Ditkin’s condition and ideals with at most countable hull in algebras of functions analytic in the unit disc

Andrzej Sołtysiak, Antoni Wawrzyńczyk

Ditkin’s condition and ideals with at most countable hull in algebras of functions analytic in the unit disc

Abstract. Agrafeuil and Zarrabi in [1] characterized all closed ideals with at most countable hull in a unital Banach algebra embedded in the classical disc algebra and satisfying certain conditions ((H1), (H2), (H3)), and the analytic Ditkin condition. We modify Ditkin’s condition and show that analogous result is true for a wider class of algebras. This is an extension of the result obtained in [1].

2010 Mathematics Subject Classification: Primary 46J20; Secondary 46J15.

Key words and phrases: Closed ideals, Banach algebras, Ditkin’s condition.

1. Introduction. In [1] and in earlier papers [4], and [5] there are introduced certain classes of Banach algebras of analytic functions in the unit disc D ⊂ C in which every closed ideal with at most countable hull has a standard form (in a sense defined by the authors).

In all mentioned cases it is assumed (in a more or less explicit form) that a considered algebra B is embedded in the algebra A ^(N

⁾ ( D) of functions analytic in D (with pointwise multiplication) and of class C ^(N

⁾ on the closed disc D for some N _B ∈ N ⁰ = N∪{0}. Moreover, it is assumed that the algebra B satisfies the analytic Ditkin condition which says the following:

for every point z 0 in the unit circle T and for every function f from the algebra which satisfies f ^(j) (z 0 ) = 0 for 0 ¬ j ¬ N B , there exists in B a sequence (σ ⁿ ) such that σ n (z 0 ) = 0 for all n and lim

n→∞ kσ ⁿ f − fk B = 0 (k · k B denotes the norm of B).

The analytic Ditkin condition is a very strong assumption which confines a class

of algebras to which obtained results can be applied. In the next section we show

a simple example of a Banach algebra B of analytic functions in the unit disc for

which N _B = 0 and in which Ditkin’s condition does not hold. Therefore none of

the results obtained in [1], [4], and [5] can be applied to describe the form of closed

ideals with at most countable hull in that algebra. On the other hand, a form of the

closed ideals in that algebra is known and all such ideals are standard (see [7]).

In the present paper we adopt arguments used in [1] to show that under modified Ditkin’s condition and suitably extended definition of a standard ideal the result analogous to the main theorem from [1] (Theorem 2.11) holds true.

A class of Banach algebras we are dealing with in the paper consists of subalge- bras B of the classical disc algebra A(D) satifying the following conditions:

(H1) The space of polynomials is a dense subset of B.

(H2) lim

n →∞ kζ ⁿ k _B

= 1 (ζ denotes the identity function z 7→ z).

(H3) There exist k ­ 0 and C > 0 such that

|1 − |λ|| ^k kfk B ¬ Ck(ζ − λ)fk B , f ∈ B, |λ| < 2.

(D) For every z 0 ∈ T there exists N(z ⁰ ) ∈ N ⁰ with the following properties:

(i) all functionals B 3 f 7→ f ^(j) (z 0 ), 0 ¬ j ¬ N(z ⁰ ), are well-defined and continuous,

(ii) there exists a sequence (ϕ n ) in the algebra B such that ϕ ⁿ (z 0 ) = 0 for all n and

n lim →∞ k(ζ − z ⁰ ) ^{N (z}

⁾⁺¹ ϕ n − (ζ − z ⁰ ) ^N(z

⁾⁺¹ k B = 0.

Remarks 1 ^◦ From (H1) it follows that the Banach algebra B has a unit 1 B = 1.

We shall assume in the sequel that k1k B = 1. Conditions (H1) and (H2) imply that the maximal ideal space of the algebra B, M(B), can be identified with D via the mapping z 7→ δ ^z , where δ z (f) = f(z). From this it follows that the algebra B is semisimple. Since the algebra A( D) is semisimple the embedding B into A(D) is continuous.

2 ^◦ Condition (D) is much easier to verify in concrete situations than the analytic Ditkin condition because we have to deal only with one given function. We call it the modified Ditkin condition. Clearly it has a local character, i.e. it depends on the degree of differentiability of functions from the algebra B at different points of the unit circle. If we assume that condition (D) holds true only at a point z 0 ∈ T, then we say that the algebra B satisfies the modified Ditkin condition at the point z ⁰ .

