SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
B \geq 4 \left[ (1 + R) \left(R + \lambda_ {2} \bar {R}\right) + \left(L + \lambda_ {2}\right) ^ {2} \right] R ^ {2} (\eta + \sqrt {\eta \bar {\lambda}}) / (\lambda \alpha \epsilon),
$$
We can again check that it suffices to take the number of iteration  $k$  as
to achieve the targe accuracy.
D Proof of Theorem 4: SVRG-MFLD
The standard argument of variance yields
$$
\sigma_ {v, k} ^ {2} \leq \max _ {1 \leq i \leq N} \frac {n - B}{B (n - 1)} \frac {1}{n} \sum_ {j = 1} ^ {n} \mathbb {E} [ \| \Delta_ {k} ^ {i, j} \| ^ {2} ],
$$
where  $\Delta_k^{i,j} = \nabla \frac{\delta\ell_j(\mu_k^{(N)})}{\delta\mu}(X_k^i) - \nabla \frac{\delta U(\mu_k^{(N)})}{\delta\mu}(X_k^i) - \nabla \frac{\delta\ell_i(\mu_s^{(N)})}{\delta\mu}(X_s^i) + \nabla \frac{\delta U(\mu_s^{(N)})}{\delta\mu}(X_s^i)$ . Here, Assumption 2 and 6 give that
$$
\| \Delta_ {k} ^ {i, j} \| ^ {2} \leq 4 (L ^ {2} W _ {2} ^ {2} (\mu_ {k} ^ {(N)}, \mu_ {s} ^ {(N)}) + L ^ {2} \| X _ {k} ^ {i} - X _ {s} ^ {i} \| ^ {2}) \leq 4 L ^ {2} \left(\frac {1}{N} \sum_ {j ^ {\prime} = 1} ^ {N} \| X _ {k} ^ {j ^ {\prime}} - X _ {s} ^ {j ^ {\prime}} \| ^ {2} + \| X _ {k} ^ {i} - X _ {s} ^ {i} \| ^ {2}\right).
$$
Taking the expectation, we have
$$
\begin{array}{l} \frac {1}{n} \sum_ {j = 1} ^ {n} \mathbb {E} [ \| \Delta_ {k} ^ {i, j} \| ^ {2} ] \leq 4 L ^ {2} \frac {1}{n} \sum_ {j = 1} ^ {n} \sum_ {l = s} ^ {k - 1} (\eta^ {2} \mathbb {E} [ \| v _ {l} ^ {i} \| ^ {2} ] + 2 \eta \lambda d) \\ \leq C_{1}L^{2}\underbrace{(k - s)}_{\leq m}(\eta^{2} + \eta \lambda), \\ \end{array}
$$
where we used that  $\mathbb{E}[\| v_k^i\|^2] \leq 2(R^2 + \lambda_2(c_r + \bar{R}^2)) \leq C_1$  by Eq. (24) with Lemma 1. Hence, we have that
$$
\sigma_ {v, k} ^ {2} \leq C _ {1} \Xi L ^ {2} m (\eta^ {2} + \eta \lambda),
$$
where  $\Xi = \frac{n - B}{B(n - 1)}$ .
$$
\tilde {\sigma} _ {v, k} ^ {2} \leq \frac {n - B}{B (n - 1)} R ^ {2} = \Xi R ^ {2},
$$
by the same argument with SGD-MFLD. We have a uniform upper bound of  $\Upsilon_{k}$  as
$$
\Upsilon_ {k} \leq \left\{ \begin{array}{l l} 4 \eta \delta_ {\eta} + \left(R + \lambda_ {2} \bar {R}\right) \left(1 + \sqrt {\frac {\lambda}{\eta}}\right) \eta^ {2} \sqrt {C _ {1} \Xi L ^ {2} m (\eta^ {2} + \eta \lambda)} \sqrt {R ^ {2} \Xi} \\ \quad + \left[ (R + \lambda_ {2} \bar {R}) R \eta^ {3} + (L + \lambda_ {2}) ^ {2} \eta^ {2} \right] \left(1 + \sqrt {\frac {\lambda}{\eta}}\right) C _ {1} \Xi L ^ {2} m (\eta^ {2} + \eta \lambda), & \text {w i t h A s s u m p t i o n 6 - (i i)}, \\ \eta C _ {1} \Xi L ^ {2} m (\eta^ {2} + \eta \lambda), & \text {w i t h o u t A s s u m p t i o n 6 - (i i)}. \end{array} \right.
$$
We again let the right hand side be  $\widetilde{\Upsilon}$ . Then, we have that
$$
\frac {1}{N} \mathbb {E} \left[ \mathcal {F} ^ {N} \left(\mu_ {k} ^ {(N)}\right) \right] - \mathcal {F} \left(\mu^ {*}\right)
$$
$$
\leq \exp (- \lambda \alpha \eta / 2) \left(\frac {1}{N} \mathbb {E} [ \mathcal {F} ^ {N} (\mu_ {k - 1} ^ {(N)}) ] - \mathcal {F} (\mu^ {*})\right) + \bar {\Upsilon} + \eta \left(\delta_ {\eta} + \frac {C _ {\lambda}}{N}\right).
