Weighted Commutant Lifting.

Transcription

1 Weighted Commutant Lifting. A. Biswas, C. Foias, A. E. Frazho. This paper presents a refined and constructive version of the commutant lifting theorem (see [14]) which includes the Treil-Volberg generalization of that theorem (see [15]). This theory is used to solve a new variant of the Sarason interpolation problem. 0 Introduction A fundamental problem in operator theory is to compute the set of all interwining liftings satisfying a certain constraint. To be precise, let T be an operator on H and T an operator on H. Let A be an operator intertwining T with T, that is, assume that A is an operator from H into H satisfying T A = AT. Recall that U on K H is a lifting of T if H is an invariant subspace for U and U H = T. Throughout P is the orthogonal projection onto H. We say that B is an intertwining lifting of A with respect to T and U if B is an operator from H into K satisfying U B = BT and P B = A. In case the operators T and U are clear from context, we will call such a B as an intertwining lifting of A. The general intertwining lifting problem is to find if possible an intertwining lifting B of A with respect to U. This problem is not always solvable. For example, let A = T = T = 1 on C 1 and let U be the matrix on C 2 defined by U = then there is no intertwining lifting of A. So one needs to add some structure on T, T and U to solve this intertwining lifting problem. In this paper we will present some new necessary and sufficient conditions for the existence of an intertwining lifting B of A. In case there exists an intertwining lifting B of A, then clearly B A.,

2 Recall that if T is an isometry and U is an isometric lifting of T, then the commutant lifting theorem ([14]) shows that there exists an intertwining lifting B of A satisfying B = A. The commutant lifting theorem provides a natural geometric framework for solving many classical and modern interpolation problems ( see [4] and [6]). If T T I and U is an isometric lifting of T, then Treil-Volberg have shown that there exists an intertwining lifting B of A also satisfying B = A (see [15]). However, the authors do not provide a method for computing an intertwining lifting of A and in fact, their proof is not constructive. In this paper, we consider the case where T is an arbitrary operator and T a contraction and let U denote an isometric lifting of T. We will show that there exists an intertwining lifting of A if and only if there exists a positive operator Q on H satisfying T QT Q A A. This result is a genuine generalization of the commutant lifting theorem and the Treil-Volberg theorem. By exploiting some of the ideas in Chapter IV of [6], we will develop two formulas for a special intertwining lifting B of A called the central intertwining lifting. Then we will show how our framework can be used to solve some weighted Sarason type interpolation problems. The solutions to these interpolation problems may turn out to be useful in Robust Control theory. Finally, we note that many of the results and techniques in monographs [4], [6] can be extended to the generalization of the commutant lifting theorem presented here. We intended to dedicate this work as an expression of our affection and admiration to the 85 th birthday of our master in Operator theory, Bela Sz.-Nagy, one of the great builders of that theory. Unfortunately, we can now only dedicate this work to his memory. 1 Preliminaries In this section, we will recall some standard results on dilation theory that will be used to construct our intertwining liftings. To begin, recall that U on K H is a dilation of an operator T on H if P U n H = T n for all integers n 0. Obviously, if U is a lifting of T, then U is also a dilation of T. Recall that U on K is a minimal isometric dilation of T, if U is an isometric dilation of T and H is cyclic for U, that is, K = U n H. n=0 If U is a minimal isometric dilation of T, then it turns out that U is also an isometric lifting of T. Obviously, if U is a minimal isometric dilation of T, then T is a contraction. On the other hand, if T is a contraction, then T admits a minimal isometric dilation. 2

3 This follows directly from the Sz.-Nagy-Schäffer model for the minimal isometric dilation of T. To develop this model, let D be the positive square root of I T T and D the closure of the range of D. Let l 2 (E) denote the Hilbert space of all square summable unilateral sequences with values in the Hilbert space E. Then the Sz.-Nagy- Schäffer minimal isometric dilation U of T is the isometry on K = H l 2 (D ) defined by U (h (f 0, f 1, f 2, )) = T h (D h, f 0, f 1, f 2, ), (1.1) where h is in H and {f j } 0 is a sequence in l 2 (D ). Clearly U in (1.1) is a lifting of T. It is easy to verify that the Sz.-Nagy-Schäffer U is indeed a minimal isometric dilation of T. The minimal isometric dilation of T is unique up to an isomorphism. To be precise, if U 1 on K 1 and U 2 on K 2 are two minimal isometric dilations of T, then there exists an unique unitary operator Φ : K 1 K 2 satisfying ΦU 1 = U 2Φ and Φ H = I. Remark 3.3 in Chapter VI of [4] shows that any isometric lifting V on G of T admits a reducing decomposition of the form V = U 1 U 2 on K 1 K 2 where U 1 is a minimal isometric dilation of T. Since U 1 is a minimal isometric dilation, there exists a unique unitary operator Φ from K = H l 2 (D ) onto K 1 satisfying ΦU = U 1Φ and Φ H = I where U is the Sz.-Nagy-Schäffer minimal isometric dilation of T. Now let W be the isometry from H l 2 (D ) into K 1 K 2 defined by W k = Φk 0. Then W is an isometry which intertwines U with V and W H = I. So if V is any isometric lifting of T, then there exists a unique isometry W intertwining the Sz.-Nagy-Schäffer minimal isometric dilation U of T with V and W H = I. An explicit formula for W is given in Proposition IV.1.3 in [6]. As before, let U be the Sz.-Nagy-Schäffer minimal isometric dilation of T. For n 1, let H n be the subspace of K be defined by H n = H ( n 1D ). The n-step lifting T n of T is defined by compressing the Sz.-Nagy-Schäffer minimal isometric dilation to H n, that is, T n is the operator on H n defined by T n = P nu H n where P n is the orthogonal projection onto H n. Corollary 2.3 in Chapter VII of [4], shows that U is the strong limit of T np n, as n approaches infinity. By consulting (1.1) it follows that a matrix representation for the n-step lifting T n of T is the n + 1 by n + 1 block 3