A function from H ^∞ is inner if the modulus of its boundary function is equal to 1 a.e. on the unit circle. If U is an inner function and f ∈ B the symbol U|f means that U divides f, i.e. there exits a function ϕ ∈ H ^∞ such that f = Uϕ (cf. [8]). A closed ideal I of the algebra B is standard, according to our definition, if there exist an inner function U and a descending family of sets H n ⊂ T, n ∈ N ⁰ , such that N (z) ­ n for every z ∈ H ⁿ and

I = {f ∈ B : U|f and f ⁽ⁿ⁾ (z) = 0 for z ∈ H ⁿ , n ∈ N 0 }.

We obtain the following theorem.

Theorem If B is a subalgebra of the disc algebra A(D) which satisfies conditions

(H1), (H2), (H3), and (D), then every closed ideal I of B with the at most countable

hull h(I) = {z ∈ D: f(z) = 0 for f ∈ I} is standard.

In the next section we give an example of an algebra B (B = A ⁽¹⁾ 1 ( D)) which satisfies all the assumptions of this theorem. The algebra A ⁽¹⁾ 1 ( D) is a very simple example of an algebra which contains functions having certain properties of differen- tiability at different boundary points of the unit disc. Algebras of this form appear in a natural way in [6] and [7] as images of convolution algebras of the Sobolev type under the Gelfand transformation. We shall deal with such algebras in some other paper (cf. [10]).

2. The theorem of Agrafeuil and Zarrabi, and the example. The follo- wing theorem is the principal result of [1] (Theorem 2.11).

Theorem 2.1 Let B be a subalgebra of the algebra A(D) which is a Banach algebra endowed with the norm k · k B . Suppose that B satisfies conditions (H1)–(H3) and the analytic Ditkin condition. Let I be a closed ideal of B such that the hull h(I) = {z ∈ D: f(z) = 0 for f ∈ I} is at most countable. Then

(1) I = {f ∈ B : U ^I |f and f ^(j) (z) = 0 for z ∈ h ^j (I), 0 ¬ j ¬ N ^B },

where N _B is the maximal nonnegative integer n such that B ⊂ A ⁽ⁿ⁾ ( D), U ^I is the greatest common inner divisor of all nonzero functions in I, and

h ^j (I) = {z ∈ T: f(z) = f ⁰ (z) = . . . = f ^(j) (z) = 0 for f ∈ I}, 0 ¬ j ¬ N ^B . Now we present the aforementioned example.

Example 2.2 Let A ⁽¹⁾ 1 ( D) be the space of functions f on D which (a) are continuous on D and analytic on D,

(b) are of class C ⁽¹⁾ on D \ {1}, (c) satisfy lim _z→1 (1 − z)f ⁰ (z) = 0.

We endowe the space A ⁽¹⁾ 1 ( D) with the norm kfk = sup

z ∈D

|f(z)| + sup

z ∈D\{1}

|(1 − z)f ⁰ (z)|.

By the maximum principle it follows that this norm coincides on A ⁽¹⁾ 1 ( D) with the norm

kfk T = sup

z ∈T |f(z)| + sup

z∈T\{1} |(1 − z)f ⁰ (z)|.

Let us verify that the latter norm is submultiplicative for all functions on T for

which the right-hand side of the formula makes sense. Suppose that f and g are

continuous functions on T and of class C ⁽¹⁾ on T \ {1}. Moreover, suppose the

property lim z →1 (1 − z)f ⁰ (z) = 0 for both functions.

Then

kfgk T = sup

z ∈T |(fg)(z)| + sup

z∈T\{1} |(1 − z)(fg) ⁰ (z)|

¬ sup

z ∈T |f(z)| sup

z ∈T |g(z)| + sup

z ∈T\{1} |(1 − z)(f ⁰ g + f g ⁰ )(z)|

¬ sup

z∈T |f(z)| sup

z∈T |g(z)| + sup

z∈T |g(z)| sup

z ∈T\{1} |(1 − z)f ⁰ (z)|

+ sup

z ∈T |f(z)| sup

z∈T\{1} |(1 − z)g ⁰ (z)|

¬ kfk T kgk T .

