SlowGuess/NeurIPS_2023_OCR huggingface dataset:
If  $\tau = \infty$ , then we have a global solution. Otherwise suppose  $\tau < \infty$ . After examining the bounds we have established in the previous step, we can see that supposing the local solution exists at some time  $T_0$ , it may be extended beyond  $T_0$  as long as the quantity
$$
\| \mu_ {t} \| _ {\mathcal {P} _ {V} (\mathbb {R} ^ {d})} = \int_ {\mathbb {R} ^ {d}} (1 + V (X (t, \boldsymbol {x}, \nu))) \nu (\mathrm {d} \boldsymbol {x})
$$
is finite at time  $T_0$ . We therefore establish an upper bound on this quantity.
$$
\begin{array}{l} \partial_ {t} \int_ {\mathbb {R} ^ {d}} (1 + V (X (t, \boldsymbol {x}, \nu))) \nu (\mathrm {d} \boldsymbol {x}) \\ = - \int_ {\mathbb {R} ^ {d}} \nabla V (X (t, \boldsymbol {x}, \nu)) ^ {\top} X (t, \boldsymbol {x}, \nu) \nu (\mathrm {d} \boldsymbol {x}) \\ - \int_ {\mathbb {R} ^ {d}} \int_ {\mathbb {R} ^ {d}} \nabla V (X (s, \boldsymbol {x}, \nu)) ^ {\top} \nabla V (X (s, \boldsymbol {x} ^ {\prime}, \nu)) X (s, \boldsymbol {x} ^ {\prime}, \nu) ^ {\top} X (s, \boldsymbol {x}, \nu) \nu (\mathrm {d} \boldsymbol {x} ^ {\prime}) \nu (\mathrm {d} \boldsymbol {x}) \\ \leq - \int_ {\mathbb {R} ^ {d}} \nabla V (X (t, \boldsymbol {x}, \nu)) ^ {\top} X (t, \boldsymbol {x}, \nu) \nu (\mathrm {d} \boldsymbol {x}) \\ \leq C _ {1, 0} \int_ {\mathbb {R} ^ {d}} (1 + V (X (t, \boldsymbol {x}, \nu))) \nu (\mathrm {d} \boldsymbol {x}). \\ \end{array}
$$
The last inequality follows from Assumption C.1. Therefore, by Gronwall's inequality we get
$$
\| \mu_ {t} \| _ {\mathcal {P} _ {V} (\mathbb {R} ^ {d})} = \int_ {\mathbb {R} ^ {d}} (1 + V (X (t, \boldsymbol {x}, \nu))) \nu (\mathrm {d} \boldsymbol {x}) \leq e ^ {C _ {1, 0} t} \int_ {\mathbb {R} ^ {d}} (1 + V (\boldsymbol {x})) \nu (\mathrm {d} \boldsymbol {x}) = e ^ {C _ {1, 0} t} \| \nu \| _ {\mathcal {P} _ {V} (\mathbb {R} ^ {d})},
$$
holds for all  $t \in [0, \tau)$ . Next we show an upper bound on  $\|X(t, \boldsymbol{x}, \nu)\|$ . We derive that
$$
\begin{array}{l} \partial_ {t} \| X (t, \boldsymbol {x}, \nu) \| ^ {2} = 2 X (t, \boldsymbol {x}, \nu) ^ {\top} \dot {X} (t, \boldsymbol {x}, \nu) \\ = - 2 X (t, \boldsymbol {x}, \nu) ^ {\top} \left(I + \int_ {\mathbb {R} ^ {d}} \nabla V \left(X (t, \boldsymbol {x} ^ {\prime}, \nu)\right) X (t, \boldsymbol {x} ^ {\prime}, \nu) ^ {\top} \nu (\mathrm {d} \boldsymbol {x} ^ {\prime})\right) X (t, \boldsymbol {x}, \nu) \\ \leq 2 \| X (t, \boldsymbol {x}, \nu) \| ^ {2} \left(1 + C _ {1, 0} \int_ {\mathbb {R} ^ {d}} (1 + V (X (t, \boldsymbol {x} ^ {\prime}, \nu))) \nu (\mathrm {d} \boldsymbol {x} ^ {\prime})\right) \\ \leq 2 \| X (t, \boldsymbol {x}, \nu) \| ^ {2} \left(1 + C _ {1, 0} e ^ {C _ {1, 0} t} \| \nu \| _ {\mathcal {P} _ {V} (\mathbb {R} ^ {d})}\right). \\ \end{array}
$$
Again the first inequality follows from Assumption C.1. Thus, by Gronwall's inequality we have
