# The Invariant Subspace Problem: examples

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?]

• Andrey, there are very general positive results, so I do not think a "concrete" candidate is known. There is a nice recent paper with good references to the state of the art on the problem: B. S. Yadav, "The Present State and Heritages of the Invariant Subspace Problem", Milan j. math. 73 (2005), 289–316. – Andrés E. Caicedo Dec 19 '10 at 18:08
• @ Andres: Thank you for the reference. – Andrey Rekalo Dec 19 '10 at 18:28
• This paper maths.leeds.ac.uk/~pmt6jrp/op_de_composition_rev.pdf gives examples of concrete operators which "all their invariant subspaces have themselves have non-trivial invariant subspaces" implies that every bounded operator on Hilbert space has an invariant subspace. Of course you might complain that the operator is concrete, but not its invariant subspaces. – David Feldman Dec 20 '10 at 8:31
• @ David: This is interesting, thanks. – Andrey Rekalo Dec 20 '10 at 10:25
• Sorry I can't find the reference - maybe an expert can supply it? Since about 1990(?), the general invariant subspace problem is known to be equivalent to a special case. Let $L^2_a(D)$ be the Bergman space of 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. – Zen Harper Jan 6 '11 at 1:22

• It should be added that in fact, if $T$ is any hypercyclic operator on a real or complex topological vector space, then any hypercyclic vector $x_0$ for $T$ has the property that $P(T)x_0$ is hypercyclic for every polynomial $P\neq 0$. – Etienne May 30 '13 at 21:47