Let $(z_n)$ be a $H^\infty$-interpolating sequence on the open complex unit disc $\mathbb D$. If $A$ is some Banach space of analytic functions on the disc, denote by $X$ the closed subspace of all the functions in $A$ that have zeros at $z_n$ for all $n \in \mathbb N$. What I'm interested in is under what conditions can we say that the annihilator of $X$ can be represented as the closed span of the point evaluation functionals $\delta_{z_n}$ on points $z_n$? It is easily seen that every $\delta_{z_n}$ must be in the annihilator but for anything else we probably need more assumptions.

I know that if $A$ is a Hardy space $H^p$ with any $1 < p < \infty$, then we have $X = BH^p$ where $B$ is the Blaschke product associated with the sequence $(z_n)$ (the sequence being interpolating is a stronger condition than it being a Blaschke sequence) and we can see the claim is true.

It seems that a similar factorization could be carried out in Bergman spaces $A^p$, $1 < p < \infty$, too but I'm not very familiar with the details. [Duren, Schuster: Bergman Spaces] (esp. p. 146) seems to say that we have a similar bounded divisor function and the claim would be true. This apparently does not apply to all zero sequences, but the sequence being interpolating is a strong enough condition to make this work. I have been unable to find any answers considering the case of standard weighted Bergman spaces $A^p_\alpha$, $-1 < \alpha < \infty$.

Would it be possible to obtain a general proof for this if we assume that $A^* = \overline{\text{span}} \{\delta_z | z \in \mathbb D\}$ holds for the dual and that $fg \in A$ for all $f \in A$ and $g \in H^\infty$? Then if the annihilator were to contain anything besides the closed span of the $\delta_{z_n}$, it would have to contain some $l \in A^*$, a (possibly infinite) linear combination of some other point evaluation functionals. I feel that it should be possible to show that this leads to a contradiction but have been unable to carry it out. We can use the interpolation property to construct a bounded function $g \in X$ that has zeros on all $z_n$ and arbitrary values on any other finite number of points (since we can add a finite number points to the sequence and have it remain interpolating) but this is not enough to guarantee that $l(g) \neq 0$. But now also $fg \in X$ for every $f \in A$ so I feel this should be enough but am unable to carry the final steps. Maybe I'm missing something obvious?

Let $(z_n)$ be a $H^\infty$-interpolating sequence on the open complex unit disc $\mathbb D$. If $A$ is some Banach space of analytic functions on the disc, denote by $X$ the closed subspace of all the functions in $A$ that have zeros at $z_n$ for all $n \in \mathbb N$. What I'm interested in is under what conditions can we say that the annihilator of $X$ can be represented as the closed span of the point evaluation functionals $\delta_{z_n}$ on points $z_n$? It is easily seen that every $\delta_{z_n}$ must be in the annihilator but for anything else we probably need more assumptions.

I know that if $A$ is a Hardy space $H^p$ with any $1 < p < \infty$, then we have $X = BH^p$ where $B$ is the Blaschke product associated with the sequence $(z_n)$ (the sequence being interpolating is a stronger condition than it being a Blaschke sequence) and we can see the claim is true.

It seems that a similar factorization could be carried out in Bergman spaces $A^p$, $1 < p < \infty$, too but I'm not very familiar with the details. [Duren, Schuster: Bergman Spaces] (esp. p. 146) seems to say that we have a similar bounded divisor function and the claim would be true. This apparently does not apply to all zero sequences, but the sequence being interpolating is a strong enough condition to make this work. I have been unable to find any answers considering the case of standard weighted Bergman spaces $A^p_\alpha$, $-1 < \alpha < \infty$.

Would it be possible to obtain a general proof for this if we assume that $A^* = \overline{\text{span}} \{\delta_z | z \in \mathbb D\}$ holds for the dual and that $fg \in A$ for all $f \in A$ and $g \in H^\infty$? Then if the annihilator were to contain anything besides the closed span of the $\delta_{z_n}$, it would have to contain some $l \in A^*$, a (possibly infinite) linear combination of some other point evaluation functionals. I feel that it should be possible to show that this leads to a contradiction but have been unable to carry it out. We can use the interpolation property to construct a bounded function $g \in X$ that has zeros on all $z_n$ and arbitrary values on any other finite number of points (since we can add a finite number points to the sequence and have it remain interpolating) but this is not enough to guarantee that $l(g) \neq 0$. But now also $fg \in X$ for every $f \in A$ so I feel this should be enough but am unable to carry the final steps. Maybe I'm missing something obvious?

