Properties of truncated Toeplitz operators

Transcription

1 Washington University in St. Louis Washington University Open Scholarship All Theses and Dissertations (ETDs January 2010 Properties of truncated Toeplitz operators Nicholas Sedlock Washington University in St. Louis Follow this and additional works at: Recommended Citation Sedlock, Nicholas, "Properties of truncated Toeplitz operators" (2010. All Theses and Dissertations (ETDs This Dissertation is brought to you for free and open access by Washington University Open Scholarship. It has been accepted for inclusion in All Theses and Dissertations (ETDs by an authorized administrator of Washington University Open Scholarship. For more information, please contact

2 WASHINGTON UNIVERSITY Department of Mathematics Dissertation Examination Committee: Richard Rochberg, Chair Albert Baernstein II Ramanath Cowsik Kabe Moen Joseph A. O Sullivan Xiang Tang PROPERTIES OF TRUNCATED TOEPLITZ OPERATORS by Nicholas Alexander Sedlock A dissertation presented to the Graduate School of Arts and Sciences of Washington University in partial fulfillment for the degree of Doctor of Philosophy May 2010 Saint Louis, Missouri

3 copyright by Nicholas Alexander Sedlock 2010 ii

4 Acknowledgments Richard Rochberg taught me almost everything I know about operator theory and functional analysis, and most of what I know about real analysis, and has patiently encouraged me in my studies these last five years. He has helped me every step of the way through the process of my research. I could never have done any of this without him. John McCarthy taught me everything else I know about operator theory and functional analysis, and provided generous financial support over the last few summers, which were some of the most productive periods of my research life. Bill Ross and Warren Wogen gave several notes, suggestions and corrections to these results, for which I give my thanks. My parents, William and Barbara, sacrificed a lot of time and money to make sure I got the best education they could provide for me, and have patiently waited for me to stop being a student and finally get a real job. Thanks to Don Knuth for writing TEX and to Leslie Lamport for writing L A TEX, thus saving me a great deal of time these last 8 months. Finally I would like to thank my wife Holly for sticking with me through my highs and lows, and giving me much needed encouragement when I was at my most discouraged. iii

5 Contents Acknowledgments iii Abstract vi 1 Introduction 1 2 Toeplitz operators on H Ku 2 and truncated Toeplitz operators The Hilbert space Ku C symmetry and Ku Some algebraic properties of truncated Toeplitz operators The H functional calculus Clark operators and Clark measures Crofoot transforms Results S u C Two conditions for the product of two TTOs to be a TTO Algebras of TTOs iv

6 4.4 Generalized shifts Bounded interpolation of TTOs α = α < Invertible TTOs of type α and their inverses α = α < Eigenvalues and eigenspaces of TTOs of type α α = α < A Other Results 49 A.1 Self-adjoint TTOs A.2 Model spaces of dimension A.3 Sums of TTOs of type α A.4 Finite rank TTOs A.5 Commutants of rank-one TTOs B u = z N 56 Bibliography 59 v

7 Abstract Let A Φ be a truncated Toeplitz operator the compression of the Hardy space Toeplitz operator T Φ to the model space H 2 uh 2, where u is a nonconstant inner function. We find a necessary and sufficient condition that the product A Φ1 A Φ2 is itself a truncated Toeplitz operator. Specifically, we show that there are algebras of truncated Toeplitz operators B α (depending on α C such that two truncated Toeplitz operators have a truncated Toeplitz operator as a product if they are both in the same B α. Some consequences of this are also discussed. vi

8 Chapter 1 Introduction Let C denote the complex plane, C the Riemann sphere, D denote the unit disc, and let T denote the unit circle. H 2 is the usual Hardy space, the subspace of L 2 (T of normalized Lebesgue measure m on T whose harmonic extensions to D are holomorphic (or, whose negative indexed Fourier coefficients are all zero. H 2 will interchangably refer to both the boundary functions and the functions on D. Let P denote the projection from L 2 (T to H 2, which is given explicitly by the Cauchy integral: (P f(λ = T f(ζ 1 λζ dm(ζ, λ D By this expression, it makes sense to think of P as an operator from L 1 (T into Hol(D, the space of holomorphic function on D, which is continuous relative to the weak topology of L 1 (T and the topology of locally uniform convergence of Hol(D. We also see that the reproducing kernel at λ D for the Hardy space is the the Szego kernel K λ := (1 λz 1. Let S denote the shift operator f zf on 1

9 H 2. Its adjoint (the backwards shift is the operator S f = f f(0 z A Toeplitz operator is the compression of a multiplication operator on L 2 (T to H 2. In other words, given Φ L 2 (T (called the symbol of the operator, T Φ = P M Φ is the operator that sends f to P (Φf for all f H 2. This operator is bounded if and only if Φ L (T, and the mapping Φ T Φ from L to the space of bounded operators on H 2 is linear and one-to-one. In the case that Φ H, the Toeplitz operator T Φ is just the multiplication operator M Φ. In [BH64], Brown and Halmos describe the algebraic properties of Toeplitz operators. Among other things, they found necessary and sufficient conditions for the product of two Toeplitz operators to itself be a Toeplitz operator, namely that either the first operator s symbol is antiholomorphic or the second operator s symbol is holomorphic. In either case, the symbol of the product is the product of the symbols (ie T Φ T Ψ = T ΦΨ. From this they derive several results about when a Toeplitz operator is invertible, unitary, or idempotent, and when the product of two Toeplitz operators is the zero operator. More recently, Sarason [Sar07] found equivalents to several of Brown and Halmos s results for truncated Toeplitz operators on the model spaces H 2 uh 2, where u is some non-constant inner function. The model spaces are the backward-shift invariant subspaces of H 2 (that they are backward shift invariant follows easily from the fact that uh 2 is clearly shift invariant. Let Ku 2 denote the space H 2 uh 2 from here forward. Let P u = P M u P M u denote the projection from L 2 to Ku. 2 Given Φ L 2 (T we then define the truncated Toeplitz operator (TTO A Φ to be the operator that sends f to P u (Φf for all f Ku. 2 Truncated Toeplitz 2

10 operators have many of the same properties as ordinary Toeplitz operators ( for example, A Φ = A Φ but there are also striking differences. For example, there are bounded truncated Toeplitz operators with unbounded symbols (though any truncated Toeplitz operator with a bounded symbol is itself bounded. Additionally, symbols are not unique: the same operator can be generated from more than one symbol, and we say that Ψ is a symbol for A Φ if A Φ = A Ψ. More background about model spaces and truncated Toeplitz operators can be found in Chapter 3. One result from the Brown-Halmos paper which does not have an equivalent in Sarason s paper is the necessary and sufficient condition for the product of two truncated Toeplitz operators to itself be a truncated Toeplitz operator. It is easy to see that the product of two bounded TTOs with holomorphic symbols is itself a TTO, but the general problem is more delicate. Our results can be found in Chapter 4. Specificaly, in Sections 4.2 and 4.3 we find two necessary and sufficient conditions for the product of two TTOs to itself be a TTO, the first of which is a condition based on the symbol of the operators in question, the second of which identifies a C -indexed family of algebras of TTOs and shows that if two TTOs have a TTO for a product, then all three operators lie in the same algebra. In Section 4.4 we show that these algebras are the commutants of certain rank-one perturbations of the compressed shift and their adjoints. In Section 4.5 we also show that the TTOs in these algebras have bounded symbols in most, but not all, cases, and in all cases find a symbol algebra for the products of TTOs. Finally, we discuss invertible TTOs in Section 4.6 and eigenvectors and eigenvalues of TTOs in Section 4.7. There are two appendices. Appendix A contains results not directly related to the main results in Chapter 4, and Appendix B applies the results of Chapter 4 to 3

11 the case u = z n in which TTOs are classical Toeplitz matrices. In what follows, I refers to the identity operator on whatever space we re considering, f, g := fg dm for all f, g T L2 (T, and f := f, f. Further, for f, g in a Hilbert space, f g represents the rank one operator f g(h := f h, g. 4

