SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\nabla_ {x} G _ {c} (\rho + \epsilon \psi , b (\rho + \epsilon \psi)) = 0.
$$
Differentiating in  $\epsilon$ , we obtain
$$
\int \delta_ {\rho} \nabla_ {x} G _ {c} [ \rho + \epsilon \psi , b (\rho + \epsilon \psi) ] \psi (z) d z + \nabla_ {x} ^ {2} G _ {c} (\rho + \epsilon \psi , b (\rho + \epsilon \psi)) \int \delta_ {\rho} b [ \rho + \epsilon \psi ] (z) \psi (z) d z = 0. \tag {22}
$$
Next, we explicitly compute all terms involved in (22). Computing the derivatives yields
$$
\nabla_ {x} G _ {c} (\rho , x) = \int \nabla_ {x} f _ {1} (z, x) \mathrm {d} \rho (z) + \int \nabla_ {x} f _ {2} (z, x) \mathrm {d} \bar {\rho} (z) + \beta (x - x _ {0})
$$
$$
\delta_ {\rho} \nabla_ {x} G _ {c} [ \rho , x ] (z) = \nabla_ {x} f _ {1} (z, x)
$$
$$
\nabla_ {x} ^ {2} G _ {c} (\rho , x) = \int \nabla_ {x} ^ {2} f _ {1} (z, x) \mathrm {d} \rho (z) + \int \nabla_ {x} ^ {2} f _ {2} (z, x) \mathrm {d} \bar {\rho} (z) + \beta \mathrm {I} _ {d}.
$$
Note that  $\nabla_x^2 G_c$  is invertible by Assumption 1, which states that  $f_{1}$  and  $f_{2}$  have positive-definite Hessians. Inverting this term and substituting these expressions into (22) for  $\epsilon = 0$  gives
$$
\begin{array}{l} \int \delta_ {\rho} b [ \rho ] (z) \psi (z) d z = - \left[ \beta I _ {d} + \int \nabla_ {x} ^ {2} f _ {1} (z, b (\rho)) d \rho (z) + \int \nabla_ {x} ^ {2} f _ {2} (z, b (\rho)) d \bar {\rho} (z) \right] ^ {- 1} \int \nabla_ {x} f _ {1} (z, b (\rho)) \psi (z) d z \\ = - \int Q (\rho) ^ {- 1} \nabla_ {x} f _ {1} (z, b (\rho)) \psi (z) d z. \\ \end{array}
$$
Finally, the lower bound on  $Q(\rho)$  follows thanks to Assumption 1.
Remark 9. If we include the additional assumption that  $f_{i} \in C^{3}(\mathbb{R}^{d} \times \mathbb{R}^{d};[0,\infty))$  for  $i = 1,2$ , then the Hessian of  $b[\rho]$  is well-defined. More precisely, the Hessian is given by
$$
\frac {\mathrm {d} ^ {2}}{\mathrm {d} \epsilon^ {2}} b [ \rho + \epsilon \psi ] | _ {\epsilon = 0} = Q (\rho) ^ {- 1} \left(\frac {\mathrm {d}}{\mathrm {d} \epsilon} Q (\rho + \epsilon \psi) | _ {\epsilon = 0} + \int \nabla_ {x} ^ {2} f _ {1} (z, b [ \rho ]) \psi (z) \mathrm {d} z\right) Q (\rho) ^ {- 1} u [ \rho , \psi ]
$$
where  $u[\rho, \psi] = \int \nabla_x f_1(z, b[\rho]) \psi(z) \, \mathrm{d}z$  and
$$
\begin{array}{l} \frac {\mathrm {d}}{\mathrm {d} \epsilon} Q _ {i j} (\rho + \epsilon \psi) | _ {\epsilon = 0} = \int \partial_ {x _ {i}} \partial_ {x _ {j}} f _ {1} (z, b [ \rho ]) \psi (z) \mathrm {d} z - \int \partial_ {x _ {i}} \partial_ {x _ {j}} \nabla_ {x} f _ {1} (z, b [ \rho ]) \psi (z) \rho (z) \mathrm {d} z Q (\rho) ^ {- 1} u [ \rho , \psi ] \\ - \int \partial_ {x _ {i}} \partial_ {x _ {j}} \nabla_ {x} f _ {2} (z, b [ \rho ]) \psi (z) \bar {\rho} (z) \mathrm {d} z Q (\rho) ^ {- 1} u [ \rho , \psi ]. \\ \end{array}
$$
Therefore, we can Taylor expand  $b[\rho]$  up to second order and control the remainder term of order  $\epsilon^2$ .
