Invertibility of a strong limit

Question

Let $H$ a Hilbert space and $A_n, A$ bounded and invertible operators such that $A_n\to A$ in the strong sense, that is for any $x\in H$, $$ ||A_n x - Ax||\to 0. $$ Is it true that $A_n^{-1}\to A^{-1}$ in the strong sense? Thank you for your help.

Answer 1

It need not be the case that $A_n^{-1} \to A_n$ in the strong operator topology.

Let $H = \ell^2(\mathbb{N})$, and for every $n\in \mathbb{N}$ define

$$A_nx = x - \frac{n}{n+1}\langle x, e_n\rangle e_n.$$

Then $A_n \to A = \operatorname{id}$ strongly, for

$$\lVert x - A_n x\rVert = \frac{n}{n+1}\lvert \langle x, e_n\rangle\rvert \to 0,$$

but we have $A_n^{-1}x = x + n\langle x, e_n\rangle e_n$ and therefore $\lVert A_n^{-1}\rVert = n+1$, so $\{ A_n^{-1} : n \in \mathbb{N}\}$ is not uniformly bounded.

Specifically, for

$$x = \sum_{n = 0}^{\infty} \frac{1}{n+1}e_n$$

we have

$$\lVert A_n^{-1} x - x\rVert = \frac{n}{n+1}\to 1.$$