SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\psi (\delta) = \int_ {0} ^ {+ \infty} \Psi \left(\alpha_ {0} t / \delta , k\right) ^ {\beta_ {0}} t \varphi (t) \mathrm {d} t \leq C _ {0} \delta^ {2}, \tag {30}
$$
$$
w i t h C _ {0} = 2 5 (2 \pi) ^ {- 1} \left(k ^ {2} + 1\right) / \left(\alpha_ {0} \beta_ {0}\right) ^ {2}.
$$
$$
\int_ {0} ^ {+ \infty} \Psi \left(\alpha_ {0} t / \delta , k\right) ^ {\beta_ {0}} t \varphi (t) \mathrm {d} t \leq (2 \pi) ^ {- 1} \int_ {0} ^ {t _ {\delta}} \Psi \left(\alpha_ {0} t / \delta , k\right) ^ {\beta_ {0}} t \mathrm {d} t + (2 \pi) ^ {- 1} \int_ {t _ {\delta}} ^ {+ \infty} \Psi \left(\alpha_ {0} t / \delta , k\right) ^ {\beta_ {0}} t \mathrm {d} t. \tag {31}
$$
4In the rest of the supplementary, we never precise the dimension  $p\in \mathbb{N}$  which can be deduced from the variable.
In addition, using that if  $u \geq 5k$  then  $\Psi(u,k) \leq \exp[-u/5]$ , we get
$$
(2 \pi) ^ {- 1} \int_ {t _ {\delta}} ^ {+ \infty} \Psi \left(\alpha_ {0} t / \delta , k\right) ^ {\beta_ {0}} t d t \leq (2 \pi) ^ {- 1} \int_ {0} ^ {+ \infty} \exp \left[ - \alpha_ {0} \beta_ {0} t / (5 \delta) \right] t d t \leq (2 \pi) ^ {- 1} 2 5 \delta^ {2} / \left(\alpha_ {0} \beta_ {0}\right) ^ {2}. \tag {32}
$$
Finally, using that for any  $u \geq 0$ ,  $\Psi(u, k) \leq 1$ , we have
$$
(2 \pi) ^ {- 1} \int_ {0} ^ {t _ {\delta}} \Psi \left(\alpha_ {0} t / \delta , k\right) ^ {\beta_ {0}} t d t \leq (2 \pi) ^ {- 1} 2 5 k ^ {2} \delta^ {2} / \alpha_ {0} ^ {2}, \tag {33}
$$
D.3 Lower bound on the inside probability and control of moments of order two and higher
Lower bound on the inside probability. We begin with the following lemma which controls the expectation of  $1 + \| Z\|$  outside of  $\mathsf{C}(x,\gamma)$ . We recall that  $\mathsf{V}$  is defined in Theorem 7-(c).
Lemma 11. Let  $\bar{\gamma} = 1$ . Let  $x \in \mathbb{V}$ ,  $Z \in \sim \mathrm{N}(0, \mathrm{Id})$  and  $\gamma \in (0, \bar{\gamma})$  then we have
$$
\left. \max  \left(\mathbb {E} \left[ \mathbb {1} _ {x + \sqrt {\gamma} Z \in \overline {{\mathcal {M}}} \cap \mathsf {C} (x, \gamma) ^ {\mathrm {c}}} \right], \mathbb {E} \left[ \langle Z, \nabla \Phi (\bar {x}) \rangle \mathbb {1} _ {x + \sqrt {\gamma} Z \in \overline {{\mathcal {M}}} \cap \mathsf {C} (x, \gamma) ^ {\mathrm {c}}} \right]\right) \leq \psi (\gamma), \right. \tag {34}
$$
with  $\psi : \mathbb{R}_+ \to \mathbb{R}_+$  such that  $\lim \sup_{t \to 0} \psi(t) / t^{1/2} < +\infty$ .
