SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Note that the constant  $C_t^{\leftarrow, (2)}$  is independent of  $h$ . We remark that when  $h \gg 1$ ,  $C_t^{\leftarrow, (1)}$  approximately behaves as follows:
Lemma E.3. When  $h\to \infty$  , we have
$$
\lim _ {\mathfrak {h} \to \infty} C _ {t} ^ {\leftarrow , (1)} = \frac {5 d (1 + M _ {t} ^ {\leftarrow})}{2} + \frac {5}{2} | \mathcal {X} _ {t} ^ {\leftarrow} | ^ {2} + \frac {(5 0 + 7 5 M _ {t} ^ {\leftarrow 2}) d M _ {t} ^ {\leftarrow}}{m _ {t} ^ {\leftarrow 2}}.
$$
We now proceed to finish the proof of Prop. 3.4. The detailed proofs of Lem. E.1, Prop. E.2, and Lem. E.3 are postponed to Appx. E.2.
Proof of Prop. 3.4. By Lem. E.1,
$$
\begin{array}{l} L (\mathfrak {h}) \stackrel {(1 3)} {=} \frac {\left(1 + \mathfrak {h} ^ {2}\right) ^ {2}}{8} \int_ {\mathbb {R} ^ {d}} \frac {\left(\Phi_ {s , T} ^ {(h ^ {\leftarrow})} (\boldsymbol {\nabla} \cdot (p _ {s} ^ {\leftarrow} E))\right) ^ {2}}{\rho_ {0}} \frac {\rho_ {0}}{p _ {0}} \tag {31} \\ \stackrel {(2 9)} {\leq} \frac {1}{2} (1 + \mathsf {h} ^ {2}) ^ {2} \int_ {\mathbb {R} ^ {d}} \frac {\left(\Phi_ {s , T} ^ {(h ^ {\leftarrow})} (\nabla \cdot (p _ {s} ^ {\leftarrow} E))\right) ^ {2}}{\rho_ {0}}. \\ \end{array}
$$
To simplify notations, let
$$
\Lambda_ {t} ^ {\leftarrow} := \Phi_ {s, t} ^ {(h ^ {\leftarrow})} \big (\nabla \cdot (p _ {s} ^ {\leftarrow} E) \big), \qquad J (\mathsf {h}, t) := \int_ {\mathbb {R} ^ {d}} \frac {\Lambda_ {t} ^ {\leftarrow 2}}{\rho_ {t} ^ {\leftarrow}}, \qquad t \in [ s, T ].
$$
After taking the time-derivative and using Lemma C.1 (in the first line below), and expressions (11) in the second line below, and integration by parts in the third line, the expression of  $\rho_t^{\leftarrow}$  (12) in the fourth line, we have
$$
\begin{array}{l} \frac {\mathrm {d}}{\mathrm {d} t} J (\mathsf {h}, t) = 2 \int_ {\mathbb {R} ^ {d}} \frac {\Lambda_ {t} ^ {\leftarrow} \mathcal {L} _ {t} ^ {(h ^ {\leftarrow})} (\Lambda_ {t} ^ {\leftarrow})}{\rho_ {t} ^ {\leftarrow}} - \int_ {\mathbb {R} ^ {d}} \frac {\Lambda_ {t} ^ {\leftarrow 2}}{(\rho_ {t} ^ {\leftarrow}) ^ {2}} \frac {\mathrm {d}}{\mathrm {d} t} \rho_ {t} ^ {\leftarrow} \\ \stackrel {(1 1)} {=} \mathrm {h} ^ {2} \int_ {\mathbb {R} ^ {d}} \frac {\Lambda_ {t} ^ {\leftarrow}}{\rho_ {t} ^ {\leftarrow}} \left(\triangle \Lambda_ {t} ^ {\leftarrow} + \boldsymbol {\nabla} \cdot \left(\nabla V _ {t} ^ {\leftarrow} \Lambda_ {t} ^ {\leftarrow}\right)\right) - \int_ {\mathbb {R} ^ {d}} \frac {\Lambda_ {t} ^ {\leftarrow 2}}{\left(\rho_ {t} ^ {\leftarrow}\right) ^ {2}} \frac {\mathrm {d}}{\mathrm {d} t} \rho_ {t} ^ {\leftarrow} \left. \right. \tag {32} \\ = - \mathsf {h} ^ {2} \int_ {\mathbb {R} ^ {d}} \nabla \frac {\Lambda_ {t} ^ {\leftarrow}}{\rho_ {t} ^ {\leftarrow}} \cdot \left(\nabla \Lambda_ {t} ^ {\leftarrow} + \nabla V _ {t} ^ {\leftarrow} \Lambda_ {t} ^ {\leftarrow}\right) - \int_ {\mathbb {R} ^ {d}} \frac {\Lambda_ {t} ^ {\leftarrow 2}}{\left(\rho_ {t} ^ {\leftarrow}\right) ^ {2}} \frac {\mathrm {d}}{\mathrm {d} t} \rho_ {t} ^ {\leftarrow} \\ = - \mathsf {h} ^ {2} \int_ {\mathbb {R} ^ {d}} \left| \nabla \frac {\Lambda_ {t} ^ {\leftarrow}}{\rho_ {t} ^ {\leftarrow}} \right| ^ {2} \rho_ {t} ^ {\leftarrow} - \int_ {\mathbb {R} ^ {d}} \frac {\Lambda_ {t} ^ {\leftarrow 2}}{\left(\rho_ {t} ^ {\leftarrow}\right) ^ {2}} \frac {\mathrm {d}}{\mathrm {d} t} \rho_ {t} ^ {\leftarrow}. \\ \end{array}