Lemma 17 (First variation of  $G_{b}$ ). The first variation of  $G_{b}$  is given by
$$
\delta_ {\rho} G _ {b} [ \rho ] (z) = h _ {1} (z) + h _ {2} (z) + \beta h _ {3} (z) - \delta_ {\rho} P [ \rho ] (z),
$$
$$
h _ {1} (z) := \frac {\delta}{\delta \rho} \left(\int f _ {1} (\tilde {z}, b (\rho)) \mathrm {d} \rho (\tilde {z})\right) (z) = \left\langle \int \nabla_ {x} f _ {1} (\tilde {z}, b (\rho)) \mathrm {d} \rho (\tilde {z}), \frac {\delta b}{\delta \rho} [ \rho ] (z) \right\rangle + f _ {1} (z, b (\rho)),
$$
$$
h _ {2} (z) := \frac {\delta}{\delta \rho} \left(\int f _ {2} (\tilde {z}, b (\rho)) \mathrm {d} \bar {\rho} (\tilde {z})\right) (z) = \left\langle \int \nabla_ {x} f _ {2} (\tilde {z}, b (\rho)) \mathrm {d} \bar {\rho} (\tilde {z}), \frac {\delta b}{\delta \rho} [ \rho ] (z) \right\rangle ,
$$
$$
h _ {3} (z) := \frac {1}{2} \frac {\delta}{\delta \rho} \| b (\rho) - x _ {0} \| ^ {2} = \left\langle b (\rho) - x _ {0}, \frac {\delta b}{\delta \rho} [ \rho ] (z) \right\rangle ,
$$
$$
\delta_ {\rho} P [ \rho ] (z) = \alpha \log (\rho (z) / \tilde {\rho} (z)) + (W * \rho) (z).
$$
Proof. We begin with general expressions for Taylor expansions of  $b: \mathcal{P}(\mathbb{R}^d) \to \mathbb{R}^d$  and  $f_i(z, b(\cdot)): \mathcal{P}(\mathbb{R}^d) \to \mathbb{R}$  for  $i = 1, 2$  around  $\rho$ . Let  $\psi \in \mathcal{T}$  with  $\mathcal{T} = \{\psi : \int \psi(z) \mathrm{d}z = 0\}$ . Then
$$
b (\rho + \epsilon \psi) = b (\rho) + \epsilon \int \frac {\delta b}{\delta \rho} [ \rho ] \left(z ^ {\prime}\right) \psi \left(z ^ {\prime}\right) d z ^ {\prime} + O \left(\epsilon^ {2}\right) \tag {23}
$$
$$
f _ {i} (z, b (\rho + \epsilon \psi)) = f _ {i} (z, b (\rho)) + \epsilon \left\langle \nabla_ {x} f _ {i} (z, b (\rho)), \int \frac {\delta b}{\delta \rho} [ \rho ] \left(z ^ {\prime}\right) \psi \left(z ^ {\prime}\right) \mathrm {d} z ^ {\prime} \right\rangle + O \left(\epsilon^ {2}\right). \tag {24}
$$
We compute explicitly each of the first variations:
(i) Using (24), we have
$$
\begin{array}{l} \int \psi (z) h _ {1} (z) \mathrm {d} z = \lim  _ {\epsilon \rightarrow 0} \frac {1}{\epsilon} \left[ \int f _ {1} (z, b (\rho + \epsilon \psi)) (\rho (z) + \epsilon \psi (z)) \mathrm {d} z - \int f _ {1} (z, b (\rho)) \rho (z) \mathrm {d} z \right] \\ = \left\langle \int \nabla_ {x} f _ {1} (z, b (\rho)) \mathrm {d} \rho (z), \int \frac {\delta b (\rho)}{\delta \rho} [ \rho ] \left(z ^ {\prime}\right) \psi \left(z ^ {\prime}\right) \mathrm {d} z ^ {\prime} \right\rangle + \int f _ {1} (z, b (\rho)) \psi (z) \mathrm {d} z \\ = \int \left\langle \int \nabla_ {x} f _ {1} (z, b (\rho)) \mathrm {d} \rho (z), \frac {\delta b (\rho)}{\delta \rho} [ \rho ] (z ^ {\prime}) \right\rangle \psi (z ^ {\prime}) \mathrm {d} z ^ {\prime} + \int f _ {1} (z, b (\rho)) \psi (z) \mathrm {d} z \\ \Rightarrow h _ {1} (z) = \left\langle \int \nabla_ {x} f _ {1} (\tilde {z}, b (\rho)) \mathrm {d} \rho (\tilde {z}), \frac {\delta b}{\delta \rho} [ \rho ] (z) \right\rangle + f _ {1} (z, b (\rho)). \\ \end{array}
