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Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators.

Invariant subspace problem: Let $T \in B(H)$. Is there a non-trivial closed $T$-invariant subspace?

Remark: This problem is known for the Banach spaces in general, but still open for an Hilbert space.

Let $(e_{n})_{n \in \mathbb{N}}$ be an orthonormal basis.
Definition : $T \in B(H)$ is banded if $\exists r \in \mathbb{N}$ such that $(Te_{n}, e_{m})\ne 0 \Rightarrow \vert n-m \vert \leq r$.

Remark: A banded operator is a thick generalization of a diagonal operator. It's also a finite sum of finite product of (orthonormal) weighted shift operators (which check obviously the ISP).

Question: Do the banded operators check the invariant subspace problem ?

Remarks:

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    $\begingroup$ @YemonChoi : the terminology "quasi-diagonalizable" is not standard, and not related to "quasidiagonal". The class of operators I defined is a thick generalization of the diagonal operators, in particular, it contains the shift. Now in section 5 of Jon Bannon's last link, a quasidiagonal operator is a sum of a block-diagonal and a compact operator, but the shift is not of this form. To avoid confusion, I will replace the name by "thick-diagonal". $\endgroup$
    – Sebastien Palcoux
    Commented Jul 27, 2013 at 20:20
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    $\begingroup$ @YemonChoi: Now a thick-diagonal operator is not a curiosity, for two reason : firstly, it's a generalization of the weight shift operators, to be more precise, it's a finite sum of finite product of weight shift operators. However, a weight shift operator checks obviously the ISP. Secondly, I don't know yet whether or not every operators are unitary equivalent to thick-diagonal operators (i.e. thick-diagonalizable, see my MSE post). $\endgroup$
    – Sebastien Palcoux
    Commented Jul 27, 2013 at 20:29
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    $\begingroup$ It is not known whether all weighted BILATERAL shifts have NTIS, and all weighted bilateral shifts are thick-diagonal. $\endgroup$
    – Bill Johnson
    Commented Jul 27, 2013 at 22:03
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    $\begingroup$ @BillJohnson : I don't understand, could you please enlighten me ? The reason of my misunderstanding is the following : a bilateral shift is an operator $T$ in $B(l^{2}(\mathbb{Z}))$ such that $Te_{n} = w_{n}e_{n+1}$, with $n \in \mathbb{Z}$, $(e_{n})_{n\in \mathbb{Z}}$ an orthonormal basis and $w_{n} \in \mathbb{C}$. Let $K_{n} = \langle e_{n}, e_{n+1}, e_{n+2}, ... \rangle $, then $\overline{K_{n}}$ is a closed, non-trivial, $T$-invariant subspace, right ? I trust you, so where is my mistake here ? $\endgroup$
    – Sebastien Palcoux
    Commented Jul 28, 2013 at 7:41
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    $\begingroup$ Here is the answer to the question in Remark. People often call such operators "banded operators." The C*-algebra on $\ell_2({\bf Z})$ generated by the banded operators (aka the uniform Roe algebra of $\bf Z$) is canonically isomorphic as $\ell_\infty({\bf Z})\rtimes {\bf Z}$, which is nuclear and has tracial states (coming from invariant means on $\bf Z$). So, any operator which is unitarily equivalent to a banded operator generates an exact C*-algebra having tracial states, which is not the case in general. $\endgroup$
    – Narutaka OZAWA
    Commented Jul 28, 2013 at 8:30

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