Strongly convergent operator sequence - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T19:17:10Z http://mathoverflow.net/feeds/question/111645 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111645/strongly-convergent-operator-sequence Strongly convergent operator sequence anton 2012-11-06T14:20:21Z 2012-11-06T20:19:03Z <p>Let $T_j$ be a sequence of compact operators on a Hilbert space $H$ which converges strongly to the identity, i.e., for each $v\in H$ the sequence $$\parallel T_jv-v\parallel$$ tends to zero. Is it true that there must exist an index $j$ such that the spectrum of $T_j$ contains a non-zero number?</p> http://mathoverflow.net/questions/111645/strongly-convergent-operator-sequence/111677#111677 Answer by Yemon Choi for Strongly convergent operator sequence Yemon Choi 2012-11-06T20:19:03Z 2012-11-06T20:19:03Z <p>I think the answer is no: there should be a sequence of quasinilpotent compact operators on $L^2[0,1]$ which converges in SOT to the identity map. This is roughly for the same reason one can have radical Banach algebras with compact multiplication and bounded approximate identities.</p> <p>The following is an outline, as I am a bit short of time and sleep right now.</p> <p>Specifically, try Volterra-type operators $T_j:L^2[0,1]\to L^2[0,1]$ $$T_j\xi(t) = \int^t_0 f_j(s)\xi(t-s)\,ds$$ where $f_j$ is something like a heat kernel or Gaussian that approaches the Dirac point mass at the origin''.</p> <p>Certainly if $f_j$ is continuous on $[0,1]$ then $T_j$ will be quasinilpotent and compact (just approximate $f_j$ with polynomials and use known properties of the classical Volterra operator).</p> <p>So I guess we just need to arrange that $\Vert T_j \xi - \xi \Vert_2 \to 0$ for any $\xi \in C[0,1]$ (then we deduce it for all $\xi \in L^2[0,1]$ by density). But now given such a $\xi$ it is uniformly continuous, so on small intervals of width $\delta$ it can only vary by $\epsilon$, so provided $f_j$ lives mostly on the interval $[0,\delta]$ and has total mass $1$ we should have $|T_j(\xi)(t)-\xi(t) | \leq 2\epsilon$ for all $t$, which certainly implies $\Vert T_j \xi - \xi \Vert_2 \leq O(\epsilon)$.</p>