let $E$ be a Banach space and $T \in \mathcal{L}(E)$
assume that there exists $S \in \mathcal{L}(E)$ such that $I – ST$ and $I-TS$ are both compact operators.
show that $T$ is a Fredholm operator.
by Fredholm's Alternative I can say that both $ST$ and $TS$ are fredholm operators, this also tells us that if such an $S$ exists then it has to be a Fredholm operator.
can anyone plot me a sketch of proof or recommends me a good reference where I can find the proof of this proposition ?
Thanks !
Best Answer
Hint: Note that $\ker(T) \subset \ker(ST)$ and that $\ker(T^*) \subset \ker((TS)^*)$.
If we want to avoid the properties of the adjoint, here's another way to see that the image of $T$ is of finite codimension.
Suppose for the purpose of contradiction that image of the map $T: E \to E$ does not have finite codimesion. Consider the quotient map $P:E \to E/\operatorname{im}(T)$; note that $P$ is not compact and that $\operatorname{im}(T) \subset \ker(P)$, which means that $PT = 0$.
Now, note that $I - TS$ is compact. It follows that $P(I - TS)$ must also be compact. However, $$ P(I - TS) = PI - (PT)S = P $$ and as we said earlier, $P$ is not compact. Thus, we have a contradiction.