$$
Then, by the Gronwall's lemma yields that
$$
\begin{array}{l} \frac {1}{N} \mathbb {E} [ \mathcal {F} ^ {N} (\mu_ {k} ^ {(N)}) ] - \mathcal {F} (\mu^ {*}) \\ \leq \exp (- \lambda \alpha \eta k / 2) \left(\frac {1}{N} \mathbb {E} [ \mathcal {F} ^ {N} (\mu_ {0} ^ {(N)}) ] - \mathcal {F} (\mu^ {*})\right) + \sum_ {l = 1} ^ {k} \exp (- \lambda \alpha \eta (k - l) / 2) \left[ \bar {\Upsilon} + \eta \left(\delta_ {\eta} + \frac {C _ {\lambda}}{N}\right) \right] \\ \leq \exp (- \lambda \alpha \eta k / 2) \left(\frac {1}{N} \mathbb {E} [ \mathcal {F} ^ {N} (\mu_ {0} ^ {(N)}) ] - \mathcal {F} (\mu^ {*})\right) + \frac {1 - \exp (- \lambda \alpha \eta k / 2)}{1 - \exp (- \lambda \alpha \eta / 2)} \left[ \bar {\Upsilon} + \eta \left(\delta_ {\eta} + \frac {C _ {\lambda}}{N}\right) \right] \\ \leq \exp (- \lambda \alpha \eta k / 2) \left(\frac {1}{N} \mathbb {E} [ \mathcal {F} ^ {N} (\mu_ {0} ^ {(N)}) ] - \mathcal {F} (\mu^ {*})\right) + \frac {4}{\lambda \alpha \eta} \bar {\Upsilon} + \frac {4}{\alpha \lambda} \left(\delta_ {\eta} + \frac {C _ {\lambda}}{N}\right), \\ \end{array}
$$
where we used  $\lambda \alpha \eta /2\leq 1 / 2$  in the last inequality.
(1) Without Assumption 6-(ii), we have
$$
\frac {1}{\lambda \alpha \eta} \bar {\Upsilon} = \frac {C _ {1}}{\lambda \alpha \eta} \Xi L ^ {2} m \eta^ {2} (\eta + \lambda) = \frac {C _ {1}}{\lambda \alpha} L ^ {2} m \eta (\eta + \lambda) \frac {(n - B)}{B (n - 1)},
$$
we obtain that
$$
\begin{array}{l} \frac {1}{N} \mathbb {E} \left[ \mathcal {F} ^ {N} \left(\mu_ {k} ^ {(N)}\right) \right] - \mathcal {F} \left(\mu^ {*}\right) \leq \epsilon + \frac {4 C _ {1} \bar {L} ^ {2} \left(\eta^ {2} + \lambda \eta\right) + 4 C _ {\lambda} / N}{\lambda \alpha} + \frac {4 C _ {1}}{\lambda \alpha} L ^ {2} m \eta (\eta + \lambda) \frac {(n - B)}{B (n - 1)} \\ = \epsilon + \frac {4 C _ {1} \bar {L} ^ {2} (\eta^ {2} + \lambda \eta) (1 + \frac {(n - B)}{B (n - 1)} m)}{\lambda \alpha} + \frac {4 C _ {\lambda}}{\lambda \alpha N} \\ \end{array}
$$
$$
k \gtrsim \frac {1}{\lambda \alpha \eta} \log (\epsilon^ {- 1}),
$$
and  $\lambda \alpha \eta /2\leq 1 / 2$  . In particular, if we set  $B\geq m$  and
$$
\eta = \frac {\alpha \epsilon}{4 \bar {L} ^ {2} C _ {1}} \wedge \frac {\sqrt {\lambda \alpha \epsilon}}{4 \bar {L} \sqrt {C _ {1}}},
$$
$$
\frac {1}{N} \mathbb {E} \left[ \mathcal {F} ^ {N} \left(\mu_ {k} ^ {(N)}\right) \right] - \mathcal {F} \left(\mu^ {*}\right) \leq \epsilon + \frac {4 C _ {\lambda}}{\alpha \lambda N},
$$
with the iteration complexity and the total gradient computation complexity:
$$
k \lesssim \left(\frac {\bar {L} ^ {2}}{\alpha \epsilon} + \frac {\bar {L}}{\sqrt {\lambda \alpha \epsilon}}\right) \frac {1}{(\lambda \alpha)} \log (\epsilon^ {- 1}),
$$
$$
B k + \frac {n k}{m} + n \lesssim \sqrt {n} k + n \lesssim \sqrt {n} \left(\frac {\bar {L} ^ {2}}{\alpha \epsilon} + \frac {\bar {L}}{\sqrt {\lambda \alpha \epsilon}}\right) \frac {1}{(\lambda \alpha)} \log (\epsilon^ {- 1}) + n,
$$
where  $m = B = \sqrt{n}$
(2) With Assumption 6-(ii), we have
$$
\frac {1}{\lambda \alpha \eta} \tilde {\Upsilon} = \frac {4}{\lambda \alpha} \delta_ {\eta} + \frac {1}{\lambda \alpha} O \left(\eta \sqrt {m (\eta^ {2} + \lambda \eta)} + \eta m (\eta^ {2} + \lambda \eta)\right) \left(1 + \sqrt {\frac {\lambda}{\eta}}\right) \frac {(n - B)}{B (n - 1)}.
