Interpolating Sequences on Uniform Algebras

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1 Interpolating Sequences on Uniform Algebras Pablo Galindo a, Mikael Lindström b, Alejandro Miralles,c a Departamento de Análisis Matemático, Universidad de Valencia Burjasot, Valencia, Spain b Department of Mathematical Sciences, University of Oulu Oulu, Finland c Instituto Universitario de Matemática Pura y Aplicada, Universidad Politécnica de Valencia Valencia, Spain Abstract We consider the problem whether a given interpolating sequence for a uniform algebra yields linear interpolation. A positive answer is obtained when we deal with dual uniform algebras. Further we prove that if the Carleson generalized condition is sufficient for a sequence to be interpolating on the algebra of bounded analytic functions on the unit ball of c 0, then it is sufficient for any dual uniform algebra. Key words: uniform algebra, interpolating sequence, analytic function, pseudohyperbolic distance 2008 MSC: Primary 46J15, Secondary 46E50 1. Background Let A be a uniform algebra and let M A denote its spectrum, i.e., its maximal ideal space endowed with the topology of the pointwise convergence. Recall that given f A, its Gelfand transform f is a continuous function on the compact space M A. Let (x n ) n=1 be a sequence of elements in M A. The sequence (x n ) is said to be an interpolating sequence for A if for any bounded sequence (α n ) n=1 C, there exists f A such that f(x n ) = α n. We look at this notion in a trivially equivalent way: Consider the restriction map R : A l defined by R(f) = ( f(x n )). Clearly R is well-defined, linear and continuous since R(f) = sup{ f(x n ) : n N} f = f. It turns out that the sequence (x n ) is interpolating for A if and only if there is a map T : l A such that R T = id l. Whenever such map T : l A is a linear operator, (x n ) is called a linear interpolating sequence for A. Supported by Project MTM (MEC-FEDER. Spain). Alejandro Miralles was also supported by the Juan de la Cierva programme (MICINN. Spain) Corresponding author addresses: galindo@uv.es (Pablo Galindo), mikael.lindstrom@oulu.fi (Mikael Lindström), almimon@csa.upv.es (Alejandro Miralles ) Preprint submitted to Elsevier October 9, 2009

2 { For any α = (α j ) l, let M α = inf f : f(x } j ) = α j, j N, f A. The constant of interpolation for (x n ) is defined by M = sup {M α : α l, α 1}. Any real number not smaller than M is a constant of interpolation for (x n ). Analogously, the notions of c 0 -(linear) interpolating sequence and its constant of interpolation are defined by simply replacing l by c 0 in the above definitions. The starting point for this research is a result of P. Beurling (see [Gr] and [C]): Theorem 1.1. Let (z j ) D be an interpolating sequence for H and M a constant of interpolation for (z j ). Then, there exists a sequence (f j ) H such that f k (z j ) = δ kj for k, j N and Some years later A. M. Davie [D] proved f j (z) M for any z D. Theorem 1.2. Let (z n ) D be a c 0 -interpolating sequence for the disk algebra A(D). Then, (z n ) is a c 0 -linear interpolating sequence for A(D). Both results lead to the following Definition 1.3. Let A be a uniform algebra and (x j ) a sequence in M A. A sequence of functions (f k ) A is a sequence of Beurling functions for (x n ) if there exists M > 0 such that f k (x j ) = δ kj for any k, j N and f j (x) M for any x M A. N. Th. Varopoulos proved (see [V] and [Gr]) a general result on uniform algebras replacing the constant of interpolation M by a worse one. Theorem 1.4. Let A be a uniform algebra on a compact set K. Consider the finite sequence {x 1, x 2,..., x n } K and let M be its constant of interpolation. For any ε > 0, there exist functions f 1, f 2,..., f n in A such that f k (x j ) = δ kj for any k, j N and such that sup x K n f j (x) M 2 + ε. This result was complemented in [GGL] by reaching the interpolation of an infinite sequence in the bidual algebra. Recall that the bidual A of a uniform algebra A is also a uniform algebra endowed with the Arens product (see [A]). 2

