Appendix A. The interpolator A

The operator A discussed in this appendix is identical to the one introduced in Appendix A of [Reference Braverman5]. We repeat some of its key properties, originally presented in [Reference Braverman5], as they are needed for the proof of Theorem 1.1. We also present several useful technical results about A that are not found in [Reference Braverman5]. Namely, Lemma A.1, Lemma A.2, Proposition A.1, and Corollary A.1.

Building on the discussion in Section 2.2.2, for a one-dimensional function $f\,:\,\delta \mathbb{Z} \to \mathbb{R}$ , we define

\begin{align*}A f(x) = \sum_{i=0}^{4} \alpha^{k(x)}_{k(x)+i}(x)f(\delta(k(x)+i)),\end{align*}

where $k(x) = \lfloor x/\delta\rfloor$ and $\alpha_{k+i}^{k}\,:\,\mathbb{R}\to \mathbb{R}$ are weights. Braverman [Reference Braverman5] described how to choose these weights to make Af(x) coincide with $f(\!\cdot\!)$ on grid points, and also to make it a differentiable function whose derivatives behave like the corresponding finite differences of $f(\!\cdot\!)$ . The idea can be applied to multidimensional grid-valued functions as well. The following result builds on Theorem A.1 of [Reference Braverman5]. We use this as an interface that contains the important properties of A without delving into the low-level details behind its construction.

Theorem A.1. Given a convex set $K \subset \mathbb{R}^{d}$ , define

\begin{align*}K_{4} = \{{\boldsymbol{x}} \in K \cap \delta \mathbb{Z}^{d} \,:\,\delta(k({\boldsymbol{x}})+ {\boldsymbol{i}}) \in K \cap \delta \mathbb{Z}^{d} \text{ for all } 0 \leq {\boldsymbol{i}} \leq 4{\boldsymbol{e}}\},\end{align*}

where $k({\boldsymbol{x}})$ by $k_{j}({\boldsymbol{x}}) = \lfloor x_j/\delta\rfloor$ . Let $\text{Conv}(K_4)$ be the convex hull of $K_4$ and for ${\boldsymbol{x}} \in \mathbb{R}^{d}$ . There exist weights $\{\alpha_{k+i}^{k} \,:\,\mathbb{R} \to \mathbb{R},\ k \in \mathbb{Z},\ i =0,1,2,3,4\}$ such that, for any $f\,:\,K \cap \delta \mathbb{Z}^{d} \to\mathbb{R}$ , the function

(A.1) \begin{align}A f({\boldsymbol{x}}) =&\ \sum_{i_d = 0}^{4} \alpha_{k_d({\boldsymbol{x}})+i_d}^{k_d({\boldsymbol{x}})}(x_d)\cdots \sum_{i_1 = 0}^{4} \alpha_{k_1({\boldsymbol{x}})+i_1}^{k_1({\boldsymbol{x}})}(x_1) f(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) \notag\\=&\ \sum_{i_1, \ldots, i_d = 0}^{4} \bigg(\prod_{j=1}^{d} \alpha_{k_j({\boldsymbol{x}}) +i_j}^{k_j({\boldsymbol{x}}) }(x_j)\bigg) f(\delta (k({\boldsymbol{x}})+{\boldsymbol{i}})) , \quad {\boldsymbol{x}} \in \text{Conv}(K_4) \end{align}

satisfies $A f({\boldsymbol{x}}) \in C^{3}(\text{Conv}(K_4))$ , where ${\boldsymbol{i}} = (i_1, \ldots, i_d)$ in (A.1). Additionally, $Af({\boldsymbol{x}})$ is infinitely differentiable almost everywhere on $\text{Conv}(K_4)$ ,

(A.2) \begin{align}A f(\delta {\boldsymbol{k}}) = f(\delta {\boldsymbol{k}}), \quad \delta {\boldsymbol{k}} \in K_{4}, \end{align}

and for $0 \leq \Vert{\boldsymbol{a}}\Vert_{1} \leq 3$ ,

(A.3)

where

(A.4)

When $\Vert{\boldsymbol{a}}\Vert_{1} = 4$ , (A.3) also holds for almost all ${\boldsymbol{x}} \in \text{Conv}(K_4)$ . Additionally, the weights $\{\alpha_{k+i}^{k} \,:\,\mathbb{R} \to \mathbb{R},\ k \in\mathbb{Z},\ i = 0,1,2,3,4\}$ are degree-7 polynomials in $(x-\delta k)/\delta$ whose coefficients do not depend on k or $\delta$ . For any $x \in\mathbb{R}$ , they satisfy

(A.5) \begin{align}\alpha_{k}^{k}(\delta k) = 1 \quad \text{ and } \quad \alpha_{k+i}^{k} (\delta k) = 0, \quad k \in \mathbb{Z},\, i = 1,2,3,4,\end{align}

(A.6) \begin{align}\sum_{i=0}^{4} \alpha^{k}_{k+i}(x) = 1, \quad k \in \mathbb{Z}, \end{align}

(A.7) \begin{align}\alpha^{k+j}_{k+j+i}(x+ \delta j) = \alpha^{k}_{k+i}(x),\quad i,j,k \in \mathbb{Z}, \end{align}

(A.8) \begin{align}\vert\alpha_{k(x) + i}^{k(x)}(x)\vert \leq 319. \end{align}

We prove (A.4) and (A.8) in Appendix A.1. The rest of Theorem A.1 is identical to Theorem A.1 of [Reference Braverman5]. We now present three useful properties of A. While Theorem A.1 only guarantees that $Af({\boldsymbol{x}})$ is thrice continuously differentiable, we need to control the fourth-order remainder term in the Taylor expansion of $Af({\boldsymbol{x}})$ . The following lemma, which would have been trivial if $Af({\boldsymbol{x}})$ were four-times continuously differentiable, helps with this. Define

(A.9) \begin{align}\mathcal{D}^{d} = \Big\{f \,:\,\mathbb{R}^{d} \to \mathbb{R} \,:\,\text{ for all ${\boldsymbol{x}}, {\boldsymbol{y}} \in \mathbb{R}^{d}$, } \vert f({\boldsymbol{x}}) - f({\boldsymbol{y}})\vert \leq \Vert{\boldsymbol{x}}-{\boldsymbol{y}}\Vert_{1} \sup_{\substack{ \min({\boldsymbol{x}},{\boldsymbol{y}}) \leq {\boldsymbol{z}} \\ {\boldsymbol{z}} \leq \max({\boldsymbol{x}},{\boldsymbol{y}}) \\ \Vert{\boldsymbol{a}}\Vert_{1} = 1}} \vert D^{{\boldsymbol{a}}} f({\boldsymbol{z}})\vert \Big\}, \end{align}

where the minimum and maximum are taken elementwise.

The second lemma is a useful identity for applying A to products of functions.

Lemma A.2. Given $f,g\,:\,\mathbb{Z}^{d} \to \mathbb{R}$ let $h({\boldsymbol{u}}) =f({\boldsymbol{u}}) g({\boldsymbol{u}})$ . There exists $\epsilon_{p}:\mathbb{R}^{d} \to \mathbb{R}$ and a constant C(d) such that

(A.10)

For our third result, let $f({\boldsymbol{x}})$ be a function defined for all ${\boldsymbol{x}} \in \mathbb{R}^{d}$ and let $f({\boldsymbol{u}})$ denote its restriction to $\delta \mathbb{Z}^{d}$ . Proposition A.1 provides an upper bound on how well $Af({\boldsymbol{x}})$ approximates $f({\boldsymbol{x}})$ . The smoother the function $f({\boldsymbol{x}})$ , the higher the exponent of $\delta$ in the error bound.

Proposition A.1. Suppose that $f \in C^{s-1}(\mathbb{R}^{d})$ for some $s \in \{1,2,3,4\}$ and that $D^{{\boldsymbol{a}}} f({\boldsymbol{x}}) \in \mathcal{D}^{d}$ when $\Vert{\boldsymbol{a}}\Vert_{1} = s-1$ . Then

(A.11)

The following corollary plays an important role later on in the proof of Lemma B.2.

Proof of Corollary A.1. If f(x) is a polynomial up to the third order, then its fourth-order derivatives are zero. The result follows from applying Proposition A.1 with $s = 4$ .

We now prove Lemmas A.1 and A.2, and Proposition A.1, in that order.

Proof of Lemma A.1. Given ${\boldsymbol{x}}, {\boldsymbol{y}} \in \mathbb{R}^{d}$ , define

\begin{align*}{\boldsymbol{x}}^{(j)} = (x_1, \ldots, x_{j-1}, y_{j}, \ldots, y_{d}), \quad 1 \leq j \leq d,\end{align*}

and note that ${\boldsymbol{x}}^{(1)} = {\boldsymbol{y}}$ and ${\boldsymbol{x}}^{(d)} = {\boldsymbol{x}}$ . Fix ${\boldsymbol{a}} \in \mathbb{Z}^{d}$ with ${\boldsymbol{a}} > 0$ and $\Vert{\boldsymbol{a}}\Vert_{1} =3$ . Then

\begin{align*}D^{{\boldsymbol{a}}} A f({\boldsymbol{x}}) - D^{{\boldsymbol{a}}} A f({\boldsymbol{y}}) =&\ D^{{\boldsymbol{a}}} A f({\boldsymbol{x}}^{(d)}) - D^{{\boldsymbol{a}}} A f({\boldsymbol{x}}^{(1)})= \sum_{j=1}^{d-1} D^{{\boldsymbol{a}}} A f({\boldsymbol{x}}^{(j+1)}) - D^{{\boldsymbol{a}}} A f({\boldsymbol{x}}^{(j)}).\end{align*}

We now show that

\begin{align*}\vert D^{{\boldsymbol{a}}} A f({\boldsymbol{x}}^{(j+1)}) - D^{{\boldsymbol{a}}} A f({\boldsymbol{x}}^{(j)})\vert \leq \vert{\boldsymbol{x}}_{j} - {\boldsymbol{y}}_{j}\vert \sup_{\substack{ \min({\boldsymbol{x}},{\boldsymbol{y}}) \leq {\boldsymbol{z}}\\ {\boldsymbol{z}} \leq \max({\boldsymbol{x}},{\boldsymbol{y}})}} \vert D^{{\boldsymbol{a}}} A f({\boldsymbol{z}})\vert.\end{align*}

Suppose that $j = d-1$ ; the argument is similar for other values of j. Note that

\begin{align*}&D^{{\boldsymbol{a}}} A f({\boldsymbol{x}})\\&\quad= \partial_{d}^{a_{d}} \cdots \partial_{1}^{a_1} \bigg(\sum_{i_d = 0}^{4} \alpha_{k_d({\boldsymbol{x}})+i_d}^{k_d({\boldsymbol{x}})}(x_d)\cdots \sum_{i_1 = 0}^{4} \alpha_{k_1({\boldsymbol{x}})+i_1}^{k_1({\boldsymbol{x}})}(x_1) f(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}}))\bigg)\\&\quad= \partial_{d}^{a_d}\sum_{i_d = 0}^{4} \alpha_{k_d(x_d)+i_d}^{k_d(x_d)}(x_d)\\& \hspace{1.3cm} \times \bigg[\partial_{d-1}^{a_{d-1}}\bigg(\sum_{i_{d-1} = 0}^{4} \alpha_{k_{d-1}(x_{d-1})+i_{d-1}}^{k_{d-1}(x_{d-1})}(x_{d-1})\cdots \partial_{1}^{a_1}\bigg(\sum_{i_1 = 0}^{4} \alpha_{k_1(x_1)+i_1}^{k_1(x_1)}(x_1) f(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) \bigg)\bigg)\bigg],\end{align*}

where the first equality follows the definition of $A f({\boldsymbol{x}})$ in (A.1). Treating $x_1, \ldots, x_{d-1}$ as fixed, let us consider the term inside the square brackets as a one-dimensional function in the dth dimension. To be precise, we define $g_{x_1,\ldots,x_{d-1}}\, :\,\delta \mathbb{Z} \to \mathbb{R}$ as

\begin{align*}&g_{x_1,\ldots,x_{d-1}}(\delta \ell)\\&\quad= \partial_{d-1}^{a_{d-1}}\bigg(\sum_{i_{d-1} = 0}^{4} \alpha_{k_{d-1}(x_{d-1})+i_{d-1}}^{k_{d-1}(x_{d-1})}(x_{d-1})\cdots \partial_{1}^{a_1}\bigg(\sum_{i_1 = 0}^{4} \alpha_{k_1(x_1)+i_1}^{k_1(x_1)}(x_1)\\& \hspace{3cm} \times f\big(\delta(k(x_1,\ldots,x_{d-1},0)+(i_1,\ldots,i_{d-1},0)) + \delta \ell \big) \bigg)\bigg), \quad \ell \in \mathbb{Z}.\end{align*}

Then

\begin{align*}D^{{\boldsymbol{a}}} A f({\boldsymbol{x}}) = \partial_{d}^{a_d}\sum_{i_d =0}^{4} \alpha_{k_d(x_d)+i_d}^{k_d(x_d)}(x_d) g_{x_1,\ldots,x_{d-1}}( \delta(k_{d}(x_d) + i_{d}) ) = \partial^{a_d} A g_{x_1,\ldots,x_{d-1}}(x_{d}).\end{align*}

Theorem A.1 states that $Ag_{x_1,\ldots,x_{d-1}}(x_d)$ is infinitely differentiable almost everywhere. Thus,

\begin{align*}&\vert D^{{\boldsymbol{a}}} A f({\boldsymbol{x}}^{(d)}) - D^{{\boldsymbol{a}}} A f({\boldsymbol{x}}^{(d-1)})\vert\\&\qquad= \vert\partial^{a_d} A g_{x_1,\ldots,x_{d-1}}(x_{d}) - \partial^{a_d} A g_{x_1,\ldots,x_{d-1}}(y_{d})\vert\\&\qquad= \vert\int_{y_d}^{x_d} \partial^{a_d+1} A g_{x_1,\ldots,x_{d-1}}(x')dx'\vert\\&\qquad\leq \vert x_d - y_d\vert \sup_{(x_d \wedge y_d) \leq z_d \leq (x_d \vee y_{d}) } \vert\partial^{a_d+1} A g_{x_1,\ldots,x_{d-1}}(z_d)\vert\\&\qquad\leq \vert x_d - y_d\vert \max_{\Vert{\boldsymbol{a}}'\Vert_{1} = \Vert{\boldsymbol{a}}\Vert_{1} + 1} \sup_{\substack{ \min({\boldsymbol{x}},{\boldsymbol{y}}) \leq {\boldsymbol{z}}\\{\boldsymbol{z}} \leq \max({\boldsymbol{x}},{\boldsymbol{y}})}} \vert D^{{\boldsymbol{a}}'} A f(z)\vert,\end{align*}

where in the last inequality we used $\partial^{a_d+1} A g_{x_1,\ldots,x_{d-1}}(x_{d}) = \partial_{d} D^{{\boldsymbol{a}}} A f({\boldsymbol{x}})$ .

