Closed ideals with countable hull in algebras of analytic functions smooth up to the boundary

Publ. Mat. 52 (2008), 19–56 CLOSED IDEALS WITH COUNTABLE HULL IN ALGEBRAS OF ANALYTIC FUNCTIONS SMOOTH UP TO THE BOUNDARY Cyril Agrafeuil and Mohamed Zarrabi Abstract We denote by Tthe unit circle and by Dthe unit disc. Let Bbe a semi-simple unital commutative Banach algebra of functions holomorphic in Dand continuous on D, endowed with the pointwise product. We assume that Bis continously imbedded in the disc algebra and satisfies the following conditions: (H1) The space of polynomials is a dense subset of B. (H2) limn→+∞kznk1/n B= 1. (H3) There exist k≥0 and C > 0 such that ˛ ˛1− |λ|˛ ˛ k‚ ‚f‚ ‚B≤C‚ ‚(z−λ)f‚ ‚B,(f∈ B,|λ|<2). When Bsatisfies in addition the analytic Ditkin condition, we give a complete characterisation of closed ideals Iof Bwith countable hull h(I), where h(I) = ˘z∈D:f(z) = 0,(f∈I)¯. Then, we apply this result to many algebras for which the structure of all closed ideals is unknown. We consider, in particular, the weighted algebras ℓ1(ω) and L1(R+, ω). 1. Introduction We denote by Tthe unit circle and by Dthe unit disc. Let nbe a nonnegative integer. We denote by an(D) the space of ntimes continuously differentiable functions on Dand holomorphic in D. We set a∞(D) = T n≥0 an(D). Notice that a(D) = a0(D) is the classical disc algebra. Let Bbe a semi-simple unital commutative Banach algebra of functions holomorphic in Dand continuous on D, endowed with the pointwise product. We assume that Bis continuously imbedded in a(D). 2000 Mathematics Subject Classification. 46J20, 46J15. Key words. Closed ideals, Banach algebras, Ditkin condition. 20 C. Agrafeuil, M. Zarrabi We denote by α:z7→ zthe identity map and we define the following conditions: (H1) The space of polynomials is a dense subset of B. (H2) limn→+∞kαnk1/n B= 1. (H3) There exist k≥0 and C > 0 such that (1) 1−|λ|k f B≤C (α−λ)f B,(f∈ B,|λ|<2). Denote by NBthe supremum of integers nsuch that B ⊂ an(D). As we will see in the next section, condition (H3) implies that NBis bounded. We say that Bsatisfies the analytic Ditkin condition if for each z0∈T and each function f∈ B such that f(k)(z0) = 0, 0 ≤k≤NB, there exists a sequence τnn≥0in B, satisfying the following properties: (A1) For all n≥0, τn(z0) = 0. (A2) lim n→+∞ (1 −τn)f B= 0. And we say that Bsatisfies the strong analytic Ditkin condition if, for each z0∈T, the sequence τnn≥0can be chosen independently of f. Denote by H∞the space of all holomorphic and bounded functions in D. For f, g ∈H∞we say that fdivides g(we write f|g) if g/f ∈H∞. Let Ibe a closed ideal of B. We denote by UIthe inner factor of I, that is the greatest inner common divisor of all nonzero functions in I(see [14, p. 85]), and we set hk(I) = nz∈D:f(z) = ···=f(k)(z) = 0,(f∈I)o(0 ≤k≤NB). The ideal Iis said to be standard if (2) I=nf∈ B :UI|fand f(k)= 0 on hk(I)∩T,0≤k≤NBo. In several Banach algebras Bsatisfying conditions (H1)–(H3), all closed ideals are standard. This is the case if Bis for example one of the following algebras: a(D), an(D) (n≥1), λs(0 < s < 1) and Hp n (n≥1, 1 < p < ∞), where λs=nf∈a(D) : f(z)−f(z′)=o(|z−z′|s),|z−z′| → 0o, and Hp n=nfholomorphic in Dand f(n)∈Hpo, Hpbeing the classical Hardy space on D(see respectively [24], [18], [19], [27]). For further examples see also [28] and [30]. Closed Ideals with Countable Hull 21 But, even in the case where NB= 0, we don’t know in general the characterisation of all closed ideals of B. For example, in A+(T), the algebra of all functions in a(D) with absolutely convergent Taylor series, it is known only when h0(I) is finite (see [15]), h0(I) is countable (see [3]) or when h0(I) is the Cantor triadic set (see [7]). And in these cases, ideals are standard. The aim of this paper is to study the structure of closed ideals with countable hull in Banach algebras satisfying conditions (H1)–(H3). The first motivation is Theorem B of [3], which is announced without proof and which concerns ideals with finite hull (see Remark 2.13). The second one is the structure of closed ideals in algebras A+ s(T) = (f∈a(D) :  f s= +∞ X n=0 |b f(n)|(1 + n)s<+∞), where sis a nonnegative real number. The characterisation of closed ideals with countable hull in A+ s(T) is known only in the case s < 1 (see [23]), which correspond to the fact that the derivation operator doesn’t act on A+ s(T). The main result of this paper (Theorem 2.11) shows