$$
The major challenge is to estimate  $\int_{\mathbb{R}^d}\frac{\Lambda_t^{\leftarrow 2}}{\left(\rho_t^{\leftarrow}\right)^2}\frac{\mathrm{d}}{\mathrm{d}t}\rho_t^{\leftarrow}$ . By Prop. E.2,
$$
\begin{array}{l} \frac {\mathrm {d}}{\mathrm {d} t} J (\mathfrak {h}, t) \leq - \left(\mathfrak {h} ^ {2} - C _ {t} ^ {\leftarrow , (2)}\right) \int_ {\mathbb {R} ^ {d}} \left| \nabla \frac {\Lambda_ {t} ^ {\leftarrow}}{\rho_ {t} ^ {\leftarrow}} \right| ^ {2} \mathrm {d} \rho_ {t} ^ {\leftarrow} + C _ {t} ^ {\leftarrow , (1)} \int_ {\mathbb {R} ^ {d}} \left(\frac {\Lambda_ {t} ^ {\leftarrow}}{\rho_ {t} ^ {\leftarrow}}\right) ^ {2} \mathrm {d} \rho_ {t} ^ {\leftarrow} \\ \stackrel {(2 8)} {\leq} \quad \left(- \left(\mathsf {h} ^ {2} - C _ {t} ^ {\leftarrow , (2)}\right) \kappa_ {t} ^ {\leftarrow} + C _ {t} ^ {\leftarrow , (1)}\right) \int_ {\mathbb {R} ^ {d}} \left(\frac {\Lambda_ {t} ^ {\leftarrow}}{\rho_ {t} ^ {\leftarrow}}\right) ^ {2}   \mathrm {d} \rho_ {t} ^ {\leftarrow}. \left. \right. \\ \end{array}
$$
To verify the condition in Prop. E.2, we can readily confirm that
$$
\int \frac {\Lambda_ {t} ^ {\leftarrow}}{\rho_ {t} ^ {\leftarrow}} \rho_ {t} ^ {\leftarrow} = \int \Lambda_ {t} ^ {\leftarrow} = \int \Phi_ {s, t} ^ {(h ^ {\leftarrow})} \big (\pmb {\nabla} \cdot (p _ {s} ^ {\leftarrow} E) \big) = \int \pmb {\nabla} \cdot (p _ {s} ^ {\leftarrow} E) = 0,
$$
where the third equality in the last equation comes the fact that the Fokker-Planck solution operator  $\Phi_{s,t}^{(h^{\leftarrow})}$  preserves the total mass. Note that we need  $h^2 - C_t^{\leftarrow, (2)} \geq 0$  in order to ensure the correct direction when applying the Poincaré inequality above, which explains the lower bound that h needs to satisfy in Prop. 3.4.
By Grownwall's inequality, for any  $t \geq s$ ,
$$
J (\mathfrak {h}, t) \leq J (h, s) \exp \left(- \int_ {s} ^ {t} \left((\mathfrak {h} ^ {2} - C _ {r} ^ {\leftarrow , (2)}\right) \kappa_ {r} ^ {\leftarrow} - C _ {r} ^ {\leftarrow , (1)}\right) \mathrm {d} r\left. \right). \tag {33}
$$
Finally, by (31), we have
$$
L (\mathfrak {h}) \leq \frac {(1 + \mathfrak {h} ^ {2}) ^ {2}}{2} J (\mathfrak {h}, T) \leq C _ {\mathfrak {h}} (1 + \mathfrak {h} ^ {2}) ^ {2} \exp \left(- \int_ {s} ^ {T} \left(\left(\mathfrak {h} ^ {2} - C _ {r} ^ {\leftarrow , (2)}\right) \kappa_ {r} ^ {\leftarrow} - C _ {r} ^ {\leftarrow , (1)}\right) \mathrm {d} r\right), \left. \right.
$$
$$
C _ {\mathrm {h}} = \frac {1}{2} J (\mathrm {h}, s) = \frac {1}{2} \int_ {\mathbb {R} ^ {d}} \frac {\left(\boldsymbol {\nabla} \cdot \left(p _ {s} ^ {\leftarrow} E\right)\right) ^ {2}}{\rho_ {s} ^ {\leftarrow}}. \tag {34}
$$
Recall that  $C_t^{\leftarrow, (2)}$  (30) is independent of  $h$ . By Lem. E.3, we already discussed that  $\lim_{h \to \infty} C_t^{\leftarrow, (1)}$  exists. When  $h \to \infty$ , we know that  $\rho_s^{\leftarrow} \rightarrow p_s^{\leftarrow}$  (11)(12), and thus,
$$
\lim  _ {\mathsf {h} \to \infty} C _ {\mathsf {h}} = \frac {1}{2} \int_ {\mathbb {R} ^ {d}} \frac {\left(\nabla \cdot \left(p _ {s} ^ {\leftarrow} E\right)\right) ^ {2}}{p _ {s} ^ {\leftarrow}}.
$$
This completes the proof of Prop. 3.4.
