Spectral Theory of Orthogonal Polynomials

Transcription

1 Spectral Theory of Orthogonal Polynomials Barry Simon IBM Professor of Mathematics and Theoretical Physics California Institute of Technology Pasadena, CA, U.S.A. Lecture 1: Introduction and Overview

2 Spectral Theory of Orthogonal Polynomials Lecture 1: Introduction and Overview Lecture 2: Szegö Theorem for OPUC Lecture 3: Three Kinds of Polynomials Asymptotics, I Lecture 4: Three Kinds of Polynomial Asymptotics, II

3 References [OPUC] B. Simon, Orthogonal Polynomials on the Unit Circle, Part 1: Classical Theory, AMS Colloquium Series 54.1, American Mathematical Society, Providence, RI, [OPUC2] B. Simon, Orthogonal Polynomials on the Unit Circle, Part 2: Spectral Theory, AMS Colloquium Series, 54.2, American Mathematical Society, Providence, RI, [SzThm] B. Simon, Szeg 's Theorem and Its Descendants: Spectral Theory for L 2 Perturbations of Orthogonal Polynomials, M. B. Porter Lectures, Princeton University Press, Princeton, NJ, 2011.

4 Spectral theory is the general theory of the relation of the fundamental parameters of an object and its spectral characteristics. Spectral characteristics means eigenvalues or scattering data or, more generally, spectral measures

5 theory Examples include Can you hear the shape of a drum? Computer tomography Isospectral manifold for the harmonic oscillator

6 theory The direct problem goes from the object to spectra. The inverse problem goes backwards. The direct problem is typically easy while the inverse problem is typically hard. For example, the domain of denition of the harmonic oscillator isospectral manifold is unknown. It is not even known if it is connected!

7 Orthogonal polynomials on the real line (OPRL) and on the unit circle (OPUC) are particularly useful because the inverse problems are easyindeed the inverse problem is the OP denition as we'll see. also enter in many applicationboth specic polynomials and the general theory.

8 Indeed, my own interest came from studying discrete Schrödinger operators on l 2 (Z) ( Hu ) n = u n+1 + u n 1 + V u n and the realization that when restricted to Z +, one had a special case of OPRL.

9 µ will be a probability measure on R. We'll always suppose that µ has bounded support [a, b] which is not a nite set of points. (We then say that µ is non-trivial.) This implies that 1, x, x 2,... are independent since P (x) 2 dµ = 0 µ is supported on the zeroes of P. Apply Gram Schmidt to 1, x,... and get monic polynomials P j (x) = x j + α j,1 x j and orthonormal (ON) polynomials p j = P j / P j

10 More generally we can do the same for any probability measure of bounded support on C. One dierence from the case of R, the linear combination of {x j } j=0 are dense in L2 (R, dµ) by Weierstrass. This may or may not be true if supp(dµ) R. If dµ = dθ/2π on D, the span of {z j } j=0 is not dense in L 2 (but is only H 2 ). Perhaps, surprisingly, we'll see later that there are measures dµ on D for which they are dense (e.g., µ purely singular). More signicantly, the argument we'll give for our recursion relation fails if supp(dµ) R.

11 Since P k is monic and {P j } k+1 j=0 span polynomials of degree at most k + 1, we have k xp k = P k+1 + B k,j P j j=0 Clearly B k,j = P j, xp k / P j 2 Now we use P j, xp k = xp j, P k (need dµ on R!!) If j < k 1, this is zero. If j = k 1, P k 1, xp k = xp k 1, P k = P k 2.

12 Thus (P 1 0); {a j } j=1, {b j} j=1 : Jacobi recursion xp N = P N+1 + b N+1 P N + a 2 NP N 1 b N R, a N = P N / P N 1 These are called Jacobi parameters. This implies P N = a N a N 1... a 1 (since P 0 = 1). This, in turn, implies p n = P n /a 1... a n obeys xp n = a n+1 p n+1 + b n+1 p n + a n p n 1

13 We have thus solved the inverse problem, i.e., µ is the spectral data and {a n, b n } n=1 are the descriptors of the object. In the orthonormal basis {p n } n=0, multiplication by x has the matrix b 1 a a 1 b 2 a J = 0 a 2 b 3 a called a Jacobi matrix.

