Let $E$ be a Banach Space and $f:Iso(E,E)\rightarrow Iso(E,E)$ be given by $f(S)=S^{-1}$. Prove that $f$ is continuos.
Hint: $f(I-T)=\sum_{n=0}^\infty T^n$ if $\lVert T\rVert<1$.
I was able to solve this without the hint as follows:
$$\lVert S^{-1}-S_o^{-1}\rVert=\lVert S^{-1}S_oS_o^{-1}-S_o^{-1} \rVert\leq \lVert S^{-1}S_o-I \rVert \lVert S_o^{-1} \rVert \leq \lVert S^{-1}S_o-S^{-1} S \rVert \lVert S_o^{-1} \rVert\Rightarrow $$ $$\lVert S^{-1}-S_o^{-1}\rVert\leq \lVert S^{-1}\rVert\lVert S_o^{-1}\rVert\lVert S-S_o\rVert$$
So we need to find a nice bound for $\lVert S^{-1} \rVert$. At first I noticed:
$$\lVert S^{-1} \rVert=\sup_{x\not=0}\frac{\lVert S^{-1}(x)\rVert}{\lVert x\rVert}=\sup_{v\not=0}\frac{\lVert S^{-1}(S(v))\rVert}{\lVert S(v)\rVert}=\sup_{v\not=0}\frac{\lVert v\rVert}{\lVert S(v)\rVert}=$$ $$\left(\inf_{v\not=0}\frac{\lVert S(v)\rVert}{\lVert v\rVert}\right)^{-1}=\frac{1}{\inf_{\lVert v\rVert =1}\lVert S(v)\rVert}$$
It is straightforward to see that for $\lVert v \rVert=1$:
$$\lVert S (v) \rVert\geq \lVert S_o(v)-(S_o-S) (v) \rVert \geq\lVert S_o(v) \rVert-\lVert S-S_o\rVert\Rightarrow$$
$$ \inf_{\lVert v\rVert=1}\lVert S (v) \rVert \geq \inf_{\lVert v\rVert=1}\lVert S_o (v) \rVert - \lVert S-S_o\rVert\Rightarrow$$
$$ \frac{1}{\lVert S^{-1} \rVert} \geq \frac{1}{\lVert S_o^{-1}\rVert} - \lVert S-S_o\rVert\Rightarrow\lVert S^{-1}\rVert\leq \frac{\lVert S_{o}^{-1}\rVert}{1-\lVert S_o-S \rVert\lVert S_o^{-1}\rVert}$$
Combining this bound for $S$ with our initial inequality yields:
$$\rVert S^{-1}- S_o^{-1} \rVert \leq \lVert S_o- S \rVert\lVert S^{-1}\rVert \lVert S_o^{-1} \rVert\leq \frac{\lVert S_{o}^{-1}\rVert^2\lVert S-S_o \rVert}{1-\lVert S_o-S \rVert\lVert S_o^{-1}\rVert}$$
And from this it is clear that if $S\in Iso(E,E)$ and $\rVert S- S_o \rVert\leq \delta(\varepsilon)\Rightarrow \rVert S^{-1}- S_o^{-1} \rVert\leq\varepsilon$.
I wonder how the hint comes into play and if my proof avoiding the hint is correct.
It suffices to show that $f$ is continuous at $I.$ Indeed, when $S\to S_0$ then $SS_0^{-1}\to I.$ By the continuity at $I,$ we get $S_0S^{-1}\to I,$ thus $S^{-1}\to S_0^{-1}.$
So it remains to prove the continuity at $I.$ Let $S\to I.$ Denote $T=I-S.$ Assume $\|T\|<1.$ Applying the hint gives $$S^{-1}-I=(I-T)^{-1}-I=\sum_{n=1}^\infty T^n$$ Thus $$\|S^{-1}-I\|\le \sum_{n=1}^\infty \|T\|^n={\|T\|\over 1-\|T\|}={\|I-S\|\over 1-\|I-S\|}$$
Remark The advantage of this approach is that it can be applied to any Banach algebra. The elements do not need to be operators acting on some space.