For a finite reversible lazy chain, the <relaxation time> is $t_{\mathrm{rel}}=1/\gamma$, where $\gamma=1-\lambda_2$ is the <spectral gap>. Its <spectral profile> is
$$
\Lambda(r)=\inf_{\pi_{\min}\leq\pi(A)\leq r}\lambda(A),
\qquad
\lambda(A)=\inf_{f\in c_0^+(A)}
\frac{\mathcal E(f,f)}{\operatorname{Var}_\pi(f)}.
$$
The variational characterization of the spectral gap is
$$
\gamma=\inf_{f\text{ nonconstant}}
\frac{\mathcal E(f,f)}{\operatorname{Var}_\pi(f)}.
$$
Every class $c_0^+(A)$ is contained in the class over which this last infimum is taken, so $\lambda(A)\geq\gamma$ and $\Lambda(r)\geq\gamma$. Therefore
$$
t_{\mathrm{rel}}=\frac1\gamma\geq\frac1{\Lambda(r)}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 215 / 3 / a / ii / Solution

{title2=$t_{\mathrm{rel}}$}

The relaxation time of a finite reversible lazy <Markov chain> is the reciprocal of its <spectral gap>: $t_{\mathrm{rel}}=(1-\lambda_2)^{-1}$.

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