Proof. Let  $\bar{r} > 0$  given by Theorem 7. First, we have that
$$
\int_ {\mathbb {R}} \int_ {\mathbb {R} ^ {d - 1}} (1 + | \alpha | + \| v \|) \mathbb {1} _ {\alpha \geq \bar {r} / \gamma^ {1 / 2}} \varphi (\alpha) \varphi (v) d \alpha d v \tag {35}
$$
$$
\leq d \int_ {\mathbb {R}} (1 + | \alpha |) \mathbb {1} _ {\alpha \geq \bar {r} / \gamma^ {1 / 2}} \varphi (\alpha) \mathrm {d} \alpha \leq d \left(\Psi \left(\bar {r} ^ {2} / \gamma , 1\right) + \exp \left[ - \bar {r} ^ {2} / (2 \gamma) \right]\right). \tag {36}
$$
Second, using Lemma 9, we have that
$$
\int_ {\mathbb {R}} \int_ {\mathbb {R} ^ {d - 1}} \mathbb {1} _ {\| v \| ^ {2} \geq (\bar {\alpha} + \sqrt {\gamma} \alpha) / (C \gamma)} \mathbb {1} _ {\bar {\alpha} + \sqrt {\gamma} \alpha \geq 0} \varphi (\alpha) \varphi (v) d \alpha d v \tag {37}
$$
$$
\leq \int_ {\mathbb {R}} \mathbb {1} _ {\bar {\alpha} + \sqrt {\gamma} \alpha \geq 0} \Psi ((\bar {\alpha} + \sqrt {\gamma} \alpha) / (C \gamma), d - 1) \varphi (\alpha) \mathrm {d} \alpha \tag {38}
$$
$$
\leq \int_ {0} ^ {+ \infty} \Psi \left(\alpha / C \gamma^ {1 / 2}, d - 1\right) \varphi \left(\alpha - \bar {\alpha} / \gamma^ {1 / 2}\right) \mathrm {d} \alpha \leq \Psi_ {1} \left(\gamma^ {1 / 2}\right). \tag {39}
$$
Second, using Lemma 10, we have that
$$
\int_ {\mathbb {R}} \int_ {\mathbb {R} ^ {d - 1}} \alpha \mathbb {1} _ {\| v \| ^ {2} \geq (\bar {\alpha} + \sqrt {\gamma} \alpha) / (C \gamma)} \mathbb {1} _ {\bar {\alpha} + \sqrt {\gamma} \alpha \geq 0} \varphi (\alpha) \varphi (v) d \alpha d v \tag {40}
$$
$$
= \int_ {\mathbb {R}} \alpha \Psi \left(\left(\bar {\alpha} + \sqrt {\gamma} \alpha\right) / (C \gamma), d - 1\right) \mathbb {1} _ {\bar {\alpha} + \sqrt {\gamma} \alpha \geq 0} \varphi (\alpha) \mathrm {d} \alpha \tag {41}
$$
$$
\leq \int_ {0} ^ {+ \infty} \Psi (\alpha / C \gamma^ {1 / 2}, d - 1) \alpha \varphi (\alpha) \mathrm {d} \alpha \leq \Psi_ {2} \left(\gamma^ {1 / 2}\right). \tag {42}
$$
Note that we have  $\lim_{\gamma \to 0} \Psi_2(\gamma^{1/2}) + \Psi_1(\gamma^{1/2}) < +\infty$ . We conclude upon combining (36), (39) and (42) with Theorem 7-(d) and the fact that  $\|\Phi(\bar{x})\| = 1$ .