4 matrx given by T D I T n = 0 0 I 0 0, (1.2) I 0 for all n 1. Notice that P nt n+1 = T np n. Moreover, I T n T n = I, where the zero operator appears n times. Using this, it readily follows that T n+1 is unitarily equivalent to the one step lifting of T n for all integers n 1. So without loss of generality, we also call T n+1 the one step lifting of T n. We say that A n a n-step intertwining lifting of A, if A n is an operator from H into H n satisfying T na n = A n T and P A n = A. Let A n be a n-step intertwining lifting of A. For j = 1, 2,, n 1, let A j be the operator form H into H j defined by A j = P j A n. Then, using P k T k+1 = T k P k, it follows that A j is also a j-step intertwining lifting of A for all j = 1, 2,, n 1. Since T j+1 is the one step lifting of T j, one can also view A j+1 as a one step intertwining lifting of A j where T j plays the role of T. For some further results on dilation theory see [11] and [4]. In the remainder of this note, we will follow the notations introduced in this section. 2 Main Results Throughout we assume that T is an operator on H, while T on H is a contraction and A is an operator from H to H intertwining T with T, that is, AT = T A. Furthermore, let V on G H be an isometric lifting of T. We begin with the following useful result. Lemma 2.1 Let Q be a positive operator on H. Assume that Q T QT and A A Q. Then, there exists a one step intertwining lifting A 1 of A satisfying A 1A 1 Q. Proof. Recall that D is the positive square root of I T T. Using AT = T A along with h H, we have (T (Q A A)T h, h) = ((T QT A T T A)h, h) ((A A A T T A)h, h) = (A D 2 Ah, h) = D Ah 2. 4

5 The inequality follows from the hypothesis that A A Q T QT. As a consequence, we obtain that, (Q A A) 1/2 T h 2 D Ah 2 for all h in H. Therefore, there exists a contraction C 1 from the space F := (Q A A) 1/2 T H into D such that C 1 (Q A A) 1/2 T = D A. So, we can extend C 1 to a contraction C mapping (Q A A) 1/2 H into D such that C F = C 1. Now let X be the operator from H into D defined by X = C(Q A A) 1/2. By construction, XT = D A. Finally, let A 1 be the operator from H into H 1 defined by A 1 = A X. We claim that A 1 is a one step intertwining lifting of A satisfying A 1A 1 Q. Using XT = D A, we obtain A 1 T = AT = T A = T 0 A = T 1 A 1. XT D A D 0 X Hence, A 1 satisfies the intertwining condition A 1 T = T 1 A 1. Since C is a contraction, for h in H, we have A 1 h 2 = Ah 2 + Xh 2 Ah 2 + (Q A A) 1/2 h 2 = (Qh, h). Thus, A 1A 1 Q. This completes the proof. Now we move on to state and prove the main result of this section. Theorem 2.1 Let V on G be an isometric lifting of T. Then the following statements are equivalent: (i) There exists an intertwining lifting B of A with respect to V ; (ii)there exists a positive operator Q 1 on H such that T Q 1 T = Q 1 A A; (iii)there exists a positive operator Q on H such that T QT Q A A. Furthermore, if (iii) holds, then there exists an intertwining lifting B of A satisfying B B Q. Proof. Let B be an intertwining lifting A. To show that (i) implies (ii), simply set Q 1 = B B. Then using V B = BT, we obtain T Q 1 T = T B BT = B V V B = B B = Q 1. Furthermore, for any h in H, we have Ah 2 = P Bh 2 Bh 2, where, as before, P denotes the orthogonal projection on H. This implies that Q 1 A A. Hence part (ii) holds. 5

6 Obviously part (ii) implies the statement in (iii). So, to complete the proof, it remains to show that (iii) implies (i). As before, let U on K be the Sz.-Nagy-Schäffer minimal isometric dilation of T. We would first construct an operator B : H K satisfying P B = A, B T = U B and B B Q. By applying Lemma 2.1, there exists a one step intertwining lifting A 1 of A satisfying A 1A 1 Q. Recall that T j+1 is the one-step lifting of T j for any integer j 1. In particular, T 2 is the one-step lifting of T 1. Again, by applying Lemma 2.1 to the operator A 1 where T 1 replaces T, there exists a one step intertwining lifting A 2 of A 1 satisfying T 2 A 2 = A 2 T and A 2A 2 Q. Obviously P A 2 = P P 1 A 2 = P A 1 = A. Therefore A 2 is a two-step intertwining lifting of A satisfying A 2A 2 Q. By continuing in this fashion, an inductive argument shows that there exists a sequence of n-step intertwining liftings A n of A satisfying A na n Q for all integers n 1. Moreover, this sequence satisfies the following nesting property: if n > j, then P j A n = A j. This nesting property implies that A n is strongly convergent to an operator B mapping H into K and B B Q. To see this, let h be in H. Then for any integer n j, we have A n h A j h 2 = (I P j )A n 2 = A n h 2 P j A n h 2 = A n h 2 A j h 2. This implies that A j h 2 A n h 2 (Qh, h). Therefore A j h 2 forms an increasing sequence bounded by (Qh, h). Thus A n h A j h converges to zero as j and n tend towards infinity. Hence A n strongly approaches an operator B. Clearly, P nb = A and B B Q. Finally, using the fact that T n strongly approaches U and T na n = A n T, it follows that U B = B T. Now, recall from our discussion in the previous section that there exists an isometry W mapping K into G satisfying W H = I, W U = V W. Taking B = W B and using W H = I, we see that Finally, notice that P B = P W B = ΦP B = A. V B = V W B = W U B = W B T = BT. This finishes the proof of (iii) implies (i). Remark 2.1. In Theorem 2.1, we may assume, without loss of generality, that the operator T is of the form cu, where U is an isometry and c is an adequate constant 6

7 satisfying c 1. To see this, assume that Theorem 2.1 holds in this case. As before, let T be an operator on H satisfying Q T QT. To obtain Theorem 2.1, set c = max {1, T }. Let U on K be the minimal isometric dilation of T/c, and let P denote the orthogonal projection on H. Then we have T P = P cu. Now let A 1 be the operator from K into H defined by A 1 = AP, and Q 1 the positive operator on K defined by Q 1 = P QP. Clearly A 1A 1 Q 1. Notice that we also have the intertwining relation T A 1 = T AP = AT P = AP cu = A 1 cu. (2.1) Hence A 1 intertwines cu with T. Moreover, using T QT Q and P cu = T P, we obtain cu Q 1 cu = cu P QP cu = P T QT P P QP = Q 1. (2.2) We may now use the (iii) implies (i) part of Theorem 2.1 with T replaced by cu, A replaced by A 1 and Q replaced by Q 1, to obtain a B 1 satisfying the properties B 1 cu = V B 1, P B 1 = A 1 and B 1B 1 Q 1. (2.3) If B = B 1 H, then we claim that B is an intertwining lifting of A with respect to T and V satisfying B B Q. Obviously, B B Q. By using (2.3) and the fact that A 1 = AP, we obtain P B = P B 1 H = A 1 H = A. (2.4) So it remains to check the intertwining relation BT = V B. To this end, notice that BT = B 1 T = B 1 cu H B 1 (cu T ) H. (2.5) We claim that B 1 (cu T ) H = 0. Since B 1B 1 Q 1 = P QP, it follows that, B 1 g 2 (QP g, P g) = 0 for all vectors g orthogonal to H. Hence B 1 (I P ) = 0. This observation, together with the fact that P (cu T ) H is zero, implies that B 1 (cu T ) H = 0. Substituting this into (2.5) yields BT = B 1 cu H = V B 1 H = V B. In other words, B intertwines T with V. Therefore B is an intertwining lifting of A satisfying B B Q, which proves our claim. By choosing Q = I, Theorem 2.1 readily yields the following result known as the commutant lifting theorem (see [14]). Corollary 2.1 Let A be a contraction intertwining T with T. Assume that T is an isometry and V is an isometric lifting of T. Then there exists a contractive intertwining lifting B of A. Remark 2.2. If Q = T QT, then the above theorem follows from the commutant lifting theorem. In this case, Q 1/2 h 2 = Q 1/2 T h 2 for all h in H. So there exists an isometry V on V = Q 1/2 H satisfying V Q 1/2 = Q 1/2 T. Because A A Q, there 7

