 Solution

Let $E_n$ and $E$ be the projection-valued measures of $A_n$ and $A$. Using the projection $P_n:\mathcal{H}\to {\mathcal{H}}_n$, defineThe scalar spectral measures converge weakly whenfor every bounded continuous function $f$ and every $v,w\in \mathcal{H}$. By the spectral theorem for normal operators on a separable Hilbert space, this is equivalent to

The assumed moment identities say precisely that this convergence holds for every monomial $f(x)=x^m$. It follows by linearity for every polynomial. For $v=w$, the $m=2$ identity givesso Markov inequality makes the positive measures $({\mu }_v^{(n)})$ tight. Higher even moments similarly control the tails of any fixed polynomial.

Given a bounded continuous $f$ and $ϵ>0$, choose $R$ so that the measure tails are uniformly small. The Weierstrass approximation theorem supplies a polynomial $p$ withMoment convergence handles $p$; tightness and a sufficiently high even moment handle the two tails. Hence $\int fd{\mu }_v^{(n)}\to \int fd{\mu }_v$. The polarization identity then gives the same conclusion for ${\mu }_{v,w}^{(n)}$. This proves weak convergence of scalar spectral measures.

The assertion fails if only $m=1$ is assumed. Let $\mathcal{H}={\ell }^2(\mathbb{N})$, let ${\mathcal{H}}_n=\mathrm{span}\{e_1,\dots ,e_n\}$, take $A=0$, and letThe reversal matrices are self-adjoint unitaries. For fixed $v,w\in {\ell }^2$,because the finite head of one vector is paired with the vanishing tail of the other. Thus the $m=1$ condition holds. However, $A_n^2=I_{{\mathcal{H}}_n}$, soin general. Taking $f(x)=x^2$ shows that the spectral measures do not converge weakly.