SlowGuess/NeurIPS_2023_OCR huggingface dataset:
where  $\tilde{f}_{x_j x_i}(x; \mathsf{M}_\varepsilon, \psi) = \frac{\partial^2 \psi(x,t)}{\partial x_j x_i}^\top \mathsf{M}_\varepsilon \frac{\partial^2 \psi(x,t)}{\partial x_j x_i}$  and  $\tilde{u}_{x_j x_i}(x) = \left\| (I - \tilde{P}) \frac{\partial^2 \psi(x,t)}{\partial x_j x_i} \right\|_{\mathcal{H}_\eta}$ .
$$
\begin{array}{l} \psi_ {i} ^ {\prime} (x, t) ^ {\top} \left(\mathsf {M} _ {\varepsilon} - \tilde {P} \mathsf {M} _ {\varepsilon} \tilde {P}\right) \psi_ {j} ^ {\prime} (x, t) \leq c _ {M} \tilde {u} _ {x _ {j}} (x) \tilde {u} _ {x _ {i}} (x) + c _ {M} ^ {1 / 2} f _ {x _ {i}} (x; \mathsf {M} _ {\varepsilon}, \psi) ^ {1 / 2} \tilde {u} _ {x _ {j}} (x) \\ + c _ {M} ^ {1 / 2} \tilde {f} _ {x _ {j}} (x; \mathrm {M} _ {\varepsilon}, \psi) ^ {1 / 2} \tilde {u} _ {x _ {i}} (x). \tag {60} \\ \end{array}
$$
Follwing similar steps of calculations as that is Theorem 8 and Theroem 2, we get
$$
\left\| \frac {\partial (\hat {p} (x , t) - \tilde {p} (x , t))}{\partial t} + \sum_ {i = i} ^ {d} \frac {\partial (\mu_ {i} (x , t) (\hat {p} (x , t) - \tilde {p} (x , t)))}{\partial x _ {i}} + (- \Delta) ^ {s} (p ^ {*} (x, t) (\hat {p} (x, t) - \tilde {p} (x, t)) \right\| _ {L ^ {2} (\tilde {X})}
$$
$$
\begin{array}{l} = P _ {1} \| \mathsf {M} _ {\varepsilon} \| C ^ {2} q _ {\eta} ^ {2} e ^ {- \frac {2 c \sigma}{h} \log \frac {c \sigma}{h}} R ^ {(d + 1) / 2} + P _ {2} C q _ {\eta} e ^ {- \frac {c \sigma}{h} \log \frac {c \sigma}{h}} \| \mathsf {M} _ {\varepsilon} \| R ^ {(d + 1) / 2} \\ + P _ {3} \left\| \mathrm {M} _ {\varepsilon} \right\| ^ {1 / 2} C q _ {\eta} e ^ {- \frac {c \sigma}{h} \log \frac {c \sigma}{h}} \| \sqrt {p (x , t)} \| _ {L ^ {2} (\tilde {x})}, \tag {61} \\ \end{array}
$$
for some constant  $P_{1}, P_{2}$  and  $P_{3}$ . Now, let us compute  $\| \mathsf{M}_{\varepsilon} \|$ . We have
$$
\| \mathsf {M} _ {\varepsilon} \| \leq T r (\mathsf {M} _ {\varepsilon}) = \hat {C} \tau^ {(d + 1) / 2} \left(1 + \varepsilon^ {\frac {2 \beta}{\beta - 2 s}} e ^ {\frac {\tilde {C} ^ {\prime}}{\tau} \varepsilon^ {- \frac {2}{\beta - 2 s}}}\right).
$$
Let us choose,  $\tau = \frac{\tilde{C}^{\prime}\varepsilon^{-\frac{2}{\beta - 2s}}}{\frac{2\beta}{\beta - 2s}\log(\frac{1 + R}{\varepsilon})}$ . Hence,
$$
\left\| \mathsf {M} _ {\varepsilon} \right\| \leq C _ {3} (1 + R) ^ {\frac {2 \beta}{\beta - 2 s}} \varepsilon^ {- \frac {d + 1}{\beta - 2 s}},
$$
For some constant  $C_3$ . Hence, then, note that  $\| p^{1/2} \|_{L^2(\mathcal{X})} = \| p \|_{L^1(\mathcal{X})}^{1/2}$ . Hence, using the argument in [RC21], we have
$$
\left\| p ^ {1 / 2} \right\| _ {L ^ {2} (\tilde {\mathcal {X}})} \leq 2 R.
