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

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

CARROLL GUILLORY (Received 20 November 1999)

Abstract.We give general 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.

2000 Mathematics Subject Classification. 46Jxx, 30Dxx.

1. Introduction. LetDdenote the open unit disk in the complex plane, andT the unit circle. ByL^∞we mean the space of essential bounded measurable functions onT with respect to the normalized Lebesgue measurement. We denote byH^∞the space of all bounded analytic functions inD. Via identification with boundary functions, H^∞can be considered as a uniformly closed subalgebra ofL^∞. Any uniformly closed subalgebra B strictly betweenH^∞and L^∞ is called a Douglas algebraB. If C is the set of all continuous functions onT, we setH^∞+C= {h+g:h∈H^∞, g∈C}. Then H^∞+Cbecomes the smallest Douglas algebra containingH^∞properly.

The function

q(z)= ∞ n=1

zn zn

z−zn

1−z¯nz (1.1)

is called a Blaschke product if_∞

n=1(1− |zn|)converges. The set {zn}is called the zero set ofq in D. Here|zn|/zn=1 is understood wheneverzn=0. We callq an interpolating Blaschke product if

infn

m:m≠n

zm−zn

1−z¯nzm

>0. (1.2)

An interpolating Blaschke productqis called sparse (or thin) if

n→∞lim

m:m≠n

zm−zn

1−z¯nzm

=1. (1.3)

The setZ(q)= {x∈M(H^∞)\D:q(x)=0}is called the zero set ofqinM(H^∞+C).

Any functionhinH^∞with|h| =1 a.e. onT is called an inner function. Since|q| =1 for any Blaschke product, Blaschke products are inner functions. LetQC=(H^∞+C)∩ H^∞∩Cand forx∈M(H^∞+C), set

Qx= y∈M

L^∞

:f (x)=f (y)∀f∈QC

. (1.4)

ThenQx is called theQC-level set forx. Forx∈M(H^∞+C), we denote byµxthe representing measure forx, and its support set by suppµx. ByH^∞[¯q]we mean the Douglas algebra generated byH^∞and the complex conjugate of the functionq. Since M(L^∞)is the Shilov boundary for every Douglas algebra, a closed setEcontained in M(L^∞)is called a peak set for a Douglas algebraBif there is a functionf inBwith f =1 onE and |f|<1 on M(L^∞)\E. A closed set E is a weak peak set forB if E is the intersection of a family of peak sets. If the setE is a weak peak set forH^∞ and we define

H_E^∞=

f∈L^∞:f|E∈H^∞|E

, (1.5)

thenH_E^∞is a Douglas algebra. For a Douglas algebraB, we defineBEsimilarly.

For an interpolating Blaschke productqwe putN(¯q)the closure of

∪

suppµx:x∈M H^∞+C

, q(x)<1

. (1.6)

ThenN(¯q)is a weak peak set forH^∞. ByN0(¯q)we denote the closure of∪{suppµx: x∈Z(q)}. In generalN0(¯q)does not equal N(¯q), but in this paperN0(¯q)=N(¯q).

Forx∈M(H^∞), we letEx= {y∈M(H^∞): suppµy=suppµx}and callEx the level set ofx. Since the sets suppµxandN(¯q)are weak peak sets forH^∞, bothH_suppµ^∞ _xand H_N(¯^∞_q)are Douglas algebras. For the interpolating Blaschke productq, set

A=

x∈M(H^∞+C)

H_suppµ^∞ _x:q(x) <1

, A0=

H_suppµ^∞ _x:x∈Z(q)

. (1.7)

Our assumptions onqthroughout imply thatH_N(¯^∞_q)=A=A0(see [8]). Forxandyin M(H^∞), the pseudohyperbolic distanceρis defined by

ρ(x, y)=supp

|h(y)|:h∞≤1, h∈H^∞, h(x)=0

. (1.8)

ForxandyinD, we have

ρ(x, y)= x−y

1−yx¯

. (1.9)

Forx∈M(H^∞), we define the Gleason partPxofxby Px=

y∈M H^∞

:ρ(x, y) <1

. (1.10)

IfPx≠{x}, thenx is said to be a nontrivial point. LetG= {x∈M(H^∞+C):Px is a nontrivial part}. We say that an interpolating Blaschke productq is of typeG if the modules is 1 on every trivial part, that is,{x∈M(H^∞+C):|q(x)|<1} ⊂G. An interpolating Blaschke productqof typeGis said to be of finite typeGif for every x∈Z(q), the setZ(q)∩Pxis a finite set.

