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

**[Invariant subspace problem][1]**: 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 ?

**Remark**: Here is a question on MSE  : [Is every operator unitary equivalent to a banded operator ?][2]  
 N. Ozawa answered "no" in its comment below (see also sections 16.3 and 16.4 of its book [here][3]).  
D. Voiculescu gave examples of quasidiagonal operators which don't generate exact $C^{*}$-algebras:  
*A note on quasidiagonal operators*, Operator Theory, 1988, 265-274.  
*Question*: Do these examples check the ISP ? 


  [1]: http://en.wikipedia.org/wiki/Invariant_subspace_problem
  [2]: http://math.stackexchange.com/questions/452908/are-all-the-operators-thick-diagonalizable
  [3]: http://www.ams.org/bookstore-getitem/item=gsm-88