TRUNCATED TOEPLITZ OPERATORS ON FINITE DIMENSIONAL SPACES

Transcription

1 TRUNCATED TOEPLITZ OPERATORS ON FINITE DIMENSIONAL SPACES JOSEPH A. CIMA, WILLIAM T. ROSS, AND WARREN R. WOGEN Abstract. In this paper, we study the matrix representations of compressions of Toeplitz operators to the finite dimensional model spaces H 2 BH 2, where B is a finite Blaschke product. In particular, we determine necessary and sufficient conditions - in terms of the matrix representation - of when a linear transformation on H 2 BH 2 is the compression of a Toeplitz operator. This result complements a related result of Sarason [6].. Introduction If H 2 is the classical Hardy space of the open unit disk D := { z < } and P is the orthogonal projection of L 2 = L 2 ( D, dθ/2π onto H 2 (see [3] for the basic definitions, one defines for ϕ L the Toeplitz operator T ϕ on H 2 by the formula T ϕ f = P (ϕf. Recently, Sarason [6] initiated a study of truncated Toeplitz operators. These are operators A ϕ defined on the model spaces K ϑ := H 2 (ϑh 2, where ϑ is an inner function, by the formula A ϕ f := P ϑ (ϕf. Here P ϑ is the orthogonal projection of L 2 onto K ϑ. In other words, A ϕ is the compression of T ϕ to K ϑ. In [6, Thm. 4.], the set T ϑ := {A ϕ : ϕ L 2 and A ϕ is bounded} is characterized as follows: A bounded operator A on K ϑ belongs to T ϑ if and only if there are functions g, g 2 K ϑ such that (. A = A zaa z + g k + k g 2, where ϑ(z ϑ(0 k(z := z and h h 2 denotes the rank-one operator h h 2 (f = f, h 2 h. Though the condition in (. determines which bounded operators on K ϑ belong to T ϑ, it is difficult to apply since it depends on the existence of the functions g, g 2 K ϑ. In this paper we will obtain, in the special case when K ϑ is finite 2000 Mathematics Subject Classification. Primary 47B35; Secondary 47B32, 30D55. Key words and phrases. Toeplitz operators, model spaces, Clark operators, matrix representations.

2 2 JOSEPH A. CIMA, WILLIAM T. ROSS, AND WARREN R. WOGEN dimensional, a more tangible condition. A finite n-dimensional model space takes the form K B, where B is a finite Blaschke product with zeros {a,, a n }. It is well known that K B consists of all functions of the form (.2 f(z = p(z n ( a jz where p is any polynomial of degree at most n. Furthermore, (.3 k λ (z := B(λB(z λz is the reproducing kernel for K B in that k λ K B for all λ D and f(λ = f, k λ f K B. In the above formula, the inner product is the L 2 inner product f, g = f(ζg(ζ dζ 2π, T where T := D. Using (.2 and interpolating, it is easy to show that given distinct points λ,, λ n D, the set {k λ,, k λn } is a basis for K B. If we assume that the zeros a,, a n of B are distinct, a (non-orthonormal basis for K B is {k a,, k an } where k aj (z = a j z. By elementary linear algebra, the complex vector space of all linear transformations on K B has dimension n 2. By Sarason [6, Thm. 7.], T B has dimension 2n. This leads to the natural question as to which linear transformations on K B belong to T B. Our first theorem is the following. Theorem.4. Let B be a finite Blaschke product of degree n with distinct zeros a,, a n and let A be any linear transformation on the n-dimensional space K B. If M A = (r i,j is the matrix representation of A with respect to the basis {k a,, k an }, then A T B if and only if ( ( B (a r,i (a a i + r,j (a j a (.5 r i,j = B, i, j n, i j. (a i a j a i Remark.6. ( Theorem.4 says that the matrix representation of a truncated Toeplitz operator is determined by the entries along the main diagonal and the first row. Notice houch matrices have dimension 2n as they should since T B has dimension 2n. (2 There is nothing special about the first row. For example, a similar result can be obtained where the representing matrix is determined by the entries along the main diagonal and the first column. (3 The proof of this theorem will also yield an algorithm for determining the symbol ϕ from the matrix entries. (4 When n = 2, the matrix ( r, r,2 r 2, r 2,2

