Moment Infinitely Divisible Weighted Shifts

Transcription

1 Moment Infinitely Divisible Weighted Shifts (joint work with Chafiq Benhida and George Exner) Raúl E. Curto GPOTS 2016 Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 1 / 47

2 Some Key Words and Phrases subnormal weighted shift Berger measure Aluthge transform Schur products other transformations moment problem completely monotone functions completely alternating functions Agler shifts infinitely divisible weighted shifts Laplace transform methods Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 2 / 47

3 Hyponormality and Subnormality L(H): algebra of operators on a Hilbert space H T L(H) is normal if T T = TT subnormal if T = N H, where N is normal and NH H (We say that N is a lifting of T, or an extension of T. hyponormal if T T TT For S,T B(H), [S,T] := ST TS. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 3 / 47

4 An n-tuple T (T 1,...,T n ) is (jointly) hyponormal if [T1,T 1] [T2,T 1] [Tn,T 1 ] [T [T1,T] :=,T 2] [T2,T 2] [Tn,T 2 ]... [T1,T n] [T2,T n] [Tn,T n ] 0. For k 1, an operator T is k-hyponormal if (T,...,T k ) is (jointly) hyponormal, i.e., [T,T] [T k,t]..... [T,T k ] [T k,t k ] 0 Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 4 / 47

5 normal subnormal k-hyponormal hyponormal (Bram-Halmos) T subnormal T is k-hyponormal for all k 1 (T,T 2,...,T k ) is hyponormal for all k 1. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 5 / 47

6 n-contractivity For n = 1,2,..., T is said to be n-contractive if ( ) n n j=0( 1) j T j T j 0. j Agler-Embry Characterization of Subnormality: Assume that T is a contraction. Then T is subnormal if and only if T is n-contractive for all n 1. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 6 / 47

7 Unilateral Weighted Shifts α {α k } k=0 l (Z + ), α k > 0 (all k 0) W α : l 2 (Z + ) l 2 (Z + ) W α e k := α k e k+1 (k 0) When α k = 1 (all k 0), W α = U +, the (unweighted) unilateral shift In general, W α = U + D α (polar decomposition) W α = sup k α k Wαe n k = α k α k+1 α k+n 1 e k+n, so Wα n n 1 = i=0 W β (i), Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 7 / 47

8 Weighted Shifts and Berger s Theorem Given a bounded sequence of positive numbers (weights) α α 0,α 1,α 2,..., the unilateral weighted shift on l 2 (Z + ) associated with α is W α e k := α k e k+1 (k 0). The moments of α are given as { γ k γ k (α) := 1 if k = 0 α α2 k 1 if k > 0 }. W α is never normal W α is hyponormal α k α k+1 (all k 0) Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 8 / 47

9 Berger Measures (Berger; Gellar-Wallen) W α is subnormal iff there exists a positive Borel measure ξ on [0, W α 2 ] such that γ k = t k dξ(t) (all k 0). ξ is the Berger measure of W α. For 0 < a < 1 we let S a := shift{a,1,1,...). The Berger measure of U + is δ 1. The Berger measure of S a is (1 a 2 )δ 0 +a 2 δ 1. The Berger measure of B + (the Bergman shift) is Lebesgue measure n+1 on the interval [0,1]; the weights of B + are α n := n+2 (n 0). Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 9 / 47

10 k-hyponormal and n-contractive U.W.S. (RC, 1990) W α is k-hyponormal iff the following Hankel moment matrices are positive for m = 0,1,2,... : γ m γ m+1 γ m+2... γ m+k γ m+1 γ m+2... γ m+k+1 γ m γ m+k γ m+k γ m+k+1... γ m+2k (An operator matrix condition is replaced by a scalar matrix condition.) (G. Exner, 2006) W α is n-contractive iff n ( ) n ( 1) j γ m+j 0, m = 0,1,... j j=0 Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 10 / 47

11 Aluthge Transform Let T be a Hilbert space operator, let P := T be its positive part, and let T = VP denote the canonical polar decomposition of T, with V a partial isometry and ker V = ker T = ker P. We define the Aluthge transform of T as AT(T) := PV P. The iterates of AT are AT n+1 := AT(AT n (T)) (n 1). The Aluthge transform has been extensive studied, in terms of algebraic, structural and spectral properties. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 11 / 47

