When are operators extended by linearity bounded? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T06:45:39Z http://mathoverflow.net/feeds/question/69843 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69843/when-are-operators-extended-by-linearity-bounded When are operators extended by linearity bounded? Adam Azzam 2011-07-09T02:07:12Z 2011-07-09T02:32:10Z <p>Greetings.</p> <p>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). </p> <p>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). </p> <p>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). </p> <p>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 </p> <p>$S(z)=\bigg\{\begin{array}{cl} v_{n+1}&amp;\text{if } z=v_{n}\text{ for some }n\in \mathbb{N}\\ 0&amp;\text{otherwise.}\end{array}$</p> <p>and extend it by linearity to a linear operator on $M$, which we'll still call $S$. It would follow then that </p> <p>$\{S^nv_{1}: n\ge 1\}=\{v_{n}: n\ge 1\}$</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/69843/when-are-operators-extended-by-linearity-bounded/69845#69845 Answer by Andreas Blass for When are operators extended by linearity bounded? Andreas Blass 2011-07-09T02:32:10Z 2011-07-09T02:32:10Z <p>The hypotheses you gave are not sufficient to make your proposed operator $S$ bounded. You could have <code>$v_n$</code> very close to <code>$v_m$</code> while <code>$v_{n+1}$</code> is far from <code>$v_{m+1}$</code>, and this could happen repeatedly with greater and greater discrepancies between the two distances.</p>