 Pre-cutoff for Markov chains

A sequence of finite irreducible Markov chains exhibits pre-cutoff if there are constants $0<A<B<\infty$ and times $t_n$ such that its worst-case total variation distance satisfiesThis only bounds the transition between two fixed multiples of a scale; cutoff requires the transition window to be negligible relative to the mixing scale.

For reversible aperiodic chains with $t_n\to \infty$, pre-cutoff implies the product condition $t_{\mathrm{rel},n}=o(t_{\mathrm{mix},n}(1/4))$. In fact the analogous Absolute relaxation time must be negligible too. To see the mechanism, the two distance limits make $t_{\mathrm{mix},n}(1/4)$ comparable to $t_n$. If $r_n$ is the largest nontrivial eigenvalue modulus, the Eigenvalue lower bound for total variation mixing gives $d_n(t)\ge r_n^t/2$. The limit at $B{t}_n$ forces $r_n^{\lfloor B{t}_n\rfloor }\to 0$, which implies $(1-r_n)t_n\to \infty$ when $t_n\to \infty$. The ordinary spectral gap is at least $1-r_n$, giving the ordinary product condition.

The diverging-scale hypothesis matters. Nonlazy simple random walk on the complete graph has $d_n(0)\to 1$ and $d_n(1)=1/n\to 0$, so the displayed pre-cutoff convention holds on the bounded scale $t_n=1$, but its Relaxation time is not negligible relative to its mixing time.