“MINIMUMIMIHHUHUHHINWHIIHNWIHIN 1O 2 '40) \ILO (D THESIS This is to certify that the thesis entitled DECOMPOSITIONS OF THE MAXIMAL IDEAL SPACE OF L°° presented by Pamela Beth Gorkin has been accepted towards fulfillment of the requirements for Ph 0 D . degree in Mathematics Wxaulehm CLALAZQA Major professor Date 5/12/82 0-7 639 MSU LIBRARIES ll/Il/l/I/l/II Ill!l/l/lll/ll/I/lll/lll/ll/I/l/I/l/I/ll/lll 3 1293 10444 0601 . RETURNING MATERIALS: Place in book drop to remove this checkout from y rrrrrrrr d. FINES Wlll be charged if book 15 returned after the date stamped below. \\ J 5 H7625 DECOMPOSITIONS OF THE MAXIMAL IDEAL SPACE OF Lan BY Pamela Beth Gorkin A DISSERTATION Submitted to Michigan State university in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOS OPHY Department of Mathematics 1982 ABSTRACT DECOMPOSITIONS OF THE MAXIMAL IDEAL SPACE or E” BY Pamela Beth Gorkin Let Lan be the Banach algebra of essentially bounded measurable functions on the unit circle and let M(fi”) denote the maximal ideal space of L”. In this paper we prove some results about M(flm). In Chapter 2 we Show the existence of one point maximal antisymmetric sets for Hai-C, thus giving the first example of a maximal antisymmetric set that equals the support set of some multiplicative linear functional on Had-C. we also Show that each Open set in a fiber contains a QC level set that is not a maximal antisymmetric set for Had-C, extending a result due to D. Sarason [21]. In Chapter 3 using facts about the maximal ideal space of HP we prove some results about closed subalgebras of a: , , a: L containing H . [21] D. Sarason, The Shilov and BishOp decompositions of Hm4-C, to appear. To my parents Anne and David Gorkin ii ACKNOWLEDGEMENTS I would like to thank my advisor, Sheldon Axler, for his guidance, encouragement and patience. He has been an excellent teacher as well as a good friend. I am very grateful to Daniel Luecking and Thomas W01ff for valuable conversations. I would also like to thank my friends ueli Daepp, Jim Hartman and Barbara MacCluer for their help and encouragement. Finally, I would like to thank Cindy Lou Smith for typing this manuscript. iii Chapter 1 Chapter 2 Chapter 3 Bibliography TABLE OF CONTENTS iv 26 38 CHAPTER 1 A complex Banach algebra B is a Banach space which is also a complex algebra such that the norm satisfies llfgll .<. llfll ll 9“ for all functions f and g in B. The space of essentially bounded Lebesgue measurable functions on the unit circle, BID, with normalized Lebesgue measure will be denoted by LP(BID,«%%) or simply L". The space L? is a Banach algebra when it is given pointwise multiplication and the essential supremum norm. Let f E L". we define f in the unit disc by . Tr f(rele) = 31;;f f(t)Pr(q-t)dt -77. 2 where Pr(e) = l-r . The extended f is a 1-2rcose+r bounded harmonic function in the Open unit disc and. as r a l the functions fr(9) = f(reie) converge to f in the weak-star tOpology on L”. The space of continuous complex valued functions on BID will be denoted by C or C(aID). We note that C is a uniformly closed subalgebra of LS, hence is also a Banach algebra. The space of bounded analytic functions on the unit disc I) will be 1 2 denoted by either H” or H9(]D). The space H” is a Banach algebra when it is given the norm Hf“ = sup |f(z)|. zen) By Fatou's Theorem, a bounded analytic function on I) has radial limits almost everywhere. By identifying each Ha function with its boundary function, Ha is isometrically isomorphic to a uniformly closed subalgebra of LP. This space will also be denoted by H9. Once we have made this identification, we can describe another closed subalgebra of L", the algebra H°°+C = {f+g :f 6 H”, g 6 C}. Sarason [19] showed that Had-C is a closed subalgebra of LP. Finally, the largest C*-subalgebra of Hmi-C will be denoted by QC. Thus QC = H°°+C n m, where the bar denotes complex conjugation. The maximal ideal space M(B) of a commutative Banach algebra B with a unit 1 is the set of multiplicative linear functionals (nonzero complex algebra homomorphisms) of B. There is a one to one correspondence between maximal ideals of B and kernels of multiplicative linear functionals on B, hence this space is identified with the space of maximal ideals in B. It is not difficult to show that for a e M(B) l¢(f)l g Hf" for all f E B and m(l) = 1. Therefore M(B) is contained in the dual space 3* of B. we give M(B) the weak-* t0pology, so a net {ma} converges to T if and only if ma(f) 4 ¢(f) for all f 6 B. With this topology, M(B) is a compact Hausdorff Space. For f 6 B, the Gelfand transform of f is the complex valued 3 function f 6 C(M(B)) defined by 2(a) = m(f) for all m 6 M(B). In the cases we are interested in here, the Gelfand transform is an isometry and we will sometimes write f for f, since the meaning will be clear from the context. we begin by mentioning some facts about M(Hm). Further information is available in [9], [10] and [11]. For each point C e I) there exists ”g E M(H‘) such that ¢g(z) = g, where 2 denotes the function f(2) = 2. In fact, the point ”g e M(H”) is uniquely determined by the condition ¢g(z) = g. Hence Q 4 cc defines an embedding of 1) into M(Hm). This embedding is a homeomorphism. By identifying g with eg, we may regard I) as an cpen subset of M(HQ). The Gelfand transform of '2 defines a map Q :MKHP) 4‘35 which is onto. Thus we write M(H") = I) U {T e