AC2 curves admit a unique minimum-action velocity field

We first prove the scalar transport estimate $\mid {\int }_0^T\text{\hspace{0.17em}⁣}\int {\mathrm{\partial }}_t\phi (t,x)d{\mu }_t(x)dt\mid \le {\int }_0^{T}s(t)\mathrm{\parallel }{\mathrm{\nabla }}_x\phi (t,\cdot ){\mathrm{\parallel }}_{L^2({\mu }_t)}dt\text{(1)}$ for $\phi \in C_c^{\mathrm{\infty }}((0,T)\times {\mathbb{R}}^d)$. Put $F(t)=\int \phi (t,x)d{\mu }_t(x)$. If $r<t$, the time increment of $\phi$ is bounded by $\mathrm{\parallel }{\mathrm{\partial }}_t\phi {\mathrm{\parallel }}_{\mathrm{\infty }}(t-r)$, and its spatial increment under an optimal coupling is bounded by $\mathrm{\parallel }\mathrm{\nabla }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}{W}_2({\mu }_r,{\mu }_t)$. The $A{C}^2$ control therefore makes F absolutely continuous. At an interior time t where the metric derivative exists, use an optimal coupling ${\pi }_h$ of ${\mu }_t,{\mu }_{t+h}$. Taylor's formula with bounded Hessian gives $\phi (t,y)-\phi (t,x)={\mathrm{\nabla }}_x\phi (t,x)\cdot (y-x)+R_t(x,y),\mathrm{\mid }{R}_t(x,y)\mathrm{\mid }\le C\mathrm{\mid }y-x{\mathrm{\mid }}^2\mathrm{.}$ Cauchy--Schwarz and [F2] yield $\mid \int (\phi (t,y)-\phi (t,x))d{\pi }_h\mid \le \mathrm{\parallel }\mathrm{\nabla }\phi (t,\cdot ){\mathrm{\parallel }}_{L^2({\mu }_t)}{W}_2({\mu }_t,{\mu }_{t+h})+C{W}_2({\mu }_t,{\mu }_{t+h}{)}^2\mathrm{.}$ After division by $\mathrm{\mid }h\mathrm{\mid }$, the remainder tends to zero and the first term tends to $s(t)\mathrm{\parallel }\mathrm{\nabla }{\phi }_t{\mathrm{\parallel }}_2$. The separate time increment converges to $\int {\mathrm{\partial }}_t\phi (t,x)d{\mu }_t(x)$ by narrow continuity and uniform convergence of the smooth time difference quotients. Hence $\mathrm{\mid }{F}^{\mathrm{\prime }}(t)-\int {\mathrm{\partial }}_t{\phi }_{t}d{\mu }_t\mathrm{\mid }\le s(t)\mathrm{\parallel }\mathrm{\nabla }{\phi }_t{\mathrm{\parallel }}_2$ almost everywhere. Since F vanishes near 0 and T, the absolutely-continuous FTC [F3] gives ${\int }_0^T{F}^{\mathrm{\prime }}=0$; integrating the pointwise derivative inequality proves (1). Holder [F4] will be used for the right side.

Let $\mathcal{G}$ be the closure in H of $\{{\mathrm{\nabla }}_x\phi :\phi \in C_c^{\mathrm{\infty }}((0,T)\times {\mathbb{R}}^d)\}$. Define $\mathrm{\ell }(\mathrm{\nabla }\phi )=-\int {\mathrm{\partial }}_t\phi dM$. Estimate (1) and [F4] show $\mathrm{\mid }\mathrm{\ell }(g)\mathrm{\mid }\le \mathrm{\parallel }s{\mathrm{\parallel }}_{L^2(0,T)}\mathrm{\parallel }g{\mathrm{\parallel }}_H$. Thus if two tests have the same gradient in H their $\mathrm{\ell }$ values agree; this defines a bounded linear functional on the gradient subspace and extends uniquely to $\mathcal{G}$. By [F1] and Riesz there is a unique $v\in \mathcal{G}$ with $\mathrm{\ell }(g)=⟨g,v{⟩}_H$ for all $g\in \mathcal{G}$, and $\mathrm{\parallel }v{\mathrm{\parallel }}_H\le \mathrm{\parallel }s{\mathrm{\parallel }}_2$. Choose a Borel representative of this $L^2(M)$ class. For every test $\phi$, $\int ({\mathrm{\partial }}_t\phi +{\mathrm{\nabla }}_x\phi \cdot v)dM=0,$ so v solves the continuity equation and has finite action.

Fix an interval I with rational endpoints in $(0,T)$ and let ${\mathcal{G}}_I$ be the closure in $L^2(M{\mathrm{\mid }}_{I\times {\mathbb{R}}^d})$ of gradients of tests compactly supported in the interior of I. The restriction $v{\mathbf{1}}_I$ belongs to ${\mathcal{G}}_I$: approximate v by global test gradients, multiply each by smooth time cutoffs converging to ${\mathbf{1}}_I$, and use [F8] dominated convergence; the time marginal is Lebesgue, so the endpoints are null. For tests supported in I, $v{\mathbf{1}}_I$ represents the restricted functional ${\mathrm{\ell }}_I$. By (1), its norm is at most $\mathrm{\parallel }s{\mathrm{\parallel }}_{L^2(I)}$. Consequently ${\int }_I\mathrm{\parallel }{v}_t{\mathrm{\parallel }}_{L^2({\mu }_t)}^{2}dt\le {\int }_{I}s(t{)}^{2}dt$ for every rational interval I. Apply [F3] to the indefinite integral of $q(t)=\mathrm{\parallel }{v}_t{\mathrm{\parallel }}_{L^2({\mu }_t)}^2-s(t{)}^2$: the interval inequalities make its values nonincreasing first on rational endpoints and then, by continuity of the indefinite integral, on all endpoints; its derivative is therefore nonpositive almost everywhere. Thus $\mathrm{\parallel }{v}_t{\mathrm{\parallel }}_{L^2({\mu }_t)}\le s(t)$ a.e. Applying [F6] to the solution v gives the reverse inequality. Hence equality holds almost everywhere.

Let w be any other Borel finite-action solution and put h=w-v. Subtracting the two weak equations gives $⟨h,\mathrm{\nabla }\phi {⟩}_H=0$ for every test. Therefore on each rational interval I, h is orthogonal to ${\mathcal{G}}_I$; step 3.1 has $v{\mathbf{1}}_I\in {\mathcal{G}}_I$, so ${\int }_I\int h_t\cdot v_{t}d{\mu }_{t}dt=0$. The scalar integrand is in $L^1(0,T)$ by Cauchy--Schwarz. Its indefinite integral has zero increments on all rational intervals and is therefore constant; [F3] makes the integrand zero almost everywhere. Thus $⟨h_t,v_t{⟩}_{L^2({\mu }_t)}=0$ a.e., and the Hilbert-space identity gives $\mathrm{\parallel }{w}_t{\mathrm{\parallel }}_{L^2({\mu }_t)}^2=\mathrm{\parallel }{v}_t+h_t{\mathrm{\parallel }}_{L^2({\mu }_t)}^2=\mathrm{\parallel }{v}_t{\mathrm{\parallel }}_{L^2({\mu }_t)}^2+\mathrm{\parallel }{h}_t{\mathrm{\parallel }}_{L^2({\mu }_t)}^2=s(t{)}^2+\mathrm{\parallel }{w}_t-v_t{\mathrm{\parallel }}_{L^2({\mu }_t)}^2\mathrm{.}$ Integrating proves the minimum value and uniqueness up to $M$-null sets.