$$
\left\| X (t, \boldsymbol {x}, \nu) \right\| \leq \exp \left(t + \left(e ^ {C _ {1, 0} t} - 1\right) \| \nu \| _ {\mathcal {P} _ {\Omega} (\mathbb {R} ^ {d})}\right) \| \boldsymbol {x} \|,
$$
for all  $t\in [0,\tau)$ . This also implies that
$$
\begin{array}{l} \| \dot {X} (t, \boldsymbol {x}, \nu) \| \\ \leq \left(1 + C _ {1, 0} e ^ {C _ {1, 0} t} \| \nu \| _ {\mathcal {P} _ {V} (\mathbb {R} ^ {d})}\right) \| X (t, \boldsymbol {x}, \nu) \| \\ \leq \left(1 + C _ {1, 0} e ^ {C _ {1, 0} t} \| \nu \| _ {\mathcal {P} _ {V} (\mathbb {R} ^ {d})}\right) \exp \left(t + \left(e ^ {C _ {1, 0} t} - 1\right) \| \nu \| _ {\mathcal {P} _ {\Omega} (\mathbb {R} ^ {d})}\right) \| \boldsymbol {x} \|, \\ \end{array}
$$
for all  $t\in [0,\tau)$
Next we extend the solution to  $t \in [0, \tau]$ . For this purpose, we first prove that for any sequence  $\{t_i\}_{i=1}^{\infty}$  such that  $0 < t_1 < t_2 < \dots < \tau$  and  $\lim_{i \to \infty} t_i = \tau$ , the sequence of functions  $\{X(t_i, \cdot, \nu)\}$  is Cauchy in  $Y$ . Then by completeness of  $Y$ , there is a limiting function for the sequence, and it is straightforward to see that such function is unique (does not rely on the choice of the sequence  $\{t_i\}_{i=1}^{\infty}$ ).
In fact, we check that for any  $i < j$
$$
\begin{array}{l} \left\| X \left(t _ {j}, \boldsymbol {x}, \nu\right) - X \left(t _ {i}, \boldsymbol {x}, \nu\right) \right\| \\ = \left(t _ {j} - t _ {i}\right) \int_ {0} ^ {1} \dot {X} \left(\lambda t _ {j} + (1 - \lambda) t _ {i}, \boldsymbol {x}, \nu\right) d \lambda \\ \leq (t_{j} - t_{i})\sup_{t\in [t_{i},t_{j}]}\| \dot{X} (t,\boldsymbol {x},\nu)| \\ \leq \left(t _ {j} - t _ {i}\right) \left(1 + C _ {1, 0} e ^ {C _ {1, 0} \tau} \| \nu \| _ {\mathcal {P} _ {V} (\mathbb {R} ^ {d})}\right) \exp \left(\tau + \left(e ^ {C _ {1, 0} \tau} - 1\right) \| \nu \| _ {\mathcal {P} _ {\Omega} (\mathbb {R} ^ {d})}\right) \| \boldsymbol {x} \|. \\ \end{array}
$$
Thus, for any  $\epsilon > 0$  there exists  $N > 0$  such that for any  $j > i > N$ , we have
$$
d _ {Y} (X (t _ {j}, \boldsymbol {x}, \nu), X (t _ {i}, \boldsymbol {x}, \nu)) = \sup  _ {\boldsymbol {x}} \frac {\| X (t _ {j} , \boldsymbol {x} , \nu) - X (t _ {i} , \boldsymbol {x} , \nu) \|}{1 + \| \boldsymbol {x} \|} <   \epsilon .
$$
In other words,  $\{X(t_i,\cdot ,\nu)\}$  is a Cauchy sequence in  $Y$
Now we know that (62) has a unique solution on  $[0, \tau]$ . Since  $\|\mu_{\tau}\|_{\mathcal{P}_V(\mathbb{R}^d)} < \infty$ , we can further find a unique solution of (62) on  $[\tau, \tau + T_0]$  for some  $T_0$  small enough, which contradicts the definition of  $\tau$ . Therefore, we conclude that  $\tau = \infty$ , and (62) has a unique global solution. Finally, thanks to the integral formulation (63)  $\dot{X}$  is continuous on  $[0, \infty) \times \mathbb{R}^d$ . The proof is complete.
