Spectral Theory of Orthogonal Polynomials

Transcription

1 Spectral Theory of Orthogonal Barry Simon IBM Professor of Mathematics and Theoretical Physics California Institute of Technology Pasadena, CA, U.S.A. Lecture 2: Szegö Theorem for OPUC for S

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

3 References for S [OPUC] B. Simon, Orthogonal on the Unit Circle, Part 1: Classical Theory, AMS Colloquium Series 54.1, American Mathematical Society, Providence, RI, [OPUC2] B. Simon, Orthogonal 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, M. B. Porter Lectures, Princeton University Press, Princeton, NJ, 2011.

4 Szeg 's Theorem as a for S Szeg 's Theorem was proven by him in 1914 as a statement about s as we discuss below. In , he rephrased it as a variational principle in OPUC. This (two-part) paper essentially invented the general theory of OPUC. In these papers, Szeg assumed dµ was purely a.c. The addition of a singular continuous part is a discovery of Verblunsky in but his work was largely ignored and he didn't get credit until about fteen years ago when, in a dierent context, Killip and Simon rediscovered his proof and then his paper.

5 Szeg 's Theorem as a for S Φ n has a variational form. Since Φ n = Proj of z n onto the orthogonal complement of {1,..., z n 1 }, Φ n = dist of z n to span of {1,..., z n 1 } = min{ P P monic, deg P = n} = min{ P P (0) = 1, deg P n} since P monic P (0) = 1. This implies Φ n+1 Φ n which is obvious from Φ n = ρ 0 ρ 1... ρ n 1 and ρ j 1.

6 Szeg 's Theorem as a Thus, clearly, lim n Φ n exists and lim Φ n = inf{ P P (0) = 1, P is a polynomial } n for S Szeg Theorem for OPUC. Let dµ = f(θ) dθ + dµ s be an arbitrary probability measure. Then (NOTE THE SQUARE) inf{ P 2 P (0) = 1, P is a polynomial } ( = exp log f(θ) dθ )

7 Szeg 's Theorem as a for S This innocuous-looking theorem will have remarkable consequences as we'll see, in part because it has multiple equivalent forms. Because f(θ) dθ <, the integral cannot diverge to +, but it can to in which case, we interpret exp( ) as 0. Indeed, by Jensen's inequality and the concavity of log, the integral is non-positive and the exponential in [0, 1] as it must be given that Φ 0 = 1. One remarkable aspect of this theorem is that dµ s doesn't enter! Before turning to the proof, we consider some equivalent forms and consequences.

8 Szeg 's Theorem as a As we've seen, Φ n = ρ 1... ρ n 1 so for S lim Φ n 2 = ( 1 αj 2) j=0 Szeg Theorem ( Version). If dµ = f(θ) dθ + dµ s, then log ( 1 α j 2) = j=0 log ( f(θ) ) dθ This is a precursor of KdV sum rules. It is clearly equivalent to the variational form.

9 Szeg 's Theorem as a for S Corollary. j=0 α j 2 < log ( f(θ) ) dθ >. A consequence of this is that dµ s can be more or less arbitrary while one still has j=0 α j 2 < ; for example, if dµs = η < 1, (1 η) dθ + dµ s = dµ has j=0 α j(µ) 2 <. This is remarkable because we'll see in a future lecture that j=0 α j < dµ is purely a.c. and ε < f(θ) < ε 1 for some ε > 0 and all θ. It is also remarkable because it is not easy to construct operators with mixed spectrum and potential decay.

