REFLEXIVITY ON HILBERT FUNCTION SPACES

In this paper we present sufficient conditions for reflexivity of any powers of the multiplication operator acting on Hilbert spaces of analytic functions on a finitely connected domain. This improves the main result of [21]. AMS Subject Classification: 47B37, 47L10


Introduction
By H(G) and H ∞ (G) we will mean respectively the set of analytic functions on a plane domain G and the set of bounded analytic functions on G. Assume that Ω is a finitely connected domain.It is well known that Ω is conformally equivalent to a circular domain.By a circular domain we mean any domain that is obtained by removing a finite number of disjoint closed subdisks from the open unit disk D. So we let Ω = D\( D1 ∪ ... ∪ DN ) where Di = {z : |z − z i | ≤ r i } (i = 1, ..., N ) are disjoint subdisks of the open unit disk D. We can choose ǫ i > 0 (i = 1, ..., N ) such that the circles and Γ 0 = {z : |z| = 1 − ǫ 0 } lying in Ω concentrate to the boundary circles of Ω so that they don't meet each other.We denote Ω i = C\ Di (i = 1, ..., N ).In [3] it is proved that where the subscript zero means that the corresponding functions vanish at ∞. Consider a Hilbert space H of functions analytic on a plane domain G, such that for each λ ∈ G the linear functional, e λ , of evaluation at λ is bounded on H. Assume further that H contains the constant functions and multiplication by the independent variable z defines a bounded linear operator M z on H.The continuity of point evaluations along with the Riesz representation theorem imply that for each λ ∈ G there is a unique function The function k λ is called the reproducing kernel for the point λ.
A complex valued function ϕ on G for which ϕf ∈ H for every f ∈ H is called a multiplier of H and the collection of all these multipliers is denoted by M(H).Each multiplier ϕ of H determines a multiplication operator M ϕ on H by M ϕ f = ϕf , f ∈ H.It is well known that each multiplier is a bounded analytic function on G (see [12]).In fact ϕ G ≤ M ϕ , where We shall use the following notation for the norm of the operator M ϕ : We also point out that if ϕ is a multiplier and λ ∈ G, then Recall that if E is a separable Banach space and A ∈ B(E), then Lat(A) is by definition the lattice of all invariant subspaces of A, and AlgLat(A) is the algebra of all operators B in B(E) such that Lat(A) ⊂ Lat(B).For the algebra B(E), the weak operator topology is the one induced by the family of seminorms p x * ,x (A) = | < Ax, x * > | where x ∈ E, x * ∈ E * and A ∈ B(E).Hence A α −→ A in the weak operator topology if and only if A α x −→ Ax weakly.Also similarly A α −→ A in the strong operator topology if and only if A α x −→ Ax in the norm topology.An operator A in B(E) is said to be reflexive if AlgLat(A) = W (A), where W (A) is the smallest subalgebra of B(E) that contains A and the identity I and is closed in the weak operator topology.

