Global drift and one-step minorization give weighted contraction and a unique invariant law

Put $w=1+\beta V$ and, for finite signed $\eta$ with $\int wd\mathrm{\mid }\eta \mathrm{\mid }<\mathrm{\infty }$, define $\mathrm{\parallel }\eta {\mathrm{\parallel }}_w={\sup }_{\mathrm{\mid }f\mathrm{\mid }\le w}\mathrm{\mid }\eta f\mathrm{\mid }$. Obtain a Hahn partition $P,N$ by [F1]. The upper bound $\mathrm{\mid }\eta f\mathrm{\mid }\le \int wd\mathrm{\mid }\eta \mathrm{\mid }$ and the test $f=w(1_P-1_N)$ give $\mathrm{\parallel }\eta {\mathrm{\parallel }}_w=\int wd\mathrm{\mid }\eta \mathrm{\mid }$. Multiplication $T\eta =w\eta$ is an isometry onto the ordinary finite-total-variation signed measures: define it by multiplying the Jordan parts, and note that the same partition is Hahn, so $\mathrm{\mid }T\eta \mathrm{\mid }=w\mathrm{\mid }\eta \mathrm{\mid }$. Conversely, for any finite signed $\zeta$, multiplication by $w^{-1}$ gives $\mathrm{\mid }{w}^{-1}\zeta \mathrm{\mid }=w^{-1}\mathrm{\mid }\zeta \mathrm{\mid }$ and is inverse to $T$. The density integration identities follow first for indicators and simple functions, then by monotone convergence. Thus [F3] makes the finite-$w$-moment signed-measure space complete. Also $\beta V\le w\le (1+\beta )V$.

Define $\eta K={\eta }^{+}K-{\eta }^{-}K$. Each positive part acts by integration of the kernel evaluations. Countable additivity follows by monotone convergence. The inequality $\mathrm{\mid }\eta K\mathrm{\mid }\le \mathrm{\mid }\eta \mathrm{\mid }K$ follows by bounding the variation sum on every countable measurable partition by the corresponding positive measure $\mathrm{\mid }\eta \mathrm{\mid }K$. The drift gives $Kw\le 1+\beta aV+\beta b\le (1+\beta b)w$, so $\eta K$ has finite $w$ moment whenever $\eta$ does. Kernel integration is Borel and finite for $\mathrm{\mid }f\mathrm{\mid }\le cw$, and $(\eta K)f=\eta (Kf)$, by the indicator identity and monotone convergence applied separately to positive and negative parts. Total mass is preserved.

The constants satisfy $0<a^{\mathrm{\prime }}<1$ and $0<\beta b<ϵ$ if $b>0$, while $\beta b=0<ϵ$ if $b=0$. Both displayed component rates are in $(0,1)$. Fix Borel $\mathrm{\mid }f\mathrm{\mid }\le w$, and write $S=V(x)+V(y)$. If $S\ge R$, then $\mathrm{\mid }Kf(x)-Kf(y)\mathrm{\mid }\le Kw(x)+Kw(y)\le 2+\beta (aS+2b)\le 2+\beta a^{\mathrm{\prime }}S$. Since $s\mapsto (2+\beta a^{\mathrm{\prime }}s)\mathrm{/}(2+\beta s)$ is decreasing for $s\ge R$, this gives $\mathrm{\mid }Kf(x)-Kf(y)\mathrm{\mid }\le r(2+\beta S)$.

If $S<R$, both states are in $D$. The residual probabilities $H_x=(K(x,\cdot )-ϵ\nu )\mathrm{/}(1-ϵ)$ are well defined there. Fixing one point of the nonempty $D$, the minorization and finite $KV$ give $ϵ\nu V\le KV(x_D)<\mathrm{\infty }$. Consequently $\nu f$ is finite, and the common term $ϵ\nu f$ cancels from $Kf(x)-Kf(y)$. Thus $\mathrm{\mid }Kf(x)-Kf(y)\mathrm{\mid }\le (1-ϵ)(H_{x}w+H_{y}w)\le 2(1-ϵ)+\beta (aS+2b)\le (2+\beta S)-2(ϵ-\beta b)\le r(2+\beta S)$. The residual evaluations are Borel on $D$, so these are legitimate kernel integrations; no selected optimal coupling is involved.

