Strongly convergent operator sequence

2012-11-06T14:20:21Z

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?

Answer by Yemon Choi for Strongly convergent operator sequence

2012-11-06T20:19:03Z

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

Strongly convergent operator sequence - MathOverflow

Strongly convergent operator sequence

Answer by Yemon Choi for Strongly convergent operator sequence