DOUGLAS ALGEBRAS WITH NO MAXIMAL SUBALGEBRA AND NO MINIMAL SUPERALGEBRA

CARROLL GUILLORY

Abstract. We give several examples of Douglas Algebras that do not have any maximal subalgebra. We find a condition on these algebras that guarantees that some do not have any minimal superalgebra. We also show that if A is the only maximal subalgebra of a Douglas algebra B, then the algebra A does not have any maximal subalgebra.

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 L^{∞} and H^{∞} is called a Douglas algebra. M(B) will denote the maximal ideal space of a Douglas algebra B. If we let X = M(L^{∞}), we can identify L^{∞} with C(X), the algebra of continuous functions on X. If C is the set of all continuous functions on T, we set

H^{∞} + C = {h + g : g ∈ C, h ∈ H^{∞}}

H^{∞} + C then 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}} is called the zero set of q in D. Here \frac{|z_{n}|}{z_{n}} = 1 is understood whenever z_{n} = 0. We call q an interpolating Blaschke product if

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

Research at MSRI is partially supported by NSF grant DMS-9022140.

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

lim_{n→∞} ∏_{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 b 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} ·

Q_{x} is called the QC-level set for x. For x ∈ M(H^{∞} + C), we denote by u_{x} the representing measure for x and its support set by supp u_{x}. By H^{∞}[q̄] we mean the Douglas algebra generated by H^{∞} and the complex conjugate of the function q. Since X is the Shilov boundary for every Douglas algebra, a closed set E contained in X is called a peak set for a Douglas algebra B if there is a function in B with f = 1 on E and |f| < 1 on X \ 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, B_{E} is similarly defined. A closed set E contained in X is called the essential set for B, denoted by ess(B), if E is the smallest set in X with the property that for f in L^{∞} with f = 0 on E, then f is in B.

For an interpolating Blaschke product q we put N(q̄) the closure of

⋃ {supp u_{x} : x ∈ M(H^{∞} + C) and |q(x)| < 1} ·

N(q̄) is a weak peak set for H^{∞} and is referred to as the nonanalytic points of q. By N_{0}(q̄) we denote the closure of

⋃ {supp u_{x} : x ∈ Z(q)} .

For an x ∈ M(H^{∞}) we let

E_{x} = {y ∈ M(H^{∞}) : supp u_{y} = supp u_{x}}

and call E_{x} the level set of x. Since the sets supp u_{x} and N(q̄) are weak peak sets for H^{∞}, both H_{supp u_{x}}^{∞} and H_{N(b̄)}^{∞} are Douglas algebras. For any interpolating Blaschke product q we set

A = ⋂_{x∈M(H^{∞}+C)\M(H^{∞}[q̄])} H_{supp u_{x}}^{∞}

and

A_{0} = ⋂_{y∈Z(q)} H_{supp u_{y}}^{∞}.

It is easy to see that A ⊆ A_{0} and it was shown in [11] that A = H_{N(q)}^{∞}. For x and y in M(H^{∞}), the pseudo-hyperbolic distance is defined by

ρ(x, y) = sup {|h(x)| : |h| ≤ 1, h ∈ H^{∞}, h(y) = 0} ·

For x and y in D we have

ρ(x, y) = |\frac{x − y}{1 − ȳx}| ·

For any x ∈ M(H^{∞}), we define the Gleason part 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. We denote by G the set of nontrivial points of M(H^{∞} + C), and for a Douglas algebra B, we set

G_{B} = G ∩ (M(H^{∞} + C) \ M(B)) ·

A point x in G_{B} is called a minimal support point of G_{B} (or simply a minimal support point of B) if there are no y ∈ G_{B} such that supp u_{y} ⊆ supp u_{x}. The set supp u_{x} is called a minimal support set for B. For Douglas algebras B and B_{0} with B_{0} ⊆ B we let Ω(B, B_{0}) be all interpolating Blaschke products q such that q̄ ∈ B but q̄ ∉ B_{0}.