12 Chapter 2 Toeplitz operators on H 2 In [BH64], Brown and Halmos provide many results about Toeplitz operators and how they behave algebraically. For example, an operator T on H 2 is a Toeplitz operator if and only if T = S T S. These results motivate our questions about TTOs as well as the work of others. If ϕ L (T, then T ϕ is bounded. For ϕ L 2 (T T ϕ is densely defined (on the polynomials. If T ϕ is bounded, however, it follows that ϕ is also bounded. Proposition Let ϕ L 2 (T such that T ϕ is a bounded operator. Then ϕ = T ϕ. Proof. Let k λ = 1 λ 2 K λ denote the normalized reproducing kernel at λ D, and note k λ 2 = k λ, k λ = (1 λ 2 K λ (λ = 1. Thus T ϕ k λ, k λ T ϕ 5

13 By the definition of the L 2 (T inner product we have T ϕ k λ, k λ = ϕ(ζ 1 λ 2 dm(ζ = ϕ(λ 1 λζ 2 where ϕ(λ is the Poisson extension of ϕ evaluated at λ. Hence ϕ(λ T ϕ for all λ and so ϕ < T ϕ. Now let ψ L (T. Then M ψ = ψ on L 2 (T and hence T ψ = P M ψ ψ. The conclusion follows. The set of all bounded Toeplitz operators is not an algebra: Proposition T Φ T Ψ is itself a Toeplitz operator if and only if either Ψ or Φ is holomorphic. In this case, T Φ T Ψ = T ΦΨ. From this several facts follow, proofs of which are in Brown and Halmos s paper: Corollary The product of two Toeplitz operators is zero if and only if one of the two operators is itself zero. 2. Suppose that T Φ is invertible. Then (T Φ 1 is a Toeplitz operator if and only if Φ is holomorphic or antiholomorphic. 3. The only unitary Toeplitz operators are the operators T c for c T. 4. T Φ is idempotent if and only if Φ = 0 or 1. A related question is when two Toeplitz operators commute: Proposition Two Toeplitz operators T Φ and T Ψ commute if and only if both Φ and Ψ are holomorphic or both are antiholomorphic, or if one is a linear function of the other. 6

14 Corollary 2. The only normal Toeplitz operators are linear functions of selfadjoint ones. 7

15 Chapter 3 K 2 u and truncated Toeplitz operators 3.1 The Hilbert space K 2 u From here forward, let u be a non-trivial inner function. K 2 u is then a reproducing kernel Hilbert space with reproducing kernels K u λ := P uk λ = 1 u(λu 1 λz Note that K u λ for λ D. is bounded for all λ and since the span of the reproducing kernels is dense in K 2 u, K u := L (T K 2 u is dense in K 2 u as well. Thus for any Φ L 2, A Φ is densely defined, since its domain contains K u. The function u is said to have an angular derivative in the sense of Carathádory (ADC at the point ζ T if u has a nontangential limit u(ζ of unit modulus at ζ and u has a nontangential limit u (ζ at ζ. It is known that u has an ADC at ζ if and only if every function in K 2 u has a nontangential limit at ζ [Sar94]. Thus there exists a reproducing kernel function K u ζ such that f, K u ζ = f(ζ. Specifically, K u ζ is the limit of K u λ as λ approaches ζ nontangentially in the disc and so Ku ζ = 1 u(ζu 1 ζz. 8

16 If u is a finite Blaschke product, both u and u are holomorphic in a domain which compactly contains D and so these boundary reproducing kernels are defined for every unimodular ζ. Just at S = T z, define S u := A z. Then Su = A z is simply the ordinary backwards shift, since Ku 2 is backwards shift invariant. H 2 = Ku uh 2 2, and by using this fact we can decompose H 2 into the direct sum of countably many disjoint subspaces generated by Ku 2 using Halmos Wandering Subspace lemma [Hal61], which states that if U is an isometry on a Hilbert space H and K = H UH, then ( ( H = U n K U n H n=0 n=0 Proposition H 2 = u n Ku 2 n=0 Proof. The operator M u is an isometry on H 2 and so we have that ( ( H 2 = u n Ku 2 u n H 2 n=0 Suppose f n=0 un H 2. If u has a zero at λ D then f has a zero of order at λ and therefore f 0. If, on the other hand, u is singular, then n=0 ( 2π e it + z u(z = ζ exp 0 e it z dµ(t 9

17 where ζ T and µ is a bounded positive singular measure. It follows that ( 2π u n e it + z (z = ζ exp 0 e it z dnµ(t where nµ is a bounded positive singular measure. Let ( 2π e it + z S(z = exp 0 e it z dν(t be the singular inner factor of f, where ν is a bounded positive singular measure. If f u n H 2, then it follows that u n divides S, which implies that ν nµ is a bounded positive singular measure. Since ν is positive and bounded and µ is positive, it follows that there is sufficienly large n such that this is not the case, and as a result f is not divisible by u n, a contradiction. Therefore it follows that n=0 un H 2 = {0} and the claim follows. 3.2 C symmetry and K 2 u In [Gar06, GP06, GP07] C-symmetric operators are introduced. Given a C-Hilbert space H and an antilinear isometric involution C on H, we say that a bounded operator T is a C-symmetric operator (CSO if T = CT C. Here by isometric we mean that Cf, Cg = g, f. C is called a conjugation operator because if we look, for example, at the space C n and define C to be pointwise complex conjugation, a bounded operator M on C n is C-symmetric if it is complex symmetric as a matrix. In L 2 (T, the operator Cf = uzf is a conjugation which bijectively maps uh 2 to zh 2 and Ku 2 to itself, and so C can be thought of as a conjugation on Ku. 2 From here on, C always refers to this operator. We will sometimes write f for Cf for 10

18 sake of readability. An operator that will come up frequently in what follows is the conjugate reproducing kernel Ku λ = u(z u(λ. By the definition of a conjugation, z λ we can see that for f K 2 u, f(λ = Ku λ, f. As it turns out, truncated Toeplitz operators are C-symmetric. The following result is Garcia and Putinar s: Lemma For Φ L 2 (T such that A Φ is bounded, CA Φ C = A Φ. Proof. Let f, g K 2 u. Then CA Φ Cf, g = Cg, A Φ Cf = uzg, Φuzf = Φf, g Necessary and sufficient conditions for the product of two CSOs to be a CSO are straightforward. Proposition Let A and B be CSOs on some Hilbert space H. Then AB is a CSO iff AB = BA iff (AB = A B. Proof. We will show 1 = 2 = 3 = 1. (1 = 2 AB = C 2 ABC 2 = C(AB C = CB A C = C 2 BC 2 AC 2 = BA. (2 = 3 (AB = (BA = A B. (3 = 1 CABC = CAC 2 BC = A B = (AB. 11

19 3.3 Some algebraic properties of truncated Toeplitz operators We will need a number of technical lemmas which can be found in [Cla72, Sar67, Sar07] which we include here for reference. The following is Theorem 4.1 in [Sar07], which gives us a Brown-Halmos-like characterization of the truncated Toeplitz operators. Fact A is a TTO iff A S u ASu = Φ K0 u + K0 u Ψ for some Φ, Ψ Ku, 2 in which case A = A Φ+Ψ, and hence if Φ L 2 (T then A Φ = 0 if and only if Φ uh 2 uh 2. Definition For two functions Φ and Ψ in L 2 (T say Φ A Ψ if A Φ = A Ψ. In what follows, when dealing with a TTO A we will usually use a symbol of the form ϕ 1 + ϕ 2 where ϕ i Ku. 2 This symbol is not unique. For example, A ϕ1 +ϕ 2 = A ϕ1 +ck0 u+ϕ 2 ck for all c C, but 0 u c(ku 0 K0 u 0 if u(0 0. The following is a necessary and sufficient condition for a TTO to be zero. Proposition Let ϕ 1, ϕ 2 K 2 u. Then A ϕ1 +ϕ 2 = 0 if and only if ϕ 1 = ck u 0 and ϕ 2 = ck u 0 for some c C. Proof. Let ϕ 1 = ck u 0 and ϕ 2 = ck u 0. Then A ϕ1 +ϕ 2 = A ck u 0 ck u 0 = A cu(zu(0 cu(zu(0 so A ϕ1 +ϕ 2 = 0. Now suppose A ϕ1 +ϕ 2 = 0. Then A S u AS u = 0 = ϕ 1 K u 0 + K u 0 ϕ 2, so ϕ 1 = ck u 0 for some c C. Hence ck u 0 K u 0 + K u 0 ϕ 2 = 0 and so ϕ 2 = ck u 0 as required. 12