3 TRUNCATED TOEPLITZ OPERATORS ON FINITE DIMENSIONAL SPACES 3 is the matrix representation of a truncated Toeplitz operator with respect to the basis {k a, k a2 } if and only if B (a r,2 = r 2, B (a 2. Although {k a,, k an } is a natural basis for K B, it is not an orthonormal one. An important orthonormal basis for K B is the Clark basis {v ζ,, v ζn } which are the normalized eigenvectors corresponding to the eigenvalues ζ j T for the Clark unitary operator U α where α T. This is formed as follows: Since B is a finite Blaschke product, it is analytic in an open neighborhood of D and hence, for each ζ T, the kernel function k ζ defines an analytic function on D. It is routine to show that k ζ K B and (.7 f(ζ = f, k ζ f K B. For each α T a routine exercise, using the fact that B never vanishes on T, will show that there are exactly n distinct points ζ,, ζ n T for which Another routine exercise will show that B(ζ j = α + B(0, j =,, n. + B(0α k ζs 2 = B (ζ s and so we form the normalized kernel functions k ζs (.8 v ζs := B (ζ s. The points ζ,, ζ n turn out to be the eigenvalues of the Clark unitary operator, (.9 U α := A z + B(0 + α B(0 2 (k 0 k 0 with corresponding eigenvectors v ζ,, v ζn. (Here and for what follows below, (.0 kλ (z := B(z B(λ. z λ One can show [6] that k λ K B for all λ D. Thus {v ζ,, v ζn } is an orthonormal basis for K B. The operators U α, first explored by Clark [2], have been well studied and generalized [, 5]. By the spectral theorem, we know that the matrix representation of U α with respect to this basis is diag(ζ,, ζ n. Our next theorem replaces the basis {k a,, k an } of kernel functions with the Clark basis {v ζ,, v ζn }. Theorem.. Suppose B is a finite Blaschke product of degree n and α T. Let A be any linear transformation on the n-dimensional space K B. If M A = (r i,j is the matrix representation of A with respect to the Clark basis {v ζ,, v ζn } corresponding to α, then A T B if and only if (.2 r i,j = B (ζ ζ j ζ i for all i, j n, i j. ( ζ j ζ i B (ζ j (ζ ζ i r,i + B (ζ i (ζ j ζ r,j Remark.3. ( Exactly as in the previous theorem, the matrix representation of a truncated Toeplitz operator is determined by the entries along the main diagonal and the first row.

4 4 JOSEPH A. CIMA, WILLIAM T. ROSS, AND WARREN R. WOGEN (2 The proof of this theorem will also yield an algorithm for determining the symbol ϕ from the matrix entries. (3 When n = 2, the matrix ( r, r,2 r 2, r 2,2 is the matrix representation of a truncated Toeplitz operator with respect to the basis {v ζ, v ζ2 } if and only if ζ r,2 = ζ 2 r 2,. If we alter the basis {v ζ,, v ζn } slightly, we get even more. Indeed, let (.4 β α := α + B(0 + B(0α, := e i 2 (arg(ζs arg(βα, e ζs := B (ζ s k ζs. Garcia and Putinar in [4] show that {e ζ,, e ζn } is an orthonormal basis which not only diagonalizes the Clark operator U α but has the additional property that the matrix representation of any truncated Toeplitz operator with respect to this basis is complex symmetric. A matrix M is complex symmetric if M = M t, where t denotes the transpose. This next theorem replaces the Clark basis {v ζ,, v ζn } with this new basis {e ζ,, e ζn }. Theorem.5. Suppose B is a finite Blaschke product of degree n and α T. Let A be any linear transformation on the n-dimensional space K B. If M A = (r i,j is the matrix representation of A with respect to the basis {e ζ,, e ζn } corresponding to α, then A T B if and only if M A is complex symmetric and (.6 r i,j = ( B (ζ w ζ j ζ i for all i, j n, i j. w j B (ζ j (ζ ζ i r,i + w i B (ζ i (ζ j ζ r,j Remark.7. When n = 2, the theorem says that any complex symmetric 2 2 matrix represents a truncated Toeplitz operator with respect to the basis {e ζ, e ζ2 }. This was previously observed by Sarason [6, 5]. In [6, 2], Sarason began a discussion on how the Clark unitary operators somehow generate the truncated Toeplitz operators (see Remark 3.3 below. In finite dimensions, we have the following result. Theorem.8. Let B be a Blaschke product of degree n and let α, α 2 T with α α 2. Then for any ϕ L 2, there are polynomials p, q of degree at most n so that (.9 A ϕ = p(u α + q(u α2. Remark.20. ( Sarason in [6, Thm. 0.] proves that p(u α T B for every polynomial p and every α T. In fact, Theorem.8 can be gleaned from the proof of Thm. 7. in [6] along with the spectral theorem for unitary operators. (2 We will see in Remark 3.3 that, in a certain sense, one can compute the polynomials p and q in (.9 from ϕ.