$$
Then, we obtain that
$$
\begin{array}{l} \frac {1}{N} \mathbb {E} [ \mathcal {F} ^ {N} (\mu_ {k} ^ {(N)}) ] - \mathcal {F} (\mu^ {*}) \\ \leq \epsilon + \frac {2 0 C _ {1} \bar {L} ^ {2} (\eta^ {2} + \lambda \eta) + 4 C _ {\lambda} / N}{\lambda \alpha} + \frac {1}{\lambda \alpha} O \left(\eta \sqrt {m (\eta^ {2} + \lambda \eta)} + \eta m (\eta^ {2} + \lambda \eta)\right) \left(1 + \sqrt {\frac {\lambda}{\eta}}\right) \frac {(n - B)}{B (n - 1)}, \\ \end{array}
$$
and  $\lambda \alpha \eta /2\leq 1 / 2$  . In particular, if we set  $B\geq \left[\sqrt{m}\frac{(\eta + \sqrt{\eta / \lambda})^2}{\eta\lambda}\lor m\frac{(\eta + \sqrt{\eta / \lambda})^3}{\eta\lambda}\right]\wedge n$  and
$$
\eta = \frac {\alpha \epsilon}{4 0 \bar {L} ^ {2} C _ {1}} \wedge \frac {\sqrt {\lambda \alpha \epsilon}}{\bar {L} \sqrt {4 0 C _ {1}}},
$$
$$
\frac {1}{N} \mathbb {E} \left[ \mathcal {F} ^ {N} \left(\mu_ {k} ^ {(N)}\right) \right] - \mathcal {F} \left(\mu^ {*}\right) \leq O (\epsilon) + \frac {4 C _ {\lambda}}{\alpha \lambda N},
$$
$$
\begin{array}{l} B k + \frac {n k}{m} + n \lesssim \max  \left\{n ^ {1 / 3} \left(1 + \sqrt {\eta / \lambda}\right) ^ {4 / 3}, \sqrt {n} \left(1 + \sqrt {\eta / \lambda}\right) ^ {3 / 2} (\sqrt {\eta \lambda}) ^ {1 / 2} \right\} k + n \\ \lesssim \max  \left\{n ^ {1 / 3} \left(1 + \sqrt {\eta / \lambda}\right) ^ {4 / 3}, \sqrt {n} \left(1 + \sqrt {\eta / \lambda}\right) ^ {3 / 2} (\sqrt {\eta \lambda}) ^ {1 / 2} \right\} \eta^ {- 1} \frac {1}{(\lambda \alpha)} \log (\epsilon^ {- 1}) + n, \\ \end{array}
$$
where  $m = \Omega (n / B) = \Omega ([n^{2 / 3}(1 + \sqrt{\eta / \lambda})^{-4 / 3}\wedge \sqrt{n} (1 + \sqrt{\eta / \lambda})^{-3 / 2}(\sqrt{\eta\lambda})^{-1 / 2})]\vee 1)$  and  $B\geq \left[n^{\frac{1}{3}}\left(1 + \sqrt{\frac{\eta}{\lambda}}\right)^{\frac{4}{3}}\lor \sqrt{n} (\eta \lambda)^{\frac{1}{4}}\left(1 + \sqrt{\frac{\eta}{\lambda}}\right)^{\frac{3}{2}}\right]\wedge n.$
E Auxiliary lemmas
E.1 Properties of the MFLD Iterates
Under Assumption 2 with Assumption 5 for SGD-MFLD or 6 for SVRG-MFLD, we can easily verify that
$$
\left\| \nabla \frac {\delta U (\mu_ {k})}{\delta \mu} \left(X _ {k} ^ {i}\right) \right\| \leq R, \mathbb {E} _ {\omega_ {k} ^ {i} \mid \mathcal {X} _ {0: k}} [ \| v _ {k} ^ {i} \| ^ {2} ] \leq 2 \left(R ^ {2} + \lambda_ {2} \left(c _ {r} + \| X _ {k} ^ {i} \| ^ {2}\right)\right). \tag {24}
$$
Lemma 1. Under Assumption 2 with Assumption 5 for SGD-MFLD or Assumption 6 for SVRG-MFLD, if  $\eta \leq \lambda_1 / (4\lambda_2)$ , we have the following uniform bound of the second moment of  $X_k^i$ :
