Optimal quadratic couplings exist and their cost is weakly lower semicontinuous

The product probability $\mu \otimes \nu$ is a coupling by [F1]. The inequality in [F5] gives its cost at most $2\int \mathrm{\mid }x{\mathrm{\mid }}^{2}d\mu +2\int \mathrm{\mid }y{\mathrm{\mid }}^{2}d\nu <\mathrm{\infty }$. Thus the coupling set is nonempty and its cost infimum $c$ is finite and nonnegative.

For each integer $n\ge 1$, by AC choose a coupling ${\pi }_n$ with $\int \mathrm{\mid }x-y{\mathrm{\mid }}^{2}d{\pi }_n\le c+1\mathrm{/}n$. Their marginals are the fixed probabilities $\mu ,\nu$. For any $\epsilon >0$, tightness of each marginal supplies compact sets $K_1,K_2$ with $\mu (K_1^c)<\epsilon \mathrm{/}2$ and $\nu (K_2^c)<\epsilon \mathrm{/}2$. Then ${\pi }_n((K_1\times K_2{)}^c)\le \mu (K_1^c)+\nu (K_2^c)<\epsilon$ for every $n$. Hence $({\pi }_n)$ is tight on the Polish space ${\mathbb{R}}^{2d}$; [F2] gives a subsequence ${\pi }_{n_j}\Rightarrow \pi$. By [F3] its marginals are $\mu ,\nu$.

For $N\ge 1$, the function $c_N(x,y)=\min \{\mathrm{\mid }x-y{\mathrm{\mid }}^2,N\}$ is bounded and continuous. Weak convergence therefore gives $\int c_{N}d\pi ={\lim }_j\int c_{N}d{\pi }_{n_j}\le {\mathrm{lim\hspace{0.17em}inf}}_j\int \mathrm{\mid }x-y{\mathrm{\mid }}^{2}d{\pi }_{n_j}\le c$. Letting $N\uparrow \mathrm{\infty }$ and using [F4] yields $\int \mathrm{\mid }x-y{\mathrm{\mid }}^{2}d\pi \le c$. Since $\pi$ is a coupling and $c$ is the infimum, equality holds; the infimum is attained.

Put $L={\mathrm{lim\hspace{0.17em}inf}}_n{W}_2({\mu }_n,{\nu }_n{)}^2$. If $L=+\mathrm{\infty }$, the asserted inequality is immediate. Otherwise choose a subsequence indexed by integers $j\ge 1$, denoted $n_j$, along which the costs converge to $L$, and choose by AC couplings ${\sigma }_j$ with costs at most $W_2({\mu }_{n_j},{\nu }_{n_j}{)}^2+1\mathrm{/}j$. The marginal sequences are tight by [F2]; choosing compact sets for each coordinate and applying the same union bound as in step 2.1 proves $({\sigma }_j)$ is tight on ${\mathbb{R}}^{2d}$. A further subsequence converges weakly to a probability $\sigma$, whose marginals are $\mu ,\nu$ by [F3]. For every $N$, bounded continuity of $c_N$ gives $\int c_{N}d\sigma \le L$. Monotone convergence [F4] then gives $\int \mathrm{\mid }x-y{\mathrm{\mid }}^{2}d\sigma \le L$. The definition of $W_2$ and the coupling property imply $W_2(\mu ,\nu {)}^2\le L$.