Spectral properties of Toeplitz+Hankel Operators

Transcription

1 Spectral properties of Toeplitz+Hankel Operators Torsten Ehrhardt University of California, Santa Cruz ICM Satellite Conference on Operator Algebras and Applications Cheongpung, Aug. 8-12, 2014

2 Overview 1. Toeplitz operators T (a): Invertibility/Fredholmness factorization theory (L -symbols) Invertibility/Fredholm theory for PC symbols Spectral theory for PC symbols 2. Toeplitz+Hankel operators T (a) + H(b): Fredholm theory for PC symbols For Special classes of Toeplitz+Hankel operators: Invertibility/Fredholmness factorization theory Invertibility theory for PC symbols Spectral theory for PC symbols 1

3 Notation T = { z C : z = 1 } unit circle Hardy spaces H p = { f L p (T) : f n = 0 for all n < 0 } H p = { f L p (T) : f n = 0 for all n > 0 } Some operators on L p (T) N, 1 < p < : P : n= f n t n n=0 f n t n Riesz projection J : f(t) t 1 f(t 1 ) flip operator L(a) : f(t) a(t)f(t) multiplication operator (a L (T) N N ) 2

4 Multiplication operator matrix representation of L(a) with respect to basis {t n } n= in Lp (T) L(a) = [a j k ] = a 0 a 1 a a 1 a 0 a 1 a 2 a 2 a 1 a 0 a a 2 a 1 a (Laurent matrix) a n = 1 2π 2π 0 a(e ix )e inx dx Fourier coefficients 3

5 Toeplitz and Hankel operators Given a L (T) N N, define T (a) = P L(a)P (H p ) N, H(a) = P L(a)JP (H p ) N acting on the vector-valued Hardy space (H p (T)) N, 1 < p <. Matrix representation with respect to standard basis {t n } n=0 in Hp : T (a) = [a j k ] H(a) = [a j+k+1 ] a 0 a 1 a 2... a 1 a 0 a 1 a 2 a 1 a a 1 a 2 a 3... a 2 a 3 a 4 a 3. a 4 a 5 4

6 PART I Toeplitz operators (classical and well known) 5

7 Toeplitz, Hankel operators: identities For a, b L (T), T (ab) = T (a)t (b) + H(a)H( b) H(ab) = T (a)h(b) + H(a)T ( b) where b(t) := b(t 1 ). Note: If a H or b H, then T (ab) = T (a)t (b) (because H(a) = 0 or H( b) = 0). T (a) lower triangular Toeplitz for a H T (b) upper triangular Toeplitz for b H 6

8 Toeplitz operators: factorization (Motivation) Assume a = a da + with a H and a + H. Then T (a) = T (a )T (d)t (a + ). If, in addition, a 1 H, d = 1, a 1 + H, then T (a) 1 = T (a + ) 1 T (a ) 1 = T (a 1 + )T (a 1 ). Note: T (a + ) 1 = T (a 1 + ) and T (a ) 1 = T (a 1 ) T (a) 1 = P L(a 1 + )P L(a 1)P = L(a 1)P L(a 1 i.e., T (a 1 ) : f a 1 + P (a 1 f) + ) 7

9 Wiener-Hopf factorization of smooth functions Consider W := { a L (T) : n= a n < } Wiener algebra W + := W H, W := W H. Assume that a W N N has a Wiener-Hopf factorization in W, i.e., with a +, a 1 + W + N N a, a 1 W N N d(t) = diag(t κ 1,..., t κ N), t T, κ 1,..., κ N Z partial indices. a(t) = a (t)d(t)a + (t) 8

10 Then: T (a) is a Fredholm operator and dim ker T (a) = κ k, κ k <0 dim ker T (a) = κ k >0 κ k. In particular: T (a) is invertible iff κ 1 = = κ N = 0. Reason: T (a) = T (a )T (d)t (a + ) with T (a ± ) invertible Moreover, for a W N N : T (a) Fredholm det a(t) 0 a(t) possesses factorization 9

