MAXIMAL SUBALGEBRA OF DOUGLAS ALGEBRA

When q is an interpolating Blaschke product, we find necessary and sufficient conditions for a subalgebra B of H[q] to be a maximal subalgebra in terms of the nonanalytic points of the noninvertible interpolating Blaschke products in B. If the set M(B) N Z(q) is not open in Z(q), we also find a condition that guarantees the xistence of a factor q0 of q in H such that B is maximal in H[]. We also give conditions that show when two arbitrary Douglas algebras A and B, with A c B have property that A is maximal in B.


I. INTRODUCTION.
Let D be the open unit disk in the complex plane and T be its boundary.Let L be the space of essentially measurable functions on T with respect to the Lebesgue measure.By H we mean the fmmily of all bounded analytic functions in D. Via identi- fication with boundary functions, H can be considered as a uniformly closed subalgebra of L A uniformly closed subalgebra B between H and L is called a Douglas algebra.
If we let C be the family of continuous functions on T, then it is well known that H+C is the smallest Douglas algebra containing H properly.For any Douglas algebra B, we denote by M(B) the space of nonzero multiplicative linear functionals on B, that is, the set of all maximal ideals in B. An algebra B 0 is said to be a maximal subalgebra of B, if B I is another algebra with the property that B 0 B I B, then either B B 0 or B I B.
An interpolating sequence {Zn}=l is a sequence in D with the property that for any bounded sequence of complex numbers { }n=l' there exists f in H such that f(z n n n for all n.A well-known condition states that a sequence {Zn}n=l is interpolating if and only if m n inf H 0. is called an interpolating Blaschke product if its zero set {Zn}n__1 For a function q in H+C, we let Z(q)= {m6M(H+C):qkm)= O} be the zero set of # in M(H+C).An inner function is a function in H of modulus almost everywhere on T. We denote by H [b] the Douglas algebra generated by H and the complex conjugate of the inner function b.
We put X M(L).Then X is the Shilow boundary for every Douglas algebra.For a point in M(H), we denote by x the representing measure on X for x and by supp x the support set for x" For a function q in L (in particular if q is an interpolating Blaschke product), we put N(q) the closure of the union set of supp x such that x eM(H+C) and Isuppx HIsupp x" Roughly speaking, N(q) is the set of nonanalytic points of q.Set QC H + C fl H + C and for x 0 in X, let Qxo {xe X: f(x)=f(x O) for f e OC}.Qxo is called the QC-level set for x 0 [9].For an inner function q, K. Izuchi has shown the following [5, Theorem l(i)].
THEOREM i.If q is an inner function that is not a finite Blaschke product, then, N(q) U {Ox; x Z(q)}. (1.4) In particular, the right side of I..4 is a closed set.Now assume that q is an interpolating Blaschke product, and let B be a Douglas algebra contained in H[].We will always assume that M(B) 8 Z(q) is not an open set in Z(q), for Izuchi has shown [6] that if B is a maximal subalgebra of H[], then M{B) fl Z(q) is not open in Z(q).We will give answers to the following two questions.When is B a maximal subalgebra of H[] or when is there a factor qo of q in H such that B is maximal in H[q-O]?These answers will be in terms of the nonanalytic points of q and the invertible inner functions of H[q] that are not invertible in B.
For a Douglas algebra B, we denote by N(B) the closure of U {supp x; x 6M(H+C)/M(B)}.
In particular N(H[]) N().In general if A and B are Douglas algebras such that A B, we put hA(B) the closure of U {supp x: x g M(A)/M(B)} and for any inner function b, hA(b) the closure of U {supp x: x g M(A), Ib(x)l i}.
It is shown in [7, Corollary 2.5] that if B c H[], then N(B) c N(), and it is not hard to show that N(q)/N(B) 2 NB(q) (in a sense the set NB(q) is generated by the nonanalytic points M(B)/M(H[]) e M( H + C)/M(H[])) 2. OUR MAIN RESULT.
We'll need the following lemma.It shows how small M(B)/M(H[]) must be if B is to be a maximal subalgebra of H[q].Let {b b is an interpolating Blaschke product with b gH[q]}, and n(B) {bOg : b 0 B}.
LEMMA i.Let q be an interpolating Blaschke product and B be a Douglas algebra contained in H[].Suppose for all bog (B), we have that NB(q) NB(b0).Then B is a maximal subalgebra of H[q].
PROOF.It suffices to show that if be (B), then B [b] H[q].Hence the only Douglas algebra between B and H[q] that contains B properly is H . Let me M(B) such that m and since NB(q) NB(b), we have that m M(H[q]) Then qlsupp __HIsupp m bl supp HIsupp m" Thus Ib(m) < 1 and we get m M(B[D]).This shows that

