SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\langle \xi , \eta \rangle := \operatorname {t r} \left(\Sigma_ {1} \Sigma_ {2}\right) + \boldsymbol {\mu} _ {1} ^ {\top} \boldsymbol {\mu} _ {2}, \tag {57}
$$
for any  $\xi, \eta \in T_{\theta}\Theta$ , where  $\xi = (\widetilde{\mu}_1, \widetilde{\Sigma}_1)$  and  $\eta = (\widetilde{\mu}_2, \widetilde{\Sigma}_2)$ , then the tangent bundle  $T\Theta$  is trivial. Since
$$
\begin{array}{c c c c} \phi : & \Theta & \to & \mathcal {P} (\mathbb {R} ^ {d}) \end{array}
$$
$$
\begin{array}{c c c} \theta & \mapsto & \rho (\cdot , \theta) \end{array}
$$
provides an immersion from  $\Theta$  to  $\mathcal{P}(\mathbb{R}^d)$ , we consider its pushforward  $\mathrm{d}\phi_{\theta}$  given by
$$
\begin{array}{l} \mathrm {d} \phi_ {\theta}: \quad T _ {\theta} \Theta \quad \rightarrow \quad T _ {\rho} \mathcal {P} (\mathbb {R} ^ {d}) \\ \xi \qquad \mapsto \quad \langle \nabla_ {\theta} \rho (\cdot , \theta), \xi \rangle . \\ \end{array}
$$
On the other hand, for any  $\Phi \in T_{\rho}^{*}\mathcal{P}(\mathbb{R}^{d})$ , the inverse canonical isomorphism of Stein metric maps it to
$$
G _ {\rho} ^ {- 1} \Phi = - \nabla \cdot \rho (\cdot , \theta) \int K (\cdot , \pmb {y}) \rho (\pmb {y}, \theta) \nabla \Phi (\pmb {y}) \mathrm {d} \pmb {y} \in T _ {\rho} \mathcal {P} (\mathbb {R} ^ {d}).
$$
$$
\langle \nabla_ {\theta} \rho (\boldsymbol {x}, \theta), \xi \rangle = - \nabla_ {\boldsymbol {x}} \cdot \rho (\boldsymbol {x}, \theta) \int K (\boldsymbol {x}, \boldsymbol {y}) \rho (\boldsymbol {y}, \theta) \nabla \Phi (\boldsymbol {y}) d \boldsymbol {y} \tag {58}
$$
Now we try to find a suitable function  $\nabla \Phi_{\xi}$  that satisfies the equation above. We compute that
$$
\begin{array}{l} \langle \nabla_ {\theta} \rho (\boldsymbol {x}, \theta), \xi \rangle \\ = \operatorname {t r} \left(\nabla_ {\Sigma} \rho (\boldsymbol {x}, \theta) \widetilde {\Sigma} _ {1}\right) + \nabla_ {\boldsymbol {\mu}} \rho (\boldsymbol {x}, \theta) ^ {\top} \widetilde {\boldsymbol {\mu}} _ {1} \\ = \left(- \frac {1}{2} \left(\operatorname {t r} \left(\Sigma^ {- 1} \widetilde {\Sigma} _ {1}\right) - (\boldsymbol {x} - \boldsymbol {\mu}) ^ {T} \Sigma^ {- 1} \widetilde {\Sigma} _ {1} \Sigma^ {- 1} (\boldsymbol {x} - \boldsymbol {\mu})\right) + \widetilde {\boldsymbol {\mu}} _ {1} ^ {\top} \Sigma^ {- 1} (\boldsymbol {x} - \boldsymbol {\mu})\right) \rho (\boldsymbol {x}, \theta). \\ \end{array}
$$
Letting  $\Psi (\pmb {x}) = \int K(\pmb {x},\pmb {y})\rho (\pmb {y},\theta)\nabla \Phi (\pmb {y})\mathrm{d}\pmb {y}$  we get
$$
\begin{array}{l} - \nabla_ {\boldsymbol {x}} \cdot \rho (\boldsymbol {x}, \theta) \int K (\boldsymbol {x}, \boldsymbol {y}) \rho (\boldsymbol {y}, \theta) \nabla \Phi (\boldsymbol {y}) d \boldsymbol {y} \\ = - \nabla_ {\boldsymbol {x}} \cdot \rho (\boldsymbol {x}, \theta) \Psi (\boldsymbol {x}) \\ = \left(\Psi (\boldsymbol {x}) ^ {\top} \Sigma^ {- 1} (\boldsymbol {x} - \boldsymbol {\mu}) - \nabla \cdot \Psi (\boldsymbol {x})\right) \rho (\boldsymbol {x}, \theta). \\ \end{array}