The following lemma allows us to give a lower bound to the quantity  $\mathbb{E}[\mathbb{1}_{x + \sqrt{\gamma} Z\in \overline{\mathcal{M}}}]$  uniformly w.r.t  $x\in \overline{\mathcal{M}}$
Lemma 12. There exists  $\bar{\gamma} >0$  such that for any  $\gamma \in (0,\bar{\gamma})$  and for any  $x\in \overline{\mathcal{M}}$ $\gamma \in (0,\bar{\gamma})$  and  $Z\sim \mathrm{N}(0,\mathrm{Id})$  we have
$$
\mathbb {E} \left[ \mathbb {1} _ {x + \sqrt {\gamma} Z \in \overline {{\mathcal {M}}}} \right] \geq 1 / 4. \tag {43}
$$
Proof. Let  $\gamma \in (0, \bar{\gamma})$ . If  $x \notin \mathsf{V}$  then  $\mathrm{B}(x, 2R) \subset \mathcal{M}$  using Theorem 7-(e) and therefore  $\mathbb{E}[\mathbb{1}_{x + \sqrt{\gamma} Z \in \overline{\mathcal{M}}}] \geq 1/4$  for  $\bar{\gamma} > 0$  small enough. Now, assume that  $x \in \mathsf{V}$ . Using Lemma 11, we have that  $\mathbb{E}[\mathbb{1}_{x + \sqrt{\gamma} Z \in \overline{\mathcal{M}} \cap \mathsf{C}(x, \gamma)^{\mathrm{c}}}] \leq \psi(\gamma)$ . In addition, using Theorem 7-(b), we have that for any  $\gamma > 0$
$$
\begin{array}{l} \mathbb {E} \left[ \mathbb {1} _ {x + \sqrt {\gamma} Z \in \overline {{\mathcal {M}}}} \right] \geq \mathbb {E} \left[ \mathbb {1} _ {x + \sqrt {\gamma} Z \in \mathsf {C} (x, \gamma)} \right] (44) \\ \geq \int_ {- \bar {\alpha} \gamma^ {- 1 / 2}} ^ {r \gamma^ {- 1 / 2}} \int_ {\nabla \Phi (\bar {x}) ^ {\perp}} \mathbb {1} _ {\| v \| ^ {2} \leq (\bar {\alpha} + \gamma^ {1 / 2} \alpha) / (C \gamma)} \varphi (\alpha) \varphi (v) d \alpha d v (45) \\ \geq \int_ {- \bar {\alpha} \gamma^ {- 1 / 2}} ^ {r \gamma^ {- 1 / 2}} (1 - \Psi ((\bar {\alpha} + \gamma^ {1 / 2} \alpha) / (C \gamma), d - 1)) \varphi (\alpha) \mathrm {d} \alpha (46) \\ \geq (1 / 2) - \Psi \left(r ^ {2} / \gamma , 1\right) - \int_ {- \bar {\alpha} \gamma^ {- 1 / 2}} ^ {+ \infty} \Psi \left(\left(\bar {\alpha} + \gamma^ {1 / 2} \alpha\right) / \left(C \gamma\right), d - 1\right) \varphi (\alpha) \mathrm {d} \alpha . (47) \\ \end{array}
$$
Hence, using Lemma 8 and Lemma 9, there exists  $\bar{\gamma} >0$  such that for any  $\gamma \in (0,\bar{\gamma})$ ,  $\Psi (r^2 /\gamma ,1) + \int_0^{+\infty}\Psi (\alpha /(C\gamma^{1 / 2}),d)\varphi (\alpha -\gamma^{1 / 2}\bar{\alpha})\mathrm{d}\alpha \leq 1 / 4$ , which concludes the proof.
Note that the result of Lemma 12 can be improved to  $1/2 - \varepsilon$  for any  $\varepsilon > 0$ . In particular, this result tells us that for  $\gamma > 0$  small enough,  $\overline{\mathcal{M}}$  looks like the hyperplane from the point of view of the Gaussian with variance  $\gamma$  centered on  $\partial \mathcal{M}$ .
Bound on moments of order two and higher. In what follows, we define for any  $\gamma > 0$ ,  $\Delta^{\gamma} : \overline{\mathcal{M}} \to \mathbb{R}_{+}$  given for any  $x \in \overline{\mathcal{M}}$  by
$$
\Delta^ {\gamma} (x) = \left(1 / \gamma\right) \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \| \sqrt {\gamma} z \| ^ {4} \varphi (z) \mathrm {d} z / \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \varphi (z) \mathrm {d} z. \tag {48}
$$
Proposition 13. We have  $\lim_{\gamma \to 0} \sup \{\Delta^\gamma(x) : x \in \overline{\mathcal{M}}\} = 0$ .
