SOME CONDITIONS ON DOUGLAS ALGEBRAS THAT IMPLY THE INVARIANCE OF THE MINIMAL ENVELOPE MAP

CARROLL GUILLORY

Abstract. We give several conditions on certain families of Douglas algebras that imply that the minimal envelope of the given algebra is the algebra itself. We also prove that the minimal envelope of the intersection of two Douglas algebras is the intersection of their minimal envelope.

1. Introduction

Let D denote the open unit disk in the complex plane, and T the unit circle. By L^{∞} we mean the space of essentially bounded measurable functions on T with respect to the normalized Lebesgue measure. We denote by H^{∞} the space of all bounded analytic functions in D. Via identification with boundary functions, H^{∞} can be considered as a uniformly closed subalgebra of L^{∞}. Any uniformly closed subalgebra B strictly between H^{∞} and L^{∞} is called a Douglas algebra. We denote by M(B) the maximal ideal space of a Douglas algebra B. If C is the set of all continuous functions on T, we set H^{∞} + C = {h + g : h ∈ H^{∞}, g ∈ C}. Then H^{∞} + C becomes the smallest Douglas algebra containing H^{∞} properly.

The function

q(z) = ∏_{n=1}^{∞} \frac{|z_{n}|}{z_{n}} \frac{z − z_{n}}{1 − z̄_{n}z}

is called a Blaschke product if ∑_{n=1}^{∞}(1 − |z_{n}|) converges. The set {z_{n}}_{n} is called the zero set of q in D. Here |z_{n}|/z_{n} = 1 is understood whenever z_{n} = 0. We call q an interpolating Blaschke product if

inf_{n} ∏_{m:m≠n} |\frac{z_{m} − z_{n}}{1 − z̄_{n}z_{m}} | > 0.

All work on this paper was done while the author was at the Mathematical Science Research Institute. The author thanks the Institute for its support during this period. Research at MSRI is supported in part by NSF grant DMS-9022140.

An interpolating Blaschke product q is called sparse (or thin) if

lim_{n→∞} ∏_{m:m≠n} |\frac{z_{m} − z_{n}}{1 − z̄_{n}z_{m}} | = 1.

The set Z(q) = {x ∈ M(H^{∞}) \ D : q(x) = 0} is called the zero set of q in M(H^{∞} + C). Any function h in H^{∞} with |h| = 1 almost everywhere on T is called an inner function. Since |q| = 1 for any Blaschke product, Blaschke products are inner functions. Let QC = (H^{∞} + C) ∩ \overline{(H^{∞} + C)} and for x ∈ M(H^{∞} + C), set

Q_{x} = {y ∈ M(L^{∞}) : f(x) = f(y) for all f ∈ QC}.

Then Q_{x} is called the QC-level set for x. For x ∈ M(H^{∞} + C), we denote by μ_{x} the representing measure for x, and its support set by supp μ_{x}. By H^{∞}[q̄] we mean the Douglas algebra generated by H^{∞} and the complex conjugate of the function q. Since M(L^{∞}) is the Shilov boundary for every Douglas algebra, a closed set E contained in M(L^{∞}) is called a peak set for a Douglas algebra B if there is a function f in B with f = 1 on E and |f| < 1 on M(L^{∞}) \ E. A closed set E is a weak peak set for B if E is the intersection of a family of peak sets. If the set E is a weak peak set for H^{∞} and we define

H_{E}^{∞} = {f ∈ L^{∞} : f|_{E} ∈ H^{∞}|_{E}},

then H_{E}^{∞} is a Douglas algebra. For a Douglas algebra B, we define B_{E} similarly. For an interpolating Blaschke product q we put N(q̄) the closure of

⋃{supp μ_{x} : x ∈ M(H^{∞} + C), |q(x)| < 1}.

Then N(q̄) is a weak peak set for H^{∞}. By N_{0}(q̄) we denote the closure of ⋃{supp μ_{x} : x ∈ Z(q)}. In general N_{0}(q̄) does not equal N(q̄), but in this paper N_{0}(q̄) = N(q̄). For x ∈ M(H^{∞}), we let E_{x} = {y ∈ M(H^{∞}) : supp μ_{y} = supp μ_{x}} and call E_{x} the level set of x. Since the sets supp μ_{x} and N(q̄) are weak peak sets for H^{∞}, both H_{supp μ_{x}}^{∞} and H_{N(q̄)}^{∞} are Douglas algebras. For the interpolating Blaschke product q, set

A = ⋂_{x∈M(H^{∞}+C)}{H_{supp μ_{x}}^{∞} : |q(x)| < 1}

and

A_{0} = ⋂{H_{supp μ_{x}}^{∞} : x ∈ Z(q)}.

