Let D be the unit disk. We show that for some relatively closed set F ⊂ D there is a function f that can be uniformly approximated on F by functions of H∞, but such that f cannot be written as f = h + g, with h ∈ H∞ and g uniformly continuous on F. This answers a question of Stray.Fil: Suarez, Fernando Daniel. Consejo Nacional de Investigaciones Científicas y Técnicas. Oficina de Coordinación Administrativa Saavedra 15. Instituto Argentino de Matemática Alberto Calderón; Argentin

Suarez, Fernando Daniel

Proceedings of the American Mathematical Society

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PROCEEDINGS OF THEAMERICAN MATHEMATICAL SOCIETYVolume 128, Number 10, Pages 3003–3007S 0002-9939(00)05577-5Article electronically published on April 28, 2000A COUNTEREXAMPLEFOR H∞ APPROXIMABLE FUNCTIONSDANIEL SUÁREZ(Communicated by Albert Baernstein II)Abstract. Let D be the unit disk. We show that for some relatively closedset F ⊂ D there is a function f that can be uniformly approximated on Fby functions of H∞, but such that f cannot be written as f = h + g, withh ∈ H∞ and g uniformly continuous on F . This answers a question of Stray.1. Introduction and preliminariesLet D denote the open unit disk and F ⊂ D be a relatively closed set. For0 < p ≤ ∞ let Ap(F ) be the space of functions that can be uniformly approximatedon F by functions of the Hardy space Hp. Also, let Cua(F ) denote the space ofuniformly continuous functions on F that are analytic on its interior. PutF̂ = {z ∈ D : |f(z)| ≤ supF|f | for all f ∈ H∞}.A. Stray proved in [4] that if 0 < p <∞, thenAp(F ) = Hp|F + Cua(F̂ )|F .(1)He also showed that A∞(F ) ⊃ Cua(F̂ ). Notice that this immediately implies theinclusion of the right member of (1) in Ap(F ) for every 0 < p ≤ ∞. The problemof whether the other inclusion holds for p =∞ is posed in [4]. The purpose of thispaper is to construct an example where the inclusion fails. That is, we will see thatthere exists a relatively closed set F ⊂ D, with F = F̂ , such that some f ∈ A∞(F )cannot be decomposed as f = h+g, with h ∈ H∞ and g ∈ Cua(F ). More generally,for the example that we construct the above decomposition is not even possible forg ∈ Cua(F ) and h in the Bloch space. In this section we fix notation and state someof the background that will be used in the construction of the example, given inSection 2. The last section is a short discussion showing that the counterexampleworks in a more general situation.The maximal ideal space M of H∞ is defined as the space of nontrivial mul-tiplicative linear functionals on H∞, provided with the weak ∗ topology. It is acompact Hausdorff space, and the Gelfand transform, f̂(x) = x(f) for f ∈ H∞ andx ∈ M, establishes an isometric morphism from H∞ into the uniform algebra ofcontinuous functions on M. Evaluations at points of the disk are in M, so D isReceived by the editors December 8, 1998.2000 Mathematics Subject Classification. Primary 30E10; Secondary 30H05.Key words and phrases. Bounded analytic functions, uniform approximation.c©2000 American Mathematical Society30033004 DANIEL SUÁREZnaturally imbedded as an open subset of M. In addition, the corona theorem [1]states that D is dense in M.Suppose that ϕ ∈ M \ D. In [3] Hoffman defined a continuous map Lϕ fromD into M \ D that generalizes the concept of Mobius transformation, Lω(z) =(ω − z)/(1 − ωz) for ω ∈ D. He showed that if (zα) is a net in D that tends to ϕ,then f ◦Lzα tends pointwise to f̂ ◦Lϕ ∈ H∞ for every f ∈ H∞. The use of nets isimposed by the fact that M is not metrizable.Let b be a Blaschke product with zero sequence {zn}. Ifδn =∏k 6=n∣∣∣∣ zn − zk1− znzk∣∣∣∣satisfies δ(b) def= inf δn > 0, then b is called an interpolating Blaschke product and{zn} an interpolating sequence. If in addition lim δn = 1 when n→∞, then b iscalled a thin product and {zn} a thin sequence. The pseudohyperbolic metric onD is ρ(z, ω) = |(ω − z)/(1− ωz)|. For z ∈ D and 0 < r < 1 we writeK(z, r) = {ω ∈ D : ρ(z, ω) < r} and ∆(z, r) = {ω ∈ D : |z − ω| < r}for the open balls of center z and radius r with respect to the pseudohyperbolicand the euclidean metric, respectively. Simple geometrical considerations [2, p. 3]show that K(z, r) = ∆(c, R), wherec =1− r21− r2|z|2 z and R =1− |z|21− r2|z|2 r.