that if Bsatisfies conditions (H1)–(H3) and the analytic Ditkin condition then every closed ideal Iof Bsuch that h0(I) is at most countable is standard. Notice that a result in the same direction was announced by Fa˘ıvyˇsevski˘ı in [9]. There is no hope to extend this result to all closed ideals of B. Indeed, when B=A+(T), J. Esterle constructed in [6] a closed ideal Isuch that h0(I) is a “thin” set, UI= 1 and I6={f∈A+(T) : f= 0 on h0(I)}. Then we apply this result (Theorem 2.11) to several concrete Banach algebras. In Section 3, we show that for every s≥0, all closed ideals in A+ s(T), with countable hull, are standard. Then, using a result of H. Hedenmalm (see [13]), we deduce a similar characterisation for a family of closed modular ideals of L1 s(R+), where L1 s(R+)=fmeasurable on R+: f 1,s =Z+∞ 0|f(t)|(1 + t)sdt<+∞. In Section 4, we give further examples of Banach algebras where Theorem 2.11 applies. Acknowledgement. We wish to thank the referee for her/his valuable comments and to have drawn our attention to the work of Fa˘ıvyˇsevski˘ı, where results in the same direction are announced. 22 C. Agrafeuil, M. Zarrabi 2. Closed ideals of B Throughout this section, Bwill denote a semi-simple unital commutative Banach algebra of functions holomorphic in Dand continuous on D, endowed with the pointwise product and continuously imbedded in a(D). Without loss of generality we assume that k1kB= 1. We recall that NBis the supremum of integers nsuch that B ⊂ an(D). We will need, in the sequel, some properties of B. Suppose that Bsatisfies conditions (H1) and (H2). We can identify the maximal ideal space of Bwith D. The identification is given by the map D∋z→δz, where δzis the point evaluation at z:δz(f) = f(z), f∈ B. Then, for all z∈D, we have g(z)≤ g B,g∈ B. It follows from the closed graph theorem that the embedding from B into aNB(D) is continuous. In particular, for all z∈Dand for all k, 0≤k≤NB, functionals χ(k) z:g7→ g(k)(z) are continuous on B. For λ∈Dand f∈a(D), we define the function Rλ(f) by Rλ(f)(z) =      f(z)−f(λ) z−λif z∈D\{λ} f′(λ) if z=λ . Clearly Rλis a linear map. Notice also that if Bsatisfies conditions (H1) and (H2) and if for every f∈ B,Rλ(f)∈ B, then the closed graph theorem asserts that Rλis a bounded operator on B. The following lemma shows that condition (H3) can be equivalently reformulated. We recall that α:z7→ zis the identity map. Lemma 2.1. Suppose that Bsatisfies conditions (H1) and (H2). Then, the following conditions are equivalent. (i) Bsatisfies condition (H3). (ii) There exists k≥0such that: (a)  αn B=O(nk), n →+∞. (b) For all λ∈D,Rλdefines a bounded linear operator on Band there exists C > 0such that  Rλ ≤C(1 −|λ|)−k(|λ|<1). Closed Ideals with Countable Hull 23 (iii) There exists k≥0such that: (c)  αn B=O(nk), n →+∞. (d) R0defines a bounded linear operator on Band  Rn 0 =O(nk), n →+∞. Proof: (i) ⇒(ii): Suppose that Bsatisfies condition (H3). We begin by showing (b). Let λ∈Dand Pbe a polynomial function. We can write P−P(λ) = (α−λ)Q, where Qis a polynomial function. So Rλ(P) = Qbelongs to B. We deduce from (H3) that 1−|λ|k Rλ(P) B≤C P−P(λ) B ≤2C P B, where kand Care nonnegative constants independent of P. Let Pbe the space of all polynomials provided with the norm of B. One has just seen that the restriction of Rλto Pis a bounded linear operator with norm not exceeding 2C1− |λ|−k. Now, since Bsatisfies (H1), Rλ|P can be extended to a bounded linear operator on Bwith norm less than 2C1− |λ|−k. Then, using the fact that functionals g7→ g(z), for z∈D, are continuous on B, it is easily seen that this extension coincides with Rλon B, which finishes the proof of part (b). Now, let us show that (a) holds. As we have observed before, the set of maximal ideal space can be identified with D. Therefore the spectrum of αis equal to D. So for all λsuch that |λ|>1, the function α−λis invertible in Band using (1), we get (3)  α−λ−1 B≤C|λ|−1−k,(1 <|λ|<2). Let r > 1, we have αn=1 2iπ Zγr λnα−λ−1dλ, where γrdenotes the circle centered in 0 with radius r. We deduce from this equality and (3) that  αn B≤Crn+1(r−1)−k(1 < r < 2). Now, it suffices to take r= 1 + 1/n to prove (a). (ii) ⇒(iii): It suffices to show that (d) holds. Let f∈ B. For λ∈D we get from part (b) of (ii),  f B= Rλ(α−λ)f B≤C1−|λ|−k (α−λ)f B. 24 C. Agrafeuil, M. Zarrabi It follows from this that for every λand λ′in D,  Rλ(f)−Rλ′(f) B ≤C21−|λ|−k1−|λ′|−k (α−λ)(α−λ′)Rλ(f)−Rλ′(f) B ≤C21−|λ|−k1−|λ′|−k (α−λ′)f−f(λ)−(α−λ)f−f(λ′) B ≤C21−|λ|−k1−|λ′|−k ×|f(λ′)−f(λ)|kαkB+|λ−λ′|kfkB+|λ′f(λ)−λf(λ′)|. Thus the map λ→Rλ(f) is continuous from Dinto B. So, for every n≥1 and r∈(0,1), the integral 1 2iπ Rγrλ−nRλ(f) dλis well defined and belongs to B. We shall now prove that (4) Rn 0(f) = 1 2iπ Zγr λ−nRλ(f) dλ, (n≥1,0< r < 1). For λand zin Dwe have Rλ(f)(z) = X m≥0 ˆ f(m)zm−λm z−λ =X m≥1 ˆ f(m) m−1 X n=0 λnzm−n−1 =X n≥0 X m≥n+1 ˆ f(m)zm−n−1 λn =X n≥0 Rn+1 0(f)(z)λn. (5) So the map λ→Rλ(f)(z) is expanded in an entire series in Dand its Taylor coefficients are Rn+1 0(f)(z)n≥0. Hence Rn 0(f)(z) = 1 2iπ Zγr λ−nRλ(f)(z) dλ, (|z|<1,0< r < 1, n ≥1). Since the linear map δz:f→f(z) is continuous on B, we get Rn 0(f)(z) = 1 2iπ Zγr λ−nRλ(f) dλ(z),(|z|<1,0< r < 1, n ≥1), Closed Ideals with Countable Hull 25 which proves (4). Therefore  Rn 0 ≤1 2πZ2π 0 r−n+1 Rreiθ (f) Bdθ≤Cr−n+1(1 −r)−k. For r= 1 −1 n, we see that (d) holds. (iii) ⇒(i): For |λ|<1, the series Pn≥0λnRn+1 0(f) is absolutely convergent in B. Then, by (5), we deduce that Rλ(f) = Pn≥0λnRn+1 0(f) and Rλ(f)∈ B. Therefore there exist constants Cand C′such that  Rλ(f) B≤X n≥0|λ|nkRn+1 0(f)kB ≤C X n≥0|λ|n(1 + n)k kfkB≤C′(1 −|λ|)−k−1kfkB. Using this for (α−λ)f, we obtain  f B= Rλ(α−λ)f B≤C′1−|λ|−k−1 (α−λ)f B. Now, let |λ|>1. The function α−λis invertible in Band we have (6)  f B≤ (α−λ)−1 B (α−λ)f B,(f∈ B). Then, it suffices to expand (α−λ)−1in series and use (c) to see that there exists a constant Csuch that (7)  (α−λ)−1 B≤X n≥0kαnkB|λ|−n−1≤C|λ|−1−k−1,(1 <|λ|<2). It follows from (6) and (7) that  f B≤C|λ|−1−k−1 (α−λ)f B,(1 <|λ|<2). Thus condition (H3) is satisfied. Remark 2.2.Suppose that Bsatisfies conditions (H1) and (H2). Condition (a) in (ii) of Lemma 2.1 can be equivalently reformulated. Assume that a∞(D)⊂ B. By the closed graph theorem the imbedding a∞(D)֒→ B is continuous. So there exist k≥0 and a constant C > 0 such that kfkB≤Ckfkak(D), whenever f∈a∞(D). In particular we have kαnkB=O(nk), n→ ∞. This implies that ak+2(D)⊂ B. Now it is easily seen that the following conditions are equivalent: (a) There exists k≥0 such that  αn B=O(nk), n→+∞. (a’) a∞(D)⊂ B. (a”) There exists k≥0 such that ak(D)⊂ B. 26 C. Agrafeuil, M. Zarrabi To show the main result of this paper we begin with the characterisation of closed ideals Isuch that UI= 1 and h0(I) is reduced to a single point. For this we need some lemmas. We denote by Ck(T) the space of ktimes continuously differentiable functions on T. Let f∈ B. If z∈D, we set kf(z) =    supnk≥0 : f∈ Ck(T) and f(k)(z) = 0oif z∈T supnk≥0 : f(k)(z) = 0oif z∈D and mf(z) = supnm≥1 : f=f1. . . fm with fi∈ B and fi(z) = 0 (1 ≤i≤m)o, with the understanding that sup ∅=−∞. We remark that kf(z)≥ min(mf(z)−1, NB). Also if z∈Dand Bsatisfies conditions (H1)–(H3), then kf(z) = mf(z)−1. Lemma 2.3. Suppose that Bsatisfies conditions (H1) and (H2). Let f∈ B,z0∈Tand kan integer with 0≤k≤NB. If kf(z0)≥k, then there exists a sequence Pmm≥0of polynomial functions such that lim m→+∞ f−(α−z0)k+1Pm B= 0. Proof: Since polynomial functions are dense in B, there exists a sequence Qmm≥0of polynomial functions such that lim m→+∞ f−Qm B= 0. For all m≥0, set Rm=Qm−Qm(z0)−Q′ m(z0)(α−z0)−···− Q(k) m(z0) k!(α−z0)k. For every integer j, 0 ≤j≤NB, the functional g7→ g(j)(z0) is continuous on B, so lim m→+∞Q(j) m(z0) = f(j)(z0) = 0. Hence, we have lim m→+∞ f−Rm B= 0. Now, since for all m≥0, the polynomial function Rmvanishes, with all its derivatives of order less or equal than k, at z0, there exists a polynomial function Pmsuch that Rm= (α−z0)k+1Pm, which concludes the proof. Closed Ideals with Countable Hull 27 Lemma 2.4. Suppose that Bsatisfies conditions (H1) and (H2). Let f∈ B,z0∈Tand kbe a nonnegative integer. If mf(z0)≥k, then there exists a sequence Pmm≥0of polynomial functions such that lim m→+∞ f−(α−z0)kPm B= 0. Proof: Assume that mf(z0)≥k. There exist functions f1,...,fkin B vanishing at z0such that f=f1...fk. It follows from Lemma 2.3 that for each i∈ {1,...,k}there exists a sequence of polynomials (Pi,m)m such that limm→+∞ fi−(α−z0)Pi,m B= 0. If for each m, we set Pm= P1,m . . . Pk,m then we have clearly limm→+∞ f−(α−z0)kPm B= 0. Lemma 2.5. Suppose that Bsatisfies conditions (H1) and (H2) and the analytic Ditkin condition. Let z0∈Tand f∈ B such that kf(z0)≥NB. Then, for all m≥1, there exists a sequence σnn≥0included in Bsuch that for each n,mσn(z0)≥m, and lim n→+∞ σnf−f B= 0. Proof: Since