**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?]

concrete, but not its invariant subspaces. $\endgroup$ – David Feldman Dec 20 '10 at 8:31Bergman spaceof analytic functions on the unit disc $D = \{ |z|<1 \}$, with squared norm the area integral $\| f \|^2 = \frac{1}{\pi}\int\int_D |f(z)|^2 dA(z)$, and the linear operator $M$ is just $(Mf)(z) = z f(z)$. Similarly to David Feldman's comment, it is not $M$ itself, but the restriction of $M$ to subspaces, which is important; but the subspaces themselves have no simple description. $\endgroup$ – Zen Harper Jan 6 '11 at 1:22