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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?]

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    $\begingroup$ 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. $\endgroup$
    – Andrés E. Caicedo
    Commented Dec 19, 2010 at 18:08
  • $\begingroup$ @ Andres: Thank you for the reference. $\endgroup$
    – Andrey Rekalo
    Commented Dec 19, 2010 at 18:28
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    $\begingroup$ 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. $\endgroup$
    – David Feldman
    Commented Dec 20, 2010 at 8:31
  • $\begingroup$ @ David: This is interesting, thanks. $\endgroup$
    – Andrey Rekalo
    Commented Dec 20, 2010 at 10:25
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    $\begingroup$ 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. $\endgroup$
    – Zen Harper
    Commented Jan 6, 2011 at 1:22

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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:

Bernard Beauzamy, An operator on a separable Hilbert space with all polynomials hypercyclic. Studia Math. 96 (1990), no. 1, 81–90. MR1055079

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).

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  • $\begingroup$ 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 $P\neq 0$. $\endgroup$
    – Etienne
    Commented May 30, 2013 at 21:47

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