A pointx∈Z(q), whereq is interpolating Blaschke product, is called a minimal element (or a minimal support point) for the Douglas algebraH^∞[¯q]if there are noy∉ M(H^∞[¯q])such that suppµy⊆suppµx; that is, ify∉M(H^∞[¯q])then either suppµy∩ suppµx=θ or suppµx⊆suppµy. We let F = {x∈G:x∈Z(q), qis finite type}.

A pointx∈M(H^∞)is called locally thin if there is an interpolating Blaschke product qsuch thatq(x)=0 and

1−Zn(α)²q

Zn(α)→1 (1.11)

wheneverZn(α)is a subnet of the zero sequence{Zn}ofq inDconverging tox. A point x ∈M(H^∞)is called locally thin if there exists a Blaschke product q which is locally thin at x. Since for locally thin points Px is nontrivial, x ∈ G. Let L= {x∈G:xis a locally thin point}, we will show thatF⊂L. Finally, we letmq= {x∈ M(H^∞+C)|xis a minimal support point ofH^∞[¯q]}.

LetBa Douglas algebra. The Bourgain algebraBb ofBrelative toL^∞is the set of f inL^∞such thatf fn+B∞→0 for every sequence{fn}inBwithfn→0 weakly in B. An algebraB is called a minimal superalgebra of an algebra A ifA⊂B and suppµy=suppµxfor allx, y∈M(A)\M(B). The minimal envelopeBmof a Douglas algebraBis defined to be the smallest Douglas algebra that contains all the minimal superalgebra of ˙B. The mapping that assigns toBthe Douglas algebraBmis called the minimal envelope map and the mapping that assigns toBthe Douglas algebraBbis called the Bourgain map.

LetEthin=[H^∞: ¯q|qis a thin interpolating Blaschke product],Efin=[H^∞: ¯b|bis of finite typeG], andEG=[H^∞: ¯ϕ|ϕis of typeG]. Then it is clear thatEthin⊆Efin⊆EG. 2. Some Douglas algebraBwith the propertyBm=B. We begin with the following theorems.

Theorem2.1. Letqbe an interpolating Blaschke product of typeG. LetB=H_N(¯^∞_q). ThenBm=B.

Proof. Letb be an interpolating Blaschke product with ¯b∉B. Then there is an m∈M(B)such thatb(m)=0. SinceB=

x∈Z(q)H^∞_suppµxthen there is ay∈Z(q)such that ¯b∉H_suppµ^∞ _y. We can assume that suppµm⊆suppµy. This implies|b(y)|<1. Let {Zn}be the zero sequence ofqinD. SinceZ(q)= {Z¯n}\{Zn}andy∈Z(q), there is a subsequence{Znβ}such thatZnβ→yasβ→ ∞. Since|b(Znβ)| → |b(y)|asβ→ ∞, there is a positive integerNand anwith 0< <1 such that|b(Znβ)|<1−for all nβ≥N. Letϕbe the factor ofqsuch thatZ(ϕ)= {Znβ}^∞_nβ≥N\{Znβ}^∞_nβ≥N. Then|b|<1 onZ(ϕ). By [12, Theorem 1], there are uncountable infinitely manyx, y∈Z(ϕ)such that suppµx∩suppµy=θ. This implies that there are infinitely manyx, y∈Z(ϕ) such thatEx∩Ey=θ and

x∈Z(ϕ)Ex⊂ {m∈M(B):|b(m)|<1}. By [9, Theorem 3]

we have thatb∉Bm. This proves our theorem.

Theorem2.2. [Efin]m=Efin.

Proof. Letqbe any interpolating Blaschke product such that ¯q∉Efin. Thenqis not of finite typeG(nor the product of a finite number of finite typeG). Hence by [10, Theorem 3.1], there arexandyinM(H^∞+C)such that|q(x)|<1,|q(y)|<1 and suppµx⊆suppµy.

Case 1. Suppose y ∈M(Efin). Then by [8, Theorem 2] there is an uncountable infinite index setI such that (a) for eachα∈I, there is an xα∈M(H^∞+C) with suppµxα⊆suppµy; (b)xα∈Z(q); and (c) forα, β∈I,α≠β, suppµxα∩suppµx_β=θ.

