5
$\begingroup$

A very natural example of a nuclear Montel space is the space $H(D)$ of all holomorphic functions on the open disc topologized by the family of seminorms

$$p_n(f)=\sup\{|f(z)|\colon |z|\leq 1-\tfrac{1}{n}\},\, n\in \mathbb N, f\in H(D) $$

I cannot find any good references concerning this space. In particular, I have two following questions:

1) Can one give examples of subspaces of $H(D)$ which are not isomorphic to it?

2) Does every copy of $H(D)$ contain further complemented one?

Thank you in advance.

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ Question 1 is missing the word "infinite-dimensional" $\endgroup$
    – Yemon Choi
    Commented Aug 14, 2011 at 23:18
  • 1
    $\begingroup$ Certainly. Then the next missing word is "closed"... $\endgroup$
    – RogersFR
    Commented Aug 14, 2011 at 23:38
  • 1
    $\begingroup$ Is question one "Does there exists one?" or "How do I write one down?" $\endgroup$
    – Helge
    Commented Aug 15, 2011 at 19:14
  • $\begingroup$ Since all the subspaces I can produce are copies of $H(D)$ I ask about the existence of other subspaces but it should be easy (I believe) to construct different ones. $\endgroup$
    – RogersFR
    Commented Aug 15, 2011 at 19:37
  • 1
    $\begingroup$ If $U$ is simply connected and not equal to the whole plane $H(U)$ is isomorphic to $H(D)$ by the Riemann mapping theorem. However, I would conjecture that this is not true if $U$ is simply connected or the whole plane. If $D\subset U$, then $H(U)$ is a subspace of $H(D)$. $\endgroup$
    – Matthias Ludewig
    Commented Jun 4, 2013 at 7:20

1 Answer 1

Reset to default
1
$\begingroup$

There is a simple and natural way to create closed subspaces of $H(D)$ which are not isomorphic to the latter, namely by considering a lacunary sequence $(\lambda_n)$ and letting $E$ be the subspace of functions whose Taylor coefficients vanish away from this sequence. As a concrete example you can take the sequence of squares of the positive integers.

Improve this answer
$\endgroup$

You must log in to answer this question.