laguerre functions have the following form (Rodrigues’ formula):

n(x) = 1

n!Ln(x) ex2, Ln(x) = ex

! d dx

"n

(xne−x, n= 0, 1, 2, . . . . (17.4) For the system {Ln : n= 0, 1, 2, . . . } of Laguerre polynomials Theorem 17.4 applies.

(7)

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], −∞ < a < b < ∞ one can take any posi- tive constant as a weight function. Thus, Lemma17.1says that the system of powers {xn : n= 0, 1, 2. . . . } is a total system of functions in the Hilbert space L2([a, b], dx).

It follows that every element f ∈ L2([a, b], dx) is the limit of a sequence of polyno- mials, in the L2-norm. Compare this with the Theorem of Stone–Weierstrass which says that every continuous function on [a, b] is the uniform limit of a sequence of polynomials.

For the special case of the interval I = [−1, 1] the Gram–Schmidt orthonormal- ization of the system of powers leads to a well-known system of polynomials.

Theorem 17.7 The system of Legendre polynomials Pn(x) = 1

2nn!

! d dx

"n

(x2− 1)n, x ∈ [−1, 1], n = 0, 1, 2, . . . (17.7) is an orthogonal basis of the Hilbert space L2([−1, 1], dx). The Legendre polynomials are normalized according to the relation

⟨Pn, Pm2 = 2

2n+ 1δnm.

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) (1− x2)y′′− 2xy+ n(n + 1)y = 0, (17.9) where y = Pn(x).

(8)

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]:

Jacobi Pnν,µ ρ(x)= (1 − x)µ, ν, µ >−1, Gegenbauer Cnλ ρ(x)= (1 − x2)λ12, λ >−1/2, 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.

Riferimenti

Documenti correlati

A data set can be explained by alternative sets of patterns, and many computational problems arise related to the choice of a particular set of patterns for a given instance.. In

Due to more complex SE spectra, the accuracy of the model describing the optical properties of silver nanoparticle arrays did not allow us to identify the surface premelting;

The temperatures shown here are: local equilibrium tem- perature T , thermodynamic non-equilibrium temperature T neq (equal to the kinetic temperature along the x axis), the

Throughout your study of Chemistry, you have explored a variety of chemical principles. Identify what you believe to be the three most important chemical principles. You should

Abstract In this paper we analyze the effects of restricted participation in a two-period general equilibrium model with incomplete financial markets and two key elements:

149 In order to simultaneously encompass both bandwagon and snob effects, in the 150 updating preference functions we will introduce a threshold effect (see Granovetter 151 1978),

Though the selection properties considered here implies for topological spaces the Lindel¨of property, we see now that they do not imply second countability: For example, consider

A: The acknowledgement of the WFD in the national legislation is expected within December 2003 followed from the identification of the Districts of Basin and of the authorities