# Criterion for a limit of invertible operators on a Banach space to be invertible

Let $A_n$ linear operators in a Banach space $B$ that have inverses. $||A_n-A|| \to 0$ for some operator $A$.

I need to prove that $A$ has an inverse operator iff the sequence $\{||A_n^{-1}||\}$ is bounded.

I am almost sure it should be solved with the Uniform boundedness principle, but I can't figure it out, neither statements.

-
A pretty standard result is that $T\mapsto T^{-1}$ is continuous on the set of invertible operators, and that implies that $T\mapsto\|T^{-1}\|$ is continuous, which in turn gives the "only if" direction. –  Jonas Meyer Jun 18 '12 at 20:15
Thank you for your answer. I'm afraid I don't know this result, hence I cannot use it (unless I prove it). In addition, I'm not sure how it gives that direction. –  Abe Jun 18 '12 at 20:17
"how it gives that direction": Convergent sequences are bounded. –  Jonas Meyer Jun 18 '12 at 20:18
Yeah, I see it now. So it's sufficient to prove that standard result. My way of thinking (of how to solve the question, not to prove the statement) was to assume $\{||A_n^{-1}||\}$ wasn't bounded, therefore there is $x$ such as $||A_n^{-1}x|| \to \infty$. Couldn't really continue from there. –  Abe Jun 18 '12 at 20:23
Forgive me if this suggestion/question is out of turn, but does this question come from a set of course exercises? If so, perhaps it could/should be tagged as homework. –  user16299 Jun 18 '12 at 20:27

Suppose that $(A_n^{-1})$ is bounded. Using the identity $a^{-1}-b^{-1}=a^{-1}(b-a)b^{-1}$ and the fact that $(A_n)$ is a Cauchy sequence, it follows that $(A_n^{-1})$ is a Cauchy sequence. Since $L(B)$ is complete, there exists an operator $T$ such that $A_n^{-1}\to T$. Taking the limit of $A_nA_n^{-1}=A_n^{-1}A_n = I$ shows that $T=A^{-1}$.
Rearranging the same identity, $(I+a^{-1}(b-a))b^{-1}=a^{-1}$. If $A$ is invertible, then $(I+A^{-1}(A_n-A))A_n^{-1}=A^{-1}$. Since $T_n:=A^{-1}(A_n-A)\to 0$, $I+T_n$ is eventually invertible, with $(I+T_n)^{-1}=\sum\limits_{k=0}^{\infty}(-T_n)^k$, and $\|(1+T_n)^{-1}\|\leq \dfrac{1}{1-\|T_n\|}\to 1$. Thus, for $n$ sufficiently large, $A_n^{-1}=(I+T_n)^{-1}A^{-1}$, and this implies that $(A_n^{-1})$ is bounded.
@Abe: I don't consider it trivial. I intentionally condensed it, seeing no reason to give a very detailed answer. How I got to it: I cannot articulate it, but for example, experience has shown that the identity in the first line is quite useful, and using $(I-a)^{-1}=\sum a^k$ is also a standard "trick". –  Jonas Meyer Jun 18 '12 at 21:18
@GastónBurrull: Because $\|A_n-A\|\to 0$ by hypothesis. Multiplication is continuous; recall the inequality $\|ST\|\leq\|S\|\|T\|$. –  Jonas Meyer Sep 2 '13 at 15:20