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 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?
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).