Bsatisfies the analytic Ditkin condition, there exists a sequence τnn≥0included in Bsatisfying conditions (A1) and (A2) associated with the function f. This gives the conclusion in the case where m= 1. Now, suppose that m≥2. We construct by induction on ka sequence τn1,...,nk(n1,...,nk)∈∪1≤k≤mNksuch that for every k∈ {1,...,m−1}and (n1, . . . , nk)∈Nkthe sequence τn1,...,nk,nn≥0 satisfies conditions (A1) and (A2) related to the function τn1,...,nkf, that is τn1,...,nk,n(z0) = 0,for every n≥0, and lim n→+∞ τn1,...,nk,nτn1,...,nkf−τn1,...,nkf B= 0. To simplify, we assume that m= 2 (the general case can be proved exactly in the same way). It is easily seen that there exist ϕ1(n)n≥0 and ϕ2(n)n≥0two increasing sequences of positive integers such that, for all n≥0,  τϕ1(n)f−f B≤1 n+ 1 and  τϕ1(n),ϕ2(n)τϕ1(n)f−τϕ1(n)f B≤1 n+ 1. Now, set σn=τϕ1(n),ϕ2(n)τϕ1(n)(n≥0). 34 C. Agrafeuil, M. Zarrabi Clearly I(U;E0,...,ENB) is a closed ideal of B. The following theorem is the main result of this paper. We recall that Bis a semi-simple unital commutative Banach algebra of functions holomorphics in Dand continuous on D, endowed with pointwise product and continuously imbedded in a(D). Theorem 2.11. Assume that Bsatisfies conditions (H1)–(H3) and the analytic Ditkin condition. Let Ibe a closed ideal of Bsuch that h0(I)is at most countable. Then Iis standard. Proof: We have obviously I⊂I0, where I0=I(UI;h0(I)∩T,..., hNB(I)∩T). Let g∈I0. To prove that g∈I, we are going to prove that I(g) = B. Notice that since I⊂I(g), we have h0(I(g)) ⊂h0(I). By Lemma 2.10, we have UI(g)= 1 and so h0(I(g)) ⊂h0(I)∩T. Let z0∈h0(I)\hNB(I)∩T. Then, z0is an isolated point of h0(I). So, by the idempotent Shilov theorem (see [10]), there exists ψ∈ B such that (ψ(z0) = 1 ψ= 0 on h0(I)\{z0}and ψ(1 −ψ)∈I. As I⊂I(ψ), we have, for all k∈ {0,...,NB},hk(I(ψ)) ⊂hk(I). Moreover, since 1 −ψ∈I(ψ) and 1 /∈I(ψ), we have h0(I(ψ)) = {z0}. And, as z0/∈hNB(I), hNB(I(ψ)) = ∅. So we deduce from Lemma 2.8 that UI(ψ)= 1. Consequently, we deduce from Lemma 2.7 that I(ψ) = nf∈ B :f(k)(z0) = 0 0≤k≤kI(ψ)(z0)o. Since kg(z0)≥kI(z0)≥kI(ψ)(z0), we get in particular that g∈I(ψ). This proves that ψ∈I(g), so that z0/∈h0(I(g)). Therefore, we have proved that h0(I(g)) ⊂hNB(I). Now, suppose that h0(I(g)) 6=∅. Since h0(I(g)) is at most countable, there exists z0an isolated point of h0(I(g)). Using once more the idempotent Shilov Theorem, there exists ϕ∈ B such that (ϕ(z0) = 1 ϕ= 0 on h0(I(g))\{z0}and ϕ(1 −ϕ)∈I(g). Let J:= I(g)(ϕ) be the division ideal of I(g) by ϕ. As z0∈hNB(I), gvanishes with all its derivatives (of order less or equal than NB) at z0. And since Bsatisfies the analytic Ditkin condition, we deduce from Closed Ideals with Countable Hull 35 Lemma 2.5 that there exists a sequence σnn≥0included in Bsuch that mσn(z0)≥NB+ 1 and (16) lim n→+∞ σnϕg −ϕg B= 0. Now, since I(g)⊂J, we deduce from Lemma 2.10 that UJ= 1. Moreover, since 1 −ϕ∈J, we have h(J)⊂ {z0}. Then, Lemma 2.7 gives the characterisation of J. In particular, by Lemma 2.4, σn∈J,n≥0. This means that, for all n≥0, σnϕg ∈I. Hence, since Iis closed, we deduce from (16) that ϕg ∈I. In other words, ϕ∈I(g), which is in contradiction with the fact that z0∈h0(I(g)). Finally, we have proved that h0(I(g)) = ∅, that is I(g) = B. Let Hbe a Banach algebra of continuous functions on T. We say that His an homogeneous Banach algebra if it satisfies the following conditions: (1) His a semi-simple and commutative Banach algebra with maximal ideal space T. (2) The set of trigonometric polynomials is a dense subset of H. (3) For all f∈ H and τ∈R, we have  fτ H= f H, where fτ(eit) = f(ei(t−τ)). Let Hbe an homogeneous algebra on T. We denote by H+the closed subalgebra of Hgenerated by {eint :n≥0}. Notice that conditions (2) and (3) implies that for every τ0∈R,  fτ−fτ0 H→0, as τ→τ0. It follows then from Theorem 2.12 of [16] that H+consists of functions in Hhaving analytic extension into D. Then, the maximal ideal space of H+is D. Recall that we denote by Cn(T) the space of ntimes continuously differentiable functions on Tand we set C∞(T) = T n≥0Cn(T). Suppose that Hcontains C∞(T), which implies in particular that His a regular algebra. Also this implies that Ck(T)⊂ Hfor some nonnegative integer k. It follows that NH, the greatest of the integers nfor which H ⊂ Cn(T), is bounded. We say that Hsatisfies the Ditkin condition if for each function f∈ H such that f(k)(1) = 0, 0 ≤k≤NH, there exists a sequence vnn≥0in Hsatisfying the following properties: (1) For all n≥0, vn= 0 on a neighbourhood of 1. (2) lim n→+∞ (1 −vn)f H= 0. We say that Hsatisfies the analytic Ditkin condition if H+satisfies the analytic Ditkin condition. 