3 The evaluation functionals at points of M A extend uniquely to be weak-star continuous multiplicative functionals on A, so we can regard M A as a subset of M A and, accordingly, we say that a sequence (x n ) M A is interpolating for A if (x n ) is interpolating for A as a subset of M A. Further results on the bidual of a uniform algebra can be found in [Ga] and [DH]. See also [GGL]. Theorem 1.5. Let A be a uniform algebra and (x n ) M A. Let M 1 such that for each finite collection {α 1, α 2,..., α n } of complex numbers of unit modulus, there exists f A satisfying f(x j ) = α j for any 1 j n and f M. Then, there is a sequence (f n ) n=1 A such that f k (x j ) = δ kj for any k, j N and f j (x) M 2 for any x M A. In the second section of the article we study the connection between interpolating sequences, linear interpolating sequences and c 0 -(linear) interpolating sequences. Roughly speaking we deduce that for dual uniform algebras, c 0 - interpolating sequences are actually linear interpolating; this applies in particular to algebras of bounded analytic functions to which we extend a result of Mujica (see below). In Section 3 we prove that if a Carleson generalized condition is sufficient for a sequence to be interpolating on the algebra of bounded analytic functions on the unit ball of c 0, then it is sufficient for any dual uniform algebra. Finally we study sequences in the maximal ideal space generalizing some results of Axler and Gorkin. 2. Linear Interpolating Sequences on Uniform algebras Our first result shows the equivalence for a sequence to be c 0 -linear interpolating and the existence of the corresponding Beurling functions. Proposition 2.1. Let A be a uniform algebra and (x n ) M A. Consider the following statements: a) (x n ) is a linear interpolating sequence for A. b) There exists a sequence (f n ) A of Beurling functions for (x n ). c) (x n ) is a c 0 -linear interpolating sequence for A. Then b) and c) are equivalent and a) implies both b) and c). Proof. Clearly, a) c). b) c). Set α = (α n ) c 0. Since sup x MA n=1 f n (x) M for any x M A, we have that n=1 α n f n (x) is also defined for any x M A. Moreover, k n=1 α nf n A converges uniformly on M A to the function n=1 α nf n since k α n f n α n f n = sup α n fn (x) n=1 n=1 3 x M A n=k+1

4 sup α n sup n k+1 x M A n=k+1 f n (x) M sup α n 0, n k+1 so n=1 α nf n belongs to A. Let R : A l be the restriction map. We define its right inverse T : c 0 A by T ((α n )) := α n f n. n=1 This is a well-defined linear operator and we have that ( ) ( R T (α) = R α n f n = α n fn (x j ) n=1 n=1 ) = (α n ) = α for all α c 0. Therefore, (x n ) is a c 0 -interpolating sequence for A. c) b). Now, there exists a linear operator T : c 0 A such that R T = id c0. Set f n := T (e n ) A. Then, (see Proposition II.D.4 in [W] for instance) there is a constant M such that n=1 u (f n) M u for every u A. Since M A lies in the unit ball of A, n=1 f n (x) M for x M A, and clearly, f n (x k ) = T (e n )(x k ) = δ nk. So, (f n ) is a sequence of Beurling functions for (x k ). In general (c) does not imply (a): Just consider the disk algebra A(D) and a convergent sequence in the unit circle. By the Rudin-Carleson Theorem, this is a c 0 -interpolating sequence for A(D) and further linear interpolating by Theorem 1.2. It is clear that this sequence is not interpolating for A(D) since this algebra is separable. Nevertheless, c 0 -linear