Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 1 / 3 / b

The Stone theorem for one-parameter unitary groups states that every strongly continuous unitary representation $U_t$ of $\mathbb{R}$ is uniquely $U_t=e^{itA}$ for a possibly unbounded self-adjoint operator $A$. Its domain isConversely every self-adjoint $A$ gives such a group by the Borel functional calculus for a normal operator.

Prove the forward direction by putting $G=iA$, initially defined as the norm derivative at zero. Its domain is dense: for $f\in C_c^{\infty }(\mathbb{R})$, the vector ${\xi }_f=\int f(t)U_t\xi dt$ has derivative $G{\xi }_f=-\int f^{\prime }(t)U_t\xi dt$. Smooth approximate identities make ${\xi }_f\to \xi$. The group law gives $U_{t}D(G)=D(G)$ and $G{U}_t=U_{t}G$ on that domain. Differentiation of preserved inner products shows that $G$ is skew-symmetric, so $A=-iG$ is symmetric.

The derivative operator is closed. Indeed, if ${\xi }_n\to \xi$ and $G{\xi }_n\to \eta$, integrate the orbit derivative to obtainTaking limits gives the same identity with $\xi ,\eta$; dividing by $t$ at zero proves $\xi \in D(G)$ and $G\xi =\eta$.

Now define bounded Laplace integralsShifting the integration variable in $U_h{R}_{\pm }\xi$ and taking the derivative at zero shows that $R_{\pm }\xi \in D(G)$ andHence $I-G$ and $I+G$ are both surjective. Since $G=iA$, this means $A+iI$ and $A-iI$ are surjective.

Here is why these range conditions establish self-adjointness rather than just symmetry. If $\eta \in D(A^{\ast })$, solve $(A-iI)\xi =(A^{\ast }-iI)\eta$. Then $\eta -\xi \in \ker (A^{\ast }-iI)=\mathrm{Ran}(A+iI{)}^{\perp }=\{0\}$. Thus $D(A^{\ast })\subseteq D(A)$, and symmetry gives equality with $A^{\ast }=A$.

The permitted functional calculus now gives $V_t=e^{itA}$. It is unitary, strongly continuous by dominated convergence against spectral measures, and differentiable with derivative $iA{V}_t\xi$ on $D(A)$. Differentiate $V_{-t}{U}_t\xi$ for $\xi \in D(A)$; the two derivative terms cancel because both groups preserve the domain and commute with their generator. Its value is consequently the constant $\xi$. Density of the domain implies $U_t=V_t$ on all of $H$.

For completeness, the converse and the exact domain follow from the same calculus. If $\xi \in D(A)$, dominated convergence applied to $|(e^{it\lambda }-1)/t|\le |\lambda |$ gives the derivative $iA\xi$. If that norm derivative exists, Fatou's lemma applied to its squared norm gives $\int {\lambda }^{2}d⟨E_A(\lambda )\xi ,\xi ⟩<\infty$, so $\xi \in D(A)$. The derivative characterization also proves uniqueness. Thus