Proof of Lemma A.2. Using (A.1) of Theorem A.1, for ${\boldsymbol{x}} \in \mathbb{R}^{d}$ ,

\begin{align*}A h({\boldsymbol{x}}) =&\ \sum_{i_1, \ldots, i_d = 0}^{4} \prod_{j=1}^{d} \alpha_{k_j({\boldsymbol{x}}) +i_j}^{k_j({\boldsymbol{x}}) }(x_j) f(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) g(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}}))\\=&\ \sum_{i_1, \ldots, i_d = 0}^{4} \prod_{j=1}^{d} \alpha_{k_j({\boldsymbol{x}}) +i_j}^{k_j({\boldsymbol{x}}) }(x_j) A f({\boldsymbol{x}}) g(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}}))\\&+ \sum_{i_1, \ldots, i_d = 0}^{4} \prod_{j=1}^{d} \alpha_{k_j({\boldsymbol{x}}) +i_j}^{k_j({\boldsymbol{x}}) }(x_j) \big(f(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) - A f({\boldsymbol{x}})\big) g(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})).\end{align*}

By the definition of A,

\begin{align*}&\sum_{i_1, \ldots, i_d = 0}^{4} \prod_{j=1}^{d} \alpha_{k_j({\boldsymbol{x}}) +i_j}^{k_j({\boldsymbol{x}}) }(x_j) A f({\boldsymbol{x}}) g(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) = Af({\boldsymbol{x}}) Ag({\boldsymbol{x}}).\end{align*}

Furthermore,

\begin{align*}& \sum_{i_1, \ldots, i_d = 0}^{4} \prod_{j=1}^{d} \alpha_{k_j({\boldsymbol{x}}) +i_j}^{k_j({\boldsymbol{x}}) }(x_j) \big(f(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) - A f({\boldsymbol{x}})\big) g(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}}))\\&\qquad = \sum_{i_1, \ldots, i_d = 0}^{4} \prod_{j=1}^{d} \alpha_{k_j({\boldsymbol{x}}) +i_j}^{k_j({\boldsymbol{x}}) }(x_j) \big(f(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) - A f({\boldsymbol{x}})\big) \big(g(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) - g(\delta k({\boldsymbol{x}}))\big),\end{align*}

because again, by the definition of A,

\begin{align*}\sum_{i_1, \ldots, i_d = 0}^{4} \prod_{j=1}^{d} \alpha_{k_j({\boldsymbol{x}}) +i_j}^{k_j({\boldsymbol{x}}) }(x_j) \big(f(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) - A f({\boldsymbol{x}})\big) g(\delta k({\boldsymbol{x}})) = g(\delta k({\boldsymbol{x}})) (Af({\boldsymbol{x}}) - A f({\boldsymbol{x}})) = 0.\end{align*}

Setting

\begin{align*}\epsilon_{p}({\boldsymbol{x}}) = \sum_{i_1, \ldots, i_d = 0}^{4} \prod_{j=1}^{d} \alpha_{k_j({\boldsymbol{x}}) +i_j}^{k_j({\boldsymbol{x}}) }(x_j) \big(f(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) - A f({\boldsymbol{x}})\big) \big(g(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) - g(\delta k({\boldsymbol{x}}))\big),\end{align*}

we have shown that $A h({\boldsymbol{x}}) = Af({\boldsymbol{x}}) A g({\boldsymbol{x}}) + \epsilon_{p}({\boldsymbol{x}})$ . Since

\begin{align*}\vert\alpha_{k_j({\boldsymbol{x}}) + i_j}^{k_j({\boldsymbol{x}})}(x_j)\vert \leq 319\end{align*}

by (A.8), it follows that

\begin{align*}\vert\epsilon_{p}({\boldsymbol{x}})\vert \leq 5^{d} 319^{d} \max_{0 \leq{\boldsymbol{i}} \leq 4{\boldsymbol{e}}} (|f(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) - A f({\boldsymbol{x}})||g(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) - g(\delta k({\boldsymbol{x}}))| ).\end{align*}

To conclude the proof, we argue that

which, when combined with the bound on $C_{A.4}(d)$ in (A.4), yields $C_{A.10}(d) \leq7975^{d}(5^{d} + dC_{A.4}(d) ) \leq 7975^{d}(5^{d} + d11404^{5} 1595^{d})$ . The first inequality follows by rewriting $g(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) - g(\delta k({\boldsymbol{x}}))$ as a telescoping series and using the fact that ${\boldsymbol{i}} \leq 4{\boldsymbol{e}}$ . Similarly,

\begin{align*}\vert f(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})) - A f({\boldsymbol{x}})\vert \leq&\ \vert f(\delta (k({\boldsymbol{x}})+{\boldsymbol{i}})) - f(\delta k({\boldsymbol{x}}))\vert + \vert f(\delta k({\boldsymbol{x}})) - A f({\boldsymbol{x}})\vert\\\leq&\ 5^{d} \max_{\substack{\Vert{\boldsymbol{a}}\Vert_{1}=1 \\ 0 \leq j \leq 4{\boldsymbol{e}} -{\boldsymbol{a}} }}\vert\Delta^{{\boldsymbol{a}}} f(\delta (k({\boldsymbol{x}}) + {\boldsymbol{j}}) \vert + \vert f(\delta k({\boldsymbol{x}})) - A f({\boldsymbol{x}})\vert,\end{align*}

and

where the last inequality is due to (A.3) of Theorem A.1.

Fix a function $f({\boldsymbol{x}})$ satisfying the conditions of Proposition A.1. The following is a key auxiliary lemma, which we prove after proving Proposition A.1.

Lemma A.3. For any $\delta < 1$ , $s \in \{1,2,3,4\}$ , and $1 \leq\Vert{\boldsymbol{a}}\Vert_{1} \leq s-1$ ,

\begin{align*}D^{{\boldsymbol{a}}} A f(\delta k({\boldsymbol{x}})) =&\ D^{{\boldsymbol{a}}} f(\delta k({\boldsymbol{x}})) + 12^{d}\delta^{s-\Vert{\boldsymbol{a}}\Vert_{1}} E^{(s)}\big([\delta k({\boldsymbol{x}}), \delta ( k({\boldsymbol{x}}) + 3 {\boldsymbol{e}})] \big), \quad {\boldsymbol{x}} \in \mathbb{R}^{d},\end{align*}

where, for any $A \subset \mathbb{R}^{d}$ , $E^{(s)}(A)$ denotes a quantity that satisfies

(A.12) \begin{align}\vert E^{(s)}(A)\vert \leq \max_{\Vert{\boldsymbol{a}}\Vert_{1} = s} \sup_{{\boldsymbol{z}} \in A } \vert D^{{\boldsymbol{a}}} f( {\boldsymbol{z}})\vert. \end{align}

Proof of Proposition A.1. Throughout the proof we let $E^{(s)}(A)$ denote a generic quantity that satisfies (A.12), but whose value may change from line to line. Expanding both $f({\boldsymbol{x}})$ and $A f({\boldsymbol{x}})$ around $\delta k({\boldsymbol{x}})$ yields

\begin{align*}A f({\boldsymbol{x}}) - f({\boldsymbol{x}}) =&\ \sum_{j = 1}^{s-1} \frac{1}{j!} \sum_{{\boldsymbol{a}} \,:\,\Vert{\boldsymbol{a}}\Vert_{1} = j} \bigg(\prod_{i=1}^{d} (x_{i} - \delta k_{i}({\boldsymbol{x}}))^{a_{i}} \bigg)\big(D^{{\boldsymbol{a}}} A f(\delta k({\boldsymbol{x}})) - D^{{\boldsymbol{a}}} f(\delta k({\boldsymbol{x}}))\big)\\&+ \frac{1}{(s-1)!} \sum_{{\boldsymbol{a}} \,:\,\Vert{\boldsymbol{a}}\Vert_{1} = s-1} \bigg(\prod_{i=1}^{d} (x_{i} - \delta k_{i}({\boldsymbol{x}}))^{a_{i}} \bigg) \big( D^{{\boldsymbol{a}}} A f(\boldsymbol{\xi}_{1}) - D^{{\boldsymbol{a}}} Af(\delta k({\boldsymbol{x}}))\big)\\&+ \frac{1}{(s-1)!} \sum_{{\boldsymbol{a}} \,:\,\Vert{\boldsymbol{a}}\Vert_{1} = s-1} \bigg(\prod_{i=1}^{d} (x_{i} - \delta k_{i}({\boldsymbol{x}}))^{a_{i}} \bigg) \big( D^{{\boldsymbol{a}}} f(\boldsymbol{\xi}_{2}) - D^{{\boldsymbol{a}}} f(\delta k({\boldsymbol{x}}))\big),\end{align*}

where $\boldsymbol{\xi}_1,\boldsymbol{\xi}_2 \in [\delta k({\boldsymbol{x}}),{\boldsymbol{x}}] \subset [\delta k({\boldsymbol{x}}) ,\delta (k({\boldsymbol{x}}) + {\boldsymbol{e}}))$ . Consider separately each of the three terms on the right-hand side. Since $\vert x_i - \delta k_i({\boldsymbol{x}})\vert \leq \delta$ , then

\begin{align*}& \sum_{j = 1}^{s-1} \frac{1}{j!} \sum_{{\boldsymbol{a}} \,:\,\Vert{\boldsymbol{a}}\Vert_{1} = j} \bigg(\prod_{i=1}^{d} (x_{i} - \delta k_{i}({\boldsymbol{x}}))^{a_{i}} \bigg)\big(D^{{\boldsymbol{a}}} A f(\delta k({\boldsymbol{x}})) - D^{{\boldsymbol{a}}} f(\delta k({\boldsymbol{x}}))\big) \\&\qquad\leq \sum_{j = 1}^{s-1} \frac{1}{j!} \sum_{{\boldsymbol{a}} \,:\,\Vert{\boldsymbol{a}}\Vert_{1} = j} \delta^{\Vert{\boldsymbol{a}}\Vert_{1}} \delta^{s-\Vert{\boldsymbol{a}}\Vert_{1}} 12^{d} E^{(s)}\big([\delta k({\boldsymbol{x}}), \delta ( k({\boldsymbol{x}}) + 3 {\boldsymbol{e}})] \big) \\&\qquad\leq \sum_{j = 1}^{s-1} \frac{1}{j!} d^{j} \delta^{s} 12^{d} E^{(s)}\big([\delta k({\boldsymbol{x}}), \delta ( k({\boldsymbol{x}}) + 3 {\boldsymbol{e}})] \big)\\&\qquad\leq 4 d^{4} 12^{d} \delta^{s} E^{(s)}\big([\delta k({\boldsymbol{x}}), \delta ( k({\boldsymbol{x}}) + 3 {\boldsymbol{e}})] \big).\end{align*}

The first inequality is due to Lemma A.3, the second is due to the fact that there are at most $d^{j}$ permutations that yield $\Vert{\boldsymbol{a}}\Vert_{1} = j$ , and the final inequality follows from $j < s \leq 4$ .

For the second term we note that

The second inequality holds because we assumed that $D^{{\boldsymbol{a}}}f({\boldsymbol{x}}) \in \mathcal{D}^{d}$ when $\Vert{\boldsymbol{a}}\Vert_{1} = s-1$ . The third holds because $\Vert\boldsymbol{\xi}_{1}-\delta k({\boldsymbol{x}})\Vert_{1} \leq d\delta$ (since $\delta k({\boldsymbol{x}}) \leq \boldsymbol{\xi}_{1} \leq{\boldsymbol{x}} < \delta(k({\boldsymbol{x}}) + {\boldsymbol{e}})$ ), and by (A.3). The second-last inequality is a consequence of the fundamental theorem of calculus: suppose that $\Delta^{{\boldsymbol{a}}}f(\delta (k({\boldsymbol{z}})+{\boldsymbol{i}}))= \Delta_{j_s} \cdots \Delta_{j_1} f(\delta(k({\boldsymbol{z}})+{\boldsymbol{i}}))$ for some $1 \leq j_1,\ldots, j_s\leq d$ . By our assumptions that $f \in C^{s-1}(\mathbb{R}^{d})$ and that $D^{{\boldsymbol{a}}} f({\boldsymbol{x}}) \in \mathcal{D}^{d}$ when $\Vert{\boldsymbol{a}}\Vert_{1} = s-1$ , it follows that

(A.13) \begin{align}&\vert\Delta_{j_s} \cdots \Delta_{j_1} f(\delta (k({\boldsymbol{z}})+{\boldsymbol{i}}))\vert \notag\\&\qquad= \Bigg|\Delta_{j_s} \int_{0}^{\delta} \cdots \int_{0}^{\delta} \partial_{j_{s-1}} \cdots \partial_{j_1} f(\delta (k({\boldsymbol{z}})+{\boldsymbol{i}}) + \gamma_{1} e_{j_1} + \cdots + \gamma_{s-1} e_{j_{s-1}}) d \gamma_{1} \cdots d\gamma_{s-1}\Bigg| \notag\\&\qquad\leq \delta^{s} \sup_{\substack{ \delta (k({\boldsymbol{z}})+{\boldsymbol{i}}) \leq {\boldsymbol{z}}' \\ {\boldsymbol{z}}' \leq \delta (k({\boldsymbol{z}})+{\boldsymbol{i}} + {\boldsymbol{a}})}} \vert D^{{\boldsymbol{a}}}f({\boldsymbol{z}}')\vert. \end{align}

Finally, for the third term, since we assumed that $D^{{\boldsymbol{a}}}f({\boldsymbol{x}}) \in \mathcal{D}^{d}$ when $\Vert{\boldsymbol{a}}\Vert_{1} = s-1$ ,

\begin{align*}&\frac{1}{(s-1)!} \sum_{{\boldsymbol{a}} \,:\,\Vert{\boldsymbol{a}}\Vert_{1} = s-1} \bigg(\prod_{i=1}^{d} (x_{i} - \delta k_{i}({\boldsymbol{x}}))^{a_{i}} \bigg) \big( D^{{\boldsymbol{a}}} f(\boldsymbol{\xi}_{2}) - D^{{\boldsymbol{a}}} f(\delta k({\boldsymbol{x}}))\big)\\&\qquad\leq d^{s-1}\delta^{s-1} \Vert\boldsymbol{\xi}_{1}-\delta k({\boldsymbol{x}})\Vert_{1} \sup_{\substack{ \delta k({\boldsymbol{x}}) \leq {\boldsymbol{z}} \\ {\boldsymbol{z}} \leq \delta ( k({\boldsymbol{x}}) + {\boldsymbol{e}}) \\ \Vert{\boldsymbol{a}}\Vert_{1} = s}} \vert D^{{\boldsymbol{a}}} A f({\boldsymbol{z}})\vert\\&\qquad\leq d^{s-1}\delta^{s-1} d \delta E^{(s)}\big([\delta k({\boldsymbol{x}}), \delta ( k({\boldsymbol{x}}) + {\boldsymbol{e}})] \big).\end{align*}

Adding up the bounds on the three terms and using the fact that $s \leq 4$ yields

To prove Lemma A.3, for $f\,:\,\mathbb{R}^{d} \to \mathbb{R}$ , $1 \leq i \leq d$ , and ${\boldsymbol{x}} \in \mathbb{R}^{d}$ , we define $\Delta_{i}^{0} f({\boldsymbol{x}}) = f({\boldsymbol{x}})$ , $\widetilde\Delta_{i}^{(0)} f({\boldsymbol{x}}) = f({\boldsymbol{x}})$ ,

(A.14) \begin{gather}\Delta_{i} f({\boldsymbol{x}}) = f({\boldsymbol{x}}+\delta {\boldsymbol{e}}_{i}) - f({\boldsymbol{x}}), \notag\\\widetilde \Delta_{i}^{(1)} f({\boldsymbol{x}})= \Big(\Delta_{i} - \frac{1}{2} \Delta_{i}^2 + \frac{1}{3} \Delta_{i}^3 \Big)f({\boldsymbol{x}}),\notag\\\widetilde \Delta_{i}^{(2)} f({\boldsymbol{x}})= \big(\Delta_{i}^2 - \Delta_{i}^3 \big) f({\boldsymbol{x}}), \quad \widetilde \Delta_{i}^{(3)} f({\boldsymbol{x}})= \Delta_{i}^3 f({\boldsymbol{x}}). \end{gather}

Lemma A.4. Given $f\,:\,\mathbb{R}^{d} \to \mathbb{R}$ and the corresponding $A f({\boldsymbol{x}})$ , for $1 \leq \Vert{\boldsymbol{a}}\Vert_{1} \leq 3$ ,

\begin{align*}D^{{\boldsymbol{a}}} A f(\delta k({\boldsymbol{x}})) =&\ \delta^{-\Vert{\boldsymbol{a}}\Vert_{1}} \widetilde \Delta^{(a_1)}_{1} \cdots \widetilde \Delta_{d}^{(a_d)} f(\delta k({\boldsymbol{x}})), \quad {\boldsymbol{x}} \in \mathbb{R}^{d}.\end{align*}

We now prove Lemmas A.3 and A.4.

