SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\left(\lambda_ {n} (\Delta \mu_ {n}) _ {i} - \lambda_ {n} ^ {*} (\Delta \mu_ {n} ^ {*}) _ {i}\right) / \mathcal {D} (G _ {n}, G _ {*, n}) \to 0, \left(\lambda_ {n} (\Delta \Sigma_ {n}) _ {u v} - \lambda_ {n} ^ {*} (\Delta \Sigma_ {n} ^ {*}) _ {u v}\right) / \mathcal {D} (G _ {n}, G _ {*, n}) \to 0
$$
for all  $1 \leq i \leq d_{1}$  and  $1 \leq u, v \leq d_{2}$  where  $(a)_i$  denotes the  $i$ -th element of vector  $a$  and  $A_{uv}$  denotes the  $(u, v)$ -th element of matrix  $A$ . Furthermore, for any  $1 \leq i, j \leq d$  ( $i$  and  $j$  can be equal),
the coefficient of  $\frac{\partial^{|\alpha|}f}{\partial\mu^{\alpha_1}\partial\Sigma^{\alpha_2}} (x|\mu_n^*,\Sigma_n^*)$  when  $(\alpha_{1})_{i} = (\alpha_{1})_{j} = 1$  and  $\alpha_{2} = 0$  leads to
$$
\left[\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left(\Delta \mu_ {n} ^ {*}\right) _ {i} \left(\Delta \mu_ {n} ^ {*}\right) _ {j} + \lambda_ {n} \left(\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}\right) _ {i} \left(\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}\right) _ {j} \right] / \mathcal {D} \left(G _ {n}, G _ {\ast , n}\right)\rightarrow 0. \tag {20}
$$
When  $i = j$ , the above limits lead to
$$
\left[ \left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) \left\{\left(\Delta \mu_ {n} ^ {*}\right) _ {i} \right\} ^ {2} + \lambda_ {n} \left\{\left(\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}\right) _ {i} \right\} ^ {2} \right] / \mathcal {D} (G _ {n}, G _ {\ast , n}) \to 0.
$$
Therefore, we would have
$$
\left[\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) \| \Delta \mu_ {n} ^ {*} \| ^ {2} + \lambda_ {n} \| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} \| ^ {2} \right] / \mathcal {D} \left(G _ {n}, G _ {* , n}\right)\rightarrow 0. \tag {21}
$$
Now, as  $\left(\lambda_n(\Delta \mu_n)_i - \lambda_n^* (\Delta \mu_n^*)_i\right) / \mathcal{D}(G_n,G_{*,n})\to 0$  we obtain that
$$
\begin{array}{l} \left(\lambda_ {n} (\Delta \mu_ {n}) _ {i} (\Delta \mu_ {n}) _ {j} - \lambda_ {n} ^ {*} (\Delta \mu_ {n} ^ {*}) _ {i} (\Delta \mu_ {n}) _ {j}\right) / \mathcal {D} (G _ {n}, G _ {\ast , n}) \quad \rightarrow \quad 0, \\ \left(\lambda_ {n} \left(\Delta \mu_ {n}\right) _ {i} \left(\Delta \mu_ {n} ^ {*}\right) _ {j} - \lambda_ {n} ^ {*} \left(\Delta \mu_ {n} ^ {*}\right) _ {i} \left(\Delta \mu_ {n} ^ {*}\right) _ {j}\right) / \mathcal {D} \left(G _ {n}, G _ {* , n}\right) \quad \rightarrow \quad 0. \tag {22} \\ \end{array}
$$
Plugging the results from (22) into (20), we ultimately achieve for any  $1 \leq i, j \leq d$  that
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left(\Delta \mu_ {n} ^ {*}\right) _ {i} \left(\Delta \mu_ {n}\right) _ {j} / \mathcal {D} \left(G _ {n}, G _ {*, n}\right)\rightarrow 0. \tag {23}
$$
Using the results from (20) and (23), we would have
$$