Our assumptions on q throughout this paper imply that H_{N(q̄)}^{∞} = A = A_{0} (see [10]). For x and y in M(H^{∞}), the pseudohyperbolic distance ρ is defined by

ρ(x, y) = supp{|h(y)| : ∥h∥_{∞} ≤ 1, h ∈ H^{∞}, h(x) = 0}.

INVARIANCE OF THE MINIMAL ENVELOPE MAP OF DOUGLAS ALGEBRAS 3

For x and y in D, we have

ρ(x, y) = |\frac{x − y}{1 − ȳx}| .

For x ∈ M(H^{∞}), we define the Gleason part P_{x} of x by

P_{x} = {y ∈ M(H^{∞}) : ρ(x, y) < 1}.

If P_{x} ≠ {x}, then x is said to be a nontrivial point. For the definition of those interpolating Blaschke products that are of type G and of finite type G, see [8].

Let B be a Douglas algebra. The Bourgain algebra B_{b} of B relative to L^{∞} is the set of f in L^{∞} such that ∥ff_{n} + B∥_{∞} → 0 for every sequence {f_{n}} in B with f_{n} → 0 weakly in B. An algebra B is called a minimal superalgebra of an algebra A if A ⊂ B and supp μ_{x} = supp μ_{y} for all x, y ∈ M(A) \ M(B). The minimal envelope B_{m} of a Douglas algebra B is defined to be the smallest Douglas algebra that contains all the minimal superalgebras of B. The mapping that assigns to B the Douglas algebra B_{m} is called the minimal envelope map.

2. Conditions for B_{m} = B

We begin with the following theorem. The case for the Bourgain algebra B_{b} has been proven in [15]. The proof used here is quite different from theirs, and can be used to show that this result also holds for the Bourgain algebras.

Theorem 1. Let A and B be Douglas algebras. Then (A ∩ B)_{m} = A_{m} ∩ B_{m}.

Proof. Since A ∩ B is contained in both A and B by Proposition 6 of [11], (A ∩ B)_{m} is contained in both A_{m} and B_{m}. Hence (A ∩ B)_{m} ⊂ A_{m} ∩ B_{m}.

To show that A_{m}∩B_{m} ⊂ (A∩B)_{m}, let ψ be an interpolating Blaschke product such that ψ̄ ∈ A_{m}∩B_{m}. We show that ψ̄ ∈ (A∩B)_{m}. By Theorem D of [11] we can assume that there is an x ∈ M(A) and a y ∈ M(B) such that {λ ∈ M(A) : |ψ(γ)| < 1} = E_{x} and {μ ∈ M(B) : |ψ(μ)| < 1} = E_{y}. This implies that M(A) = M(A[ψ̄]) ∪ E_{x} and M(B) = M(B[ψ̄]) ∪ E_{y}. Thus

\begin{aligned}M(A ∩ B) & = M(A) ∪ M(B) = M(A[ψ̄]) ∪ M(B[ψ̄]) ∪ E_{x} ∪ E_{y} \\ & = M(A[ψ̄] ∩ B[ψ̄]) ∪ E_{x} ∪ E_{y} = M((A ∩ B)[ψ̄]) ∪ E_{x} ∪ E_{y}. \end{aligned}

Hence {ω ∈ M(A ∩ B) : |ψ(ω)| < 1} = E_{x} ∪ E_{y}. By Theorem 3 of [11] we have ψ̄ ∈ (A ∩ B)_{m}. Our theorem follows.

The following corollaries are immediate consequences of the theorem.

Corollary 1. Let A and B be Douglas algebras with A = A_{m} and B = B_{m}. Then A ∩ B = (A ∩ B)_{m}.

4 CARROLL GUILLORY

Corollary 2. Let B_{0} be a Douglas algebra with (B_{0})_{m} = B_{0} and B be a Douglas algebra such that there is an interpolating Blaschke product q with {λ ∈ M(B) : |q(λ)| = 1} ⊂ M(B_{0}). If A has the property that A_{m} = A, then B[q̄]∩A = (B[q̄]∩A)_{m}. In particular, H^{∞}[φ̄] ∩ A = (H^{∞}[q̄] ∩ A)_{m} for every interpolating Blaschke product φ.