$$
(ii) Similarly, using again (24),
$$
\begin{array}{l} \int \psi (z) h _ {2} (z) d z = \lim  _ {\epsilon \rightarrow 0} \frac {1}{\epsilon} \left[ \int f _ {2} (z, b (\rho + \epsilon \psi)) d \bar {\rho} (z) - \int f _ {2} (z, b (\rho)) \bar {\rho} (z) d z \right] \\ = \int \left\langle \int \nabla_ {x} f _ {2} (\tilde {z}, b (\rho)) d \bar {\rho} (\tilde {z}), \frac {\delta b}{\delta \rho} [ \rho ] (z) \right\rangle \psi (z) d z \\ \Rightarrow h _ {2} (z) = \left\langle \int \nabla_ {x} f _ {2} (\tilde {z}, b (\rho)) \mathrm {d} \bar {\rho} (\tilde {z}), \frac {\delta b}{\delta \rho} [ \rho ] (z) \right\rangle . \\ \end{array}
$$
(iii) Finally, from (23) it follows that
$$
\begin{array}{l} \int \psi (z) h _ {3} (z) \mathrm {d} z = \lim  _ {\epsilon \rightarrow 0} \frac {1}{2 \epsilon} \left[ \langle b (\rho + \epsilon \psi) - x _ {0}, b (\rho + \epsilon \psi) - x _ {0} \rangle - \langle b (\rho) - x _ {0}, b (\rho) - x _ {0} \rangle \right] \\ = \int \left\langle b (\rho) - x _ {0}, \frac {\delta b}{\delta \rho} [ \rho ] (z) \right\rangle \psi (z) d z \\ \Rightarrow h _ {3} (z) = \left\langle b (\rho) - x _ {0}, \frac {\delta b}{\delta \rho} [ \rho ] (z) \right\rangle . \\ \end{array}
$$
Finally, the expression for  $\delta_{\rho}P[\rho ]$  follows by direct computation
Lemma 18. Denote  $G_{b}(\rho) \coloneqq G_{c}(\rho, b(\rho))$  with  $b(\rho)$  given by (5). Then  $\delta_{\rho}G_{b}[\rho] = \delta_{\rho}G_{c}[\rho]|_{x=b(\rho)}$ .
Proof. We start by computing  $\delta_{\rho}G_{c}(\cdot ,x)[\rho ](z)$  for any  $z,x\in \mathbb{R}^d$
$$
\delta_ {\rho} G _ {c} (\cdot , x) [ \rho ] (z) = f _ {1} (z, x) - \delta_ {\rho} P [ \rho ] (z). \tag {25}
$$
Next, we compute  $\delta_{\rho}G_{b}$ . Using Lemma 17, the first variation of  $G_{b}$  is given by
$$
\begin{array}{l} \delta_ {\rho} G _ {b} [ \rho ] (z) = h _ {1} (z) + h _ {2} (z) + \beta h _ {3} (z) - \delta_ {\rho} P [ \rho ] (z) \\ = - \left\langle \left[ \int \nabla_ {x} f _ {1} (\tilde {z}, b (\rho)) \mathrm {d} \rho (\tilde {z}) + \int \nabla_ {x} f _ {2} (\tilde {z}, b (\rho)) \mathrm {d} \bar {\rho} (\tilde {z}) + \beta (b (\rho) - x _ {0}) \right], \delta_ {\rho} b [ \rho ] (z) \right\rangle \\ + f _ {1} (z, b (\rho)) - \delta_ {\rho} P [ \rho ] (z). \\ \end{array}
$$
$$
\nabla_ {x} G _ {c} (\rho , x) = \int \nabla_ {x} f _ {1} (\tilde {z}, x) \mathrm {d} \rho (\tilde {z}) + \int \nabla_ {x} f _ {2} (\tilde {z}, x) \mathrm {d} \bar {\rho} (\tilde {z}) + \beta (x - x _ {0}), \tag {26}
$$
and by the definition of the best response  $b(\rho)$ , we have  $\nabla_{x}G_{x}(\rho ,x)|_{x = b(\rho)} = 0$ . Substituting into the expression for  $\delta_{\rho}G_{b}$  and using (25), we obtain
$$
\delta_ {\rho} G _ {b} [ \rho ] (z) = f _ {1} (z, b (\rho)) - \delta_ {\rho} P [ \rho ] (z) = \delta_ {\rho} G _ {c} (\cdot , x) [ \rho ] (z) \Bigg | _ {x = b (\rho)}.