interpolating sequences are always linear interpolating when we deal with the bidual A : Proposition 2.2. Let A be a uniform algebra and (x n ) M A. If (x n ) is a c 0 -linear interpolating sequence for A, then (x n ) is linear interpolating for A. Proof. Since (x n ) is c 0 -linear interpolating, then there exists a linear operator T : c 0 A such that R T = id c0. Then, for α = (α n ) c 0, we have that T (α)(x n ) = α n for any n N. Consider the second adjoint T : l A and fix α l. We have that T (x n ) l 1 and, considering the sections α k = (α 1,..., α k, 0,... ) c 0, we have that the sequence (α k ) w(l, l 1 )-converges to α. Therefore, it follows that T (α), x n = α, T (x n ) = lim k (α 1,..., α k, 0,... ), T (x n ) = lim k (α 1,..., α k, 0,... ), T (x n ) = lim k T ((α 1,..., α k, 0,... )), x n = α n. Hence, for any α l, we have that T (α)(x n ) = α n for any n N. Next we consider dual uniform algebras A = X, that is uniform algebras isomorphically isometric to the dual of some Banach space X. We prove that c 0 -interpolating sequences (x n ) M A X for A are also linear interpolating. 4

5 Lemma 2.3. Let A be a dual uniform algebra A = X for some Banach space X and consider (x n ) M A X a c 0 -linear interpolating sequence for A. Then (x n ) is linear interpolating for A. Proof. Since (x n ) is c 0 -linear interpolating, there is a linear operator T : c 0 A such that R T = id c0. Then, as above, the sequence of Beurling functions (T (e n )) = (f n ) defines a w(a, A )-Cauchy series in A, n=1 f n, and there is a constant M such that n=1 u (f n) M u for every u A. Hence, for each sequence α = (α j ) l, the series n=1 α nf n is a w(a, A )-Cauchy series in A and u (α n f n ) M u α for every u A. n=1 Therefore, n=1 α nf n is a w(a, X) Cauchy series, thus convergent since the ball in A of radius M α is w(a, X) compact. Hence, n=1 α nf n A. In addition, we have that n=1 α nf n M α. Define the map T : l A by T ((α j )) := α jf j. It is linear and, since each x k belongs to X, we obtain that α j f j (x k ) = α j fj (x k ) = α k. Thus (x n ) is linear interpolating for A. Theorem 2.4. Let A = X be a dual uniform algebra and (x n ) M A X. The following statements are equivalent: a) Every finite subset of (x n ) is interpolating for A and there exists a constant of interpolation independent of the number of interpolated terms. b) (x n ) is c 0 -interpolating for A. c) (x n ) is linear interpolating for A. Proof. It is clear that c) b). b) a). Let (x n ) be a c 0 -interpolating sequence. Then, there is a map T : c 0 A such that R T = id c0. Consider the set B = R 1 (c 0 ). Since R is continuous, we have that B is a closed subspace of A and, therefore, it is a Banach subspace of A. Moreover, the mapping R B : B c 0 is surjective since for all α c 0 we have that T (α) B and R(T (α)) = α. In consequence, R B is an open mapping and the quotient mapping R B : B/ ker ( R B ) c0 has a continuous linear inverse. Consider M = R B 1 > 0. We have that sup {inf{ f : f A, ( f(x n )) n = α}} α 1 5

6 sup {inf{ f : f B, ( f(x n )) n = α}} = M. α 1 Now, for every finite subset {x 1,..., x n }, it is clear that the constant of interpolation M n := sup {inf{ f : f A, f(x j ) = η j, j = 1, 2,..., n}} (η j) 1 satisfies the inequality M n M. a) c) We will use Theorem 1.4 and a normal families reasoning. In addition, we will use some ideas from Theorem 1.5. Let (ɛ n ) be a null sequence of positive numbers. By Theorem 1.4, we have that for each ɛ n there are functions f n 1,..., f n n in A such that f n j (x k) = δ jk for j, k = 1,..., n and such that sup x M A n f j n(x) M 2 + ɛ n. Since A = X, its unit ball is a w(a, X)-relatively compact set, thus from the sequence (fj n) n j we may obtain a net (fj α ) w(a, X)-convergent to