36 C. Agrafeuil, M. Zarrabi Corollary 2.12. Let Hbe an homogeneous Banach algebra on Tthat contains C∞(T)and satisfies the analytic Ditkin condition. Let Ibe a closed ideal of H+such that h0(I)is at most countable. Then Iis standard. Proof: By Theorem 2.11 it suffices to check that H+satisfies conditions (H1)–(H3). This is clearly the case for (H1) and (H2). It remains to check condition (H3). As we have observed before there exists a nonnegative integer ksuch that Ck(T)⊂ H. By the closed Graph theorem this imbedding is continuous. So there exists a constant Csuch that  f B≤C f Ck(T), f ∈ Ck(T). In particular we have  αn =O(|n|k), |n| → ∞. So there exists a constant Csuch that for every λ∈D, k(α−λ)−1kH≤X n≥0kα−n−1kH|λ|n≤C(|1−|λ|)−k−1. So, for every λ∈Dand f∈ H+,Rλ(f)∈ H+and  Rλ(f) H≤ k(α−λ)−1kHkf−f(λ)kH≤2C(|1−|λ|)−k−1kfkH. It follows now from Lemma 2.1 that H+satisfy condition (H3). Remark 2.13.1) Let Hbe an homogeneous Banach algebra on Tthat contains C∞(T) and such that H+satisfies the analytic Ditkin condition. It is announced in [3] that if Iis a closed ideal of H+such that h0(I) is a finite set then I=IH∩UIH∞(D), where IHis the closed ideal of Hgenerated by I. 2) Let Hbe an homogeneous Banach algebra on Tsuch that kαnkH=O(nk) (n→+∞),for some k≥0, and lim n→+∞ log kα−nkH √n= 0. We can show that if Iis a closed ideal of Hsuch that ˜ h0(I) is countable then I={f∈ H :f(k)= 0 on ˜ hk(I),(0 ≤k≤NH)}, where ˜ hk(I) = {z∈T:f(z) = ···=f(k)(z) = 0,(f∈I)}. The proof of this result uses similar arguments as in the case of Theorem 2.11 with some simplications, since there is no problem in this case with the inner factor. Closed Ideals with Countable Hull 37 3. Closed ideals in ℓ1(ω) and L1(R+, ω) Let sbe a nonnegative real number. We set ωs(t) = (1 + t)s,t≥0. Denote by ℓ1(ωs) the set of all complex sequences x= (xn)n≥0such that kxnks:= X n≥0|xn|ωs(n)<+∞. Clearly ℓ1(ωs), equipped with convolution product and norm k ks, is a semi-simple unitary commutative Banach algebra. The set of maximal ideals of ℓ1(ωs) can be identified with D. Considering the Gelfand transfom x→ˆx, where ˆx(z) = Pn≥0xnzn,z∈D, we see that ℓ1(ωs) is isometrically isomorphic to the Beurling weighted Banach algebra A+ s(T) =   f∈a(D) : kfks:= X n≥0|ˆ f(n)|ωs(n)<+∞  . So we will just be interested by closed ideals of A+ s(T). The structure of closed ideals Iof A+ s(T) such that h0(I) is countable, is known only in the cases when s < 1 (see [23], [3] and [7]) and when s≥1 and UI= 1 (see [1]). Here we give a complete characterisation of such ideals. With notation of the above section, we have A+ s(T) = As(T)+where As(T) = (f∈ C(T) : kfks:= X n∈Z|ˆ f(n)|(1 + |n|)s<+∞). Clearly As(T) is an homogeneous Banach algebra on Tthat contain C∞(T). Moreover by Proposition 2.4 of [1], A+ s(T) satisfies the analytic Ditkin condition. Then we deduce immediately from Corollary 2.12 the following result. Theorem 3.1. Let sbe a nonnegative real number, and Ibe a closed ideal of A+ s(T)such that h0(I)is at most countable. Then Iis standard. Now we turn to the Banach algebra L1(R+, ωs)=fmeasurable on R+: f 1,s =Z+∞ 0|f(t)|ωs(t) dt<+∞. L1(R+, ωs),kk1,s, equipped with the convolution product, is a Banach algebra. We denote by Pthe open half plane {z∈C: Re(z)>0}. For f∈L1(R+), we define the Laplace transform L(f) by L(f)(z) = Z+∞ 0 f(t)e−tz dt, (z∈P). 38 C. Agrafeuil, M. Zarrabi Let Jbe a closed ideal of L1(R+, ωs). We set hk(J)=nz∈P:L(f)(z)=···=L(f)(k)(z)=0,(f∈J)o,(0≤k≤[s]), where [s] is the integer such that [s]≤s < [s] + 1. And we denote by TJits inner factor, that is the greatest common divisor of all Laplace transform of nonzero functions in J. Also we set βJ= inff∈Jβf, where βf= sup{a:f= 0 on [0, a] (almost everywhere)}. Notice that J=L1(R+, ωs) if and only if h0(I) = ∅and βJ= 0 ([22], [11]). We say that Jis modular if L1(R+, ωs)/J admits a unit. It is well-known that Jis a closed modular ideal of L1(R+, ωs) if and only if h0(J) is compact and βJ= 0 (in case s= 0, see for example [4, Proposition 2.1]). When s= 0, V. P. Gurari˘ı showed in [11] that if Jis a closed ideal of L1(R+) = L1(R+, ω0) such that h0(J) is at most countable, then (17) J=L−1L(J)∞, where L(J)∞is the closure of