Using Lemma I.2 with $\delta = \frac{\bar{\epsilon}_2}{16}$, we deduce that for all $x \in \mathcal{X}_j^{\text{near}}(r)$, $\eta_{\alpha, \beta}(x) \le 1 - \frac{15}{32}\bar{\epsilon}_2 \mathfrak{d}_g(x, x_j^0)^2$.

With the same reasoning, for all $x \in \mathcal{G}_{x_j^0}(\mathcal{X}_j^{\text{near}}(r))$ we can obtain $\left\| D_2[\eta_{\alpha^j, \beta^j}](x) - K_{\text{norm}}^{(02)}(x_j^0, x) \right\|_x \le \frac{\bar{\epsilon}_2}{16}$.

So from Lemma I.2, for all $x \in \mathcal{X}_j^{\text{near}}(r)$ we have $|1 - \eta_{\alpha^j, \beta^j}(x)| \le \frac{B_{02} + \bar{\epsilon}_2/16}{2} \mathfrak{d}_g(x, x_j^0)^2$. We used that $B_{02} \ge \bar{\epsilon}_2$. We also have, for $i \ne j$, and $x \in \mathcal{G}_{x_i^0}(\mathcal{X}_i^{\text{near}}(r))$,

$$ \begin{aligned} \|D_2[\eta_{\alpha^j, \beta^j}](x)\|_x &= \left\| \alpha_i^j K_{\text{norm}}^{(02)}(x_i^0, x) + \sum_{l \ne i} \alpha_l^j K_{\text{norm}}^{(02)}(x_l^0, x) + [\beta_i^j] K_{\text{norm}}^{(12)}(x_i^0, x) + \sum_{l \ne i} [\beta_l^j] K_{\text{norm}}^{(12)}(x_l^0, x) \right\|_x , \\ &\le |\alpha_i^j| B_{02} + \|\alpha^j\|_{\infty} h + \max_i \left\| \beta_i^j \right\|_{x_i^0} (B_{12} + h), \\ &\le \frac{2h}{1-2h} B_{02} + \frac{h}{1-2h} + 4h(B_{12} + h), \\ &\le \frac{\bar{\epsilon}_2}{16}. \end{aligned} $$

Lemma I.2 ensures that for all $x \in \mathcal{X}_i^{\text{near}}(r)$, $|\eta_{\alpha^j, \beta^j}(x)| \le \frac{\bar{\epsilon}_2}{16} \mathfrak{d}_g(x, x_i^0)^2$. This concludes the proof. $\square$

**Norm of the certificates** We use bounds on the norm of the certificates to get the controls in estimation and in prediction ($c_p$ in Assumption 1). The construction of certificates presented in Lemma I.1 allows us to obtain the bounds displayed in the following proposition.

**Proposition I.1.** The certificates constructed in Lemma I.1 verify, for all $x \in \mathbb{R}^d \times [u_{\min}, +\infty)^d$, for all $j \in \{1, \dots, s\}$,

$$ \eta_{\alpha, \beta}(x) = \langle \Psi \delta_x, p_{\alpha, \beta} \rangle_{\mathbb{L}} \quad \text{and} \quad \eta_{\alpha^j, \beta^j}(x) = \langle \Psi \delta_x, p_{\alpha^j, \beta^j} \rangle_{\mathbb{L}} \quad \text{with} \quad \|p_{\alpha, \beta}\|_{\mathbb{L}} \le \sqrt{2s}, \quad \|p_{\alpha^j, \beta^j}\|_{\mathbb{L}} \le \sqrt{2}. $$

*Proof.* From Lemma 5.1, for all $j \in \{1, \dots, s\}$,

$$ p_{\alpha, \beta} = \sum_{i=1}^{s} \alpha_i \Psi \delta_{x_i^0} + \sum_{i=1}^{s} \beta_i \nabla_x (\Psi \delta_{x_i^0}) $$

and

$$ p_{\alpha^j, \beta^j} = \sum_{i=1}^{s} \alpha_i^j \Psi \delta_{x_i^0} + \sum_{i=1}^{s} \beta_i^j \nabla_x (\Psi \delta_{x_i^0}). $$

So recalling (63) and defining

$$ D_{\mathfrak{g}} := \left( \begin{array}{c} \mathrm{Id}_s \\ \mathfrak{g}_{x_1^0}^{-\frac{1}{2}} \\ \vdots \\ \mathfrak{g}_{x_s^0}^{-\frac{1}{2}} \end{array} \right) \in \mathbb{R}^{s(d+1) \times s(d+1)}, \quad \tilde{\Upsilon} := D_{\mathfrak{g}} \Upsilon D_{\mathfrak{g}}, $$

using the results of [Poon et al., 2023, p. 268] we have $\|p_{\alpha, \beta}\|_{\mathbb{L}}^2 = \begin{pmatrix} \alpha \\ \beta \end{pmatrix}^T \Upsilon \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \mathbf{u}_s^T \tilde{\Upsilon}^{-1} \mathbf{u}_s$. We can apply [Poon et al., 2023, Lemma 3] that gives $\|\tilde{\Upsilon}^{-1}\|_2 \le 2$. So $\|p_{\alpha, \beta}\|_{\mathbb{L}}^2 \le \|\tilde{\Upsilon}^{-1}\|_2 \|\mathbf{u}_s\|_2^2 \le 2s$. We repeat the same reasoning to bound $\|p_{\alpha^j, \beta^j}\|_{\mathbb{L}}^2$: we have $\|p_{\alpha^j, \beta^j}\|_{\mathbb{L}}^2 = (\mathbf{u}_s^j)^T \tilde{\Upsilon}^{-1} \mathbf{u}_s^j \le \|\tilde{\Upsilon}^{-1}\|_2 \|\mathbf{u}_s^j\|_2^2 \le 2$. $\square$

The combination of Theorem I.1 and Proposition I.1 concludes the proof of Theorem 5.2.
J Proof of Theorem 5.3

**Lemma J.1.** Let $x, x' \in \mathbb{R}^d \times [u_{\min}, +\infty)^d$. We can establish that

$$
\begin{align*}
\| K_{\text{norm}}^{(00)}(x, x') \|_{x, x'} &= |K_{\text{norm}}(x, x')|, \\
\| K_{\text{norm}}^{(10)}(x, x') \|_{x, x'} &= \|\mathfrak{g}_x^{-1/2} \nabla_1 K_{\text{norm}}(x, x')\|_2, \\
\| K_{\text{norm}}^{(11)}(x, x') \|_{x, x'} &= \|\mathfrak{g}_x^{-1/2} \nabla_1 \nabla_2 K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2}\|_2, \\
\| K_{\text{norm}}^{(02)}(x, x') \|_{x, x'} &= \|\mathfrak{g}_{x'}^{-1/2} H_2^g K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2}\|_2, \\
\| K_{\text{norm}}^{(12)}(x, x') \|_{x, x'} &\le \sqrt{2d} \max_{k=1, \dots, d} \left\{ \begin{array}{l} \| \mathfrak{g}_{t_k t_k}^{-1/2} \mathfrak{g}_{x'}^{-1/2} \partial_{t_k} H_2^g K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2} \|_2, \\ \| \mathfrak{g}_{u_k u_k}^{-1/2} \mathfrak{g}_{x'}^{-1/2} \partial_{u_k} H_2^g K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2} \|_2 \end{array} \right\}.
\end{align*}
$$

*Proof.* To get the simplified expressions for the operator norms (Definition 5.1), we use that $\|v\|_x^2 := v^T g_x v$. The result for $\|K_{\text{norm}}^{(10)}(x, x')\|_{x, x'}$, $\|K_{\text{norm}}^{(11)}(x, x')\|_{x, x'}$, $\|K_{\text{norm}}^{(02)}(x, x')\|_{x, x'}$ is stated by [Poon et al., 2023, Equation (27)].

To deal with $\|K_{\text{norm}}^{(12)}(x, x')\|_{x, x'}$, we also use that our metric is diagonal. Denoting $\tilde{q} = \sqrt{\mathfrak{g}_x q}$, $\tilde{V}_1 = \sqrt{\mathfrak{g}_{x'} V_1}$, $\tilde{V}_2 = \sqrt{\mathfrak{g}_{x'} V_2}$, we have

$$
\begin{align*}
\| K_{\text{norm}}^{(12)}(x, x') \|_{x, x'} &= \sup_{\substack{\|V_1\|_{x'} \le 1 \\ \|q=(q_1, \dots, q_{2d})\|_x \le 1}} \sum_{k=1}^d q_k V_1^T \partial_{t_k} H_2^g K_{\text{norm}}(x, x') V_2 + \sum_{k=1}^d q_{k+d} V_1^T \partial_{u_k} H_2^g K_{\text{norm}}(x, x') V_2, \\
&= \sup_{\|\tilde{V}_1\|_2, \|\tilde{V}_2\|_2 \le 1, \|\tilde{q}\|_2 \le 1} \left( \sum_{k=1}^d \tilde{q}_k \tilde{V}_1^T \mathfrak{g}_{t_k t_k}^{-1/2} \mathfrak{g}_{x'}^{-1/2} \partial_{t_k} H_2^g K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2} \tilde{V}_2 \right. \\
&\qquad \left. + \sum_{k=1}^d \tilde{q}_{k+d} \tilde{V}_1^T \mathfrak{g}_{u_k u_k}^{-1/2} \mathfrak{g}_{x'}^{-1/2} \partial_{u_k} H_2^g K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2} \tilde{V}_2 \right), \\
&\le \sqrt{2d} \max_{k=1, \dots, d} \left\{ \begin{aligned}
& \sup_{\|\tilde{V}_1\|_2, \|\tilde{V}_2\|_2 \le 1} \left| \tilde{V}_1^T \mathfrak{g}_{t_k t_k}^{-1/2} \mathfrak{g}_{x'}^{-1/2} \partial_{t_k} H_2^g K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2} \tilde{V}_2 \right|, \\
& \sup_{\|\tilde{V}_1\|_2, \|\tilde{V}_2\|_2 \le 1} \left| \tilde{V}_1^T \mathfrak{g}_{u_k u_k}^{-1/2} \mathfrak{g}_{x'}^{-1/2} \partial_{u_k} H_2^g K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2} \tilde{V}_2 \right|
\end{aligned} \right\}
\end{align*}
$$

giving the desired result.

□

**Global controls** To check the first condition of Definition 5.2, we give global controls for the Riemannian derivatives of the kernel. We take advantage of the metric used, which allows us to have uniform bounds on $\mathbb{R}^d \times [u_{\min}, +\infty)^d$.

**Lemma J.2 (Global controls).** For $(i, j) \in \{0, 1\} \times \{0, 1, 2\}$ we have

$$
\sup_{x, x' \in \mathbb{R}^d \times [u_{\min}, +\infty)^d} \| K_{\text{norm}}^{(ij)}(x, x') \|_{x, x'} \le B_{ij}
$$

with

$$
\begin{align*}
B_{00} &= 1, \\
B_{10} &= B_{01} = \sqrt{2d}, \\
B_{11} &= 2d, \\
B_{02} &= B_{20} = \sqrt{4d^2 + 10d}, \\
B_{12} &= B_{21} = \sqrt{2dB_{02}}.
\end{align*}
$$
Proof. The bounds are based on the expressions given in Lemma J.1. We make use of the relation between the Frobenius norm and the 2-norm:

$$ \forall M = (m_{ij})_{\substack{1 \le i \le n \\ 1 \le j \le m}} \in \mathbb{R}^{n \times m}, \quad \|M\|_2 \le \sqrt{\sum_{\substack{1 \le i \le n \\ 1 \le j \le m}} m_{ij}^2} \le \sqrt{nm \max_{i,j} m_{ij}^2}. $$

Let $x, x' \in \mathbb{R}^d \times [u_{\min}, +\infty)^d$.

**B00:** Using Cauchy-Schwarz and since the kernel is normalized, we have

$$ |K_{\text{norm}}(x, x')| = |\langle \Psi \delta_x, \Psi \delta_{x'} \rangle_{\mathbb{L}}| \le \sup_{x \in \mathbb{R}^d \times [u_{\min}, +\infty)^d} \|\Psi \delta_x\|_{\mathbb{L}}^2 = \sup_{x \in \mathbb{R}^d \times [u_{\min}, +\infty)^d} K_{\text{norm}}(x, x) = 1. $$

**B10:** We use that $\mathfrak{g}_x$ is diagonal, and name its diagonal elements by $\text{diag}(\mathfrak{g}_{t_1 t_1}, \dots, \mathfrak{g}_{t_d t_d}, \mathfrak{g}_{u_1 u_1}, \dots, \mathfrak{g}_{u_d u_d}) = \mathfrak{g}_x$. Remark that $\nabla_1 K_{\text{norm}}(x, x') = \langle \nabla \Psi \delta_x, \Psi \delta_{x'} \rangle_{\mathbb{L}}$ along with, for all $k \in \{1, \dots, d\}$, $\|\partial_{t_k} \Psi \delta_x\|_{\mathbb{L}} = \mathfrak{g}_{t_k t_k}^{1/2}$ and $\|\partial_{u_k} \Psi \delta_x\|_{\mathbb{L}} = \mathfrak{g}_{u_k u_k}^{1/2}$. We use Cauchy-Schwarz to write

$$ \begin{aligned} \| \mathfrak{g}_x^{-1/2} \nabla_1 K_{\text{norm}}(x, x') \|_2 &= \sqrt{\sum_{k=1}^d \mathfrak{g}_{t_k t_k}^{-1} \langle \partial_{t_k} \Psi \delta_x, \Psi \delta_{x'} \rangle_{\mathbb{L}}^2 + \sum_{k=1}^d \mathfrak{g}_{u_k u_k}^{-1} \langle \partial_{u_k} \Psi \delta_x, \Psi \delta_{x'} \rangle_{\mathbb{L}}^2}, \\ &\le \sqrt{\sum_{k=1}^d \| \Psi \delta_{x'} \|_{\mathbb{L}}^2 + \sum_{k=1}^d \| \Psi \delta_{x'} \|_{\mathbb{L}}^2}, \\ &\le \sqrt{2d}. \end{aligned} $$

**B11:** We use the same reasoning. With $\mathfrak{g}_x^{-1/2} \nabla_1 \nabla_2 K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2} = (m_{ij})_{1 \le i,j \le 2d}$, we have $m_{ij}$ of the form $\mathfrak{g}_{b_k b_k}^{-1/2} \mathfrak{g}_{b_l b_l}^{-1/2} \langle \partial_{b_k} \Psi \delta_x, \partial_{b_l} \Psi \delta_{x'} \rangle_{\mathbb{L}}$ where $b_m$ stands for $u_m$ or $t_m$, $1 \le m \le d$. So

$$ m_{ij}^2 \le \mathfrak{g}_{b_k b_k}^{-1} \mathfrak{g}_{b_l b_l}^{-1} \| \partial_{b_k} \Psi \delta_x \|_{\mathbb{L}}^2 \| \partial_{b_l} \Psi \delta_{x'} \|_{\mathbb{L}}^2 \le 1, $$

from which we conclude that $\|M\|_2 \le \sqrt{4d^2} = 2d$.

