Strong convergence rates of probabilistic integrators for ordinary differential equations

Abstract

Probabilistic integration of a continuous dynamical system is a way of systematically introducing discretisation error, at scales no larger than errors introduced by standard numerical discretisation, in order to enable thorough exploration of possible responses of the system to inputs. It is thus a potentially useful approach in a number of applications such as forward uncertainty quantification, inverse problems, and data assimilation. We extend the convergence analysis of probabilistic integrators for deterministic ordinary differential equations, as proposed by Conrad et al. (Stat Comput 27(4):1065–1082, 2017. https://doi.org/10.1007/s11222-016-9671-0), to establish mean-square convergence in the uniform norm on discrete- or continuous-time solutions under relaxed regularity assumptions on the driving vector fields and their induced flows. Specifically, we show that randomised high-order integrators for globally Lipschitz flows and randomised Euler integrators for dissipative vector fields with polynomially bounded local Lipschitz constants all have the same mean-square convergence rate as their deterministic counterparts, provided that the variance of the integration noise is not of higher order than the corresponding deterministic integrator. These and similar results are proven for probabilistic integrators where the random perturbations may be state-dependent, non-Gaussian, or non-centred random variables.

A Proofs

Proof of Lemma 2.1

Assertion (2.3) holds immediately for $n=1$, so let $n\in {\mathbb {N}}\setminus \{1\}$, and recall the binomial formula: for $x,y\in {\mathbb {R}}$ and $n\in {\mathbb {N}}\setminus \{1\}$,

$$\begin{aligned} (x + y)^n = \sum ^n_{k=0} \left( {\begin{array}{c} n \\ k \end{array}}\right) x^{k} y^{n - k} = x^n + y^n + \sum ^{n-1}_{k=1} \left( {\begin{array}{c} n \\ k \end{array}}\right) x^{k} y^{n-k}. \end{aligned}$$

Fix $\delta >0$. By (2.1), for any $1\le k\le n-1$,

$$\begin{aligned} x^{k} y^{n-k}&\le \delta \frac{k}{n}x^{n} + \frac{1}{\delta ^{k/(n-k)}}\frac{n-k}{n}y^n \\&\le \delta \frac{k}{n}x^{n} + \frac{1}{\delta ^{n-1}}\frac{n-k}{n}y^n, \end{aligned}$$

where the second inequality follows from $-\tfrac{k}{n-k}\ge -(n-1)$. Therefore,

$$\begin{aligned} (x + y)^n&\le x^n \left( 1 + \delta \sum ^{n-1}_{k=1}\left( {\begin{array}{c} n \\ k \end{array}}\right) \frac{k}{n} \right) \nonumber \\&\quad +\, y^n \left( 1 + \frac{1}{\delta ^{n-1}}\sum ^{n-1}_{k=1}\left( {\begin{array}{c} n \\ k \end{array}}\right) \frac{n-k}{n} \right) , \end{aligned}$$

(A.1)

and the proof is complete upon observing that

$$\begin{aligned} \sum ^{n-1}_{k=1}\left( {\begin{array}{c} n \\ k \end{array}}\right) \frac{k}{n}&=\sum ^{n-1}_{k=1} \left( {\begin{array}{c} n-1 \\ k-1 \end{array}}\right) = \sum ^{n-1}_{j=0} \left( {\begin{array}{c} n - 1 \\ j \end{array}}\right) - \left( {\begin{array}{c} n - 1 \\ n - 1 \end{array}}\right) \\&= (1 + 1)^{n-1} -1 \le 2^{n-1} \end{aligned}$$

and bounding the other binomial sum in a similar way. $\square$

Proof of Theorem 3.4

By (3.3),

$$\begin{aligned} \Vert e_{k + 1} \Vert ^{2} =&\left\| \bigl ( \varPhi ^{\tau }(u_{k}) - \varPhi ^{\tau }(U_{k}) \bigr ) - \bigl ( \varPsi ^{\tau }(U_{k}) - \varPhi ^{\tau }(U_{k}) \bigr ) \right\| ^{2} \\&+ \Vert \xi _{k}(\tau ) \Vert ^{2} + 2 \langle \varPhi ^{\tau }(u_{k}) - \varPsi ^{\tau }(U_{k}) , \xi _{k}(\tau ) \rangle . \end{aligned}$$

By (2.2) with $\delta = \tau$, by Assumptions 3.1 and 3.2, and using that $\tau <\tau ^*\le 1$,

$$\begin{aligned}&\left\| \varPhi ^{\tau }(u_{k})-\varPsi ^{\tau }(U_{k}) \right\| ^2\nonumber \\&\quad = \left\| \bigl ( \varPhi ^{\tau }(u_{k}) - \varPhi ^{\tau }(U_{k}) \bigr ) - \bigl ( \varPsi ^{\tau }(U_{k}) - \varPhi ^{\tau }(U_{k}) \bigr ) \right\| ^{2} \nonumber \\&\quad \le (1 + \tau ) \Vert \varPhi ^{\tau }(u_{k}) - \varPhi ^{\tau }(U_{k}) \Vert ^{2} \ \nonumber \\&\qquad + (1 + \tau ^{-1}) \Vert \varPsi ^{\tau }(U_{k}) - \varPhi ^{\tau }(U_{k}) \Vert ^{2} \nonumber \\&\quad \le (1 + \tau ) (1 + C_\varPhi \tau )^{2} \Vert e_{k} \Vert ^{2} + 2C_\varPsi ^2 \tau ^{1 + 2 q}. \end{aligned}$$

(A.2)

Observe that $[(1 + \tau )(1 + C_\varPhi \tau )^2-1]\tau ^{-1}$ equals a quadratic polynomial in $\tau$ with coefficients $a_0$, $a_1$, and $a_2$. Calculating these coefficients and defining

$$\begin{aligned} C_1=C_1(C_\varPhi ,\tau ^*):=1 + 2C_\varPhi + C_\varPhi (2 + C_\varPhi )\tau ^*+ C_\varPhi ^2(\tau ^*)^2 \end{aligned}$$

(A.3)

then yields that $[(1 + \tau )(1 + C_\varPhi \tau )^2-1]\tau ^{-1}\le C_1$ for all $0<\tau <\tau ^*$.

