[step:Sum the Euler-Lagrange identities into a discrete weak formulation]Let $\varphi\in C_c^\infty([0,\infty)\times\mathbb R^n)$. Choose $T_\varphi>0$ such that
\begin{align*}
\operatorname{supp}\varphi\subset[0,T_\varphi]\times K_\varphi
\end{align*}
for some compact set $K_\varphi\subset\mathbb R^n$. For each $k\ge0$, define the spatial test function $\varphi_k^\tau:\mathbb R^n\to\mathbb R$ by
\begin{align*}
\varphi_k^\tau(x)=\varphi(k\tau,x).
\end{align*}
This map belongs to $C_c^\infty(\mathbb R^n;\mathbb R)$.
Apply the previous step with $\psi=\varphi_{k+1}^\tau$ and choose, for each $k$, an optimal plan $\pi_k^\tau\in\Pi(\rho_{k+1}^\tau,\rho_k^\tau)$. Let $K_\tau$ be the largest integer such that $K_\tau\tau\le T_\varphi+\tau$. Summing the one-step identities for $0\le k\le K_\tau$ gives a transport sum
\begin{align*}
\frac{1}{\tau}\sum_{k=0}^{K_\tau}\int_{\mathbb R^n\times\mathbb R^n}\left(\varphi((k+1)\tau,x)-\varphi((k+1)\tau,y)\right)\,d\pi_k^\tau(x,y).
\end{align*}
By Taylor's theorem in the spatial variable,
\begin{align*}
\varphi((k+1)\tau,x)-\varphi((k+1)\tau,y)=\nabla\varphi((k+1)\tau,x)\cdot(x-y)+r_{k,\tau}(x,y),
\end{align*}
where
\begin{align*}
|r_{k,\tau}(x,y)|\le \frac{1}{2}\|D_x^2\varphi\|_\infty |x-y|^2.
\end{align*}
Since $\pi_k^\tau$ is optimal, the summed spatial remainder satisfies
\begin{align*}
\left|E_\tau^{\mathrm{space}}\right|\le \frac{1}{2\tau}\|D_x^2\varphi\|_\infty\sum_{k=0}^{K_\tau}W_2^2(\rho_{k+1}^\tau,\rho_k^\tau).
\end{align*}
This is combined with the one-step remainder $R_{k,\tau}[\varphi_{k+1}^\tau]$ from the Euler-Lagrange identity; after multiplication by $\tau$ in the summed weak formulation, the total error is bounded by a constant multiple of $\sum_{k=0}^{K_\tau}W_2^2(\rho_{k+1}^\tau,\rho_k^\tau)$ and therefore tends to $0$.
The zeroth-order transport terms telescope after adding and subtracting $\varphi(k\tau,\cdot)$ against $\rho_{k+1}^\tau$:
\begin{align*}
\sum_{k=0}^{K_\tau}\left(\int_{\mathbb R^n}\varphi((k+1)\tau,x)\,d\rho_{k+1}^\tau(x)-\int_{\mathbb R^n}\varphi((k+1)\tau,y)\,d\rho_k^\tau(y)\right)
\end{align*}
\begin{align*}
= -\int_{\mathbb R^n}\varphi(0,x)\,d\rho_0(x)-\sum_{k=0}^{K_\tau}\int_{\mathbb R^n}\left(\varphi((k+1)\tau,x)-\varphi(k\tau,x)\right)\,d\rho_{k+1}^\tau(x)+o(1).
\end{align*}
The terminal term is $o(1)$ because $\varphi(t,\cdot)=0$ for $t>T_\varphi$ and $(K_\tau+1)\tau>T_\varphi$ for small $\tau$. Taylor's theorem in time gives
\begin{align*}
\varphi((k+1)\tau,x)-\varphi(k\tau,x)=\int_{k\tau}^{(k+1)\tau}\partial_t\varphi(r,x)\,d\mathcal L^1(r),
\end{align*}
and the uniform continuity of $\partial_t\varphi$ on its compact support converts the sum into
\begin{align*}
\int_0^\infty\int_{\mathbb R^n}\partial_t\varphi(t,x)\,d\bar\rho_\tau(t)(x)\,d\mathcal L^1(t)+o(1).
\end{align*}
Consequently the discrete weak formulation is
\begin{align*}
\int_0^\infty\int_{\mathbb R^n}\left(\partial_t\varphi(t,x)+\Delta\varphi(t,x)-\nabla V(x)\cdot\nabla\varphi(t,x)\right)\,d\bar\rho_\tau(t)(x)\,d\mathcal L^1(t)+\int_{\mathbb R^n}\varphi(0,x)\,d\rho_0(x)=o(1).
\end{align*}[/step]