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. Lectures 11 & 12: Selected Additional, I, II

2 Spectral Theory of Orthogonal Polynomials Lecture 9: Fuchsian Groups and Finite Gaps, I Lecture 10: Fuchsian Groups and Finite Gaps, II Lecture 11: Selected Additional, I Lecture 11: Selected Additional, 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 and Its Descendants: Spectral Theory for L 2 Perturbations of Orthogonal Polynomials, M. B. Porter Lectures, Princeton University Press, Princeton, NJ, 2011.

4 and Essential Spectrum, [SzThm], Ÿ7.3 Reectionless and, [SzThm], Ÿ7.47.6, [SzThm], Chap. 8, [OPUC1], Chap. 4 Clock Behavior and Universality, [SzThm], Ÿ2.16, 2.17, 3.11, 3.12, 5.11 Toda Flows, [SzThm], Chap. 6 Regular Measures, [SzThm], Ÿ5.9 Ratio Asymptotics and Weak Limits, [OPUC2], Ÿ9.59.8

5 Right limits were introduced and studied by LastSimon in two papers [Inv. Math. 135 (1999), ; J. Anal. Math. 98 (2006), ]. Given a set of bounded (one-sided) Jacobi parameters {a n, b n } n=1, we say a two-sided sequence {ã n, b n } n= is a right limit if there exists m j so that for all l, as j, a mj +l ã l ; b mj +l b l The set of all right limits is denoted R(J). By compactness, it is non-empty and a closed subset of [ R, R] in the product topology (R = sup n a n + b n ).

6 We say λ σ,pp ( J) if and only if u L, u 0, with Ju = λu. By a Weyl sequence argument, σ,pp ( J) σ( J). LastSimon proved (σ,pp (J) is from [SzThm] book). σ ess (J) = Jr R(J)σ(J r ) = Jr R(J)σ,pp (J r ) That σ(j r ) σ ess (J) is easy. If u n has nite support with (J r λ)u < ε, then with u (j) n = u mj +n, we have lim j (J λ)u (j) ε and u (j) 0 weakly. The other parts are not hard but somewhat involved. LastSimon also proved Σ ac (J) Jr Σ ac (J r ).

7 Reectionless A reectionless Jacobi matrix on e R is a whole line matrix obeying n, G nn (x + i0) ir for a.e. x e. It is sucient that this holds for three successive n's. By the reection principle (dierent reection!), we have that if (a, b) e, then G nn is continuous on (a, b), so J has purely a.c. spectrum on (a, b).

8 Reectionless Given J, a whole line Jacobi matrix, J + and J are the Jacobi matrices with parameters {a n, b n } n=1 and {a n, b n+1 } n=1 and J with parameters {a n 1, b n } n=1. J +/ are the two pieces we get by setting a 0 = 0 and J + and J by setting a 0 = a 1 = 0 = b 0. m ±, m are their m-functions. The u ± formulae for G and m yield G 00 (z) = ( a 2 0m + (z) m (z) 1) 1 G 00 (z) = ( z b 0 + a 2 1 m + a 2 0m +) 1

9 Reectionless A basic fact is. J is reectionless if and only if for a.e. x e m + (x + i0) = ( a 2 0m (x + i0) ) 1 a 2 0 m+ (x + i0) = b 0 x a 2 1 m(x + i0) These conditions imply G 00 (x + i0) is pure imaginary since Re ( a 2 0 m+ (x + i0) (m (x + i0)) 1) = 0.

10 Reectionless To get full result, one shows that Weyl solutions for J + and J, call them u + n and u n, have a.c. boundary values (since m does). Re G nn (x + i0) for n = 0, ±1 is equivalent to (all at x + i0) Im(u + n u n ) = 0 for n = ±1 and W (= Wronskian) has Re W = 0 and that this is equivalent to u n = u + n.

11 Reectionless Torus Suppose now e = l+1 j=1 [α j, β j ] is a nite gap set. Let J be reectionless on e with σ(j) = e. Let m + (z) be the m-function for the half-line operator J +. Since m + is real in the gaps and on (, α 1 ) and (β l+1, ), we have that m + (x + i0) = m + (x i0) = [a 2 0m (x i0)] 1 This implies that m + dened on S + can be analytically continued to S by dening it to be (a 2 0 m (z)) 1 on S. Since m has a zero at, (m ) 1 has a pole there.