20 Proposition I = A 1 = A K u 0 = A K u 0 Proof. Let f K 2 u. Then A 1 f = P (1 f = f, so the first equality holds. Now A 1 A K u 0 = A 1 1+u(0u(z = A u(0u(z = 0, so the second equality holds. Finally, this implies that A K u 0 is self-adjoint, and so the third equality holds. Corollary 3. Let ϕ and ψ be in H 2 such that A ϕ+ψ is bounded, and let c C. Then A ϕ+ψ+ck u 0 = A ϕ+ψ+c = A ϕ+ψ+ck u 0. Fact For λ D, S uk u λ = λk u λ u(λ K u 0, S u Ku λ = λ K u λ u(λku 0 2. For nonzero λ D, S u Kλ u = 1 λ (Ku λ K0 u, Su K λ u = 1 ( Ku λ λ K 0 u 3. These equalities all hold for λ T such that u has an ADC at λ. Fact I S u S u = K u 0 K u 0 2. I S us u = K u 0 K u 0 The only compact TTO in H 2 is the zero operator. In K 2 u, however, there are many finite rank TTOs. Fact

21 1. Let λ D. Then K λ u Ku λ is a TTO. If λ D, then K u λ Ku λ = A u z λ 2. Let λ T such that u has an ADC at λ. Then K u λ K u λ = A K u λ +K u λ 1 3. Let λ D (or let λ T such that u has an ADC at λ. Then n 1 ( ( n 1 d j Ku λ j dλ j j=0 dn j 1 K u λ dλ n j 1 is a TTO. If λ D, then n 1 ( ( n 1 d j Ku λ j dλ j j=0 dn j 1 K u λ dλ n j 1 = A (n 1!u (z λ n 3.4 The H functional calculus Definition A TTO A is of holomorphic type if there is a function ϕ K 2 u such that A = A ϕ. TTOs of anti-holomorphic type are therefore the adjoints of TTOs of holomorphic type. Corollary 4. Let ϕ, ψ K 2 u. Then A ϕ+ψ is of holomorphic type if and only if ψ = ck u 0 for some c C. type. The product of two TTOs of holomorphic type is itself a TTO of holomorphic 14

22 Proposition Let ϕ, ψ H 2 such that A ϕ, A ψ are bounded. Then A ϕ A ψ = A Pu[ϕP uψ], and so A ϕ A ψ = A Pu[ϕP uψ]. Proof. We proceed using Fact A ϕ A ψ S u A ϕ A ψ S u = A ϕ A ψ (I S u S u = A ϕ A ψ (K u 0 K u 0 = (A ϕ A ψ K u 0 K u 0 = (A ϕ P u ψ K u 0 = (P u [ϕp u ψ] K u 0 Thus the TTOs of holomorphic type form an algebra. It turns out that this algebra is precisely the commutant of the compressed shift S u. Details are laid out in [Sar67, Saw09]. We reproduce those results here for reference. Fact Let ϕ H. Then 1. A ϕ ϕ. 2. The map ϕ A ϕ is linear and multiplicative. 3. If ψ is the greatest common inner divisor of u and the inner factor ofϕ, then ker A ϕ = u ψ H2 uh 2 so in particular A ϕ = 0 if and only if ϕ uh 2 and A ϕ is injective if and only if the inner factor of ϕ and u are relatively prime. 15

23 This is Sarason s characterization of the commutant of S u. Fact Let A be a bounded operator on K 2 u that commutes with S u. Then there is a bounded function ϕ H such that A = A ϕ and A = ϕ. Recall that the spectrum of an operator A on a Hilbert space H is the set {λ C : (λi A is not invertible on H}. Fact λ is in the spectrum of A ϕ if and only if inf z D ( u(z + ϕ(z λ = Clark operators and Clark measures In [Cla72], Clark unitary operators and Clark measures are introduced. For reference we include a summary of these results based on their treatment in [CMR98], slightly modified to better coincide with our main results. Definition Let α T. Define U α = S u + unitary operator. α 1 αu(0 Ku 0 K 0 u. Then U α is a The spectral theorem says that U α is therefore unitarily equivalent to M z on L 2 (T, µ α for some Borel µ α. We call µ α a Clark measure. The Lebesgue decomposition of µ α with respect to m has no continuous part, so µ α = (µ α d + (µ α s where (µ α d is discrete and (µ α s is continuous and singular. Fact Let E α be the subset of T where u(ζ = α and let E be the subset of E α where u does not have an angular derivative. Then (µ α d = ζ e α\e δ ζ u (ζ 16

24 If u is a finite Blaschke product, then µ α = (µ α d. From this one can deduce that the eigenvalues of U α are the ζ T such that u(ζ = α and u (ζ, and that their associated eigenvectors are the functions K u ζ. Define V α : L 2 (T, µ α Hol(D by letting V α (K λ = Kλ u /(1 αu(λ and extending to the whole domain. Then the following is true. Fact V α M z = U α V α 3.6 Crofoot transforms Let α D. Define u α = u α 1 αu. Then T α = T (1 α 2 1/2 (1 αu is a bounded operator from H 2 to itself. More is true, however. The following is due to Crofoot [Cro94], which discusses the idea of invertible maps between model spaces more generally. Fact T α is an unitary map from K 2 u α onto K 2 u. 17

25 Chapter 4 Results 4.1 S u C In what follows, the operator S u C will feature prominently. Note that since S u is C-symmetric, we have that S u C = CS u, and that therefore, since S u = S restricted to K 2 u we can explicitly compute S u Cf. Proposition Let f K 2 u. 1. S u C span(k u 0 is an antilinear involutive isometry. 2. S u Cf = u(f f(0. 3. P u ( uf = Su f. 4. If u(0 = 0, Ker S u C = sp (K u 0. If u(0 0, Ker S u C = {0}. 5. If S u f = K u 0, then u(0 0 and f = ck u 0 for some c C. 6. If S u f = ck u 0, then f sp (K u 0. Proof. 18

26 1. Let f Ku 2 with f(0 = 0. Then S u f(0 = S u Ku 0, f = 0 and so S u C maps K 2 u sp (K u 0 to itself. Antilinearity is obvious. Let f, g sp (K u 0. Then S u CS u C = S u S u = I K u 0 K u 0, so if f K u 0, S u CS u Cf = f and so S u C is an involution. Finally, S u f, Su g = CSuf, CSug = Sug, Suf = S u Sug, f = g, f 2. S u Cf = CS uf = C(z[f f(0] = u(f f(0. 3. uf = S u f + uf(0. 4. Follows from Fact If u(0 = 0, then ker S u C = C(uzH 2 uh 2. Thus elements of ker S u C are of the form g where g is a holomorphic function. Hence g is constant, which means ker S u C = sp (K u 0. If u(0 0, then ker S u = {0} and the conclusion follows. 5. First note that if u(0 = 0 then S u f(0 = S u f, K u 0 = f, S uk0 u = f, 0 = 0 for all f K 2 u. So since K u 0 (0 = 1, u(0 0. Thus it remains to show that if S u f = K u 0, then f = ck u 0. Since S u C( K u 0 /u(0 = K u 0 by Fact we have f + Ku 0 u(0 Ker S uc = sp (K u 0, and the conclusion follows. 6. Follows from (4 and (5. Proposition Let Φ = ϕ 1 + ϕ 2, ϕ i K 2 u such that A Φ is bounded. Then A Φ K u 0 = ϕ 1 + ϕ 2 (0K u 0 u(0s u ϕ 2 and A Φ Ku 0 = ϕ 2 + ϕ 1 (0 K u 0 u(0s uϕ 1. 19