It is easy to see that the space A ⁽¹⁾ 1 ( D) is a unital algebra with respect to the pointwise multiplication. Moreover it is complete with respect to the norm k · k T . Hence A ⁽¹⁾ 1 ( D) is a Banach algebra continuously embedded in the disc algebra A(D).

The algebras similar to A ⁽¹⁾ 1 ( D) were studied in [6] and [7].

We need the following properties of this algebra.

Proposition 2.3 (i) The space of polynomials is dense in A ⁽¹⁾ 1 ( D).

(ii) lim _n→∞ kζ ⁿ k

= 1.

(iii) |1 − |λ|| ² kfk ¬ 3k(ζ − λ)fk for all f ∈ A ⁽¹⁾ 1 ( D), |λ| < 2.

Proof (i) As proved in [7], the space A ⁽¹⁾ 1 ( D) is isomorphic (as a Banach space) under the mapping f 7→ (1 − ζ)f with a closed ideal of the algebra A ⁽¹⁾ ( D) of functions analytic in D and of class C ⁽¹⁾ on the closed disc. More exactly,

(1 − ζ)A ⁽¹⁾ 1 ( D) = I ¹ = {g ∈ A ⁽¹⁾ ( D): g(1) = 0}.

This implies (i).

(ii) By definition of the norm

kζ ⁿ k = kζ ⁿ k T = 1 + n sup

z∈T\{1} |(1 − z)z ^{n −1} | = 2n + 1, which gives (ii).

(iii) For |λ| = 1 inequality (iii) is obvious. Let |λ| 6= 1. We have kfk = kfk T =

(ζ − λ)f 1 ζ − λ

_T ¬ k(ζ − λ)fk T

1

ζ − λ _T . Since

1

ζ − λ

_T = sup

z ∈T

    1

z − λ  

  + sup _z∈T\{1}

    1 − z

(z − λ) ²  

  ¬ 1

|1 − |λ|| + 2

|1 − |λ|| ²

¬ 3

|1 − |λ|| ²

property (iii) is proved. Notice that we have used the norm k · k T for the function 1

ζ − λ not belonging to the algebra A ⁽¹⁾ 1 ( D). 

Hence the algebra A ⁽¹⁾ 1 ( D) satisfies conditions (H1), (H2), and (H3). The number N _B = 0 for B = A ⁽¹⁾ 1 ( D) since there are elements of A ⁽¹⁾ 1 ( D) which are not derivable at 1 (for example the functions ϕ n (z) = (1 − z)

defined by an appropiate branch of logarithm).

The set I ₋₁ ¹ = {f ∈ A ⁽¹⁾ 1 ( D): f(−1) = f ⁰ (−1) = 0} is obviously a closed ideal of A ⁽¹⁾ 1 ( D) with the one-point hull h(I −1 ¹ ) = {−1}. However it does not have form (1) from Theorem 2.1.

We see that in this case Theorem 2.1 does not work. The reason for that is that the algebra A ⁽¹⁾ 1 ( D) does not satisfy the analytic Ditkin condition.

3. The shift operator acting on the quotient algebra B/I. From now on we shall denote the norm k · k B of a given Banach algebra B simply by the symbol k · k. As proved in [1] (Lemma 2.1) condition (H3) defining the space B can be formulated in a different way.

For λ ∈ D let us define the operator L ^λ acting on the algebra A( D) by the formula

L λ (f)(z) =

 



f (z) − f(λ)

z − λ if z 6= λ, f ⁰ (λ) if z = λ.

Theorem 3.1 Suppose that a Banach algebra B satisfies conditions (H1) and (H2).

Then the following statements are equivalent.

(i) B satisfies condition (H3).

(ii) There exists k ­ 0 such that (a) kζ ⁿ k = O(n ^k ) as n → ∞;

(b) for all λ ∈ D the operator L ^λ is bounded on B and there is C > 0 such that

kL ^λ k ¬ C(1 − |λ|) ^−k (|λ| < 1).