10 Szeg 's Theorem and for S Given {c n } n=, the corresponding N N matrix T N (c) has the form c 0 c 1... c N 1 c 1 c 0... c N..... c N+1 c N+2... c 0 i.e., ( T N (c) ) ij = c j i. If µ is a measure, we set c j = e ijθ dµ(θ) and write (µ is called the symbol) D N (µ) = det ( T N+1 (µ) )

11 Szeg 's Theorem and Notice that in the L 2 (dµ) inner product, ( TN ) kj = eikθ, e ijθ = z k, z j Writing Φ N = z N + l.o. and using sums of rows and columns, one sees that D N (µ) = det ( Φ j, Φ k ) 0 j, k N = Φ 0 2 Φ N 2 for S

12 Szeg 's Theorem and Since Φ j, one sees that Thus, lim D N(µ) 1/N+1 = lim Φ N 2 N N Form of Szeg 's Theorem. For any µ, lim N 1 N + 1 log D N(µ) = log f(θ) dθ for S

13 Szeg 's Theorem and Aside: It is known that if dµ s = 0 and for S and then log ( f(θ) ) n= n L n 2 < n=1 log D N (µ) = (N + 1) L 0 + L n e inθ n L n 2 + o(1) n=1 This is the Strong Szeg Theorem. [OPUC1], Chap. 6 has many proofs of this.

14 When are in L 2 ( D, dµ)? By Weierstrass' Theorem, for any µ of compact support on R, the polynomials in x are dense in L 2 (R, dµ). But this is not true for D. Indeed, if dµ = dθ, the closure of the polynomials are those functions in L 2 whose negative Fourier coecient e inθ f(e iθ ) dθ = 0 for n 1. On the other hand, we'll see soon that if supp(dµ) D, the polynomials are dense. for S

15 When are in L 2 ( D, dµ)? for S Theorem (Kolmogorov-Krein). If dµ = f dθ + dµ s, then the polynomials in z are dense in L 2 ( D, dµ) if and only if log f(e iθ ) dθ =. They found this because this density result was relevant to their theory of prediction for stochastic processes. Given Szeg 's Theorem, the proof is almost trivial for inf P z 1 P 2 L 2 = inf P 1 zp 2 L 2 = inf Q Q(0)=1 Q 2 L = exp ( 2 log f dθ )

16 When are in L 2 ( D, dµ)? for S So z 1 closure of polys log f dθ =. Thus, if the integral is nite, z 1 / closure of polys and thus, polynomials are not dense. On the other hand, if z 1 = lim P n, then z 2 = lim n P n [ limm P m ] so all polynomials in z and z 1 are in closure of polys and they are dense (by Weierstrass' other density theory). Krein used this to show (see [SzThm], p. 319) that on R, if dρ = F dx + dρ ν, then {e iαx } α 0 are dense in L 2 log F (x) dx =. This, in turn, implies that if 1+x 2 x n dρ(x) <, the moment problem is indeterminate if the integral is nite, for example, dρ(x) = e x α dx, α < 1

17 Strategy of the Proof As with all good proofs of equalities, we'll prove two inequalities. We'll use entropy term for exp [ log f dθ ] for reasons that will become clear soon. The proof that lim n Φ n 2 is an upper bound will be variational. We'll show that for any polynomial with P (0) = 1, we have P 2 entropy term. for S

18 Strategy of the Proof The lower bound on the entropy term will come from the fact that µ entropy term is weakly upper-semicontinuous (usc), i.e., µ n µ S(µ) lim sup S(µ n ). We'll prove that S(µ) = j=0( 1 αj 2) 1/2 for BernsteinSzego measures by direct calculation and then use this and usc to get the other inequality. for S

19 for S Lemma. For any polynomial P, with P (0) 0, we have that log P (e iθ ) dθ log P (0) Remark. One proof notes that log ( P (z) ) is subharmonic. Proof. If {z j } m j=1 are zeros in D, let Q(z) = m j=1 1 z j z z z j Then log Q(z) is analytic in D, so P (z)

20 log Q(0) = lim r 1 = log Q(re iθ ) dθ = log P (e iθ ) dθ log Q(e iθ ) dθ But, Q(0) = m j=1 z j 1 P (0) P (0). for S

21 for S For any polynomial, P, with P (0) 0, dµ = f dθ + dµ s, we have P (e iθ ) 2 dµ(θ) (by Jensen) = ( exp by the Lemma. Thus inf P P (0)=1 P (e iθ ) 2 f(θ) dθ exp [ 2 log P (e iθ ) + log ( f(θ) )] dθ 2 log ( P (e iθ ) dθ ) ( exp log f dθ ) ) ( P (0) 2 exp log f(θ) dθ ) P (e iθ ) 2 dµ exp ( log ( f(θ) )) dθ

