Let $X$ be a Banach space and let $\mathrm{Iso}(X)$ be its group of isometries, i.e., the set of surjective linear maps $T: X \to X$ with $\|Tx\| = \|x\|$.


**Q: Is $\mathrm{Iso}(X)$ a topological group under the strong topology?**


While it is easy to show that multiplication is continuous, it is not clear to me how to show that inversion is continuous. I did not find a reference in the literature for this statement.


If $X = H$ is a separable Hilbert space, then $\mathrm{Iso(X)} = \mathrm{U}(H)$, the unitary group of $H$, and the statement that this is a topological group is well-established. However, the proof (that I know) uses the fact that the weak and the strong operator topology agree on $\mathrm{U}(H)$ and that the inverse is just given by $u \mapsto u^\*$, which is continuous in the weak operator topology.



Group of isometries of Banach spaces a topological group?
That the inverse is continuous for the strong topology would actually be true for any *bounded* subgroup of $GL(X)$, the invertible operators on $X$.


Firstly, as translation is continuous, it suffices to consider continuity at the identity.


Now let $(T\_i)$ be a *bounded* net of invertibles converging strong to $I$, and with $(T\_i^{-1})$ also bounded, say a common bound of $K$. Then, for $x\in X$,
$$ \|T\_i^{-1}(x) - x\| = \|T\_i^{-1}(x - T\_i(x))\| \leq K \|x-T\_i(x)\| \rightarrow 0 $$
by assumption that $T\_i\rightarrow I$ strongly. Thus also $T\_i^{-1}\rightarrow I$ strongly.