E.2 Proof of Lem. E.1, Prop. E.2, and Lem. E.3
Proof of Lem. E.1.
(i) From  $m_0 > 1$  in Assumption 3.3,  $x \mapsto U_0(x) - x^2 / 2$  is strongly convex and thus is surely bounded from below by some  $c_U \in \mathbb{R}$ .
(ii) The second result is a classical result by Bakry-Emery criterion [2, 3], as  $V_{t}^{\leftarrow}$  (12) is strongly convex with Hessian lower bound  $\kappa_{t}^{\leftarrow}$  (16).
(iii) To prove the third one, note that
$$
\begin{array}{l} Z _ {V _ {0}} := \int_ {\mathbb {R} ^ {d}} e ^ {- V _ {0} \stackrel {(1 1)} {=}} \int_ {\mathbb {R} ^ {d}} e ^ {- (1 + h ^ {- 2}) U _ {0} (x) + h ^ {- 2} | x | ^ {2} / 2} d x \\ = \int_ {\mathbb {R} ^ {d}} e ^ {- U _ {0} (x)} e ^ {- \mathsf {h} ^ {- 2} (U _ {0} (x) - | x | ^ {2} / 2)} \mathrm {d} x \\ \leq e ^ {- \mathsf {h} ^ {- 2} c _ {U}} \int_ {\mathbb {R} ^ {d}} e ^ {- U _ {0}} = e ^ {- \mathsf {h} ^ {- 2} c _ {U}}. \\ \end{array}
$$
To prove the lower bound,
$$
Z _ {V _ {0}} \geq \int_ {\mathbb {R} ^ {d}} e ^ {- (1 + h ^ {- 2}) U _ {0} (x)} d x = e ^ {- (1 + h ^ {- 2}) U _ {0} (\mathcal {X} _ {0})} \int_ {\mathbb {R} ^ {d}} e ^ {- (1 + h ^ {- 2}) \left(U _ {0} (x) - U _ {0} (\mathcal {X} _ {0})\right)} d x. \tag {35}
$$
Recall from the beginning of this Appendix,  $\mathcal{X}_0$  is defined as the global minimum of  $x\mapsto U_0(x)$ . Then,  $U_{0}(x)\geq U_{0}(\mathcal{X}_{0})$  for any  $x$  and we know  $\mathsf{h}\mapsto e^{-(1 + \mathsf{h}^{-2})\left(U_0(x) - U_0(\mathcal{X}_0)\right)}$  is monotone increasing. By monotone convergence theorem,
$$
\begin{array}{l} \lim  _ {\mathfrak {h} \rightarrow \infty} \int_ {\mathbb {R} ^ {d}} e ^ {- (1 + \mathfrak {h} ^ {- 2}) \left(U _ {0} (x) - U _ {0} (\mathcal {X} _ {0})\right)} d x = \int_ {\mathbb {R} ^ {d}} \lim  _ {\mathfrak {h} \rightarrow \infty} e ^ {- (1 + \mathfrak {h} ^ {- 2}) \left(U _ {0} (x) - U _ {0} (\mathcal {X} _ {0})\right)} d x \\ = \int_ {\mathbb {R} ^ {d}} e ^ {- U _ {0} (x) + U _ {0} (\mathcal {X} _ {0})} d x = e ^ {U _ {0} (\mathcal {X} _ {0})}. \\ \end{array}
$$
Thus, the limit of the right-hand side of (35) is 1 when  $\mathsf{h} \to \infty$ . Hence, the lower bound of  $Z_{V_0}$  follows immediately. Finally, when  $\mathsf{h} \geq \mathsf{h}_0(\delta)$ ,
$$
\frac {\rho_ {0}}{p _ {0}} (x) = \frac {e ^ {- (1 + h ^ {- 2}) U _ {0} (x) + h ^ {- 2} | x | ^ {2} / 2}}{Z _ {V _ {0}} e ^ {- U _ {0}}} \leq \frac {e ^ {- h ^ {- 2} c _ {U}}}{1 - \delta}. \tag {36}
$$
(iv) When we pick  $\delta = 1 / 2$  and choose  $h$  as specified, we can immediately obtain (29).
Before we prove Prop. E.2, we need the following three lemmas.
