3
$\begingroup$

$\DeclareMathOperator\C{C}\DeclareMathOperator\HC{HC}$Definitions:


Let $T:X\rightarrow X$ be a bonded linear operator on a separable (infinite-dimensional) Banach space and define the sets:

  • $ \HC(T)\triangleq \left\{ x \in X:\, \overline{\{T^n(x)\}_{n \in \mathbb{N}}} = X \right\}, $
  • $ \C(T)\triangleq \left\{ x \in X:\, \overline{\operatorname{span}\{T^n(x)\}_{n \in \mathbb{N}}} = X \right\}. $

*When $\HC(T)\neq \emptyset$, then $T$ is called hypercylic.*

Facts:

It is known (see Theorem 8.13 of Grosse-Erdmann and Manguillot - Linear chaos) that every separable Banach space admits a hypercyclic operator $T$. Moreover, by the Birkhoff's Transitivity Theorem, it is known that for such an operator $T$, the set $\HC(T)$ is a dense $G_{\delta}$ subset of $X$ not containing the origin.

Question:

Suppose that $T$ is hypercyclic. What do we known about $\C(T)$ other than the fact that $\HC(T)\subseteq \C(T)$ (and in particular $\C(T)\cup\{0\}$ contains a dense linear subspace)?

More specifically: How much "bigger" is $\C(T)$ than $\HC(T)$?

-I leave the interpretation of "bigger" up to the reader.

$\endgroup$
2
  • 1
    $\begingroup$ You asked how much bigger is $\operatorname{HC}(T)$ than $\operatorname C(T)$, but I'm pretty sure it's the other way around. I edited accordingly. Also, that Amazon link is awesome. It's the 2011th edition! $\endgroup$
    – LSpice
    Commented Jun 1, 2020 at 21:56
  • 1
    $\begingroup$ Thanks for pointing that out. Also that's hilarious, I didn't catch that... they must spend a lot of time making versions haha $\endgroup$
    – ABIM
    Commented Jun 1, 2020 at 22:15

0

You must log in to answer this question.