# There exists a continuous inverse of $(\text{id}-A)$ in the set $(\text{id}-A)(X)$.

Exercise :

Let $$X$$ be a Banach space and $$A \in \mathcal{L}_c(X)$$ (means that $$A$$ is a compact operator). Suppose that $$(\text{id}-A)$$ is $$"1-1"$$. Show that the operator $$(\text{id}-A)$$ has a continuous inverse over the set $$V=(\text{id}-A)(X)$$.

Question :

As I fail to comprehend the question fully into functional analysis terms, how would one show that there exists a continuous inverse ? What is sufficient to prove that ?

Since $$(\text{id}-A)$$ is injective, there would exists an inverse if it was also surjective as well, but I can't seem to find a way around that. Also for the continuity, what's the catch ? Does showing that $$(\text{id}-A)$$ is an isometry has anything to do with that ?

Any hint on how to approach or elaboration will be the most appreciated since I cannot get a grasp on an intuition.

• This depends a little bit on your background, e.g. do you know basic Fredholm theory? – Klaus Mar 17 at 19:11
• @Klaus Thanks from replying. I am afraid that we have only mentioned what a Fredholm operator is and what its index is. After that, we have neither elaborated nor used it anywhere. – Rebellos Mar 17 at 19:12
• Basically you need that the index is invariant under compact perturbations. This would imply surjectivity. – Klaus Mar 17 at 19:16
• @Klaus I am at loss on how to elaborate over that, literally never used any of these. What about continuity ? Anyway, I would be very welcome if you could elaborate. – Rebellos Mar 17 at 19:17
• The inverse of any bounded invertible operator is bounded again (bounded inverse theorem). Boundedness is equivalent to continuity. – Klaus Mar 17 at 19:20