Sincey ∈M(Efin) we have thatxα ∈M(Efin) for all α∈I. Thus

α∈IExα ⊂ {m∈ M(Efin)|q(m) <1}. By [9, Theorem 3],q∉[Efin]m.

Case2. Ify∉M(Efin). Then again for eachα∈I, we have thatxα∈M(Efin). This follows from [11, Theorem 9] which states that every invertible interpolating Blaschke

product inEfinis the product of a finite number of interpolating Blaschke product of finite typeG. So ifbis invertible inEfin, and|b(y)|<1, then|b(xα)| =1 for allα∉I by [9, Theorem 3.1]. Thusxα∈M(Efin)for allα∈I, and so

α∈IExα⊂ {m∈M(Efin)|

|q(y)|<1}. This implies that ¯q∉[Efin]m. ThereforeEfin=[Efin]m. Theorem2.3. Ethin=[Ethin]m.

Proof. Letϕ be an interpolating Blaschke product such that ¯ϕ∉[Ethin]. Since Ethin⊆Efin, we have that[Ethin]m⊆[Efin]m=Efin. We can assume that ¯ϕ∈Efin. We will show that there is an infinite setF⊆Z(ϕ)such that for anyx, m∈F, suppµx∩ suppµm=θ and

x∈F ⊂ {w∈M(Ethin)| |ϕ(w)|<1}. By [9, Theorem 3.1] this will show that ¯ϕ∉[Ethin]m. Hence will show that[Ethin]m=Ethin. Since ¯ϕ∉Ethinthere is anx0∈M(Ethin)such thatϕ(x0)=0. Let{Zn}be the zero sequence ofϕin ˙D. Since ϕis interpolating andZ(ϕ)= {Z¯n}\{Zn}, there exists a subsequence{Znβ}such that Znβ→x0.We will show that the set{Z¯nB}\{Znh} ⊂M(Ethin). Letqbe any interpolating Blachke product such that ¯q∈Ethin(thenqis the product of a finite number of thin interpolating Blaschke product). Sincex0∈M(Ethin), we have that|q(x0)| =1. Thus

|q(Znβ)| →1 asβ→ ∞. This shows that|q| =1 on{Z¯nβ}\{Znβ}. Since this is true for anyqwith ¯q∈Ethin, we have

Z¯nβ

\ Znβ

⊂M Ethin

. (2.1)

Since{Z¯nβ}\{Znβ}is homeomorphic to the compactification of the integer, we have that {Z¯nβ}\{Znβ}is uncountably infinite and by [12, Theorem 1], there are infinite manyx, y∈ {Z¯nβ}\{Znβ}such that suppµ∩suppµy =θ. Thus ifF = {Z¯nβ}\{Znβ},

then we have

m∈F

Em⊂ w∈M

Ethin

|ϕ(w) <1

. (2.2)

Hence ¯ϕ∉[Ethin]m. This proves our theorem.

We are unable to give solutions to the following two problems.

Problem2.4. Letqbe an interpolating Blaschke product that is not of typeG. Is it true that(H_N(¯^∞_q))m=H_N(¯^∞_q)?

Let B be any Douglas algebra,B≠H^∞+C, and letW be a weak peak set forB.

Mortini and Younis have shown that the restricted algebraH_W^∞has the property that [H_W^∞]m=[H^∞_W]b=H_W^∞, if W is a weak peak set forH^∞.

Problem 2.5. What conditions must be imposed on the Douglas algebraB such thatBW =[BW]m? For example, ifB=H^∞[¯q], where qis an interpolating Blaschke product of typeG, andWis a weak peak set forB, then isBW=[BW]m?

Problem2.6. Is[EG]m=EG?

We also have the following theorem. The case for the Bourgain algebraBbhas been proven in [15]. The proof used here is quite different from theirs, and can be used to show that his result also holds for the Bourgain algebras.

Theorem2.7. LetAandBbe Douglas algebras. Then(A∩B)m=Am∩Bm.

Proof. SinceA∩Bis contained in bothAandBby [9, Proposition 6],(A∩B)mis contained in bothAmandBm. Hence(A∩B)m⊂Am∩Bm.