8 exists a contraction C from V into H such that A = CQ 1/2. Using AT = T A and V Q 1/2 = Q 1/2 T, it follows that C intertwines V with T. By applying the commutant lifting theorem to C, V and T, there exists a contractive intertwining lifting B o of C satisfying V B o = B o V, where V on G is an isometric lifting of T. Now let B be the operator from H into K defined by B = B o Q 1/2. By employing V Q 1/2 = Q 1/2 T, we see that B is an intertwining lifting of A satisfying B B Q. This proves Theorem 2.1 when Q = T QT. Moreover, if B is an intertwining lifting of A satisfying B B Q, then B admits a unique factorization of the form B = B o Q 1/2 where B 0 is a contraction from V into H. Using V Q 1/2 = Q 1/2 T, it follows that this B o is also a contractive intertwining lifting of C with respect to V and V. Therefore the relation B = B o Q 1/2 provides a one to one correspondence between the set of all intertwining liftings B of A satisfying B B Q and the set of all contractive intertwining liftings B o of C. From Theorem 2.1 with Q = I, we obtain the following result due to S. Treil and A. Volberg (see [15]): Corollary 2.2 Let A be a contraction intertwining T with T. Assume that T T I and U is an isometric lifting of T. Then there exists a contractive intertwining lifting B of A. Remark 2.3. In Remark 2.2 we showed that if Q = T QT, then Theorem 2.1 is a consequence of the commutant lifting theorem. Next, we will present an example inspired by [12] and [13], to show that one cannot prove Theorem 2.1 by a similar application of the Treil-Volberg theorem. First, let us establish some notations. For any Hilbert space Y, L 2 (Y) denotes the Lebesgue space of all Y valued, strongly measurable square integrable functions on the unit circle T = {e ıt : 0 t < 2π} and by H 2 (Y) the Hardy (sub)space of all h L 2 (Y) with Fourier expansions of the form a n e ınt. Recall that such an h may also be viewed as the analytic function n=0 h(z) = a n z n for z D = {z C : z < 1} (for details, see Chapter IX of [4] n=0 or Chapter V of [11]). Also for two separable Hilbert spaces U and Y, we denote by L (U, Y) the set of all essentially bounded, strongly measurable functions defined on T with values bounded linear operators from U into Y. Also, H (U, Y) is the subspace of L (U, Y) formed by the functions with Fourier expansions of the form A n e ınt, where A n, n N, are bounded linear operators from U into Y. Such n=0 a function can also be identified with its extension A n z n for z D, which is a bounded operator-valued analytic function on D (see Chapter IX of [4] or Chapter V of [11] for details). 8 n=0

9 Let M be the diagonal operator on l 2 defined by M = 1 1/n and let E be a separable Hilbert space. Notice that M is a self-adjoint contraction whose kernel is zero. Let L (E, E) be the set of all strongly Lebesgue measurable, essentially bounded functions on the unit circle in the complex plane whose values are linear operators on E. Let Θ and be the constant functions in L (E, E) defined by Θ(e ıt ) = M and (e ıt ) = (1 Θ(e ıt ) Θ(e ıt )) 1/2 for 0 t 2π. Clearly, Θ is a contractive analytic two sided outer function. Let T 1 on H be the completely nonunitay contraction in the functional model with the characteristic function Θ; see Ch VI in [11] for details. To be specific, recall that the space H is given by H = [H 2 (E) L 2 (E)] {Θu u : u H 2 (E)}. The contraction T 1 = P (S Ŝ) H where P is the orthogonal projection onto H, and S is the canonical unilateral shift on H 2 (E), while Ŝ is the canonical bilateral shift on L 2 (E). Since M is self-adjoint, the characteristic function for T1 coincides with the characteristic function of T 1. According to Theorem VI.3.4 in [11], we have T1 = ΨT 1 Ψ for some unitary operator Ψ. Because the kernel of Θ(0) = M is trivial, Theorem VI.4.1 in [11] shows that 0 is not an eigenvalue of T 1. Hence both T 1 and T1 are one to one and their ranges are dense in H. Let R be the invariant subspace for Ŝ defined by R = L2 (E), and R the operator on R defined by R = Ŝ R. Recall that U = S R on K = H2 (E) R is the minimal isometric dilation of T 1. In particular, P (S R) = T 1 P. Let f be the function in L (T) be defined by f(e ıt ) = e ıt/2 where 0 t < 2π. Let H be the contraction on H defined by H = P f(r)p R H, where P R is the orthogonal projection onto R. Then the adjoint of H is computed by H = P f(r) P R H = P Rf(R)P R H = P (S R)(0 f(r))p R H = T 1 H. (2.6) Hence H = T 1 H, and thus H = H T 1 = T 1 HT 1. This readily implies that H = T 1 HT 1 = T 1 HΨT 1 Ψ. (2.7) We claim that the range of H is dense in H, that is, HH = H. To prove this, iterate (2.7) to obtain H = T1 n HT1 n, for all integers n 1. Recall that the range of T1 is dense in H. Hence the previous relation implies that HH M where M = 0 T 1 n HH. Using P (S R) n = T1 n P for any positive integer n, we obtain T n 1 H = P (S R) n (0 f(r))p R H = P P R (0 f(r))(s R) n H. So h H is orthogonal to M if and only if f(r) P R h is orthogonal to (S R) n H for all integers n 0. Because S R is a minimal isometric dilation of T 1, it follows 9