Combining the preceding estimates yields

$$\begin{aligned} \Vert e_{k + 1} \Vert ^{2} - \Vert e_{k} \Vert ^{2}&\le C_1 \tau \Vert e_{k} \Vert ^{2} + 2C_\varPsi ^2 \tau ^{1 + 2 q} + \Vert \xi _{k}(\tau ) \Vert ^{2} \nonumber \\&\quad +\, 2 \langle \varPhi ^{\tau }(u_{k}) - \varPsi ^{\tau }(U_{k}) , \xi _{k}(\tau ) \rangle . \end{aligned}$$

(A.4)

Using (A.4) in the telescoping sum

$$\begin{aligned} \Vert e_{k} \Vert ^{2} - \Vert e_{0} \Vert ^{2} = \sum _{j = 0}^{k - 1} \bigl ( \Vert e_{j + 1} \Vert ^{2} - \Vert e_{j} \Vert ^{2} \bigr ), \end{aligned}$$

the fact that $e_0=u_0-U_0=0$ and $K=T/\tau$, we obtain

$$\begin{aligned} \Vert e_{k} \Vert ^2&\le \sum ^{k-1}_{j=0}\biggr [C_1\tau \Vert e_j \Vert ^2 + C_\varPsi \tau ^{1 + 2q} + \Vert \xi _j(\tau ) \Vert ^2 \\&\quad + 2\left\langle \varPhi ^\tau (u_k)-\varPsi ^\tau (U_k) , \xi _k(\tau ) \right\rangle \biggr ] \\&\le C_1\tau \sum ^{k-1}_{j=0} \Vert e_j \Vert ^2 + \sum ^{K-1}_{j=0}\left( 2C_\varPsi ^2\tau ^{1 + 2q} + \Vert \xi _j(\tau ) \Vert ^2\right) \\&\quad + 2\left\| \sum ^{k-1}_{j=0}\left\langle \varPhi ^\tau (u_k)-\varPsi ^\tau (U_k) , \xi _k(\tau ) \right\rangle \right\| \\&\le C_1\tau \sum ^{k-1}_{j=0} \Vert e_j \Vert ^2 + 2TC_\varPsi ^2\tau ^{2q} + \sum ^{K-1}_{j=0}\Vert \xi _j(\tau ) \Vert ^2 \\&\quad + 2\left\| \sum ^{k-1}_{j=0}\left\langle \varPhi ^\tau (u_k)-\varPsi ^\tau (U_k) , \xi _k(\tau ) \right\rangle \right\| . \end{aligned}$$

It follows from the last inequality that

$$\begin{aligned} \max _{\ell \le k}\Vert e_{\ell } \Vert ^2&\le C_1\tau \sum ^{k-1}_{j=0} \Vert e_j \Vert ^2 + 2TC_\varPsi ^2\tau ^{2q} + \sum ^{K-1}_{j=0}\Vert \xi _j(\tau ) \Vert ^2 \\&\quad + 2\max _{\ell \le k}\left\| \sum ^{\ell -1}_{j=0}\left\langle \varPhi ^\tau (u_k)-\varPsi ^\tau (U_k) , \xi _k(\tau ) \right\rangle \right\| . \end{aligned}$$

Now replace $\Vert e_j \Vert ^2$ on the right-hand side with $\max _{\ell \le j}\Vert e_\ell \Vert ^2$ and take expectations of both sides of the inequality. Since Assumption 3.3 holds with $R=2$,

$$\begin{aligned} {\mathbb {E}}\left[ \sum _{j = 0}^{K - 1}\Vert \xi _{j}(\tau ) \Vert ^{2} \right] \le \frac{T}{\tau } (C_{\xi ,R}\tau ^{p + 1/2})^2=T C_{\xi ,R}^2\tau ^{2p}. \end{aligned}$$

Next, define for every $k\in [K]$ the $\sigma$-algebra ${\mathcal {F}}_{j}$ generated by $\xi _{0}(\tau ),\ldots ,\xi _{j}(\tau )$ Then, the sequence $({\mathcal {F}}_{j})_{j\in [K]}$ forms a filtration. Define $(M_k)_{k\in [K]}$ by

$$\begin{aligned} M_k:=\sum ^k_{j=0}\left\langle \varPhi ^\tau (u_j)-\varPsi ^\tau (U_j) , \xi _j(\tau ) \right\rangle . \end{aligned}$$

We want to show that this process is a martingale with respect to $({\mathcal {F}}_{j})_{j\in [K]}$. By (1.6), $U_j$ is measurable with respect to ${\mathcal {F}}_{j-1}$, so $M_k$ is measurable with respect to ${\mathcal {F}}_k$. Hence $(M_k)_{k\in [K]}$ is adapted with respect to $({\mathcal {F}}_k)_{k\in [K]}$. Observe that

$$\begin{aligned} {\mathbb {E}}\left[ \Vert M_k \Vert \right]&\le \sum ^k_{j=0}{\mathbb {E}}\left[ \left\| \left\langle \varPhi ^\tau (u_j)-\varPsi ^\tau (U_j) , \xi _j(\tau ) \right\rangle \right\| \right] \\&\le \sum ^k_{j=0}\left( {\mathbb {E}}\left[ \Vert \varPhi ^\tau (u_j)-\varPsi ^\tau (U_j) \Vert ^2\right] + {\mathbb {E}}\left[ \Vert \xi _j(\tau ) \Vert ^2\right] \right) \\&\le 2\sum ^k_{j=0}\left( \Vert \varPhi ^\tau (u_j) \Vert ^2 + {\mathbb {E}}\left[ \Vert \varPsi ^\tau (U_j) \Vert ^2+\Vert \xi _j(\tau ) \Vert ^2\right] \right) . \end{aligned}$$

