Measures, orthogonal polynomials, and continued fractions Michael Anshelevich November 7, 2008
MEASURES AND ORTHOGONAL POLYNOMIALS. µ a positive measure on R. A linear functional µ[p ] = P (x) dµ(x), µ : R[x] R. Positive: µ[p (x) 2 ] 0. Inner product P, Q = µ[p Q]. Gram-Schmidt {, x, x 2, x 3,... } monic orthogonal polynomials {P 0 =, P, P 2, P 3,...}.
Theorem. (Favard, Stone, etc.) For some β i R, γ i 0 xp n = P n+ + β n P n + γ n P n. 2nd order recursion relation. Two independent solutions {P n, Q n }. Initial conditions P = 0, P 0 =, P = x β 0, Q 0 = 0, Q =, Q 2 = x β. Exercise. [ Pn (x) P n (y) Q n (x) = (I µ) x y ]. 2
µ { (β0, β, β 2,...) (γ, γ 2, γ 3,...) } equivalent. More explicit relation? If know { (β0, β, β 2,...) (γ, γ 2, γ 3,...) }, how to recover µ? Without going through {P n }. Cauchy transform G µ (z)= z x dµ(x) = µ [ z x ] = µ[] z + µ[x] z 2 + µ[x2 ] z 3 + µ[x3 ] z 4 +... 3
G µ (z) = µ[] z + µ[x] z 2 + µ[x2 ] z 3 + µ[x3 ] z 4 +.... Theorem. Also G µ (z) = z β 0 z β Same coefficients as in the recursion. γ γ 2 z β 2 γ 3 z... Note: µ 0 all γ 0, no determinants. Proof II. Flajolet (980): lattice paths. 4
Proof I. Look at G(z) = z β 0 z β γ z β 2 γ 2 γ 3... z β n γ n H G = polynomial polynomial. Claim. G = Q n γ n Q n H P n γ n P n H, where {P, Q} with recursion {β i, γ i }. 5
Proof. By induction. Assume and say Then G = G= z β 0 γ H 0 = Q γ Q 0 H 0 P γ P 0 H 0. G = Q n γ n Q n H P n γ n P n H H = z β n γ n+ K. Q n γ n Q n z β n γ n+ K P n γ n P n z β n γ n+ K = zq n β n Q n γ n+ Q n K γ n Q n zp n β n P n γ n+ P n K γ n P n = Q n+ γ n+ Q n K P n+ γ n+ P n K. 6
FINITE CONTINUED FRACTIONS. Let G n = G cut off at level n. G n (z) = z β 0 z β γ z β 2 γ 2 γ 3... H = 0. z β n 0 G n = Q n γ n Q n H P n γ n P n H = Q n P n. 7
G n = Q n P n. P n monic, n real roots G n (z)= P n (x) = n i= (x x i ). z x dµ n(x) = Q n(z) (z xi ) = (partial fractions) = a i z x i. G n = Cauchy transform of µ n = a i δ xi, x i = roots of P n, a i = Q n(x i ) P n(x i ). 8
G n (z) = z β 0 z β γ z β 2 γ 2 γ 3... G n approximate G. z β n 0 G n = G µn, µ n = n i= a i δ xi. Do µ n approximate µ? Yes. In fact, µ n [P (x)] = µ[p (x)] for deg P 2n. 9
GAUSSIAN QUADRATURE. Want to evaluate f(x) dµ(x) Riemann sums. n i= a i f(x i ). Want P (x) dµ(x) = n i= a i P (x i ) for P of low degree. How to choose a i, x i? Answer: take x i = roots of P n. Choose a i so that x k dµ(x) = a i x k i, k = 0,,..., n (n equations, n unknowns). Our a i = Q n(x i ) P n(x i ) work. 0
Proof. Lagrange interpolation: for any P with deg P < n, Note so µ [ Pn (x) µ[p (x)] = x x i P (x) = n ] i= = µ n i= P (x i )P n (x) P n (x i)(x x i ). [ Pn (x) P n (x i ) x x i P (x i ) P n (x i) Q n(x i ) = n ] i= = Q n (x i ) Q n (x i ) P n (x i) δ x i [P ] µ[p (x)] = n i= a i δ xi [P (x)] µ[p (x)] = µ n [P (x)] for deg P < n. In fact, same x i, a i work for k = n, n +,..., 2n.