11 Generalized Factorization Generalized or Φ-factorization in L p of a (L ) N N : with a + (H q ) N N, a 1 + (Hp ) N N, a (H p ) N N, a 1 (Hq ) N N, a(t) = a (t)d(t)a + (t) d(t) = diag(t κ 1,..., t κ N), t T, κ 1,..., κ N Z partial indices, the mapping f a 1 + P (a + f) ( ) extends to a bounded linear operator on (H p ) N. (Here 1 < p <, 1/p + 1/q = 1.) 10

12 Remarks: existence of factorization, factors and part. indices depend on p partial indices are unique, up to change of order factors a ± are unique up to simple modification condition ( ) is equivalent to an A p -condition or Hunt-Muckenhoupt-Wheeden condition for N = 1 and, e.g., functions a P C: factorization can be constructed explicitly for N > 1, except in special cases, no explicit procedure for the construction of the factorization (or determination of the partial indices) is known. (P C = set of piecewise continuous functions on T) 11

13 Construction in the scalar case (illustration) Construction of factorization in case a W : Assume a(t) 0 and wind(a) = κ. Then with a(t) = a (t)t κ a + (t) a + = exp(p b), a = exp((i P )b), and b(t) = log [ a(t)t κ]. 12

14 Theorem: (Gohberg, Krupnik) Let a (L ) N N. Then T (a) is a Fredholm operator on (H p ) N if and only if a admits a Φ-factorization in L p. In this case, dim ker T (a) = κ k <0 κ k, dim ker T (a) = κ k >0 κ k. Hence: T (a) is invertible iff, in addition, κ 1 = = κ N = 0. 13

15 Toeplitz operators: Fredholm theory for P C-symbols Theorem: Let a P C N N. Then T (a) is Fredholm on (H p ) N iff a(t + 0) and a(t 0) are invertible matrices for all t T, for each t T, the arguments of the eigenvalues of the matrix a(t 0)a(t + 0) 1 are all different from 2π/p + 2πZ, i.e., 1 2π arg EV [ k a(t 0)a(t + 0) 1 ] / 1 p + Z, 1 k N. Here: one-sided limits of a P C N N at t T: a(t ± 0) = lim ε +0 a(te±iε ) 14

16 Essential spectrum of T (a), a P C: λ belongs to the essential spectrum of T (a) (acting on H p ) T (a λ) is not Fredholm on H p for some t T: a(t + 0) λ = 0 or a(t 0) λ = 0 or ( ) 1 a(t 0) λ 2π arg a(t + 0) λ 1 p + Z 15

17 Corollary: The essential spectrum of T (a) with a P C is sp ess (T (a) = R(a) t J A 1/p [a(t 0), a(t + 0)]. Here: R(a) = essential range of the function a P C L, A 1/p [u, v] = { z C : 1 2π arg u z v z 1 p + Z} line segment (p = 2) or certain circular arc (p 2) connecting the points u and v, J = { t T : a(t + 0) a(t 0)} set of jumps of a 16

18 i i Example i- 1+ i -1- i i i i+ -i- 1- -i -1+ -i i image of some a P C Fig. 1: image of sp ess (T (a)); p = 2 Fig. 2: p =1.3 17

19 i i i i p = 2 p = 2.4 i i i i p = 2.8 p =

20 Spectrum of T (a), a P C Coburn s lemma: If a L (T), a 0, then T (a) or T (a) has a trivial kernel. Corollary: Let a L (T). Then: T (a) is invertible T (a) is Fredholm and has Fredholm index zero. For a P C, the spectrum of T (a) can be described geometrically: sp(t (a)) = im(a #,p ) { λ / im(a #,p ) : wind(a #,p ; λ) 0 } 19