$$
We choose  $\nabla \Phi_{\xi}(\pmb{x}) = S_1(\pmb{x} - \pmb{\mu}) + \pmb{b}_1$ , where  $S_1 \in \operatorname{Sym}(d, \mathbb{R})$  and  $\pmb{b}_1 \in \mathbb{R}^d$  will be determined later. Note that  $S_1$  needs to be symmetric because the gradient field is curl-free. We derive that
$$
\begin{array}{l} \Psi (\boldsymbol {x}) = \int (\boldsymbol {x} ^ {\top} \boldsymbol {y} + 1) \rho (\boldsymbol {y}, \theta) \nabla \Phi_ {\xi} (\boldsymbol {y}) d \boldsymbol {y} \\ = \int (\boldsymbol {x} ^ {\top} \boldsymbol {y} + 1) \rho (\boldsymbol {y}, \theta) (S _ {1} (\boldsymbol {y} - \boldsymbol {\mu}) + \boldsymbol {b} _ {1}) d \boldsymbol {y} \\ = \int (\boldsymbol {x} ^ {\top} (\boldsymbol {y} + \boldsymbol {\mu}) + 1) \rho (\boldsymbol {y} + \boldsymbol {\mu}, \theta) (S _ {1} \boldsymbol {y} + \boldsymbol {b} _ {1}) d \boldsymbol {y} \\ = S _ {1} \Sigma \boldsymbol {x} + (\boldsymbol {x} ^ {\top} \boldsymbol {\mu} + 1) \boldsymbol {b} _ {1} \\ = \left(S _ {1} \Sigma + \boldsymbol {b} _ {1} \boldsymbol {\mu} ^ {\top}\right) \boldsymbol {x} + \boldsymbol {b} _ {1}. \\ \end{array}
$$
By comparison of the coefficients, we need
$$
\left\{ \begin{array}{l} \left(S _ {1} \Sigma + \boldsymbol {b} _ {1} \boldsymbol {\mu} ^ {\top}\right) ^ {\top} \Sigma^ {- 1} + \Sigma^ {- 1} \left(S _ {1} \Sigma + \boldsymbol {b} _ {1} \boldsymbol {\mu} ^ {\top}\right) = \Sigma^ {- 1} \widetilde {\Sigma} _ {1} \Sigma^ {- 1}, \\ \operatorname {t r} \left(S _ {1} \Sigma + \boldsymbol {b} _ {1} \boldsymbol {\mu} ^ {\top}\right) = \frac {1}{2} \operatorname {t r} \left(\Sigma^ {- 1} \widetilde {\Sigma} _ {1}\right), \\ S _ {1} \Sigma \boldsymbol {\mu} + \left(1 + \boldsymbol {\mu} ^ {\top} \boldsymbol {\mu}\right) \boldsymbol {b} _ {1} = \widetilde {\boldsymbol {\mu}} _ {1}. \end{array} \right. \tag {59}
$$
Note that the Lyapunov equation
$$
P X + X P = Q,
$$
$$
P = \Sigma \left(I - \frac {1}{1 + \boldsymbol {\mu} ^ {\top} \boldsymbol {\mu}} \boldsymbol {\mu} \boldsymbol {\mu} ^ {\top}\right) \Sigma ,
$$
$$
Q = \widetilde {\Sigma} _ {1} - \frac {1}{1 + \boldsymbol {\mu} ^ {\top} \boldsymbol {\mu}} (\Sigma \boldsymbol {\mu} \widetilde {\boldsymbol {\mu}} _ {1} ^ {\top} + \widetilde {\boldsymbol {\mu}} _ {1} \boldsymbol {\mu} ^ {\top} \Sigma),
$$
has a unique solution  $X = S_{1}\in \operatorname {Sym}(d,\mathbb{R})$  . Together with
$$
\boldsymbol {b} _ {1} = \frac {1}{1 + \boldsymbol {\mu} ^ {\top} \boldsymbol {\mu}} (\widetilde {\boldsymbol {\mu}} _ {1} - S _ {1} \Sigma \boldsymbol {\mu}),
$$
we find that (59) holds with the unique solution described above. Now from the calculations above it is straightforward to check that (59) is equivalent to the following equation
$$
G _ {\theta} ^ {- 1} \left(\boldsymbol {b} _ {1}, \frac {1}{2} S _ {1}\right) = \left(\widetilde {\boldsymbol {\mu}} _ {1}, \widetilde {\Sigma} _ {1}\right),
$$
where  $G_{\theta}^{-1}$  is the map defined in (29). Thus, the existence and uniqueness of the solution indicates that  $G_{\theta}$  is an isomorphism.
