REFLEXIVITY, FACTORIZATION, AND HANKEL OPERATORS

EDWARD AZOFF, RUBÉN A. MARTÍNEZ-AVENDAÑO, AND JAMES SOLAZZO

Abstract. The space Hank of Hankel operators acting on the Hardy space H^{2} is a module over H^{∞}. There is a natural correspondence between weak* closed submodules of Hank and individual inner functions, and we apply work of V. Kapustin on Jordan models to characterize which submodules are reflexive in terms of the canonical factorization of these functions. We also prove that reflexivity of any weak* closed subspace of Hank is equivalent to reflexivity of the largest H^{∞} module it contains. Analogous results are obtained in the finite dimensional and “semi-infinite” dimensional settings.

1. Introduction

In [7], A. Beurling set up a correspondence between invariant subspaces of the unilateral shift and inner functions. Applications in invariant subspace theory proper have been most successful for operators enjoying a symbolic calculus over the Hardy space H^{∞}. In [26], D. Sarason showed the algebra AT of analytic Toeplitz operators to be reflexive in the sense that any operator sharing its invariant subspaces must in fact belong to AT . More recently, H^{∞} functional calculi played a key role in Scott Brown’s proof of the intransitivity of subnormal operators [8] and in joint work by H. Bercovici, S. Brown, and C. Pearcy establishing intransitivity of each operator of norm one whose spectrum contains the unit circle [3].

A. Loginov and V. Shulman, in [18], generalized the notion of reflexivity to operator spaces which are not necessarily closed under composition. The reflexive closure of a linear space L of Hilbert space operators, denoted ref L, consists of those operators B which belong to L locally in the sense that Bx belongs to the norm closure of Lx for each vector x in the underlying Hilbert space. The space L is reflexive if it coincides with its reflexive closure; at the opposite extreme, L is transitive if every operator on the underlying Hilbert space belongs to its reflexive closure. The concept of reflexivity of subspaces played an important spoiler role in Warren Wogen’s celebrated paper [28] in which he imbedded recalcitrant linear operators in singly–generated operator algebras, thereby solving several long standing questions concerning reflexivity of individual operators.

Hankel (Toeplitz) matrices are defined by the property that their entries only depend on the sum (respectively difference) of row and column number, i.e., they are “constant on cross-diagonals (diagonals)”. The terms also refer to bounded Hilbert space operators whose matrices relative to “standard” orthonormal bases take this form. The best choice to make in the separable infinite-dimensional case is the Hardy space H^{2}, equipped with its usual basis consisting of powers of z. Via the Nehari map, the space Hank of all Hankel operators on H^{2} is linearly isometrically isomorphic and weak* homeomorphic to L^{∞}/(zH^{∞}), the Banach space dual of H^{1}.

In [2], the first author and M. Ptak showed that each weak* closed linear space of (not necessarily analytic) Toeplitz operators is either reflexive or transitive and there are many of each type. By way of contrast, the first result of the present paper implies that no proper weak* closed subspace of Hank can be transitive.

Proposition 1.1. Hank is transitive, but its subspaces are “relatively reflexive” in the sense that each weak* closed subspace K of Hank satisfies Hank ∩ ref K = K.

Both Toeplitz and Hankel operators have L^{∞} symbols. Proposition 1.1 reflects the crucial difference: for Hankel operators, the Hilbert space inner product

⟨H_{ϕ}f, g⟩ = \frac{1}{2π} ∫ ϕfg^{∗}

involves the analytic function g^{∗} : z ↦ \overline{g(\overline{z})}. This means that studying reflexivity in Hank relies on cataloguing the different ways to factor an H^{1} function as a product of H^{2} functions, as opposed to the H^{2} · \overline{H^{2}} factorizations which underlie [2].

We next take up reflexivity of subspaces of Hank. Since reflexive closures are always weak* closed, all subspaces of Hank discussed below will be assumed to be closed in that topology. Given a subset N of H^{1}, we define its companion space by

C(N) := {H_{ϕ} : ∫ ϕu = 0 for each u ∈ N} . (1)

We will see that every weak* closed subspace of Hank is a companion space.

We will concentrate on two special companion spaces associated with an individual element u of H^{1}: the companion space C_{u} of the singleton set {u}, and the companion space M_{u} of the set uH^{∞}. Every hyperspace of Hank takes the former form, and Beurling’s Theorem will show that M_{u} represents the most general (weak* closed) submodule of Hank.

Here is our main result. The proof, given in Section 5, does not depend on Section 4; it is accomplished by carefully studying a preadjoint of the Nehari map which sends trace class operators to H^{1} functions.

Theorem 1.2. Let N ⊂ H^{1}, take W := C(N) to be the corresponding subspace of Hank, and set θ as the greatest common divisor of the inner factors of the members of N. Then ref W = W + ref M_{θ}.

To gauge the strength of this theorem, consider the special case when N consists of a single outer function ω and thus W = C_{ω}. In this case θ = 1 and M_{θ} = {0} is trivially reflexive, hence the theorem above gives ref C_{ω} = C_{ω}. So even though C_{ω} is a hyperplane in the transitive space Hank, the theorem still guarantees its reflexivity. More generally, Theorem 1.2 leads to equivalence of the first three conditions in the following result.

Theorem 1.3. Let N ⊂ H^{1}, take W := C(N) to be the corresponding subspace of Hank, and set θ as the greatest common divisor of the inner factors of the members of N. Then the following are equivalent:

(1) W is reflexive.

(2) C_{θ} is reflexive.

(3) M_{θ} is reflexive.

(4) S(θ) is reflexive.

(5) θ = βσ where β is a Blaschke product having no repeated roots while σ is a singular inner function whose associated measure vanishes on Beurling-Carleson sets.

In particular, reflexivity of W is equivalent to reflexivity of the largest H^{∞} module it contains.

The equivalence (3) iff (4) is established in Section 4. In connection with Condition (4), recall that if θ is an inner function and S is the unilateral shift operator on H^{2} (i.e., multiplication by the independent variable z), the Jordan model operator S(θ) is defined to be the compression of S to the space H^{2} ⊖ θH^{2}. The equivalence of (4) and (5) is due to V. Kapustin [16, 17].

In Section 6, we determine the finite-rank members of ref M_{θ}. These are absent when θ is a totally nonatomic inner function, but plentiful in the opposite case that the measure associated with the singular factor of θ is supported on a countable set. This leads to a concrete computation of ref M_{θ} in the latter situation.

All Hankel and Toeplitz operators are compressions of multiplication operators. So far, we have been discussing the so called “infinite” case of Hankel operators in B(H^{2}). We now describe our results in the “semi-infinite” and “finite” cases, covered in the last two sections of the paper. (This less technical material can be read immediately after Section 2.) One of the original motivations of the present paper was the use of such settings to gain insight on the Toeplitz spaces studied in [2]. In fact, the simple device of reversing the order of a finite orthonormal basis shows reflexivity questions for spaces of Hankel and Toeplitz operators are entirely equivalent in these settings.

Write P_{n} for the set of all polynomials with degree less than or equal to n, thought of as a subspace of H^{2}. Each of these spaces (including H^{2}) is equipped with a standard orthonormal basis, and a bounded operator acting between two of them is considered to be Hankel if its matrix relative to the corresponding standard bases is constant on skew diagonals. We write Hank(P_{n}, H^{2}) for the space of Hankel operators in B(P_{n}, H^{2}) (the semi-infinite case) and Hank(P_{n}, P_{m}) for the Hankel operators in B(P_{n}, P_{m}) (the finite case). These operators have symbols as well, taken from \overline{H^{2}} in the semi-infinite case, and from \overline{P_{m+n}} in the finite case. Companion spaces are defined as in Display (1), with N ranging over the subsets of the corresponding preduals, H^{2} and P_{n+m} respectively.

Theorem 1.4. Suppose n > 0. Let N ⊂ H^{2}, take M := C(N) to be the corresponding subspace of Hank(P_{n}, H^{2}), and set γ as the greatest common divisor of the inner factors of the members of N. Then the following are equivalent:

(1) M is reflexive.

(2) C_{γ} is reflexive.

(3) γ is a Blaschke product without repeated roots.

The criteria in the next theorem are quite computable; one can check whether r(z) has any repeated roots by applying the Euclidean algorithm to r and r′.

Theorem 1.5. Let N be a set of polynomials in P_{m+n}, take M := C(N) to be the corresponding subspace of Hank(P_{n}, P_{m}), and set r = gcd N. Then the following are equivalent:

(1) M is reflexive.

(2) M is contained in a reflexive hyperplane.

(3) Some member of N has degree ≥ m + n − 1 and r(z) has no repeated roots.

Note the contrast between the middle conditions of the last two theorems: the hyperplane C_{γ} of 1.4 need not contain M; on the other hand, reflexivity of M in 1.5 does not imply reflexivity of C_{r}. Proofs techniques are also different: 1.4 takes advantage of the existence of a strictly cyclic vector for Hank(P_{n}, H^{2}); that tool is not available for 1.5 and we rely on a Theorem of Bertini instead.

2. Preliminaries

Hankel and Toeplitz Operators. We will concentrate on operators on the Hardy–Hilbert space H^{2} (see [12, 13, 15] for further information on H^{2}), but we will use the letter H whenever we make a definition that is valid for general Hilbert space. We think of H^{2} both as a space of analytic functions and as a subspace of L^{2} := L^{2}(𝕋). We also think of the space L^{∞} as a subset of L^{2} and we define the subspace H^{∞} of L^{∞} as L^{∞} ∩ H^{2}. The dual space of H^{1} is L^{∞}/(zH^{∞}) (see, for example, [12, p. 161]), under the pairing

⟨ϕ + zH^{∞}, u⟩ := \frac{1}{2π} ∫_{−π}^{π} ϕ(e^{iθ})u(e^{iθ}) dθ, ϕ + zH^{∞} ∈ L^{∞}/(zH^{∞}), u ∈ H^{1}.

The (forward) shift operator S is defined on H^{2} as multiplication by the independent variable z. This has the effect of shifting the Fourier coefficients forward by one place (hence the name). Its adjoint, S^{∗}, is usually called the backward shift since it shifts the Fourier coefficients one place back. The flip operator J : L^{2} → L^{2}, defined by (Jf)(z) := f(\overline{z}), interchanges analytic and coanalytic functions. If f is a function in L^{2}, we also define f^{∗} by f^{∗}(z) = \overline{f(\overline{z})}; note that f^{∗} ∈ H^{2} if and only f ∈ H^{2}.

The main objects of study of this paper are Hankel operators. Most of the following basic facts concerning them can be found in [22, 23, 24].

We will say that an operator H is Hankel if, with respect to a canonical basis, the matrix of H has the property that all diagonals perpendicular to the main one (usually called “skew–diagonals”) are constant.

The canonical basis for H^{2} is {z^{n}}_{n=0}^{∞}, and thus H ∈ B(H^{2}) is a Hankel operator if and only if it satisfies the equation S^{∗}H = HS. We will denote the space of all bounded Hankel operators on H^{2} by Hank.

Given ϕ ∈ L^{∞}, define the operator H_{ϕ} : H^{2} ⟶ H^{2} by

H_{ϕ}f = P J(ϕf),

where P denotes the orthogonal projection from L^{2} onto H^{2} and J is the flip operator defined above. Since ϕ ∈ L^{∞} we have that H_{ϕ} ∈ B(H^{2}); also (H_{ϕ})^{∗} = H_{ϕ^{∗}}. It is easy to see that each such H_{ϕ} is a Hankel operator. The converse is a classical result of Nehari [20].

Theorem 2.1. The Nehari map Γ^{∗} : L^{∞}/(zH^{∞}) → B(H^{2}), defined by

Γ^{∗}(ϕ + zH^{∞}) = H_{ϕ}, ϕ ∈ L^{∞},

is an isometry with range Hank.

Much of what we do in this article ultimately depends on the following basic computation.

Proposition 2.2. Given a symbol ϕ ∈ L^{∞} and vectors f, g ∈ H^{2}, we have

⟨H_{ϕ}f, g⟩ = \frac{1}{2π} ∫_{−π}^{π} ϕ(e^{iθ})f(e^{iθ})g^{∗}(e^{iθ})dθ.

Proof. By definition, we have ⟨H_{ϕ}f, g⟩ = ⟨P J(ϕf), g⟩ = ⟨J(ϕf), g⟩, which coincides with the integral \frac{1}{2π} ∫_{−π}^{π} ϕ(e^{−it})f(e^{−it})\overline{g(e^{it}})dt. The desired result follows by making the substitution θ = −t and recalling the definition of g^{∗}. □

Recall that an operator T ∈ B(H^{2}) is called a Toeplitz operator if S^{∗}T S = T. This means that the matrix representation of T with respect to the canonical basis has the property that the diagonals parallel to the main one are constant.

For each ϕ ∈ L^{∞} define the operator T_{ϕ} : H^{2} ⟶ H^{2} as

T_{ϕ}f = P (ϕf),

where P is the orthogonal projection of L^{2} onto H^{2}. Since ϕ ∈ L^{∞} we have that T_{ϕ} ∈ B(H^{2}). It is easy to see that the operator thus defined is a Toeplitz operator and one can show that all Toeplitz operators arise in this form.

The function ϕ above is called the symbol of the Toeplitz operator T_{ϕ}. Notice that in this case the symbols are unique. We shall need the fact that T_{ϕ}^{∗} = T\overline{ϕ} for any ϕ ∈ L^{∞}. When ϕ ∈ H^{∞}, we will often write ϕ(S) for the Toeplitz operator T_{ϕ} and ϕ(S^{∗}) for T_{J(ϕ)}. These conventions, which are a special case of the H^{∞} functional calculus for completely nonunitary contractions, will improve the appearance of many formulas in the sequel. (No corresponding convention is used for Hankel operators because powers of Hankel operators are usually not Hankel.) In particular, the shift and its adjoint are both examples of Toeplitz operators: S = T_{z} and S^{∗} = T\overline{z}. For all these basic facts about Toeplitz operators, consult [12].