27 Proof. ( ] A Φ K0 u = P u [(ϕ 1 + ϕ 2 1 u(0u ] = P u [ϕ 1 + ϕ 2 u(0ϕ 1 u u(0ϕ 2 u = ϕ 1 + P u (ϕ 2 (0 u(0p u (uϕ 2 = ϕ 1 + ϕ 2 (0K u 0 u(0s u ϕ 2 The second equation follows from the first. 4.2 Two conditions for the product of two TTOs to be a TTO We are interested in the case that A Φ A Ψ is a TTO. Here is a necessary and sufficient condition for this to be true. Lemma Let Φ = ϕ 1 + ϕ 2 and Ψ = ψ 1 + ψ 2 where ϕ i, ψ i K 2 u such that A Φ, A Ψ are bounded. Then A Φ A Ψ is a TTO if and only if ϕ 1 ψ 2 (S u ϕ 2 (S u ψ1 = Φ 0 K u 0 + K u 0 Ψ 0 for some Φ 0, Ψ 0 K 2 u. Proof. In what follows, Φ 0 and Ψ 0 represent functions in Ku 2 that can be different from use to use. By Fact 3.3.1, A Φ A Ψ is a TTO if and only if A Φ A Ψ S u A Φ A Ψ Su = Φ 0 K0 u + K0 u Ψ 0. It suffices to show that A Φ A Ψ S u A Φ A Ψ Su = ϕ 1 ψ 2 (S u ϕ 2 (S u ψ1 + Φ 0 K0 u + K0 u Ψ 0. By Fact 3.3.6, I = SuS u + K 0 u K 0 u, and 20

28 by Fact we have that S u A Φ Su = A Φ ϕ 1 K0 u K0 u ϕ 2 and so A Φ A Ψ S u A Φ A Ψ Su = A Φ A Ψ (S u A Φ Ku0 (S u A Ψ Ku0 (A Φ ϕ 1 K0 u K0 u ϕ 2 (A Ψ ψ 1 K0 u K0 u ψ 2 For sake of readability, we simplify each term in the right hand side separately. First the second term. S u A Φ Ku 0 = S u ( ϕ 2 + ϕ 1 (0 K u 0 u(0s uϕ 1 = S u ϕ 2 u(0ϕ 1 (0K u 0 u(0s u S uϕ 1 = S u ϕ 2 u(0ϕ 1 (0K u 0 u(0ϕ 1 + u(0 (K u 0 K u 0 ϕ 1 = S u ϕ 2 u(0ϕ 1 (0K u 0 u(0ϕ 1 + u(0ϕ 1 (0K u 0 = S u ϕ 2 u(0ϕ 1 The first equality is by Proposition Therefore, (S u A Φ Ku0 (S u A Ψ Ku0 = (S u ϕ 2 u(0ϕ 1 (S u ψ1 u(0ψ 2 ] = S u ϕ 2 S u ψ1 u(0 [ϕ 1 S u ψ1 u(0 [S u ϕ 2 ψ 2 ] + u(0 2 [ϕ 1 ψ 2 ] Next, the third term. 21

29 (A Φ ϕ 1 K u 0 K u 0 ϕ 2 (A Ψ ψ 1 K u 0 K u 0 ψ 2 = A Φ A Ψ (A Φ ψ 1 K u 0 (A Φ K u 0 ψ 2 ϕ 1 (A Ψ K u 0 + ψ 1 (0 (ϕ 1 K u 0 + ( 1 u(0 2 ϕ 1 ψ 2 K u 0 (A Ψ ϕ 2 + ψ 1, ϕ 2 (K u 0 K u 0 + ϕ 2 (0 (K u 0 ψ 2 = A Φ A Ψ (A Φ ψ 1 K u 0 + ψ 1 (0 (ϕ 1 K u 0 K u 0 (A Ψ ϕ 2 + ψ 1, ϕ 2 (K0 u K0 u + ϕ 2 (0 (K0 u ψ 2 + ( 1 u(0 2 ϕ 1 ψ 2 (ϕ 1 + ϕ 2 (0K u0 u(0s u ϕ 2 ψ 2 ϕ 1 (ψ 2 + ψ 1 (0K u0 u(0s u ψ1 Grouping the F K u 0 and K u 0 G terms together, we get A Φ A Ψ + [ ψ 1, ϕ 2 K u 0 A Φ ψ 1 ] K u 0 K0 u (A Ψ ϕ 2 ( 1 + u(0 2 ϕ 1 ψ 2 + u(0 (S u ϕ 2 ψ 2 + u(0 (ϕ 1 S u ψ1 By combining the expanded terms together, we get A Φ A Ψ S u A Φ A Ψ S u = ϕ 1 ψ 2 S u ϕ 2 S u ψ1 + [A Φ ψ 1 ψ 1, ϕ 2 K u 0 ] K u 0 + K u 0 (A Ψ ϕ 2 and the result follows. In fact, we have found the symbol of the product of two TTOs in the event 22

30 that their product is a TTO. Proposition If Φ = ϕ 1 +ϕ 2 and Ψ = ψ 1 +ψ 2 where ϕ i, ψ i Ku 2 and A Φ, A Ψ are bounded, and ϕ 1 ψ 2 (S u ϕ 2 (S u ψ1 = Φ 0 K0 u +K0 u Ψ 0 for some Φ 0, Ψ 0 Ku, 2 then A Φ A Ψ is the TTO with symbol A Φ ψ 1 ψ 1, ϕ 2 K0 u + A Ψ ϕ 2 + Φ 0 + Ψ 0 Lemma Let A Φ A Ψ be a TTO. If one of the two operators is of holomorphic (resp. antiholomorphic type, then either that operator is actually ci or the other operator is also of holomorphic (resp. antiholomorphic type. Proof. Since A Φ A Ψ be a TTO, A Φ and A Ψ commute, and by taking adjoints we have that A Φ A Ψ is a TTO as well. Thus without loss of generality we suppose A Φ is of holomorphic type. We will show that either A Φ = ci or A Ψ is of holomorphic type. Let ϕ, ψ 1, ψ 2 K 2 u such that Φ A ϕ and Ψ A ψ 1 + ψ 2. Then A φ A ψ1 +ψ 2 is a TTO and so by Lemma ϕ ψ 2 = Φ 0 K u 0 + K u 0 Ψ 0 for some Φ 0, Ψ 0 K 2 u. So either Φ 0 = c 1 K u 0 or Ψ 0 = c 2 K u 0. If Φ 0 = c 1 K u 0, then it follows that ϕ = c 3 K u 0 and A Φ = ci. Similarly, if Ψ 0 = c 2 K u 0, then it follows that ψ 2 = c 4 K u 0 and so A Ψ is of holomorphic type. Definition For α C, B α := { } A ϕ+αsu ϕ+c ϕ K2 u, c C with B understood to mean the vector space {A ϕ+c ϕ Ku, 2 c C}. Note that this makes B 0 the vector space of TTOs of holomorphic type and B the vector space of TTOs of antiholomorphic type. An operator is of type α if it is in B α. That B α is a vector space for α C \ {0} is straightforward to see since the operator f S u f is C-linear and additive. The following is a useful alternative symbol for the operators in B α. Proposition If A Φ is of type α, then there exists ϕ 0 Ku 2 and c C such that ϕ 0 (0 = 0 and A Φ = A ϕ0 +αs u ϕ 0 +c 23

31 Proof. By definition, Φ A ϕ + αs u ϕ + c 1 for some ϕ K 2 u and c 1 C. Rewrite ϕ = ϕ 0 + c 2 K u 0, where ϕ K 2 u, ϕ 0 (0 = 0 and c 2 C. Then S u ϕ = S u C(ϕ 0 + c 2 K u 0 = S u ϕ 0 + c 2 S u Ku 0 = S u ϕ 0 c 2 u(0k u 0 Then by Proposition the result follows. Definition Let ϕ K 2 u and c C. Define B α ϕ+c := A ϕ+c+αcs (ϕ+c = A ϕ+αsu ϕ+c = Bα ϕ + ci for α C \ {0}. ϕ + c is the B α -symbol of the operator B if B = B α ϕ+c. Proposition Let B be of type α for some α C. Then B is of type 1/α using the convention 1/0 = and 1/ = 0. Proof. If α = 0 or this is obvious, so assume α C \ {0}. There exists ϕ K 2 u with ϕ(0 = 0 and c C such that B = B α ϕ+c = A ϕ+c+αsu ϕ. Now S u S u ϕ = ϕ since ϕ(0 = 0. So let χ = αs u ϕ. It follows that ϕ = 1 α S u χ and so B = A ϕ+c+αsu ϕ = A χ+c+ 1 Su χ α = B 1/α χ+c B1/α Proposition Bϕ α = ci if and only if ϕ sp (K0 u. Specifically, BK α = 0 ( u 1 αu(0 I, and so if 1 αu(0, then any TTO of type α can be written in 24

