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

Question

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.

Answer 1

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 $n\geq n_0$.

Answer 2

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 $$(I-T)^{-1}=I+T+T^2+\dots.$$