(iii) There exists k ­ 0 such that (c) kζ ⁿ k = O(n ^k ) as n → ∞;

(d) L 0 defines a bounded operator on B and

kL ⁿ 0 k = O(n ^k ) as n → ∞.

Remark Notice that the operator L λ is a left inverse of the operator S − λ, where S is defined on the algebra B by Sf = ζf.

Let I be a closed ideal of a Banach algebra B. We denote by U ^I the greatest common inner divisor of all nonzero elements of I. Let S be the operator on the algebra B/I defined by S[f] = [ζf]. The following theorem was in fact proved in [1]

although never formulated explicitly. We give the proof for completeness.

Theorem 3.2 Let a Banach algebra B satisfy conditions (H1), (H2), and (H3).

Let I be a closed ideal of the algebra B with h(I) ⊂ T and let U ^I = 1. Then the operator S has the following properties:

kS ⁿ k = O(n ^k ) as n → ∞ (the integer k is the same as in (H3)), kS ⁻ⁿ k = O(exp ε √

n) as n → ∞, for every ε > 0.

Proof The property kS ⁿ k = O(n ^k ), n → ∞, follows immediately by Theorem 3.1 (iii) (d).

Notice that σ(S) = σ B/I ([ζ]). Let µ ∈ σ B/I ([ζ]). Then µ = ψ([ζ]) for some ψ ∈ M(B/I). The functional e ψ : B 3 f 7→ ψ([f]) belongs to M(B) hence it is of the form δ z for some z ∈ D. Then µ = e ψ(ζ) = δ z (ζ) = z. Since ψ vanishes on I we have µ ∈ h(I). The other inclusion is obvious. Thus we have σ(S) = h(I). In particular, the operator S − λ is invertible for λ ∈ D.

The relation (z −λ)L ^λ (f)(z) = f(z) −f(λ) leads to the formula [ζ −λ][L ^λ (f)] = [f] − f(λ), i.e. (S − λ)[L ^λ (f)] = [f] − f(λ).

For f ∈ I it follows (S − λ)[L ^λ (f)] = −f(λ) and

k − f(λ)(S − λ) ⁻¹ k ¬ kL ^λ (f)k ¬ kL ^λ kkfk ¬ C kfk (1 − |λ|) ^k , according to Theorem 3.1 (ii).

Every function f ∈ A(D) factorizes in the form f = BV G, where B is the corresponding Blaschke product, V is the singular inner factor of the form V (z) = R

T z+ζ

z −ζ dν(ζ) and G is the outer factor.

Lemma 5 (c) from [2] asserts that for every mapping ϕ analytic in D with values in some Banach space A and for every analytic function f which is not identically zero on D with the singular inner factor R

T z+ζ

z −ζ dν(ζ) we have kϕ(z)k = O K ^1+|z| 2

exp ^2ν( _1−|z| ^T)+ε

as |z| → 1−, for every ε > 0, where K(r) = sup _{z ∈T} kf(rz)ϕ(rz)k.

We appply this result to the resolvent λ 7→ (S − λ) ⁻¹ . For every ε > 0 it holds k(S − λ) ⁻¹ k = O (1 − |λ|) ^−k exp ^2ν( _1−|λ| ^T)+ε

as |λ| → 1 − .

As stated in [1] (Lemma 2.6; see also [3] Lemma 1.3) under the condition that the greatest common inner divisor of nonzero elements of I is equal to 1, there is a sequence f n ∈ I such that the associated singular measures ν ⁿ satisfy ν n ( T) → 0.

This leads to

k(S − λ) ⁻¹ k = O exp _1−|λ| ^ε

as |λ| → 1−, for every ε > 0.

Now, Lemma 2 (b) from [2] implies that kS ⁻ⁿ k = O(exp ε √ n) as n → ∞. 

We need the following result obtained by Atzmon in [2], Corollary 1.

Theorem 3.3 If T is an invertible Banach space operator with a finite spectrum σ(T ) = {z ¹ , . . . , z p } which satisfies the following conditions:

kT ⁿ k = O(n ^k ) as n → ∞ for some integer k ­ 0, kT ⁻ⁿ k = O(exp ε √

n) as n → ∞ for every ε > 0, then p(T ) = 0, where p denotes the polynomial p(z) = Q p

j=1 (z − z ^j ) ^k+1 . From Theorems 3.2 and 3.3 we immediately obtain the following:

Corollary 3.4 Let a Banach algebra B satisfy conditions (H1), (H2), and (H3).