\begin{array}{l} \frac {\lambda_ {n} (\Delta \mu_ {n}) _ {i} (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) _ {j}}{\mathcal {D} (G _ {n} , G _ {*, n})} \to \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) (\Delta \mu_ {n}) _ {i} (\Delta \mu_ {n} ^ {*}) _ {j}}{\mathcal {D} (G _ {n} , G _ {*, n})} \to 0, \\ \frac {\lambda_ {n} ^ {*} (\Delta \mu_ {n} ^ {*}) _ {i} (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) _ {j}}{\mathcal {D} (G _ {n} , G _ {* , n})} \rightarrow \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) (\Delta \mu_ {n} ^ {*}) _ {i} (\Delta \mu_ {n}) _ {j}}{\mathcal {D} (G _ {n} , G _ {* , n})} \rightarrow 0 \\ \end{array}
$$
for any  $1 \leq i, j \leq d$ . Therefore, it leads to
$$
\begin{array}{l} \frac {\sum_ {1 \leq i , j \leq d} \lambda_ {n} | (\Delta \mu_ {n}) _ {i} | | (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) _ {j} |}{\mathcal {D} (G _ {n} , G _ {*, n})} = \frac {\lambda_ {n} \sum_ {1 \leq i \leq d} | (\Delta \mu_ {n}) _ {i} | \sum_ {1 \leq i \leq d} | (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) _ {i} |}{\mathcal {D} (G _ {n} , G _ {*, n})} \rightarrow 0, \\ \frac {\sum_ {1 \leq i , j \leq d} \lambda_ {n} ^ {*} | (\Delta \mu_ {n} ^ {*}) _ {i} | | (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) _ {j} |}{\mathcal {D} (G _ {n} , G _ {*, n})} = \frac {\lambda_ {n} ^ {*} \sum_ {1 \leq i \leq d} | (\Delta \mu_ {n}) _ {i} ^ {*} | \sum_ {1 \leq i \leq d} | (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) _ {i} |}{\mathcal {D} (G _ {n} , G _ {*, n})} \to 0. \\ \end{array}
$$
The above results mean that
$$
\lambda_ {n} \| \Delta \mu_ {n} \| \| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} \| / \mathcal {D} (G _ {n}, G _ {* , n}) \rightarrow 0, \lambda_ {n} ^ {*} \| \Delta \mu_ {n} ^ {*} \| \| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} \| / \mathcal {D} (G _ {n}, G _ {* , n}) \rightarrow 0. \tag {24}
$$
By applying the above argument with the coefficients of  $\frac{\partial^{|\alpha|}f}{\partial\mu^{\alpha_1}\partial\Sigma^{\alpha_2}} (x|\mu_n^*,\Sigma_n^*)$  when  $\alpha_{1} = 0$  and  $(\alpha_{2})_{u_{1}v_{1}} = (\alpha_{2})_{u_{2}v_{2}} = 1$  for any two pairs  $(u_{1},v_{1}),(u_{2},v_{2})$  (not necessarily distinct) such that  $1\leq u_{1},u_{2},v_{1},v_{2}\leq d$  or  $(\alpha_{1})_{i} = 1$  and  $(\alpha_{2})_{uv} = 1$  for any  $1\leq i\leq d$  and  $1\leq u,v\leq d$ , we respectively obtain that
$$
\begin{array}{l} \left[\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) \| \Delta \Sigma_ {n} ^ {*} \| ^ {2} + \lambda_ {n} \| \Delta \Sigma_ {n} - \Delta \Sigma_ {n} ^ {*} \| ^ {2} \right] / \mathcal {D} \left(G _ {n}, G _ {*}, n\right)\rightarrow 0, \\ \lambda_ {n} \| \Delta \Sigma_ {n} \| \| \Delta \Sigma_ {n} - \Delta \Sigma_ {n} ^ {*} \| / \mathcal {D} (G _ {n}, G _ {* * n}) \rightarrow 0, \lambda_ {n} ^ {*} \| \Delta \Sigma_ {n} ^ {*} \| \| \Delta \Sigma_ {n} - \Delta \Sigma_ {n} ^ {*} \| / \mathcal {D} (G _ {n}, G _ {* * n}) \rightarrow 0, \\ \end{array}
$$
$$
\lambda_ {n} \| \Delta \mu_ {n} \| \| \Delta \Sigma_ {n} - \Delta \Sigma_ {n} ^ {*} \| / \mathcal {D} \left(G _ {n}, G _ {* , n}\right)\rightarrow 0, \lambda_ {n} ^ {*} \| \Delta \mu_ {n} ^ {*} \| \| \Delta \Sigma_ {n} - \Delta \Sigma_ {n} ^ {*} \| / \mathcal {D} \left(G _ {n}, G _ {* , n}\right)\rightarrow 0. \tag {25}