32 the form B α ϕ, ϕ K 2 u. More generally, B α ϕ+c = A (ϕ+c(1+αu for any ϕ K 2 u and c C. Proof. Say B α ϕ ci = A ϕ+αsu ϕ ck u 0 = 0 Then by Proposition 3.3.3, ϕ ck0 u = c 2 K0 u. In the other direction, BK α = 0 ( u A K u = A 0 +αs u Ku K u 0 0 αu(0k = 1 αu(0 I by Proposition Finally, A 0 u K u 0 (1+αu = A = A (1 u(0u(1+αu 1 u(0u+αu αu(0 (1 = αu(0 I and the result follows. Write ϕ = ϕ 0 + c 1 K u 0. B α ϕ 0 +c 1 K u 0 +c = B α ϕ 0 + B α c 1 K u 0 + Bα c = A ϕ0 +αs u ϕ 0 + c 1 B α K u 0 + ci = A ϕ0 +αuϕ 0 + A c1 K u 0 (1+αu + A c(1+αu The conclusion follows. The following is a method for determining when a TTO is of type α. Proposition Let A := A ϕ1 +ϕ 2 be bounded, where ϕ i K 2 u. 1. If α C, then A is of type α if and only if αs u ϕ 1 ϕ 2 sp (K u A is of type if and only if ϕ 1 sp (K u 0. Proof. 1. Let A ϕ1 +ϕ 2 be of type α. Then there is some ϕ K 2 u and c C such that A ϕ1 +ϕ 2 = A ϕ+ck u 0 +αs u ϕ. Thus we have A ϕ 1 ϕ ck u 0 +ϕ 2 αs u ϕ = 0. So by Proposition we have that ϕ 1 ϕ sp (K u 0 and that ϕ 2 αs u ϕ 25

33 sp (K u 0. So then by Fact 3.3.5, we have that S u ϕ 1 S u ϕ sp (K u 0 and so αs u ϕ 1 αs u ϕ ϕ 2 + αs u ϕ = αs u ϕ 1 ϕ 2 sp (K u 0. Now suppose that αs u ϕ 1 ϕ 2 sp (K u 0. Then ϕ 2 = αs u ϕ 1 + ck u 0 for some c C and thus A ϕ1 +ϕ 2 = A ϕ1 +αs u ϕ 1 +ck u 0 is of type α. 2. A is of type if and only if ϕ 1 + ϕ 2 A ψ for some ψ K 2 u, which is true if and only if ϕ 1 = P u (ψ ϕ 2 A ψ(0 ϕ 2 (0 which is true if and only if ϕ 1 sp (K u 0. The following is a generalization of Lemma Lemma Let A Φ A Ψ be a TTO and let α C. If one of the operators in the product is of type α, then either it is a constant multiple of the identity, or the other is of type α as well. Proof. Since A Φ A Ψ is a TTO, A Φ A Ψ = A Ψ A Φ and we assume wlog that A Φ is of type α. If α {0, } then the conclusion follows from Lemma 4.2.3, so assume α C, α 0. So Φ A ϕ 0 +αs u ϕ 0 +ck u 0 and Ψ A ψ 1 +ψ 2 for some ϕ 0, ψ 1, ψ 2 K 2 u, ϕ 0 (0 = 0, c C. By Lemma 4.2.1, there exists Φ 0, Ψ 0 K 2 u such that ( Φ 0 K0 u + K0 u Ψ 0 = (ϕ 0 + ck0 u ψ 2 S u(αsu ϕ 0 (S u ψ1 = ϕ 0 ψ 2 + ck0 u ψ 2 ϕ 0 (αs u ψ1 = ϕ 0 (ψ 2 αs u ψ1 + ck0 u ψ 2 So we have that ϕ 0 (ψ 2 αs u ψ1 = Φ 0 K0 u +K0 u Ψ 1. So either Φ 0 and K0 u are linearly dependent or Ψ 1 and K u 0 are. If Φ 0 and K u 0 are linearly dependent, then 26

34 Φ 0 = c 1 K u 0 which means ϕ 0 = c 2 K u 0, but this and ϕ 0 (0 = 0 then imply that c 2 = 0, and so ϕ 0 = 0 and A Φ = ci. Otherwise, Ψ 1 = c 3 K u 0 and so ψ 2 αs u ψ1 = c 4 K u 0, which means A Ψ is of type α by Proposition We now present our main result. Theorem Let Φ, Ψ A (Ku. 2 Then A Φ A Ψ is a TTO if and only if one of two (not mutually exclusive cases holds: Trivial case: Either A Φ or A Ψ is equal to ci for some c C. Non-trivial case: A Φ and A Ψ are both of type α for some α C. Proof. In what follows we will use the fact that if Φ and Ψ are functions such that A Φ A Ψ is a TTO, then for any complex constants c 1, c 2 A Φ+c1 A Ψ+c2 is also a TTO. First we prove the sufficiency of both cases. In the trivial case, if either A Φ or A Ψ is equal to ci, then A Φ A Ψ is clearly a TTO. In the non-trivial case, if α = 0 or, the product is clearly a TTO, so assume α C \ {0}. A Φ = B α ϕ+c 1 and A Ψ = B α ψ+c 2 for some ϕ, ψ K 2 u such that ϕ(0 = ψ(0 = 0 and c 1, c 2 C. It follows from Propositions that B α ϕ+c 1 B α ψ+c 2 is a TTO if and only if B α ϕb α ψ is as well. By the fact that S u C = CS u and Fact 3.3.6, we have that ( [( ] α ϕ S u ψ α S u Su ϕ S u ψ = α (ϕ S u Suϕ S u ψ ( = [(K0 u K0 u ϕ] αs u ψ [ ] = K0 u αϕ(0s u ψ = 0 So by Lemma and the earlier discussion, it follows that the product of the operators in the non-trivial case is a TTO. In the other direction, suppose A Φ A Ψ is a TTO. By Lemma it suffices 27

35 to show that one of A Φ and A Ψ is of type α for some α which we can do with Proposition There exists ϕ i, ψ i Ku 2 such that we may assume wlog that Φ = ϕ 1 + ϕ 2 and that Ψ = ψ 1 + ψ 2. Then it follows by Lemma that ϕ 1 ψ 2 (S u ϕ 2 (S u ψ1 = Φ 0 K u 0 + K u 0 Ψ 0 holds for some Φ 0, Ψ 0 in K 2 u. This can happen in one of five ways: 1. ϕ 1 ψ 2 (S u ϕ 2 (S u ψ1 = 0 2. ϕ 1 ψ 2 (S u ϕ 2 (S u ψ1 = c(k u 0 K u 0 ; c C \ {0} 3. ϕ 1 ψ 2 (S u ϕ 2 (S u ψ1 = Φ 0 K u 0 ; Φ 0 ck u 0 4. ϕ 1 ψ 2 (S u ϕ 2 (S u ψ1 = K u 0 Ψ 0 ; Ψ 0 ck u 0 5. ϕ 1 ψ 2 (S u ϕ 2 (S u ψ1 = Φ 0 K u 0 + K u 0 Ψ 0, Φ 0 ; Ψ 0 ck u 0 First consider the case that one of ϕ 1, ψ 2, S u ϕ 2, S u ψ1 is in sp (K0 u. If S u ϕ 2 (resp. S u ψ1 equals ck0 u, then ϕ 2 (resp. ψ 1 is also a constant multiple of K0 u, and thus A Φ is of holomorphic type (resp. A Ψ is of antiholomorphic type. Similarly, if ϕ 1 (resp. ψ 2 equals ck0 u, then A Φ is of antiholomorphic type (resp. A Ψ is of holomorphic type. In what follows, c and c i represent complex constants that may change from paragraph to paragraph. Case 1: We have ϕ 1 ψ 2 = (S u ϕ 2 (S u ψ1, which means that ψ 2 and S u ψ1 are linearly dependent. Both ψ 2 and S u ψ1 are non-zero, so ψ 2 = αs u ψ1 for α 0 and it follows that A Ψ is of type α. 28

