Greetings.


Suppose that $H$ is a separable infinite-dimensional Hilbert space and that $M$ is an infinite 
dimensional closed subspace of $H$. Suppose that {$v\_{n}: n\ge 1$} is an infinite linearly independent subset of $M$ that is bounded and bounded away from zero (in the case I'm considering, the collection {$v\_{n}: n\ge 1$} is somewhere dense in M, so the set of norms is dense "between" its infimum and supremum). 


In general, I'm trying to determine if there necessarily exists an bounded linear operator $S$ on $M$ and some element $x\in M$ with {$S^{n}x: n\ge 1$} $= $ {$v\_{n}: n\ge 1$} (or {$S^{n}x: n\ge 1$} $\subseteq $ {$v\_{n}: n\ge 1$} and {$S^{n}x: n\ge 1$} is somewhere dense). 


I'm familiar with the results of S. Grivaux, who shows that every countable dense linearly independent subset of a separable Banach space is the orbit of a bounded linear operator. But I don't want to assume that {$v\_{n}: n\ge 1$} is dense (also, extending it to a dense linearly independent set doesn't help because we get the wrong inclusion). 


What I attempted in this. Extend {$v\_{n}: n\ge 1$} to a basis for $M$, call it $B$. Define a map $S$ on the basis 


$S(z)=\bigg\{\begin{array}{cl} v\_{n+1}&\text{if } z=v\_{n}\text{ for some }n\in \mathbb{N}\\\ 0&\text{otherwise.}\end{array}$


and extend it by linearity to a linear operator on $M$, which we'll still call $S$. It would follow then that 


$\{S^nv\_{1}: n\ge 1\}=\{v\_{n}: n\ge 1\}$


But I'm unsure if this operator derived from the extension is a bounded operator. Intuitively, I believe that this operator is bounded, since {$v\_{n}: n\ge 1$} is bounded and bounded away from zero. However, the proof is eluding me.


EDIT: There are some hypothesis/background I find irrelevant, and they're omitted. So if anyone would like any more information, please let me know.



When are operators extended by linearity bounded?
The hypotheses you gave are not sufficient to make your proposed operator $S$ bounded. You could have $v\_n$ very close to $v\_m$ while $v\_{n+1}$ is far from $v\_{m+1}$, and this could happen repeatedly with greater and greater discrepancies between the two distances.