$$
Combining the results from (21), (24), and (25) leads to
$$
1 = \mathcal {D} (G _ {n}, G _ {*}, n) / \mathcal {D} (G _ {n}, G _ {*}, n) \rightarrow 0,
$$
which is a contradiction. As a consequence, not all the coefficients of  $\frac{\partial^{|\alpha|}f}{\partial\mu^{\alpha_1}\partial\Sigma^{\alpha_2}} (x|\mu_n^*,\Sigma_n^*)$  go to 0 as  $1\leq |\alpha |\leq 2$ . Follow the argument of Proposition C.1, by denoting  $m_{n}$  to be the maximum of the absolute values of the coefficients of  $\frac{\partial^{|\alpha|}f}{\partial\mu^{\alpha_1}\partial\Sigma^{\alpha_2}} (x|\mu_n^*,\Sigma_n^*)$  we achieve for all  $x$  that
$$
\frac {1}{m _ {n}} \frac {p _ {G _ {n}} (x) - p _ {G _ {* , n}} (x)}{W _ {2} ^ {2} \left(G _ {n} , G _ {* , n}\right)} \rightarrow \sum_ {| \alpha | = 1} ^ {2} \tau_ {\alpha} \frac {\partial^ {| \alpha |} f}{\partial \mu^ {\alpha_ {1}} \partial \Sigma^ {\alpha_ {2}}} \left(x | \mu_ {0}, \Sigma_ {0}\right) = 0
$$
where  $\tau_{\alpha} \in \mathbb{R}$  are some coefficients such that not all of them are 0. Due to the second order identifiability condition of  $f$ , the above equation implies that  $\tau_{\alpha} = 0$  for all  $\alpha$  such that  $|\alpha| = 2$ , which is a contradiction. As a consequence, Case 1 cannot happen.
Case 2: Exactly one of  $A_{n}$  and  $B_{n}$  goes to 0, i.e., there exists at least one component among  $(\mu_n,\Sigma_n)$  and  $(\mu_n^*,\Sigma_n^*)$  that does not converge to  $(\mu_0,\Sigma_0)$  as  $n\to \infty$ . Due to the symmetry of  $A_{n}$  and  $B_{n}$ , we assume without loss of generality that  $A_{n}\not\to 0$  and  $B_{n}\to 0$ , which is equivalent to  $(\mu_n,\Sigma_n)\rightarrow (\mu ',\Sigma ')\neq (\mu_0,\Sigma_0)$  while  $(\mu_n^*,\Sigma_n^*)\to (\mu_0,\Sigma_0)$  as  $n\to \infty$ . We denote
$$
\mathcal {D} ^ {\prime} \left(G _ {n}, G _ {* n}\right) = \left| \lambda_ {n} ^ {*} - \lambda_ {n} \right| B _ {n} + \lambda_ {n} A _ {n} + \lambda_ {n} ^ {*} B _ {n}.
$$
Since  $[p_{G_n}(x) - p_{G_{*,n}}(x)] / \mathcal{D}(G_n,G_{*,n})\to 0$  , we achieve that  $[p_{G_n}(x) - p_{G_{*,n}}(x)] / \mathcal{D}'(G_n,G_{*,n})$ $\rightarrow 0$  for all  $x$  as  $\mathcal{D}(G_n,G_{*,n})\lesssim \mathcal{D}'(G_n,G_{*,n})$  . By means of Taylor expansion up to the first order, we have
$$
\begin{array}{l} \frac {p _ {G _ {n}} (x) - p _ {G _ {* , n}} (x)}{\mathcal {D} ^ {\prime} (G _ {n} , G _ {* , n})} = \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) [ f (x | \mu_ {0} , \Sigma_ {0}) - f (x | \mu_ {n} ^ {*} , \Sigma_ {n} ^ {*}) ] + \lambda_ {n} f (x | \mu_ {n} , \Sigma_ {n}) - \lambda_ {n} f (x | \mu_ {n} ^ {*} , \Sigma_ {n} ^ {*})}{\mathcal {D} ^ {\prime} (G _ {n} , G _ {* , n})} \\ = \frac {\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) \left(\sum_ {| \alpha | = 1} \frac {(- \Delta \mu_ {n} ^ {*}) ^ {\alpha_ {1}} (- \Delta \Sigma_ {n} ^ {*}) ^ {\alpha_ {2}}}{\alpha !} \frac {\partial f}{\partial \mu^ {\alpha_ {1}} \partial \Sigma^ {\alpha_ {2}}} \left(x \mid \mu_ {n} ^ {*}, \Sigma_ {n} ^ {*}\right) + R _ {1} ^ {\prime} (x)\right)}{\mathcal {D} ^ {\prime} \left(G _ {n} , G _ {*, n}\right)} \\ + \frac {\lambda_ {n} f (x | \mu_ {n} , \Sigma_ {n}) - \lambda_ {n} f (x | \mu_ {n} ^ {*} , \Sigma_ {n} ^ {*})}{\mathcal {D} ^ {\prime} \left(G _ {n} , G _ {*, n}\right)} \\ \end{array}