M(Hé) :|¢(z)| = l]. The Corona Theorem [5] states that I) is dense in M(Hm). we shall also be interested in M(Lm). Since La is a C*-algebra, L? is isometrically isomorphic (via the Gelfand transform) to C(M(LQ)). In [9] and [11] it is shown that M(L”) is a totally disconnected, in fact, extremally disconnected, compact Hausdorff space. For these and other relevant facts about the tepology of M(L”) the reader is referred to [9] and [11]. For each m E M(H”), there is a unique positive Borel measure uw on M(Lm) such that ¢(f) = f a fdp for all f 6 H M(L ) If m e M(Had-C) the closed support of n¢ is denoted supp pm, or simply supp e. Let B denote a closed subalgebra of Lan containing the constant functions which separates the points of M(La). A closed subset S g M(Lm) is called a peak set for B if there is a function f e B such that 2(a) = 1 for w E S and |f(¢)| < l for W e M(L”) ~ S. The function f is said to be a peaking function for S. A closed subset S of M(L”) is called a weak peak set for B if it is the intersection of peak sets. If S is a weak peak set for B, then the restriction algebra B [S is a Banach algebra [9, p.57]. Let B denote a closed subalgebra of Lco containing the function 2. For k G Bib we let Ml(B) = [m 6 M(B)::¢(z) = l}. we call MA(B) the B-fiber over x.- It is not hard to show that M(L”) = LJ MX(L”) and ' keen M(Ha) = I) U lJ {e 6 M(Hm) :¢(z) = l}. Furthermore, 1661) M(H°°+C) = U [ca 6 M(H‘”) :cp(Z) = x} = M(H‘”) ~ In. X651) The LP-fiber over A is a weak peak set for H”, hence for Hmdbc. NOte that z is constant on each fiber. Therefore each polynomial is also constant on each fiber. By the Stone-Weierstrass Theorem, we see that any continuous function f is constant on each fiber and its value on the fiber over A is simply f(l). Therefore Hm4-C IMX(L°) = H°°lM,‘(L°°). 5 There are several other important decompositions of M(L”). The first that we shall discuss here is the Shilov decomposition of M(Lm). For t 6 M(L”) we let E = {m E M(L") :¢(q) = ¢(q) for all q 6 QC}. We call W Ew the QC level set correSponding to W- For t E M(QC) we may also write Et = hp 6 M(L”) :cp(CI) = th) for all q e QC} - In this case we call Et the QC level set corresponding to t. Each QC level set is contained in some La fiber. we have the following theorem of Shilov [22]: Theorem 1.1. Let f E L". If for each QC level set, E there exists a function g 6 H? such that W, f |E¢ = 9 ‘EW , then f e Ha4-c, The second decomposition is BishOp's decomposition of M(La). Before stating BishOp's theorem, we need to define the notion of an antisymmetric set. A set S E M(L”) is called an antisymmetric set for HQ+C if whenever f E Hal-C and f |S is real valued, then f |S is constant. A maximal antisymmetric set for Hm4-C is a weak peak set for Ha4-C. Note that each antisymmetric set is contained in some QC level set and it is easy to see that if {Sc} denotes the set of maximal antisymmetric sets for Hm4-C, then M(Lm) = lJ SO. Bish0p's theorem [3] says the following: a 6 Theorem 1.2. Let {Sc} denote the maximal antisymmetric sets for HP-PC. If f 6 LG is such that for each maximal antisymmetric set Sc there exists geH with f|Sa=g|Sa, then fEH+C. Sarason [21] has given an example of a QC level set that is not an antisymmetric set for Had-C. Thus BishOp's decomposition for M(L”) is strictly finer than Shilov's decomposition for M(L”). The third theorem along these lines is due to Sarason [20]. Theorem 1.3. Let f E L”. If for each m E M(Hmi-C) there exists 9 E H? such that f [supp m = g lsupp e, then f e H°°+C. The relationship of Sarason's theorem to the others is not clear. However,one can say something about how Sarason's theorem relates to Bish0p's theorem. Each support set is an antisymmetric set for Had-C. Thus Sarason's theorem is a refinement of Bish0p's theorem. In [20] and [21] Sarason asked for the precise relation between support sets and sets of antisymmetry for Hwi-C. Is every maximal antisymmetric set for Had-C the support set of a multiplicative linear functional on Hm4-C? This question is still cpen. In fact it was unknown whether any maximal antisymmetric set equals the support set of a multiplicative linear functional on Ha4-C. we shall show the existence of a maximal antisymmetric set consisting of a single point. The relation of this result to Sarason's question is indicated below. Let a e M(L”). Then point evaluation at w is a positive measure and is the unique positive measure satisfying a(f) = f a fdpm for all f e H“. The closed M(L ) support of this measure is [m]. Thus a maximal anti- symmetric set for Hmi-C consisting of a single point must be a support set. We will give many examples of one point maximal anitsymmetric sets and will show that many of these are contained in QC level sets consisting of more than one point. In Chapter 3 we use some of these results to obtain information about closed subalgebras of L" containing H . CHAPTER 2 The following theorem is the main result of this Chapter . Theorem 2.1. Let {In} be a sequence of distinct points of a 2D 5' [1] with "n .. 1. Let [n :- Mxn(L¢) and \l E {\VniMu' ) n M1(LQ). Then {[1] is a maximal anti- symmetric set for Hai-C. An unpublished result of K. Hoffman Shows that any point of M(La) in the closure of a sequence of points from distinct L? fibers is a maximal support set. Our proof is independent of this fact, although Hoffman's result follows easily from Theorem 2.1. In order to prove Theorem 2.1, we need the result given below. Theorem 2.2. Let {In} be a sequence of distinct points of 62D ~ {1] such that An 4 1. Let {In} be a sequence of intervals of 32D with '3 fl lJ I = Q and 1“main!