**B02:** Here, we denote $(m_{ij})_{1 \le i,j \le 2d} = \mathfrak{g}_{x'}^{-1/2} H_2^\mathfrak{g} K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2}$. Recall that

$$ H_2^\mathfrak{g} K_{\text{norm}}(x, x') = \nabla_2^2 K_{\text{norm}}(x, x') - \sum_{k=1}^d \Gamma^{t'_k} \partial_{t'_k} K_{\text{norm}}(x, x') - \sum_{k=1}^d \Gamma^{u'_k} \partial_{u'_k} K_{\text{norm}}(x, x'). $$

Using that $\Gamma^{b_k b_l b_m} \ne 0$ implies that $b_k = b_l = t_k, b_m = u_k$ or $b_k = b_m = t_k, b_l = u_k$ or $b_k = u_m, b_l = b_m = t_k$ or $b_k = b_l = b_m = u_k$, we can treat the 4d terms $m_{ii}, m_{i2i}, m_{2ii}, m_{2i2i}$ ($i \in \{1, \dots, d\}$) separately from the rest of the matrix.

The other terms are of the form

$$ \mathfrak{g}_{b'_k b'_k}^{-1/2} \mathfrak{g}_{b'_l b'_l}^{-1/2} \partial_{b'_k b'_l} K_{\text{norm}}(x, x') $$

where $k \ne l$ (i.e. the derivatives are associated with different dimensions). By abuse of notation, we denote $K_{\text{norm}}(x_m, x'_m)$ an evaluation of the kernel in dimension $d=1$ at points $x_m = (t_m, u_m), x'_m = (t'_m, u'_m)$, and we consider in a similar way $\Psi \delta_{x_m}$. Remark that $\mathfrak{g}_{b_m b_m} = \|\partial_{b_m} \Psi \delta_{x_m}\|_{\mathbb{L}}^2$. As $K_{\text{norm}}(x, x') = \prod_{m=1}^d K_{\text{norm}}(x_m, x'_m)$, for $k \ne l$ we have

$$ \partial_{b'_k b'_l} K_{\text{norm}}(x, x') = \partial_{b'_k} K_{\text{norm}}(x_k, x'_k) \partial_{b'_l} K_{\text{norm}}(x_l, x'_l) \prod_{m \notin \{k,l\}} K_{\text{norm}}(x_m, x'_m). \quad (65) $$

So using Cauchy-Schwarz, for $m_{ij}$ such that $i \ne j$ and $i \ne 2j, j \ne 2i$,

$$ m_{ij}^2 \le \mathfrak{g}_{b'_k b'_k}^{-1} \mathfrak{g}_{b'_l b'_l}^{-1} \left( \partial_{b'_k} K_{\text{norm}}(x_k, x'_k) \right)^2 \left( \partial_{b'_l} K_{\text{norm}}(x_l, x'_l) \right)^2 \prod_{m \notin \{k,l\}} K_{\text{norm}}(x_m, x'_m)^2 \le 1. $$

This quantity bounds $(2d)^2 - 4d$ squared coefficients of $M$.

The terms $m_{ii}, m_{i2i}$ and $m_{2ii}, m_{2i2i}$ ($i \in \{1, \dots, d\}$) are of the form

$$ \mathfrak{g}_{t'_k t'_k}^{-1} \left( \partial_{t'_k t'_k} K_{\text{norm}}(x, x') - \Gamma^{u'_k t'_k t'_k} \partial_{u'_k} K_{\text{norm}}(x, x') \right) \quad (66) $$

$$ \text{ or } \mathfrak{g}_{t'_k t'_k}^{-1/2} \mathfrak{g}_{u'_k u'_k}^{-1/2} \left( \partial_{t'_k u'_k} K_{\text{norm}}(x, x') - \Gamma^{t'_k t'_k u'_k} \partial_{t'_k} K_{\text{norm}}(x, x') \right) \quad (67) $$

$$ \text{ or } \mathfrak{g}_{u'_k u'_k}^{-1} \left( \partial_{u'_k u'_k} K_{\text{norm}}(x, x') - \Gamma^{u'_k u'_k u'_k} \partial_{u'_k} K_{\text{norm}}(x, x') \right). \quad (68) $$
These forms concern respectively $d$, $2d$, $d$ coefficients. We can again reduce the problem to dimension 1 using the decomposition of the kernel:

$$
\begin{align}
\partial_{t'_k t'_k} K_{\text{norm}}(x, x') - \Gamma^{u'_k} t'_k t'_k \partial_{u'_k} K_{\text{norm}}(x, x') &= \left( \partial_{t'_k t'_k} K_{\text{norm}}(x_k, x'_k) - \Gamma^{u'_k} t'_k t'_k \partial_{u'_k} K_{\text{norm}}(x_k, x'_k) \right) \prod_{l \neq k} K_{\text{norm}}(x_l, x'_l), \nonumber \\
\partial_{t'_k u'_k} K_{\text{norm}}(x, x') - \Gamma^{t'_k} t'_k u'_k \partial_{t'_k} K_{\text{norm}}(x, x') &= \left( \partial_{t'_k u'_k} K_{\text{norm}}(x_k, x'_k) - \Gamma^{t'_k} t'_k u'_k \partial_{t'_k} K_{\text{norm}}(x_k, x'_k) \right) \prod_{l \neq k} K_{\text{norm}}(x_l, x'_l), \nonumber \\
\partial_{u'_k u'_k} K_{\text{norm}}(x, x') - \Gamma^{u'_k} u'_k u'_k \partial_{u'_k} K_{\text{norm}}(x, x') &= \left( \partial_{u'_k u'_k} K_{\text{norm}}(x_k, x'_k) - \Gamma^{u'_k} u'_k u'_k \partial_{u'_k} K_{\text{norm}}(x_k, x'_k) \right) \prod_{l \neq k} K_{\text{norm}}(x_l, x'_l). \tag{69}
\end{align}
$$

So for the first form (66),

$$
\begin{align*}
m_{ii}^2 &= \mathfrak{g}_{t'_k t'_k}^{-2} \langle \Psi \delta_x, \partial_{t'_k t'_k} \Psi \delta_{x'} - \Gamma^{u'_k} t'_k t'_k \partial_{u'_k} \Psi \delta_{x'} \rangle_{\mathbb{L}}^2, \\
&\le \mathfrak{g}_{t'_k t'_k}^{-2} \left\| \partial_{t'_k t'_k} \Psi \delta_{x'_k} - \Gamma^{u'_k} t'_k t'_k \partial_{u'_k} \Psi \delta_{x'_k} \right\|_{\mathbb{L}}^2, \\
&= \mathfrak{g}_{t'_k t'_k}^{-2} \left( \partial_{t_k t'_k t'_k} K_{\text{norm}}(x'_k, x'_k) + \Gamma^{u'_k 2} t'_k t'_k \partial_{u'_k} K_{\text{norm}}(x'_k, x'_k) - 2\Gamma^{u'_k} t'_k t'_k \partial_{t_k t'_k} K_{\text{norm}}(x'_k, x'_k) \right), \\
&= 1
\end{align*}
$$

where $\partial_{t_k} K_{\text{norm}}$ (resp. $\partial_{t'_k} K_{\text{norm}}$) denotes a derivative w.r.t. the first (resp. the second) variable of the kernel. We can calculate this quantity, which is constant equal to 1 (see [Giard, 2025, Section IV]). In a similar way, for the second form (67) we have

$$
\begin{align*}
m_{2i2i}^2, m_{2ii}^2 &\le \mathfrak{g}_{t'_k t'_k}^{-1} \mathfrak{g}_{u'_k u'_k}^{-1} \left\| \partial_{t'_k u'_k} \Psi \delta_{x'_k} - \Gamma^{u'_k} t'_k u'_k \partial_{t'_k} \Psi \delta_{x'_k} \right\|_{\mathbb{L}}^2, \\
&= \mathfrak{g}_{t'_k t'_k}^{-1} \mathfrak{g}_{u'_k u'_k}^{-1} \left( \partial_{t_k u'_k} K_{\text{norm}}(x'_k, x'_k) + \Gamma^{t'_k 2} t'_k u'_k \partial_{t_k t'_k} K_{\text{norm}}(x'_k, x'_k) - 2\Gamma^{t'_k} t'_k u'_k \partial_{t_k u'_k} K_{\text{norm}}(x'_k, x'_k) \right), \\
&= 3.
\end{align*}
$$

For the third one (68),

$$
\begin{align*}
m_{2i2i}^2 &\le \mathfrak{g}_{u'_k u'_k}^{-2} \left( \partial_{u'_k u'_k} K_{\text{norm}}(x'_k, x'_k) + \Gamma^{u'_k 2} u'_k u'_k \partial_{u'_k} K_{\text{norm}}(x'_k, x'_k) - 2\Gamma^{u'_k} u'_k u'_k \partial_{u'_k} K_{\text{norm}}(x'_k, x'_k) \right), \\
&= 7.
\end{align*}
$$

$$
\text{Hence } \|M\|_2 \le \sqrt{4d^2 - 4d + d + 2 \times 3d + 7d} = \sqrt{4d^2 + 10d}.
$$

B$_{12}$: Lemma J.1 gives

$$
\left\| K_{\text{norm}}^{(12)}(x, x') \right\|_{x,x'} \le \sqrt{2d} \max_{t_k, u_k, k=1,\dots,d} \left\{ \left\| \mathfrak{g}_{t_k t_k}^{-1/2} \mathfrak{g}_{x'}^{-1/2} \partial_{t_k} H_2^{\mathfrak{g}} K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2} \right\|_2, \left\| \mathfrak{g}_{u_k u_k}^{-1/2} \mathfrak{g}_{x'}^{-1/2} \partial_{u_k} H_2^{\mathfrak{g}} K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2} \right\|_2 \right\}.
$$

For any $k \in \{1, \dots, d\}$, $b_k = t_k$ or $b_k = u_k$, the coefficients of the matrix $M_k = \mathfrak{g}_{b_k b_k}^{-1/2} \mathfrak{g}_{x'}^{-1/2} \partial_{b_k} H_2^{\mathfrak{g}} K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2}$ are of the form

$$
\mathfrak{g}_{b_m b_m}^{-1/2} \mathfrak{g}_{b'_k b'_k}^{-1/2} \mathfrak{g}_{b'_k b'_k}^{-1/2} \partial_{b_m b'_k b'_k} K_{\text{norm}}(x, x') \quad \text{where } k \neq l
$$

$$
\text{or } \mathfrak{g}_{b_m b_m}^{-1/2} \mathfrak{g}_{t'_k t'_k}^{-1} \left( \partial_{b_m t'_k t'_k} K_{\text{norm}}(x, x') - \Gamma^{u'_k} t'_k t'_k \partial_{b_m u'_k} K_{\text{norm}}(x, x') \right)
$$

$$
\text{or } \mathfrak{g}_{b_m b_m}^{-1/2} \mathfrak{g}_{t'_k t'_k}^{-1/2} \mathfrak{g}_{u'_k u'_k}^{-1/2} \left( \partial_{b_m t'_k u'_k} K_{\text{norm}}(x, x') - \Gamma^{t'_k} t'_k u'_k \partial_{b_m t'_k} K_{\text{norm}}(x, x') \right)
$$

$$
\text{or } \mathfrak{g}_{b_m b_m}^{-1/2} \mathfrak{g}_{u'_k u'_k}^{-1} \left( \partial_{b_m u'_k u'_k} K_{\text{norm}}(x, x') - \Gamma^{u'_k} u'_k u'_k \partial_{b_m u'_k} K_{\text{norm}}(x, x') \right).
$$

These forms correspond respectively to $(2d)^2 - 4d$, $d$, $2d$, $d$ coefficients. We can bound them with the same arguments as before. We obtain $\|M_k\|_2 \le B_{02}$, so $\|K_{\text{norm}}^{(12)}(x, x')\|_{x,x'} \le \sqrt{2dB_{02}}$. $\square$

Controls when $d(x, x')$ is small

**Lemma J.3** (Curvature constants in dimension $d=1$). Let $r>0$. Let $x, x' \in \mathbb{R} \times [u_{\min}, +\infty)$. If $d(x, x') \ge r$, then $K_{\text{norm}}(x, x') \le 1 - \bar{\epsilon}_0(r)$ where $\bar{\epsilon}_0(r) \le 1 - e^{-r^2/2}$. Moreover, if $d(x, x') \le r$ where $r \le 0.32$, then

$$
-v^T H_2^{\mathfrak{g}} K_{\text{norm}}(x, x') v \ge \bar{\epsilon}_2(r) \|v\|_{x'}^2 \quad \forall v \in \mathbb{R}^2
$$

$$
\text{where } \bar{\epsilon}_2(r) \le e^{-r^2/2} |G(r)| \quad (G(r) \text{ defined by (76) below}).
$$
**Proof.** Obtaining $\bar{\varepsilon}_0$: By definition of the semi-distance (see (21)), $d(x, x') \ge r$ implies that $K_{\text{norm}}(x, x') \le e^{-r^2/2}$. Obtaining $\bar{\varepsilon}_2$, general overview: We use that

$$ -v^T H_2^g K_{\text{norm}}(x, x') v \ge \bar{\varepsilon}_2 \|v\|_{x'}^2, \quad \forall v \in \mathbb{R}^2 \iff -v^T g_{x'}^{-1/2} H_2^g K_{\text{norm}}(x, x') g_{x'}^{-1/2} v \ge \bar{\varepsilon}_2 \|v\|_2^2, \quad \forall v \in \mathbb{R}^2. $$

Defining $\tilde{H}^{02}(x, x') := K_{\text{norm}}(x, x')^{-1} g_{x'}^{-1/2} H_2^g K_{\text{norm}}(x, x') g_{x'}^{-1/2}$, we have (see [Giard, 2025, Section V.1])

$$ \tilde{H}^{02}(x, x') = \begin{pmatrix} \tilde{H}_{t't'}^{02}(x, x') & \tilde{H}_{t'u'}^{02}(x, x') \\ \tilde{H}_{t'u'}^{02}(x, x') & \tilde{H}_{u'u'}^{02}(x, x') \end{pmatrix} \quad \text{where} $$

$$ \tilde{H}_{t't'}^{02}(x, x') = -1, $$

$$ \tilde{H}_{t'u'}^{02}(x, x') = \frac{(t-t')^3(2u'^2 + \tau^2)^{3/2}}{\sqrt{2}(u^2 + u'^2 + \tau^2)^3} - \frac{3(t-t')(2u'^2 + \tau^2)^{1/2}(u'^2 - u^2)}{\sqrt{2}(u^2 + u'^2 + \tau^2)^2}, $$

$$ \tilde{H}_{u'u'}^{02}(x, x') = \frac{(t-t')^4(2u'^2 + \tau^2)^2}{2(u^2 + u'^2 + \tau^2)^4} + \frac{3(t-t')^2(2u'^2 + \tau^2)(u^2 - u'^2)}{(u^2 + u'^2 + \tau^2)^3} + \frac{(u^2 - u'^2)^2}{2(u^2 + u'^2 + \tau^2)^2} - \frac{(2u^2 + \tau^2)(2u'^2 + \tau^2)}{(u^2 + u'^2 + \tau^2)^2}. \quad (70) $$

If the maximal eigenvalue $\lambda$ of $\tilde{H}^{02}(x, x')$ is smaller than some $c < 0$, as $K_{\text{norm}}(x, x') \ge e^{-r^2/2}$,

$$ -v^T H_2^g K_{\text{norm}}(x, x') v \ge -\lambda K_{\text{norm}}(x, x') \|v\|_{\mathbb{L}}^2 \ge -c e^{-r^2/2} \|v\|_{\mathbb{L}}^2 $$

and we can take $\bar{\varepsilon}_2(r) = -e^{-r^2/2}c$. It remains to bound $\lambda$. We use that

$$ \lambda \le \max\{\tilde{H}_{t't'}^{02}(x, x'), \tilde{H}_{u'u'}^{02}(x, x')\} + |\tilde{H}_{t'u'}^{02}(x, x')|. $$