To show thatAm∩Bm⊂(A∩B)m, letϕbe an interpolating Blaschke product such that ¯ϕ∈Am∩Bm. We show that ¯ϕ∈(A∩B)m. By [9, Theorem D] we can assume that there is anx∈M(A)and a y∈M(B)such that{λ∈M(A):|ψ(γ)|<1} =Ex and {µ∈M(B):|ψ(µ)|<1} =Ey. Thus

M(A∩B)=M(A)∪M(B)=M

A[ψ]∩B[¯ ψ]¯

∪Ex∪Ey

=M

(A∩B)[ψ]¯

∪Ex∪Ey. (2.3)

Hence{w∈(A∩B)m:|ψ(w)|<1} =Ex∪Ey. By [9, Theorem 3] we have ¯ψ∈(A∩B)m. Then our theorem follows.

The following corollaries are immediate consequences of the theorem.

Corollary2.8. LetAandBbe Douglas algebras withA=AmandB=Bm. Then A∩B=(A∩B)m.

Corollary 2.9. LetB0 be a Douglas algebra with (B0)m=B0 andB be a Dou- glas algebra such that there is an interpolating Blaschke product q with M(B0)∩ M(H^∞[¯q])=M(B)∩M(H^∞[¯q]). IfAis a Douglas algebra with the propertyAm=A, then(B[ϕ]¯ ∩A)m=B[¯q]∩A. In particular,H^∞[ϕ]¯ ∩A=(H^∞[ϕ]¯ ∩A)m for every interpolating Blaschke productϕ.

Proof. Our hypothesis implies thatB0[q]¯ =B[¯q], hence[B[¯q]]m=B[¯q]. SoB[¯q]∩ A=(B[¯q]∩A)mfollows from our theorem. By [2, Theorem 1] we haveH^∞_m=H^∞+C, so by our theoremH_m^∞[¯q]∩A=[H^∞+C][¯q]∩A.

Theorem 2.10. LetB0 be a Douglas algebra such that(B0)m=B0. LetB be any other Douglas algebra such that there is an interpolating Blaschke product ϕ with M(B0)∩M(H^∞[ϕ])¯ =M(B)∩M(H^∞[ϕ]). For an interpolating Blaschke product¯ qset Bx=B[ϕ]¯ ∩H_suppµ^∞ _x for eachx∈Z(q). PutBe= ∩{Bx:x∈Z(q)}. Then ifqis of type G, then(Be)m=Be.

Proof. We show thatBe=B[ϕ]¯ ∩HN(¯^∞q). By an unpublished result of D. Sarason,

M Be

=M





x∈Z(q)

Bx



=

x∈Z(q)

M Bx

=

x∈Z(q)

M

B[ϕ]¯ ∩Hsuppµ^∞ x

=

x∈Z(q)

M B[ϕ]¯

∩M

Hsuppµ^∞ _x

=M B[ϕ]¯

∪

x∈Z(q)

MHsuppµ^∞ _x

(2.4)

Now, if q is of type G, [7, Proposition 1] and [10, Theorem 3.2(i)] implies x∈Z(q)H_suppµ^∞ _x=H_N(^∞_q)¯. Thus

x∈Z(q)

M

Hsuppµ^∞ _x

=M H_N(¯^∞_q)

, (2.5)

and we get

M Be

=M B[ϕ]¯

∪M H_N(¯^∞_q)

=M

B[ϕ]¯ ∩HN(^∞q)¯

. (2.6)

So by the Chang-Marshall theorem [1,13] we haveBe=B[ϕ]¯ ∩H_N(^∞_q)¯.

The conditionM(B0)∩M(H^∞[ϕ])¯ =M(B)∩M(H^∞[ϕ]), implies that¯ B[ϕ]¯ =B0[ϕ].¯ Hence[B[ϕ]]¯ m=[B0[ϕ]]¯ m=[B0]m[ϕ]¯ =B0[ϕ]¯ =B[ϕ], where the middle equality¯ follows from [9, Theorem 4]. ByTheorem 2.1(and its proof) we have(H_N(¯^∞_q))m=H_N(¯^∞_q) if ¯qis of typeG. Thus

Be

m=

B[ϕ]¯ ∩H_N(¯^∞_q)

m=B[ϕ]¯ ∩H_N(¯^∞_q)=Be. (2.7) This proves our theorem.