“ Kn 6 In. Then there exists q 6 QC satisfying: (1) q is continuous except at l = l : 43]!“ O (2) [arg q(xn)-n| < for all n ; (3) lar9q(>.)l<%; for xeaJD~UIn The proof will be in two parts. The first part is Lemma 2.3 given below. In what follows we let u denote the harmonic conjugate of u. The Space of continuous real valued functions with continuous first derivatives will be 1 1 Re If 116ch continuous function [11, p.79]. denoted by C then u is a real valued Lemma 2.3. Let I be an Open interval contained in 52D and let w E I. Then given 6 > O and x0 6 n9', there exists u e Ciz with ”qun < e, |u(z)| < e for zeaD~I and u(w)=xo. Proof: By choosing 5 > O sufficiently small and rotating, we may assume that w = l and I = {elez-26<:e<:26]. l with R x “VH0° < l. |V(Z)| g,l for z E BID ~ I and 6(1) =-1? , l v It is enough to show that there exists v E C for then u = satisfies Hun” < e, |u(z)| < ('3 6(1) for z 6 52D ~ I and 3(1) = l0. xna l l-[— It is not hard to show that lim ———TE—— = 1n k for den - X X k > O. we use this fact below. To find v, let 6 > O + Zvl and x0 6 It be given. Choose k so that 1n k > 6 etan:2 Choose an odd integer m satisfying 10 1 - 2wk l (i)m[1- (3%)m] > ——9—5- , (ii) Elfi< a and (iii) cos(7lfi) > -2- . e tan-5 r . I 0 if z 6 52D ~'§ Let v(z) = < and extend v o 3.3 g'w, v(e- Writing v(e) for v(e 3(0) Hence Suppose z = e19 ¢ of v 1 U -% tan 92-) (mt)m if z = e1t 6 [e'l'S 1 1 = k7<8 319- 2? k e ' Xo 6(0) >‘1r . Since the (closed) support I. I 2 we have 11 11' We) V(e+t) -V(6-t) Q1; l ‘ S I-w 2 tan-E- 2n = .l V(6+t)-V(e-t)| gg+ l A6+t)-v(6-t_1 g; .3 I 2v ‘5 27 [tlgp 2 tan 2 5<1t13F 2 tan 2 _ vte+t) -V(e-t)l _d_r-_ HVH, - I 2tan£ |21r$tan§_$ égltlgr 2 2 Therefore |v(z)| g_l if z 6 52D ~ I, as desired. Proof of Theorem 2.2: Given intervals In with In n lJn Im = ¢, 1n e In and In a 1 choose functions . 1 ~ 1 un 6 C11 with Hun“a < 2n+3 , lun(z)| < 25:3. for ~ a: z 6 62D ~ In and un(xn) = (2ni-l)v. Let u = SEE un. Then u 6 CR and since the map T :L2 4 L2 defined by ~ A: Q ~ 2 T(f) = f is continuous, u = Z) un in L norm. Since n=1 1 ~ m . each un e Clt' un 6 Clt’ It is easy to see that [#2: un]m converge uniformly to u on compact subsets of 62D ~ [1]. Hence a is continuous except possibly at 1 = 1. Let q = e1“. Then q = eu+1u e"‘1 6 Hmi-C and 'q = e-u":Lu eu E Hmi-C. Therefore q E QC. For any n we have 12 n) 1260. m m ‘7‘ ifiun) , arg e ‘ -1r| = [arg e larg qun) -4] i[(2n+1)1r+ Z Emunll = [arg e m7ln -1r| i 23 {'10. ) m n _ main .1. - larg(-e )-1r[ < 4» and if xeam~uin, then 123 u (A) m m 1 |arg q().)| = |arg e |< Z . Before we present the proof of Theorem 2.1 we prove a corollary of Theorem 2.2 that will be used frequently. Corollary 2.4. Let t EM1(QC) and [An] bea sequence of distinct points of a D ~ [1} such that t is in the M(QC) closure of a sequence of points {tn}, C (D where tn 6 “kn“; ) and )‘n 4 1. Then Et _C_ It: Mln(L ). Proof: Suppose cp e M(Lm) ~ U 145‘ (La). If n n cp 6 NHL”) ~ M1(L°°), then cp 6 ML”) ~ E . Therefore we d- may assume that T e M1(Lm). Since M(Lm) has a basis of clOpen sets (sets that are both closed and Open), we can .____.5_ find a clopen set F g M(L‘”) with cp e F g M(L‘”) ~ U Mx (L ). n n For each n, M). (La) 53 M(Lm) ~ F and therefore n a U “9' 6 ML”) : lcp’(z)-). I >-]‘} 2F. Since F is compact, m=l n m N a 1 there exists N such that f] [cp’ E M(L ) : [cp’(z) -1 | SEE} m=l n 13 is contained in M(L") ~ F. Thus there exists an interval . . . 09 Q ~ In with kn E In satisfying M1(L ) g_M(L ) F for all 1 E 3;. By choosing In sufficiently small we may assume that -I_n U I = . Note that (*)U{ (Lmlzlé-I—LE M(L”) ~ F. Thus there is a QC function q satisfying conditions (1) - (3) of Theorem 2.2. For any n and any t 6 M1 (La) we have, by (l) and n (2) of Theorem 2.2, that [arg q(¢)-wi g_% . Passing to M(QC) ‘we have larg q(t)-nl g_%- for any t 6 M1 (La). n Therefore for any w’ 6 E we have larg q(¢')-—w| $_%u t To see that m e M(L‘) ~ Et we shall Show that |arg q(¢)|‘g_%u Choose 6 > O and let Fe = [n E M(LQ) :|arg T(q)-arg n(q)| < 6]. Then Fe n F is an Open set in M(LQ) containing m. We claim that there exists x0 7! 1 such that Mk (L°°) n F n Fe 9‘ (l). 0 To establish this it is enough to show the following: Claim M1(Lm) has no interior in M(L”). Proof of claim: Suppose F is a c10pen subset of M(L‘”) with F _c_ M1(L°°). Then XE e C(M(L°°)). Hence there exists a measurable set E _c:_ a D of positive measure such A that XF = x If f is a nonconstant function in Ha E. such that f is continuous at 1, then f is constant on Ml(L°). Therefore f is constant on F. Hence f is constant on E. Since E has positive measure, f 14 must be a constant function. This contradiction establishes the claim. Choose *0 E a I) satisfying 10 # l and a — MXO(L ) n F n Fe #’¢. By (*), 10 E a I) lJ In' Hence 1 m |arg q(xo)| $.35 Let 13,0 6 MXO(L ) n F 0 Fe' Then 1 l . |arg $6,0(q)| 3-3“ Therefore |arg m(q)| g_Z~+e. Since 6 ‘was arbitrary |arg ¢(q)| S_%u Therefore T E M(Lm) ~ Et' so M(L”) ~ (U MX(LP)) E M(L") " Et ‘which implies the result. Proof of Theorem 2.1. Choose m e M(L”) ‘with T #'w such that T and t are in the same QC level set. If no such w exists, then ET = [w] and hence