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Let $T:X\to X$ be a bounded operator on Banach space $X$ such that $A= \{x: \{T^n(x)\}_{n=0}^\infty \text{ is unbounded } \}$ be residual in $X$.

In my research, the set of vector $x\in X$ with bounded orbit is closed i.e. $A$ is open set in $X$. Clearly orbit of $x=0$ is bounded. Is there a non-zero vector $x\in X$ that its orbit is bounded?

user479859
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1 Answers1

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$T$ could be $\lambda Id$ with $|\lambda|>1\implies A=X\backslash\{0\}$.

edit: I answer to this: "it seems that if the set of non-zero vector in $X$ with unbounded orbit is open and residual, then the set of non-zero vector with bounded orbit is empty.".

claim: Let $X$ a infinite Hilbert space then every normal and compact operator $T$ (with an eigenvalues of less than one length ) has a non-zero vector with bounded orbit and $A$ is open and residual.

proof: By the spectral theorem there exists an O.N. base $(e_j)$ such that $Te_j=\lambda_je_j$, now its easy to see that $v\in A^c$ (i.e. has a bounded orbit) if and only if $\langle v,e_j \rangle=0$ for every $e_j$ s.t. $|\lambda_j|>1$.

By the infinity dimension, the eigenvalues goes to zero so there exists $e_j$ s.t. $|e_j|\le1$ and these vector stays in $A^c$.

Now we have only to prove that $A$ is open and residual:

I'll prove that $A^c$ is closed. Because A is a metrizable space i can work with the classical sequences. Let $v^k\in A^c$ that converges to some $v\in X$ then: $$\langle v,e_j\rangle=\lim \langle v^k,e_j\rangle=\lim 0=0 \;\text{ for every for every $e_j$ s.t. $|\lambda_j|>1$} $$ $$\implies v\in A^c\implies A^c\text{ is closed.}$$

Now, because $A$ is open, we only need to prove that its dense. By the hypothesis exists $e_j$ with $|\lambda_j|>1$ and if $v\in A^c$ then $v+\varepsilon e_j\in A$ for every small $\varepsilon>0\implies A$ is dense.

Bongo
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  • Thanks. But $A= X\setminus {0}$ means that there is no non-zero vector with bounded orbit. Is it true? – user479859 Aug 09 '23 at 14:18
  • Yes, in fact you need additional hypothesis for operator $T$ (if you want a non-zero vector with bounded orbit). – Bongo Aug 09 '23 at 14:25
  • Is there such hypothesis? I think the set of non-zero vector with bounded orbit may be empty set. – user479859 Aug 09 '23 at 14:28
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    If $T$ is a compact and normal then its diagonalizable . Now if your space is infinite dimensional surely the eigenvalues $(\lambda_j)$goes to zero $\implies$ the eigenvector associated with the eigenvalues s.t. $|\lambda_j|<1$ will have bounded orbit. – Bongo Aug 09 '23 at 14:35
  • In the case of $T$ is compact and normal on infinite dimensional space $X$, is it true that the set of vectors with unbounded orbit is an open and residual in $X$? . – user479859 Aug 09 '23 at 14:56
  • do you want an answer for this particular case ? (note that now $X$ is an Hilbert space) – Bongo Aug 09 '23 at 15:26
  • My research is dynamical system and I does have few information about operators on infinite -dimensional space. Some example that I saw in this subject, it seems that if the set of non-zero vector in $X$ with unbounded orbit is open and residual, then the set of non-zero vector with bounded orbit is empty. I am so thankful if you help me to know that it is true or not. – user479859 Aug 09 '23 at 15:41
  • I edit my answer. There are some settings where your last statement is false. – Bongo Aug 09 '23 at 16:35