= Solution
Let $B=2\mathbb Z_{2n}^d$. The chain induced on $B$ by the lazy walk on $A$ has uniform stationary distribution, and an excursion from one even vertex can return there or reach only one of its $2d$ even lattice neighbors at displacement $\pm2e_i$. Its transition conductances are bounded above by constants depending only on $d$. Testing its Dirichlet quotient with
$$
f(2x)=\cos(\pi x_1/n)
$$
therefore gives
$$
\gamma_B\leq\frac{c_d}{n^2}.
$$
Indeed neighboring values differ by $O(n^{-1})$, while the variance of $f$ is bounded below uniformly. The supplied trace-chain theorem gives $\gamma_B\geq\gamma^{(2n)}(A)$, and hence
$$
\gamma^{(2n)}(A)\leq\frac{c_d}{n^2}.
$$
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