Lemma E.4. Under Assumption 3.3, we have
$$
| \partial_ {t} U _ {t} (x) | \leq \frac {d (1 + M _ {t})}{2} + \frac {| x | ^ {2}}{4} + \frac {3 M _ {t} ^ {2}}{4} | x - \mathcal {X} _ {t} | ^ {2}.
$$
Proof. Since  $p_t$  follows the Fokker-Planck equation  $\partial_t p_t = \nabla \cdot \left( \frac{1}{2} x p_t \right) + \frac{1}{2} \triangle p_t$ , then  $U_t = -\log p_t$  satisfies  $\partial_t U_t(x) = -\frac{1}{2} \left( d - x \cdot \nabla U_t(x) - \triangle U_t(x) + |\nabla U_t(x)|^2 \right)$ . Then by Assumption 3.3,
$$
\begin{array}{l} \left| \partial_ {t} U _ {t} (x) \right| \leq \frac {d}{2} + \frac {1}{2} | x \cdot \nabla U _ {t} (x) | + \frac {1}{2} \triangle U _ {t} (x) + \frac {1}{2} | \nabla U _ {t} (x) | ^ {2} \quad (\text {t r i a n g l e i n e q u a l i t y}) \\ \leq \frac {d}{2} + \frac {1}{4} \left(| x | ^ {2} + | \nabla U _ {t} (x) | ^ {2}\right) + \frac {1}{2} \triangle U _ {t} (x) + \frac {1}{2} | \nabla U _ {t} (x) | ^ {2} \quad (\text {C a u c h y i n e q u a l i t y}) \\ \leq \frac {d}{2} + \frac {1}{4} | x | ^ {2} + \frac {3}{4} | \nabla U _ {t} (x) - \nabla U _ {t} (\mathcal {X} _ {t}) | ^ {2} + \frac {d M _ {t}}{2} \quad (\text {b y} \nabla U _ {t} (\mathcal {X} _ {t}) \equiv 0) \\ \leq \frac {d (1 + M _ {t})}{2} + \frac {| x | ^ {2}}{4} + \frac {3 M _ {t} ^ {2}}{4} | x - \mathcal {X} _ {t} | ^ {2}. \\ \end{array}
$$
Recall that  $\mathcal{X}_t$  is defined as the global minimum of  $U_{t}$  and thus  $\nabla U_t(\mathcal{X}_t) = 0$ .
Lemma E.5. When  $h \geq \frac{1}{2}$ ,
$$
\frac {\left| \frac {\mathrm {d}}{\mathrm {d} t} \rho_ {t} ^ {\leftarrow} \right|}{\rho_ {t} ^ {\leftarrow}} (x) \leq \xi_ {t} ^ {\leftarrow} + \varsigma_ {t} ^ {\leftarrow} | x - \mathcal {X} _ {t} ^ {\leftarrow} | ^ {2}, \tag {37}
$$
$$
\left\{ \begin{array}{l} \xi_ {t} ^ {\leftarrow} := \left| \partial_ {t} \log Z _ {V _ {t} ^ {\leftarrow}} \right| + \frac {5 d (1 + M _ {t} ^ {\leftarrow})}{2} + \frac {5}{2} | \mathcal {X} _ {t} ^ {\leftarrow} | ^ {2}, \\ \varsigma_ {t} ^ {\leftarrow} := 5 \big (\frac {1}{2} + \frac {3 M _ {t} ^ {\leftarrow - 2}}{4} \big). \end{array} \right.
$$
Proof. By direct calculation from the definitions of  $\rho_t^{\leftarrow}$  in (12) and  $V_{t}^{\leftarrow}$  in (11),
$$
\partial_ {t} \rho_ {t} ^ {\leftarrow} = - \left(\partial_ {t} \log Z _ {V _ {t} ^ {\leftarrow}}\right) \rho_ {t} ^ {\leftarrow} - \left(1 + h ^ {- 2}\right) \rho_ {t} ^ {\leftarrow} \partial_ {t} U _ {t} ^ {\leftarrow}.
