Is every operator unitary equivalent to a banded operator? Let $H$ be an infinite dimensional separable Hilbert space and $B(H)$ the algebra of bounded operators. **Definition** : Let $(e_{n})_{n \in \mathbb{N}}$ be an orthonormal basis. $T \in B(H)$ is **banded** if $\exists r \in \mathbb{N}$ such that $ (Te_{n}, e_{m})\ne 0 \Rightarrow \vert n-m \vert \leq r$. > Is every operator unitary equivalent to a banded operator ? **Remark** : A _banded_ operator is a thick generalization of a diagonal operator. It's also a finite sum of finite product of weight shift operators.

N. Ozawa answered "no" here :
<< The $C^{*}$-algebra on $ℓ^{2}(\mathbb{Z})$ generated by the banded operators (aka the uniform Roe algebra of $\mathbb{Z}$) is canonically isomorphic as $ℓ^{∞}(\mathbb{Z})⋊\mathbb{Z}$, which is nuclear and has tracial states (coming from invariant means on $\mathbb{Z}$). So, any operator which is unitarily equivalent to a banded operator generates an exact $C^{*}$-algebra having tracial states, which is not the case in general >>.

For more details, see the sections 16.3 and 16.4 of his book with Nate Brown.

D. Voiculescu gave the first example of quasidiagonal operators which don't generate exact $C^{*}$-algebras:
_A note on quasidiagonal operators_ , Operator Theory, 1988, 265-274.