Proof of Lemma A.3. Fix $s \in \{1,2,3,4\}$ and ${\boldsymbol{a}}$ with $1 \leq \Vert{\boldsymbol{a}}\Vert_{1} \leq s-1$ . For any $A \subset \mathbb{R}^{d}$ , we write $E_{i}^{(s)}(A)$ to denote any quantity satisfying

\begin{align*}\vert E_{i}^{(s)}(A)\vert \leq \sup_{ {\boldsymbol{z}} \in A} \vert\partial_{i}^{s} f({\boldsymbol{z}})\vert,\end{align*}

and recall from (A.12) that $E^{(s)}(A)$ denotes a quantity satisfying

\begin{align*}\vert E^{(s)}(A)\vert \leq \max_{\Vert{\boldsymbol{a}}\Vert_{1} = s} \sup_{{\boldsymbol{z}} \in A } \vert D^{{\boldsymbol{a}}} f( {\boldsymbol{z}})\vert.\end{align*}

Suppose we have shown that, for $1 \leq i \leq d$ and $1 \leq j \leq s-1$ ,

(A.15) \begin{align}\widetilde \Delta_{i}^{(j)} f({\boldsymbol{x}}) =&\, \delta^{j}\partial_{i}^{j} f({\boldsymbol{x}}) + 5 \delta^{s}E_{i}^{(s)}([{\boldsymbol{x}}, {\boldsymbol{x}} + 3 \delta{\boldsymbol{e}}_{i}]). \end{align}

Combining (A.15) with Lemma A.4 yields

\begin{align*}&D^{{\boldsymbol{a}}} A f(\delta k({\boldsymbol{x}}))\\&\quad= \delta^{-\Vert{\boldsymbol{a}}\Vert_{1}} \widetilde \Delta^{(a_1)}_{1} \cdots \widetilde \Delta_{d}^{(a_d)} f(\delta k({\boldsymbol{x}})) \notag\\&\quad= \delta^{-\Vert{\boldsymbol{a}}\Vert_{1}} \widetilde\Delta^{(a_1)}_{1} \cdots \widetilde \Delta_{d-1}^{(a_{d-1})} (\delta^{a_d} \partial_{d}^{a_d} f(\delta k({\boldsymbol{x}})) + 5\delta^{s} E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta(k({\boldsymbol{x}}) + 3 {\boldsymbol{e}}_{d})]) ) \notag\\&\quad= \delta^{-\Vert{\boldsymbol{a}}\Vert_{1}} \delta^{a_d} \widetilde\Delta^{(a_1)}_{1} \cdots \widetilde \Delta_{d-1}^{(a_{d-1})}\partial_{d}^{a_d} f(\delta k({\boldsymbol{x}})) \\&\qquad+ 5\delta^{s-\Vert{\boldsymbol{a}}\Vert_{1}} \widetilde \Delta^{(a_1)}_{1}\cdots \widetilde \Delta_{d-1}^{(a_{d-1})} E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) + 3{\boldsymbol{e}}_{d})])\\&\quad= \delta^{-\Vert{\boldsymbol{a}}\Vert_{1}} \delta^{a_d}\widetilde\Delta^{(a_1)}_{1} \cdots \widetilde \Delta_{d-2}^{(a_{d-2})} (\delta^{a_{d-1}} \partial_{d-1}^{a_{d-1}} \partial_{d}^{a_d} f(\delta k({\boldsymbol{x}})) + 5 \delta^{s} E_{d-1}^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) + 3{\boldsymbol{e}}_{d-1})]) )\\&\qquad + 5 \delta^{s-\Vert{\boldsymbol{a}}\Vert_{1}} \widetilde\Delta^{(a_1)}_{1} \cdots \widetilde \Delta_{d-1}^{(a_{d-1})}E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) + 3{\boldsymbol{e}}_{d})]).\end{align*}

Continuing recursively, we arrive at

(A.16) \begin{align}D^{{\boldsymbol{a}}} A f(\delta k({\boldsymbol{x}})) =&\ \delta^{-\Vert{\boldsymbol{a}}\Vert_{1}} \delta^{a_1} \cdots \delta^{a_d} D^{{\boldsymbol{a}}} f(\delta k({\boldsymbol{x}})) \notag\\&+ 5 \delta^{s-\Vert{\boldsymbol{a}}\Vert_{1}} \delta^{a_d} \cdots\delta^{a_2} E_{1}^{(s)}([\delta k({\boldsymbol{x}}), \delta(k({\boldsymbol{x}}) + 3 {\boldsymbol{e}}_{1})]) \notag\\&+ \cdots \notag\\&+ 5 \delta^{s-\Vert{\boldsymbol{a}}\Vert_{1}} \widetilde\Delta^{(a_1)}_{1} \cdots \widetilde \Delta_{d-1}^{(a_{d-1})}E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) + 3{\boldsymbol{e}}_{d})]) \end{align}

We now argue that, for any $1 \leq j \leq d-1$ ,

(A.17) \begin{align}\widetilde \Delta^{(a_1)}_{1} \cdots \widetilde \Delta_{j-1}^{(a_{j-1})}E_{j}^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) + 3{\boldsymbol{e}}_{j})]) = 12^{j-1}E^{(s)}([\delta k({\boldsymbol{x}}),\delta (k({\boldsymbol{x}}) + 3 {\boldsymbol{e}})]), \end{align}

because applying (A.17) to (A.16) and using the fact that $\delta < 1$ yields

\begin{align*}D^{{\boldsymbol{a}}} A f(\delta k({\boldsymbol{x}})) =&\ D^{{\boldsymbol{a}}} f(\delta k({\boldsymbol{x}})) + 5\delta^{s-\Vert{\boldsymbol{a}}\Vert_{1}} \Big(\sum_{j=1}^{d-1} 12^{j-1}\Big) E^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) + 3{\boldsymbol{e}})])\\=&\ D^{{\boldsymbol{a}}} f(\delta k({\boldsymbol{x}})) + 12^{d}\delta^{s-\Vert{\boldsymbol{a}}\Vert_{1}} E^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) + 3 {\boldsymbol{e}})])\end{align*}

where the second equality follows from $5 \sum_{j=1}^{d-1} 12^{j-1} \leq12^{d}$ . To prove (A.17), we set $j = d$ and evaluate $\widetilde \Delta^{(a_1)}_{1} \cdots \widetilde \Delta_{d-1}^{(a_{d-1})}E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) + 3{\boldsymbol{e}}_{d})])$ . We remind the reader that, for any $1 \leq \ell\leq d-1$ ,

\begin{align*}&\Delta_{\ell} E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta(k({\boldsymbol{x}}) + 3 {\boldsymbol{e}}_{d})])\\&\qquad= E_{d}^{(s)}([\delta k({\boldsymbol{x}}+\delta{\boldsymbol{e}}_{\ell}), \delta (k({\boldsymbol{x}}+\delta{\boldsymbol{e}}_{\ell}) + 3 {\boldsymbol{e}}_{d})]) - E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) + 3{\boldsymbol{e}}_{d})])\\&\qquad= 2 E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta(k({\boldsymbol{x}}) +{\boldsymbol{e}}_{\ell}+ 3{\boldsymbol{e}}_{d})]),\end{align*}

where in the second equality we used the fact that $k({\boldsymbol{x}}+\delta {\boldsymbol{e}}_{\ell}) = k({\boldsymbol{x}})+{\boldsymbol{e}}_{\ell}$ , which follows from the definition of $k({\boldsymbol{x}})$ . Similarly,

\begin{align*}\Delta_{\ell}^2 E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta(k({\boldsymbol{x}}) + 3 {\boldsymbol{e}}_{d})]) =&\ 4 E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) +2{\boldsymbol{e}}_{\ell}+3{\boldsymbol{e}}_{d})]),\\\Delta_{\ell}^3 E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta(k({\boldsymbol{x}}) + 3 {\boldsymbol{e}}_{d})]) =&\ 8 E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) +3{\boldsymbol{e}}_{\ell}+3{\boldsymbol{e}}_{d})]).\end{align*}

These expressions, together with the definition of $\widetilde\Delta_{d-1}^{(a_{d-1})} f({\boldsymbol{x}})$ in (A.14) and the fact that $0 \leq a_{d-1} \leq 3$ , imply that

\begin{align*}\widetilde \Delta_{d-1}^{(a_{d-1})} E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) + 3{\boldsymbol{e}}_{d})]) = 12 E_{d}^{(s)}([\delta k({\boldsymbol{x}}),\delta (k({\boldsymbol{x}}) +3{\boldsymbol{e}}_{d-1}+3{\boldsymbol{e}}_{d})])\end{align*}

and, similarly,

\begin{align*}\widetilde \Delta^{(a_1)}_{1} \cdots \widetilde \Delta_{d-1}^{(a_{d-1})}E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) + 3{\boldsymbol{e}}_{d})]) =&\ 12^{d-1} E_{d}^{(s)}([\delta k({\boldsymbol{x}}), \delta (k({\boldsymbol{x}}) + 3 {\boldsymbol{e}})])\\=&\ 12^{d-1} E^{(s)}([\delta k({\boldsymbol{x}}), \delta(k({\boldsymbol{x}}) + 3 {\boldsymbol{e}})]),\end{align*}

proving (A.17).

It remains to prove (A.15). Suppose that $s = 4$ , Taylor expansion yields

(A.18) \begin{gather}\Delta_{i} f({\boldsymbol{x}}) = \delta \partial_{i} f({\boldsymbol{x}}) +\frac{1}{2}\delta^2 \partial_{i}^{2} f({\boldsymbol{x}}) +\frac{1}{6}\delta^3 \partial_{i}^{3} f({\boldsymbol{x}}) + \frac{1}{24}\delta^{4} E_{i}^{(4)}([{\boldsymbol{x}}, {\boldsymbol{x}} + \delta{\boldsymbol{e}}_{i}]), \notag\\\Delta_{i}^2 f({\boldsymbol{x}}) = \delta^2 \partial_{i}^2f({\boldsymbol{x}}) + \delta^3 \partial_{i}^3 f({\boldsymbol{x}}) +\frac{2}{3} \delta^4 E_{i}^{(4)}([{\boldsymbol{x}}, {\boldsymbol{x}} + 2\delta {\boldsymbol{e}}_{i}]), \notag\\\Delta_{i}^3 f({\boldsymbol{x}}) =\delta^3 \partial_{i}^3f({\boldsymbol{x}}) + \frac{17}{6} \delta^4 E_{i}^{(4)}([{\boldsymbol{x}},{\boldsymbol{x}} + 3\delta {\boldsymbol{e}}_{i}]), \end{gather}

The second equality is obtained by applying $\Delta_{i}$ to both sides of the first equality and observing that

\begin{align*}\Delta_{i} E_{i}^{(4)}([{\boldsymbol{x}}, {\boldsymbol{x}} + \delta{\boldsymbol{e}}_{i}]) = E_{i}^{(4)}([{\boldsymbol{x}}+ \delta{\boldsymbol{e}}_{i}, {\boldsymbol{x}} + 2\delta {\boldsymbol{e}}_{i}]) -E_{i}^{(4)}([{\boldsymbol{x}}, {\boldsymbol{x}} + \delta{\boldsymbol{e}}_{i}]) = 2E_{i}^{(4)}([{\boldsymbol{x}}, {\boldsymbol{x}} +2\delta {\boldsymbol{e}}_{i}]).\end{align*}

Similarly, the third equality is obtained by applying $\Delta_{i}$ to both sides of the second equality. To prove (A.15), we combine (A.18) with the definition of $\widetilde \Delta_{i}^{(j)}f({\boldsymbol{x}})$ to obtain

\begin{align*}\widetilde \Delta_{i}^{(1)} f({\boldsymbol{x}}) &=\ \delta \partial_{i}f({\boldsymbol{x}}) + \frac{1}{24} \delta^{4}E_{i}^{(4)}([{\boldsymbol{x}}, {\boldsymbol{x}} + \delta{\boldsymbol{e}}_{i}]) - \frac{1}{2}\times\frac{2}{3} \delta^4E_{i}^{(4)}([{\boldsymbol{x}}, {\boldsymbol{x}} + 2 \delta{\boldsymbol{e}}_{i}])\\&\hskip4.5mm+ \frac{1}{3}\times\frac{17}{6} \delta^4E_{i}^{(4)}([{\boldsymbol{x}}, {\boldsymbol{x}} + 3\delta{\boldsymbol{e}}_{i}])\\&= \delta \partial_{i} f({\boldsymbol{x}}) +2\delta^4E_{i}^{(4)}([{\boldsymbol{x}}, {\boldsymbol{x}} + 3\delta{\boldsymbol{e}}_{i}]),\end{align*}

and

\begin{align*}\widetilde \Delta_{i}^{(2)} f({\boldsymbol{x}})=&\ \delta^2 \partial_{i}^2f({\boldsymbol{x}}) + 4\delta^4E_{i}^{(4)}([{\boldsymbol{x}},{\boldsymbol{x}} + 3\delta {\boldsymbol{e}}_{i}]),\\\widetilde \Delta_{i}^{(3)} f({\boldsymbol{x}})=&\ \delta^3 \partial_{i}^3f({\boldsymbol{x}}) + 3\delta^4E_{i}^{(4)}([{\boldsymbol{x}},{\boldsymbol{x}} + 3\delta {\boldsymbol{e}}_{i}]).\end{align*}

For completeness, we outline the main steps needed to prove (A.15) when $s<4$ . Suppose that $s = 3$ , in which case Taylor expansion yields

\begin{gather*}\Delta_{i} f({\boldsymbol{x}}) = \delta \partial_{i} f({\boldsymbol{x}}) +\frac{1}{2}\delta^2\partial_{i}^{2} f({\boldsymbol{x}}) + \frac{1}{6}\delta^{3} E_{i}^{(3)}([{\boldsymbol{x}}, {\boldsymbol{x}} + \delta{\boldsymbol{e}}_{i}]),\\\Delta_{i}^2 f({\boldsymbol{x}}) = \delta^2 \partial_{i}^2f({\boldsymbol{x}}) + \frac{4}{3} \delta^3 E_{i}^{(3)}([{\boldsymbol{x}},{\boldsymbol{x}} + 2\delta {\boldsymbol{e}}_{i}]),\\\Delta_{i}^3 f({\boldsymbol{x}}) = \frac{11}{3} \delta^3E_{i}^{(3)}([{\boldsymbol{x}}, {\boldsymbol{x}} + 3\delta{\boldsymbol{e}}_{i}]),\end{gather*}

which implies that

\begin{gather*}\widetilde \Delta_{i}^{(1)} f({\boldsymbol{x}}) = \delta \partial_{i}f({\boldsymbol{x}}) + 3\delta^3E_{i}^{(3)}([{\boldsymbol{x}},{\boldsymbol{x}} + 3\delta {\boldsymbol{e}}_{i}]),\\\widetilde \Delta_{i}^{(2)} f({\boldsymbol{x}})= \delta^2 \partial_{i}^2f({\boldsymbol{x}}) + 5 \delta^3E_{i}^{(3)}([{\boldsymbol{x}},{\boldsymbol{x}} + 3\delta {\boldsymbol{e}}_{i}]).\end{gather*}