We denote by Ω(B) the set of all interpolating Blaschke products q with q̄ ∈ B. Let B be a Douglas algebra. The Bourgain algebra B_{b} of B relative to L^{∞} is the set of those elements of L^{∞}, f, such that ∥ff_{n} + B∥_{∞} → 0 for every sequence {f_{n}} in B with f_{n} → 0 weakly. The minimal envelop B_{m} of a Douglas algebra B is defined to be the smallest Douglas algebra which contains all minimal superalgebra of B. An algebra A is called a minimal superalgebra of B if for all x, y ∈ M(B) \ M(A), x ≠ y implies supp u_{x} = supp u_{y}.

A major part of this paper was done while the author was at the Mathematical Sciences Research Institute. The author thanks the Institute for its support and also thanks Pamela Gorkin for the many valuable discussions.

2. A Condition for Douglas Algebras to Have Equal Essential Sets

Consider the Douglas algebras A and A_{0} defined above. In [11] some conditions were given when A ⊊ A_{0} but yet ess(A) = ess(A_{0}). This happened because ess(A) = N(q̄) and ess(A_{0}) = N_{0}(q̄) (this is not hard to show). Theorem 1 of [11] gives conditions when ess(A) ≠ ess(A_{0}). The conditions found in Theorem 5 of [11] are far more complicated then those found in Theorem 1 below. Yet ess(A) = ess(A_{0}) in that Theorem (Theorem 5 of [11]) and also satisfies the condition in Theorem 1.

Theorem 1. Let B_{0} be a subalgebra of a Douglas algebra B with

ess(B_{0}) ≠ X.

If for every x ∈ M(B_{0}) \ M(B) we have ess(H_{supp u_{x}}^{∞}) = ess(B_{supp u_{x}}), then ess(B) = ess(B_{0}).

Proof: We note that ess(H_{supp u_{x}}^{∞}) = supp u_{x}. Hence if ess(H_{supp u_{x}}^{∞}) is contained in ess(B) for every x ∈ M(B_{0}) \ M(B), then supp u_{y} ⊂ ess(B) for every y ∈ M(B_{0}) and so we get ess(B_{0}) ⊂ ess(B). Since B_{0} ⊂ B we have that ess(B) ⊂ ess(B_{0}), and we get ess(B) = ess(B_{0}).

Corollary 1. Let B_{0} be a maximal subalgebra of a Douglas algebra B. Then ess(B_{0}) = ess(B).

Since M(B_{0}) = M(B) ∪ E_{x} for some x ∈ M(B_{0}) \ M(B), we have that if z and y are in M(B_{0}) \ M(B), then supp u_{y} = supp u_{x} = supp u_{z}. Now ess(B_{supp u_{x}}) = ess(H_{supp u_{x}}^{∞}) since the set

⋃ {supp u_{y} : y ∈ M(B) ∩ M(H_{supp u_{x}}^{∞})}

is dense in supp u_{x} (because x is a minimal support point of B). Thus ess(B) = ess(B_{0}).

Corollary 2. Let A be any Douglas algebra and q be an interpolating Blaschke product with q̄ ∉ A. Then ess(A) = ess(A[q̄]).

Proof: Let B = A[q̄] and let x ∈ M(A) \ M(B). We need the following two facts:

i) M(A) = {x ∈ M(H^{∞} + C) : A_{supp u_{x}} = H_{supp u_{x}}^{∞}} and

ii) If A ⊂ B, then A_{supp u_{x}} ⊂ B_{supp u_{x}}.

We use i) and ii) to show that B_{supp u_{x}} = A_{supp u_{x}}[b̄]. Since A ⊂ B by ii) M(B_{supp u_{x}}) ⊂ M(A_{supp u_{x}}). By the Chang–Marshall Theorem [1,16] it suffices to show that M(B_{supp u_{x}}) = M(A_{supp u_{x}}[q̄]). Let y ∈ M(B_{supp u_{x}}). Since M(B_{supp u_{x}}) ⊂ M(A_{supp u_{x}}) we have that y ∈ M(A_{supp u_{x}}) and |q(y)| = 1 (since M(B) = {y ∈ M(A) : |q(y)| = 1}). Hence y ∈ M(A_{supp u_{x}}[q̄]). Therefore M(B_{supp u_{x}}) ⊂ M(A_{supp u_{x}}[q̄]).

