Show that $f(T)=T^{-1}$ is a homeomorphism on Aut(V)

Question

I have proven part i and ii.

(i) I essentially used the fact that if $||s||<1$ then $I-S$ is invertible.

(ii) Consider $B(T,||T^{-1}||^{-1}) $

But im stuck at iii. I cant find the right $\delta$ that would allow me to show that the mapping is continuous. Any ideas?

Answer 1

When solving (i), you probably found the formula $$(T-S)^{-1}=(T(I-T^{-1}S))^{-1}=(I-T^{-1}S)^{-1}T^{-1}=\left(\sum_{n=0}^\infty (T^{-1}S)^n\right)T^{-1}$$

Subtracting from $T^{-1}$ and taking the norm:

$$\Vert T^{-1}-(T-S)^{-1}\Vert\leq\sum_{n=1}^\infty(\Vert T^{-1}\Vert\Vert S\Vert)^n=\frac{\Vert T^{-1}\Vert\Vert S\Vert}{1-\Vert T^{-1}\Vert\Vert S\Vert}$$

If $T-S$ is very close to $T$, then $\Vert S\Vert$ is very small, so $\Vert T^{-1}\Vert\Vert S\Vert$ is also very close to $0$, and the rightmost term above is also very close to $0$ (because $x\mapsto x/(1-x)$ is continuous).

Therefore $f\colon T\mapsto T^{-1}$ is continuous on $Aut(V)$. But $f\circ f=I$, i.e., $f$ has a continuous inverse (itself), so $f$ is a homeomorphism.