Momentum operator under the Fourier transform

The multiplier $m(\xi )=2\pi \xi$ is real, finite and measurable. Apply [F1] with the specified unitary $U={\mathcal{F}}_2$ from [F2]. The domain condition $mUf\in L^2$ is equivalent to $\xi Uf\in L^2$ since $2\pi \mathrm{\ne }0$. Hence [F1] gives exactly the stated self-adjoint operator and the strongly continuous group $e^{itP}={\mathcal{F}}_2^{-1}{M}_{e^{2\pi it\xi }}{\mathcal{F}}_2$, including both directions of its derivative-domain criterion.

If $f\in \mathcal{S}$, [F3] gives $\mathcal{F}(-i{f}^{\mathrm{\prime }})=2\pi \xi \widehat{f}$. The left side is a Schwartz function and hence is in $L^2$; by [F6] and [F2], this proves $f\in D(P)$ and $Pf=-i{f}^{\mathrm{\prime }}$. Also [F4] with $a=-t$ gives $\mathcal{F}(f(\cdot +t))=e^{2\pi it\xi }\widehat{f}$. By [F6] and step 1.1 it follows that $e^{itP}f=f(\cdot +t)$ for Schwartz $f$, with the plus sign appropriate to $e^{itP}$.

For any $f\in L^2$, use [F5] and countable choice to take $f_j\in \mathcal{S}$ tending to $f$. Both $e^{itP}$ and $f\mapsto f(\cdot +t)$ are isometries, respectively by step 1.1 and [F7]. Therefore the norm of their difference on $f$ is at most $2\mathrm{\parallel }f-f_j{\mathrm{\parallel }}_2$, since it is zero on $f_j$ by step 2.1. Let $j\to \mathrm{\infty }$. This proves the group identity on every class. The exact domain and derivative assertion remain those proved in step 1.1; no unspecified self-adjoint extension or general functional calculus is used.