One way of thinking of the following well-known and easy to prove fact is that analytic Toeplitz operators implement the H^{∞}-module structure on Hank. This makes the Nehari map a module isomorphism onto Hank.

Proposition 2.3. If ψ ∈ L^{∞} and ϕ ∈ H^{∞} then ϕ(S^{∗})H_{ψ} = H_{ψϕ} = H_{ψ}ϕ(S).

Duality. Assume that X is a Banach space, and that X^{∗} is its dual space, where the duality is implemented by the bilinear form

⟨·, ·⟩ : X^{∗} × X ⟶ ℂ.

In such a setting, the annihilator of a subset L of X is defined by

L^{⊥} := {a ∈ X^{∗} : ⟨a, t⟩ = 0 for all t ∈ L},

while the preannihilator of a subset M of X^{∗} is defined by

M_{⊥} := {t ∈ X : ⟨a, t⟩ = 0 for all a ∈ M}.

From topological vector space theory, we know that the map L ↦ L^{⊥} sets up a one-to-one correspondence between the collection of norm-closed linear manifolds in X and the collection of weak* closed linear manifolds in X^{∗}; the inverse map is implemented by taking preannihilators. In fact, we will reserve the term subspace to refer to these two types of linear manifolds, i.e., either norm closed in X or weak* closed in X^{∗}. Moreover, if M is a subset of X or X^{∗}, we write [M] for the subspace (closed in the appropriate topology) generated by M; given another subset L ⊂ M, we say L is total in M if [L] = [M].

Reflexivity. Our main use of duality will involve the spaces B(H) and Tr(H), where B(H) is the full algebra of bounded linear operators acting on a separable Hilbert space H, while Tr(H) is its ideal of trace class operators under the trace norm [27]. The space B(H) is the dual of Tr(H) under the pairing

⟨A, T ⟩ := tr (AT), A ∈ B(H), T ∈ Tr(H).

The corresponding weak* topology on B(H) is sometimes called the ultraweak operator topology.

The rank-one operators play an important role in what follows. Given f, g ∈ H, the operator f ⊗ g is defined by (f ⊗ g)h = ⟨h, g⟩ f. Every operator of rank one or less takes this form, and each operator of rank n can be written as a sum of n such operators. The collection of operators having rank at most n is denoted F_{n}(H) and the collection of all operators of finite rank is denoted F(H); reference to the underlying Hilbert space is usually omitted. If A and B are bounded operators on H, we have that A(f ⊗ g)B = (Af) ⊗ (B^{∗}g) and

⟨B, f ⊗ g⟩ = tr (B(f ⊗ g)) = ⟨Bf, g⟩ .

The first ⟨·, ·⟩ in this equation reflects the duality between B(H) and Tr(H) (without adjunction), while the second ⟨·, ·⟩ above refers to the usual inner product (involving conjugation) on H.

Whenever we write a series of the form ∑_{n=1}^{∞} f_{n} ⊗ g_{n}, it will be assumed that ∑ ∥f_{n}∥∥g_{n}∥ < ∞. Such series converge in the trace norm topology and every trace class operator can be represented in this way. The preceding display generalizes to

⟨B, ∑ f_{n} ⊗ g_{n}⟩ = ∑ ⟨Bf_{n}, g_{n}⟩ , B ∈ B(H), ∑ f_{n} ⊗ g_{n} ∈ Tr(H).

We now recall the following definitions, along with some of their basic properties. All of it can be found in [1]. Historical perspective can also be gained from [5], [11] (especially Section 22) and [25].

Definition 2.4. Let L be a linear manifold in B(H).

(1) The reflexive closure of L is given by

ref L := \{B ∈ B(H) : Bf ∈ \overline{Lf} for all f ∈ H\}

(2) The space L is transitive if ref L exhausts the whole space of operators.

(3) The space L is reflexive if L = ref L.

(4) A single operator B ∈ B(H) is reflexive if the weak* closed algebra generated by B and the identity operator is a reflexive linear manifold.

(5) We say L is elementary if L_{⊥} + F_{1} exhausts the predual space Tr(H).

We always have L ⊂ ref L ⊂ B(H), with (2) and (3) representing extreme possibilities. Note that A ∉ ref L if and only if there are vectors f, g ∈ H with g ⊥ Lf, but ⟨A, f ⊗ g⟩ = ⟨Af, g⟩ ≠ 0. This provides the useful characterization

ref L = (L_{⊥} ∩ F_{1})^{⊥}.

In particular, ref L is always weak* closed and L is transitive if and only if L_{⊥} contains no rank-one operators; for weak* closed L, reflexivity is tantamount to having L_{⊥} ∩ F_{1} total in L_{⊥}.

More generally, for a positive integer k, a weak* closed subspace L is said to be k-reflexive if L_{⊥} ∩ F_{k} total in L_{⊥}. There is a fleeting reference to 2-reflexivity in Theorem 3.2, but otherwise, this concept will not be needed in this paper.

Concept (5) above goes by the name Property ℵ_{1} in [5]. We will find the third part of the following Proposition especially useful.

Proposition 2.5. Let L be an elementary subspace of B(H) and suppose K is a subspace of L.

(1) K = L ∩ ref K

(2) No proper subspace of L can be transitive.

(3) If ref K ⊂ L then K is reflexive.

Proof. Since every operator space is contained in its reflexive closure, we only have to be concerned with the inclusion ⊃ of (1). So assume a ∉ K, but a ∈ L. Then there is a trace class operator t ∈ K_{⊥} with ⟨a, t⟩ ≠ 0. Because L is elementary, we may even assume t ∈ F_{1}, which means a ∉ ref K, and (1) is established. (2) and (3) are immediate consequences of (1). □

We also need a few general facts concerning reflexivity.

Lemma 2.6. Suppose X, Y are operators, while T is a linear space of operators. (1) Then X(ref T )Y ⊂ ref (XT Y).

(2) If Y XA = A for each A ∈ T , then

(a) Xref T = ref (XT ).

(b) Y ref (XT ) = ref T .

(c) T is reflexive if and only if XT is reflexive.

Proof. Given C ∈ ref T and f in the underlying Hilbert space, we have

XCY f ∈ X[\overline{T Y f}] ⊂ \overline{XT Y f},

so XCY ∈ ref (XT Y). Thus X(ref T )Y ⊂ ref (XT Y) and (1) is established. In particular, (I − XY)ref (XT ) ⊂ ref ((X − XY X)T ) = {0}, so

ref (XT ) = XY ref (XT ) ⊂ Xref (Y XT ) = Xref T ⊂ ref (XT ),

and ref (XT ) = Xref T , giving (2a). Multiplying the last equation by Y on the left also gives Y ref (XT ) = Y Xref T = ref T , and (2b) is established. Finally, (2c) follows immediately from (2a) and (2b). □

The reader primarily interested in the semi-infinite or finite cases should now skip to Section 7 or Section 8 respectively.

3. Cosymbols and companion spaces

The cosymbol map. The Nehari map Γ^{∗} of Theorem 2.1 is weak* to weak* continuous. By topological vector space theory, we therefore know it is dual to a map Γ : Tr(H^{2}) → H^{1}. We refer to Γ as the cosymbol map. It will allow us to investigate various reflexivity questions in a function theoretic setting.

Proposition 3.1. The cosymbol map Γ has the following properties.

(1) Γ is given by the explicit formula Γ(∑ f_{n} ⊗ g_{n}) = ∑ f_{n}g_{n}^{∗}.

(2) Γ is a contraction.

(3) Γ is surjective. In fact, given u ∈ H^{1}, there is an operator f ⊗ g of rank at most one satisfying Γ(f ⊗ g) = u and ∥f ⊗ g∥_{Tr(H^{2})} = ∥u∥_{1}.

(4) The kernel of Γ is Hank_{⊥}.

Proof. Given f, g ∈ H^{2} and ϕ ∈ L^{∞}, we have

⟨ϕ + zH^{∞}, Γ(f ⊗ g)⟩ = ⟨H_{ϕ}, f ⊗ g⟩ =< H_{ϕ}f, g >=< ϕ + zH^{∞}, fg^{∗} >,

where the first equality comes from the definition of dual operator, while the third equality is the basic Proposition 2.2. This means Γ(f ⊗ g) = fg^{∗}, from which (1) follows by linearity and continuity. (2) is true since ∥Γ^{∗}∥ = 1 by the Nehari Theorem.

It is well known that any function u ∈ H^{1} can be factored as u = fg^{∗} with f, g ∈ H^{2} and ∥f∥_{2} = ∥g∥_{2} = \sqrt{∥u∥_{1}}. It then follows that Γ(f ⊗ g) = u and ∥f ⊗ g∥_{Tr(H^{2})} = ∥u∥_{1}, which establishes (3).

Nehari’s Theorem tells us that range(Γ^{∗}) = Hank, whence kernel(Γ) = Hank_{⊥} by general topological vector space theory. □

The full space Hank. We now use Proposition 3.1 to obtain reflexivity properties of the full space of Hankel operators on H^{2}. Theorem 3.2 was proven in [19].

Theorem 3.2. The space Hank is transitive, but it is 2-reflexive and elementary.

Proof. From Proposition 3.1, we know that Hank_{⊥} is the kernel of the cosymbol map Γ. In particular, f ⊗ g ∈ Hank_{⊥} implies fg^{∗} = 0, which only occurs when f = 0 or g = 0. Thus Hank_{⊥} has no rank-one members, which makes Hank transitive.

For 2-reflexivity, suppose A ∈ (Hank_{⊥} ∩ F_{2})^{⊥}. For each f, g ∈ H^{2}, we have Γ(f ⊗ zg − zf ⊗ g) = 0, which implies f ⊗ zg − zf ⊗ g ∈ Hank_{⊥} ∩ F_{2} since ker Γ = Hank_{⊥}. Hence,

⟨(S^{∗}A − AS)f, g⟩ = ⟨A, f ⊗ zg − zf ⊗ g⟩ = 0,

implying S^{∗}A = AS. Thus A ∈ Hank.

To see that Hank is elementary, let T ∈ Tr(H^{2}). Set u := Γ(T), and choose f, g ∈ H^{2} as in Part (3) of Proposition 3.1. Then

T = (T − f ⊗ g) + f ⊗ g ∈ Hank_{⊥} + F_{1}.

Thus Hank_{⊥} + F_{1} exhausts Tr(H^{2}) and Hank is elementary by definition. (In fact we have shown that Hank_{⊥} + Ball(F_{1}) covers the unit ball of Tr(H^{2}) which means that Hank has the stronger Property ℵ_{1}(1) in the terminology of [5].) □

The following is a special case of Proposition 2.5. The third assertion will be used every time we show a subspace of Hank to be reflexive. Part (2) was obtained in [19].

Corollary 3.3. Let K be a subspace of Hank.

(1) K = Hank ∩ ref K.

(2) No proper subspace of Hank can be transitive.

(3) ref K ⊂ Hank if and only if K is reflexive.

Proof of Proposition 1.1: Observe that Proposition 1.1 is just the first assertion of Theorem 3.2 and Part (1) of Corollary 3.3. □

Companion spaces. Proposition 3.4 will enable us to move freely between subspaces of Hank and H^{1}.

For notational clarity, we will use the symbol ± to denote annihilators (and preannihilators) relative to the duality between L^{∞}/(zH^{∞}) and H^{1} mentioned at the beginning of the previous section while continuing to use the standard ⊥ notation for annihilators relating to the duality between B(H^{2}) and Tr(H^{2}).

One further piece of notation will prove convenient for us. When M is a subspace of a Banach space X, we will write LAT(M) for the lattice of all subspaces of M and COLAT(M) for the lattice of all subspaces of X which contain M. The same conventions are adopted for (weak* closed) subspaces of X^{∗}.

Proposition 3.4. For each subset K of H^{1}, we have (Γ^{−1}(K))^{⊥} = Γ^{∗}(K^{±}). Moreover, the following is a commutative diagram of lattice isomorphisms and antiisomorphisms.

LAT(H^{1})  ⟶^{±} LAT(L^{∞}/(zH^{∞}))
Γ^{−1}↓  ↓_{Γ^{∗}}
COLAT(Hank_{⊥}) ⟶^{⊥} LAT(Hank)

Proof. Given ϕ + zH^{∞} ∈ K^{±} and t ∈ Γ^{−1}(K), we have ⟨Γ^{∗}(ϕ + zH^{∞}), t⟩ = ⟨ϕ + zH^{∞}, Γ(t)⟩ = 0, so (Γ^{−1}(K))^{⊥} ⊃ Γ^{∗}(K^{±}). To get the opposite inclusion, take A ∈ (Γ^{−1}(K))^{⊥}. Since Γ^{−1}(K) ⊃ kernel Γ, we see that A ∈ (Hank_{⊥})^{⊥} = Hank. Since the range of the Nehari map Γ^{∗} exhausts Hank, we can write A = Γ^{∗}(ϕ + zH^{∞}) for some ϕ ∈ L^{∞}. Let f ∈ K. By surjectivity of Γ, we have f = Γ(t) for some t ∈ Γ^{−1}(K), and

< ϕ + zH^{∞}, f >=< A, t >= 0.

This shows that ϕ + zH^{∞} ∈ K^{±}, whence A ∈ Γ^{∗}(K^{±}) as desired.

Since any M ∈ LAT(L^{∞}/zH^{∞}) can be expressed as M = K^{±} for K = M_{±}, we see that Γ^{∗} maps LAT(L^{∞}/zH^{∞}) into LAT(Hank), so all the maps in the diagram are well-defined, and the diagram commutes.

