Metric derivatives of AC2 Wasserstein curves

Let $D=\mathbb{Q}\cap [0,T]$ together with $0,T$. For each $q\in D$, set $f_q(t)=W_2({\mu }_t,{\mu }_q)$. The reverse triangle inequality and the assumed control give $\mathrm{\mid }{f}_q(t)-f_q(s)\mathrm{\mid }\le {\int }_s^{t}g(r)dr$, so $f_q$ is absolutely continuous. By [F1], it has a derivative almost everywhere, and because $g\in L^1$ by Holder, at common differentiability points of $f_q$ and $G(t)={\int }_0^{t}g(r)dr$, difference quotients give $\mathrm{\mid }{f}_q^{\mathrm{\prime }}(t)\mathrm{\mid }\le g(t)$. Since $D$ is countable, outside one null set all these derivatives exist and are bounded by $g$. Define $m(t)={\sup }_{q\in D}\mathrm{\mid }{f}_q^{\mathrm{\prime }}(t)\mathrm{\mid }$ there and set $m=0$ on the null set. Difference quotients along $h=1\mathrm{/}n$ give measurable versions of the scalar derivatives, so $m$ is measurable; also $0\le m\le g$ almost everywhere and hence $m\in L^2(0,T)$.

For any $s,t$, continuity of the curve and density of $D$ imply $W_2({\mu }_s,{\mu }_t)={\sup }_{q\in D}\mathrm{\mid }{f}_q(t)-f_q(s)\mathrm{\mid }\mathrm{.}$ The inequality from right to left is the reverse triangle inequality. For the other direction take $q_j\in D$ with $q_j\to s$; then $f_{q_j}(s)\to 0$ and $f_{q_j}(t)\to W_2({\mu }_t,{\mu }_s)$. For each q, [F1] and step 1.1 give $\mathrm{\mid }{f}_q(t)-f_q(s)\mathrm{\mid }\le {\int }_s^t\mathrm{\mid }{f}_q^{\mathrm{\prime }}(r)\mathrm{\mid }dr\le {\int }_s^{t}m(r)dr$. Taking the supremum proves $W_2({\mu }_s,{\mu }_t)\le {\int }_s^{t}m(r)dr$.

The function $H(t)={\int }_0^{t}m(r)dr$ is absolutely continuous, and [F1] gives $H^{\mathrm{\prime }}(t)=m(t)$ almost everywhere. At a time t where $H$ is differentiable and every $f_q$ is differentiable, step 2.1 gives ${\mathrm{lim\hspace{0.17em}sup}}_{h\to 0}\frac{W_2({\mu }_{t+h},{\mu }_t)}{\mathrm{\mid }h\mathrm{\mid }}\le m(t),$ while $W_2({\mu }_{t+h},{\mu }_t)\ge \mathrm{\mid }{f}_q(t+h)-f_q(t)\mathrm{\mid }$ gives a lower limit at least $\mathrm{\mid }{f}_q^{\mathrm{\prime }}(t)\mathrm{\mid }$ for every q. Taking the supremum yields a lower limit at least $m(t)$. Thus the metric derivative exists and equals $m(t)$ almost everywhere. The interval bound and square integrability follow from step 2.1 and $m\le g$.

Let h be any other nonnegative $L^2$ upper control. Repeating the scalar difference-quotient bound of step 1.1 with h gives $\mathrm{\mid }{f}_q^{\mathrm{\prime }}(t)\mathrm{\mid }\le h(t)$ almost everywhere for each q. Countability of D lets these inequalities hold off a single null set, so $m(t)\le h(t)$ there. By step 3.1, $\mathrm{\mid }{\dot{\mu }}_t\mathrm{\mid }=m(t)$ almost everywhere, proving leastness.