Proof of Theorem C.2. Given  $\nu$ , let  $X(t,x,\nu)$  be the mean field characteristic flow defined in (62), and let  $\rho_t = X(t,\cdot,\nu)_{\#}\nu$ . Note that (2) can be rewritten as
$$
\dot {\rho} _ {t} + \nabla \cdot \left(\rho_ {t} U [ \rho_ {t} ]\right) = 0,
$$
where  $U[\rho ]$  is the vector field given by
$$
U [ \rho ] (\boldsymbol {x}) = - \boldsymbol {x} - \int_ {\mathbb {R} ^ {d}} \nabla V (\boldsymbol {y}) \boldsymbol {y} ^ {\top} \boldsymbol {x} \nu (\mathrm {d} \boldsymbol {y}) = - \boldsymbol {x} - \int_ {\mathbb {R} ^ {d}} \rho (\boldsymbol {y}) \nabla V (\boldsymbol {y}) \boldsymbol {y} ^ {\top} \boldsymbol {x} \mathrm {d} \boldsymbol {y}.
$$
Then  $\rho_{t}$  is a weak solution to (2) in the sense that
$$
\sup  _ {t \in [ 0, T ]} \| \rho_ {t} \| _ {\mathcal {P} _ {V}} <   \infty , \quad \forall T > 0,
$$
$$
\int_ {0} ^ {\infty} \int_ {\mathbb {R} ^ {d}} \left(\dot {\phi} (t, \boldsymbol {x}) + \nabla \phi (t, \boldsymbol {x}) ^ {\top} U [ \rho_ {t} ] (\boldsymbol {x})\right) \rho_ {t} (\mathrm {d} \boldsymbol {x}) \mathrm {d} t + \int_ {\mathbb {R} ^ {d}} \phi (0, \boldsymbol {x}) \nu (\mathrm {d} \boldsymbol {x}) = 0
$$
holds for all  $\phi \in \mathcal{C}_0^\infty ([0,\infty)\times \mathbb{R}^d)$ . This is either directly checked or follows immediately from Theorem 5.34 in [83].
By Lemma H.2 there exists some constant  $C_1$  such that
$$
\| \rho_ {t} \| _ {\mathcal {P} _ {V}} \leq \| \nu \| _ {\mathcal {P} _ {V}}.
$$
Suppose that  $\nu \in \mathcal{P}^p (\mathbb{R}^d)\cap \mathcal{P}_V(\mathbb{R}^d)$ . As shown in the proof of Lemma H.2, the map  $X(t,\boldsymbol {x},\nu)$  is an element of the space  $Y$  with  $d_{Y}(X(t,\boldsymbol {x},\nu),\boldsymbol {x})\leq C_{1}e^{C_{1}t}$ . Therefore, since
$$
\| X (t, \boldsymbol {x}, \nu) \| ^ {p} \leq 2 ^ {p} \| \boldsymbol {x} \| ^ {p} + 2 ^ {p} (1 + \| \boldsymbol {x} \|) ^ {p} d _ {Y} (X (t, \boldsymbol {x}, \nu), \boldsymbol {x}) ^ {p},
$$
$$
\| \rho_ {t} \| _ {\mathcal {P} ^ {p}} = \int_ {\mathbb {R} ^ {d}} \| \boldsymbol {y} \| ^ {p} \rho_ {t} (\mathrm {d} \boldsymbol {y}) = \int_ {\mathbb {R} ^ {d}} \| X (t, \boldsymbol {x}, \nu) \| ^ {p} \nu (\mathrm {d} \boldsymbol {x}) \leq e ^ {C _ {2} t} \| \nu \| _ {\mathcal {P} ^ {p}}
$$
for some constant  $C_2 > 0$  and  $\rho_t\in \mathcal{P}^p (\mathbb{R}^d)\cap \mathcal{P}_V(\mathbb{R}^d)$
We now explain that the uniqueness of the weak solution follows from the uniqueness of the mean field characteristic flow. Suppose  $q \in \mathcal{C}([0,T],\mathcal{P}_V(\mathbb{R}^d))$  is another weak solution to (2). By definition of the weak solution, the vector field  $(t,\pmb {x})\mapsto U[q_t](\pmb {x})$  is bounded over  $[0,T]\times \mathbb{R}^d$ , continuous in  $t$  and Lipschitz continuous in  $\pmb{x}$ . Then we can define a continuous family of maps  $\widetilde{X} (t,\cdot ,\nu)$  by
$$
\dot {\widetilde {X}} = U [ q _ {t} ] (\widetilde {X})
$$
$$
\widetilde {X} (0, \boldsymbol {x}, \nu) = \boldsymbol {x}
$$
And the measure  $\widetilde{q}_t = \widetilde{X}(t, \cdot, \nu)_{\#} \nu$  is a weak solution to the transport equation
$$
\dot {\widetilde {q}} _ {t} + \nabla \cdot (\widetilde {q} _ {t} U [ q _ {t} ] (\boldsymbol {x})) = 0
$$
with initial condition  $\widetilde{q}_0 = \nu = q_0$ . Uniqueness of the solution to this linear equation implies that  $\widetilde{q}_t = q_t$ . Thus, we have  $\widetilde{X}(t,\cdot,\nu)_{\#}\nu = q_t$ . In other words,  $\widetilde{X}(t,\boldsymbol{x},\nu)$  is the mean field characteristic flow for  $\nu$ . Uniqueness of the characteristic flow implies that  $\widetilde{X} = X$ , and hence  $q_t = \rho_t$ . Thus, we conclude that the weak solution is unique.