Similarly, we let
$$
\nabla \Phi_ {\eta} (\boldsymbol {x}) = S _ {2} (\boldsymbol {x} - \boldsymbol {\mu}) + \boldsymbol {b} _ {2},
$$
where  $X = S_{2}\in \operatorname {Sym}(d,\mathbb{R})$  is the unique solution of the Lyapunov equation
$$
Q = \tilde {\Sigma} _ {2} - \frac {1}{1 + \boldsymbol {\mu} ^ {\top} \boldsymbol {\mu}} (\Sigma \boldsymbol {\mu} \tilde {\boldsymbol {\mu}} _ {2} ^ {\top} + \tilde {\boldsymbol {\mu}} _ {2} \boldsymbol {\mu} ^ {\top} \Sigma),
$$
$$
\boldsymbol {b} _ {2} = \frac {1}{1 + \boldsymbol {\mu} ^ {\top} \boldsymbol {\mu}} (\widetilde {\boldsymbol {\mu}} _ {2} - S _ {2} \Sigma \boldsymbol {\mu}).
$$
Now we compute the Riemannian tensor
$$
\begin{array}{l} g _ {\theta} (\xi , \eta) = g _ {\rho} (\xi , \eta) = \int \Phi_ {\xi} (\boldsymbol {x}) G _ {\rho} ^ {- 1} \Phi_ {\eta} (\boldsymbol {x}) d \boldsymbol {x} \\ = \int (\nabla \Phi_ {\xi} (\boldsymbol {x})) ^ {\top} \rho (\boldsymbol {x}, \theta) \int (\boldsymbol {x} ^ {\top} \boldsymbol {y} + 1) \rho (\boldsymbol {y}, \theta) \nabla \Phi_ {\eta} (\boldsymbol {y}) d \boldsymbol {y} d \boldsymbol {x} \\ = \int \rho (\boldsymbol {x}, \theta) \left(S _ {1} (\boldsymbol {x} - \boldsymbol {\mu}) + \boldsymbol {b} _ {1}\right) ^ {\top} \left(\left(S _ {2} \Sigma + \boldsymbol {b} _ {2} \boldsymbol {\mu} ^ {\top}\right) \boldsymbol {x} + \boldsymbol {b} _ {2}\right) d \boldsymbol {x} \\ = \operatorname {t r} \left(S _ {1} \left(S _ {2} \Sigma + \boldsymbol {b} _ {2} \boldsymbol {\mu} ^ {\top}\right) \Sigma\right) + \boldsymbol {b} _ {1} ^ {\top} \left(S _ {2} \Sigma \boldsymbol {\mu} + \left(1 + \boldsymbol {\mu} ^ {\top} \boldsymbol {\mu}\right) \boldsymbol {b} _ {2}\right) \\ = \operatorname {t r} \left(S _ {1} S _ {2} \Sigma^ {2}\right) + \left(\boldsymbol {b} _ {1} ^ {\top} S _ {2} + \boldsymbol {b} _ {2} ^ {\top} S _ {1}\right) \Sigma \boldsymbol {\mu} + \left(1 + \boldsymbol {\mu} ^ {\top} \boldsymbol {\mu}\right) \boldsymbol {b} _ {1} ^ {\top} \boldsymbol {b} _ {2}. \\ \end{array}
$$
Finally, we show that  $G_{\theta}$  is indeed the canonical isomorphism corresponding to  $g_{\theta}(\cdot, \cdot)$ . We check that
$$
\begin{array}{l} g _ {\theta} (\xi , \eta) = \operatorname {t r} \left(S _ {1} \left(S _ {2} \Sigma + \boldsymbol {b} _ {2} \boldsymbol {\mu} ^ {\top}\right) \Sigma\right) + \boldsymbol {b} _ {1} ^ {\top} \left(S _ {2} \Sigma \boldsymbol {\mu} + \left(1 + \boldsymbol {\mu} ^ {\top} \boldsymbol {\mu}\right) \boldsymbol {b} _ {2}\right) \\ = \frac {1}{2} \operatorname {t r} \left(S _ {1} \tilde {\Sigma} _ {2}\right) + \boldsymbol {b} _ {1} ^ {\top} \tilde {\boldsymbol {\mu}} _ {2} = \langle G _ {\theta} (\xi), \eta \rangle . \\ \end{array}
$$
Proof of Theorem B.7. Similar to the proof of Theorem B.3, we define the inner product on  $T_{\theta}\Theta$  by
$$
\langle \xi , \eta \rangle := \operatorname {t r} \left(\Sigma_ {1} \Sigma_ {2}\right) + \boldsymbol {\mu} _ {1} ^ {\top} \boldsymbol {\mu} _ {2},
$$
for any  $\xi, \eta \in T_{\theta}\Theta$ , where  $\xi = (\widetilde{\mu}_1, \widetilde{\Sigma}_1)$  and  $\eta = (\widetilde{\mu}_2, \widetilde{\Sigma}_2)$ . The map
$$
\begin{array}{c c c} \theta & \mapsto & \rho (\cdot , \theta), \end{array}
$$
where  $\rho (\cdot ,\Sigma)$  denotes the density of  $\mathcal{N}(\pmb {\mu},\Sigma)$ , provides an immersion from  $\Theta$  to  $\mathcal{P}(\mathbb{R}^d)$ . We consider its pushforward  $\mathrm{d}\phi_{\theta}$  given by
$$
\mathrm {d} \phi_ {\theta}: \quad T _ {\theta} \Theta \quad \rightarrow \quad T _ {\rho} \mathcal {P} (\mathbb {R} ^ {d})
$$
$$
\xi \qquad \mapsto \quad \langle \nabla_ {\theta} \rho (\cdot , \theta), \xi \rangle .