an element F j A. Since x k X, it follows that F j (x k ) = lim α f α j (x k ) = δ jk for any j, k N. Fix m 1 and let a 1,..., a m be complex numbers of unit modulus. For any n m and all x M A we have m m a j f n n j (x) f j n(x) f j n(x) M 2 + ɛ n. From this, we have for any n m, m a jfj n M 2 + ɛ n. Thus for all u B X, we have u, m a jfj n > M 2 + ɛ n, and by passing to the w(a, X) limit, we obtain u, m a jf j > M 2, hence m a jf j M 2. Therefore, we obtain m a j Fj (x) M 2 for any x M A. This is true in particular for a j = F j (x) / F j (x) if F j (x) 0 and, therefore, we obtain m F j (x) M 2 for all x M A. By Proposition 2.1, we obtain that (x n ) is a c 0 -linear interpolating sequence for A and the proof is completed by applying Lemma 2.3. The following corollary proves that, under some assumptions, uniform algebras A are dual algebras A = X and we can choose the predual X to contain the interpolating sequence (x n ). Corollary 2.5. Let A be a closed subalgebra of l (Y ) for some set Y whose points are separated by A. Suppose that the limit of any bounded net of functions in A that converges pointwise on Y also belongs to A. If (x n ) is a c 0 -interpolating sequence for A, then it is linear interpolating for A. 6

7 Proof. Set Y 1 = Y {x n : n N} M A, so A is also a closed subalgebra of l (Y 1 ). Since Y 1 satisfies the same assumptions as Y, the condition on pointwise bounded limits guarantees by the Krein-Schmulian theorem that A is a weak* closed subspace of l (Y 1 ). Thus A is the dual of the Banach space X := l 1 (Y 1 )/A, where A = {x l 1 (Y 1 ) : x, f = 0 for all f A} and every y Y 1 is identified with the characteristic function δ y. Therefore, (x n ) X. Now, it suffices to apply Theorem 2.4. Let E be a complex Banach space with open unit ball B E. Given an open set U E, a function f : U C is said to be analytic if it is Fréchet differentiable. The space H (U) denotes the set {f : U C : f is analytic and bounded}. We denote by A u (U) the set {f : U C : f is analytic and uniformly continuous}. Both spaces are uniform algebras endowed with the sup-norm f = sup{ f(x) : x U}. J. Mujica proved that if (x n ) U is an interpolating sequence for H (U), then it is also linear interpolating [M]. This result is extended by Corollary 2.5 to any c 0 -interpolating sequence since H (U) is a closed subalgebra of l (U) fulfilling the assumptions by Montel s Theorem. Therefore we may state: Proposition 2.6. Let A = H (U) and (x n ) M A. Then, the following conditions are equivalent, a) There exists a sequence (f n ) of Beurling functions for (x n ). b) The sequence (x n ) is c 0 -interpolating for H (U). c) The sequence (x n ) is linear interpolating for H (U). We close the section by emphasizing that the assumption of the algebra being a dual space is crucial to obtain linear interpolation from bare interpolation. Indeed, A. M. Davie proved [D] that, when we deal with the algebra A u (2B c0 ), there exists an example of a c 0 -interpolating sequence on its spectrum M A = 2B l (see Proposition 21 in [GL1]) which is not c 0 -linear interpolating. We provide a somehow different proof of this result as an application of results on composition operators. Recall that the Bartle-Graves selection Theorem states that if E and F are Banach spaces and T : E F is a surjective linear operator, then there exists a continuous function g : F E such that T g is the identity map on F. Example 2.7. Let A be the algebra A u (2B c0 ). There exists a c 0 -interpolating sequence for A which does not admit linear interpolating subsequences. Proof. Let {f j } be a dense sequence in the unit ball of c 0, chosen in c 00, that is, f j (n) = 0 for n large enough depending on j. Define x i B l by x i (j) = f j (i). As we have mentioned above, the spectrum of A u (2B c0 ) is given by M A = 2B l = {(z n ) : z n 2}. 7