Edit: I have found out that a divisor function does indeed exist in all standard weighted Bergman spaces with my conditions: (http://www.ams.org/mathscinet-getitem?mr=427650)

Let $(z_n)$ be a $H^\infty$-interpolating sequence on the open complex unit disc $\mathbb D$. If $A$ is some Banach space of analytic functions on the disc, denote by $X$ the closed subspace of all the functions in $A$ that have zeros at $z_n$ for all $n \in \mathbb N$. What I'm interested in is under what conditions can we say that the annihilator of $X$ can be represented as the closed span of the point evaluation functionals $\delta_{z_n}$ on points $z_n$? It is easily seen that every $\delta_{z_n}$ must be in the annihilator but for anything else we probably need more assumptions.

I know that if $A$ is a Hardy space $H^p$ with any $1 < p < \infty$, then we have $X = BH^p$ where $B$ is the Blaschke product associated with the sequence $(z_n)$ (the sequence being interpolating is a stronger condition than it being a Blaschke sequence) and we can see the claim is true.

It seems that a similar factorization could be carried out in Bergman spaces $A^p$, $1 < p < \infty$, too but I'm not very familiar with the details. [Duren, Schuster: Bergman Spaces] (esp. p. 146) seems to say that we have a similar bounded divisor function and the claim would be true. This apparently does not apply to all zero sequences, but the sequence being interpolating is a strong enough condition to make this work. I have been unable to find any answers considering the case of standard weighted Bergman spaces $A^p_\alpha$, $-1 < \alpha < \infty$.

Would it be possible to obtain a general proof for this if we assume that $A^* = \overline{\text{span}} \{\delta_z | z \in \mathbb D\}$ holds for the dual and that $fg \in A$ for all $f \in A$ and $g \in H^\infty$? Then if the annihilator were to contain anything besides the closed span of the $\delta_{z_n}$, it would have to contain some $l \in A^*$, a (possibly infinite) linear combination of some other point evaluation functionals. I feel that it should be possible to show that this leads to a contradiction but have been unable to carry it out. We can use the interpolation property to construct a bounded function $g \in X$ that has zeros on all $z_n$ and arbitrary values on any other finite number of points (since we can add a finite number points to the sequence and have it remain interpolating) but this is not enough to guarantee that $l(g) = 0$. But now also $fg \in X$ for every $f \in A$ so I feel this should be enough but am unable to carry the final steps. Maybe I'm missing something obvious?

Let $(z_n)$ be a $H^\infty$-interpolating sequence on the open complex unit disc $\mathbb D$. If $A$ is some Banach space of analytic functions on the disc, denote by $X$ the closed subspace of all the functions in $A$ that have zeros at $z_n$ for all $n \in \mathbb N$. What I'm interested in is under what conditions can we say that the annihilator of $X$ can be represented as the closed span of the point evaluation functionals $\delta_{z_n}$ on points $z_n$? It is easily seen that every $\delta_{z_n}$ must be in the annihilator but for anything else we probably need more assumptions.

I know that if $A$ is a Hardy space $H^p$ with any $1 < p < \infty$, then we have $X = BH^p$ where $B$ is the Blaschke product associated with the sequence $(z_n)$ (the sequence being interpolating is a stronger condition than it being a Blaschke sequence) and we can see the claim is true.

It seems that a similar factorization could be carried out in Bergman spaces $A^p$, $1 < p < \infty$, too but I'm not very familiar with the details. [Duren, Schuster: Bergman Spaces] (esp. p. 146) seems to say that we have a similar bounded divisor function and the claim would be true. This apparently does not apply to all zero sequences, but the sequence being interpolating is a strong enough condition to make this work. I have been unable to find any answers considering the case of standard weighted Bergman spaces $A^p_\alpha$, $-1 < \alpha < \infty$.

Would it be possible to obtain a general proof for this if we assume that $A^* = \overline{\text{span}} \{\delta_z | z \in \mathbb D\}$ holds for the dual and that $fg \in A$ for all $f \in A$ and $g \in H^\infty$? Then if the annihilator were to contain anything besides the closed span of the $\delta_{z_n}$, it would have to contain some $l \in A^*$, a (possibly infinite) linear combination of some other point evaluation functionals. I feel that it should be possible to show that this leads to a contradiction but have been unable to carry it out. We can use the interpolation property to construct a bounded function $g \in X$ that has zeros on all $z_n$ and arbitrary values on any other finite number of points (since we can add a finite number points to the sequence and have it remain interpolating) but this is not enough to guarantee that $l(g) \neq 0$. But now also $fg \in X$ for every $f \in A$ so I feel this should be enough but am unable to carry the final steps. Maybe I'm missing something obvious?