SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
1 = \frac {\operatorname* {m a x} \left\{\left| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} \right| ^ {2} , \left| \Delta v _ {n} - \Delta v _ {n} ^ {*} \right|\right\}}{M _ {n} ^ {2}} \rightarrow 0,
$$
which is a contradiction. Therefore, we would only have  $M_{n} \in \max \left\{|\Delta \mu_{n}^{*}|, |\Delta v_{n}^{*}|^{1/2}\right\}$ . For the simplicity of the proof, we only consider the setting when  $M_{n} = |\Delta \mu_{n}^{*}|$  for all  $n$  (by subsequence argument). The setting that  $M_{n} = |\Delta v_{n}^{*}|^{1/2}$  for all  $n$  can be argued in the similar fashion. Now, if we have
$$
\max \left\{\left| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} \right|, \left| \Delta v _ {n} - \Delta v _ {n} ^ {*} \right| ^ {1 / 2} \right\} / M _ {n} \not \to 0,
$$
then by dividing the numerator and denominator of  $H_{i}$  with  $\lambda_{n}\left(\max \left\{\left|\Delta \mu_{n} - \Delta \mu_{n}^{*}\right|, \left|\Delta v_{n} - \Delta v_{n}^{*}\right|^{1 / 2}\right\}\right)^{i}$  as  $1 \leq i \leq 2$ , we would achieve
$$
1 = \frac {\max \left\{\left| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} \right| ^ {2} , \left| \Delta v _ {n} - \Delta v _ {n} ^ {*} \right| \right\}}{\max \left\{\left| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} \right| ^ {2} , \left| \Delta v _ {n} - \Delta v _ {n} ^ {*} \right| \right\}} \to 0,
$$
a contradiction. Therefore, we must have
$$
\left. \right. \max  \left\{\left| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} \right|, \left| \Delta v _ {n} - \Delta v _ {n} ^ {*} \right| ^ {1 / 2} \right\} / M _ {n} \neq 0 \tag {26}
$$
as  $n\to \infty$  . Now, we further divide the argument under that setting of  $M_{n}$  into two small cases
Case 1.2.1:  $(\lambda_n^* - \lambda_n)|\Delta \mu_n^*|^4 \leq \lambda_n|\Delta \mu_n - \Delta \mu_n^*|^4$  for all  $n$  (by subsequence argument). Since  $M_n = |\Delta \mu_n^*|$ , we would have  $(\lambda_n^* - \lambda_n)|\Delta \mu_n^*|^i \leq \lambda_n|\Delta \mu_n - \Delta \mu_n^*|^i$  for all  $n$  and  $1 \leq l \leq 4$ . From here, we obtain that
$$
\frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta \mu_ {n} ^ {*} | ^ {4}}{\lambda_ {n} | \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} |} \leq \frac {\lambda_ {n} | \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} | ^ {4}}{\lambda_ {n} | \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} |} \to 0,
$$
$$
\frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta v _ {n} ^ {*} | ^ {2}}{\lambda_ {n} | \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} |} \leq \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta \mu_ {n} ^ {*} | ^ {4}}{\lambda_ {n} | \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} |} \to 0.
$$
If  $|\Delta \mu_n - \Delta \mu_n^*| / |\Delta v_n - \Delta v_n^*|^{1/2} \nrightarrow 0$ , by diving both the numerator and the denominator of  $H_1$  by  $\lambda_n |\Delta \mu_n - \Delta \mu_n^*|$  and given that the new denominator goes to 0, the new numerator must converge to 0, i.e. we have
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) \Delta \mu_ {n} ^ {*} / \left\{\lambda_ {n} \left(\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}\right)\right\}\rightarrow - 1.
