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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.