the maximal antisymmetric set containing w, 5*, satisfies SW = {W} and we are done. we assume then that such a m exists. Since m #'w, there exists a c10pen set F with a e F and w e M(L”) ~ F. Thus passing to a subsequence of Mn} if necessary, we may assume that U: _c_:_ M(L”) ~ F. By a theorem of Axler [l] for each, n we can find fn e H”+c with “full“ = 1 such that Wn‘fn)‘ = 1 and n(fn) = O for all n E F. using an idea of Sarason.we let Gn denote the Open ellipse with major axis {-1.1} and minor axis [-i/n,i/n]. Let Tn denote a conformal mapping of the Open unit disc 1) onto Gn such that Tn(0) = 0 and by [18, p.309] we may assume Tn e C. Choose zn ell) 'with |zn| > 321 , Tn(zn) real and Tn(zn) > 321 . By multiplying fn by a constant of modulus one, we may assume that |zn|wn(fn) = zn. 15 Since H""+C|Mk (La) = HWIMx (La), there exists an Hm function whose restriction to (L ) is f L ). “in n M'kn Multiplying that function by a suitable peaking function forMx (I?) ‘we obtain a function 9n 6 H" such that n 1 a _ a: ”gnu“n < TE;T and gn |M1n(L ) - fn |Min(L ). Thus Tn<:(|zn|gn) 6 HS. Let n 6 M(L”). we claim that n(Tn<:|zn|gn) = Tn(|zn|n(gn)). To see this, note that Tn is a uniform limit of polynomials with [[th < 1, then n(Tnof) = nu? pm'n(f)) = pm,n' lim p (n(f)) = T (n(f)). Therefore for each n we have m m,n n ¢n(Tn"‘znlgn) = Tn(‘zn1wn(gn)) _ _ .JE. - Tn(lzn|¢n(fn)) - Tn(zn) > n+1 - If T E F n M1 (La) for some n, then n T(Tn<:|zn|gn) = Tn([zn|T(fn)) = Tn(0) = 0 . For each Xn' choose intervals In centered at kn with I n {J I = ¢ where the Lebesgue measure of I , n min m n . . l [In|, satisfies |In| < 2n+4 and leaJD"(UIn). Let n 0(In) = {z 6 JD : |z-xn| < E511] and let hn be a peaking function for M1 (Lm). By raising hn to a sufficiently n l 2n+4 large power we may assume that th |i§ " CHInlllm < Let Kh be a Mgbius transformation such that Kn(l) = o and Kn(xn) = 1. Let hn = hn(Tn..|zn[gn)Kn. Then 16 (l) in ‘Mkn(L ) = (Tncalzn‘gn)l Mln(Lm) for all n and ‘- 1 (2) “in | 1D ~ (3(InllLJD 27);: . m a: Let Lm= 2 1n and L= Z 1. n=1 n=l n' It is easy to see that Lm converges to L uniformly on compact subsets of 35-{l}. Furthermore, HLmu‘g,2 and thus L E H°(In . To see that L [Et is real valued, let 6 > 0 be given. Choose N such that Z '%<% . Let I be an n=N 2 cpen interval of 52D containing 1 such that max K. IQ<-§-—. Then M.|IHG<-§- j=1.2,...,N. 19,911 31 n 3,. l , 3,. Choose in the QC level set correSponding to w, EW. “‘0 Let V = {n E M(La) : |n(L)- ¢O(L)| < %} n xU M)‘(L¢). Then GI V is an cpen set about 10' By Corollary 2.4, there exists an integer m satisfying m > max(N,-%) and such that on Q , van(L);!<[>. Let chEVnMx(L).SlnceZLn m m n converges uniformly on ‘fm ' we have N |Im cpo(L)| = |Im ETOUnH = [nil]- Im cpo(Ln)+Im comm) + 23 Im m (L ) rde 0 n‘ n¢m N S n§1 |Im (p0(14n)‘ + |Im cpo(Tm o (‘zm‘gm))‘+ + Z Imcp (z ) n=N+1‘ O n ‘ nan 17 N -§— 0 SnEJ, 3N+ lIm(Tm (qpo(|zm[gm))l+ 1 + Z) ———- =N+1 2n+4 nfm Therefore |Im wo(L)| < %§-. Since 6 was arbitrary, ¢O(L) must be real valued. Recall that we chose T to be a point in .M(L°) with T #’¢ such that m and w are in the same QC level set and F was a clOpen subset of M(L?) with T e F and t 6 {tn} (L ) g M(L‘”) ~ F. Since in E MKL?) :[n(L)-¢(L)| <-%} is an cpen subset containing t, there exists n with n12 7 and w 6 M1 (La) such n n that |uyn(L)-¢(L)| <-é—. Thus 1 l1’11”")I = Wn("'n)+m§5lll>n ‘VnumH 23%]:- 323- Z: 5 Therefore IW(L)[.2-%. To determine ¢(L), note that U = [n e M(Lm) :|n(L)-¢(L)| <‘%] n F is an Open set in M(L") containing o. By Corollary 2.4 there exists m such that Mxm(L°°) n Ual’ q). Let cpm e Mxm(L°) n U. Then we have 1 |c"111(1‘) 1 = Wmum) + ngm c"m“n’ l S 1ch(‘¢'m)| + ngmlcpmun) l 30+; 18 3 . Therefore |¢(L)|lg.§- and ¢(L) #’¢(L). The maximal antisymmetric set S containing w is W contained in E so L [S is real valued. Thus L IS *0 is constant. Therefore c Q S. Since T was an arbitrary point of E distinct from t, S = {w} and the proof W is complete. we will show that many of the points that are in the M(L”) closure of a sequence of points from distinct L? fibers are contained in QC level sets consisting of more than one point. we will also show that not every QC level set contains such a point. Before proceeding to the proofs of these results, some related results must be presented. we call a sequence [23.)]:1 of distinct points in I) Q an interpolating sequence if whenever {wj}j=l is a bounded sequence of complex numbers, there exists a function for all j. It is well known in“ 'thf.=. 6 ‘w1 (zj) 'wJ [4] that a sequence { is an interpolating sequence zj} if and only if there exists a constant 5 > 0 such that i=1 -2, H :3£:T_J. 2:6 > O for k = l,2,3,... . A Blaschke J#k 1"szk C product with zeroes {zj}j-l _<:_ ]D is a function b e H°(]D) of the form -'2. z-z. b(z)=).1'ITz—1|-—_—-1 for 261D j j l-z.z 3 where Ill = 1 and Z)(l-|zn|) < m. If the zeroes of n 19 b form an interpolating sequence, b is called an interpolating Blaschke product. A Blaschke product is an inner function, so that |b(eie)| = l for almost all e. A great deal of information has been obtained about interpolating Blaschke products. In [14] P. Jones showed that the interpolating Blaschke products separate the points of M(Hm). We will use the theorem stated below. The proof of this theorem is given in [11, p.205]. Theorem 2.5. Let [2.] a be an interpolating 3 j=1 sequence and let b be the Blaschke product with zeroes a '-——-s-'M(H°) {2,}. . Then {2,}. is homeomorphic to the J J=l J j=1 Stone-Cech compactification of [zj].