Using the assumption that $X\in L^2_{{\mathbb {P}}} \implies \varPsi ^\tau (X)\in L^2_{{\mathbb {P}}}$, (1.6), Assumption 3.3, and the fact that $U_0=u_0$ is fixed, it follows that $U_j$ and $\varPsi ^\tau (U_j)$ belong to $L^2_{{\mathbb {P}}}$; thus $M_k$ belongs to $L^1_{{\mathbb {P}}}$ for every $k\in [K]$. We now use the assumption that ${\mathbb {E}}[\xi _j(\tau )]=0$ for every $j\in [K]$, and that the $(\xi _k(\tau ))_{k\in [K]}$ are mutually independent, in order to establish the martingale property:

$$\begin{aligned}&{\mathbb {E}}\left[ M_k-M_{k-1}\vert {\mathcal {F}}_{k-1}\right] \\&\quad = {\mathbb {E}}\left[ \left\langle \varPhi ^\tau (u_k) - \varPsi ^\tau (U_k) , \xi _k(\tau ) \right\rangle \big \vert {\mathcal {F}}_{k-1}\right] , \end{aligned}$$

and the right-hand side vanishes since $U_k$ is measurable with respect to ${\mathcal {F}}_{k-1}$ as noted earlier. Since $(M_k)_{k\in [K]}$ is a martingale, we may apply the Burkholder–Davis–Gundy inequality (Peškir 1996, Equation 2.2). Letting $[Y]_{\ell }$ denote the quadratic variation up to time $\ell$ of a process $Y_{k}$, we have

$$\begin{aligned}&{\mathbb {E}}\left[ \max _{k \le \ell } \left\| \sum _{j = 0}^{k - 1} \langle \varPhi ^{\tau } (u_{j}) - \varPsi ^{\tau } (U_{j}) , \xi _{j}(\tau ) \rangle \right\| \right] \\&\quad \le 3{\mathbb {E}}\Bigl [ [ \langle \varPhi ^{\tau } (u_{\bullet }) - \varPsi ^{\tau } (U_{\bullet }) , \xi _{\bullet }(\tau ) \rangle ]_{\ell - 1}^{1/2} \Bigr ]\le 3 {\mathbb {E}}\left[ ab \right] \end{aligned}$$

where we define $b:=\sqrt{ \sum _{j = 1}^{\ell - 1} \Vert \xi _{j}(\tau ) - \xi _{j - 1}(\tau ) \Vert ^{2} }$ and $a:=\sqrt{\max _{j \le \ell } \Vert \varPhi ^{\tau } (u_{j}) - \varPsi ^{\tau } (U_{j}) \Vert ^2}$. Using (2.1) with the same a and b, $r=r^*=2$, and $\delta =[6(1 + \tau )(1 + C_\varPhi \tau )^2]^{-1}$, and using (A.2), it follows that

$$\begin{aligned}&3 {\mathbb {E}}\Biggl [\left( \max _{j \le \ell } \Vert \varPhi ^{\tau } (u_{j}) - \varPsi ^{\tau } (U_{j}) \Vert \right) \sqrt{ \sum _{j = 1}^{\ell - 1} \Vert \xi _{j}(\tau ) - \xi _{j - 1}(\tau ) \Vert ^{2} } \Biggr ] \\&\quad \le \frac{1}{4}\left( {\mathbb {E}}\left[ \max _{j\le \ell } \Vert e_j \Vert ^2\right] + 2C_\varPsi ^2\tau ^{1 + 2q}\right) \\&\quad \quad + 9(1 + \tau ^*)(1 + C_\varPhi \tau ^*)^2\sum ^{\ell -1}_{j=1} {\mathbb {E}}\left[ \Vert \xi _{j}(\tau )-\xi _{j-1}(\tau ) \Vert ^2\right] \\&\quad \le \frac{1}{4}\left( {\mathbb {E}}\left[ \max _{j\le \ell } \Vert e_j \Vert ^2\right] + 2C_\varPsi ^2\tau ^{1 + 2q}\right) \\&\qquad +18(1 + \tau ^*)(1 + C_\varPhi \tau ^*)^2\sum ^{\ell -1}_{j=1} {\mathbb {E}}\left[ \Vert \xi _{j}(\tau ) \Vert ^2\right] \end{aligned}$$

where we applied (2.2) with $\delta =1$, $r=r^*=2$, $a=\xi _{j}(\tau )$ and $b=\xi _{j-1}(\tau )$ to obtain the last inequality. Thus by Assumption 3.3 and by using $\ell -1\le K=T/\tau$,

$$\begin{aligned}&2 {\mathbb {E}}\left[ \max _{k \le \ell } \left\| \sum _{j = 0}^{\ell - 1} \langle \varPhi ^{\tau } (u_{j}) - \varPsi ^{\tau } (U_{j}) , \xi _{j}(\tau ) \rangle \right\| \right] \\&\quad \le \frac{1}{2}\left( {\mathbb {E}}\left[ \max _{k\le \ell } \Vert e_{k} \Vert ^2 \right] + 2C_\varPsi ^2 \tau ^{1 + 2q}\right) \\&\qquad + 36(1 + \tau ^*)(1 + C_\varPhi \tau ^*)^2\sum ^{\ell -1}_{j=0} {\mathbb {E}}\left[ \Vert \xi _{j}(\tau ) \Vert ^2 \right] \\&\quad \le \frac{1}{2}\left( {\mathbb {E}}\left[ \max _{k\le \ell } \Vert e_{k} \Vert ^2 \right] + 2C_\varPsi ^2 \tau ^{1 + 2q}\right) \\&\qquad + 36(1 + \tau ^*)(1 + C_\varPhi \tau ^*)^2T\tau ^{-1} \left( C_{\xi ,R}\tau ^{p + 1/2}\right) ^2 \end{aligned}$$

Combining the preceding estimates, we obtain

$$\begin{aligned}&{\mathbb {E}}\biggl [ \max _{\ell \le k} \Vert e_{\ell } \Vert ^{2} \biggr ] \\&\quad \le \tau C_1 \sum _{j = 0}^{k - 1} {\mathbb {E}}\biggl [ \max _{\ell \le j} \Vert e_{\ell } \Vert ^{2} \biggr ] + 2TC_\varPsi ^2\tau ^{2q} + TC_{\xi ,R}^2\tau ^{2p} \\&\qquad + \frac{1}{2}\left( {\mathbb {E}}\left[ \max _{k\le \ell } \Vert e_{k} \Vert ^2 \right] + 2C_\varPsi ^2 \tau ^{1 + 2q}\right) \\&\qquad + 36(1 + \tau ^*)(1 + C_\varPhi \tau ^*)^2T\tau ^{-1} \left( C_{\xi ,R}\tau ^{p + 1/2}\right) ^2, \end{aligned}$$

and by rearranging terms and using that $\tau <\tau ^*\le 1$, we obtain

$$\begin{aligned}&{\mathbb {E}}\biggl [ \max _{\ell \le k} \Vert e_{\ell } \Vert ^{2} \biggr ] \\&\quad \le 2\tau C_1 \sum _{j = 0}^{k - 1} {\mathbb {E}}\biggl [ \max _{\ell \le j} \Vert e_{\ell } \Vert ^{2} \biggr ]+ 4(1 + T)C_\varPsi ^2\tau ^{2q} \\&\qquad + 2TC_{\xi ,R}^2(1 + 36(1 + \tau ^*)(1 + C_\varPhi \tau ^*)^2)\tau ^{2p}. \end{aligned}$$