Now suppose y ∉ M(B_{supp u_{x}}). If y ∉ M(A_{supp u_{x}}), then y ∉ A_{supp u_{x}}[b̄] and we have nothing to prove. So we can assume that y ∈ M(A_{supp u_{x}}). Now since y ∉ M(B_{supp u_{x}}) implies that |q(y)| < 1 and y ∈ M(A_{supp u_{x}}). Hence y ∉ M(A_{supp u_{x}}[q̄]). Thus M(A_{supp u_{x}}[q̄]) ⊂ M(B_{supp u_{x}}). By the Chang–Marshall Theorem we have that M(A_{supp u_{x}}[q̄]) = M(B_{supp u_{x}}). Hence A_{supp u_{x}}[q̄] = B_{supp u_{x}}. Thus we have

ess(B_{supp u_{x}}) = ess (A_{supp u_{x}}[q̄])
= ess (H_{supp u_{x}}^{∞}[q̄]) by i)

So it suffices to show that

ess (H_{supp u_{x}}^{∞}[q̄]) = ess (H_{supp u_{x}}^{∞}) .

To do this set B_{1} = H_{supp u_{x}}^{∞} and

E = ⋃ {supp u_{y} : supp u_{y} ⊂ supp u_{x}, |q(y)| = 1} .

Assume that Ē, the closure of E, is properly contained in supp u_{x}. Put

B_{2} = H̄_{E}^{∞} = \overline{\{f ∈ L^{∞} : f|_{E} ∈ H^{∞}|_{E}\}}.

By [2, page 39] M(B_{2}) = {m ∈ M(H^{∞}) : supp u_{m} ⊆ Ē}∪M(L^{∞}). Since Ē ⊊ supp u_{x} we have that B_{1} ⊊ B_{2}. Therefore M(B_{2}) ⊊ M(B_{1}) and so there is a nontrivial point x_{0} ∈ M(B_{1}) \ M(B_{2}) [5, Proposition 4.1] such that (a) supp u_{x_{0}} ⊊ supp u_{x}, (b) supp u_{x_{0}} ⊄ Ē [otherwise x_{0} ∈ M(B_{2})], (c) |q(x_{0})| < 1 and (d) supp u_{x_{0}} ∩ E = ∅. In (a) supp u_{x_{0}} ⊊ supp u_{x} for if supp u_{x_{0}} = supp u_{x} for all x_{0} ∈ M(B_{1}) \ M(B_{2}), then B_{1} is a maximal subalgebra of B_{2}, so by Corollary 1 we have Ē = supp u_{x} = ess(B_{1}) (in fact B_{2} has no maximal subalgebras). The fact that |q(x_{0})| < 1 implies that q̄ ∉ H_{supp u_{x_{0}}}^{∞}.

Therefore there is a y_{0} ∈ Z(q) such that supp u_{y_{0}} ⊆ supp u_{x_{0}}. By Theorem 2 of [10] there is a z_{0} ∈ Z(q) such that supp u_{z_{0}} is a minimal support set for H^{∞}[q̄] that is contained in supp u_{x_{0}} ⊆ supp u_{x}. Since q is an interpolating Blaschke product supp u_{z_{0}} is not trivial. By [5, Theorem 4.2] there is m ∈ M(H^{∞}+C) so that supp u_{m} is nontrivial and supp u_{m} ⊊ supp u_{z_{0}} ⊆ supp u_{x_{0}}. Since supp u_{z_{0}} is a minimal support set for H^{∞}[q̄] we have that |q(m)| = 1. Thus supp u_{x_{0}} ∩ E ≠ ∅. This contradicts (d). So Ē = supp u_{x} and ess(H_{supp u_{x}}^{∞}) = ess(H_{supp u_{x}[q̄]}^{∞}) and we get ess(B_{supp u_{x}}) = ess(H_{supp u_{x}}^{∞}) for every x ∈ M(A) \ M(B), so Corollary 2 follows from our Theorem.