°°l and every zero 3: of b in M(H”) ~ I) is in the .M(H”) closure of { 13.; z. 3 The following lemma was proven by K. Clancey and JuA. Gosselin [7]. Lemma 2.6. Let u be an inner function. If t 6 M(QC) ‘with u IEt invertible in HQ |Et' then u lEt is constant. In fact {t E M(QC) :u [St is constant] = {t e M(QC) :u |Et is invertible in H" |E is an Open t} set in M(QC). This was also proven by R.G. Douglas in [8]. using Lemma 2.6 together with the following result of D.E. Marshall [16, p.15] , we prove a similar result about characteristic 20 functions. In what follows H°[f] denotes the closed subalgebra of L? generated by H0° and f. Theorem 2.7. Let XE be a nonconstant characteristic function in LS. Then there is an inner function u such that H“ [XE] = 3‘15] Theorem 2.8. Let XE be a nonconstant characteristic function in L“. If x |E 6 Ha IE for some QC level E t0 t0 set Et , then XE |Et is constant. 0 O Sarason has given a proof of Theorem 2.8. Since his proof is unpublished we include a proof below. Our proof is different from Sarason's: his did not use Theorem 2.7. Proof: By Theorem 2.7 there exists an inner function u such that H¢[XE] = Haffi]. Therefore M(H°°[XE]) = M(H°°[T1']). Thus M(H°° [ 1i:t ) _<:_ M(H°°[XE]) = 0 .M(H"[E]). Hence |¢(u)| = l for all T E M(Ha |Et)° Therefore u lEt is invertible in H9 lEt' By Lemma 2.6 there exists 0 open in M(QC) containing t. with 0 3|}: EHmlE forall tee. Let qEQC with q(t0)=l, t t q(s) = O for s e M(QC) ~ 0’ and O g_q g 1. Choose 1] e M(H"+C). If supp w 5 Es and s e M(QC) ~ 0, then q |supp m = O and therefore qu |supp t = 0. If supp t 5 Et and t 6 G, then '11 l supp l” 6 Ho |supp x]; and hence XE |supp W e H" |supp w. By Theorem 1.3 21 QXE E Hmi-C- Since qXE is real valued, qu 6 QC. A Thus qu [Eto is constant. Since q(m) = 1 for all T 6 Et we must have XE [Et constant, as desired. 0 0 By the Shilov Idempotent Theorem we obtain the corollary below, answering a question of R.G. Douglas in [8]. Corollary 2.9. If t E M(QC). then M(Ho[Et) is connected. It is a consequence of the following result of T. wolff [25] that any function f c L? is constant on \ some QC level set. Theorem 2.10. Let f E L”. There exists an outer function q E QC n H” such that qf 6 QC. we will show that for any 1 e 61D and any clOpen set F contained in M1(LQ) there exists a QC level set contained in F. we will make frequent use of the following [21] and [24]. Theorem 2.11. Let f and g be functions in LS. If for each T e M(Hmd-C) either f [supp T e H" [supp T or g [supp T 6 H? [supp e, then for each QC level set E either f [Et 6 H? [E co t or g [Et 6 H [Et' t This theorem together with the following unpublished result of K. Hoffman provides us with much more information 22 about the points in M(L”) that are in the closure of . . a . a sequence of points from distinct L fibers. a Theorem 2.12. Let [zj]. be an interpolating J=1 sequence such that Z-Z lim :1 “=1. n‘kn mn -Ez 9* 1 mn G If m e {zni (H ) and T E M(Hai-C), then supp T is a maximal support. Theorem 2.13. Let E be a nonempty clOpen subset of M1(L°). Then E contains a QC level set that is not a maximal antisymmetric set for Hoi-C. We remark that Sarason [21] has shown that each L6° fiber contains a QC level set that is not a maximal anti- symmetric set for Hai-C. Some of the ideas used to prove Theorem 2.13 are similar to techniques communicated by T. Wolff (private communication). Proof: Let F be a clopen subset of M(L”) such that E = F n M1(LO). Then there exists a measurable subset G A of 52D of positive measure such that XF = XG‘ Let [A ] is n be a sequence of distinct points of a D with 1n=e n 4 1 is and lim xG(re n) r41 = 1. We claim that there exists an interpolating sequence {2m} with the following prOperties: 2-2 (1) lim n J—J =1 n4cn mn l-E'z # m n 23 (2) {2m} is the disjoint union of interpolating . 16 an _ n sequences {2mm}n=1 such that zm,n - rm'ne for suitable choices of rm,n and (3) XG(Zm) 4 l as m 4 m , we construct such a sequence as follows: Let 21 = 21 1 = is ' . 1 _ _ 162 r1,le where O < rl,l < 1. Choose 22 - 22,1 — rz'le 1 22-21 '2 such that XG(22) >-§ and -—-1:——- > e . Choose l-z z . 2 1 161 1 - rl'ze such that xG(23) > l-§- and - 1 ie 3+j ._ _ _ 2 2 j - 1,2. We choose 24 - 22,2 — r2'2e l 1 24-2. 24+j satisfying xG(z4) > l-Z- and “"TZTTL ‘2 e j = 1,2,3. l-z.z j 4 We continue to choose zn satisfying (2) such that -; z -z. n+j 'xG(zn) > l-% and 13-2—1- )e 2 for j< n. It -z.z j n is not hard to see that (1) and (3) also hold. °° my“). Then cpn e M)‘ (Hm). Let n Let' wn E {zmrn}m:l flM Hfic .. :90 6 ‘pn ( ) n M1(H ). By Theorem 2.12 we have that for each n supp mn is a maximal support set. Let b be the interpolating Blaschke product with zeroes {2m}. Choose u 6 M(Hmi-C). If 45 [supp ¢ ¢ Ha [supp h, then [¢(b)[ < 1. Since b [supp W is not invertible in 24 H"D [supp w, there exists 7 E M(Hm [supp ¢) = {n E M(Hm) : supp Tl g supp u] with T(b) = 0. By Theorem 2.5, T E‘TE;T and hence by Theorem 2.12, supp T is a maximal support set. Therefore supp r = supp t. Since xG(zm) 4 l as m 4 m, we have T(XG) = 1. Thus supp 'r _c:_ G, so supp (i _C_ G. Thus for any h 6 M(Hm+C) either '5 [supp W 6 H? [supp t or XF [supp w E H? |supp t. By Theorem 2.11 on each QC level set E we have t b [Et 6 H? [Et or XF [Et 6 H [Et. ConSider the QC level set Et containing supp ”0‘ Since 0 b [supp $0 ¢ H9 [supp $0 we must have XF [Eto E H [Et0. By Theorem 2.8 XF [Et is constant. By (3) above, 0 we must have E c.F. to.