Theorem2.11. Let{qn}be a sequence of interpolating Blaschke product that are not invertible in a Douglas algebraA. Suppose for eachnthere is a Douglas algebraAn

with[An]m=AnandM(A)∩M(H^∞[¯qn])=M(An)∩M(H^∞[¯qn]). LetB=_∞

n=1A[¯qn].

ThenBm=B.

Proof. SinceB⊆A[¯qn]forn=1,2, . . . ,by [9, Proposition 6] we haveBm⊆(A[¯qn])m

for alln. SinceM(A)∩M(H^∞[¯q])=M(An)∩M(H^∞[¯q]), we haveA[q¯n]=An[¯qn]. So using [9, Theorem 4] we have

A

¯ qn

m=

An[¯qn]

m= An

m

q¯n

=An

q¯n

=A

¯ qn

. (2.8)

ThusBm⊆_∞

n=1[A[¯qn]]m=_∞

n=1A[¯qn]=B.

Corollary2.12. IfB=_∞

n=1H^∞[¯qn], thenBm=B.

Before we prove our next theorem, we have the following related lemmas.Lemma 2.13shows that the setF⊂L.

Lemma2.13. Letqbe an interpolating Blaschke product that is of finite typeG. Then eachx∈Z(q)is a locally thin point. That isF⊂L.

Proof. Sinceqis of finite typeG, the setZ(q)∩Pxis a finite set, hence there is a factorq0ofqsuch thatZ(q)∩Px= {x}. HenceH_suppµ^∞ _x[¯q0]is a minimal superalgebra ofH_suppµ^∞ _x. HenceH_supp^∞ _µx⊂H_suppµ^∞ _x[¯q0]. By [14, Theorem 5] we have thatxis a locally thin point.

Lemma2.14. LetB=H^∞_suppµy, whereyis a trivial point. IfB⊂Bm, then there is an interpolating Blaschke productqand anx∈Z(q)such thatsuppµx=suppµy.

Proof. IfB⊂Bmthen, by [8, Theorem D], there is an interpolating Blaschke prod- uctqsuch thatB[¯q]is a minimal superalgebra ofB. Hence suppµyis a minimal sup- port set ofH^∞[¯q]. By [6, Theorem 2] there is anx∈Z(q)such that suppµx=suppµy.

Theorem2.15. Letqbe an interpolating Blaschke product and setm˜q= {y∈mq: suppµy=suppµt, tis a trivial point}. SetT=

y∈m˜qH_suppµ^∞ _y. ThenTb=T.

Proof. SupposeT≠Tb. Then by [9, Theorem C] there is an interpolating Blaschke productCsuch thatZ(C)∩M(T)={y}. By [7, Theorem 1] we haveM(T )=M(T [C])∪¯

Py andyis a minimal support point ofH^∞[C]. By [14, Proposition 6],¯ (H_suppµ^∞ _y)b= H_suppµ^∞ _y[C]. So by [14, Theorem 5],¯ y is a locally thin point. We show that this is not the case by showing that there isx∈m˜qwith suppµx=suppµy. By the remark following the proof of [14, Theorem 5], this will lead to a contradiction. Suppose that suppµy≠suppµx for allx∈m˜q. Then for eachx∈m˜qone of the following must occur. (i) suppµx⊂suppµy(properly contained in); (ii) suppµx∩suppµy=θ; or (iii) suppµy⊂suppµx(properly contained in). We show that none of these can happen.

Suppose (i) is true. Then|C(x)| =1 which implies that ¯C∈H_suppµ^∞ _x, sincey is a minimal support point ofC. Thus (i) cannot hold for anyx∈m˜q. If (ii) holds for all x∈m˜q, then again ¯C∈T. Hence neither (i) or (ii) can hold for allx∈m˜q.

Now if (iii) holds for somex∈m˜q, then by [8, Theorem 2] there is an uncountable set Γ ⊂Z(C)such that suppµα⊆suppµx (properly contained in) for allα∈Γ and suppµα∩suppµβ=θfor allα, β∈Γ andα≠β. Sincex∈M(T )we haveα∈M(T ), henceΓ⊂M(T )∩Z(C). This implies thatZ(C)∩M(T )≠{y}. This leads to a contradic- tion. Thus (iii) cannot hold for anyx∈m˜q. So if there is ay∈M(T )∩Z(C)such that T [C]¯ is a minimal superalgebra forT, then there is ax∈m˜qwith suppµy=suppµx. This implies thatxis a locally thin point, which is impossible. ThusTb=T.