By the discrete Grönwall inequality (Theorem 2.1) with $x_k:={\mathbb {E}}[ \max _{\ell \le k} \Vert e_{\ell } \Vert ^{2} ]$ and constant $\alpha _k$ and $\beta _j=2\tau C_1$, and by using that $K=T/\tau$, we obtain

$$\begin{aligned}&{\mathbb {E}}\biggl [ \max _{\ell \in [K]} \Vert e_{\ell } \Vert ^{2} \biggr ]\\&\quad \le \exp ( 2 T C_1 )\left[ 4(1 + T)C_\varPsi ^2\tau ^{2q}\right. \\&\qquad \left. + \,2TC_{\xi ,R}^2(1 + 36(1 + \tau ^*)(1 + C_\varPhi \tau ^*)^2)\tau ^{2p}\right] . \end{aligned}$$

This establishes (3.4). $\square$

Proof of Theorem 3.5

Let $0\le k\le K-1$ and $n \in {\mathbb {N}}$. By applying the triangle inequality, (2.3), Assumptions 3.1 and 3.2, and by using that $1 + \tau 2^{n-1}\le 1 + 2^{n-1}$ (since $\tau \le 1$),

$$\begin{aligned}&\Vert e_{k + 1} \Vert ^n \\&\quad \le \left( \Vert \varPhi ^\tau (u_k)-\varPsi ^\tau (U_k) \Vert + \Vert \xi _k(\tau ) \Vert \right) ^n \\&\quad \le (1 + \tau 2^{n-1})\Vert \varPhi ^\tau (u_k)-\varPsi ^\tau (U_k) \Vert ^n \\&\qquad + (1 + (2/\tau )^{n-1})\Vert \xi _k(\tau ) \Vert ^n \\&\quad \le (1 + \tau 2^{n-1})\left( (1 + \tau 2^{n-1}) \Vert \varPhi ^\tau (u_k)-\varPhi ^\tau (U_k) \Vert ^n\right. \\&\qquad +\left. (1 + (2/\tau )^{n-1})\Vert \varPhi ^\tau (U_k)-\varPsi ^\tau (U_k) \Vert ^n\right) \\&\qquad + (1 + (2/\tau )^{n-1})\Vert \xi _k(\tau ) \Vert ^n \\&\quad \le (1 + \tau 2^{n-1})^2(1 + \tau C_\varPhi )^n\Vert e_k \Vert ^n \\&\qquad + \left( 1 + (2/\tau )^{n-1}\right) \left( (1 + 2^{n-1})C_\varPsi ^n\tau ^{n(q + 1)} + \Vert \xi _k(\tau ) \Vert ^n\right) . \end{aligned}$$

Observe that, since $2^{n-1}$ and $C_\varPhi$ are non-negative, and since $0<\tau <\tau ^*$,

$$\begin{aligned} C_\varPhi (n,\tau ):=\left[ (1 + \tau 2^{n-1})^2(1 + \tau C_\varPhi )^n-1\right] \tau ^{-1}. \end{aligned}$$

(A.5)

Note that $C_\varPhi (n,\tau )\le C_\varPhi (n,\tau ^*)$.

Since $n\ge 1$ implies that $1 + (2/\tau )^{n-1}\le 2^n\tau ^{1-n}$, we have

$$\begin{aligned}&\Vert e_{k + 1} \Vert ^n-\Vert e_{k} \Vert ^n \\&\quad \le C_\varPhi (m,\tau ^*)\tau \Vert e_{k} \Vert ^{m} + \tau ^{1-n}(1 + 2^{n-1})^2C^n_\varPsi \tau ^{n(q + 1)} \\&\qquad + \tau ^{1-n}(1 + 2^{n-1})\Vert \xi _k(\tau ) \Vert ^n \\&\quad \le C_\varPhi (m,\tau ^*)\tau \Vert e_k \Vert ^n + \tau ^{1-n}(4 C_\varPsi \tau ^{q + 1})^n \\&\qquad + \tau ^{1-n}(2\Vert \xi _k(\tau ) \Vert )^n. \end{aligned}$$

Decomposing $\Vert e_{k + 1} \Vert ^n-\Vert e_0 \Vert ^n$ as a telescoping sum, using that $e_0=u_0-U_0=0$, using the non-negativity of the summands on the right-hand side of the last inequality, and using the relation $\Vert e_\ell \Vert ^n\le \max _{j\le \ell }\Vert e_j \Vert ^n$, we obtain

$$\begin{aligned} \max _{\ell \le k + 1}\Vert e_\ell \Vert ^n&\le \left( \tau ^{1-n}\sum ^{K-1}_{k=0}\left( (4 C_\varPsi \tau ^{q + 1})^n + \left( 2\Vert \xi _k(\tau ) \Vert \right) ^n\right) \right) \\&\quad + C_\varPhi (n,\tau ^*)\tau \sum ^k_{\ell =0}\max _{j\le \ell }\Vert e_j \Vert ^n. \end{aligned}$$

Using that $K=T\tau$ and Grönwall’s inequality (Theorem 2.1),

$$\begin{aligned}&\max _{\ell \in [K]}\Vert e_\ell \Vert ^n \nonumber \\&\quad \le (4 C_\varPsi \tau ^{q})^n T\exp \left( TC_\varPhi (n,\tau ^*)\right) \nonumber \\&\qquad + \left( \tau ^{1-n}2^n\sum ^{K-1}_{k=0}\Vert \xi _k(\tau ) \Vert ^n\right) \exp \left( TC_\varPhi (n,\tau ^*)\right) . \end{aligned}$$

(A.6)

Taking expectations, using (3.2) with $w=n$ and $v=1$, and using that $K=T/\tau$ yields

$$\begin{aligned} {\mathbb {E}}\left[ \max _{\ell \in [K]}\Vert e_\ell \Vert ^n\right]&\le (4 C_\varPsi \tau ^{q})^n T\exp \left( TC_\varPhi (n,\tau ^*)\right) \\&\quad + 2^n TC^n_{\xi ,R}\tau ^{n(p-1/2)}\exp \left( TC_\varPhi (n,\tau ^*)\right) . \end{aligned}$$

