Convergence of a sequence of inverse operators

Question

Let $T_n$ be a sequence of invertible bounded linear operators between Banach spaces converging to $T$. I naively thought that the sequence of inverses $T_n^{-1}$ may converge to $T^{-1}$. On the other hand, however, I vaguely remember that I have been taught that a side condition such as $\Vert T_n^{-1}\Vert < M$, where $M$ is indepedent of $n$, may be needed for $T_n^{-1}$ to converge to $T^{-1}$. I couldn't find a reference to the memory. Is it really possible that $T_n^{-1}$ does not converge to $T^{-1}$ in general?

Answer 1

Let $c_n$ be some sequence converging to $0$ and define $T_n (x) := c_n\cdot x$. Then $T_n \to 0$ in the operator topology. We have $T_n^{-1}(x) := \frac{1}{c_n}x$ which does not converge in the operator topology (we have $\|T_n^{-1}\| = \frac{1}{c_n}$).

Answer 2

There are some details missing regarding type of convergence. Consider norm-convergence: If you assume $T_n$ invertible and $\|T^{-1}_n\|\leq M$ then for $\|T-T_n\|<1/M$, $T$ is also invertible and you get a von neumann series by developping $$ T^{-1}= (T_n + (T-T_n))^{-1} = (I + T_n^{-1} (T-T_n))^{-1} T_n^{-1}$$ Sometimes you may use the identity $$ T_n^{-1} - T^{-1} = T_n^{-1} (T - T_n) T^{-1} $$ to get better estimates. If you don't have any bounds on $T_n^{-1}$ nor $T^{-1}$ then it becomes more difficult...