We will provide control of the three terms in the right-hand side of the previous inequality. Basic inequalities: Using that $d(x, x') \le r$, we have

$$ \frac{|t-t'|}{\sqrt{u^2 + u'^2 + \tau^2}} \le r \quad (71) $$

and

$$ \frac{u^2 + u'^2 + \tau^2}{\sqrt{2u^2 + \tau^2}\sqrt{2u'^2 + \tau^2}} \le e^{r^2}. $$

Using (35), we get

$$ \frac{|u^2 - u'^2|}{u^2 + u'^2 + \tau^2} \le \sqrt{e^{2r^2} - 1}. \quad (72) $$

From (36) we also have

$$ \frac{2u'^2 + \tau^2}{u^2 + u'^2 + \tau^2} \le e^{r^2} + \sqrt{e^{2r^2} - 1}. \quad (73) $$

Control of $\tilde{H}_{t'u'}^{02}$: Using (71), (72) and (73), we get from (70) that

$$ \begin{aligned} |\tilde{H}_{t'u'}^{02}(x, x')| & \le r^3 \frac{(2u'^2 + \tau^2)^{3/2}}{\sqrt{2}(u^2 + u'^2 + \tau^2)^{3/2}} + 3r \frac{(2u'^2 + \tau^2)^{1/2}|u^2 - u'^2|}{\sqrt{2}(u^2 + u'^2 + \tau^2)^{3/2}}, \\ & \le \frac{1}{\sqrt{2}} r^3 (e^{r^2} + \sqrt{e^{2r^2} - 1})^{3/2} + \frac{3}{\sqrt{2}} r \sqrt{e^{2r^2} - 1} \sqrt{e^{r^2} + \sqrt{e^{2r^2} - 1}}. \end{aligned} \quad (74) $$

Control of $\tilde{H}_{u'u'}^{02}$: As $\frac{\sqrt{2u^2+\tau^2}\sqrt{2u'^2+\tau^2}}{u^2+u'^2+\tau^2} \ge e^{-r^2}$,

$$ \begin{aligned} \tilde{H}_{u'u'}^{02}(x, x') & \le r^4 \frac{(2u'^2 + \tau^2)^2}{2(u^2 + u'^2 + \tau^2)^2} + 3r^2 \frac{(2u'^2 + \tau^2)|u^2 - u'^2|}{(u^2 + u'^2 + \tau^2)^2} + \frac{(u^2 - u'^2)^2}{2(u^2 + u'^2 + \tau^2)^2} - \frac{(2u^2 + \tau^2)(2u'^2 + \tau^2)}{(u^2 + u'^2 + \tau^2)^2}, \\ & \le \frac{1}{2} r^4 (e^{r^2} + \sqrt{e^{2r^2} - 1})^2 + 3r^2 \sqrt{e^{2r^2} - 1}(e^{r^2} + \sqrt{e^{2r^2} - 1}) + \frac{1}{2}(e^{2r^2} - 1) - e^{-2r^2}. \end{aligned} \quad (75) $$

where we have used (71), (72) and (73) again.

Conclusion: The previous bound is greater than $\tilde{H}_{t'u'}^{02}(x, x') = -1$ (because $-e^{-2r^2} \ge -1$). It comes that

$$ \begin{aligned} \lambda & \le \frac{1}{\sqrt{2}} r^3 (e^{r^2} + \sqrt{e^{2r^2} - 1})^{3/2} + \frac{3}{\sqrt{2}} r \sqrt{e^{2r^2} - 1} \sqrt{e^{r^2} + \sqrt{e^{2r^2} - 1}} \\ & \quad + \frac{1}{2} r^4 (e^{r^2} + \sqrt{e^{2r^2} - 1})^2 + 3r^2 \sqrt{e^{2r^2} - 1}(e^{r^2} + \sqrt{e^{2r^2} - 1}) + \frac{1}{2}(e^{2r^2} - 1) - e^{-2r^2} =: G(r). \end{aligned} \quad (76) $$
The function $G$ is non-decreasing on $\mathbb{R}^+$ as a sum of non-decreasing functions. It is negative for $r \le 0.32$ (see [Giard, 2025, Section V.2]). Then $\bar{\varepsilon}_2(r)$ can be chosen as $-e^{-r^2/2}G(r) = e^{-r^2/2}|G(r)|$ (or smaller) for $r \in [0, 0.32]$. $\square$

**Lemma J.4** (Curvature constants in dimension $d \ge 1$). Let $r \ge 0$. Let $x, x' \in \mathbb{R}^d \times [u_{\min}, +\infty)^d$. If $d(x, x') \ge r$, then $K_{\text{norm}}(x, x') \le 1 - \bar{\varepsilon}_0(r)$ where $\bar{\varepsilon}_0(r) \le 1 - e^{-r^2/2}$. Moreover, if $d(x, x') \le r$ where $r = \frac{r_0}{\sqrt{d}}$ with $0 < r_0 \le 0.32$, then

$$-vH_2^g K_{\text{norm}}(x, x')v \ge \bar{\varepsilon}_2(r) \|v\|_{x'}^2 \quad \forall v \in \mathbb{R}^2$$

where $\bar{\varepsilon}_2(r) \le e^{-r_0^2/2}|G(r_0)|$ ($G(r_0)$ defined by (76)).

*Proof.* To get $\bar{\varepsilon}_0(r)$, remark again that $d(x, x') \ge r$ implies that $K_{\text{norm}}(x, x') \le e^{-r^2/2}$ (see (21)).

**Reduction to dimension 1:** For $\bar{\varepsilon}_2(r)$, we use the same reasoning as in Lemma J.3.

We denote $\tilde{H}^{02}(x, x') := K_{\text{norm}}(x, x')^{-1}g_{x'}^{-1/2}H_2^g K_{\text{norm}}(x, x')g_{x'}^{-1/2}$. Its maximum eigenvalue $\lambda$ is smaller than $\max_{1 \le i \le 2d}\{|\tilde{H}_{ii}^{02}(x, x') + \sum_{j \ne i}|\tilde{H}_{ij}^{02}(x, x')|\}$. We denote $\tilde{H}_{ij}^{02}(x, x') = \tilde{H}_{b'_k b'_k}^{02}(x, x')$ to specify that this coefficient corresponds to the derivatives w.r.t. $b'_k, b'_l$ where $1 \le k, l \le d$ ($b$ can be $u$ or $t$).

We then remark that we can reduce the problem to dimension 1. First, according to (21), $d(x, x') \le r$ implies that $d(x_k, x'_k) \le r$ for all $k = 1, \dots, d$. Using (65), we get for $l \ne k$

$$\begin{aligned} |\tilde{H}_{b'_k b'_k}^{02}(x, x')| & \le |g_{b'_k b'_k}^{-1/2} g_{b'_l b'_l}^{-1/2} K_{\text{norm}}(x_k, x'_k)^{-1} K_{\text{norm}}(x_l, x'_l)^{-1} \partial_{b'_k} K_{\text{norm}}(x_k, x'_k) \partial_{b'_l} K_{\text{norm}}(x_l, x'_l)|, \\ & = |g_{b'_k b'_k}^{-1/2} K_{\text{norm}}(x_k, x'_k)^{-1} \partial_{b'_k} K_{\text{norm}}(x_k, x'_k)| |g_{b'_l b'_l}^{-1/2} K_{\text{norm}}(x_l, x'_l)^{-1} \partial_{b'_l} K_{\text{norm}}(x_l, x'_l)|. \end{aligned} \quad (77)$$

For $l = k$, using (69) we have

$$\tilde{H}_{t'_k t'_k}^{02}(x, x') \le \tilde{H}_{t'_k t'_k}^{02}(x_k, x'_k) = -1, \quad \tilde{H}_{u'_k u'_k}^{02}(x, x') \le \tilde{H}_{u'_k u'_k}^{02}(x_k, x'_k), \quad |\tilde{H}_{t'_k t'_k}^{02}(x, x')| \le |\tilde{H}_{t'_k t'_k}^{02}(x_k, x'_k)|. \quad (78)$$

**Bounds for $\tilde{H}_{b'_k b'_k}^{02}(x, x')$:** We use (78). From (74) and (75), we have

$$\tilde{H}_{t'_k t'_k}^{02}(x, x') \lor \tilde{H}_{u'_k u'_k}^{02}(x, x') \le \frac{1}{2}r^4(e^{r^2} + \sqrt{e^{2r^2} - 1})^2 + 3r^2\sqrt{e^{2r^2} - 1}(e^{r^2} + \sqrt{e^{2r^2} - 1}) + \frac{1}{2}(e^{2r^2} - 1) - e^{-2r^2}$$

and

$$|\tilde{H}_{t'_k u'_k}^{02}(x, x')| \le \frac{1}{\sqrt{2}}r^3(e^{r^2} + \sqrt{e^{2r^2} - 1})^{3/2} + \frac{3}{\sqrt{2}}r\sqrt{e^{2r^2} - 1}\sqrt{e^{r^2} + \sqrt{e^{2r^2} - 1}}.$$

**Bounds for $\tilde{H}_{b'_k b'_k}^{02}(x, x')$, $l \ne k$:** See [Giard, 2025, Section V.3]. We calculate

$$g_{x_k}^{-1/2} K_{\text{norm}}(x_k, x'_k)^{-1} \nabla_1 K_{\text{norm}}(x_k, x'_k) = \left( \frac{-(t_k - t'_k)\sqrt{2u_k^2 + \tau^2}}{u_k^2 + u'_k^2 + \tau^2} \frac{(t_k - t'_k)^2(2u_k^2 + \tau^2)}{\sqrt{2}(u_k^2 + u'_k^2 + \tau^2)^2} + \frac{u_k^2 - u_k^2}{\sqrt{2}(u_k^2 + u'_k^2 + \tau^2)} \right).$$

With (71), (72) and (73), it follows using (77) that for $l \ne k$,

$$\begin{aligned} |\tilde{H}_{t'_k t'_k}^{02}(x, x')| & \le \frac{|t'_k - t_k|\sqrt{2u_k^2 + \tau^2}}{u_k^2 + u'_k^2 + \tau^2} \frac{|t_l - t'_l|\sqrt{2u_l^2 + \tau^2}}{u_l^2 + u'_l^2 + \tau^2}, \\ & \le r^2(e^{r^2} + \sqrt{e^{2r^2} - 1}) \end{aligned}$$

and

$$\begin{aligned} |\tilde{H}_{u'_k u'_k}^{02}(x, x')| & \le \frac{|t'_l - t_l|\sqrt{2u_l^2 + \tau^2}}{u_l^2 + u'_l^2 + \tau^2} \left( \frac{(t_k - t'_k)^2(2u_k^2 + \tau^2)}{\sqrt{2}(u_k^2 + u'_k^2 + \tau^2)^2} + \frac{|u_k^2 - u'_k^2|}{\sqrt{2}(u_k^2 + u'_k^2 + \tau^2)} \right), \\ & \le r\sqrt{e^{r^2} + \sqrt{e^{2r^2} - 1}} \times \frac{1}{\sqrt{2}}(r^2(e^{r^2} + \sqrt{e^{2r^2} - 1}) + \sqrt{e^{2r^2} - 1}) \end{aligned}$$

along with

$$|\tilde{H}_{u'_k u'_k}^{02}(x, x')| \le \frac{1}{2} \left( r^2(e^{r^2} + \sqrt{e^{2r^2} - 1}) + \sqrt{e^{2r^2} - 1} \right)^2.$$
**Conclusion:** The bound found for $|\tilde{H}_{u_k u_l}^{02}(x, x')|$ is greater than the one found for $|\tilde{H}_{t_k t_l}^{02}(x, x')|$. In fact, using that $e^{r^2} \ge 1 + r^2$,

$$ 
\begin{aligned}
\frac{1}{2} \left( r^2(e^{r^2} + \sqrt{e^{2r^2} - 1}) + \sqrt{e^{2r^2} - 1} \right)^2 &\ge r^2(e^{r^2} + \sqrt{e^{2r^2} - 1})\sqrt{e^{2r^2} - 1} + \frac{1}{2}(e^{2r^2} - 1), \\
&\ge r^2\sqrt{e^{2r^2} - 1} + \frac{1}{2}(2r^2 + 1)(e^{2r^2} - 1), \\
&\ge r^2\sqrt{e^{2r^2} - 1} + \frac{1}{2}(2r^2 + 1)(e^{r^2}(1 + r^2) - 1), \\
&\ge r^2\sqrt{e^{2r^2} - 1} + r^2e^{r^2} + \frac{1}{2}(r^2 + 1)e^{r^2} - \frac{1}{2}(2r^2 + 1), \\
&\ge r^2\sqrt{e^{2r^2} - 1} + r^2e^{r^2}.
\end{aligned}
$$

Hence

$$ 
\begin{aligned}
\lambda &\le \frac{1}{2}r^4(e^{r^2} + \sqrt{e^{2r^2} - 1})^2 + 3r^2\sqrt{e^{2r^2} - 1}(e^{r^2} + \sqrt{e^{2r^2} - 1}) + \frac{1}{2}(e^{2r^2} - 1) - e^{-2r^2} \\
&\quad + (d-1)\frac{1}{2}\left(r^2(e^{r^2} + \sqrt{e^{2r^2} - 1}) + \sqrt{e^{2r^2} - 1}\right)^2 \\
&\quad + (d-1)r\sqrt{e^{r^2} + \sqrt{e^{2r^2} - 1}} \times \frac{1}{\sqrt{2}}\left(r^2(e^{r^2} + \sqrt{e^{2r^2} - 1}) + \sqrt{e^{2r^2} - 1}\right) \\
&\quad + \frac{1}{\sqrt{2}}r^3(e^{r^2} + \sqrt{e^{2r^2} - 1})^{3/2} + \frac{3}{\sqrt{2}}r\sqrt{e^{2r^2} - 1}\sqrt{e^{r^2} + \sqrt{e^{2r^2} - 1}}, \\
&= -e^{-2r^2} + 2r^2\sqrt{e^{2r^2} - 1}(e^{r^2} + \sqrt{e^{2r^2} - 1}) + \frac{d}{2}\left(r^2(e^{r^2} + \sqrt{e^{2r^2} - 1}) + \sqrt{e^{2r^2} - 1}\right)^2 \\
&\quad + \frac{dr}{\sqrt{2}}\sqrt{e^{r^2} + \sqrt{e^{2r^2} - 1}}\left(r^2(e^{r^2} + \sqrt{e^{2r^2} - 1}) + \sqrt{e^{2r^2} - 1}\right) \\
&\quad + \sqrt{2r}\sqrt{e^{2r^2} - 1}\sqrt{e^{r^2} + \sqrt{e^{2r^2} - 1}} =: G_d(r).
\end{aligned}
$$

Furthermore, $r \in \mathbb{R}^+ \mapsto \sqrt{e^{2r^2} - 1}$ is convex. In fact, for $r > 0$, using that $e^{2r^2} \ge 1 + 2r^2$, we have

$$ \frac{\partial^2}{\partial r^2} \sqrt{e^{2r^2} - 1} = \frac{2e^{2r^2}(2r^2e^{2r^2} - 4r^2 + e^{2r^2} - 1)}{(e^{2r^2} - 1)^{3/2}} \ge \frac{2e^{2r^2}(2r^2 - 4r^2 + 1 + 2r^2 - 1)}{(e^{2r^2} - 1)^{3/2}} \ge 0. $$