- Let {mnc} be a subnet of {en} such that who 4 ”0' Choose [n 6 supp en. Then some subnet of {an ] converges. d we may assume without loss of generality that 1n 4 to. a Therefore to [QC = no [QC. Hence [0 6 Et . By Theorem 2.1, O {to} is a maximal antisymmetric set for H°4-C. Since $0 6 M(Hm) ~ M(LQ), supp m0 consists of more than one point. Since [0 6 Eto and supp mo E Eto , Eto is not a maximal antisymmetric set for Hm4-C. One may ask whether every QC level set contains a point that is in the closure of a sequence of points from distinct L” fibers. The answer to this question is no, as we shall see. We first state a lemma due to Sarason that will appear in [21]. 25 [Lemma 2.14. Let b be an inner function. If m E M(Had-C) and b [supp T is nonconstant, then b(supp cp) = a 13. z ] a be an interpolating n n=l Example 2.15. Let [ sequence with 2n 4 1‘ and let T e [zn]H(H ), ¢ 6 M(Hwi-C). Let EV denote the QC level set containing supp T- Then B does not contain a point which is in W the closure of a sequence Of points from distinct fibers. Proof: Suppose not, that is, suppose mo 6 E1 and $0 6 [muffllL ) 'where ”n 6 M1 (Lo) and xn #’xm for n n #’m. Let b be the interpolating Blaschke product with zeroes [Zn] a . By assumption 2 4 1. Thus b n=l n is continuous at 1 for l # 1. Hence b [Mx(Lm) is constant for 1 7! 1. Let U = {¢EM(L°°) : [(;(b) -cpo(b)[<-]2'-}. Then U is an Open set in M(Lm) containing m0. 1.:(L°° ) Therefore e0 6 {¢n.:¢n E U] By Corollary 2.4, = e . .. M(L‘”) Ed, Ech g u {Mxn(L ) .cpn e Nan(L ) n U] . Hence Ecp 9 TI. If b [ Ecp were nonconstant, then by Lemma 2.14 0 O we have b(Em ) = 52D. Therefore there exists n e E o c"o with n(b) = -¢O(b). Hence n é'fi. Thus b [Ecp is constant. 0 Therefore [¢(b)[ = l and hence w cannot be in the closure Of the zeroes Of b. This contradiction implies the result. CHAPTER 3 In this chapter we prove some related results about closed subalgebras of LCD containing H”. A closed subalgebra B Of Lon containing H“" is called a Douglas algebra. Thus qu-C is a Douglas algebra: it is the smallest closed subalgebra of Lon prOperly containing ff’ EHDp.376]. S.-Y. A. Chang and D.E. Marshall proved the following theorem about Douglas algebras [6], [15]. Theorem 3.1. (Chang-Marshall Theorem) Let B be a closed subalgebra Of L” containing HF. Then B is generated as an algebra by HP and the set E where B = [b- : b is an interpolating Blaschke product and '5 6 B}. This theorem was proven in two stages. Chang showed that if B1 is a Douglas algebra and B2 is any other closed subalgebra Of LP containing H? with M(Bl) = M(BZ)' then B1 = 32‘ Marshall completed the proof by showing that if B is a closed subalgebra Of L” containing HP, then there is a set 6 Of interpolating Blaschke products such that M(B) = M(H‘”[E : b e 6]) where Hm[b : b E 6] denotes the algebra generated by H” and [B : b 6 B}. Marshall's half Of the proof used the result stated below [26]. 26 27 Theorem 3.2. Let u be an inner function. For each O, O < a < 1, there exists an interpolating Blaschke product (:a such that l . (1) [ca(z)[ $16- If [u(z)[ g a (z 6 JD) and (2) There exists Ba < 1 such that if ca(z) = 0, then [u(2)[ 3 Ba (z e D). The proof Of this theorem yields more information than what was given above. In fact, the proof shows the following: Theorem 3.3. Let {uY] be a family Of inner functions. For any a, O < O < 1, there is an interpolating Blaschke product Ca such that (1) [ca(2)|g-f15 if 83p luY(z)l g a (z e D) and (2) There exists Ba < 1 such that if ca(z) = 0, then 53p [uY(z)[ g Ba (2 6 D). Using this form Of Marshall's result we prove a theorem that was proven by D. Sarason (unpublished). Our proof is different than Sarason's proof. Theorem 3.4. Let {BY} be a family Of closed subalgebras Of Lon containing Ha. Then M(flB)=L_J—M_(—B—)'. YY Y Y 28 If Hco E:BY' then the continuous map F :MKBY) 4 M(Ha), defined by restricting each multiplicative linear functional on BY tO H”, is a homeomorphism. Accordingly, we think Of M(By) as a closed subset Of M(H”). we also note that if u is an inner function, then M(H°[E]) = {T 6 M(H") :[¢(u)[ = 1]. These remarks will be used in the proof Of Theorem 3.4. Before we proceed with the proof Of Theorem 3.4 we prove a lemma that will be used in the proof Of Theorem 3.4. Lemma 3.5. Let cpeM(H°°+C) ~ u M(BY) and let (2n) be a sequence Of points Of I) such that cp E [zni _c._' M(Hm) ~ U M(BY). Then for each y, there exists Y a Blaschke product bY such that 'S§ 6 BY and 33p [bY(zn)[ < l. lggggf: Suppose not. Then there exists Y such that whenever bY is a Blaschke product invertible in BY, we must have sup [bY(zn)[ = 1. For any Blaschke product b let Ab = [the M(Hai-C) :[V(b)| = 1] n {W 6 M(Hai-C) :¢ 6 TE;7]. Then Ab is a closed subset Of M(H”+C) and if 36 BY, then Ab #4). We claim that n [Ab :b is a Blaschke product and [5'6 By} #’¢. b By compactness it is enough to show that if b b l' 2'°'°’ n n are Blaschke products and b1,b2,...,bn 6 BY' then jCH Abj5¥¢. n - Since bj 6 BY for j = l,2,...,n, we have n bj 6B .Therefore .