Geometric strong mixing gives an Lp Poisson solution

Invariance [F7], extended with [F6], gives $\pi P\mathrm{\mid }h\mathrm{\mid }=\pi \mathrm{\mid }h\mathrm{\mid }$ for integrable $h$, so the absolute kernel integral is finite off a measurable $\pi$-null set. A change of representative on a $\pi$-null set $A$ does not change its image class, since $\pi P{1}_A=\pi (A)=0$ implies $P{1}_A=0$ almost everywhere. For $h\in L^p$, [F3] gives $h\in L^2$, and [F2] identifies $Ph(X_0)$ with $\mathbb{E}[h(X_1)\mid X_0]$. Conditional contraction [F3] and stationarity give $\mathrm{\parallel }Ph{\mathrm{\parallel }}_p\le \mathrm{\parallel }h{\mathrm{\parallel }}_p$. The bounded iteration identity in [F2] extends to these $L^p$ classes by replacing $h$ with its bounded clipped truncations: their $L^p$ error tends to zero because the integrable tail $\mathrm{\mid }h{\mathrm{\mid }}^p{1}_{\{\mathrm{\mid }h\mathrm{\mid }>K\}}$ tends to zero, as follows from [F6] applied to its complementary truncations; conditional and kernel contractions pass this error to zero. Thus $P$ is a well-defined linear contraction on $L^p$, its iterates agree with the kernel iterates, and $P^{n}h(X_0)=\mathbb{E}[h(X_n)\mid X_0]$. This also proves $P^{n}f\in L^p$ before any covariance estimate is used.

Fix $n\ge 1$, put $Z=P^{n}f$, and let $p^{\mathrm{\prime }}=p\mathrm{/}(p-1)$ and $W_K(x)=\mathrm{sgn}(Z(x))\mathrm{\mid }Z(x){\mathrm{\mid }}^{p-1}{1}_{\{\mathrm{\mid }Z(x)\mathrm{\mid }\le K\}}$ for integers $K\ge 1$. This function is bounded, belongs to $L^2$, and satisfies $\mathrm{\parallel }{W}_K{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}\le \mathrm{\parallel }Z{\mathrm{\parallel }}_p^{p-1}$. The centered $f(X_n)$ is in $L^2$ by [F3]. Step 1.1 and bounded conditional multiplication give $\mathbb{E}[W_K(X_0)f(X_n)]=\pi (W_K{P}^{n}f)=\pi (\mathrm{\mid }Z{\mathrm{\mid }}^p{1}_{\{\mathrm{\mid }Z\mathrm{\mid }\le K\}})$; since $\pi f=0$, this expectation is the covariance of $W_K(X_0)$ and $f(X_n)$. Bounded conditional multiplication here follows directly from the defining conditional expectation event identity, first for simple $W_K$ and then by uniform bounded approximation. Thus [F1] is applied only to square-integrable variables, even though $p^{\mathrm{\prime }}<2$.

Put $a={\alpha }_X(n)$. The quantile identity in [F1] gives $\mathrm{\parallel }{Q}_{W_K(X_0)}{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}=\mathrm{\parallel }{W}_K{\mathrm{\parallel }}_{p^{\mathrm{\prime }}}$ and $\mathrm{\parallel }{Q}_{f(X_n)}{\mathrm{\parallel }}_q=\mathrm{\parallel }f{\mathrm{\parallel }}_q$. Since $1\mathrm{/}{p}^{\mathrm{\prime }}+1\mathrm{/}q=1-\theta$, apply the product inequality [F4] to these two quantile functions in $L^{1\mathrm{/}(1-\theta )}(0,1)$, and then Hölder with $1_{(0,a)}$, whose $L^{1\mathrm{/}\theta }$ norm is $a^{\theta }$. Therefore [F1] and step 2.1 give $\pi (\mathrm{\mid }Z{\mathrm{\mid }}^p{1}_{\{\mathrm{\mid }Z\mathrm{\mid }\le K\}})\le 4a^{\theta }\mathrm{\parallel }Z{\mathrm{\parallel }}_p^{p-1}\mathrm{\parallel }f{\mathrm{\parallel }}_q$. If $a=0$ the quantile integral and the right-hand side are zero. Let $K\to \mathrm{\infty }$ using [F6]. If $\mathrm{\parallel }Z{\mathrm{\parallel }}_p=0$ the required bound is immediate; otherwise divide by $\mathrm{\parallel }Z{\mathrm{\parallel }}_p^{p-1}$ to obtain $\mathrm{\parallel }{P}^{n}f{\mathrm{\parallel }}_p\le 4{\alpha }_X(n{)}^{\theta }\mathrm{\parallel }f{\mathrm{\parallel }}_q$. The restriction $p<q$ ensures $\theta >0$ and makes all Hölder exponents just used valid.

The geometric bound makes ${\sum }_{n\ge 1}{\alpha }_X(n{)}^{\theta }\le C^{\theta }{\sum }_{n\ge 1}{r}^{n\theta }<\mathrm{\infty }$, including $C=0$. By step 3.1, [F4] makes $g_m={\sum }_{n=0}^m{P}^{n}f$ a Cauchy sequence in $L^p$. Completeness [F5] gives $g_m\to g\in L^p$. The contraction in step 1.1 gives $P{g}_m\to Pg$, while $g_m-P{g}_m=f-P^{m+1}f\to f$ in $L^p$ by step 3.1. Thus $g-Pg=f$ as almost-everywhere classes. Choose the finite measurable representative supplied by [F5], setting it to zero on its measurable null exceptional set if necessary, and use the version convention from step 1.1 for $Pg$. These choices make the asserted almost-everywhere identity precise on the original sigma-algebra without completing the measure.

Apply [F1] directly to $f(X_0),f(X_n)$, both square-integrable. Their quantiles have common $L^q(0,1)$ norm $\mathrm{\parallel }f{\mathrm{\parallel }}_q$. Two applications of [F4], with reciprocal exponents $1\mathrm{/}q,1\mathrm{/}q,1-2\mathrm{/}q$, yield $\mathrm{\mid }{c}_n\mathrm{\mid }\le 4{\alpha }_X(n{)}^{1-2\mathrm{/}q}\mathrm{\parallel }f{\mathrm{\parallel }}_q^2$. The exponent is positive because $q>2$; the geometric bound therefore gives ${\sum }_{n\ge 1}\mathrm{\mid }{c}_n\mathrm{\mid }<\mathrm{\infty }$. At a zero mixing coefficient the quantile integral is zero, so no division by that coefficient is needed.

Centering, stationarity and [F8] give $n^{-1}{\mathbb{E}}_{\pi }({\sum }_{k=0}^{n-1}f(X_k){)}^2=\pi (f^2)+2{\sum }_{h=1}^{n-1}(1-h\mathrm{/}n)c_h$. To pass to its limit, choose any finite $H$; the terms $h\le H$ converge individually to $c_h$, while the absolute value of either remaining weighted or limiting tail is at most ${\sum }_{h>H}\mathrm{\mid }{c}_h\mathrm{\mid }$, which tends to zero as $H\to \mathrm{\infty }$ by step 4.2. The limit is the claimed finite covariance sum $v$. Each left-hand side is nonnegative, hence $v\ge 0$. When $f=0$ almost everywhere all bounds and the choice $g=0$ hold directly; no claim of uniqueness of $g$ is made. The strict exponent range and all zero-mixing cases were handled in steps 3.1, 4.1, 4.2. AC is used through the Markov/conditional and completeness suppliers, not as an imported covariance or Poisson theorem.