$$
On the other hand, for any  $\Phi \in T_{\rho}^{*}\mathcal{P}(\mathbb{R}^{d})$ , the inverse canonical isomorphism of regularized Stein metric maps it to
$$
G _ {\rho} ^ {- 1} \Phi = - \nabla \cdot \left( \right.\rho \big ((1 - \nu) \mathcal {T} _ {K, \rho} + \nu I \big) ^ {- 1} \mathcal {T} _ {K, \rho} (\nabla \Phi) \big) \in T _ {\rho} \mathcal {P} (\mathbb {R} ^ {d}).
$$
$$
\langle \nabla_ {\theta} \rho (\boldsymbol {x}, \theta), \xi \rangle = - \nabla_ {\boldsymbol {x}} \cdot \left(\rho (\boldsymbol {x}, \theta) \left((1 - \nu) \mathcal {T} _ {K, \rho} + \nu I\right) ^ {- 1} \mathcal {T} _ {K, \rho} (\nabla \Phi (\boldsymbol {x}))\right) \tag {60}
$$
Letting  $\Psi (\pmb {x}) = ((1 - \nu)\mathcal{T}_{K,\rho} + \nu I)^{-1}\mathcal{T}_{K,\rho}$ $(\nabla \Phi (\pmb {x}))$  , we get that the RHS of (60) is equal to
$$
- \nabla_ {\boldsymbol {x}} \cdot \rho (\boldsymbol {x}, \Sigma) \Psi (\boldsymbol {x}) = \left(\Psi (\boldsymbol {x}) ^ {\top} \Sigma^ {- 1} (\boldsymbol {x} - \boldsymbol {\mu}) - \nabla \cdot \Psi (\boldsymbol {x})\right) \rho (\boldsymbol {x}, \Sigma).
$$
$$
\begin{array}{l} \mathcal {T} _ {K, \rho} (\nabla \Phi_ {\xi} (\boldsymbol {x})) = \int ((\boldsymbol {x} - \boldsymbol {\mu}) ^ {\top} (\boldsymbol {y} - \boldsymbol {\mu}) + 1) \rho (\boldsymbol {y}, \theta) \nabla \Phi_ {\xi} (\boldsymbol {y}) \mathrm {d} \boldsymbol {y} \\ = \int ((\boldsymbol {x} - \boldsymbol {\mu}) ^ {\top} (\boldsymbol {y} - \boldsymbol {\mu}) + 1) \rho (\boldsymbol {y}, \theta) (S _ {1} (\boldsymbol {y} - \boldsymbol {\mu}) + \boldsymbol {b} _ {1}) d \boldsymbol {y} \\ = S _ {1} \Sigma (\boldsymbol {x} - \boldsymbol {\mu}) + \boldsymbol {b} _ {1}. \\ \end{array}
$$
Thus, we know  $\Psi(\pmb{x}) = S_1\Sigma((1 - \nu)\Sigma + \nu I)^{-1}\pmb{x} + \pmb{b}_1$ . By comparison of the coefficients, we need  $\pmb{b}_1 = \widetilde{\pmb{\mu}}_1$  and
$$
\left\{ \begin{array}{l} \Sigma \left((1 - \nu) \Sigma + \nu I\right) ^ {- 1} S _ {1} \Sigma^ {- 1} + \Sigma^ {- 1} S _ {1} \Sigma \left((1 - \nu) \Sigma + \nu I\right) ^ {- 1} = \Sigma^ {- 1} \widetilde {\Sigma} _ {1} \Sigma^ {- 1}, \\ \operatorname {t r} \left(S _ {1} \Sigma \left((1 - \nu) \Sigma + \nu I\right) ^ {- 1}\right) = \frac {1}{2} \operatorname {t r} \left(\Sigma^ {- 1} \widetilde {\Sigma} _ {1}\right). \end{array} \right. \tag {61}
$$
Note that the first equation is equivalent to the following Lyapunov equation
$$
\Sigma^ {2} \left(\left(1 - \nu\right) \Sigma + \nu I\right) ^ {- 1} X + X \Sigma^ {2} \left(\left(1 - \nu\right) \Sigma + \nu I\right) ^ {- 1} = \widetilde {\Sigma} _ {1},
$$
which has a unique solution  $X = S_{1} \in \operatorname{Sym}(d, \mathbb{R})$ , and the second equation is automatically satisfied once we have the first one. Now from the calculations above it is straightforward to see that
$$
G _ {\theta} ^ {- 1} (\boldsymbol {b} _ {1}, S _ {1}) = \left(\boldsymbol {b} _ {1}, 2 \left(\left((1 - \nu) \Sigma + \nu I\right) ^ {- 1} \Sigma^ {2} S _ {1} + S _ {1} \left(\left(1 - \nu\right) \Sigma + \nu I\right) ^ {- 1} \Sigma^ {2}\right)\right).