When $s = 2$ ,

\begin{gather*}\Delta_{i} f({\boldsymbol{x}}) = \delta\partial_{i} f({\boldsymbol{x}}) +\frac{1}{2}\delta^2 E_{i}^{(2)}([{\boldsymbol{x}}, {\boldsymbol{x}} +\delta {\boldsymbol{e}}_{i}]),\\\Delta_{i}^2 f({\boldsymbol{x}}) = 2\delta^2 E_{i}^{(2)}([{\boldsymbol{x}},{\boldsymbol{x}} + 2\delta {\boldsymbol{e}}_{i}]), \quad \text{ and } \quad\Delta_{i}^3 f({\boldsymbol{x}}) = 4\delta^2 E_{i}^{(2)}([{\boldsymbol{x}},{\boldsymbol{x}} + 3\delta {\boldsymbol{e}}_{i}]),\end{gather*}

implying that

\begin{align*}\widetilde \Delta_{i}^{(1)} f({\boldsymbol{x}}) =&\ \delta \partial_{i}f({\boldsymbol{x}}) + 3\delta^3E_{i}^{(2)}([{\boldsymbol{x}},{\boldsymbol{x}} + 3\delta {\boldsymbol{e}}_{i}]).\end{align*}

Proof of Lemma A.4. Let us first consider the case when $d = 1$ . According to the original definition of A f(x) in Theorem 1 of [Reference Braverman5],

\begin{align*}A f(x) = P_{k(x)}(x), \quad x \in \mathbb{R},\end{align*}

where

(A.19) \begin{align}P_{k}(x) =&\ f(\delta k) + \Big(\frac{x-\delta k}{\delta} \Big)(\Delta - \frac{1}{2}\Delta^2 + \frac{1}{3}\Delta^3) f(\delta k) \notag\\&+ \frac{1}{2} \Big(\frac{x-\delta k}{\delta} \Big)^2\big(\Delta^2 - \Delta^3\big) f(\delta k) + \frac{1}{6} \Big(\frac{x-\delta k}{\delta} \Big)^3 \Delta^3 f(\delta k)\notag\\& -\frac{23}{3} \Big(\frac{x-\delta k}{\delta} \Big)^4 \Delta^4 f(\delta k) +\frac{41}{2} \Big(\frac{x-\delta k}{\delta} \Big)^5\Delta^4 f(\delta k) \notag\\&- \frac{55}{3} \Big(\frac{x-\delta k}{\delta} \Big)^6\Delta^4 f(\delta k) +\frac{11}{2} \Big(\frac{x-\delta k}{\delta} \Big)^7\Delta^4 f(\delta k) , \quad x \in \mathbb{R}. \end{align}

We can also write A f(x) as the weighted sum

\begin{align*}A f(x) = \sum_{i=0}^{4} \alpha^{k(x)}_{k(x)+i}(x)f(\delta(k(x)+i)),\end{align*}

where the weights $\alpha^{k}_{k+i}(x)$ are defined by the corresponding polynomial $P_{k}(x)$ . Now $A f \in C^{3}(\mathbb{R})$ by Theorem A.1, meaning that $\partial^{j} A f(\delta k(x))$ equals the corresponding right derivative of $P_{k(x)}(x)$ at $x =\delta k(x)$ for $1 \leq j \leq 3$ . The result follows immediately by inspecting the derivatives of (A.20).

When $d > 1$ , the proof is similar. Let us write ${\boldsymbol{k}}$ instead of $k({\boldsymbol{x}})$ for notational convenience. Since $Af({\boldsymbol{x}}) \in C^{3}(\mathbb{R}^{d})$ by Theorem A.1, we fix ${\boldsymbol{a}}$ with $1 \leq\Vert{\boldsymbol{a}}\Vert_{1} \leq 3$ and consider

\begin{align*}D^{{\boldsymbol{a}}} A f({\boldsymbol{x}}) =&\ D^{{\boldsymbol{a}}}\Bigg( \sum_{i_d = 0}^{4} \alpha_{k_d+i_d}^{k_d}(x_d)\cdots \sum_{i_1 = 0}^{4} \alpha_{k_1+i_1}^{k_1}(x_1) f(\delta({\boldsymbol{k}}+{\boldsymbol{i}}))\Bigg) \\=&\ \Bigg(\sum_{i_d = 0}^{4} \partial_{d}^{a_d}\alpha_{k_d+i_d}^{k_d}(x_d)\cdots \bigg(\sum_{i_2 = 0}^{4} \partial_{2}^{a_2}\alpha_{k_2+i_2}^{k_2}(x_2)\bigg(\sum_{i_1 = 0}^{4} \partial_{1}^{a_1}\alpha_{k_1+i_1}^{k_1}(x_1) f(\delta({\boldsymbol{k}}+{\boldsymbol{i}})) \bigg)\bigg)\Bigg).\end{align*}

The first equality follows from (A.1) of Theorem A.1. If we think of the innermost sum as a one-dimensional function in $x_1$ only, it follows that

\begin{align*}\sum_{i_1 = 0}^{4} \partial_{1}^{a_1}\alpha_{k_1+i_1}^{k_1}(\delta k_1) f(\delta({\boldsymbol{k}}+{\boldsymbol{i}})) = \delta^{-a_1}\widetilde \Delta^{(a_1)}_{1} f\big(\delta ({\boldsymbol{k}}+(0,i_2,\ldots,i_{d}))\big).\end{align*}

Proceeding identically along each of the remaining dimensions yields the result.

Before concluding this section, we prove Lemmas 2.4 and 2.5.

Proof of Lemma 2.4. Fix $h \in \mathcal{M}_{4}$ and note that

\begin{align*}\vert\mathbb{E} h({\boldsymbol{V}}) - \mathbb{E} h({\boldsymbol{V}}')\vert \leq&\ \vert\mathbb{E} h({\boldsymbol{V}}) - \mathbb{E} A h({\boldsymbol{V}}')\vert + \vert\mathbb{E} A h({\boldsymbol{V}}') - \mathbb{E} h({\boldsymbol{V}}')\vert,\end{align*}

where $A h(\!\cdot\!)$ is understood to be the operator A applied to the restriction of $h({\boldsymbol{x}})$ to $\delta \mathbb{Z}^{d}$ . Proposition A.1 and the fact that $h \in\mathcal{M}_{4}$ imply that

Furthermore, (A.13) implies that the restriction of $h({\boldsymbol{x}})$ to $\delta \mathbb{Z}^{d}$ belongs to $\mathcal{M}_{\mathrm{disc},4}$ and, therefore,

\begin{align*}\vert\mathbb{E} h({\boldsymbol{V}}) - \mathbb{E} A h({\boldsymbol{V}}')\vert \leq \sup_{\substack{h \in \mathcal{M}_{\mathrm{disc},4} }} \vert\mathbb{E} h({\boldsymbol{V}}) - \mathbb{E} Ah({\boldsymbol{V}}')\vert.\end{align*}

Proof of Lemma 2.5. Since $h \in \mathcal{M}_{\mathrm{disc},4}$ implies that $\vert h({\boldsymbol{u}} + \delta {\boldsymbol{e}}_{i}) - h({\boldsymbol{u}})\vert \leq \delta$ for all $1 \leq i \leq K-1$ , it follows that

\begin{align*}\vert h({\boldsymbol{u}})\vert \leq \vert h(0)\vert + \Vert{\boldsymbol{u}}\Vert_{1}, \quad {\boldsymbol{u}} \in \delta \mathbb{Z}^{K-1}.\end{align*}

Furthermore, (A.3) of Theorem A.1 and the fact that $h \in \mathcal{M}_{\mathrm{disc},4}$ imply that the first-order partial derivatives of $A h({\boldsymbol{x}})$ are bounded by $C_{A.4}(d)$ , and therefore,

where the last inequality follows from $A h(0) = h(0)$ and ${\boldsymbol{Z}} \in \bar \nabla^{K}$ . To bound $\vert\mathcal{G}_{{\boldsymbol{Z}}} A f_h({\boldsymbol{Z}})\vert$ , we recall that

\begin{align*}\mathcal{G}_{{\boldsymbol{Z}}} A f_h({\boldsymbol{x}}) = \frac{1}{2} \sum_{i=1}^{K-1} (\beta_{i} - s x_{i}) \partial_{i} A f_h({\boldsymbol{x}}) + \frac{1}{2} \sum_{i,j = 1}^{K-1} x_{i}(\delta_{ij} - x_{j}) \partial_{i}\partial_{j} A f_h({\boldsymbol{x}}), \quad {\boldsymbol{x}} \in \bar \nabla^{K}.\end{align*}

Note that $\vert\beta_{i} - s x_{i}\vert \leq s$ , $\vert x_{i}(\delta_{ij}-x_{j})\vert \leq 1$ , and that (A.3) of Theorem A.1 implies that

Recalling from (B.7) that $\Vert f_h\Vert \leq \delta^{-1} K B_{1}(f_h)$ , we combine the bounds to arrive at

where in the last inequality we used $\delta \leq 1$ .

Appendix A. Explicit constants in Theorem A.1

Proof of (A.4) and (A.8). We show that

which implies the result. Consider first the case when $d = 1$ ; we want a bound on A f(x) and its first four derivatives. For any $f\,:\,\mathbb{Z}\to \mathbb{R}$ , (A.1) of [Reference Braverman5] states that

\begin{align*}A f(x) = \sum_{i=0}^{4} \alpha^{k(x)}_{k(x)+i}(x)f(\delta(k(x)+i)) = P_{k(x)}(x),\end{align*}

where

(A.20) \begin{align}P_{k}(x) =&\ f(\delta k) + \Big(\frac{x-\delta k}{\delta} \Big)(\Delta - \frac{1}{2}\Delta^2 + \frac{1}{3}\Delta^3) f(\delta k) \notag\\&+ \frac{1}{2} \Big(\frac{x-\delta k}{\delta} \Big)^2\big(\Delta^2 - \Delta^3\big) f(\delta k) + \frac{1}{6} \Big(\frac{x-\delta k}{\delta} \Big)^3 \Delta^3 f(\delta k)\notag\\& -\frac{23}{3} \Big(\frac{x-\delta k}{\delta} \Big)^4 \Delta^4 f(\delta k) +\frac{41}{2} \Big(\frac{x-\delta k}{\delta} \Big)^5\Delta^4 f(\delta k) \notag\\&- \frac{55}{3} \Big(\frac{x-\delta k}{\delta} \Big)^6\Delta^4 f(\delta k) +\frac{11}{2} \Big(\frac{x-\delta k}{\delta} \Big)^7\Delta^4 f(\delta k) , \quad x \in \mathbb{R}. \end{align}

To bound $\vert A f(x)\vert$ , we need to bound each of the terms on the right-hand side of (A.20). For instance,

\begin{align*}(\Delta - \frac{1}{2}\Delta^2 + \frac{1}{3}\Delta^3) f(\delta k) =& - f(\delta k) + f(\delta (k+1))\\&- \frac{1}{2}\big( f(\delta k) - 2 f(\delta (k+1)) + f(\delta (k+2))\big)\\&+ \frac{1}{3}\big( \!-f(\delta k) +3 f(\delta (k+1)) -3 f(\delta (k+2)) + f(\delta (k+3))\big),\end{align*}

implying that

\begin{align*}\vert(\Delta - \frac{1}{2}\Delta^2 + \frac{1}{3}\Delta^3) f(\delta k)\vert \leq 3 \times 5 \max_{0 \leq i \leq 4 } \vert f(k(x)+i)\vert.\end{align*}

Proceeding similarly with each of the terms on the right-hand side of (A.20), we see that $\vert A f(x)\vert$ is bounded by

\begin{align*}& \Big(1 + 3\Big(\frac{x-\delta k(x)}{\delta}\Big) + \frac{5}{2} \Big(\frac{x-\delta k(x)}{\delta}\Big)^2 + \frac{1}{2} \Big(\frac{x-\delta k(x)}{\delta}\Big)^3 + \frac{23}{3} \times 6 \Big(\frac{x-\delta k(x)}{\delta}\Big)^4\\&+ \frac{41}{2} \times 6 \Big(\frac{x-\delta k(x)}{\delta}\Big)^5 + \frac{55}{3} \times 6\Big(\frac{x-\delta (x)}{\delta}\Big)^6 + \frac{11}{2} \times 6 \Big(\frac{x-\delta k(x)}{\delta}\Big)^7 \Big) \max_{0 \leq i \leq 4 } \vert f(k(x)+i)\vert\\&\qquad\leq 319 \times 5 \max_{0 \leq i \leq 4 } \vert f(k(x)+i)\vert.\end{align*}

Note that (A.8) follows from a similar argument because $\alpha_{k+i}^{k}(x)$ is the coefficient corresponding to $f(\delta(k+i))$ . To bound $\vert \partial A f(x)\vert$ , we differentiate (A.20) once to get

\begin{align*}\partial P_{k}(x) =&\ \frac{1}{\delta} \bigg( (1 - \frac{1}{2}\Delta + \frac{1}{3}\Delta^2) \Delta f(\delta k) \notag\\&\hskip5.5mm+ \frac{1}{2} \times 2 \Big(\frac{x-\delta k}{\delta} \Big)\big(\Delta - \Delta^2\big) ( \Delta f(\delta k)) + \frac{1}{6} \times 3 \Big(\frac{x-\delta k}{\delta} \Big)^2 \Delta^2 ( \Delta f(\delta k))\notag\\&\hskip5.5mm -\frac{23}{3} \times 4 \Big(\frac{x-\delta k}{\delta} \Big)^3 \Delta^3 ( \Delta f(\delta k)) +\frac{41}{2} \times 5 \Big(\frac{x-\delta k}{\delta} \Big)^4\Delta^3 ( \Delta f(\delta k)) \notag\\&\hskip5.5mm- \frac{55}{3} \times 6 \Big(\frac{x-\delta k}{\delta} \Big)^5\Delta^3 ( \Delta f(\delta k)) +\frac{11}{2} \times 7 \Big(\frac{x-\delta k}{\delta} \Big)^6\Delta^3 ( \Delta f(\delta k)) \bigg) , \quad x \in \mathbb{R}.\end{align*}

Bounding the right-hand side the same way as before yields

\begin{align*}\vert \partial A f(x)\vert \leq&\ \frac{1}{\delta} \Big( (1+1/2+2/3) + (1+2) \Big(\frac{x-\delta k(x)}{\delta}\Big) + \frac{3}{6}\times 2 \Big(\frac{x-\delta k(x)}{\delta}\Big)^2\\&+ \frac{23}{3} \times 4 \times 3 \Big(\frac{x-\delta k(x)}{\delta}\Big)^3 + \frac{41}{2}\times 5 \times 3 \Big(\frac{x-\delta k(x)}{\delta}\Big)^4\\&+ \frac{55}{3}\times 6 \times 3\Big(\frac{x-\delta (x)}{\delta}\Big)^5 + \frac{11}{2} \times 7 \times 3\Big(\frac{x-\delta k(x)}{\delta}\Big)^6 \Big) \times \max_{0 \leq i \leq 3} \vert\Delta f(k(x)+i)\vert\\\leq&\ \frac{1}{\delta} 852 \times 4 \max_{0 \leq i \leq 3} \vert\Delta f(k(x)+i)\vert.\end{align*}

