Behavior of orbits under small perturbations

Question

Perhaps this question is too easy for mathoverflow, at least this is how it seems, but I got no answer on stackexchange.

Suppose $T$ is a bounded linear operator on $l_2$ and $x\in l_2$ is a vector such that the orbit $(T^{n}x)$ is linearly independent.

Can one find an $\epsilon>0$ such that for all $||y-x||<\epsilon$ the orbit $(T^{n}y)$ is also linearly independent?

This is probably not true in general. Does then a weaker requirement hold: that for all $\epsilon>0$ we can find a $y\in l_2$ with $||y-x||<\epsilon$ and the orbit $(T^{n}x)$ is linearly independent?

What if linearly independent is replace by not linearly independent or by basic sequence?

Answer 1

The first version does not work, as you suspected. For example, if $Te_n= (1/n)e_{n+1}$ for $n\ge 0$ on $H=\ell^2(\mathbb Z)$, then the $T^n e_0$, $n\ge 0$ are linearly independent. However, we can easily arrange matters in such a way that $T^N(e_0+\delta e_{-1})=0$ for a small $\delta>0$: $T$ will map $e_{-1}$ to $e_{-2}$ and then to $e_{-3}$ etc. We wait until $T^Ne_0$ has the same size as $\delta$, and then we finally map $e_{-N}$ to $-e_N$.

Then we just continue in this way: we define $Te_j$ for $j<-N$ such that $T^{M} (e_0+\delta' e_{-N-1})=0$ for an even smaller $\delta'>0$ etc.

The second claim is true. In fact, the set of vectors $y$ for which $\{T^ny\}$ is l.i. is a dense $G_{\delta}$ set. This follows from the observation that $$ A_N = \{ y\in H: y,Ty,\ldots , T^N y \:\:\textrm{are l.i.}\} $$ is a dense open set. It's easy to see that $A_N$ is open, and to prove that it's dense, suppose that we had a $y\in H$ and an $N\ge 1$ such that $y+z\notin A_N$ for all $\|z\|<r$, for some $r>0$. Then, if $N$ was minimal with this property, we have that $$ T^Nz \in L(y,Ty, \ldots, T^{N-1}y, z, Tz, \ldots T^{N-1} z ) \quad\quad\quad\quad (1) $$ for all $z$ of small norm. However, since $T$ is linear, (1) then holds for all $z\in H$. Moreover, $T^n z$ for $n\ge N$ lies in the same (at most) $2N$ dimensional space. This contradicts our assumption that there is an $x\in\bigcap A_n$.

Answer 2

For any chaotic operator the first statement is wrong because the set of points with dense orbits and the set of points with periodic orbits are both dense (this is the definition of a chaotic linear operator). It remains to note that dense orbits must be linearly independent: Otherwise you find some non-trivial linear combination $\sum_{k=0}^n \alpha_k T^k(x)=0$ which easily implies that the orbit is contained in the linear span of $\lbrace x, T(x),\ldots,T^n(x)\rbrace$ which is closed.

There are many chaotic operators, e.g., Rolewicz' weighted backward shift on $\ell^2$, $(x_n)_{n\in\mathbb N} \mapsto (\lambda x_{n+1})_{n\in\mathbb N}$ for $|\lambda|>1$.