5 TRUNCATED TOEPLITZ OPERATORS ON FINITE DIMENSIONAL SPACES 5 For a given ϕ L 2, decompose ϕ as Now write A ϕ as 2. Proof of Theorem.4 ϕ = ψ + ψ 2 + η + η 2, ψ, ψ 2 K B, η, η 2 BH 2. A ϕ = A ψ+ψ 2 + A η+η 2 and notice from [6, Thm. 3.] that the second term on the right is zero. Thus } (2. {A ψ+ψ2 : ψ, ψ 2 K B = T B. We are assuming that the zeros a,, a n of B are distinct and so the functions kaj (z = B(z z a j, form a basis for K B and ( B(z kaj (z = z a j j =,, n, j =,, n form a basis for K B. From the above discussion and (2., T B consists of A ϕ, where ( B(ζ B(ζ (2.2 ϕ(ζ = c j + d j ζ a j ζ a j and c j, d j are arbitrary complex numbers. Combine this with the identity (2.3 k λ k λ = A B. z λ [6, Thm. 7.] and its adjoint to see that T B consists of operators of the form (2.4 c j k aj k aj + d j kaj k aj, where c j, d j are complex numbers. In a moment, we will find the matrix representation of the above operator with respect to the basis {k a,, k an }. Before doing this, we need a few formulas. Using the reproducing property of k aj and the definitions of k aj (.3 and k aj (.0 we obtain { 0, if i j; (2.5 k ai, k aj = B (a j, if i = j. and k ai, k aj =. a j a i We know, since {k a,, k an } is a basis for K B, that kaj = h s (a j k as s= for some complex constants h s (a j. Using (2.5 one can compute h s (a j and get (2.6 kaj = k B as. (a s a s a j s=

6 6 JOSEPH A. CIMA, WILLIAM T. ROSS, AND WARREN R. WOGEN We are now ready for the proof in Theorem.4. Let A ϕ be of the form in (2.4 and let (b s,p s,p n = M Aϕ be the matrix representation of A ϕ with respect to the basis {k a,, k an }. We want to show that ( ( B (a b,s (a a s + b,p (a p a (2.7 b s,p = B, s, p n, s p. (a s a p a s A computation with (2.4, (2.5, and (2.6 will show that A ϕ k ap = c p B (a p k ap + d j k B as. (a s ( a s a j ( a p a j Thus s= b s,p = c p B (a p δ s,p + B (a s The identities in (2.7 follow from the formula d j ( a s a j ( a p a j. ( a s a j ( a p a j = a s a p a s a s a j + a p a p a s a p a j. One direction of the proof is now complete. For the other direction, let V denote the set of all matrices satisfying (.5. These identities show that each M = (r i,j V is determined uniquely by its entries along the diagonal and the first row. Furthermore, M is a linear function of these entries. It follows that V is a 2n dimensional vector space. We have already shown via (2.7 that V := {M Aϕ : A ϕ T B } V and, from Sarason s theorem, V has dimension 2n. Therefore, V = V. The proof is now complete. Remark 2.8. ( Note that {D,, D n, R 2,, R n } is an explicit basis for V. Here D k = diag(0,,, 0,, 0 and R k is the matrix satisfying (.5 with r,k =, r,j = 0 if j k, and r j,j = 0 for all j. For example, if n = 3, then D = 0 0 0, D 2 = 0 0, D 3 = 0 0 0, R 2 = R 3 = 0 0 B (a B (a (a 2 a B (a (a 2 a 3B (a B (a B (a 3 (a a 3B (a (a 2 a 3B (a 3 0 (a a 2B (a (a 3 a 2B (a 2 (a 3 a B (a (a 3 a 2B (a 2 In the above, denotes complex conjugation of all the entries of the matrix (not the conjugate transpose.

