= Solution
Let $E_n$ and $E$ be the <projection-valued measures> of $A_n$ and $A$. Using the projection $P_n:\mathcal H\to\mathcal H_n$, define
$$
\mu^{(n)}_{v,w}(B)
=\langle E_n(B)P_nv,P_nw\rangle.
$$
The <scalar spectral measures> converge weakly when
$$
\boxed{
\int_{\mathbb R}f\,d\mu^{(n)}_{v,w}
\longrightarrow
\int_{\mathbb R}f\,d\mu_{v,w}}
$$
for 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
$$
\langle f(A_n)P_nv,P_nw\rangle
\longrightarrow
\langle f(A)v,w\rangle.
$$
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 gives
$$
\int x^2\,d\mu^{(n)}_v(x)\longrightarrow
\int x^2\,d\mu_v(x),
$$
so <Markov inequality> makes the positive measures $(\mu^{(n)}_v)$ tight. Higher even moments similarly control the tails of any fixed polynomial.
Given a bounded continuous $f$ and $\epsilon>0$, choose $R$ so that the measure tails are uniformly small. The <Weierstrass approximation theorem> supplies a polynomial $p$ with
$$
\sup_{|x|\leq R}|f(x)-p(x)|<\epsilon.
$$
Moment convergence handles $p$; tightness and a sufficiently high even moment handle the two tails. Hence $\int f\,d\mu_v^{(n)}\to\int f\,d\mu_v$. 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 $\mathcal H_n=\operatorname{span}\{e_1,\ldots,e_n\}$, take $A=0$, and let
$$
A_ne_j=e_{n+1-j}
\qquad(1\leq j\leq n).
$$
The reversal matrices are self-adjoint unitaries. For fixed $v,w\in\ell^2$,
$$
\langle A_nP_nv,P_nw\rangle\longrightarrow0,
$$
because the finite head of one vector is paired with the vanishing tail of the other. Thus the $m=1$ condition holds. However, $A_n^2=I_{\mathcal H_n}$, so
$$
\langle A_n^2P_nv,P_nw\rangle
\longrightarrow\langle v,w\rangle
\ne0=\langle A^2v,w\rangle
$$
in general. Taking $f(x)=x^2$ shows that the spectral measures do not converge weakly.
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