Similarly,

\begin{align*}\partial^{2} P_{k}(x) =&\ \frac{1}{\delta^2} \bigg( \big(1 - \Delta\big) ( \Delta^2 f(\delta k)) + \Big(\frac{x-\delta k}{\delta} \Big) \Delta ( \Delta^2 f(\delta k))\notag\\& -23 \times 4 \Big(\frac{x-\delta k}{\delta} \Big)^2 \Delta^2 ( \Delta^2 f(\delta k)) +81\times 5 \Big(\frac{x-\delta k}{\delta} \Big)^3\Delta^2 ( \Delta^2 f(\delta k)) \notag\\&- \frac{55}{3} \times 6 \times 5 \Big(\frac{x-\delta k}{\delta} \Big)^4\Delta^2 ( \Delta^2 f(\delta k)) \\&+\frac{11}{2} \times 7 \times 6 \Big(\frac{x-\delta k}{\delta} \Big)^5\Delta^2 ( \Delta^2 f(\delta k)) \bigg) , \quad x \in \mathbb{R},\end{align*}

\begin{align*}\partial^{3} P_{k}(x) =&\ \frac{1}{\delta^3} \bigg( \Delta^3 f(\delta k) -23\! \times 4 \times 2 \Big(\frac{x-\delta k}{\delta} \Big) \Delta ( \Delta^3 f(\delta k)) +81\!\times 5\times 3 \Big(\frac{x-\delta k}{\delta} \Big)^2\Delta ( \Delta^3 f(\delta k)) \notag\\&- \frac{55}{3} \times 6 \times 5 \times 4 \Big(\frac{x-\delta k}{\delta} \Big)^3\Delta ( \Delta^3 f(\delta k)) \\&+\frac{11}{2} \times 7 \times 6 \times 5 \Big(\frac{x-\delta k}{\delta} \Big)^4\Delta ( \Delta^3 f(\delta k)) \bigg) , \quad x \in \mathbb{R},\end{align*}

and

\begin{align*}\partial^{4} P_{k}(x) =&\ \frac{1}{\delta^4} \bigg( -23 \times 4 \times 2 \Delta^4 f(\delta k) +81\times 5\times 3 \times 2 \Big(\frac{x-\delta k}{\delta} \Big) \Delta^4 f(\delta k) \notag\\&- \frac{55}{3} \times 6 \times 5 \times 4 \times 3 \Big(\frac{x-\delta k}{\delta} \Big)^2 \Delta^4 f(\delta k) \\&+\frac{11}{2} \times 7 \times 6 \times 5 \times 4 \Big(\frac{x-\delta k}{\delta} \Big)^4 \Delta^4 f(\delta k) \bigg) , \quad x \in \mathbb{R},\end{align*}

yield the bounds

\begin{gather*}\vert \partial^2 A f(x)\vert \leq \frac{1}{\delta^2} 2559 \times 3 \max_{0 \leq i \leq 2} \vert\Delta^2 f(k(x)+i)\vert,\\\vert \partial^3 A f(x)\vert \leq \frac{1}{\delta^3} 4775 \times 2 \max_{0 \leq i \leq 1} \vert\Delta^3 f(k(x)+i)\vert,\\\vert \partial^4 A f(x)\vert \leq \frac{1}{\delta^4} 11404 \vert\Delta^4 f(\delta k(x))\vert.\end{gather*}

When $d > 1$ , we recall from (A.1) that

\begin{align*}A f({\boldsymbol{x}}) =&\ \sum_{i_d = 0}^{4} \alpha_{k_d({\boldsymbol{x}})+i_d}^{k_d({\boldsymbol{x}})}(x_d)\cdots \sum_{i_1 = 0}^{4} \alpha_{k_1({\boldsymbol{x}})+i_1}^{k_1({\boldsymbol{x}})}(x_1) f(\delta(k({\boldsymbol{x}})+{\boldsymbol{i}})).\end{align*}

The result follows by applying the bounds from the $d=1$ case along each dimension.

Appendix B. Generator expansion and Stein factors

This appendix is dedicated to proving Lemmas 2.2 and 2.3. Recall that, for ${\boldsymbol{x}} \in\mathbb{R}^{K-1}$ , we define $k({\boldsymbol{x}})$ by $k_{j}({\boldsymbol{x}}) = \lfloor x_j/\delta\rfloor$ , that $\delta = 1/N$ , $\nabla^{K}_{N}$ is defined in (2.7), and that $\bar\nabla^{K}_{N} = \text{Conv}(\nabla^{K}_{N})$ . We may assume without loss of generality that $100K^2 < N$ , which implies that $\nabla^{K}_{N} \neq\emptyset$ , because the claim in Lemma 2.2 holds trivially for all $0 < N \leq 100K^{2}$ (since this covers a finite number of N). We claim that

(B.1) \begin{align}\text{ if ${\boldsymbol{x}} \in \bar \nabla^{K}_{N}$ then $\delta(k({\boldsymbol{x}}) + {\boldsymbol{i}}) \in \nabla^{K}$ for all $0 \leq{\boldsymbol{i}} \leq 10{\boldsymbol{e}}$}. \end{align}

We argue this as follows. Since $0 \leq \delta k({\boldsymbol{x}}) \leq {\boldsymbol{x}}$ , then $\delta k({\boldsymbol{x}}) \in \nabla^{K}_{N}$ for any ${\boldsymbol{x}} \in \bar \nabla^{K}_{N}$ , because $ \delta \sum_{j=1}^{K-1} k_j({\boldsymbol{x}}) \leq \sum_{j=1}^{K-1} x_j$ . Thus, for all $0 \leq i \leq 10{\boldsymbol{e}}$ ,

\begin{align*}\delta \sum_{j=1}^{K-1} (k_j({\boldsymbol{x}}) + i_j) \leq 1 - 10K/\sqrt{N} + 10K/N < 1,\end{align*}

implying (B.1). Combining (B.1) with (A.1) of Theorem A.1 implies that $A (\mathcal{G}_{{\boldsymbol{U}}} A f_h)({\boldsymbol{x}})$ is well defined if ${\boldsymbol{x}} \in \bar \nabla^{K}_{N}$ . To derive an expression for $A(\mathcal{G}_{{\boldsymbol{U}}} A f_h)({\boldsymbol{x}})$ , we define

\begin{gather*}b_{i}({\boldsymbol{u}}) = \mathbb{E}_{u}(U_i(1) - u_i),\\a_{ij}({\boldsymbol{u}}) = \mathbb{E}_{u}(U_i(1) - u_i)(U_j(1) - u_j),\\c_{ijk}({\boldsymbol{u}}) = \mathbb{E}_{u}(U_i(1) - u_i)(U_j(1) - u_j)(U_{k}(1)-u_{k}),\\\bar d_{ijk\ell}({\boldsymbol{u}}) = \mathbb{E}_{u} \vert(U_i(1) -u_i)(U_j(1) - u_j)(U_{k}(1)-u_{k})(U_{\ell}(1)-u_{\ell})\vert, \quad{\boldsymbol{u}} \in \nabla^{K}.\end{gather*}

Since $A f_h \in C^{3}(\mathbb{R}^{K-1})$ , by Theorem A.1, we know that, for any ${\boldsymbol{u}}\in \nabla^{K}$ ,

\begin{align*}\mathcal{G}_{{\boldsymbol{U}}} (A f_h)({\boldsymbol{u}}) &= \mathbb{E}_{{\boldsymbol{u}}} A f_h({\boldsymbol{U}}(1)) - A f_h({\boldsymbol{u}})\\&= \sum_{i=1}^{K-1} b_{i}({\boldsymbol{u}}) \partial_{i} A f_h({\boldsymbol{u}}) + \frac{1}{2}\sum_{i,j=1}^{K-1} a_{ij}({\boldsymbol{u}}) \partial_{i}\partial_{j} A f_h({\boldsymbol{u}})\\& \hskip4mm + \frac{1}{6} \sum_{i,j,k=1}^{K-1} c_{ijk}({\boldsymbol{u}}) \partial_{i}\partial_{j}\partial_{k} A f_h({\boldsymbol{u}}) + \epsilon({\boldsymbol{u}}),\end{align*}

where

\begin{align*}\epsilon({\boldsymbol{u}}) =&\ \frac{1}{6} \sum_{i,j,k=1}^{K-1}\mathbb{E}_{{\boldsymbol{u}}} \Big( (U_i(1) - u_i)(U_j(1) -u_j)(U_{k}(1)-u_{k}) \big(\partial_{i}\partial_{j}\partial_{k}Af_h(\boldsymbol{\xi}) - \partial_{i}\partial_{j}\partial_{k}Af_h({\boldsymbol{u}})\big)\Big),\end{align*}

and $\boldsymbol{\xi}$ is between ${\boldsymbol{u}}$ and ${\boldsymbol{U}}(1)$ . Since A is a linear operator, it follows that, for any ${\boldsymbol{x}} \in \bar \nabla^{K}_{N}$ ,

(B.2) \begin{align}A (\mathcal{G}_{{\boldsymbol{U}}} A f_h)({\boldsymbol{x}}) &= \sum_{i=1}^{K-1} A( b_{i}\partial_{i} A f_h)({\boldsymbol{x}}) + \frac{1}{2}\sum_{i,j =1}^{K-1} A ( a_{ij}\partial_{i}\partial_{j} A f_h)({\boldsymbol{x}}) \notag\\&\hskip4mm + \frac{1}{6} \sum_{i,j,k =1}^{K-1} A ( c_{ijk}\partial_{i}\partial_{j}\partial_{k} A f_h)({\boldsymbol{x}}) + A \epsilon({\boldsymbol{x}}). \end{align}

We present several lemmas needed to analyze the right-hand side. Recall assumption (1.3) or that $p_i = \beta_i/(2N) = \delta\beta_{i}/2$ and, subsequently, $\Sigma = \delta s/2$ . Also recall the definition of $B_i(\!\cdot\!)$ from (2.1).

Lemma B.1. Fix ${\boldsymbol{u}} \in \nabla^{K}$ and let $\bar u_{i} = u_{i}\Sigma - p_{i}$ . Then for $1 \leq i,j \leq K-1$ ,

\begin{gather*}b_{i}({\boldsymbol{u}}) = - \bar u_{i} \quad \text{ and } \quad a_{ij}({\boldsymbol{u}}) = \bar u_{i}\bar u_{j} + \frac{1}{N}\big(u_{i} - \bar u_{i} \big) \big( \delta_{ij} - ( u_{j} - \bar u_{j})\big),\\B_{1}(b_{i}) \leq \delta^2 s \quad \text{ and } \quad B_{1}(a_{ij}) \leq \delta^2 5(1+s)^2.\end{gather*}

Furthermore, for any $1 \leq i,j,k,\ell \leq K-1$ ,

\begin{align*}&\vert c_{ijk}({\boldsymbol{u}})\vert \leq 20\delta^2(1+s)^3 \quad \text{ and } \quad \bar d_{ijk\ell}({\boldsymbol{u}}) \leq \delta^2(2+s)^4, \quad {\boldsymbol{u}} \in \nabla^{K}.\end{align*}

The proof of Lemma B.1 is left to later in this appendix as it essentially has many elementary but tedious moment calculations. Since $b_{i}({\boldsymbol{u}})$ and $a_{ij}({\boldsymbol{u}})$ are polynomials in ${\boldsymbol{u}}$ , we let $b_{i}({\boldsymbol{x}})$ and $a_{ij}({\boldsymbol{x}})$ be their natural extensions to $\bar\nabla^{K}$ . Furthermore, Corollary A.1 implies that $Ab_{i}({\boldsymbol{x}}) = b_{i}({\boldsymbol{x}})$ and $Aa_{ij}({\boldsymbol{x}}) = a_{ij}({\boldsymbol{x}})$ .

Lemma B.2. For any ${\boldsymbol{x}} \in \bar \nabla^{K}_{N}$ ,

\begin{gather*}A( b_{i}\partial_{i} A f_h)({\boldsymbol{x}}) = b_i({\boldsymbol{x}}) \partial_{i} A f_{h}({\boldsymbol{x}}) + \tilde \epsilon_{i}({\boldsymbol{x}}), \\A( a_{ij}\partial_{i} A f_h)({\boldsymbol{x}}) = a_{ij}({\boldsymbol{x}})\partial_{i}\partial_{j} A f_{h}({\boldsymbol{x}}) + \tilde \epsilon_{ij}({\boldsymbol{x}}),\\A ( c_{ijk}\partial_{i}\partial_{j}\partial_{k} A f_h)({\boldsymbol{x}}) = \tilde \epsilon_{ijk}({\boldsymbol{x}}),\end{gather*}

where

(B.3)

(B.4)

(B.5)

Furthermore,

(B.6)

Proof of Lemma 2.2. Assumption (1.3) states that $p_{i} = \beta_{i}/(2N)$ , which implies that $\Sigma = s/(2N)$ . Thus, $\mathcal{G}_{{\boldsymbol{Z}}}f({\boldsymbol{x}})$ , which is defined in (1.8), satisfies

\begin{align*}\delta\mathcal{G}_{{\boldsymbol{Z}}} f({\boldsymbol{x}}) =&\ \sum_{i=1}^{K-1} (p_{i} - \Sigma x_{i}) \partial_{i} f({\boldsymbol{x}}) + \frac{1}{2N} \sum_{i,j = 1}^{K-1} x_{i}(\delta_{ij} - x_{j}) \partial_{i}\partial_{j} f({\boldsymbol{x}})\\=&\ \sum_{i=1}^{K-1} b_{i}({\boldsymbol{x}}) \partial_{i} f({\boldsymbol{x}}) + \frac{1}{2N} \sum_{i,j = 1}^{K-1} x_{i}(\delta_{ij} - x_{j}) \partial_{i}\partial_{j} f({\boldsymbol{x}}), \quad {\boldsymbol{x}} \in \bar \nabla^{K},\end{align*}

where the second equality is due to Lemma B.1. Combining Lemma B.2 with (B.2) yields

\begin{align*}A (\mathcal{G}_{U} A f_h)({\boldsymbol{x}}) &= \sum_{i=1}^{K-1} b_{i}({\boldsymbol{x}})\partial_{i} A f_h ({\boldsymbol{x}}) + \frac{1}{2}\sum_{i,j =1}^{K-1} a_{ij}({\boldsymbol{x}})\partial_{i}\partial_{j} A f_h ({\boldsymbol{x}})\\&\hskip4mm+ \sum_{i=1}^{K-1} \tilde \epsilon_{i}({\boldsymbol{x}}) + \frac{1}{2}\sum_{i,j =1}^{K-1} \tilde \epsilon_{ij}({\boldsymbol{x}}) + \frac{1}{6} \sum_{i,j,k =1}^{K-1} \tilde \epsilon_{ijk}({\boldsymbol{x}}) + A \epsilon({\boldsymbol{x}}).\end{align*}

Let us compare the first two terms on the right-hand side to $\delta \mathcal{G}_{{\boldsymbol{Z}}} A f_h({\boldsymbol{x}})$ . Set $\bar x_{i} = x_{i} \Sigma - p_{i}$ and observe by Lemma B.1 that

\begin{align*}a_{ij}({\boldsymbol{x}}) =&\ \bar x_{i}\bar x_{j} + \frac{1}{N}\big(x_{i} - \bar x_{i} \big) \big( \delta_{ij} - ( x_{j} - \bar x_{j})\big)\\=&\ \frac{1}{N} x_{i} (\delta_{ij} - x_{j}) - \frac{1}{N}\Big( \bar x_{i}(\delta_{ij} - x_{j}) - x_{i} \bar x_{j} - \bar x_{i} \bar x_{j} \Big) + \bar x_{i}\bar x_{j}.\end{align*}

Since $\delta = 1/N$ , $\vert\bar x_{i}\vert \leq \Sigma \leq \delta s$ , and $\vert x_i\vert \leq 1$ , it follows that

Combining this bound with the bounds on $\vert\tilde \epsilon_{i}({\boldsymbol{x}})\vert, \ldots, \vert A \epsilon({\boldsymbol{x}})\vert$ from Lemma B.2 yields

We first state and prove an auxiliary lemma, followed by the proof of Lemma B.2.

