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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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    @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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    @GastónBurrull: Because $\|A_n-A\|\to 0$ by hypothesis. Multiplication is continuous; recall the inequality $\|ST\|\leq\|S\|\|T\|$.2013-09-02