[proofplan]
We compute the first variation of $\mathcal E$ at a smooth positive density by perturbing the density and differentiating the entropy and potential terms. This gives the formal variational derivative $\log\rho_t+1+V$, up to the usual additive constant on the probability simplex. Substituting this derivative into the stated Otto gradient-flow convention reduces the equation to a pointwise differential identity, and positivity of $\rho_t$ gives $\rho_t\nabla\log\rho_t=\nabla\rho_t$. Taking the divergence then yields the Fokker-Planck equation.
[/proofplan]
[step:Compute the first variation of the entropy plus potential energy]
Fix $t\in(0,T)$. Let $\eta:\mathbb R^n\to\mathbb R$ be a smooth rapidly decaying function such that
\begin{align*}
\int_{\mathbb R^n}\eta(x)\,d\mathcal L^n(x)=0.
\end{align*}
For $|\varepsilon|$ sufficiently small, define the perturbed density $r_\varepsilon:\mathbb R^n\to(0,\infty)$ by
\begin{align*}
r_\varepsilon(x)\coloneqq \rho_t(x)+\varepsilon\eta(x).
\end{align*}
The zero-mass condition on $\eta$ gives
\begin{align*}
\int_{\mathbb R^n}r_\varepsilon(x)\,d\mathcal L^n(x)=1,
\end{align*}
so this is a formal tangent perturbation inside the space of probability densities.
By the assumed smoothness and decay, differentiating under the integral sign is justified. Since the derivative of $s\mapsto s\log s$ on $(0,\infty)$ is $s\mapsto \log s+1$, we obtain
\begin{align*}
\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}\int_{\mathbb R^n}r_\varepsilon(x)\log r_\varepsilon(x)\,d\mathcal L^n(x)=\int_{\mathbb R^n}(\log\rho_t(x)+1)\eta(x)\,d\mathcal L^n(x).
\end{align*}
Similarly, linearity of the potential energy gives
\begin{align*}
\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}\int_{\mathbb R^n}V(x)r_\varepsilon(x)\,d\mathcal L^n(x)=\int_{\mathbb R^n}V(x)\eta(x)\,d\mathcal L^n(x).
\end{align*}
Adding the two identities gives
\begin{align*}
\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}\mathcal E[r_\varepsilon]=\int_{\mathbb R^n}(\log\rho_t(x)+1+V(x))\eta(x)\,d\mathcal L^n(x).
\end{align*}
Thus the formal first variation may be represented by
\begin{align*}
\frac{\delta\mathcal E}{\delta\rho}(\rho_t)(x)=\log\rho_t(x)+1+V(x).
\end{align*}
On the probability simplex this representative is determined only up to an additive constant, but the gradient-flow equation uses its spatial gradient, so this ambiguity has no effect.
[guided]
Fix a time $t\in(0,T)$. To identify the formal Wasserstein gradient, we first identify the first variation of the energy at the density $\rho_t$. A tangent perturbation to the space of probability densities must preserve total mass to first order, so we choose a smooth rapidly decaying function $\eta:\mathbb R^n\to\mathbb R$ satisfying
\begin{align*}
\int_{\mathbb R^n}\eta(x)\,d\mathcal L^n(x)=0.
\end{align*}
For $|\varepsilon|$ small, define $r_\varepsilon:\mathbb R^n\to(0,\infty)$ by
\begin{align*}
r_\varepsilon(x)\coloneqq \rho_t(x)+\varepsilon\eta(x).
\end{align*}
The zero integral of $\eta$ gives
\begin{align*}
\int_{\mathbb R^n}r_\varepsilon(x)\,d\mathcal L^n(x)=\int_{\mathbb R^n}\rho_t(x)\,d\mathcal L^n(x)+\varepsilon\int_{\mathbb R^n}\eta(x)\,d\mathcal L^n(x)=1.
\end{align*}
Thus $r_\varepsilon$ is a formal probability-density perturbation of $\rho_t$.
