On the proof of Heisenberg's uncertainty principle

@1: Yes, $S$ and $T$ are (unbounded) operators defined on suitable subspaces of $L^2$ going into $L^2$. You probably want either the Schwartz space or $H^1$ / its fourier transform respectively as domain for $T$ and $S$.

@3: If one of the integrals on the RHS of theorem 1 is infinite, there is really nothing to prove. If the other integral is zero, then $f=0$ and the inequaltiy is true. If the other integral is non-zero, then the RHS is $+\infty$ and the inequality is also true. Therefore you can assume right away and w.l.o.g. that both integrals are finite, i.e. not only $f\in L^2$, but $(x-a)f(x) \in L^2$ and $(\xi-b)\hat{f}(\xi)\in L^2$ as well so that $x f(x)\in L^2$ and $\xi \hat{f}(\xi)\in L^2$. Now since Fourier transform exchanges multiplication by $x$ with differentiation (up to some $\pm i$), this means that $f'(x)\in L^2$ as well. One has to be a bit careful here because this is only a weak derivative, but that doesn't change anything relevant.

@2: And this is the reason you can apply Plancherel's theorem: You assume wlog that all the relevant functions are in $L^2$.