We mention here that Corollary 2 was proved in [20, Theorem 2] by another method.

There are algebras B_{0} and B that satisfy the hypothesis of Theorem 1 and are not of the form B = B_{0}[q̄] for any interpolating Blaschke product (if B_{0} ⊆ B). To see this let Γ be the collection of sparse Blaschke products and B be the Douglas algebra [H^{∞} : q̄; q ∈ Γ]. Let q_{0} be any element in Γ and put B_{0} = H^{∞}[q̄]. Then B_{0} ⊂ B. By a Theorem of Hedenmaln’s [13] we have that if b is a Blaschke product such that b̄ ∈ B, then b = b_{1} ⋯ b_{n} where each b_{i}, i = 1, …, n, is a sparse Blaschke product. Hence if x ∈ M(B_{0}) \ M(B), then x is the zero of some sparse Blaschke product. So x is a minimal support point of B for every x ∈ M(B_{0}) \ M(B). This implies that ess(B_{supp u_{x}}) = ess(H_{supp u_{x}}^{∞}). So, by Theorem 1, we have ess(B) = ess(B_{0}) (Theorem 2 below shows that H_{supp u_{x}}^{∞} is a maximal subalgebra of B_{supp u_{x}}.) Now suppose there is a Blaschke product q ∈ Ω(B, B_{0}) with B = B_{0}[q̄]. Again by Hedenmaln’s Theorem we have q = q_{1} ⋯ q_{n} with each q_{i} a sparse Blaschke product. Let Q be any infinite sparse Blaschke product such that |Q| = 1 on ⋃_{x∈Z(q)} P_{x}. Then there is a m ∈ M(H^{∞} + C) such that Q(m) = 0 but m ∉ ⋃_{x∈Z(q)} P_{x}. Thus |q(m)| = 1 and so we get that m ∈ M(B_{0}[q̄]). Thus Q̄ ∉ B_{0}[q̄] and yet Q̄ ∈ B. This implies that B_{0}[q̄] ⊊ B, which is a contradiction.

3. Maximal Subalgebras That Have No Maximal Subalgebra

We begin by extending Proposition 1 of [9]. There the authors showed that if x ∈ Z(q) with q a sparse Blaschke product, then the algebra H_{supp u_{x}}^{∞} is a maximal subalgebra of H_{supp u_{x}}^{∞}[q̄]. Below we show that this is true for a larger class.

Theorem 2. Let A be any Douglas algebra with maximal subalgebra and x be a minimal support point of G_{A}. Then H_{supp u_{x}}^{∞} is a maximal subalgebra of A_{supp u_{x}} and A_{supp u_{x}} = H_{supp u_{x}}^{∞}[q̄] for some q ∈ Ω(A).

Proof: Let B_{0} = H_{supp u_{x}}^{∞} and B = A_{supp u_{x}}. Suppose x is a minimal support set for G_{A}. Then we have for any interpolating Blaschke product ψ ∈ Ω(A) with |ψ(x)| < 1 and any y ∈ M(H^{∞} + C) with supp u_{y} ⊊ supp u_{x}, |ψ(y)| = 1. Thus if ψ_{0} ∈ Ω(B) and |ψ_{0}(x)| < 1 there is a ψ ∈ Ω(A) such that ψ|_{supp u_{x}} = ψ_{0}|_{supp u_{x}}. This implies that |ψ_{0}(y)| = 1 for every such y. Hence x is a minimal support point for G_{B}. Note that this implies that M(B) = M(B_{0}) \ E_{x} where E_{x} is the level set for x. Hence M(B_{0}) = M(B) ∪ E_{x}, so by Theorem 1 of [10], B_{0} is a maximal subalgebra of B. Let q be any element in Ω(A) with q(x) = 0. Then q ∈ Ω(B, B_{0}) and we have that B = B_{0}[q̄].

Theorem 3. A Douglas algebra A has no maximal subalgebra if and only if H_{supp u_{x}}^{∞} is not a maximal subalgebra of A_{supp u_{x}} for every x ∈ G_{A}.