$$
However, since we have  $|(\Delta \mu_n - \Delta \mu_n^*) / \Delta \mu_n^* \to 0$ , the above result would imply that
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left| \Delta \mu_ {n} ^ {*} \right| ^ {4} / \left\{ \right.\lambda_ {n} \mid \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} \left. \right| ^ {4} \} \rightarrow \infty ,
$$
which is a contradiction to the assumption of Case 1.2.1.1. As a consequence, we must have  $|\Delta \mu_n - \Delta \mu_n^*| / |\Delta v_n - \Delta v_n^*|^{1/2} \to 0$ . Now, we also have that
$$
\frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta \mu_ {n} ^ {*} | ^ {4}}{\lambda_ {n} | \Delta v _ {n} - \Delta v _ {n} ^ {*} | ^ {i / 2}} \lesssim \frac {\lambda_ {n} | \Delta v _ {n} - \Delta v _ {n} ^ {*} | ^ {2}}{\lambda_ {n} | \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} | ^ {i / 2}} \to 0,
$$
$$
\frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta v _ {n} ^ {*} | ^ {4}}{\lambda_ {n} | \Delta v _ {n} - \Delta v _ {n} ^ {*} | ^ {i / 2}} \leq \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta \mu_ {n} ^ {*} | ^ {4}}{\lambda_ {n} | \Delta v _ {n} - \Delta v _ {n} ^ {*} | ^ {i / 2}} \to 0.
$$
for all  $1 \leq i \leq 3$ . Without loss of generality, we assume that  $\Delta v_{n} - \Delta v_{n}^{*} > 0$  for all  $n$ . We denote  $(- \Delta \mu_{n}^{*}) = q_{1}^{n} (\Delta v_{n} - \Delta v_{n}^{*})$  and  $\Delta v_{n}^{*} = q_{2}^{n} (\Delta v_{n} - \Delta v_{n}^{*})$  for all  $n$ . From the result of (26), we would have  $|q_{1}^{n}| \to \infty$ . Given the above results, by dividing the numerators and the denominators of  $H_{\beta}$  by  $\lambda_{n} (\Delta v_{n} - \Delta v_{n}^{*})^{\beta / 2}$  for any  $1 \leq \beta \leq 3$ , we would have the new denominators go to 0. Therefore, all the new numerators of these  $H_{\beta}$  also go to 0, i.e. we achieve the following system of limits
$$
\frac {\lambda_ {n} ^ {*} - \lambda_ {n}}{\lambda_ {n}} q _ {1} ^ {n} \rightarrow 0, \frac {\lambda_ {n} ^ {*} - \lambda_ {n}}{\lambda_ {n}} \left\{(q _ {1} ^ {n}) ^ {2} + q _ {2} ^ {n} \right\} + 1 \rightarrow 0, \frac {\lambda_ {n} ^ {*} - \lambda_ {n}}{\lambda_ {n}} \left(\frac {(q _ {1} ^ {n}) ^ {3}}{6} + \frac {q _ {1} ^ {n} q _ {2} ^ {n}}{2}\right)\rightarrow 0.
$$
Since  $|q_1^n| \to \infty$ , the last limit in the above system implies that  $(\lambda_n^* - \lambda_n)\left(\frac{(q_1^n)^2}{3} + q_2^n\right) / \lambda_n \to 0$ . Combining this result with the second limit in the above system yields that  $(\lambda_n^* - \lambda_n)(q_1^n)^2 / \lambda_n + 3/2 \to 0$ , which cannot happen. Therefore, Case 1.2.1 does not hold.