To complete the proof, it suffices to show that three of the lattice maps in the diagram are bijective. Like all annihilator maps, the horizontal maps ±, ⊥ are lattice anti-isomorphisms. Write π for the natural projection map from Tr(H^{2}) to Tr(H^{2})/Hank_{⊥}. Then π induces a lattice isomorphism between COLAT(Hank_{⊥}) and LAT(Tr(H^{2})/Hank_{⊥}). Also, Γ∘π^{−1} is a Banach space isomorphism and hence sets up a lattice isomorphism between LAT(Tr(H^{2})/Hank_{⊥}) and LAT(H^{2}). Thus Γ = (Γ ∘ π^{−1}) ∘ π provides a lattice isomorphism, and the vertical map Γ^{−1} is our third lattice bijection. □

Definition 3.5. The companion space to a subset K of H^{1} is defined as

C(K) := {H_{ϕ} ∈ Hank : ∫ ϕu = 0 for each u ∈ K}.

For an individual u ∈ H^{1}, we write C_{u} := C({u}) and M_{u} := C(uH^{∞}).

Thus, in duality notation, we have C(K) = Γ^{∗}(K^{±}), which coincides with (Γ^{−1}(K))^{⊥} by the last proposition.

Here are some observations on companion spaces.

Proposition 3.6. Let u ∈ H^{1} and K ⊂ H^{1}.

(1) C(K) only depends on the closed linear span [K] of K.

(2) C sets up a lattice anti-isomorphism between the subspaces of H^{1} and subspaces of Hank.

(3) C_{u} is the most general hyperspace of Hank.

(4) M_{u} is the most general submodule of Hank; it only depends on the inner factor of u.

(5) Let s be the greatest common divisor of all inner factors of members of [K]. Then M_{s} is the largest submodule of C(K).

(6) In order for a rank-one operator f ⊗ g to belong to [C(K)]_{⊥} it is necessary and sufficient that fg^{∗} ∈ [K].

Proof. We will be referring to the diagram of Proposition 3.4.

(1) We have C(K) = Γ^{∗}(K^{±}) and K^{±} = [K]^{±}.

(2) This follows from the bijectivity of the lattice maps in the diagram.

(3) The Nehari map Γ^{∗} preserves codimension (since it is a vector space isomorphism), while taking annihilators interchanges codimension with dimension.

(4) Given ϕ + zH^{∞} ∈ L^{∞}/(zH^{∞}), u ∈ H^{1}, and f ∈ H^{∞}, we have

⟨f(ϕ + zH^{∞}), u⟩ = ⟨ϕ + zH^{∞}, fu⟩ .

It follows that N^{±} is a submodule of L^{∞}/(zH^{∞}) whenever N is a submodule of H^{1}; conversely, when K is a submodule of L^{∞}/(zH^{∞}), its preannihilator K_{±} must be a submodule of H^{1}. Notice that Γ^{∗} is an H^{∞}-module isomorphism and thus M is a submodule of Hank if and only if Γ^{∗−1}(M) is a submodule of L^{∞}/(zH^{∞}), and this happens if and only if (Γ^{∗−1}(M))_{±} is a submodule of H^{1}. But Beurling’s Theorem tells us that every submodule of H^{1} is singly-generated and of the form [uH^{∞}]. Also, it only depends on the inner factor of u. Thus (Γ^{∗−1}(M))_{±} = [uH^{∞}] for some u ∈ H^{1} and since M_{u} = C(uH^{∞}) = Γ^{∗}([uH^{∞}]^{±}), it follows that M must equal M_{u}.

(5) The largest submodule M of C(K) is {H_{ϕ} : H_{fϕ} ∈ C(K) for all f ∈ H^{∞}}. The submodule C^{−1}(M) equals [H^{∞}K] which by Beurling’s Theorem equals sH^{∞}.

(6) By Part (1) we know that C(K) = C([K]) = (Γ^{−1}([K]))^{⊥}, so [C(K)]_{⊥} = Γ^{−1}([K]) by continuity. But f ⊗ g belongs to the latter space if and only if fg^{∗} = Γ(f ⊗ g) ∈ [K].

□

Parts (4) and (5) of the last proposition motivate the following definition.

Definition 3.7. Let M be a subspace of Hank. The greatest common divisor of all inner factors of members of C^{−1}(M) is called the seed of M and denoted seed (M).

Corollary 3.8. Let W be a subspace of Hank with seed θ.

(1) M_{θ} is the largest H^{∞}-submodule of Hank contained in W.

(2) The rank-one operator f ⊗ g belongs to W_{⊥} if and only if fg^{∗} ∈ C^{−1}(W).

(3) The rank-one operator f ⊗ g belongs to [M_{θ}]_{⊥} if and only if θ divides fg^{∗}.

Proof. Part (1) follows by applying Proposition 3.6(5) with K := C^{−1}(W). To obtain part (2), apply Proposition 3.6(6) to K := C^{−1}(W). For (3), apply Proposition 3.6(6) to K := θH^{∞} to see that f ⊗ g ∈ [M_{θ}]_{⊥} if and

only if fg^{∗} ∈ [θH^{∞}]. But the latter condition is equivalent to fg^{∗} being divisible by θ. □

4. Reflexivity of submodules of Hank

In Theorem 4.5, we establish the equivalence of Conditions (3) and (4) of Theorem 1.3, which we then combine with Kapustin’s result to the effect that Conditions (4) and (5) of Theorem 1.3 are also equivalent. Theorem 4.5 is a two-way street which can be used to apply the results of Section 6 below to Jordan models.

Definition 4.1. Let θ be an inner function. The subspace H^{2} ⊖ θH^{2} is denoted by H(θ). The Jordan model S(θ) : H^{2} ⟶ H^{2} is defined by

S(θ) := P_{H(θ)}S.

Here P_{H(θ)} is the orthogonal projection from H^{2} to H(θ).

Note that since θH^{2} is invariant under the shift S, we actually have S(θ) = P_{H(θ)}S = P_{H(θ)}SP_{H(θ)}. In fact, the usual convention is to restrict the domain and codomain of S(θ) to H(θ) (the resulting operator being called the compression of the shift S to H(θ)) but it simplifies our notation to have all these operators acting on the same Hilbert space H^{2}. Jordan models have been studied extensively (see, for example, [4, 21]) and it is well known that they are related to Hankel operators. There is a natural H^{∞} functional calculus for S(θ). Proofs of Theorem 4.3 can be found, for example, in [4, p. 41] and [21, p. 230].

Definition 4.2. For every ϕ ∈ H^{∞}, the operator ϕ(S(θ)) on H^{2} is defined by ϕ(S(θ)) := P_{H(θ)}ϕ(S).

Theorem 4.3. Let θ be an inner function. Then the weak* closed algebra generated by S(θ) is

A_{S(θ)} := {ϕ(S(θ)) : ϕ ∈ H^{∞}} .

The identity of this algebra is P_{H(θ)}; when domains and codomains are restricted to H(θ), the algebra A_{S(θ)} becomes the commutant of S(θ).

The following variant of the Commutant Lifting Theorem (see e.g. [21, p. 230]) provides our basic tool for relating spaces of Jordan models and Hankel operators.

Lemma 4.4. Let θ be an inner function and write X := H_{z\overline{θ}}.

(1) X^{∗}X = P_{H(θ)}, i.e., X is a partial isometry with initial space H(θ).

(2) For every function ϕ ∈ H^{∞} we have H_{z\overline{θ}ϕ} = Xϕ(S(θ)).

(3) M_{θ} = XA_{S(θ)}.

Proof. (1) can be found on Page 34 of [24]; it can also be checked directly.

H_{z\overline{θ}}ϕ(S) = H_{z\overline{θ}ϕ} by Proposition 2.3. Since H_{z\overline{θ}} = H_{z\overline{θ}}P_{H(θ)} by Part (1), we have

H_{z\overline{θ}ϕ} = H_{z\overline{θ}}ϕ(S) = H_{z\overline{θ}}P_{H(θ)}ϕ(S) = H_{z\overline{θ}}ϕ(S(θ)),

establishing (2).

For (3), first apply the definition to note that the Hankel operator H_{ψ} is in M_{θ} if and only if ∫ ψθf = 0 for all f ∈ H^{∞}, i.e., if and only if ψθ ∈ zH^{∞} which is in turn equivalent to ψ being of the form z\overline{θ}ϕ for some function ϕ ∈ H^{∞}.

That is, a Hankel operator is in M_{θ} if and only if it has the form H_{z\overline{θ}ϕ} for ϕ ∈ H^{∞}. As ϕ ranges through H^{∞}, Part (2) implies that the corresponding Hankel operators H_{z\overline{θ}ϕ} exhaust M_{θ}, while ϕ(S(θ)) traces out A_{S(θ)} by Theorem 4.3. □

Theorem 4.5. Let θ be an inner function. Then:

(1) ref M_{θ} = H_{z\overline{θ}}ref A_{S(θ)}.

(2) ref A_{S(θ)} = (H_{z\overline{θ}})^{∗}ref M_{θ}.

(3) M_{θ} is a reflexive module of Hankel operators if and only if S(θ) is a reflexive Jordan model.

Proof. Take X = H_{z\overline{θ}} and Y = X^{∗}, and write T := A_{S(θ)}. The preceding lemma shows that the hypothesis of Lemma 2.6(2) is satisfied and hence the conclusions of the current Theorem follow. □

Kapustin’s Theorem. We close this section with some detail concerning the equivalence of parts (4) and (5) of Theorem 1.3.

Definition 4.6. Let E be a closed subset of (−π, π]. We say that E is a Beurling-Carleson set (a BC-set) if E has zero Lebesgue measure and the sum ∑ ℓ_{k} log \frac{2π}{ℓ_{k}} is finite. Here {ℓ_{k}} consists of the lengths of the intervals making up the complement of E.

We say that a measure μ is BC-vanishing if μ(E) = 0 for every BC-set E.

Note in particular that singleton point sets are BC and thus BC-vanishing measures must be totally nonatomic.

Here is the main result of [16, 17]

Theorem 4.7 (Kapustin). Let θ be an inner function and write μ for the representing measure of the singular factor of θ. The following statements are equivalent.

(1) The function θ has no multiple zeros in 𝔻 and μ(E) = 0 for every BC-set E.

(2) The algebra H^{∞}/θH^{∞} is the weak*-closed linear span of its idempotent members.

(3) The Jordan model S(θ) is reflexive.

Regarding the proof of Theorem 1.3: That (3) ⟺ (4) is the content of Theorem 4.5, while (4) ⟺ (5) follows from Theorem 4.7. □

5. Reflexivity of subspaces of Hank

In this section, we complete the proofs of Theorems 1.2 and 1.3. When ω is outer, M_{ω} = {0} is the only submodule of C_{ω}. The 0-module is trivially reflexive, so our first job is establishing reflexivity of the hyperspace C_{ω}.

Proposition 5.1. If ω is outer then C_{ω} is reflexive.

Proof. Let B ∈ ref C_{ω} = ((C_{ω})_{⊥} ∩ F_{1})^{⊥}. Factor ω = ω_{1}ω_{2}^{∗} as the product of two outer H^{2} functions. Given a real number a ∈ (−1, 1), define functions f, g ∈ H^{2} by f(z) := 1 − az and g(z) := \frac{1}{1−az}. Then fg = 1 so (gω_{1})(f^{∗}ω_{2})^{∗} = ω. In view of Corollary 3.8(2), we have (gω_{1}) ⊗ (f^{∗}ω_{2}) ∈ (C_{ω})_{⊥} and hence

⟨(I − aS^{∗})B(I − aS)^{−1}ω_{1}, ω_{2}⟩ = ⟨Bgω_{1}, f^{∗}ω_{2}⟩ = 0, −1 < a < 1.

Expanding this expression as a power series in a yields

(B + ∑_{n=0}^{∞} a^{n+1}(BS − S^{∗}B)S^{n}) ω_{1} ⊥ ω_{2}, −1 < a < 1.

Since the orthogonal complement of ω_{2} is linear and (norm) closed, repeatedly differentiating and evaluating at a = 0 shows the coefficient of each power of a must belong to this space as well:

⟨(BS − S^{∗}B)S^{n}ω_{1}, ω_{2}⟩ = 0 n = 0, 1, ….

Replacing f by f(z) := (1 − az) exp(bz^{m}) and g by g(z) := \frac{exp(−bz^{m})}{1−az} in the above argument, the last display becomes

⟨exp(b(S^{∗})^{m})(BS − S^{∗}B)S^{n} exp(−bS^{m})ω_{1}, ω_{2}⟩ = 0, −1 < b < 1; m, n ∈ ℕ.

Differentiating the last expression once with respect to b and then setting b = 0 yields

⟨(S^{∗})^{m}(BS − S^{∗}B)S^{n}ω_{1}, ω_{2}⟩ − ⟨(BS − S^{∗}B)S^{m+n}(ω_{1}), ω_{2}⟩ = 0.

Since we already know the second inner product in this display is zero, we conclude that

(BS − S^{∗}B)z^{n}ω_{1} ⊥ z^{m}ω_{2}, m, n ∈ ℕ.

Since ω_{1}, ω_{2} are outer, this means BS − S^{∗}B = 0, so B ∈ Hank. An appeal to Corollary 3.3(3) completes the proof. □

Our next task is to show more generally that reflexivity of C_{θω} does not depend on the outer factor ω. The plan is to use Toeplitz operators to build up the class of inner functions θ for which this is true. Proposition 5.1 got us started with the case θ = 1.

Proposition 5.2. Let u ∈ H^{1} and ϕ, ψ ∈ H^{∞}.

(1) If ψϕ divides u, then B ∈ ref C_{u} implies ϕ(S^{∗})Bψ(S) ∈ ref C_{u/(ψϕ)}.

(2) If ψϕ divides u, then B ∈ ref M_{u} implies ϕ(S^{∗})Bψ(S) ∈ ref M_{u/(ψϕ)}.
(3) B ∈ ref M_{u} implies ϕ(S^{∗})Bψ(S) ∈ ref M_{u}.

