SlowGuess/NeurIPS_2023_OCR huggingface dataset:
where  $\stackrel{\mathrm{tr}}{=}^{}$  denotes equal in trace. Now if  $C_t, \Sigma_t \in S_\epsilon$  then
$$
\begin{array}{l} \operatorname {t r} \left(\left(C _ {t} - \Sigma_ {t}\right) \left(C _ {t} + \Sigma_ {t}\right) \left(C _ {t} - \Sigma_ {t}\right) Q ^ {- 1}\right) = \operatorname {t r} \left(Q ^ {- 1 / 2} \left(C _ {t} - \Sigma_ {t}\right) \left(C _ {t} + \Sigma_ {t}\right) \left(C _ {t} - \Sigma_ {t}\right) Q ^ {- 1 / 2}\right) \\ \geq \operatorname {t r} \left(2 (1 - \epsilon) Q ^ {- 1 / 2} \left(C _ {t} - \Sigma_ {t}\right) Q \left(C _ {t} - \Sigma_ {t}\right) Q ^ {- 1 / 2}\right) \stackrel {\diamond} {\geq} 2 (1 - \epsilon) \operatorname {t r} \left(\left(C _ {t} - \Sigma_ {t}\right) ^ {2}\right), \tag {89} \\ \end{array}
$$
$$
\begin{array}{l} \operatorname {t r} \left(\left(C _ {t} - \Sigma_ {t}\right) ^ {2} \left(C _ {t} + \Sigma_ {t}\right) Q ^ {- 1}\right) - 2 (1 - \epsilon) \operatorname {t r} \left(\left(C _ {t} - \Sigma_ {t}\right) ^ {2}\right) \\ = \operatorname {t r} \left(\left(C _ {t} - \Sigma_ {t}\right) ^ {2} \left(C _ {t} + \Sigma_ {t} - 2 (1 - \epsilon) Q\right) Q ^ {- 1}\right) \\ = \frac {1}{2} \operatorname {t r} \left(\left(C _ {t} - \Sigma_ {t}\right) \left(\left(C _ {t} + \Sigma_ {t} - 2 (1 - \epsilon) Q\right) Q ^ {- 1} + Q ^ {- 1} \left(C _ {t} + \Sigma_ {t} - 2 (1 - \epsilon) Q\right)\right) \left(C _ {t} - \Sigma_ {t}\right)\right) \geq 0. \tag {90} \\ \end{array}
$$
Note that  $\diamond$  is not trivially true. We show it as a lemma: Suppose  $A$  is a symmetric matrix and  $B$  is a positive definite symmetric matrix. Then  $\operatorname{tr}(ABAB^{-1}) \geq \operatorname{tr}(A^2)$ .
In fact, we write  $B = P\Lambda P^{\top}$  where  $P$  is an orthogonal matrix and  $\Lambda = \mathrm{diag}\{\lambda_2,\dots ,\lambda_d\}$  is a diagonal matrix. Then
$$
\operatorname {t r} \left(A B A B ^ {- 1}\right) = \operatorname {t r} \left(A P \Lambda P ^ {\top} A P \Lambda^ {- 1} P ^ {\top}\right) = \left\| \Lambda^ {1 / 2} P ^ {\top} A P \Lambda^ {- 1 / 2} \right\| _ {F} ^ {2}.
$$
Denoting  $P^{\top}AP = (a_{ij})$  we get
$$
\begin{array}{l} \left\| \Lambda^ {1 / 2} P ^ {\top} A P \Lambda^ {- 1 / 2} \right\| _ {F} ^ {2} = \sum_ {i, j = 1} ^ {d} \left(\frac {\sqrt {\lambda_ {i}}}{\sqrt {\lambda_ {j}}} a _ {i j}\right) ^ {2} \\ = \frac {1}{2} \sum_ {i, j = 1} ^ {d} \left(\frac {\lambda_ {i}}{\lambda_ {j}} a _ {i j} ^ {2} + \frac {\lambda_ {j}}{\lambda_ {i}} a _ {j i} ^ {2}\right) \geq \sum_ {i, j = 1} ^ {d} a _ {i j} ^ {2} = \operatorname {t r} (A ^ {2}). \\ \end{array}
$$
Thus, substituting (90) and (89) into (88) we get
$$
\frac {\mathrm {d}}{\mathrm {d} t} \| C _ {t} - \Sigma_ {t} \| _ {F} ^ {2} = - 2 \operatorname {t r} \left((C _ {t} - \Sigma_ {t}) (\dot {C} _ {t} - \dot {\Sigma} _ {t})\right) \leq - 4 (1 - 2 \epsilon) \| C _ {t} - \Sigma_ {t} \| _ {F} ^ {2} \leq 0.
$$
Suppose at time  $t_0$  both  $C_{t_0}$  and  $\Sigma_{t_0}$  lie in  $\mathcal{S}_{\epsilon}$ . Then for any  $t \geq t_0$  we have
$$
\left\| C _ {t} - \Sigma_ {t} \right\| _ {F} \leq e ^ {- 2 (1 - 2 \epsilon) (t - t _ {0})} \left\| C _ {t _ {0}} - \Sigma_ {t _ {0}} \right\| _ {F}.
$$