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Let $B$ be a separable Banach space and let $L:B\rightarrow B$ be a hypercyclic operator; here I use the definition of hypercyclicity given implicitly by Birkhoff's Transitivity Theorem: continuous linear operator for which every two distinct non-empty open subsets $U$ and $V$ there is some $N\in \mathbb{N}$ such that $L^N(U)\cap V \neq \emptyset$.

Also, let $b \in B$ be given and be such that $L^n(b)\neq L^m(b)$ for any $n,m \in \mathbb{N}$ with $n\neq m$.

Is there an iterative procedure/algorithm for constructing a sequence $\{b_n\}_{n \in \mathbb{N}}$, using only $b$, $L$, and the vector space structure on $B$ such that $\lim\limits_{n\to\infty} b_n$ converges in $B$ to a hypercyclic vector? I'm looking especially for something constructive.

Some special cases of such constructions are:

Are there other known procedures or broader (constructive/algorithmic) methods?

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    $\begingroup$ Definitely some of them but usually not all of them. In what particular terms do you want us to describe them? I mean, take the usual example of twice the backward shift in $\ell^2$. What answer do you expect here? $\endgroup$
    – fedja
    Commented Jul 28, 2020 at 16:21
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    $\begingroup$ Another "what can be said" question ... these are not very satisfying. $\endgroup$
    – Nik Weaver
    Commented Jul 28, 2020 at 16:33
  • $\begingroup$ @NikWeaver I made the question much more precise now. I'm looking for an algorithm to construct a hypercylic vector. $\endgroup$
    – ABIM
    Commented Jul 28, 2020 at 18:05
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    $\begingroup$ You probably should make your question more precise in a constructive sense. In constructive math, the statement "$L$ is a hypercyclic operator" should be equivalent to having an explicit construction of a hypercyclic vector, because existence implies explicit constructibility. This renders your question difficult to answer in a meaningful way. For example, a more meaningful constructive question would be: `Is there a constructive proof of Kitai's Criterion?'. $\endgroup$
    – Franka Waaldijk
    Commented Jul 31, 2020 at 7:50
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    $\begingroup$ In construcive mathematics it often makes sense to use an apartness relation instead of inequality. In a normed space that would be $\| x - y \| > 0$. For instance, if you know $x \neq 0$ you cannot yet normalize $x$ to $x / \| x \|$, but if you know $\| x \| > 0$ then you can. $\endgroup$
    – Andrej Bauer
    Commented Jul 31, 2020 at 16:19

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