Uniform ergodicity gives no uniform practical mixing time

The invariant law is $\pi =(1\mathrm{/}2,1\mathrm{/}2)$ by symmetry, and each state has positive holding probability $1-\eta$. Every entry is positive, so it is irreducible by [F4], and each state has positive holding probability, so it is aperiodic. Moreover every row dominates $2\eta \pi$, because its two entries are at least $\eta$. This is a whole-space minorization with $m=1$, $ϵ=2\eta$.

The theorem [F3] gives ${\sup }_x\mathrm{\parallel }{P}_{\eta }^n(x,\cdot )-\pi {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}\le (1-2\eta {)}^n$, so each fixed chain is uniformly geometrically ergodic.

The nonconstant eigenvector $(1,-1)$ has eigenvalue $1-2\eta$. Direct multiplication, or induction, gives $\mathrm{\parallel }{P}_{\eta }^n(0,\cdot )-\pi {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}=\frac{1}{2}(1-2\eta {)}^n\mathrm{.}$ For every fixed $n$ this tends to $1\mathrm{/}2$ as $\eta \downarrow 0$. Hence no finite $n$ makes the distance smaller than (for example) $1\mathrm{/}4$ uniformly over the family. Each chain is uniformly ergodic, but the mixing time depends on $\eta$.