If $Q$ is the generator of a well-behaved continuous-time Markov process on a finite state space and $p$ is the invariant distribution, the corresponding Dirichlet form is $\mathcal{D}_Q(f) := \frac{1}{2} \sum_{j,k} p_j Q_{jk} (f_j - f_k)^2$. Write $Var_p(f)$ for the variance of $f$. In studies of convergence it is of great interest to find the infimum of $\mathcal{D}_Q/Var_p$ (where the denominator is restricted to be nonzero). W/l/o/g it suffices to consider $f$ for which $\sum_j p_j f_j = 0$, so that $Var_p(f) = \sum_j p_j f_j^2$.

If $Q^{(m)} = c_m Q$ for $c_m > 0$ and $Q^\otimes = \sum_m I^{\otimes(m-1)} \otimes Q^{(m)} \otimes I^{\otimes(N-m)}$, it can be shown that $\mathcal{D}_{Q^\otimes}(f^{\otimes N})/Var_{p^{\otimes N}}(f^{\otimes N}) = N \langle c \rangle \mathcal{D}_Q(f)/Var_{p}(f)$, where the arithmetic mean is indicated. Which is great and all, but:

Is the infimum of $\mathcal{D}_{Q^\otimes}/Var_{p^{\otimes > N}}$ actually attained for an argument of the form $f^{\otimes N}$?

I feel like the answer should be trivial but for some reason I'm not seeing it.

UPDATE: Given my answer below, I will award the bounty to the best answer that deals with some combination of finding/addressing any errors in this answer and reconciling the implied result for mixing times of tuples of Markov processes with this paper (PDF) of Ycart and coworkers

viz., if the cutoff and mixing times are equivalent as claimed, why do they appear to grow differently with $N$?

For an $N$-tuple of identical processes, the result (Lemma 2 and Theorem 3) of Ycart et al. seems to indicate that the ($L^2$) mixing time should go as $\frac{1}{2} \log N$ times the mixing time of a single process. The calculations here seem to indicate that the mixing time goes as $1/N$ times the mixing time of a single process. Clearly I am mistaken somewhere.