Let I be a closed ideal of B such that h(I) = {z 0 } ⊂ T and U ^I = 1 . Let S be the shift operator acting on B/I. Then (S − z 0 ) ^k+1 = 0.

Remark From Corollary 3.4 it follows that for a closed ideal I such that h(I) = {z ⁰ } ⊂ T and U ^I = 1 the range of the operator (S − z ⁰ ) ^k+1 is contained in I and therefore (ζ − z ⁰ ) ^k+1 B ⊂ I.

4. Consequences of the modified Ditkin condition.

Proposition 4.1 Let a Banach algebra B satisfy conditions (H1), (H2), and the modified Ditkin condition at a point z 0 ∈ T. Let I be a closed ideal of B. Suppose that there exists m ¬ N(z ⁰ ) such that f ^(j) (z 0 ) = 0 for all 0 ¬ j ¬ m and f ∈ I.

Then

I ⊂ (ζ − z ⁰ ) ^m+1 B.

Proof Let g ∈ I. Since polynomials are dense in B we take a sequence of poly- nomials P n (z) = P M

l=0 a n,l (z − z 0 ) ^l which approximates g in B. The functionals f 7→ f ^(j) (z 0 ), 0 ¬ j ¬ m, are continuous, hence a ^n,j → 0 as n → ∞ for the same set of j ⁰ s. The polynomials Q n (z) = P M

l=m+1 a n,l (z − z ⁰ ) ^l which are elements of the space (ζ − z ⁰ ) ^m+1 B, also converge to g. The proof follows. 

For z ₀ ∈ T and p ∈ N 0 we denote

I _z ^p

= {f ∈ B : f(z ⁰ ) = f ⁰ (z 0 ) = . . . = f ^(p) (z 0 ) = 0}.

Theorem 4.2 Let a Banach algebra B satisfy conditions (H1), (H2), and the mo- dified Ditkin condition at a point z 0 ∈ T. For every integer p ­ N(z ⁰ ) the following equalities hold true:

(ζ − z 0 ) ^p+1 B = (ζ − z 0 ) ^{N (z}

⁾⁺¹ B = I z ^N

^(z

⁾ .

Proof It is enough to prove that I z ^N(z

⁾ ⊂ (ζ − z 0 ) ^p+1 B. Take any ε > 0 and f ∈ I ^{z N}

^(z

⁾ . By Proposition 4.1 we get ϕ 1 ∈ B such that

(2) kf − (ζ − z ⁰ ) ^{N (z}

⁾⁺¹ ϕ 1 k < ¹ 3 ε.

From the modified Ditkin condition at the point z 0 we get a sequence (σ n ) in I _z ⁰

such that (ζ − z ⁰ ) ^N(z

⁾⁺¹ σ n → (ζ − z ⁰ ) ^{N (z}

⁾⁺¹ as n → ∞. By an easy in- duction we obtain for every r ∈ N subsequences σ ^j

(n)
, σ j

(n)
, . . . , σ j

(n)
such that (ζ − z ⁰ ) ^{N (z}

⁾⁺¹ σ j

(n) σ j

(n) . . . σ j

(n) → (ζ − z ⁰ ) ^{N (z}

⁾⁺¹ as n → ∞. Let τ n = σ j

(n) σ j

(n) . . . σ j

(n) . Then τ n ∈ (ζ − z ⁰ ) ^r B for all n and

(3) k(ζ − z 0 ) ^{N (z}

⁾⁺¹ ϕ 1 τ n − (ζ − z 0 ) ^N(z

⁾⁺¹ ϕ 1 k < ¹ ₃ ε for n large enough. Now, for the given τ n we take ϕ 2 ∈ B such that

kτ ⁿ − (ζ − z ⁰ ) ^r ϕ 2 k < ¹ 3 ε k(ζ − z ⁰ ) ^{N (z}

⁾⁺¹ ϕ 1 k ⁻¹ . (We may assume that ϕ 1 6= 0.) This implies

(4) k(ζ − z ⁰ ) ^N(z

⁾⁺¹ ϕ 1 τ n − (ζ − z ⁰ ) ^N(z

⁾⁺¹ (ζ − z ⁰ ) ^r ϕ 1 ϕ 2 k < ¹ 3 ε.