Let $\eta$ have zero mass and finite $w$ norm. If $\eta \mathrm{\ne }0$, its Jordan masses agree at some $m>0$; put $p={\eta }^{+}\mathrm{/}m$ and $q={\eta }^{-}\mathrm{/}m$. They concentrate on disjoint Hahn sets. Integration of steps 3.1 and 4.1 against $p\otimes q$ gives, for every $\mathrm{\mid }f\mathrm{\mid }\le w$, $\mathrm{\mid }(\eta K)f\mathrm{\mid }=m\mathrm{\mid }\int (Kf(x)-Kf(y))p(dx)q(dy)\mathrm{\mid }\le mr\int (w(x)+w(y))p(dx)q(dy)=r\mathrm{\parallel }\eta {\mathrm{\parallel }}_w$. Every integral is finite by the finite moments and step 2.1. For $\eta =0$ the inequality is immediate. Taking the supremum and iterating, using preservation of mass and moments, proves $\mathrm{\parallel }\eta K^n{\mathrm{\parallel }}_w\le r^n\mathrm{\parallel }\eta {\mathrm{\parallel }}_w$. Norm comparison in step 1.1 yields $\mathrm{\parallel }\eta K^n{\mathrm{\parallel }}_V\le C{r}^n\mathrm{\parallel }\eta {\mathrm{\parallel }}_V$, including differences of the stated probabilities.

Fix $x_0\in {\mathbb{R}}^d$ and set ${\mu }_n={\delta }_{x_0}{K}^n$. The difference ${\mu }_1-{\mu }_0$ has finite $w$ norm, and step 5.1 gives $\mathrm{\parallel }{\mu }_{n+1}-{\mu }_n{\mathrm{\parallel }}_w\le r^n\mathrm{\parallel }{\mu }_1-{\mu }_0{\mathrm{\parallel }}_w$. The geometric tail sum makes $({\mu }_n)$ Cauchy, so step 1.1 gives a signed-measure limit $\pi$ with finite $w$ norm. Norm convergence implies $\pi (A)={\lim }_n{\mu }_n(A)\ge 0$ for every Borel $A$ and $\pi ({\mathbb{R}}^d)=1$. Thus $\pi$ is a probability and $\pi V<\mathrm{\infty }$. Step 5.1 gives $\mathrm{\parallel }{\mu }_{n}K-\pi K{\mathrm{\parallel }}_w\le r\mathrm{\parallel }{\mu }_n-\pi {\mathrm{\parallel }}_w\to 0$; since ${\mu }_{n}K={\mu }_{n+1}\to \pi$, it follows that $\pi K=\pi$. Finally drift iteration gives ${\mu }_{n}V\le a^{n}V(x_0)+b(1-a^n)\mathrm{/}(1-a)$. Because $\beta V\le w$, norm convergence gives ${\mu }_{n}V\to \pi V$, whence $\pi V\le b\mathrm{/}(1-a)$.

Apply step 5.1 to ${\delta }_x-\pi$. Its $V$ norm is at most $V(x)+\pi V$, giving the pointwise asserted bound. If $\lambda$ is any invariant probability, without a moment assumption, then for each Borel $A$ the pointwise convergence $K^n(x,A)\to \pi (A)$ and $0\le K^n(x,A)\le 1$ give $\lambda (A)=\int K^n(x,A)\lambda (dx)\to \pi (A)$ by dominated convergence. Thus $\lambda =\pi$. Together with steps 5.1 and 6.1 this proves all assertions.