Notes on Certain Star-Shift Invariant Subspaces - CiteSeerX

Computational Methods and Function Theory Volume 4 (2004), No. 2, 461–474

Notes on Certain Star-Shift Invariant Subspaces John R. Akeroyd and Kristi Karber (Communicated by Laurent Baratchart) Abstract. Our work addresses the question: for which (infinite) Blaschke products B does the star-shift invariant subspace KB := H 2 (D) BH 2 (D) contain a (non-trivial) function with a non-trivial singular inner factor? In the case that the zeros of B have only finitely many accumulation points w1 , w2 , . . . , wn in T, a recent paper shows that, for an affirmative answer, there necessarily exist k, 1 ≤ k ≤ n, and a subsequence of the zeros of B that converges tangentially to wk on “both sides” of wk . One of the results in this article improves upon this theorem. And, currently, the only examples of Blaschke products in the literature that are shown to yield an affirmative answer are those that have a proper factor b that satisfies b(D) 6= D. We produce many examples here that have no such factor. Keywords. Invariant subspace, backward shift, inner function. 2000 MSC. 30H05, 47B38, 49J20.

1. Preliminaries Let D denote the unit disk {z : |z| < 1}, let T denote the unit circle {z : |z| = 1} and let m denote normalized Lebesgue measure on T. For 0 < p < ∞, the Hardy space H p (D) is the set of functions f that are analytic in D such that Z p kf kp := sup |f (rζ)|p dm(ζ) < ∞. 0 0} and D− the lower half-disk {z ∈ D : Im(z) < 0}. If {ξk }∞ k=1 is a sequence in D that has just one iθ accumulation point e in T, then let {ξj+ }j denote the “subsequence” of {ξk }∞ k=1 that is contained in eiθ · D+ and {ξj− }j the “subsequence” that is contained in eiθ · D− . Notice that if there is a chord of T with one endpoint at eiθ such that {ξk }∞ k=1 is contained in just one of the regions bounded by this chord and T, then either {|eiθ − ξj+ |/(1 − |ξj+ |)}j or {|eiθ − ξj− |/(1 − |ξj− |)}j is bounded. And so, if {ξk }∞ is a Blaschke sequence in D, then this “chordal condition” implies that k=1P P either j |eiθ − ξj+ | or j |eiθ − ξj− | converges. Yet it is quite possible for one or both of these series to be convergent without the chordal condition holding. And so the following theorem is indeed an improvement upon Theorem 2.1. Theorem 2.2. Let B be an infinite Blaschke product whose zeros {zn }∞ n=1 accumulate only at finitely many points w1 , w2 , . . . , wm ∈ T. For s = 1, 2, . . . , m, let ∞ {zk,s }∞ < ρ}, k=1 be the subsequence of {zn }n=1 that is contained in D∩{z P: |z −ws | + where ρ = min{1, |ws − wt |/2 : s 6= t}. If, for each s, either j |ws − zj,s | or P − j |ws − zj,s | converges, then KB does not contain a function of the form f Sµ , where f ∈ H 2 (D), f 6≡ 0, and Sµ is a (non-trivial) singular inner function.