Proof. Given H_{η} ∈ C_{u}, we have ∫ ηu = 0 by definition. Proposition 2.3 yields ϕ(S^{∗})H_{η}ψ(S) = H_{ϕηψ}. Since we already know ∫ (ϕηψ)(u/(ψϕ)) = 0, we conclude that ϕ(S^{∗})C_{u}ψ(S) ⊂ C_{u/(ψϕ)} Applying Lemma 2.6(1), we then get

ϕ(S^{∗})(ref C_{u})ψ(S) ⊂ ref (ϕ(S^{∗})C_{u}ψ(S)) ⊂ ref (C_{u/(ψϕ)}),

which establishes the first assertion of the proposition. The second assertion can be proved in the same manner. The third assertion follows from the fact that, since M_{u} is an H^{∞}-module, ϕ(S^{∗})M_{u}ψ(S) ⊂ M_{u}. Now apply Lemma 2.6(1) as above. □

Corollary 5.3. Suppose F is a family of inner functions having a least common multiple v.

(1) ref M_{v} is the (weak*) closed linear span of {ref M_{u} : u ∈ F}.

(2) In order that M_{v} be reflexive it is necessary and sufficient that M_{u} be reflexive for each u ∈ F.

(3) In order that C_{v} be reflexive it is sufficient that C_{u} be reflexive for each u ∈ F.

Proof. Write W for the (weak*) closed linear span of {ref M_{u} : u ∈ F}. Since M_{v} ⊃ M_{u} for each u ∈ F, we at least have W ⊂ ref M_{v} of (1). To get the opposite inclusion, let B ∈ ref M_{v}. Then for each u ∈ F and each p ∈ \frac{v}{u}H^{∞}, Bp(S) ∈ ref M_{u} ⊂ W by parts (3) and (2) of the previous proposition. Since gcd\{\frac{v}{u} : u ∈ F\} = 1, we conclude by Beurling’s Theorem that the constant function 1 belongs to the (weak*) closed linear span of \{\frac{v}{u}H^{∞} : u ∈ F\}, whence B ∈ W since W is a (weak*) closed subspace. This establishes (1).

To prove (2) observe first that, since M_{v} ⊃ M_{u}, reflexivity of M_{v} implies reflexivity of M_{u} for all u ∈ F, by Corollary 3.3(3). For the opposite direction, observe that since M_{v} is the (weak*) closed linear span of {M_{u} : u ∈ F}, reflexivity of M_{v} follows from part (1).

For (3), suppose each C_{u} is reflexive and let B ∈ ref C_{v}. Then the previous proposition yields B (\frac{v}{u}) (S) ∈ ref C_{u} ⊂ Hank for each u ∈ F. Define

I := {ψ ∈ H^{∞} : Bψ(S) ∈ Hank} ,

which is a (weak*) closed ideal. It then follows that \frac{v}{u} ∈ I for all u ∈ F. Since gcd\{\frac{v}{u} : u ∈ F\} = 1, Beurling’s Theorem tells us the constant function 1 is in I, whence B ∈ Hank. □

The span of any two distinct hyperplanes in Hank will contain all of Hank so (1) fails rather spectacularly when M is replaced by C. However, the converse of (3) is an immediate consequence of (2) and Corollary 5.11 below. Similar comments apply to the following dual of the last Corollary.

Corollary 5.4. Suppose F is a family of inner functions and write v for its greatest common divisor. Then ref M_{v} is the intersection of {ref M_{u} : u ∈ F}.

Proof. Write W for the intersection of {ref M_{u} : u ∈ F}. Since M_{v} ⊂ M_{u} for each u ∈ F, we at least have ref M_{v} ⊂ W. To get the opposite inclusion, let B ∈ ref W. Then for each u ∈ F, we have B \frac{u}{v}(S) ∈ \frac{u}{v}(S)ref M_{u} ⊂ ref M_{v}. Define

I := {ψ ∈ H^{∞} : Bψ(S) ∈ ref M_{v}},

which is a (weak*) closed ideal. Since gcd\{\frac{u}{v} : u ∈ F\} = 1, Beurling’s Theorem allows us to conclude that B ∈ ref M_{v} as well. □

Consistency. Recall that, by Proposition 2.3 above, if it is known a priori that A ∈ Hank, then ϕ(S^{∗})A and Aϕ(S) are both Hankel and they agree. If A ∉ Hank, it still might turn out that one of ϕ(S^{∗})A or Aϕ(S) is Hankel, but this doesn’t guarantee that they’re both Hankel and even when they are, they might not agree. For simple examples, take ϕ(z) = z with A = 1 ⊗ z^{2} and A = 1 ⊗ z. The goal of the next three lemmas is to show that this pathology doesn’t arise when we apply Proposition 5.2.

The rank-one Hankel operators are well-known. For a ∈ 𝔻, define the operator R_{a} by

R_{a} := \frac{1}{1 − az} ⊗ \frac{1}{1 − āz} .

It is easily seen that R_{a} is the Hankel operator H_{\frac{1}{1−az̄}} and that, in fact, every rankone Hankel operator is a scalar multiple of some R_{a}. Note that ∫ \frac{u}{1−az̄} = u(a) for every u ∈ H^{1}. Thus R_{a} ∈ C_{u} if and only if u(a) = 0. In particular, C_{u} (and hence M_{u}) has no rank-one members when the Blaschke factor of u is trivial.

Lemma 5.5. Suppose β is a Möbius function, while ω is outer. Then C_{βω} is reflexive and β(S^{∗})A = Aβ(S) for each A ∈ ref C_{βω}.

Proof. Choose a in the unit disc so that β(z) = \frac{z−a}{1−\overline{a}z}. Given A ∈ ref C_{βω}, Proposition 5.2 tells us that β(S^{∗})A and Aβ(S) both belong to ref C_{ω} = C_{ω} ⊂ Hank. Thus β(S^{∗})(β(S^{∗})A−Aβ(S)) = 0 (by Proposition 2.3) and the range of β(S^{∗})A−Aβ(S) is contained in the one-dimensional kernel of β(S^{∗}). Since C_{ω} has no rank-one members, we conclude β(S^{∗})A − Aβ(S) = 0 and the second conclusion of the Lemma is established.

The remaining conclusion comes from the computation

(S^{∗} − aI)A = (I − \overline{a}S^{∗})β(S^{∗})A
= (I − \overline{a}S^{∗})Aβ(S)
= Aβ(S)(1 − \overline{a}S)
= A(S − aI),

where the third equality follows from Aβ(S) being a Hankel operator. This equality implies S^{∗}A = AS, which puts A ∈ Hank. □

The next proof does not directly deal with reflexivity.

Lemma 5.6. Suppose A ∈ ref C_{σω} where σ is a singular inner function, while ω is outer. Then σ(S^{∗})A = Aσ(S).

Proof. Recall that one can define arbitrary powers of singular inner functions. For each positive integer n and integer k, 0 ≤ k ≤ n, Proposition 5.2 yields

σ^{\frac{k}{n}} (S^{∗})Aσ^{\frac{n−k}{n}} (S) ∈ ref C_{ω} = C_{ω}.

In particular, these operators are all in Hank and hence intertwine σ^{\frac{1}{n}} (S^{∗}) and σ^{\frac{1}{n}} (S) (by Proposition 2.3): σ^{\frac{k+1}{n}} (S^{∗})Aσ^{\frac{n−k}{n}} (S) = σ^{\frac{k}{n}} (S^{∗})Aσ^{\frac{n+1−k}{n}} (S). Applying these equations in turn with k = 1, 2, …, n − 1, we conclude

σ^{\frac{1}{n}} (S^{∗})Aσ(S) = σ(S^{∗})Aσ^{\frac{1}{n}} (S)

whence taking the limit as n → ∞ yields the desired result. □

Lemma 5.7. Suppose A ∈ ref C_{θω} where θ is an inner function, while ω is outer. Then θ(S^{∗})A = Aθ(S).

Proof. Set F := {θ inner: the conclusion of the Lemma holds for each outer ω}.

The preceding two lemmas imply that F contains all Möbius transformations and singular inner functions. We next check that F is closed under multiplication. Let θ, η ∈ F, with ω is outer, and A ∈ ref C_{θηω}. Then θ(S^{∗})A ∈ ref C_{ηω}, so

η(S^{∗})θ(S^{∗})A = θ(S^{∗})Aη(S),

since η ∈ F. But Aη(S) ∈ ref C_{θω}, so

θ(S^{∗})Aη(S) = Aη(S)θ(S),

since θ ∈ F. Combining these equations gives (θη)(S^{∗})A = A(θη)(S), as desired.

Suppose now that β = ∏ β_{n} is a Blaschke product, that ω is outer, and that A ∈ ref C_{βω}. Write ρ_{n} := ∏_{k=1}^{n} β_{k} and τ_{n} := \frac{β}{ρ_{n}}. For each n, we have τ_{n}(S^{∗})A ∈ C_{ρ_{n}ω}, so

ρ_{n}(S^{∗})τ_{n}(S^{∗})A = τ_{n}(S^{∗})Aρ_{n}(S)

since ρ_{n} ∈ F. Taking the limit on n, we see that β ∈ F.

The result for general θ now follows from the canonical factorization of inner functions. □

Main results.

Proposition 5.8. Let W be a weak* closed subspace of Hank with seed θ. Given B ∈ ref W, there is an individual Hankel operator H_{ϕ} such that v(S^{∗})Bu(S) = Bθ(S) = H_{ϕ} whenever u, v are inner functions satisfying uv = θ.

Proof. Suppose θ = uv is any factorization of θ with u and v inner functions. Fix an element of C^{−1}(W) with inner-outer factorization ψω, so that B ∈ C_{ψω} and set η := \frac{ψ}{θ}. Then Bu(S) ∈ ref C_{vηω} by Proposition 5.2. Lemma 5.7 tells us that

η(S^{∗})v(S^{∗}) (Bu(S)) = (Bu(S)) v(S)η(S) = Bθ(S)η(S),

and the latter operator belongs to ref C_{ω} ⊂ Hank. Define

I := {η ∈ H^{∞} : η(S^{∗})v(S^{∗})Bu(S) = Bθ(S)η(S) ∈ Hank}

which is a (weak*) closed ideal. By definition of seed,

gcd \{\frac{ψ}{θ} : ψ inner part of functions in C^{−1}(W)\} = 1,

whence we conclude by Beurling’s Theorem that v(S^{∗})Bu(S) = Bθ(S) and that this operator in is Hank. The proof is completed by taking H_{ϕ} = Bθ(S). □

Proposition 5.9. Let W be a weak* closed subspace of Hank with seed θ. Then the following are equivalent for an operator B on H^{2}.

(1) B ∈ ref M_{θ}.

(2) B ∈ ref W and Bθ(S) = 0.

(3) B satisfies v(S^{∗})Bu(S) = 0 whenever u, v are inner functions with uv = θ.

Proof. Assuming (1), we get B ∈ ref W because M_{θ} ⊂ W. Also, Proposition 5.2 tells us that Bθ(S) ∈ M_{1} = {0} yielding (2).

That (2) implies (3) follows from the preceding proposition.

Finally assume (3) and suppose f ⊗ g ∈ (M_{θ})_{⊥} ∩ F_{1}. To complete the proof, we must show ⟨Bf, g⟩ = 0. Corollary 3.8 tells us θ | fg^{∗}.

Take u to be the greatest common divisor of θ and the inner factor of f, and set v := \frac{θ}{u}. Then v must divide g^{∗}. Thus

⟨Bf, g⟩ = ⟨B (u\frac{f}{u} ) , v^{∗} \frac{g}{v^{∗}} ⟩ = ⟨v(S^{∗})Bu(S) (\frac{f}{u} ) , \frac{g}{v^{∗}} ⟩ = 0.

Thus B ⊥ f ⊗ g and the proof is complete. □

Theorem 5.10. Let W be a weak* closed subspace of Hank with seed θ. Then

ref W = W + ref M_{θ}.

Proof. Clearly, the right-hand side is contained in the left-hand side. For the reverse inclusion, suppose A ∈ ref W, choose H_{ϕ} as in Proposition 5.8 and set B := A − H_{ϕ\overline{θ}}. Then B satisfies the third condition of Proposition 5.9 so B ∈ ref M_{θ} ⊂ ref W. In particular, H_{ϕ\overline{θ}} = A − B ∈ ref W and since Hank ∩ ref W = W, we have that H_{ϕ\overline{θ}} ∈ W. Thus A = H_{ϕ\overline{θ}} + B ∈ W + ref M_{θ}. □

Proof of Theorem 1.2: Observe that this is just the previous theorem. □

Corollary 5.11. Let N ⊂ H^{1}, take W := C(N) to be the corresponding subspace of Hank, and set θ as the greatest common divisor of the inner factors of the members of N. Then the following are equivalent:

(1) W is reflexive.

(2) C_{θ} is reflexive.

(3) M_{θ} is reflexive. In particular, reflexivity of W is equivalent to reflexivity of the largest H^{∞}-module it contains.

Proof. Proposition 3.6 implies that C^{−1}(W) is the closed linear span of N, whence θ is in fact the seed of W. Theorem 5.10 tells us that ref M_{θ} ⊂ Hank if and only if ref W ⊂ Hank, so the equivalence of (1) and (3) follows by Corollary 3.3. Applying this general result to the special case N = {θ} then yields (2) ⟺ (3). The “in particular” assertion then follows from Corollary 3.8. □

Observe that the missing converse of Corollary 5.3 (3) follows immediately from the above corollary.

Proof of Theorem 1.3: The last corollary establishes (1) ⟺ (2) ⟺ (3); That (3) ⟺ (4) is the content of Theorem 4.5, while (4) ⟺ (5) follows from Kapustin’s Theorem 4.7. □

6. Finite-rank members of reflexive closures

Recall that given a singular inner function σ there is an associated finite positive Borel measure μ on (−π, π] which is singular with respect to Lebesgue measure and such that, except for a constant factor of modulus 1,

σ(z) = exp (−\frac{1}{2π} ∫_{−π}^{π} \frac{e^{it} + z}{e^{it} − z} dμ(t)) . (2)

The function σ is said to be totally atomic if the associated measure is supported on a countable set, and totally nonatomic if that measure has no atoms. Observe that any singular inner function σ can then be written as σ = γη, where γ is totally atomic and η is totally nonatomic.

