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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.

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  • $\begingroup$ This depends a little bit on your background, e.g. do you know basic Fredholm theory? $\endgroup$ – Klaus Mar 17 at 19:11
  • $\begingroup$ @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. $\endgroup$ – Rebellos Mar 17 at 19:12
  • $\begingroup$ Basically you need that the index is invariant under compact perturbations. This would imply surjectivity. $\endgroup$ – Klaus Mar 17 at 19:16
  • $\begingroup$ @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. $\endgroup$ – Rebellos Mar 17 at 19:17
  • $\begingroup$ The inverse of any bounded invertible operator is bounded again (bounded inverse theorem). Boundedness is equivalent to continuity. $\endgroup$ – Klaus Mar 17 at 19:20

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