CLOSED IDEALS WITH COUNTABLE HULL IN ALGEBRAS OF ANALYTIC FUNCTIONS SMOOTH UP TO THE BOUNDARY

We denote by T the unit circle and by D the unit disc. Let B be a semi-simple unital commutative Banach algebra of functions holomorphic in D and continuous on D, endowed with the pointwise product. We assume that B is continously imbedded in the disc algebra and satisfies the following conditions: (H1) The space of polynomials is a dense subset of B. (H2) limn→+∞ ‖z‖ 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 B satisfies in addition the analytic Ditkin condition, we give a complete characterisation of closed ideals I of B with 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 `(ω) and L(R, ω).


Introduction
We denote by T the unit circle and by D the unit disc. Let n be a nonnegative integer. We denote by a n (D) the space of n times continuously differentiable functions on D and holomorphic in D. We set a ∞ (D) = n≥0 a n (D). Notice that a(D) = a 0 (D) is the classical disc algebra.
Let B be a semi-simple unital commutative Banach algebra of functions holomorphic in D and continuous on D, endowed with the pointwise product. We assume that B is continuously imbedded in a(D). We denote by α : z → z the identity map and we define the following conditions: (H1) The space of polynomials is a dense subset of B. (H2) lim n→+∞ α n 1/n B = 1. (H3) There exist k ≥ 0 and C > 0 such that Denote by N B the supremum of integers n such that B ⊂ a n (D). As we will see in the next section, condition (H3) implies that N B is bounded.
We say that B satisfies the analytic Ditkin condition if for each z 0 ∈ T and each function f ∈ B such that f (k) (z 0 ) = 0, 0 ≤ k ≤ N B , there exists a sequence τ n n≥0 in B, satisfying the following properties: (A1) For all n ≥ 0, τ n (z 0 ) = 0.
And we say that B satisfies the strong analytic Ditkin condition if, for each z 0 ∈ T, the sequence τ n n≥0 can 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 f divides g (we write f |g) if g/f ∈ H ∞ . Let I be a closed ideal of B. We denote by U I the inner factor of I, that is the greatest inner common divisor of all nonzero functions in I (see [14] p.85), and we set The ideal I is said to be standard if In several Banach algebras B satisfying condition (H1)-(H3), all closed ideals are standard. This is the case if B is for example one of the following algebras : a(D), a n (D) (n ≥ 1), λ s (0 < s < 1) and H p n (n ≥ 1, 1 < p < ∞), where and H p n = f holomorphic in D and f (n) ∈ H p , H p being the classical Hardy space on D (see respectively [24], [18], [19], [27]). For further examples see also [28] and [31].
But, even in the case where N B = 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 h 0 (I) is finite (see [15]), h 0 (I) is countable (see [3]) or when h 0 (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 where s is 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 B satisfies conditions (H1)-(H3) and the analytic Ditkin condition then every closed ideal I of B such that h 0 (I) is at most countable is standard. Notice that a result in the same direction was announced by Faivishevskii 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 I such that h 0 (I) is a "thin" set, U I = 1 and I = {f ∈ A + (T) : f = 0 on h 0 (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 L 1 s (R + ), where In section 4, we give further examples of Banach algebras where Theorem 2.11 applies.
Acknowledgement. We wish to thank the referee for his valuable comments and to have drawn our attention to the work of Faivishevskii, where results in the same direction are announced.

Closed ideals of B
Throughout this section, B will denote a semi-simple unital commutative Banach algebra of functions holomorphic in D and continuous on D, endowed with the pointwise product and continuously imbedded in a(D). Without loss of generality we assume that 1 B = 1. We recall that N B is the supremum of integers n such that B ⊂ a n (D).
We will need, in the sequel, some properties of B.