Lemma B.3. For any ${\boldsymbol{u}} \in \nabla^{K}$ and any integer $M \geq0$ ,

\begin{align*}\mathbb{P}_{{\boldsymbol{u}}}(U_{K}(1) \leq M/N) \leq \frac{(M+1)N^{M}}{( \Sigma - p_{K})^{M}} ( 1- u_{K}(1-\Sigma))^{N}.\end{align*}

Proof of Lemma B.3. Recall that from (1.1), $N U_{K}(1)|u_{K} \sim{\rm Binomial}(N, u_{K} - \bar u_{K})$ , where $\bar u_{K} = -\Sigma u_{K} + p_{K}$ . For any $0 \leq j \leq M$ ,

\begin{align*}\mathbb{P}_{{\boldsymbol{u}}}(U_{K}(1) = j/N) =&\ \frac{N!}{j!(N-j)!} (u_{K}(1-\Sigma) + p_{K})^{j} ( 1- u_{K}(1-\Sigma) - p_{K})^{N-j}\\\leq&\ N^{j} ( 1- u_{K}(1-\Sigma) - p_{K})^{N} \frac{1}{( 1- u_{K}(1-\Sigma) - p_{K})^{j}}\\\leq&\ N^{M} ( 1- u_{K}(1-\Sigma))^{N} \frac{1}{( \Sigma - p_{K})^{j}}\\\leq&\ \frac{N^{M}}{( \Sigma - p_{K})^{M}} ( 1- u_{K}(1-\Sigma))^{N},\end{align*}

where the second inequality follows from $1- u_{K}(1-\Sigma) - p_{K} \geq \Sigma - p_{K} $ since $u_{K} \leq 1$ . Thus,

\begin{align*}\mathbb{P}_{{\boldsymbol{u}}}(U_{K}(1) \leq M/N) = \sum_{j=0}^{M} \mathbb{P}_{{\boldsymbol{u}}}(U_{K}(1) = j/N) \leq \frac{(M+1)N^{M}}{( \Sigma - p_{K})^{M}} ( 1- u_{K}(1-\Sigma) )^{N}.\end{align*}

Proof of Lemma B.2. Let us prove (B.3). Note that $A b_i({\boldsymbol{x}}) = b_i({\boldsymbol{x}})$ by Corollary A.1. Lemma A.2 then gives us

\begin{align*}A(b_{i}\partial_{i} A f_h)({\boldsymbol{x}}) =&\ A b_{i}(x) A (\partial_{i} A f_{h}) ({\boldsymbol{x}}) + \epsilon_{i}({\boldsymbol{x}})\\=&\ b_{i}({\boldsymbol{x}}) \partial_{i} A f_{h} ({\boldsymbol{x}}) + \epsilon_{i}({\boldsymbol{x}}) + b_{i}({\boldsymbol{x}}) \big(A (\partial_{i} A f_{h}) ({\boldsymbol{x}}) - \partial_{i} A f_{h}({\boldsymbol{x}}) \big),\end{align*}

where

The constant $C_{A.10}(K)$ is as in Lemma A.2 and the second inequality is due to the bound on $B_{1}(b_i)$ from Lemma B.1. Now

where the second-last inequality is due to (A.3) of Theorem A.1. Thus,

To complete the proof of (B.3), note that

The first inequality follows from $\vert b_i\vert({\boldsymbol{u}}) \leq\Sigma \leq \delta s$ ; see Lemma B.1. The second inequality follows from Proposition A.1 with $f({\boldsymbol{x}}) =\partial_{i} A f_h({\boldsymbol{x}})$ and $s = 3$ there, and the third inequality follows from (A.3) of Theorem A.1. The proof of (B.4) is similar to that of (B.3). Lemma A.2 and the fact that $Aa_{ij}({\boldsymbol{x}}) = a_{ij}({\boldsymbol{x}})$ imply that

\begin{align*}A(a_{ij}\partial_{i}\partial_{j} A f_h)({\boldsymbol{x}}) =&\ A a_{ij}({\boldsymbol{x}}) A (\partial_{i}\partial_{j} A f_{h}) (x) + \epsilon_{ij}({\boldsymbol{x}})\\=&\ a_{ij}(x) \partial_{i}\partial_{j} A f_{h} ({\boldsymbol{x}}) + \epsilon_{ij}({\boldsymbol{x}}) + a_{ij}({\boldsymbol{x}}) \big(A (\partial_{i}\partial_{j} A f_{h}) ({\boldsymbol{x}}) - \partial_{i}\partial_{j} A f_{h}({\boldsymbol{x}}) \big),\end{align*}

where

We know that $B_{1}(a_{ij}) \leq \delta^2 5(1+s)^2$ by Lemma B.1, so that repeating the arguments used to bound $\epsilon_{i}({\boldsymbol{x}})$ yields

Furthermore, since $\vert a_{ij}({\boldsymbol{x}})\vert \leq \Sigma^{2} + \delta \leq \delta(\delta s^2 + 1)$ ,

which proves (B.4). Let us prove (B.5). Note that

where the first inequality is due to (A.8) and Lemma B.1, and the second inequality is due to (A.3) of Theorem A.1, and the final inequality is due to (B.1). Lastly, to prove (B.6), we bound $\vert A \epsilon({\boldsymbol{x}})\vert$ by bounding $\epsilon({\boldsymbol{u}}), {\boldsymbol{u}} \in \nabla^{K}$ and using (A.8).

From (A.1) of Theorem A.1, we see that $A \epsilon({\boldsymbol{x}})$ depends only on $\epsilon(\delta (k({\boldsymbol{x}})+{\boldsymbol{i}}))$ for $0 \leq {\boldsymbol{i}} \leq 4{\boldsymbol{e}}$ , so going forward we fix ${\boldsymbol{u}} = \delta (k({\boldsymbol{x}})+{\boldsymbol{i}})$ . Recall that

\begin{align*}\epsilon({\boldsymbol{u}}) =&\ \frac{1}{6}\sum_{i,j,k=1}^{K-1}\mathbb{E}_{{\boldsymbol{u}}} \Big( (U_i(1) -u_i)(U_j(1) - u_j)(U_{k}(1)-u_{k})\big(\partial_{i}\partial_{j}\partial_{k}A f_h(\boldsymbol{\xi}) -\partial_{i}\partial_{j}\partial_{k}A f_h({\boldsymbol{u}})\big)\Big),\end{align*}

where $\boldsymbol{\xi}$ is between ${\boldsymbol{u}}$ and ${\boldsymbol{U}}(1)$ . We know that $\partial_{i}\partial_{j}\partial_{k}A f_h \in \mathcal{D}^{K-1}$ by Lemma A.1, implying that

\begin{align*}\vert\partial_{i}\partial_{j}\partial_{k}A f_h(\boldsymbol{\xi}) - \partial_{i}\partial_{j}\partial_{k}A f_h({\boldsymbol{u}})\vert \leq&\ \Vert{\boldsymbol{U}}(1)-{\boldsymbol{u}}\Vert_{1} \sup_{\substack{ \min({\boldsymbol{u}},{\boldsymbol{U}}(1)) \leq {\boldsymbol{z}} \\ {\boldsymbol{z}} \leq \max({\boldsymbol{u}},{\boldsymbol{U}}(1)) \\ \Vert{\boldsymbol{a}}\Vert_{1} = 4}} \vert D^{{\boldsymbol{a}}} Af_h({\boldsymbol{z}})\vert.\end{align*}

Consider separately the cases when $U_{K}(1) > 4K/N$ and $U_{K}(1) \leq4K/N$ . If $U_{K}(1) > 4K/N$ then

For the last inequality, we used the facts that $U(1) + \delta {\boldsymbol{a}} \in \nabla^{K}$ and ${\boldsymbol{u}} + \delta {\boldsymbol{a}} \in \nabla^{K}$ for $\Vert a\Vert_{1} \leq 4$ , because $U_{K}(1) > 4K/N$ and ${\boldsymbol{x}} \in \bar \nabla^{K}_{N}$ , respectively. If $U_{K}(1) \leq 4K/N$ , we use the coarse bound of

The last inequality is due to the fact that

(B.7) \begin{align}\Vert f_h\Vert \leq K N B_1(f_h) = K \delta^{-1} B_1(f_h), \end{align}

which we verify at the end of the proof. Combining the bounds on $\vert D^{{\boldsymbol{a}}} Af_h({\boldsymbol{z}})\vert$ with the definition of $\epsilon({\boldsymbol{u}})$ yields

In the second-last inequality we used the coarse bound of $\vert U_i(1) -u_i\vert \leq 1$ to bound the second term, and in the last inequality we used the bound on $\bar d_{ijk\ell}({\boldsymbol{u}})$ from Lemma B.1. Furthermore, ${\boldsymbol{x}} \in \bar\nabla^{K}_{N}$ implies that, for any $0 \leq {\boldsymbol{i}} \leq4{\boldsymbol{e}}$ ,

\begin{align*}1 - \sum_{j=1}^{K-1}\delta(k_j({\boldsymbol{x}})+i_j) \geq 1 - \sum_{j=1}^{K-1}\delta k_j({\boldsymbol{x}}) - 4K/N \geq 10K\sqrt{N}/N - 4K/N \geq 10K(\sqrt{N}-1)/N,\end{align*}

and, therefore, $( 1- u_{K}(1-\Sigma))^{N}$ on the right-hand side is bounded by $( 1- (10K(\sqrt{N}-1)/N) (1-\Sigma))^{N}$ .

Finally, (A.8) yields the upper bound of

\begin{align*}\vert A \epsilon({\boldsymbol{x}})\vert =&\ \bigg| \sum_{i_1, \ldots, i_{K-1} = 0}^{4} \bigg(\prod_{j=1}^{d} \alpha_{k_j({\boldsymbol{x}}) +i_j}^{k_j({\boldsymbol{x}}) }(x_j)\bigg) \epsilon(\delta (k({\boldsymbol{x}}) + {\boldsymbol{i}})) \bigg| \leq 5^{K}319^{K} \max_{0 \leq {\boldsymbol{i}} \leq 4{\boldsymbol{e}}} \vert\epsilon(\delta (k({\boldsymbol{x}}) + {\boldsymbol{i}}))\vert,\end{align*}

which proves (B.6). It remains to prove (B.7). Since we chose $f_h({\boldsymbol{u}}) = 0$ for ${\boldsymbol{u}}$ outside of $\nabla^{K}$ , we need to show that

\begin{align*}\vert f_h({\boldsymbol{u}})\vert \leq N K B_{1}(f_h), \quad{\boldsymbol{u}} \in \nabla^{K}.\end{align*}

Letting $\pi({\boldsymbol{u}}) = \mathbb{P}({\boldsymbol{U}} = {\boldsymbol{u}})$ , we have

\begin{align*}f_h({\boldsymbol{u}}) = \sum_{n=0}^{\infty} \big( \mathbb{E}_{u} h({\boldsymbol{U}}(n)) - \mathbb{E} h({\boldsymbol{U}}) \big) =&\ \sum_{n=0}^{\infty} \sum_{{\boldsymbol{u}}' \in \nabla^{K}} \pi({\boldsymbol{u}}') \big( \mathbb{E}_{{\boldsymbol{u}}} h({\boldsymbol{U}}(n)) - \mathbb{E}_{{\boldsymbol{u}}'} h({\boldsymbol{U}}(n)) \big)\\=&\ \sum_{{\boldsymbol{u}}' \in \nabla^{K}} \sum_{n=0}^{\infty}\pi({\boldsymbol{u}}') \big( \mathbb{E}_{{\boldsymbol{u}}} h({\boldsymbol{U}}(n)) - \mathbb{E}_{{\boldsymbol{u}}'} h({\boldsymbol{U}}(n)) \big),\end{align*}

where the interchange is justified by the Fubini–Tonelli theorem because $\{{\boldsymbol{U}}(n)\}$ is geometrically ergodic. Since

\begin{align*}\Delta_{i} f_h({\boldsymbol{u}}) = \sum_{n=0}^{\infty} \big( \mathbb{E}_{{\boldsymbol{u}} + \delta {\boldsymbol{e}}_{i}} h(U(n)) - \mathbb{E}_{{\boldsymbol{u}}} h({\boldsymbol{U}}(n)) \big),\end{align*}

it follows that

\begin{align*}\sum_{n=0}^{\infty} \vert\mathbb{E}_{{\boldsymbol{u}}} h({\boldsymbol{U}}(n)) - \mathbb{E}_{{\boldsymbol{u}}'} h({\boldsymbol{U}}(n)) \vert \leq \Vert({\boldsymbol{u}} - {\boldsymbol{u}}')/\delta\Vert_{1} B_{1}(f_h) \leq N K B_{1}(f_h),\end{align*}

proving (B.7).

We now prove Lemma B.1. Recall that

\begin{align*}( (NU_{1}(1),\ldots, NU_{K}(1)) | U(0) = {\boldsymbol{u}} ) \sim \text{Multinomial}(N, \big(u_{1} - \bar u_{1}, \ldots,u_{K} - \bar u_{K}) ),\end{align*}

where $\bar u_{i} = u_{i}\Sigma - p_{i}$ . We also require the following result about the moments of the multinomial distribution.

Lemma B.4. Let $X \sim {\rm Multinomial}(N, (p_1, \ldots, p_k))$ . Then for all $i \neq j \neq k$ ,

\begin{gather*}\mathbb{E}(X_i) = N p_i, \qquad\mathbb{E}(X_i^2) = N(N-1)p_i^2 + Np_i, \qquad\mathbb{E}(X_i X_j) = N(N-1)p_ip_j,\\\mathbb{E}(X_iX_jX_k) = N(N-1)(N-2)p_ip_jp_k,\\\mathbb{E}(X_i^2X_j) = N(N-1)(N-2) p_i^2p_j + N(N-1)p_ip_j,\\\mathbb{E}(X_i^3) = N(N-1)(N-2)p_i^3 + 3N(N-1)p_i^2 - 2Np_i.\end{gather*}

Proof of Lemma B.1. For convenience, we write $\mathbb{E}(\!\cdot\!)$ instead of $\mathbb{E}_{u}(\!\cdot\!)$ and $U_{i}$ instead of $U_{i}(1)$ ; e.g. we write $\mathbb{E} U_{i}$ instead of $\mathbb{E}_{u} U_{i}(1)$ . Since $\mathbb{E} N U_i = N(u_{i} - \bar u_{i})$ , it follows from Lemma B.4 that

\begin{align*}b_{i}({\boldsymbol{u}}) = \mathbb{E}(U_i - u_i) = -\bar u_{i}.\end{align*}

Next, we observe that

\begin{align*}a_{ij}({\boldsymbol{u}}) = \mathbb{E}(U_i - u_i)(U_j - u_j) =&\ \mathbb{E}( U_{i} U_{j}) - u_{i} \mathbb{E} U_{j}- u_{j} \mathbb{E} U_{i} + u_{i} u_{j}.\end{align*}

Lemma B.4 implies that $\mathbb{E} U_{j} = (u_{j} - \bar u_{j})$ and

\begin{align*}\mathbb{E} ( U_{i} U_{j}) =&\ (1-1/N)(u_i - \bar u_i) (u_j - \bar u_{j})\\=&\ u_i u_j - u_i \bar u_j - u_j \bar u_i + \bar u_i \bar u_j- \frac{1}{N}(u_i - \bar u_i) (u_j - \bar u_{j}),\end{align*}

from which it follows that

(B.8) \begin{align}a_{ij}({\boldsymbol{u}}) =&\ \bar u_i \bar u_j-\frac{1}{N}(u_i - \bar u_i) (u_j - \bar u_{j}). \end{align}