The following summarizes the main results of this section.

Corollary 6.1. Let θ be a nonconstant inner function.

(1) If θ is a totally nonatomic singular inner function, then ref M_{θ} has no nonzero finite-rank members.

(2) In all other cases, ref M_{θ} has rank-one members.

(3) In particular, when the singular factor of θ is totally atomic, then ref M_{θ} is the weak* closed linear span of its rank-one members.

The negative result (1) reflects the depth of Kapustin’s work [16, 17]. In connection with (3), the relevant rank-one operators will be explicitly exhibited. As an application, we will compute ref W when the the singular factor of the seed of W is totally atomic, and when the seed is a Blaschke product.

The following proposition does not extend to operators of rank larger than one. Indeed, each hyperspace C_{u} has members of every rank greater than one, while M_{ψ} reduces to {0} for outer ψ, Proposition 6.2. Let W be a subspace of Hank and write θ for its seed.

(1) W and M_{θ} have the same rank-one members.

(2) ref W and ref M_{θ} also have the same rank-one members.

Proof. Suppose B ∈ F_{1} ∩ ref W. Let u ∈ C^{−1}(W) have inner-outer factorization u = ϕω so that B ∈ C_{u}. Then Bϕ(S) ∈ ref C_{ω} = C_{ω}. But C_{ω} has no rank-one members so Bϕ(S) = 0. Define

I := {ψ ∈ H^{∞} : Bψ(S) = 0},

which is a (weak*) closed ideal. Since θ is the greatest common divisor of the functions in I another application of Beurling’s Theorem allows us to conclude that Bθ(S) = 0, whence B ∈ ref M_{θ} by Proposition 5.9. Thus we have shown F_{1} ∩ ref W ⊂ F_{1} ∩ ref M_{θ}. But M_{θ} ⊂ W, so F_{1} ∩ ref W = F_{1} ∩ ref M_{θ}, proving (2). Intersecting both sides of the last equality with Hank gives (1) by Corollary 3.3(1). □

Pairwise divisors. The following concept relates our question to arithmetic in H^{∞}. Thanks to Jeremy Praissman for the nomenclature.

Definition 6.3. Suppose f, g, and θ are nonconstant inner functions. Then we say that (f, g) is a pairwise divisor for θ if whenever θ = uv is a factorization involving inner functions, then either f divides u or g divides v.

Proposition 6.4. Suppose (f, g) is a pairwise divisor of θ while B is an operator on H^{2} satisfying f(S^{∗})B = Bg(S) = 0. Then B belongs to the (weak*) closed linear span of the rank-one members of ref M_{θ}.

Proof. Suppose θ = uv = vu is a factorization involving inner functions. If f divides v the hypothesis gives f(S^{∗})B = 0 and hence v(S^{∗})B = 0. Otherwise g divides u and hence Bu(S) = 0. In either case, v(S^{∗})Bu(S) = 0, so B ∈ ref M_{θ} by Proposition 5.9.

Now write X := (f(S^{∗}))^{∗}f(S^{∗}) and notice that X is an orthogonal projection. Observe that kernel f(S^{∗}) = (f^{∗}H^{2})^{⊥} is nontrivial since f is nonconstant. Fix an orthonormal basis {x_{n}} of kernel f(S^{∗}). Then I −X is projection onto kernel f(S^{∗}) and thus the series ∑_{n} x_{n} ⊗ x_{n} converges weak* to I − X. By hypothesis, XB = 0, whence

B = (I − X)B = ∑_{n}(x_{n} ⊗ x_{n})B.

For each n, f(S^{∗})(x_{n} ⊗x_{n})B = 0 since x_{n} ∈ kernel f(S^{∗}). Also, since Bg(S) = 0 we have that (x_{n} ⊗ x_{n})Bg(S) = 0 by orthogonality of the sequence {x_{n}}. Therefore, by the preceeding paragraph, each term in the above display belongs to ref M_{θ} and hence B is in the (weak*) closed linear span of the operators (x_{n} ⊗ x_{n})B ∈ ref M_{θ}. □

Proposition 6.4 has a partial converse.

Proposition 6.5. Suppose B ∈ ref M_{θ} has finite rank and its rank is minimal among all nonzero members of ref M_{θ}. Then B has rank one and there is a pairwise divisor (f, g) of θ satisfying f(S^{∗})B = Bg(S) = 0.

Proof. Take J := {h ∈ H^{∞} : Bh(S) = 0}. Then J is a (weak*) closed ideal in H^{∞}, whence Beurling’s Theorem provides a nonconstant inner function g such that J = gH^{∞}. Thus the inner members of J coincide with the inner multiples of g.

Similarly, there is a nonconstant inner function f such that, for inner h, h(S^{∗})B = 0 if and only if f divides h. In particular, f(S^{∗})B = Bg(S) = 0. Now suppose θ = uv is any inner factorization of θ. By Proposition 5.9, u(S^{∗})Bv(S) = 0. Now if Bv(S) = 0, then g divides v. Otherwise, the rank minimality hypothesis implies range B = range Bv(S) which implies range u(S^{∗})B = range u(S^{∗})Bv(S) = {0}, whence u(S^{∗})B = 0 and f divides u. This completes the proof that (f, g) is a pairwise divisor for θ. Proposition 6.4 then expresses B as a linear combination of rank-one members of ref M_{θ}, whence the minimality assumption implies B itself has rank one. □

Corollary 6.6. The following are equivalent for inner θ and nonzero x, y ∈ H^{2}. (1) x ⊗ y^{∗} ∈ ref M_{θ}.

(2) f(S^{∗})x = g(S^{∗})y = 0 for some pairwise divisor (f, g) of θ.

Proof. Assuming (1), choose (f, g) from Proposition 6.5. Then 0 = f(S^{∗})(x⊗y^{∗}) = (f(S^{∗})x)⊗y^{∗} implies f(S^{∗})x = 0. Also 0 = (x⊗y^{∗})(g(S)) = x⊗((g(S))^{∗}(y^{∗})). This means 0 = (g(S))^{∗}(y^{∗}) which in turn equals (g(S^{∗})(y))^{∗} and we have established (2).

Conversely if (2) holds, then 0 = f(S^{∗})(x ⊗ y^{∗}) = (x ⊗ y^{∗})g(S), and (1) follows by Proposition 6.4. □

Example 6.7. Take θ(z) = z^{n}, and suppose i, j are nonnegative integers with i + j ≤ n − 1. Then z^{i} ⊗ z^{j} ∈ ref M_{θ}.

Proof. (z^{i+1}, z^{j+1}) is a pairwise divisor for θ. Apply the Corollary 6.6 with x = z^{i} and y = z^{j}. □

Application to ref M_{θ}. It is possible to characterize all pairwise divisors.

Proposition 6.8. Let θ be a nonconstant inner function. Then up to scalar multiples, all pairwise divisors of θ take one of the following two forms.

(1) (β^{i}, β^{j}) where β is is a Möbius transformation, i and j are positive integers, and β^{i+j−1} divides θ.

(2) (α^{r}, α^{1−r}) where α is a singular inner function whose measure is supported at a single point, 0 < r < 1, and α divides θ.

Proof. That the displayed pairs work follows from uniqueness of the canonical Blaschke-singular factorization of inner functions.

For the converse, let (f, g) be a pairwise divisor for θ. The factorization θ = θ · 1 = 1 · θ forces f and g to divide θ. Suppose first that f(a) = 0 for some a ∈ 𝔻 and write β for the corresponding Möbius transformation. There is a highest power of β which divides θ which thus factors as θ = ρβ^{n} for some ρ which is relatively prime to β. Since f does not divide ρ, we conclude that g = β^{j} for some 1 ≤ j ≤ n. But we can also factor θ = β^{n−j+1}[β^{j−1}ρ]. Since g doesn’t divide the bracketed factor, we conclude that that f = β^{i} for some i ≤ n − j + 1 so our pairwise divisor takes the form (1).

It remains to consider the case where f and g are singular inner functions. Since any Blaschke factor of θ is irrelevant to questions of divisibility by f and g, we may as well assume that θ is singular inner as well. Write μ for the measure on the unit circle associated to θ as in Equation (2). Then inner divisors of θ correspond to measures which are less than or equal to μ setwise. For inner divisors h and k of θ, we write h ≪ k and h ⊥ k when the corresponding measures are related in these ways. Applying the Jordan Decomposition Theorem, we can factor θ = ρσ where ρ ⊥ f and σ ≪ f. Since f does not divide ρ, we conclude g must divide σ so in fact g ≪ σ ≪ f. By symmetry f and g must be mutually absolutely continuous. But now, if h is any nontrivial divisor of f, then (h, g) is still a pairwise divisor of θ whence h ≈ g ≈ f. In other words, the measure associated to f is mutually absolutely continuous with every smaller measure, and this forces the measures of f and g to be supported at a single common point of 𝕋. Take α := fg to see that (f, g) takes the form (2). □

Corollary 6.9. Let σ be a totally nonatomic singular inner function. Then σ does not have a pairwise divisor.

Proof. This is an immediate consequence of Proposition 6.8. □

Proposition 6.10. Suppose θ is a power of a Möbius function. Then ref M_{θ} is the (weak*) closed span of its rank-one members.

Proof. Express θ = β^{n} where β is the Möbius function β(z) = \frac{z−a}{1−āz} and write V for the isometry β(S). Let B ∈ ref M_{θ}. Since BV^{n} = 0, we have

B = ∑_{j=0}^{n−1} BV^{j}(I − V V^{∗})(V^{∗})^{j}.

Fix j and write C := BV^{j}(I − V V^{∗})(V^{∗})^{j}. By Proposition 5.9 we have that β^{n−j}(S^{∗})C = 0. Also, since (I − V V^{∗})V = 0 we have Cβ^{j+1}(S) = 0. Since (β^{n−j}, β^{j+1}) is a pairwise divisor for β^{n}, by Proposition 6.4, we conclude that each term in the preceding display is in the (weak*) closed linear span of the rank-one members of ref M_{θ} and the proof is complete. □

Proposition 6.11. Suppose θ is a singular inner function whose measure is supported at a single point. Then ref M_{θ} is the (weak*) closed linear span of its rank-one members.

Proof. We must show F_{1} ∩ ref M_{θ} is total in ref M_{θ}. Let B ∈ ref M_{θ}.

Let V be the isometry θ(S). Observe that for each natural number n, since BV = 0, we have

BV ^{\frac{1}{n}} = ∑_{k=1}^{n−1} BV ^{\frac{1}{n}} V ^{\frac{k−1}{n}} (I − V ^{\frac{1}{n}} (V^{∗})^{\frac{1}{n}} ) (V^{∗})^{\frac{k−1}{n}}

Fix k and write C := BV ^{\frac{k}{n}} (I − V ^{\frac{1}{n}} (V^{∗})^{\frac{1}{n}} ) (V^{∗})^{\frac{k−1}{n}}. By Proposition 5.9, we have

θ^{1− \frac{k}{n}} (S^{∗})C = 0. Also, since (I − V ^{\frac{1}{n}} (V^{∗})^{\frac{1}{n}})V ^{\frac{1}{n}} = 0, it follows that CV ^{\frac{k}{n}} = 0. Since (θ^{1− \frac{k}{n}} , θ^{\frac{k}{n}}) is a pairwise divisor for θ, Proposition 6.4 tells us that C belongs to the (weak*) closed span of F_{1}∩ref M_{θ}. Thus each term in the preceeding display belongs to the (weak*) closed linear span of F_{1} ∩ ref M_{θ} and hence BV ^{\frac{1}{n}} belongs to the (weak*) closed linear span of F_{1} ∩ ref M_{θ}. Taking the limit as n → ∞ yields B ∈ [F_{1} ∩ ref M_{θ}] as desired. □

All the pieces are now in place for the proof of Corollary 6.1.

Theorem 6.12. Let θ be an inner function having factorization θ = γσ, where γ is an inner function with singular factor totally atomic and σ is a totally nonatomic singular inner function. Then

[F_{1} ∩ ref M_{θ}] = [F ∩ ref M_{θ}] = ref M_{γ}.

Proof. The inclusion [F_{1} ∩ ref M_{θ}] ⊂ [F ∩ ref M_{θ}] is obvious.

Suppose B ∈ F ∩ ref M_{θ}. Then Bγ(S) ∈ F ∩ ref M_{σ}. But Corollary 6.9 tells us σ has no pairwise divisors, whence Proposition 6.5 implies ref M_{σ} has no nonzero members of finite rank. Thus Bγ(S) = 0. On the other hand, Bσ(S) ∈ ref M_{γ}. Since γ and σ are relatively prime, there exists H^{∞} functions h and k such that 1 = γh+σk and hence B = Bγ(S)h(S)+Bσ(S)k(S) = Bσ(S)k(S). By Proposition 5.2(3), we can then conclude that B ∈ ref M_{γ}. Thus we have established the second inclusion [F ∩ ref M_{θ}] ⊂ ref M_{γ}.

For the remaining inclusion, express γ = Π_{n}γ_{n} where each γ_{n} is either a power of a Möbius function or a singular inner function whose measure is supported at a single point; no two γ_{n}’s should have a common zero or be supported at the same point of 𝕋. Applying Propositions 6.10 and 6.11, we learn that for each n,

ref M_{γ_{n}} = [F_{1} ∩ ref M_{γ_{n}}] ⊂ [F_{1} ∩ ref M_{θ}].