For P k, n k 2n, P k (x) = A(x)P n (x) + B(x), deg A, B n. µ[p k ] =, P k = 0. To show: µ n [P k ] = 0. µ[ap n ] = A, P n = 0 deg A < n µ[b] = 0 µ n [B] = 0. Finally, so µ n [P k ] = 0. µ n [AP n ] = a i A(x i )P n (x i ) = 0, So µ n µ, G n G, and therefore G µ = G. 2
If know {β i, γ i }, can find µ? Usually hard: G µ (z) = z β 0 an infinite expression. z β γ γ 2 z β 2 γ 3 z... Class of explicit examples. Semicircle law: 2πt 4t x 2 [ 2 t,2 t] dx. 2.0.5 y.0 0.5 0.0 3 2 0 x 2 3 3
Semicircle law: 2πt 4t x 2 [ 2 t,2 t] dx. 2.0.5 y.0 0.5 0.0 3 2 0 x 2 3 Marchenko-Pastur distributions: 2π 4t (x t) 2 x [+t 2 t,+t+2 t] (x) dx + max( t, 0)δ 0..0.0 0.75 0.75 y 0.5 y 0.5 0.25 0.25 0.0 0 2 x 3 4 0.0 0 2 x 3 4 4
Semicircular, Marchenko-Pastur orthogonal polynomials satisfy n=0 P n (x)u n = A(u) B(u)x. In general: free Meixner distributions 2πt 4(t + c) (x b) 2 + (b/t)x + (c/t 2 )x 2 [b 2 t+c,b+2 t+c] dx +0,, 2 atoms. AC support an interval polynomial polynomial limited atoms outside of the AC support Not SC part 5
PERIODIC CONTINUED FRACTIONS. β i+n = β i, γ i+n = γ i. H = G in G(z) = z β 0 z β γ z β 2 γ 2 γ 3... z β n γ n H G = Q n γ n Q n G P n γ n P n G. γ n P n G 2 (γ n Q n + P n )G + Q n = 0. Quadratic equation! D = (γ n Q n + P n ) 2 4γ n Q n P n. 6
G = (γ nq n + P n ) D 2γ n P n = Stieltjes inversion formula: z x dµ(x). dµ(x) > 0 if dµ(x) = π lim y 0 Im G(x + iy). D(x) ir, i.e. D < 0. D degree 2n, n intervals for D 0. No SC. atoms: roots of P n, at most (n ). Recall D = (γ n Q n + P n ) 2 4γ n Q n P n. So if P n (a) = 0, then D(a) 0. Atoms outside of the AC support. 7
Eventually constant continued fractions (n = ) β i = β, γ i = γ for i N. polynomial on one interval. polynomial If β = 0, γ = (Peherstorfer?) Bernstein-Szegő class 4 x 2 polynomial on [ 2, 2]. Weyl s Theorem. If β i 0, γ i, then σ ess (µ) = [ 2, 2]. Denisov-Rakhmanov Theorem. If σ ess (µ) = AC support of µ = [ 2, 2], then β i 0, γ i. 8
Eventually periodic polynomial polynomial on n intervals. polynomial polynomial on n intervals eventually periodic. Weyl: if {β i, γ i } asymptotically periodic, same essential spectrum as for actually periodic. Converse false. Last, Simon: if {β i, γ i } approaches the isospectral torus of a periodic sequence, same essential spectrum. Damanik, Killip, Simon: converse true. 9
QUESTIONS. Connection between random matrices and (eventually) periodic continued fractions (Pastur). Multivariate (non-commutative) orthogonal polynomials, states, continued fractions. All exist. Continued fractions matricial. Formulas for states with periodic continued fractions. Free Meixner states known; constant after step 2. If not formulas for states, description of their operator algebras. Connection between multi-matrix models and states with multivariate (eventually) periodic continued fractions. 20