Operator Theory and Related Topics, Université de Lille 1, 2010 An application of generalized spectral operators

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1 Operator Theory and Related Topics, Université de Lille 1, 2010 An application of generalized spectral operators Definition (Dunford) spectral operator : U B(X) (X = complex Banach space) spectral measure E : Bor(C) {projectors of X} s.t. UE(B) = E(B)U, σ(u E(B)X) B B Bor(C) Example U =scalar : U = C λde λ Theorem (Dunford) U spectral U = scalar+quasinilpotent Remark if B = closed, then E(B) = X U (B) where X U (B) := {x X : σ U (x) B} σ U (x) = C \ {λ 0 : x = x(λ) X analytic, λ λ 0, s.t. (λi U)x(λ) x} 1

2 Definition (Foiaş) Y = spectral maximal space of U : Y X closed, linear, invariant, s.t. if Z is another closed, linear, invariant subspace with σ(u, Z) σ(u, Y ), then Z Y U decomposable : for every finite open covering (G i ) i of σ(u) there is a set (Y i ) i of spectral maximal spaces s.t. σ(u, Y i ) G i and X = i Y i Proposition (Foias) Dunford, Bishop,... U spectral U decomposable Definition (Foias, Sz-Nagy) T i B(X i ) i = 1, 2 T 1 is a quasiaffine transformation of T 2 : A B(X 1, X 2 ) injective with dense range s.t. T 1 = A 1 T 2 A. Write T 1 < T 2. T 1, T 2 are quasisimilar : T 1 < T 2, T 2 < T 1. hyperinvariant subspace : invariant under the commutant 2

3 (ρ k ) k 0 ρ k > 0, ρ 0 = 1 C := {f C : c = c f s.t. f (k) (x) c k+1 ρ k x, k} C is quasianalytic : f (k) (x 0 ) = 0 k f 0 Theorem (Denjoy-Carleman) C quasianalytic n 1 L n = where L n = inf k n ρ 1/k k non-quasianalytic f, g 0 s.t. fg 0 3

4 Given ρ n 1 (n Z) s.t. ρ n+m ρ n ρ m ; lim n ρ 1/ n n = 1, the space of all continuous functions f(e it ) = n c ne int with f := c n ρ n < n Z becomes a Banach algebra A (ρn ) If Beurling s condition lnρ n n < n Z is verified (example: ρ n = n n ρ, 0 < ρ < 1), the algebra A (ρn ) is regular. In particular, A (ρn ) contains funct. f, g 0 s.t. fg 0. Definition (Colojoară; Foiaş) Let U B(H) be invertible. Set ρ n = U n and A U := A (ρn ) n. We call U A U -unitary if A U is regular and there exists a continuous morphism of algebras E : A U f f(u) B(H) taking any polynomial n c nz n into n c nu n ( U decomposable; U = E(z)) 4

5 Theorem (Vasilescu; Colojoara, Foias) If the Banach algebra A generated by the range of E is inverse closed, then: the space M of maximal ideals of A is σ(e(z)); the Gelfand representation A Â satisfies: ˆ E(f) = f σ(e(z)), σ(e(f)) = f(σ(e(z))). Theorem (Wermer) If U is A U -unitary, it has invariant subspaces f(u)g(u) = (fg)(u) = 0, f(u) 0, g(u) 0; H 0 := ker f(u) UH 0 H 0 : f(u)h = 0 f(u)uh = Uf(U)h = 0 5

6 Theorem 1 (Colojoară, Foiaş, Sz.-Nagy) Let U, V, T B(X) s.t. U < T < V, where U = A U -unitary, V = A V -unitary and σ(u) a single point. Then T has hyperinvariant subspaces. Application (Bishop type operators) Fix α (0, 1) \ Q and define T α : L 2 [0, 1] L 2 [0, 1] by (T α h)(x) = x h({x + α}), x [0, 1] where {y} = the fractional part of y (y = n + s with n Z and s [0, 1); {y} := s). Equivalently, (T α h)(e 2πix ) = xh(e 2πi(x+α) ) on the space L 2 of the unit circle. Bishop: T α = candidate for an operator without invariant subspaces 6

