Zero-variance geometric Markov sums have a uniform strong rate

Apply [F1] to obtain $g\in L^p(\pi )$ with $f=g-Pg$ almost everywhere; choose its finite measurable version as in that supplier. By [F3], $g\in L^2$. Stationarity makes the Poisson identity hold at every $X_k$ simultaneously off a countable union of null events. By [F2], on this common event $f(X_k)=g(X_k)-g(X_{k+1})+D_{k+1}$. Summing gives $S_n:={\sum }_{k=0}^{n-1}f(X_k)=R_n+M_n$, with $R_n=g(X_0)-g(X_n)$ and $M_n={\sum }_{k=0}^{n-1}{D}_{k+1}$. No ergodic theorem or central limit theorem has been used.

Put ${\tau }^2={\mathbb{E}}_{\pi }{D}_1^2$. Stationarity of transition pairs gives $\mathbb{E}{D}_j^2={\tau }^2$ for every $j\ge 1$. If $i<j$, then $D_i$ is ${\mathcal{F}}_{j-1}$-measurable, its product with $D_j$ is integrable by [F3], and [F2]–[F3] give $\mathbb{E}[D_i{D}_j]=\mathbb{E}[D_i\mathbb{E}(D_j\mid {\mathcal{F}}_{j-1})]=0$. Thus $\mathbb{E}{M}_n^2=n{\tau }^2$. The inequality $(a-b{)}^2\le 2a^2+2b^2$ and stationarity give $\mathbb{E}{R}_n^2\le 4\mathrm{\parallel }g{\mathrm{\parallel }}_2^2$. Cauchy–Schwarz [F3] and step 1.1 imply $\mathrm{\mid }{n}^{-1}\mathbb{E}{S}_n^2-{\tau }^2\mathrm{\mid }\le 4\mathrm{\parallel }g{\mathrm{\parallel }}_2^2\mathrm{/}n+4\mathrm{\parallel }g{\mathrm{\parallel }}_2\tau \mathrm{/}\sqrt{n}\to 0$. By [F1], the same normalized second moments tend to $v=0$, so ${\tau }^2=0$. A nonnegative variable of zero expectation is zero almost surely; applying this to each $D_j^2$ and taking the countable intersection shows that all $D_j$ vanish simultaneously. Hence $S_n=g(X_0)-g(X_n)$ for every $n$ on one probability-one event.

Fix $\epsilon >0$. Pointwise, the number of positive integers $k$ with $\mathrm{\mid }g(x){\mathrm{\mid }}^p>{\epsilon }^{p}k$ is at most ${\epsilon }^{-p}\mathrm{\mid }g(x){\mathrm{\mid }}^p$. Stationarity and [F4], applied first to finite sums and then to their increasing limit, therefore give ${\sum }_{k\ge 1}{\mathbb{P}}_{\pi }(\mathrm{\mid }g(X_k)\mathrm{\mid }>\epsilon k^{1\mathrm{/}p})={\sum }_{k\ge 1}\pi (\mathrm{\mid }g{\mathrm{\mid }}^p>{\epsilon }^{p}k)\le {\epsilon }^{-p}\pi \mathrm{\mid }g{\mathrm{\mid }}^p<\mathrm{\infty }$. Borel–Cantelli [F4], intersected over positive rational $\epsilon$, yields $g(X_k)=o(k^{1\mathrm{/}p})$ almost surely. On that event, for every $\epsilon >0$ the sufficiently late terms obey $\mathrm{\mid }g(X_k)\mathrm{\mid }\le \epsilon k^{1\mathrm{/}p}\le \epsilon n^{1\mathrm{/}p}$ for $k\le n$, while the maximum of the finitely many earlier terms divided by $n^{1\mathrm{/}p}$ tends to zero. Thus ${\max }_{k\le n}\mathrm{\mid }g(X_k)\mathrm{\mid }=o(n^{1\mathrm{/}p})$. Step 2.1 gives ${\max }_{0\le j\le n}\mathrm{\mid }{S}_j\mathrm{\mid }\le \mathrm{\mid }g(X_0)\mathrm{\mid }+{\max }_{j\le n}\mathrm{\mid }g(X_j)\mathrm{\mid }=o(n^{1\mathrm{/}p})$, proving the stationary conclusion.