Similarly,

\begin{align*} a_{ii}({\boldsymbol{u}}) = \mathbb{E}(U_i - u_i)^2 =&\ \mathbb{E}( U_{i}^2 ) - 2 u_{i} \mathbb{E} U_{i} + u_{i}^2 = \mathbb{E}( U_{i}^2 ) - 2 u_{i} (u_{i} - \bar u_{i}) + u_{i}^2.\end{align*}

By Lemma B.4, we know that

\begin{align*}\mathbb{E} ( U_{i}^2 ) =&\ (1-1/N)(u_i - \bar u_i)^2 + \frac{1}{N} (u_i-\bar u_i)\\=&\ (u_i^2 - 2u_i \bar u_i + \bar u_i^2) + \frac{1}{N} \big( (u_i - \bar u_i) - (u_i - \bar u_i)^2\big),\end{align*}

and we conclude that

(B.9) \begin{align}a_{ii}(u) =&\ \bar u_i^2 + \frac{1}{N} \big((u_i - \bar u_i) - (u_i - \bar u_i)^2\big). \end{align}

The form of $b_i({\boldsymbol{u}})$ immediately implies that $B_{1}(b_i) = \delta \Sigma \leq \delta^2 s$ . Furthermore, one can check that

\begin{align*}a_{ii}({\boldsymbol{u}})=&\ u_i^2\big( \Sigma^2 -(1/N) (1-\Sigma)^2\big) + u_i \big(\! -2\Sigma p_i + (1/N) (1-\Sigma)(1-2p_i) \big)\\& + p_i^2+(1/N)p_i(1-p_i),\end{align*}

from which it follows that

\begin{align*}\vert\Delta_{i} a_{ii}({\boldsymbol{u}})\vert =&\ \big|(2u_i \delta + \delta^2) \big( \Sigma^2 -(1/N) (1-\Sigma)^2\big) + \delta \big(\! -2\Sigma p_i + (1/N) (1-\Sigma)(1-2p_i) \big)\big|\\\leq&\ \delta^2 5(1+s)^2.\end{align*}

The bound is crude but simple, and follows from the facts that $u_i \leq1$ , $\delta = 1/N \leq 1$ , and $p_i \leq \Sigma \leq \delta s$ . Similarly, one can show that

\begin{align*}\vert\Delta_{i} a_{ij}({\boldsymbol{u}})\vert, \vert\Delta_{j} a_{ij}({\boldsymbol{u}})\vert \leq \delta^2 5(1+s)^2, \quad i \neq j,\end{align*}

implying the bound on $B_{1}(a_{ij})$ .

To prove $\vert c_{ijk}({\boldsymbol{u}})\vert \leq 20\delta^2(1+s)^3$ , we compute and then bound $c_{iii}({\boldsymbol{u}})$ , $c_{iij}({\boldsymbol{u}})$ , and $c_{ijk}({\boldsymbol{u}})$ , for $i \neq j \neq k$ . We begin with

\begin{align*}c_{iii}({\boldsymbol{u}}) =&\ \mathbb{E} (U_i-u_i)^{3} = \mathbb{E} U_{i}(U_{i}-u_{i})^{2} - u_{i} \mathbb{E}(U_i - u_i)^2\\=&\ \mathbb{E}(U_{i}^{3} - 2 u_{i} U_{i}^{2} + u_{i}^{2} U_{i}) - u_{i} \mathbb{E}(U_i - u_i)^2.\end{align*}

From (B.9), we know that

\begin{align*}- u_{i} \mathbb{E}(U_i - u_i)^2 =&\ -u_{i} \bar u_i^2 - \frac{1}{N} u_{i}\big((u_i - \bar u_i) - (u_i - \bar u_i)^2 \big).\end{align*}

For the remaining terms, we use Lemma B.4 to see that

\begin{gather*}\begin{aligned}\mathbb{E} U_{i}^{3} &= (1-1/N)(1-2/N)(u_i - \bar u_i )^3 + \frac{3}{N} (1-1/N)(u_i - \bar u_i )^2 - \frac{2}{N^2} (u_i - \bar u_i)\\&= (u_i - \bar u_i )^3 - \frac{3}{N}(u_i - \bar u_i )^3 + \frac{2}{N^2}(u_i - \bar u_i )^3 + \frac{3}{N} (u_i - \bar u_i )^2 - \frac{3}{N^2}(u_i - \bar u_i )^2 - \frac{2}{N^2}(u_i - \bar u_i ),\end{aligned}\\\begin{aligned}-2u_i \mathbb{E} U_{i}^2 &= -2u_i \Big( (1-1/N) (u_i - \bar u_i )^2 + \frac{1}{N} (u_i - \bar u_i ) \Big)\\&= -2 u_i (u_i - \bar u_i )^2 + 2 u_i \frac{1}{N} (u_i - \bar u_i )^2 - 2u_i \frac{1}{N} (u_i - \bar u_i ),\end{aligned}\\u_i^2 \mathbb{E} U_i = u_i^2 (u_i - \bar u_i).\end{gather*}

We begin with the terms with a $1/N^2$ factor. Since $0 \leq u_i - \bar u_i \leq 1$ , it follows that

\begin{align*}\left|\frac{2}{N^2}(u_i - \bar u_i)^3 - \frac{3}{N^2}(u_i - \bar u_i)^2 - \frac{2}{N^2}(u_i - \bar u_i) \right| \leq \frac{5}{N^2}.\end{align*}

Let us now consider only those containing $1/N$ in front of them. Namely,

\begin{align*}& - \frac{3}{N}(u_i - \bar u_i )^3+ \frac{3}{N} (u_i - \bar u_i )^2+ 2 u_i \frac{1}{N} (u_i - \bar u_i )^2 - 2u_i \frac{1}{N} (u_i - \bar u_i )- \frac{1}{N} u_{i}\big((u_i - \bar u_i) - (u_i - \bar u_i)^2\big)\\&\qquad= \frac{3}{N} \Big( \!- (u_i - \bar u_i )^3+ (u_i - \bar u_i )^2+ u_i (u_i - \bar u_i )^2 - u_i (u_i - \bar u_i ) \Big)\\&\qquad= \frac{3}{N} \Big( \bar u_i (u_i - \bar u_i )^2 + (u_i - \bar u_i )^2 - u_i (u_i - \bar u_i ) \Big)\\&\qquad= \frac{3}{N} \Big( \bar u_i (u_i - \bar u_i )^2 - \bar u_i (u_i - \bar u_i ) \Big)\\&\qquad= \frac{3}{N} \bar u_i (u_i - \bar u_i ) \Big( (u_i - \bar u_i ) - 1\Big),\end{align*}

and note that

\begin{align*}\Big|\frac{3}{N} \bar u_i (u_i - \bar u_i ) \Big( (u_i - \bar u_i ) - 1\Big)\Big| \leq \frac{ 3\Sigma}{4N} = \frac{ 3s}{8 N^2}.\end{align*}

Lastly, we collect all the terms without $1/N$ or $1/N^2$ in front of them. Their sum equals

\begin{align*}&(u_i - \bar u_i )^3 -2 u_i (u_i - \bar u_i )^2 + u_i^2 (u_i - \bar u_i)-u_{i} \bar u_i^2\\&\qquad= u_i^3 - 3\bar u_i u_i^2 + 3 \bar u_i^2 u_i - \bar u_i^3 - 2u_i^3 + 4 u_i^2 \bar u_i - 2 u_i \bar u_i^2 + u_i^3 - u_i^2 \bar u_i - u_i \bar u_i^2\\&\qquad= - 3\bar u_i u_i^2 + 3 \bar u_i^2 u_i - \bar u_i^3 + 4 u_i^2 \bar u_i - 2 u_i \bar u_i^2 - u_i^2 \bar u_i - u_i \bar u_i^2\\&\qquad= 3 \bar u_i^2 u_i - \bar u_i^3 - 2 u_i \bar u_i^2 - u_i \bar u_i^2\\&\qquad= - \bar u_i^3.\end{align*}

Since $\vert\bar u_{i}^{3}\vert \leq \Sigma^3 = s^3/(8N^3)$ , we have shown that $\vert c_{iii}(u)\vert \leq 5/N^2 + 3s/(8N^2) + s^3/(8N^3) \leq 20 \delta^2(1+s)^3$ . Next, we bound

\begin{align*}c_{iij}({\boldsymbol{u}}) &= \mathbb{E}(U_i-u_i)^2(U_j-u_j)\\&= \mathbb{E} U_i(U_i-u_i)(U_j-u_j) - u_i\mathbb{E}(U_i-u_i)(U_j-u_j)\\&= \mathbb{E} U_{i}^{2} U_{j} - u_{j} \mathbb{E} U_{i}^2 - u_{i}\mathbb{E} U_{i}U_{j} + u_{i}u_{j} \mathbb{E} U_{i} - u_i\mathbb{E}(U_i-u_i)(U_j-u_j).\end{align*}

From (B.8), we know that

\begin{align*}- u_i\mathbb{E}(U_i-u_i)(U_j-u_j) =&\ -u_i \Big( \bar u_i \bar u_j- \frac{1}{N} (u_i - \bar u_i) (u_j - \bar u_{j}) \Big),\end{align*}

and, for the rest of the terms, we use Lemma B.4 to obtain

\begin{gather*}\begin{aligned}\mathbb{E} U_{i}^{2} U_{j} =&\ (1-1/N)(1-2/N) (u_i - \bar u_i)^2 (u_j - \bar u_j) + \frac{1}{N}(1-1/N) (u_i - \bar u_i) (u_j - \bar u_j)\\=&\ (u_i - \bar u_i)^2 (u_j - \bar u_j) - \frac{3}{N} (u_i - \bar u_i)^2 (u_j - \bar u_j)+ \frac{2}{N^2} (u_i - \bar u_i)^2 (u_j - \bar u_j)\\&+ \frac{1}{N} (u_i - \bar u_i) (u_j - \bar u_j) - \frac{1}{N^2} (u_i - \bar u_i) (u_j - \bar u_j),\\- u_{j} \mathbb{E} U_{i}^2 =&\ -u_{j} \Big( (1-1/N) (u_i - \bar u_i)^2 + \frac{1}{N} (u_i - \bar u_i) \Big)\\=&\ -u_{j} \Big( (u_i - \bar u_i)^2 - \frac{1}{N } (u_i - \bar u_i)^2 + \frac{1}{N} (u_i - \bar u_i) \Big),\end{aligned}\\- u_{i}\mathbb{E} U_{i}U_{j} = - u_{i} \Big( (u_i - \bar u_i) (u_j - \bar u_j) - \frac{1}{N}(u_i - \bar u_i) (u_j - \bar u_j) \Big),\\u_{i}u_{j} \mathbb{E} U_{i} = u_{i}u_{j}(u_i - \bar u_i).\end{gather*}

Considering the $1/N^2$ terms,

\begin{align*}\left|\frac{2}{N^2}(u_i - \bar u_i)^2 (u_j - \bar u_j) - \frac{1}{N^2}(u_i- \bar u_i) (u_j - \bar u_j)\right| \leq \frac{1}{N^2}.\end{align*}

Collecting all terms with $1/N$ in front, the result equals

\begin{align*}& \frac{1}{N} (- 3 (u_i - \bar u_i)^2 (u_j - \bar u_j)+ (u_i - \bar u_i) (u_j - \bar u_j) + u_j (u_i - \bar u_i)^2 - u_j (u_i - \bar u_i)\\& \hskip4mm + u_{i} (u_i - \bar u_i) (u_j - \bar u_j) + u_{i} (u_i - \bar u_i) (u_j - \bar u_j) )\\&\qquad= \frac{1}{N}(u_i - \bar u_i) (- 3 (u_i - \bar u_i) (u_j - \bar u_j)+ (u_j - \bar u_j) + u_j (u_i - \bar u_i) - u_j + 2 u_{i} (u_j - \bar u_j) )\\&\qquad= \frac{1}{N}(u_i - \bar u_i) (- 3 u_i u_j + 3 u_i \bar u_j + 3 u_j \bar u_i - 3 \bar u_i \bar u_j \\&\qquad\qquad\qquad\qquad+ (u_j - \bar u_j) + u_j u_i - u_j \bar u_i - u_j + 2 u_{i} u_j - 2 u_{i} \bar u_j )\\&\qquad= \frac{1}{N}(u_i - \bar u_i) ( 3 u_i \bar u_j + 3 u_j \bar u_i - 3 \bar u_i \bar u_j - \bar u_j - u_j \bar u_i - 2 u_{i} \bar u_j ),\end{align*}

the absolute value of which can be bounded by $15 \Sigma/N$ , because $\bar u_{i}, \bar u_{j} \leq \Sigma$ . Next, collecting all terms without $1/N^2$ or $1/N$ yields

\begin{align*}&(u_i - \bar u_i)^2 (u_j - \bar u_j)- u_{j} (u_i - \bar u_i)^2- u_{i} (u_i - \bar u_i) (u_j - \bar u_j) +u_{i}u_{j}(u_i - \bar u_i)-u_i \bar u_i \bar u_j \\&\qquad= - \bar u_j (u_i - \bar u_i)^2 - u_{i} (u_i - \bar u_i) (u_j - \bar u_j) +u_{i}u_{j}(u_i - \bar u_i)-u_i \bar u_i \bar u_j\\&\qquad= - \bar u_j (u_i - \bar u_i)^2 +u_{i} \bar u_j (u_i - \bar u_i) -u_i \bar u_i \bar u_j\\&\qquad= - \bar u_j ( (u_i - \bar u_i)^2 - u_{i} (u_i - \bar u_i) +u_i \bar u_i )\\&\qquad= - \bar u_j ( - \bar u_i (u_i - \bar u_i) +u_i \bar u_i )\\&\qquad= - \bar u_j \bar u_i^2,\end{align*}

which is bounded by $\Sigma^{3}$ . Thus, we have shown that

\begin{align*}\vert c_{iij}({\boldsymbol{u}})\vert \leq \frac{1}{N^2} + \frac{15 \Sigma}{N} + \Sigma^3 \leq 20 \delta^2(1+s)^3.\end{align*}

Lastly, we consider

\begin{align*}c_{ijk}({\boldsymbol{u}}) =&\ \mathbb{E}(U_i-u_i)(U_j-u_j) (U_k-u_k)\\=&\ \mathbb{E} U_{k} (U_i-u_i)(U_j-u_j) - u_{k} \mathbb{E} (U_i-u_i)(U_j-u_j)\\=&\ \mathbb{E} U_{k} U_{i} U_{j} - u_{j} \mathbb{E} U_{k} U_{i} - u_{i}\mathbb{E} U_{k} U_{j} + u_{i} u_{j} \mathbb{E} U_{k} - u_{k} \mathbb{E}(U_i-u_i)(U_j-u_j).\end{align*}

From (B.8) we have

\begin{align*}- u_{k} \mathbb{E} (U_i-u_i)(U_j-u_j) =&\ - u_{k} \Big( \bar u_i \bar u_j- \frac{1}{N} (u_i - \bar u_i) (u_j - \bar u_{j}) \Big),\end{align*}

and from Lemma B.4 we have

\begin{align*}\mathbb{E} U_{k} U_{i} U_{j} =&\ \Big( 1 - \frac{3}{N} + \frac{2}{N^2} \Big) (u_{i} - \bar u_{i})(u_{j} - \bar u_{j})(u_{k} - \bar u_{k}),\\- u_{j} \mathbb{E} U_{k} U_{i} =&\ - u_{j} \Big( (u_k - \bar u_k) (u_i - \bar u_i) - \frac{1}{N}(u_k - \bar u_k) (u_i - \bar u_i) \Big),\\- u_{i} \mathbb{E} U_{k} U_{j} =&\ - u_{i} \Big( (u_k - \bar u_k) (u_j - \bar u_j) - \frac{1}{N}(u_k - \bar u_k) (u_j - \bar u_j) \Big).\end{align*}