$$
The proofs of Theorems B.4 and B.6 are similar to that of Theorems B.3 and B.7. In particular, other proofs of Theorem B.6 can also be found in literature, for example, see [77, 60]. Finally, Corollary B.5 is the direct corollary of Theorem B.3 or Theorem B.4.
H Proofs for Section C
Given a probability measure  $\mu$  and a Borel-measurable map  $f$ , we denote by  $f_{\#}\mu$  the pushforward of the measure  $\mu$  under the map  $f$ .
Definition H.1 (Mean field characteristic flow). Given a probability measure  $\nu$ , we say that the map
$$
X (t, \pmb {x}, \nu): [ 0, \infty) \times \mathbb {R} ^ {d} \rightarrow \mathbb {R} ^ {d}
$$
is a mean field characteristic flow associated to the particle system (13) or to the mean field PDE (2) if  $X$  is  $\mathcal{C}^1$  in time and solves the following problem
$$
\dot {X} (t, \boldsymbol {x}, \nu) = - (\nabla K * \mu_ {t}) (X (t, \boldsymbol {x}, \nu)) - (K * (\mu_ {t} \nabla V)) (X (t, \boldsymbol {x}, \nu))
$$
$$
\mu_ {t} = X (t, \cdot , \nu) _ {\#} \nu \tag {62}
$$
$$
X (0, \boldsymbol {x}, \nu) = \boldsymbol {x}
$$
Note that here  $X(t,\cdot ,\nu)_{\#}\nu$  is the push-forward of  $\nu$  under the map  $\pmb {x}\mapsto X(t,\cdot ,\nu)$ , and  $\{X(t,\cdot ,\nu)\}_{t\geq 0,\nu}$  can be regarded as a family of maps from  $\mathbb{R}^d$  to  $\mathbb{R}^d$ , parameterized by  $t$  and  $\nu$ . In the lemma below we show that the mean field characteristic flow (62) is well-defined.
Lemma H.2 (Solution of the mean field characteristic flow). Assume the conditions of Assumption C.1 hold, and  $\nu \in \mathcal{P}_V(\mathbb{R}^d)$ . For any  $T > 0$ , there exists a unique solution  $X(\cdot ,\cdot ,\nu)\in \mathcal{C}^{1}([0,T],Y)$  to the problem (62), where  $Y$  is the function space given by
$$
Y := \left\{u \in \mathcal {C} (\mathbb {R} ^ {d}, \mathbb {R} ^ {d}): \sup  _ {\boldsymbol {x} \in \mathbb {R} ^ {d}} \frac {\| u (\boldsymbol {x}) - \boldsymbol {x} \|}{1 + \| \boldsymbol {x} \|} <   \infty \right\}.
$$
Moreover, the measure  $\mu_t = X(t,\cdot,\nu)_{\#}\nu$  satisfies
$$
\| \mu_ {t} \| _ {\mathcal {P} _ {V}} \leq e ^ {C t} \| \nu \| _ {\mathcal {P} _ {V}},
$$
for some constant  $C$  that is independent of  $\nu$ .
Proof. We prove the lemma in two steps. First we show local well-posedness of the mean field characteristic flow. Second we extend the local solution to  $t \in [0, \infty)$ .