12 Reectionless The only other possible poles are in the gaps. (a 0 m (z)) 1 = b 0 z a 2 1 m(z), so (a2 0 m ) 1 has a pole m does. G 00 = (z b 0 + a 2 1 m + a2 0 m+ ) 1 shows that if m or m + have poles at x 0 (β j, α j+1 ), then G 00 has a zero there. Since G 00 is monotone (and bounded) on (β j, α j+1 ), it has at most one zero there. If it has a zero at x, either m + or m or both have a pole there.

13 Reectionless If m + and m both have poles there, one can show the Weyl solutions agree and vanish at n = 0 which implies the whole line problem has an eigenvalue at x, contrary to our hypothesis that σ(j) = e. Thus, at a zero of G 00 in a gap m + has a pole at precisely one of the points z ± with π(z ± ) = x 0. A further analysis shows that if G 00 has no zeros in a given gap, it vanishes at one end or the other where one has a simple pole in the local coordinates.

14 Reectionless We thus see if J is reectionless on e and σ(j) = e, m + is a minimal Herglotz function. Conversely, if m + is a minimal Herglotz function, (a 2 0 m+ (τ(z))) 1 denes a J (where a 2 0 is picked so (a 2 0 m+ (τ(z))) 1 z 1 near + ) and J +, a 0, J t together into a reectionless J. We have thus proven:. There is a 1-1 correspondence between the isospectral torus on e (dened as minimal Herglotz functions) and reectionless J's.

15 Remling proved the following remarkable theorem: Remlings. If J is a half-line Jacobi matrix and e Σ ac (J), then any right limit, J r, is reectionless on e. There is an attractive intuition: J r repeats innitely often far from each other. If there were any reection, a wave would get trapped by the innitely many reections but by RiemannLebesgue, a.c. wave packets get out to.

16 Alas, no one has a proof using this intuition. The only proof we have is original proof which takes 20 dense pages in [SzThm] and which I can follow but don't understand!

17 DenisovRakhmanovRemling (Rakhmanov [Math. Sb. 32 (1977), ; 46 (1983), ]). Let dµ = f dθ 2π + dµ s on D with f(θ) > 0 for Lebesgue a.e. θ. Then α n (dµ) 0. (Denisov [Proc. A.M.S. 130 (2004), ). Let dµ = fdx + dµ s ; ess supp(dµ) = [ 2, 2]; f(x) > 0 for a.e. x [ 2, 2]. Then a n (dµ) 1, b n (dµ) 0. (DamanikKillipSimon, Ann. Math. 171 (2010), ). Let e be a nite gap set with each band having rational harmonic measure. Let dµ = fdx + dµ s ; ess(dµ) = e; f(x) > 0 for a.e. x e. Then, as l, {a n l, b n l } n=1 (periodic) isospectral torus for e. approaches the two-sided

18 DenisovRakhmanovRemling (Remling [Ann. Math 174 (2011), ]). The last theorem holds for any nite gap set, e. Remarks. 1. This was conjectured by DamanikKillipSimon. 2. This implies the Denisov result and also Rakhmanov once one has an extension of Remlin to OPUC which was accomplished by Breuer, Ryckman, Zinchenko [Comm. Math. Phys. 292 (2009), 128]. The is an immediate consequence of and the analysis of J's reectionless on e with σ(j) = e.

19 (DamanikKillipSimon [Ann. Math. 171 (2010), ]). Let be the discriminant associated to a nite gap set e R with rational harmonic measures ( is degree p, measures are q j /p with no common factor for p {q j } j=1 l+1 ). Let J be a whole line Jacobi matrix and S : l 2 l 2 by (Su) n = u n 1. Then (J) = S p + S p J isospectral torus for e DKS called this the magic formula.