$$
\mathbb {E} [ \| X _ {k} ^ {i} \| ^ {2} ] \leq \mathbb {E} [ \| X _ {0} ^ {i} \| ^ {2} ] + \frac {2}{\lambda_ {1}} \left[ \left(\frac {\lambda_ {1}}{8 \lambda_ {2}} + \frac {1}{2 \lambda_ {1}}\right) (R ^ {2} + \lambda_ {2} c _ {r}) + \lambda d \right],
$$
for any  $k\geq 1$
Proof. By the update rule of  $X_{k}^{i}$  and Assumption 1, we have
$$
\begin{array}{l} \mathbb {E} [ \| X _ {k + 1} ^ {i} \| ^ {2} ] = \mathbb {E} [ \| X _ {k} ^ {i} \| ^ {2} ] - 2 \mathbb {E} [ \langle X _ {k} ^ {i}, \eta v _ {k} ^ {i} + \sqrt {2 \eta \lambda} \xi_ {k} ^ {i} \rangle ] + \mathbb {E} [ \| \eta v _ {k} ^ {i} + \sqrt {2 \eta \lambda} \xi_ {k} ^ {i} \| ^ {2} ] \\ = \mathbb {E} [ \| X _ {k} ^ {i} \| ^ {2} ] - 2 \eta \mathbb {E} \left[ \left\langle X _ {k} ^ {i}, \nabla \frac {\delta U (\mu_ {k})}{\delta \mu} (X _ {k} ^ {i}) + \nabla r (X _ {k} ^ {i}) \right\rangle \right] + \mathbb {E} [ \eta^ {2} \| v _ {k} ^ {i} \| ^ {2} ] + 2 \eta \lambda d \\ \leq \mathbb {E} [ \| X _ {k} ^ {i} \| ^ {2} ] + 2 \eta R \mathbb {E} [ \| X _ {k} ^ {i} \| ] - 2 \eta \lambda_ {1} \mathbb {E} [ \| X _ {k} ^ {i} \| ^ {2} ] + 2 \eta^ {2} (R ^ {2} + \lambda_ {2} (c _ {r} + \| X _ {k} ^ {i} \| ^ {2})) + 2 \eta \lambda d \\ \leq \left(1 - 2 \eta \lambda_ {1} + 2 \eta^ {2} \lambda_ {2}\right) \mathbb {E} [ \| X _ {k} ^ {i} \| ^ {2} ] + 2 \eta R \mathbb {E} [ \| X _ {k} ^ {i} \| ] + 2 \eta (\eta (R ^ {2} + \lambda_ {2} c _ {r}) + \lambda d) \\ \leq \left(1 - \eta \lambda_ {1}\right) \mathbb {E} \left[ \| X _ {k} ^ {i} \| ^ {2} \right] + 2 \eta \left(\eta \left(R ^ {2} + \lambda_ {2} c _ {r}\right) + \lambda d + R ^ {2} / \lambda_ {1}\right) \\ \left(\because 2 R \mathbb {E} [ \| X _ {k} ^ {i} \| ] \leq \lambda_ {1} \mathbb {E} [ \| X _ {k} ^ {i} \| ^ {2} ] / 2 + 2 R ^ {2} / \lambda_ {1}\right). \\ \end{array}
$$
where we used the assumption  $\eta \leq \lambda_1 / (4\lambda_2)$ . Then, by the Gronwall lemma, it holds that
$$
\begin{array}{l} \mathbb {E} [ \| X _ {k} ^ {i} \| ^ {2} ] \leq (1 - \eta \lambda_ {1}) ^ {k} \mathbb {E} [ \| X _ {0} ^ {i} \| ^ {2} ] + \frac {1 - (1 - \eta \lambda_ {1}) ^ {k}}{\eta \lambda_ {1}} \eta [ \eta (R ^ {2} + \lambda_ {2} c _ {r}) + \lambda d + R ^ {2} / \lambda_ {1} ] \\ \leq \mathbb {E} [ \| X _ {0} ^ {i} \| ^ {2} ] + \frac {1}{\lambda_ {1}} \left[ \eta \left(R ^ {2} + \lambda_ {2} c _ {r}\right) + \lambda d + R ^ {2} / \left(\lambda_ {1}\right) \right]. \\ \end{array}
$$
This with the assumption  $\eta \leq \lambda_1 / (4\lambda_2)$  yields the assertion.