For the all-start part, define the original path event $E=\{(x_k{)}_{k\ge 0}:n^{-1\mathrm{/}p}{\sum }_{k=0}^{n-1}f(x_k)\to 0\}$. It is product-measurable, since the condition is a countable combination of inequalities for finite measurable sums. It is invariant under deleting the first coordinate: if $s_n={\sum }_{k=0}^{n-1}f(x_k)$, the shifted sum is $s_{n+1}-f(x_0)$, and the reverse identity is $s_n=f(x_0)+(s_n-f(x_0))$ with the second term the shifted sum of length $n-1$. Since $f(x_0)$ is finite and $((n+1)\mathrm{/}n{)}^{1\mathrm{/}p}$ and $((n-1)\mathrm{/}n{)}^{1\mathrm{/}p}$ tend to one, either normalized convergence implies the other. Iterating gives invariance under deleting any fixed finite prefix. Also, on every real path, $s_n=o(n^{1\mathrm{/}p})$ is equivalent to ${\max }_{j\le n}\mathrm{\mid }{s}_j\mathrm{\mid }=o(n^{1\mathrm{/}p})$: the reverse direction is immediate, and the forward direction follows by discarding a finite initial segment and bounding all later $\mathrm{\mid }{s}_j\mathrm{\mid }$ by $\epsilon j^{1\mathrm{/}p}$. Thus it suffices to transfer the event $E$, rather than any Poisson identity outside a stationary full-measure set.

Let $h(x)={\mathbb{P}}_x(E)$. By [F5], $h$ is measurable and $0\le h\le 1$. The canonical stationary chain has the same path law as the stationary chain already considered, so step 3.1 gives $\pi (h)={\mathbb{P}}_{\pi }(E)=1$; this equality also follows by applying [F5] at time zero and taking expectations. For every state $x$ and integer $n\ge 1$, prefix-deletion invariance and [F5] give $h(x)={\mathbb{E}}_x{1}_E={\mathbb{E}}_x[h(X_n)]=P^{n}h(x)$. These are identities for each canonical starting state, not simultaneous versions of a conditional expectation over an uncountable family of starts.

For any probability measures $\mu ,\nu$ and measurable $0\le h\le 1$, let $h_m=2^{-m}{\sum }_{j=1}^{2^m}{1}_{\{h\ge j{2}^{-m}\}}$. Then $\mathrm{\mid }{h}_m-h\mathrm{\mid }\le 2^{-m}$ and [F6] gives $\mathrm{\mid }\mu (h_m)-\nu (h_m)\mathrm{\mid }\le \mathrm{\parallel }\mu -\nu {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}$. Letting $m\to \mathrm{\infty }$ proves the same bound for $h$. Under the additional pointwise total-variation assumption, apply it to $\mu =P^n(x,\cdot )$ and $\nu =\pi$. Step 5.1 gives $h(x)=P^{n}h(x)\to \pi (h)=1$ for every $x$, hence $h(x)=1$ everywhere. For any initial law $\mu$, [F5] at time zero gives ${\mathbb{P}}_{\mu }(E)=\int hd\mu =1$. The deterministic equivalence in step 4.1 transfers the uniform maximal rate as well. No moment or support condition on $\mu$ is needed.

Finally suppose the everywhere common geometric bound in the Statement holds. Since its envelope is finite at every state, it implies $P^n(x,\cdot )\to \pi$ in total variation for every $x$. By [F5] the canonical invariant-initial-law chain is stationary. The stronger arbitrary-measurable-space clause of [F7] supplies a finite integrable geometric envelope, and its mixing implication supplies the stationary geometric strong-mixing bound needed by [F1]. Therefore the proof in steps 1.1, 2.1, 3.1, 4.1, 5.1, 6.1 applies to the stated $f$ and zero covariance sum under every initial law. The zero function, a one-state chain, empty partial sums and zero martingale variance are all included; an empty state space admits no invariant probability and does not satisfy the hypotheses. Strict $2<p<q$ is used by [F1] and the $L^2$ comparison. AC is inherited precisely through [F1], [F2], [F3], [F5] and [F7]. This completes both the stationary result and the asserted all-start sufficient conditions.