Proof. The hypothesis {λ ∈ M(B) : |q(λ)| = 1} ⊂ M(B_{0}) implies that B_{0}[q̄] = B[q̄], hence [B[q̄]]_{m} = B[q̄]. So B[q̄] ∩ A = (B[q̄] ∩ A)_{m} follows from our theorem. By Theorem 1 of [2] we have H_{m}^{∞} = H^{∞}+C, so {λ ∈ M(H^{∞}) : |ψ(λ)| = 1} ⊂ M(H^{∞}+C) for every interpolating Blaschke product. The second part of the corollary follows.

Theorem 2. Let B_{0} be a Douglas algebra such that (B_{0})_{m} = B_{0}. Let B be any other Douglas algebra such that there is an interpolating Blaschke product φ with {λ ∈ M(B) : |φ(λ)| = 1} ⊂ M(B_{0}). For an interpolating Blaschke product q set B_{x} = B[φ̄] ∩ H_{supp μ_{x}}^{∞} for each x ∈ Z(q). Put B_{e} = ⋂{B_{x} : x ∈ Z(q)}.

(i) If q is of finite type G, then (B_{e})_{b} = B_{e}. (ii) If q is the product of a finite number of sparse interpolating Blaschke products,

then (B_{e})_{m} = B_{e}.

Proof. We show that B_{e} = B[φ̄] ∩ H_{N(q̄)}^{∞}. By, an unpublished result of D. Sarason,

M(B_{e}) = M( ⋂_{x∈Z(q)} B_{x}) = \overline{⋃_{x∈Z(q)} M(B_{x})}

= \overline{⋃_{x∈Z(q)} M(B[φ̄] ∩ H_{supp μ_{x}}^{∞})} = \overline{⋃_{x∈Z(q)}(M(B[φ̄]) ∪ M(H_{supp μ_{x}}^{∞}))} ,

M(B_{e}) = M(B[φ̄]) ∪ \overline{⋃_{x∈Z(q)} M(H_{supp μ_{x}}^{∞})} .

Now, if q is of finite type g, Proposition 1 of [9] and Theorem 3.2(i) of [8] imply ⋂_{x∈Z(q)} H_{supp μ_{x}}^{∞} = H_{N(q̄)}^{∞}. Thus

\overline{⋃_{x∈Z(q)} M(H_{supp μ_{x}}^{∞})} = M(H_{N(q̄)}^{∞})

and we get

\begin{aligned}M(B_{e}) & = M(B[φ̄]) ∪ M(H_{N(q̄)}^{∞}) \\ & = M(B[φ̄] ∩ H_{N(q̄)}^{∞}). \end{aligned}

So by the Chang–Marshall Theorem [1, 13] we have B_{e} = B[φ̄] ∩ H_{supp μ_{x}}^{∞}.

The condition {λ ∈ M(B) : |φ(λ)| = 1} ⊂ M(B_{0}) implies that B[φ̄] = B_{0}[φ̄]. Hence [B[φ̄]]_{m} = [B_{0}[φ̄]]_{m} = [B_{0}]_{m}[φ̄] = B_{0}[φ̄] = B[φ̄], where the middle equality follows from Theorem 4 of [11]. By theorem 1 of [7] (and its proof) we have (H_{N(q̄)}^{∞})_{b} = H_{N(q̄)}^{∞} if q is of finite type G. Thus

(B_{e})_{b} = (B[φ̄] ∩ H_{N(q̄)}^{∞})_{b} = B[φ̄] ∩ H_{N(q̄)}^{∞} = B_{e}.

This proves (i).

Now if q is the product of a finite number of sparse Blaschke products, then again by Theorem 1 of [7] we have (H_{N(q̄)}^{∞})_{m} = H_{N(q̄)}^{∞}, and so (B_{e})_{m} = B_{e}. This proves (ii).

Corollary 3. Suppose the hypothesis of Theorem 2 holds if φ = q. Then B_{e} = B, and:

(i) B = (B)_{b} if q is of finite type G; (ii) B = B_{m} is the product of a finite number of sparse Blaschke products.

Theorem 3. Let B be a Douglas algebra and let E be a peak set for B. Then (B_{E})_{m} = B_{E}.

Proof. Since E is a peak set for B, there is a function f ∈ B such that f = 1 on E and |f| < 1 on M(L^{∞}) \ E. By [3, p. 39] we have

M(B_{E}) = {x ∈ M(B) : supp μ_{x} ⊆ E} ∪ M(L^{∞}) = {x ∈ M(B) : f(x) = 1} ∪ M(L^{∞}).