10 that P R h = 0. Since Θ is outer, for any nonzero h in H, the quantity T1 n h does not approach zero, as n approaches infinity; see Proposition VI.3.5 in [11]. According to Proposition II.3.1 in [11], we also have P R h = lim T1 n h. This implies that P R h = 0 is equivalent to h being zero. This proves that HH = H. Finally, from (2.6) and HH = H, it follows that H has dense range. Therefore ker H = {0} where ker denotes the kernel. Now let Q be the positive operator on H defined by Q = H H. Set T = T1 and let A be the operator on H defined by A = HΨH. Using (2.7) along with the fact that T 1 and H are contractions, it is easy to verify that T QT Q A A. By employing (2.7) once again, we have for all h H AT h 2 = T 1 HΨT 1 HT 1 h 2 = T 1 HΨHh 2 Ah 2. (2.8) Relation (2.8) allows us to define a contraction T on H such that T A = AT. Since A = HΨH and the range of H is dense in H, it follows that the range of A is also dense in H. So the contraction T is uniquely determined by the relation T A = AT. We claim that there is no operator V 1 such that V 1 H = HT. We proceed by contradiction. Suppose there exists a V 1 satisfying V 1 H = HT. Then, using H = T 1 HT1 and recalling that T = T1, we obtain V 1 T 1 HT = HT. Because the range of both T and H are dense in H, we conclude that V 1 T 1 = I. This also implies that (T 1 V 1 I)T 1 = 0. Recall that the range of T 1 is also dense in H. Therefore T 1 V 1 = I, and thus T 1 is invertible. However, according to Theorem VI.4.1 in [11], T 1 is invertible if and only if Θ(0) = M is invertible. Obviously M is not invertible. Hence T 1 is not invertible. This is a contradiction. The contradiction stemmed from assuming the existence of an operator V 1 satisfying V 1 H = HT. To obtain our counter-example, let T, T, A and Q be the operators on H above. The relation A A Q implies that there exists a contraction C on H satisfying A = CQ 1/2. To apply Treil-Volberg theorem to deduce (iii) implies (i) part of Theorem 2.1, we must obtain an operator V on H such that T C = CV and V V I. (2.9) Recall that H is one to one and the range of H is dense in H. So H admits a polar decomposition of the form H = ΓQ 1/2 where Γ is a unitary operator. Using Γ H = Q 1/2, we obtain CΓ H = CQ 1/2 = A = HΨH. Because the range of H is dense in H, we have H = CΓ Ψ. In particular, C is one to one. If T C = CV, then CΓ HT = AT = T A = T CΓ H = CV Γ H. 10

11 Since C is one to one, Γ HT = V Γ H. Hence HT = V 1 H where V 1 = ΓV Γ. Therefore, if there exists an operator V satisfying (2.9), then there exists an operator V 1 satisfying V 1 H = HT, which is impossible. So in this case there exists no V satisfying (2.9), and one cannot readily apply the Treil-Volberg theorem to obtain an intertwining lifting. However, Theorem 2.1 shows that there exists an intertwining lifting B of A satisfying B B Q. 3 General Intertwining Lifting Problem Let T and T be two operators on H and H respectively and let A : H H be an operator intertwining T and T, that is, A satisfies the relation AT = T A. Recall that an operator L on L is a lifting for an operator T on H, if H is an invariant subspace for L and L H = T or, equivalently, P L = T P where, as before, P denotes the orthogonal projection on H. An operator B : H L is termed an intertwining lifting of A with respect to T and L if it satisfies BT = L B and P B = A. (3.1) As noted in the introduction, an intertwining lifting B may not always exist. Then the general intertwining lifting problem is to find necessary and sufficient conditions for the existence of an intertwining lifting and to give an explicit description of all intertwining liftings provided those conditions are satisfied. We first provide a sufficient condition for the existence of an intertwining lifting in case T is a contraction. This result is a generalization of Corollary 4.6 in [2]. Proposition 3.1 Let A be an operator intertwining T with T. Assume that T is a contraction and L on L is a contractive lifting of T. Then the following statements are equivalent: (a)there exists a positive operator Q 1 on H such that T Q 1 T = Q 1 A A; (b)there exists a positive operator Q on H such that T QT Q A A. If either (a) or (b) holds, then there exists an operator Y from H into L intertwining T with L satisfying A = P Y and Y Y Q. Moreover, conditions (a) and (b) are necessary if A admits an intertwining lifting for all contractive liftings L of T. Proof. Clearly part (a) trivially implies (b). If (b) holds, then we claim that exists an operator Y intertwining T with L satisfying A = P Y and Y Y Q. To establish this fact, let U be the minimal isometric dilation of L. Then, U is an isometric lifting of T. By employing Theorem 2.1, there exists an intertwining 11

12 lifting B of A satisfying B B Q. Let Y be the operator from H into L defined by Y = P L B where P L is the orthogonal projection onto L. Using P L U = L P L, along with the fact that B intertwines T with U, it follows that L Y = P L U B = P L BT = Y T. Hence Y intertwines T with L. Moreover, P Y = P P L B = A. Because Y = P L B we also have Y Y Q which proves our claim, that is, there exists an operator Y intertwining T with L satisfying A = P Y and Y Y Q. Finally, notice that if we choose Q 1 = B B, then part(a) holds. This follows from the fact that U B = BT and U is an isometry. To complete the proof, assume now that A admits an intertwining lifting for all contractive liftings L of T. Then, in particular, A has an intertwining lifting when L is the minimal isometric dilation of T. By virtue of Theorem 2.1, we can conclude that both conditions (a) and (b) hold. Remark 3.1. In Proposition 3.1, neither condition (a) nor (b) is not necessary for the existence of an intertwining lifting of A. For example, let H = H 2 = H 2 (C) and T = S where S is the canonical unilateral shift on H 2. If there is a positive operator Q satisfying T QT Q A A, then by iteration, we obtain T n QT n Q A A for all positive integers n. Since T n 0 strongly, as n approaches infinity, the above inequality implies that Q = A = 0. However, there are examples with A nonzero and T = S, such that A admits an intertwining lifting. For instance, take A = T = T = S. Let Ŝ be the canonical bilateral shift on L2 = L 2 (C) and set L = Ŝ H 2. Let L on L be the compression of Ŝ to L, that is, L = P L Ŝ L. Clearly L is a lifting of T. If Q = I, then obviously Q A A. Now let Y be the operator from H 2 into L defined by Y = L H 2. Then Y intertwines T with L and A = P Y, although, the condition T QT Q in part (b) of the previous proposition is not satisfied. From Proposition 3.1, we may deduce the following result as a corollary. Corollary 3.1 Let T on H and T on H be two operators and let L on L H be a lifting of T. Furthermore, let A be an operator intertwining T with T, that is, A satisfies AT = T A. If there is a positive operator Q on H satisfying A A Q and T QT L 2 Q, then A admits an intertwining lifting B with respect to T and L satisfying B B Q. Proof. Since the case L = 0 is trivial, we assume that L 0. Let us denote by c = 1/ L and let T 1 = ct, T 1 = ct and L 1 = cl. Then, clearly, L 1 is a lifting of T 1 and A intertwines T 1 with T 1. Furthermore, we have, T 1 QT 1 = c 2 T QT c 2 L 2 Q = Q. (3.2) 12

