The Gaussian coupling bound is exact for a point mass

The density in [F1] is the product of $d$ one-dimensional normal densities translated by $a$. Each coordinate has mean $a_j$ and variance $c^2ϵ$, so its second moment about $a$ is $d{c}^2ϵ$; both laws therefore have finite second moments.

If $\pi$ is any coupling of ${\mu }_{ϵ}$ and ${\delta }_a$, its second marginal assigns mass one to $\{a\}$, hence $\pi ({\mathbb{R}}^d\times \{a\})=1$. Its quadratic cost is therefore $\int \mathrm{\mid }x-y{\mathrm{\mid }}^2\pi (dx,dy)=\int \mathrm{\mid }x-a{\mathrm{\mid }}^2{\mu }_{ϵ}(dx)=d{c}^2ϵ\mathrm{.}$ There is at least one coupling: the continuous map $x\mapsto (x,a)$ pushes ${\mu }_{ϵ}$ forward to a Borel probability on ${\mathbb{R}}^d\times {\mathbb{R}}^d$ with marginals ${\mu }_{ϵ}$ and ${\delta }_a$ by [F4]. Thus the infimum over all couplings equals this common cost. Taking the square root gives the stated equality. The endpoint bound [F2] has the same value because $v(ϵ)=c^2ϵ$.