Global unitary evolution at fixed finite regulator

Parseval gives ${\sum }_n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2=\mathrm{\parallel }f{\mathrm{\parallel }}^2$. Since $\mathrm{\mid }{e}^{-it{\lambda }_n}\mathrm{\mid }=1$, the finite partial sums defining $U(t)f$ are Cauchy in $H_N$, so converge; Parseval gives $\mathrm{\parallel }U(t)f{\mathrm{\parallel }}^2={\sum }_n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2=\mathrm{\parallel }f{\mathrm{\parallel }}^2$. The coefficient formula shows $U(-t)U(t)f=f=U(t)U(-t)f$ and $U(s)U(t)f=U(s+t)f$, first on finite sums and then on all of $H_N$ by norm convergence and the norm bound. Thus $U(t)$ is a surjective isometry, hence unitary, and $U(0)=I$.

Fix $s\in \mathbb{R}$ and $f\in H_N$. Given $\epsilon >0$, choose a finite set $F\subset \mathbb{N}$ with ${\sum }_{n\mathrm{\notin }F}\mathrm{\mid }{c}_n{\mathrm{\mid }}^2<{\epsilon }^2\mathrm{/}64$, possible by convergence of the nonnegative series in [F2]. The norm of the difference on the complement is at most $2({\sum }_{n\mathrm{\notin }F}\mathrm{\mid }{c}_n{\mathrm{\mid }}^2{)}^{1\mathrm{/}2}<\epsilon \mathrm{/}4$. On the finite set $F$, each scalar $e^{-it{\lambda }_n}$ is continuous in $t$, so the finite partial difference has norm below $3\epsilon \mathrm{/}4$ for $t$ sufficiently close to $s$. The triangle inequality proves $\mathrm{\parallel }U(t)f-U(s)f\mathrm{\parallel }<\epsilon$, establishing strong continuity.

Let $f\in D(h)$, so ${\sum }_n{\lambda }_n^2\mathrm{\mid }{c}_n{\mathrm{\mid }}^2<\mathrm{\infty }$. Since $\mathrm{\mid }{e}^{-it{\lambda }_n}\mathrm{\mid }=1$, $U(t)f$ has the same finite weighted coefficient sum and belongs to $D(h)$. For $r\mathrm{\ne }0$, $\frac{U(t+r)f-U(t)f}{r}={\sum }_n{e}^{-it{\lambda }_n}\frac{e^{-ir{\lambda }_n}-1}{r}{c}_n{\psi }_n\mathrm{.}$ The scalar integral identity $e^{-ir\lambda }-1={\int }_0^r(-i\lambda )e^{-is\lambda }ds$ gives $\mathrm{\mid }(e^{-ir\lambda }-1)\mathrm{/}r\mathrm{\mid }\le \mathrm{\mid }\lambda \mathrm{\mid }$. Thus the difference quotient coefficients are dominated in square sum by ${\lambda }_n^2\mathrm{\mid }{c}_n{\mathrm{\mid }}^2$. On each finite set they converge to $-i{\lambda }_n{e}^{-it{\lambda }_n}{c}_n$; choosing a finite set with small tail in that dominating series and then taking $r\to 0$ proves convergence in $H_N$ to $-i{\sum }_n{\lambda }_n{e}^{-it{\lambda }_n}{c}_n{\psi }_n=-ihU(t)f$.

Conversely, suppose the norm derivative at $t=0$ exists and equals $g\in H_N$. Continuity of the coefficient functional $v\mapsto ⟨v,{\psi }_n⟩$ from [F3] makes its value on the difference quotient converge to $-i{\lambda }_n{c}_n$. Hence $⟨g,{\psi }_n⟩=-i{\lambda }_n{c}_n$ for every $n$. Parseval for $g$ implies ${\sum }_n{\lambda }_n^2\mathrm{\mid }{c}_n{\mathrm{\mid }}^2=\mathrm{\parallel }g{\mathrm{\parallel }}^2<\mathrm{\infty }$, so $f\in D(h)$ by [F1]. If the derivative exists at any other $t$, the same coefficient argument multiplies each coefficient by a unit phase and yields the identical summability condition. Thus the differentiability domain is exactly $D(h)$ at every $t$.