21 PART II Toeplitz+Hankel operators 20

22 Toeplitz+Hankel operators The goal is to develop, if possible, a Fredholm and invertibility theory (more generally: spectral theory) for Toeplitz + Hankel operators a, b L (T). T (a) + H(b), Questions: Is there relation to factorization theory? Can one establish explicit invertibility criteria, say, for a, b P C? Is there a geometric charaterization of the spectrum? 21

23 Fredholm theory for T (a) + H(b) with a, b P C N N One can establish explicit criteria for the Fredholmness of T (a) + H(b) with a, b P C N N. The conditions can be expressed in terms of a(t ± 0) and b(t ± 0) exclusively, similar as, but more complicated than in the Toeplitz case. Moreover, available: Fredholm index of T (a) + H(b) essential spectrum of T (a) + H(b) Difficult (open) problem: invertibility and spectrum 22

24 Equivalence after extension Notion: A L(X) and B L(Y ) are called equivalent after extension if there exist Banach spaces X 1 and Y 1 and invertible operators E : X X 1 Y Y 1, F : Y Y 1 X X 1 such that ( A 0 ) ( B 0 ) E 0 I X1 F = 0 I Y1. In this case: A invertible iff B invertible, A Fredholm iff B Fredholm, dim ker A = dim ker B, etc. 23

25 T+H operators block T-operators Let a, b L (T) and assume a 1 L (T). Then ( ) T (a) + H(b) 0 acting of H p H p 0 T (a) H(b) is equivalent after extension to T (Φ) acting of H p H p with Φ = ( a b 0 1 ) ( 1 0 b ã ) 1 = ( a bã 1 b bã 1 bã 1 ã 1 ) Remarks: Φ is triangular if aã = b b (important special case) Triangular matrix functions can (under certain conditions) be factored explicitly. Special case: a = b, i.e., T (a) + H(a). 24

26 Special Toeplitz+Hankel operators: T (a) + H(a) [Basor/E. 2004] 25

27 Identities for T (a) + H(a) Let us consider M(a) := T (a) + H(a). Recall T (ab) = T (a)t (b) + H(a)H( b) H(ab) = T (a)h(b) + H(a)T ( b) to derive the identity M(ab) = M(a)M(b) + H(a)M( b b) Hence M(ab) = M(a)M(b) if a H or b = b (i.e., b(t) = b(t 1 )). 26

28 This suggest to consider the asymmetric factorization: where a, a 1 H a 0 = ã 0 and a 0, a 1 0 L In this case: a = a da 0 M(a) = M(a )M(d)M(a 0 ) where M(a ) = T (a ) is invertible: T (a ) 1 = T (a 1 ) M(a 0 ) is invertible: M(a 0 ) 1 = M(a 1 0 ) 27

29 Trivial observation: even symbols Suppose a L (T) is even. Then M(a) is Fredholm (invertible) iff a 1 L (T). Hence sp(m(a)) = sp ess (M(a)) = R(a). Consequence: (Essential) spectrum of M(a) can be disconnected. (contrasting the case of T (a)) 28

30 Asymmetric factorization vs. Wiener-Hopf factorization (Heuristic) Assume a = a da 0 with a ±1 H and a ±1 0 L and a 0 = ã 0. Then i.e., aã 1 = a da 0 ã 1 0 d 1 ã 1 aã 1 = a rã 1 with middle factor r = d d 1 and ã ±1 H. = a d d 1 ã 1 29

31 Asymmetric factorization and Fredholmness of M(a) We say that a L possesses an asymmetric Φ-factorization in L p if with a(t) = a (t)t κ a 0 (t) (1 + t 1 )a (t) H p, (1 t 1 )a 1 (t) Hq, 1 t a 0 (t) L q even, 1 + t a 1 0 (t) Lp even, κ Z (index), the mapping f a 1 0 (I + J)P (a 1 extends to a bounded linear operator on H p. f) ( ) Here L p even := { b L p (T) : b = b } and 1 < p <, 1/p + 1/q = 1. 30