Suppose that B satisfies conditions (H1) and (H2). We can identify the maximal ideal space of B with D. The identification is given by the map where δ z is the point evaluation at z : Then, for all z ∈ D, we have It follows from the closed graph theorem that the embedding from B into a N B (D) is continuous. In particular, for all z ∈ D and for all k, 0 ≤ k ≤ N B , functionals For λ ∈ D and f ∈ a(D), we define the function R λ (f ) by Clearly R λ is a linear map. Notice also that if B satisfies condition (H1) and (H2) and if for every f ∈ B, R λ (f ) ∈ B, then the closed graph theorem asserts that R λ is a bounded operator on B.
The following Lemma shows that condition (H3) can be equivalently reformulated. We recall that α : z → z is the identity map.
(ii) There exists k ≥ 0 such that: (a) α n B = O(n k ), n → +∞. (b) For all λ ∈ D, R λ defines a bounded linear operator on B and there exists C > 0 such that (iii) There exists k ≥ 0 such that: Suppose that B satisfies condition (H3). We begin by showing (b). Let λ ∈ D and P be a polynomial function. We can write P − P (λ) = (α − λ)Q, where Q is a polynomial function. So R λ (P ) = Q belongs to B. We deduce from (H3) that where k and C are nonnegative constants independent of P . Let P be the space of all polynomials provided with the norm of B. One has just seen that the restriction of R λ to P is a bounded linear operator with norm not exceeding 2C 1 − |λ| −k . Now, since B satisfies (H1), R λ|P can be extended to a bounded linear operator on B with norm less than 2C 1 − |λ| −k . Then, using the fact that functionals g → g(z), for z ∈ D, are continuous on B, it is easily seen that this extension coincides with R λ on 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 α is equal to D. So for all λ such that |λ| > 1, the function α − λ is invertible in B and using (1), we get Let r > 1, we have α n = 1 2iπ γr λ n α − λ −1 dλ, where γ r denotes the circle centered in 0 with radius r. We deduce from this equality and (3) that . 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), It follows from this that for every λ and λ in D, Thus the map λ → R λ (f ) is continuous from D into B. So, for every n ≥ 1 and r ∈ (0, 1), the integral 1 2iπ γr λ −n R λ (f )dλ is well defined and belongs to B. We shall now prove that For λ and z in D we have So the map λ → R λ (f )(z) is expanded in an entire series in D and its Taylor coefficients are R n+1 Since the linear map δ z : f → f (z) is continuous on B, we get which proves (4). Therefore For r = 1 − 1 n , we see that (d) holds. (iii) ⇒ (i) : For |λ| < 1, the series n≥0 λ n R n+1 0 (f ) is absolutely convergent in B. Then, by (5), we deduce that R λ (f ) = n≥0 λ n R n+1 0 (f ) and R λ (f ) ∈ B. Therefore there exist constants C and C such that Using this for (α − λ)f , we obtain Then, it suffices to expand (α − λ) −1 in series and use (c) to see that there exists a constant C such that It follows from (6) and (7) that Thus condition (H3) is satisfied. To show the main result of this paper we begin with the characterisation of closed ideals I such that U I = 1 and h 0 (I) is reduced to a single point. For this we need some lemmas. We denote by C k (T) the space of k times continuously differentiable functions on T. Let f ∈ B. If z ∈ D, we set with the understanding that sup ∅ = −∞. We remark that Also if z ∈ D and B satisfies conditions (H1)-(H3), then k f (z) = m f (z) − 1. For all m ≥ 0, set Now, since for all m ≥ 0, the polynomial function R m vanishes, with all its derivatives of order less or equal than k, at z 0 , there exists a polynomial function P m such that which concludes the proof. Proof. Since B satisfies the analytic Ditkin condition, there exists a sequence τ n n≥0 included in B satisfying 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 k a sequence τ n 1 ,...,n k (n 1 ,...,n k )∈∪ 1≤k≤m N k such that for every k ∈ {1, . . . m−1} and (n 1 , . . . , n k ) ∈ N k the sequence τ n 1 ,...,n k ,n n≥0 satisfies conditions (A1) and (A2) related to the function τ n 1 ,...,n k f , that is τ n 1 ,...,n k ,n (z 0 ) = 0, for every n ≥ 0, and lim n→+∞ τ n 1 ,...,n k ,n τ n 1 ,...,n k f − τ n 1 ,...,n k f 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 ϕ 1 (n) n≥0 and ϕ 2 (n) n≥0 two increasing sequences of positive integers such that, for all n ≥ 0, Then, as And since for all n ≥ 0, Let f ∈ a(D). We denote by U f its inner factor. We have is the Blaschke product constructed with zeros of f in D, and S f its singular inner factor, associated with a singular measure µ f . Let I be a nonzero closed ideal of B and z ∈ D. We set and (what have already been defined in the introduction) We recall that U I stands for the inner factor of I, that is the greatest common divisor of all nonzero functions in I (see [14] p.85). Let B I be the Blaschke product constructed with elements of h 0 (I) ∩ D (taking multiplicity into account). Then U I = B I S I , where S I is the singular inner function associated with U I . We denote by µ I the singular positive measure which defines S I .
