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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. – 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
@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.) – user16299 Jun 18 '12 at 20:29
up vote 2 down vote accepted

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! – Abe Jun 18 '12 at 20:52
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! – Abe Jun 18 '12 at 21:11
@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
Indeed. I knew the second "trick", but not the first one. I'll keep it in mind. Thanks a lot for your help! – Abe Jun 18 '12 at 21:23
@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

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