7 TRUNCATED TOEPLITZ OPERATORS ON FINITE DIMENSIONAL SPACES 7 (2 If P j := B (a j k a j k aj, notice from the above proof that Pj 2 = P j, P j = I, P j P l = δ j,l P j, T B = span{p j, Pj : j =,, n}. Similar identities hold for P j = B (a j k aj k aj. These identities exhibit the linear dependence of the set of 2n operators {P j, Pj : j =,, n}. A little work will show, for example, that the set {P j, Pl : j = 2,, n; l =,, n} forms a basis for T B consisting of rank one idempotents. (3 Using similar techniques, one can compute A ϕ from (2.4 with respect to the basis { k a,, k an }. In this case, the b s,p entry of this matrix is b s,p = d p B (a p δ s,p + c j B (a s ( a s a j ( a p a j and the necessary and sufficient condition for a matrix (r s,p to represent (with respect to the basis { k a,, k an } something from T B is ( r s,p = B (a r,s (a a s + r,p (a p a B, s, p n, s p. (a s a p a s 3. Proof of Theorem.8 The following lemma can be gleaned from [6, Thm. 7.]. We include a proof here. Lemma 3.. Suppose w,, w 2n are distinct points of T. Then the rank-one operators k w k w,, k w2n k w2n are linearly independent. Proof. Suppose c,, c 2n are complex constants such that (3.2 2n c j k wj k wj = 0. Since k w,, k wn are linearly independent, there is a g K B such that k w, g =, k wj, g = 0, j = 2,, n. Apply to this g the operator on the left hand side of (3.2 to see that c k w + 2n j=n+ c j g, k wj k wj = 0. However, the vectors k w, k wn+,, k w2n are linearly independent and so c = 0. Now take an appropriate g to show that c 2 = 0 and so on.

8 8 JOSEPH A. CIMA, WILLIAM T. ROSS, AND WARREN R. WOGEN Here is the proof of Theorem.8. Let α, α 2 T with α α 2. Let ζ,, ζ n and η,, η n be the points in T so that B(ζ j = β α := α + B(0 + B(0α, B(η j = β α2 := α 2 + B(0 + B(0α 2, j =,, n. Notice that the points ζ,, ζ n, η,, η n are distinct and recall that {v ζ,, v ζn } is a orthonormal basis for K B of eigenvectors of U α. In a similar way, {v η,, v ηn } is a orthonormal basis for K B of eigenvectors of U α2. Let P ζj := v ζj v ζj, P ηj := v ηj v ηj, j =,, n and observe that these operators are the orthogonal projections onto the eigenspaces spanned by k ζj (respectively k ηj. In [6, Thm. 5.] it was shown, for any ζ T, that k ζ k ζ = A kζ +k ζ and so these projections P ζj, P ηj also belong to T B. Furthermore, by the spectral theorem for unitary operators, we have, for any analytic polynomials p and q, p(u α = p(ζ j v ζj v ζj, q(u α2 = q(η j v ηj v ηj. and so p(u α, q(u α2 T B. Then, to show that it suffices to prove that T B = {(U α i, (U α2 j, i, j n}, T B = {P ζj, P ηj : j =,, n}, which follows directly from Lemma 3. and the fact that T B has dimension 2n. Remark 3.3. ( Theorem.8 says that any A ϕ takes the form p(u α +q(u α2 for some polynomials p and q. We remark here that we can determine p and q from the symbol ϕ provided it is chosen in a particular way. To see how to do this, notice in the proof of Theorem.8, how we have shown that {kζj k ζj, k ηj k ηj : j =,, n} = T B. In fact any 2n of these will form a basis for T B. But since k ζ k ζ = A kζ +k ζ, every operator in T B can be written as A ϕ where ϕ = c j (k ζj + k ζj + d j (k ηj + k ηj. Choose polynomials p and q of degree at most n for which p(ζ j = B (ζ j c j, q(η j = B (η j d j, j =,, n. Then we have A ϕ = p(u α + q(u α2.

9 TRUNCATED TOEPLITZ OPERATORS ON FINITE DIMENSIONAL SPACES 9 Indeed, from the spectral theorem, p(u α = p(ζ j v ζj v ζj, q(u α2 = q(η j v ηj v ηj. The result now follows. (2 Sarason [6, 2] began a discussion on how the Clark unitary operators generate T ϑ for a general inner function ϑ. He used the Clark theory and some recent results of Aleksandrov and Poltoratski to prove, for a bounded Borel function ϕ and an inner function ϑ, the following integral formula: (3.4 A ϕ = T ϕ(u α dα 2π, where the above integral is understood in the weak sense, i.e., A ϕ f, g = ϕ(u α f, g dα 2π, f, g K ϑ. T When ϕ L 2 (not necessarily bounded, there is also a version of this formula, although it must be interpreted in a very special way. Sarason also proves that T ϑ is closed in the weak operator topology. Is it the case that (3.5 T ϑ := {q(u α : q is a trigonometric polynomial; α T}? In the above, is the closed linear span in the weak operator topology. This is certainly true when ϑ is a finite Blaschke product (Theorem.8. In order to prove (3.5, it suffices, by means of (3.4, to prove that {A ϕ : ϕ L } is dense (weak operator topology in T ϑ. As mentioned earlier, it is unknown whether or not the above set is actually equal to T ϑ. 4. Proof of Theorem. and Theorem.5 Fix α, α 2 T with α α 2. Let {ζ j, η j : j =,, n} be the points of T for which B(ζ j = β α, B(η j = β α2, j =,, n. We know from Remark 3.3 that any member of T B takes the form c j k ζj k ζj + d j k ηj k ηj for some complex constants c j, d j. Let e ζs = v ζs, where (4. = e i 2 (arg(ζs arg(βα. To prove Theorem.5, let us compute c j k ζj k ζj + d j k ηj k ηj e ζp, e ζs, the matrix representation of this operator with respect to the basis {e ζ,, e ζn }