Um M(B[b]) c M(H[q]
), and B is maximal in H[q].Using Theorem i above, it is not hard to show directly that N(B[b]) N(q).However, by Proposition 4.1 of [7], this condition is not sufficient We let E NB(q).This can be a very complicated set.For example, it can contain supp x where x belongs to a trivial Gleason part or a Gleason part where lql i, but yet q 0 on this part [see 3].So for B to be maximal in H[q], E must be as simple as possible.To see how simple, we set A(B) , then there are interpolating Blaschke products a 0 and a 1 in (B) such that NB(q) n N(a 0) a N(a 1) .Thus we get B B[a0] c '[q].To see this, just note that both NB(q) N(a 0) e 0 and NB(q) N(a 1) e since a 0 and a belong to B (B).Since their intersection is empty, there is an Xlg N(B) such that HIsupp x Thus NB[0](q) N(q), which implies that B[a 0] c H[q]. a01supp Xl Obviously, B B[a0] so B cannot be maximal in H[q] unless E 0 .We_now state.PROPOSITION i. LetB be a DouNlas algebra properly contained in H[q], and suppose Then the following statements are equivalent: PROOF.We prove the following: Suppose (i) holds.We will show that NB(q) NB(b) for all b e (B).Using Lemma i, this will prove that B is a maximal subalgebra of H H'[q], hence N(B) N(B[b]) N(q).Now N(q) N(B) U NB(q), so by the above equality we have that NB(q) N(B[b]).Thus, if x e M(B) such that -qlsupp : H'Isupp x implies that supp xN(B[b]) Thus blsupp H'Isup p x and NB(q) NB(b).We have (i) (ii).
Next suppose that (ii) holds.It is clear that E 0 c E 0 E. We must show that E01E0 and EIE 0 are empty sets.First we show that E01E0 is empty.Suppose not.
e H Isup p x and Then there is an xeM(B) and a b 0e A (B) such that b01supp x supp -x E0" It is clear by Theorem 1 that supp x N N( 0) .Consider the algebra B[b0].Since bog A(B), B B[b0].Since supp x 5 N(q) and supp x N(b0)' we have that Ib0(x) I, so we have supp x N(q)/NB[0](q)" This implies that B[b 0] which is a contradiction.So E 0 E 0. Now we show that E/E 0 is empty.Again suppose not.Hence there is a y M(B) such that supp y E, but supp by E 0. There is a b A(b) such that supp y N(b). (since supp y N(q)/NB[](q)), which is a contradiction.
So we get E 0 E.This shows that (ii) (iii).
It is trivial that if (iii) holds, E 0 NB(q).
If (iv) holds and b is any interpolating Blaschke product in &](B), then by (iv) NB(q) c_ NB(b so by Lemma i, B is a maximal subalgebra of H[q]. Finally, suppose (ii) holds.We are going to show that N(B) N(q).Suppose not.
Then N(B) c N(q).By Theorem there is a Q-C level set Q with N(B) N Q .Put H'[q] and N(B 0) N(q).Since N(B) 0 Q , we also have B B 0 (because N(B) N(B0)).This implies that B is not a maximal sub- algebra of H[ q], which is a contradiction Thus N(B) N(q).Now suppose we have that E 0 5 E 0 E (E 0 is possible).
When is there a factor q0 of q in H such that B is a maximal subalgebra of H[q0 (B H[q0 is not possible)?To answer this question, let 0 {q0:qq0 e H }, and 0 (B) {q0 g 0 B H[q0]}.
Set F N(q0).Suppose F N(q0) for some factor q0 of q in H Then