8 It is clear that each x i belongs to the spectrum and the sequence (x i ) converges to 0 there since the Gelfand topology coincides with the pointwise topology and x i (j) = f j (i) 0 when i. Consider the restriction map R : A u (2B c0 ) c which is defined by R(f) = ( f(x i )) i=1. For any j N, let z j A u (2B c0 ) be the coordinate functions defined by z j (x) = x j. We have that z j (x i ) = x i (j) = f j (i), so then R maps the unit ball of A u (2B c0 ) onto a dense set of B c0. In consequence, the mapping R : A u (2B c0 ) c is onto by the open mapping Theorem and therefore, by the Bartle-Graves selection Theorem, there exists a map T : c A u (2B c0 ) such that R T = Id c. By taking the restriction map T : c 0 A u (2B c0 ), we obtain that (x i ) is a c 0 -interpolating sequence. We show that (x i ) has no linear interpolating subsequences. Consider the natural embedding ι : B c0 2B c0 and the composition operator C ι : H (2B c0 ) H (B c0 ) defined by C ι (f) = f Bc0. This operator is completely continuous according to Proposition 10 in [GLR]. Observe that the restriction C ι : A u (2B c0 ) A u (B c0 ) is still completely continuous and the adjoint C ι restricted to the spectrum B l is the canonical embedding B l 2B l. Indeed, if δ x B l, the homomorphism Cι (δ x ) coincides with δ x since both coincide on the linear functionals on c 0 (i.e., on l 1 ) and consequently on the dense subspace of finite type polynomials P f (c 0 ). Suppose that (x i ) has a c 0 -linear interpolating subsequence. Without loss of generality, we can assume that (x i ) itself is c 0 -linear interpolating. Then there exist a linear operator T : c 0 A u (2B c0 ) and a sequence (F k ) A u (2B c0 ) such that F k (x i ) = δ ki and further, F j (x) M for all x B l. This means that F j is weakly Cauchy, so C ι(f j ) = F j Bc0 is a Cauchy series in A u (B c0 ). Therefore, (F k Bc0 ) k is a null sequence there. However, this is not possible since F k Bc0 = C ι (F k ) x k, C ι (F k ) = C ι (x k ), F k = F k (x k ) = Interpolating sequences and the pseudohyperbolic metric Recall that for a given uniform algebra A, the pseudohyperbolic distance for x, y M A, is given by ρ A (x, y) = sup{ρ(f(x), f(y)) : f A, f 1}. In the particular case that A = H, it is well-known that ρ(z, w) = z w 1 zw for z, w D. 8

9 It is also well-known (see [AGL]) that the expression for the pseudohyperbolic distance for H (B c0 ) is given by x(n) y(n) ρ(x, y) = sup for x, y B n N 1 x(n)y(n) c0. (3.1) A sequence (x n ) M A is said to be (ɛ)-hyperbolically separated if there is ɛ > 0, such that ρ A (x n, x m ) ɛ for all n m. Remark 3.1. Let (x n ) M A. If (x n ) is c 0 -interpolating for A with interpolation constant M, then (x n ) is ( 1 M ) -hyperbolically separated. This necessary condition is very far from being sufficient: Let {q n } be the subset of D of complex numbers with rational argument in [0, 2π] and take A as the disc algebra. Clearly {q n } is a hyperbolically separated sequence since ρ(q n, q m ) = 1 if n m. It is not c 0 -interpolating since for the sequence ( 1 n ) if there were f A such that f(q n ) = 1 n, then we would have got for every ɛ > 0 some n 0 such that f(q n ) < ɛ for n > n 0. Therefore, f ɛ because {q n : n > n 0 } is dense in the Shilov boundary D. So we would have shown f = 0. The most well-known result on interpolating sequences for H is due to L. Carleson and states that a necessary and sufficient condition for a sequence (z n ) D to be interpolating for H is the existence of δ > 0 such that z k z j 1 z k z j > δ for any j N. k j Such Carleson condition can be stated for general uniform algebras: Definition 3.2. Let A be a uniform