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

**Definition** :  $T \in B(H)$ is **quasi-diagonalizable** if $\exists r \in \mathbb{N}$ and $(e_{n})_{n \in \mathbb{N}}$ orthonormal basis :  
$$
 \forall n , m \in \mathbb{N} \ , \ (T.e_{n}, e_{m})\ne 0 \Rightarrow \vert n-m \vert \leq r
$$

**[Invariant subspace problem (ISP)][1]**: $\forall T \in B(H)$, is there a non-trivial, closed, $T$-invariant subspace $K \subset H$ (i.e., $T(K) \subset K = \overline{K}$ with $K \ne \{0\}$,  $H$)?   


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



> Is it true that the quasi-diagonalizable operators check the ISP?

**Remark**: I post  the following question on MSE : [Are all the operators quasi-diagonalizable?][2]  
It's probably 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-quasi-diagonalizable