L(J) in A0(P), the algebra of all functions fwhich are continuous on P, holomorphic in P, and such that f(z)→0 as |z| → +∞. The closed ideals of A0(P) are given by the Beurling-Rudin theorem, thanks to a conformal transform between P and D. So, if h0(J) is at most countable, we have J=nf∈L1(R+) : TJ| L(f) and L(f) = 0 on h0(J)∩iRo. Using a result of H. Hedenmalm, we will deduce from Theorem 3.1 a similar result concerning closed ideals of L1(R+, ωs), for any nonnegative real number s. If Iis a closed ideal of A+ s(T), we denote by π=πIthe canonical surjection from A+ s(T) onto A+ s(T)/I. If h0(I)⊂D\[−1,0], we can define the function z7→ π(α)z, which is an entire function of finite exponential type. Furthermore, we have  π(α)t s≤sup 0≤u≤1 π(α)u s(1 + [t])s ≤sup 0≤u≤1 π(α)u s(1 + t)s(t≥0). Then, we define ΦI:L1(R+, s)→A+ s(T)/I by ΦI(f) = Z+∞ 0 f(t)π(α)tdt. Closed Ideals with Countable Hull 39 ΦIis a continuous homomorphism. We set I=nIclosed ideals of A+ s(T) : h0(I)⊂D\[−1,0]o and J=nJclosed modular ideals of L1(R+, s) : h0(J)⊂[0,+∞)×(−π, π)o. The following result, due to H. Hedenmalm ([13, Theorem 4.3 and Remark 4.5]), is essential for our purpose, since it gives a one-to-one correspondence between Iand J(see also [4, Theorem 2.2]). Theorem 3.2 ([13]).The mapping Θ: I7→ ker ΦIdefines a bijection between Iand J. Futhermore, if J∈ J, we have Θ−1(J) = (f∈A+ s(T) : +∞ X n=0 b f(n)δn!∗g∈J, g∈L1(R+, ωs)), where δnis the Dirac measure at n. Until the end of this paper, X∗will denote the dual of a normed space X, and the duality will be denoted by hf, gi(f∈X, g ∈X∗). Let Ibe a closed ideal of A+ s(T) and π=πIthe canonical surjection from A+ s(T) onto A+ s(T)/I. Let 0 ≤k≤[s] and w∈hk(I). The functional χ(k) w:g7→ g(k)(w) is continuous on A+ s(T) and we have I⊂ ker χ(k) w. So there exists ˜χ(k) w∈A+ s(T)/I∗such that hπ(g),˜χ(k) wi=hg, χ(k) wi(g∈A+ s(T)). Let Ψ be an analytic function on a neighbourhood of h0(I). We use the Dunford-Schwarz functional calculus to define Ψ(π(α)) by Ψ(π(α)) = 1 2iπ Z∂Ω Ψ(ξ)ξ−π(α)−1dξ, where ∂Ω is the boundary of a suitable open neighborhood Ω of h0(I) in which Ψ is holomorphic. Then, we have the following lemma. Lemma 3.3. Let Ibe a closed ideal of A+ s(T). Let Ψbe an analytic function on a neighbourhood of h0(I). Then hΨ(π(α)),˜χ(k) wi= Ψ(k)(w) (0 ≤k≤[s], w ∈hk(I)). Proof: We have Ψ(π(α)) = 1 2iπ Z∂Ω Ψ(ξ)ξ−π(α)−1dξ. 40 C. Agrafeuil, M. Zarrabi Since ˜χ(k) wis a continuous functional, we have (18) hΨ(π(α)),˜χ(k) wi=1 2iπ Z∂Ω Ψ(ξ)ξ−π(α)−1,˜χ(k) wdξ. Hence, the result will follow from the Cauchy integral formulae, if we prove that ξ−π(α)−1,˜χ(k) w=k!(ξ−w)−k−1. Let ξ∈∂Ω and g∈A+ s(T) such that π(g)ξ−π(α)= 1. Since, w∈hk(I), we have the following relations: g(w)(ξ−w) = 1 and g(j)(w)(ξ−w)−jg(j−1)(w) = 0,(1 ≤j≤k). So, we deduce easily from these equalities that g(k)(w) = k!(ξ−w)−k−1. Then, we have ξ−π(α)−1,˜χ(k) w=hπ(g),˜χ(k) wi =g(k)(w) =k!(ξ−w)−k−1. Then the result follows from the above equality, (18) and the Cauchy integral formulae. We also need the two following lemmas. Lemma 3.4. Let I∈ I and J= Θ(I). Then, we have hk(I) = e−hk(J),(0 ≤k≤[s]). Proof: Let 0 ≤k≤[s], w∈hk(I) and f∈J. We have ΦI(f) = 0 and in particular, (19) Z+∞ 0 f(t)π(α)tdt, ˜χ(j) w= 0 (0 ≤j≤k). Now, applying Lemma 3.3 with Ψ(z) = zt(t≥0), we get, for all 0 ≤ j≤k, Z+∞ 0 f(t)π(α)tdt, ˜χ(j) w=Z+∞ 0 f(t)hπ(α)t,˜χ(j) widt =Z+∞ 0 f(t)t(t−1) ...(t−j+ 1)wt−jdt, (20) Closed Ideals with Countable Hull 41 with the understanding that t(t−1) ...(t−j−1) = 1 if j= 0. So, we easily deduce from (19) and (20) that, for all 0 ≤j≤k, L(f)(j)(−log w) = (−1)jZ+∞ 0 f(t)tjwtdt= 0. This proves that −log w∈hk(J), and so hk(I)⊂e−hk(J). Now, let w∈hk(J) and f∈I. For g∈L1(R+, ωs) we set Fg= +∞ P n=0 b f(n)δn∗g. We have L(Fg)(z)= +∞ X n=0 b f(n)Z+∞ 0 (δn∗g)(t)e−tz dt =f(e−z)L(g)(z),z∈P, g ∈L1(R+, ωs). (21) According to Theorem 3.2, we have Fg∈J, for all g∈L1(R+, ωs). It follows that (22) L(Fg)(j)(w) = 0,0≤j≤k, g ∈L1(R+, ωs). Now, we easily deduce from (21) and (22) that f(j)(e−w) = 0,(0 ≤j≤k). This proves that e−hk(J)⊂hk(I), and concludes the proof of this lemma. Lemma 3.5. Let J, J′∈ J. Then, we have TJ=TJ′⇐⇒ UΘ−1(J)=UΘ−1(J′). Proof: Let J∈ J and set I= Θ−1(J). We begin to prove that (23) J0= Θ0(I0), where J0(resp. I0) is the closure of Jin L1(R+) (resp. A+(T)) and where Θ0is the map Θ corresponding to the case s= 0. Clearly J0and I0are closed ideals. Moreover we have TJ0=TJand UI0=UI. By Theorem 3.2 we have I=(f∈A+ s(T) : +∞ X n=0 b f(n)δn!