Is it true thatF=L? That is, every locally thin point is the zero of an interpolating Blaschke product of finite typeG.

Problem 2.16. What ifF is properly contained inL. Suppose x∈L\F. Is it true that suppµxis maximal?

BothBb and Bmare generated by a special type of minimal superalgebras (deter- mined by the character of the minimal support point).

Under certain conditions we can determine the Bourgain algebra and the minimal envelope algebras of maximal subalgebras of a Douglas algebraB(hereBwill have a maximal subalgebra).

Theorem2.17. LetBbe a Douglas algebra such thatBb[q]¯ =B[¯q]orBm[¯q]=B[¯q]

for some interpolating Blaschke productq. LetAxbe the maximal subalgebra ofB[¯q]

for whichxis the corresponding minimal support point ofB[¯q]. Then either (i) Ax⊂(Ax)band(Ax)b=(Ax)m=B[¯q], or

(ii) Ax=(Ax)band(Ax)m=B[¯q].

Proof. By [14, Theorem 3], [5, Theorem 3], and [9, Theorem 4], our hypothesis implies thatB[¯q]=(B[¯q])b=(B[¯q])m.

Now letAxbe any maximal subalgebra ofB[¯q]associated with the minimal support pointx. First we assume thatxis a locally thin point (see [14, Theorem 5]). Then the maximal ideal space ofB[¯q]andAxare related by the equation

M Ax

=M B[¯q]

∪Px. (2.9)

ThenB[¯q]is a minimal superalgebra ofAx, henceB[¯q]⊂(Ax)b. Now using [14, The- orem 3] again, we get

B[¯q]= B[¯q]

b= Ax[¯q]

b= Ax

b[q]¯ = Ax

b since ¯q∈ Ax

b. (2.10)

SimilarlyB[¯q]=(Ax)m=(Ax)bifxis a locally thin point. This proves (i).

Now suppose thatxis not a locally thin point (e.g.,Pxis not a homeomorphic disk, see [3]). ThenM(Ax)andM(B[¯q])are related by

M Ax

=M B[¯q]

∪Ex (2.11)

with Px ⊆Ex (or Z(q)∩Px has infinitely many points). Using [9, Theorem 4] we have

B[¯q]= B[¯q]

m= Ax[¯q]

m

= Ax

m[¯q] by [6, Theorem 4]

= Ax

m since ¯q∈ Ax

m.

(2.12)

Now, for any Douglas algebraAwe have thatA⊆Ab⊆Am. Thus Ax⊆

Ax

b⊆B[¯q]= Ax

m. (2.13)

SinceAxis a maximal subalgebra ofB[¯q], we have thatAx=(Ax)bifxis not locally thin. This proves (ii).

Corollary2.18. Letqbe any interpolating Blaschke product and setB=H^∞[¯q].

LetAxbe any maximal subalgebra ofBthat corresponds to the minimal support point ofB. Then either

(i) Ax⊂(Ax)band(Ax)b=(Ax)m=B, or (ii) Ax=(Ax)band(Ax)m=B.

Corollary2.19. LetAbe any interpolating Douglas algebra such thatA=Abor A=Am, and letqbe any interpolating Blaschke product such thatq¯∉A. SetB=A[¯q]

and letBxbe any maximal subalgebra ofBcorresponding to the minimal support point ofA[¯q]. Then either

(i) Bx⊂(Bx)band(Bx)b=(Bx)m=B, or (ii) Bx=(Bx)band(Bx)m=B.

Theorem 2.20. LetB be a Douglas algebra that has a maximal subalgebra Ax, wherexis the minimal support point ofBcorresponding toAx. Then(Ax)=Bm.

