The topology of W2 is narrow convergence plus uniform second-moment tails

Suppose $W_2({\mu }_n,\mu )\to 0$. For each $n$ choose an optimal coupling ${\pi }_n$ and regard its coordinates as $X_n,Y$ with laws ${\mu }_n,\mu$. The cost gives $\int \mathrm{\mid }{X}_n-Y{\mathrm{\mid }}^{2}d{\pi }_n\to 0$. For bounded continuous $\phi$, choose a ball $B(0,K)$ with $\mu (B(0,K{)}^c)$ small. On the event $\mathrm{\mid }Y\mathrm{\mid }\le K$ and $\mathrm{\mid }{X}_n-Y\mathrm{\mid }\le \delta <1$, both points lie in the compact ball $\bar{B}(0,K+1)$, where $\phi$ is uniformly continuous. The complement has probability at most $\mu (B(0,K{)}^c)+{\delta }^{-2}\int \mathrm{\mid }{X}_n-Y{\mathrm{\mid }}^{2}d{\pi }_n$. First choose $K$, then $\delta$, and then $n$; the boundedness of $\phi$ shows $\int \phi d{\mu }_n-\int \phi d\mu \to 0$. Thus ${\mu }_n\Rightarrow \mu$.

Conversely suppose narrow convergence and the stated uniform tail condition. Fix $\epsilon >0$. Choose $R$ so that the second-moment tails outside $Q=[-R,R{]}^d$ of all ${\mu }_n$ and of $\mu$ are less than $\epsilon$; the tail of $\mu$ vanishes because $\mu \in {\mathcal{P}}_2$. Enlarge R slightly if necessary so its coordinate faces have zero $\mu$-mass. Partition each coordinate interval $[-R,R]$ by finitely many levels that are not atoms of the corresponding one-dimensional marginal of $\mu$, with mesh small enough that each resulting rectangular cell has diameter at most $\epsilon$. Such levels exist because each probability has at most countably many atoms. Together with $Q^c$, these cells form a finite Borel partition $(C_i{)}_{i=0}^N$ whose boundaries have $\mu$-measure zero. By [F2], $a_{n,i}:={\mu }_n(C_i)\to a_i:=\mu (C_i)$ for each $i$.

Write $Z_n=X_n-Y$ and $a_n=\int \mathrm{\mid }{Z}_n{\mathrm{\mid }}^{2}d{\pi }_n\to 0$. For $R>K$, on $\{\mathrm{\mid }{X}_n\mathrm{\mid }>R\}$ either $\mathrm{\mid }Y\mathrm{\mid }>K$ or $\mathrm{\mid }{Z}_n\mathrm{\mid }>R-K$. Since $\mathrm{\mid }{X}_n{\mathrm{\mid }}^2\le 2\mathrm{\mid }Y{\mathrm{\mid }}^2+2\mathrm{\mid }{Z}_n{\mathrm{\mid }}^2$, for every $n$, ${\int }_{\mathrm{\mid }{X}_n\mathrm{\mid }>R}\mathrm{\mid }{X}_n{\mathrm{\mid }}^{2}d{\pi }_n\le 2{\int }_{\mathrm{\mid }Y\mathrm{\mid }>K}\mathrm{\mid }Y{\mathrm{\mid }}^{2}d{\pi }_n+2K^2\frac{a_n}{(R-K{)}^2}+2a_n\mathrm{.}$ Given a target $\eta >0$, first choose K so the first term is below $\eta \mathrm{/}4$; then choose N so $a_n<\eta \mathrm{/}8$ for $n\ge N$; then choose R>K so the middle term is below $\eta \mathrm{/}8$ for all $n\ge N$. The last term is below $\eta \mathrm{/}4$. Finally enlarge R so the finitely many tails for $n<N$ are below $\eta$. This proves the required uniform tail limit.

For each bounded cell $C_i\subset Q$, match mass $r_{n,i}=\min (a_{n,i},a_i)$. If $r_{n,i}>0$, take $r_{n,i}$ times the product of the probability measures ${\mu }_n{\mathrm{\mid }}_{C_i}\mathrm{/}{a}_{n,i}$ and $\mu {\mathrm{\mid }}_{C_i}\mathrm{/}{a}_i$; if $r_{n,i}=0$, use the zero measure. This gives a coupling piece supported on $C_i\times C_i$ with cost at most ${\epsilon }^2{r}_{n,i}$. Subtract its two marginals from ${\mu }_n$ and $\mu$. The residual measures have equal mass. If that mass is positive, their normalized product, multiplied by that mass, couples them; if it is zero there is no residual piece. The sum is a coupling ${\pi }_n$ of ${\mu }_n,\mu$. Its matched cost is at most ${\epsilon }^2$. For the residual coupling, $\mathrm{\mid }x-y{\mathrm{\mid }}^2\le 2\mathrm{\mid }x{\mathrm{\mid }}^2+2\mathrm{\mid }y{\mathrm{\mid }}^2$. Its residual second moments are bounded by the two tails outside $Q$ plus $d{R}^2{\sum }_{i=1}^N\mathrm{\mid }{a}_{n,i}-a_i\mathrm{\mid }$ on each side: inside a bounded cell, unmatched mass is exactly the positive part of the difference of its two cell masses. Therefore $W_2({\mu }_n,\mu {)}^2\le \int \mathrm{\mid }x-y{\mathrm{\mid }}^{2}d{\pi }_n\le {\epsilon }^2+4\epsilon +4d{R}^2{\sum }_{i=1}^N\mathrm{\mid }{a}_{n,i}-a_i\mathrm{\mid }\mathrm{.}$ For this fixed finite partition the sum tends to zero, so ${\mathrm{lim\hspace{0.17em}sup}}_n{W}_2({\mu }_n,\mu {)}^2\le {\epsilon }^2+4\epsilon$. Letting $\epsilon \downarrow 0$ proves $W_2({\mu }_n,\mu )\to 0$.

For the witness, the mass outside a ball of radius $R$ is at most $n^{-2}$ for indices $n>R$ and is zero for indices $n\le R$; hence the family is tight. For bounded continuous $\phi$, $\int \phi d{\mu }_n=(1-n^{-2})\phi (0)+n^{-2}\phi (n{e}_1)\to \phi (0)$, proving narrow convergence. Every coupling with second marginal ${\delta }_0$ is concentrated on ${\mathbb{R}}^d\times \{0\}$, so its cost is forced to be $\int \mathrm{\mid }x{\mathrm{\mid }}^{2}d{\mu }_n=1$. The asserted failure of W2 convergence follows.