The inverse map on $\mathcal{l_i(H,H)}$, is continuous

Question

Let $\mathcal{H}$ a Hilbert space, and $\mathcal{l_i(H,H)}=\{f\in \mathcal{F(H,H)}, f \text{ is a linear continuous invertible map on } \mathcal{H}\}$

I want to prove that $\psi : \mathcal{l_i(H,H)} \rightarrow \mathcal{l_i(H,H)}$, $T \rightarrow T^{-1}$. Is continuous .

First of all, I think that $\psi$ is not linear.

So I supposed that : if $(T_n)_n \in \mathcal{l_i(H,H)} $, convergent to $T$.

Anad I want to prove that $(T_n^{-1})_n \in \mathcal{l_i(H,H)} $, convergent to $T^{-1}$.

and since $\mathcal{l_i(H,H)} $ is a Banach space, we just have to prove that it's a cauchy sequence, but unfortunately I can't.

Answer 1

You might know the following basic theorem:

Theorem (Neumann): If $\|A\|<1$, then $I-A$ is invertible and $\|(I-A)^{-1}\|\le\frac 1{1-\|A\|}$.

Now, let $T$ be invertible and let $S$ be as close to $T$ such that $\|S-T\| < \|T^{-1}\|^{-1}$. Then, setting $A = I-T^{-1}S = T^{-1}(T-S)$, we have $$S = TT^{-1}S = T[I-(I-T^{-1}S)] = T(I-A)$$ and since $\|A\|=\|T^{-1}(T-S)\|\le\|T^{-1}\|\|S-T\| < 1$, we see that $S$ is invertible. Moreover, $$ \|S^{-1}\| = \|(I-A)^{-1}T^{-1}\|\le\frac{\|T^{-1}\|}{1-\|A\|} = \frac{\|T^{-1}\|}{1-\|T^{-1}(T-S)\|}\le\frac{\|T^{-1}\|}{1-\|T^{-1}\|\|S-T\|}. $$ Now, assume that even $\|S-T\|\le\frac 12\|T^{-1}\|^{-1}$. Then $\|S^{-1}\|\le 2\|T^{-1}\|$ and hence $$ \|S^{-1}-T^{-1}\| = \|S^{-1}(T-S)T^{-1}\|\le\|S^{-1}\|\|T^{-1}\|\|S-T\|\le 2\|T^{-1}\|^2\|S-T\|, $$ showing (in particular) the continuity of $T\mapsto T^{-1}$.