Main Results
The operator M z has been the focus of attention for several decades and many of its properties have been studied (e.g.[2,12]).In [1o] Sarason proved that normal operators are reflexive.It was shown by J. Deddens (see [4]) that every isometry is reflexive.Also, R. Olin and J. Thomson (see [8]) have shown that subnormal operators are reflexive.H. Bercovici, C. Foias, J. Langsam, and C. Pearcy (see [1]) have shown that (BCP)-operators are reflexive.The reflexive operators on a finite dimensional space were characterized by J. Deddens and P. A. Fillmore (see [5]).In [7,11,14 -21] some sufficient conditions for the reflexivity of multiplication operators on some function spaces have been investigated.Also, reflexivity of canonical models were studied in [6].In [17] it is proved that if M z is invertible on the space L p (β), then it is reflexive.In this article we would like to give some sufficient conditions so that the powers of the operator M z , acting on a Hilbert space of analytic functions on a finitely connected domain, becomes reflexive.This extends the main result of the paper [21].For a good source of reflexivity see [9].
From now on, let Ω be a finitely connected domain in the complex plane and suppose that the Hilbert space H under consideration satisfy the following axioms: Axiom 1. H is a subspace of the space of all analytic functions on Ω.
Axiom 2. For each λ ∈ Ω, the linear functional of evaluation at λ, e(λ), is bounded on H. Axiom 3. The sequence {f k } k is an orthogonal basis for H where f k (z) = z k for all integers k.
For h ∈ H(D) ∩ M(H) and w ∈ ∂D, define h w by h w (z) = h(wz).Thus ĥw (n) = w n ĥ(n) for all n.Note that since |w| = 1, we have Also, we say that Furthermore, we assume that H holds in the following axiom: For the proof of the main result we will need the following lemmas.
Thus for all m, M ϕ w f m −→ M ϕ f m as w −→ 1.Since M ϕ w = M ϕ < ∞ for all w ∈ ∂D, indeed M ϕ w −→ M ϕ in the strong operator topology.
(ii) First note that the strong operator continuity of ϕ w allows us to define This completes the proof.
Throughout this paper we suppose that M z is bounded on H.In the following by H(G) and H ∞ (G) we will mean respectively the set of analytic functions on a plane domain G and the set of bounded analytic functions on G.We should prove that the operators L and M φ(j)f j have the same matrix entries with respect to the orthogonal basis {f j } j .We have which is equal to φ(n − m) f n 2 whenever n = m + j, and is 0 else.On the other hand we note that which is equal to φ(n − m) f n 2 whenever n = m + j, and is 0 else.Hence L = M φ(j)f j .Since {f j } j is a basis for H, the proof is complete.
Note that K n ≥ 0 and For all n ≥ 0, P n (ϕ) ∈ H(D) ∩ M(H) and by Lemma 2.1 (ii), we get Put r n = P n (ϕ) and note that M rn is represented by the matrix whose (i,j)-th entry is Hence lim for all base elements f j and f i in H.By the boundedness of the sequence {M rn }, we have M rn → M ϕ in the weak operator topology.This completes the proof.
Then by the proof of Lemma 2.3, we get M rn → M ϕ in the weak operator topology where r n = P n (ϕ).Since r n is a polynomial and M rn = r n (M z ), we conclude that M ϕ ∈ W (M z ).
Theorem 2.5.For all k ≥ 1, the operator M z k is reflexive on H.
First, we note that convergence in H implies uniform convergence on compact subsets of Ω.For this let K be a compact subset of Ω and consider the family of bounded linear functionals {e λ : So by the Principle of Uniform Boundedness Theorem, the family {e λ : λ ∈ K} is bounded.Put c = sup{ e λ : λ ∈ K} and let a sequence {f n } n converges to f in H. Then we have Hence convergence in H implies uniform convergence on compact subsets of Ω. Set Note that L 0 is a subspace of H.To see that L 0 is closed, let {g k } be a sequence in L 0 such that g k → h in H. Since Γ 0 is a compact subset of Ω, it is now easy to see that h ∈ L 0 and so L 0 is closed in H. Also, clearly L 0 is invariant under M z and contains the constants.Let k ∈ N and note that W (M z k ) ⊂ AlgLat(M z k ).On the other hand, let A ∈ AlgLat(M z k ).Since Lat(M z ) ⊂ Lat(M z k ), thus we have Lat(M z ) ⊂ Lat(A).This implies that A ∈ AlgLat(M z ).Note that since M * z e(λ) = λe(λ) for all λ in Ω, the one dimensional span of e(λ) is invariant under M * z .Therefore it is invariant under A * and we can write A * e(λ) = ϕ(λ)e(λ), λ ∈ Ω.So < Af, e(λ) >=< f, A * e(λ) >= ϕ(λ)f (λ) for all f ∈ H and λ ∈ Ω.This implies that A = M ϕ and ϕ ∈ M(H), hence ϕ ∈ H ∞ (Ω).Since L 0 ∈ Lat(M z ), we have AL 0 ⊂ L 0 , so A1 = ϕ ∈ L 0 .By applying the Cauchy integral formula we can write ϕ = ϕ 0 + ϕ 1 + ... + ϕ N where ϕ 0 ∈ H(D) and ϕ i ∈ H 0 (Ω i ) (i=1,...,N) (here H 0 (Ω i ) denotes the space of all functions in H(Ω i ) that vanishes at ∞). Set g = ϕ 1 + ... + ϕ N .Therefore g is analytic in ext(Γ ′ 0 ) the unbounded component of C \Γ ′ 0 where the circle Γ ′ 0 is chosen sufficiently close to Γ 0 with smaller radius so that Γ 0 lies in ext(Γ ′ 0 ).Now, we can write From this it follows that ĝ(n) = 0 for all integers n ≤ −1 and so g(z) = 0, z ∈ ext(Γ ′ 0 ).Hence g ≡ 0 which implies that ϕ = ϕ 0 ∈ H(D).Thus ϕ ∈ H(D) ∩ M(H) and so by Lemma 2.3, there exists a sequence of polynomials {r n } such that M rn → M ϕ in the weak operator topology.Now, we use a similar method used in the proof of the main theorem in [21]: let M k be the closed linear span of the set {f nk : n ≥ 0} (recall that f i (z) = z i for all i).We have M z k f nk = f (n+1)k ∈ M k for all n ≥ 0. Thus M k ∈ Lat(M z k ), and so Hence φ(i) = 0 for all i = nk, n ≥ 0. Now, by a consequence of the particular construction of r n used in Lemma 2.3, each r n should be a polynomial in z k , i.e., r n (z) = q n (z k ) for some polynomial q n .Thus M rn = r n (M z ) = q n (M z k ) → A in the weak operator topology.Hence A ∈ W (M z k ).Thus M z k is reflexive and so the proof is complete.
In the following we give an example of a Hilbert space for which the axioms 1 through 4, hold.

Lemma 2 . 1 .
Let ϕ ∈ H(D) ∩ M(H).Then: (i) If w −→ 1, then M ϕw −→ M ϕ in the strong operator topology.(ii) If g is a continuous complex valued function on ∂D and dλ = |dw|/2π is the normalized Lebesgue measure on ∂D, then the operator ∂D ϕ w g(w)dλ defined by

Lemma 2 . 2 .
Let ϕ ∈ H(D) ∩ M(H) and let p be a polynomial.Then ϕ * p ∈ M(H) and ∂D ϕ w p( w)dλ = M ϕ * p where (ϕ * p)(z) = i φ(i)p(i)f i .Proof.It is enough to consider the case p = f j .Define the operator L by L = ∂D ϕ w f j ( w)dλ.