Let I be any interpolating Blaschke product such that Z(I) ∩ M(B_{E}) ≠ ∅. We will show that there is an uncountable set Γ ⊂ Z(I) ∩ M(B_{E}) such that E_{α} ≠ E_{β} for distinct α, β ∈ Γ. Let {z_{n}} be the zero sequence of I in D and y ∈ Z(I) ∩ M(B_{E}). Since y ∈ Z(I), there is a subsequence {z_{n_{k}}} of {z_{n}} such that z_{n_{k}} → y. Since y ∈ M(B_{E}), we have f(z_{n_{k}}) → 1. Let I_{1} be the factor of I such that {z_{n_{k}}} is the zero sequence of I_{1}. Then

Z(I_{1}) = \overline{\{z_{n_{k}}\}} \ \{z_{n_{k}}\}

and since Z(I_{1}) is equivalent to the Čech compactification of the integers, Z(I_{1}) is an uncountable set. Since f(z_{n_{k}}) → 1, we have that f = 1 on Z(I_{1}). Thus Z(I_{1}) ⊂ Z(I) ∩ M(B_{E}), and so Z(I) ∩ M(B_{E}) is an infinite set. To show that Γ exists, take a subsequence {z_{n_{k_{0}}}} of {z_{n_{k}}} such that {z_{n_{k_{0}}}} is a sparse Blaschke sequence in D (see [4]). Let I_{0} be the sparse Blaschke product whose zero sequence in D is {z_{n_{k_{0}}}}. Then, by Theorem 1 of [12], we have

N(I_{0}) = ⋃_{x∈Z(I_{0})} Q_{x}.

Since Z(I_{0}) is uncountable, Lemma 4 of [12] says that Q_{x} ≠ Q_{y} for distinct x, y ∈ Z(I_{0}). This implies that supp μ_{x} ∩ supp μ_{y} = ∅ and E_{x} ≠ E_{y}. Set Γ = Z(I_{0}). Then

⋃_{α∈Γ} E_{α} ⊆ {λ ∈ M(B_{E}) : |I(λ)| < 1}.

By Theorem 3 of [11], Ī ∉ (B_{E})_{m}. This shows that (B_{E})_{m} = B_{E}.

Theorem 4. Let A be a Douglas algebra and let {q_{n}} be a sequence of interpolating Blaschke products such that \overline{q_{n}} ∉ A for every n. Let B = A[\overline{q_{n}}; n = 1, 2, 3 …]. Then B_{m} = B.

Proof. Since M(B) = {λ ∈ M(A) : |q_{n(λ)}| = 1, n = 1, 2, … }, we can define the function F on M(H^{∞}) as

F (x) = ∑_{n=1}^{∞} \frac{1}{2^{n}} |q_{n}(x)|.

Then F = 1 on M(B) and |F | < 1 on M(H^{∞}) \ M(B). Let C be any interpolating Blaschke product such that Z(C) ∩ M(B) ≠ ∅. Let {z_{n}} be the zero sequence of C in D. As in the proof of Theorem 3 we can find a sparse subsequence {z_{n_{k}}} of {z_{n}} such that F = 1 on \overline{\{z_{n_{k}}}\} \ \{z_{n_{k}}\}. If we set C_{1} to be the factor of C whose zero sequence in D is {z_{n_{k}}}, then as in the proof of Theorem 3 we can show that E_{α} ≠ E_{β} for distinct α, β ∈ Z(C_{1}), and

⋃_{α∈Z(C_{1})} E_{α} ⊂ {λ ∈ M(B) : |Cλ| < 1}.

So, by Theorem 3 of [11], we have C̄ ∉ B_{m}. This implies that B_{m} = B.

Theorem 5. Let {q_{n}} be a sequence of interpolating Blaschke products not invertible in a Douglas algebra A. Suppose that for each n there is a Douglas algebra A_{n} with (A_{n})_{m} = A_{n} and

{λ ∈ M(A) : |q_{n}(λ)| = 1} ⊂ M(A_{n}).

Let B = ⋂_{n=1}^{∞} A[q̄_{n}]. Then B_{m} = B.

Proof. Since B ⊆ A[q̄_{n}] for n = 1, 2 …, by Proposition 6 of [11] we have B_{m} ⊆ (A[q̄_{n}])_{m}. Hence

B_{m} ⊆ ⋂_{n=1}^{∞}(A[\overline{q}_{n}])_{m}.