Rearranging the above produces the desired inequality. $\square$

Proof of Corollary 3.1

Let $m\in {\mathbb {N}}$ be arbitrary. Using (A.6), and applying (2.4) twice, we obtain

$$\begin{aligned} \max _{\ell \in [K]}\Vert e_\ell \Vert ^{nm}&\le 2^{m-1}e^{(TC_\varPhi (n,\tau ^*))^m}\biggr [\left( (4 C_\varPsi \tau ^q)^{n}T\right) ^m \\&\quad +\left. (\tau ^{1-n}2^{n})^m\left( \sum ^{K-1}_{k=0}\Vert \xi _k(\tau ) \Vert ^{n}\right) ^m\right] . \end{aligned}$$

Taking expectations and using (3.2) with $w=n$ and $v=m$, we obtain

$$\begin{aligned}&{\mathbb {E}}\left[ \max _{\ell \in [K]}\Vert e_\ell \Vert ^{nm}\right] \\&\quad \le 2^{m-1}\exp (TC_\varPhi (n,\tau ^*))^m\left( \left( (4 C_\varPsi \tau ^q)^{n}T\right) ^m \right. \\&\qquad +\left. \left( \tau ^{1-n}2^n\right) ^m \left( TC^n_{\xi ,R}\tau ^{n(p + 1/2)-1}\right) ^m\right) . \end{aligned}$$

The conclusion follows by the series expansion of the exponential and the dominated convergence theorem. $\square$

Proof of Theorem 4.2

Recall that the solution map $\varPhi ^\tau$ of the initial value problem (1.1) satisfies

$$\begin{aligned} \varPhi ^\tau (a) :=a + \int _0^\tau f(\varPhi ^{t}(a)) \, \mathrm {d}t. \end{aligned}$$

For any $\tau >0$ and $a,b\in {\mathbb {R}}^{d}$, Assumption 4.1 and the integral Grönwall–Bellman inequality yield

$$\begin{aligned}&\Vert \varPhi ^{\tau }(a)-\varPhi ^{\tau }(b) \Vert \\&\quad = \left\| a-b + \int _0^{\tau } f(\varPhi ^{t}(a))-f(\varPhi ^{t}(b))\, \mathrm {d}t \right\| \\&\quad \le \left\| a-b \right\| \\&\qquad + D\int _0^{\tau } (1 + \Vert \varPhi ^{t}(a) \Vert ^s + \Vert \varPhi ^{t}(b) \Vert ^s)\Vert \varPhi ^{t}(a)-\varPhi ^{t}(b) \Vert \,\mathrm {d}t \\&\quad \le \left\| a-b \right\| \exp \left( D\int _0^{\tau } (1 + \Vert \varPhi ^{t}(a) \Vert ^s + \Vert \varPhi ^{t}(b) \Vert ^s)\, \mathrm {d}t\right) . \end{aligned}$$

Given the boundedness hypothesis on the $(\xi _k(\tau ))_{k\in [K]}$, we may define a finite constant $C>0$ that does not depend on $\tau$ or k, such that

$$\begin{aligned} \Vert \varPhi ^{\tau }(u_{k}) - \varPhi ^{\tau }(U_{k}) \Vert&\le \Vert e_{k} \Vert \exp \left( D\tau (1 + 2C)\right) \\&\le \Vert e_{k} \Vert (1 + C'\tau ). \end{aligned}$$

The rest of the proof follows in a similar manner to that of Theorem 3.5. $\square$

Proof of Lemma 4.1

In what follows, we shall omit the dependence of all random variables on $\omega$, with the understanding that $\omega$ is arbitrary. Let $n\in [K]$, where $K = T/\tau \in {\mathbb {N}}$. From (1.6) we have, by (2.1),

$$\begin{aligned} \Vert U_{n + 1} \Vert ^2 \le (1 + \tau ) \left\| \varPsi ^{\tau }(U_n) \right\| ^2 + (1 + \tau ^{-1}) \Vert \xi _n(\tau ) \Vert ^2 . \end{aligned}$$

(A.7)

Taking the inner product of (4.3) with $\varPsi ^{\tau }(U_n)$, we obtain by (4.4)

$$\begin{aligned}&\left\| \varPsi ^{\tau }(U_n) \right\| ^2\\&\quad = \langle \varPsi ^{\tau }(U_n) , U_n \rangle + \tau \langle f(\varPsi ^{\tau }(U_n)) , \varPsi ^{\tau }(U_n) \rangle \\&\quad \le \frac{1}{2} \left( \left\| \varPsi ^{\tau }(U_n) \right\| ^2 + \Vert U_n \Vert ^2\right) + \tau \left( \alpha + \beta \left\| \varPsi ^{\tau }(U_n) \right\| ^2 \right) \\&\quad = \left\| \varPsi ^{\tau }(U_n) \right\| ^2 \left( \frac{1}{2} + \beta \tau \right) + \frac{1}{2} \Vert U_n \Vert ^2 + \alpha \tau . \end{aligned}$$

Thus,

$$\begin{aligned} \left\| \varPsi ^{\tau }(U_n) \right\| ^2&\le \frac{1}{1-2\beta \tau } \left( \Vert U_n \Vert ^2 + 2\alpha \tau \right) \nonumber \\&\le \frac{1}{1-2 \vert \beta \vert \tau } \left( \Vert U_n \Vert ^2 + 2\alpha \tau \right) , \end{aligned}$$

(A.8)

where we used the inequality $1-2\vert \beta \vert \tau \le 1 + 2\beta \tau$ for the second inequality. Then, (A.7) and (A.8) yield

$$\begin{aligned} \Vert U_n \Vert ^2 \le \frac{1 + \tau }{1-2\vert \beta \vert \tau } \left( \Vert U_{n-1} \Vert ^2 + 2\alpha \tau \right) + \frac{1 + \tau }{\tau } \Vert \xi _{n-1}(\tau ) \Vert ^2 . \end{aligned}$$

(A.9)

Let $c_1(\tau ):=\tfrac{1+2\vert \beta \vert }{1-2\vert \beta \vert \tau }$ and $c_2(\tau ):=\tfrac{2\alpha }{1-2\vert \beta \vert \tau }$. By (A.9), it follows that

$$\begin{aligned}&\Vert U_n \Vert ^2 - \Vert U_{n-1} \Vert ^2 \\&\quad \le \tau c_1(\tau ) \Vert U_{n-1} \Vert ^2 + (1 + \tau ) \left( \tau c_2(\tau ) + \tau ^{-1} \Vert \xi _{n-1}(\tau ) \Vert ^2 \right) . \end{aligned}$$