12 For instance, (i) T = AT(T) T is quasinormal; (ii) (Aluthge, 1990) If 0 < p < 1 2 and T is p-hyponormal, then AT(T) is (p )-hyponormal; (iii) (Jung, Ko & Pearcy, 2000) If AT(T) has a n.i.s., then T has a n.i.s. (iv) (Kim-Ko, 2005; Kimura, 2004) T has property (β) if and only if AT(T) has property (β); and (v) (Ando, 2005) (T λ) 1 (AT(T) λ) 1 (λ / σ(t)). Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 12 / 47

13 On the other hand, G. Exner (IWOTA 2006 Lecture): subnormality is not preserved under AT. Concretely, Exner proved that the Aluthge transform of the weighted shift in the following example is not subnormal. Example (RC, Y. Poon and J. Yoon, 2005) Let 1 α α n := 2, if n = 0, 2 n n +1, if n 1 Then W α is subnormal, with 3-atomic Berger measure µ = 1 3 (δ 0 +δ 1/2 +δ 1 ). Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 13 / 47

14 Note that the Aluthge transform of a weighted shift is again a weighted shift. Concretely, the weights of AT(W α ) are α0 α 1, α 1 α 2, α 2 α 3, α 3 α 4,. Define W α := shift ( α 0, α 1, α 2, ). Then AT(W α ) is the Schur product of W α and its restriction to the subspace {e 1,e 2, }. Thus, a sufficient condition for the subnormality of AT(W α ) is the subnormality of W α. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 14 / 47

15 To compare W α and AT(W α ), we assume that W α is subnormal. This guarantees that both W α and AT(W α ) are subnormal. Denote the Berger measures of W α and AT(W α ) by µ and AT(µ), resp. Problem How is AT(µ) related to µ? In Exner s example, is the fact that card supp µ = 3 of significance, since it is an odd number, and the Aluthge transform of W α is related to Wα? 2 Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 15 / 47

16 Other Transformations Given W α and 0 < p < 1, consider the weighted shift W α p with weights α (p) i := (α i ) p (i 0). If W α is subnormal, with Berger measure µ, under what conditions is W α p subnormal? If it is, how is the Berger measure of W α p related to µ? When is W α p subnormal for all 0 < p < 1? Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 16 / 47

17 Agler Shifts For j = 2,3,..., the j-th Agler shift A j is given by α j := 1 j, 2 j +1, 3 j +2,... It is well known that A j is subnormal, with Berger measure dµ j (t) = (j 1)(1 t) j 2 dt. Clearly, A 2 is the Bergman shift, and the remaining Agler shifts are the upper row shifts of the Drury-Arveson 2-variable weighted shift, which incidentally is a spherical complete hyperexpansion. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 17 / 47

18 Weight Diagram of the Drury-Arveson Shift T 1 T 2 (0,1) (0,2) (0,3) (0,0) (1,0) (2,0) (3,0) δ dt (1 t)dt (1 t) 2 dt Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 18 / 47

19 Powers and Aluthge transforms of the Agler shifts are again subnormal, via an approach based on completely monotone functions. Definition A function f : R + R + \{0} is completely monotone if its derivatives alternate in sign: f (2j) 0 for j 1, and f (2j+1) le0 for j 0. Proposition f is completely monotone if and only if f = L(µ) for some positive measure µ, where L denotes the Laplace transform. Fact: If a completely monotone function f interpolates the moments of a shift, i.e., f(m) = γ m for all m 0, then the shift is subnormal. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 19 / 47

20 Theorem (Exner, 2009) For j = 2,3,..., let A j be the j-th Agler shift. Any p-th power transformation (p > 0) of A j is subnormal, as is any n-th iterated Aluthge transform of A j. However, the proof (which uses monotone function theory) offers no information about the Berger measure of the resulting shift. Using the difference operator we can rephrase the Agler-Embry conditions for shifts. 0 ϕ := ϕ, ( ϕ)(n) := ϕ(n) ϕ(n+1), for all m 1. m ϕ := m 1 ϕ, Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 20 / 47