$$
Hence, by Lem. E.4 and  $h \geq \frac{1}{2}$ ,
$$
\begin{array}{l} \frac {\left| \partial_ {t} \rho_ {t} ^ {\leftarrow} \right|}{\rho_ {t} ^ {\leftarrow}} (x) \leq \left| \partial_ {t} \log Z _ {V _ {t} ^ {\leftarrow}} \right| + (1 + h ^ {- 2}) \left(\frac {d \left(1 + M _ {t} ^ {\leftarrow}\right)}{2} + \frac {\left| x \right| ^ {2}}{4} + \frac {3 M _ {t} ^ {\leftarrow 2}}{4} \left| x - \mathcal {X} _ {t} ^ {\leftarrow} \right| ^ {2}\right)\left. \right. \\ \leq \left(\left| \partial_ {t} \log Z _ {V _ {t} ^ {\leftarrow}} \right| + \frac {5 d \left(1 + M _ {t} ^ {\leftarrow}\right)}{2}\right) \\ + 5 \left(\frac {\left| \mathcal {X} _ {t} ^ {\leftarrow} \right| ^ {2}}{2} + \frac {\left| x - \mathcal {X} _ {t} ^ {\leftarrow} \right| ^ {2}}{2} + \frac {3 M _ {t} ^ {\leftarrow 2}}{4} \left| x - \mathcal {X} _ {t} ^ {\leftarrow} \right| ^ {2}\right) \\ = \xi_ {t} ^ {\leftarrow} + 5 \left(\frac {1}{2} + \frac {3 M _ {t} ^ {\leftarrow 2}}{4}\right) | x - \mathcal {X} _ {t} ^ {\leftarrow} | ^ {2}. \\ \end{array}
$$
Lemma E.6. Assume that  $\rho \propto e^{-V}$  and  $\varphi$  decays fast enough such that
$$
\int_ {B _ {0} (R)} \nabla \cdot (\varphi^ {2} \nabla \rho) \rightarrow 0,
$$
as the radius  $R \to \infty$  ( $B_0(R)$  is the ball centered at 0 with radius  $R$ ), e.g.,  $\varphi \in C_0^2(\mathbb{R}^d)$ . Then
$$
\int_ {\mathbb {R} ^ {d}} \left| \nabla V \right| ^ {2} \varphi^ {2} \rho \leq 4 \int_ {\mathbb {R} ^ {d}} \left| \nabla \varphi \right| ^ {2} \rho + 2 \int_ {\mathbb {R} ^ {d}} \varphi^ {2} \triangle V \rho . \tag {38}
$$
Proof. The identity  $\nabla \cdot (\varphi^2\nabla \rho) = \nabla \varphi^2\cdot \nabla \rho +\varphi^2\triangle \rho$  and the trivial facts  $\nabla \rho = -\rho \nabla V,\triangle \rho = (|\nabla V|^2 -\triangle V)\rho$  show that
$$
\begin{array}{l} 0 = \int \nabla \varphi^ {2} \cdot \nabla \rho + \varphi^ {2} \triangle \rho \\ = - 2 \int (\nabla V \cdot \nabla \varphi) \varphi \rho + \int \varphi^ {2} (- \triangle V + | \nabla V | ^ {2}) \rho \\ \geq - \frac {1}{2} \int | \nabla V | ^ {2} \varphi^ {2} \rho - 2 \int | \nabla \varphi | ^ {2} \rho - \int \varphi^ {2} \triangle V \rho + \int | \nabla V | ^ {2} \varphi^ {2} \rho \\ = \frac {1}{2} \int | \nabla V | ^ {2} \varphi^ {2} \rho - 2 \int | \nabla \varphi | ^ {2} \rho - \int \varphi^ {2} \triangle V \rho . \\ \end{array}
$$
$$
\int | \nabla V | ^ {2} \varphi^ {2} \rho \leq 4 \int | \nabla \varphi | ^ {2} \rho + 2 \int \varphi^ {2} \triangle V \rho .
$$
Proof of Prop. E.2. The above three lemmas show that
$$
\begin{array}{l} \left| \int_ {\mathbb {R} ^ {d}} \varphi^ {2} \frac {\mathrm {d}}{\mathrm {d} t} \rho_ {t} ^ {\leftarrow} \right| \stackrel {(3 7)} {\leq} \int_ {\mathbb {R} ^ {d}} \varphi^ {2} \left(\xi_ {t} ^ {\leftarrow} + \varsigma_ {t} ^ {\leftarrow} | x - \mathcal {X} _ {t} ^ {\leftarrow} | ^ {2}\right) \rho_ {t} ^ {\leftarrow} \\ = \xi_ {t} ^ {\leftarrow} \int_ {\mathbb {R} ^ {d}} \varphi^ {2} \rho_ {t} ^ {\leftarrow} + \varsigma_ {t} ^ {\leftarrow} \int_ {\mathbb {R} ^ {d}} \varphi^ {2} | x - \mathcal {Y} _ {t} ^ {\leftarrow} + \mathcal {Y} _ {t} ^ {\leftarrow} - \mathcal {X} _ {t} ^ {\leftarrow} | ^ {2} \rho_ {t} ^ {\leftarrow} \\ \end{array}
$$