13 Thus, we may invoke Proposition 3.1 to conclude that there exists an intertwining lifting B of A with respect to T 1 and L 1 such that B B Q, that is an operator satisfying P B = A, BT 1 = L 1B and B B Q where P acting on L denotes the orthogonal projection on H. Clearly, this B is an intertwining lifting of A with respect to T and L as well, and this proves the corollary. Remark 3.2. We note that the statement in Corollary 3.1 remains unchanged when the operators T, T and L are multiplied by a constant. This multiplication also permits the reduction of the general intertwining lifting problem to the case when the operators T, T and L are contractions. To see this, set 1/c = max {1, T, T, L } and notice that cl is a lifting of ct and AcT = ct A. An operator B satisfies (3.1) if and only if it satisfies BcT = cl B and P B = A. (3.3) So, in the general intertwining lifting problem, we may assume, without loss of generality, that T and T are contractions and L is a contractive lifting of T. Furthermore, using Proposition 3.2 in [8], and by taking adjoints of all the operators mentioned therein, it follows that this problem can be reduced to the case where L is a co-isometry. 4 The Central Intertwining Lifting In this section, we will follow some of the ideas in Chapter IV of [6] to present an explicit expression for a special intertwining lifting B of A satisfying B B Q. This B reduces to the central intertwining lifting of A when T is an isometry and Q = γi. As before, assume that Q is a positive operator on H and A A Q. Furthermore, A is an operator intertwining T with T where T is a contraction and T is an operator satisfying T QT Q. Recall that we denote by D the positive square root of I T T and by D the closure of the range of D. Now let D A be the positive square root of Q A A and D A be the closed range of D A. Let ω be the operator from F : = (Q A A) 1/2 T H into D D A H defined by D A ωd A T = D A. (4.1) (T QT Q) 1/2 We claim that ω is an isometry. To see this notice that AT = T A gives T (Q A A)T = T QT Q + Q A A + A A A T T A. (4.2) 13

14 So for all h in H, we have D A T h 2 = ((T QT Q)h, h) + D A h 2 + D Ah 2. By rearranging terms, this readily implies that ω in (4.1) is an isometry. Let Π be the contraction from D D A H onto D and Π A be the contraction from D D A H onto D A defined by Π (d d A h) = d and Π A (d d A h) = d A. (4.3) where d d A h is in D D A H. Finally, it is useful to observe that Π ωd A T = D A and Π A ωd A T = D A. (4.4) By following the results in Chapter IV of [6], we are ready to define the central intertwining lifting of A in our setting. To this end, let U be the Sz.-Nagy- Schäffer minimal isometric dilation of T. Let P F be the orthogonal projection onto F. Let B Q be the linear map from H into H l 2 (D ) defined by A Π ωp F D A Π B Q = ωp F Π A ωp F D A Π ωp F (Π A ωp F ) 2. (4.5) D A Π ωp F (Π A ωp F ) 3 D A. This B Q is called the central intertwining lifting of A with respect to T, the Sz.-Nagy- Schäffer minimal isometric dilation U and Q. The following theorem shows that B Q is an intertwining lifting of A with respect to T and U satisfying B Q B Q Q. Recall that if V is any isometric lifting of T, then there exists a unique isometry W intertwining the Sz.-Nagy-Schäffer minimal isometric dilation U of T with V and W H = I. So if V is any isometric lifting of T, then B = W B Q is called the central intertwining lifting of A with respect to T, V and Q. Since B Q is an intertwining lifting of A with respect to T and U, it follows that B is an intertwining lifting of A with respect to T and V and B B Q. This fact is established in the following result. Theorem 4.1 Let Q be a positive operator on H and T an operator on H satisfying T QT Q. Let A be an operator intertwining T with a contraction T on H and assume that A A Q. Finally, let V be any isometric lifting of T, and W the 14

15 unique isometry intertwining the Sz.-Nagy-Schäffer minimal isometric dilation U of T with V. Then B = W B Q is an intertwining lifting of A with respect to T and V and B B Q. Proof. Without of generality, we can assume that V = U is the Sz.-Nagy- Schäffer minimal isometric dilation of T. Using (4.4) and the form of the Sz.-Nagy- Schäffer minimal isometric dilation in (1.1), it follows that U B Q = B Q T provided that B Q is a bounded operator. Obviously P B Q = A. Hence B Q is an intertwining lifting of A. To complete the proof, it remains to show that B Q is a bounded operator and that B Q B Q Q. To this end, let h be in H. Since Π f 2 f 2 Π A f 2 for all f in D D A H, we have n Π ωp F (Π A ωp F ) j D A h 2 j=0 n ωp F (Π A ωp F ) j D A h 2 j=0 n (Π A ωp F ) j P F D A h 2 j=0 n Π A ωp F (Π A ωp F ) j D A h 2 j=0 n (Π A ωp F ) j+1 P F D A h 2 j=0 P F D A h 2 (Π A ωp F ) n+1 P F D A h 2 P F D A h 2. Since this holds for all integers n 0, the operator B Q is bounded. Moreover, by letting n approach infinity in the above inequality, we obtain B Q h 2 = Ah 2 + Π ωp F (Π A ωp F ) j D A h 2 j=0 Ah 2 + P F D A h 2 Ah 2 + D A h 2 = (Qh, h). Therefore BQ B Q Q. This completes the proof. If T is an isometry and Q = γi, then B Q is precisely the central intertwining lifting of A defined in Chapter IV of [6]. Now assume that Q A A is strictly positive. By following some of the ideas in Chapter IV of [6], we will present a slightly more explicit formula for the central solution B Q. If Q A A is strictly positive, then equation (4.1) or (4.2) shows that T (Q A A)T is invertible. Recall that if X is any operator such that X X is invertible, then the range of X is closed. Moreover, the orthogonal projection onto the range of X is given by P = X(X X) 1 X. To verify this simply notice that P = P 2 = P, and P and X have the same range. Let X = D A T, and recall that F is the closure of the range of D A T. Since X X = T (Q A A)T is invertible, 15