21 For example, ( ϕ)(n) = ϕ(n) ϕ(n+1), ( 2 ϕ)(n) = ϕ(n) 2ϕ(n+1)+ϕ(n+2), ( 3 ϕ)(n) = ϕ(n) 3ϕ(n+1)+3ϕ(n+2) ϕ(n+3), and so on. A sequence ϕ is said to be completely monotone if ( m ϕ)(n) 0 for all m,n 0. It follows that the Agler-Embry conditions for subnormality of W α entail that the moment sequence (γ n ) is completely monotone. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 21 / 47

22 A sequence ψ is said to be completely alternating if ( m ψ)(n) 0 for all m,n 0. Similarly, given an integer k 1, a sequence ψ is said to be k-alternating if ( k ψ)(n) 0 for all n 0. Proposition ψ is completely alternating if and only if ϕ := e tψ is completely monotone for all t > 0. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 22 / 47

23 Theorem ψ is completely alternating if and only if it has an associated Lévy-Khintchin measure; that is, ψ(n) = a+bn+ 1 0 (1 t n+1 ) dµ(t), where µ 0. In the case of the Agler shifts A j, the sequence of weights squared is 1 0 (1 t n+1 )(j 1)t j 2 dt. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 23 / 47

24 CA: completely alternating sequences CA + : completely alternating sequences with all positive terms LogCA: positive-term sequences (x n ) such that (lnx n ) CA. These sequence sets have been studied by A. Athavale, V.M. Sholapurkar and A. Ranjekar, among others. CA and CA + are closed under sums, the addition of constants, and (positive) scalar multiples. Log CA is closed under (positive) scalar multiplication, products, and positive powers. ϕ CA + ln(1+ϕ) CA + ϕ CA + ϕ p CA + Since ( n+1 n+2 ) CA n+1 + we have ( n+2 ) CA +. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 24 / 47

25 Proposition (Benhida, RC and Exner, 2015) CA + LogCA. Proposition (Benhida, RC and Exner, 2015) LogCA CA. Proof. ( ) ( n+1 n+2 )3 The sequence is log completely alternating but not completely ( ) alternating. For, A 2 (with weight sequence ) has weights squared n+1 n+2 completely alternating. Therefore the weights squared are log completely alternating, so clearly any power of them is log completely alternating. ( ) On the other hand, one checks, using Mathematica, that ( n+1 n+2 )3 is not completely alternating. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 25 / 47

26 Moment Infinitely Divisible (MID) Weighted Shifts Definition W α is MID if the shift with weight sequence (αn) t corresponds to a subnormal weighted shift, for every t > 0. Theorem (Benhida, RC and Exner, 2015) Let W α be a weighted shift with (positive) weight sequence α = (α n ). Then W α is moment infinitely divisible if and only if (i) α n 1, n = 0,1,..., and (ii) (αn) 2 is log completely alternating; equivalently, (lnα n ) CA. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 26 / 47

27 Corollary (i) All Agler shifts are MID (ii) Also, all S(a,b,c,d) shifts, and their subshifts, are MID. Here S(a,b,c,d) is the shift defined by RC, Y.T. Poon and J. Yoon, with weights, where a,b,c,d > 0 and ad > bc. an+b cn+d Corollary Suppose W α is a weighted shift. 1 If W α is MID, let W α be the rank-one perturbation consisting only of decreasing the zeroth weight of W α (while leaving it positive). Then W α is MID; 2 If the weighted shift with weight sequence e α = (e α 0,e α 1,...) is MID, then W α is MID. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 27 / 47

28 We now improve a result of C. Cowen and J. Long, who established the subnormality of a certain weighted shift of importance in considering Toeplitz operators (existence of non-normal, non-analytic subnormal Toeplitz operators). Corollary If 0 < p < 1, the weighted shift with weights α n := (1 p 2n+2 ) 1 2 for n = 0,1,... is subnormal and moment infinitely divisible. Further, if p 1,p 2,...,p k is a collection of values in (0,1), the weighted shift with weights α n := k i=1 (1 p2n+2 i ) 1 2 for n = 0,1,... is subnormal and MID. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 28 / 47