Let $r_0 > 0$. As $r := \frac{r_0}{\sqrt{d}} \le r_0$, $\sqrt{e^{2r_0^2} - 1} \le \frac{\sqrt{e^{2r_0^2} - 1}}{\sqrt{d}}$ and we deduce that $G_d(r) \le G_1(r_0) = G(r_0)$ (see (76)). If $r_0 \le 0.32$, this quantity is negative (proof of Lemma J.3). We can take $\bar{\varepsilon}_2(\frac{r_0}{\sqrt{d}}) = -e^{-\frac{r_0^2}{2d}}G(r_0)$. The choice of the dependence on $d$ for $r = \frac{r_0}{\sqrt{d}}$ is intended to compensate for the term $d(e^{2r^2} - 1)$ appearing in $G_d(r)$. $\square$

**Controls when $d(x, x')$ is large** The constraint $\tau \le u_{\min}$ is used in the following lemma.

**Lemma J.5** (Bounds under a large separation, in dimension $d = 1$). Let $\Delta > 0$. Let $x, x' \in \mathbb{R} \times [u_{\min}, +\infty)$. Assume that $\tau \le u_{\min}$. If $d(x, x') \ge \Delta$, then

$$ \max_{(i,j) \in \{0,1\} \times \{0,1,2\}} \left\| K_{\text{norm}}^{(ij)}(x, x') \right\|_{x,x'} \le \sqrt{2} 153.05 e^{-\frac{\Delta^2}{4}}. $$

*Proof.* The calculations in this proof are presented in [Giard, 2025, Sections VI.1 and VI.2]. Let $x, x' \in \mathbb{R} \times [u_{\min}, +\infty)$ such that $d(x, x') \ge \Delta$. The following bounds on the operator norms are based on the simplified expressions from Lemma J.1. We will use the inequalities

$$ \frac{\sqrt{2u^2 + \tau^2}\sqrt{2u'^2 + \tau'^2}}{u'^2 + u^2 + \tau^2} \le \frac{|u'^2 - u^2|}{u'^2 + u^2 + \tau^2} \le 1, \quad \frac{2u^2 + \tau^2}{u'^2 + u^2 + \tau^2} \le 2, \quad \frac{2u^2 + \tau^2}{u^2} \le 3, \quad (79) $$

which holds since $\tau \le u_{\min}$. As it holds $y^q e^{-\frac{u^2}{4}} \le (\frac{2q}{e})^{q/2}$ for all $y \ge 0$, $q \ge 1$, we also have that for all $x, x' \in \mathbb{R} \times [u_{\min}, +\infty)$,

$$ 
\begin{aligned}
&\forall q \ge 1, \quad \frac{|t - t'|^q}{(u'^2 + u^2 + \tau^2)^{\frac{q}{2}}} \sqrt{K_{\text{norm}}(x, x')} \le \left(\frac{2q}{e}\right)^{q/2} \frac{(2u^2 + \tau^2)^{1/8}(2u'^2 + \tau'^2)^{1/8}}{(u'^2 + u^2 + \tau^2)^{\frac{1}{4}}}, \\
&\le \left(\frac{2q}{e}\right)^{q/2}. \qquad (80)
\end{aligned}
$$
We will also use that $\sqrt{K_{\text{norm}}(x, x')} \le e^{-\Delta^2/4}$ for $d(x, x') \ge \Delta$ (see (21)).

In what follows, we bound the 2-norm of a matrix $M = (m_{ij})_{ij}$ by its Frobenius norm $\sqrt{\sum_{i,j} m_{ij}^2}$.

$\|K_{\text{norm}}^{(00)}(x, x')\|$: We have $|K_{\text{norm}}(x, x')| \le e^{-\Delta^2/2}$, which is an immediate consequence of (21).

$\|K_{\text{norm}}^{(10)}(x, x')\|_x$: First,

$$ \begin{pmatrix} H_t^{10}(x, x') \\ H_u^{10}(x, x') \end{pmatrix} = H^{10}(x, x') := \sqrt{K_{\text{norm}}(x, x')}^{-1} \mathfrak{g}_x^{-1/2} \nabla_1 K_{\text{norm}}(x, x'), \\ = \sqrt{K_{\text{norm}}(x, x')} \begin{pmatrix} \frac{(t-t')\sqrt{2u^2+\tau^2}}{u^2+u'^2+\tau^2} \\ \frac{(t-t')^2(2u^2+\tau^2)}{\sqrt{2}(u^2+u'^2+\tau^2)^2} + \frac{u'^2-u^2}{\sqrt{2}(u^2+u'^2+\tau^2)} \end{pmatrix}. $$

Using (80) and (79) along with $\sqrt{K_{\text{norm}}(x, x')} \le 1$, it comes

$$ \begin{gathered} |H_t^{10}(x, x')| \le \frac{2}{\sqrt{e}} \\ \text{and} \quad |H_u^{10}(x, x')| \le \left( \frac{4\sqrt{2}}{e} + \frac{1}{\sqrt{2}} \right), \end{gathered} \qquad (81) $$

hence $\|H^{10}(x, x')\|_2 \le \sqrt{\frac{4}{e} + \left(\frac{4\sqrt{2}}{e} + \frac{1}{\sqrt{2}}\right)^2}$ and

$$ \|K_{\text{norm}}^{(10)}(x, x')\|_x = \sqrt{K_{\text{norm}}(x, x')} \|H^{10}(x, x')\|_2 \le e^{-\Delta^2/4} \sqrt{\frac{4}{e} + \left(\frac{4\sqrt{2}}{e} + \frac{1}{\sqrt{2}}\right)^2}. $$

$\|K_{\text{norm}}^{(11)}(x, x')\|_{x,x'}$: We have

$$ H^{11}(x, x') := \sqrt{K_{\text{norm}}(x, x')}^{-1} \mathfrak{g}_x^{-1/2} \nabla_1 \nabla_2 K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2} = \sqrt{K_{\text{norm}}(x, x')} \begin{pmatrix} \tilde{H}_{tt'}^{11}(x, x') & \tilde{H}_{tu'}^{11}(x, x') \\ \tilde{H}_{tu'}^{11}(x, x') & \tilde{H}_{uu'}^{11}(x, x') \end{pmatrix} $$

where

$$ \tilde{H}_{tt'}^{11}(x, x') = \frac{\sqrt{2u^2 + \tau^2}\sqrt{2u'^2 + \tau^2}}{u^2 + u'^2 + \tau^2} - \frac{(t-t')^2\sqrt{2u^2 + \tau^2}\sqrt{2u'^2 + \tau^2}}{(u^2 + u'^2 + \tau^2)^2}, $$

$$ \tilde{H}_{tu'}^{11}(x, x') = \frac{-(t-t')^3 u'(2u^2 + \tau^2)\sqrt{2u'^2 + \tau^2}}{\sqrt{2}u(u^2 + u'^2 + \tau^2)^3} + \frac{\sqrt{2}(t-t')u'(2u^2 + \tau^2)\sqrt{2u'^2 + \tau^2}}{u(u^2 + u'^2 + \tau^2)^2} \\ + \frac{(t-t')u'(u'^2 - u^2)(2u^2 + \tau^2)}{\sqrt{2}u\sqrt{2u'^2 + \tau^2}(u^2 + u'^2 + \tau^2)^2}, $$

$$ \tilde{H}_{tu'}^{11}(x, x') = \tilde{H}_{tu'}^{11}(x', x), $$

$$ \tilde{H}_{uu'}^{11}(x, x') = \frac{(t-t')^4(2u^2 + \tau^2)(2u'^2 + \tau^2)}{2(u^2 + u'^2 + \tau^2)^4} + \frac{-2(t-t')^2(2u^2 + \tau^2)(2u'^2 + \tau^2)}{(u^2 + u'^2 + \tau^2)^3} + \frac{(t-t')^2(u'^2 - u^2)^2}{(u^2 + u'^2 + \tau^2)^3} \\ + \frac{(2u^2 + \tau^2)(2u'^2 + \tau^2)}{(u^2 + u'^2 + \tau^2)^2} - \frac{(u^2 - u'^2)^2}{2(u^2 + u'^2 + \tau^2)^2}. $$

The constraint $\tau \le u_{\min}$ is used here, to control $\frac{\sqrt{2u^2+\tau^2}}{u}$ in $\tilde{H}_{tu'}^{11}, \tilde{H}_{tu'}^{11}$. Using again (80), (79) and the normalization $\sqrt{K_{\text{norm}}(x, x')} \le 1$, it comes

$$ \begin{align} |H_{tt'}^{11}(x, x')| &\le \left(1 + \frac{4}{e}\right), \nonumber \\ |H_{uu'}^{11}(x, x')| &\vee |H_{tu'}^{11}(x, x')| \le \left(\frac{\sqrt{3}}{\sqrt{2}}\left(\frac{6}{e}\right)^{3/2} + \frac{2\sqrt{3}}{\sqrt{e}} + \frac{\sqrt{3}}{\sqrt{e}}\right), \tag{82} \\ |H_{uu'}^{11}(x, x')| &\le \left(\frac{1}{2}\left(\frac{8}{e}\right)^2 + \frac{8}{e} + \frac{4}{e} + 1 + \frac{1}{2}\right). \nonumber \end{align} $$
Hence $\|H^{(11)}(x, x')\|_2 \le \sqrt{(1 + \frac{4}{e})^2 + 2(\frac{\sqrt{3}}{\sqrt{2}}(\frac{6}{e})^{3/2} + \frac{3\sqrt{3}}{\sqrt{e}})^2 + (\frac{32}{e^2} + \frac{12}{e} + \frac{3}{2})^2}$ and

$$ \|K_{\text{norm}}^{(11)}(x, x')\|_{x,x'} \le e^{-\Delta^2/4} \sqrt{\left(1 + \frac{4}{e}\right)^2 + 2\left(\frac{\sqrt{3}}{\sqrt{2}}\left(\frac{6}{e}\right)^{3/2} + \frac{3\sqrt{3}}{\sqrt{e}}\right)^2 + \left(\frac{32}{e^2} + \frac{12}{e} + \frac{3}{2}\right)^2} $$

$\|K_{\text{norm}}^{(02)}(x, x')\|_{x'}$: We use (70) to get the expression of

$$ H^{02}(x, x') := \sqrt{K_{\text{norm}}(x, x')} \tilde{H}^{02}(x, x') = \sqrt{K_{\text{norm}}(x, x')}^{-1} \mathfrak{g}_{x'}^{-1/2} H_2^{\mathfrak{g}} K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2}. $$

Using the same techniques as before, we find $\|H^{02}(x, x')\|_2 \le \sqrt{1 + 2\left(3\frac{\sqrt{2}}{\sqrt{e}} + 2\left(\frac{6}{e}\right)^{3/2}\right)^2 + \left(\frac{128}{e^2} + \frac{24}{e} + \frac{3}{2}\right)^2}$ so

$$ \|K_{\text{norm}}^{(02)}(x, x')\|_{x'} \le e^{-\Delta^2/4} \sqrt{1 + 2\left(3\frac{\sqrt{2}}{\sqrt{e}} + 2\left(\frac{6}{e}\right)^{3/2}\right)^2 + \left(\frac{128}{e^2} + \frac{24}{e} + \frac{3}{2}\right)^2} $$

$\|K_{\text{norm}}^{(12)}(x, x')\|_{x,x'}$: We denote

$$ H^{12,1}(x, x') = \sqrt{K_{\text{norm}}(x, x')}^{-1} \mathfrak{g}_{tt}^{-1/2} \mathfrak{g}_{x'}^{-1/2} \partial_t H_2^{\mathfrak{g}} K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2} $$

and

$$ H^{12,2}(x, x') = \sqrt{K_{\text{norm}}(x, x')}^{-1} \mathfrak{g}_{uu}^{-1/2} \mathfrak{g}_{x'}^{-1/2} \partial_u H_2^{\mathfrak{g}} K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2}. $$

We find $\tilde{H}^{12,1} = \sqrt{K_{\text{norm}}(x, x')} \begin{pmatrix} \tilde{H}_{t't'}^{12,1}(x, x') & \tilde{H}_{t'u'}^{12,1}(x, x') \\ \tilde{H}_{t'u'}^{12,1}(x, x') & \tilde{H}_{u'u'}^{12,1}(x, x') \end{pmatrix}$ with

$$ \tilde{H}_{t't'}^{12,1}(x, x') = \frac{(t - t')\sqrt{2u^2 + \tau^2}}{u^2 + u'^2 + \tau^2}, $$

$$ \tilde{H}_{t'u'}^{12,1}(x, x') = \frac{3(t - t')^2 \sqrt{2u^2 + \tau^2} (2u'^2 + \tau^2)^{3/2}}{\sqrt{2}(u^2 + u'^2 + \tau^2)^3} - \frac{(t - t')^4 \sqrt{2u^2 + \tau^2} (2u'^2 + \tau^2)^{3/2}}{\sqrt{2}(u^2 + u'^2 + \tau^2)^4} \\ + \frac{3(t - t')^2 \sqrt{2u^2 + \tau^2} \sqrt{2u'^2 + \tau^2} (u'^2 - u^2)}{\sqrt{2}(u^2 + u'^2 + \tau^2)^3} + \frac{3\sqrt{2u^2 + \tau^2} \sqrt{2u'^2 + \tau^2} (u^2 - u'^2)}{\sqrt{2}(u^2 + u'^2 + \tau^2)^2} $$

and

$$ \tilde{H}_{u'u'}^{12,1}(x, x') = \frac{-(t-t')^5(2u'^2 + \tau^2)^2 \sqrt{2u^2 + \tau^2}}{2(u^2 + u'^2 + \tau^2)^5} + \frac{2(t-t')^3(2u'^2 + \tau^2)^2 \sqrt{2u^2 + \tau^2}}{(u^2 + u'^2 + \tau^2)^4} \\ + \frac{3(t-t')^3(u'^2 - u^2)(2u'^2 + \tau^2)\sqrt{2u^2 + \tau^2}}{(u^2 + u'^2 + \tau^2)^4} - \frac{(t-t')\sqrt{2u^2 + \tau^2}(u^2 - u'^2)^2}{2(u^2 + u'^2 + \tau^2)^3} \\ + \frac{(t-t')(2u'^2 + \tau^2)(2u^2 + \tau^2)^{3/2}}{(u^2 + u'^2 + \tau^2)^3} + \frac{6(t-t')(2u'^2 + \tau^2)(u^2 - u'^2)\sqrt{2u^2 + \tau^2}}{(u^2 + u'^2 + \tau^2)^3}. $$