$$
Lemma 19 (Uniform boundedness of the best response). Let Assumption 1 hold. Then for any  $\rho \in \mathcal{P}(\mathbb{R}^d)$ , we have
$$
\left\| b (\rho) \right\| ^ {2} \leq \left\| x _ {0} \right\| ^ {2} + \frac {2 (a _ {1} + a _ {2})}{\beta}.
$$
Proof. By definition of the best response  $b(\rho)$ , we have
$$
\int \nabla_ {x} f _ {1} (z, b (\rho)) \mathrm {d} \rho_ {t} + \int \nabla_ {x} f _ {2} (z, b (\rho)) \mathrm {d} \bar {\rho} (z) + \beta (b (\rho) - x _ {0}) = 0.
$$
To show that that  $b(\rho)$  is uniformly bounded, we take the inner product of the above expression with  $b(\rho)$  itself
$$
\beta \| b (\rho) \| ^ {2} = \beta x _ {0} \cdot b (\rho) - \int \nabla_ {x} f _ {1} (z, b (\rho)) \cdot b (\rho) d \rho (z) - \int \nabla_ {x} f _ {2} (z, b (\rho)) \cdot b (\rho) d \bar {\rho} (z).
$$
Using Assumption 1 to bound the two integrals, together with using Young's inequality to bound the first term on the right-hand side, we obtain
$$
\beta \| b (\rho) \| ^ {2} \leq \frac {\beta}{2} \| x _ {0} \| ^ {2} + \frac {\beta}{2} \| b (\rho) \| + a _ {1} + a _ {2},
$$
which concludes the proof after rearranging terms.
Lemma 20 (Upper semi-continuity). Let Assumptions 1-3 hold. The functional  $G_{c}:\mathcal{P}(\mathbb{R}^{d})\times \mathbb{R}^{d}\to [-\infty , + \infty ]$  is upper semi-continuous when  $\mathcal{P}(\mathbb{R}^d)\times \mathbb{R}^d$  is endowed with the product topology of the weak-\* topology and the Euclidean topology. Moreover, the functional  $G_{b}:\mathcal{P}(\mathbb{R}^{d})\to [-\infty , + \infty ]$  is upper semi-continuous with respect to the weak-\* topology.
Proof. The functional  $G_{c}:\mathcal{P}(\mathbb{R}^{d})\times \mathbb{R}^{d}\to [-\infty , + \infty ]$  is continuous in the second variable thanks to Assumption 1. Similarly,  $\int f_1(z,x)\mathrm{d}\rho (z) + \int f_2(z,x)\mathrm{d}\bar{\rho} (z)$  is continuous in  $\rho$  thanks to [San15, Proposition 7.1] using the continuity of  $f_{1}$  and  $f_{2}$ . Further,  $-P$  is upper semi-continuous using [Pos75] and [San15, Proposition 7.2] thanks to Assumptions 2 and 3. This concludes the continuity properties for  $G_{c}$ .
The upper semi-continuity of  $G_{b}$  then follows from a direct application of a version of Berge's maximum theorem [AB06, Lemma 16.30]. Let  $R := \| x_0\|^2 + \frac{2(a_1 + a_2)}{\beta} > 0$ . We define  $\varphi : (\mathcal{P}(\mathbb{R}^d), W_2) \to \mathbb{R}^d$  as the correspondence that maps any  $\rho \in \mathcal{P}(\mathbb{R}^d)$  to the closed ball  $\overline{B_R(0)} \subset \mathbb{R}^d$ . Then the graph of  $\varphi$  is  $\operatorname{Gr} \varphi = \mathcal{P}(\mathbb{R}^d) \times \{\overline{B_R(0)}\}$ . With this definition of  $\varphi$ , the range of  $\varphi$  is compact and  $\varphi$  is continuous with respect to weak- $*$  convergence, and so it is in particular upper hemicontinuous. Thanks to Lemma 19, the best response function  $b(\rho)$  is always contained in  $\overline{B_R(0)}$  for any choice of  $\rho \in \mathcal{P}(\mathbb{R}^d)$ . As a result, maximizing  $-G_c(\rho, x)$  in  $x$  over  $\mathbb{R}^d$  for a fixed  $\rho \in \mathcal{P}(\mathbb{R}^d)$  reduces to maximizing it over  $\overline{B_R(0)}$ . Using the notation introduced above, we can restrict  $G_c$  to  $G_c : \operatorname{Gr} \varphi \to \mathbb{R}$  and write
$$
G _ {b} (\rho) := \max  _ {\hat {x} \in \varphi (\rho)} - G _ {c} (\rho , \hat {x}).
$$
Because  $G_{c}(\rho, x)$  is upper semi-continuous when  $\mathcal{P}(\mathbb{R}^2) \times \mathbb{R}^d$  is endowed with the product topology of the weak- $*$  topology and the Euclidean topology, [AB06, Lemma 16.30] guarantees that  $G_{b}(\cdot)$  is upper semi-continuous in the weak- $*$  topology.