The following Lemma appears in [7] in the particular case where B = A + (T) . Since B ⊂ a(D), the proof uses exactly same arguments, so we state it without proof. Suppose that B satisfies conditions (H1)-(H3). Let I be a closed ideal of B such that h 0 (I) ⊂ T. Denote by π I the canonical surjection from B onto B/I. Let T be the operator on B/I defined by where α : z → z is the identity map; T I is a bounded linear operator on B/I such that Sp (T I ) = h 0 (I). Let .
By Lemma 2.1, there exist nonnegative contants C and k such that Denote by ν f (resp. ν I ) the discrete part of the singular measure µ f (resp. m I ). Then, we deduce from (9) and from Lemma 5c of [2] that, for all ε > 0, Now, the following Lemma gives the characterisation of closed ideals I of B such that U I = 1 and h 0 (I) = {z 0 }, z 0 ∈ T. Then Proof. To simplify notations, we put k I = k I (z 0 ). It follows from Lemma 2.3 that where the closure is taken for the norm in B. Now, we are interested in showing the other inclusion. By Lemma 2.1, there exists a nonnegative integer k such that α n = O(n k ), n → ∞. The operator T defined on B/I by On the other hand, we deduce from (10) that, for all ε > 0, We obtain by Lemma 2 of [2] that, for all ε > 0, Then, it follows from Corollary 1 of [2] that (T − z 0 ) k+1 = 0, which means that Denote by k 0 the smallest of nonnegative integers p such that (α − z 0 ) p+1 ∈ I. We are going to prove that k 0 ≤ k I . Since B satisfies the analytic Ditkin condition, we deduce from Lemma 2.5 that there exists a sequence σ n n≥0 included in B such that m σn (z 0 ) ≥ k + 1 and Since m σn (z 0 ) ≥ k + 1, we deduce from Lemma 2.4 that there exists a sequence P n,k k≥0 of functions included in B such that, for all n ≥ 0, Since (α − z 0 ) k+1 ∈ I, it follows from this equality that, for all n ≥ 0, σ n ∈ I. Then, we deduce from (12) We (13), we get Now, using the definition of k 0 , one can easily prove that the family (π I (α)−z 0 ) k 0≤k≤k 0 is linearly independent. So we deduce from (14) that for all k ∈ {0, . . . , k 0 }, f (k) (z 0 ) = 0. Hence, for all f ∈ I, k f (z 0 ) ≥ k 0 , and so k I ≥ k 0 . Consequently, (α − z 0 ) k I +1 ∈ I, which proves that Finally, we have proved that (the second equality follows from Lemma 2.3).
Lemma 2.8. Let µ be a singular positive measure on T, and S µ the singular inner function associated with µ. Let p be a nonnegative integer and f ∈ a p (D) such that S µ f . Then, we have where Supp µ denotes the support of the measure µ.