Using the same ideas as before to bound the coefficients, we get

$$ \|H^{12,1}(x, x')\|_2 \le \sqrt{\frac{4}{e} + 2\left(\frac{18\sqrt{2}}{e} + \sqrt{2}\frac{64}{e^2} + \frac{3}{\sqrt{2}}\right)^2 + \left(\sqrt{2}\left(\frac{10}{e}\right)^{5/2} + 7\sqrt{2}\left(\frac{6}{e}\right)^{3/2} + \frac{15}{\sqrt{e}}\right)^2} $$
We also have $\tilde{H}^{12,2} = \sqrt{K_{\text{norm}}(x, x')} \begin{pmatrix} \tilde{H}_{t't'}^{12,1}(x, x') & \tilde{H}_{t'u'}^{12,1}(x, x') \\ \tilde{H}_{t'u'}^{12,1}(x, x') & \tilde{H}_{u'u'}^{12,1}(x, x') \end{pmatrix}$ with

$$ \tilde{H}_{t't'}^{12,1}(x, x') = \frac{-(t-t')^2(2u^2 + \tau^2)}{\sqrt{2}(u^2 + u'^2 + \tau^2)^2} + \frac{u^2 - u'^2}{\sqrt{2}(u^2 + u'^2 + \tau^2)}, $$

$$ \tilde{H}_{t'u'}^{12,1}(x, x') = \frac{(t-t')^5(2u'^2 + \tau^2)^{3/2}(2u^2 + \tau^2)}{2(u^2 + u'^2 + \tau^2)^5} + \frac{(t-t')^3(2u'^2 + \tau^2)^{1/2}(2u^2 + \tau^2)(u^2 - u'^2)}{(u^2 + u'^2 + \tau^2)^4} \\ - \frac{4(t-t')^3(2u'^2 + \tau^2)^{3/2}(2u^2 + \tau^2)}{(u^2 + u'^2 + \tau^2)^4} + \frac{(t-t')^3(2u'^2 + \tau^2)^{1/2}}{(u^2 + u'^2 + \tau^2)^2} \\ + \frac{3(t-t')(2u'^2 + \tau^2)^{1/2}(2u^2 + \tau^2)^2}{(u^2 + u'^2 + \tau^2)^3} + \frac{21(t-t')(2u'^2 + \tau^2)^{1/2}(2u^2 + \tau^2)(u'^2 - u^2)}{2(u^2 + u'^2 + \tau^2)^3} \\ + \frac{3(t-t')(2u'^2 + \tau^2)^{1/2}(u^2 - u'^2)}{2(u^2 + u'^2 + \tau^2)^2} $$

and

$$ \tilde{H}_{u'u'}^{12,1}(x, x') = \frac{\sqrt{2}(t-t')^6(2u'^2 + \tau^2)^2(2u^2 + \tau^2)}{4(u^2 + u'^2 + \tau^2)^6} + \frac{5\sqrt{2}(t-t')^4(2u'^2 + \tau^2)(2u^2 + \tau^2)(u^2 - u'^2)}{4(u^2 + u'^2 + \tau^2)^5} \\ - \frac{2\sqrt{2}(t-t')^4(2u'^2 + \tau^2)^2(2u^2 + \tau^2)}{(u^2 + u'^2 + \tau^2)^5} + \frac{\sqrt{2}(t-t')^4(2u'^2 + \tau^2)(u^2 - u'^2)^2}{2(u^2 + u'^2 + \tau^2)^5} \\ - \frac{5\sqrt{2}(t-t')^2(2u'^2 + \tau^2)(u^2 - u'^2)^2}{4(u^2 + u'^2 + \tau^2)^4} + \frac{5\sqrt{2}(t-t')^2(2u^2 + \tau^2)(2u'^2 + \tau^2)^2}{2(u^2 + u'^2 + \tau^2)^4} \\ + \frac{\sqrt{2}(t-t')^2(u^2 - u'^2)^3}{2(u^2 + u'^2 + \tau^2)^4} - \frac{7\sqrt{2}(t-t')^2(2u^2 + \tau^2)(2u'^2 + \tau^2)(u^2 - u'^2)}{(u^2 + u'^2 + \tau^2)^4} \\ + \frac{7\sqrt{2}(2u'^2 + \tau^2)(2u^2 + \tau^2)(u^2 - u'^2)}{2(u^2 + u'^2 + \tau^2)^3} - \frac{\sqrt{2}(u^2 - u'^2)^3}{4(u^2 + u'^2 + \tau^2)^3}. $$

We get

$$ \|H^{12,2}(x, x')\|_2 \le \sqrt{ \left( \left( \frac{4\sqrt{2}}{e} + 1 \right)^2 + 2 \left( \frac{1}{\sqrt{2}} \left( \frac{10}{e} \right)^{5/2} + 6\sqrt{2} \left( \frac{6}{e} \right)^{3/2} + \frac{36}{\sqrt{e}} \right)^2 \right) + \left( \frac{432\sqrt{2}}{e^3} + \frac{272\sqrt{2}}{e^2} + \frac{60\sqrt{2}}{e} + \frac{15\sqrt{2}}{4} \right)^2 } \le 153.05. $$

As this bound is greater than the one found for $\|H^{12,1}(x, x')\|_2$, we deduce that

$$ \|K_{\text{norm}}^{(12)}(x, x')\|_{x,x'} \le \sqrt{2} 153.05 e^{-\Delta^2/4}. $$

This bound is the largest we obtained for $\|K_{\text{norm}}^{(ij)}(x, x')\|_{x,x'}$, hence it is an upper bound for all the derivatives investigated by this lemma. □

**Lemma J.6** (Bounds under a large separation, in dimension $d \ge 1$). Let $\Delta > 0$. Let $x, x' \in \mathbb{R}^d \times [u_{\min}, +\infty)^d$. Assume that $\tau \le u_{\min}$. If $d(x, x') \ge \Delta$, then

$$ \max_{(i,j) \in \{0,1\} \times \{0,1,2\}} \left\| K_{\text{norm}}^{(ij)}(x, x') \right\|_{x,x'} \le \sqrt{2d}(170.5 + 25.78d)e^{-\Delta^2/4}. $$

*Proof.* Let $x, x' \in \mathbb{R}^d \times [u_{\min}, +\infty)^d$ such that $d(x, x') \ge \Delta$. We use the decomposition of the kernel $K_{\text{norm}}(x, x') = \prod_{k=1}^d K_{\text{norm}}(x_k, x'_k)$ to reuse the results proven in dimension 1 (see Lemma J.5). As in Lemma J.4, we use the expressions for the operator norms given in Lemma J.1 and we bound the 2-norm of a matrix $M = (m_{ij})_{ij}$ by $\sqrt{\sum_{i,j} m_{ij}^2}$. The calculations in this proof are presented in [Giard, 2025, Section VI.3].

$$ \|K_{\text{norm}}^{(00)}(x, x')\|: \text{ According to (21), we still have } |K_{\text{norm}}^{(00)}(x, x')| \le e^{-\Delta^2/2}. $$

$$ \|K_{\text{norm}}^{(10)}(x, x')\|_x: \text{ We recall that the notation } \partial_{b_k} K_{\text{norm}} \text{ (resp. } \partial_{b'_k} K_{\text{norm}}) \text{ refers to a derivative w.r.t. to the} $$
first (resp. the second) variable of the kernel, where $b_k$ is $t_k$ or $u_k$. The decomposition of the kernel gives $\partial_{b_k} K_{\text{norm}}(x, x') = \partial_{b_k} K_{\text{norm}}(x_k, x'_k) \prod_{l \neq k}^{d} K_{\text{norm}}(x_l, x'_l)$. Using (80) for $K_{\text{norm}}(x_k, x'_k)$, we get that

$$ \forall q \ge 1, \quad \frac{|t_k - t'_k|^q}{(u_k^2 + u_k^2 + \tau^2)^{q/2}} \sqrt{K_{\text{norm}}(x_k, x'_k)} \le \left(\frac{2q}{e}\right)^{q/2}. $$

We have

$$ |\mathfrak{g}_{b_k b_k}^{-1/2} \partial_{b_k} K_{\text{norm}}(x, x')| \le \sqrt{\prod_{l=1}^{d} K_{\text{norm}}(x_l, x'_l)} |\sqrt{K_{\text{norm}}(x_k, x'_k)}^{-1} \mathfrak{g}_{b_k b_k}^{-1/2} \partial_{b_k} K_{\text{norm}}(x_k, x'_k)|. $$

The term in the right-hand side has already been dealt with in Lemma J.5: (81) gives a bound for

$$ |\sqrt{K_{\text{norm}}(x_k, x'_k)}^{-1} \mathfrak{g}_{b_k b_k}^{-1/2} \partial_{b_k} K_{\text{norm}}(x_k, x'_k)| = |H_{b_k}^{10}(x_k, x'_k)|, $$

that we denote by $B_{H_{b_k}^{10}}$ (note that this bound does not depend on $\Delta$). So

$$ |\mathfrak{g}_{b_k b_k}^{-1/2} \partial_{b_k} K_{\text{norm}}(x, x')| \le e^{-\Delta^2/4} B_{H_{b_k}^{10}}. $$

$$ \text{Hence } \left\| K_{\text{norm}}^{(10)}(x, x') \right\|_{x} \le \sqrt{d} \sqrt{B_{H_{t_k}^{10}}^2 + B_{H_{u_k}^{10}}^2} e^{-\Delta^2/4} \le 3.05 \sqrt{d} e^{-\Delta^2/4}. $$

$\|K_{\text{norm}}^{(11)}(x, x')\|_{x,x'}$: We use in the same way as for $B_{H_{b_k}^{(10)}}$ the notation $B_{H_{b_k,b'_k}^{(11)}}$, denoting the bound obtained in Lemma J.5 for $\mathfrak{g}_{x_k}^{-1/2} \partial_{b_k,b'_k} K_{\text{norm}}(x_k, x'_k) \mathfrak{g}_{x'_k}^{-1/2} \sqrt{K_{\text{norm}}(x_k, x'_k)}^{-1} = |H_{b_k b'_k}^{11}(x_k, x'_k)|$ (see (82)). For $k \neq l$,

$$ \begin{aligned} |\mathfrak{g}_{b_k b_k}^{-1/2} \partial_{b_k b'_k} K_{\text{norm}}(x_k, x'_k) \mathfrak{g}_{b'_k b'_k}^{-1/2}| & \le \sqrt{\prod_{m=1}^{d} K_{\text{norm}}(x_m, x'_m)} |\sqrt{K_{\text{norm}}(x_k, x'_k)}^{-1} \mathfrak{g}_{b_k b_k}^{-1/2} \partial_{b_k} K_{\text{norm}}(x_k, x'_k)| \\ & \quad \times |\sqrt{K_{\text{norm}}(x_l, x'_l)}^{-1} \mathfrak{g}_{b'_k b'_k}^{-1/2} \partial_{b'_k} K_{\text{norm}}(x_l, x'_l)|, \\ & \le e^{-\Delta^2/4} B_{H_{b_k}^{10}} B_{H_{b'_k}^{01}} = e^{-\Delta^2/4} B_{H_{b_k}^{10}} B_{H_{b'_k}^{10}}. \end{aligned} $$

For $k=l$, we have in the same way

$$ |\mathfrak{g}_{l_k l_k}^{-1/2} \partial_{l_k l'_k} K_{\text{norm}}(x_k, x'_k) \mathfrak{g}_{l'_k l'_k}^{-1/2}| \le e^{-\Delta^2/4} B_{H_{l'_k, l'_k}^{11}}, $$

$$ |\mathfrak{g}_{u_k u_k}^{-1/2} \partial_{u_k u'_k} K_{\text{norm}}(x_k, x'_k) \mathfrak{g}_{u'_k u'_k}^{-1/2}| \le e^{-\Delta^2/4} B_{H_{u'_k, u'_k}^{11}}, $$

$$ |\mathfrak{g}_{u_k u_k}^{-1/2} \partial_{u_k u'_k} K_{\text{norm}}(x_k, x'_k) \mathfrak{g}_{u'_k u'_k}^{-1/2}| \le e^{-\Delta^2/4} B_{H_{u'_k, u'_k}^{11}}. $$

So

$$ \begin{aligned} \left\| K_{\text{norm}}^{(11)}(x, x') \right\|_{x,x'} & \le e^{-\Delta^2/4} \sqrt{d(d-1)B_{H_{t_k}^{10}}^4 + d(d-1)B_{H_{u_k}^{10}}^4 + 2d(d-1)B_{H_{t_k}^{10}}^2 B_{H_{u_k}^{10}}^2} \\ & \quad + 2dB_{H_{t_k, u'_k}^{11}}^2 + dB_{H_{t_k, t'_k}^{11}}^2 + dB_{H_{u_k, u'_k}^{11}}^2, \\ & \le (9.25d + 6.95)e^{-\Delta^2/4}. \end{aligned} $$

$\|K_{\text{norm}}^{(02)}(x, x')\|_{x'}$: With the same arguments, the terms of $\mathfrak{g}_{x'}^{-1/2} H_2^2 K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2}$ corresponding to derivatives taken in $b'_k, b'_l$ with $k \neq l$ can be bounded by $e^{-\Delta^2/4} B_{H_{b_k}^{10}} B_{H_{b'_k}^{10}}$. We bound the terms corresponding to $k=l$ with $e^{-\Delta^2/4} B_{H_{t_k}^{02}}$ or $e^{-\Delta^2/4} B_{H_{t'_k}^{02}}$ or $e^{-\Delta^2/4} B_{H_{u'_k}^{02}}$, where $B_{H_{b'_k b'_k}^{02}}$ denotes again the bound obtained in Lemma J.5. We get

$$ \begin{aligned} \left\| K_{\text{norm}}^{(02)}(x, x') \right\|_{x'} & \le e^{-\Delta^2/4} \sqrt{d(d-1)B_{H_{t_k}^{10}}^4 + d(d-1)B_{H_{u_k}^{10}}^4 + 2d(d-1)B_{H_{t_k}^{10}}^2 B_{H_{u_k}^{10}}^2} \\ & \quad + 2dB_{H_{t_k, u'_k}^{02}}^2 + dB_{H_{t_k, t'_k}^{02}}^2 + dB_{H_{u_k, u'_k}^{02}}^2, \\ & \le (9.25d + 45.9)e^{-\Delta^2/4}. \end{aligned} $$
$\|K_{\text{norm}}^{(12)}(x, x')\|_{x, x'}$: We denote $M_{b_k} = \mathfrak{g}_{b_k b_k}^{-1/2} \mathfrak{g}_{x'}^{-1/2} \partial_{b_k} H_2^\mathfrak{g} K_{\text{norm}}(x, x') \mathfrak{g}_{x'}^{-1/2}$. The terms at positions $b'_l, b'_m$ where $l \neq m$ and $l, m \neq k$ can be bounded by $e^{-\Delta^2/4} B_{H_{b_k}^{10}} B_{H_{b_l}^{10}} B_{H_{b_m}^{10}}$. The terms at positions $b'_l, b'_k$ or $b'_k, b'_l$ where $l \neq k$ can be bounded by $e^{-\Delta^2/4} B_{H_{b_k, t'_k}^{11}} B_{H_{b_l}^{10}}$ or $e^{-\Delta^2/4} B_{H_{b_k, u'_k}^{11}} B_{H_{b_l}^{10}}$. The terms at positions $b'_l, b'_m$ where $l = m \neq k$ can be bounded by $e^{-\Delta^2/4} B_{H_{b_k}^{10}} B_{H_{b'_l b'_m}^{02}}$. The terms at positions $b'_l, b'_m$ where $l = m = k$ can be bounded by $e^{-\Delta^2/4} B_{H_{t'_k, t'_k}^{12, b_k}}$ or $e^{-\Delta^2/4} B_{H_{t'_k, u'_k}^{12, b_k}}$ or $e^{-\Delta^2/4} B_{H_{u'_k, u'_k}^{12, b_k}}$. Writing $M_{b_k} = (m_{b'_l b'_m})_{b'_l, b'_m} \in \{t_1, \dots, t_d, u_1, \dots, u_d\}$, we use the following bound for its 2-norm: $\|M_{b_k}\|_2 \le \max_l \sqrt{m_{t'_l t'_l}^2 + m_{t'_l u'_l}^2 + m_{u'_l t'_l}^2 + m_{u'_l u'_l}^2} + \sqrt{\sum_{l \neq m} m_{b'_k b'_l}^2}$. It comes that