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Proof. Our proof is a modification of the second of the two proofs of Theorem 2.1 given in [1]. We suppose, to the contrary, that there exists f Sµ in KB , where f ∈ H 2 (D), f 6≡ 0 and Sµ is a non-trivial singular inner function. Appealing to equation (1.1) and Remark 1.2, and composing with an appropriately chosen M¨obius transformation from D onto itself, we can reduce to the case: (i) 1 is an accumulation point of {zn }∞ n=1 , and in fact is the only such accumuiθ lation point in {e : −π/2 ≤ θ ≤ π/2}; (ii) f is an outer function in H 2 (D) and µ is a point mass at 1; so,   z+1 Sµ (z) = Sc (z) := exp c , z−1 where c is some positive constant. P + | converges. Without loss of generality we may further assume that j |1 − zj,1 We now introduce some specific regions whose geometries are related to each other in a useful way. Let U = {z : |z − (1 + i)| < 1} denote the disk of radius 1 and center 1 + i. Let Γ be the chord of the unit circle T that has endpoints i and 1, and let W be the chordal region bounded by Γ and {eiθ : 0 ≤ θ ≤ π/2}; we let W ∗ = {1/¯ z : z ∈ W }. Notice that     n πo 1 i 1 ∗ iθ (∂W ) \ e : 0 < θ < = z : z − + =√ \ D. 2 2 2 2 √ √ √ Let ∆ = {z : |z−(1/2+i/(2 3))| < 1/ 3}. Evidently, i/ 3 and 1 are antipodal ¯ ∪ W ∗ . Indeed, the angle formed by ∂W ∗ at 1 ¯ ⊆D points of the circle ∂∆, and ∆ is greater than that formed by ∂(∆ \ D) at 1; see Figure 1.

U W∗ W ∆ D

Figure 1.

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P + Now by our assumption that j |1 − zj,1 | converges, the zeros of B that are contained in W , and hence, the poles of B that are contained in W ∗ , form a Blaschke sequence with respect to the disk U . Let b be the Blaschke product for the disk U whose zeros are the poles of B in W ∗ , taking multiplicity into account. Factor B as the product B1 B2 , where B1 and B2 are Blaschke products in D and the zeros of B1 are precisely those of B in W . Claim. |b(1/¯ z )| ≤ |B1 (z)| for all z in W . To establish this claim we need only show that if a ∈ W and if ϕa and ψa are M¨obius transformations from D and U (respectively) onto the unit disk D such that ϕa (a) = ψa (1/¯ a) = 0, then |ψa (1/¯ z )| ≤ |ϕa (z)| whenever z ∈ ∂W . This is abundantly clear if z ∈ T ∩ ∂W , and so the question only remains open for z in Γ. The technical difficulties are reduced if we transplant the problem under a M¨obius transformation that sends i to ∞ and 1 to 0. Under such a transformation the problem reduces to showing that if α = c + id, where 0 < c < d, then |z + α ¯ | ≤ |z − α ¯ | whenever z = x + ix and x > 0. This is easily verified, and so our claim holds. Recalling equation (1.1), we can find a positive constant M such that, for z in W ∗ \ {poles of B1 }, |B(z)| |f (z)Sc (z)| ≤ M . |z| − 1 Our claim then gives us: |f (z)b(z)Sc (z)| ≤ M

|B2 (z)| , |z| − 1

once again, for z in W ∗ \ {poles of B1 }. And hence, for all such z, (2.1) (2.2)

M |B2 (z)| , |S2c/3 (z)|(|z| − 1) M |B2 (z)| |f (z)b(z)S2c/3 (z)| ≤ . |Sc/3 (z)|(|z| − 1) |f (z)b(z)Sc/3 (z)| ≤

We now argue that inequalities (2.1) and (2.2), along with [1], Lemma 2.2, imply that both F1 := f bSc/3 and F2 := f bS2c/3 are in N (∆) — the Nevanlinna class of ∆. We first note that, by the remarks immediately following equation (1.1) and by our definition of b, both F1 and F2 have analytic continuations√to ∆; in fact, to all of D ∪ {eiθ√ : 0 < θ < π/2} ∪ W ∗ . For 0 < r < 1/ 3, let ∆r = {z : |z − (1/2 + i/(2 3))| < r}, let γ√r = ∂∆r and let σr denote normalized Lebesgue measure on γr . If r is near 1/ 3, then γr is “partitioned” into four relatively open subarcs: γr,1 , γr,2 , γr,3 and γr,4 , where γr,1 and γr,3 are contained in D ∩ U , γr,2 is contained in W ∗ and γr,4 is contained in D \ U¯ . Straightforward

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calculations show that σr |γr,1 + σr |γr,3 is a Carleson measure (for H 2 (D)) with Carleson constant C independent of r. Hence, Z Z 2 |Fk | dσr ≤ |f |2 dσr ≤ Ckf k2H 2 (D) for k = 1, 2. γr,1 ∪γr,3

