MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 7 down vote accepted

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.

The following is an outline, as I am a bit short of time and sleep right now.

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''.

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).

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 $\Vert T_j \xi - \xi \Vert_2 \leq O(\epsilon)$.

share|cite|improve this answer
Yes, you are right, Yemon. If you are willing to theorem quote, use the Erdos Density Theorem to get norm one finite rank operators that leave every subspace of the Volterra nest invariant and converge strongly to the identity; necessarily these are nilpotent operators (because the Volterra nest is continuous). – Bill Johnson Nov 7 '12 at 1:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.