Fix  $r > 0$ , and we define
$$
Y _ {r} := \left\{u \in Y: \sup  _ {\boldsymbol {x} \in \mathbb {R} ^ {d}} \frac {\| u (\boldsymbol {x}) - \boldsymbol {x} \|}{1 + \| \boldsymbol {x} \|} \leq r \right\}.
$$
We show that there exists  $T_0 > 0$  such that (62) has a unique solution  $X(t, \boldsymbol{x}, \nu)$  on  $t \in [0, T_0]$ , and the solution is in the following function class
$$
S _ {r} := \mathcal {C} \left(\left[ 0, T _ {0} \right], Y _ {r}\right),
$$
which is a complete metric space equipped with the uniform metric
$$
d _ {S} (u, v) := \sup  _ {t \in [ 0, T _ {0} ]} d _ {Y} (u (t, \cdot), v (t, \cdot)), \quad d _ {Y} (u, v) := \sup  _ {\boldsymbol {x} \in \mathbb {R} ^ {d}} \frac {\| u (\boldsymbol {x}) - v (\boldsymbol {x}) \|}{1 + \| \boldsymbol {x} \|}.
$$
Now we check the integral formulation of (62) given by
$$
\begin{array}{l} X (t, \boldsymbol {x}, \nu) = \boldsymbol {x} - \int_ {0} ^ {t} \int_ {\mathbb {R} ^ {d}} \nabla_ {2} K \left(X (s, \boldsymbol {x}, \nu), X (s, \boldsymbol {x} ^ {\prime}, \nu)\right) \nu (\mathrm {d} \boldsymbol {x} ^ {\prime}) \mathrm {d} s \\ - \int_ {0} ^ {t} \int_ {\mathbb {R} ^ {d}} K (X (s, \boldsymbol {x}, \nu), X (s, \boldsymbol {x} ^ {\prime}, \nu)) \nabla V (X (s, \boldsymbol {x} ^ {\prime}, \nu)) \nu (\mathrm {d} \boldsymbol {x} ^ {\prime}) \mathrm {d} s \\ = \boldsymbol {x} - \int_ {0} ^ {t} X (s, \boldsymbol {x}, \nu) \mathrm {d} s - \int_ {0} ^ {t} \int_ {\mathbb {R} ^ {d}} \nabla V \left(X \left(s, \boldsymbol {x} ^ {\prime}, \nu\right)\right) X \left(s, \boldsymbol {x} ^ {\prime}, \nu\right) ^ {\top} X (s, \boldsymbol {x}, \nu) \nu (\mathrm {d} \boldsymbol {x} ^ {\prime}) \mathrm {d} s. \tag {63} \\ \end{array}
$$
Let us define the operator  $\mathcal{F}:u(t,\cdot)\mapsto \mathcal{F}(u)(t,\cdot)$  by
$$
\mathcal {F} (u) (t, \boldsymbol {x}) := \boldsymbol {x} - \int_ {0} ^ {t} u (s, \boldsymbol {x})   \mathrm {d} s - \int_ {0} ^ {t} \int_ {\mathbb {R} ^ {d}} \nabla V \left(u (s, \boldsymbol {x} ^ {\prime})\right) u (s, \boldsymbol {x} ^ {\prime}) ^ {\top} u (s, \boldsymbol {x}) \nu \left(\mathrm {d} \boldsymbol {x} ^ {\prime}\right)   \mathrm {d} s.
$$
We aim to show that  $\mathcal{F}$  is a contraction map in  $S_{r}$ , and thus, it has a unique fixed point. For this purpose we first prove that  $\mathcal{F}$  maps  $S_{r}$  into  $S_{r}$ . It is straightforward to check that  $(t,\pmb {x})\mapsto \mathcal{F}(u)(t,\pmb {x})$ . We now need to establish a bound on  $\| \mathcal{F}(u)(t,\pmb {x}) - \pmb {x}\|$ . If  $u\in S_r$ , then for any  $s\in [0,T_0]$  and  $\pmb {x}\in \mathbb{R}^d$
$$
\left\| u (s, \boldsymbol {x}) \right\| \leq \left\| \boldsymbol {x} \right\| + \left\| u (s, \boldsymbol {x}) - \boldsymbol {x} \right\| \leq (r + 1) \left\| \boldsymbol {x} \right\| + r.