$$ \|M_{t_k}\|_2 \le e^{-\Delta^2/4} \left( \max \left\{ \sqrt{B_{H_{t_k}^{10}}^2 B_{t'_k t'_k}^{02} + 2B_{H_{t_k}^{10}}^2 B_{t'_k u'_k}^{02} + B_{H_{t_k}^{10}}^2 B_{u'_k u'_k}^{02}}, \sqrt{2B_{H_{t'_k, u'_k}^{12,1}}^2 + B_{H_{t'_k, t'_k}^{12,1}}^2 + B_{H_{u'_k, u'_k}^{12,1}}^2} \right\} \right. \\ \left. + \sqrt{ (d^2 - 3d + 2)B_{H_{t_k}^{10}}^2 (B_{H_{t_k}^{10}}^4 + 2B_{H_{t_k}^{10}}^2 B_{H_{u_k}^{10}}^2 + B_{H_{u_k}^{10}}^4) } \right. \\ \left. + 2(d-1) (B_{H_{t'_k, t'_k}^{11}}^2 B_{H_{t_k}^{10}}^2 + B_{H_{t'_k, t'_k}^{11}}^2 B_{H_{u_k}^{10}}^2 + B_{H_{t'_k, u'_k}^{11}}^2 B_{H_{t_k}^{10}}^2 + B_{H_{t'_k, u'_k}^{11}}^2 B_{H_{u_k}^{10}}^2) \right) $$

and

$$ \|M_{u_k}\|_2 \le e^{-\Delta^2/4} \left( \max \left\{ \sqrt{B_{H_{u_k}^{10}}^2 B_{t'_k t'_k}^{02} + 2B_{H_{u_k}^{10}}^2 B_{t'_k u'_k}^{02} + B_{H_{u_k}^{10}}^2 B_{u'_k u'_k}^{02}}, \sqrt{2B_{H_{t'_k, u'_k}^{12,2}}^2 + B_{H_{t'_k, t'_k}^{12,2}}^2 + B_{H_{u'_k, u'_k}^{12,2}}^2} \right\} \right. \\ \left. + \sqrt{ (d^2 - 3d + 2)B_{H_{u_k}^{10}}^2 (B_{H_{t_k}^{10}}^4 + 2B_{H_{t_k}^{10}}^2 B_{H_{u_k}^{10}}^2 + B_{H_{u_k}^{10}}^4) } \right. \\ \left. + 2(d-1) (B_{H_{u'_k, t'_k}^{11}}^2 B_{H_{t_k}^{10}}^2 + B_{H_{u'_k, t'_k}^{11}}^2 B_{H_{u_k}^{10}}^2 + B_{H_{u'_k, u'_k}^{11}}^2 B_{H_{t_k}^{10}}^2 + B_{H_{u'_k, u'_k}^{11}}^2 B_{H_{u_k}^{10}}^2) \right), \\ \le e^{-\Delta^2/4}(170.5 + 25.78d). $$

As this last bound is greater than the previous one, $\|K_{\text{norm}}^{(12)}(x, x')\|_{x, x'} \le \sqrt{2d}(170.5 + 25.78d)e^{-\Delta^2/4}$. $\square$

### Choice of $r$, $\Delta$ for the LPC

**Proposition J.1** ($K_{\text{norm}}$ satisfies the LPC, $d=1$). Let $s \ge 2$. Assume that $\mathcal{X} \subset \mathbb{R} \times [u_{\min}, u_{\max}]$ and that $\tau \le u_{\min}$. Then $K_{\text{norm}}$ satisfies the LPC (see Definition 5.2) with parameters $s$, $\Delta(s) = 2\sqrt{13.88 + \ln(s-1)}$, $r = 0.3025$, $\bar{\varepsilon}_2(0.3025) = 0.13139$, $\bar{\varepsilon}_0(0.3025) = 0.04472$.

*Proof.* To establish that $K_{\text{norm}}$ satisfies the LPC, we first determine the size of the near regions $r$, giving the constraint on the minimal separation $\Delta$ (see Definition 5.2).

We want to pick $r$ such that $\bar{\varepsilon}_0(r)$, $\bar{\varepsilon}_2(r)$ exist and $\frac{1}{64} \min \left\{ \frac{\bar{\varepsilon}_0(r)}{B_0}, \frac{\bar{\varepsilon}_2(r)}{B_2} \right\}$ is maximal, using Lemmas J.3 and J.2. This allows us to get the smallest $\Delta$ (see item 3 of Definition 5.2). Graphically, we choose $r = 0.3025$ (see [Giard, 2025, Section VII.1]). We can take $\bar{\varepsilon}_2(0.3025) = 0.13139$ and $\bar{\varepsilon}_0(0.3025) = 0.04472$. Then

$$ \frac{1}{64} \min \left( \frac{0.04472}{B_0}, \frac{0.13139}{B_2} \right) \ge 0.000204618 := c_r $$

(see [Giard, 2025, Section VII.1]).

Let $s \ge 2$. The minimal separation $\Delta(s)$ must satisfy item 3 of Definition 5.2. Using Lemma J.5, it suffices that

$$ (s-1)\sqrt{2}153.05e^{-\frac{\Delta(s)}{4}} \le c_r, \\ \text{i.e. that } \Delta(s) \ge 2\sqrt{c_{\Delta} + \ln(s-1)} \text{ where } c_{\Delta} = 13.88 \ge \ln\left(\frac{\sqrt{2}153.05}{c_r}\right) \text{ (see [Giard, 2025, Section VII.1])}. \square $$

**Proposition J.2** ($K_{\text{norm}}$ satisfies the LPC, $d \ge 1$). Let $s \ge 2$. Assume that $\mathcal{X} \subset \mathbb{R}^d \times [u_{\min}, u_{\max}]^d$ and that $\tau \le u_{\min}$. Then $K_{\text{norm}}$ satisfies the LPC with parameters $s$, $r = \frac{0.3025}{\sqrt{d}}$, $\bar{\varepsilon}_2(r) = 0.13139$, $\bar{\varepsilon}_0(r) = \frac{0.0894}{2d}$, $\Delta(s) = 2\sqrt{11.9 + 3\ln(d+6.62) + \ln(s-1)}$.

*Proof.* We take Proposition J.1 as a starting point. We make use of Lemmas J.2, J.4 and J.6. Setting $r = \frac{0.3025}{\sqrt{d}}$, we can take $\bar{\varepsilon}_2(r) = 0.13139$ (as in dimension $d=1$) because $d \in \mathbb{N}^* \mapsto e^{-\frac{r^2}{2d}}|G(r_0)|$ with $r_0 = 0.3025$ is non-decreasing.
Furthermore, for $r \le 0.3025$ we have $1 - e^{-r^2/2} \ge 0.977 r^2$. In fact, for $0 < c < 1$, denoting

$$ \Box_c : r \in \mathbb{R}^+ \mapsto 1 - e^{-r^2/2} - c \frac{r^2}{2}, $$

we have $\frac{\partial}{\partial r}\Box_c(r) = r(e^{-r^2/2} - c)$ so $\Box_c$ is increasing then decreasing. Then remark that $\Box_c(0) = 0$ and $\Box_{0.977}(0.3025) \ge 0$. So we can take $\bar{\varepsilon}_0(r) = \frac{0.0894}{2d} \le 0.977 \frac{0.3025^2}{2d}$ (see [Giard, 2025, Section VII.2]).

Moreover, $\frac{\bar{\varepsilon}_0(r)}{1+B_{00}+B_{10}} \le \frac{\bar{\varepsilon}_2(r)}{1+B_{02}+B_{12}}$ (see [Giard, 2025, Section VII.2]). So

$$ \frac{1}{64} \min \left( \frac{\bar{\varepsilon}_0(r)}{1+B_{00}+B_{10}}, \frac{\bar{\varepsilon}_2(r)}{1+B_{02}+B_{12}} \right) \ge \frac{\varepsilon_0(r)}{64(1+B_{00}+B_{10})} \ge \frac{0.0894}{64(2+\sqrt{2d})} =: c_{d,r}. $$

The minimal separation should verify $(s-1)\sqrt{2d}(170.5 + 25.78d)e^{-\Delta(s)^2/4} \le c_{d,r}$, i.e.

$$ \Delta(s) \ge 2\sqrt{\ln\left(\frac{64}{0.0894}\right) + \ln\left((170.5 + 25.78d)2d(2 + \sqrt{2d})\sqrt{2d}\right) + \ln(s-1)}. $$

It suffices that $\Delta(s) \ge 2\sqrt{11.9 + 3\ln(d+6.62) + \ln(s-1)}$ (see [Giard, 2025, Section VII.2]). $\square$

# K Proofs of Section 6

## K.1 Proof of Lemma 6.1

We construct $\eta_{\text{NDSC}}$ in the same way as in Theorem 5.2, but we do not track the constants and dependence on $d$. We first determine $r_{\text{NDSC}}$ such that $\eta_{\text{NDSC}}$ can satisfy (NDSC). Then we prove that $K_{\text{norm}}$ verifies the LPC (Definition 5.2) with $r = r_{\text{NDSC}}$ and a sufficiently large minimal separation, showing the existence of the non-degenerate certificate $\eta_{\text{NDSC}}$.

Choice of $r_{\text{NDSC}}$: Given $r_{\text{NDSC}} > 0$, according to Lemma 1.2 it suffices that

$$ 1 - \frac{B_{02} + \bar{\varepsilon}_2(r_{\text{NDSC}})/16}{2} \mathfrak{D}_g(x, x_j^0)^2 > -1 \quad \forall x \in \mathcal{X}_j^{\text{near}}(r_{\text{NDSC}}) \quad (83) $$

to have $\eta_{\text{NDSC}} > -1$ (the existence of this certificate is proven later). Lemma J.2 gives $B_{02} = \sqrt{4d^2 + 10d}$. To bound $\mathfrak{D}_g(x, x_j^0)^2$, we use Lemma H.7. If $d(x, x_j^0) < \sqrt{2}$, we have $\mathfrak{D}_g(x, x_j^0)^2 \le dF(d(x, x_j^0))$ where $F$ is defined by (62) in the appendix. We remark that $F$ is continuous on $[0, \sqrt{2}]$ and that $F(0) = 0$. Furthermore, if $r \le \frac{0.32}{\sqrt{d}}$ we can take $\bar{\varepsilon}_2(r) = e^{-\frac{0.32^2}{2}}|G(0.32)|$ (see Lemma J.4). This quantity does not depend on $r$. So there exists $0 < r_{\text{NDSC}} \le \frac{0.32}{\sqrt{d}}$ (that depends only on $d$) such that (83) is satisfied.

LPC and existence of $\eta_{\text{NDSC}}$: It remains to show the existence of $\eta_{\text{NDSC}}$, a global non-degenerate certificate of the form (18) satisfying Definition 3.2 where the near regions are of radius $r_{\text{NDSC}}$. We first prove the LPC with this choice of radius, choosing $\Delta$ large enough, and then make use of Theorem 5.2.

The curvature constants $\bar{\varepsilon}_0(r_{\text{NDSC}})$ and $\bar{\varepsilon}_2(r_{\text{NDSC}})$ are given by Lemma J.4. The separation $\Delta(s)$ should satisfy

$$ (s-1)\sqrt{2}153.05e^{-\frac{\Delta(s)^2}{4}} \le \frac{1}{64} \min \left( \frac{\bar{\varepsilon}_0(r_{\text{NDSC}})}{B_0}, \frac{\bar{\varepsilon}_2(r_{\text{NDSC}})}{B_2} \right). $$

Such a $\Delta(s)$ exists as $e^{-\frac{\Delta(s)^2}{4}} \xrightarrow{\Delta(s) \to +\infty} 0$. It depends only on $d$ (through $\bar{\varepsilon}_0(r_{\text{NDSC}})$ and $\bar{\varepsilon}_2(r_{\text{NDSC}})$) and $s$. Then $K_{\text{norm}}$ satisfies the LPC (Definition 5.2) with these parameters.

Finally, Theorem 5.2 applies. The minimal separation $\Delta_{\text{NDSC}}$ follows from Lemma 5.3. It depends on $d$, $u_{\text{min}}$, $u_{\text{max}}$, $\tau$, $r_{\text{NDSC}}$, $\Delta(s)$. Under this separation, we can construct a non-degenerate certificate $\eta_{\text{NDSC}}$ verifying $\eta_{\text{NDSC}} > -1$. We can also construct local non-degenerate certificates $\eta_{j,\text{NDSC}}$ for $j = 1, \dots, s$. Negative definiteness of $\nabla^2\eta_{\text{NDSC}}(x_j^0)$: Let $j \in \{1, \dots, s\}$. To show that $\nabla^2\eta_{\text{NDSC}}(x_j^0)$ is negative-definite, first remark that $\nabla^2\eta_{\text{NDSC}}(x_j^0) = D_2[\eta_{\text{NDSC}}](x_j^0)$ because $\nabla\eta_{\text{NDSC}}(x_j^0) = 0$ (equality of the Riemannian and Euclidean hessians, see Definition 5.1). Then, for $v \in \mathbb{R}^2$ such that $\|v\|_{x_j^0} = 1$, using the control on the near regions (proof of Theorem 5.2) we deduce that

$$ D_2[\eta_{\text{NDSC}}](x_j^0)[v, v] \le K_{\text{norm}}^{(02)}(x_j^0, x_j^0)[v, v] + \left\| D_2[\eta_{\text{NDSC}}](x_j^0) - K_{\text{norm}}^{(02)}(x_j^0, x_j^0) \right\|_{x_j^0} \le -\bar{\varepsilon}_2 + \frac{\bar{\varepsilon}_2}{16} = -\frac{\bar{\varepsilon}_2}{15}. $$

So for all $v \in \mathbb{R}^{2d} \setminus \{0\}$,

$$ v^T \nabla^2 \eta_{\text{NDSC}}(x_j^0) v \le - \|v\|_{x_j^0} \frac{\bar{\varepsilon}_2}{15} < 0. $$
## K.2 Proof of Theorem 6.1

**Lemma K.1.** Under the hypothesis of Lemma 6.1, $G_{S_0}$ is full-rank, where

$$G_{S_0} : (\alpha_1, \dots, \alpha_s), (\beta_1, \dots, \beta_s) \in \mathbb{R}^s \times \mathbb{R}^{2d \times s} \mapsto \sum_{j=1}^{s} \alpha_j \Psi \delta_{x_j^0} + \sum_{j=1}^{s} \beta_j^T \nabla_x (\Psi \delta_{x_j^0}). \quad (84)$$

Moreover, $B := \max_{i=0,1,2} \sup_{x \in \mathbb{R}^d \times [u_{\min}, +\infty)^d} \|D_i \Psi(\delta_x)\|_x < \infty$ ($D_i$ and the associated norm defined in Definition 5.1).

**Proof.** $G_{S_0}$ is full-rank: We used the invertibility of $\Upsilon$ (see (63)) to construct $\eta_{\text{NDSC}}$: we already know that

$$\sum_{j=1}^{s} \alpha_j K_{\text{norm}}(x_j^0, \cdot) + \sum_{j=1}^{s} \beta_j^T \nabla_1 K_{\text{norm}}(x_j^0, \cdot) = 0 \quad \Rightarrow \quad \alpha_j = 0, \beta_j = 0_{\mathbb{R}^2} \quad \forall j = 1, \dots, s.$$

**$B < \infty$:** Remark that $\sup_{x \in \mathbb{R}^d \times [u_{\min}, +\infty)^d} \|D_i \Psi(\delta_x)\|_x \le \sup_{x \in \mathbb{R}^d \times [u_{\min}, +\infty)^d} \sqrt{\|K_{\text{norm}}^{(ii)}(x, x)\|_{x,x}}$. To deal with $i = 0, 1$, remark that $\sqrt{\|K_{\text{norm}}^{(ii)}(x, x)\|_{x,x}} \le \sqrt{B_{ii}}$ with $B_{00}, B_{11}$ defined in Lemma J.2.