and finally
We have the following.
COROLLARY I. Let q0 be a factor of q i__qn H such that F N(q0) and assume F 0 .
If any of the followin conditions hold: Then B is a maxima subalgebra of H 0 H[q0 where rio(B).qo The fact that F N(q0) for some q0 fl0 (B) implies that H 0 H[0 and our corollary follows from Proposition i.
We now consider this question for the genral Douglas algebras.Let A and B be Douglas algebras such that A c B and there is an inner function q with B c A[q].
When this occurs we say that A is near B. It is well known that if B L and A is any Douglas algebra properly contained in B, then A is not near B, that is, B S A[q] for any inner function q.In fact L is not countably generated over any Douglas algebra A [I0].So by the results of C. Sundberg [i0] any Douglas algebra B which is countably generated over A is also near it.
The following result comes from [2, Lemma 5] and gives equivalent conditions for two Douglas algebras to be near each other [see II, Theorem for a similar result].
THEOREM 2. Le___!tA and B be DouRlas alRebras with H +C A B and be an inner function.Then the following statements are equivalent.
PROOF.Assume (i) holds; we show that B c_ A. Let b be any interpolating Blaschke product for which is in B. If x is in ZA(b), we show that x is also in ZA(q).Now x is in M(A) and b(x) 0 implies that x is not in M(B), since b is in B.
So by (i) we have that x must be in ZA(q).Thus ZA(b) c_ ZA(q), and by Theorem of [A] we have b is in A. Now let f be any function in B. By the Chang Marshall Theorem [1,8] there is a sequence of functions {hn} in H and a sequence of inter- polating Blaschke products {bn} with bn B for all n, such that hnbn f" It is trivial that if (iii) holds, NA(B) c_ NA(b for all We are done. In Proposltion 4.1 of [7] Izuchi constructed a family of Douglas algebras B contained in H' [q] with the property that N(B) N().By Proposition I, we have that this family is a family of maximal subalgebras of H[q].
Finally we close this paper with the following question that I have been uable to answer.
QUESTION i. Recall that if q is an interpolating Blaschke product, then N(q) N(B) U NB(q) for any Douglas algebra with B H[q].Does there exist a Douglas algebra Bo -c H[q] with N B (q) N(q).O

Call for Papers
As a multidisciplinary field, financial engineering is becoming increasingly important in today's economic and financial world, especially in areas such as portfolio management, asset valuation and prediction, fraud detection, and credit risk management.For example, in a credit risk context, the recently approved Basel II guidelines advise financial institutions to build comprehensible credit risk models in order to optimize their capital allocation policy.Computational methods are being intensively studied and applied to improve the quality of the financial decisions that need to be made.Until now, computational methods and models are central to the analysis of economic and financial decisions.However, more and more researchers have found that the financial environment is not ruled by mathematical distributions or statistical models.In such situations, some attempts have also been made to develop financial engineering models using intelligent computing approaches.For example, an artificial neural network (ANN) is a nonparametric estimation technique which does not make any distributional assumptions regarding the underlying asset.Instead, ANN approach develops a model using sets of unknown parameters and lets the optimization routine seek the best fitting parameters to obtain the desired results.The main aim of this special issue is not to merely illustrate the superior performance of a new intelligent computational method, but also to demonstrate how it can be used effectively in a financial engineering environment to improve and facilitate financial decision making.In this sense, the submissions should especially address how the results of estimated computational models (e.g., ANN, support vector machines, evolutionary algorithm, and fuzzy models) can be used to develop intelligent, easy-to-use, and/or comprehensible computational systems (e.g., decision support systems, agent-based system, and web-based systems) This special issue will include (but not be limited to) the following topics: • Computational methods: artificial intelligence, neural networks, evolutionary algorithms, fuzzy inference, hybrid learning, ensemble learning, cooperative learning, multiagent learning [q].Let b e (B) and consider the Douglas algebra B[b].We have B B[b] B and B[b] H[q] I; I is an interpolating Blaschke product with le H [q] and IIQe HIQ].By Proposition 4.1 of [7], we have B 0 q0e0(B)    B c_ H[q0 ].So q0 is our possible candidate.Blaschke product with c e H[q0]}, But h (b) f belongs to A since b s in A for all n.This proves (ii) n i n Since this is true for any b e(B,A), we have NA(B) _D W Thus W NA(B) if A is maximal in B.

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Application fields: asset valuation and prediction, asset allocation and portfolio selection, bankruptcy prediction, fraud detection, credit risk management • Implementation aspects: decision support systems, expert systems, information systems, intelligent agents, web service, monitoring, deployment, implementation