Lastly, we show the regularity result: If  $\nu$  has a density  $\rho_0(\pmb{x})\geq 0$ , then  $\rho_{t}$  also has a density. Furthermore, if  $\rho_0\in \mathcal{H}^k (\mathbb{R}^d)$  for some  $k$ , then we have  $\rho_t\in \mathcal{H}^k (\mathbb{R}^d)$ .
Note that we have already proven that  $\rho_t\in \mathcal{C}\left([0,T],\mathcal{P}_V\right)$  and
$$
\| \rho_ {t} \| _ {\mathcal {P} _ {V}} \leq e ^ {C t} \| \rho_ {0} \| _ {\mathcal {P} _ {V}}, \quad t \geq 0.
$$
$$
\begin{array}{l} U [ \rho ] (\boldsymbol {x}) = - \boldsymbol {x} - \int_ {\mathbb {R} ^ {d}} \nabla V (\boldsymbol {y}) \boldsymbol {y} ^ {\top} \boldsymbol {x} d \mu (\boldsymbol {y}) \\ = - \boldsymbol {x} + \int_ {\mathbb {R} ^ {d}} V (\boldsymbol {y}) \boldsymbol {x} \mathrm {d} \mu (\boldsymbol {y}) \\ = \int_ {\mathbb {R} ^ {d}} (V (\boldsymbol {y}) - 1) \mathrm {d} \mu (\boldsymbol {y}) \boldsymbol {x}, \\ \end{array}
$$
$$
| U [ \rho_ {t} ] (\boldsymbol {x}) | \leq e ^ {C t} \| \rho_ {0} \| _ {\mathcal {P} _ {V}} \| \boldsymbol {x} \|,
$$
$$
\| \nabla U [ \rho_ {t} ] (\pmb {x}) \| \leq e ^ {C t} \| \rho_ {0} \| _ {\mathcal {P} _ {V}},
$$
$$
D _ {\boldsymbol {x}} ^ {j} U [ \rho_ {t} ] (\boldsymbol {x}) = 0 \text {f o r} j = 2, \dots , k + 1.
$$
Thus,  $U(t,\pmb {x})\coloneqq U[\rho_t](\pmb {x})\in \mathcal{C}\left([0,T],\mathcal{C}_b^{k + 1}(\mathbb{R}^d)\right)$  where  $\mathcal{C}_b^{k + 1}(\mathbb{R}^d)$  is the space of continuous functions with bounded  $(k + 1)$ -th order derivatives. Let  $\Phi_t(\pmb {x}) = X(t,\pmb {x},\nu)$  denote the characteristic flow. Since  $\Phi_t$  satisfies the ODE system
$$
\partial_ {t} \Phi_ {t} (\boldsymbol {x}) = U [ \rho_ {t} ] (\Phi_ {t} (\boldsymbol {x})),
$$
from the regularity theory of ODE systems (see Chapter 2 of [79]) we know that both the map  $\pmb{x} \mapsto \Phi_t$  and its inverse  $\Phi_t^{-1}$  are  $\mathcal{C}^k$ . Therefore, if  $\rho_0$  has a density, then  $\rho_t$  also has a density and it is given by
$$
\rho_ {t} (\pmb {x}) = \left(\Phi_ {t}\right) _ {\#} \rho_ {0} = \rho_ {0} (\Phi_ {t} ^ {- 1} (\pmb {x})) \exp \left(- \int_ {0} ^ {t} (\nabla_ {\pmb {x}} \cdot U [ \rho_ {s} ]) (\Phi_ {s} \circ \Phi_ {t} ^ {- 1} (\pmb {x})) \mathrm {d} s\right).