Cauchy ineq.  $\leq \left(\xi_{t}^{\leftarrow} + 2\varsigma_{t}^{\leftarrow}|\mathcal{V}_{t}^{\leftarrow} - \mathcal{X}_{t}^{\leftarrow}|^{2}\right)\int_{\mathbb{R}^{d}}\varphi^{2}\rho_{t}^{\leftarrow} + 2\varsigma_{t}^{\leftarrow}\int_{\mathbb{R}^{d}}\varphi^{2}|x - \mathcal{Y}_{t}^{\leftarrow}|^{2}\rho_{t}^{\leftarrow}$
$$
\begin{array}{l} \stackrel {(1 4)} {\leq} \left(\xi_ {t} ^ {\leftarrow} + 2 \varsigma_ {t} ^ {\leftarrow} \left| \mathcal {Y} _ {t} ^ {\leftarrow} - \mathcal {X} _ {t} ^ {\leftarrow} \right| ^ {2}\right) \int_ {\mathbb {R} ^ {d}} \varphi^ {2} \rho_ {t} ^ {\leftarrow} + \frac {2 \varsigma_ {t} ^ {\leftarrow}}{m _ {t} ^ {\leftarrow 2}} \int_ {\mathbb {R} ^ {d}} \varphi^ {2} \left| \nabla V _ {t} ^ {\leftarrow} (x) - \nabla V _ {t} ^ {\leftarrow} \left(\mathcal {Y} _ {t} ^ {\leftarrow}\right)\right| ^ {2} \rho_ {t} ^ {\leftarrow} \left. \right.\left. \right. \\ = \left(\xi_ {t} ^ {\leftarrow} + 2 \varsigma_ {t} ^ {\leftarrow} | \mathcal {Y} _ {t} ^ {\leftarrow} - \mathcal {X} _ {t} ^ {\leftarrow} | ^ {2}\right) \int_ {\mathbb {R} ^ {d}} \varphi^ {2} \rho_ {t} ^ {\leftarrow} + \frac {2 \varsigma_ {t} ^ {\leftarrow}}{m _ {t} ^ {\leftarrow 2}} \int_ {\mathbb {R} ^ {d}} \varphi^ {2} | \nabla V _ {t} ^ {\leftarrow} (x) | ^ {2} \rho_ {t} ^ {\leftarrow} \\ \stackrel {(3 8)} {\leq} \left(\xi_ {t} ^ {\leftarrow} + 2 \varsigma_ {t} ^ {\leftarrow} | \mathcal {Y} _ {t} ^ {\leftarrow} - \mathcal {X} _ {t} ^ {\leftarrow} | ^ {2}\right) \int_ {\mathbb {R} ^ {d}} \varphi^ {2} \rho_ {t} ^ {\leftarrow} + \frac {2 \varsigma_ {t} ^ {\leftarrow}}{m _ {t} ^ {\leftarrow 2}} \big (4 \int_ {\mathbb {R} ^ {d}} | \nabla \varphi | ^ {2} \rho_ {t} ^ {\leftarrow} + 2 \int_ {\mathbb {R} ^ {d}} \varphi^ {2} \triangle V _ {t} ^ {\leftarrow} \rho_ {t} ^ {\leftarrow} \big) \\ \stackrel {(1 4)} {\leq} \left(\xi_ {t} ^ {\leftarrow} + 2 \varsigma_ {t} ^ {\leftarrow} \left| \mathcal {Y} _ {t} ^ {\leftarrow} - \mathcal {X} _ {t} ^ {\leftarrow} \right| ^ {2}\right) \int_ {\mathbb {R} ^ {d}} \varphi^ {2} \rho_ {t} ^ {\leftarrow} + \frac {2 \varsigma_ {t} ^ {\leftarrow}}{m _ {t} ^ {\leftarrow 2}} \left(4 \int_ {\mathbb {R} ^ {d}} \left| \nabla \varphi \right| ^ {2} \rho_ {t} ^ {\leftarrow} + 1 0 M _ {t} ^ {\leftarrow} d \int_ {\mathbb {R} ^ {d}} \varphi^ {2} \rho_ {t} ^ {\leftarrow}\right)\left. \right) \\ = \left(\xi_ {t} ^ {\leftarrow} + 2 \varsigma_ {t} ^ {\leftarrow} | \mathcal {Y} _ {t} ^ {\leftarrow} - \mathcal {X} _ {t} ^ {\leftarrow} | ^ {2} + \frac {2 0 M _ {t} ^ {\leftarrow} d \varsigma_ {t} ^ {\leftarrow}}{m _ {t} ^ {\leftarrow 2}}\right) \int_ {\mathbb {R} ^ {d}} \varphi^ {2} \rho_ {t} ^ {\leftarrow} + \frac {8 \varsigma_ {t} ^ {\leftarrow}}{m _ {t} ^ {\leftarrow 2}} \int_ {\mathbb {R} ^ {d}} | \nabla \varphi | ^ {2} \rho_ {t} ^ {\leftarrow}. \\ \end{array}
$$
To get the fifth line, we used the definition of  $\mathcal{V}_t^{\leftarrow}\colon \mathcal{V}_t^{\leftarrow}$  is the global minimum of  $V_{t}^{\leftarrow}$  so that  $\nabla V_t^{\leftarrow}(\mathcal{V}_t^{\leftarrow})\equiv 0$  . To get the second last line, we used the fact that
$$
\triangle V _ {t} ^ {\leftarrow} = \left(1 + h ^ {- 2}\right) \triangle U _ {t} ^ {\leftarrow} - h ^ {- 2} \leq \left(1 + h ^ {- 2}\right) d M _ {t} ^ {\leftarrow} - h ^ {- 2} \stackrel {(\text {b y} h \geq 1 / 2)} {\leq} 5 d M _ {t} ^ {\leftarrow}.