Proof. Let  $\bar{\gamma} > 0$  given by Lemma 12. Let  $x \in \overline{\mathcal{M}}$  and  $\gamma \in (0, \bar{\gamma})$ . We have using Lemma 12
$$
\int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \varphi (z) \mathrm {d} z \geq 1 / 4. \tag {49}
$$
$$
(1 / \gamma) \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \| \sqrt {\gamma} z \| ^ {4} \varphi (z) d z \leq 3 \gamma d ^ {2}. \tag {50}
$$
Therefore, we get that for any  $\gamma \in (0,\bar{\gamma})$ $\Delta^{\gamma}(x)\leq 12\gamma d^{2}$  , which concludes the proof.
In what follows, we define for any  $\gamma > 0$ ,  $\dot{\Sigma}^{\gamma}: \overline{\mathcal{M}} \to \mathrm{S}_d^+(\mathbb{R})$  given for any  $x \in \overline{\mathcal{M}}$  by
$$
\hat {\Sigma} ^ {\gamma} (x) = \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} z \otimes z \varphi (z) \mathrm {d} z / \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \varphi (z) \mathrm {d} z. \tag {51}
$$
Proposition 14. There exists  $\bar{\gamma} >0$  such that for any  $x\in \overline{\mathcal{M}}$  and  $\gamma \in (0,\bar{\gamma})$  we have
$$
\| \hat {\Sigma} ^ {\gamma} (x) \| \leq 4 d. \tag {52}
$$
Proof. Let  $x \in \overline{\mathcal{M}}$  and  $\bar{\gamma} > 0$  given by Lemma 12. For any  $\gamma \in (0, \bar{\gamma})$ , we have using Lemma 12
$$
\int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \varphi (z) \mathrm {d} z \geq 1 / 4. \tag {53}
$$
$$
\int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \| z \| ^ {2} \varphi (z) \mathrm {d} z \leq d, \tag {54}
$$
D.4 Properties of large drift terms
Finally, we define for any  $\gamma > 0$ ,  $\hat{b}^{\gamma} : \overline{\mathcal{M}} \to \mathbb{R}^{d}$  given for any  $x \in \overline{\mathcal{M}}$  by
$$
\hat {b} ^ {\gamma} (x) = \gamma^ {- 1 / 2} \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} z \varphi (z) \mathrm {d} z / \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \varphi (z) \mathrm {d} z. \tag {55}
$$
First, we show away from the boundary the drift  $\hat{b}^{\gamma}$  converges to zero.
Proposition 15. There exists  $\bar{\gamma} >0$  such that for any  $\gamma \in (0,\bar{\gamma})$ ,  $r > 0$  and  $x\in \overline{\mathcal{M}}$  such that  $d(x,\partial \mathcal{M})\geq r$  we have  $\| \hat{b}^{\gamma}(x)\| \leq 2d\Psi (r / \gamma ,d)^{1 / 2} / \gamma^{1 / 2}$
Proof. Let  $x \in \overline{\mathcal{M}}$  and  $\bar{\gamma} > 0$  given by Lemma 12. For any  $\gamma \in (0, \bar{\gamma})$  we have using Lemma 12
$$
\int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \varphi (z) \mathrm {d} z \geq 1 / 4. \tag {56}
$$
$$
\begin{array}{l} \left\| \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} z \varphi (z) \mathrm {d} z \right\| \leq \left\| \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {\| z \| \leq r / \gamma^ {1 / 2}} z \varphi (z) \mathrm {d} z \right\| + \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {\| z \| \geq r / \gamma^ {1 / 2}} \| z \| \varphi (z) \mathrm {d} z (57) \\ \leq 2 \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {\| z \| \geq r / \gamma^ {1 / 2}} \| z \| \varphi (z) \mathrm {d} z \leq 2 d \Psi (r / \gamma , d) ^ {1 / 2} / \gamma^ {1 / 2}, (58) \\ \end{array}
$$
We have the following corollary.
Corollary 16. There exists  $\bar{\gamma} >0$  such that for any  $\delta >0$  there exists  $M_{\delta} > 0$  such that for any  $\gamma \in (0,\bar{\gamma})$  and  $x\in \overline{\mathcal{M}}$ ,  $\| \hat{b}^{\gamma}(x)\| \geq M_{\delta}$ , then  $\Phi (x)\leq \delta$ .