γr,1 ∪γr,3

Concerning γr,2 , we recall inequalities (2.1) and (2.2) above and appeal to [1], Lemma 2.2, to obtain: Z log+ |Fk | dσr γr,2

is bounded independent of r, for k = 1, 2. And we find the same to be true with γr,4 in place of γr,2 by applying [1], Lemma 2.2, once again; this time without the assistance of other inequalities. Combining these we find that Z log+ |Fk | dσr γr

is bounded independent of r, for k = 1, 2. Therefore, both F1 and F2 are in N (∆) from which it follows that Sc/3 ∈ N (∆). But this is false! We have arrived at a contradiction of our leading supposition and so our proof is complete. Remark 2.3. In [5], Theorem 5 (i), K. Dyakonov shows that if Sµ is a singular inner function, α ∈ D and bµ,α (z) :=

Sµ (z) − α 1−α ¯ Sµ (z)

is a Blaschke product, then there is an outer function f in H ∞ (D) (namely, 1/(1 − α ¯ Sµ )) such that f Sµ ∈ Kzbµ,α . In the case that µ = cδ{1} (a positively weighted point mass at 1), bc,α := bcδ{1} ,α is a Blaschke product for all α in D, except for α = 0. And for any non-zero α in D, the zeros of bc,α are contained in ∞ some circle in D that is tangent P to T+at 1; letP{zn }n=1− be the sequence of zeros of bc,α . By Theorem 2.2, both j |1−zj,1 | and j |1−zj,1 | must diverge, and indeed they do. But for small c, these series diverge slowly and thus indicate a certain “sharpness” in Theorem 2.2. If, however, the zeros of a Blaschke product B lie on a curve in D that is tangent to T at 1 of high (exponential) order of tangency, then the problem seems to change character; in this case, KB is much less likely to contain a non-trivial function of the form f Sc ; cf. [5, Theorem 5 (ii)].

3. Examples whose Blaschke factors satisfy b(D) = D As was observed in [1], Section 6, a Blaschke product b has the property that b(D) 6= D if and only if b has the form b(z) =

Sµ (z) − α , 1−α ¯ Sµ (z)

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where Sµ is some singular inner function and α ∈ D. So far in the literature, the only Blaschke products B that have been shown to have the property that KB contains a (non-trivial) function with a non-trivial singular inner factor are those that have a proper factor b such that b(D) 6= D; cf. [1, Theorem 6.1] and [6, Lemma 1]. The question has been posed as to whether or not these are the only Blaschke products with this property; cf. [1, Question 6.2]. Our results in this section answer this question with a resounding “No”. We begin with a general “perturbation” result. The convergence condition in the hypothesis of this result is the same as that of [1, Theorem 6.10], yet our goal and methods here are quite different. Theorem 3.1. Let B be a Blaschke product such that KB contains a function of the form f Sµ , where 0 6≡ f ∈ H 2 (D) and Sµ is a non-trivial singular inner ∞ function. Let {an }∞ n=1 be the zeros of B listed according to multiplicity. If {αn }n=1 is any sequence in D such that ∞ X |an − αn | n=1

