Question. Is there a concrete example of a bounded linear operator on a Hilbert space for which it is not known if it has a non-trivial closed invariant subspace?
[Added 24.01.2011: According to Bernard Beauzamy (Introduction to Operator Theory and Invariant Subspaces, Elsevier (1988), p. 345),
the operator which is "closest" to a counter-example is the one built by the present author: it has one hypercyclic point $x_0$, and for every polynomial $p$ with complex coefficients, $p(T)x_0$ is also hypercyclic. Therefore, the operator has a vector space of hypercyclic points (thus solving a question raised by P. Halmos), but it may still have points which are not cyclic at all, thus having Invariant Subspaces.
Beauzamy refers to his manuscript "The orbits of a linear operator". I have not been able to find an electronic version of this manuscript (or paper) online. Does anyone know where one may find a description of the example? Is it presently known whether the operator in Beauzamy's example has an invariant subspace?]
Best Answer
It seems likely that the author of the question has found the reference in the meantime. I will provide it here for the sake of completeness.
The article containing the construction of the operator described in Bernard Beauzamy (Introduction to Operator Theory and Invariant Subspaces, Elsevier (1988), p. 345 can be found here:
Direct link to the document:
http://matwbn.icm.edu.pl/ksiazki/sm/sm96/sm9618.pdf
It seems still not to be known whether this bounded operator on a separable Hilbert space admits a non-trivial closed invariant subspace. Note that there are quite a number of articles containing a reference to the above article by Beauzamy (cf. subscription-only databases MathSciNet and ISI Web of Knowledge).