Proof. Let  $\bar{\gamma} > 0$  given by Lemma 12. Let  $f: \mathbb{R}_+ \to \mathbb{R}_+$  given for any  $r > 0$  by  $f(r) = \sup \{\gamma > 0: \Psi(r / \gamma, 1)^{1/2} / \gamma^{1/2}\}$ . We have that  $f$  is non-increasing and  $\lim_{r \to 0} f(r) = +\infty$ . Let  $\delta > 0$  and  $M_{\delta} = 2df(\delta / C)$  with  $C = \sup \{\|\nabla \Phi(x)\| : x \in \overline{\mathcal{M}}\}$ . Let  $\gamma \in (0, \bar{\gamma})$  and  $x \in \overline{\mathcal{M}}$  such that  $\|\hat{b}^\gamma(x)\| \geq M_\delta$  then using Proposition 15 we have that  $d(x, \partial \mathcal{M}) \leq \delta / C$ . Let  $\bar{x} \in \partial \mathcal{M}$  such that  $\|x - \bar{x}\| = d(x, \partial \mathcal{M})$ . We have
$$
\Phi (x) \leq \Phi (\bar {x}) + \int_ {0} ^ {1} \left\langle \nabla \Phi (\bar {x} + t (x - \bar {x})), x - \bar {x} \right\rangle \mathrm {d} t \leq \delta , \tag {59}
$$
For ease of notation, for any  $\gamma > 0$ , we define  $\bar{b}^{\gamma} = \gamma^{1/2}\hat{b}^{\gamma}$ , the renormalized version of the drift. First, we have the following result which will ensure that the drift projected on the normal component does not vanish.
Lemma 17. There exists  $\bar{\gamma} >0$  such that for any  $\gamma \in (0,\bar{\gamma})$  and  $x\in \mathsf{V}$  we have
$$
\langle \bar {b} ^ {\gamma} (x), \nabla \Phi (\bar {x}) \rangle \geq \| \bar {b} ^ {\gamma} (x) \| - \psi (\gamma), \tag {60}
$$
with  $\psi : \mathbb{R}_+ \to \mathbb{R}_+$  such that  $\lim \sup_{\gamma \to 0} \psi(\gamma) / \sqrt{\gamma} < +\infty$ .
$$
\int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \varphi (z) \mathrm {d} z \geq 1 / 4. \tag {61}
$$
$$
\begin{array}{l} \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \langle z, \nabla \Phi (\bar {x}) \rangle \varphi (z) \mathrm {d} z \geq \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in C (x, \gamma)} \langle z, \nabla \Phi (\bar {x}) \rangle \varphi (z) \mathrm {d} z (62) \\ - \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M} \cap \mathrm {C} (x, \gamma) ^ {\mathrm {c}}} \langle z, \nabla \Phi (\bar {x}) \rangle \varphi (z). (63) \\ \end{array}
$$
Using Lemma 11, we get that
$$
\int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \langle z, \nabla \Phi (\bar {x}) \rangle \varphi (z) \mathrm {d} z \geq \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathsf {C} (x, \gamma)} \langle z, \nabla \Phi (\bar {x}) \rangle \varphi (z) \mathrm {d} z - \psi (\gamma). \tag {64}
$$
Let  $\{e_i\}_{i=1}^{d-1}$  a basis of  $\nabla \Phi(\bar{x})^\perp$ . Using Theorem 7-(b), we have that for any  $i \in \{1, \ldots, d-1\}$
$$
