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In this question the disc algebra $\mathcal{A}(\mathbb{D})$ is the Banach algebra of all holomorphic functions on the unit open disc $\mathbb{D} \subset \mathbb{C}$ which have a continuous extension to the closure $\bar{D}$ of the open disc.

Questions: Is there a complete classification of all Banach subspaces of the disc algebra $\mathcal{A}(\mathbb{D})$ which are invariants under the differentiation operator? Is there a complete classification of such Banach subspaces for which the differentiation operator is a bounded operator?

Is there a complete classification of all Banach sub algebras of $\mathcal{A}(\mathbb{D})$ which are invariant under the differentiation operator?

Having these questions in my mind, I arrived at this question:

Is there a holomorphic function on open unit disc with this property?

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    $\begingroup$ When you say "Banach subspace" do you simply mean a closed subspace of ${\mathcal A}({\mathbb D})$? I think you will have problems finding infinite-dimensional examples, simply because differentiation increases the supremum norm. $\endgroup$
    – Yemon Choi
    Commented Apr 20, 2019 at 14:45
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    $\begingroup$ this book may be of interest archive.org/details/banachspacesofan032699mbp $\endgroup$
    – Pietro Majer
    Commented Apr 21, 2019 at 11:06
  • $\begingroup$ @PietroMajer Many thanks for this refetence. $\endgroup$
    – Ali Taghavi
    Commented Apr 21, 2019 at 19:07
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    $\begingroup$ @PietroMajer I'm not sure Hoffman's book will help that much: my recollection is that its main focus is on Hardy spaces (including $H^\infty$ to be fair). But perhaps there is something there which enables progress on Ali's question $\endgroup$
    – Yemon Choi
    Commented Apr 21, 2019 at 23:53
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    $\begingroup$ @AliTaghavi I mean that you will have problems defining d/dz on all of your Banach space, even before you seek to show boundedness. Recall that $d/dz$ is only densely-defined as an operator from ${\mathcal A}({\mathbb D})$ to itself. One again: could you please clarify what you mean by "Banach subspace" in this context? $\endgroup$
    – Yemon Choi
    Commented Apr 21, 2019 at 23:55

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