Proof: Suppose A has no maximal subalgebra and let x ∈ G_{A}. Since x is not minimal support set of G_{A} there is a y ∈ G_{A} with supp u_{y} ⊊ supp u_{x}, and a ψ ∈ Ω(A) such that |ψ(y)| < 1. Since ψ̄ ∉ H_{supp u_{y}}^{∞}, we can assume that ψ(y) = 0. Hence y ∉ M(A_{supp u_{x}}). By Lemma 4 of [8] there is a ψ_{0} ∈ Ω(A_{supp u_{x}}) such that |ψ_{0}(y)| = 1 and ψ_{0}(x) = 0. Then we have H_{supp u_{x}}^{∞} ⊊ H_{supp u_{x}}^{∞}[ψ̄_{0}] ⊊ A_{supp u_{x}}. So H_{supp u_{x}}^{∞} is not a maximal subalgebra of A_{supp u_{x}}.

Suppose that for all x ∈ G_{A}, H_{supp u_{x}}^{∞} is not a maximal subalgebra of A_{supp u_{x}}. Then there is an algebra B with H_{supp u_{x}}^{∞} ⊊ B ⊆ A_{supp u_{x}}. Thus we can find a y ∈ M(H^{∞} + C) such that supp u_{y} ⊊ supp u_{x} and y ∈ M(B) \ M(A_{supp u_{x}}). This implies that there is an interpolating Blaschke product q with q̄ ∈ B ⊂ A_{supp u_{x}} such that |q(y)| = 1 and |q(x)| < 1. Hence there is a q_{0} ∈ Ω(A) with q_{0}|_{supp u_{x}} = q|_{supp u_{x}}. So |q_{0}(y)| = 1 and |q_{0}(x)| < 1. This implies that x is not a minimal support point of G_{A} for every x ∈ G_{A}. So by Theorem 1 of [10] A has no maximal subalgebra.

Proposition 1. Let x ∈ M(H^{∞})\M(L^{∞}). Then H_{supp u_{x}}^{∞} has no maximal subalgebra.

Proof: Now ess(H_{supp u_{x}}^{∞}) = supp u_{x}. Hence if y ∈ G_{H_{supp ux}^{∞}}, then supp u_{y}∩supp u_{x} = ∅. Hence if A is a subalgebra of H_{supp u_{x}}^{∞}, then there is a y ∈ M(A)\M(H_{supp u_{x}}^{∞}) with supp u_{y} ∩ supp u_{x} = ∅. Hence ess(A) ⊇ supp u_{y} ∪ supp u_{x} ⊋ supp u_{x} = ess(H_{supp u_{x}}^{∞}). By Corollary 1, A is not a maximal subalgebra of H_{supp u_{x}}^{∞}. Proposition 2. Let A be a Douglas algebra that has only one maximal subalgebra A_{0}. Then A_{0} has no maximal subalgebra.

Proof: Suppose there is a subalgebra B_{0} ⊆ A_{0} such that B_{0} is a maximal subalgebra of A_{0}. Then by Theorem 1 of [10] there is an x_{0} ∈ G_{A_{0}} such that

M(B_{0}) = M(A_{0}) ∪ E_{x_{0}}  (1)

Since A_{0} is a maximal subalgebra of A there is an x ∈ G_{A} ∩ M(B_{0}) such that

M(A_{0}) = M(A) ∪ E_{x}  (2)

