Let $H$ be a Hilbert space and suppose that $\{T_n\}$ is a sequence of invertible operators in $B(X)$ which converges to $T \in B(X)$. Suppose also that $\|T_n^{−1}\|<1$ for all $n \in \mathbb{N}$. Show that $T$ is invertible.
My attempt:
Thm : Let $H$ be a real Hilbert space. Let $A : H \to H$ be a bounded linear operator which is strictly positive definite, so that $\exists \ \beta >0 \ \ s.t. \langle A x,x\rangle \ge \beta \|x\|^2$. Then, for every $f \in H$, there exists a unique $u = A^{−1}f \in H$ such that $Au = f$ (i.e. $A$ is invertible) . The inverse operator $A^{−1}$ satisfies $\|A^{−1}\|\le \frac{1}{\beta}$.
$(1)$ Since $T_n$ is is invertible for all $n \in \mathbb{N}$, $\exists \ \beta >0 \ \ s.t. \langle T_n x,x\rangle \ge \beta \|x\|^2$ for all $n \in \mathbb{N}$. Since $\|T_n^{−1}\|<1 \implies \beta =1$.
$(2)$ Since $T_n \in B(X) , \ \forall n\in \mathbb{N}$ By uniform boundedness principle they're uniformly bounded and the limit $T$ is also linear and bounded.
$(3)$ Since $T,T_n \in B(X) , \ \forall n\in \mathbb{N}$ , there exists a subsequence $T_{nk}\rightharpoonup T$ i.e. $\langle T_{nk}x,x\rangle \to \langle Tx,x\rangle $, as $k\to \infty, \ \forall x \in H$.
So from part (1) we have the $ \langle Tx,x\rangle = \lim_{k\to \infty} \langle T_{nk}x,x\rangle \ge \|x\|^2$.
It is not true that invertibility of $T_n$ gives you the inequality you stated in (1). So this method completely breaks down.
Note that $\|I-T_n^{-1}T\|=\|T_n^{-1}T_n-T_n^{-1}T\|\leq \|T_n^{-1}\|\|T_n-T\| <1$ for $n$ sufficiently large. From the following well known fact it follows that $T_n^{-1}T$ is invertible and hence $T=T_n T_n^{-1}T$ is also invertible.
Lemma: If $S$ is a bounded operator with $\|S\| <1$ then $I-S$ is invertible
The lemma is proved by showing that the series $\sum S^{n}$ converges in operator norm and its sum is the inverse of $I-S$.