= Solution
Part c gives the integral representation
$$
\mathcal E_K(f)
=\frac12\int\pi(dx)K(x,dy)(f(y)-f(x))^2.
$$
The integrand vanishes on the diagonal $y=x$. Thus the assumed off-diagonal <Peskun ordering> implies
$$
\mathcal E_K(f)\geq\mathcal E_Q(f)
$$
for every $f\in L^2(\pi)$.
On the mean-zero subspace, the variational characterization of the <spectral gap> of a positive reversible kernel is
$$
\operatorname{gap}(K)
=\inf_{\pi(f)=0,\ f\ne0}
\frac{\mathcal E_K(f)}{\operatorname{Var}_\pi(f)}.
$$
It follows immediately that
$$
\operatorname{gap}(K)\geq\operatorname{gap}(Q).
$$
Equivalently, the energy inequality says $K\preceq Q$ in the <Löwner order> on $L^2_0(\pi)$. Positivity permits the <operator monotonicity of the square root> and hence $K^{1/2}\preceq Q^{1/2}$; the <projection-valued measure>[spectral representations] in the question identify the top spectral values and give the same gap inequality.
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