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I have several related questions on Analytic functions and Hyperfunction as topological vector spaces (I am mainly interested in questions 4,6,10):

For an open set $U\subset \mathbb C^n$ we can consider the space of analytic functions on U as a topological vector space with the topology induced from the space of smooth functions.

  1. Is this the standard topology considered on this space?
  2. Am I right that this is a nuclear Frechet space (as it is a closed subspace of nuclear Frechet space)?

If $Z\subset \mathbb C^n$ is a closed set, then the space of analytic functions on $Z$ is a direct limit of the spaces of analytic functions on open sets containing  $Z$. This defines a topology on the space of analytic functions on $Z$.

  1. Is this the standard topology considered on this space?
  2. Is this space nuclear?
  3. Is the space of real analytic functions on $\mathbb R^n$  Nuclear? This should follow from the previous question.

Questions on Hyperfunction:

  1. Is it true that the space of compactly supported Hyperfunction on $\mathbb R^n$ is the dual of the space of real analytic functions on $\mathbb R^n$?
  2. If yes, Is the natural topology to consider on the space of compactly supported Hyperfunction on $\mathbb R^n$ is the strong dual topology to the one on the space of real analytic functions on $\mathbb R^n$?
  3. is the space of compactly supported Hyperfunction on $\mathbb R^n$ is nuclear? This should follow from the previous questions.
  4. What is the natural topology to consider on  the space of Hyperfunction on $\mathbb R^n$? (see also: Topology on space of hyperfunctions)
  5. Is this space nuclear?

Thanks a lot!

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    $\begingroup$ I'm not that much of a purist when it comes to asking multiple questions in one, but 10 sure seems like a lot. Regardless, at least for 1. this is the same as the uniform convergence on compact spaces, which is surely used a lot (by differentiating the Cauchy's integral formula you can bound any derivative from the function itself). I have no idea what a nuclear space or a hyperfunction is, so can't help with other questions. $\endgroup$ Commented Nov 1, 2024 at 8:46
  • $\begingroup$ Thank you. I have added a remark about the questions I am mainly interested in. The questions are related so I prefer to have all of them for the context. $\endgroup$ Commented Nov 1, 2024 at 20:39
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    $\begingroup$ When $Z$ is a compact subset of $\mathbb C^n$, the ring of overconvergent functions on $Z$ should be nuclear. In condensed mathematics, this is Prop 9.13 of [Clausen–Scholze, Condensed mathematics and complex geometry], and their argument probably applies to classical functional analysis: first reduce to the case that $Z$ is a closed polydisc, then it is a sequential colimit along nuclear operators. $\endgroup$ Commented Nov 1, 2024 at 21:49
  • $\begingroup$ 1. At least in the case $U \subseteq \mathbb C$ open, the usual topology of $A(U)$ is: uniform convergence on compact sets. It is a Fréchet space. The uniform limit on compact sets of analytic functions is automatically analytic. $\endgroup$ Commented Nov 2, 2024 at 14:41

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