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.