Proof. By hypothesis, there exists g ∈ H ∞ (D) such that S µ g = f . Now, if λ ∈ T, we deduce from ( [21], p. 63) that Hence, an induction on k proves that f (k) (λ) = 0 for all k ∈ {0, . . . , p}. This concludes the proof. Lemma 2.9. Assume that B satisfies conditions (H1)-(H3). Let f ∈ B and let ψ be an infinitely differentiable function on T. Then the Cauchy transform of ψf |T defined by Proof. Let z ∈ D. Cauchy formulae shows that for n ≥ 0, and that for n < 0, Since ψ is infinitely differentiable, it follows from Lemma 2.1, that the series converges. So n≥0ψ (n)z n f + n<0ψ (n)R −n 0 (f ) belongs to B and by (15), it is equal to C(ψf ). This finishes the proof.
Let I be a closed ideal of B and g ∈ B. We denote by I(g) the division ideal of I by g, that is Clearly I(g) is a closed ideal of B that contains I.
The following lemma is known in the case B = A + (T) (see Lemma 1.2 and Lemma 1.5 of [7]). The proof is based on a similar argument used in [7] and on Lemma 2.9 established below. We include the proof to see how Lemma 2.9 operates here. Proof. Notice that I ⊂ I(g), which implies that h 0 (I(g)) ⊂ h 0 (I). Let λ ∈ h 0 (I) ∩ D. Let f ∈ I such that k f (λ) = k I (λ). Definẽ Hencef g ∈ I, which means thatf ∈ I(g). Now, by construction,f (λ) = 0, which proves that λ / ∈ h 0 (I(g)). This shows that h 0 (I(g)) ⊂ h 0 (I) ∩ T, and consequently, B I(g) = 1. Since h 0 (I) is at most countable, it suffices to show that µ I(g) ({z}) = 0 for all z ∈ h 0 (I) ∩ T, to prove that S I(g) = 1. Let z 0 ∈ h 0 (I) ∩ T and let ϕ be a nonzero outer function in a ∞ (D) ⊂ B such that ϕ (k) (z 0 ) = 0, k ≥ 0. Let It is easily seen, by Taylor formulae, that ψ is infinitely differentiable on T.
Let U be an inner function and E 0 , . . . , E N B be closed subsets of T such that E N B ⊂ . . . ⊂ E 0 . We set .
Let J := I(g) (ϕ) be the division ideal of I(g) by ϕ. As z 0 ∈ h N B (I), g vanishes with all its derivatives (of order less or equal than N B ) at z 0 . And since B satisfies the analytic Ditkin condition, we deduce from Lemma 2.5 that there exists a sequence σ n n≥0 included in B such that m σn (z 0 ) ≥ N B + 1 and lim n→+∞ σ n ϕg − ϕg B = 0.
Now, since I(g) ⊂ J, we deduce from Lemma 2.10 that U J = 1. Moreover, since 1 − ϕ ∈ J, we have h(J) ⊂ {z 0 }. Then, Lemma 2.7 gives the characterisation of J. In particular, by Lemma 2.4, σ n ∈ J, n ≥ 0. This means that, for all n ≥ 0, σ n ϕg ∈ I. Hence, since I is closed, we deduce from (16) that ϕg ∈ I. In other words, ϕ ∈ I(g), which is in contradiction with the fact that z 0 ∈ h 0 (I(g)). Finally, we have proved that h 0 (I(g)) = ∅, that is I(g) = B.
Let H be a Banach algebra of continuous functions on T. We say that H is an homogeneous Banach algebra if it satisfies the following conditions: (1) H is a semi-simple and commutative Banach algebra with maximal ideal space T.
(2) The set of trigonometric polynomials is a dense subset of H.
Let H be an homogeneous algebra on T. We denote by H + the closed subalgebra of H generated by {e int : n ≥ 0}.
Notice that conditions (2) and (3) implies that for every τ 0 ∈ R, f τ − f τ 0 H → 0, as τ → τ 0 . It follows then from Theorem 2.12 of [16] that H + consists of functions in H having analytic extension into D. Then, the maximal ideal space of H + is D.
Recall that we denote by C n (T) the space of n times continuously differentiable functions on T and we set C ∞ (T) = n≥0 C n (T).
