**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*][1],  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?]


  [1]: http://books.google.com.ua/books?id=u4ejT0G6TrwC&printsec=frontcover&dq=Introduction+to+operator+theory+and+invariant+subspaces&source=bl&ots=whYpI3zpUE&sig=XhnbtpmvtrbyrLU85I1Mbo5HSP8&hl=en&ei=mW89Tc6sDMecOvXPtJ8L&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBIQ6AEwAA#v=onepage&q&f=false