Herglotz spectral measure and the summable-covariance density

For any finite integers $h_1,\dots ,h_m$ and complex numbers $c_1,\dots ,c_m$, finite second moments make ${\sum }_j{c}_j(X_{h_j}-\mu )$ square-integrable. Expanding its squared modulus and using stationarity gives ${\sum }_{j,k=1}^m{c}_j\overline{c_k}\gamma (h_j-h_k)=\mathbb{E}{\mid {\sum }_{j=1}^m{c}_j(X_{h_j}-\mu )\mid }^2\ge 0.$ Thus $\gamma$ is positive definite on the additive group $\mathbb{Z}$ by [F1]–[F3].

Now assume absolute summability and define $f(\lambda )=(2\pi {)}^{-1}{\sum }_{h\in \mathbb{Z}}\gamma (h)e^{-ih\lambda }$. For every $N$, the tail is bounded uniformly in $\lambda$ by $(2\pi {)}^{-1}{\sum }_{\mathrm{\mid }h\mathrm{\mid }>N}\mathrm{\mid }\gamma (h)\mathrm{\mid }$, which tends to zero; finite partial sums are continuous, so the series converges uniformly to a continuous function $f$ by [F14].

For each $N\ge 0$, set $f_N(\lambda ):=\frac{1}{2\pi }{\sum }_{\mathrm{\mid }h\mathrm{\mid }\le N}(1-\frac{\mathrm{\mid }h\mathrm{\mid }}{N+1})\gamma (h)e^{-ih\lambda }\mathrm{.}$ Using $c_j=e^{-ij\lambda }$ in [F3] for $j=0,\dots ,N$ and grouping pairs by $h=j-k$ gives $f_N(\lambda )=\frac{1}{2\pi (N+1)}{\sum }_{j,k=0}^N{e}^{-i(j-k)\lambda }\gamma (j-k)\ge 0.$ Moreover, $2\pi \mathrm{\mid }{f}_N(\lambda )-f(\lambda )\mathrm{\mid }\le {\sum }_{\mathrm{\mid }h\mathrm{\mid }>N}\mathrm{\mid }\gamma (h)\mathrm{\mid }+{\sum }_{\mathrm{\mid }h\mathrm{\mid }\le N}\frac{\mathrm{\mid }h\mathrm{\mid }}{N+1}\mathrm{\mid }\gamma (h)\mathrm{\mid },$ and the right side tends to zero: first make the absolutely summable tail small, then let $N\to \mathrm{\infty }$ on the remaining finite sum. Thus $f_N\to f$ uniformly, so $f$ is real and nonnegative.

Give $\mathbb{Z}$ the discrete topology. Every singleton is compact and open, distinct integers have disjoint open singleton neighborhoods, and maps from a discrete space are continuous; hence $\mathbb{Z}$ is a locally compact Hausdorff abelian topological group and $\gamma$ is continuous. By [F4] and step 1.1, there is a unique finite positive Radon measure $M$ on $\widehat{\mathbb{Z}}$ with $\gamma (h)={\int }_{\widehat{\mathbb{Z}}}\chi (h)M(d\chi ),M(\widehat{\mathbb{Z}})=\gamma (0)\mathrm{.}$ Use the homeomorphism $\mathrm{\Psi }:\mathbb{T}\to \widehat{\mathbb{Z}}$ from [F5] and define $F(B):=M(\mathrm{\Psi }(B))$ for Borel $B\subseteq \mathbb{T}$. Images of Borel sets are Borel by [F7], so this is a finite positive Borel measure. It is finite on compact sets because $\mathrm{\Psi }$ maps compact sets to compact sets. For Borel $E\subseteq \mathbb{T}$, outer regularity of $M$ gives an open $V\supseteq \mathrm{\Psi }(E)$ with $M(V)<M(\mathrm{\Psi }(E))+\epsilon$; then $U={\mathrm{\Psi }}^{-1}(V)$ is open, contains $E$, and $F(U)=M(V)$. For open $O\subseteq \mathbb{T}$, inner regularity gives compact $C\subseteq \mathrm{\Psi }(O)$ with $M(C)>M(\mathrm{\Psi }(O))-\epsilon$; then $K={\mathrm{\Psi }}^{-1}(C)$ is compact in $O$ and $F(K)=M(C)$. Thus $F$ is Radon by [F6]. For any nonnegative Borel $q$ on $\widehat{\mathbb{Z}}$, ${\int }_{\mathbb{T}}q(\mathrm{\Psi }(z))F(dz)={\int }_{\widehat{\mathbb{Z}}}q(\chi )M(d\chi )\mathrm{.}$ Indeed the identity holds for indicators by the definition of $F$, hence for nonnegative simple functions by finite addition, and for nonnegative Borel functions by increasing simple approximation and monotone convergence [F8]. It also holds for bounded complex Borel $q$ by applying that argument to its positive and negative real and imaginary parts; these are integrable by [F9] because both measures are finite. Taking $q(\chi )=\chi (h)$ and using [F5] proves $\gamma (h)={\int }_{\mathbb{T}}{e}^{ih\lambda }F(d\lambda )$. Also $F(\mathbb{T})=M(\widehat{\mathbb{Z}})=\gamma (0)$. If another finite positive Radon measure $G$ on $\mathbb{T}$ represents $\gamma$, let $\tilde{G}(E):=G({\mathrm{\Psi }}^{-1}(E))$ for Borel $E\subseteq \widehat{\mathbb{Z}}$. It is finite on compact sets because ${\mathrm{\Psi }}^{-1}$ maps compact sets to compact sets. For Borel $E$, outer regularity of $G$ gives open $U\supseteq {\mathrm{\Psi }}^{-1}(E)$ with $G(U)<G({\mathrm{\Psi }}^{-1}(E))+\epsilon$; then $\mathrm{\Psi }(U)$ is open, contains $E$, and $\tilde{G}(\mathrm{\Psi }(U))=G(U)$. For open $V\subseteq \widehat{\mathbb{Z}}$, inner regularity gives compact $K\subseteq {\mathrm{\Psi }}^{-1}(V)$ with $G(K)>G({\mathrm{\Psi }}^{-1}(V))-\epsilon$; then $\mathrm{\Psi }(K)$ is compact in $V$ and $\tilde{G}(\mathrm{\Psi }(K))=G(K)$. Thus $\tilde{G}$ is Radon by [F6] and [F7]. The same indicator/simple/monotone argument gives ${\int }_{\widehat{\mathbb{Z}}}q(\chi )\tilde{G}(d\chi )={\int }_{\mathbb{T}}q(\mathrm{\Psi }(z))G(dz)$ for bounded Borel $q$. Taking $q(\chi )=\chi (h)$ shows $\tilde{G}$ represents $\gamma$; uniqueness in [F4] gives $\tilde{G}=M$. Hence $G(B)=\tilde{G}(\mathrm{\Psi }(B))=M(\mathrm{\Psi }(B))=F(B)$ for every Borel $B\subseteq \mathbb{T}$. The integrands $\chi \mapsto \chi (h)$ are Borel by [F15], so the representing integrals are defined.