$$
By Assumption C.1, we have that
$$
\begin{array}{l} \left\| \nabla V (u (s, \boldsymbol {x} ^ {\prime})) u (s, \boldsymbol {x} ^ {\prime}) ^ {\top} \right\| = \| \nabla V (u (s, \boldsymbol {x} ^ {\prime})) \| \cdot \| u (s, \boldsymbol {x} ^ {\prime}) \| \\ \leq (r + 1) \left(1 + \| \boldsymbol {x} ^ {\prime} \|\right) \| \nabla V \left(u (s, \boldsymbol {x} ^ {\prime})\right) \| \\ \leq (r + 1) C _ {r + 1, r} \left(1 + V \left(\boldsymbol {x} ^ {\prime}\right)\right). \\ \end{array}
$$
$$
\begin{array}{l} \left\| \mathcal {F} (u) (t, \pmb {x}) - \pmb {x} \right\| \\ \leq t \big ((r + 1) \| \boldsymbol {x} \| + r \big) + t \big ((r + 1) \| \boldsymbol {x} \| + r \big) (r + 1) C _ {r + 1, r} \int (1 + V (\boldsymbol {x} ^ {\prime})) \nu (\mathrm {d} \boldsymbol {x} ^ {\prime}) \\ \leq \widetilde {C} _ {r} t (1 + \| \boldsymbol {x} \|), \\ \end{array}
$$
for some constant  $\widetilde{C}_r$ , where we used the assumption that  $\nu \in \mathcal{P}_V(\mathbb{R}^d)$ . Therefore, choosing  $T_0 \leq r / \widetilde{C}_r$  we get
$$
\sup  _ {t \in [ 0, T _ {0} ]} \sup  _ {\boldsymbol {x} \in \mathbb {R} ^ {d}} \frac {\| \mathcal {F} (u) (t , \boldsymbol {x}) - \boldsymbol {x} \|}{1 + \| \boldsymbol {x} \|} \leq \widetilde {C} _ {r} T _ {0} \leq r,
$$
which shows that  $\mathcal{F}$  maps from  $S_{r}$  to  $S_{r}$  for sufficiently small  $T_0$ . Next, we prove that  $\mathcal{F}$  is indeed a contraction map. If  $u,v\in S_r$ , then for any  $t\in [0,T_0]$  and  $\pmb {x}\in \mathbb{R}^{d}$
$$
\begin{array}{l} \left\| \mathcal {F} (u) (t, \boldsymbol {x}) - \mathcal {F} (v) (t, \boldsymbol {x}) \right\| \\ \leq \int_ {0} ^ {t} \| u (s, \boldsymbol {x}) - v (s, \boldsymbol {x}) \| d s + \int_ {0} ^ {t} \int_ {\mathbb {R} ^ {d}} \| \nabla V (u (s, \boldsymbol {x} ^ {\prime})) u (s, \boldsymbol {x} ^ {\prime}) ^ {\top} \| \nu (d x ^ {\prime}) \| u (s, \boldsymbol {x}) - v (s, \boldsymbol {x}) \| d s \\ + \int_ {0} ^ {t} \int_ {\mathbb {R} ^ {d}} \| \nabla V (u (s, \boldsymbol {x} ^ {\prime})) \| \cdot \| u (s, \boldsymbol {x} ^ {\prime}) - v (s, \boldsymbol {x} ^ {\prime}) \| \nu (\mathrm {d} \boldsymbol {x} ^ {\prime}) \| v (s, \boldsymbol {x}) \| \mathrm {d} s \\ + \int_ {0} ^ {t} \int_ {\mathbb {R} ^ {d}} \| \nabla V (u (s, \boldsymbol {x} ^ {\prime})) - \nabla V (v (s, \boldsymbol {x} ^ {\prime})) \| \cdot \| v (s, \boldsymbol {x} ^ {\prime}) \| \nu (\mathrm {d} \boldsymbol {x} ^ {\prime}) \| v (s, \boldsymbol {x}) \| \mathrm {d} s =: \mathrm {I} + \mathrm {I I} + \mathrm {I I I} + \mathrm {I V}. \\ \end{array}
$$
Term I can be upper-bounded by
$$
I / (1 + \| \boldsymbol {x} \|) \leq \int_ {0} ^ {t} \frac {\left\| u (s , \boldsymbol {x}) - v (s , \boldsymbol {x}) \right\|}{1 + \| \boldsymbol {x} \|} d s \leq t d _ {S} (u, v). \tag {64}
$$
Similarly we have
$$
\left. \mathbf {I I} / (1 + \| \boldsymbol {x} \|) \leq t d _ {S} (u, v) (r + 1) C _ {r + 1, r} \int (1 + V (\boldsymbol {x} ^ {\prime})) \nu (\mathrm {d} \boldsymbol {x} ^ {\prime}). \right. \tag {65}
$$
For the third term, we apply Assumption C.1 and get
$$