36 Case 2: We have ϕ 1 ψ 2 (S u ϕ 2 (S u ψ1 = c(k0 u K0 u ; c 0. So either ϕ 1 and S u ϕ 2 are linearly dependent or S u ψ1 and ψ 2 are. In the latter case, we again get that A Ψ is of type α for some α 0 Assume instead that ϕ 1 = c 1 S u ϕ 2 for c 1 0. It follows that S u ϕ 2 = c 2 K0 u which means A Φ is of holomorphic type. Case 3: We have ϕ 1 ψ 2 (S u ϕ 2 (S u ψ1 = Φ 0 K0 u ; Φ 0 ck0 u. So either ϕ 1 and S u ϕ 2 are linearly dependent or S u ψ1 and ψ 2 are. In the latter case, we again get that A Ψ is of type α for some α 0. Assume instead that ϕ 1 = c 1 S u ϕ 2 for c 1 0. Then S u ϕ 2 (c 1 ψ 2 S u ψ1 = Φ 0 K0 u, so c 1 ψ 2 S u ψ1 = c 2 K0 u, c 2 0. So by Proposition A Ψ is of type α for some α. Case 4: We have ϕ 1 ψ 2 (S u ϕ 2 (S u ψ1 = K0 u Ψ 0 ; Ψ 0 ck0 u. So either ϕ 1 and S u ϕ 2 are linearly dependent or S u ψ1 and ψ 2 are. If ϕ 1 and S u ϕ 2 are linearly dependent, then it follows that ϕ 1 = c 1 K u 0 and hence Φ is of type. Otherwise, there exists α 0 such that ψ 2 = αs u ψ1 which means A Ψ is of type α. Case 5: We have ϕ 1 ψ 2 (S u ϕ 2 (S u ψ1 = Φ 0 K u 0 +K u 0 Ψ 0 ; Φ 0, Ψ 0 ck u 0. There exists f K 2 u such that f(0 = 0 and f, Φ 0 = 1. Then we have K0 u = (Ψ 0 K0 u + K0 u Φ 0 f = (ψ 2 ϕ 1 f (S u ψ1 S u ϕ 2 f = ψ 2 f, ϕ 1 S u ψ1 f, S u ϕ 2 If f, ϕ 1 = 0, then ck u 0 = S u ψ1, and so A Ψ is of type. Similarly, if f, S u ϕ 2 = 0, then ck u 0 = ψ 2 and A Ψ is of type 0. So we can assume that ψ 2 = αs u ψ1 + ck u 0 for some α 0. Thus A Ψ is of type α by Proposition Example Let λ D and consider the rank one TTO A = K u λ Ku λ. A simple computation shows that ( Ku λ Ku λ 2 = u (λ K u λ Ku λ so it follows that 29

37 K u λ Ku λ is of type α for some α C. Furthermore, in the event that u (λ = 1, A is idempotent, and in the event that u (λ = 0, A is nilpotent. Fact says that the function u/(z λ is a symbol for A, and since u/(z λ A K u λ + u(λ/(z λ A K u λ + u(λzk λ A K u λ + u(λs uk u λ A is of type u(λ. Now instead suppose that ζ T such that u has an ADC at ζ, and consider A = Kζ u Ku ζ. Again it is clear that A2 is a scalar multiple of A and hence A is of type α for some α. Since A is self-adjoint, it follows that α is unimodular. A simple computation shows that K u ζ = ζu(ζku ζ so S u Ku ζ = ζ K u ζ u(ζku 0 = u(ζ ( K u ζ K u 0 Thus K u ζ 1 A u(ζs u Ku ζ and so by Fact K u ζ + u(ζs u K u ζ is a symbol for A, which is therefore of type u(ζ. Example Given ϕ, ψ K u, A ϕ(1+αu and A ψ(1+αu are bounded operators (because their symbols are bounded and we have that A ϕ(1+αu A ψ(1+αu = A ϕψ + A α 2 u 2 ϕψ + A ϕ A αsu ψ + A αsu ϕ A ψ We know that A ϕ A αsu ψ+a αsu ϕ A ψ is a TTO, so we find its symbol using Fact ( A ϕ A ψ S αsu u A ϕ A αsu ψs u = A ϕ (I S u Su A αsu = A ψ ϕ K0 u A αsu ψk 0 u = α ϕ S u ψ 30

38 On the other hand, A αsu ϕ A ψ S u A αsu ϕ A ψs u ( =A αsu ϕ A ψ S u A αsu ϕ S us u A ψ Su + S u A αsu ϕ Ku 0 K 0 u ( A αsu ϕ Ku 0 αs u ϕ A ψ S u =A αsu ϕ A ψ (A ψ ψ K0 u αs u Suϕ S u ψ =P u (αuϕψ K0 u + K0 u P u (uαϕ (ψ ψ(0 α uϕψ, 1 (K0 u K0 u αϕ S u ψ + αϕ(0k u 0 S u ψ The sum of both terms is therefore P u (αuϕψ K u 0 + K u 0 P u (αuϕψ αûϕψ(0ku 0 K u 0 which is equal to A αuϕψ S u A αuϕψ S u and so A ϕ(1+αu A ψ(1+αu = A ϕψ(1+αu+α 2 u 2 Now since ϕ and ψ are in K u, their product is in H 2. Consider the function ϕψ P u (ϕψ = uh for some h H 2. For g H 2 we have h, ug = uh, u 2 g = ϕψ, u 2 g = uϕuψ, g = 0 and so ϕψ = h 1 + uh 2 for some h 1, h 2 K 2 u. So ϕψ(1 + αu + α 2 u 2 = (h 1 + uh 2 (1 + αu + α 2 u 2 A h 1 (1 + αu + αh 2 (1 + αu and so A ϕψ(1+αu+α 2 u 2 is of type α, and equals B α h 1 +αh 2. 31

39 If u is a finite Blaschke product, Ku 2 is finite dimensional and equal to Ku. Given ϕ, ψ Ku 2 and c 1, c 2 C, we have then that Bϕ+c α 1 Bψ+c α 2 = BϕB α ψ α + c 1Bψ α + c 2 Bϕ α +c 1 c 2 is of type α since all the terms on the right hand side are. We therefore have a symbol calculus for the multiplication of TTOs in the case that Ku 2 is finite dimensional. 4.3 Algebras of TTOs The families B 0 and B are actually algebras by Proposition Example showed that in certain cases, the product of two TTOs of type α is itself of type α. We will now show that for all α C, B α is an algebra. Theorem Let ϕ, ψ K 2 u such that ϕ(0 = ψ(0 = 0, let α C and let c 1, c 2 C. Then B α ϕ+c 1 B α ψ+c 2 is of type α and so B α is an algebra. Proof. B α ϕ+c 1 B α ψ+c 2 = B α ϕb α ψ + c 1 B α ψ + c 2 B α ϕ + c 1 c 2 so it suffices that B α ϕb α ψ be a TTO of type α. As mentioned above, if α = 0 or this follows from Proposition Suppose then that α C\{0}. By Proposition we can find the symbol of the product. We have B α ϕb α ψ = A ϕ+αsu ϕ A ψ+αs u ψ and since ϕ ( ( αs u ψ S u αsu ϕ S u ψ = α [(I Su Su ϕ] S u ψ = 0 32