Proposition 21. Let  $\alpha, \beta > 0$  and assume Assumptions 1-4 hold with the parameters satisfying  $\alpha \tilde{\lambda} > \Lambda_1$ . Fix  $\rho_0, \rho_1 \in \mathcal{P}(\mathbb{R}^d)$ . Along any geodesic  $(\rho_s)_{s \in [0,1]} \in \mathcal{P}_2(\mathbb{R}^d)$  connecting  $\rho_0$  to  $\rho_1$ , we have for all  $s \in [0,1]$
$$
\frac {\mathrm {d} ^ {2}}{\mathrm {d} s ^ {2}} G _ {b} \left(\rho_ {s}\right) \leq - \lambda_ {b} W _ {1} \left(\rho_ {0}, \rho_ {1}\right) ^ {2}, \quad \lambda_ {b} := \alpha \tilde {\lambda} - \Lambda_ {1},. \tag {27}
$$
As a result, the functional  $G_{b}:\mathcal{P}_{2}(\mathbb{R}^{d})\to [-\infty , + \infty ]$  is uniformly displacement concave with constant  $\lambda_{b} > 0$ .
Proof. Consider any  $\rho_0, \rho_1 \in \mathcal{P}_2(\mathbb{R}^d)$ . Then any  $W_{2}$ -geodesic  $(\rho_s)_{s \in [0,1]}$  connecting  $\rho_0$  with  $\rho_1$  solves the following system of geodesic equations:
$$
\left\{ \begin{array}{l} \partial_ {s} \rho_ {s} + \operatorname {d i v} \left(\rho_ {s} v _ {s}\right) = 0, \\ \partial_ {s} \left(\rho_ {s} v _ {s}\right) + \operatorname {d i v} \left(\rho_ {s} v _ {s} \otimes v _ {s}\right) = 0, \end{array} \right. \tag {28}
$$
where  $\rho_s: \mathbb{R}^d \to \mathbb{R}$  and  $v_s: \mathbb{R}^d \mapsto \mathbb{R}^d$ . The first derivative of  $G_b$  along geodesics can be computed explicitly as
$$
\begin{array}{l} \frac {\mathrm {d}}{\mathrm {d} s} G _ {b} \left(\rho_ {s}\right) = \int \nabla_ {z} f _ {1} \left(z, b \left(\rho_ {s}\right)\right) \cdot v _ {s} (z) \rho_ {s} (z) \mathrm {d} z - \frac {\mathrm {d}}{\mathrm {d} s} P \left(\rho_ {s}\right) \\ + \left\langle \left[ \int \nabla_ {x} f _ {1} (z, x) \mathrm {d} \rho_ {s} (z) + \int \nabla_ {x} f _ {2} (z, x) \mathrm {d} \bar {\rho} (z) + \beta (x - x _ {0}) \right] \Big | _ {x = b (\rho_ {s})}, \frac {\mathrm {d}}{\mathrm {d} s} b (\rho_ {s}) \right\rangle . \\ \end{array}
$$
The left-hand side of the inner product is zero by definition of the best response  $b(\rho_s)$  to  $\rho_s$ , see (26). Therefore
$$
\frac {\mathrm {d}}{\mathrm {d} s} G _ {b} \left(\rho_ {s}\right) = \int \nabla_ {z} f _ {1} \left(z, b \left(\rho_ {s}\right)\right) \cdot v _ {s} (z) \rho_ {s} (z) \mathrm {d} z - \frac {\mathrm {d}}{\mathrm {d} s} P \left(\rho_ {s}\right).