Lemma 2. Let  $\bar{R}^2 \coloneqq \mathbb{E}[||X_0^i ||^2] + \frac{1}{\lambda_1}\left[\left(\frac{\lambda_1}{4\lambda_2} +\frac{1}{\lambda_1}\right)(R^2 +\lambda_2c_r) + \lambda d\right]$ . Under Assumptions 1 and 2, if  $\eta \leq \lambda_1 / (4\lambda_2)$  and we define
$$
\delta_ {\eta} = 8 \left[ R ^ {2} + \lambda_ {2} \left(c _ {r} + \bar {R} ^ {2}\right) + d \right] \bar {L} ^ {2} \left(\eta^ {2} + \lambda \eta\right),
$$
$$
\mathbb {E} _ {\widetilde {\mathcal {X}} _ {t}, \widetilde {\mathcal {X}} _ {0}} \left[ \left\| \nabla \frac {\delta F (\tilde {\mu} _ {0} ^ {(N)})}{\delta \mu} (\widetilde {X} _ {0} ^ {i}) - \nabla \frac {\delta F (\tilde {\mu} _ {t} ^ {(N)})}{\delta \mu} (\widetilde {X} _ {t} ^ {i}) \right\| ^ {2} \right] \leq \delta_ {\eta}.
$$
Proof. The proof is basically a mean field generalization of Nitanda et al. (2022). By Assumptions 1 and 2, the first two terms in the right hand side can be bounded as
$$
\begin{array}{l} \left\| \nabla \frac {\delta F (\tilde {\mu} _ {0} ^ {(N)})}{\delta \mu} \left(\widetilde {X} _ {0} ^ {i}\right) - \nabla \frac {\delta F \left(\tilde {\mu} _ {t} ^ {(N)}\right)}{\delta \mu} \left(\widetilde {X} _ {t} ^ {i}\right) \right\| \\ = \left\| \nabla \frac {\delta U \left(\tilde {\mu} _ {0} ^ {(N)}\right)}{\delta \mu} \left(\widetilde {X} _ {0} ^ {i}\right) + \nabla r \left(\widetilde {X} _ {0} ^ {i}\right) - \nabla \frac {\delta U \left(\tilde {\mu} _ {t} ^ {(N)}\right)}{\delta \mu} \left(\widetilde {X} _ {t} ^ {i}\right) - \nabla r \left(\widetilde {X} _ {t} ^ {i}\right) \right\| \\ \leq L \left(W _ {2} \left(\tilde {\mu} _ {0} ^ {(N)}, \tilde {\mu} _ {t} ^ {(N)}\right) + \| \widetilde {X} _ {0} ^ {i} - \widetilde {X} _ {t} ^ {i} \|\right) + \lambda_ {2} \| \widetilde {X} _ {t} ^ {i} - \hat {X} _ {t} ^ {i} \|. \tag {25} \\ \end{array}
$$
Therefore, the right hand side can be further bounded as
$$
\begin{array}{l} \left\| \nabla \frac {\delta F (\tilde {\mu} _ {0} ^ {(N)})}{\delta \mu} (\widetilde {X} _ {0} ^ {i}) - \nabla \frac {\delta F (\tilde {\mu} _ {t} ^ {(N)})}{\delta \mu} (\widetilde {X} _ {t} ^ {i}) \right\| ^ {2} \\ \leq \left(L + \lambda_ {2}\right) ^ {2} \left(W _ {2} \left(\tilde {\mu} _ {0} ^ {(N)}, \tilde {\mu} _ {t} ^ {(N)}\right) + \| \tilde {X} _ {0} ^ {i} - \tilde {X} _ {t} ^ {i} \|\right) ^ {2} \\ \leq 2 \bar {L} ^ {2} \frac {1}{N} \sum_ {i = 1} ^ {N} \| \widetilde {X} _ {0} ^ {i} - \widetilde {X} _ {t} ^ {i} \| ^ {2} + 2 \bar {L} ^ {2} \| \widetilde {X} _ {0} ^ {i} - \widetilde {X} _ {t} ^ {i} \| ^ {2} \\ \leq 2 \bar {L} ^ {2} \frac {1}{N} \sum_ {i = 1} ^ {N} \left\| t v _ {k} ^ {i} - \sqrt {2 t \lambda} \xi_ {k} ^ {i} \right\| ^ {2} + 2 \bar {L} ^ {2} \| t v _ {k} ^ {i} - \sqrt {2 t \lambda} \xi_ {k} ^ {i} \| ^ {2}. \\ \end{array}
$$
Then, by taking the expectation, it holds that
$$
\begin{array}{l} \mathbb {E} _ {\widetilde {\mathcal {X}} _ {t}, \widetilde {\mathcal {X}} _ {0}} \left[ \left\| \nabla \frac {\delta F (\tilde {\mu} _ {0} ^ {(N)})}{\delta \mu} (\widetilde {X} _ {0} ^ {i}) - \nabla \frac {\delta F (\tilde {\mu} _ {t} ^ {(N)})}{\delta \mu} (\widetilde {X} _ {t} ^ {i}) \right\| ^ {2} \right] \\ \leq 2 \bar {L} ^ {2} \frac {1}{N} \sum_ {i = 1} ^ {N} \left(t ^ {2} \mathbb {E} \left[ \| v _ {k} ^ {i} \| ^ {2} \right] + 2 t \lambda d\right) + 2 \bar {L} ^ {2} \left(t ^ {2} \mathbb {E} \left[ \| v _ {k} ^ {i} \| ^ {2} \right] + 2 t \lambda d\right) \\ \leq 4 \bar {L} ^ {2} \left[ t ^ {2} 2 \left(R ^ {2} + \lambda_ {2} \left(c _ {r} + \bar {R} ^ {2}\right)\right) + 2 t \lambda d \right] \quad (\because \text {L e m m a} 1) \\ = 8 \bar {L} ^ {2} \left[ t ^ {2} \left(R ^ {2} + \lambda_ {2} \left(c _ {r} + \bar {R} ^ {2}\right)\right) + t \lambda d \right] \\ \leq 8 \left[ R ^ {2} + \lambda_ {2} \left(c _ {r} + \bar {R} ^ {2}\right) + d \right] \bar {L} ^ {2} \left(t ^ {2} + t \lambda\right). \\ \end{array}
$$
Then, by noticing  $t \leq \eta$ , we obtain the assertion.