By (1) and (2) we have that M(B_{0}) = M(A) ∪ E_{x} ∪ E_{x_{0}}. Since x ∈ M(A_{0}) we have that supp u_{x} ≠ supp u_{x_{0}}. Also since x_{0} ∉ E_{x} and x_{0} ∈ G_{A_{0}} implies that supp u_{x_{0}} ⊄ supp u_{x} (otherwise x_{0} ∈ M(A) by (2). We show that x_{0} is a minimal support point of G_{A}, and hence get a contradiction. Let y ∈ M(H^{∞} + C) such that supp u_{y} ⊊ supp u_{x_{0}}. Since x_{0} is a minimal support point of G_{A_{0}} we have that y ∈ M(A_{0}) = M(A) ∪ E_{x}. If y ∈ M(A) then we are done. So we can assume y ∈ E_{x}. If y ∈ E_{x}, then supp u_{y} = supp u_{x}, so we have that supp u_{x} ⊊ supp u_{x_{0}}. Since x ∉ M(A) there is an interpolating Blaschke product q with q̄ ∈ A and such that q(x) = 0. By Theorem 2 of [11] there is an uncountable set U of Z(q) such that (a) supp u_{m} ⊊ supp u_{x_{0}} for all m ∈ U and (b) supp u_{m} ∩ supp u_{k} for all m, k ∈ U, m ≠ k. By (1) each such m ∈ U is in M(A_{0}). Since for all m ∈ U (except if m = x) we have supp u_{x} ∩ supp u_{m} = ∅, hence by (2), m ∈ M(A). But q̄ ∈ A and U ⊂ Z(q) ∩ M(A). This is a contradiction, and we get y ∉ E_{x}. So y ∈ M(A) and since supp u_{x_{0}} ≠ supp u_{x} we have that x_{0} is a minimal support point of G_{A}. This is a contradiction. So A_{0} has no maximal subalgebra.

Note that Proposition 1 follows from Proposition 2 if x is a minimal support point for some interpolating Blaschke product.

Let q be an interpolating Blaschke product. We consider the algebra H_{N(q̄)}^{∞}. Certainly H_{N(q̄)}^{∞} is not known to be a maximal subalgebra of any Douglas algebra, but does have some of the same properties of H_{supp u_{x}}^{∞}. For example Proposition 3. The algebra H_{N(b̄)}^{∞} has no maximal subalgebra.

Proof: Set B = H_{N(q̄)}^{∞}. Let x ∈ G_{B} and suppose x is a minimal support point for G_{B}. Then if y ∈ M(H^{∞} + C) such that supp u_{y} ⊊ supp u_{x} we have that y ∈ M(B). By [2, page 39] we must have that supp u_{y} ⊆ N(q̄) = ess(B). Thus we have that ess(B) ∩ supp u_{x} = N(q̄) ∩ supp u_{x} ≠ ∅. By [14, Theorem 1] N(q̄) = ⋃_{x∈Z(q)} Q_{x}. So there is an x_{0} ∈ Z(q) such that supp u_{x} ∩ Q_{x_{0}} ≠ ∅. By the definition of Q_{x_{0}}, we have that supp u_{x} ⊂ Q_{x_{0}}. By [2, page 39] this implies that x ∈ M(B), which is a contradiction. So if x ∈ G_{B}, then supp u_{x} ∩ N(q̄) = ∅, which implies that x is not a minimal support point for G_{B}. B has no maximal subalgebra.

4. Minimal Superalgebras of H_{supp u_{x}}^{∞}

We will compute the Bourgain algebras and the Minimal Envelopes of the Douglas algebra H_{supp u_{x}}^{∞} for any x ∈ M(H^{∞} + C). We have the following

Theorem 4. Let x ∈ M(H^{∞}+C)\M(L^{∞} such that |q(x)| < 1 for some interpolating Blaschke product q and set B = H_{S_{supp ux}}^{∞}. Then

i) Either B_{b} = B or B_{b} = B[ψ̄] for some interpolating Blaschke product ψ. ii) Either B_{m} = B_{b} = B or B_{m} = B[ψ̄] for some interpolating Blaschke product ψ.