=1 n 3 An #4). Since An 5 0 Ab , we see that j:]_3 J 29 n (1 Ab #’¢. Let n 6 0 [Ab :b is a Blaschke product i=1 3 and [3'6 BY]. If IE; 6 BY' then [n(b)[ = 1. Therefore n e M(BY). But 11 6 [2n] 5 M(B") ~ U M(BY) and this Y contradiction implies the result. Proof Of Theorem 3.4. Clearly U M(B ) E M( 0 BY). Y Y Y Suppose cp eM(H°°+C) ~ U M(BY). Thus M(H”) ~ U M(BY) is Y Y an Open subset Of M(H") containing m. The Corona Theorem [5] asserts that the Open unit disk I) is a dense subset Of M(HQ). In fact, if W 6 M(Hmi-C), then t is in the M(H”) closure Of a sequence in I) [12]. Thus there exists a sequence Of points [zn] c_:_ :D such that T E'TE;TM(HQ). Without loss Of generality we may assume that cp 6 [2n] _c_:_ M(H") ~ U M(BY). By Lemma 3.5 Y for each y there exists a Blaschke product b h th t b ' 1 and ‘5’ B . ra' ' Y suc a 33p [ Y(zn)| < Y 6 Y By iSing bY to a sufficiently large power, we may assume that sup [by(zn)[ g.%u By Theorem 3.3 there exists an inter- n polating Blaschke product c satisfying (1) and (2) Of Theorem 3.3 with a =-%. Let [wh] denote the zero sequence Of c. By (2) Of Theorem 3.3, there exists B < 1 such that if C(z) = 0, then sup[bY(z)[ g_B. Hence Y sup [bY(wh)[ g.B for all n. By (1) Of Theorem 3.3, Y [C(Zn)[ < f%- for all n. we claim that 30 and [b] [T(Cll < 1 . . 1 Since [C(Zn)[ 3-1-5 for all n and cp E [zni , [m(c)[.ggf% . Thus [b] above holds. TO establish [a], suppose [a] does not hold. Then there exists Y0 such that E'¢ BY . Therefore there exists n E M(BY ) ‘with 0 O q(c) = 0. By Theorem 2.5, n G [wh[. By the remarks made above, sup [bY(wh)['g.B for all n. Therefore Y .— [n(bYO)I g.B < 1. Since b E B . n cannot belong to Y Y O O M(BY ) which is a contradiction. Therefore [a] holds. 0 New [a] and [b] imply that m 6 M(H”) ~ M((W BY) and Y hence M((W B ) =lJ M(B ) as desired. Y Y Y Y Theorem 3.1 implies that if f e L” ~ H”, then E”[£] is generated by H? and the complex conjugates Of countably many interpolating Blaschke products. C. Sundberg [23] has shown that L" is not countably generated as an algebra over H”. 'We prove an extension Of Sundberg's result below. Theorem 3.6. Let B be a closed subalgebra of Lon containing HSA-C. If there exists an Open set 0- in M(H°°+C) such that «p 7! o n M(B) g M(L‘”), then B is not countably generated as an algebra over Ha. Proof: Suppose B is generated by H“0 together with countably many L? functions. By Theorem 3.1 31 and remarks made above we see that B must be generated by H“ together with the complex conjugates of countably many interpolating Blaschke products {bj :j 6 ll]. Following a method Of Axler [l] we construct a single Blaschke product b such that M(Hmfbl) g M(B). TO do this, let {zn,j}n:1 denote the zeroes Of bj' Since the zeroes Of bj form a Blaschke sequence, there exists an integer Nj such that Z (l-[zn j[) <—lj- . Let b denote the ' 2 nzn. 3 a Blaschke product With zeroes J31 [zn,j]nj2N . If cp e M(H°°+C) ~ M(B), then there exists j() 6 N such that [cp(bj )[ < 1. Therefore bj [supp T is not invertible O O in H? [supp m and hence there exists T 6 M(Hé [supp m) such that T(b. ) = 0. By Theorem 2.5 T 6 [z i: and hence T(b) = 0. Thus b [supp T is not constant. Since supp T E supp m, we see that b [supp T is non- constant. Therefore [m(b)[ < l and c9 6 M(H°°+C) ~ M(H°°['13]). Thus M(B) _O_ M(Hafii] ). Let n E G n M(L”). Since M(La) has no interior in M(Ha-l-C), we can choose a net {no} 55 M(H”+C) ~ M(Lm) with n E Tfigj. Since n 6 0- we may assume that {ha} 5:0. By a result Of D.J. Newman [17], there exists a Blaschke product c such that if T 6 M(HQT-c) and |e(h)[ < 1. then e(c) = 0. Since 0 n M(B) _c_: M(L‘”), and M(H°[B]) g M(B), we have 0 n M(H"[S’]) g M(L‘”). For each a, n0 6 O and no 6 M(Hmi-C) ~ M(Lm). Hence 32 [na(b)[ < 1. Therefore na(c) = O for all a. Thus q(c) = 0. However n E M(La) so [n(C)[ = l. which is a contradiction. Therefore B cannot be countably generated over Hm. Many examples can be given Of Douglas algebras that are generated by Ha together with one L? function. For example, Hai-C = Hafz]. Donald Marshall showed in [16] that if f is a simple function, then there is an inter- polating Blaschke product b such that Half] = Han-5] . It is not known which algebras have the form Hmffi] for a single interpolating Blaschke product b. we use the following result Of Sarason [21] to give two examples Of Douglas algebras that are not generated by H" and the complex conjugate Of just one Blaschke product. The first is an example Of a Douglas algebra which is generated by H? and one function f 6 L7, but is not generated by Han and the complex conjugate Of one inner function. Theorem 3.7. (Sarason [21]) There exists a function f e H" such that f [M1(L°) is real valued and f is nonconstant on some QC level set contained in M1(L°). Example 3.8. Let f be the function given in Theorem 3.7. Then Hmff] does not equal Héffi] for any inner function u. 33 .gggggz Suppose Hoff] = Haffi] for some inner function '3. By assumption f [M1(L°) is real valued, hence f must be constant on the support set Of every multiplicative linear functional T E M1(H”). Thus M1(H°°) g M(H°[-f-]) = M(HQ [11]). Therefore for each m e M1(H"), u [supp m is constant. we Claim that there exists an Open interval I g a D such that l e I and such that for each 1 6 I and each T E MX(H°) we have 3 [supp m 6 HQ [supp m. If not, there exists a sequence Of points {in} with in 4 l, and wn 6 M1n(H¢) with ¢n(u) = 0. By compactness there exists m e E;;TM(H°)rlM1(H°) with ¢(u) = 0. .Since u |supp T is constant, this is a contradiction. Therefore there exists an Open interval I g a D such that l 6 I and u [ supp q; is constant for all cpEMX(H°°_) whenever I E I. Let E EMl(L°°) bea t QC level set such that f [Et is nonconstant. Choose a continuous function g such that the (closed) support Of g is contained in I and 9(1) = 1. If m 6 MX(HS) and 1 e I, then 9 |supp m is constant and u [supp ¢ is constant. Thus f |supp T is constant. Therefore gf [supp m 6 Ha [supp T- If 1 E 62D ~ I, then gf [supp m is identically zero. By Theorem 1.3 gf e Hmi-C. Since ET 6 Hmi-C, we have g? E QC. Therefore gf [Et is constant. Hence If [Et is constant, which is a contradiction. Hence Huff] does not equal H°f51 for any inner function u. 34 The second example shows that the intersection Of two algebras of the form H?