16 the range of D A T is closed. Hence F is the range of D A T. Moreover, the orthogonal projection onto F is given by P F = D A T (T QT T A AT ) 1 T D A. (4.6) Now let T A be the operator on H defined by T A = (Q A A)T (T QT T A AT ) 1. (4.7) By using the definition of ω in (4.1) along with the above formula for the orthogonal projection for P F, we have Π A ωp F = D A (T QT T A AT ) 1 T D A = D A T AD 1 A. Hence, Π A ωp F = D A T A D 1 A. In other words, T A is similar to the contraction Π AωP F. By employing Π A ωp F D A = D A T A and the formula for P F in (4.6), we obtain for all integers n 0, Π ωp F (Π A ωp F ) n D A = Π ωp F D A T n A = D AT n+1 A. Substituting this into our formula for the central intertwining B Q for A in (4.5), yields A D AT A B Q = D ATA 2. (4.8) D ATA 3. Summing up this analysis we readily obtain the following result. Corollary 4.1 Let Q be a positive operator on H and T an operator on H satisfying T QT Q. Let A be an operator intertwining T with a contraction T on H and assume that A A Q. Finally, let V be any isometric lifting of T, and W the unique isometry intertwining the Sz.-Nagy-Schäffer minimal isometric dilation U of T with V. If Q A A is strictly positive, then the central intertwining lifting of A with respect to T, V and Q is given by W B Q where B Q is computed according to (4.8). To complete this section we will follow some of the results in Chapter IV of [6] to present a quotient formula for computing the central intertwining lifting of A when T is a unilateral shift. First let us recall some standard definitions. Let 16

17 U, Y be two Hilbert spaces and let F be a function in L (U, Y), then M F is the multiplication operator from L 2 (U) into L 2 (Y) defined by M F g = F g where g is in L 2 (U). Throughout T F denotes the Toeplitz operator from H 2 (U) into H 2 (Y) defined by T F = P + M F H 2 (U), where P + is the orthogonal projection onto H 2 (Y). Notice that if F is a function in H (U, Y), then T F = M F H 2 (U). Recall that Ŝ is the canonical bilateral shift on L 2 (U) if Ŝ(f)(eıt ) = e ıt f(e ıt ) where f is in L 2 (U). The operator S is the canonical unilateral shift on H 2 (U) if S = Ŝ H2 (U) where Ŝ is the canonical bilateral shift on L 2 (U). Let Θ be an analytic function whose values are bounded operators on U and assume that Θu is in H 2 (U) for all u in U. Then we say that Θ is outer if the closed linear span of {S n ΘU : n = 0, 1, 2,...} is all of H 2 (U). Finally, Π o denotes the operator from H 2 (U) onto U defined by Π o h = h(0) where h is in H 2 (U). The following result provides us with a method for computing the centeral intertwining lifting when T is a unilateral shift. Corollary 4.2 Let Q be a positive operator on H and T a canonical unilateral shift on H 2 (U) satisfying T QT Q. Let S be the canonical unilateral shift on H 2 (Y). Let H be an invariant subspace for S and T the contraction on H defined by T = S H. Let A be an operator intertwining T with T and assume that Q A A is strictly positive. Finally, let N and R be the operator valued analytic functions defined by N(λ)u = (AD 2 A Π ou)(λ) and R(λ)u = (D 2 A Π ou)(λ) (u U). (4.9) Then R is an outer function. Moreover, if F (λ) = N(λ)R(λ) 1, then F is a well defined function in H (U, Y). Futhermore, the central intertwining lifting B of A with respect to T, S and Q is given by B = T F. Proof. Let B be the central intertwining lifting of A with respect to S and Q. Since S B = BT where T and S are canonical unilateral shifts on the appropriate H 2 spaces, there exists a unique function F in H (U, Y) satisfying T F = B. Recall that B = W B Q where W is an isometry satisfying W H = I. Because T u = 0 for all u in U, we have T A D 2 A u = 0; see (4.7). So by consulting the formula for B Q in (4.8), we obtain F Ru = T F D 2 A u = W B QD 2 A u = W AD 2 A u = Nu. Hence F (λ)r(λ) = N(λ) for all λ in the open unit disc. By adjusting the proof of Lemma IV.4.3 in [6] to our setting, we see that R is an outer function. Therefore F = NR 1. This completes the proof. A minor modification of the previous proof readily yields the following result. Corollary 4.3 Let Q be a positive operator on H and T a canonical unilateral shift on H 2 (U) satisfying T QT Q. Let Ŝ be the canonical bilateral shift on L2 (Y). 17

18 Let H be an invariant subspace for Ŝ and T the contraction on H defined by T = Ŝ H. Let A be an operator intertwining T with T and assume that Q A A is strictly positive. Finally, let N and R be the operator valued functions defined by N(e ıt )u = (AD 2 A Π ou)(e ıt ) and R(e ıt )u = (D 2 A Π ou)(e ıt ) (u U). (4.10) Then R is an outer function. Moreover, if F = NR 1, then F is a well defined function in L (U, Y). Futhermore, the central intertwining lifting B of A with respect to T, Ŝ and Q is given by B = M F H 2 (U). 5 Weighted Sarason Interpolation In this section, and henceforth throughout, we will denote L 2 = L 2 (C), H 2 = H 2 (C), L = L (C, C) and H = H (C, C). We will use some of the above results to solve the following weighted Sarason interpolation problem: Given a function f in H, an inner function m and an outer function θ in H, find if possible a function h in H such that f mh θ a.e. on the unit circle. (5.1) We say that h is a solution to the weighted Sarason interpolation problem if h is a function in H satisfying (5.1). Theorem 5.1 Let H be the subspace of H 2 defined by H = H 2 mh 2. Let A be the operator from H 2 into H defined by A = P T f where P is the orthogonal projection onto H. Then there exists a solution to the weighted Sarason interpolation problem if and only if A A T θ T θ. Proof. The proof is similar to the solution of the Sarason problem in Section II.4 of [6]. For completeness, a proof is given. Let T be the canonical unilateral shift on H 2. Clearly mh 2 is an invariant subspace for T. Hence H is an invariant subspace for T. Let T be the contraction on H defined by T = T H. Then T is an isometric lifting of T. Recall that P is the orthogonal projection onto H. Using P T = T P, it follows that A intertwines T with T. In this setting, V = T, and thus B is an intertwining lifting of A if B is an operator on H 2 commuting with T and satisfying P B = A. We claim that B is an intertwining lifting of A if and only if B = T g where g = f mh for some h in H. If B = T g where g = f mh for some h in H, then for v in H 2, we have P Bv = P T g v = P (f mh)v = P fv = Av. 18