$$
Moreover, since  $\rho$  satisfies
$$
\dot {\rho} _ {t} = - \nabla \cdot \left(\rho_ {t} U [ \rho_ {t} ]\right)
$$
with the vector field  $U(t, \boldsymbol{x}) \in \mathcal{C}\left([0, T], \mathcal{C}_b^{k+1}(\mathbb{R}^d)\right)$ , it follows from Lemma 2.8 of [44] that
$$
\rho \in \mathcal {C} ([ 0, T ], \mathcal {H} ^ {k} (\mathbb {R} ^ {d}))
$$
for any  $T\geq 0$
Proof of Theorem C.3. We show that the particle system (13) is well-posed and that the empirical measure is a weak solution to the mean field PDE. We introduce the function
$$
H _ {N} (X _ {t}) = \frac {1}{N} \sum_ {i = 1} ^ {N} V (\boldsymbol {x} _ {i} ^ {(t)}) + 1.
$$
Since  $V$  is  $\mathcal{C}^1$  hence locally Lipschitz, by Picard-Lindelöf theorem the problem (13) has a unique solution up to some time  $T_0 > 0$ . Intuitively we only need to show that the solution does not blow up at any finite time. We claim that for some constant  $C$ ,
$$
H _ {N} \left(X _ {t}\right) \leq H _ {N} \left(X _ {0}\right) \cdot e ^ {C t}. \tag {68}
$$
To establish this, we first differentiate  $V(x_{i}(t))$  with respect to  $t$  and sum over  $i$ :
$$
\begin{array}{l} \partial_ {t} \left(\frac {1}{N} \sum_ {i = 1} ^ {N} V \left(\boldsymbol {x} _ {i} ^ {(t)}\right)\right) \\ = - \frac {1}{N} \sum_ {i = 1} ^ {N} \nabla V \left(\boldsymbol {x} _ {i} ^ {(t)}\right) ^ {\top} \boldsymbol {x} _ {i} ^ {(t)} - \frac {1}{N ^ {2}} \sum_ {i, j = 1} ^ {N} \boldsymbol {x} _ {i} ^ {(t)} ^ {\top} \boldsymbol {x} _ {j} ^ {(t)} \nabla V \left(\boldsymbol {x} _ {i} ^ {(t)}\right) ^ {\top} \nabla V \left(\boldsymbol {x} _ {j} ^ {(t)}\right) \\ \leq C _ {1, 0} \left(\frac {1}{N} \sum_ {i = 1} ^ {N} V \left(\boldsymbol {x} _ {i} ^ {(t)}\right) + 1\right). \\ \end{array}
$$
Note that here we have used Assumption C.1. By Gronwall's inequality, (68) holds.
Now to be rigorous, once again we define
$$
\tau := \sup  \left\{t \in \mathbb {R} ^ {+} \cup \{\infty \}: (1 3) \text {h a s a (u n i q u e) s o l u t i o n o n} [ 0, t) \right\}.
$$
If  $\tau <  \infty$  , we define
$$
\boldsymbol {x} _ {i} ^ {(\tau)} := \lim  _ {t \nearrow \tau^ {-}} \boldsymbol {x} _ {i} ^ {(t)}.
$$
Then (13) has a unique solution on  $[0, \tau]$ . Again by Picard-Lindelöf theorem, there exists some  $\epsilon > 0$  such that (13) has a unique solution on  $[\tau, \tau + \epsilon]$ , which contradicts the definition of  $\tau$ . Thus, we conclude that  $\tau = \infty$ , which means that there is a global unique solution to (13).
Having established the well-posedness of the finite particle system, it now follows from the definition of the characteristic flow  $X\left(t,\pmb {x},\mu_0^N\right)$  that
$$
\boldsymbol {x} _ {i} ^ {(t)} = X \left(t, \boldsymbol {x} _ {i} ^ {(t)}, \mu_ {0} ^ {N}\right)
$$
$$
\mu_ {t} ^ {N} (\mathrm {d} \boldsymbol {x}) = \left(X \left(t, \cdot , \mu_ {0} ^ {N}\right)\right) _ {\#} \mu_ {0} ^ {N}.
$$
Similar to the proof of Theorem C.2, we conclude that  $\mu_t^N$  is a weak solution to the mean field PDE (2).