Using the telescoping sum

$$\begin{aligned} \Vert U_n \Vert ^2 = \Vert U_0 \Vert ^2 + \sum ^n_{i=1}\left( \Vert U_i \Vert ^2-\Vert U_{i-1} \Vert ^2\right) \end{aligned}$$

it follows that

$$\begin{aligned}&\Vert U_n \Vert ^2 \le \Vert U_0 \Vert ^2 + \sum ^n_{i=1} \left[ \tau c_1(\tau )\Vert U_{i-1} \Vert ^2 \right. \\&\quad +\left. (1 + \tau ) \left( \tau c_2(\tau ) + \tau ^{-1}\Vert \xi _{i-1}(\tau ) \Vert ^2 \right) \right] . \end{aligned}$$

Since $n \le K :=T/\tau$, and since the right-hand side of the inequality above is non-negative,

$$\begin{aligned} \Vert U_n \Vert ^2&\le \Vert U_0 \Vert ^2 + (1 + \tau ) \left( Tc_2(\tau ) + \tau ^{-1}\sum ^{T/\tau }_{i=1}\Vert \xi _{i-1}(\tau ) \Vert ^2\right) \\&\quad + \tau c_1(\tau )\sum ^{n-1}_{i=0}\Vert U_i \Vert ^2. \end{aligned}$$

Applying the Grönwall inequality (Theorem 2.1), yields, for all $n\in [K]$,

$$\begin{aligned}&\max _{i\in [K]} \Vert U_i \Vert ^2 \\&\quad \le \left[ \Vert U_0 \Vert ^2 + (1 + \tau ) \left( Tc_2 + \frac{1}{\tau }\sum ^{T/\tau }_{i=1} \Vert \xi _{i-1}(\tau ) \Vert ^2 \right) \right] \exp \left( Tc_1 \right) \\&\quad \le (1 + \tau ) \left[ \Vert U_0 \Vert ^2 + Tc_2 + \frac{1}{\tau } \sum ^{T/\tau }_{i=1} \Vert \xi _{i-1}(\tau ) \Vert ^2\right] \exp \left( Tc_1\right) \\&\quad \le C_2 \left( 1 + \tau ^{-1}\sum ^{T/\tau }_{i=1}\Vert \xi _{i-1}(\tau ) \Vert ^2 \right) , \end{aligned}$$

where we define, for $\tau '$ as in Assumption 4.4, the scalar

$$\begin{aligned} C_2=(1 + \tau ')\max \left\{ 1, \Vert U_0 \Vert ^2 + Tc_2(\tau ')\right\} \exp \left( Tc_1(\tau ')\right) . \end{aligned}$$

(A.10)

This yields (4.5) for $n=1$. By applying (2.4), we obtain (4.5) for arbitrary $n\in {\mathbb {N}}$. $\square$

Proof of Proposition 4.2

Recall that in Assumption 4.1, we assume $f \in C^1({\mathbb {R}}^{d}; {\mathbb {R}}^{d})$. Therefore, Taylor’s theorem applied to the function $t\mapsto \varPhi ^t(a)$ yields

$$\begin{aligned} \varPhi ^\tau (a)=a + \tau f(a) + \tau R^\tau (a), \end{aligned}$$

(A.11)

where $R^\tau (a)\rightarrow 0$ as $\tau \rightarrow 0$. Then, by (4.1a), (4.3), and (2.4),

$$\begin{aligned}&\Vert \varPsi ^\tau (U_k)-\varPhi ^\tau (U_k) \Vert ^{2n} \nonumber \\&\quad \le 2^{2n-1}\tau ^{2n} \left( \Vert f(\varPsi ^\tau (U_k))- f(U_k) \Vert ^{2n}+ \Vert R^\tau (U_k) \Vert ^{2n}\right) . \end{aligned}$$

(A.12)

By (4.1a), (4.3), (4.2), and (2.4) with the fact that $C_\varPhi \ge 1$ in Assumption 4.1, we obtain

$$\begin{aligned}&\left\| f(U_{k})-f(\varPsi ^{\tau }(U_{k})) \right\| ^{2n} \\&\quad \le C_\varPhi ^{2n} \left( 1 + \Vert U_{k} \Vert ^{s} + \left\| \varPsi ^{\tau }(U_{k}) \right\| ^{s} \right) ^{2n} \left\| U_{k}-\varPsi ^{\tau }(U_{k}) \right\| ^{2n} \\&\quad = C_\varPhi ^{2n} \left( 1 + \Vert U_{k} \Vert ^{s} + \left\| \varPsi ^{\tau }(U_{k}) \right\| ^{s}\right) ^{2n} \left\| \tau f(\varPsi ^{\tau }(U_{k})) \right\| ^{2n} \\&\quad \le \tau ^{2n} C_\varPhi ^{2n}\left( 1 + \Vert U_{k} \Vert ^{s} + \left\| \varPsi ^{\tau }(U_{k}) \right\| ^{s}\right) ^{2n} \\&\qquad \times \left( C_\varPhi \left( 1 + \left\| \varPsi ^{\tau }(U_{k}) \right\| ^s\right) \Vert \varPsi ^\tau (U_k) \Vert + \Vert f(0) \Vert \right) ^{2n} \\&\quad \le \tau ^{2n} 3^{2(2n-1)}C_\varPhi ^{4n}\left( 1 + \Vert U_{k} \Vert ^{2ns} + \left\| \varPsi ^{\tau }(U_{k}) \right\| ^{2ns}\right) \\&\qquad \times \left( \Vert \varPsi ^\tau (U_k) \Vert ^{2n} + \left\| \varPsi ^{\tau }(U_{k}) \right\| ^{2n(s + 1)} + \Vert f(0) \Vert ^{2n}\right) . \end{aligned}$$