But Proposition 5.3(1) tells us that ref M_{γ} is the closed linear span of {ref M_{γ_{n}}}. Thus ref M_{γ} ⊂ [F_{1} ∩ ref M_{θ}] and the cycle is complete. □

Proof of Corollary 6.1: Observe that, for θ inner, M_{θ} ≠ {0} if and only if θ is nonconstant. Then, we only need to apply the previous theorem with γ = 1, γ nonconstant, and σ = 1 to get Parts (1), (2), and (3) respectively. □

We close this subsection with a refinement of Corollary 6.1(1).

Proposition 6.13. Let W be a weak*-closed subspace of Hank whose seed θ is a totally nonatomic singular inner function. Then all finite-rank members of ref W belong to W itself.

Proof. Let A ∈ ref W have finite rank. Apply Proposition 5.8 to find an individual Hankel operator H_{ϕ} such that

v(S^{∗})Au(S) = Aθ(S) = H_{ϕ}

whenever u, v are inner functions satisfying uv = θ. Since H_{ϕ} has finite rank, the vectors H_{ϕ}1, H_{ϕ}z, H_{ϕ}z^{2}, … are linearly dependent and there is a polynomial p ≢ 0 such that 0 = H_{ϕ}p = H_{pϕ}1. Since 1 is a separating vector for Hank, it follows that H_{pϕ} = 0 and thus Aθ(S)p(S) = 0. Multiplying the last display through by p(S) yields

v(S^{∗})Ap(S)u(S) = Ap(S)θ(S) = 0

whenever uv = θ. But now Proposition 5.9 yields Ap(S) ∈ ref M_{θ}. In view of Corollary 6.1(1), this means Ap(S) = 0, and we have already observed that Aθ(S) ∈ Hank. Since p and θ are relatively prime, there exist H^{∞} functions h and k such that 1 = ph+θk. Hence A = Ap(S)h(S)+Aθ(S)k(S) = Aθ(S)k(S) ∈ Hank and we get A ∈ Hank ∩ ref W = W as desired. □

Computing reflexive closures. Theorem 5.10 can be used to compute reflexive closures of nonreflexive spaces. For example, (matrices of) operators in W := M_{z^{k}} are supported on the top k skew diagonals and hence the same is true for members of ref W. On the other hand, Example 6.7 shows that all matrices supported on this triangle do in fact belong to ref W. Since there are precisely k independent Hankel operators supported on that triangle, the codimension of W in ref W (which quantifies how nonreflexive W is) is \frac{k(k−1)}{2}. Theorem 5.10 now tells us how to compute the reflexive closures of all Hankel spaces having seed z^{k}. The following result generalizes this analysis.

Corollary 6.14. Suppose θ is an inner function whose singular factor is totally atomic and set

ℜ := {x ⊗ y^{∗} : f(S^{∗})x = g(S^{∗})y = 0 for some pairwise divisor (f, g) of θ}.

(1) ref M_{θ} is the weak* closed linear span of ℜ.

(2) If W is any weak* closed subspace of Hank with seed θ, then ref W = W + \overline{span ℜ}.

Proof. For (1), combine Corollaries 6.1(3) and 6.6.

For (2), apply Theorem 5.10 to the result of Part (1). □

Of course, the description of ℜ can hardly be considered concrete, though Proposition 6.8 helps. The set C given in Corollary 6.15 is is more satisfying in this respect.

Corollary 6.15. Suppose θ is a Blaschke product and set

C := \{\frac{z^{i−1}}{(1 − az)^{i}} ⊗ \frac{z^{j−1}}{(1 − \overline{a}z)^{j}} : i, j ∈ ℕ, (\frac{z − a}{1 − \overline{a}z} )^{i+j−1} divides θ\} .

(1) ref M_{θ} is the (weak*) closed linear span of C.

(2) If W is any (weak*) closed subspace of Hank with seed θ, then ref W = W + \overline{span C}.

Proof. In this case, all pairwise divisors of θ take the form of Proposition 6.8(1). Thus the set ℜ of the last corollary becomes

ℜ := {x ⊗ y^{∗} : β^{i}(S^{∗})x = β^{j}(S^{∗})y = 0 for some Möbius β with β^{i+j−1} dividing θ}.

But observe that, if β is the Möbius function β(z) = \frac{z−a}{1−\overline{a}z}, then

kernel(β^{n}(S^{∗})) = ((β^{∗})^{n}H^{2})^{⊥} = span \{\frac{z^{k}}{(1 − az)^{k+1}} : k = 0, …, n − 1\} ,

and thus the span of ℜ equals the span of C. □

Corollary 6.16. If θ is an inner function having a multiple root or a nontrivial atomic inner factor, then ref M_{θ} has a non-Hankel member of rank one.

Proof. Since ϕ|θ implies ref M_{ϕ} ⊂ ref M_{θ}, it suffices to consider the cases when (1) θ = β^{2} for β a Möbius transformation vanishing at a ∈ 𝔻 and

(2) θ = α where α is a singular inner function whose measure is supported at a single point in T.

In Case (1) Corollary 6.15 tells us \frac{1}{1−az} ⊗ \frac{z}{(1−\overline{a}z)^{2}} ∈ ref M_{θ} and we know this operator does not belong to Hank because it is not a scalar multiple of R_{a}.

In Case (2), Corollary 6.1(2) tells us ref M_{θ} has a rank-one element which can’t be Hankel since M_{θ} itself has no rank-one elements (because θ has no roots). □

Relation to earlier papers. A main goal of our work to this point has been to identify precisely which (weak* closed) subspaces of Hank are reflexive. The following summarizes the results we have obtained.

Theorem 6.17. Let W be a weak* closed subspace of Hank, and write θ for its seed, i.e., M_{θ} is the largest submodule of Hank contained in W. Consider the canonical factorization θ = βασ, where β is a Blaschke product, α is a totally atomic singular inner function and σ is a totally nonatomic singular inner function. Let μ the singular measure corresponding to the singular inner function σ. Then,

(1) W is reflexive if and only if M_{θ} is reflexive if and only if C_{θ} is reflexive if and only if S(θ) is reflexive.

(2) M_{θ} is reflexive if and only if M_{β}, M_{α} and M_{σ} are reflexive.

(3) M_{β} is reflexive if and only if β has no repeated roots.

(4) M_{α} is reflexive if and only if α = 1.

(5) M_{σ} is reflexive if and only if μ is a BC-vanishing measure.

Earlier proofs of the Jordan model analogues of Parts (3) and (4) can be found in [29]. (5) depends strongly on Kapustin’s deep work, and is unlikely to be simplified. On the other hand, the proofs we have given of everything except (5) are independent of existing papers on Jordan models:

• (1) follows from Corollary 5.11 and Theorem 4.5,

• (2) is Corollary 5.3(2),

• the ‘if’ implication of (3) comes from Lemma 5.5 and Corollary 5.3(2),

• the ‘if’ implication of (4) comes from Corollary 5.1,

• the ‘only if’ implications of (3) and (4) result from Corollary 6.16. In particular, we were able to use operators of rank one to get the negative

content of (3) and (4). Proposition 6.13 explains why such an approach cannot help with (5).

7. Semi–infinite Dimensional

Let us turn our attention to the so-called semi–infinite dimensional case. First, for each n ∈ ℕ, let P_{n} denote the set of all polynomials with degree less than or equal to n, thought of as a subspace of H^{2}. The space B(P_{n}, H^{2}) is dual to the trace class Tr(H^{2}, P_{n}) via the bilinear form

⟨A, T ⟩ = trace(AT), A ∈ B(P_{n}, H^{2}), T ∈ Tr(H^{2}, P_{n}).

For n ∈ ℕ and ϕ ∈ \overline{H^{2}} define H_{ϕ} : P_{n} ⟶ H^{2} by H_{ϕ}f = P J(ϕf), where P denotes the orthogonal projection onto H^{2} and J is the flip operator J(f)(z) := f(\overline{z}). Thus, H_{ϕ} is an operator in B(P_{n}, H^{2}). It is easy to check that the matrix of H_{ϕ} with respect to the canonical bases {z^{k}}_{k=0}^{n} and {z^{k}}_{k=0}^{∞}, for P_{n} and H^{2} respectively, has constant skew–diagonals. Conversely, it is easy to see that all Hankel operators are of the form H_{ϕ}, as above.

Since ∥H_{ϕ}z^{j}∥ ≤ ∥ϕ∥ for j = 0 … n with equality for j = 0, it follows that

∥H_{ϕ}∥ ≤ \sqrt{n + 1} ∥ϕ∥ = \sqrt{n + 1} ∥H_{ϕ}(1)∥, ϕ ∈ \overline{H^{2}}. (3)

We denote the set of all Hankel operators in B(P_{n}, H^{2}) as Hank(P_{n}, H^{2}).

For each H_{ϕ} ∈ Hank(P_{n}, H^{2}), f ∈ P_{n} and g ∈ H^{2}, a straightforward calculation shows that

⟨H_{ϕ}, f ⊗ g⟩ = ⟨H_{ϕ}f, g⟩ = \frac{1}{2π} ∫_{−π}^{π} ϕ(e^{iθ})f(e^{iθ})g^{∗}(e^{iθ})dθ.

Every element T of Tr(H^{2}, P_{n}) has a unique expression T = ∑_{j=0}^{n} z^{j} ⊗ g_{j}(z), and we define its cosymbol Γ(T) := ∑_{j=0}^{n} z^{j}g_{j}^{∗}(z). The preceding display then generalizes to

⟨H_{ϕ}, T ⟩ = \frac{1}{2π} ∫ ϕΓ(T), ϕ ∈ \overline{H^{2}}, T ∈ Tr(H^{2}, P_{n}). (4)

Proposition 7.1. (1) For each g ∈ H^{2}, we have Γ(1 ⊗ g^{∗}) = g.

(2) Γ is bounded and has range H^{2}.

(3) The kernel of Γ is the preannihilator of Hank(P_{n}, H^{2})

Proof. (1) is a matter of definition and (2) is easy to check. (3) follows from Equation (4). □

Theorem 7.2. Let n ∈ ℕ. The space Hank(P_{n}, H^{2}) is transitive and elementary.

Proof. Suppose f ⊗ g ∈ Hank(P_{n}, H^{2})_{⊥}. Then Equation (4) yields ∫ ϕfg^{∗} = 0 for all ϕ ∈ \overline{H^{2}}. But this means fg^{∗} ≡ 0 whence f = 0 or g = 0. Thus Hank(P_{n}, H^{2})_{⊥} has no rank-one members and Hank(P_{n}, H^{2}) is transitive.

Given T ∈ Tr(H^{2}, P_{n}), we can write T = 1 ⊗ (Γ(T))^{∗} + [T − 1 ⊗ (Γ(T))^{∗}]. Since the bracketed expression is in the kernel of Γ, we have Tr(H^{2}, P_{n}) = F_{1} + Hank(P_{n}, H^{2})_{⊥} and Hank(P_{n}, H^{2}) is elementary. □

In particular, an appropriate version of Proposition 2.5 (in our semi-infinite dimensional setting) tells us no proper subspace of Hank(P_{n}, H^{2}) is transitive.

We now study subspaces of Hank(P_{n}, H^{2}). Because P_{n} is finite-dimensional, every norm closed linear manifold in B(P_{n}, H^{2}) is automatically weak* closed, so we need not worry about the latter topology. Equation (3) shows that the symbol map is is a Banach space isomorphism between \overline{H^{2}} and Hank(P_{n}, H^{2}). Moreover, taking perps relative to the bilinear form

⟨f, g⟩ = \frac{1}{2π} ∫ fg, f ∈ \overline{H^{2}}, g ∈ H^{2}

sets up a one-to-one correspondence between subspaces of \overline{H^{2}} and subspaces of H^{2}. Thus the companion spaces of the following definition are all subspaces of Hank(P_{n}, H^{2}), and every subspace of Hank(P_{n}, H^{2}) is accounted for in this way.

Definition 7.3. The companion space to a subset N of H^{2} is defined as

C^{n}(N) := {H_{ϕ} ∈ Hank(P_{n}, H^{2}) : ∫ ϕu = 0 for each u ∈ N}.

For an individual u ∈ H^{2}, we write C_{u}^{n} := C^{n}({u}).

In particular, C_{u}^{n} is the most general hyperplane in Hank(P_{n}, H^{2}).

Lemma 7.4. Let n ∈ ℕ, f ∈ P_{n}, g ∈ H^{2} and N ⊂ H^{2}.

(1) In order that a trace class operator T belong to (C^{n}(N))_{⊥} it is necessary and sufficient that Γ(T) belong to the closed linear span [N] of N.

(2) In particular, f ⊗ g ∈ (C^{n}(N))_{⊥} if and only if fg^{∗} belongs to [N].

Proof. For necessity in (1), assume Γ(T) ∈ [N] and let H_{ϕ} ∈ C^{n}(N). By definition, the last statement implies ∫ ϕu = 0 for all u ∈ N. Thus ∫ ϕΓ(T) = 0 as well, whence ⟨H_{ϕ}, T ⟩ = 0 by Equation 4. Thus T ∈ (C^{n}(N))_{⊥} as required.

For the converse, suppose Γ(T) ∉ [N] and choose ϕ ∈ \overline{H^{2}} with ∫ ϕu = 0 for all u ∈ N but ∫ ϕΓ(T) ≠ 0. The last two conditions respectively mean H_{ϕ} ∈ C^{n}(N), but ⟨H_{ϕ}, T ⟩ ≠ 0 and we have shown that T ∉ (C^{n}(N))_{⊥}.

(2) is a special case of (1) since Γ(f ⊗ g) = fg^{∗}. □

Corollary 7.5. Let n ∈ ℕ, f ∈ P_{n}, g ∈ H^{2} and u ∈ H^{2}. Then f ⊗ g ∈ (C_{u}^{n})_{⊥} if and only if there exists λ ∈ ℂ with fg^{∗} = λu.

Note also that B(P_{0}, H^{2}) is reflexive and elementary (thus every subspace is reflexive [1, Proposition 2.10]). The main result of this section is the analog of Theorem 1.3 in the semi–infinite dimensional setting.