Finally from (2), (3), and (4) we get

kf − (ζ − z ⁰ ) ^N(z

^)+1+r ϕ 1 ϕ 2 k < ε

which ends the proof. _

Immediately we obtain the following result which if fundamental for what fol- lows.

Theorem 4.3 Let a Banach algebra B satisfies conditions (H1), (H2), (H3), and the modified Ditkin condition at a point z 0 ∈ T. Let I be a closed ideal of B such that h(I) = {z 0 }. Suppose that U ^I = 1 and let

m I = max{m: f(z 0 ) = f ⁰ (z ₀ ) = . . . = f ^(m) (z ₀ ) = 0 for all f ∈ I } < ∞.

Then

I = I _z ^m

= {f ∈ B : f(z ⁰ ) = f ⁰ (z 0 ) = . . . = f ^(m

⁾ (z 0 ) = 0}.

Proof By Corollary 3.4 and Proposition 4.1 we obtain the inclusions (ζ − z ⁰ ) ^k+1 B ⊂ I ⊂ (ζ − z ⁰ ) ^m

⁺¹ B.

Let p ­ 0 be the smallest integer such that (ζ −z 0 ) ^p+1 B ⊂ I. Obviously m ^I ¬ p.

It suffices to show that p ¬ m ^I . We can assume that p ¬ N(z 0 ) since, if p >

N (z 0 ) then Theorem 4.2 implies

(ζ − z ⁰ ) ^N(z

⁾⁺¹ B = (ζ − z ⁰ ) ^p+1 B ⊂ I.

Let f ∈ I. Define g(z) = P p

j=0 f

(z

)

j! (z − z ⁰ ) ^j . The function h = f − g satisfies h ^(j) (z 0 ) = 0, 0 ¬ j ¬ p, and by Proposition 4.1 h = lim ⁿ →∞ h n , where h n ∈ (ζ − z ⁰ ) ^p+1 B ⊂ I. It follows that h ∈ I and we have 0 = [h] = [f] − [g] = −[g] in the space B/I. This proves that P p

j=0 f

(z

)

j! [(ζ − z 0 ) ^j ] = 0. However the functions [(ζ − z 0 ) ^j ], 0 ¬ j ¬ p, are linearly independent in B/I by the definition of p.

All coefficients ^f

_j! ^(z

⁾ are equal to zero and we conclude that f ^(j) (z 0 ) = 0 for

0 ¬ j ¬ p. This proves that p = m ^I and ends the proof. _

Remark Theorem 4.2 states that in the algebras we are interested in the ideals of the form I z ^N

^(z

⁾ are minimal between ideals satisfying h ⁰ (I) = {z ⁰ } and U ^I = 1.

Suppose a Banach algebra B satisfies conditions (H1), (H2), and for some N ∈ N ⁰ and z 0 ∈ T the functionals f 7→ f ^(j) (z 0 ), 0 ¬ j ¬ N, are continuous. If the ideal I z ^N

is minimal, then the modified Ditkin condition at the point z 0 is satisfied.

Indeed, the ideal (ζ − z ⁰ ) ^N+1 I _z ⁰

is contained in I _z ^N

. The latter is the minimal closed ideal with h(I) = {z 0 } and U ^I = 1. Therefore (ζ − z 0 ) ^N+1 I _z ⁰

is dense in I _z ^N

. In particular, we can approximate the function (ζ −z 0 ) ^{N +1} by sequences of the form (ζ − z 0 ) ^N+1 σ n , where σ n (z ₀ ) = 0. This means that it is necessary that the algebra satisfies the modified Ditkin condition at the point z 0 if we want the ideal I _z ^N

to be minimal.

5. The division ideal and ideals with at most countable hull. Let I be an ideal of B and let g ∈ B. The division ideal I(g) is defined by the formula

I(g) = {f ∈ B : fg ∈ I}.

