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Suppose we wish to prove the following.

Let $X$ be a Banach space and let $(\mathbf\Lambda_n)_{n=1}^\infty \subset \mathcal L(X)$ be a sequence of invertible bounded linear operators on $X$ that converges in operator norm to some invertible map $\mathbf\Lambda$. Then $\mathbf\Lambda_n^{-1}$ converges in operator norm to $\mathbf\Lambda^{-1}$.

So I started as follows. By factoring $\mathbf\Lambda_n^{-1}$ from the left and $\mathbf\Lambda^{-1}$ from the right and using the submultiplicativity of the operator norm, we see that $$ \Vert \mathbf\Lambda_n^{-1} - \mathbf\Lambda^{-1} \Vert = \Vert \mathbf\Lambda_n^{-1}(\mathbf\Lambda - \mathbf\Lambda_n)\mathbf\Lambda^{-1} \Vert \leq \Vert \mathbf\Lambda_n^{-1} \Vert \Vert \mathbf\Lambda - \mathbf\Lambda_n\Vert \Vert\mathbf\Lambda^{-1} \Vert. \quad\quad (1) $$ Now we know by assumption that $$ \Vert \mathbf\Lambda - \mathbf\Lambda_n\Vert \to 0 , \quad n \to \infty, $$ so it would be nice if we could prove that the other two norms in $(1)$ are bounded uniformly in $n$.

The Banach isomorphism theorem allows us to conclude that $\mathbf\Lambda^{-1}$ is bounded, i.e. $$\Vert\mathbf\Lambda^{-1} \Vert \leq c$$ but also that each $\mathbf\Lambda_n^{-1}$ is bounded, only this time the bound might depend on $n$, i.e. $$\Vert\mathbf\Lambda_n^{-1} \Vert \leq c_n. $$ This is somewhat unfortunate, since we would like to let $n\to\infty$ in $(1)$ to reach the desired conclusion. Is it possible to achieve a uniform bound for $\Vert\mathbf\Lambda_n^{-1} \Vert$?

At this point I would like to emphasize that the limit $\mathbf\Lambda$ is assumed to be invertible. I found posts such as [1] and [2], where certain examples for which this is not possible are given, but it seems that it was not assumed that the limit $\mathbf\Lambda$ is invertible in any of the presented examples.

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Here is a direct argument. Since $T$ is invertible, it is bounded below: there exists $c>0$ such that $\|Tx\|\geq c\|x\|$ for all $x$. You can take, for instance, $c=\|T^{-1}\|^{-1}$.

Because $T_n\to T$, there exists $n_0$ such that, for all $n\geq n_0$, $\|T_n-T\|\leq c/2$. Then $$ \|T_nx\|=\|Tx-(T-T_n)x\|\geq \|Tx\|-\|(T-T_n)x\|\geq c\|x\|-\tfrac c2\|x\|=\tfrac c2\,\|x\| $$ for all $n\geq n_0$ and all $x$. In particular, for fixed $n$ and $x$ replace $x$ with $T_n^{-1}x$ above to get $$ \|T_n^{-1}x\|\leq\tfrac2c\,\|x\|. $$ Thus $$\|T_n^{-1}\|\leq\tfrac2c$$ for all $n\geq n_0$.

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  • $\begingroup$ Wonderful, thanks! $\endgroup$ – MisterRiemann Dec 17 '18 at 19:26
  • $\begingroup$ Glad I could help. $\endgroup$ – Martin Argerami Dec 17 '18 at 20:28
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Hints, in easier-to-type notation: First, it's enough to consider the case $T_n\to I$, where $I$ is the identity, because... . And for that case, note that if $||T||<1$ then $$(I-T)^{-1}=I+T+T^2+\dots.$$

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  • $\begingroup$ Thank you for your answer. I'm not seeing how I can use this to directly prove that $\Vert T_n^{-1} \Vert$ is uniformly bounded though. I feel that your hints can be used to prove the proposition, from which one may deduce the wanted uniform boundedness as well. Am I missing something? (P.S. the $\mathbf\Lambda$'s become easier to type once one gets used to them, and they look much more aesthetically pleasing than $T$ in my opinion. :-) ). $\endgroup$ – MisterRiemann Dec 17 '18 at 18:14
  • $\begingroup$ I think the answer proves that if $T$ is close to $I$, then $\|T^{-1}-I\| \leq C\|T-I\|$. $\endgroup$ – Mindlack Dec 17 '18 at 18:48

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