From (A.8) and (2.4), it holds that for any n and r such that $nr\ge 1$,

$$\begin{aligned} \Vert \varPsi ^\tau (U_k) \Vert ^{2nr}&\le \frac{2^{nr-1}}{(1-2\vert \beta \vert \tau )^{nr}}\left( \Vert U_k \Vert ^{2nr} + (2\alpha \tau )^{nr}\right) \\&\le \frac{2^{nr-1}}{(1-2\vert \beta \vert \tau ')^{nr}}\left( \Vert U_k \Vert ^{2nr} + (2\alpha \tau ')^{nr}\right) , \end{aligned}$$

for $\tau '$ in Assumption 4.4. Applying the second inequality for the appropriate values of r and computing exponents yields that, for the polynomials $\pi _1$, $\pi _2$ and $\pi$ defined on ${\mathbb {R}}$ by

$$\begin{aligned} \pi _1(x)&:=\left( 1 + x^{ns} + \frac{2^{ns-1}}{(1-2\vert \beta \vert \tau ')^{ns}} \left( x^{ns} + \left( 2\alpha \tau '\right) ^{ns}\right) \right) \\ \pi _2(x)&:=\frac{2^{n(s + 1)-1}}{(1-2\vert \beta \vert \tau ')^{n(s + 1)}} \\&\qquad \times \,\left( x^{n} + \left( 2\alpha \tau '\right) ^{n} + x^{n(s + 1)}+ \left( 2\alpha \tau '\right) ^{n(s + 1)} + \Vert f(0) \Vert ^{2n}\right) \end{aligned}$$

and $\pi (x):=\pi _1(x)\pi _2(x)$, it follows from Lemma 4.1 that

$$\begin{aligned}&\left\| f(U_{k})-f(\varPsi ^{\tau }(U_{k})) \right\| ^{2n} \\&\quad \le \tau ^{2n}3^{2(2n-1)}C^{4n}_\varPhi \pi \left( \Vert U_k \Vert ^2\right) \\&\quad \le \tau ^{2n}3^{2(2n-1)}C^{4n}_\varPhi \pi \left( \max _{i\in [K]}\Vert U_i \Vert ^2\right) . \end{aligned}$$

Taking expectations, applying Proposition 4.1, and using that $\tau <\tau '$ to bound the right-hand side of inequality (4.6) in Proposition 4.1, we may define some $C_3=C_3(\alpha ,\beta ,C_\varPhi ,\tau ',n)$ that does not depend on k or $\tau$, such that

$$\begin{aligned}&{\mathbb {E}}\left[ \left\| f(U_{k})-f(\varPsi ^{\tau }(U_{k})) \right\| ^{2n}\right] \nonumber \\&\quad \le \tau ^{2n} 3^{2(2n-1)}C_\varPhi ^{4n}{\mathbb {E}}\left[ \pi \left( \max _{i\in [K]}\Vert U_i \Vert ^2\right) \right] =:\tau ^{2n}C_3. \end{aligned}$$

(A.13)

By Proposition 4.1, the finiteness of $C_3$ follows from the hypothesis $R\ge 2n(2s + 1)$ and the observation that $\pi _1(x^2)$ and $\pi _2(x^2)$ have degree ns and $n(s + 1)$ in $x^2$, respectively.

Now it remains to show that $\Vert R^\tau (U_k) \Vert ^{2n}\in L^1_{{\mathbb {P}}}$. From (4.1a), (4.1b), and (A.11), we obtain

$$\begin{aligned}&\tau \Vert R^\tau (a)-R^\tau (b) \Vert \nonumber \\&\quad =\Vert \varPhi ^\tau (a)-a-\tau f(a)-\varPhi ^\tau (b)-b-\tau f(b) \Vert \nonumber \\&\quad \le \Vert \varPhi ^\tau (a)-\varPhi ^\tau (b) \Vert +\Vert a-b \Vert +\tau \Vert f(a)-f(b) \Vert \nonumber \\&\quad = 2\left( 1 + \tau C_\varPhi \left( 1 + \Vert a \Vert ^s + \Vert b \Vert ^s\right) \Vert a-b \Vert \right) \end{aligned}$$

(A.14)

By the triangle inequality and (A.14),

$$\begin{aligned}&\tau \Vert R^\tau (U_k) \Vert&\\&\quad \le \tau \left( \Vert R^\tau (0) \Vert + 2\left( 1 + \tau C_\varPhi (1 + \Vert U_k \Vert ^s)\Vert U_k \Vert \right) \right) \\&\quad \le \tau \left( \Vert R^\tau (0) \Vert + 2\left( 1 + \tau C_\varPhi (\Vert U_k \Vert + \Vert U_k \Vert ^{s + 1})\right) \right) \\&\quad \le \tau \biggr ( \Vert R^\tau (0) \Vert + 2 \\&\qquad \left. +\, 2C_\varPhi \left( \max _{k\in [K]}\Vert U_k \Vert + \max _{k\in [K]}\Vert U_k \Vert ^{s + 1}\right) \right) . \end{aligned}$$

Then by applying (2.4) and Proposition 4.1 with the hypothesis that $R\ge 2n(2s + 1)\ge 2n(s + 1)$, and using the bound $\tau <\tau '$, it follows that we may define a positive scalar $C_4$ that does not depend on k or $\tau$, such that

$$\begin{aligned}&{\mathbb {E}}\left[ \tau ^{2n}\Vert R^\tau (U_k) \Vert ^{2n} \right] \nonumber \\&\quad \le \tau ^{2n}{\mathbb {E}}\biggr [\biggr ( \Vert R^\tau (0) \Vert + 2 \nonumber \\&\qquad + \,\left. \left. 2C_\varPhi \left( \max _{k\in [K]}\Vert U_k \Vert + \max _{k\in [K]}\Vert U_k \Vert ^{s + 1}\right) \right) ^{2n}\right] \nonumber \\&\quad =:\tau ^{2n}C_4. \end{aligned}$$

(A.15)

Therefore, with $C_3$ and $C_4$ as in (A.13) and (A.15) above, (A.12) yields

$$\begin{aligned} {\mathbb {E}}\left[ \Vert \varPsi ^\tau (U_k)-\varPhi ^\tau (U_k) \Vert ^{2n}\right]&\le 2^{2n-1}\tau ^{4n}\left( C_3 + C_4\right) \nonumber \\&=:C_\varPsi \tau ^{4n} \end{aligned}$$

(A.16)

as desired. $\square$

The proof below makes clear that we make absolutely no effort to find optimal constants.