40 the symbol is A ϕ+αsu ϕ ψ ψ, αs u ϕ K u 0 + A ψ+αsu ψαs u ϕ Now S u Ku 0 sp (K u 0 so it suffices to show that αs u C ( A ϕ(1+αu ψ A ψ(1+αu αs u ϕ = ck u 0 for some c C. Because both sides of this equation are in K 2 u and the reproducing kernels Kλ u are dense in K2 u it suffices to show that the above equation holds ( pointwise for all λ D. So let λ D. Note that K0 u (λ = 1 u(0u(λ = Kλ u(0. αs u C ( A ϕ(1+αu ψ ( (λ = α S u A ϕ(1+αu ψ (λ = α S u P u ( (ϕ(1 + αu uzψ, K u λ = α uzϕψ(1 + αu, S K u λ = α uzϕψ(1 + αu, z (K u λ K u λ(0 = α uϕψ(1 + αu, Kλ u Kλ(0 u = ψ(1 + αuαuϕ, Kλ u K u 0 (λ uϕψ(1 + αu, 1 = A ψ(1+αu αs u ϕ, Kλ u K u0 (λ uϕψ(1 + αu, 1 ( = A ψ(1+αu αs u ϕ (λ K u0 (λ uϕψ(1 + αu, 1 Note that uϕψ(1 + αu, 1 does not depend on λ and is thus constant, since α, ϕ and ψ are fixed. By Proposition we see that B α is an algebra of commuting operators. If α = 1 then B α is an algebra of commuting normal operators by Proposition

41 4.4 Generalized shifts Definition Let α D. Then Su α α := S u + 1 u(0α Ku 0 K 0 u Note that Su 0 = S u. These are the generalized shift operators and were defined by Sarason in [Sar07]. They are the sum of two TTOs and hence are all TTOs themselves. If α is unimodular, then S α u is in fact one of the Clark unitary operators as defined in Section 3.5. The assumption that α 1 ensures that 1 u(0α is non-zero. Lemma Let α D. Then S α u is of type α. Specifically, S α u = 1 1 u(0α Bα S uk u 0 + αu (0K u 0 1 u(0α Proof. Su α α = S u + 1 u(0α Ku 0 K 0 u = A SuK u 0 + α 1 u(0α A u z = A SuK 0 u+ α ( Ku 1 u(0α 0 +u(0z = A SuK u 0 = 1 ( 1 αu(0+αu(0 + α K u 1 αu(0 1 u(0α 0 1 u(0α A S uk u 0 +α K u 0 34

42 Which is of type α. The symbol of B α ϕ is ϕ + αs u ϕ. S u C ( S u K u 0 + αu (0K u 0 1 u(0α A S u S u K u 0 + αu (0S u Ku 0 1 αu(0 A K 0 u u (0K0 u αu (0u(0K0 u 1 αu(0 A K 0 u u (0 1 αu(0 Ku 0 Therefore the symbol of B α S uk u 0 + αu (0K u 0 1 u(0α is ( S u K0 u + αu (0K0 u 1 u(0α + α K 0 u u (0 1 αu(0 Ku A 0 S u K0 u + α K 0 u The result follows. Lemma Let A be a bounded operator on K 2 u and let α D. Then AS α u = S α u A if and only if A is of type α. Proof. If A is of type α, then ASu α = Su α A by Proposition and Theorem To prove the other direction, assume ASu α = Su α A. The first corollary of Theorem 10.1 in [Sar07] implies that A is then a TTO, and hence C-symmetric. Using the equations and S α u A = S u A + ASu α α = AS u + 1 αu(0 (AKu 0 K 0 u α 1 αu(0 Ku 0 (A Ku0 = S u A + α 1 αu(0 Ku 0 u (ÃK 0 35

43 we can compute the symbol of A using Fact A S u AS u = A AS u S u α 1 αu(0 AKu 0 S u Ku 0 + α 1 αu(0 Ku 0 S u ÃK u 0 = AK0 u K0 u + u(0α 1 αu(0 AKu 0 K0 u α + 1 αu(0 Ku 0 S u ÃK0 u ( = AKu 0 1 αu(0 Ku 0 + K0 u AK0 u αs u C 1 αu(0 And so the symbol of A is 1 1 αu(0 Bα AK. 0 u ( AK0 u + αs AK u 1 αu(0 uc 0 1 αu(0 which is the symbol of Corollary 5. If A is of type α, α 1, then A = 1 1 αu(0 Bα AK u 0. This result yields another (simpler proof of Theorem By taking adjoints if necessary, assume that A Φ and A Ψ are both of type α D. Then it follows that S α u commutes with both A Φ and A Ψ, and thus commutes with their product. Therefore, their product is of type α as well. Therefore, for any α C, B α is a weakly closed algebra. Theorem Let A n be TTOs of type α D such that A n converges to A in the weak operator topology. Then A is of type α. Proof. Let f, g K 2 u. Then S α u Af, g = Af, S α u g = lim n A n f, S α u g = lim n A n S α u f, g = AS α u f, g. 4.5 Bounded interpolation of TTOs Recall the following result due to Sarason [Sar67]: 36

44 Fact Let A be a bounded operator that commutes with S u. Then there exists a function ϕ H such that A = ϕ and A = A ϕ. Hence every bounded operator of type 0 has a bounded symbol. In this section we show that any operator of type α D has a bounded symbol. By taking adjoints, it follows that if α is not unimodular, then any operator of type α has a bounded symbol. In the event that α is unimodular, we will show that there are TTOs of type α with no bounded symbol α = 1 The case of α = 1 is indirectly dealt with in [Sar07, BCF + 09] and we collect those results here. There are TTOs of unimodular type without a bounded symbol under certain conditions. Specifically, in [BCF + 09] the following is proven: Fact Suppose that u is an inner function with an ADC at ζ T but such that K u ζ L p (T for some p > 2. Then K u ζ Ku ζ is a bounded TTO with no bounded symbol. They also give some conditions on the zeroes of u to ensure that K u ζ Lp (T. Example shows that K u ζ Ku ζ is of type u(ζ, and hence it is an example of a TTO of unimodular type without a bounded symbol. If, however, we weaken what we mean by bounded symbol we can find a bounded symbol for any TTO of unimodular type. Specifically, we change the measure with respect to which we take the sup norm of a function. Let α be unimodular, and fixed for the rest of this section. An operator is of type α if and only if it commutes with S α u. S α u is unitarily equivalent to M z on the space L 2 (T, µ α where µ α is the Clark measure associated with S α u. The 37

45 commutant of M z is the space of multiplication operators induced by L (µ α and so by using the unitary equivalence, every operator of type α is equal to Φ(Su α where Φ L (µ α. In this sense we can think about Φ as a bounded symbol for the operator. This gives us a symbol calculus of sorts for operators of type α: given Φ, Ψ bounded µ α almost everywhere, the product of M Φ and M Ψ is clearly M ΦΨ where ΦΨ is itself bounded µ α almost everywhere. Hence Φ(Su α Ψ(Su α = ΦΨ(Su α. We can use this symbol calculus to precisely describe the unitary TTOs on a given model space. Proposition Let A be a TTO. Then A is unitary if and only if it is equal to Φ(Su α for some α T and some Φ L (T, µ α such that Φ = 1 µ α almost everywhere. Specifically, any unitary TTO is of unimodular type, and commutes with the Clark unitary operator of the same type. Proof. If A is unitary then AA = I, which means that A and A must both be of the same type α. Thus α = α 1 which implies that α is of unimodular type. So A = Φ(S α u for some Φ L (T, µ α. Then I = AA = Φ(S α u Φ(S α u = Φ 2 (S α u which implies that Φ = 1 µ α almost everywhere. The other direction is obvious α < 1 Recall that u α = u α 1 αu for α D. In what follows, we will be dealing with TTOs on both K 2 u and K 2 u α. Let A u Φ refer to a TTO on K2 u and A uα Φ a TTO on K2 u α. We first consider operators of the form A u ϕ/(1 αu for ϕ H2. Proposition A u ϕ/(1 αu = Bα ϕ for ϕ K 2 u and α D. 38

