Gaussian smoothing converges weakly at the endpoint

The coefficients $b(t,x)=0$ and $\sigma (t,x)=g(t)I_d$ are jointly Borel, globally Lipschitz in $x$ with constant zero, and satisfy $\mathrm{\mid }b\mathrm{\mid }+\mathrm{\parallel }\sigma {\mathrm{\parallel }}_F=\sqrt{d}g(t)\le \sqrt{d}{g}_{\max }$. The initial support gives $\mathbb{E}\mathrm{\mid }{X}_0{\mathrm{\mid }}^2\le R^2$, and the complete usual filtered space, adapted Brownian increments, and independence of $X_0$ are hypotheses of the statement. Thus [F3] supplies the unique continuous strong solution. By [F2], $X_{ϵ}-X_0={\int }_0^{ϵ}g(r)d{W}_r$. The characteristic-function identity in [F1] identifies this vector with $N_d(0,v(ϵ)I_d)$, whose coordinate variances sum to $dv(ϵ)$. Thus $\mathbb{E}\mathrm{\mid }{X}_{ϵ}-X_0{\mathrm{\mid }}^2=dv(ϵ)$. Also $v(ϵ)\le g_{\max }^{}ϵ\to 0$. For every $a>0$, $\mathrm{\mid }{X}_{ϵ}-X_0{\mathrm{\mid }}^2\ge a^2{\mathbf{1}}_{\{\mathrm{\mid }{X}_{ϵ}-X_0\mathrm{\mid }>a\}}$; taking expectations gives $P(\mathrm{\mid }{X}_{ϵ}-X_0\mathrm{\mid }>a)\le dv(ϵ)\mathrm{/}{a}^2\to 0$.

The support condition gives $\mathbb{E}\mathrm{\mid }{X}_0{\mathrm{\mid }}^2\le R^2$, and step 1.1 gives $\mathbb{E}\mathrm{\mid }{X}_{ϵ}{\mathrm{\mid }}^2\le 2R^2+2dv(T)<\mathrm{\infty }$. Hence both laws lie in the quadratic Wasserstein space. The measurable pair $(X_{ϵ},X_0)$ has a Borel law on ${\mathbb{R}}^d\times {\mathbb{R}}^d$ by [F8]; by [F5] its marginals are ${\mu }_{ϵ}$ and $\mu$, and [F4] identifies ${\mu }_{ϵ}$ as the smoothed law from the definition. Thus it is a coupling. Its cost is $\mathbb{E}\mathrm{\mid }{X}_{ϵ}-X_0{\mathrm{\mid }}^2=dv(ϵ)$. Taking the infimum in [F7] proves $W_2({\mu }_{ϵ},\mu )\le \sqrt{dv(ϵ)}$.

Fix bounded continuous $f$ and let $K=\bar{B}(0,R+1)$. Its restriction to compact $K$ is uniformly continuous; define ${\omega }_K(\delta )=\sup \{\mathrm{\mid }f(x)-f(y)\mathrm{\mid }:x,y\in K,\mathrm{\mid }x-y\mathrm{\mid }\le \delta \}$, which tends to zero as $\delta \downarrow 0$. For $0<\delta <1$, on the event $\mathrm{\mid }{X}_{ϵ}-X_0\mathrm{\mid }\le \delta$, both vectors belong to $K$, so their $f$-values differ by at most ${\omega }_K(\delta )$. On the complementary event the difference is at most $2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$. Step 1.1 and Markov's inequality therefore give $\mid \int fd{\mu }_{ϵ}-\int fd\mu \mid \le {\omega }_K(\delta )+2\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}\frac{dv(ϵ)}{{\delta }^2}\mathrm{.}$ For fixed $\delta$ the second term tends to zero as $ϵ\downarrow 0$; then letting $\delta \downarrow 0$ proves convergence for every such $f$. This is weak convergence by [F6]. The estimates establish only the endpoint statements in the theorem.