20 Recall for J isospectral torus, (E(θ)) = 2 cos θ where E(θ) are eigenvalues of Floquet solutions with u n+p = e iθ u n. In a suitable spectral representation S p is multiplication by e iθ and J by E(θ) so that the above shows that (J) = S p + S p.

21 For the converse, if e (J) = S p + S p, then [J, e (J)] so [J, S p + S p ]. Using that J has nite width, this implies (an argument of Naiman [Soviet Math. Dokl. 3 (1962), ]) [J, S p ] = 0, i.e., J has periodic p. Thus, J (J) e (J) = 0. Again, using nite width of J, p(j) = 0 for a polynomial p = 0. Thus J = e J isospectral torus.

22 To use this idea, note that in general, if J is a Jacobi matrix, (J) is not tri-diagonal but 2p + 1 diagonal (since (J l ) kj 0 k j l) which can be thought of as p p block tridiagonal. The DKS strategy is to extend results on approach to [ 2, 2] to the matrix-valued case.

23 For example, to get the extension of Denisov, if J has Σ ac (J) = e, then (J) has multiplicity p a.c. spectrum on [ 2, 2]. By a matrix extension of Denisov, this implies all right limits of (J) are S p + S p. By the Magic, this proves any right limit J r obeys (J r ) = S p + S p, so J r isospectral torus.

24 DKS are thus able to prove DRR for periodic (also follows from Remling) Shohat-Nevai for perturbations of isospectral torus (also follows from CSZ) LiebThirring and so Nevai conjecture for perturbations of periodic (also follows from FrankSimon) KillipSimon for perturbations of periodic with all gaps open (only known proof!)

25 In 2003, Cantero, Moral, and Velázquez [Linear Alg. Appl. 362 (2003), 2951] found the right matrix representation for OPUC 82 years after Szeg invented OPUC! Its usefullness for spectral theory of OPUC was found by GolinskiiSimon and presented in Section 4.3 of [OPUC]. Just as Jacobi matrices come from general self-adjoint operators with cyclic vector, CMV matrices come from unitary matrices with cyclic vector and in this form (without the OPUC connection) they were found in the numerical linear algebra community in 1991 by Bunse-GerstnerElsner [Linear Alg. Appl. 154 (1991), ].

26 For OPUC, orthonormalized z n may not give a basis for L 2 ( D, dµ). CMV had the idea of orthonormalizing, 1, z, z 1, z 2,... which always gives a basis! Remarkably, the resulting basis, {χ j } j 0 can be expressed in terms of the ϕ n 's and ϕ n's. If σ n = χ 2n, τ n = χ 2n 1 ; n = 0, 1, 2,... (n 1 for =) then σ n = z n ϕ 2n, τ n = z n+1 ϕ 2n 1

27 If instead, one orthonormalizes, 1, z, z 1, z 2,..., one gets an ON basis, {x j } j=0 and if s n = x 2n, n = 0, 1, 2,..., t n = x 2n 1, n = 1, 2,..., then s n = z n ϕ 2n, t n = z n ϕ 2n 1 One denes the CMV and alternate CMV matrices by C ij (dµ) = χ i, zχ j ; Cij (dµ) = x i, zx j

28 One denes L ij (dµ) = x i, χ j, m ij (dµ) = χ i, zx j so one has the LM factorization If Θ j = ( ᾱj ρ j ρ i α j ) C = LM, then L = Θ 0 Θ 2 Θ 4... and M = Θ 1 Θ 3... C = ML

29 Thus, C is 5-diagonal with structures of 4 2 blocks. Matrix elements are quadratic in α's and ρ's, explicitly ᾱ 0 ᾱ 1 ρ 0 ρ 1 ρ ρ 0 ᾱ 1 α 0 ρ 1 α C = 0 ᾱ 2 ρ 1 ᾱ 2 α 1 ᾱ 3 ρ 2 ρ 3 ρ ρ 2 ρ 1 ρ 2 α 1 ᾱ 3 α 2 ρ 3 α ᾱ 4 ρ 3 ᾱ 4 α