algebra and (x n ) a sequence in M A. The sequence (x n ) is said to satisfy the generalized Carleson condition if there exists δ > 0 such that ρ A (x k, x j ) > δ for all j N. (3.2) k j B. Berndtsson [B] proved that the generalized Carleson condition is sufficient for a sequence in the unit ball B n of C n to be interpolating for H (B n ); this was extended to any Hilbert space in [GGL1]. In [BCL] the authors showed that the analogous result also holds for the polidisc. In addition, they noticed that if the constant of interpolation was independent of the dimension n, then for arbitrary uniform algebras A and finite sequences (x k ) N k=1 M A satisfying the generalized Carleson condition, we would obtain that (x k ) N k=1 is an interpolating sequence for A with the constant of interpolation depending only on δ, and not on the number of points N in the sequence. Such kind of dependence of the constant of interpolation plays a decisive role in the description of the spectra of some composition operators, see [GM]. Following this idea, we have 9

10 Proposition 3.3. Suppose that any sequence (x n ) B c0 satisfying the generalized Carleson condition 3.2 is interpolating for H (B c0 ) with interpolation constant depending only on δ. Then, for any dual uniform algebra A = X, all sequences (x n ) X M A satisfying the generalized Carleson condition 3.2 are linear interpolating sequences for A with constant of interpolation depending only on δ. Proof. Assume there is δ > 0 such that j k ρ A(x j, x k ) > δ for all k. For k, j N there exists f k,j A such that f k,j (x k ) = 0, f k,j 1 and ( ρ A (x j, x k ) f j,k (x j ) 1 1 ) 2 j+k ρ A (x j, x k ). Fix n N and define ϕ : M A B c0 according to ϕ(u) := (u(f 1,1 ), u(f 1,2 ), u(f 2,1 ),..., u(f n,n ), 0, 0,...). Then, by the expression for the pseudohyperbolic distance 3.1 for c 0, we have that ρ H (B c0 )(ϕ(u), ϕ(v)) = max j,k {ρ(u(f j,k), v(f j,k )}. Thus Hence r s ρ H (B c0 )(ϕ(x r ), ϕ(x s )) = max j,k {ρ(f j,k(x r ), f j,k (x s ))}) ρ(f s,r (x r ), f s,r (x s )) n ρ H (B c0 )(ϕ(x r ), ϕ(x s )) r s ( 1 1 ) n 2 r+s ρ A (x r, x s ) r s ( 1 1 ) 2 r+s ρ A (x r, x s ). n r s r s ( 1 1 ) 2 r+s ρ A (x r, x s ) ( 1 1 ) 2 r+s δ r=1 (1 12 r ) δ which is greater than Cδ since the product converges. Then, the hypothesis guarantees that the finite sequences (ϕ(x j )) n are interpolating sequences for H (B c0 ) with interpolation constant M depending only on δ. Therefore, for given (α j ) n C such that α j 1 for any j = 1,..., n, there is F H (B c0 ) such that F M and (F ϕ)(x j ) = F (ϕ(x j )) = α j. Since F actually depends only of a finite number of variables, it turns out that F ϕ A. Hence (x j ) n interpolates (α j) n by means of (F ϕ) and F ϕ M. By Theorem 2.4, the sequence (x n ) is interpolating for A and, in addition, the constant of interpolation is bounded by M 2. Notice that for finite dimensional Banach spaces E, interpolating sequences (x n ) for H (B E ) satisfy that x n converge to 1. If we deal with infinite dimensional Banach spaces, we can extend this result as follows, 10

11 Proposition 3.4. Consider the following statements: a) All polynomials on E are weakly continuous on bounded sets. b) If (x n ) B E is an interpolating sequence for H (B E ), then x n converges to 1. c) The space E does not contain copies of l 1. Then a) implies b) and b) implies c). In addition, if all polynomials on E are weakly sequentially continuous, then c) implies a) and, therefore, the three statements are equivalent. Proof. b) c) If E contains a copy of l 1, then it is well-known that l 2 is a quotient of E. Let q : E l 2 be the quotient map and denote by e n the n th unit vector of the canonical basis of l 2. Let (y n ) E be a bounded sequence such that q(y n ) = e n for