\begin{array}{l} \left\| \nabla V (u (s, \boldsymbol {x} ^ {\prime})) \right\| \cdot \left\| u (s, \boldsymbol {x} ^ {\prime}) - v (s, \boldsymbol {x} ^ {\prime}) \right\| \\ = (1 + \| \boldsymbol {x} ^ {\prime} \|) \| \nabla V (u (s, \boldsymbol {x} ^ {\prime})) \| \cdot \frac {\| u (s , \boldsymbol {x} ^ {\prime}) - v (s , \boldsymbol {x} ^ {\prime}) \|}{1 + \| \boldsymbol {x} ^ {\prime} \|} \\ \leq (r + 1) C _ {r + 1, r} \left(1 + V \left(\boldsymbol {x} ^ {\prime}\right)\right) d _ {S} (u, v). \\ \end{array}
$$
$$
\begin{array}{l} \operatorname {I I I} / (1 + \| \boldsymbol {x} \|) \leq (r + 1) C _ {r + 1, r} \int (1 + V (\boldsymbol {x} ^ {\prime})) \nu (\mathrm {d} \boldsymbol {x} ^ {\prime}) \int_ {0} ^ {t} \frac {\| v (s , \boldsymbol {x}) \|}{1 + \| \boldsymbol {x} \|} \mathrm {d} s \\ \leq t d _ {S} (u, v) (r + 1) ^ {2} C _ {r + 1, r} \int (1 + V (\boldsymbol {x} ^ {\prime})) \nu (\mathrm {d} \boldsymbol {x} ^ {\prime}). \tag {66} \\ \end{array}
$$
Finally applying Assumption C.1 once again, we have
$$
\begin{array}{l} \left\| \nabla V \left(u (s, \boldsymbol {x} ^ {\prime})\right) - \nabla V \left(v (s, \boldsymbol {x} ^ {\prime})\right) \right\| \cdot \left\| v (s, \boldsymbol {x} ^ {\prime}) \right\| \\ = (1 + \| \boldsymbol {x} ^ {\prime} \|) \| \nabla V (u (s, \boldsymbol {x} ^ {\prime})) - \nabla V (v (s, \boldsymbol {x} ^ {\prime})) \| \cdot \frac {\| v (s , \boldsymbol {x} ^ {\prime}) \|}{1 + \| \boldsymbol {x} ^ {\prime} \|} \\ \leq (r + 1) \left(1 + \| \boldsymbol {x} ^ {\prime} \|\right) \max _ {\lambda \in [ 0, 1 ]} \left\| \nabla^ {2} V \left(\lambda u (s, \boldsymbol {x} ^ {\prime}) + (1 - \lambda) v (s, \boldsymbol {x} ^ {\prime})\right) \right\| \cdot \left\| u (s, \boldsymbol {x} ^ {\prime}) - v (s, \boldsymbol {x} ^ {\prime}) \right\| \\ = (r + 1) \left(1 + \| \boldsymbol {x} ^ {\prime} \|\right) ^ {2} \max  _ {\lambda \in [ 0, 1 ]} \left\| \nabla^ {2} V \left(\lambda u (s, \boldsymbol {x} ^ {\prime}) + (1 - \lambda) v (s, \boldsymbol {x} ^ {\prime})\right) \right\| \frac {\left\| u (s , \boldsymbol {x} ^ {\prime}) - v (s , \boldsymbol {x} ^ {\prime}) \right\|}{1 + \| \boldsymbol {x} ^ {\prime} \|} \\ \leq (r + 1) C _ {r + 1, r} \left(1 + V \left(\boldsymbol {x} ^ {\prime}\right)\right) d _ {S} (u, v), \\ \end{array}
$$
where in  $(\ast)$  we have used the fact that  $\lambda u + (1 - \lambda)v \in S_r$ , and thus,  $\lambda u + (1 - \lambda)v$  also satisfies the inequality (3.3), which enables us to apply Assumption C.1.
Thus, Term IV is also bounded from above by (the same as the upper bound of Term III)
$$
\left. \mathbf {I V} / (1 + \| \boldsymbol {x} \|) \leq t d _ {S} (u, v) (r + 1) ^ {2} C _ {r + 1, r} \int (1 + V (\boldsymbol {x} ^ {\prime})) \nu (\mathrm {d} \boldsymbol {x} ^ {\prime}). \right. \tag {67}
$$
Now combining (64)-(67), we conclude that  $\mathcal{F}$  is a contraction on  $S_{r}$  when  $T_{0}$  is small enough. By the contraction mapping theorem,  $\mathcal{F}$  has a unique fixed point  $X(\cdot ,\cdot ,\nu)\in S_r$  , which solves (63). After defining  $\mu_t = X(t,\cdot ,\nu)_{\#}\nu$  , one sees that  $X(t,x,\nu)$  solves (62) in the small time interval  $[0,T_0]$
Now we proceed to the second step of extending the local solution. Define
$$
\tau := \sup  \left\{t \in \mathbb {R} ^ {+} \cup \{\infty \}: (6 2) \text {h a s a (u n i q u e) s o l u t i o n o n} [ 0, t) \right\}.
$$