E.2 Wasserstein Distance Bound
Recall that  $W_{2}(\mu, \nu)$  is the 2-Wasserstein distance between  $\mu$  and  $\nu$ . We let  $D(\mu, \nu) = \int \log \left( \frac{\nu}{\mu} \right) \mathrm{d}\nu$  be the KL-divergence between  $\mu$  and  $\nu$ .
Lemma 3. Under Assumptions 1 and 3, it holds that
$$
W _ {2} ^ {2} (\mu^ {(N)}, \mu^ {* N}) \leq \frac {2}{\lambda \alpha} (\mathcal {F} ^ {N} (\mu^ {(N)}) - N \mathcal {F} (\mu^ {*})).
$$
Proof. By Assumption 3,  $\mu^{*}$  satisfies the LSI condition with a constant  $\alpha >0$ . Then, it is known that its tensor product  $\mu^{*N}$  also satisfies the LSI condition with the same constant  $\alpha$  (see, for example, Proposition 5.2.7 of Bakry et al. (2014)). Then, Otto-Villani theorem Otto and Villani (2000) yields the Talagrand's inequality of the tensorized measure  $\mu^{*N}$ :
$$
W _ {2} ^ {2} (\mu^ {(N)}, \mu^ {* N}) \leq \frac {2}{\alpha} D (\mu^ {(N)}, \mu^ {* N}).
$$
Moreover, the proof of Theorem 2.11 of Chen et al. (2022) yields that, under Assumption 1, it holds that
$$
D \left(\mu^ {(N)}, \mu^ {* N}\right) \leq \frac {1}{\lambda} \left(\mathscr {F} ^ {N} \left(\mu^ {(N)}\right) - N \mathscr {F} \left(\mu^ {*}\right)\right). \tag {26}
$$
Lemma 4. Suppose that  $|h_x(z) - h_{x'}(z)| \leq L\| x - x'\|$  ( $\forall x, x' \in \mathbb{R}^d$ ) and let  $V_{\mu^*} \coloneqq \operatorname{Var}_{\mu^*}(f_{\mu^*}) = \int (f_{\mu^*}(z) - h_x(z))^2 \, \mathrm{d}\mu^*(x)$ , then it holds that
$$
\mathbb {E} _ {\mathcal {X} _ {k} \sim \mu_ {k} ^ {(N)}} [ (f _ {\mu_ {\mathcal {X} _ {k}}} (z) - f _ {\mu^ {*}} (z)) ^ {2} ] \leq \frac {2 L ^ {2}}{N} W _ {2} ^ {2} (\mu_ {k} ^ {(N)}, \mu^ {* N}) + \frac {2}{N} V _ {\mu^ {*}}.
$$
Proof. Consider a coupling  $\gamma$  of  $\mu_{\mathcal{X}_k}^{(N)}$  and  $\mu^{*N}$  and let  $(\mathcal{X}_k,\mathcal{X}_*) = ((X_k^i)_{i = 1}^N,(X_i^i)_{i = 1}^N)$  be a random variable obeying the law  $\gamma$ . Then,
$$
\begin{array}{l} \left(f _ {\mu_ {\mathcal {X} _ {k}}} (z) - f _ {\mu^ {*}} (z)\right) ^ {2} = \left(\frac {1}{N} \sum_ {i = 1} ^ {N} h _ {X _ {k} ^ {i}} (z) - \int h _ {x} (z) \mathrm {d} \mu^ {*} (x)\right) ^ {2} \\ = \left(\frac {1}{N} \sum_ {i = 1} ^ {N} \left(h _ {X _ {k} ^ {i}} (z) - h _ {X _ {*} ^ {i}} (z)\right) + \frac {1}{N} \sum_ {i = 1} ^ {N} h _ {X _ {*} ^ {i}} (z) - \int h _ {x} (z) d \mu^ {*} (x)\right) ^ {2} \\ \leq 2 \left(\frac {1}{N} \sum_ {i = 1} ^ {N} \left(h _ {X _ {k} ^ {i}} (z) - h _ {X _ {*} ^ {i}} (z)\right)\right) ^ {2} + 2 \left(\frac {1}{N} \sum_ {i = 1} ^ {N} h _ {X _ {*} ^ {i}} (z) - \int h _ {x} (z) \mathrm {d} \mu^ {*} (x)\right) ^ {2} \\ \leq 2 \left(\frac {1}{N} \sum_ {i = 1} ^ {N} L \| X _ {k} ^ {i} - X _ {*} ^ {i} \|\right) ^ {2} + 2 \left(\frac {1}{N} \sum_ {i = 1} ^ {N} h _ {X _ {*} ^ {i}} (z) - \int h _ {x} (z) \mathrm {d} \mu^ {*} (x)\right) ^ {2} \\ \leq 2 L ^ {2} \frac {1}{N} \sum_ {i = 1} ^ {N} \| X _ {k} ^ {i} - X _ {*} ^ {i} \| ^ {2} + 2 \left(\frac {1}{N} \sum_ {i = 1} ^ {N} h _ {X _ {*} ^ {i}} (z) - \int h _ {x} (z) \mathrm {d} \mu^ {*} (x)\right) ^ {2}. \\ \end{array}
$$
Then, by taking the expectation of the both side with respect to  $(\mathcal{X}_k,\mathcal{X}_*)$ , we have that
$$
\begin{array}{l} \mathbb {E} _ {\mathcal {X} _ {k} \sim \mu_ {k} ^ {(N)}} \left[ \left(f _ {\mu_ {\mathcal {X} _ {k}}} (z) - f _ {\mu^ {*}} (z)\right) ^ {2} \right] \\ \leq 2 L ^ {2} \mathbb {E} _ {\gamma} \left[ \frac {1}{N} \sum_ {i = 1} ^ {N} \| X _ {k} ^ {i} - X _ {*} ^ {i} \| ^ {2} \right] + \frac {2}{N} V _ {\mu^ {*}}. \\ \end{array}
$$
Hence, taking the infimum of the coupling  $\gamma$  yields that
Remark 2. By the self-consistent condition of  $\mu^{*}$ , we can see that  $\mu^{*}$  is sub-Gaussian. Therefore,  $V_{\mu^{*}} < \infty$  is always satisfied.
E.3 Logarithmic Sobolev Inequality
Lemma 5. Under Assumptions 1 and 2,  $\mu^{*}$  and  $p_{\mathcal{X}}$  satisfy the LSI condition with a constant
$$
\alpha \geq \frac {\lambda_ {1}}{2 \lambda} \exp \left(- 4 \frac {R ^ {2}}{\lambda_ {1} \lambda} \sqrt {2 d / \pi}\right) \vee \left\{\frac {4 \lambda}{\lambda_ {1}} + \left(\frac {R}{\lambda_ {1}} + \sqrt {\frac {2 \lambda}{\lambda_ {1}}}\right) ^ {2} e ^ {\frac {R ^ {2}}{2 \lambda_ {1} \lambda}} \left[ 2 + d + \frac {d}{2} \log \left(\frac {\lambda_ {2}}{\lambda_ {1}}\right) + 4 \frac {R ^ {2}}{\lambda_ {1} \lambda} \right] \right\} ^ {- 1}.
$$
Proof. In the following, we give two lower bounds of  $\alpha$ . By taking the maximum of the two, we obtain the assertion.
(1) By Assumption 2,  $\frac{\delta U(\mu)}{\delta\mu}$  is Lipschitz continuous with the Lipschitz constant  $R$ , which implies that  $\frac{1}{\lambda}\frac{\delta U(\mu)}{\delta\mu}$  is  $R / \lambda$ -Lipschitz continuous. Hence,  $\mu^{*}$  and  $p_{\mathcal{X}}$  is a Lipshitz perturbation of  $\nu(x) \propto \exp(-\lambda^{-1}r(x))$ . Since  $\lambda^{-1}\nabla \nabla^{\top}r(x) \succeq \frac{\lambda_1}{\lambda}I$  by Assumption 1, Miclo's trick (Lemma 2.1 of Bardet et al. (2018)) yields that  $\mu^{*}$  and  $p_{\mathcal{X}}$  satisfy the LSI with a constant
$$
\alpha \geq \frac {\lambda_ {1}}{2 \lambda} \exp \left(- 4 \frac {\lambda}{\lambda_ {1}} \left(\frac {R}{\lambda}\right) ^ {2} \sqrt {2 d / \pi}\right) = \frac {\lambda_ {1}}{2 \lambda} \exp \left(- 4 \frac {R ^ {2}}{\lambda_ {1} \lambda} \sqrt {2 d / \pi}\right).