1 − |an |

converges, then {αn }∞ n=1 is a Blaschke sequence and the Blaschke product B built 2 with zero sequence {αn }∞ n=1 has the property that there exists F 6≡ 0 in H (D) such that F Sµ ∈ KB . P∞ Proof. Since {an }∞ n=1 is a Blaschke sequence in D and n=1 |an − αn |/(1 − |an |) is also a Blaschke sequence. And converges, it is straightforward that {αn }∞ n=1 similarly it follows that if An := |αn |¯ an /(|an |¯ αn ) for all values of n except for Q those finitely many for which either an or αn are 0, then n An converges to a non-zero constant A. Now, by our hypothesis, there is a non-trivial singular inner function Sµ and a function f 6≡ 0 in H 2 (D) such that f Sµ ∈ KB . Recalling Discussion 1.4, we can reduce to the case that (3.1) f¯S¯µ B = b0 f m a.e. on T, where f is an outer function in H 2 (D) and b0 is a Blaschke product such that b0 (0) = 0. Our overarching strategy here is to convert factors of B into factors of B while including other terms that help maintain equality in (3.1). Let an be a zero of B; we may assume that an 6= αn . We describe various tactics for the three basic cases under the heading: an 6= 0 or αn 6= 0. If an 6= 0, then −¯ an z − an B(z) = Bnˆ (z) |an | 1 − a ¯n z and so, multiplying both sides of equation (3.1) by ζ¯ (ζ ∈ T), we obtain −¯ an ¯ 0 f (1 − a f¯S¯µ (1 − a ¯n ζ) Bnˆ = ζb ¯n ζ) |an |

m a.e.

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And therefore,     −¯ a 1 − a ¯ ζ 1 − a ¯ ζ n n n ¯ ¯ 0f (1 − αn ζ) Bnˆ = ζb (1 − α ¯ n ζ) f¯S¯µ 1−α ¯nζ |an | 1−α ¯nζ

469

m a.e.

Multiplying both sides by ζ we obtain (m a.e.)      1 − a ¯ ζ ζ − α −¯ a 1 − a ¯ ζ n n n n f¯S¯µ Bnˆ = b0 f . 1−α ¯nζ 1−α ¯ n ζ |an | 1−α ¯nζ And hence, if αn 6= 0 (and an 6= 0), then       1−a ¯n ζ −¯ αn ζ − αn 1−a ¯n ζ ¯ ¯ f Sµ An Bnˆ = b0 f m a.e. 1−α ¯ n ζ |αn | 1 − α ¯nζ 1−α ¯nζ If an 6= 0 and yet αn = 0, then    −¯ an ζ − an ¯ ¯ Bnˆ = b0 f f Sµ |an | 1−a ¯n ζ and therefore,   −¯ a n f¯S¯µ (1 − a ¯n ζ) ζBnˆ = b0 f (1 − a ¯n ζ) m a.e. |an | And if an = 0, yet αn 6= 0, then  2  −¯ αn ζ − αn ¯ ¯ f Sµ ζBnˆ = b0 f m a.e. |αn | 1 − α ¯nζ From this we obtain        1 −¯ α ζ − α −α 1 n n n f¯S¯µ Bnˆ = b0 f m a.e. 1−α ¯nζ |αn | 1−α ¯nζ |αn | 1−α ¯nζ Employing these tactics as we inductively work through the factors of B, we arrive at an equation of the form g¯S¯µ B = b0 h m a.e., where g and h are outer functions in H 2 (D), provided ∞ Y 1−a ¯n z n=1

1−α ¯nz

¯ to a non-zero function. And this occurs if converges uniformly on D ∞ X |an − αn | n=1

1 − |αn |

converges, which is equivalent to the convergence of ∞ X |an − αn | . 1 − |an | n=1

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Remark 3.2. One can use Theorem 3.1 to generate a variety of Blaschke products B such that B has no factor that satisfies b(D) 6= D, and yet KB contains functions of the form f Sµ , where 0 6≡ f ∈ H 2 (D) and Sµ is a non-trivial singular inner function. For instance, if S(z) := exp((z + 1)/(z − 1)), we know that z 7→

S(z) − a 1−a ¯S(z)

is a Blaschke product (provided 0 6= a ∈ D). And hence, by [5, Theorem 5 (i)], if B(z) := z(S(z) − a)/(1 − a ¯S(z)) and f (z) := 1/(1 − a ¯S(z)) (where 0 6= a ∈ D), ∞ then f S ∈ KB . Let {an }n=1 be an enumeration of the zeros of B listed according to multiplicity and let {αn }∞ n=1 be any sequence in D such that ∞ X |an − αn | n=1

