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

The Stone theorem for the circle group says that a strongly continuous unitary representation $U$ of $\mathbb{T}=\{z:|z|=1\}$ has an orthogonal decompositionEquivalently, $U_z={\sum }_n{z}^n{P}_n$ strongly, with mutually orthogonal projections $P_n$ summing strongly to the identity. Define these projections using normalized Haar measure:Haar invariance gives $U_w{P}_n=w^n{P}_n$. Changing $z$ to $z^{-1}$ in the adjoint integral proves $P_n^{\ast }=P_n$. Applying the defining integral to $P_m\xi$ gives $P_n{P}_m={\delta }_{nm}{P}_m$ by orthogonality of the characters $z^k$. Thus $P_n$ are orthogonal projections, and their ranges are exactly the indicated character subspaces.

It remains to prove completeness of these subspaces. The Fejér kernel has the finite expansionIt is nonnegative, has normalized integral one, and its integral outside any fixed neighborhood of the identity tends to zero, since the denominator in the usual geometric-sum formula is bounded away from zero there. Strong continuity therefore givessplit the integral into that neighborhood, where $\Vert U_z\xi -\xi \Vert$ is small, and its complement, where it is bounded by $2\Vert \xi \Vert$. The integral on the left is ${\sum }_{|n|<N}(1-|n|/N)P_n\xi$. Hence the closed span of all $H_n$ is $H$. Orthogonality then gives the unweighted strong sum and the stated representation formula.

Conversely, any orthogonal decomposition indexed by the integers gives this representation; continuity follows first on finite sums and then on all vectors by the uniform unitary norm bound. With $z=e^{it}$ its generator isThis diagonal operator is self-adjoint and $U_{e^{it}}=e^{itA}$. Thus the circle theorem is the integer-spectrum version of the one-parameter theorem.

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