7 More generally, where ϕ L [0, 1] T α,ϕ h (x) = ϕ(x)h(e 2πi(x+α) ) 7

8 Davie, Blecher, MacDonald, Flattot, Chalendar, Partington Remark If α Q, then T α has invariant subspaces Proposition (Davie) If α Q, T α has no eigenvectors Theorem 2 (Davie 70 ) For almost every irrational α, the operator T α has hyperinvariant subspaces T := et α ( the spectral radius r(t) of T is 1) Generally, we have (MacDonald) r(t) = lim n T n r(t ϕ,α ) = e 1 0 ln ϕ(x) dx good estimates of T ±n existence of invariant subspaces (if T is invertible) Wermer, Atzmon... Proof of Davie s theorem (6 steps) 1) extend T to a space L of (classes of) Lebesgue measurable functions defined a.e. on [0, 1], so that T 1 exists, T n f(x) = F n (x)f({x αn}) 8

9 F n (x) = e n {x α} {x nα} 2) write [0, 1] = t E t with E t = {x : {x nα} are bounded from below } 3) n there are p, q relatively prime s.t. α p q < 1 q 2, q n (Dirichlet); for almost all α, we can take q n 1/4 (Levy: x p n(x) q n (x) continuous fractions; lim n 1 n lnq n(x) = universal constant) 4) {x nα} {x n p q } estimates for F n(x) on E t (by Stirling s fomula) T ±n L 2 (E t ) n nρ 5) L := {f : f L < } where for suitable constants c t, f L := k c tk E tk n Z T n f (x)e n ρ 2 dx 9

10 Using 4) we obtain f L ct. f L 2 Let T denote T acting on L. Then by Beurling s condition T is A T-unitary; σ( T) = unit circle Let A : L 2 L be the inclusion L 2 L; then A is bounded with dense range Also, AT = TA, that is, T = A 1 TA; Then T is a quasiaffine transformation of T, T < T Properties: B < C C < B ; if B is A B -unitary B is A B -unitary Now T also is a Bishop-type operator; then similarly T < T for another A T -unitary operator T. Hence T < T = T with T = A T -unitary. Thus T < T < T. 6) Apply Colojoara-Foias Theorem 1 10

11 Proposition Let α (0, 1) be an arbitrary irrational number. Let ϕ(x) = e f(x) for 0 < x < 1, where f(x) = n k= n c ke 2kπix is an arbitrary trigonometric polynomial. Then T α,ϕ defined by T α,ϕ h (x) = ϕ(x)h(e 2πi(x+α) ) has hyperinvariant subspaces. 11

12 Definition The index indα of α is the supremum of all l > 0 s.t. for any k > 0 there exist p, q with α p q < k q l Liouville: if indα = α is transcendent Dirichlet: for all irrational α we have indα 2 Roth: if α is algebraic irrational ind α = 2 Jarnik: almost all α (0, 1) have finite index More generally, consider (Th)(x) = ϕ(x)h(e 2πi(x+α) ) on L p Davie: for all irrational α (0, 1) with indα < and ϕ(x) = x, on L 2 MacDonald: for all irrational α with indα < and ln ϕ well-approximable by step functions of intervals indα < and ln ϕ L p piecewise monotone with p > ind α indα < and ϕ analytic in a neighbourhood of [0, 1], on L p Flattot: for ϕ(x) = x s and a larger class of α s, on L 2 Chalendar & Partington: for products of Bishop ops. MacDonald: T n 12

13 Idea for more general cases: T n f(x) = F n (x)f({x αn}), and estimating F n (x) is equivalent to estimate 1 n lnf n(x) = 1 n [F(x)+F(τ(x))+ +F(τn 1 (x))] F dx a n where F(x) := lnx and τ(z) := ze 2πiα, by a suitable sequence of constants a n 0 (a n := 1 n ε for example). Therefore: good hypotheses on α and ϕ should lead to uniform estimates of the speed of convergence in (this particular case of) Birkhoff s ergodic theorem Speed of convergence in the ergodic theorem: Let F := ln ϕ and write F(e 2πix ) = k c ke 2πi kx Let τ act on the unit circle by rotation of angle 2πα (Weyl automorphism) We seek for: 1 n n 1 j=1 F(τ j (x)) Fdx a n ( 0 as n ) 13

14 Kowada: if F L 1 with c k a n = O(1/n) ct k indα+1+ε, we may take Kachurovsky: estimates in terms of the measure E U ( )F, F where F L 2 and E U is the spectral measure of the unitary Uh = h τ; for example: if c k ct k 2+ε, we may take a n = O(1/n ǫ ) Sinai, Fomin, Cornfeld: if α is well approximable by rationals, then τ is well approximable by periodic automorphisms ( ergodicity) 14