\int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathrm {C} (x, \gamma)} \langle z, e _ {i} \rangle \varphi (z) \mathrm {d} z = \int_ {- \bar {\alpha} / \gamma^ {1 / 2}} ^ {r / \gamma^ {1 / 2}} \int_ {\nabla \Phi (\bar {x}) ^ {\perp}} \mathbb {1} _ {\| v \| ^ {2} \leq (\gamma^ {1 / 2} \alpha + \bar {\alpha}) / \gamma} \langle v, e _ {i} \rangle \varphi (v) \varphi (\alpha) \mathrm {d} v \mathrm {d} \alpha . \tag {65}
$$
Hence, combining this result and the Cauchy-Schwarz inequality we have for any  $i \in \{1, \dots, d - 1\}$
$$
\begin{array}{l} \left(\int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathbb {C} (x, \gamma)} \langle z, e _ {i} \rangle \varphi (z) \mathrm {d} z\right) ^ {2} = \left(\int_ {- \bar {\alpha} / \gamma^ {1 / 2}} ^ {r / \gamma^ {1 / 2}} \int_ {\nabla \Phi (\bar {x}) ^ {\perp}} \mathbb {1} _ {\| v \| ^ {2} \geq (\gamma^ {1 / 2} \alpha + \bar {\alpha}) / \gamma} \langle v, e _ {i} \rangle \varphi (v) \varphi (\alpha) \mathrm {d} v \mathrm {d} \alpha\right) ^ {2} (66) \\ \leq \int_ {\nabla \Phi (\bar {x}) ^ {\perp}} \left\langle v, e _ {i} \right\rangle^ {2} \varphi (v) d v \left(\int_ {- \bar {\alpha} / \gamma^ {1 / 2}} ^ {r / \gamma^ {1 / 2}} \Psi \left(\left(\bar {\alpha} + \alpha \gamma^ {1 / 2}\right) / \gamma , d - 1\right) ^ {1 / 2} \varphi (\alpha) d \alpha\right) ^ {2} (67) \\ \leq \left(\int_ {- \bar {\alpha} / \gamma^ {1 / 2}} ^ {r / \gamma^ {1 / 2}} \Psi \left(\left(\bar {\alpha} + \alpha \gamma^ {1 / 2}\right) / \gamma , d - 1\right) ^ {1 / 2} \varphi (\alpha) \mathrm {d} \alpha\right) ^ {2}. (68) \\ \end{array}
$$
Hence, using Lemma 9, we get that
$$
\sum_ {i = 1} ^ {d - 1} \left(\int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathrm {C} (x, \gamma)} \langle z, e _ {i} \rangle \varphi (z) \mathrm {d} z\right) ^ {2} \leq (d - 1) \psi^ {2} (\gamma), \tag {69}
$$
with  $\psi$  given by Lemma 9 with  $\beta_0 = 1 / 2$ . Therefore, we get that
$$
\begin{array}{l} \left(\int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in C (x, \gamma)} \langle z, \nabla \Phi (\bar {z}) \rangle \varphi (z) \mathrm {d} z\right) ^ {2} (70) \\ = \left(\int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \varphi (z) \mathrm {d} z\right) ^ {2} \| \bar {b} ^ {\gamma} (x) \| ^ {2} - \sum_ {i = 1} ^ {d - 1} \left(\int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathrm {C} (x, \gamma)} \langle z, e _ {i} \rangle \varphi (z) \mathrm {d} z\right) ^ {2} (71) \\ \geq \left(\int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \varphi (z) \mathrm {d} z\right) ^ {2} \| \bar {b} ^ {\gamma} (x) \| ^ {2} - \psi (\gamma) ^ {2}. (72) \\ \end{array}
$$
We conclude the proof upon using that for any  $a, b \geq 0$ ,  $(a + b)^{1/2} \leq a^{1/2} + b^{1/2}$  and (61).
We are now ready to state the following lower bound on the drift.
