The Hilbert lift induces the localized Wasserstein continuity equation

Since $m_t$ is the image of $X_t$ under Bochner integration, [F3] and the Lipschitz bound on $\mathrm{\Phi }$ show $t\mapsto m_t$ is continuous. The full shifted selection $S(a,u)$ is jointly Borel on $\mathcal{H}\times Q_r$ by [F2]; extending $v_t=-S(a(t),u)$ by zero outside the closed set $Q_r$ gives a Borel field on $[0,T]\times {\mathbb{R}}^d$. Here $a(t)=\mathrm{\nabla }R(m_t)$ is continuous. By [F1] and [F2], a representative of the Hilbert derivative satisfies $\dot{X}(t,s)=v_t(X(t,s))$ for almost every $(t,s)$. Its squared integral is ${\int }_0^T\int \mathrm{\mid }{v}_t(u){\mathrm{\mid }}^{2}d{\mu }_t(u)dt={\int }_0^T\mathrm{\parallel }\dot{X}(t){\mathrm{\parallel }}_K^{2}dt<\mathrm{\infty }$.

For $\phi \in C_c^{\mathrm{\infty }}((0,T)\times {\mathbb{R}}^d)$, the scalar chain rule along the almost-everywhere absolutely continuous label paths gives ${\displaystyle \frac{d}{dt}\phi (t,X(t,s))={\mathrm{\partial }}_t\phi (t,X(t,s))+{\mathrm{\nabla }}_x\phi (t,X(t,s))\cdot v_t(X(t,s))}$ for almost every $(t,s)$. The right side is integrable: the test derivatives are bounded, ${\mu }_t$ is a probability, and [F1] gives finite $L^1$ velocity action on $[0,T]$. Integrating in $s$ and $t$, Fubini and the vanishing of $\phi$ near $t=0,T$ give the weak continuity-equation identity for ${\mu }_t$.

The coupling $(X(s),X(t){)}_{\mathrm{\#}}{\mu }_0$ gives $W_2({\mu }_s,{\mu }_t{)}^2\le \mathrm{\parallel }X(s)-X(t){\mathrm{\parallel }}_K^2$. Cauchy–Schwarz in time and [F1] imply $W_2({\mu }_s,{\mu }_t)\le {\int }_s^t\mathrm{\parallel }\dot{X}(q){\mathrm{\parallel }}_{K}dq$ for $s<t$. Thus $\mu \in A{C}^2$ on $[0,T]$. Alternatively [F4] applies to the continuity equation and the action bound from step 1.1.

Fix a time $t$ and put $f_t(u)=⟨\mathrm{\nabla }R(m_t),\mathrm{\Phi }(u)⟩+V(u)+{\iota }_{Q_r}(u)$. On $Q_r$ this function is bounded, lower semicontinuous and semiconvex, with a nonempty subdifferential and minimum section $g_t=-v_t$ bounded uniformly on compact time intervals by [F2]. For small $\epsilon >0$ let $J_{\epsilon }x$ minimize $f_t(y)+\mathrm{\mid }x-y{\mathrm{\mid }}^2\mathrm{/}(2\epsilon )$ and let $f_{t,\epsilon }$ be the minimum value. The resolvent proof in [F2] gives $J_{\epsilon }$ Lipschitz. Comparing the envelope at $x+h$ using $J_{\epsilon }x$, and at $x$ using $J_{\epsilon }(x+h)$, bounds the difference from $⟨(x-J_{\epsilon }x)\mathrm{/}\epsilon ,h⟩$ by $O(\mathrm{\mid }h{\mathrm{\mid }}^2)$ locally. Thus $f_{t,\epsilon }$ is $C^1$ and its gradient is $(x-J_{\epsilon }x)\mathrm{/}\epsilon$. The monotonicity estimate from [F2] gives $\mathrm{\mid }\mathrm{\nabla }{f}_{t,\epsilon }(x)\mathrm{\mid }\le \mathrm{\mid }{g}_t(x)\mathrm{\mid }\mathrm{/}(1+\epsilon {\alpha }_t)$ on $Q_r$, with ${\alpha }_t$ a semiconvexity coefficient; the same proof gives pointwise convergence to $g_t(x)$. Dominated convergence yields convergence in $L^2({\mu }_t)$. Each gradient belongs to the closure of gradients of compactly supported smooth functions: multiply the $C^1$ envelope by smooth cutoffs equal to one on $B_N$, with derivative $O(1\mathrm{/}N)$, and then mollify. On $Q_r$ the envelope stays between the fixed lower bound and upper bound of $f_t$, so the cutoff derivative term tends to zero in $L^2({\mu }_t)$; the other tail tends to zero by the bound just proved. Mollification converges uniformly in $C^1$ on each compact set. Therefore $g_t$ belongs to the spatial gradient closure in $L^2({\mu }_t)$.

Let $w$ be a minimum-action field for $\mu$ as in the AC² minimal-velocity theorem. Subtracting its continuity equation from that of $v$ and using time tensor tests shows $\int (v_t-w_t)\cdot \mathrm{\nabla }\psi d{\mu }_t=0$ for almost every $t$, simultaneously for a countable set of compactly supported smooth $\psi$ dense in $C^1$ on each compact ball. Such a set is obtained by rational polynomial approximations and fixed smooth cutoffs; extend the equality to every compactly supported smooth test by uniform gradient approximation. By step 2.3, $v_t$ belongs to the $L^2({\mu }_t)$ closure of these gradients. Hence $⟨v_t-w_t,v_t{⟩}_{L^2({\mu }_t)}=0$, and therefore $\mathrm{\parallel }{w}_t{\mathrm{\parallel }}_{L^2({\mu }_t)}^2=\mathrm{\parallel }{v}_t{\mathrm{\parallel }}_{L^2({\mu }_t)}^2+\mathrm{\parallel }{w}_t-v_t{\mathrm{\parallel }}_{L^2({\mu }_t)}^2$ almost everywhere. Thus $v$ is a minimum-action field; [F6] identifies its action with ${\int }_0^T\mathrm{\mid }{\dot{\mu }}_t{\mathrm{\mid }}_{W_2}^{2}dt$. The constraint is included: for any absolutely continuous label path in $Q_r$, at a two-sided differentiability time and every $n\in N_{Q_r}(X_t)$, the inequalities $n\cdot (X_{t+h}-X_t)\le 0$ for both signs of $h$ give $n\cdot {\dot{X}}_t=0$. Thus normal-cone terms contribute no velocity pairing, and the constrained minimum cannot be replaced by an unconstrained selection.