any n N. Let C > 0 be such that y n C 2 and set x n = yn C. We check that (x n ) is a c 0 -interpolating sequence. Indeed, take (α n ) c 0 and let P P ( 2 E) be the polynomial defined by P (x) = C 2 α n x 2 n for x = (x n ) l 2. n=1 The polynomial P q : E C also belongs to P ( 2 E) and ( ) q(yj ) (P q)(x j ) = P = P (e j) C C 2 = α j. Then, (x j ) is c 0 -interpolating but x n 1 2. This contradicts b). a) b) Suppose that the conclusion fails. We can suppose, passing to subsequences if necessary, that there exists an interpolating sequence (x n ) such that x j r < 1 for all j. By the assumptions on E, every f H (B E ) is weakly uniformly continuous on rb E. Hence every f H (B E ) extends to an analytic function ˆf on B E which is weak-star continuous on rb E. By Rosenthal s l 1 Theorem, either (x j ) has a subsequence equivalent to the unit basis of l 1 or it has a weak Cauchy subsequence. The first alternative cannot hold since otherwise we could find, in the same way we have done in (b) (c), a polynomial which would be non weakly continuous on the unit ball. So we can suppose that (x j ) is a weak Cauchy sequence. If we consider (x j ) as a sequence in E, we get that (x j ) is a w -convergent sequence in E since B E is w -compact. Let x j x in (rb E, w(e, E )). Then f(x j ) ˆf(x) for all f H (B E ). On the other hand, since (x j ) is c 0 -interpolating, it is interpolating by Theorem 2.4, which contradicts the former assertion. It is easy that c) a) under the assumption that all polynomials are weakly sequentially continuous since if E does not contain copies of l 1, all weakly 11

12 sequentially continuous polynomials are weakly continuous on bounded sets (see [Di]). Recall (see [ACG]) that for E = l p, 1 < p < +, any sequence (λ j e j ) with 0 < inf λ j and λ j < 1 is c 0 -interpolating for H (B E ), so (c) may not imply either (b) or (a) in the above proposition if the assumption on polynomials is not satisfied. Let us also remark that for (c) to imply (a) it suffices that E enjoys the Dunford-Pettis property. In [CGG], several examples of spaces satisfying that all polynomials are weakly sequentially continuous and lacking the Dunford-Pettis property are exhibited; let us mention among them the dual S of the Schreier space and the predual d (w) of some Lorentz sequence space d(w; 1). 4. Sequences in the maximal ideal space It is well-known that two open pseudohyperbolic balls in M A of radius 1 either are disjoint or coincide. These balls are called the Gleason parts of A. For a discussion of Gleason parts, see Chapter VI of [Ga]. The following notion was introduced in [GGL] in close connection with interpolating sequences. A subset B M A is said to be hyperbolically bounded if it is contained in a finite union of pseudohyperbolic balls whose radii are strictly less than 1. Theorem 4.1. [GGL] Let A be a uniform algebra. Let E be a subset of M A which is not hyperbolically bounded. Then, for each ε > 0 there are a sequence (x k ) E and a sequence of Beurling functions (F j ) in A satisfying F k (x j ) = δ k,j and k=1 F k(x) 1 + ε for any x M A. From this result we obtain, Proposition 4.2. Let A = X and suppose that (x n ) X M A is not hyperbolically bounded. Then, for each ɛ > 0 there is a subsequence (x nk ) which is (linear) interpolating for A with constant of interpolation M < 1 + ɛ. Proof. If (x n ) is not hyperbolically bounded, then by Theorem 4.1 we find a subsequence (x nk ) which is (linear) interpolating for A with constant of interpolation M < 1 + ɛ. Actually, (x nk ) is interpolating for A. Indeed, set (η j ) l and F A so that F (x nk ) = η k. Select a bounded net (f α ) in A w(a, A ) convergent to F. As the bounded sets in A are w(a, X) relatively compact, there is a subnet of (f α ) w(a, X) convergent to some g A, which we denote the same. Then g(x nk ) = lim α f α (x nk ) = F (x nk ), which completes the proof. Proposition 4.3. If A is a dual space, A = X, and (x n ) X M A converges to x M A, then there is 0 < r < 1 such that ρ A (x n, x) < r for n big enough. 12