Case 1.2.2:  $(\lambda_{n}^{*} - \lambda_{n})|\Delta \mu_{n}^{*}|^{4} > \lambda_{n}|\Delta \mu_{n} - \Delta \mu_{n}^{*}|^{4}$  for all  $n$  (by subsequence argument). If  $(\lambda_{n}^{*} - \lambda_{n})|\Delta \mu_{n}^{*}|^{4} \leq \lambda_{n}|\Delta v_{n} - \Delta v_{n}^{*}|^{2}$  for all  $n$ , the by using the same argument as that of Case 1.2.1, we quickly achieve the contradiction. Therefore, we must have  $(\lambda_{n}^{*} - \lambda_{n})|\Delta \mu_{n}^{*}|^{4} > \lambda_{n}|\Delta v_{n} - \Delta v_{n}^{*}|^{2}$ . Denote  $(\Delta \mu_{n} - \Delta \mu_{n}^{*}) = m_{1}^{n}(-\Delta \mu_{n}^{*}), (-\Delta v_{n}^{*}) = m_{2}^{n}(\Delta \mu_{n}^{*})^{2},$  and  $(\Delta v_{n} - \Delta v_{n}^{*}) = m_{3}^{n}(\Delta \mu_{n}^{*})^{2}$ . Since  $M_{n} = |\Delta \mu_{n}^{*}|$ , we would have  $|m_i^n| \leq 1$  for all  $1 \leq i \leq 3$ . Denote  $m_i^n \to m_i$  for all  $1 \leq i \leq 3$  (by subsequence argument). The results of (26) lead to  $m_1 = m_3 = 0$ . Now by dividing both the numerator and denominator of  $H_{\beta}$  by  $(\lambda_n^* -\lambda_n)(-\Delta \mu_n^*)^\beta$  for any  $1 \leq \beta \leq 4$ , as the new denominators of  $H_{|\beta |}$  do not go to  $\infty$ , we would also achieve that the new numerators of  $H_{|\beta |}$  go to 0, i.e. the following system of limits hold
$$
1 + \frac {\lambda_ {n} ^ {*} - \lambda_ {n}}{\lambda_ {n}} m _ {1} ^ {n} \rightarrow 0, \left[ 1 + \frac {\lambda_ {n} ^ {*} - \lambda_ {n}}{\lambda_ {n}} (m _ {1} ^ {n}) ^ {2} \right] + m _ {2} ^ {n} + \frac {\lambda_ {n} ^ {*} - \lambda_ {n}}{\lambda_ {n}} m _ {3} ^ {n} \rightarrow 0,
$$
$$
\left(1 + \frac {\lambda_ {n} ^ {*} - \lambda_ {n}}{\lambda_ {n}} (m _ {1} ^ {n}) ^ {3}\right) / 6 + \left(m _ {2} ^ {n} + \frac {\lambda_ {n} ^ {*} - \lambda_ {n}}{\lambda_ {n}} m _ {1} ^ {n} m _ {3} ^ {n}\right) / 2 \rightarrow 0,
$$
$$
\left(1 + \frac {\lambda_ {n} ^ {*} - \lambda_ {n}}{\lambda_ {n}} (m _ {1} ^ {n}) ^ {4}\right) / 2 4 + \left(m _ {2} ^ {n} + \frac {\lambda_ {n} ^ {*} - \lambda_ {n}}{\lambda_ {n}} (m _ {3} ^ {n}) ^ {2}\right) / 4 + \left((m _ {2} ^ {n}) ^ {2} + \frac {\lambda_ {n} ^ {*} - \lambda_ {n}}{\lambda_ {n}} (m _ {3} ^ {n}) ^ {2}\right) / 8 \rightarrow 0.
$$
Combining with  $m_1^n \to 0$ , the first and third limit of the above system of limits imply that  $m_2 = -1/3$ . From here, the second and fourth limit yields that  $1/6 + m_2 + m_2^2/2 = 0$ , which is a contradiction. Therefore, Case 1.2.2 cannot hold.
Case 2:  $\left(\lambda_{n}(|\Delta \mu_{n}|^{2} + |\Delta v_{n}|) + \lambda_{n}^{*}(|\Delta \mu_{n}^{*}|^{2} + |\Delta v_{n}^{*}|)\right) / \left\{\lambda_{n}(|\mu_{n} - \mu_{n}^{*}|^{2} + |v_{n} - v_{n}^{*}|)\right\} \to \infty .$
$$
\begin{array}{l} \overline {{\mathcal {Q}}} (G _ {n}, G _ {*, n}) = (\lambda_ {n} ^ {*} - \lambda_ {n}) (| \Delta \mu_ {n} | ^ {2} + | \Delta v _ {n} |) (| \Delta \mu_ {n} ^ {*} | ^ {2} + | \Delta v _ {n} ^ {*} |) + \left(\lambda_ {n} (| \Delta \mu_ {n} | ^ {2} + | \Delta v _ {n} |\right) \\ + \lambda_ {n} ^ {*} (| \Delta \mu_ {n} ^ {*} | ^ {2} + | \Delta v _ {n} ^ {*} |)) \left(| \mu_ {n} - \mu_ {n} ^ {*} | ^ {2} + | v _ {n} - v _ {n} ^ {*} |\right). \\ \end{array}