The combined $1/N^2$ terms can be bounded by $2/N^2$ . Collecting terms with $1/N$ in front yields

\begin{align*}& \frac{1}{N} ( -3(u_{i} - \bar u_{i})(u_{j} - \bar u_{j})(u_{k} - \bar u_{k}) + u_{j} (u_k - \bar u_k) (u_i - \bar u_i)\\& \hskip4mm + u_{i} (u_k - \bar u_k) (u_j - \bar u_j) + u_k (u_i - \bar u_i) (u_j - \bar u_{j}) )\\&\qquad= \frac{1}{N} ( \bar u_j (u_{i} - \bar u_{i})(u_{k} - \bar u_{k}) + \bar u_i (u_{j} - \bar u_{j})(u_{k} - \bar u_{k})+ \bar u_k (u_{j} - \bar u_{j})(u_{i} - \bar u_{i}) ),\end{align*}

and this term can be bounded by $3 \Sigma /N$ . Similarly, collecting all terms without $1/N^2$ and $1/N$ yields

\begin{align*}&(u_{i} - \bar u_{i})(u_{j} - \bar u_{j})(u_{k} - \bar u_{k}) - u_{j} (u_k - \bar u_k) (u_i - \bar u_i) - u_{i} (u_k - \bar u_k) (u_j - \bar u_j)\\&+ u_{i} u_{j} (u_k - \bar u_k) - u_{k} \bar u_i \bar u_j\\&\qquad= - \bar u_{j} (u_{i} - \bar u_{i}) (u_{k} - \bar u_{k})- u_{i} (u_k - \bar u_k) (u_j - \bar u_j) + u_{i} u_{j} (u_k - \bar u_k) - u_{k} \bar u_i \bar u_j\\&\qquad= - \bar u_{j} ( u_{i} u_k - u_k \bar u_i - u_i \bar u_k + \bar u_i \bar u_k) - u_{i} (u_k u_j -u_k \bar u_j - u_j \bar u_k + \bar u_j \bar u_k) + u_{i} u_{j} u_k\\&\hskip11mm- u_{i} u_{j} \bar u_k - u_{k} \bar u_i \bar u_j\\&\qquad= - \bar u_{j} ( u_{i} u_k - u_k \bar u_i - u_i \bar u_k + \bar u_i \bar u_k) - u_{i} ( -u_k \bar u_j + \bar u_j \bar u_k) - u_{k} \bar u_i \bar u_j\\&\qquad= - \bar u_{j} ( - u_k \bar u_i - u_i \bar u_k + \bar u_i \bar u_k) - u_{i}\bar u_j \bar u_k - u_{k} \bar u_i \bar u_j,\end{align*}

and this term can be bounded by $5\Sigma^2$ . This completes the bound for

\begin{align*}\vert c_{ijk}({\boldsymbol{u}})\vert \leq \frac{2}{N^2} + \frac{3\Sigma}{N} + 5 \Sigma^3 \leq 20\delta^2 (1+s)^3.\end{align*}

Finally, we show that

\begin{align*}\bar d_{ijk\ell}({\boldsymbol{u}}) \leq \left( \frac{2}{\sqrt N} + \Sigma\right)^4 \leq \delta^2(2+s)^4. \end{align*}

Using the Cauchy–Schwarz inequality twice,

\begin{align*}&\mathbb{E}\vert(U_i - u_i)(U_j - u_j)(U_k - u_k)(U_{\ell} - u_{\ell})\vert\\&\qquad\leq \left[\mathbb{E}(U_i - u_i)^4\right]^\frac14 \left[\mathbb{E}_u(U_j - u_j)^4\right]^\frac14 \left[\mathbb{E}(U_k - u_k)^4\right]^\frac14 \left[\mathbb{E}_u(U_{\ell} - u_{\ell})^4\right]^\frac14.\end{align*}

Recalling that $N U_i | u_i \sim \text{Bin}( N, u_i - \bar u_{i})$ and then using Minkowski’s inequality,

\begin{align*}\left[\mathbb{E}(U_i - u_i)^4\right]^\frac14 &= \frac{1}{N} \left[\mathbb{E}_u ( NU_i - N(u_i - \bar u_{i}) - N \bar u_{i})^4 \right]^\frac14\\& \leq \frac{1}{N} \left[ \mathbb{E}_u ( NU_i - N(u_i- \bar u_{i}))^4 \right]^\frac14 + \frac{1}{N} N \bar u_{i}.\end{align*}

Noting that, for $Y \sim \text{Bin}(n,p)$ ,

\begin{align*}\mathbb{E}(Y - np)^4 &= 3(np(1-p))^2 + np(1-p)(1-6p(1-p))\\&\leq 3(np(1-p)^2 + np(1-p) \leq 4n^2,\end{align*}

we conclude that

\begin{align*}\left[\mathbb{E}(U_i - u_i)^4\right]^\frac14 &\leq \frac{1}{N}\Big( (4N^2)^\frac14 + N \bar u_{i}\Big) \leq \frac{2}{{\sqrt N}\,} + \Sigma,\end{align*}

and the final bound is now clear.

Finally we complete this appendix with the proof of the Stein factor bounds in Lemma 2.3.

Proof of Lemma 2.3. For notational clarity, given ${\boldsymbol{a}}$ such that $\|{\boldsymbol{a}}\| = k$ , decompose ${\boldsymbol{a}}$ into its components such that ${\boldsymbol{a}} = {\boldsymbol{a}}_1 + \ldots +{\boldsymbol{a}}_k$ , where ${\boldsymbol{a}}_i$ is a standard basis unit vector. Furthermore, for the remainder of this proof, let ${\boldsymbol{U}}_{\boldsymbol{u}}(t)$ denote the process ${\boldsymbol{U}}(t)$ started at ${\boldsymbol{U}}(0) = {\boldsymbol{u}}$ . We first start with the case where $\| {\boldsymbol{a}} \| = 1$ . Starting from (2.5), given $h \in \mathcal{M}_{\mathrm{disc},4}(C)$ ,

(B.10) \begin{align}\vert\Delta^{{\boldsymbol{a}}_1} f_h({\boldsymbol{u}})\vert = \biggl\vert\sum_{t=0}^\infty \left[\mathbb{E} h ({\boldsymbol{U}}_{{\boldsymbol{u}} + \delta{\boldsymbol{a}}_1}(t)) - \mathbb{E} h ({\boldsymbol{U}}_{\boldsymbol{u}}(t)) \right] \biggr\vert \leq \sum_{t=0}^\infty C \mathbb{E} \| {\boldsymbol{U}}_{{\boldsymbol{u}} + \delta {\boldsymbol{a}}_1 }(t) - {\boldsymbol{U}}_{{\boldsymbol{u}}}(t) \|_{1}. \end{align}

We couple the two processes ${\boldsymbol{U}}_{{\boldsymbol{u}} + \delta{\boldsymbol{a}}_1}(t)$ and ${\boldsymbol{U}}_{\boldsymbol{u}}(t)$ in the following manner. Index the parents so that the types for individuals in both processes match except for the one entry where the first process will have an individual of type depending on ${\boldsymbol{a}}_1$ and the second process has an individual of type K. (Recall we reserve the final type K to be the remainder.) Given the current generation, to generate the next generation, take a random sample of size N from the indices $\{1,\ldots N\}$ , and use this common random sample to choose parents for the offspring of both processes. Given mutation is parent independent, we also couple the mutations identically across both processes in the obvious manner. Figure 1 illustrates the joint evolution of ${\boldsymbol{U}}_{{\boldsymbol{u}} + \delta {\boldsymbol{a}}_1}(t)$ and ${\boldsymbol{U}}_{\boldsymbol{u}}(t)$ .

Figure 1. An illustration of our coupling at times $t = 0,1,2$ with population size $N = 6$ and $K= 5$ types. Each gene type is colour and number coded; e.g. type 1 is green, type 5 is red, etc. An ‘M’ next to a row represents a mutation, while arrows represent parental relationships; e.g. rows two and three (from the bottom) in the middle plot are children of the first row in the leftmost plot, while row six of the middle plot mutated. Coupling occurs at time $t = 2$ since rows two and three in the middle plot have no children.

Given the starting configuration at time 0, let ${\boldsymbol{V}}(t) =(V_1(t), \ldots, V_N(t))$ denote the process that tracks the ancestry of the original configuration. That is, $V_j(t)$ tracks the number of individuals at time t that trace their ancestry directly back to individual j at time 0. Note that if a mutation occurs, it is removed from this process, hence, $\|{\boldsymbol{V}}(0)\|_1 = N$ , and $\lim_{t \to\infty} {\boldsymbol{V}}(t) = 0$ . For readers familiar with coalescent theory, this is analogous to a coalescent process looking forwards in time.

Without loss of generality, we can therefore set

\begin{align*}\| {\boldsymbol{U}}_{{\boldsymbol{u}} + \delta{\boldsymbol{a}}_1 }(t) - {\boldsymbol{U}}_{{\boldsymbol{u}}}(t) \|_{1} = \delta V_1(t).\end{align*}

In the context of Figure 1, $V_1(t)$ is tracking the number of replicates at time t of the pair (5,1) from the final row in the processes at time 0. Therefore, in this particular realisation, $V_1(0) =1, V_1(1) = 2, V_1(2) = 0$ . Given $V_1(t-1)$ , $V_1(t)$ will be made up of the number of times one of the $V_1(t-1)$ individuals are chosen, which will occur with probability $\delta V_1(t-1)$ , who also do not mutate. Hence, recalling that $\Sigma = \sum_{i=1}^K p_k$ denotes the probability of any mutation occurring,

\begin{align*}V_1(t) | V_1(t-1) \sim \text{Bin}\Big(N, \delta V_1(t-1) (1-\Sigma) \Big).\end{align*}

Observing that $\mathbb{E}[V_1(1)] = \mathbb{E}[ \mathbb{E}[V_1(1) | V_1(0)]] =\delta\mathbb{E}[V_1(0)](1-\Sigma) = \delta(1-\Sigma)$ , and then applying this recursively we conclude that $\mathbb{E} [V_1(t)] = \delta(1-\Sigma)^t$ for all integers $t \geq 0$ . Therefore, following on from (B.10), for $h\in \mathcal M_{\mathrm{disc},4}(C)$ ,

\begin{align*}\vert\Delta^{{\boldsymbol{a}}_1}f_h({\boldsymbol{u}})\vert \leq \sum_{t=0}^\infty C \delta \mathbb{E} V_1(t) =\sum_{t=0}^\infty C \delta(1-\Sigma)^t= \frac{C\delta}{\Sigma}.\end{align*}

For the second-order difference where $\|{\boldsymbol{a}}\|=2$ , for $h \in \mathcal M_{\mathrm{disc},4}(C)$ ,

(B.11) \begin{align}\vert\Delta^{{\boldsymbol{a}}} f_h({\boldsymbol{u}}) \vert&= \biggl\vert \sum_{t=0}^\infty \mathbb{E} \left[ h ({\boldsymbol{U}}_{{\boldsymbol{u}} + \delta {\boldsymbol{a}}_1 +\delta {\boldsymbol{a}}_2}(t)) - h ({\boldsymbol{U}}_{{\boldsymbol{u}} + \delta {\boldsymbol{a}}_1}(t)) - h ({\boldsymbol{U}}_{{\boldsymbol{u}} + \delta {\boldsymbol{a}}_2}(t)) + h ({\boldsymbol{U}}_{\boldsymbol{u}}(t)) \right] \biggr\vert\notag\\&\leq C \sum_{t=0}^\infty \mathbb{E} \left[\| {\boldsymbol{U}}_{{\boldsymbol{u}} + {\boldsymbol{a}}_1 + {\boldsymbol{a}}_2}(t)- {\boldsymbol{U}}_{{\boldsymbol{u}} + {\boldsymbol{a}}_1}(t)\|_1 \| {\boldsymbol{U}}_{{\boldsymbol{u}} + {\boldsymbol{a}}_2}(t) - {\boldsymbol{U}}_{{\boldsymbol{u}}}(t)\|_1\right], \end{align}

where the inequality is due to the fact that, in general,

(B.12) \begin{align}\vert f({\boldsymbol{x}}+{\boldsymbol{a}}+ {\boldsymbol{b}}) - f({\boldsymbol{x}}+{\boldsymbol{a}}) - f({\boldsymbol{x}}+{\boldsymbol{b}}) - f({\boldsymbol{x}})\vert \leq \|{\boldsymbol{a}}\|_1\|{\boldsymbol{b}}\|_1 \max_{\Vert{\boldsymbol{a}}'\Vert=2} \Vert\Delta^{{\boldsymbol{a}}'}f\Vert. \end{align}

For one-dimensional functions $f\,:\,\mathbb{Z} \to \mathbb{R}$ , inequality (B.12) follows from

\begin{align*}&f(x + a + b) - f(x + a) -f(x+b) + f(x)\\&\qquad= \sum_{i=0}^{b-1} \sum_{j=1}^{a-1} \big(f(x +j+ i) - 2 f(x +j-1+ i)+ f(x +j-1+ i -1)\big), \quad x,a,b \in \mathbb{Z}.\end{align*}

A similar idea can be used to justify (B.12) for multidimensional functions.

We couple the four processes on the right-hand side in an analogous manner to the first difference; i.e. parent selection and mutation are coupled to be identical for all four processes. Figure 2 illustrates their evolution. Similar to the first difference, we need to keep track of any differences between the four processes, which can be achieved by tracking genealogies using the process ${\boldsymbol{V}}(t)$ where we set $(V_1(t),V_2(t))$ to jointly track the propagation of the initial two rows in Figure 2.

Figure 2. An illustration of our second-order difference coupling at times $t = 0,1,2$ with population size $N = 6$ and $K= 5$ types.

Therefore, we can without loss of generality set

\begin{align*}( \| {\boldsymbol{U}}_{{\boldsymbol{u}} + {\boldsymbol{a}}_1 + {\boldsymbol{a}}_2}(t)- {\boldsymbol{U}}_{{\boldsymbol{u}} + {\boldsymbol{a}}_1}(t)\|_1, \| {\boldsymbol{U}}_{{\boldsymbol{u}} + {\boldsymbol{a}}_2}(t) - {\boldsymbol{U}}_{{\boldsymbol{u}}}(t)\|_1 ) = (\delta V_1(t), \delta V_2(t)).\end{align*}

Analogously to the first difference, we can set $V_1(t)$ and $V_2(t)$ jointly such that

\begin{align*}( (V_1(t), \!V_2(t)) \vert V_1(t-1), \!V_2(t-1)) \sim \text{Multinomial}\left(N, \!\left\{ \delta V_1(t-1) (1-\Sigma), \delta V_2(t-1) (1-\!\Sigma)\right\}\right)\!.\end{align*}

Given $V_1(0) = V_2(0) = 1$ , we can recursively show, with the help of Lemma B.4, that $\mathbb{E}[N^2 V_1(t) V_2(t)] =N(N-1)\delta^2(1-\Sigma)^{2t}$ . Hence, continuing from (B.11),

\begin{align*}\vert\Delta^{{\boldsymbol{a}}} f_h({\boldsymbol{u}})\vert\leq \sum_{t=0}^\infty C \delta^2\mathbb{E} V_1(t)V_2(t) \leq \sum_{t=0}^\infty C\delta^2 (1-\Sigma)^{2t}= \frac{C\delta^2}{(1-(1-\Sigma)^2)}.\end{align*}

We omit the proof of the bounds when $\Vert{\boldsymbol{a}}\Vert = 3,4$ , as they follow from exactly the same proof methodology.