$$
where  $R_1'(x)$  is Taylor remainder that satisfies  $(\lambda_n^* - \lambda_n)|R_1'(x)| / \mathcal{D}'(G_n, G_{*,n}) = O(B_n^{\gamma'}) \to 0$  for some positive number  $\gamma' > 0$ . Since  $(\mu_n, \Sigma_n)$  and  $(\mu_n^*, \Sigma_n^*)$  do not have the same limit, they will be different when  $n$  is large enough, i.e.,  $n \geq M'$  for some value of  $M'$ . Now, as  $n \geq M'$ ,  $[p_{G_n}(x) - p_{G_{*,n}}(x)] / \mathcal{D}'(G_n, G_{*,n})$  becomes a linear combination of  $\frac{\partial f}{\partial \mu^{\alpha_1} \partial \Sigma^{\alpha_2}}(x|\mu_n^*, \Sigma_n^*)$  for all  $|\alpha| \leq 1$  and  $f(x|\mu_n, \Sigma_n)$ . If all of the coefficients of these terms go to 0, we would have  $\lambda_n / \mathcal{D}'(G_n, G_{*,n}) \to 0$ ,  $(\lambda_n^* - \lambda_n)(-\Delta \mu_n^*)_i / \mathcal{D}'(G_n, G_{*,n}) \to 0$ , and  $(\lambda_n^* - \lambda_n)(-\Delta \Sigma_n^*)_{uv} / \mathcal{D}'(G_n, G_{*,n}) \to 0$  for all  $1 \leq i \leq d_1$  and  $1 \leq u, v \leq d_2$ . It would imply that  $(\lambda_n^* - \lambda_n)B_n / \mathcal{D}'(G_n, G_{*,n}) \to 0$ ,  $\lambda_n A_n / \mathcal{D}'(G_n, G_{*,n}) \to 0$ , and  $\lambda_n B_n / \mathcal{D}'(G_n, G_{*,n}) \to 0$ . These results lead to
$$
1 = \left(\left| \lambda_ {n} ^ {*} - \lambda_ {n} \right| B _ {n} + \lambda_ {n} A _ {n} + \lambda_ {n} ^ {*} B _ {n}\right) / \mathcal {D} ^ {\prime} \left(G _ {n}, G _ {* , n}\right)\rightarrow 0,
$$
a contradiction. Therefore, not all the coefficients of  $\frac{\partial f}{\partial\mu^{\alpha_1}\partial\Sigma^{\alpha_2}} (x|\mu_n^*,\Sigma_n^*)$  and  $f(x|\mu_n,\Sigma_n)$  go to 0. By defining  $m_{n}^{\prime}$  to be the maximum of these coefficients, we achieve for all  $x$  that
$$
\frac {1}{m _ {n} ^ {\prime}} \frac {p _ {G _ {n}} (x) - p _ {G _ {* , n}} (x)}{\mathcal {D} ^ {\prime} \left(G _ {n} , G _ {* , n}\right)} \rightarrow \eta^ {\prime} f (x | \mu_ {0}, \Sigma_ {0}) + \sum_ {| \alpha | = 0} ^ {1} \tau_ {\alpha} ^ {\prime} \frac {\partial^ {| \alpha |} f}{\partial \mu^ {\alpha_ {1}} \partial \Sigma^ {\alpha_ {2}}} (x | \mu^ {\prime}, \Sigma^ {\prime}) = 0,
$$
where  $\eta^{\prime}$  and  $\tau_{\alpha}^{\prime}$  are coefficients such that not all of them are 0, which is a contradiction to the first order identifiability of  $f$ . As a consequence, Case 2 cannot hold.
Case 3: Both  $A_{n}$  and  $B_{n}$  do not go to 0, i.e.,  $(\mu_n,\Sigma_n)$  and  $(\mu_n^*,\Sigma_n^*)$  do not converge to  $(\mu_0,\Sigma_0)$  as  $n\to \infty$ . Since  $\mathcal{D}_n(G_n,G_{*,n})\lesssim \mathcal{K}(G_n,G_{*,n}) = |\lambda_n - \lambda_n^* | + (\lambda_n + \lambda_n^*)C_n$  and  $[p_{G_n}(x) - p_{G_{*,n}}(x)] / \mathcal{D}(G_n,G_{*,n})\to 0$ , we achieve that  $[p_{G_n}(x) - p_{G_{*,n}}(x)] / \mathcal{K}(G_n,G_{*,n})\to 0$  for all  $x$ . From here, by using the same argument as that of the proof of Proposition C.1, we also reach the contradiction. Therefore, Case 3 cannot happen.