$$
We will demonstrate that  $\mathcal{Q}(G_n, G_{*,n}) \asymp \overline{\mathcal{Q}}(G_n, G_{*,n})$ . In fact, from the above formulation of  $\overline{\mathcal{Q}}(G_n, G_{*,n})$ , we would have that
$$
\begin{array}{l} \overline {{\mathcal {Q}}} (G _ {n}, G _ {*}, n) \leq 2 (\lambda_ {n} ^ {*} - \lambda_ {n}) (| \Delta \mu_ {n} ^ {*} | ^ {2} + | \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} | ^ {2} + | \Delta v _ {n} ^ {*} | + | \Delta v _ {n} - \Delta v _ {n} ^ {*} |) (| \Delta \mu_ {n} ^ {*} | ^ {2} + | \Delta v _ {n} ^ {*} |) \\ + \quad 2 \left(\lambda_ {n} \left(| \Delta \mu_ {n} | ^ {2} + | \Delta v _ {n} |\right) + \lambda_ {n} ^ {*} \left(| \Delta \mu_ {n} ^ {*} | ^ {2} + | \Delta v _ {n} ^ {*} |\right)\right) \left(| \mu_ {n} - \mu_ {n} ^ {*} | ^ {2} + | v _ {n} - v _ {n} ^ {*} |\right) \\ \leq 2 \mathcal {Q} \left(G _ {n}, G _ {* , n}\right) \\ \end{array}
$$
where the first inequality is due to the triangle inequality and basic inequality  $(a + b)^2 \leq 2(a^2 + b^2)$  and the second inequality is due to the following result
$$
\left. \left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) \left(| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} | ^ {2} + | \Delta v _ {n} - \Delta v _ {n} ^ {*} |\right) \leq \lambda_ {n} ^ {*} \left(| \mu_ {n} - \mu_ {n} ^ {*} | ^ {2} + | v _ {n} - v _ {n} ^ {*} |\right) \right.
$$
On the other hand, we also have that
$$
\begin{array}{l} 2 \bar {\mathcal {Q}} (G _ {n}, G _ {* n}) \geq (\lambda_ {n} ^ {*} - \lambda_ {n}) \left(\left| \Delta \mu_ {n} ^ {*} \right| ^ {2} + \left| \Delta v _ {n} ^ {*} \right|\right) \left(\left| \Delta \mu_ {n} \right| ^ {2} + \left| \Delta v _ {n} \right| + \left| \mu_ {n} - \mu_ {n} ^ {*} \right| ^ {2} + \left| v _ {n} - v _ {n} ^ {*} \right|\right) \\ + \left. \left(\lambda_ {n} \left(\left| \Delta \mu_ {n} \right| ^ {2} + \left| \Delta v _ {n} \right|\right) + \lambda_ {n} ^ {*} \left(\left| \Delta \mu_ {n} ^ {*} \right| ^ {2} + \left| \Delta v _ {n} ^ {*} \right|\right)\right) \left(\left| \mu_ {n} - \mu_ {n} ^ {*} \right| ^ {2} + \left| v _ {n} - v _ {n} ^ {*} \right|\right) \right. \\ \geq \mathcal {Q} (G _ {n}, G _ {* n}) / 2 \\ \end{array}
$$
where the last inequality is due to triangle inequality and basic inequality  $(a + b)^2 \leq 2(a^2 + b^2)$ . Therefore, we conclude that  $\mathcal{Q}(G_n, G_{*,n}) \asymp \overline{\mathcal{Q}}(G_n, G_{*,n})$ . Now, since  $H_\beta \to 0$  for all  $1 \leq \beta \leq 8$ , we would have that
$$
F _ {\beta} = \frac {\sum_ {\alpha_ {1} , \alpha_ {2}} \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) (- \Delta \mu_ {n} ^ {*}) ^ {\alpha_ {1}} (- \Delta v _ {n} ^ {*}) ^ {\alpha_ {2}} + \lambda_ {n} (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) ^ {\alpha_ {1}} (\Delta v _ {n} - \Delta v _ {n} ^ {*}) ^ {\alpha_ {2}}}{2 ^ {| \alpha_ {2} |} \alpha_ {1} ! \alpha_ {2} !}}{\overline {{\mathcal {Q}}} (G _ {n} , G _ {*, n})} \to 0.