30 Consequences of CMV New proof of Verblunsky's Trace class theory: α j α j < C(α) C( α) l 1 Σ ac (α) = Σ ac ( α) j=0 Weyl : α j β j 0 σ ess (α) = σ ess (β) New proof of Geronimus Relatins [KillipNenciu, IMRN 50 (2004), ] Haar on CUE induced measures on Verblunsky coecients [KillipNenciu, IMRN 50 (2004), ]

31 We've discussed the DOS but that only tells part of the story of the distribution of zeros for OPs. In the next two slides, I'll show you the zeros for two dierent sets of Verblunsky coecients. One has the α's iidrvunfortunately uniform in a real interval (hence the complex conjugate symmetry) rather than uniform on some circle. I'm not sure of the DOS. The other is α n = ( 3 4 )n+1 where the DOS is known (a result of Mhaskar-Sa) to be uniform on the circle of radius 3/4.

32 Here is the random casewhich looks irregular: In the rotation symmetry case, it is known to be Poisson [Stoiciu, JAT 139 (2006) 2964; DaviesSimon, JAT 141 (2006), ]

33 And here is the exponential decay casewhich looks regular: In this case, the zeros are asymptotically equally spacedwhich I called clock spacing.

34 I want to focus on the clock-spacing case, but for OPRL, not OPUC. When a n 1, b n 0, the DOS will be the equilibrium measure, ρ e (x)dx for e = [ 2, 2], i.e., ρ e (x) = π 1 (4 x 2 ) 1/2 so clock only means locally equally spaced. If x (j) n (E 0 ) is the zeros of p n (x) with x ( 2) n < x ( 1) n < x (0) n < E 0 < x (1) n... then clock spacing is n [ x (j+1) n x (j) ] n 1/ρ(E0 )

35 The earliest general clock-spacing results are due to Erd sturan [Ann. Math. 44 (1940), ] and very general results are in LastSimon [Comm. Pure Appl. Math. 61 (2008), ]. But I want to focus on a wonderful approach of Lubinskyactually two approaches, both based on the CD kernel [Lubinsky, Ann. Math. 170 (2009), ; J. Anal. Math. 106 (2008), ]

36 With his rst method, Simon [J. Math. Anal. 105 (2008), ] and Totik [Ark. Mat. 47 (2009), ] obtained clock space for fairly general a.c. dµ on sets e with e int = e and AvilaLastSimon [Anal. PDE 3 (2010), 81108], using Lubinsky's second method, even have some results for a.c. spectrum on positive measure Cantor sets. Here, we'll focus on the underlying method.

37 and Universality A critical tool is the CD (for Christoel Darboux) kernel K n (x, y) = n p n (x)p n (y) j=0 which is the integral kernel in L 2 (R, dµ) of the projection into polynomials of degree n or smaller. (CD ). For x y K n (x, y) = a ( n+1 pn+1 (x)p n (y) p n+1 (y)p n (x) ) x y

38 and Universality Proof. If L n (x, y) = a n+1 ( pn+1 (x)p n (y) p n+1 (y)p n (x) ) recursion relation for xp n (x) times p n (y) minus recursion relation for yp n (y) times p n (x) says (x y)p n (x)p n (y) = L n (x, y) L n 1 (x, y) This plus induction says that CP formula holds.

39 and Universality Universality on [ 2, 2] = e says, for any compact interval I [ 2, 2], 1 n + 1 K n(x n, x n ) ρ e(x ) w(x ) if say n x x A uniformly in x I and x n 's with A xed. K n (x + a n, x + b n ) K n (x, x ) uniformly in x I, a A, b = A. sin( πρ e (x )(b a) ) πρ e (x )(b a)

40 and Universality (FreudLevin ). Universality clock spacing. Proof. By CD formula, if p n (x) = 0, y x, then p n (y) = 0 K n (x, y) = 0 (since p n (x) = 0 p n+1 (x) 0). Universality controls zeros of K n for n large. Explicity K n (x, y) 0 if x y with 0 α < 1 and n large. α ρ e(x) Zeros of sin says at least one zero near zero + above says no more than 1. k ρ and e(x )

41 and Universality Lubinsky provided two methods of proving universality. See his papers or Sections 3.11 and 3.12 of [SzThm].