1 − |an |

converges but such that no circle in D that is tangent to T at 1 contains more than a finite number of αn 0 s. Let B be the Blaschke product built with zero ∞ sequence {αn }∞ n=1 . Since {αn }n=1 necessarily has only one accumulation point in T, namely 1, the only factors of B that could satisfy b(D) 6= D are those of the form bc,α , as introduced in Remark 2.3. But each of these have infinitely many zeros on some circle in D that is tangent to T at 1. Therefore, B cannot have a factor b that satisfies b(D) 6= D. Yet, by Theorem 3.1, KB contains a non-trivial function that has S as a factor. We conclude this paper with two specific examples of Blaschke products B that have no factor b that satisfies b(D) 6= D, but have the property that KB contains (non-trivial) functions with non-trivial singular inner factors. There is nothing that special about these particular examples except that they allow us to illustrate certain methods. Example 3.3. For z in D, let g(z) = S(z)(3 − z) − 1, where, as before,   z+1 S(z) := exp . z−1 Trivially, g ∈ H ∞ (D), g is analytic across T \ {1} and g(x) → −1 as x in [0, 1) increases to 1. Therefore, g has no singular inner factor, and so g = hB0 , where h is an outer function in H ∞ (D) and B0 is a Blaschke product. We intend to show that there is an outer function f in H 2 (D) such that f S ∈ KB , where B(z) := zB0 (z). Now since hH 2 (D) is dense in H 2 (D), this is equivalent to showing that there is an outer function f in H 2 (D) such that Z f (ζ)S(ζ)g(ζ)ζ n dm(ζ) = 0 for n = 1, 2, . . . . T

For ζ in T, f (ζ)S(ζ)g(ζ) = f (ζ)(3 − ζ) − f (ζ)S(ζ).

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Therefore, we look for an outer function f in H 2 (D) such that (f · (3 − ζ) − f S) ⊥ H02 (D), where H02 (D) := {f ∈ H 2 (D) : f (0) = 0}. But notice that (for ζ in T), f (ζ) · (3 − ζ) = 3f (ζ) −

f (ζ) − f (0) f (0) − ζ ζ

and f (0)/ζ ⊥ H02 (D). Moreover, f (z) − f (0) − f S, z which we want orthogonal to H02 (D), is (necessarily) in H 2 (D). It follows that 3f (z) −

f (ζ) − f (0) − f (ζ)S(ζ) ζ is (necessarily) constant for m a.e. ζ in T. Now since f is an outer function, we may assume that f (0) = 1. And so the problem reduces to a search for an outer function f in H 2 (D) that satisfies 3f (ζ) −

f (ζ) − 1 − f (ζ)S(ζ) = M ζ (in T), for some constant M . Equivalently, 3f (ζ) −

m a.e.

f (ζ)(3ζ − (1 + ζS(ζ))) = M ζ − 1

m a.e.

Notice that z 7→ [3z − (1 + zS(z))] is bounded away from 0 for z (in D) near T and, by Rouch´e’s Theorem, this function has just one zero in D and this zero does not occur at z = 0. Therefore, we may choose M so that the zero of z 7→ M z − 1 coincides with the zero of z 7→ [3z − (1 + zS(z))] in D. With this choice of M , Mz − 1 f (z) := 3z − (1 + zS(z)) is an outer function in H ∞ (D) and one easily verifies that Z f (ζ)S(ζ)g(ζ)ζ n dm(ζ) = 0 for n = 1, 2, 3, . . . . T

Therefore, f S ∈ KB . Among other things, this implies that B0 (the Blaschke factor of g) is an infinite Blaschke product. Furthermore, notice that the zeros of g in D occur symmetrically with respect to R and 1 is the only possible accumulation point of these in T. Composing g with ϕ(w) := (w + 1)/(w − 1) — a standard M¨obius transformation from {w : Re(w) < 0} onto D — one finds that the zeros of g in D accumulate at 1 on a curve γ in D that is tangent to T at 1 with order of tangency like that of a circle. And, moreover, any circle in D that is tangent to T at 1 intersects γ at no more than two points. From this and observations like those made in Remark 3.2, it follows that B can have no factor b that satisfies b(D) 6= D.