$$
Similar to Case 1, under Case 2 we also consider two distinct settings of  $\lambda_n^* / \lambda_n$
Case 2.1:  $\lambda_n^* /\lambda_n\neq \infty$  . Under this case, we denote
$$
M _ {n} ^ {\prime} := \max  \left\{\left| \Delta \mu_ {n} \right| ^ {2}, \left| \Delta v _ {n} \right|, \left| \Delta \mu_ {n} ^ {*} \right| ^ {2}, \left| \Delta v _ {n} ^ {*} \right| \right\}.
$$
From the assumption of Case 2, we would have
$$
\left| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} \right| ^ {2} / M _ {n} ^ {\prime} \rightarrow 0, \left| \Delta v _ {n} - \Delta v _ {n} ^ {*} \right| / M _ {n} ^ {\prime} \rightarrow 0. \tag {27}
$$
Due to the symmetry between  $\left( |\Delta \mu_n|^2, |\Delta v_n| \right)$  and  $\left( |\Delta \mu_n^*|^2, |\Delta v_n^*| \right)$ , we assume without loss of generality that  $M_n' \in \max \left\{ |\Delta \mu_n|^2, |\Delta v_n| \right\}$ . Under that assumption, we have two distinct cases
Case 2.1.1:  $M_n' = |\Delta \mu_n|^2$  for all  $n$  (by the subsequence argument). From (27), we have  $|\Delta \mu_n - \Delta \mu_n^*| / |\Delta \mu_n| \to 0$ , i.e.,  $\Delta \mu_n / \Delta \mu_n^* \to 1$ . To be able to utilize the assumptions of Case 2, we will need to study the formulations of  $F_\beta$  more deeply. In fact, when  $\beta = 1$  simple calculation yields
$$
A _ {1} := \left(\lambda_ {n} \Delta \mu_ {n} - \lambda_ {n} ^ {*} \Delta \mu_ {n} ^ {*}\right) / \overline {{\mathcal {Q}}} \left(G _ {n}, G _ {*, n}\right)\rightarrow 0.
$$
When  $\beta = 2$ , we have
$$
F _ {2} = \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) (\Delta \mu_ {n} ^ {*}) ^ {2} + \lambda_ {n} (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) ^ {2} + (\lambda_ {n} ^ {*} - \lambda_ {n}) (- \Delta v _ {n} ^ {*}) + \lambda_ {n} (\Delta v _ {n} - \Delta v _ {n} ^ {*})}{\overline {{Q}} (G _ {n} , G _ {* , n})} \to 0.
$$
Combining with the result of  $A_{1}$ , it is clear that
$$
\frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) (\Delta \mu_ {n} ^ {*}) ^ {2} + \lambda_ {n} (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) ^ {2}}{\overline {{\mathcal {Q}}} (G _ {n} , G _ {\ast , n})} \to \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) \Delta \mu_ {n} \Delta \mu_ {n} ^ {*}}{\overline {{\mathcal {Q}}} (G _ {n} , G _ {\ast , n})}.
$$
Combining the above result with  $F_{2}\to 0$  , we would have
$$
A _ {2} := \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) \Delta \mu_ {n} \Delta \mu_ {n} ^ {*} + \lambda_ {n} \Delta v _ {n} - \lambda_ {n} ^ {*} \Delta v _ {n} ^ {*}}{\overline {{\mathcal {Q}}} (G _ {n} , G _ {* , n})} \to 0.