$$
Differentiating a second time, using (28) and integration by parts, we obtain
$$
\frac {\mathrm {d} ^ {2}}{\mathrm {d} s ^ {2}} G _ {b} (\rho_ {s}) = L _ {1} (\rho_ {s}) + L _ {2} (\rho_ {s}) - \frac {\mathrm {d} ^ {2}}{\mathrm {d} s ^ {2}} P (\rho_ {s}),
$$
$$
L _ {1} \left(\rho_ {s}\right) := \int \nabla_ {z} ^ {2} f _ {1} (z, b \left(\rho_ {s}\right)) \cdot \left(v _ {s} \otimes v _ {s}\right) \rho_ {s} d z = \int \left\langle v _ {s}, \nabla_ {z} ^ {2} f _ {1} (z, b \left(\rho_ {s}\right)) \cdot v _ {s} \right\rangle \rho_ {s} d z,
$$
$$
L _ {2} \left(\rho_ {s}\right) := \int \frac {\mathrm {d}}{\mathrm {d} s} b \left(\rho_ {s}\right) \cdot \nabla_ {x} \nabla_ {z} f _ {1} (z, b \left(\rho_ {s}\right)) \cdot v _ {s} (z) \rho_ {s} (z) \mathrm {d} z.
$$
From (16), we have that
$$
\frac {\mathrm {d} ^ {2}}{\mathrm {d} s ^ {2}} \tilde {P} (\rho_ {s}) \geq \alpha \tilde {\lambda} W _ {2} (\rho_ {0}, \rho_ {1}) ^ {2},
$$
and thanks to Assumption 4 it follows that
$$
L _ {1} (s) \leq \Lambda_ {1} W _ {2} \left(\rho_ {0}, \rho_ {1}\right) ^ {2}.
$$
This leaves  $L_{2}$  to bound; we first consider the term  $\frac{\mathrm{d}}{\mathrm{d}s} b(\rho_s)$ :
$$
\begin{array}{l} \frac {\mathrm {d}}{\mathrm {d} s} b \left(\rho_ {s}\right) = \int \delta_ {\rho} b \left[ \rho_ {s} \right] (\tilde {z}) \partial_ {s} \rho_ {s} (\mathrm {d} \tilde {z}) = - \int \delta_ {\rho} b \left[ \rho_ {s} \right] (\tilde {z}) \operatorname {d i v} \left(\rho_ {s} v _ {s}\right) \mathrm {d} \tilde {z} \\ = \int \nabla_ {z} \delta_ {\rho} b [ \rho_ {s} ] (\tilde {z}) \cdot v _ {s} (\tilde {z}) \mathrm {d} \rho_ {s} (\tilde {z}). \\ \end{array}
$$
Defining  $u(\rho_s)\in \mathbb{R}^d$  by
$$
u \left(\rho_ {s}\right) := \int \nabla_ {x} \nabla_ {z} f _ {1} (z, b \left(\rho_ {s}\right)) \cdot v _ {s} (z) d \rho_ {s} (z),
$$
using the results from Lemma 16 for  $\nabla_{z}\delta_{\rho}b[\rho_{s}]$ , Assumption 1 and the fact that  $Q(\rho)$  is constant in  $z$  and  $x$ , we have
$$
\begin{array}{l} L _ {2} (\rho_ {s}) = - \iint \left[ Q (\rho_ {s}) ^ {- 1} \nabla_ {x} \nabla_ {z} f _ {1} (\tilde {z}, b (\rho_ {s})) \cdot v _ {s} (\tilde {z}) \right] \cdot \nabla_ {x} \nabla_ {z} f _ {1} (z, b (\rho_ {s})) \cdot v _ {s} (z) d \rho_ {s} (z) d \rho_ {s} (\tilde {z}) \\ = - \left\langle u \left(\rho_ {s}\right), Q \left(\rho_ {s}\right) ^ {- 1} u \left(\rho_ {s}\right) \right\rangle \leq 0 \\ \end{array}
$$
Combining all terms together, we obtain
$$
\frac {\mathrm {d} ^ {2}}{\mathrm {d} s ^ {2}} G _ {b} \left(\rho_ {s}\right) \leq - \left(\alpha \tilde {\lambda} - \Lambda_ {1}\right) W _ {2} \left(\rho_ {0}, \rho_ {1}\right) ^ {2}.
