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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.

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    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.2012-06-18
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    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.2012-06-18
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    "how it gives that direction": Convergent sequences are bounded.2012-06-18
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    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.2012-06-18
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    @JonasMeyer I can't remember right now if this result is *needed* to prove the continuity of inversion in a Banach algebra, but I have a vague feeling it can be proved without using that result. (However, I am under-caffeinated and could well be misremembering, I haven't sat down to check the details.)2012-06-18

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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.

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    That looks good. I still have to go over the details, especially in the second direction, but it will certainly help me. Thanks!2012-06-18
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    OK, I looked more closely and it "works". There are just some minor things to note, probably so minor it looks trivial to you. Anyway, could you hint me for how you got to the solution? Thank you!2012-06-18
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    @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".2012-06-18
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    Indeed. I knew the second "trick", but not the first one. I'll keep it in mind. Thanks a lot for your help!2012-06-18
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    Why $T_n\to 0$? Is $A^{-1}$ continuous at $0$?2013-09-02
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    @GastónBurrull: Because $\|A_n-A\|\to 0$ by hypothesis. Multiplication is continuous; recall the inequality $\|ST\|\leq\|S\|\|T\|$.2013-09-02