Theorem 7.6. Suppose n ∈ ℕ and u ∈ H^{2}. Then C_{u}^{n} is reflexive if and only if the inner factor of u is a Blaschke factor with no repeated roots.

The following lemma depends on the fact that 1 is a strictly separating vector for Hank(P_{n}, H^{2}) (Equation (3)). Whence this argument cannot be used in the infinite dimensional setting or the finite dimensional setting!

Lemma 7.7. Let M be a subspace of Hank(P_{n}, H^{2}). The following statements are equivalent:

(1) The space M is reflexive,

(2) If A1 = 0, and A ∈ ref M, then A = 0.

Proof. Assume that (1) holds, A1 = 0, and A ∈ ref M. Since A ∈ M we have that A is Hankel, whence A = 0 since 1 is a separating vector.

Now assume that (2) holds and that A ∈ ref M. We must show that A ∈ M. Since A ∈ ref M we have that A1 ∈ \overline{M(1)}, by definition of reflexivity. This amounts to saying that there exist a sequence of operators {A_{n}} in M with ∥A_{n}1−A1∥ → 0. This implies that the sequence {A_{n}1} is Cauchy, and since 1 is a strictly separating vector, that {A_{n}} is also a Cauchy sequence. Since A_{n} ∈ M and M is closed, it follows that there exists an operator A_{0} ∈ M ⊂ ref M with A_{n} → A_{0}. Hence, we have that (A − A_{0})1 = 0 and A − A_{0} ∈ ref M. The hypothesis implies that A − A_{0} = 0. Therefore, A = A_{0} ∈ M. □

Our technique for showing hyperplanes of Hankel operators to be nonreflexive is to exhibit non-Hankel operators in their reflexive closure. We should remind the reader that we use P_{n} to denote the operator P_{n} : H^{2} ⟶ P_{n}, the orthogonal projection onto the polynomials of degree at most n.

Proposition 7.8. Let n ∈ ℕ and fix a ∈ 𝔻. Then the non-Hankel rank-one operator \frac{z}{(1−az)^{2}} ⊗ P_{n} \frac{1}{1−\overline{a}z} belongs to the reflexive closure of C^{n}((z − a)^{2}H^{2}).

Proof. Suppose f ∈ P_{n} and g ∈ H^{2} with f ⊗ g ∈ (C^{n}((z − a)^{2}H^{2}))_{⊥}. Then, by Lemma 7.4, fg^{∗} has a repeated root at a. Notice that

\begin{aligned}⟨\frac{z}{(1 − az)^{2}} ⊗ P_{n} \frac{1}{1 − \overline{a}z} , f ⊗ g⟩ & = ⟨f, P_{n} \frac{1}{1 − \overline{a}z} ⟩ ⟨\frac{z}{(1 − az)^{2}} , g⟩ \\ & = ⟨f, \frac{1}{1 − \overline{a}z} ⟩ ⟨g^{∗}, \frac{z}{(1 − \overline{a}z)^{2}} ⟩ \end{aligned}

and since \frac{1}{1−\overline{a}z} acts as a reproducing kernel in P_{n} while \frac{z}{(1−\overline{a}z)^{2}} is a reproducing kernel for the first derivative in H^{2}, we then have that

⟨\frac{1}{1 − az} ⊗ P_{n} \frac{z}{(1 − \overline{a}z)^{2}} , f ⊗ g⟩ = f(a)g^{∗′}(a).

But if fg^{∗} has a repeated root at a then either f or g^{∗′} has a root at a, which means the previous equation equals zero. □

Proposition 7.9. Let n > 0. Suppose σ is a nontrivial singular inner function. Choose r to be a nonzero member of (σ^{∗}H^{2})^{⊥} (necessarily nonconstant). Then the non-Hankel operator r ⊗ 1 belongs to the reflexive closure of C^{n}(σH^{2}).

Proof. Suppose f ∈ P_{n} and g ∈ H^{2} with f ⊗ g ∈ (C^{n}(σH^{2}))_{⊥}. Then σ divides fg^{∗} by Lemma 7.4. Since f ∈ P_{n}, the singular inner function σ must divide g^{∗}. Since σ^{∗} divides g, it follows that g ∈ σ^{∗}H^{2} and thus ⟨r, g⟩ = 0. Notice then that ⟨r ⊗ 1, f ⊗ g⟩ = ⟨f, 1⟩ ⟨r, g⟩ = 0. That r ⊗ 1 is not Hankel is clear since r is not constant (look at its matrix). □

The following proposition isolates the function theory needed in applying Lemma 7.7 in the proof of Theorem 7.11. If a ∈ 𝔻 \ \{0\} let b_{a}(z) = \frac{|a|}{a} \frac{a−z}{1−\overline{a}z}, and if a = 0 let b_{a}(z) = z.

Proposition 7.10. Let F ⊂ H^{2} and consider the following subspace of H^{2}

N := [\{\frac{u(z)}{1 − bz} \}_{b∈𝔻} , \{\frac{u(z)}{z − c}\}_{u(c)=0} : u ∈ F] .

Suppose β := gcd{inner factor of u : u ∈ F} is a Blaschke product with no repeated roots. Then N = H^{2}.

Proof. First observe that F ⊂ N. For each u ∈ F we have that

[\{\frac{u(z)}{1 − bz} \}_{b∈𝔻}] = [\{z^{n}u(z)\}_{n∈ℕ}]

and

z \frac{u(z)}{z − c} = u(z) + c \frac{u(z)}{z − c} whenever u(c) = 0.

Hence, N ⊂ H^{2} is invariant under multiplication by z and by Beurling’s Theorem N = ϕH^{2} for some inner function ϕ. Since ϕ must divide each u ∈ F, we see it must divide β, and since ϕ divides \frac{u(z)}{z−c} for each root c of β, we conclude that ϕ has no roots at all and hence must be constant. □

Theorem 7.11. Let K be a subset of H^{2}, and take γ to be the greatest common divisor of the inner factors of members of K. Then the following are equivalent.

(1) The companion space M of K is reflexive.

(2) γ is a Blaschke product without repeated roots.

Proof. Suppose first that γ has a repeated root a. Then K ⊂ (z − a)^{2}H^{2} so C^{n}((z − a)^{2}H^{2}) ⊂ M. By Proposition 7.8, we know C^{n}((z − a)^{2}H^{2}) is nonreflexive, so the larger space M cannot be reflexive either. Similarly, if γ has a non trivial singular inner factor σ, then M contains the nonreflexive space C^{n}(σH^{2}) and hence is nonreflexive as well.

For the converse, assume γ is a Blaschke product with no repeated roots and let A ∈ ref M. Assume that A1 = Az = ⋯ = Az^{k} = 0 for some fixed k, 0 ≤ k < n. We will show that Az^{k+1} = 0. Then by induction we will get that, if A1 = 0, then A = 0. Whence M will be reflexive by Lemma 7.7.

Given a, b ∈ 𝔻, and u ∈ K, we have by Lemma 7.4 that

(1 − az)^{k}(1 − bz) ⊗ \frac{u^{∗}(z)}{(1 − \overline{a}z)^{k}(1 − \overline{b}z)} ∈ M_{⊥}

and, since A ∈ ref M, it follows that

A(1 − az)^{k}(1 − bz) ⊥ \frac{u^{∗}(z)}{(1 − \overline{a}z)^{k}(1 − \overline{b}z)}.

The induction hypothesis then implies, for nonzero a, b ∈ 𝔻, that

Az^{k+1} ⊥ \frac{u^{∗}(z)}{(1 − \overline{a}z)^{k}(1 − \overline{b}z)}.

But since ∥\frac{1}{(1−\overline{a}z)^{k}} − 1∥_{∞} → 0 as a → 0, we have that

∥\frac{u^{∗}(z)}{(1 − \overline{a}z)^{k}(1 − \overline{b}z)} − \frac{u^{∗}(z)}{(1 − \overline{b}z)}∥_{2} → 0, as a → 0.

Hence,

Az^{k+1} ⊥ \frac{u^{∗}(z)}{1 − \overline{b}z}

for all b ∈ 𝔻 (the case b = 0 obtained by taking the limit as b → 0.)

Using similar reasoning, but starting with the fact that, for c ∈ 𝔻 with u(c) = 0, the operator

(1 − az)^{k}(z − c) ⊗ \frac{u^{∗}(z)}{(1 − \overline{a}z)^{k}(z − \overline{c})} ∈ M_{⊥}

for all a ∈ 𝔻, we obtain that Az^{k+1} ⊥ \frac{u^{∗}(z)}{z − c̄} whenever u(c) = 0. Thus the vector

Az^{k+1} is perpendicular to the space N, as defined in Proposition 7.10. But N = H^{2} by that Proposition and thus Az^{k+1} = 0.

By induction A = 0 and therefore M is reflexive. □

Proof of Theorem 7.6: Set K := {u} in Theorem 7.11. □

Proof of Theorem 1.4: The equivalency of conditions (2) and (3) is just Theorem 7.6. Now, choose K ⊂ H^{2} such that M = C^{n}(K) in Theorem 7.11 to obtain the equivalency of (1) and (3). □

8. Finite Dimensional

We now consider the finite dimensional setting. For each n ∈ ℕ we let P_{n} := \{p ∈ ℂ[z] : deg p ≤ n\}. Given m, n ∈ ℕ and ϕ ∈ span\{1, \overline{z}, …, \overline{z}^{m+n}\} define H_{ϕ} : P_{n} → P_{m} as

H_{ϕ}f = P_{m}J(ϕf),

where P_{m} denotes the orthogonal projection onto P_{m}, and J is the flip operator J(p)(z) := p(\overline{z}). Thus, H_{ϕ} is an operator in B(P_{n}, P_{m}) whose matrix relative to the standard (polynomial) bases has constant skew–diagonals, and clearly all Hankel operators arise in this manner. The full space of Hankel operators in B(P_{n}, P_{m}) will be denoted by Hank(P_{n}, P_{m}). In this case there is a one-to-one relation between a Hankel operator and its symbol.

Notice also that for each H_{ϕ} ∈ Hank(P_{n}, P_{m}), f ∈ P_{n} and g ∈ P_{m}, we have that

(5)  ⟨H_{ϕ}f, g⟩ = \frac{1}{2π} ∫_{−π}^{π} ϕ(e^{iθ})f(e^{iθ})g^{∗}(e^{iθ})dt,

Again, ⟨H_{ϕ}f, g⟩ only depends on the product fg^{∗} and not on the individual factors f and g^{∗}.

The following theorem is proved for the n × n case in [1, Example 3.4]. Using the exact same argument (or adapting arguments from the previous sections), one can show the following.

Theorem 8.1. Let n, m ∈ ℕ.

(1) Hank(P_{n}, P_{m}) has dimension n + m + 1.

(2) Hank(P_{n}, P_{m}) is transitive.

(3) Hank(P_{n}, P_{m}) is elementary.

In [1, Proposition 4.7], it is shown that if a subspace M of B(P_{n}, P_{m}) is transitive, then dim M ≥ n + m + 1. Therefore, no proper subspace of Hank(P_{n}, P_{m}) can be transitive; the same conclusion follows from a slight modification of Part 2 of Corollary 3.3.

Definition 8.2. Let n, m ∈ ℕ. For a polynomial u ∈ P_{m+n} we define C_{u}^{m,n} in Hank(P_{n}, P_{m}) as

C_{u}^{m,n} := \{H_{ϕ} ∈ Hank(P_{n}, P_{m}) : \frac{1}{2π} ∫_{−π}^{π} ϕ(e^{iθ})u(e^{iθ})dθ = 0\} .

Since, dim Hank(P_{n}, P_{m}) = dim P_{m+n}, this is the most general hyperplane in Hank(P_{n}, P_{m}).

Lemma 8.3. Let m, n ∈ ℕ, f ∈ P_{n} and g ∈ P_{m}. Then, f ⊗ g ∈ (C_{u}^{m,n})_{⊥} if and only if fg^{∗} = λu for some λ ∈ ℂ

Proof. Assume fg^{∗} = λu for some λ ∈ ℂ and let H_{ϕ} ∈ C_{u}^{m,n}. By definition, the last statement implies ∫ ϕu = 0. Thus ∫ ϕfg^{∗} = 0 as well, whence ⟨H_{ϕ}f, g⟩ = 0 by Equation (5). Thus f ⊗ g ∈ (C_{u}^{m,n})_{⊥} as required.

Conversely, assume fg^{∗} ∉ [u]. As in Lemma 7.4, choose ϕ ∈ span\{1, \overline{z}, …, \overline{z}^{m+n}\} such that ∫ ϕu = 0 but ∫ ϕfg^{∗} ≠ 0. It then follows that while H_{ϕ} ∈ C_{u}^{m,n} also ⟨H_{ϕ}, f ⊗ g⟩ ≠ 0, and hence f ⊗ g ∉ (C_{u}^{m,n})_{⊥}. □

Proposition 8.4. Fix m, n ∈ ℕ and let u ∈ P_{m+n}. Then (1) if deg u < m + n − 1, then C_{u}^{m,n} is not reflexive,

(2) if u has a repeated root, then C_{u}^{m,n} is not reflexive.

Proof. Let f ∈ P_{n} and g ∈ P_{m} with f ⊗ g ∈ (C_{u}^{m,n})_{⊥} and thus fg^{∗} = λu.

Let us show (1). Observe that the rank-one operator z^{m−1} ⊗ z^{n} is not a Hankel operator. However, if the degree of the polynomial u is strictly less than m + n − 1 and fg^{∗} = λu, then either

(a) the degree of f equals n, which implies that the degree g is less than or equal to m − 2,

(b) or the degree of f is strictly less than n. In either case, we have that f ⊗ g is orthogonal to z^{m−1} ⊗ z^{n}. That is, the reflexive closure of C_{u}^{m,n} contains a non-Hankel operator.