To show that B_{m} = B it suffices to show that (A[q̄_{n}])_{m} = A[q̄_{n}] for all n. Since {λ ∈ M(A) : |q_{n}(λ)| = 1} ⊂ M(A_{n}), we have A[q̄_{n}] = A_{n}[q̄_{n}]. So using Theorem 4 of

INVARIANCE OF THE MINIMAL ENVELOPE MAP OF DOUGLAS ALGEBRAS 7

[11] we have

(A[q̄_{n}])_{m} = (A_{n}[q̄_{n}])_{m} = (A_{n})_{m}[q̄_{n}] = A_{n}[q̄_{n}] = A[q̄_{n}].

Thus B_{m} ⊆ ⋂_{n=1}^{∞}(A[q̄_{n}])_{m} = ⋂_{n=1}^{∞} A[q̄_{n}] = B.

Corollary 4. If B = ⋂_{n=1}^{∞} H^{∞}[q̄_{n}], then B_{m} = B.

Theorem 6. Let B be a Douglas algebra. Then B ⊂ B_{m} if and only if there exists a point x ∈ M(B) whose level set E_{x} is an open subset in M(B).

Proof. Let B ⊂ B_{m}. By Theorem D of [11] there exists an interpolating Blaschke product φ such that {λ ∈ M(B) : |φ(λ)| < 1} = E_{x} for some x ∈ M(B). By Lemma 9(ii) of [11], E_{x} is an open subset of M(B).

Suppose that B = B_{m}. Then for every y ∈ M(B) and every interpolating Blaschke product q with q(y) = 0 the set Z(q) ∩ M(B) contains an infinite subset Γ such that E_{α} ≠ E_{β} for all distinct α, β ∈ Γ, and

⋃_{α∈Γ} E_{α} = {λ ∈ M(B) : |q(λ)| < 1}.

We’ll show that there is a subset \{x_{σ}\}_{σ∈A} of Γ such that y ∈ \overline{\{x_{σ}\}}_{σ∈A} \ \{y\}, that is,

E_{y} ⊆ \overline{\{E_{α}\}}_{α∈\{Γ\\{y\}\}}.

Suppose to the contrary that E_{y} ⊄ \overline{\{E_{α}\}}_{α∈\{Γ\\{y\}\}}. Then \overline{\{E_{α}\}}_{α∈\{Γ\\{y\}\}} and \{y\} are two closed disjoint subsets of M(H^{∞}). Choose neighborhoods V of \overline{\{E_{α}\}}_{α∈\{Γ\\{y\}\}} and U of {y} such that Ū ∩ V̄ = ∅. Let q_{1} be the the factor of q with zeros in U ∩ D. Then we see that Z(q_{1}) ∩ M(B) ⊂ E_{y} and {λ ∈ M(B) : |q_{1}(λ)| < 1} = E_{y}. Thus q̄_{1} ∈ B_{m} \ B, which is a contradiction. Hence

E_{y} ⊂ \overline{\{E_{α}\}}_{α∈\{Γ\\{y\}\}}.

Thus there is a subset {x_{σ}}_{σ∈A} ⊂ Γ such that x_{σ} → y, and x_{σ} ∉ E_{y} for all σ ∈ A. This implies that M(B) \ E_{y} is not a closed subset of M(H^{∞}). Thus E_{y} is not an open subset of M(B). This proves our theorem.

I have been unable to answer the following two questions, which I close with.

Question 1. Find an x ∈ M(H^{∞} + C) \ M(L^{∞}) such that H_{supp μ_{x}}^{∞} = (H_{supp μ_{x}}^{∞})_{m} (if such an x exists).

Question 2. Find a Douglas algebra B such that B ⊊ B_{b} ⊊ B_{m} (if such a Douglas algebra exists).

An immediate consequence of Theorem 6 above and Theorem 7 of [14], related to Question 2, is the following:

Theorem 7. Let B be a Douglas algebra. Then B = B_{b} ⊊ B_{m} if and only if for all x ∈ M(B) the set P_{x} is not open in M(B) but for some x ∈ M(B) the set E_{x} is an open subset in M(B).

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Department of Mathematics, University of Southwestern Louisiana, Lafayette, LA 70504

Current address: Mathematical Science Research Institute, 100 Centennial Drive, Berkeley, CA 94720