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Example 3.4. In this example we consider g(z) := S(z) − (1 − z), where S is as before. Once again g ∈ H ∞ (D) and g is analytic across T \ {1}. So the only possible singular inner factors of g are of the form   z+1 Sc (z) := exp c , z−1 where c is some positive constant. However, if g were to have such a factor, then this would imply that (1 − z) has a non-trivial singular inner factor, which we know is false. Therefore, g has no (non-trivial) singular inner factor. Let B0 denote the Blaschke factor of g and let B(z) = zB0 (z). As in Example 3.3, we compose g with ϕ(w) := (w + 1)/(w − 1) and find that B has no factor b that satisfies b(D) 6= D. Indeed, the zeros of B0 lie on a curve in D that is symmetric with respect to R and tangent to T at 1, though with a lower order of tangency than that of a circle. In contrast with the result of Example 3.3, we argue here that KB contains no (non-trivial) function with S as a factor. If KB contained a function of the form f S, where 0 6≡ f ∈ H 2 (D), then, by Remark 1.2, we may assume that f is an outer function and proceed as in Example 3.3 (necessarily) to the equation f (ζ)(ζ + (1 − ζ)S(ζ)) = M ζ + 1

m a.e.

on T for some constant M . Under the conformal mapping ϕ(w) := (w+1)/(w−1), the zeros of (z + (1 − z)S(z)) in D correspond to solutions of the equation 2ew = 1 + w in {w : Re(w) < 0}. And a solution to this latter equation occurs when both 2eRe(w) = |1 + w| and Im(w) = arg(1 + w). Notice that 2eRe(w) = |1 + w| traces out a Jordan arc γ in {w : Re(w) ≤ 0} that is symmetric with respect to R√and that intersects the imaginary axis iR at pre√ cisely two points, namely, i 3 and −i 3. Let w = u + iv, where u and v are in R. Since u 7→ 4e2u − (1 + u)2 − v 2 is (strictly) increasing, any horizontal line {w : Im(w) = v√ 0 } intersects √ γ at most once. From this it follows that Im(w) decreases from 3 to −√ 3 as one traverses γ counterclockwise. Since arg(1 + w) < Im(w) at w = i 3, but as one traverses γ counterclockwise in {w : Re(w) ≤ 0 and Im(w) > 0}, Im(w) tends to 0 while arg(1 + w) increases to π (observe that cos(arg(1 + w)) = (1 + u)/(2eu ) for w in γ), there is necessarily a unique zero w0 of 2ew − (1 + w) in {w : Re(w) < 0 and Im(w) > 0}. Taking derivatives, we find that w0 is a zero of multiplicity 1. Therefore, 2ew − (1 + w) has precisely two zeros in {w : Re(w) < 0} each of multiplicity 1, namely w0 and w¯0 . Hence, (ζ + (1 − ζ)S(ζ)) has precisely two (simple) zeros in D; call these zeros z0 and z¯0 . Therefore, one cannot find an outer function f in H 2 (D) such that f (ζ)(ζ + (1 − ζ)S(ζ)) = M ζ + 1 m a.e. It follows that there is no (non-trivial) function in KB that has S as a factor. However, we now proceed to show that for any constant c, 0 < c < 1, there is a non-zero function f in H 2 (D) (and hence an outer function f in H 2 (D)) such