Suppose that H contains C ∞ (T), which implies in particular that H is a regular algebra. Also this implies that C k (T) ⊂ H for some nonnegative integer k. It follows that N H , the greatest of the integers n for which H ⊂ C n (T), is bounded. We say that H satisfies the Ditkin condition if for each function f ∈ H such that f (k) (1) = 0, 0 ≤ k ≤ N H , there exists a sequence v n n≥0 in H satisfying the following properties: (1) For all n ≥ 0, v n = 0 on a neighbourhood of 1.
We say that H satisfies the analytic Ditkin condition if H + satisfies the analytic Ditkin condition. Corollary 2.12. Let H be an homogeneous Banach algebra on T that contains C ∞ (T) and satisfies the analytic Ditkin condition. Let I be a closed ideal of H + such that h 0 (I) is at most countable. Then I is 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 k such that C k (T) ⊂ H. By the closed Graph theorem this imbedding is continuous. So there exists a constant C such that In particular we have α n = O(|n| k ), |n| → ∞. So there exists a constant C such that for every λ ∈ D, So, for every λ ∈ D and f ∈ H + , R λ (f ) ∈ H + and It follows now from Lemma 2.1 that H + satisfy condition (H3).
Remark 2.13. 1) Let H be an homogeneous Banach algebra on T that contains C ∞ (T) and such that H + satisfies the analytic Ditkin condition. It is announced in [3] that if I is a closed ideal of H + such that h 0 (I) is a finite set then where I H is the closed ideal of H generated by I. 3. Closed ideals in 1 (ω) and L 1 (R + , ω)

2) Let H be an homogeneous Banach algebra on T such that
Let s be a nonnegative real number. We set ω s (t) = (1 + t) s , t ≥ 0. Denote by 1 (ω s ) the set of all complex sequences x = (x n ) n≥0 such that x n s := n≥0 |x n |ω s (n) < +∞.
Clearly 1 (ω s ), equipped with convolution product and norm s , is a semi-simple unitary commutative Banach algebra. The set of maximal ideals of 1 (ω s ) can be identified with D. Considering the Gelfand transfom x →x, wherex(z) = n≥0 x n z n , z ∈ D, we see that 1 (ω s ) is isometrically isomorphic to the Beurling weighted Banach algebra So we will just be interested by closed ideals of A + s (T). The structure of closed ideals I of A + s (T) such that h 0 (I) is countable, is known only in the cases when s < 1 (see [23], [3] and [7]) and when s ≥ 1 and U I = 1 (see [1]). Here we give a complete characterisation of such ideals.
With notation of the above section, we have A + Clearly A s (T) is an homogeneous Banach algebra on T that 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 s be a nonnegative real number, and I be a closed ideal of A + s (T) such that h 0 (I) is at most countable. Then I is standard.

Now we turn to the Banach algebra
1,s , equipped with the convolution product, is a Banach algebra. We denote by P the open half plane {z ∈ C : Re(z) > 0}. For f ∈ L 1 (R + ), we define the Laplace transform L(f ) by Let J be a closed ideal of L 1 (R + , ω s ). We set Notice that J = L 1 (R + , ω s ) if and only if h 0 (I) = ∅ and β J = 0 ( [22], [11]). We say that J is modular if L 1 (R + , ω s )/J admits a unit.
It is well-known that J is a closed modular ideal of L 1 (R + , ω s ) if and only if h 0 (J) is compact and β J = 0 (in case s = 0, see for example [4], Proposition 2.1). When s = 0, V. P. Gurarii showed in [11] that if J is a closed ideal of L 1 (R + ) = L 1 (R + , ω 0 ) such that h 0 (J) is at most countable, then where L(J) ∞ is the closure of L(J) in A 0 (P ), the algebra of all functions f which are continuous on P , holomorphic in P , and such that f (z) → 0 as |z| → +∞. The closed ideals of A 0 (P ) are given by the Beurling-Rudin theorem, thanks to a conformal transform between P and D. So, if h 0 (J) is at most countable, we have Using a result of H. Hedenmalm, we will deduce from Theorem 3.1 a similar result concerning closed ideals of L 1 (R + , ω s ), for any nonnegative real number s.