$$
Now, we have two small cases
Case 2.1.1.1:  $\Delta v_{n} / (\Delta \mu_{n})^{2}\to 0$  as  $n\to \infty$ . From (27), since we have  $|\Delta v_{n} - \Delta v_{n}^{*}| / (\Delta \mu_{n})^{2}\to 0$ , it implies that  $\Delta v_n^* /(\Delta \mu_n)^2\rightarrow 0$ . Since  $\Delta \mu_{n} / \Delta \mu_{n}^{*}\rightarrow 1$ , we also have that  $\Delta v_{n}^{*} / (\Delta \mu_{n}^{*})^{2}\rightarrow 0$ . Now, from the formulations of  $\overline{\mathcal{Q}}(G_n,G_{*,n})$  we have
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left| \Delta v _ {n} \Delta v _ {n} ^ {*} \right| / \bar {\mathcal {Q}} \left(G _ {n}, G _ {* n}\right) \leq \left| \Delta v _ {n} \right| / \left| \Delta \mu_ {n} \right| ^ {2} \rightarrow 0,
$$
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left| \Delta v _ {n} \left(\Delta \mu_ {n} ^ {*}\right) ^ {2} \right| / \overline {{\mathcal {Q}}} \left(G _ {n}, G _ {\ast , n}\right) \leq \left| \Delta v _ {n} \right| / \left| \Delta \mu_ {n} \right| ^ {2} \rightarrow 0,
$$
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left| \Delta v _ {n} ^ {*} \left(\Delta \mu_ {n}\right) ^ {2} \right| / \overline {{\mathcal {Q}}} \left(G _ {n}, G _ {* , n}\right) \leq \left| \Delta v _ {n} ^ {*} \right| / \left| \Delta \mu_ {n} ^ {*} \right| ^ {2} \rightarrow 0. \tag {28}
$$
From the result that  $A_{2}\rightarrow 0$ , by multiplying  $A_{2}$  with  $\Delta \mu_n\Delta \mu_n^*$ , we would also have that
$$
\frac {\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left(\Delta \mu_ {n} \Delta \mu_ {n} ^ {*}\right) ^ {2} + \left(\lambda_ {n} \Delta v _ {n} - \lambda_ {n} ^ {*} \Delta v _ {n} ^ {*}\right) \Delta \mu_ {n} \Delta \mu_ {n} ^ {*}}{\bar {Q} \left(G _ {n} , G _ {* , n}\right)} \rightarrow 0. \tag {29}
$$
As  $\lambda_n^* / \lambda_n \not\to \infty$ , we have two distinct settings of  $\lambda_n^* / \lambda_n$
Case 2.1.1.1:  $\lambda_n^* / \lambda_n \nrightarrow 1$ . Using the result from (28) and the fact that  $\Delta \mu_n / \Delta \mu_n^* \to 1$ , we would obtain that
$$
\left(\lambda_ {n} \Delta v _ {n} - \lambda_ {n} ^ {*} \Delta v _ {n} ^ {*}\right) \Delta \mu_ {n} \Delta \mu_ {n} ^ {*} / \overline {{\mathcal {Q}}} \left(G _ {n}, G _ {*, n}\right)\rightarrow 0.
$$
Combining the above result with (29), it leads to
$$
\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right)\left(\Delta \mu_ {n} \Delta \mu_ {n} ^ {*}\right) ^ {2} / \bar {\mathcal {Q}} \left(G _ {n}, G _ {*, n}\right)\rightarrow 0. \tag {30}
$$
Combining (28) and (30), we would achieve that
$$
\frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) (| \Delta \mu_ {n} | ^ {2} + | \Delta v _ {n} |) (| \Delta \mu_ {n} ^ {*} | ^ {2} + | \Delta v _ {n} ^ {*} |)}{\overline {{\mathcal {Q}}} (G _ {n} , G _ {* , n})} \to 0.
$$
From the formulation of  $\overline{\mathcal{Q}}(G_n, G_{*,n})$ , the above limit implies that
$$
E := \frac {\left(\lambda_ {n} (| \Delta \mu_ {n} | ^ {2} + | \Delta v _ {n} |) + \lambda_ {n} ^ {*} | (\Delta \mu_ {n} ^ {*} | ^ {2} + | \Delta v _ {n} ^ {*} |\right) \left(| \mu_ {n} - \mu_ {n} ^ {*} | ^ {2} + | v _ {n} - v _ {n} ^ {*} |\right)}{\overline {{\mathcal {Q}}} (G _ {n} , G _ {* , n})} \to 1.
$$
Due to the previous assumptions, we obtain that
$$
E \lesssim \frac {\operatorname* {m a x} \left\{\lambda_ {n} (\Delta \mu_ {n}) ^ {2} (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) ^ {2} , \lambda_ {n} (\Delta \mu_ {n}) ^ {2} (\Delta v _ {n} - \Delta v _ {n} ^ {*}) \right\}}{\overline {{\mathcal {Q}}} (G _ {n} , G _ {*, n})}.