19 Obviously B commutes with T. Hence B = T g is an intertwining lifting of A. On the other hand, if B is an intertwining lifting of A, then B commutes with the unilateral shift T. So there exists a function g in H satisfying B = T g. Because P B = A, it follows that P (f g)v = 0 for all v in H 2. Thus f g is contained in mh 2. In other words, f g = mh for some h in H 2. Since h = (f g)/m where f and g are functions in H, the function h must be in H. Therefore g = f mh for some h in H, which proves our claim. To complete the proof notice that g is a function in H satisfying g θ a.e. on the unit circle if and only if Tg T g Tθ T θ. Now let h be a solution to the weighted Sarason problem, that is, let h be a function in H satisfying (5.1). If g = f mh, then B = T g is an intertwining lifting of A satisfying B B = Tg T g Tθ T θ. This readily implies that A A Tθ T θ. If A A Tθ T θ, then Theorem 2.1 or 4.1 with Q = Tθ T θ shows that there exists an intertwining lifting B of A satisfying B B Tθ T θ. Moreover, according to our previous analysis B = T g where g = f mh for some h in H. Hence Tg T g = B B Tθ T θ. Therefore g θ a.e. on the unit circle and thus h is a solution to the weighted Sarason interpolation problem. This completes the proof. Remark 5.1. If Tθ T θ A A is strictly positive, then Corollary 4.2 can be used to compute a solution to the weighted Sarason problem. To see this, let n = AD 2 A 1 and r = D 2 A 1 where 1 is the constant function one in H2. According to Corollary 4.2, the central intertwining lifting B of A is given by B = T g where g = n/r. By our previous discussion g = f mh for some h in H. Therefore a solution to the weighted Sarason interpolation problem is given by h = (rf n)/rm. Corollary 5.1 Let A be the operator from H 2 into H = H 2 mh 2 defined by A = P T f. Then there exists a solution to the weighted Sarason interpolation problem if and only if A admits a factorization of the form A = CT θ where C is a contraction from H 2 into H. In this case, C intertwines T with T where T is the canonical unilateral shift on H 2 and T = P T H. Proof. If A = CT θ where C is a contraction, then obviously A A T θ T θ. The previous theorem shows that there exists a solution to the weighted Sarason interpolation problem. On the other hand, if there exists a solution to the weighted Sarason problem, then A A T θ T θ. Because θ is outer, the closure of the range of T θ is all of H 2. So there exists a contraction C from H 2 into H satisfying A = CT θ. Because T commutes with T θ, it follows that C intertwines T with T. This completes the proof. Now assume that A = CT θ with C a contraction and recall that T is an isometric dilation of T. In view of the fact that T commutes with T θ and that θ is outer, it is easy to see that C intertwines T with T. Hence, B o is an intertwining lifting of 19

20 C with respect to T if and only if B o is an operator on H 2 commuting with T and P B o = C. Since B o commutes with T there exists a function q in H satisfying B o = T q and B o = q. Motivated by this we call a function q an interpolant for C if q is a function in H and T q is an intertwining lifting of C. Obviously, the relation B o = T q yields a one to one correspordence between the set of all intertwining lifting B o of C and the set of all interpolants q of C. In particular, B o is a contractive intertwining liftings of C if and only if B o = T q where q is an interpolant for C satisfying q 1. If Q = T θ T θ, then Q 1/2 = ΩT θ where Ω is a unitary operator on H 2. By adjusting Remark 2.2 to this setting, it follows that the set of all intertwining liftings B of A satisfying B B T θ T θ are given by B = B o T θ where B o is a contractive intertwining lifting of C. In other words, the set of all solutions h to the weighted Sarason interpolation problem is given by qθ = f mh where q is an interpolant for C satisfying q 1. Finally, it is noted that there exists a unique solution to the weighted Sarason interpolation problem if and only if C has only one contractive intertwining lifting. Now let us use the functional calculus in Chapter III of [11] to gain some further insight into the weighted Sarason interpolation problem with data f, m and θ. To this end, notice that T is a C o contraction whose minimal function is m. Corollary 5.2 Let T be the unilateral shift on H 2 and T the compression of T to H = H 2 mh 2. Then there exists a solution to the weighted Sarason interpolation problem if and only if f(t ) f(t ) θ(t ) θ(t ). (5.2) Proof. Let u be any function in H. Then using P T = T P, we have P T u = P u(t ) = u(t )P. (5.3) In particular, this implies that A = f(t )P. If (5.2) holds, then there exists a contraction Γ on H satisfying f(t ) = Γθ(T ). Using P T θ = θ(t )P, we have A = f(t )P = Γθ(T )P = ΓP T θ. Obviously C = ΓP is a contraction. Hence A admits factorization of the form A = CT θ where C is a contraction. So if (5.2) holds, then there exists a solution to the weighted Sarason interpolation problem by Corollary 5.1. On the other hand, if there exists a solution to the weighted Sarason interpolation problem, then A admits factorization of the form A = CT θ where C is a contraction. Moreover, C intertwines T with T. We claim that mh 2 ker C, where ker denotes the kernel. If v is in H 2, then Cmθv = CT θ mv = Amv = f(t )P mv = 0. 20