32 Theorem (Basor/ E. 2004): Let a L. Then M(a) = T (a) + H(a) is Fredholm in H p iff a admits an asymmetric Φ-factorization in H p. In this case dim ker M(a) = min{0, κ}, dim ker M(a) = max{0, κ} In particular: M(a) is invertible iff κ = 0. Remark: Condition ( ) is equivalent to an A p -condition for the function σ defined on [ 1, 1] by σ(cos x) = a 1 0 (eix ) (1 + cos x)1/2q (1 cos x) 1/2p 31

33 Asymmetric vs. Antisymmetric factorization Assume a, a 1 L. Then: a admits an asymmetric factorization in L p, a(t) = a (t) t κ a 0 (t), (1 + t 1 )a (t) H p, (1 t 1 )a 1 (t) Hq, 1 t a 0 (t) L q even, 1 + t a 1 0 (t) Lp even, aã 1 admits an anti-symmetric factorization in L p, a(t)ã 1 (t) = a (t) t 2κ ã 1 (t), (1 + t 1 )a (t) H p, (1 t 1 )a 1 (t) Hq. 32

34 Fredholmness of M(a), a P C Theorem. Let a P C. Then M(a) = T (a) + H(a) is Fredholm on H p iff a(t ± 0) 0 for all t T and 1 2π for all t T, Im(t) > π arg a(1 0) a(1 + 0) / 1 2p + Z arg a( 1 0) a( 1 + 0) / 1 2p Z 1 2π arg a(t 0)a( t 0) a(t + 0)a( t + 0) / 1 p + Z 33

35 Essential spectrum of M(a) Corollary: The essential spectrum of T (a) + H(a) on H p with a P C is equal to sp ess (T (a) =R(a) A 1/2p [a(1 0), a(1 + 0)] A 1/2p+1/2 [a( 1 0), a( 1 + 0)] t J H 1/p [a(t 0), a(t + 0), a( t 0), a( t + 0)]. where H 1/p (u, v, x, y) = { λ C : 1 2π arg (u λ)(x λ) (v λ)(y λ) 1 p + Z } 34

36 -i Example Fig. 3: p = i- 1+ i -1- i i i i+ -i- 1- -i -1+ -i -i image of some a P C Fig. 1: image of sp ess (M(a)); p = 2 Fig. 5: p =2Fig. 2: p =1.3 35

37 Fig. Fig. 1: image 3: p = of Fig. Fig. 2: p 4: =1.3 p = i -i -i Fig. p = 1: 2 image of Fig. Fig. 3: p 5: = p =2 -i -i p = Fig : p =1.3 Fig. Fig. 4: p 6: =1.34 p =5 i i+ -1 -i i -i-1+ -i -i i 1+ i i- i i i i i i i i i i i- i i -1- i i -i p = Fig. 1: Fig. image 3: p of= i p = 1.3 Fig. 2: Fig. p =1.3 4: p =

38 Spectrum of M(a), a P C For a P C and M(a) Fredholm on H p, the Fredholm index of M(a) can be determined geometrically. An analogue of Coburn s lemma holds: Theorem: Let a L, a 1 L. Then M(a) or M(a) has a trivial kernel. Corollary: Let a L. Then M(a) is invertible iff M(a) is Fredholm and Fredholm index of T (a) is zero. For a P C, the spectrum of M(a) can be described geometrically. 37

39 Generalizations Same methods as for T (a) + H(a) = (a j k + a j+k+1 ) also work for operators T (a) H(a) = (a j k a j+k+1 ) and T (a) + H(at) = (a j k + a j+k ) and T (a) H(at 1 ) = (a j k a j+k+2 ) (Results are similar) 38