$$
Proof of Lem. E.3. Note that  $\varsigma_t^{\leftarrow}$  does not depend on  $h$ . When  $h \to \infty$ , as  $V_t^{\leftarrow} \rightarrow U_t^{\leftarrow}$  (11), we know  $Z_{V_t^{\leftarrow}}$  converges to  $Z_{U_t^{\leftarrow}} \equiv 1$  by the definition of  $p_t \coloneqq e^{-U_t}$ , we immediately know that
$$
\lim  _ {\mathsf {h} \to \infty} \left| \partial_ {t} \log Z _ {V _ {t} ^ {\leftarrow}} \right| = 0.
$$
$$
\lim  _ {\mathfrak {h} \to \infty} \xi_ {t} ^ {\leftarrow} = \frac {5 d \left(1 + M _ {t} ^ {\leftarrow}\right)}{2} + \frac {5}{2} | \mathcal {X} _ {t} ^ {\leftarrow} | ^ {2}.
$$
Moreover,  $\mathcal{Y}_t\to \mathcal{X}_t$  as  $h\rightarrow \infty$  , we have
$$
\begin{array}{l} \lim  _ {\mathfrak {h} \to \infty} C _ {t} ^ {\leftarrow , (1)} = \lim  _ {\mathfrak {h} \to \infty} \xi_ {t} ^ {\leftarrow} + 2 \lim  _ {\mathfrak {h} \to \infty} \varsigma_ {t} ^ {\leftarrow} | \mathcal {Y} _ {t} ^ {\leftarrow} - \mathcal {X} _ {t} ^ {\leftarrow} | ^ {2} + \frac {2 0 d M _ {t} ^ {\leftarrow} \varsigma_ {t} ^ {\leftarrow}}{m _ {t} ^ {\leftarrow 2}}, \\ = \frac {5 d (1 + M _ {t} ^ {\leftarrow})}{2} + \frac {5}{2} | \mathcal {X} _ {t} ^ {\leftarrow} | ^ {2} + \frac {(5 0 + 7 5 M _ {t} ^ {\leftarrow 2}) d M _ {t} ^ {\leftarrow}}{m _ {t} ^ {\leftarrow 2}}. \\ \end{array}
$$
E.3 Proof of Prop. 3.5
Case I:  $h = 0$ . For this ODE case, (8) becomes
$$
\mathcal {L} _ {t} ^ {(0)} (\mu) (x) = \frac {1}{2} \nabla \cdot \left(\nabla \left(U _ {t} ^ {\leftarrow} (x) - | x | ^ {2} / 2\right) \mu (x)\right), \tag {39}
$$
and the formula of leading order term  $L(0)$  is
$$
L (0) = \frac {1}{2} \int_ {\mathbb {R} ^ {d}} \frac {v _ {T} ^ {2} (x)}{p _ {0} (x)} \mathrm {d} x, \qquad v _ {T} \stackrel {(1 0)} {=} - \frac {1}{2} \int_ {T - a} ^ {T} \Phi_ {t, T} ^ {(0)} (\nabla \cdot (p _ {t} ^ {\leftarrow} E)) \mathrm {d} t,
$$
since we perturb the  $\mathcal{E}_t^{\leftarrow}$  only when  $t\approx T$  , that is,  $\mathcal{E}_t^{\leftarrow}(x) = \mathbb{I}_{t\in [T - a,T]}E(x)$  for a small positive  $a$  Then we know that
$$
v _ {T} \sim - \frac {1}{2} \int_ {T - a} ^ {T} \nabla \cdot (p _ {0} E)) \mathrm {d} t = - \frac {a}{2} \nabla \cdot (p _ {0} E).
$$
Recall that  $p_0 = p_T^{\leftarrow} \approx p_t^{\leftarrow}$  when  $t \approx T$ ; moreover,  $\Phi_{t,T}^{(0)} \approx \Phi_{T,T}^{(0)} \equiv \mathrm{Id}$ . Therefore, the leading order term of  $L(0)$  when  $a \ll 1$  is
$$
L (0) \sim \frac {a ^ {2}}{8} \int_ {\mathbb {R} ^ {d}} \frac {(\pmb {\nabla} \cdot (p _ {0} E)) ^ {2}}{p _ {0}}.