(2)The following well-known result of Hoffman can be found in [3] or [2, pp. 404-405].Lemma 1. Let b be a Blaschke product with zero sequence {zn}. If δ = δ(b) > 0,then there are 0 < ε(δ), r(δ) < 1 such that(i) a(z) = (b(z)−ω)/(1−ωb(z)) is an interpolating Blaschke product when |ω| <ε(δ),(ii) {z ∈ D : |b(z)| < ε(δ)} =⋃n Vn, where Vn ⊂ K(zn, r(δ)), and(iii) b(Vn) = ∆(0, ε(δ)) for every n.2. The exampleLet X ⊂ D be any subset. We can think of X as contained in M or in thecomplex plane C. For G = M or C we write closGX for the closure of X in thespace G.Let b be a thin product with zero sequence 0 ≤ zn < 1 (for instance zn = 1−n−n)and put δ = δ(b). We fix some ε with 0 < ε < ε(δ), where ε(δ) is given by Lemma1. Consider the setsΩε = {ω ∈ R : ε/2 ≤ |ω| ≤ ε}andE = {x ∈ M : b̂(x) ∈ Ωε}.Then F def= E∩D is a relatively closed subset of D such that F̂ = F . In fact, supposethat z ∈ D is not in F . Then b(z) 6∈ Ωε and since Ωε is polynomially convex, thereis a polynomial p such that |p(b(z))| > supΩε |p| = supF |p ◦ b|. Since p ◦ b ∈ H∞,then z 6∈ F̂ .A COUNTEREXAMPLE FOR H∞ APPROXIMABLE FUNCTIONS 3005Furthermore, we claim that E = closMF . The continuity of b̂ on M obviouslyimplies that E ⊃ closMF . For the other inclusion take x ∈ E. Therefore b̂(x) ∈Ωε ⊂ {z ∈ D : |z| < ε(δ)}, and (i) of Lemma 1 tells us thata(z) = (b(z)− b̂(x))/(1 − b̂(x)b(z))is an interpolating Blaschke product. If {ωn} denotes the zero sequence of a andZ(â) = {y ∈ M : â(y) = 0}, then the fact that a is interpolating easily yieldsclosM{ωn} = Z(â) (see [2, VII, Ex. 4]). Besides, we have b(ωn) = b̂(x) ∈ Ωε forall n, which means that {ωn} ⊂ F . Therefore x ∈ Z(â) = closM{ωn} ⊂ closMF ,and the inclusion is proved.Let ϕ ∈ closM{zn} \ D. Since {zn} is a thin sequence, a result of Hoffman[3, pp. 106-107] asserts that b̂ ◦ Lϕ(z) = λϕz for every z ∈ D, where λϕ ∈ C hasmodulus 1. Moreover, if (zα) is a subnet of {zn} that tends to ϕ, then b◦Lzα→b̂◦Lϕpointwise, and since all the functions b◦Lzα take real values on the interval (−1, 1),then so does b̂◦Lϕ. Therefore λϕ = 1 or −1 (depending on ϕ). It can be shown thatboth possibilities occur, but we are not going to use this fact here. The symmetryof Ωε with respect to the origin implies that b̂◦Lϕ(z) = λϕz ∈ Ωε for every z ∈ Ωε.Consequently Lϕ(Ωε) ⊂ E, and since Lϕ(D) ⊂M\ D, thenLϕ(Ωε) ⊂ E \ D for every ϕ ∈ closM{zn} \ D.(3)Since Ωε is polynomially convex and 0 6∈ Ωε, then Runge’s theorem asserts that thefunction 1/ω is the uniform limit on Ωε of some sequence of polynomials {pn(ω)}.Therefore pn((b(z)) converges uniformly to 1/b(z) on F , showing that 1/b ∈ A∞(F ).We are going to prove that 1/b cannot be decomposed as in (1) with p = ∞.Suppose otherwise that there exist h ∈ H∞ and a function f uniformly continuouson F such that 1/b = h + f on F . By a simple geometrical argument, the lasttwo items of Lemma 1 and (2) imply that closCF = F ∪ {1}. Since f is uniformlycontinuous on F , then it can be extended to some continuous function on F ∪ {1}.That is, there is c ∈ C such that lim f(z) = c when z→1 with z ∈ F . Therefore1/b(z)− (h(z) + c)→0 when z→1 with z ∈ F . Multiplying this expression by b andwriting h1 = h+ c we obtain1− b(z)h1(z)→0 when z→1 with z ∈ F.