Finally, we show Theorem C.5.
Proof of Theorem C.5. Recall that  $p = q' = q / (q - 1)$ . By assumption  $\| \nu_i \|_{\mathcal{P}^p} \leq R < \infty$  and the fact that  $\mathcal{P}^p(\mathbb{R}^d) \subset \mathcal{P}_V(\mathbb{R}^d)$ , we know that there exists  $C > 0$  depending on  $R$  such that
$$
\| \nu_ {i} \| _ {\mathcal {P} _ {V}} \leq C <   \infty .
$$
From the proof of Theorem C.2 and Definition H.1, we know that the weak solutions  $\mu_{i,t}$  take the form
$$
\mu_ {i, t} = (X (t, \cdot , \nu_ {i})) _ {\#}, \quad i = 1, 2.
$$
Now we bound  $\mathcal{W}_p^p (\mu_{1,t},\mu_{2,t})$  using  $\mathcal{W}_p^p (\nu_1,\nu_2)$ . Let  $\pi^0$  be a coupling between  $\nu_{1}$  and  $\nu_{2}$ . For  $\delta >0$  define  $\phi_{\delta}(\boldsymbol {x})\coloneqq \frac{1}{p} (\| \boldsymbol {x}\| +\delta)^{p / 2}$  to be an approximation to  $\frac{1}{p}\| \pmb {x}\| ^p$ . Given any two points  $\pmb{x}_1,\pmb{x}_2\in \mathbb{R}^d$ , we have that
$$
\begin{array}{l} \partial_ {t} \phi_ {\delta} (X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2})) \\ = - \nabla \phi_ {\delta} \left(X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2})\right) ^ {\top} \\ \left\{\left(X \left(t, \boldsymbol {x} _ {1}, \nu_ {1}\right) - X \left(t, \boldsymbol {x} _ {2}, \nu_ {2}\right)\right) \right. \\ + \left(\int_ {\mathbb {R} ^ {2 d}} \nabla V \left(X \left(t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1}\right)\right) X \left(t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1}\right) ^ {\top} X \left(t, \boldsymbol {x} _ {1}, \nu_ {1}\right) \nu_ {1} \left(\mathrm {d} \boldsymbol {x} _ {1} ^ {\prime}\right) \right. \\ \left. \left. - \int_ {\mathbb {R} ^ {2 d}} \nabla V \left(X \left(t, \boldsymbol {x} _ {2} ^ {\prime}, \nu_ {2}\right)\right) X \left(t, \boldsymbol {x} _ {2} ^ {\prime}, \nu_ {2}\right) ^ {\top} X \left(t, \boldsymbol {x} _ {2} ^ {\prime}, \nu_ {2}\right) \nu_ {2} \left(\mathrm {d} \boldsymbol {x} _ {2} ^ {\prime}\right)\right) \right\} \\ = - \nabla \phi_ {\delta} \left(X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2})\right) ^ {\top} \\ \left\{(X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2})) \right. \\ + \int_ {\mathbb {R} ^ {2 d}} \nabla V (X (t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1})) X (t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1}) ^ {\top} (X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2})) \pi^ {0} (\mathrm {d} \boldsymbol {x} _ {1} ^ {\prime} \mathrm {d} \boldsymbol {x} _ {2} ^ {\prime}) \\ + \int_ {\mathbb {R} ^ {2 d}} \nabla V (X (t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1})) (X (t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2} ^ {\prime}, \nu_ {2})) ^ {\top} X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \pi^ {0} (\mathrm {d} \boldsymbol {x} _ {1} ^ {\prime} \mathrm {d} \boldsymbol {x} _ {2} ^ {\prime}) \\ \left. + \int_ {\mathbb {R} ^ {2 d}} \left(\nabla V \left(X (t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1})\right) - \nabla V \left(X (t, \boldsymbol {x} _ {2} ^ {\prime}, \nu_ {2})\right)\right) X (t, \boldsymbol {x} _ {2} ^ {\prime}, \nu_ {2}) ^ {\top} X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \pi^ {0} (\mathrm {d} \boldsymbol {x} _ {1} ^ {\prime} \mathrm {d} \boldsymbol {x} _ {2} ^ {\prime}) \right\} \\ =: I _ {1} + I _ {2} + I _ {3} + I _ {4}. \\ \end{array}
$$
Below we bound  $I_{i}$  individually. First, noticing that
$$
\left\| \nabla \phi_ {\delta} (\boldsymbol {x}) \right\| = \left\| \left(\| \boldsymbol {x} \| ^ {2} + \delta\right) ^ {p / 2 - 1} \boldsymbol {x} \right\| \leq \| \boldsymbol {x} \| ^ {p - 1}, \tag {69}
$$
$$