Now compute the derivative of each part of $\mathcal E$. The entropy integrand is the scalar function $s\mapsto s\log s$ on $(0,\infty)$, whose derivative is $s\mapsto \log s+1$. Since the theorem assumes precisely the smoothness and decay needed to differentiate under the integral sign, the entropy term satisfies
\begin{align*}
\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}\int_{\mathbb R^n}r_\varepsilon(x)\log r_\varepsilon(x)\,d\mathcal L^n(x)=\int_{\mathbb R^n}(\log\rho_t(x)+1)\eta(x)\,d\mathcal L^n(x).
\end{align*}
The potential energy is linear in the density, so its derivative in the direction $\eta$ is
\begin{align*}
\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}\int_{\mathbb R^n}V(x)r_\varepsilon(x)\,d\mathcal L^n(x)=\int_{\mathbb R^n}V(x)\eta(x)\,d\mathcal L^n(x).
\end{align*}
Adding the entropy and potential contributions yields
\begin{align*}
\frac{d}{d\varepsilon}\bigg|_{\varepsilon=0}\mathcal E[r_\varepsilon]=\int_{\mathbb R^n}(\log\rho_t(x)+1+V(x))\eta(x)\,d\mathcal L^n(x).
\end{align*}
This is exactly the defining formal identity for the first variation, so we may take
\begin{align*}
\frac{\delta\mathcal E}{\delta\rho}(\rho_t)(x)=\log\rho_t(x)+1+V(x).
\end{align*}
Because tangent perturbations have zero total mass, adding a constant to this expression would not change the first variation. This is harmless here: the Wasserstein gradient-flow convention differentiates the first variation in space, and the spatial gradient of a constant is zero.
[/guided]
[/step]
[step:Insert the first variation into the formal Wasserstein gradient-flow convention]
By the convention in the theorem statement, the formal Wasserstein gradient-flow equation is
\begin{align*}
\partial_t\rho_t=\nabla\cdot\left(\rho_t\nabla\frac{\delta\mathcal E}{\delta\rho}(\rho_t)\right).
\end{align*}
Using the representative computed above gives, for every $(t,x)\in(0,T)\times\mathbb R^n$,
\begin{align*}
\nabla\frac{\delta\mathcal E}{\delta\rho}(\rho_t)(x)=\nabla\log\rho_t(x)+\nabla V(x),
\end{align*}
because the spatial gradient of the constant $1$ is zero. Therefore the formal equation becomes
\begin{align*}
\partial_t\rho_t=\nabla\cdot(\rho_t\nabla\log\rho_t)+\nabla\cdot(\rho_t\nabla V).
\end{align*}
[/step]
[step:Simplify the logarithmic entropy term to the Laplacian]
Since $\rho_t(x)>0$ for every $x\in\mathbb R^n$ and $\rho_t$ is smooth, the chain rule gives
\begin{align*}
\nabla\log\rho_t(x)=\frac{\nabla\rho_t(x)}{\rho_t(x)}.
\end{align*}
Multiplying by $\rho_t(x)$ gives the pointwise identity
\begin{align*}
\rho_t(x)\nabla\log\rho_t(x)=\nabla\rho_t(x).
\end{align*}
Taking the divergence of both sides yields
\begin{align*}
\nabla\cdot(\rho_t\nabla\log\rho_t)=\nabla\cdot(\nabla\rho_t)=\Delta\rho_t.
\end{align*}
[/step]
[step:Identify the resulting equation as the Fokker-Planck equation]
Substituting
\begin{align*}
\nabla\cdot(\rho_t\nabla\log\rho_t)=\Delta\rho_t
\end{align*}
into the formal gradient-flow equation gives
\begin{align*}
\partial_t\rho_t=\Delta\rho_t+\nabla\cdot(\rho_t\nabla V).
\end{align*}
This is exactly the stated Fokker-Planck equation on $(0,T)\times\mathbb R^n$. The proof is only a formal Otto-calculus computation, and it uses the decay hypothesis only to justify the differentiations and integrations implicit in the first-variation calculation.
[/step]