Proposition 18. There exist  $\bar{\gamma} >0$ ,  $M\geq 0$  and  $c > 0$  such that for any  $x\in \overline{\mathcal{M}}$  and  $\gamma \in (0,\bar{\gamma})$  if  $\| \hat{b}^{\gamma}(x)\| \geq M$  then  $x\in \mathsf{V}$  and
$$
\min  \left(\langle \hat {b} ^ {\gamma} (x), \nabla \Phi (x) \rangle , \langle \hat {b} ^ {\gamma} (x), \nabla \Phi (\bar {x}) \rangle\right) \geq c \| \hat {b} ^ {\gamma} (x) \|. \tag {73}
$$
Proof. Let  $\bar{\gamma} > 0$  given by Lemma 12 and  $M_0 = 4\sup \{\psi(\gamma) / \gamma^{1/2} : \gamma \in (0, \bar{\gamma}]\}$ . In addition, let  $c = 1/4$ . Using Proposition 15 and Theorem 7-(e), there exists  $M_1 \geq 0$  such that for any any  $x \in \overline{\mathcal{M}}$ , if  $\|\hat{b}^\gamma(x)\| \geq M_1$  then  $x \in \mathsf{V}$  and  $x = \bar{x} + \alpha \nabla \Phi(\bar{x})$  with  $\alpha \leq 1/(4C)$  and  $C = \sup \{\|\nabla^2 \Phi(x)\| : x \in \overline{\mathcal{M}}\}$ . We denote  $M = \max(M_0, M_1)$ . Let  $\gamma \in (0, \bar{\gamma})$  and  $x \in \overline{\mathcal{M}}$  such that  $\|\hat{b}^\gamma(x)\| \geq M$ . Using Lemma 17, we have that
$$
\langle \hat {b} ^ {\gamma} (x), \nabla \Phi (\bar {x}) \rangle \geq \| \hat {b} ^ {\gamma} (x) \| - \psi (\gamma) / \gamma^ {1 / 2}. \tag {74}
$$
Using that  $\psi (\gamma) / \gamma^{1 / 2}\leq M / 2\leq \| \hat{b}^{\gamma}(x)\| /2$  , we have
$$
\langle \hat {b} ^ {\gamma} (x), \nabla \Phi (\bar {x}) \rangle \geq (1 / 2) \| \hat {b} ^ {\gamma} (x) \|. \tag {75}
$$
Since  $\| x - \bar{x} \| \leq \alpha \leq 1 / (4C)$  we have  $\langle \hat{b}^{\gamma}(x), \nabla \Phi(x) \rangle \geq (1/2 - C\alpha) \| \hat{b}^{\gamma}(x) \| \geq \| \hat{b}^{\gamma}(x) \| / 4$ , which concludes the proof.
D.5 Convergence on compact sets
In this section, we show the convergence of the drift and diffusion matrix on compact sets. We recall that  $\mathcal{M}$  does not include its boundary  $\partial \mathcal{M}$ .
Proposition 19. For any compact set  $\mathsf{K} \subset \mathcal{M}$  and  $\varepsilon > 0$ , there exists  $\bar{\gamma} > 0$  such that for any  $\gamma \in (0, \bar{\gamma})$  we have for any  $x \in \mathsf{K}$
$$
\left\| \hat {b} ^ {\gamma} (x) \right\| \leq \varepsilon , \quad \left\| \hat {\Sigma} ^ {\gamma} (x) - \operatorname {I d} \right\| \leq \varepsilon . \tag {76}
$$
Proof. Let  $\mathsf{K} \subset \mathcal{M}$  be a compact set and  $\gamma > 0$ . Since  $\mathsf{K} \cap \partial \mathcal{M} = \emptyset$ , there exists  $r > 0$  such that for any  $x \in \mathsf{K}$ ,  $d(x, \partial \mathcal{M}) > r$ . Therefore, we have that for any  $x \in \mathsf{K}$
$$
\left\| \hat {b} ^ {\gamma} (x) \right\| = \gamma^ {- 1 / 2} \| \int_ {x + \sqrt {\gamma} z \in \mathcal {M}} z \varphi (z) \mathrm {d} z \| / \int_ {x + \sqrt {\gamma} z \in \mathcal {M}} \varphi (z) \mathrm {d} z. \tag {77}
$$
In addition, using the Cauchy-Schwarz inequality we have
$$
\begin{array}{l} \left\| \int_ {x + \sqrt {\gamma} z \in \mathcal {M}} z \varphi (z) \mathrm {d} z \right\| \leq \left\| \int_ {\mathbb {R} ^ {d}} z \varphi (z) \mathrm {d} z \right\| + \int_ {\mathcal {M} ^ {\mathrm {c}}} \| z \| \varphi (z) \mathrm {d} z (78) \\ \leq \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {\| z \| \geq r / \gamma^ {1 / 2}} \| z \| \varphi (z) d z \leq \sqrt {d} \Psi \left(r ^ {2} / \gamma , d\right) ^ {1 / 2}. (79) \\ \end{array}