22 for S One can also get a variational upper bound to complete the proof. The idea is to consider the function ( e iθ + z D(z) = exp e iθ z ) dθ log(f(θ)) 4π Formally, and we'll see later that D is actually in H 2 (D) and has boundary values, D(e iθ ) = lim r D(re iθ ) exists for a.e. θ and D(e iθ ) 2 = f(θ). If dµ s = 0, we have P (z) = D(0)/D(z) has P (0) = 0 and P (z) 2 dµ = D(0) 2 f(θ) 2[ f(θ) dθ ] = D(0) 2 = exp ( log(f(0)) dθ )

23 P isn't a polynomial but one can approximate by polynomials. Handling dµ s is a separate issue, but it can be done (see [OPUC1], Section 2.5 and [SzThm], Section 2.12). for S

24 The Suppose α j = 0 for j N. Then, we've seen that for S Thus, dµ = f(θ) dθ, f(θ) = ϕ N(e iθ ) 2 log f(θ) = 2 log ϕ N(e iθ ) = log Φ N 2 2 log Φ N(e iθ ) Since Φ N (z) is analytic in a nbhd of D, so is log ( Φ N (z)), so dθ log Φ N(e iθ ) = log Φ N(0) = 0 Thus, log f(θ) dθ N 1 = log Φ N 2 ( = log 1 αj 2) 1/2 proving Szeg 's Theorem in this case. j=0

25 The Given two prob. measures on D, we dene their relative entropy by { if µ is not ν-a.c. S(µ ν) = log ( dµ ) dν dµ if µ is ν-a.c. For example, S(gdν dν) = g log(g)dν Usually ν is xed and we vary µ. for S

26 The for S We claim that ( dθ S f dθ ) + dµ s = log ( f(θ) ) dθ For µ is ν-a.c. i f(θ) 0 for dθ -a.e. θ. If f(θ) = 0 on a positive Lebesgue measure set, the integral is, so both sides are. If f(θ) 0 for a.e. θ, dµ dν = f 1 χ S where χ S is a set with dµ s (S) = 0 and S = 1. Clearly log ( dµ ) = dν log ( f(θ) ) dθ

27 for S Here is a basic fact which we'll make plausible but not formally prove (but see Section 2.2 of [SzThm]). Theorem. Let E( D) be the continuous strictly positive functions on D. Then S(µ ν) = inf S(f; µ, ν) where f E( D) S(f; µ, ν) = f(x)dν(x) 1 + log ( f(x) ) dµ for S Proof. If dµ = gdν with g E( D), then S(g; gdν, ν) = 1 1 log ( g(x) ) dµ = S(gdν ν) By an approximation argument (and control of dµ s ) one obtains S(µ ν) inf S

28 for S for S Let's prove S(f; µ, ν) S(µ ν) in case dµ s = 0 so so that where Then S(f; µ, ν) = dν = g 1 dµ Q g(x) ( f(x) ) dµ(x) Q b (x) = xb 1 1 log x Q b (x) = b 1 x 1, Q b (x) = x 2 0 so Q b is convex, Q b (b) = 0, so Q b(x) Q b (b), i.e., Thus Q b (x) log(b) S(f; µ, ν) log ( g(x) ) dµ(x) = S(µ ν)

29 for S For each xed f in E( D), S(f; µ, ν) is linear and weakly continuous so the inf is concave and weakly usc, i.e. Theorem. S(µ ν) is jointly converse and jointly weakly usc in µ and ν. Corollary. Dene Sz(µ) = log f dθ Then µ Sz(µ) is weakly usc. if dµ = f dθ + dµ s. for S

30 The end of the proof Let µ have Verblunsky coecients, {α n } n=0. Let µ n be the approximation. We've proven above that By weak usc Sz(µ n ) = n 1 j=0 ρ 2 j for S Sz(µ) lim Sz(µ n ) = ρ 2 j j=0 which is the other inequality that we needed to prove.