29 Example The weighted shift with weights α n := ln(n+1), n = 0,1,..., n or the shift with weights the square roots of these, is MID and subnormal. This is because this sequence is completely alternating (and it increases to the Euler constant γ ). Corollary If W α is moment infinitely divisible then so is AT(W α ). Proof. Since (lnα n ) is completely alternating, so is (ln α n α n+1 ). Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 29 / 47

30 Mean and Cesàro Transforms Definition Let T VP be the polar decomposition of T. The mean transform is T := VP+PV 2. Theorem (S.H. Lee, W.Y. Lee and J. Yoon, 2014) The mean transform of B (2) + is subnormal. Proposition Suppose W α has (α 2 n) CA, and let α n := α 2 n +α2 n+1 2. Then W α and Ŵ α are MID and hence subnormal. Moreover, if (α n ) CA then the Ŵα is MID and hence subnormal. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 30 / 47

31 Corollary The mean transform of B + and the mean transform of B (2) + hence subnormal. are MID, Proposition If the sequence (α n ) is completely alternating then its Cesàro transform (C(α n )) is completely alternating. If the sequence (α n ) is log completely alternating, its geometric Cesàro transform (GC(α n )) is log completely alternating. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 31 / 47

32 Proposition If (γ n ) is a completely monotone sequence then so is (e γn 1 ), and therefore the shift with these moments is subnormal. Proposition If (γ n ) is an infinitely divisible sequence then so is (e γn 1 ), and therefore the shift with these moments is moment infinitely divisible. Corollary If the sequence (α n ) is completely alternating, then the shift with weights e αn is moment infinitely divisible. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 32 / 47

33 The Square Root Problem Given a subnormal weighted shift W α, under what conditions is its square root W α subnormal? Equivalently, by the uniqueness in the Hamburger moment problem: ( 2 t n dµ(t) = t dν(t)) n, n = 0,1,2,.... (1) Thus, the Square Root Problem can be stated as follows: Given a probability measure µ (supported on a compact interval in R + ), does there exist ν satisfying (1)? If ν exists, can one find it? Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 33 / 47

34 A Basic Case Problem What is the Berger measure for the square root of the Bergman shift, B+, with weight sequence , 4 3, 4 4,...? Equivalently, what is the measure ν with the following moment matching equations: ( t n dt = t n dν(t)) 2 (n = 0,1,2,...)? Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 34 / 47

35 Associated Moment Problem for Other Transformations For 0 < p < 1 consider the moment matching equations ( t n dµ(t) = t n dν(t)) p, n = 0,1,2,.... (2) One then tries to find one measure, given the other. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 35 / 47

36 A Result About the Support Let us recast the moment matching equations in terms of product measures: For each n, t n dµ(t) = = = = ( ( ) 2 t n dν(t) ) ( s n dν(s) s n t n dν(s)dν(t) s n t n d(ν ν). ) t n dν(t) Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 36 / 47

37 If we define p : R R R by p(x,y) := xy, we obtain that µ = (ν ν) p 1. Theorem (RC & G. Exner, 2015; J. Stochel & J.B. Stochel, 2012) Suppose that µ and ν are measures supported in some compact subset of R + satisfying (1), and let p(x,y) xy be as above. Then (1) µ = (ν ν) p 1 (2) supp(µ) = (supp(ν)) 2. When all of the above happens, we will write µ = ν 2. This result allows one to decipher the above mentioned case of a 3-atomic Berger measure. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 37 / 47

38 Observe that in this situation µ(e) = (ν ν)(p 1 (E)) = (ν ν)({(s,t) : st E}). One can give a geometrical interpretation, using the following picture (in the case when E = [a,b] ) (0,1) a b (0,0) (1,0) Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 38 / 47

39 Absolutely Continuous Measures Assume we wish to solve the Square Problem µ = ν 2, and we know that ν is absolutely continuous with respect to Lebesgue measure on [0, 1]. Write dν(t) g(t)dt, with g the Radon-Nikodym derivative in L 1. It appears reasonable to hope that µ will also be absolutely continuous, and to pursue its Radon-Nikodym derivative, call it f. Now, for 0 < a < 1, we have (almost everywhere) f(a) = lim n µ([a,a+1/n]). 1/n Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 39 / 47