14 Since b n = xp 2 n 1(x) dµ, a n = supp(µ) [ R, R] b n R, a n R. xp n 1 (x)p n (x) dµ ( Conversely, if sup n an + b n ) = α <, J is a bounded matrix of norm at most 3α. In that case, the spectral theorem implies there is a measure dµ so that (1, 0,...) t, J l (1, 0,...) t = x l dµ(x) If one uses Gram-Schmidt to orthonormalize {J l (1, 0,...) t } l=0, one nds µ has Jacobi matrix exactly given by J.

15 We have thus proven (his paper was in 1935; really due to Stieltjes in 1894 or to Stone in 1932)..There is a oneone correspondence between bounded Jacobi parameters {a n, b n } n=1 [ (0, ) R ] and non-trivial probability measures, µ, of bounded support via: µ {a n, b n } (OP recursion) {a n, b n } µ (Spectral Theorem) There are also results for µ's with unbounded support so long as x n dµ <. In this case, {a n, b n } µ may not be unique because J may not be essentially self-adjoint on vectors of nite support.

16 Let dµ be a non-trivial probability measure on D. As in the OPRL case, we use Gram-Schmidt to dene monic, Φ n (z) and ON OP's ϕ n (z). In the OPRL case, if z is multiplication by the underlying variable, z = z. This is replaced by z z = 1. In the OPRL case, P n+1 xp n {1, x 1,..., x n 2 }.

17 In the OPUC case, Φ n+1 zφ n {z,..., z n }, since if j 1. zφ, z j = Φ, z j 1 In the OPRL case, we used deg P = n and P {1, x,..., x n 2 } P = c 1 P n + c 2 P n 1. In the OPUC case, we want to characterize deg P = n, P {z, z 2,..., z n }.

18 Dene on degree n polynomials to themselves by ( ) 1 Q (z) = z n Q z (bad but standard notation!) or n n Q(z) = c j z j Q (z) = c n j z j j=0 j=0 Then, is unitary and so for deg Q = n Q {1,..., z n 1 } Q = c Φ n is equivalent to Q {z,..., z n } Q = c Φ n

19 Thus, we see, there are parameters {α n } n=0 (called Verblunsky ) so that Φ n+1 (z) = zφ n α n Φ n(z) This is the Szeg Recursion (History: Szeg and Geronimus in 1939; Verblunsky in ) Applying for deg n + 1 polynomials to this yields Φ n+1(z) = Φ n(z) α n zφ n The strange looking ᾱ n rather than say +α n is to have the α n be the Schur parameter of the Schur function of µ (Geronimus); also the Verblunsky parameterization then agrees with α n. These are discussed in [OPUC1].

20 Φ n monic constant term in Φ n is 1 Φ n(0) = 1. This plus Φ n+1 = zφ n ᾱ n Φ n(z) implies Φ n+1 (0) = α n i.e., Φ n determines α n 1.

21 For OPRL, we saw P n+1 / P n = a n+1. We are looking for the analog for OPUC. Szeg Recursion Φ n+1 + ᾱ n Φ n = zφ n Φ n+1 Φ n Φ n α n 2 Φ n 2 = zφ n 2 Multiplication by z unitary plus antiunitary Φ n+1 2 = ρ 2 n Φ n 2 ; ρ 2 n = 1 α n 2 which implies α n < 1 (i.e., α n D) and Φ n = ρ n 1 ρ 0

22 ( ϕn+1 ϕ n+1 ) = A n (z) ( ϕn ϕ n ) x; A n = ρ 1 n ( z ) ᾱn α n z 1 det A n 0 if z 0, so we can get ϕ n (Φ n ) from ϕ n+1 (Φ n+1 ) by zφ n = ρ 2 [ n Φn+1 + ᾱ n Φ n+1] Φ n = ρ 2 n [ Φn+1 + α n Φ n+1 ]

23 We see that Φ n+1 determines α n, so by induction and inverse recursion, Theorem. If two measures have the same Φ n, they have the same {Φ j } n 1 j=0 and {α j} n 1 j=0.