$$
By combining the results from (28) and (30), we can verify that
$$
\frac {\lambda_ {n} (\Delta \mu_ {n}) ^ {2} (\Delta \mu_ {n} - \Delta \mu_ {n} ^ {*}) ^ {2}}{\overline {{\mathcal {Q}}} (G _ {n} , G _ {* , n})} \rightarrow \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) \left(- (\Delta \mu_ {n}) ^ {2} (\Delta \mu_ {n} ^ {*}) ^ {2} + (\Delta \mu_ {n}) ^ {3} \Delta \mu_ {n} ^ {*}\right)}{\overline {{\mathcal {Q}}} (G _ {n} , G _ {* , n})} \rightarrow 0,
$$
$$
\frac {\lambda_ {n} (\Delta \mu_ {n}) ^ {2} (\Delta v _ {n} - \Delta v _ {n} ^ {*})}{\overline {{\mathcal {Q}}} (G _ {n} , G _ {* , n})} \rightarrow \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) (\Delta \mu_ {n}) ^ {2} \Delta v _ {n} ^ {*}}{\overline {{\mathcal {Q}}} (G _ {n} , G _ {* , n})} \rightarrow 0.
$$
Therefore, we achieve  $E \to 0$ , which is a contradiction. As a consequence, Case 2.1.1.1 cannot happen.
Case 2.1.1.1.2:  $\lambda_n^* /\lambda_n\to 1$  . Under this case, if we have
$$
\max \left\{\frac {\lambda_ {n} | \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} | ^ {2}}{(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta \mu_ {n} | ^ {2}}, \frac {\lambda_ {n} | \Delta v _ {n} - \Delta v _ {n} ^ {*} |}{(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta \mu_ {n} | ^ {2}} \right\} \to \infty ,
$$
then we will achieve that
$$
(\lambda_ {n} ^ {*} - \lambda_ {n}) (\Delta \mu_ {n} \Delta \mu_ {n} ^ {*}) ^ {2} / \overline {{\mathcal {Q}}} (G _ {n}, G _ {\ast , n}) \leq \min \biggl \{\frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta \mu_ {n} ^ {*} | ^ {2}}{\lambda_ {n} | \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} | ^ {2}}, \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta \mu_ {n} ^ {*} | ^ {2}}{\lambda_ {n} | \Delta v _ {n} - \Delta v _ {n} ^ {*} |} \biggr \} \to 0.
$$
From here, by using the same argument as that of Case 2.1.1.1.1, we will obtain  $E \to 0$ , which is a contradiction. Therefore, we would have that
$$
\left. \max  \left\{\frac {\lambda_ {n} \left| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} \right| ^ {2}}{\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) \left| \Delta \mu_ {n} \right| ^ {2}}, \frac {\lambda_ {n} \left| \Delta v _ {n} - \Delta v _ {n} ^ {*} \right|}{\left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) \left| \Delta \mu_ {n} \right| ^ {2}} \right\} \not \rightarrow \infty . \tag {31}
$$
With that assumption, it leads to  $\overline{\mathcal{Q}}(G_n, G_{*,n}) \asymp (\lambda_n^* - \lambda_n)(\Delta \mu_n^*)^2 (\Delta \mu_n)^2 \asymp (\lambda_n^* - \lambda_n)(\Delta \mu_n^*)^4$  as  $\Delta \mu_n^* / \Delta \mu_n \to 1$ . Now, we denote  $\Delta \mu_n - \Delta \mu_n^* = \tau_1^n \Delta \mu_n^*$  and  $\Delta v_n - \Delta v_n^* = \tau_2^n (\Delta \mu_n^*)^2$ . From the assumption of Case 2.1.1.1, we would have that  $\tau_1^n \to 0$  and  $\tau_2^n \to 0$ . By dividing both the numerator and the denominator of  $F_3$  by  $(\lambda_n^* - \lambda_n)(\Delta \mu_n^*)^3$ , as the new denominators of  $F_3$  goes to 0, we also obtain the numerator of this term goes to 0, i.e., the following holds
$$
\left\{- 1 + \frac {\lambda_ {n}}{\lambda_ {n} ^ {*} - \lambda_ {n}} (\tau_ {1} ^ {n}) ^ {3} \right\} / 6 + \frac {\lambda_ {n}}{2 (\lambda_ {n} ^ {*} - \lambda_ {n})} \tau_ {1} ^ {n} \tau_ {2} ^ {n} \rightarrow 0.