Put $s_N(\lambda ):=(2\pi {)}^{-1}{\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}\gamma (k)e^{-ik\lambda }$. Absolute summability gives $\mathrm{\parallel }{s}_N-f{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$. For $N\ge \mathrm{\mid }h\mathrm{\mid }$, the finite sum and [F11] give ${\int }_{\mathbb{T}}{e}^{ih\lambda }{s}_N(\lambda )d\lambda =\frac{1}{2\pi }{\sum }_{\mathrm{\mid }k\mathrm{\mid }\le N}\gamma (k){\int }_{\mathbb{T}}{e}^{i(h-k)\lambda }d\lambda =\gamma (h)\mathrm{.}$ If $q_N=e^{ih\lambda }(f-s_N)$ and ${\tau }_N=\mathrm{\parallel }f-s_N{\mathrm{\parallel }}_{\mathrm{\infty }}$, then $\mathrm{\mid }\mathrm{Re}{q}_N\mathrm{\mid },\mathrm{\mid }\mathrm{Im}{q}_N\mathrm{\mid }\le {\tau }_N$. By [F9], [F13], and $d\lambda (\mathbb{T})=2\pi$, the integrals of the real and imaginary parts each lie between $-2\pi {\tau }_N$ and $2\pi {\tau }_N$: subtracting either part from the constant ${\tau }_N$ or adding it to that constant gives a nonnegative function, so monotonicity and linearity apply. The complex integral is the real-part integral plus $i$ times the imaginary-part integral [F9], hence its modulus is at most the sum of the two absolute values, $4\pi {\tau }_N$. This tends to zero. Hence ${\int }_{\mathbb{T}}{e}^{ih\lambda }f(\lambda )d\lambda =\gamma (h)\mathrm{.}$ By [F12], $\nu (B):={\int }_{B}fd\lambda$ is a measure; it is finite and positive since $0\le f\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $d\lambda (\mathbb{T})=2\pi$. It is Radon: for a Borel $E$, outer regularity of $d\lambda$ [F10] gives open $U\supseteq E$ with $$\begin{gathered} d\lambda (U \\ setminusE)<\epsilon \mathrm{/}(\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}+1) \end{gathered}$$, hence $$\begin{gathered} \nu (U \\ setminusE)<\epsilon \end{gathered}$$. Countable additivity [F12] then yields $\nu (U)<\nu (E)+\epsilon$. For open $O$, inner regularity gives compact $K\subseteq O$ with $$\begin{gathered} d\lambda (O \\ setminusK)<\epsilon \mathrm{/}(\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}+1) \end{gathered}$$, hence $$\begin{gathered} \nu (O \\ setminusK)<\epsilon \end{gathered}$$ and $\nu (O)<\nu (K)+\epsilon$ by [F12]. These are the regularity clauses in [F6]. Thus $\nu$ is a finite positive Radon measure representing $\gamma$, and uniqueness in [F4] gives $F=fd\lambda$, with the stated density.