$$
(2) We can easily check that  $V(x) = \frac{r(x)}{\lambda}$  and  $H(x) = \frac{1}{\lambda} \frac{\delta U(\tilde{\mu})}{\delta \mu}(x)$ , for appropriately chosen  $\tilde{\mu}$ , satisfies the conditions in Lemma 6 with  $c_1 = \frac{\lambda_1}{\lambda}$ ,  $c_2 = \frac{\lambda_2}{\lambda}$ ,  $c_V = c_r$ , and  $\bar{L} = \frac{R}{\lambda}$ . Hence,  $\mu^*$  and  $p_{\mathcal{X}}$  satisfy the LSI with a constant  $\alpha$  such that
$$
\alpha \geq \left\{\frac {4 \lambda}{\lambda_ {1}} + \left(\frac {L}{\lambda_ {1}} + \sqrt {\frac {2 \lambda}{\lambda_ {1}}}\right) ^ {2} e ^ {\frac {R ^ {2}}{2 \lambda_ {1} \lambda}} \left[ 2 + d + \frac {d}{2} \log \left(\frac {\lambda_ {2}}{\lambda_ {1}}\right) + 4 \frac {R ^ {2}}{\lambda_ {1} \lambda} \right] \right\} ^ {- 1}.
$$
Lemma 6 (Log Sobolev inequality with Lipschitz perturbation). Let  $\nu(x) \propto \exp(-V(x))$  where  $\nabla \nabla^{\top} V(x) \succeq c_1 I$ ,  $x^{\top} \nabla V(x) \geq c_1 \|x\|^2$  and  $0 \leq V(x) \leq c_2 (c_V + \|x\|^2)$  with  $c_1, c_2, c_V > 0$ , and let  $H: \mathbb{R}^d \to \mathbb{R}$  is a Lipschitz continuous function with the Lipschitz constant  $\bar{L}$ . Suppose that  $\mu \in \mathcal{P}$  is given by  $\mu(x) \propto \exp(-H(x)) \nu(x)$ . Then,  $\mu$  satisfies the LSI with a constant  $\alpha$  such that
$$
\alpha \geq \left\{\frac {4}{c _ {1}} + \left(\frac {\bar {L}}{c _ {1}} + \sqrt {\frac {2}{c _ {1}}}\right) ^ {2} e ^ {\frac {\bar {L} ^ {2}}{2 c _ {1}}} \left[ 2 + d + \frac {d}{2} \log \left(\frac {c _ {2}}{c _ {1}}\right) + 4 \frac {\bar {L} ^ {2}}{c _ {1}} \right] \right\} ^ {- 1}.
$$
Proof. Since  $\nu$  is a strongly log-concave distribution, the Bakry-Emery argument (Bakry and Emery, 1985b) yields that it satisfies the LSI condition with a constant  $\alpha' = c_1$  by the assumption  $\nabla \nabla^\top V(x) \succeq c_1 I$ .
Next, we evaluate the second moment of  $\nu$  because it is required in the following analysis. Since we know that  $\nu$  is the stationary distribution of the SDE  $\mathrm{d}X_{t} = -\nabla V(X_{t})\mathrm{d} + \sqrt{2}\mathrm{d}W_{t}$ , its corresponding infinitesimal generator gives that
$$
\frac {\mathrm {d}}{\mathrm {d} t} \mathbb {E} [ \| X _ {t} \| ^ {2} ] = \mathbb {E} [ - 2 X _ {t} ^ {\top} \nabla V (X _ {t}) ] + 2 d \leq - 2 c _ {1} \mathbb {E} [ \| X _ {t} \| ^ {2} ] + 2 d.
$$
Then, by the Gronwall lemma, we have
$$
\begin{array}{l} \mathbb {E} \left[ \| X _ {t} \| ^ {2} \right] \leq \exp (- 2 c _ {1} t) \mathbb {E} \left[ \| X _ {0} \| ^ {2} \right] + \int_ {0} ^ {t} \exp (- 2 c _ {1} (t - s))) \mathrm {d} s 2 d \\ = \exp (- 2 c _ {1} t) \mathbb {E} [ \| X _ {0} \| ^ {2} ] + (1 - \exp (- 2 c _ {1} t)) \frac {d}{c _ {1}}. \\ \end{array}
$$
Hence, by taking  $X_0 = 0$  (a.s.), we see that  $\lim \sup_t\mathbb{E}[\| X_t\|^2 ]\leq d / c_1$  that yields  $\mathbb{E}_{\nu}[\| X^2 \| ]\leq d / c_1$ . A distribution  $\tilde{\mu}$  satisfies the Poincaré inequality with a constant  $\tilde{\alpha}$  if it holds that
$$
\mathbb {E} _ {\tilde {\mu}} [ (f - \mathbb {E} _ {\tilde {\mu}} [ f ]) ^ {2} ] \leq \frac {1}{\tilde {\alpha}} \mathbb {E} _ {\tilde {\mu}} [ \| \nabla f \| ^ {2} ],
$$
for all bounded  $f$  with bounded derivatives. By Example (3) in Section 7.1 of Cattiaux and Guillin (2014),  $\mu$  satisfies the Poincaré inequality with a constant  $\tilde{\alpha}$  such that
$$
\frac {1}{\tilde {\alpha}} \leq \frac {1}{2} \left(\frac {2 \bar {L}}{c _ {1}} + \sqrt {\frac {8}{c _ {1}}}\right) ^ {2} e ^ {\frac {\bar {L} ^ {2}}{2 c c _ {1}}}.
$$