∗g∈J, (g∈L1(R+, ωs))). 42 C. Agrafeuil, M. Zarrabi Since, for every g∈L1(R+, ωs), the map f→+∞ P n=0 b f(n)δn∗gis continuous from A+(T) in L1(R+), we get I0⊂(f∈A+(T) : +∞ X n=0 b f(n)δn!∗g∈J0,(g∈L1(R+, ωs))). Also since, for every f∈A+(T), the map g→+∞ P n=0 b f(n)δn∗gis continuous from L1(R+) in itself and since L1(R+, ωs) is dense in L1(R+), we obtain I0⊂(f∈A+(T) : +∞ X n=0 b f(n)δn!∗g∈J0,(g∈L1(R+))). This means that I0⊂Θ−1 0(J0). Since Θ0is increasing with respect to inclusion order we have Θ0(I0)⊂J0. It remains to prove the other inclusion. Denote by ψthe homomorphism from A+ s(T)/I into A+(T)/I0defined by ψπI(f)=πI0(f). Since ψis an homomorphism, it is easily seen that Φ0 I0|A+ s(T)=ψ◦ΦI, where Φ0 I0is the map ΦIcorresponding to s= 0 and I=I0. Hence, we have ker ΦI⊂ker Φ0 I0∩A+ s(T)⊂ker Φ0 I0. In other words, we have J⊂Θ0(I0). Since Θ0(I0) is closed in L1(R+), we get J0⊂Θ0(I0). Finally, we have proved (23). Let J′∈ J,I′= Θ−1(J′) and J′ 0(resp. I′ 0) be the closure of J′ (resp. I′) in L1(R+) (resp. A+(T)). We have J′ 0= Θ0(I0). Denote by H∞(D) (resp. H∞(P)) the space of bounded holomorphic function on D(resp. P). Denote also by A0(P) the space of holomorphic functions on P, continuous on P, and vanishing at infity. It follows from Lemma 3.7 of [4] and from (23) that Θ0UI0H∞(D)∩A+(T)=L−1TJ0H∞(P)∩A0(P) and Θ0UI′ 0H∞(D)∩A+(T)=L−1TJ′ 0H∞(P)∩A0(P). Thus we see that UI0=UI′ 0if and only if TJ0=TJ′ 0. This finishes the proof since UI0=UI,UI′ 0=UI′,TJ0=TJand TJ′ 0=TJ′. Closed Ideals with Countable Hull 43 Theorem 3.6. Let sbe a nonnegative real number, and Jbe a closed ideal of L1(R+, ωs)such that h0 +(J)is compact and at most countable. Then J=nf∈L1(R+, ωs) : TJ| L(f)and L(f)(k)= 0 on hk(J)∩iR,(0 ≤k≤[s])o. Proof: We first suppose that Jis modular. For a real number a6= 0 and f∈L1(R+, ωs) we set Ta(f)(x) = af(ax), x∈R;Tais an isomorphism of L1(R+, ωs). So using this transformation we may assume without loss of generality that h0 +(J) is contained in [0,+∞)×(−π, π). Set J0=nf∈L1(R+, ωs) : TJ| L(f) and L(f)(k)= 0 on hk(J),(0 ≤k≤[s])o. We also set I= Θ−1(J) and I0= Θ−1(J0). By definition of J0, we have TJ=TJ0. So, we deduce from Lemma 3.5 that (24) SI=SI0. Furthermore, for all 0 ≤k≤[s], we have hk(J) = hk(J0). So we deduce from Lemma 3.4 that (25) hk(I) = hk(I0),(0 ≤k≤[s]). It follows then from (24), (25) and Theorem 3.1 that I=I0. Since Θ is a bijection, we have J=J0. Suppose now that Jis not modular. We set J1=δ−βI∗J;J1is a closed ideal and we have βJ1= 0, TJ(z) = e−βIzTJ1(z), (z∈P) and hk(J1) = hk(J), (0 ≤k≤[s]). So J1is modular and applying the previous result to J1, we obtain the desired equality for J. 4. Further examples In this section, we give three further examples of algebras where, to our knowledge, the structure of closed ideals with countable hull has not been already given. For these algebras, the most difficult part is to prove that they satisfy the analytic Ditkin condition. Indeed, the fact that they satisfy conditions (H1)–(H3) is quite easy to verify and so, left to the reader. 50 C. Agrafeuil, M. Zarrabi Let β≥0 be a real number and j≥0 an integer. Then there exists a constant C > 0 such that for every real number xwith 0 ≤x < 1, (34) +∞ X k=j (1 + k)βxk≤Cxj 1−xmax j+ 1,1 1−xβ . Applying this inequality for β=sp and x=n n+1 sp−p/p′ and using the fact that n+1 nj−1≤n+1 njmin 1,j n, we obtain (35)  (en−1)f p. +∞ X j=0 min 1,j nsp |ˆ f(j)|pmax j+ 1, n + 1sp + (n+1)−p +∞ X j=0 (1 + j)sp|ˆ f(j)|p j−1 X k=0 n n+ 1k(sp−p/p′) (1+k)p/p′ !. Using again (34), we obtain j−1 X k=0 n n+ 1k(sp−p/p′) (1 + k)p/p′≤ +∞ X k=0 n n+ 1k(sp−p/p′) (1 + k)p/p′ .(n+ 1)p/p′+1 = (n+ 1)p. It follows now from (35) and the Lebesgue dominated convergence theorem that lim n→∞ (en−1)f = 0. Remark 4.3.For s= 1 + 1/p′, we have NFℓp(N,s)= 0. But in this case we have  (en−1)(α−1) p=1 (n+ 1)p+1 np(n+ 1)p +∞ X k=1 n n+ 1kp (1 + k)αp which implies that  (en−1)(α−1) 90,as n→0. We get this with the help of the following inequality: 1 (1 −x)β+1 ≤1 + (1 + β)X k≥1 kβxk,(β≥0,0≤x < 1). Closed Ideals with Countable Hull 51 4.3. The algebra λω.Let wbe a nonnegative nondecreasing subadditive function on (0,+∞) such that w(0+) = 0. Let pbe a nonnegative integer. We define the Banach algebra λ(p) w=nf∈ap(D) : f(p)(z)−f(p)(z′)=ow(|z−z′|),|z−z′| → 0o, equipped with the norm  f λw,p= f Cp(T)+supz,z′∈Tf(p)(z)−f(p)(z′) w(|z−z′|). We set λ(0) w=λw. When