In sum, we achieve the conclusion of the proposition.
C.3 Proof of Theorem 3.6
For the simplicity of proof argument, we will only consider the univariate setting of Gaussian kernel, i.e., when both  $\mu$  and  $\Sigma = \sigma^2$  are scalars. The argument for the multivariate setting of Gaussian kernel can be argued in the rather similar fashion, which is omitted. Throughout this proof, we denote  $v\coloneqq \sigma^2$ . Now, according to the proof argument of Proposition C.1 and Proposition C.3, to achieve the conclusion of the theorem it suffices to demonstrate the following result:
Proposition C.4. Given  $\overline{G} = (\overline{\lambda},\overline{\mu},\overline{v})$  such that  $\overline{\lambda}\in [0,1]$  and  $(\overline{\mu},\overline{v})$  can be identical to  $(\mu_0,v_0)$ . Then, the following holds
(a) If  $(\mu_0, v_0) \neq (\overline{\mu}, \overline{v})$  and  $\overline{\lambda} > 0$ , then
(b) If  $(\mu_0, v_0) \equiv (\overline{\mu}, \overline{v})$  or  $(\mu_0, v_0) \neq (\overline{\mu}, \overline{v})$  and  $\overline{\lambda} = 0$ , then
$$
\lim  _ {\epsilon \rightarrow 0} \inf  _ {G, G _ {*}} \left\{\frac {\| p _ {G} - p _ {G _ {*}} \| _ {\infty}}{\mathcal {Q} (G , G _ {*})}: \mathcal {Q} (G, \overline {{G}}) \vee \mathcal {Q} (G _ {*}, \overline {{G}}) \leq \epsilon \right\} > 0.
$$
Proof. We will only provide the proof for part (b) since the proofs for part (a) can be argued in similar fashion as that of Proposition C.1. Assume that the conclusion of Proposition C.4 does not hold. It implies that we can find two sequences  $G_{n} = (\lambda_{n},\mu_{n},v_{n})$  and  $G_{*,n} = (\lambda_{n}^{*},\mu_{n}^{*},v_{n}^{*})$  such that  $\mathcal{Q}(G_n,\overline{G})\to 0$ ,  $\mathcal{Q}(G_{*,n},\overline{G})\to 0$ , and  $\| p_{G_n} - p_{G_{*,n}}\|_{\infty} / \mathcal{Q}(G_n,G_{*,n})\to 0$  as  $n\to \infty$ . Due to the symmetry between  $\lambda_{n}$  and  $\lambda_{n}^{*}$ , we can assume without loss of generality that  $\lambda_{n}^{*}\geq \lambda_{n}$ . Therefore, we achieve that
$$
\begin{array}{l} \mathcal {Q} (G _ {n}, G _ {*, n}) = (\lambda_ {n} ^ {*} - \lambda_ {n}) (| \Delta \mu_ {n} ^ {*} | ^ {4} + | \Delta v _ {n} ^ {*} | ^ {2}) + \bigg (\lambda_ {n} (| \Delta \mu_ {n} | ^ {2} + | \Delta v _ {n} |) + \lambda_ {n} ^ {*} (| \Delta \mu_ {n} ^ {*} | ^ {2} + | \Delta v _ {n} ^ {*} |) \bigg) \times \\ \times \left(\left| \mu_ {n} - \mu_ {n} ^ {*} \right| ^ {2} + \left| v _ {n} - v _ {n} ^ {*} \right|\right). \\ \end{array}
$$
In this proof, we only consider the scenario when  $\| (\Delta \mu_n,\Delta v_n)\| \to 0$  and  $\| (\Delta \mu_n^*,\Delta v_n^*)\| \to 0$  since the arguments for other settings of these two terms are similar to those of Case 2 and Case 3 in the proof of Proposition C.3. As being indicated in Section 3.2.2, the univariate Gaussian kernel contains the partial differential equation structure  $\frac{\partial^2f}{\partial\mu^2} (x|\mu ,v) = 2\frac{\partial f}{\partial v} (x|\mu ,v)$  for all  $\mu \in \Theta$  and  $v\in \Omega$ . Therefore, for any  $\alpha = (\alpha_{1},\alpha_{2})$  we can check that