$$
Using Lemma 8 and Lemma 12, there exists  $\bar{\gamma}_0 > 0$  such that for any  $\gamma \in (0,\bar{\gamma}_0)$  we have that for any  $x\in K$
$$
\left\| \hat {b} ^ {\gamma} (x) \right\| \leq 4 d \Psi \left(r ^ {2} / \gamma , 1\right) ^ {1 / 2} / \gamma^ {1 / 2} \leq \varepsilon , \tag {80}
$$
which concludes the first part of the proof. Similarly, we have that for any  $x \in \mathsf{K}$
$$
\begin{array}{l} \left\| \int_ {x + \sqrt {\gamma} z \in \mathcal {M}} (z \otimes z - \operatorname {I d}) \varphi (z) \mathrm {d} z \right\| \leq \left\| \int_ {\mathbb {R} ^ {d}} (z \otimes z - \operatorname {I d}) \varphi (z) \mathrm {d} z \right\| + \int_ {\mathcal {M} ^ {\mathrm {c}}} \| z \| \varphi (z) \mathrm {d} z (81) \\ \leq \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {\| z \| \geq r / \gamma^ {1 / 2}} \| z \otimes z - \operatorname {I d} \| \varphi (z) \mathrm {d} z (82) \\ \leq \sqrt {2} (1 + 3 d ^ {2}) ^ {1 / 2} \Psi \left(r ^ {2} / \gamma , d\right) ^ {1 / 2}. (83) \\ \end{array}
$$
Using Lemma 8 and Lemma 12, there exists  $\bar{\gamma}_1 > 0$  such that for any  $\gamma \in (0,\bar{\gamma}_1)$ , we have that for any  $x\in K$
$$
\left\| \hat {\Sigma} ^ {\gamma} (x) - \operatorname {I d} \right\| \leq 4 \sqrt {2} \left(1 + 3 d ^ {2}\right) ^ {1 / 2} \Psi \left(r ^ {2} / \gamma , 1\right) ^ {1 / 2} \leq \varepsilon , \tag {84}
$$
which concludes the proof upon letting  $\bar{\gamma} = \min (\bar{\gamma}_0,\bar{\gamma}_1)$
D.6 Convergence on the boundary
Finally, we investigate the behavior at the boundary of the diffusion matrix and the drift. First, we show that there is a lower bound to the diffusion matrix near the boundary. Second, we show that the renormalized drift converges to the outward normal.
Proposition 20. There exist  $c > 0$  and  $\bar{\gamma} > 0$  such that for any  $\gamma \in (0, \bar{\gamma})$ ,  $u \in \mathbb{R}^d$  and  $x \in \mathsf{V}$  we have
$$
\langle u, \hat {\Sigma} ^ {\gamma} (x) u \rangle \geq c \| u \| ^ {2}. \tag {85}
$$
In particular, there exist  $r,\varepsilon >0$  such that for any  $\gamma \in (0,\bar{\gamma})$  and  $x\in \overline{\mathcal{M}}$  with  $d(x,\partial \mathcal{M})\leq r$
$$
\langle \nabla \Phi (x), \hat {\Sigma} ^ {\gamma} (x) \nabla \Phi (x) \rangle \geq \varepsilon . \tag {86}
$$
Proof. First, we show (85). Let  $x \in \mathsf{V}$ . We have for any  $u \in \mathbb{R}^d$
$$
\begin{array}{l} \langle u, \hat {\Sigma} ^ {\gamma} (x) u \rangle = \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \langle z, u \rangle^ {2} \varphi (z) \mathrm {d} z / \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in \mathcal {M}} \mathrm {d} z (87) \\ \geq \int_ {\mathbb {R} ^ {d}} \mathbb {1} _ {x + \sqrt {\gamma} z \in C (x, \gamma)} \langle z, u \rangle^ {2} \varphi (z) \mathrm {d} z. (88) \\ \end{array}
$$
For any  $u\in \mathbb{R}^d$  , let  $\alpha_{u} = \langle u,\nabla \Phi (\bar{x})\rangle$  . Using Theorem 7-(b) we have for any  $u\in \mathbb{R}^d$