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that f Sc ∈ KB . For any such constant c we note that, for f 6≡ 0 in H 2 (D) and m a.e. ζ in T, f (ζ)Sc (ζ)g(ζ) = f (ζ)S1−c (ζ) − (1 − ζ)f (ζ)Sc (ζ). Mimicking the methods of Example 3.3, we hope to obtain (f (ζ)S1−c (ζ) − (1 − ζ)f (ζ)Sc (ζ)) ⊥ H02 (D); that is, S1−c (ζ)(f (ζ) − (1 − ζ)f (ζ)S(ζ)) ⊥ H02 (D). Or, in other words, (f (ζ) − (1 − ζ)f (ζ)S(ζ)) ⊥ S1−c · H02 (D); which in turn is equivalent to   f (ζ)S(ζ) − f (0)S(0) f (ζ) − f (ζ)S(ζ) + ∈ H 2 (D) S1−c · H02 (D). ζ Since we may assume that f (0)S(0) = 1 (see Remark 1.2), the problem reduces to finding a function h in H 2 (D) S1−c · H02 (D) so that the equation f (ζ)(ζ + (1 − ζ)S(ζ)) = ζh(ζ) + 1 is satisfied m a.e. by some non-trivial function f in H 2 (D). First notice that if α, β and d are constants and 0 < d ≤ 1 − c, then (αSd + β) ∈ H 2 (D) S1−c · H02 (D). And therefore, zh(z) + 1 can have the form αzSd (z) + βz + 1. We argue that α and β can be chosen so that αzSd (z) + βz + 1 has a zero at both z0 and z¯0 . Now, by adjusting d if necessary, we may assume that Sd (z0 ) 6∈ R. Moreover, for non-zero α, (−βz − 1)/(αz) can have the form A + C/z for any prescribed real constants A and C (provided z 6= 0). And real constants A and C can be found so that A + C/z equals Sd at z0 and hence at z¯0 . Therefore, we can find h in H 2 (D) S1−c · H02 (D) so that zh(z) + 1 has a zero at both z0 and z¯0 . With this choice of h, we define f on D \ {z0 , z¯0 } by f (z) =

zh(z) + 1 . z + (1 − z)S(z)

Since the only zeros in D of the denominator of f are z0 and z¯0 , and these are both simple zeros, f has analytic continuation to D. And our (earlier) analysis of the zeros of z + (1 − z)S(z) shows that the denominator of f is bounded away from 0 for z (in D) near T. Therefore, 0 6≡ f ∈ H ∞ (D) and, indeed, f Sc ∈ KB . Acknowledgement. The first author takes this opportunity to express gratitude to John B. Conway for his guidance and encouragement over the years. We also thank the referee for helpful suggestions leading to an improvement of exposition.

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J. R. Akeroyd and K. Karber

CMFT

References 1. J. R. Akeroyd, D. Khavinson and H. S. Shapiro, Weak compactness in certain star-shift invariant subspaces, J. Funct. Anal. 202 (2003), 98–122. 2. J. A. Cima and W. T. Ross, The Backward Shift on the Hardy Space, Mathematical Surveys and Monographs, 79. American Mathematical Society, Providence, RI, 2000. 3. R. G. Douglas, H. S. Shapiro and A. L. Shields, Cyclic vectors and invariant subspaces for the backward shift operator, Ann. Inst. Fourier 20 (1970), 37–76. 4. P. Duren, Theory of H p Spaces, Academic Press, New York, 1970. 5. K. M. Dyakonov, Kernels of T¨ oplitz operators via Bourgain’s factorization theorem, J. Funct. Anal. 170 (2000), 93–106. 6. K. M. Dyakonov, Zero sets and multiplier theorems for star-invariant subspaces, J. Anal. Math. 86 (2002), 247–269. 7. J. B. Garnett, Bounded Analytic Functions, Academic Press, New York, 1982. 8. K. Hoffman, Banach Spaces of Analytic Functions, Prentice-Hall, Englewood Cliffs, N.J., 1962. 9. N. K. Nikolskii, Treatise on the Shift Operator, Springer-Verlag, Berlin, 1986. John R. Akeroyd E-mail: [email protected] Address: Department of Mathematics, University of Arkansas, Fayetteville, AR 72701, U.S.A. Kristi Karber E-mail: [email protected] Address: Department of Mathematics, University of Arkansas, Fayetteville, AR 72701, U.S.A.