(4)Suppose that x ∈ E \ D and let (zα) be a net in F that converges to x. It is clearthat if we look at (zα) as a net in the topological space C, then zα→1. It followsfrom (4) that 1 − b̂(x)ĥ1(x) = lim(1 − b(zα)h1(zα)) = 0. That is, 1 − b̂ĥ1 ≡ 0 onE \ D. Therefore, for every ϕ ∈ closM{zn} \ D the inclusion of (3) implies that1− b̂ĥ1 ≡ 0 on Lϕ(Ωε). So, writing gϕ = ĥ1 ◦ Lϕ ∈ H∞, we get0 = 1− (b̂ ◦ Lϕ(z))(ĥ1 ◦ Lϕ(z)) = 1− λϕzgϕ(z) for z ∈ Ωε.Since the only analytic function that vanishes on Ωε is the trivial function, then1− λϕzgϕ(z) = 0 for every z ∈ D, which is clearly impossible (take z = 0).3. More general impossibilitiesA further analysis of our example with the aid of additional theory will rule outmore general decompositions of the same type for 1/b. The arguments are outlinedbelow.3006 DANIEL SUÁREZSuppose that K ⊂ M is a closed set and V ⊂ M is an open neighborhood ofK. Let h ∈ H∞(V ∩ D). The corona theorem easily implies that closM(V ∩ D) isa neighborhood of K. In addition, h can be continuously extended to some openneighborhood of K in M (see [5, Thm. 3.2]).Proposition 2. Let ε(δ), r(δ) as in Lemma 1 and assume the hypotheses of theexample. Then there is no decomposition of the form1/b = h+ f on F,(5)with f ∈ Cua(F ) and h ∈ H∞(⋃K(zn, r(δ))).Proof. Suppose that h ∈ H∞(⋃K(zn, r(δ))) and let η with ε < η < ε(δ). Theset V = {x ∈ M : |b̂(x)| < ε(δ)} is an open neighborhood of the closed setK = {x ∈ M : |b̂(x)| ≤ η}. By Lemma 1 V ∩ D ⊂⋃K(zn, r(δ)), and consequentlythe comments preceding the proposition imply that h can be extended to somecontinuous function on K, say h̃.Since |b(ω)| < η for every ω ∈ K(zn, η) = Lzn(∆(0, η)), then Lϕ(∆(0, η)) ⊂ Kfor every ϕ ∈ closM{zn}. So, h̃ is continuous on{Lϕ(∆(0, η)) : ϕ ∈ closM{zn}}.Let ϕ ∈ closM{zn}\D and (zα) be a subnet of {zn} that tends to ϕ. The continuityof h̃ on the above set implies that h ◦ Lzα(z) tends pointwise to the (necessarilyanalytic) function h̃ ◦ Lϕ(z) for |z| < η.These facts allow us to repeat almost word-by-word the arguments of the previoussection. In fact, if (5) holds with f ∈ Cua(F ) and h ∈ H∞(⋃K(zn, r(δ))), thenby the same reasons as in Section 2 there is c ∈ C such that the function gϕ(z) =(h̃+ c)◦Lϕ(z) is analytic on ∆(0, η) and 1−λϕzgϕ(z) = 0 for z ∈ Ωε (observe thatΩε ⊂ ∆(0, η)). Again, this is impossible.We recall that the Bloch space B consists of the analytic functions f on D suchthat‖f‖Bdef= supz∈D(1− |z|2)|f ′(z)| <∞.It is well known (see [6, p. 81]) that every h ∈ B is uniformly continuous withrespect to the metric ρ. An immediate consequence of Proposition 2 is that thedecomposition (5) is not possible with f ∈ Cua(F ) and h ∈ B. To see this supposethat there is such decomposition. Then h is bounded on F , and since by Lemma1 ρ(zn, F ) < r(δ) < 1 for every n, the uniform continuity of h with respect to ρimplies that h is bounded on {zn}. By the same reason h must be bounded on⋃K(zn, r(δ)). That is, h ∈ H∞(⋃K(zn, r(δ))), which contradicts the proposition.AcknowledgementI am grateful to Arne Stray for pointing out that the counterexample works formore general spaces than H∞.References[1] L. Carleson, Interpolations by bounded analytic functions and the corona theorem, Ann. ofMath. 76 (1962), 547-559. MR 25:5186[2] J. B. Garnett, “Bounded Analytic Functions”, Academic Press, New York (1981).MR 83g:30037A COUNTEREXAMPLE FOR H∞ APPROXIMABLE FUNCTIONS 3007[3] K. Hoffman, Bounded analytic functions and Gleason parts, Ann. of Math. 86 (1967), 74-111.MR 35:5945[4] A. Stray, Mergelyan type theorems for some function spaces, Publicacions Matemàtiques 39(1995), 61-69. MR 96g:30067[5] F. D. Suárez, Čech cohomology and covering dimension for the H∞ maximal ideal space, J.Funct. Anal. 123 (1994), 233-263. MR 95g:46100[6] K. Zhu, “Operator Theory in Function Spaces”, Marcel Dekker, New York and Basel (1990).MR 92c:47031Departamento de Matemática, Facultad de Cs. Exactas y Naturales, UBA, Pab. I,Ciudad Universitaria, (1428) Núñez, Capital Federal, ArgentinaCurrent address: Departamento de Análisis Matemático, Universidad de La Laguna, 38271 LaLaguna, Tenerife, SpainE-mail address: dsuarez@dm.uba.ar

A counterexample for H ∞ approximable functions

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A counterexample for H ∞ approximable functions

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