Proof. By [9, Theorem 4] we have that(Ax)m⊆Bm sinceAx⊆B. SinceAx is a maximal subalgebra ofBthere is anx0∈M(A)\M(B)and an interpolating Blaschke productψ0such that

M Ax

=M A[ψ¯0]

∪Ex₀=M(B)∪Ex₀. (2.14) So by [9, Theorem D] we haveB⊆(Ax)m. IfB0is another superalgebra containingAx, then there is somey0∈M(Ax)such thatM(Ax)=M(B0)∪Ey0. Hence we have that B0⊆(Ax)mandy0∈M(B), otherwiseEy₀=Ex₀. To show thatBm⊆(Ax)m, letψbe any interpolating Blaschke product such that ¯ψ∈Bm. Then by [9, Theorem D] we can assume that

λ∈M(B):|ψ(λ)|<1

=Ex (2.15)

for somex∈M(B). But m∈M

Ax

:ψ(m) <1

=Ex∪ m∈M

Ax

:|ψ(m)|<1

∩Ex₀. (2.16) The set on the right-hand side is eitherEx orEx∪Ex0. Hence by [9, Theorem 4] we have that ¯ψ∈(Am)m.

For(Ax)bwe have the following special result if we assume an additional assump- tion.

Theorem 2.21. LetB be a Douglas algebra that has a maximal subalgebra Ax, wherexis the minimal support point ofB corresponding toAx. Assume thatPx is a nonhomeomorphic disk. Then(Ax)b⊆Bb.

Proof. SinceB⊆Bb and(Ax)b⊆Bb, it suffices to show thatB(Ax)b. SinceAx

is a minimal subalgebra ofBcorresponding tox, by [7, Theorem 1] we have M

Ax

=M(B)∪Ex. (2.17)

Note thatPx⊂Ex. Hence ifψis any interpolating Blaschke product with ¯ψ∈B, but ψ¯∉Ax. Then we have, by [5, Corollary 1.5], the setM(Ax)∩Z(ψ)⊇Px∩Z(ψ)is an infinite set. Then ¯ψ∉(Ax)b. HenceB(Ax)b. (See [4, Theorem 2].)

The following two propositions on minimal support points seem to indicate that the sets given in them are smaller in some sense than the set in the following two well-known facts.

Fact1. LetBbe any Douglas algebra. Then an interpolating Blaschke productqis invertible inBif and only ifZ(q)∩M(B)=θ.

Fact2. For any Douglas algebraBwe have B=

x∈M(B)

H_suppµ^∞ _x. (2.18)

Proposition2.22. An interpolating Blaschke productqis invertible in a Douglas algebraBif and only ifM(B)∩mq=θ.

Proof. We have that mq⊆ Z(q), hence if mq∩M(B)≠θ then ¯q ∉B. (See [7, Theorem 2].)

To prove the converse, suppose that ¯q∉B. Then byFact 1,Z(q)∩M(B)≠θ. There is ay0∈Z(q)such that suppµy₀⊆suppµxfor anyx∈Z(q)∩M(B)andy0∈mq. (See [7, Theorem 2].) Sincex∈M(B)we have thatM(H_suppµ^∞ _x)⊂M(B). SinceM(H_suppµ^∞ _x)⊂ M(L^∞)∪ {λ∈M(H^∞+C): suppµλ⊆suppµx}, we have thaty0∈M(B). Hencemq∩ M(B)≠θ.

Let B be any Douglas algebra and set mq(B)=mq∩M(B). Let MB= ∪{mq(B): qis an interpolating Blaschke product, ¯q∉B}. LetΓ(B)= {xα}α∈Λbe the family of all minimal support points forMBsuch that suppµα∩suppµβ=θifα≠β. Then

Proposition2.23. B=

x_α∈Γ(B)H_suppµ^∞ _α.

Proof. Ifx∈MB, then there is anxα∈Γ(B)such that suppµxα=suppµx. So it suffices to show that

B=

x∈MB

H^∞_suppµx. (2.19)

SetB0=

x∈MBH_suppµ^∞ _x. Since MB⊆M(B), byFact 2 we have thatB⊆B0. Suppose B⊆B0. Then by the Chang-Marshall theorem [1,13] there is an interpolating Blaschke productqsuch that ¯q∈B0but ¯q∉B. Hence there is ay∈(B)such thatq(y)=0.

Hence ¯q∉H_suppµ^∞ _y

0, so ¯qcannot be invertible inB0. ThusB0=B.

Acknowledgement. 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.

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Carroll Guillory: Department of Mathematics, University of Louisiana at Lafayette, Lafayette, LA70504, USA

E-mail address:[email protected]