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 **thick-diagonal** if $\exists r \in \mathbb{N}$ such that $(Te_{n}, e_{m})\ne 0 \Rightarrow \vert n-m \vert \leq r$.   

**Remark**: A *thick-diagonal* operator is a thick generalization of a diagonal operator.  It's also a finite sum of finite product of weight shift operators. Now, a weight shift operator checks obviously the ISP.


>**Question**: Do the thick-diagonal operators check the invariant subspace problem ?

**Remark**: I post  the following question on MSE : [Are all the operators thick-diagonalizable?][2]  
It's perhaps false, but of course, if it's true, the question here would be equivalent to the ISP, and I would be interested to know if this way is (in your opinion) promising, and ever known (reference).  



  [1]: http://en.wikipedia.org/wiki/Invariant_subspace_problem
  [2]: http://math.stackexchange.com/questions/452908/are-all-the-operators-thick-diagonalizable