[proofplan] The proof is an equality-case analysis of the Benamou-Brenier lower bound. For every admissible finite-action pair, the dynamic Wasserstein estimate gives a length bound controlled by the $L^2(\mu_t)$ norm of the velocity, and Cauchy-Schwarz converts this into the action lower bound $W_2^2(\mu_0,\mu_1)$. A minimizing pair must make both inequalities equalities, which forces the velocity norm to be constant in time and the curve to realize the Wasserstein distance proportionally on every subinterval. Conversely, if the curve is a constant-speed geodesic and the velocity realizes the metric derivative a.e., then its action equals $W_2^2(\mu_0,\mu_1)$, hence it is minimizing by the Benamou-Brenier formula. [/proofplan] [step:Define the speed function and record the dynamic estimate] Define \begin{align*} b:[0,1]\to[0,\infty] \end{align*} by \begin{align*} b(t):=\left(\int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x)\right)^{1/2} \end{align*} for every $t$ for which $v_t\in L^2(\mu_t;\mathbb R^n)$, and redefine $b(t):=0$ on the remaining $\mathcal L^1$-null set. Since $(\mu_t,v_t)$ has finite action, $b\in L^2(0,1)$ and hence $b\in L^1(0,1)$. We use the standard dynamic estimate for solutions of the continuity equation in Wasserstein space: for every $0\le s\le t\le 1$, \begin{align*} W_2(\mu_s,\mu_t)\le\int_s^t b(r)\,d\mathcal L^1(r). \end{align*} This estimate applies because $(\mu_t,v_t)$ is narrowly continuous, solves the continuity equation distributionally, and has finite quadratic action. In particular, $(\mu_t)$ is absolutely continuous as a curve in $(\mathcal P_2(\mathbb R^n),W_2)$. By the metric-derivative characterization for absolutely continuous metric curves [citetheorem:10022], the metric derivative $|\mu'|(t)$ exists for $\mathcal L^1$-a.e. $t$ and satisfies \begin{align*} |\mu'|(t)\le b(t) \end{align*} for $\mathcal L^1$-a.e. $t\in[0,1]$. [guided] The first object to isolate is the time-dependent size of the velocity field. Define \begin{align*} b:[0,1]\to[0,\infty] \end{align*} by \begin{align*} b(t):=\left(\int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x)\right)^{1/2} \end{align*} at times where $v_t$ is an element of $L^2(\mu_t;\mathbb R^n)$, and set $b(t):=0$ on the exceptional null set. The finite-action hypothesis says exactly that \begin{align*} \int_0^1 b(t)^2\,d\mathcal L^1(t)<\infty, \end{align*} so $b\in L^2(0,1)$. Since the time interval has finite $\mathcal L^1$-measure, Cauchy-Schwarz also gives $b\in L^1(0,1)$. The continuity equation connects this Eulerian velocity size to Wasserstein motion. The standard dynamic estimate for admissible solutions of \begin{align*} \partial_t\mu_t+\operatorname{div}(v_t\mu_t)=0 \end{align*} states that for every $0\le s\le t\le 1$, \begin{align*} W_2(\mu_s,\mu_t)\le\int_s^t b(r)\,d\mathcal L^1(r). \end{align*} Its hypotheses are precisely the ones in the statement: the curve is narrowly continuous, the vector field is Borel, the continuity equation holds in the distributional sense, and the quadratic action is finite. This estimate says that $b$ is an admissible metric upper gradient for the curve $(\mu_t)$. Now apply the metric-derivative theorem for absolutely continuous curves in complete metric spaces [citetheorem:10022], with the [complete metric space](/page/Complete%20Metric%20Space) taken to be $(\mathcal P_2(\mathbb R^n),W_2)$. The estimate above proves absolute continuity of $(\mu_t)$ and gives an admissible control function $b$. Therefore the metric derivative $|\mu'|(t)$ exists for $\mathcal L^1$-a.e. $t$ and is bounded above by every such control function. Hence \begin{align*} |\mu'|(t)\le b(t) \end{align*} for $\mathcal