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:

• Here is a question on MSE : Is every operator unitary equivalent to a banded operator ?
  N. Ozawa answered "no" in its comment below (see also sections 16.3 and 16.4 of its book here).
  D. Voiculescu gave examples of quasidiagonal operators which don't generate exact $C^{*}$-algebras:
  A note on quasidiagonal operators, Operator Theory, 1988, 265-274.

• An ISP counter-example is necessarily irreducible, but the Voiculescu examples are not irreducible, so the more relevant question is the following (MSE):
  Is every irreducible operator unitary equivalent to a banded operator?

• See also the MSE question : Does an irreducible operator generate an exact $C^{∗}$-algebra?.
  C. Eckhardt and M. Argerami answered "no". The first counter-example is given by the full group $C^{*}$-algebra $C^{*}(\mathbb{F}_{2})$ which is primitive, singly-generated and non-exact.