most recent 30 from http://mathoverflow.net 2013-06-19T19:50:03Z http://mathoverflow.net/feeds/question/1685 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1685/range-of-a-certain-linear-operator Alessandro S 2009-10-21T17:01:04Z 2009-10-21T20:37:58Z

<p>Consider the following hermitian form on the sobolev space H^1(I), of an interval I: g(u,v):= \int_I (du/dt dv/dt - \rho(t) u v)dt, where \rho is a nice bounded function on I. Riesz representation theorem gives us a bounded linear operator A on the Hilbert space H^1(I) such that (Au,v) = g(u,v), where (,) is the inner product of H^1(I). The question is: can you find sufficient conditions on \rho for A to have a dense range?</p>

http://mathoverflow.net/questions/1685/range-of-a-certain-linear-operator/1716#1716 Yemon Choi 2009-10-21T20:37:58Z 2009-10-21T20:37:58Z

<p>This looks a bit like it could be a H/W exercise, but having started to type something up I might as well give most of it. I'm assuming all your functions are real-valued, since otherwise your form isn't hermitian.</p> <p>Start by noting that a hermitian, bounded linear operator on Hilbert space has dense range if and only if it's injective. (Let $v\in H$: then $v$ is orthogonal to $Tu$ for all $u$ if and only if $Tv=T^*v$ is orthogonal to all $u$, i.e. if and only if $Tv=0$.)</p> <p>So your operator $A$ has dense range if and only if it is injective. In particular (again, assuming that we're talking about real-valued H^1(I) here), if \rho is negative a.e. and not identically zero, then the only solution of $Au=0$ is $u=0$ (just consider $g(u,u)$ ). </p> <p>Conversely, suppose that \rho is positive on some sub-interval $[a,b]$. Then I think we can find $u\in H^1(I)$ which is supported on $[a,b]$ and is not identically zero, such that $(du/dt)^2 - \rho(t)u(t)^2=0$ almost everywhere. (Hint: just try doing it!)</p> <p>Depending on how nice your function \rho is supposed to be, that almost answers your question. If you're merely requiring it to be integrable then I'd have to think a bit more on this.</p>