3
$\begingroup$

An operator $T$ on a separable Hilbert space $H$ is called unicellular if any two closed invariant subspaces $M$ and $N$ are comparable; that is either $M\subseteq N$ or $N\subseteq M$. There are many examples of (compact) unicellular operators, for example, the Volterra operator.

I am looking at a weaker property. Namely, the existence of a closed nontrivial invariant subspace $M$ that is comparable to all the other closed, invariant subspaces. Obviously, any closed invariant subspace of a unicellular operator satisfies this property. On the other hand, any diagonal operator will not be unicellular and won't satisfy this weaker property either.

In particular, I am interested in compact operators that do not have eigenvalues. My questions are:

  1. Are there examples of compact operators without eigenvalues having this weaker property, but are not unicellular? (I suspect the answer is Yes)

  2. Does any compact operator without eigenvalues have this weaker property? (I suspect the answer is No)

Edit: The question is edited for clarity.

Improve this question
$\endgroup$
3
  • $\begingroup$ Maybe I'm being a bit naive, but how can I see that the Volterra operator is unicellular? $\endgroup$
    – Jochen Glueck
    Mar 23, 2022 at 8:21
  • 1
    $\begingroup$ It is not trivial (as far as I know) that its invariant subspaces are only the obvious ones, the space of functions that vanish almost everywhere on $[0,a]$ for $a<1$. The references I could find are old. Donoghue, 1957, The lattice of invariant subspaces of a completely continuous quasinilpotent transformation, and Sarason, 1965, A remark on the Volterra operator. $\endgroup$
    – Markus
    Mar 23, 2022 at 14:39
  • $\begingroup$ Thanks for your reply! $\endgroup$
    – Jochen Glueck
    Mar 23, 2022 at 15:08

2 Answers 2

Reset to default
1
$\begingroup$

$V\oplus V\in B(L^2(0,1)\oplus L^2(0,1))$, with $V$ being the Volterra operator, is an operator without eigenvalues that does not have your property. The invariant subspaces are $L^2(0,a)\oplus L^2(0,b)$, and for any such space $M=A\oplus B$, we can find another one $N=C\oplus D$, with $C\subsetneq A$, $D\supsetneq B$, unless $M=0, H$.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ I think the invariant subspaces are of the form $L^2(a,1)$, but your argument stands nonetheless. $\endgroup$
    – Markus
    Mar 24, 2022 at 21:13
  • 1
    $\begingroup$ @Markus: You are right if we use what is probably the standard definition of $V$; my spaces work for $(Vf)(x)=\int_x^1 f(t)\, dt$ (which is really $V^*$ for the other $V$). $\endgroup$
    – Christian Remling
    Mar 24, 2022 at 21:24
1
$\begingroup$

The weaker version of unicellularity does not hold; consider the compact operator $T$ defined on a separable Hilbert space $H$ with orthogonal basis $(e_n)$ by $Te_n=\frac1ne_n$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Of course, I wanted to exclude diagonal operators. I will edit my question, for clarity. $\endgroup$
    – Markus
    Mar 23, 2022 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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