$$
Remark 10. Under some additional assumptions on the functions  $f_{1}$  and  $f_{2}$ , we can obtain an improved convergence rate. In particular, assume that for all  $z, x \in \mathbb{R}^d$ ,
- there exists a constant  $\Lambda_2 \geq \lambda_2 \geq 0$  such that  $\nabla_x^2 f_2(z,x) \preceq \Lambda_2 \mathrm{I}_d$ ;
- there exists a constant  $\sigma \geq 0$  such that  $\| \nabla_x\nabla_zf_1(z,x)\| \geq \sigma$
Then we have  $-Q(\rho_s)^{-1} \preceq -1 / (\beta + \Lambda_1 + \Lambda_2) \mathrm{I}_d$ . Using Lemma 16, we then obtain a stronger bound on  $L_2$  as follows:
$$
\begin{array}{l} L _ {2} \left(\rho_ {s}\right) \leq - \frac {1}{\beta + \Lambda_ {1} + \Lambda_ {2}} \| u \left(\rho_ {s}\right) \| ^ {2} \leq - \frac {1}{\beta + \Lambda_ {1} + \Lambda_ {2}} \int \left\| \nabla_ {x} \nabla_ {z} f _ {1} (z, b \left(\rho_ {s}\right)) \right\| ^ {2} d \rho_ {s} (z) \int \| v _ {s} (z) \| ^ {2} d \rho_ {s} (z) \\ \leq - \frac {\sigma^ {2}}{\beta + \Lambda_ {1} + \Lambda_ {2}} W _ {2} \left(\rho_ {0}, \rho_ {1}\right) ^ {2}. \\ \end{array}
$$
This means we can improve the convergence rate in (27) to  $\lambda_{b} := \alpha \tilde{\lambda} + \frac{\sigma^{2}}{\beta + \Lambda_{1} + \Lambda_{2}} - \Lambda_{1}$ .
Proposition 22 (Ground state). Let Assumptions 1-4 hold for  $\alpha \tilde{\lambda} > \Lambda_1 \geq 0$  and  $\beta > 0$ . Then there exists a unique maximizer  $\rho_*$  for the functional  $G_b$  over  $\mathcal{P}(\mathbb{R}^d)$ , and it satisfies  $\rho_* \in \mathcal{P}_2(\mathbb{R}^d) \cap L^1(\mathbb{R}^d)$  and  $\rho_*$  is absolutely continuous with respect to  $\tilde{\rho}$ .
Proof. Uniqueness of the maximizer (if it exists) is guaranteed by the uniform concavity provided by Lemma 21. To show existence of a maximizer, we use the direct method in the calculus of variations, requiring the following key properties for  $G_{b}$ : (1) boundedness from above, (2) upper semi-continuity, and (3) tightness of any minimizing sequence. To show (1), note that  $\nabla_z^2(f_1(z,x) + \alpha \log \tilde{\rho}(z)) \preceq -(\alpha \tilde{\lambda} - \Lambda_1) I_d$  for all  $z, x \in \mathbb{R}^d \times \mathbb{R}^d$  by Assumptions 2 and 4, and so
$$
f _ {1} (z, x) + \alpha \log \tilde {\rho} (z) \leq c _ {0} (x) - \frac {\left(\alpha \tilde {\lambda} - \Lambda_ {1}\right)}{4} | z | ^ {2} \quad \forall (z, x) \in \mathbb {R} ^ {d} \times \mathbb {R} ^ {d} \tag {29}
$$
with  $c_{0}(x) \coloneqq f_{1}(0, x) + \alpha \log \tilde{\rho}(0) + \frac{1}{\alpha \tilde{\lambda} - \Lambda_{1}} \left\| \nabla_{z}\left[f_{1}(0, x) + \alpha \log \tilde{\rho}(0)\right]\right\|^{2}$ . Therefore,
$$
\begin{array}{l} G _ {b} (\rho) = \int \left[ f _ {1} (z, b (\rho)) + \alpha \log \tilde {\rho} (z) \right] \mathrm {d} \rho (z) + \int f _ {2} (z, b (\rho)) \mathrm {d} \bar {\rho} (z) + \frac {\beta}{2} \| b (\rho) - x _ {0} \| ^ {2} \\ - \alpha \int \rho \log \rho - \int \rho W * \rho \\ \leq c _ {0} (b (\rho)) + \int f _ {2} (z, b (\rho)) \mathrm {d} \bar {\rho} (z) + \frac {\beta}{2} \| b (\rho) - x _ {0} \| ^ {2}. \\ \end{array}
$$
To estimate each of the remaining terms on the right-hand side, denote  $R \coloneqq \| x_0\|^2 + \frac{2(a_1 + a_2)}{\beta}$  and recall that  $\| b(\rho)\| \leq R$  for any  $\rho \in \mathcal{P}(\mathbb{R}^d)$  thanks to Lemma 19. By continuity of  $f_{1}$  and  $\log \tilde{\rho}$ , there exists a constant  $c_{1} \in \mathbb{R}$  such that
$$
\sup  _ {x \in B _ {R} (0)} c _ {0} (x) = \sup  _ {x \in B _ {R} (0)} \left[ f _ {1} (0, x) + \alpha \log \tilde {\rho} (0) + \frac {1}{\alpha \tilde {\lambda} - \Lambda_ {1}} \| \nabla_ {z} \left(f _ {1} (0, x) + \alpha \log \tilde {\rho} (0)\right) \| ^ {2} \right] \leq c _ {1}. \tag {30}
$$
The second term is controlled by  $c_{2}$  thanks to Assumption 4. And the third term can be bounded directly to obtain
$$
G _ {b} (\rho) \leq c _ {1} + c _ {2} + \beta \left(R ^ {2} + \| x _ {0} \| ^ {2}\right).
