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Markov chain Monte Carlo asymptotic variance for a stationary Markov chain and $\psi \in L^2(\pi )$ iswhenever the limit and series exist. If a reversible Markov chain has positive $L_0^2(\pi )$ spectral gap $\gamma$, then the spectral theorem for normal operators on a separable Hilbert space gives

Let $L_n=\mathrm{span}\{e_1,\dots ,e_n\}$ and let $P_n$ be its orthogonal projection. For $z\in \mathbb{C}$, defineBecause $A$ is unitary,The displayed finite Hermitian matrix uses only finitely many matrix entries of $A$, and its least eigenvalue can be approximated by an arithmetic algorithm in the SCI hierarchy.

As $L_n$ increases densely,A unitary operator is normal, so the spectral theorem for normal operators on a separable Hilbert space identifies the limit asThe functions ${\gamma }_n$ are continuous and decrease to a continuous function on the compact unit circle. The Dini theorem therefore gives uniform convergence there.

Take successively finer rational meshes $G_n$ around $\mathbb{T}$. From the finitely computed values of ${\gamma }_n$, retain the mesh minima in the comparison neighborhoods whose radii are ${\gamma }_n(z)$; equivalently, use the standard local-minimum construction for a decreasing approximation to a distance function. Call the resulting finite set ${\Gamma }_n(A)$. Uniform convergence and the shrinking mesh implyin Hausdorff distance. Every operation at stage $n$ is finite and arithmetic, so $({\Gamma }_n)$ is the required one-limit sequence.

There is no finite-stage certificate that the whole output has the correct Hausdorff error: the convergence of ${\gamma }_n$ has no uniform computable rate over all unitary operators, and unseen matrix entries can still reveal a missing spectral component. A small computed residual can certify that an individual output point lies near the spectrum, but it cannot verify that ${\Gamma }_n(A)$ covers all of $\mathrm{Sp}(A)$. Thus the full finite-stage output is not verifiable without additional information.

A projection-valued measure on the Borel sets of $\mathbb{C}$ is a map $E$ into the orthogonal projections on a separable Hilbert space $\mathcal{H}$ such thatand for pairwise disjoint $B_j$,for every $v\in \mathcal{H}$, with convergence in norm.

The spectral theorem for normal operators on a separable Hilbert space states that a bounded normal operator $A$ has a unique projection-valued measure supported on $\mathrm{Sp}(A)$ for whichMore generally, the Borel functional calculus for a normal operator is

For $v,w\in \mathcal{H}$, the scalar spectral measures areIf $A$ is self-adjoint, its spectrum and hence the support of $E$ lie in $\mathbb{R}$. Moreover,so ${\mu }_v$ is a positive measure, andThe paper prints total mass $\Vert v\Vert$; with the standard definition it is $\Vert v{\Vert }^2$, so the unsquared norm is a typographical error.

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.

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.

Every bounded positive operator $T$ has a unique positive operator $T^{1/2}$ satisfying $(T^{1/2}{)}^2=T$. The spectral theorem for normal operators on a separable Hilbert space constructs it by applying the scalar square-root function to the spectrum.