$$
\frac {\partial^ {| \alpha |} f}{\partial \mu^ {\alpha_ {1}} \partial v ^ {\alpha_ {2}}} (x | \mu , v) = \frac {1}{2 ^ {\alpha_ {2}}} \frac {\partial^ {\beta} f}{\partial \mu^ {\beta}} (x | \mu , v)
$$
where  $\beta = \alpha_{1} + 2\alpha_{2}$ . Now, by means of Taylor expansion up to the fourth order, we obtain
$$
\begin{array}{l} \frac {p _ {G _ {n}} (x) - p _ {G _ {* , n}} (x)}{\mathcal {Q} (G _ {n} , G _ {* , n})} = \frac {\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) \left(\sum_ {| \alpha | = 1} ^ {4} \frac {(- \Delta \mu_ {n} ^ {*}) ^ {\alpha_ {1}} (- \Delta v _ {n} ^ {*}) ^ {\alpha_ {2}}}{\alpha_ {1} ! \alpha_ {2} !} \frac {\partial^ {| \alpha |} f}{\partial \mu^ {\alpha_ {1}} \partial v ^ {\alpha_ {2}}} (x | \mu_ {n} ^ {*}, v _ {n} ^ {*}) + R _ {1} (x)\right)}{\mathcal {Q} (G _ {n} , G _ {* , n})} \\ + \frac {\lambda_ {n} \left(\sum_ {| \alpha | = 1} ^ {4} \frac {\left(\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}\right) ^ {\alpha_ {1}} \left(\Delta v _ {n} - \Delta v _ {n} ^ {*}\right) ^ {\alpha_ {2}}}{\alpha_ {1} ! \alpha_ {2} !} \frac {\partial^ {| \alpha |} f}{\partial \mu^ {\alpha_ {1}} \partial v ^ {\alpha_ {2}}} \left(x | \mu_ {n} ^ {*} , v _ {n} ^ {*}\right) + R _ {2} (x)\right)}{\mathcal {Q} \left(G _ {n} , G _ {* , n}\right)} \\ = \sum_ {\beta = 1} ^ {8} \sum_ {\alpha_ {1}, \alpha_ {2}} \frac {\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) \left(- \Delta \mu_ {n} ^ {*}\right) ^ {\alpha_ {1}} \left(- \Delta v _ {n} ^ {*}\right) ^ {\alpha_ {2}} + \lambda_ {n} \left(\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}\right) ^ {\alpha_ {1}} \left(\Delta v _ {n} - \Delta v _ {n} ^ {*}\right) ^ {\alpha_ {2}}}{2 ^ {\alpha_ {2}} \alpha_ {1} ! \alpha_ {2} ! \mathcal {Q} \left(G _ {n} , G _ {* , n}\right)} \\ \times \frac {\partial^ {\beta} f}{\partial \mu^ {\beta}} (x | \mu_ {n} ^ {*}, v _ {n} ^ {*}) + \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) R _ {1} (x) + \lambda_ {n} R _ {2} (x)}{\mathcal {Q} (G _ {n} , G _ {*, n})} \\ \end{array}
$$
where  $R_{1}(x), R_{2}(x)$  are Taylor remainders and the range of  $\alpha_{1}, \alpha_{2}$  in the summation of the second equality satisfies  $\beta = \alpha_{1} + 2\alpha_{2}$ . As Gaussian kernel admits fourth-order uniform Lipschitz condition, it is clear that
$$
\frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | R _ {1} (x) | + \lambda_ {n} | R _ {2} (x) |}{\mathcal {Q} (G _ {n} , G _ {* , n})} = \mathcal {O} (\| (\Delta \mu_ {n} ^ {*}, \Delta v _ {n} ^ {*}) \| ^ {\gamma} + \| (\mu_ {n}, v _ {n}) - (\mu_ {n} ^ {*}, v _ {n} ^ {*}) \| ^ {\gamma}) \rightarrow 0
$$
as  $n\to \infty$  for some  $\gamma >0$ . Therefore, we can consider  $[p_{G_n}(x) - p_{G_{*,n}}(x)] / \mathcal{Q}(G_n,G_{*,n})$  as a linear combination of  $\frac{\partial^{\beta}f}{\partial\mu^{\beta}} (x|\mu_n^*,v_n^*)$  for  $1\leq \beta \leq 8$ . If all of the coefficients of these terms go to 0, then we obtain
$$