$$
This concludes the proof of (1). Statement (2) was shown in Lemma 20. Then we obtain a minimizing sequence  $(\rho_{n})\in \mathcal{P}(\mathbb{R}^{d})$  which is in the closed unit ball of  $C_0(\mathbb{R}^d)^*$  and so the Banach-Anaoglu theorem [Rud91, Theorem 3.15] there exists a limit  $\rho_{*}$  in the Radon measures and a subsequence (not relabeled) such that  $\rho_{n}\stackrel {*}{\to}\rho_{*}$ . In fact,  $\rho_{*}$  is absolutely continuous with respect to  $\tilde{\rho}$  as otherwise  $G_{b}(\rho_{*}) = -\infty$ , which contradicts that  $G_{b}(\cdot) > - \infty$  somewhere. We conclude that  $\rho_{*}\in L^{1}(\mathbb{R}^{d})$  since  $\tilde{\rho}\in L^{1}(\mathbb{R}^{d})$  by Assumption 2. To ensure  $\rho_{*}\in \mathcal{P}(\mathbb{R}^{d})$ , we require (3) tightness of the minimizing sequence  $(\rho_{n})$ . By Markov's inequality [Gho02] it is sufficient to establish a uniform bound on the second moments:
$$
\int \| z \| ^ {2} \mathrm {d} \rho_ {n} (z) <   C \quad \forall n \in \mathrm {N}. \tag {31}
$$
To see this we proceed in a similar way as in the proof of Proposition 10. Defining
$$
K (\rho) := - \int \left[ f _ {1} (z, b (\rho)) + \alpha \log \tilde {\rho} (z) \right] \mathrm {d} \rho (z) + \alpha \int \rho \log \rho \mathrm {d} z + \frac {1}{2} \int \rho W * \rho \mathrm {d} z,
$$
we have  $K(\rho) = -G_b(\rho) + \int f_2(z, b(\rho)) \mathrm{d}\bar{\rho}(z) + \frac{\beta}{2} \|b(\rho) - x_0\|^2$ . Then using again the bound on  $b(\rho)$  from Lemma 19,
$$
\begin{array}{l} K (\rho) \leq - G _ {b} (\rho) + \sup  _ {x \in B _ {R} (0)} \int f _ {2} (z, x) \mathrm {d} \bar {\rho} (z) + \beta \left(R ^ {2} + \| x _ {0} \| ^ {2}\right) \\ \leq - G _ {b} (\rho) + c _ {2} + \beta \left(R ^ {2} + \left\| x _ {0} \right\| ^ {2}\right), \\ \end{array}
$$
where the last inequality is thanks to Assumption 4. Hence, using the estimates (29) and (30) from above, and noting that the sequence  $(\rho_n)$  is minimizing  $(-G_b)$ , we have
$$
\begin{array}{l} \frac {(\alpha \tilde {\lambda} - \Lambda_ {1})}{4} \int \| z \| ^ {2} \mathrm {d} \rho_ {n} (z) \leq c _ {0} (b (\rho_ {n})) - \int [ f _ {1} (z, b (\rho_ {n})) + \alpha \log \tilde {\rho} (z) ] \mathrm {d} \rho_ {n} (z) \\ \leq c _ {1} + K \left(\rho_ {n}\right) \leq c _ {1} - G _ {b} \left(\rho_ {n}\right) + c _ {2} + \beta \left(R ^ {2} + \left\| x _ {0} \right\| ^ {2}\right) \\ \leq c _ {1} - G _ {b} \left(\rho_ {1}\right) + c _ {2} + \beta \left(R ^ {2} + \left\| x _ {0} \right\| ^ {2}\right) <   \infty . \\ \end{array}
$$
which uniformly bounds the second moments of  $(\rho_n)$ . This concludes the proof for the estimate (31) and also ensures that  $\rho_{*}\in \mathcal{P}_{2}(\mathbb{R}^{d})$ .
Corollary 23. Any maximizer  $\rho_{*}$  of  $G_{b}$  is a steady state for equation (5) according to Definition 9, and satisfies  $\mathrm{supp}(\rho_{*}) = \mathrm{supp}(\tilde{\rho})$ .