m-step Dobrushin certificate for a stationary probability

The transpose $K=P^{\mathsf{T}}$ is a finite row transition matrix by [F1]. By [F5] it has an invariant probability row vector, and [F4] transposes this to a stationary column probability $\pi$ for P.

If ${\pi }_1,{\pi }_2$ are stationary column probabilities, then $z={\pi }_1-{\pi }_2$ lies in H and $P^{m}z=z$. By [F3], $\mathrm{\parallel }z{\mathrm{\parallel }}_1=\mathrm{\parallel }{P}^{m}z{\mathrm{\parallel }}_1\le q\mathrm{\parallel }z{\mathrm{\parallel }}_1$; since $q<1$, this forces $z=0$. Thus $\pi$ is unique.

Let $e=x-\pi$. Both vectors have coordinate sum one, so $e\in H$. Since $P^m\pi =\pi$, reverse triangle inequality and [F3] give $\mathrm{\parallel }{P}^{m}x-x{\mathrm{\parallel }}_1=\mathrm{\parallel }{P}^{m}e-e{\mathrm{\parallel }}_1\ge \mathrm{\parallel }e{\mathrm{\parallel }}_1-\mathrm{\parallel }{P}^{m}e{\mathrm{\parallel }}_1\ge (1-q)\mathrm{\parallel }e{\mathrm{\parallel }}_1$. Dividing by $1-q>0$ proves the first bound.

The telescoping identity $P^{m}x-x={\sum }_{k=0}^{m-1}{P}^k(Px-x)$ follows by cancellation of adjacent terms. Applying [F2] repeatedly gives $\mathrm{\parallel }{P}^k(Px-x){\mathrm{\parallel }}_1\le \mathrm{\parallel }Px-x{\mathrm{\parallel }}_1$ for each k; the triangle inequality therefore yields $\mathrm{\parallel }{P}^{m}x-x{\mathrm{\parallel }}_1\le m\mathrm{\parallel }Px-x{\mathrm{\parallel }}_1$.

Combine steps 2.2 and 3.1. The proof does not require irreducibility or aperiodicity once the stated m-step contraction is certified; if $q\ge 1$, the division step is unavailable and no bound is asserted.