[Bi] and H°[Sé], with b1 and b interpolating Blaschke products is not necessarily 2 Of the form HS[E] where c is an inner function. we remark that if b1 and b2 satisfy lim max([b1(z)[,[b2(z)[) = 1, IZl"1 then it has been shown [2] that H" ['51] n Hafiz] = H°°+C. Our example shows that it is possible to have lim max([b1(z)[,[b2(z)[) = 1 (note that [z[ 4 1 has been 241 replaced by z 4 l) and such that Ha[5i] n H°[Sé] is not Of the form H°[E] where u is an inner function. The second example is similar to the first. we will use the result Of D. Marshall [15] stated below. Theorem 3.9. Let S be a set contained in M(La) and containing more than one point. Then there exists an interpolating Blaschke product b which is nonconstant on S. Lemma 3.10. Let Et be a QC level set containing more than one point. Then U {S : S _c_ Et] S nontrivial S maximal antisymmetric is dense in Et' Proof: If this is not the case, then there exists a clopen subset F of Bt with In E_<:_E . U {S:SEEt] t S nontrivial S maximal antisymmetric Then M(H‘” [Et) = hp 6 M(H‘”) :supp Cp 5 F} u (E:t ~ F). 35 Therefore for each T E M(Hco [Et), XF [supp m E H? [supp m. Thus XF 6 Ha [Et' This implies M(HaD [Et) is disconnected, contradicting Corollary 2.9. Egampge 3.11. There exist interpolating Blaschke products b1 and b2 such that for every antisymmetric_ set S ELM1(L°), either bl [S is constant or b2 [3 is constant, but on some QC level set E ‘E;M1(L°) neither t b1 [Et is constant nor b2 [Et is constant. Proof: By Theorem 3.7 there exists f e H0° real valued on M1(L°°) and a QC level set E EM1(L°°) such t that f [Et is nonconstant. Since the nontrivial maximal antisymmetric sets contained in Et are dense in Et’ f must assume different values on two maximal antisymmetric sets S1 and 82 contained in Ft. Without loss Of generality we may assume that f(Sl) = 0 and f(82) = 1. Let 01 = [cp e M(H‘”) : (um -1| >1) and c2 = to e M(H‘”) : lens) -1[ < 31;}. Then Mm” |sj) goj for j = 1,2. By Theorem 3.9, there exist interpolating Blaschke products b1 and b2 such that b1 [S1 is nonconstant and b2 [S2 is nonconstant. Therefore there exists ejemsmlsj) j=1,2 with cpj(bj)=0 j=1.2. Let Q Q {zn,l}n_1 and {zn,2}n=1 denote the zero sequences of b1 and b2, respectively. Without loss Of generality we may or") 7—1401“) assume that zn,l E 01 and zn'2 E 02. Let S be a maximal antisymmetric set contained in Ml(L°). Then either M(Hm [ S) E 0: . in which case 36 b1 [S is constant (since for any blaschke product either b(S) = 52D or b [S is constant) or M(H0° [5) n 01 #’¢. Since f E H? and f [M1(Lm) is real valued, f [S is constant. Therefore we must have M(H°° [S) E 01. Hence b2 [S is constant. Clearly neither bl nor b2 is constant on Et and we are done. we remark that Example 3.9 shows that Theorem 2.11 does not generalize to the case where f [supp m 6 H? [supp m or g [supp T 6 Ha [supp T for all m E M1(H"4-C). Example 3.12. There exist interpolating Blaschke products b1 and b such that Hé[hi] n Hmffié] is not 2 generated by Han and the complex conjugate of a single inner function. Proof: Choose b1, b2, Et' ml and T2 as in Example 3.11. Suppose there is an inner function u such that 11°51] 0 11°52] = H°°[Ti] Then by Example 3.11 M1(H°°) 2 ms” [751]) U M(H”['152] ) EM(H°°[E]). As was shown in Example 3.8, there exists an Open interval I g; a 13 containing 1, such that Mxm‘”) _<_:_ ma" [3]) = M(H°°['b'l] n Hafiz—2]) for all 1 E I. By Theorem 3.4, M101“) 5; M(H‘”[Bl]) u M(H°[Bz]) for x e I. Let g be a continuous function such that supp g E_I and 9(1) = 1. Let cp€M(H.+C). If eeMxm‘”) and >. €6D~I, then 37 9-51 [ supp :9 and 932 [ supp cp are both identically zero. If e e MX(HQ) and x e I then cp e M(H°°[Bll) UM(H°°[‘152]). Therefore either bl [supp T is constant or b2 [supp m is constant. Hence for any cp 6 M(H°°+C) either gBi [supp m 6 HS [supp m or gfié [supp m e H" [supp m. By Theorem 2.11, we have either gEi [Et 6 H? [Et or gSé [Et 6 H9 [Et. Since 9 [M1(LP) is identically 1, . ._ a - 5’ . . either b1 [Et 6 H [Et or b2 [Et 6 H [Et which is a contradiction. 10. 11. 12. 13. BIBLIOGRAPHY S. Axler, Factorization of La functions, Ann. of Math. 106 (1977), 567-572. S. Axler, S.-Y. A. Chang and D. Sarason, Products Of Toeplitz Operators, Integral Equations and Operator Theory 1 (1978). 285-309. E. BishOp, A generalization Of the Stone-weierstrass theorem, Pacific J. Math. 11 (1961), 777-783. L. Carleson, An interpolation problem for bounded analytic functions, Amer. J. Math. 80 (1958), 921-930. , Interpolations by bounded analytic functions and the corona problem, Ann. Of Math. 76 (1962), 547-559. S.-Y. A. 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