Proof: We will use Theorem 2 of [4] which says that for any interpolating Blaschke product ψ with ψ̄ ∈ B_{b}, we have that the set Z(ψ) ∩ M(B) is a finite set and the fact that

M(B) = M (L^{∞}) ∪ {m ∈ M(H^{∞}) : supp u_{m} ⊆ supp u_{x}}  (3)

We claim that if ψ is an interpolating Blaschke product such that ψ̄ ∈ B_{b}, then Z(ψ) ∩ M(B) ⊂ E_{x}, the level set of x. Suppose not. Then there is an x_{0} ∈ Z(ψ) ∩ M(B) such that supp u_{x_{0}} ⊊ supp u_{x}. By Theorem 2 of [11] there is an uncountable set Γ of Z(ψ) such that (a) supp u_{γ} ⊊ supp u_{x} for all γ ∈ Γ and (b) supp u_{x_{m}}∩supp u_{γ} = ∅ for all m, γ ∈ Γ, m ≠ γ. By (a) and (3) each γ ∈ M(B) and so Γ ⊂ Z(ψ) ∩ M(B). This implies that the set Z(ψ) ∩ M(B) is infinite. This is a contradiction. Hence if x_{0} ∈ Z(ψ) ∩ M(B) then supp u_{x_{0}} = supp u_{x}, so we get Z(ψ) ∩ M(B) ⊂ E_{x}. There are two possibilities (1) the set Z(ψ) ∩ M(B) = ∅ for which ψ̄ ∈ B, so B_{b} ⊆ B. This gives us the case when B_{b} = B. (2) If Z(ψ)∩M(B) ≠ ∅ but finite. Then the algebra B[ψ̄] ⊆ B_{b} To show that B_{b} = B[ψ̄], let ψ_{0} be other interpolating Blaschke product with ψ̄_{0} ∈ B_{b}. Since both sets Z(ψ_{0})∩M(B) and Z(ψ)∩M(B) are contained in E_{x}, we have that M(B) \ M(B[ψ̄_{0}]) = M(B) \ M(B[ψ̄]) = E_{x}. Thus M(B[ψ̄_{0}]) = M(B[ψ̄]) and by the Chang–Marshall Theorem [1, 16] we have B[ψ̄_{0}] = B[ψ̄], since this is true for all ψ, ψ_{0} we have by Theorem C of [12], B_{b} = B[ψ̄] for any such ψ B[ψ̄] is a minimal superalgebra of B.) This proves (i).

To prove (ii) let ψ̄ ∈ B_{m}. Then, by Theorem 3 of [12] there is a finite set {x_{1}, …, x_{n}} ⊂ Z(ψ) ∩ M(B) such that {u ∈ M(B) : |ψ(u)| < 1} = E_{x_{1}} ∪ ⋯ ∪ E_{x_{n}}. Again we claim that E_{x_{1}} = E_{x_{2}} = ⋯ = E_{x_{n}} = E_{x}. Suppose that E_{x_{1}} ≠ E_{x_{2}}. Then supp u_{x_{1}} ≠ supp u_{x_{2}}. By (3) either supp u_{x_{1}} ⊊ supp u_{x} or supp u_{x_{1}} ⊊ supp u_{x} or both. Let us suppose that supp u_{x_{1}} ⊊ supp u_{x}. Then by Theorem 2 of [11] there is an uncountable set Γ such that E_{α} ≠ E_{β} for all α, β ∈ Γ and ⋃_{α∈Γ} E_{α} ⊂ {u ∈ M(B) : |ψ(u)| < 1}. This contradicts Theorem 3 of [12]. Thus E_{x_{1}} = E_{x_{2}} = ⋯ = E_{x_{n}} = E_{x}. As before we have that for ψ̄ ∈ B_{m}, Z(ψ) ∩ M(B) ⊂ E_{x} and B_{m} = B[ψ̄] if Z(ψ) ∩ M(B) ≠ ∅. This proves ii).

Corollary 3. i) Let x ∈ M(H^{∞} + C) \ M(L^{∞}) and B = H_{S_{ux}}^{∞}. Then B ⊂ B_{m} if and only if x is a minimal support point of H^{∞}[ψ̄] for some interpolating Blaschke product ψ. ii) B = B_{b} = B_{m} if and only if x is not a minimal support point of H^{∞}[ψ̄] for any

interpolating Blaschke product.

Theorem 4 i) has also appeared in [17].

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University of Southwestern Louisiana, Lafayette, Louisiana 70504 Current address: Mathematical Sciences Research Institute, 1000 Centennial Drive, Berkeley,

CA 94720 E-mail address: cjg@usl.edu