21 Because θ is outer, this implies that CmH 2 = {0}. Thus mh 2 ker C. Therefore C = CP and f(t )P = A = CP T θ = CP θ(t )P. Since CP is a contraction, (5.2)holds. This completes the proof. In many applications f, m and θ are rational functions. In this case, H is finite dimensional, and the spectrum of T is simply the zeros of m. Since θ is a rational outer function, θ(t ) is invertible. Obviously (5.2) holds if and only if f(t )θ(t ) 1 is a contraction. So if f, m and θ are rational, then there exists a solution to the weighted Sarason interpolation problem if and only if f(t )θ(t ) 1 is a contraction. Moreover, in this case, the contraction C satisfying A = CT θ is given by C = f(t )θ(t ) 1 P. To verify this simply notice that f(t )θ(t ) 1 P T θ = A. Finally, it is noted that in the rational case one can use the the results in Chapter VI of [6] to compute the set of all interpolants q of C satisfying q 1, and thus compute the set of all solutions to the weighted Sarason interpolation problem. Let us conclude this section with the following result. Corollary 5.3 Assume that f, m and θ are rational functions. Then there exists a unique solution to the weighted Sarason interpolation problem if and only if θ(t ) θ(t ) f(t ) f(t ) (5.4) is positive and singular. In this case, if h is the unique function in H solving the weighted Sarason interpolation problem, then f mh = θ on the unit circle. Proof. By Corollary 5.2 and Corollary 5.1, we can assume that there exists a solution to the weighted Sarason interpolation problem, and A = CT θ where C is the contraction given by C = f(t )θ(t ) 1 P. Recall that B is an intertwining liftings of A satisfying B B Q if and only if B = B o T θ where B o is a contractive intertwining lifting of C. In particular, there is a unique solution to the weighted Sarason interpolation problem if and only if there is only one contractive intertwining lifting of C. It is well known that there is a unique contractive intertwining lifting of C if and only if the kernel of I C C is nonzero; see Corollary IV.2.7 in [6]. However, the kernel of I C C is nonzero if and only if the operator in (5.4) is positive and singular. So there exists a unique solution to the weighted Sarason interpolation problem if and only if operator in (5.4) is positive and singular. To complete the proof assume that there is a unique solution to the weighted Sarason interpolation problem. Then C is a contraction, the kernel of I C C is nonzero, and there exists a unique contractive intertwining lifting B o of C. Corollary IV.2.7 in [6] also shows that B o = T q where q is a rational Blaschke product. Therefore, there is only one intertwining lifting B of A satisfying B B Q and this 21

22 B is given by B = T q T θ. Recall that B = T g where g = f mh for some h in H. This h must be the unique solution to the weighted Sarason interpolation problem. Obviously f mh = qθ. Because q is inner, f mh = θ on the unit circle. This completes the proof. Remark 5.2. As before, let f, m and θ be rational functions. Assume that there exists a solution to the weighted Sarason interpolation problem, or equivalently, A = CT θ where C is a contraction. Then there exists a unique solution to the weighted Sarason interpolation problem if and only if zero is an eigenvalue of I C C. If there exists a unique solution to the weighted Sarason interpolation problem, then zero is also in the spectrum of Tθ T θ A A. To see this, observe that Tθ T θ A A = Tθ (I C C)T θ. If zero is not in the spectrum of Tθ T θ A A, then Tθ T θ A A is invertible, and thus Tθ T θ is invertible. This implies that I C C is invertible, which contradicts the fact that zero is an eigenvalue of I C C. Therefore, if there exists a unique solution to the weighted Sarason interpolation problem, then zero is in the spectrum of Tθ T θ A A. However, zero may not be an eigenvalue of Tθ T θ A A. For example, if f 1, m(λ) λ and θ = 1 m, then T = 0. Hence f(t ) = 1 = θ(t ). According to Corollary 5.3, there is a unique solution to to the weighted Sarason interpolation problem. In fact, h 1 is the only solution. In this case, zero is not an eigenvalue of Tθ T θ A A. Remark 5.3. If zero is in the spectrum of Tθ T θ A A, then it does not necessarily imply that there is a unique solution to the weighted Sarason problem. For example, if f 0, m(λ) λ and θ = 1 m, then A = 0. Obviously zero is in the spectrum of Tθ T θ. Hence zero is in the spectrum of Tθ T θ A A. In this case f(t ) = 0 and θ(t ) = 1, and thus there exists infinitely many solutions to the weighted Sarason problem. 6 Hyper-weighted Sarason Interpolation The previous weighted Sarason interpolation problem can be solved by using the commutant lifting theorem. In this section we will present a generalization of this interpolation problem, which is not readily solved by the commutant lifting theorem. Then we will use Theorem 2.1 or Theorem 4.1 to solve this interpolation problem. To this end, let a be a function in L and H a the Hankel operator from H 2 into K 2 = L 2 H 2 defined by H a = P M a H 2 where P is the orthogonal projection onto K 2. Now consider the following interpolation problem which we will call the hyper-weighted Sarason interpolation problem : Given a function f in H, an inner function m and a function a in L, find if possible a function h in H such that (f mh)v 2 + H a v 2 av 2 (for all v H 2 ). (6.1) 22

23 We say that h is a solution to the hyper-weighted Sarason interpolation problem in (6.1) if h is a function in H satisfying (6.1). Notice that if a = θ is an outer function, then H a = 0 and this problem reduces to the weighted Sarason problem considered in (5.1). Moreover, it is easy to construct examples where a is in K 2 and the interpolation problem in (6.1) does not have a solution. For example, let a = e ıt b where b is an inner function. If h is a solution to (6.1), then choosing v = b yields f mh 2 0. So in this case, there exists a solution to the interpolation problem in (6.1) if and only if f is in mh 2. To obtain necessary and sufficent conditions for the existence of a solution to the hyper-weighted Sarason interpolation problem, recall that T a is the Toeplitz operator on H 2 defined by T a = P + M a H 2 where P + is the orthogonal projection onto H 2. Theorem 6.1 Let H be the subspace of H 2 defined by H = H 2 mh 2. Let A be the operator from H 2 into H defined by A = P T f where P is the orthogonal projection onto H. Then there exists a solution to the hyper-weighted Sarason interpolation problem in (6.1) if and only if A A T a T a. Proof. Let h be any function in H 2 and set g = f mh. Then the inequality in (6.1) is equivalent to T g v 2 av 2 P av 2 = T a v 2. Hence there exists a solution to the interpolation problem in (6.1) if and only if T g v T a v for all v in H 2, or equivalently, Tg T g Ta T a. As in the proof of Theorem 5.1, let T be the canonical unilateral shift on H 2 and T the compression of T to H = H 2 mh 2. Then B is an intertwining lifting of A if and only if B = T g where g = f mh for some h in H. So if h is a a solution to the interpolation problem in (6.1), then B = T g is an intertwining lifting of A where g = f mh. Using A = P B, we have A A B B = Tg T g Ta T a. Now assume that A A Ta T a, and set Q = Ta T a. Then T QT Q (see Proposition 6.2 below). According to Theorem 2.1 or 4.1, there exists an intertwining lifting B of A satisfying B B Q = Ta T a. Thus, B = T g where g = f mh for some h in H. Hence Tg T g = B B Ta T a. Therefore (6.1) holds and h is a solution to the hyper-weighted Sarason interpolation problem. This completes the proof. In the proof of the preceeding theorem we used the fact that T QT Q for Q = Ta T a. In fact, one has the more general result: Proposition 6.2 Let S be the canonical unilateral shift on H 2 (E) where E is a seperable Hilbert space, and let F be a function in L (E, E). If Q = T F T F, then S QS Q and equality holds if and only if F H (E, E). 23