Theorem 12.4 -- mixing is at least the relaxation time

Let $P$ be an irreducible, aperiodic Markov chain on a finite state space $V$, reversible with respect to its stationary distribution $\pi$ (detailed balance: $\pi(x)P(x,y)=\pi(y)P(y,x)$). Among the eigenvalues of $P$ — the real $\lambda$ admitting a nonzero $f$ with $Pf=\lambda f$ — let $\lambda_\star$ be the largest absolute value of an eigenvalue different from $1$; the absolute spectral gap is $\gamma_\star=1-\lambda_\star$ and the relaxation time is $t_{\mathrm{rel}}=1/\gamma_\star$. For a tolerance $\varepsilon$, the mixing time $t_{\mathrm{mix}}(\varepsilon)$ is the first $t$ with $\max_x\|P^t(x,\cdot)-\pi\|_{TV}\le\varepsilon$, where $\|\mu-\nu\|_{TV}=\max_A|\mu(A)-\nu(A)|$.

The theorem (Theorem 12.4 of Levin–Peres–Wilmer) asserts: for every $0<\varepsilon<1$,

A chain cannot mix faster than it relaxes: an eigenfunction with eigenvalue close to $1$ decays like $\lambda_\star^t$ and remains a visible witness against stationarity for about $t_{\mathrm{rel}}$ steps. Together with the companion upper bound $t_{\mathrm{mix}}(\varepsilon)\le\log(1/\varepsilon\pi_{\min})\,t_{\mathrm{rel}}+1$, this sandwiches the mixing time between $t_{\mathrm{rel}}$ and $t_{\mathrm{rel}}\log(1/\pi_{\min})$ for reversible chains.