46 2. If ϕ H 2, then A u ϕ/(1 αu = Au ϕ. Specifically, A u (1 αu 1 = I. 3. S α u = A u z/(1 αu. Proof. 1. Since we can compute But since uϕ zh 2 A u ϕ/(1 αu = Bα ϕ. 1 1 αu = (αu n n=0 ϕ 1 αu = ϕ(αu n n=0 it follows that n=0 ϕ(αun A ϕ(1 + αu and so 2. ϕ/(1 αu A ϕ + αuϕ/(1 αu A ϕ, since uϕ/(1 αu uh 2. ( 3. First note that Su α 1 = B α 1 αu(0 S uk0 u + αu (0I, so it suffices to show that (1 αu(0a u z/(1 αu = Bα S uk u 0 +αu (0I. Since z = S u K u 0 +up (uz, A u z/(1 αu = B α S uk u 0 + A up (uz/(1 αu. Now up (uz/(1 αu A αp (uz/(1 αu, and since K u 0 = (u u(0 z, P (uz = K u 0 (0 u(0z = u (0 u(0z and so A u z/(1 αu = Bα S uk u 0 + αu (0I + αu(0a u z/(1 αu. The result follows. Recall T α = M (1 α 2 1/2 (1 αu is an unitary map from K 2 u α onto K 2 u. Note that T α 1 = M (1 α 2 1/2 (1 αu 1. Lemma Let ϕ H 2 and α D. Then T α A uα ϕ T 1 α = A u ϕ/(1 αu and T α A uα ϕ T α 1 = A u ϕ/(1 αu. Therefore Auα ϕ and A u ϕ/(1 αu if ψ H 2, then A u ϕ/(1 αu = Au ψ/(1 αu if and only if u α (ϕ ψ. have the same norm, and 39

47 Proof. It suffices to show that the equalities hold on K u, so let f K u. Then ( fϕ A u ϕ/(1 αuf = P u = P 1 αu ( ( fϕ ufϕ up 1 αu 1 αu On the other hand, ( fϕ T α A uα ϕ Tα 1 f = (1 αu P uα 1 αu [ ( ] fϕ = (1 αu 1 αu u uα fϕ αp 1 αu ( ufϕ = fϕ (u αp 1 αu ( ( αufϕ ufϕ = fϕ + P up 1 αu 1 αu ( ( fϕ ufϕ = P up 1 αu 1 αu Similarly, ( fϕ Aϕ/(1 αuf u = P u 1 αu ( ( ϕf uϕf = P up 1 αu 1 αu ( ϕf = (1 αup 1 αu 40

48 and on the other hand, ( fϕ T α A uα ϕ T α 1 f = (1 αu P uα 1 αu [ ( ( ] ϕf uα ϕf = (1 αu P u α P 1 αu 1 αu [ ( ( ] ϕf uϕf = (1 αu P u α P 1 αu 1 αu ( ϕf = (1 αup 1 αu Theorem Let A be an bounded operator on K 2 u and let α D. Then A is of type α if and only if there is a function ϕ H 2 such that A = A u ϕ/(1 αu. In either case, there is a function ψ H such that ψ = A and A = A u ψ/(1 αu and therefore every operator of type α has a bounded symbol. Further, if ϕ, ψ are in H then A u ϕ/(1 αu Au ψ/(1 αu = Au ϕψ/(1 αu. Proof. Let B = Tα 1 AT α. Then AA u z/(1 αu = A u z/(1 αua if and only if BA uα z = T 1 α AA u z/(1 αut α = T 1 A u z/(1 αuat α = A uα B α z But this is true if and only if B = A uα ϕ for some ϕ H 2 which is true if and only if A = A u ϕ/(1 αu for some ϕ H2, hence the first claim holds. To prove the second claim, note that Fact implies that there is a function ψ H such that A uα ϕ = A uα ψ and Auα ϕ = ψ. By Lemma it therefore follows that 41

49 A = A u ψ/(1 αu. Since T α is unitary, A = ψ. To prove the last claim, compute A u ϕ/(1 αua u ψ/(1 αu = Tα 1 A uα ϕ A uα ψ T α = Tα 1 A uα ϕψ T α = A u ϕψ/(1 αu If α > 1 and A is of type α, then A is of type 1/α D, and so the above results can be applied to A to get similar results for A. Specifically, A has a bounded symbol. Thus for all α such that α = 1, any operator of type α has a bounded symbol. Since any operator of type α has a B α -symbol in Ku, 2 we might want to figure out the B α -symbol of the operator A ϕ/(1 αu in the event α D. We can achieve this by looking at the decomposition of H 2 induced by the Wandering Subspace lemma by Proposition Proposition Let ϕ = n=0 un ϕ n where ϕ n K 2 u for all n N. Then A ϕ 1 αu = A n=0 α n ϕn 1 αu Proof. Let f, g K u. It suffices to show that A ϕ f = A 1 αu n=0 α n ϕn f 1 αu 42

50 A simple calculation yields ϕ A ϕ f, g = 1 αu 1 αu, fg u n ϕ n = 1 αu, fg = n=0 n=0 u n ϕ n 1 αu f, g Since (1 αu 1 = m αm u m, for each n we have u n ϕ n 1 αu A u n ϕ n A ϕ n A ϕ n α n m=0 m=n α m u m α m u m n m=0 A αn ϕ n 1 αu α m u m and so the conclusion follows. 4.6 Invertible TTOs of type α and their inverses. We begin with a theorem that follows from the above. Theorem Let A be an invertible TTO. Then A 1 is a TTO if and only if A is of type α for some α C. As a result, A 1 is also of type α. Proof. If A 1 is a TTO, then both A and A 1 are of type α for some α C by Theorem since their product is I = A K u 0. If A is of type α, either α 1 or A is of type β = 1/α 1. In the first case, we have that AS α u = S α u A, so 43

51 A 1 S α u = A 1 S α u AA 1 = A 1 AS α u A 1 = S α u A 1 and A 1 is a TTO of type α by Lemma In the second case, we have that A is an invertible TTO of type β where β 1, so its inverse is a TTO of type β as well. By taking adjoints again, the result follows. This raises the question of when a TTO of type α is invertible in the first place. We consider two cases when α = 1, and when α < α = 1 We again consider the picture in L 2 (T, µ α, where a TTO of type α becomes the multiplication operator M Φ where Φ L (T, µ α. It is easy to see precisely when this operator is invertible. Proposition Let Φ L (T, µ α. Then M Φ is invertible if and only if there exists δ > 0 such that Φ δ µ α almost everywhere, and its inverse is M 1/Φ. For a more concrete example, assume u is a finite Blaschke product of degree n. Then by Fact it follows that µ α = n j=1 δ ζj u (ζ j where ζ j are the n distinct zeroes of u α. Thus it follows that M Φ is invertible if and only if Φ is non-zero on the set where u = α α < 1 Fact gives necessary and sufficient conditions for an operator of holomorphic type to be invertible. The following result is a generalization. 44

52 Proposition Let α D and let ϕ H. Then A u ϕ/(1 αu is invertible if and only if inf z D ( u α (z + ϕ(z > 0 Proof. A u ϕ/(1 αu is invertible if and only if Auα ϕ only if inf z D ( u α (z + ϕ(z > 0 by Fact is invertible, which is true if and Again we consider the case that u is a finite Blaschke product. Proposition Let α D and ϕ H, and let u be a finite Blaschke product. Then A uα ϕ is invertible if and only if ϕ(ζ 0 for all ζ such that u α = 0. Proof. From our assumptions, we have that u α is a finite Blaschke product. Suppose there exists ζ D such that u α (ζ = ϕ(ζ = 0. Then clearly A uα ϕ is not invertible. If, on the other hand, inf z D ( u α (z + ϕ(z = 0 then since u α (z is bounded away from zero near T it follows that there exists some ζ D such that u α (ζ = ϕ(ζ = Eigenvalues and eigenspaces of TTOs of type α α = 1 In this section we assume α is in D and so once again a TTO of type α can be thought of as Φ(Su α = V α M Φ for some Φ L (T, µ α. Since V α is unitary, the eigenvalues of Φ(Su α are the same as the eigenvalues of M Φ and the associated eigenspaces are related by V α. Proposition Let {Φ = λ} denote {ζ T : Φ(ζ = λ} and let {Φ λ} denote its complement. Then the eigenvalues of Φ(Su α are the λ C such that µ α ({Φ = λ} > 0. The associated eigenspace is {V α f : f L 2 (T, µ α and f {Φ λ} = 0}. 45