Obviously I ⊂ I(g) and if I is closed, then I(g) is closed. This implies that h(I(g)) ⊂ h(I).

In Lemmas 2.10, and 2.8 in [1] the following results were proved.

Theorem 5.1 Let B be a Banach algebra which satisfies conditions (H1), (H2), (H3), and the modified Ditkin condition (D).

(a) If I is a closed ideal of B such that h(I) is at most countable and g ∈ B is divisible by U I , then U I(g) = 1.

(b) If f ∈ B is divisible by an inner function S associated to a singular measure µ, then f ^(j) (z) = 0 for all z ∈ supp µ and for every 0 ¬ j ¬ N(z).

In this section we describe all closed ideals of B with at most countable hull.

For a closed ideal I of B and an integer j ∈ N ⁰ we define

h ^j (I) = {z ∈ T: N(z) ­ j, f(z) = f ⁰ (z) = . . . = f ^(j) (z) = 0 for all f ∈ I }.

The collection {h ^j (I)},  ∈ N ⁰ , is a descending family of subsets of the unit circle.

Definition 5.2 For any descending family of sets H n ⊂ T, n ∈ N ⁰ , such that N (z) ­ n for all z ∈ H ⁿ and for a given inner function U we define

I(U ; H n , n ∈ N ⁰ ) = {f ∈ B : U|f and f ⁽ⁿ⁾ (z) = 0 for z ∈ H n , n ∈ N ⁰ }.

Ideals of B which are of this form are called standard.

Now we prove the main result of the paper.

Theorem 5.3 Let B be a Banach algebra which satisfies conditions (H1), (H2), (H3), and the modified Ditkin condition (D). If I is a closed ideal of B and h(I) is at most countable, then I is a standard ideal:

I = I(U I ; h ^j (I), j ∈ N ⁰ ).

Proof Obviously I ⊂ I(U ^I ; h ^j (I), j ∈ N ⁰ ). Hence it remains to prove that every g ∈ I(U ^I ; h ^j (I), j ∈ N ⁰ ) belongs to I. It will be done if we prove that the division ideal I(g) is equal to B. Since in a unital Banach algebra B the condition h(I) = ∅ implies I = B it suffices to show that U ^I(g) = 1 and h ⁰ (I(g)) = ∅.

The first property is immediate by Theorem 5.1 (a).

For every j ­ 0 we have h ^j (I(g)) ⊂ h ^j (I) since I ⊂ I(g). Let z 0 ∈ h ⁰ (I(g)). If z ₀ is not an isolated point in h ⁰ (I), then for every f ∈ I and 0 ¬ j ¬ N(z 0 ) we have f ^(j) (z ₀ ) = 0, hence z ₀ ∈ h ^{N (z}

⁾ (I). Now, every z ₀ ∈ h ⁰ (I) such that z ₀ 6∈ h ^N(z

⁾ (I) is an isolated point of h ⁰ (I).

By the Shilov idempotent theorem (see [11]) applied to the algebra B/I there exists ψ ∈ B which vanishes on h ⁰ (I) \ {z ⁰ }, satisfies ψ(z ⁰ ) = 1, and ψ(1 − ψ) ∈ I.

Let us consider the ideal I(ψ). Again, we have h ⁰ (I(ψ)) ⊂ h ⁰ (I), but 1 − ψ, which is an element of I(ψ), never vanishes on h ⁰ (I) \ {z ⁰ }, so h ⁰ (I(ψ)) contains at most z 0 . Theorem 5.1 (b) implies that U _I(ψ) = 1, since z 0 6∈ h ^{N (z}

⁾ (I).

By Theorem 4.3 we obtain that the ideal I(ψ) is of the form I(ψ) = {f ∈ B : f ^(j) (z 0 ) = 0, 0 ¬ j ¬ m},

for some m ∈ N ⁰ which obviously is less or equal than N(z 0 ). In particular, g ∈ I(ψ) which implies ψ ∈ I(g), and finally z ⁰ 6∈ h ⁰ (I(g)).

Thus we have proved that h ⁰ (I(g)) contains only elements of h ⁰ (I) which satisfy z 0 ∈ h ^N(z

⁾ (I). It means that for z ₀ ∈ h ⁰ (I(g)) and for all f ∈ I(U ^I ; h ^j (I), j ∈ N 0 ) it holds

f (z 0 ) = f ⁰ (z ₀ ) = . . . = f ^(N(z

⁾⁾ = 0.