I _ {1} \leq \| X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \| ^ {p}. \tag {70}
$$
Next we bound  $I_{2}$ :
$$
\begin{array}{l} I _ {2} \leq \| X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \| ^ {p} \left\| \int_ {\mathbb {R} ^ {2 d}} \nabla V \left(X \left(t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1}\right)\right) X \left(t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1}\right) ^ {\top} \nu_ {1} \left(\mathrm {d} \boldsymbol {x} _ {1} ^ {\prime}\right) \right\| \\ \leq \left\| X \left(t, \boldsymbol {x} _ {1}, \nu_ {1}\right) - X \left(t, \boldsymbol {x} _ {2}, \nu_ {2}\right) \right\| ^ {p}. (71) \\ \left(\int_ {\mathbb {R} ^ {2 d}} \| \nabla V (X (t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1})) \| ^ {q} \nu_ {1} (\mathrm {d} \boldsymbol {x} _ {1} ^ {\prime})\right) ^ {1 / q} \left(\int_ {\mathbb {R} ^ {2 d}} \| X (t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1}) \| ^ {p} \nu_ {1} (\mathrm {d} \boldsymbol {x} _ {1} ^ {\prime})\right) ^ {1 / p} \\ \stackrel {b} {\leq} \| X (t, \pmb {x} _ {1}, \nu_ {1}) - X (t, \pmb {x} _ {2}, \nu_ {2}) \| ^ {p} C _ {V} ^ {1 / q} \| \mu_ {1, t} \| _ {\mathcal {P} _ {V}} ^ {1 / q} \cdot \| \mu_ {1, t} \| _ {\mathcal {P} _ {V}} \\ \leq C _ {V} ^ {1 / q} e ^ {\left(C _ {1} / q + C _ {2}\right) t} \| \nu_ {1} \| _ {\mathcal {P} _ {V}} ^ {1 / q} \cdot \| \nu_ {1} \| _ {\mathcal {P} ^ {p}} \| X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \| ^ {p}. (72) \\ \end{array}
$$
Here we have applied Hölder's inequality in  $a$  and Theorem C.2 in  $c$ . The inequality  $b$  is due to Assumption C.4 and the definitions of the  $\mathcal{P}_V$ -norm and  $\mathcal{P}^p$ -norm.
Similarly for  $I_{3}$  we use Hölder's inequality again and get
$$
\begin{array}{l} I _ {3} \leq \| X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \| ^ {p - 1} \cdot \| X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \| \cdot \\ \int_ {\mathbb {R} ^ {2 d}} \| \nabla V (X (t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1})) \| \cdot \| X (t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2} ^ {\prime}, \nu_ {2}) \| \pi^ {0} (\mathrm {d} \boldsymbol {x} _ {1} ^ {\prime} \mathrm {d} \boldsymbol {x} _ {2} ^ {\prime}) \\ \leq \left\| X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \right\| ^ {p - 1} \cdot \left\| X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \right\| \cdot \\ \left(\int_ {\mathbb {R} ^ {2 d}} \| \nabla V (X (t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1})) \| ^ {q} \nu_ {1} (\mathrm {d} \boldsymbol {x} _ {1} ^ {\prime})\right) ^ {1 / q} \left(\int_ {\mathbb {R} ^ {2 d}} \| X (t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2} ^ {\prime}, \nu_ {2}) \| ^ {p} \pi^ {0} (\mathrm {d} \boldsymbol {x} _ {1} ^ {\prime} \mathrm {d} \boldsymbol {x} _ {2} ^ {\prime})\right) ^ {1 / p} \\ \leq C _ {V} ^ {1 / q} e ^ {(C _ {1} / q + C _ {2}) t} \| \nu_ {1} \| _ {\mathcal {P} _ {V}} ^ {1 / q} \| X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \| ^ {p - 1} \| X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \| \cdot \\ \left(\int_ {\mathbb {R} ^ {2 d}} \left\| X \left(t, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1}\right) - X \left(t, \boldsymbol {x} _ {2} ^ {\prime}, \nu_ {2}\right) \right\| ^ {p} \pi^ {0} \left(\mathrm {d} \boldsymbol {x} _ {1} ^ {\prime} \mathrm {d} \boldsymbol {x} _ {2} ^ {\prime}\right)\right) ^ {1 / p}. \tag {73} \\ \end{array}
$$
Finally we proceed to bound  $I_4$ . An application of the intermediate value theorem to the difference of  $\nabla V$  yields that
Note that to get  $a$  we have applied (43) of Assumption C.4 and  $b$  is implied by Theorem C.2.
