The product condition is necessary for cutoff

Consider a family of chains $P^{(n)}$ on finite state spaces, each irreducible and aperiodic, each reversible with respect to its stationary distribution $\pi_n$ (detailed balance $\pi_n(x)P^{(n)}(x,y)=\pi_n(y)P^{(n)}(y,x)$). For each chain: $d_n(t)=\max_x\|P^{(n)t}(x,\cdot)-\pi_n\|_{TV}$ is the worst-case total variation distance, $t^{(n)}_{\mathrm{mix}}(\varepsilon)=\min\{t:d_n(t)\le\varepsilon\}$ with $t^{(n)}_{\mathrm{mix}}=t^{(n)}_{\mathrm{mix}}(1/4)$; among the eigenvalues (real $\lambda$ with $P f=\lambda f$, $f\ne0$), $\lambda_\star$ is the largest absolute value of an eigenvalue $\ne1$, and $t^{(n)}_{\mathrm{rel}}=(1-\lambda_\star)^{-1}$ is the relaxation time (Mission VII). The family has a cutoff when $t^{(n)}_{\mathrm{mix}}(\varepsilon)/t^{(n)}_{\mathrm{mix}}(1-\varepsilon)\to1$ for every $0<\varepsilon<1$.

The theorem (Proposition 18.4 of Levin–Peres–Wilmer) asserts: if the mixing times grow to infinity but stay comparable to the relaxation times — $t^{(n)}_{\mathrm{mix}}\le C\,t^{(n)}_{\mathrm{rel}}$ for a fixed constant $C$ — then the family has no cutoff.

The product condition $t_{\mathrm{rel}}=o(t_{\mathrm{mix}})$ is therefore necessary for cutoff. The reason: the relaxation-time lower bound of Mission VII gives $t_{\mathrm{mix}}(\varepsilon)\ge(t_{\mathrm{rel}}-1)\log(1/2\varepsilon)$, so if $t_{\mathrm{rel}}$ is proportional to $t_{\mathrm{mix}}$, shrinking $\varepsilon$ inflates $t_{\mathrm{mix}}(\varepsilon)/t_{\mathrm{mix}}$ beyond any bound — an eigenfunction decaying only like $\lambda_\star^t$ keeps the collapse from being abrupt. Whether the product condition is also sufficient for reversible families was a famous question of Peres; it fails in general, which makes this necessary direction the definitive elementary statement.