In particular this is true for the function g.

If h ⁰ (I(g)) 6= ∅, then it contains an isolated point z ⁰ . Again, by the Shilov idempotent theorem there exists ϕ ∈ B which vanishes on h ⁰ (I(g)) \ {z ⁰ }, satisfies ϕ(z 0 ) = 1, and ϕ(1 − ϕ) ∈ I(g).

Finally, consider the ideal J = (I(g))(ϕ). Obviously U J = 1 and since 1 − ϕ ∈ J it follows that h ⁰ (J) = {z ⁰ }. Theorem 4.3 implies that

J = {f ∈ B : f(z 0 ) = f ⁰ (z ₀ ) = . . . = f ^(M) (z ₀ ) = 0}

for some M ¬ N(z 0 ). The function g, and consequently also the product ϕg, belong to the ideal I z ^N(z

⁾ and according to Theorem 4.2 there exists in the latter space an approximative unit (ϕ n ). We have

n→∞ lim kϕ ⁿ ϕg − ϕgk B = 0.

The functions ϕ n belong to I z ^N

^(z

⁾ ⊂ J, hence ϕ ⁿ ϕ ∈ I(g) and ϕ ⁿ ϕg ∈ I. This implies ϕg ∈ I and consequently ϕ ∈ I(g). This is a contradiction, because ϕ(z ⁰ ) = 1, while z ₀ ∈ h ⁰ (I(g)).

In this way we have proved that I(g) = B and the proof is finished. 

6. Final remarks. Now we go back to the algebra B = A ⁽¹⁾ 1 ( D) from Example

2.2. We have seen that A ⁽¹⁾ 1 ( D) does not satisfy the analytic Ditkin condition. Now

we prove that the modified Ditkin condition (D) holds true in that algebra.

First we show that this condition is satisfied at the point 1. Notice that N(1) = 0.

We take the following sequence of functions in A ⁽¹⁾ 1 ( D) : ϕ n (z) = (1 − z)

.

(i) The functions ϕ n (z) = (1−z)

are bounded on D and tend almost uniformly to 1 on D\{1}. For |z−1| < 1 we have |ϕ n (z)| < 1. Take any ε with 0 < ε < 1. Then there exists n ₀ such that for all n ­ n 0 and |z − 1| ­ ¹ ₂ ε we have |ϕ ⁿ (z) − 1| < ¹ ₄ ε.

Since |z − 1| ¬ 2 on D we get

|(z − 1)(ϕ ⁿ (z) − 1)| < ε

for all z ∈ D and n ­ n ⁰ . This means that (z − 1)ϕ ⁿ (z) → z − 1 uniformly on D.

(ii) Since

(1 − z) ((z − 1) (ϕ ⁿ (z) − 1)) ⁰ = (1 − z) 1 + n ¹

ϕ n (z) − 1
,

similar argument as in (i) shows that (1 − z) ((z − 1) (ϕ ⁿ (z) − 1)) ⁰ → 0 uniformly on D.

From (i) and (ii) it follows that k(ζ − 1)ϕ ⁿ − (ζ − 1)k → 0 as n → ∞, i.e. the modified Ditkin condition is satisfied at the point 1.

If |z ⁰ | = 1 and z ⁰ 6= 1, then N(z ⁰ ) = 1. We take the functions ψ n (z) =

z−z

z − ( ¹⁺

) ^z

. Elementary calculations show that k(ζ − z ⁰ ) ² ψ n − (ζ − z ⁰ ) ² k → 0 as n → ∞.

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Andrzej Sołtysiak

Faculty of Mathematics and Computer Science, Adam Mickiewicz University ul. Umultowska 87, 61–614 Poznań, Poland

E-mail: asoltys@amu.edu.pl Antoni Wawrzyńczyk

Departamento de Matem´ aticas, Universidad Autónoma Metropolitana-Iztapalapa AP 55-534, 09340 M´exico D. F., M´exico

E-mail: awaw@xanum.uam.mx

(Received: 28.01.2012)