Let $X$, $Y$ be Banach spaces, $D(T)$ dense in $X$ and $D(S)$ dense in $Y$. Furthermore let $T:D(T)\to D(S)$ and $S:D(S)\to D(T)$ be linear bounded operators with $ST = id_{D(T)}$ and $TS = id_{D(S)}$.
If $\tilde{T}$ and $\tilde{S}$ are the bounded continuations of $T$ and $S$ on $X$ resp. $Y$, is it true that $\tilde{T} = \tilde{S}^{-1}$ and $\tilde{S} = \tilde{T}^{-1}$?
I have a proof in mind (see answer below) but the proof seems too simply for me. Can someone confirm that my proof is right, or if not, point out the mistake?
My current proof:
Let $f\in X$ and $(f_n)_n\subset D(T)$ a sequence with $f_n\to f$ for $n\to\infty$, then $$ \tilde{S}\tilde{T} f = \tilde{S}\tilde{T} \lim f_n = \lim \tilde{S}\tilde{T} f_n $$ holds by continuity of the extensions. Now use $f_n \in D(T)$ and $T f_n\in D(S)$ to deduce $$ \lim\tilde{S}\tilde{T} f_n = \lim ST f_n = \lim f_n = f $$ using $ST = id_{D(T)}$ in the second step.
This shows $\tilde{S}\tilde{T} = id_X$ and by symmetry $\tilde{T}\tilde{S} = id_Y$ follows as well; thus finishing the proof.