To show (2), suppose that u has a repeated root a, that is (z − a)^{2}|u which implies that either (z − a)^{2} | f or (z − a) | g^{∗}. Let us define

x := ∑_{k=0}^{m} a^{k}z^{k} and y := ∑_{k=0}^{n−1}(k + 1)\overline{a}^{k}z^{k+1}.

Direct computation confirms that x acts as a reproducing kernel for evaluation at \overline{a} while y acts as a reproducing kernel for derivative evaluation at a:

⟨u, x⟩ = u(\overline{a}), u ∈ P_{m}; ⟨v, y⟩ = v′(a), v ∈ P_{n}.

We will show that f ⊗ g is orthogonal to the non-Hankel rank-one operator x ⊗ y (to see that x ⊗ y is a non-Hankel operator just observe that ⟨(x ⊗ y)1, z⟩ = 0 but ⟨(x ⊗ y)z, 1⟩ = 1 and thus its matrix is not constant on the skew-diagonals).

Thus we have ⟨x ⊗ y, f ⊗ g⟩ = ⟨x, g⟩ ⟨f, y⟩ = g^{∗}(a)f′(a) = 0. Hence the non-Hankel operator x ⊗ y is in the reflexive closure of C_{u}^{m,n} and we have that C_{u}^{m,n} is not reflexive. □

The converse of Proposition 8.4 is the main result of this section.

Theorem 8.5. Fix m, n ∈ ℕ and let u ∈ P_{m+n}. If the polynomial u has no repeated roots and deg u ≥ m + n − 1, then C_{u}^{m,n} is reflexive.

The operators in the next definition can be thought of as “analytic Toeplitz operators acting between different spaces”. Given ϕ ∈ P_{1} we let

T_{ϕ} : P_{n−1} ⟶ P_{n},

be the linear operator defined by T_{ϕ}(f) = ϕf.

Proposition 8.6. Let B ∈ B(P_{n}, P_{m}) where n ≥ 2. The following are equivalent: (1) B ∈ Hank(P_{n}, P_{m}),

(2) BT_{ϕ} ∈ Hank(P_{n−1}, P_{m}) for all ϕ ∈ P_{1},
(3) BT_{1}, BT_{z} ∈ Hank(P_{n−1}, P_{m}).

Proof. (1) implies (2) implies (3) is clear. It remains to prove that (3) implies (1). It suffices to show that each entry of the matrix associated to the operator B, with respect to standard orthonormal bases for P_{m} and P_{n}, is a Hankel matrix. For 1 ≤ k ≤ n − 1 and 1 ≤ l ≤ m we have that

⟨Bz^{k}, z^{l−1}⟩ = ⟨(BT_{1})z^{k}, z^{l−1}⟩
= ⟨(BT_{1})z^{k−1}, z^{l}⟩
= ⟨Bz^{k−1}, z^{l}⟩ .

Now, for k = n and 1 ≤ l ≤ m we have that

⟨Bz^{k}, z^{l−1}⟩ = ⟨(BT_{z})z^{k−1}, z^{l−1}⟩
= ⟨(BT_{z})z^{k−2}, z^{l}⟩
= ⟨Bz^{k−1}, z^{l}⟩ .

Therefore the associated matrix of B is a Hankel matrix which is equivalent to B ∈ Hank(P_{n}, P_{m}). □

Proposition 8.7. Let u ∈ P_{m+n−1} and ϕ ∈ P_{1}. Then C_{uϕ}^{m,n}T_{ϕ} ⊂ C_{u}^{m,n−1}.

Proof. Let H_{ψ} ∈ C_{uϕ}^{m,n}. Then ∫ ψuϕ = 0. Since H_{ψϕ} = H_{ψ}T_{ϕ} and ∫ (ψϕ)u = 0 we have that H_{ψ}T_{ϕ} ∈ C_{u}^{m,n−1} as desired. □

Before proving Theorem 8.5, we show a special case of it.

Proposition 8.8. C_{u}^{1,1} is reflexive if u has no repeated roots and deg u ≥ 1.

Proof. C_{u}^{1,1} is 2-dimensional, so dim((C_{u}^{1,1})_{⊥}) = 4 − 2 is 2, as well. Hence showing C_{u}^{1,1} is reflexive amounts to finding two linearly independent rank-one members of (C_{u}^{1,1})_{⊥}.

First suppose that deg u = 2, that is u(z) = (z − a)(z − b) for distinct values a, b ∈ ℂ. Then by Lemma 8.3 we have that

(z − a) ⊗ (z − \overline{b}) (z − b) ⊗ (z − \overline{a})

are linearly independent rank-one members of (C_{u}^{1,1})_{⊥}.

Now suppose that deg u = 1, that is u(z) = (z − a) for some a ∈ ℂ. Then we have that, as before,

(z − a) ⊗ 1 1 ⊗ (z − \overline{a})

are linearly independent rank-one members of (C_{u}^{1,1})_{⊥}. □

Proof of Theorem 8.5: Consider the following statement:

P (m, n) : if u ∈ P_{m+n} has no repeated roots and deg u ≥ m + n − 1,
then C_{u}^{m,n} is reflexive.

Let m, n ∈ ℕ and assume that P (m, n) is true. Now let u ∈ P_{m+n+1} be such that deg u ≥ m + n and u has no repeated roots, let B ∈ ref C_{u}^{m,n+1} and let a ≠ b be two distinct roots of u. Then

BT_{z−a} ∈ (ref C_{u}^{m,n+1}) T_{z−a} ⊂ ref (C_{u}^{m,n+1}T_{z−a})
= Cm,n_{\frac{u}{z−a}}
⊂ Hank(P_{n}, P_{m}),

where we are using the fact that if M is a subspace of operators and T an operator, then (ref M) T ⊂ ref (M T), (which can be found in [2] Lemma 4.5), combined with Proposition 8.7 and the reflexivity of Cm,n_{\frac{u}{z−a}}.

Similarly, BT_{z−b} ∈ Hank(P_{n}, P_{m}). Since 1, z ∈ span {z − a, z − b} we have that BT_{1}, BT_{z} ∈ Hank(P_{n}, P_{m}). Hence by Proposition 8.6, the operator B ∈ Hank(P_{n+1}, P_{m}). Thus C_{u}^{m,n+1} is reflexive and P (m, n + 1) is true. So we have that

∀ m, n ∈ ℕ [P (m, n) ⟹ P (m, n + 1)]. (6)

By Proposition 8.8 we have that P (1, 1) is true and (6) implies that for all n ∈ ℕ that P (1, n) is true. Now observe that (C_{u}^{m,n})^{∗} = C_{u^{∗}}^{n,m}. Thus by symmetry we have that for all m ∈ ℕ that P (m, 1) is true. Applying (6) again we get that for all m, n ∈ ℕ that P (m, n) is true. □

We will need the following classical result.

Theorem 8.9 (Bertini’s Theorem). Suppose p(z) and q(z) are polynomials (over ℂ) with no common roots. Then p(z) + tq(z) has a repeated root for at most finitely many t ∈ ℂ.

Proof. If a is a repeated root of the polynomial p(z) + tq(z) for some t ∈ ℂ, then p(a)q′(a) − p′(a)q(a) = 0. Thus there are at most deg (p) + deg (q) − 1 numbers a satisfying this equation and thus only a finite number of a’s can be repeated roots for p(z) + tq(z), regardless of t.

If the same a is a repeated root for the polynomials p(z) + tq(z) and p(z) + sq(z) for some s, t ∈ ℂ such that s ≠ t, then, p and q share a common root, which contradicts the hypothesis. Hence, once an element of the finite set described above is a repeated root of p(z) + tq(z), it cannot be a repeated root of any other p(z) + sq(z), t ≠ s. Hence, the polynomial p(z) + tq(z) has a repeated root for at most finitely many t ∈ ℂ. □

Proposition 8.10. Let F ⊂ ℂ be a finite nonempty set and suppose k > 1. If w_{1}, …, w_{k} are relatively prime polynomials, i. e. gcd(w_{1}, …, w_{k}) = 1, then there exist t_{2}, …, t_{k} ∈ ℂ such that the polynomial

w_{1} + t_{2}w_{2} + ⋯ + t_{k}w_{k}

has no repeated roots, no roots in F, and degree ≥ deg(w_{1}).

Proof. Let F ⊂ ℂ be a finite nonempty set. Let w_{1}, w_{2} be polynomials which are relatively prime.

If for some a ∈ F we can choose s ≠ t ∈ ℂ such that a is a root of both the polynomials w_{1} + sw_{2} and w_{1} + tw_{2} then a would be a common root of w_{1} and w_{2}, contradicting our assumption that w_{1} and w_{2} are relatively prime polynomials. It follows from the finiteness of F that there can only be finitely many t for which w_{1} + tw_{2} has a root in F. In view of Bertini’s Theorem, for all but finitely many t ∈ ℂ, the polynomial w_{1} + tw_{2} has no repeated roots as well as no roots in F. Also, there can be at most one t for which deg(w_{1} + tw_{2}) < deg(w_{1}).

Now assume k > 2 and that the Proposition holds for relatively prime (k − 1)- tuples of polynomials. Suppose we are given relatively prime polynomials w_{1}, ..., w_{k}. Set g := gcd(w_{1}, …, w_{k−1}), and take F′ the set of roots of w_{k}. Applying (part of) the inductive hypothesis, we obtain t_{2}, ..., t_{k−1} ∈ ℂ such that the polynomial \frac{w_{1}}{g} +t_{2} \frac{w_{2}}{g} +⋯+t_{k−1} \frac{w_{k−1}}{g} has no roots in F′ and its degree is at least deg \frac{w_{1}}{g}. Since g has no roots in common with w_{k}, we conclude that w_{1} + t_{2}w_{2} + ⋯ + t_{k−1}w_{k−1} is relatively prime to w_{k}. An application of the first paragraph of the proof completes the inductive argument. □

In view of the analogous statement to Proposition 2.5 in our context, any reflexive subspace of a reflexive hyperspace of Hank(P_{n}, P_{m}) will also be reflexive. The following generalized version of Theorem 8.5 shows that conversely every reflexive subspace of Hank(P_{n}, P_{m}) is contained in a reflexive hyperspace.

Corollary 8.11. Fix m, n ∈ ℕ, let u_{1}, …, u_{k} ∈ P_{m+n}, and set r = gcd(u_{1}, …, u_{k}). Then the following are equivalent:

(1) ⋂_{j=1}^{k} C_{u_{j}}^{m,n} is reflexive.
(2) ⋂_{j=1}^{k} C_{u_{j}}^{m,n} is contained in a reflexive hyperspace of Hank(P_{n}, P_{m}).

(3) Some u_{i} has degree at least m + n − 1 and r(z) has no repeated roots.

Proof. We will first show that (1) implies (3). Suppose that r has a repeated root and and let a ∈ ℂ be such a root. Consider the polynomials k_{0}, k_{1} ∈ P_{n+m} defined to have the property that

\frac{1}{2π} ∫_{−π}^{π} \overline{k_{0}(e^{iθ})}p(e^{iθ})dθ = p(a) and \frac{1}{2π} ∫_{−π}^{π} \overline{k_{1}(e^{iθ})}p(e^{iθ})dθ = p′(a)

(these polynomials are easily seen to exist: see the proof of Proposition 8.4).

Since (z − a)^{2} divides u_{1}, u_{2}, …, u_{k}, we have that for all j, H\overline{k_{0}} and H\overline{k_{1}} belong to C_{u_{j}}^{m,n}. Thus

span \{H\overline{k_{0}}, H\overline{k_{1}}\} ⊂ ⋂_{j=1}^{k} C_{u_{j}}^{m,n},

whence ref (span \{H\overline{k_{0}}, H\overline{k_{1}}\}) is contained in ref ( ⋂_{j=1}^{k} C_{u_{j}}^{m,n}). But one can easily

check that the non-Hankel operator k_{0} ⊗ k_{1} ∈ ref (span \{H\overline{k_{0}}, H\overline{k_{1}}\}). Therefore

ref ( ⋂_{j=1}^{k} C_{u_{j}}) contains a non-Hankel member, so ⋂_{j=1}^{k} C_{u_{j}}^{m,n} is not reflexive.

If every u_{i} has degree less than m + n − 1, then one can check, in a similar

way to the arguments above, that ref ( ⋂_{j=1}^{k} C_{u_{j}}) contains the non-Hankel operator

z^{m−1} ⊗ z^{n} and ⋂_{j=1}^{k} C_{u_{j}}^{m,n} is not reflexive.

We will now show that (3) implies (2). When k = 1, we have r = u_{1} and it suffices to apply Theorem 8.5. Thus assume k > 2 and for definiteness that u_{1} has degree at least m + n − 1.

Let r = gcd(u_{1}, ..., u_{k}) and suppose that r has no repeated roots. Next, let v_{j} = u_{j}/r for j = 1, ..., k. Thus, by Proposition 8.10 we can choose t_{2}, ..., t_{k} ∈ ℂ such that

v_{1} + t_{2}v_{2} + ⋯ + t_{k}v_{k}

has no repeated roots, no roots in common with r and its degree is at least that of v_{1}. Multiplying through by r, we see that

u_{1} + t_{2}u_{2} + ⋯ + t_{k}u_{k}

has no repeated roots and has degree at least m + n − 1.

Thus Theorem 8.5 tells us C_{u_{1}+t_{2}u_{2}+···+t_{k}u_{k}}^{m,n} is reflexive.

Finally (2) implies (1) since Hank(P_{n}, P_{m}) is elementary. □

Proof of Theorem 1.5: Observe that Corollary 8.11 is just Theorem 1.5. □

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Department of Mathematics, University of Georgia, Athens, GA 30602-7403

E-mail address: azoff@math.uga.edu

Centro de Investigación en Matemáticas, Universidad Autónoma del Estado de Hidalgo, Pachuca, Hidalgo, 42184, Mexico

E-mail address: rubenma@uaeh.edu.mx

Department of Mathematics, University of Georgia, Athens, GA 30602-7403

E-mail address: solazzo@math.uga.edu