$$
Term  $R$  comes from the fact that for each time instance  $p(x,t)$  is a density and  $t\in (0 < R)$ . Now,
$$
\begin{array}{l} \left\| \frac {\partial (\hat {p} (x , t) - \tilde {p} (x , t))}{\partial t} + \sum_ {i = i} ^ {d} \frac {\partial \left(\mu_ {i} (x , t) (\hat {p} (x , t) - \tilde {p} (x , t))\right)}{\partial x _ {i}} + (- \Delta) ^ {s} \left(p ^ {\star} (x, t) (\hat {p} (x, t) - \tilde {p} (x, t)) \right. \right\| _ {L ^ {2} (\hat {X})} \\ = P _ {1} C _ {3} (1 + R) ^ {\frac {2 \beta}{\beta - 2 s}} \varepsilon^ {- \frac {d + 1}{\beta - 2 s}} C ^ {2} q _ {\eta} ^ {2} e ^ {\frac {- 2 c \sigma}{h} \log \frac {c \sigma}{h}} R ^ {(d + 1) / 2} + P _ {2} C q _ {\eta} e ^ {- \frac {c \sigma}{h} \log \frac {c \sigma}{h}} C _ {3} (1 + R) ^ {\frac {2 \beta}{\beta - 2 s}} \varepsilon^ {- \frac {d + 1}{\beta - 2 s}} R ^ {(d + 1) / 2} \\ + P _ {3} C _ {3} ^ {1 / 2} R (1 + R) ^ {\frac {\beta}{\beta - 2 s}} \varepsilon^ {- \frac {d + 1}{2 (\beta - 2 s)}} C q _ {\eta} e ^ {- \frac {c \sigma}{h} \log \frac {c \sigma}{h}} \| \sqrt {p (x , t)} \| _ {L ^ {2} (\tilde {X})}. \\ \end{array}
$$
By choosing  $h = c\sigma / s$  with  $s = \max(C', (1 + \frac{d + 1}{2(\beta - 2s)})\log \frac{1}{\varepsilon} + (1 + \frac{d}{2})\log(1 + R) + \log(\hat{C}) + e)$ , for some big enough  $\hat{C}$ . Since  $s \geq e$ , then  $\log s \geq 1$ , so
$$
C e ^ {- \frac {c \sigma}{h} \log \frac {c \sigma}{h}} = C e ^ {- s \log s} \leq C e ^ {- s} \leq C ^ {\prime} (1 + R) ^ {- d / 2} \varepsilon^ {1 + \frac {d + 1}{2 (\beta - 2 s)}}.
$$
Hence, combining everything, we have from triangles inequality,
$$
\begin{array}{l} \left\| \frac {\partial \left(p ^ {\star} (x , t) - \tilde {p} (x , t)\right)}{\partial t} + \sum_ {i = i} ^ {d} \frac {\partial \left(\mu_ {i} (x , t) \left(p ^ {\star} (x , t) - \tilde {p} (x , t)\right)\right)}{\partial x _ {i}} + (- \Delta) ^ {s} \left(p ^ {\star} (x, t) - \tilde {p} (x, t)\right) \right\| _ {L ^ {2} (\tilde {\mathcal {X}})} \\ = O (\varepsilon) \\ \end{array}
$$
$$
m \leq \left(1 + \frac {2 R \sqrt {d + 1}}{h}\right) ^ {d + 1}.
$$
Then, since  $h \geq c\sigma / (C_4 \log \frac{C_5 \log(1 + R)}{\varepsilon})$  with  $C_4 = 1 + d / \min(2(\beta - 2s), 2)$  and  $C_5 = (\hat{C})^{1/C_4}$  and since  $\sigma = \min(R, 1/\sqrt{\tau})$ , then  $R / \sigma = \max(1, R\sqrt{\tau}) \leq 1 + \sqrt{\tilde{C}'} \varepsilon^{-1/(\beta - 2)} (\log \frac{1 + R}{\varepsilon})^{-1/2}$ , so after final calculation, we have
$$
m ^ {\frac {1}{d + 1}} \leq 1 + 2 R \sqrt {d + 1} / h \leq C _ {8} + C _ {9} \log \frac {1 + R}{\varepsilon} + C _ {1 0} \varepsilon^ {- \frac {1}{\beta - 2 s}} \left(\log \frac {1 + R}{\varepsilon}\right) ^ {1 / 2},
$$