L^1$-a.e. $t\in[0,1]$. [/guided] [/step] [step:Derive the Benamou-Brenier lower bound from length and Cauchy-Schwarz] Let \begin{align*} D:=W_2(\mu_0,\mu_1). \end{align*} Using the dynamic estimate with $s=0$ and $t=1$ gives \begin{align*} D\le\int_0^1 b(r)\,d\mathcal L^1(r). \end{align*} Applying the [Cauchy-Schwarz inequality](/theorems/432) on the [measure space](/page/Measure%20Space) $([0,1],\mathcal B([0,1]),\mathcal L^1)$ to the functions $b$ and $1$, we obtain \begin{align*} \int_0^1 b(r)\,d\mathcal L^1(r)\le\left(\int_0^1 b(r)^2\,d\mathcal L^1(r)\right)^{1/2}. \end{align*} Therefore every admissible finite-action pair satisfies \begin{align*} \mathcal A(\mu,v)=\int_0^1 b(r)^2\,d\mathcal L^1(r)\ge D^2. \end{align*} The Benamou-Brenier formula asserts that the infimum of $\mathcal A$ over admissible finite-action pairs joining $\mu_0$ to $\mu_1$ is exactly $D^2$. [/step] [step:Use equality in the lower-bound chain for a minimizing pair] Assume first that $(\mu_t,v_t)$ is a Benamou-Brenier minimizer. By the previous step and the Benamou-Brenier formula, \begin{align*} \int_0^1 b(r)^2\,d\mathcal L^1(r)=D^2. \end{align*} The chain of inequalities \begin{align*} D\le\int_0^1 b(r)\,d\mathcal L^1(r)\le\left(\int_0^1 b(r)^2\,d\mathcal L^1(r)\right)^{1/2}=D \end{align*} forces \begin{align*} \int_0^1 b(r)\,d\mathcal L^1(r)=D. \end{align*} Equality in Cauchy-Schwarz for the pair of functions $b$ and $1$ implies that $b$ is proportional to $1$ in $L^2(0,1)$. Since the common value of the $L^1$ norm of $b$ is $D$, the proportionality constant is $D$, and therefore \begin{align*} b(t)=D \end{align*} for $\mathcal L^1$-a.e. $t\in[0,1]$. [/step] [step:Prove that the minimizing curve realizes every subinterval distance proportionally] For $0\le s\le t\le 1$, the dynamic estimate and the previous step give \begin{align*} W_2(\mu_s,\mu_t)\le\int_s^t b(r)\,d\mathcal L^1(r)=D(t-s). \end{align*} On the other hand, applying this upper bound to the intervals $[0,s]$, $[s,t]$, and $[t,1]$, and then using the triangle inequality in $(\mathcal P_2(\mathbb R^n),W_2)$, gives \begin{align*} D=W_2(\mu_0,\mu_1)\le W_2(\mu_0,\mu_s)+W_2(\mu_s,\mu_t)+W_2(\mu_t,\mu_1)\le Ds+W_2(\mu_s,\mu_t)+D(1-t). \end{align*} Hence \begin{align*} W_2(\mu_s,\mu_t)\ge D(t-s). \end{align*} Combining the upper and lower bounds yields \begin{align*} W_2(\mu_s,\mu_t)=D(t-s) \end{align*} for all $0\le s\le t\le 1$. Thus $(\mu_t)$ is a constant-speed $W_2$-geodesic from $\mu_0$ to $\mu_1$. Since a constant-speed geodesic with speed $D$ has metric derivative $|\mu'|(t)=D$ for $\mathcal L^1$-a.e. $t$, and since $b(t)=D$ for $\mathcal L^1$-a.e. $t$, we conclude that \begin{align*} \int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x)=b(t)^2=|\mu'|^2(t) \end{align*} for $\mathcal L^1$-a.e. $t\in[0,1]$. [/step] [step:Compute the action of a constant-speed geodesic with metric-speed velocity] Conversely, assume that $(\mu_t)$ is a constant-speed $W_2$-geodesic from $\mu_0$ to $\mu_1$ and that \begin{align*} \int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x)=|\mu'|^2(t) \end{align*} for $\mathcal L^1$-a.e. $t$. Let $D:=W_2(\mu_0,\mu_1)$. Since $(\mu_t)$ is a constant-speed geodesic on the interval $[0,1]$, its metric derivative satisfies \begin{align*} |\mu'|(t)=D \end{align*} for $\mathcal L^1$-a.e. $t\in[0,1]$. Therefore \begin{align*} \mathcal A(\mu,v)=\int_0^1\int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x)\,d\mathcal L^1(t)=\int_0^1 D^2\,d\mathcal L^1(t)=D^2. \end{align*} By the Benamou-Brenier formula, $D^2$ is the minimum possible action among all admissible finite-action pairs joining $\mu_0$ to $\mu_1$. Hence $(\mu_t,v_t)$ is a Benamou-Brenier minimizer. This proves both implications. [/step]