L _ {\beta} = \frac {\sum_ {\alpha_ {1} , \alpha_ {2}} \frac {\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) (- \Delta \mu_ {n} ^ {*}) ^ {\alpha_ {1}} (- \Delta v _ {n} ^ {*}) ^ {\alpha_ {2}} + \lambda_ {n} (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) ^ {\alpha_ {1}} (\Delta v _ {n} - \Delta v _ {n} ^ {*}) ^ {\alpha_ {2}}}{2 ^ {| \alpha_ {2} |} \alpha_ {1} ! \alpha_ {2} !}}{\mathcal {Q} (G _ {n} , G _ {*, n})} \rightarrow 0
$$
for any  $1 \leq \beta \leq 8$ . Now, we divide our argument with  $L_{\beta}$  into two key cases
Case 1:  $\left(\lambda_n(|\Delta \mu_n|^2 + |\Delta v_n|) + \lambda_n^* |(\Delta \mu_n^*|^2 + |\Delta v_n^*)\right) / \left\{\lambda_n(|\mu_n - \mu_n^*|^2 + |v_n - v_n^*)\right\} \not\to \infty.$
It implies that as  $n$  is large enough, we would have
$$
Q \left(G _ {n}, G _ {* , n}\right) \lesssim \left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) \left(\left| \Delta \mu_ {n} ^ {*} \right| ^ {4} + \left| \Delta v _ {n} ^ {*} \right| ^ {2}\right) + \lambda_ {n} \left(\left| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} \right| ^ {4} + \left| \Delta v _ {n} - \Delta v _ {n} ^ {*} \right| ^ {2}\right).
$$
Combining the above result with  $L_{\beta}\rightarrow 0$  for all  $1\leq \beta \leq 8$  , we get
$$
H _ {\beta} = \frac {\sum_ {\alpha_ {1} , \alpha_ {2}} \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) (- \Delta \mu_ {n} ^ {*}) ^ {\alpha_ {1}} (- \Delta v _ {n} ^ {*}) ^ {\alpha_ {2}} + \lambda_ {n} (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) ^ {\alpha_ {1}} (\Delta v _ {n} - \Delta v _ {n} ^ {*}) ^ {\alpha_ {2}}}{2 ^ {| \alpha_ {2} |} \alpha_ {1} ! \alpha_ {2} !}}{\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) (| \Delta \mu_ {n} ^ {*} | ^ {4} + | \Delta v _ {n} ^ {*} | ^ {2}) + \lambda_ {n} (| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} | ^ {4} + | \Delta v _ {n} - \Delta v _ {n} ^ {*} | ^ {2})} \to 0
$$
Note that, when the denominator of the above limits is  $(\lambda_n^* - \lambda_n)(|\Delta \mu_n^*|^4 + |\Delta v_n^*|^4) + \lambda_n(|\Delta \mu_n - \Delta \mu_n^*|^4 + |\Delta v_n - \Delta v_n^*|^4)$ , the technique for studying the above system of limits with this denominator has been considered in Proposition 2.3 in [21]. However, since the current denominator of  $H_{\beta}$  strongly dominates by the previous denominator, we must develop a more sophisticated control of  $H_{\beta}$  as  $1 \leq \beta \leq 8$  to obtain a concrete understanding of their limits. Due to the symmetry between  $\lambda_n^* - \lambda_n$  and  $\lambda_n$ , we assume without loss of generality that  $\lambda_n^* - \lambda_n \leq \lambda_n$  for all  $n$  (by the subsequence argument). We have two possibilities regarding  $\lambda_n$  and  $\lambda_n^*$
Case 1.1:  $(\lambda_n^* - \lambda_n) / \lambda_n \neq 0$  as  $n \to \infty$ . Under that setting, we define  $p_n = \max \left\{\lambda_n^* - \lambda_n, \lambda_n\right\}$  and
$$
M _ {n} = \max  \left\{| \Delta \mu_ {n} ^ {*} |, | \Delta \mu_ {n} ^ {*} - \Delta \mu_ {n} |, | \Delta v _ {n} ^ {*} | ^ {1 / 2}, | \Delta v _ {n} ^ {*} - \Delta v _ {n} | ^ {1 / 2} \right\}
$$