We follow the same approach as in the proof of Lemma J.2 to obtain an upper bound $B_{22}$ on $\|K_{\text{norm}}^{(22)}(x, x')\|_{x,x'}$ with $(\mathfrak{g}_x^{-1/2} H^\mathfrak{g} \Psi \delta_x \mathfrak{g}_x^{-1/2})_{k,l}$ representing the coordinate at position $(k, l)$, we have

$$
\begin{aligned}
\| K_{\text{norm}}^{(22)}(x, x') \|_{x, x'} &= \sup_{\substack{\|\tilde{V}\|_2 \le 1, \|\tilde{Q}\|_2 \le 1 \\ \|\tilde{V}'\|_2 \le 1, \|\tilde{Q}'\|_2 \le 1}} \left\langle \tilde{V}^T \mathfrak{g}_x^{-1/2} H^\mathfrak{g} \Psi \delta_x \mathfrak{g}_x^{-1/2} \tilde{Q}, \tilde{V}'^T \mathfrak{g}_{x'}^{-1/2} H^\mathfrak{g} \Psi \delta_{x'} \mathfrak{g}_{x'}^{-1/2} \tilde{Q}' \right\rangle_{\mathbb{L}}, \\
&= \sup_{\substack{\|\tilde{V}\|_2 \le 1, \|\tilde{Q}\|_2 \le 1 \\ \|\tilde{V}'\|_2 \le 1, \|\tilde{Q}'\|_2 \le 1}} \sum_{k,l,k',l'} V_k V_{k'} Q_l Q_{l'} \left\langle (\mathfrak{g}_x^{-1/2} H^\mathfrak{g} \Psi \delta_x \mathfrak{g}_x^{-1/2})_{k,l}, (\mathfrak{g}_{x'}^{-1/2} H^\mathfrak{g} \Psi \delta_{x'} \mathfrak{g}_{x'}^{-1/2})_{k',l'} \right\rangle_{\mathbb{L}}, \\
&\le \sqrt{\sum_{k,l,k',l'} \left\langle (\mathfrak{g}_x^{-1/2} H^\mathfrak{g} \Psi \delta_x \mathfrak{g}_x^{-1/2})_{k,l}, (\mathfrak{g}_{x'}^{-1/2} H^\mathfrak{g} \Psi \delta_{x'} \mathfrak{g}_{x'}^{-1/2})_{k',l'} \right\rangle_{\mathbb{L}}^2}, \\
&\le 4d^2 \sup_{k,l,k',l',x \in \mathbb{R}^d \times [u_{\min}, +\infty)^d} \left\| \left\langle (\mathfrak{g}_x^{-1/2} H^\mathfrak{g} \Psi \delta_x \mathfrak{g}_x^{-1/2})_{k,l}, (\mathfrak{g}_{x'}^{-1/2} H^\mathfrak{g} \Psi \delta_{x'} \mathfrak{g}_{x'}^{-1/2})_{k',l'} \right\rangle_{\mathbb{L}} \right\|, \\
&\le 4d^2 \sup_{k,l} \left\| (\mathfrak{g}_x^{-1/2} H^\mathfrak{g} \Psi \delta_x \mathfrak{g}_x^{-1/2})_{k,l} \right\|_{\mathbb{L}}^2, \\
&\le 4d^2 \sup_{k,l} \mathfrak{g}_{b_k b_l}^{-1} \mathfrak{g}_{b_k b_l}^{-1} \|(H^\mathfrak{g} \Psi \delta_x)_{b_k b_l} \|_{\mathbb{L}}^2.
\end{aligned}
$$

The terms $\mathfrak{g}_{b_k b_k}^{-1} \mathfrak{g}_{b_l b_l}^{-1} \|(H^\mathfrak{g} \Psi \delta_x)_{b_k b_l} \|_{\mathbb{L}}^2$ take one of the following forms ($b$ represents $t$ or $u$, while $k, l$ refer to the dimension in the following):

$$
\begin{aligned}
& \mathfrak{g}_{b_k b_k}^{-1} \mathfrak{g}_{b_l b_l}^{-1} \| \partial_{b_k b_l} \Psi \delta_x \|_{\mathbb{L}}^2 \quad \text{with } k \neq l, \\
& \mathfrak{g}_{t_k t_k}^{-2} \| \partial_{t_k t_k} \Psi \delta_x - \Gamma^{t_k} t_k u_k \partial_{t_k} \Psi \delta_x \|_{\mathbb{L}}^2, \\
& \mathfrak{g}_{t_k t_k}^{-1} \mathfrak{g}_{u_k u_k}^{-1} \| \partial_{t_k u_k} \Psi \delta_x - \Gamma^{t_k} t_k u_k \partial_{t_k} \Psi \delta_x \|_{\mathbb{L}}^2, \\
& \mathfrak{g}_{u_k u_k}^{-2} \| \partial_{u_k u_k} \Psi \delta_x - \Gamma^{u_k} u_k u_k \partial_{t_k} \Psi \delta_x \|_{\mathbb{L}}^2.
\end{aligned}
$$

From the proof of Lemma J.2, we know that these terms are respectively bounded by 1, 1, 3, 7. Hence $B_{22} \le 28d^2$, and $B \le 7\sqrt{2d}$. $\square$

Below we present the main steps of the proof of Theorem 6.1, adapted from [Duval and Peyré, 2015, Lemma 2, Proposition 7] and [Poon, 2019, Section 4.3]. We work under the assumptions of Lemma 6.1: there exists $\eta_{\text{NDSC}}$ of the form (18) that verifies $|\eta_{\text{NDSC}}| \le 1$, $|\eta_{\text{NDSC}}(x)| = 1 \iff x \in \{x_j^0\}_j$ and $\nabla^2 \eta_{\text{NDSC}}(x_j^0) < 0$ for all $j = 1, \dots, s$ (Non Degenerate Source Condition, or NDSC).
**Dual, noisy and noiseless problems** We denote $\mathcal{D}_{\kappa,b}$ the dual problem with regularization $\kappa$ and noise $b$.
More precisely:

$$ \arg\max_{p \in \mathcal{L}, \|\Psi^* p\|_{\infty} \le 1} \langle \Psi \mu_{\omega}^0, p \rangle_{\mathcal{L}} \quad (\mathcal{D}_{0,0}) $$

is the dual problem of

$$ \arg\min_{\mu \in \mathcal{M}(\mathcal{X})} \|\mu\|_{\text{TV}} \text{ such that } \Psi\mu = \Psi\mu_{\omega}^{0}. \quad (\mathcal{P}_{0,0}) $$

For $\kappa > 0$, the dual problem is

$$ \arg\min_{p \in \mathcal{L}, \|\Psi^* p\|_{\infty} \le 1} \left\| \frac{y}{\kappa} - p \right\|_{\mathcal{L}}^2, \quad (\mathcal{D}_{\kappa,b}) $$

where we observe $y = \Psi\mu_{\omega}^0 + b$ (i.e. $b = \Gamma_n$). Its solution is unique. It is the dual problem of

$$ \arg\min_{\mu \in \mathcal{M}(\mathcal{X})} \frac{1}{2} \|y - \Psi\mu\|_{\mathcal{L}}^2 + \kappa \|\mu\|_{\text{TV}}. \quad (\mathcal{P}_{\kappa,b}) $$

We use the notation $p_{\kappa,b}$ for the solution of the dual problem, and $\eta_{\kappa,b} = \Psi^* p_{\kappa,b}$. For $\kappa = 0$, we choose the particular dual solution defined by $(p_{0,0})$.

The dual and primal solutions are connected via the subdifferential of the TV-norm: we recall that for $\eta \in \mathcal{C}(\mathcal{X})$ and $\mu \in \mathcal{M}(\mathcal{X})$,

$$ \eta \in \partial \|\mu\|_{\text{TV}} \iff (\|\eta\|_{\infty} \le 1, \supp(\mu^{-}) \subset \{\eta = -1\}, \supp(\mu^{+}) \subset \{\eta = 1\}). $$

Strong duality ensures that for $\mu_{\kappa,b}$ a solution of the primal problem, $\eta_{\kappa,b} \in \partial \|\mu_{\kappa,b}\|_{\text{TV}}$. Moreover, for $\kappa > 0$ we have $p_{\kappa,b} = -\frac{1}{\kappa}(\Psi\mu_{\kappa,b} - \Psi\mu_{\omega}^0 - b)$.

**Lemma K.2.** Under the assumptions of Lemma 6.1, $\mu_{\omega}^0$ is the unique solution of $(\mathcal{P}_{0,0})$. Moreover, $\eta_{0,0} := \eta_{\text{NDSC}}|_{\mathcal{X}} = \Psi^* p_{0,0}$ where $p_{0,0}$ is the solution of $(\mathcal{D}_{0,0})$ with minimal norm, i.e.

$$ p_{0,0} = \arg\min_{p \in \mathcal{L}} \{\|p\|_{\mathcal{L}} : \Psi^* p \in \partial \|\mu_{\omega}^0\|_{\text{TV}}\}. $$

*Proof.* As $\eta_{\text{NDSC}}|_{\mathcal{X}} \in \partial \|\mu_{\omega}^0\|_{\text{TV}}$, we deduce that $p_{0,0}$ is a solution of $(\mathcal{D}_{0,0})$, linked to any solution $\mu_{0,0}$ of $(\mathcal{P}_{0,0})$ by $\Psi^* p_{0,0} = \eta_{\text{NDSC}}|_{\mathcal{X}} \in \partial \|\mu_{0,0}\|_{\text{TV}}$. So $\supp(\mu_{0,0}) \subset \{|\eta_{\text{NDSC}}|_{\mathcal{X}}| = 1\} = \{x_j^0\}_{j=1}^s$. Using the injectivity of $G_{S_0}$ (Lemma K.1), we have $\mu_{0,0} = \mu_{\omega}^0$.

Moreover, as $\eta_{\text{NDSC}}$ (which is of the form (18)) verifies the NDSC, [Duval and Peyré, 2015, Proposition 7] shows that $p_{0,0}$ is of minimal norm. $\square$

**Lemma K.3** (Convergence of the dual solutions). It holds that $\|p_{\kappa,0} - p_{0,0}\|_{\mathcal{L}} \xrightarrow{\kappa \to 0} 0$.

$$ \text{Moreover, } \|p_{\kappa,b} - p_{\kappa,0}\|_{\mathcal{L}} \le \frac{\|b\|_{\mathcal{L}}}{\kappa}. $$

*Proof.* $\|p_{\kappa,0} - p_{0,0}\|_{\mathcal{L}} \xrightarrow{\kappa \to 0} 0$: Using that $p_{\kappa,0}$ is the solution of $(\mathcal{D}_{\kappa,b})$ for $b = 0$ and $p_{0,0}$ is a solution of $(\mathcal{D}_{0,0})$, writing $y = \Psi\mu_{\omega}^0$ we have

$$ \langle y, p_{\kappa,0} \rangle_{\mathcal{L}} - \frac{\kappa}{2} \|p_{\kappa,0}\|^2 \ge \langle y, p_{0,0} \rangle_{\mathcal{L}} - \frac{\kappa}{2} \|p_{0,0}\|^2 \quad \text{and} \quad \langle y, p_{0,0} \rangle_{\mathcal{L}} \ge \langle y, p_{\kappa,0} \rangle_{\mathcal{L}}. $$

We deduce that $\|p_{\kappa,0}\|_{\mathcal{L}}$ is bounded by $\|p_{0,0}\|_{\mathcal{L}}$. Given $\kappa_n \to 0$, as the closed unit ball of a Hilbert space is weakly sequentially compact, we can extract $p_{\kappa_{n_k},0}$ that converges weakly towards $p^* \in \mathcal{L}$.

We show that $p^* = p_{0,0}$. We have $\langle y, p^* \rangle_{\mathcal{L}} \ge \langle y, p_{0,0} \rangle_{\mathcal{L}}$ so

$$ \langle \mu_{\omega}^{0}, \Psi^{*} p^{*} \rangle \ge \langle \mu_{\omega}^{0}, \Psi^{*} p_{0,0} \rangle = \|\mu_{\omega}^{0}\|_{\text{TV}}. $$

We also have that $\Psi^*$ is weakly continuous from $\mathcal{L}$ to $\mathcal{C}(\mathcal{X})$ so

$$ \| \Psi^* p^* \|_{\infty} \le \| \Psi^* p_{\kappa_{n_k},0} \|_{\infty} = 1, $$

hence $p^* \in \partial \|\mu_{\omega}^0\|_{\text{TV}}$. Finally,

$$ \|p_{0,0}\|_{\mathcal{L}} \ge \liminf \|p_{\kappa_{n_k},0}\|_{\mathcal{L}} \ge \|p^*\|_{\mathcal{L}} $$

as if $h_n \to h$ in $\mathcal{L}$, then $\|h\|_{\mathcal{L}}^2 \le \liminf \|h_n\|_{\mathcal{L}}^2$. Hence $p^* = p_{0,0}$.

Then $p_{\kappa_{n_k},0} \to p_{0,0}$ strongly because $\|p_{\kappa_{n_k},0}\|_{\mathcal{L}} \to \|p_{0,0}\|_{\mathcal{L}}$ and $p_{\kappa_{n_k},0} \to p_{0,0}$. Finally, $p_{\kappa,0} \xrightarrow{n_k \to \infty} p_{0,0}$ strongly because any convergent subsequence of $(p_{\kappa,0})_\kappa$ has limit $p_{0,0}$.
$\|p_{\kappa,b} - p_{\kappa,0}\|_{\mathbb{L}} \le \frac{\|b\|_{\mathbb{L}}}{\kappa}$: The mapping $P: \frac{y}{\kappa} \to p_{\kappa,b}$ is the projection onto a close convex set so it is non expansive: for $h_1, h_2 \in \mathbb{L}$,

$$ \langle h_1 - P(h_1), P(h_2) - P(h_1) \rangle \le 0 \quad \text{and} \quad \langle h_2 - P(h_2), P(h_1) - P(h_2) \rangle \le 0. $$

Subtracting the two and applying Cauchy-Schwarz leads to

$$ \langle P(h_1) - P(h_2), P(h_1) - P(h_2) \rangle \le \langle h_1 - h_2, P(h_1) - P(h_2) \rangle \le \|h_1 - h_2\| \|P(h_1) - P(h_2)\|. $$

Hence

$$ \|p_{\kappa,b} - p_{\kappa,0}\| \le \left\| \frac{\Psi \mu_{\omega}^0 + b - \Psi \mu_{\omega}^0}{\kappa} \right\| = \frac{\|b\|}{\kappa}. $$

□

**Proposition K.1** (Sparsity of the solution). Under the assumptions of Lemma 6.1, there exists $\tilde{\kappa}_0 > 0$, $\gamma_0$ such that for all $\kappa \le \tilde{\kappa}_0$ and $b$ such that $\|b\|_{\mathbb{L}} \le \gamma_0 \kappa$, the solution $\mu_{\kappa,b}$ of $(P_{\kappa,b})$ is a discrete measure and has at most 1 particle in each $X_j^{\text{near}}(r_{\text{NDSC}})$.

