Uniformly bounding a sequence $(\mathbf\Lambda_n^{-1})_{n=1}^\infty$ of inverses of bounded linear operators

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.

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 $$n\geq n_0$$.
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 $$(I-T)^{-1}=I+T+T^2+\dots.$$
• 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. :-) ). – MSDG Dec 17 '18 at 18:14
• I think the answer proves that if $T$ is close to $I$, then $\|T^{-1}-I\| \leq C\|T-I\|$. – Mindlack Dec 17 '18 at 18:48