24 A similar argument to the one that led to α n < 1 yields Theorem. All zeros of Φ n lie in D. Proof. Φ n (z 0 ) = 0 Φ n = (z z 0 )p, deg p = n 1 zp = Φ n + z 0 p and p Φ n p 2 = Φ n 2 + z 0 2 p 2 z 0 < 1 Corollary. All zeros of Φ n(z) lie in C \ D.

25 Here is a second proof that only uses. By induction, suppose that all zeros of Φ n are in D. Then, for β < 1 zφ n + βφ n 0 on D since zφ n (z) = Φ n(z) on D. ( 1 z = z) If Φ (β) n+1 = zφ n + βφ n, then at β = 0, all zeros of Φ (β) in D. n+1 are As β varies in D, all zeros of Φ (β) n+1 are trapped in D. QED.

26 We are heading towards a proof that any {α n } n=0 D are the Verblunsky of a measure on D (analog of ). It will depend on Theorem ( measures). Let {α (0) j } n 1 j=0 Dn. Let ϕ n (z) be the normalized degree n polynomial obtained by. Let dµ n (θ) = dθ 2π ϕ n (e iθ ) 2 Then dµ n has Verblunsky α j (dµ n ) = { α (0) j j = 0,..., n 1 0 j n

27 The rst step of the proof is to show that k, l, n with k < n + l z k z l ϕ n (z)dµ n (θ) = 0 z=e iθ For z D ϕ n (z) = ϕ n ( 1 z ) = z n ϕ n(z). Thus the integral above is z k z l ϕ n (z) z n ϕ n (z) ϕ n(z) dz 2πiz = 1 2π z l+n k 1 dz ϕ n(z) is zero since [ ϕ n(z) ] 1 is analytic on a neighborhood of D and l + n k 1 0.

28 Thus, z l ϕ n is a multiple of the OP's for dµ n. Since z l ϕ n 2 dµ = 1, we see that ϕ n+k (z; dµ) = z k ϕ n (z); k > 0. As we saw, Φ n determines {α j } n 1 j=0 and Φ j by inverse and Φ j+1 (0) = α j. This shows that { ϕ j (x) j = 0,..., n ϕ j (z; dµ) = z j n ϕ n (z) j = n, n + 1,... implying the claimed result.

29 Given {α j } j=0 D, we can form dµ n as above. Via Φj (e iθ )dµ(e iθ ) = 0, {Φ j } n j=0 determines { z j dµ} n j=0 inductively (actually they determine more moments). Thus z j dµ n = z j dµ m j min(n, m) and z j dµ n = ( z j dµ n ). Thus, dµ n has a weak limit dµ. Clearly, α j (dµ ) = α j. We have thus proven Verblunsky's Theorem. µ {α j (µ)} j=0 sets up a 11 correspondence between non-trivial probability measures on D and D.

30 Simon [CRM Proc. and Lecture Notes 42 (2007), ] has proven an analog of the approximation for OPRL (the analog for Schrödinger operators is due to Carmona; hence the name): Let dρ be a probability measure on R with x n dρ < for all n. Let {p n } n=0 be its orthonormal polynomials and {a n, b n } n=1 its Jacobi parameters. Let dν n (x) = dx/ [ π(a 2 np 2 n(x) + p 2 n 1(x)) ] Then, for l = 0,..., 2n 2, x l dν n = x l dρ. If the moment problem for dρ is determinate, then dν n dρ weakly.

31 One important consequence of this result is Theorem. If I R is an interval and for all x I and some c > 0, we have that c a 2 np 2 n(x) + p 2 n 1(x) c 1 then dρ I has a.c. part and no singular spectrum. Similarly, for I D and µ a probability measure c ϕ n (z) c 1 all z I implies dµ I has a.c. part and no singular spectrum. Remark. A much stronger result is known (see e.g., Simon [Proc AMS 124 (1996), 3361]); I can be any set and c can be x-dependent.