Wasserstein metric and continuity-equation foundations — Examples

For $s,t\in [0,T]$, any coupling of ${\delta }_{a+sb}$ and ${\delta }_{a+tb}$ is concentrated at $((a+sb),(a+tb))$, hence [F1] gives $W_2({\mu }_s,{\mu }_t)=\mathrm{\mid }t-s\mathrm{\mid }\mathrm{\mid }b\mathrm{\mid }$. For every bounded continuous f, $f(a+tb)$ is continuous, so the curve is narrowly continuous. Thus it is $A{C}^2$ with constant metric derivative $\mathrm{\mid }b\mathrm{\mid }$ (the endpoint difference quotient is exactly $\mathrm{\mid }b\mathrm{\mid }$ at every interior time).

For $\phi \in C_c^{\mathrm{\infty }}((0,T)\times {\mathbb{R}}^d)$, the chain rule gives $\frac{d}{dt}\phi (t,a+tb)={\mathrm{\partial }}_t\phi (t,a+tb)+{\mathrm{\nabla }}_x\phi (t,a+tb)\cdot b\mathrm{.}$ The left side integrates to zero over $[0,T]$ because the test vanishes near both time endpoints; this is exactly the weak continuity-equation identity for ${\mu }_t={\delta }_{a+tb}$ and $v_t=b$. Its action is ${\int }_0^T\mathrm{\mid }b{\mathrm{\mid }}^{2}dt=T\mathrm{\mid }b{\mathrm{\mid }}^2$. By [F3] and step 1.1, this equals ${\int }_0^T\mathrm{\mid }{\dot{\mu }}_t{\mathrm{\mid }}^{2}dt$, the minimum action.

Write ${\mu }_{m,\sigma }=\mathrm{Law}(m+\sigma \xi )$. Let $X,Y$ be any coupling of ${\mu }_{m,\sigma }$ and ${\mu }_{n,\tau }$, and set $U=X-m$, $V=Y-n$. Since $EU=EV=0$, expansion gives $E\mathrm{\mid }X-Y{\mathrm{\mid }}^2=\mathrm{\mid }m-n{\mathrm{\mid }}^2+E\mathrm{\mid }U{\mathrm{\mid }}^2+E\mathrm{\mid }V{\mathrm{\mid }}^2-2E(U\cdot V)\mathrm{.}$ By [F1] and [F2], this is at least $\mathrm{\mid }m-n{\mathrm{\mid }}^2+d({\sigma }^2+{\tau }^2-2\sigma \tau )=\mathrm{\mid }m-n{\mathrm{\mid }}^2+d(\sigma -\tau {)}^2$. Conversely, couple the two laws by $X=m+\sigma \xi$, $Y=n+\tau \xi$ using the same standard Gaussian vector. The moments in [F1] give $E\mathrm{\mid }X-Y{\mathrm{\mid }}^2=\mathrm{\mid }m-n{\mathrm{\mid }}^2+d(\sigma -\tau {)}^2$. Taking the infimum proves $W_2({\mu }_{m,\sigma },{\mu }_{n,\tau }{)}^2=\mathrm{\mid }m-n{\mathrm{\mid }}^2+d(\sigma -\tau {)}^2\mathrm{.}$

Apply step 1.1 to $(m_s,{\sigma }_s)$ and $(m_t,{\sigma }_t)$. Their differences are $(t-s)(m_1-m_0)$ and $(t-s)({\sigma }_1-{\sigma }_0)$, so $W_2({\mu }_s,{\mu }_t)=\mathrm{\mid }t-s\mathrm{\mid }\sqrt{\mathrm{\mid }{m}_1-m_0{\mathrm{\mid }}^2+d({\sigma }_1-{\sigma }_0{)}^2}\mathrm{.}$ This proves the constant-speed geodesic claim and identifies its speed.

For each $z$, define ${\gamma }_z(t)=m_t+{\sigma }_{t}z$. Direct differentiation gives ${\dot{\gamma }}_z=m_1-m_0+({\sigma }_1-{\sigma }_0)z=v_t({\gamma }_z(t))$. For any bounded continuous test f, dominated convergence [F4] gives $\int fd{\mu }_t=E[f(m_t+{\sigma }_t\xi )]$ continuously in t, so the curve is narrowly continuous. For a compactly supported smooth test $\phi$, the chain rule [F6] gives $\frac{d}{dt}\phi (t,{\gamma }_{\xi }(t))={\mathrm{\partial }}_t\phi (t,{\gamma }_{\xi }(t))+\mathrm{\nabla }\phi (t,{\gamma }_{\xi }(t))\cdot v_t({\gamma }_{\xi }(t))$. Integrate over $t$ and the Gaussian law. The FTC [F6] and the test vanishing near both time endpoints make the time integral zero; its derivative is integrable since $v_t({\gamma }_{\xi }(t))$ is affine in $\xi$ with finite first moment by [F2, F3]. The resulting identity is the weak continuity equation for ${\mu }_t$. Also $\int \mathrm{\mid }{v}_t(x){\mathrm{\mid }}^{2}d{\mu }_t(x)=E\mathrm{\mid }{m}_1-m_0+({\sigma }_1-{\sigma }_0)\xi {\mathrm{\mid }}^2=\mathrm{\mid }{m}_1-m_0{\mathrm{\mid }}^2+d({\sigma }_1-{\sigma }_0{)}^2,$ using the zero coordinate means and unit variances in [F2]. Its time integral is the squared-distance value from step 1.1, so [F4] makes this velocity minimum-action.