If I is a closed ideal of A + s (T), we denote by π = π I the canonical surjection from A + s (T) onto A + s (T)/I. If h 0 (I) ⊂ D\[−1, 0], we can define the function z → π(α) z , which is an entire function of finite exponential type. Furthermore, we have Then, we define Φ I : Φ I is a continuous homomorphism. We set 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 I and J (see also [4], Theorem 2.2).

Theorem 3.2. ([13])
The mapping Θ : I → ker Φ I defines a bijection between I and J . Futhermore, if J ∈ J , we have where δ n is 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 < f, g > (f ∈ X, g ∈ X * ).
Let I be a closed ideal of A + s (T) and π = π I the canonical surjection from A + ). Let Ψ be an analytic function on a neighbourhood of h 0 (I). We use the Dunford-Schwarz functional calculus to define Ψ(π(α)) by where ∂Ω is the boundary of a suitable open neighborhood Ω of h 0 (I) in which Ψ is holomorphic. Then, we have the following Lemma.
Lemma 3.3. Let I be a closed ideal of A + s (T). Let Ψ be an analytic function on a neighbourhood of h 0 (I). Then Proof. We have w is a continuous functional, we have Hence, the result will follow from the Cauchy integral formulae, if we prove that Let ξ ∈ ∂Ω and g ∈ A + s (T) such that π(g) ξ − π(α) = 1. Since, w ∈ h k (I), we have the following relations: So, we deduce easily from these equalities that Then, we have Then the result follows from the above equality, (18) and the Cauchy integral formulae.
We also need the two following lemmas. Proof. Let 0 ≤ k ≤ [s], w ∈ h k (I) and f ∈ J. We have Φ I (f ) = 0 and in particular, Now, applying Lemma 3.3 with Ψ(z) = z t (t ≥ 0), we get, for all 0 ≤ j ≤ k, (20) 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, This proves that − log w ∈ h k (J), and so According to Theorem 3.2, we have F g ∈ J, for all g ∈ L 1 (R + , ω s ). It follows that Now, we easily deduce from (21) and (22) that This proves that e −h k (J) ⊂ h k (I), and concludes the proof of this lemma. Lemma 3.5. Let J, J ∈ J . Then, we have Proof. Let J ∈ J and set I = Θ −1 (J). We begin to prove that where J 0 (resp. I 0 ) is the closure of J in L 1 (R + ) (resp. A + (T)) and where Θ 0 is the map Θ corresponding to the case s = 0. Clearly J 0 and I 0 are closed ideals. Moreover we have T J 0 = T J and U I 0 = U I . By Theorem 3.2 we have f (n)δ n * g ∈ J, (g ∈ L 1 (R + , ω s )) .
Since, for every g ∈ L 1 (R + , ω s ), the map f → +∞ n=0 f (n)δ n * g is continuous from Also since, for every f ∈ A + (T), the map g → +∞ n=0 f (n)δ n * g is continuous from This means that I 0 ⊂ Θ −1 0 (J 0 ). Since Θ 0 is increasing with respect to inclusion order we have Θ 0 (I 0 ) ⊂ J 0 .
It remains to prove the other inclusion. Denote by ψ the homomorphism from A + s (T)/I into A + (T)/I 0 defined by ψ π I (f ) = π I 0 (f ). Since ψ is an homomorphism, it is easily seen that Φ 0 where Φ 0 I 0 is the map Φ I corresponding to s = 0 and I = I 0 . Hence, we have ker Φ I ⊂ ker Φ 0 In other words, we have J ⊂ Θ 0 (I 0 ). Since Θ 0 (I 0 ) is closed in L 1 (R + ), we get J 0 ⊂ Θ 0 (I 0 ).