10 0 JOSEPH A. CIMA, WILLIAM T. ROSS, AND WARREN R. WOGEN Since {e ζ,, e ζn } is an orthonormal basis for K B, every f K B Fourier expansion f(z = f, e ζs e ζs (z = B (ζ s f(ζ se ζs (z and so s= (4.2 f, g = First notice that e ζs (ζ q = s= s= f(ζ s g(ζ s B, f, g K B. (ζ s { w s B (ζ s k ws B (ζ ζ s (ζ q = s, if s = q; 0, if s q. From the above inner product formula in (4.2, we have (kζj k ζj e ζp, e ζs = n q= ((k ζj k ζj e ζp (ζ q e ζs (ζ q B (ζ q has the = ((k ζj k ζj e ζp (ζ s B (ζ s = B (ζ s e ζ p, k ζj k ζj (ζ s w p = B (ζ s B (ζ p k ζ p (ζ j k ζj (ζ s In a similar way, = { B (ζ s, if s = p = j; 0, otherwise. (k ηj k ηj e ζp, e ζs = B (ζ s B (ζ p k ζ p (η j k ηj (ζ s w p w p = B (ζ s B (ζ p = B (ζ s w p B(ζ p B(η j B(η j B(ζ s ζ p η j η j ζ s β α β α2 β α2 β α B (ζ p ζ p η j η j ζ s = β α2 β α 2 w s B (ζ s B (ζ p ζ p η j η j ζ s w p = β α2 β α 2 w s w p ( η j ζ p B (ζ s B (ζ p ζ p η j η j ζ s ( η j ζ p = η j β α2 β α 2 w s w p ζ p B (ζ s B (ζ p η j ζ s η j ζ p The definition of w p from (4. yields the identity ζ p = β α w p 2. Use this identity to manipulate the last line of the above expression to β α η j β α2 β α 2 w s w p. B (ζ s B (ζ p η j ζ s η j ζ p

11 TRUNCATED TOEPLITZ OPERATORS ON FINITE DIMENSIONAL SPACES Putting this all together, we get c j k ζj k ζj + d j k ηj k ηj e ζp, e ζs n = c p B (ζ p δ s,p β α β α2 β α 2 w η j d j p. B (ζ s B (ζ p η j ζ s η j ζ p Using the partial fraction decomposition = η j ζ s η j ζ p ζ s ζ p ( η j ζ s η j ζ p one can verify the identities in (.6. Thus the matrix representation of any truncated Toeplitz operator - with respect to the basis {e ζ,, e ζn } - satisfies the conditions in (.6. The proof of the converse is nearly the same as the proof of the converse in Theorem.4. Using similar calculations as in the proof of Theorem.5, one proves that c j k ζj k ζj + d j k ηj k ηj v ζp, v ζs n = c p B (ζ p δ s,p β α β α2 β α 2 ζ η j d j p. B (ζ s B (ζ p η j ζ s η j ζ p Now follow the rest of the proof of Theorem.5 to prove Theorem.. References. J. A. Cima, A. L. Matheson, and W. T. Ross, The Cauchy transform, Mathematical Surveys and Monographs, vol. 25, American Mathematical Society, Providence, RI, D. N. Clark, One dimensional perturbations of restricted shifts, J. Analyse Math. 25 (972, P. L. Duren, Theory of H p spaces, Academic Press, New York, S. R. Garcia and M. Putinar, Complex symmetric operators and applications, Trans. Amer. Math. Soc. 358 (2006, no. 3, (electronic. 5. D. Sarason, Sub-Hardy Hilbert spaces in the unit disk, University of Arkansas Lecture Notes in the Mathematical Sciences, 0, John Wiley & Sons Inc., New York, , Algebraic properties of truncated Toeplitz operators, Operators and Matrices (2007, no. 4, Department of Mathematics, University of North Carolina, Chapel Hill, North Carolina address: cima@ .unc.edu Department of Mathematics and Computer Science, University of Richmond, Richmond, Virginia address: wross@richmond.edu Department of Mathematics, University of North Carolina, Chapel Hill, North Carolina address: wrw@ .unc.edu,