Additionally, we let  $(\lambda_n^* - \lambda_n) / p_n \to c_1^2$ ,  $\lambda_n / p_n \to c_2^2$ ,  $\Delta \mu_n^* / M_n \to -a_1$ ,  $(\Delta \mu_n^* - \Delta \mu_n) / M_n \to a_2$ ,  $\Delta v_n^* / M_n^2 \to -2b_1$ , and  $(\Delta v_n - \Delta v_n^*) / M_n^2 \to 2b_2$ . From here, at least one among  $a_1, a_2, b_1, b_2$  and both  $c_1, c_2$  are different from 0. Now, by dividing both the numerators and the denominators of  $H_\beta$  as  $1 \leq \beta \leq 4$  by  $p_n M_n^\beta$ , we achieve the following system of polynomial equations
$$
\begin{array}{l} c _ {1} ^ {2} a _ {1} + c _ {2} ^ {2} a _ {2} = 0 \\ \frac {1}{2} \left(c _ {1} ^ {2} a _ {1} ^ {2} + c _ {2} ^ {2} a _ {2} ^ {2}\right) + c _ {1} ^ {2} b _ {1} + c _ {2} ^ {2} b _ {2} = 0 \\ \frac {1}{3 !} \left(c _ {1} ^ {2} a _ {1} ^ {3} + c _ {2} ^ {2} a _ {2} ^ {3}\right) + c _ {1} ^ {2} a _ {1} b _ {1} + c _ {2} ^ {2} a _ {2} b _ {2} = 0 \\ \frac {1}{4 !} \left(c _ {1} ^ {2} a _ {1} ^ {4} + c _ {2} ^ {2} a _ {2} ^ {4}\right) + \frac {1}{2 !} \left(c _ {1} ^ {2} a _ {1} ^ {2} b _ {1} + c _ {2} ^ {2} a _ {2} ^ {2} b _ {2}\right) + \frac {1}{2 !} \left(c _ {1} ^ {2} b _ {1} ^ {2} + c _ {2} ^ {2} b _ {2} ^ {2}\right) = 0, \\ \end{array}
$$
As being indicated in Proposition 2.1 in [21], this system will only admits the trivial solution, i.e.,  $a_1 = a_2 = b_1 = b_2 = 0$ , which is a contradiction. Therefore, Case 1.1 cannot happen.
Case 1.2:  $(\lambda_n^* - \lambda_n) / \lambda_n \to 0$ , i.e.,  $\lambda_n^* / \lambda_n \to 1$ , as  $n \to \infty$ . Under that setting, if  $M_n \in \max \left\{ |\Delta \mu_n - \Delta \mu_n^*|, |\Delta v_n - \Delta v_n^*|^{1/2} \right\}$ , then we have
$$
\lambda_ {n} M _ {n} ^ {4} = \max \Biggl \{(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta \mu_ {n} ^ {*} | ^ {4}, (\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} | ^ {4}, \lambda_ {n} | \Delta v _ {n} ^ {*} | ^ {2}, \lambda_ {n} | \Delta v _ {n} - \Delta v _ {n} ^ {*} | ^ {2} \Biggr \}.
$$
By dividing both the numerator and the denominator of  $H_{1}$  by  $\lambda_{n}M_{n}$ , given that the new denominator of  $H_{1}$  goes to 0, its new numerator also goes to 0, i.e., we obtain
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left(- \Delta \mu_ {n} ^ {*}\right) / \left\{\lambda_ {n} M _ {n} \right\} + \left(\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}\right) / M _ {n} \rightarrow 0.
$$
Since  $(\lambda_n^* - \lambda_n) / \lambda_n \to 0$  and  $|\Delta \mu_n^*| \leq M_n$ , we have  $(\lambda_n^* - \lambda_n)(-\Delta \mu_n^*) / \{\lambda_n M_n\} \to 0$ . Therefore, we have  $(\Delta \mu_n - \Delta \mu_n^*) / M_n \to 0$ . With the previous results, by dividing both the numerator and the denominator of  $H_2$  by  $\lambda_n M_n^2$  and given that the new denominator goes to 0, we have
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left(- \Delta v _ {n} ^ {*}\right) / \left\{\lambda_ {n} M _ {n} ^ {2} \right\} + \left(\Delta v _ {n} - \Delta v _ {n} ^ {*}\right) / M _ {n} ^ {2} \rightarrow 0.
$$
As  $(\lambda_n^* - \lambda_n)(-\Delta v_n^*) / \{\lambda_n M_n^2\} \to 0$  (due to the assumption of  $M_{n}$ ), we get  $(\Delta v_{n} - \Delta v_{n}^{*}) / M_{n}^{2}\to 0$ . These results imply that