Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 358 / 3

The spectral theorem for normal operators on a separable Hilbert space states that a normal operator $A$ has a unique projection-valued measure $E$ on its spectrum such thatFor bounded $A$ this integral acts on all of $H$. For an unbounded normal operator,Equivalently, $A$ is unitarily equivalent to multiplication by a measurable function on a direct sum of $L^2$ spaces.

By the spectral theorem, the operator in parentheses acts at spectral value $t\in \mathbb{R}$ byThe Poisson kernel converges to $1$ for $t\in (a,b)$, to $1/2$ at $t=a,b$, and to $0$ outside $[a,b]$. It is uniformly bounded, so dominated convergence in the spectral measure of a normal operator gives the strong limitApplying this operator to $v$ proves the claimed Stone formula.

Let $p(x)=x^3-x$. Since $A$ is multiplication by $p$, its spectral projection is multiplication by $1_{p^{-1}(S)}$. HenceSplit $[-1,1]$ at the two critical points $\pm 1/\sqrt{3}$. On each resulting interval, $p$ is monotone, so one-dimensional change of variables shows that the measure is absolutely continuous. For almost every $t$,The density vanishes outsideIts inverse-square-root singularities at the two critical values are locally integrable, so they do not create singular spectral measure.