40 Still special, but more general operators: T (a) + H(b) with aã = b b 39

41 Fredholmness of T (a) + H(b), a, b P C Theorem: Let a, b P C satisfy aã = b b and assume (wlog) a 1, b 1 P C. Let 1 < p <, 1/p + 1/q = 1. Put c = a b = b ã, d = ã b = b a. Then T (a) + H(b) is Fredholm on H p if and only if 1 2π arg c (1) / p + Z, 1 2π arg d (1) / q + Z, 1 2π arg c ( 1) / 1 2p + Z, 1 2π arg d ( 1) / 1 2q + Z, 1 2π arg ( c (τ) c + (τ) for each τ T, Im(τ) > 0. ) / 1 p + Z, 1 2π arg Notation: c ± (τ) = c(τ ± 0), d ± (τ) = d(τ ± 0). ( d (τ) d + (τ) ) / 1 q + Z, 40

42 Geometric Interpretation The function c P C satisfies c c = 1. Construct a curve c #,p as follows: consider the image of c(e ix ), 0 < x < π curve with jumps fill in the arc A 1/2+1/2p (1, c + (1)) (if 1 c + (1)) fill in the arcs A 1/p (c (τ), c + (τ)) (whenever c (τ) c + (τ), Im(τ) > 0) fill in the arc A 1/2p (c ( 1), 1) (if 1 c (1)) This gives a closed oriented curve c #,p. Similar construction to obtain a closed oriented curve d #,q. 41

43 Under the assumptions on a, b as in the theorem: T (a) + H(b) is Fredholm on H p iff 0 does not lie on any of the curves c #,p and d #,q. Moreover,. ind(t (a) + H(b)) = wind(d #,q ) wind(c #,p ) 42

44 Example: Consider function c P C, c c = 1 whose image c(e ix ), 0 < x < π is as follows: i c i 0 c i 0 c 1 0 c i 43

45 i c i 0 i i c i 0 c 1 0 c i c(e ix ), 0 < x < π p = 2 i p = 1.5 i i i i p = 4/3 i p = 1.16 i p = 1.13 i 44

46 Factorization results: Theorem (Basor/ E. 2013): Let a, b L satisfy aã = b b, and assume a 1, b 1 L. Define the auxiliary functions c = a b = b ã, d = ã b = b a. Now assume that c and d have factorizations of the form with n, m Z and 1 c = c + t 2n c +, d = d +t 2m d < p <, 1/p + 1/q = 1. (1 + t)c + H q, (1 t)c 1 + Hp (1 + t)d + H p, (1 t)d 1 + Hq 45

47 Now also assume that T (a) + H(b) are Fredholm operators on H p. Then dim ker(t (a) + H(b)) = 0 if n > 0, m 0 n if n 0, m 0 dim ker A n,m if n > 0, m > 0 m n if n 0, m > 0, dim ker(t (a) + H(b)) = Therein, in case n > 0, m > 0, 0 if m > 0, n 0 m if m 0, n 0 dim ker(a n,m ) T if m > 0, n > 0 n m if m 0, n > 0. and A n,m := [ ] n 1 m 1 ρ i j + ρ i+j i=0 j=0. ρ := t m n (1 + t)(1 + t 1 ) c + d + b 1 L 1 (T). In particular, the Fredholm index of T (a) + H(b) is equal to m n. 46

48 Remarks: Factorizations of c, d exist if T (a) + H(b) is Fredholm and a, b P C. They can be constructed explicitly. n = wind(c #,p ), m = wind(d #,q ) In general (a, b / P C) such factorizations need not exist, even if T (a) + H(b) is invertible!!! Corollary: Assume all of the above (in particular Fredholmness of T (a) + H(b)). Then T (a) + H(b) is invertible on H p iff n = m = 0, or n = m > 0 and A n,n is an invertible matrix. 47

49 What s missing...? Nonetheless, we have no description of spectrum of yet!!! T (a) + H(b) with aã = b b, a, b P C Reason: T (a) + H(b) λi = T (a λ) + H(b λ), but (ã λ)(a λ) ( b λ)(b λ) in general. 48

50 Thank you! 49