$$
D _ {p} (\pi) (s) := \left(\int_ {\mathbb {R} ^ {2 d}} \| X (s, \boldsymbol {x} _ {1} ^ {\prime}, \nu_ {1}) - X (s, \boldsymbol {x} _ {2} ^ {\prime}, \nu_ {2}) \| ^ {p} \pi (\mathrm {d} \boldsymbol {x} _ {1} ^ {\prime} \mathrm {d} \boldsymbol {x} _ {2} ^ {\prime})\right) ^ {1 / p},
$$
combining (70), (71), (73) and (74) we obtain that
$$
\begin{array}{l} \partial_ {t} \phi_ {\delta} (X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2})) \\ \leq 4 C _ {V} ^ {1 / q} e ^ {(C _ {1} / q + C _ {2}) t} R ^ {(q + 1) / q} \| X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \| ^ {p - 1}. \\ \left(\| X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \| + D _ {p} (\pi^ {0}) (t) \| X (t, \boldsymbol {x} _ {2}, \nu_ {2}) \|\right). \\ \end{array}
$$
Now integrating the inequality above with respect to the coupling  $\pi^0 (\mathrm{d}\pmb {x}_1,\mathrm{d}\pmb {x}_2)$  using the fact that
$$
\begin{array}{l} \int_ {\mathbb {R} ^ {2 d}} \| X (s, \boldsymbol {x} _ {1}, \nu_ {1}) - X (s, \boldsymbol {x} _ {2}, \nu_ {2}) \| ^ {p - 1} \| X (s, \boldsymbol {x} _ {2}, \nu_ {2}) \| \pi^ {0} (\mathrm {d} \boldsymbol {x} _ {1} \mathrm {d} \boldsymbol {x} _ {2}) \\ \leq \left(\int_ {\mathbb {R} ^ {2 d}} \| X (s, \boldsymbol {x} _ {1}, \nu_ {1}) - X (s, \boldsymbol {x} _ {2}, \nu_ {2}) \| ^ {p} \pi^ {0} (\mathrm {d} \boldsymbol {x} _ {1} \mathrm {d} \boldsymbol {x} _ {2})\right) ^ {(p - 1) / p} \left(\int_ {\mathbb {R} ^ {2 d}} \| X (s, \boldsymbol {x} _ {2}, \nu_ {2}) \| ^ {p} \nu_ {2} (\boldsymbol {x} _ {2})\right) ^ {1 / p} \\ \leq e ^ {C _ {2} t} R D _ {p} ^ {p - 1} \left(\pi^ {0}\right) (s). \\ \end{array}
$$
$$
\partial_ {t} \phi_ {\delta} \left(X (t, \pmb {x} _ {1}, \nu_ {1}) - X (t, \pmb {x} _ {2}, \nu_ {2})\right) \leq 8 C _ {V} ^ {1 / q} e ^ {(C _ {1} / q + 2 C _ {2}) t} R ^ {(2 q + 1) / q} D _ {p} ^ {p} (\pi^ {0}) (t) \leq C _ {V, R} e ^ {C T} D _ {p} ^ {p} (\pi^ {0}) (t).
$$
Integrating  $t$  we get
$$
\phi_ {\delta} \left(X (t, \boldsymbol {x} _ {1}, \nu_ {1}) - X (t, \boldsymbol {x} _ {2}, \nu_ {2})\right) \leq \phi_ {\delta} (\boldsymbol {x} _ {1} - \boldsymbol {x} _ {2}) + C _ {V, R} e ^ {C T} \int_ {0} ^ {t} D _ {p} ^ {p} (\pi^ {0}) (s) d s.
$$
Finally letting  $\delta \to 0$  yields
$$
D _ {p} ^ {p} (\pi^ {0}) (t) \leq D _ {p} ^ {p} (\pi^ {0}) (0) + C _ {V, R} e ^ {C T} \int_ {0} ^ {t} D _ {p} ^ {p} (\pi^ {0}) (s) d s.
$$