40 µ([a,a+1/n]) = = = p 1 (E) a+1/n 1 a 1 + a+1/n a/x 1 (a+1/n)/x a a/x 1dν dν 1 g(y)g(x)dy dx + a/x (a+1/n)/x 1 g(y)g(x)dy dx 1 g(y)g(x)dy dx, (where we have used that g vanishes outside [0,1]). Then with f the Radon-Nikodym derivative of µ, we have: Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 40 / 47

41 µ([a,a+1/n]) f(a) = lim n 1/n ( ) 1 1 = lim n 1/n a = = 1 a 1 a lim n a/x (a+1/n)/x a/x (a+1/n)/x 1 nx g( a x )g(x)1 dx (a.e.). x 1 g(y)g(x)dy dx 1 g(y)dy 1 x g(x)dx Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 41 / 47

42 We thus have: Proposition (RC & G. Exner, 2015) Let µ = ν 2 and assume that dν(t) = g(t)dt, with g L 1 ([0,1]). Then dµ(t) = f(t)dt, where f(a) = 1 a g( a x )g(x)1 x dx (a.e.). Remark For future use, we note that, since g vanishes outside [0,1], one has 1 a g( a 1 x )g(x)1 x dx = g( a x )g(x)1 dx. (3) x This displays the integral as a convolution g g with respect to Haar measure on the multiplicative semigroup (0, 1). 0 Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 42 / 47

43 As a consequence, we get a concrete expression for f if g is a polynomial. Theorem (RC & G. Exner, 2015) Suppose that g is a polynomial, say g(x) n i=0 a ix i, positive on [0,1] and inducing a probability measure dν(t) = g(t)χ (0,1) (t)dt. Then µ = ν 2 is absolutely continuous with respect to Lebesgue measure and with Radon-Nikodym derivative f χ 0,1) where f is given by f(x) = + n 1 ( ) 1 x i+1 n i 1 a j a j+i+1 x j + i +1 i=0 lnx 2 i= n 1 n ai 2 x i i=0 j=0 ( ) 1 x i+1 n i +1 j= i 1 a j a j+i+1 x j. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 43 / 47

44 Corollary Let ν be the measure on [0,1] given by dν(t) = 1dt; i.e., ν is Lebesgue measure on [0,1]. Then ν 2 is given by lntdt. In particular, (1dt) 2 is singular at the origin. Corollary The Aluthge transform of the weighted shift associated with the Berger measure lntdt is A 3, the third Agler shift; that is, AT(A (2) 2 ) = A 3. This result does not hold for other Agler shifts. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 44 / 47

45 Laplace Transform Method Theorem (A. Athavale, 2013) The square root measure for 1dt (on [0,1]) is 1 π ( lnt) ( 1/2) dt. The key to this and allied results is the Laplace transform and the movement of the moment problem from the interval [0,1] to the interval [0, ). Recall that the Laplace transform of a function h is H L{h} where H(s) := 0 e st h(t)dt. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 45 / 47

46 Now suppose that F := L{g}. Then F(s +1) = L{e t g(t)}(s) First Shifting Theorem = = = e st e t g(t)dt u s+11 u f(u)du (u := e t,f(u) := g( lnu)) u s f(u)du. Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 46 / 47

47 Theorem (RC & G. Exner, 2015) The q-th power of 1dt (on [0,1]) for q > 0 is f(u)du = 1 Γ(q) ( lnu)q 1 du (where Γ denotes the classical Gamma function). Alternatively, the Berger ( ) q measure of the weighted shift whose moment sequence is 1 Γ(q) ( lnu)q 1 du. Corollary (Special Case: q = 1 2 ) du = Γ( 1 2 du = π ( lnu) 1 2 du. 2 )( lnu) 1 n+1 n=1 is Raúl E. Curto (Urbana, IL, 5/26/2016) Infinitely Divisible Shifts 47 / 47