$$
Case II:  $\mathsf{h} \to \infty$ . For the SDE case,
$$
\begin{array}{l} v _ {T} \stackrel {(1 0)} {=} - \frac {1 + h ^ {2}}{2} \int_ {0} ^ {T} \Phi_ {t, T} ^ {(h)} \left(\boldsymbol {\nabla} \cdot \left(p _ {t} ^ {\leftarrow} \mathcal {E} _ {t} ^ {\leftarrow}\right)\right) d t \\ \sim - \frac {1 + h ^ {2}}{2} \int_ {T - a} ^ {T} \Phi_ {t, T} ^ {(h)} (\boldsymbol {\nabla} \cdot (p _ {0} E)) d t \quad (b y a \ll 1) \\ \sim - \frac {1 + h ^ {2}}{2} \int_ {T - a} ^ {T} e ^ {(T - t) \frac {h ^ {2}}{2} \mathcal {K} _ {T} ^ {\leftarrow}} (\boldsymbol {\nabla} \cdot (p _ {0} E)) d t \quad (\text {b y} a \ll 1) \\ = - \frac {1 + h ^ {2}}{2} \int_ {0} ^ {a} e ^ {t \frac {h ^ {2}}{2} \mathcal {K} _ {T} ^ {\leftarrow}} (\nabla \cdot (p _ {0} E)) d t, \\ \end{array}
$$
$$
\mathcal {K} _ {t} ^ {\leftarrow} (\mu) := \triangle \mu + \boldsymbol {\nabla} \cdot \left(\nabla V _ {t} ^ {\leftarrow} \mu\right). \tag {40}
$$
Then when we further assume  $h \gg 1$ ,
$$
\begin{array}{l} \frac {1}{2} \int_ {\mathbb {R} ^ {d}} \frac {(v _ {T}) ^ {2}}{p _ {0}} \lesssim \frac {1}{2} \int_ {\mathbb {R} ^ {d}} \frac {(v _ {T}) ^ {2}}{\rho_ {0}} \quad (\text {b y} \\ \lesssim \frac {(1 + h ^ {2}) ^ {2}}{8} \left(\frac {1 - e ^ {- a \frac {h ^ {2}}{2} \kappa_ {0}}}{\frac {h ^ {2}}{2} \kappa_ {0}}\right) ^ {2} \int_ {\mathbb {R} ^ {d}} \frac {(\boldsymbol {\nabla} \cdot (p _ {0} E)) ^ {2}}{\rho_ {0}} \quad (\text {b y L e m . D . 2}) \\ \sim \frac {(1 + h ^ {2}) ^ {2}}{h ^ {4}} \frac {\left(1 - e ^ {- a \frac {h ^ {2}}{2} \kappa_ {0}}\right) ^ {2}}{2 \kappa_ {0} ^ {2}} \int_ {\mathbb {R} ^ {d}} \frac {(\boldsymbol {\nabla} \cdot (p _ {0} E)) ^ {2}}{\rho_ {0}} \\ \sim \frac {\left(1 - e ^ {- a \frac {h ^ {2}}{2} \kappa_ {0}}\right) ^ {2}}{2 \kappa_ {0} ^ {2}} \int_ {\mathbb {R} ^ {d}} \frac {(\pmb {\nabla} \cdot (p _ {0} E)) ^ {2}}{\rho_ {0}} \\ \sim \frac {\left(1 - e ^ {- a \frac {\mathrm {h} ^ {2}}{2} \kappa_ {0}}\right) ^ {2}}{2 \kappa_ {0} ^ {2}} \int_ {\mathbb {R} ^ {d}} \frac {(\nabla \cdot (p _ {0} E)) ^ {2}}{p _ {0}}. \\ \end{array}
$$
To apply Lem. D.2 in the second line, we remark that  $\mathcal{K}_T^{\leftarrow}(\mu) = \triangle \mu +\nabla \cdot (\nabla V_T^{\leftarrow}\mu)$ ,  $V_{T}^{\leftarrow}$  has Hessian lower bound  $\kappa_T^\leftarrow \equiv \kappa_0$  (16).
E.4 Discussion on the pulse-shape error
Though the error of the score function is time-dependent, we can always divide the error function into a linear combination of step functions:
$$
\mathcal {E} _ {t} ^ {\leftarrow} (x) \approx \sum_ {j} \phi_ {j} (x) \chi_ {[ t _ {j}, t _ {j + 1} ]} (t),
$$
where  $\chi$  is the indicator function and  $\phi_j$  is the function value of  $\mathcal{E}_t^{\leftarrow}$  on the time interval  $[t_j, t_{j+1}]$ . Without loss of generality, assume that the time-discretization is uniform, i.e.,  $t_{j+1} - t_j = \Delta t$  for any  $j$ . If we don't worry about ill-behaved functions, this decomposition can be made arbitrarily accurate by choosing a smaller time interval; it is not difficult to make this approximation mathematically rigorous. It is hard to assume that this error function at time e.g.,  $t = 0.1$  has some subtle connection
with its error at e.g.,  $t = 0.8$  in general; how the error function exactly looks like depends on a vast amount of hyper-parameters in training. Therefore, we might as well treat each  $\phi_j$  as "mostly independent" in order to handle the worst case situation. If we consider how each  $\phi_j$  contributes to the final sample generation error (quantified via KL divergence), then we might as well study each one independently. We remark that this is simply a reasonable assumption in order to treat generic error types.
More specifically, recall that in Prop. 3.2,  $v_{T}$  linearly depends on each  $\phi_{j}$