Some conditions on Douglas algebras that imply the invariance of the minimal envelope map

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.


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 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 z m − z n 1 −z 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.
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) ∩ (H ∞ + C) and for x ∈ M(H ∞ + C), set 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 ∞ [q] 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 E is a Douglas algebra. For a Douglas algebra B, we define B E similarly. For an interpolating Blaschke product q we put N(q) the closure of Then N(q) is a weak peak set for H ∞ . By N 0 (q) we denote the closure of {supp µ x : x ∈ Z(q)}. In general N 0 (q) does not equal N(q), but in this paper N 0 (q) = N(q). 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(q) are weak peak sets for H ∞ , both H ∞ supp µx and H ∞ N (q) are Douglas algebras. For the interpolating Blaschke product q, Our assumptions on q throughout this paper imply that H ∞ N (q) = A = A 0 (see [10]). For x and y in M(H ∞ ), the pseudohyperbolic distance ρ is defined by 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 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.

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
Proof. Since A ∩ B is contained in both A and B by Proposition 6 of [11], To show that A m ∩B m ⊂ (A∩B) m , let ψ be an interpolating Blaschke product such thatψ ∈ A m ∩B m . We show thatψ ∈ (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 The following corollaries are immediate consequences of the theorem.    [11]. By theorem 1 of [7] (and its proof) we have (

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 (q) ) m = H ∞ N (q) , and so (B e ) m = B e . This proves (ii). 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 and since Z(I 1 ) is equivalent to theČech 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 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 By Theorem 3 of [11],Ī ∈ (B E ) m . This shows that (B E ) m = B E . Thus 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 We'll show that there is a subset {x σ } σ∈A of Γ such that y ∈ {x σ } σ∈A \ {y}, that is, Suppose to the contrary that E y ⊂ {E α } α∈{Γ\{y}} . Then {E α } α∈{Γ\{y}} and {y} are two closed disjoint subsets of M(H ∞ ). Choose neighborhoods V of {E α } α∈{Γ\{y}} and U of {y} such thatŪ ∩V = ∅. 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 q 1 ∈ B m \ B, which is a contradiction. Hence 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 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: