TITLE: Decomposing operators into the sum of a quasinilpotent and something else
QUESTION [7 upvotes]: I seem to remember some result of the following sort:

Alleged Theorem. Every bounded operator on a separable complex Hilbert space can be decomposed as the sum of a normal operator and a quasinilpotent operator.

However, I can't remember if this is the correct statement of the result. Is the theorem as stated obviously true? obviously false? Is there a related result with "normal" replaced by some other class of operators?
I googled it and found this paper, which is similar to the above stated theorem, but it places spectral restrictions on the class of bounded operators for which it works.

REPLY [4 votes]: Are you thinking of the Weyl-von Neumann-Berg Theorem?
Every normal operator $N$ on a separable Hilbert space can be decomposed as $N=D+K$, the sum of a a diagonal normal operator $D$ and a compact operator $K$?
Or the Brown-Douglas-Fillmore Theorem?
An operator $T \in B(H)$ is a compact perturbation of a normal operator if and only if $T$ is essentially normal and the Fredholm index $\operatorname{ind} (T-\lambda)=0$ for every $\lambda$ not in the essential spectrum of $T$.
See this Encyclopedia of Mathematics page for further details .
Or a theorem due to West?
Every Riesz operator is the sum of a quasinilpotent operator and a compact operator.