$$
From (31), we have that  $\lambda_{n}(\tau_{1}^{n})^{2} / (\lambda_{n}^{*} - \lambda_{n})\not\to\infty$  and  $\lambda_{n}\tau_{2}^{n} / (\lambda_{n}^{*} - \lambda_{n})\not\to\infty$ . Therefore, since  $\tau_{1}^{n}\rightarrow 0$  and  $\tau_2^n\rightarrow 0$ , we would achieve that  $\lambda_{n}(\tau_{1}^{n})^{3} / (\lambda_{n}^{*} - \lambda_{n})\rightarrow 0$  and  $\lambda_{n}\tau_{1}^{n}\tau_{2}^{n} / (\lambda_{n}^{*} - \lambda_{n})\rightarrow 0$ . By plugging these results to the above limit, it implies that  $-1 / 6 = 0$ , which is a contradiction. As a consequence, Case 2.1.1.1.2 cannot hold.
Case 2.1.1.2:  $\Delta v_{n} / (\Delta \mu_{n})^{2}\not\to 0$  as  $n\to \infty$ . Under that case, we will only consider the setting that  $\lambda_n^* /\lambda_n\rightarrow 1$  as the argument for other settings of that ratio can be argued in the similar fashion. Since we have  $|\Delta v_{n} - \bar{\Delta} v_{n}^{*}| / (\Delta \mu_{n})^{2}\rightarrow 0$ , it leads to  $\Delta v_{n}^{*} / (\Delta \mu_{n})^{2}\not\to 0$ . Combining with  $\Delta \mu_n / \Delta \mu_n^*\rightarrow 1$ , it implies that as  $n$  is large enough we would have
$$
\max  \left\{\left(\Delta \mu_ {n}\right) ^ {2}, \left(\Delta \mu_ {n} ^ {*}\right) ^ {2} \right\} \lesssim \min  \left\{\left| \Delta v _ {n} \right|, \left| \Delta v _ {n} ^ {*} \right| \right\}. \tag {32}
$$
According to the formulation of  $\overline{\mathcal{Q}}(G_n, G_{*,n})$ , we achieve
$$
\begin{array}{l} \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta v _ {n} \Delta v _ {n} ^ {*} |}{\overline {{Q}} (G _ {n} , G _ {* , n})} \leq \min  \left\{\frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta v _ {n} ^ {*} |}{\lambda_ {n} | \Delta v _ {n} - \Delta v _ {n} ^ {*} |}, \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta v _ {n} |}{\lambda_ {n} ^ {*} | \Delta v _ {n} - \Delta v _ {n} ^ {*} |}, \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta v _ {n} ^ {*} |}{\lambda_ {n} | \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} | ^ {2}}, \right. \\ \left. \frac {(\lambda_ {n} ^ {*} - \lambda_ {n}) | \Delta v _ {n} |}{\lambda_ {n} ^ {*} | \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} | ^ {2}} \right\} = B. \\ \end{array}
$$
If we have  $B \to 0$ , we would get  $\frac{(\lambda_n^* - \lambda_n)|\Delta v_n\Delta v_n^*|}{\overline{\mathcal{Q}}(G_n, G_{*,n})} \to 0$ . Combining with (32), we can check that all the results in (28), (30), and (31) hold. With similar arguments as Case 2.1.1.1, we achieve  $\overline{\mathcal{Q}}(G_n, G_{*,n}) / \overline{\mathcal{Q}}(G_n, G_{*,n}) \to 0$ , a contradiction. Therefore, we must have  $B \not\to 0$ . It implies that as  $n$  is large enough we must have
$$
\max  \left\{\lambda_ {n}, \lambda_ {n} ^ {*} \right\} \max  \left\{| \Delta \mu_ {n} - \Delta \mu_ {n} ^ {*} | ^ {2}, | \Delta v _ {n} - \Delta v _ {n} ^ {*} | \right\} \lesssim \left(\lambda_ {n} ^ {*} - \lambda_ {n}\right) \min  \left\{| \Delta v _ {n} |, | \Delta v _ {n} ^ {*} | \right\}.
$$