Denote by H ∞ (D) (resp. H ∞ (P )) the space of bounded holomorphic function on D (resp. P ). Denote also by A 0 (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 Thus we see that U I 0 = U I 0 if and only if T J 0 = T J 0 . This finishes the proof since Theorem 3.6. Let s be a nonnegative real number, and J be a closed ideal of L 1 (R + , ω s ) such that h 0 + (J) is compact and at most countable. Then Proof. We first suppose that J is modular. For a real number a = 0 and f ∈ L 1 (R + , ω s ) we set T a (f )(x) = af (ax), x ∈ R; T a is an isomorphism of L 1 (R + , ω s ). So using this transformation we may assume without loss of generality that h 0 We also set I = Θ −1 (J) and I 0 = Θ −1 (J 0 ). By definition of J 0 , we have T J = T J 0 . So, we deduce from Lemma 3.5 that Furthermore, for all 0 ≤ k ≤ [s], we have h k (J) = h k (J 0 ). So we deduce from Lemma 3.4 that It follows then from (24), (25) and Theorem 3.1 that I = I 0 . Since Θ is a bijection, we have J = J 0 . Suppose now that J is not modular. We set J 1 = δ −β I * J; J 1 is a closed ideal and we have β J 1 = 0, T J (z) = e −β I z T J 1 (z), (z ∈ P ) and h k (J 1 ) = h k (J), (0 ≤ k ≤ [s]). So J 1 is modular and applying the previous result to J 1 , we obtain the desired equality for J.

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 condition (H1)-(H3) is quite easy to verify and so, left to the reader.
Recall that we denote α : z → z the identity map. We will need the following sequence e n n≥1 defined by Notice that e n ∞ ≤ 2 and lim n→∞ e n f − f ∞ = 0 for every function f ∈ a(D) such that f (1) = 0. Here . ∞ denotes the supremum norm on D.

4.1.
The algebra H 1 m . Let 1 ≤ p < +∞ and m ≥ 1 an integer. We define H p m to be the algebra of all functions f which are analytic in D and such that f (m) ∈ H p , equipped with the norm |L| ≤ C max z∈L ρ(z), for any arc L ⊂ T, where ρ(z) = dist z, h m−1 (I) , C > 0 and γ > 0 are certain constants and |L| is the lengh of L. Clearly these conditions are not satisfied by all closed ideals with countable hull. Proof. It is easily seen that N H 1 m = m − 1. We will suppose, without loss of generality, that z 0 = 1. We begin to show the result for m = 1. So, let f ∈ H 1 1 such that f (1) = 0, we are going to prove that lim n→+∞ e n − 1 f 1,1 = 0.
We have Since f ∈ H 1 (D) and f (1) = 0, we have Consequently, Now, an easy calculation and Cauchy formulae show that 1 n So, using the dominated convergence theorem of Lebesgue, we deduce from the above inequality, the fact that f ∈ H 1 (D) and (27) that Moreover it is easily seen that lim n→+∞ e n − 1 f (0) = 0.
So we deduce from (28) and (29) that (26) is satisfied. Now, suppose that m ≥ 2. For all n ≥ 1, set τ n = e m n and let f ∈ H 1 m such that f (1) = .

4.2.
The algebra F p (N, s). Let 1 ≤ p < +∞ and s > 0 be real numbers. We denote by F p (N, s) the space of all continuous functions on T such that By the Hölder inequality, we obtain We have clearly v k ≤ 1 sp −1 (n + 1). We write with the understanding that " .
Since p − sp > −1, it follows from the last inequality that u k ≤ e p +1−sp (n + 1) + e p +1−sp (n + 1) − 1 e p +1−sp (n + 1) n + 1 n k(p +1−sp ) , where the notation A B means that there exists a constant C > 0, independent of n, k and f such that A ≤ CB.
(see [28], [31], [19]). For α > 0, we set There is no characterisation of closed ideals of the algebras λ wα (see the introduction of [31]). For this reason we consider here this algebra. Let us introduce the following condition : there exists C > 0, such that w(y) w(x) ≤ C y x , 0 < x ≤ y.
This condition is satisfied by concave functions and in particular by functions w α , α > 0.
Proof. It is easily checked that B is 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.