p≥1, all closed ideals of λ(p) ware standard, but when p= 0, this holds if w(x) = O(xα), x→0, for some α∈(0,1] or if wsatisfies the Zygmund condition, that is Zδ 0 w(x) xdx=o(w(δ)), δ →0 (see [28], [30], [19]). For α > 0, we set wα(x) = 1 |log 1 x|+ 1α, x > 0. There is no characterisation of closed ideals of the algebras λwα(see the introduction of [30]). For this reason we consider here this algebra. Let us introduce the following condition: (36) There exists C > 0,such that w(y) w(x)≤Cy x,0< x ≤y. This condition is satisfied by concave functions and in particular by functions wα,α > 0. Lemma 4.4. Assume that wsatisfies condition (36). Then λwsatisfies the strong analytic Ditkin condition. Proof: We have Nλw= 0. Let f∈λwsuch that f(1) = 0, we have to prove that (37) lim n→+∞sup z,z′∈T(en−1)f(z)−(en−1)f(z′) w(|z−z′|)= 0. For z, z′∈Tand n≥1, set An(z, z′) = (en−1)f(z)−(en−1)f(z′) w(|z−z′|). Let ε > 0. Since f∈λw, there exists η > 0 such that (38) |z−z′| ≤ η=⇒f(z)−f(z′) w(|z−z′|)≤ε. 52 C. Agrafeuil, M. Zarrabi Let z, z′∈T. We will distinguish two cases: Case 1: |z−1|>η 2and |z′−1|>η 2. An easy calculation shows that An(z, z′) = Bn(z, z′)+Cn(z, z′), where Bn(z, z′) = (en(z)−1)f(z)−f(z′) w(|z−z′|) (39) and Cn(z, z′) = f(z′) w(|z−z′|)en(z)−en(z′).(40) Since lim n→+∞sup |ξ−1|>η 2en(ξ)−1= 0, it is easily seen that there exists n1≥0 such that, for all n≥n1,|Bn(z, z′)| ≤ ε. On the other hand side we have en(z)−en(z′) = z−z′ n(z−1−1/n)(z′−1−1/n), which implies that en(z)−en(z′)≤4|z−z′| nη2. Moreover condition (36), implies that |z−z′| w(|z−z′|)≤C′, where C′=2 Cw(2) . So there exists n2≥0 such that, for all n≥n2,|Cn(z, z′)| ≤ ε. So, in this case, there exists n3= max{n1, n2}such that, for all n≥n3,|An(z, z′)| ≤ 2ε. Case 2: |z−1| ≤ η 2or |z′−1| ≤ η 2. Without loss of generality, we can suppose that |z′−1| ≤ η 2. If |z−z′| ≤ η, we deduce from (38) that |Bn(z, z′)| ≤ 3ε. If |z−z′|> η, then |z−1|>η 2, and we can find, as in the previous case, n4≥0 such that, for all n≥n4,|Bn(z, z′)| ≤ ε. Now, we have to estimate Cn(z, z′). Since |z′−1| ≤ η 2and f(1) = 0, we have by (38), |f(z′)| ≤ εω(|z′−1|), and so (41) Cn(z, z′)≤εω(|z′−1|) ω(|z−z′|)|z−z′| n|z−1−1/n||z′−1−1/n|. Closed Ideals with Countable Hull 53 Suppose that |z′−1| ≥ |z−z′|. By condition (36), we have w(|z′−1|) w(|z−z′|)≤ C|z′−1| |z−z′|, so that Cn(z, z′)≤Cε |z′−1| n|z−1−1/n||z′−1−1/n| ≤Cε. Suppose that |z′−1|<|z−z′|. Then, since wis nondecreasing, w(|z′−1|) w(|z−z′|)≤1. Furthermore, a standard computation shows that |z−z′| n|z−1−1/n||z′−1−1/n|≤ |en(z)|+|en(z′)| ≤ 2, (n≥1 and z, z′∈T). So we deduce from (41) that |Cn(z, z′)| ≤ 2ε. In this case, for all n≥n4, we have |An(z, z′)| ≤ C1ε, where C1= 3 + max{2, C}. Finally, if we set n0= max{n3, n4}, we have |An(z, z′)| ≤ C1ε, (n≥n0and z, z′∈T), which proves (37). Theorem 4.5. Let Bbe one of the following algebras: (a) H1 m, for m≥1an integer. (b) Fℓp(N, s), for 1< p < +∞and 0< s −1 p′<1. (c) λω, for ωsatisfying condition (36). Let Ibe a closed ideal of Bsuch that h0(I)is at most countable. Then Iis standard. Proof: It is easily checked that Bis continuously imbedded in a(D) and satisfies conditions (H1)–(H3). Now the theorem follows immediately from Lemmas 4.1, 4.2, 4.4, and Theorem 2.11. References [1] C. Agrafeuil, Id´eaux ferm´es de certaines alg`ebres de Beurling et application aux op´erateurs `a spectre d´enombrable, Studia Math. 167(2) (2005), 133–151. [2] A. 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Shirokov, Standard ideals of the algebra H1 n, (Russian), Funktsional. Anal. i Prilozhen. 13(1) (1979), 86–87; translation in: Functional Anal. Appl. 13(1) (1979), 73–74. [30] N. A. Shirokov, Closed ideals of algebras of type Bα pq, (Russian), Izv. Akad. Nauk SSSR Ser. Mat. 46(6) (1982), 1316–1332, 1344; translation in: Math. USSR-Izv. 21 (1983), 585–600. [31] G. E. ˇ Silov, Homogeneous rings of functions, Amer. Math. Soc. Transl. 92 (1953), 65 pp. [32] B. A. Taylor and D. L. Williams, Ideals in rings of analytic functions with smooth boundary values, Canad. J. Math. 22 (1970), 1266–1283. Cyril Agrafeuil: 59, rue du Moulin-Vert 75014 Paris France E-mail address:[email protected] Mohamed Zarrabi: Universit´e Bordeaux 1 Laboratoire Bordelais d’Analyse et de G´eom´etrie (UMR 5467) 351, cours de la Lib´eration 33405 Talence Cedex France E-mail address:[email protected] Primera versi´o rebuda el 13 de febrer de 2006, darrera versi´o rebuda el 19 de setembre de 2007.