W2 convergence implies weak convergence of homogeneous projections

Let $(X_n,Y_n)$ have joint law ${\pi }_n$ and write $D_n=X_n-Y_n$. AC supplies the sequence of couplings in [A1]. Then $\mathbb{E}\mathrm{\mid }{D}_n{\mathrm{\mid }}^2\to 0$, and ${\sup }_n\mathbb{E}\mathrm{\mid }{X}_n{\mathrm{\mid }}^2<\mathrm{\infty }$ by $\mathrm{\mid }{X}_n{\mathrm{\mid }}^2\le 2\mathrm{\mid }{Y}_n{\mathrm{\mid }}^2+2\mathrm{\mid }{D}_n{\mathrm{\mid }}^2$. For bounded continuous $\psi$, the function $H(w,\theta )=w\psi (\theta )$ is continuous on the closed Euclidean subspace $\mathbb{R}\times \mathrm{\Theta }$ and has at most linear growth. Multiplying $H$ by a continuous radial cutoff of compact support in that closed subspace gives a uniformly continuous function, so convergence in probability of $D_n$ gives convergence of its expectations. The truncation error is uniformly small because $\mathbb{E}[\mathrm{\mid }{X}_n\mathrm{\mid }{1}_{\mathrm{\mid }{X}_n\mathrm{\mid }>R}]\le R^{-1}\mathbb{E}\mathrm{\mid }{X}_n{\mathrm{\mid }}^2$ and the same bound holds for $Y_n$. Hence $\mathbb{E}H(X_n)-\mathbb{E}H(Y_n)\to 0$. By [F2], this is weak convergence of $h^1({\mu }_n)$ against every such $\psi$.

Fix continuous $f$ on ${\mathbb{S}}^{d-1}$ and set ${\mathrm{\Psi }}_f(0)=0$, ${\mathrm{\Psi }}_f(u)=\mathrm{\mid }u{\mathrm{\mid }}^{2}f(u\mathrm{/}\mathrm{\mid }u\mathrm{\mid })$ for $u\mathrm{\ne }0$. This function is continuous and $\mathrm{\mid }{\mathrm{\Psi }}_f(u)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\mid }u{\mathrm{\mid }}^2$. The quadratic moments are asymptotically uniformly integrable: if $\mathrm{\mid }{X}_n\mathrm{\mid }>R$, then either $\mathrm{\mid }{Y}_n\mathrm{\mid }>R\mathrm{/}2$ or $\mathrm{\mid }{D}_n\mathrm{\mid }>R\mathrm{/}2$. Using [A1], $\mathrm{\mid }{X}_n{\mathrm{\mid }}^2\le 2\mathrm{\mid }{Y}_n{\mathrm{\mid }}^2+2\mathrm{\mid }{D}_n{\mathrm{\mid }}^2$, and splitting the term $\mathrm{\mid }{Y}_n{\mathrm{\mid }}^2{\mathbf{1}}_{\mathrm{\mid }{D}_n\mathrm{\mid }>R\mathrm{/}2}$ at $\mathrm{\mid }{Y}_n\mathrm{\mid }=R\mathrm{/}2$ gives $\mathbb{E}[\mathrm{\mid }{X}_n{\mathrm{\mid }}^2{\mathbf{1}}_{\mathrm{\mid }{X}_n\mathrm{\mid }>R}]\le 4\mathbb{E}\mathrm{\mid }{D}_n{\mathrm{\mid }}^2+4\mathbb{E}[\mathrm{\mid }{Y}_n{\mathrm{\mid }}^2{\mathbf{1}}_{\mathrm{\mid }{Y}_n\mathrm{\mid }>R\mathrm{/}2}]\mathrm{.}$ The right side tends uniformly to zero in the limit $n\to \mathrm{\infty }$ followed by $R\to \mathrm{\infty }$; the same estimate applies to $Y_n=Y$. A compactly supported continuous truncation of ${\mathrm{\Psi }}_f$ has convergent integrals because it is uniformly continuous and $D_n\to 0$ in probability. The asymptotic tail bound makes the truncation error tend to zero. Hence $\int {\mathrm{\Psi }}_{f}d{\mu }_n\to \int {\mathrm{\Psi }}_{f}d\mu$, which is weak convergence of $h^2({\mu }_n)$ by [F2].