Uniform density ratios bound the Dobrushin coefficient

Fix $x,x^{\mathrm{\prime }}\in S$ and write $p=k_x$, $q=k_{x^{\mathrm{\prime }}}$. The assumed ordered-pair bound gives $p\le Cq$ almost everywhere, so $q\ge p\mathrm{/}C$. Where $p\le q$, $\min (p,q)=p\ge p\mathrm{/}C$; where $p>q$, the same inequality $p\le Cq$ gives $\min (p,q)=q\ge p\mathrm{/}C$. Hence $\int \min (p,q)d\lambda \ge C^{-1}\int pd\lambda =C^{-1}$.

Let $A_{+}=\{z:p(z)\ge q(z)\}$. For every measurable $A$, $\mathrm{\mid }{\int }_A(p-q)d\lambda \mathrm{\mid }\le \int (p-q{)}^{+}d\lambda$. Since $\int (p-q)=0$, the positive and negative parts have equal integral, and $\int (p-q{)}^{+}d\lambda =1-\int \min (p,q)d\lambda$; equality is attained at $A_{+}$. Thus [F2] gives $\parallel K(x,\cdot )-K(x^{\mathrm{\prime }},\cdot ){\parallel }_{\mathrm{T}\mathrm{V}}=1-\int \min (p,q)d\lambda \le 1-C^{-1}$ by step 1.1.

The bound in step 2.1 holds for every pair $x,x^{\mathrm{\prime }}$, so taking their supremum in [F1] proves $\delta (K)\le 1-C^{-1}$.