I was reading the paper Three decades of the Lomonosov invariant subspace theorem by Kubrusly and I found the following statement (see Section 3):
for every operator $T$ there exists a nonzero compact operator $K$ such that $\operatorname{rank}(TK-KT) \leq 2$ (reason: this always holds when $K$ is a rank $1$ operator).
It is clear why the reason implies the statement, but how do we prove that there is always a compact $K$ with rank one such that $\operatorname{rank}(TK-KT) \leq 2$, whichever is $T \in B(X)$ (for some Banach space $X$)?
Any suggestion is appreciated, thanks!
EDIT: I have found a proof for $T$ having finite rank, and I think that it also holds for operators with infinite rank. We know that: $$ \operatorname{rank}(AB) \leq \min(\operatorname{rank}(A), \operatorname{rank}(B)) $$ and $$\operatorname{rank}(A+B) \leq \operatorname{rank}(A) + \operatorname{rank}(B)$$ Since, by assumption, $\operatorname{rank}(K)=1$, we have: $$ 0 \leq \operatorname{rank}(TK),\operatorname{rank}(KT) \leq 1 $$ Thus, from subadditivity we see that $$ \operatorname{rank}(TK-KT) \leq 2 $$ for every compact $K$ with rank $1$. My question is: does this work even in the infinite case? I think that the above properties also hold in the infinite rank case, which would imply this statement, but I'm not sure about this.
Let $0\neq u \in X,0\neq \phi \in E'$ and $K:= u \otimes \phi$ that is $$Kx=\phi(x)u \; \forall\; x \in X.$$ Then $K$ is a rank-one operator, hence compact. In fact, every rank one operator is of this form.
Since $\operatorname{range} KT \subseteq \operatorname{range} K,$ therefore $\operatorname{rank} KT \leq 1.$ Also $$TK=Tu \otimes \phi$$ and hence $\operatorname{rank} TK \leq 1.$ It follows that $$\operatorname{rank}(TK-KT)\leq 2.$$