13 Proof. The sequence (x n ) cannot have, as a convergent sequence, interpolating subsequences. Thus, the above proposition shows that (x n ) is hyperbolically bounded. So there is 0 < r < 1 and a finite number of disjoint closed balls for the ρ A distance of radius r, B 1,..., B m such that (x n ) B 1 B m. Since ρ A as a supremum of continuous functions in M A is lower semi-continuous on M A, such balls are closed in M A. Then only one of such balls contains an infinite number of elements in (x n ), since, otherwise, x would be in the closure of, at least, two of them. The corollary below extends Corollary 9 of [AG] to A = H (B E ) for arbitrary E; in particular if (z n ) B E converges in M A, then (z n ) lies strictly inside B E. However, it does not necessarily converge to some z B E as we will see later. Corollary 4.4. Let A be a uniform algebra on Y such that the limit of any bounded net of functions in A that converges pointwise on Y also belongs to A. If (x n ) converges to x M A, then there is 0 < r < 1 such that ρ A (x n, x) < r for n large enough. Proof. According to Corollary 2.5, it turns out that A is the dual space of some X with (x n ) X. Now, simply apply the above proposition. By using the same argument of Axler and Gorkin in [AG] Corollary 10 we may also extend their result: Each path connected component of the spectrum of H (B E ) is contained in a Gleason part. Whereas in the maximal ideal space of H convergent sequences are norm convergent (see [AG] Cor. 13 or [GL] Prop. 17), this may fail in H (B E ) if E is infinite dimensional. Indeed: Set E = c 0 and z n = ( 1 2,. n).., 1 2, 0,... ). Then (z n) is w(l, l 1 )-convergent to z = ( 1 2,..., 1 2,... ), and, clearly, not norm convergent. According to Davie and Gamelin [DG], for all f H (B E ) with Taylor series f = P m, the series Pm converges in B E to a function f H (B E ), where P is the Aron-Berner extension of P to E. Since polynomials on c 0 are weakly uniformly continuous on bounded sets, their Aron-Berner extensions are w(l, l 1 ) continuous on bounded sets. Thus it turns out that for any n- homogeneous polynomial P, lim P (z n ) = P (z). Since the partial sums of the Taylor series approximate f on 1 2 B c 0, one checks that lim f(z n ) = f(z) for all f H (B c0 ), that is, δ zn δ z in the maximal ideal space. However (δ zn ) is not a Cauchy sequence in H (B c0 ) because then (z n ) would be a Cauchy sequence in B c0. References [A] R. Arens, Operation induced in function classes, Monatsh. Math. 55 (1951),

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15 [GLR] P. Galindo, M. Lindström and R. Ryan, Weakly compact composition operators between algebras of bounded analytic functions, Proc. Amer. Math. Soc. 128 (1999), [GM] P. Galindo and A. Miralles, Spectra of non-power compact composition operators on H spaces, Integr. equ. oper. theory. (to appear) [Ga] T. W. Gamelin, Uniform algebras on plane sets, Approximation Theory, Academic Press (1973), [Gr] J.B. Garnett, Bounded analytic functions, Academic Press, New York- London (1981). [M] J. Mujica, Linearization of holomorphic mappings on infinite dimensional spaces, Rev. Unión Mat. Argentina 37 (1991), [V] N. Th. Varopoulos, Ensembles pics et esembles d interpolation pour les algebras uniformes, C. R. Acad. Sci. Paris, Sér. A 272 (1971), [W] P. Wojtaszczyk, Banach spaces for analysts, Cambridge studies in advanced mathematics 25, Cambridge University Press, Cambridge (1991). 15