*Proof.* We recall the definitions of the covariant derivatives and the associated norms in Definition 5.1.

Using NDSC, the fact that $\mathcal{X}$ is compact and that $\text{diam}(X_j^{\text{near}}(r_\epsilon)) \underset{r_\epsilon \to 0}{\to} 0$, we deduce that there exists $\epsilon > 0$, $r_\epsilon \in (0, r_{\text{NDSC}}]$ such that $|\eta_{0,0}(x)| < 1 - \epsilon$ for all $x \notin \bigcup_j X_j^{\text{near}}(r_\epsilon)$ and $H^g \eta_{0,0}(x) < -\epsilon I$ for all $x \in \bigcup_j \mathcal{G}_{x_j^0}(X_j^{\text{near}}(r_\epsilon))$ (because $H^g \eta_{0,0}(x_j^0) = \nabla^2 \eta_{0,0}(x_j^0)$ is negative-definite).

In light of Lemma K.3, we can choose $\tilde{\kappa}_0$ such that for all $\kappa \le \tilde{\kappa}_0$, $B \|p_{\kappa,0} - p_{0,0}\|_{\mathbb{L}} \le \frac{\epsilon}{4}$ (where $B$ is defined in Lemma K.1). We assume that $\|b\|_{\mathbb{L}} \le \gamma_0 \kappa$ with $\gamma_0 = \frac{\epsilon}{4B}$. Using Cauchy-Schwarz we get that for all $x \in \mathbb{R}^d \times [u_{\min}, +\infty)^d$, $i = 0, 1, 2$,

$$ \begin{aligned} \|D_i \eta_{\kappa,b}(x) - D_i \eta_{0,0}(x)\|_{x} &\le \|p_{\kappa,b} - p_{0,0}\|_{\mathbb{L}} \|D_i \Psi \delta_x\|_{x}, \\ &\le B \left( \frac{\|b\|_{\mathbb{L}}}{\kappa} + \|p_{\kappa,0} - p_{0,0}\|_{\mathbb{L}} \right), \\ &\le \frac{\epsilon}{2}. \end{aligned} $$

So $|\eta_{\kappa,b}(x)| < 1$ for all $x \notin \bigcup_j X_j^{\text{near}}(r_\epsilon)$ and $H^g \eta_{\kappa,b}(x)$ is negative-definite for all $x \in \bigcup_j \mathcal{G}_{x_j^0}(X_j^{\text{near}}(r_\epsilon))$. Hence there exists at most 1 point $\bar{x}$ in each $X_j^{\text{near}}(r_\epsilon)$ such that $\eta_{\kappa,b}(\bar{x}) = 1$. In fact for such a point, $\nabla \eta_{\kappa,b}(\bar{x}) = 0$ because $|\eta_{\kappa,b}| \le 1$. So for $x \in X_j^{\text{near}}(r_\epsilon)$, denoting $\gamma$ the geodesic connecting $\bar{x}$ to $x$, we have

$$ \eta_{\kappa,b}(x) = 1 + \int_{0}^{1} (1-y)\dot{\gamma}(y)^{T} H^{g} \eta_{\kappa,b}(x) \dot{\gamma}(y) dy < 1. $$

As $\eta_{\kappa,b} \in \partial \|\mu_{\kappa,b}\|_{\text{TV}}$, we deduce that $\mu_{\kappa,b}$ is a discrete measure and that it has at most 1 particle in each $X_j^{\text{near}}(r_\epsilon)$. □

**Lemma K.4.** Under the assumptions of Lemma 6.1, there exists $\kappa_0 > 0$, $\gamma_0$ such that for all $\kappa \le \kappa_0$ and $b$ such that $\|b\|_{\mathbb{L}} \le \gamma_0 \kappa$, $\mu_{\kappa,b}$ is a discrete measure and has exactly 1 particle in each $X_j^{\text{near}}(r_{\text{NDSC}})$.

*Proof.* We take Proposition K.1 as a starting point. We show that $\mu_{\kappa,b}(X_j^{\text{near}}(r_\epsilon)) > 0$ for $\|b\|_{\mathbb{L}} \le \gamma_0 \kappa$ and $\kappa \le \kappa_0$ with $\kappa_0 \le \tilde{\kappa}_0$ small enough.

It suffices to show the weak* convergence of $\mu_{\kappa,b}$ towards $\mu_\omega^0$ as $\kappa \to 0$ and for $\|b\|_{\mathbb{L}} \le \gamma_0 \kappa$ (this result is also stated in [Duval and Peyré, 2015, Proposition 4]). We have

$$ \frac{1}{2} \| \Psi \mu_{\omega}^{0} + b - \Psi \mu_{\kappa,b} \|_{\mathbb{L}}^{2} + \kappa \| \mu_{\kappa,b} \|_{\text{TV}} \le \frac{1}{2} \| b \|_{\mathbb{L}}^{2} + \kappa \| \mu_{\omega}^{0} \|_{\text{TV}} $$

since $\mu_{\kappa,b}$ minimizes

$$ J_{\kappa,b}: \mu \mapsto \frac{1}{2} \| \Psi \mu_{\omega}^{0} + b - \Psi \mu \|_{\mathbb{L}}^{2} + \kappa \| \mu \|_{\text{TV}}. $$

Dividing by $\kappa$, it follows that $\|\mu_{\kappa,b}\|_{\text{TV}}$ is bounded by $\frac{\gamma_0^2}{2} \kappa + \|\mu_\omega^0\|_{\text{TV}}$. So we can extract $\mu_k$ a weak* convergent subsequence of $\mu_{\kappa,b}$ as $\kappa, b \to 0$ with $\|b\|_{\mathbb{L}} \le \gamma_0 \kappa$. Let $\mu^*$ be its limit. By the lower semi-continuity of the TV norm for the weak* convergence, $\|\mu^*\|_{\text{TV}} \le \liminf \|\mu_k\|_{\text{TV}} \le \|\mu_\omega^0\|_{\text{TV}}$. To establish $\mu^* = \mu_\omega^0$, it remains to show that $\Psi \mu^* = \Psi \mu_\omega^0$ (because $\mu_\omega^0$ is the only solution of $(P_{0,0})$, see Lemma K.2). Recall that $\|p_{\kappa,b} - p_{\kappa,0}\|_{\mathbb{L}} \le \frac{\|b\|_{\mathbb{L}}}{\kappa}$
and that $p_{\kappa,b} = -\frac{1}{\kappa}(\Psi\mu_{\kappa,b} - \Psi\mu_{\omega}^0 - b)$, giving $\|\Psi\mu_{\kappa,b} - b - \Psi\hat{\mu}_{\kappa,0}\|_{\mathbb{L}} \le \|b\|$. As $\|\Psi\mu_{\kappa,0} - \Psi\mu_{\omega}^0\|_{\mathbb{L}}^2 \le 2\kappa \|\mu_{\omega}^0\|_{\text{TV}}$ (because $J_{\kappa,0}(\mu_{\kappa,0}) \le J_{\kappa,0}(\mu_{\omega}^0)$), it comes

$$ \begin{aligned} \|\Psi\mu_{\kappa,b} - \Psi\mu_{\omega}^0\|_{\mathbb{L}} &\le \|\Psi\mu_{\kappa,b} - \Psi\hat{\mu}_{\kappa,0}\|_{\mathbb{L}} + \|\Psi\mu_{\kappa,0} - \Psi\mu_{\omega}^0\|_{\mathbb{L}}, \\ &\le 2\|b\|_{\mathbb{L}} + \sqrt{2\kappa \|\mu_{\omega}^0\|_{\text{TV}}}. \end{aligned} $$

As $\Psi$ is weak* to weak continuous from $M(\mathcal{X})$ to $\mathbb{L}$, for all $p \in \mathbb{L}$ we have

$$ |\langle p, \Psi(\mu_k - \mu_{\omega}^0) \rangle_{\mathbb{L}}| \to |\langle p, \Psi(\mu^* - \mu_{\omega}^0) \rangle_{\mathbb{L}}| $$

Applying Cauchy-Schwarz, we have $|\langle p, \Psi(\mu_k - \mu_{\omega}^0) \rangle_{\mathbb{L}}| \to 0$ from which we conclude that $\langle p, \Psi(\mu^* - \mu_{\omega}^0) \rangle_{\mathbb{L}} = 0$. So $\Psi\mu^* = \Psi\mu_{\omega}^0$, and $\mu_{\kappa,b}$ converges towards $\mu_{\omega}^0$ as $\kappa \to 0$ and $\|b\|_{\mathbb{L}} \le \gamma_0\kappa$ for the weak* topology. This concludes the proof. $\square$

## K.3 Proof of Corollary 6.1

If $n \ge \frac{c_{\kappa}^2}{\kappa_0^2(2\pi)^{d/2}\tau^{d/2}}$, choosing $\kappa = \frac{c_{\kappa}}{(2\pi)^{d/4}\tau^{d/2}\sqrt{n}}$ we have $\kappa \le \kappa_0$. Moreover, it holds that $\|\Gamma_n\| \le \gamma_0\kappa$ with probability greater than $1 - C_{\Gamma}e^{-\left(\frac{\gamma_0 c_{\kappa}}{C_{\Gamma}}\right)^2}$ (c.f. Lemma 3.1 where we took $\rho = \frac{\gamma_0^2 c_{\kappa}^2}{C_{\Gamma}^2}$). Theorem 6.1 applies: $\mu_{n,\omega}^* = \sum_{j=1}^s \omega_j^* \delta_{x_j^*}$ where $\omega_j^* > 0$ and $x_j^* \in \mathcal{X}_j^{\text{near}}(r_{\text{NDSC}})$ for all $j = 1, \dots, s$.

From (28), using $\|\Gamma_n\|_{\mathbb{L}} \le \gamma_0\kappa$ and $\|p_{\text{NDSC}}\|_{\mathbb{L}} \le \sqrt{2s}$ (see Lemma 6.1), we have

$$ \begin{aligned} D_{\eta_{\text{NDSC}}}(\mu_{n,\omega}^*, \mu_{\omega}^0) &\le \frac{\kappa}{2}(\gamma_0 + \|p_{\text{NDSC}}\|_{\mathbb{L}})^2, \\ &\le \frac{c_{\kappa}}{2(2\pi)^{d/4}\tau^{d/2}\sqrt{n}}(\gamma_0 + \sqrt{2s})^2. \end{aligned} \quad (85) $$

Bound for $|\omega_j^0 - \omega_j^*|$: Recalling (32), as $\mu_{n,\omega}^*(\mathcal{X}^{\text{far}}(r_{\text{NDSC}})) = 0$ we have

$$ \begin{aligned} |\omega_j^0 - \omega_j^*| &\le \|p_{j,\text{NDSC}}\|_{\mathbb{L}} (2 \|\Gamma_n\|_{\mathbb{L}} + 2\kappa \|p_{\text{NDSC}}\|_{\mathbb{L}}) + \frac{\tilde{\varepsilon}_{2,\text{NDSC}}}{\varepsilon_{2,\text{NDSC}}} D_{\eta_{\text{NDSC}}} (\mu_{n,\omega}^*, \mu_{\omega}^0), \\ &\le 2\kappa \|p_{j,\text{NDSC}}\|_{\mathbb{L}} (\gamma_0 + \|p_{\text{NDSC}}\|_{\mathbb{L}}) + \frac{\tilde{\varepsilon}_{2,\text{NDSC}}}{\varepsilon_{2,\text{NDSC}}} D_{\eta_{\text{NDSC}}} (\mu_{n,\omega}^*, \mu_{\omega}^0), \\ &\le \frac{c_{\kappa}}{(2\pi)^{d/4}\tau^{d/2}\sqrt{n}} \left( 2\sqrt{2}(\gamma_0 + \sqrt{2s}) + \frac{1}{2}(\gamma_0 + \sqrt{2s})^2 \frac{\tilde{\varepsilon}_{2,\text{NDSC}}}{\varepsilon_{2,\text{NDSC}}} \right). \end{aligned} $$

The second factor of this product only depends on $\mathcal{X}$, $\tau$ and $\mu^0$.

Control of $d(x_j^*, x_j^0)$: Using (85) together with (31), we get

$$ \sum_{j=1}^{s} \omega_j^* \delta_{\mathfrak{g}}(x_j^*, x_j^0)^2 \le \frac{1}{\varepsilon_{2,\text{NDSC}}} \frac{c_{\kappa}}{2(2\pi)^{d/4}\tau^{d/2}\sqrt{n}} (\gamma_0 + \sqrt{2s})^2. $$

To control the proximity between $x_j^0$ and $x_j^*$ with the semi-distance, we use Lemma H.5. It comes that

$$ \sum_{j=1}^{s} \omega_j^* d(x_j^*, x_j^0)^2 \le \frac{\tilde{\varepsilon}_{3,\text{NDSC}}}{\varepsilon_{2,\text{NDSC}}} \frac{c_{\kappa}}{2(2\pi)^{d/4}\tau^{d/2}\sqrt{n}} (\gamma_0 + \sqrt{2s})^2 $$

where $\tilde{\varepsilon}_{3,\text{NDSC}}$ only depends on $r_{\text{NDSC}}$. As $\omega_j^* \ge \omega_j^0 - |\omega_j^0 - \omega_j^*|$, if $|\omega_j^0 - \omega_j^*| < \omega_j^0$ then

$$ \begin{aligned} d(x_j^*, x_j^0)^2 &\le \frac{1}{\omega_j^0 - |\omega_j^0 - \omega_j^*|} \omega_j^* d(x_j^*, x_j^0)^2, \\ &\le \frac{1}{\omega_j^0 - |\omega_j^0 - \omega_j^*|} \frac{\tilde{\varepsilon}_{3,\text{NDSC}}}{\varepsilon_{2,\text{NDSC}}} \frac{c_{\kappa}}{2(2\pi)^{d/4}\tau^{d/2}\sqrt{n}} (\gamma_0 + \sqrt{2s})^2. \end{aligned} $$

Recalling (23) and the condition $\frac{n}{c_{\kappa}^2} \ge \frac{1}{\kappa_0^2(2\pi)^{d/2}\tau^{d/2}}$, if $c_{\kappa}$ is chosen as $c_{\kappa,n} = \frac{o}{n \to +\infty}(\sqrt{n})$, there exists $n_0 \in \mathbb{N}$ depending on $(c_{\kappa,n})_n, \mu^0, \mathcal{X}, \tau$ and $\tilde{\varepsilon}_0 > 0$ that depends on $\mu^0, \mathcal{X}, \tau$ (through $\gamma_0, \omega_j^0, s$ etc.) such that for all $n \ge n_0$, $|\omega_j^0 - \omega_j^*| < \omega_j^0$ and $\frac{\tilde{\varepsilon}_{3,\text{NDSC}}}{\varepsilon_{2,\text{NDSC}}} \frac{(\gamma_0 + \sqrt{2s})^2}{2(2\pi)^{d/4}\tau^{d/2}(\omega_j^0 - |\omega_j^0 - \omega_j^*|)} \le \tilde{\varepsilon}_0$ with probability at least $1 - C_{\Gamma}e^{-\left(\frac{\gamma_0 c_{\kappa,n}}{C_{\Gamma}}\right)^2}$. Then $d(x_j^*, x_j^0)^2 \le \tilde{\varepsilon}_0 \frac{c_{\kappa,n}}{\sqrt{n}}$.