Proof of Theorem 4.5

Let $n\in {\mathbb {N}}$. By (2.3)

$$\begin{aligned}&\Vert e_{k + 1} \Vert ^{2n} \\&\quad \le (1 + \tau 2^{2n-1})\left[ (1 + \tau 2^{2n-1})\Vert \varPhi ^\tau (u_k)-\varPhi ^\tau (U_k) \Vert ^{2n}\right. \\&\qquad \left. + (1 + (2/\tau )^{2n-1})\Vert \varPhi ^\tau (U_k)-\varPsi ^\tau (U_k) \Vert ^{2n}\right] \\&\qquad + (1 + (2/\tau )^{2n-1}\Vert \xi _k(\tau ) \Vert ^{2n}. \end{aligned}$$

Since $\tau \le 1$ and $n\ge 1$, it holds that $1 + \tau ^{1-2n} 2^{2n-1}\le \tau ^{1-2n}(1 + 2^{2n-1})$ and $1 + \tau 2^{2n-1}\le 1 + 2^{2n-1}$. Using these inequalities, (2.3), and (4.1b) in the preceding inequality, we obtain

$$\begin{aligned}&\Vert e_{k + 1} \Vert ^{2n} \\&\quad \le (1 + \tau 2^{2n-1})^2\left( 1 + \tau C_\varPhi (1 + \Vert u_k \Vert ^s + \Vert U_k \Vert ^s)\right) ^{2n}\Vert e_k \Vert ^{2n} \\&\qquad + (1 + 2^{2n-1})^2\tau ^{1-2n} \\&\qquad \times \left( \Vert \varPhi ^\tau (U_k)-\varPsi ^\tau (U_k) \Vert ^{2n} + \Vert \xi _k(\tau ) \Vert ^{2n}\right) . \end{aligned}$$

Using (2.3) again, we obtain

$$\begin{aligned}&\left( 1 + \tau C_\varPhi (1 + \Vert u_k \Vert ^s + \Vert U_k \Vert ^s)\right) ^{2n}\Vert e_k \Vert ^{2n} \\&\quad \le \left[ \left( 1 + \tau C_\varPhi \left( 1 + \max _{\ell \in [K]}\Vert u_\ell \Vert ^s\right) \right) \Vert e_k \Vert \right. \\&\qquad +\left. \tau C_\varPhi \max _{\ell \in [K]}\Vert U_\ell \Vert ^s\Vert e_k \Vert \right] ^{2n} \\&\quad \le (1 + \tau 2^{2n-1})\left( 1 + \tau C_\varPhi \left( 1 + \max _{\ell \in [K]}\Vert u_\ell \Vert ^s\right) \right) ^{2n}\Vert e_k \Vert ^{2n} \\&\qquad + (1 + \tau ^{1-2n}2^{2n-1})(\tau C_\varPhi )^{2n}\max _{\ell \in [K]}\Vert U_\ell \Vert ^{2ns}\Vert e_k \Vert ^{2n} \end{aligned}$$

so that by defining

$$\begin{aligned} C_5=C_5 (n,s,C_\varPhi ,u):=\max \{2^{2n-1},C_\varPhi (1 + \Vert u \Vert ^s_\infty )\} \end{aligned}$$

we have

$$\begin{aligned}&\left( 1 + \tau C_\varPhi (1 + \Vert u_k \Vert ^s + \Vert U_k \Vert ^s)\right) ^{2n}\Vert e_k \Vert ^{2n} \\&\quad \le \left( 1 + \tau C_5\right) ^{2n + 1}\Vert e_k \Vert ^{2n} \\&\qquad + \tau ^{1-2n}(1 + 2^{2n-1})\left( \tau C_\varPhi \max _{\ell \in [K]}\Vert U_\ell \Vert ^{s}\Vert e_k \Vert \right) ^{2n} \end{aligned}$$

and, therefore,

$$\begin{aligned}&\Vert e_{k + 1} \Vert ^{2n}-\Vert e_k \Vert ^{2n} \\&\quad \le \left[ \left( 1 + \tau C_5\right) ^{2n + 1}-1\right] \tau ^{-1}\tau \Vert e_k \Vert ^{2n} \\&\qquad + \tau ^{1-2n}(1 + 2^{2n-1})\left( \tau C_\varPhi \max _{\ell \in [K]}\Vert U_\ell \Vert ^{s}\Vert e_k \Vert \right) ^{2n} \\&\qquad + (1 + 2^{2n-1})^2\tau ^{1-2n} \\&\qquad \times \left( \Vert \varPhi ^\tau (U_k)-\varPsi ^\tau (U_k) \Vert ^{2n} + \Vert \xi _k(\tau ) \Vert ^{2n}\right) . \end{aligned}$$

By non-negativity of $C_5$, it follows that $[(1 + \tau C_5)^{2n + 1}-1]\tau ^{-1}$ is a polynomial of degree 2n in $\tau$ with positive coefficients. In particular, if we recall the definition of $C_5$ and define $C_6$ by

$$\begin{aligned}&C_6(C_\varPhi ,n,s,(u_t)_{0\le t\le T},\tau ') \nonumber \\&\quad :=\left[ \left( 1 + \tau ' \max \{2^{2n-1},C_\varPhi (1 + \Vert u \Vert ^s_{\infty }\}\right) ^{2n + 1}-1\right] (\tau ')^{-1} , \end{aligned}$$

(A.17)

then $C_6$ does not depend on $\tau$, $[(1 + \tau C_5)^{2n + 1}-1]\tau ^{-1}\le C_6$ for all $0<\tau <\tau '$, and

$$\begin{aligned}&\Vert e_{k + 1} \Vert ^{2n}-\Vert e_k \Vert ^{2n} \\&\quad \le C_6\tau \Vert e_k \Vert ^{2n} \\&\qquad + \tau ^{1-2n}(1 + 2^{2n-1})\left( \tau C_\varPhi \max _{\ell \in [K]}\Vert U_\ell \Vert ^{s}\Vert e_k \Vert \right) ^{2n} \\&\qquad + (1 + 2^{2n-1})^2\tau ^{1-2n} \\&\qquad \times \left( \Vert \varPhi ^\tau (U_k)-\varPsi ^\tau (U_k) \Vert ^{2n} + \Vert \xi _k(\tau ) \Vert ^{2n}\right) . \end{aligned}$$

By the telescoping sum associated to $\Vert e_{k + 1} \Vert ^{2n}-\Vert e_k \Vert ^{2n}$, the fact that $e_0=0$, the bound $1 + 2^{2n-1}\le 2^{2n}$, the non-negativity of the terms on the right-hand side of the inequality above, and the bound $\Vert e_j \Vert \le \max _{\ell \le j}\Vert e_\ell \Vert$, we obtain

$$\begin{aligned}&\max _{\ell \le k + 1}\Vert e_{\ell } \Vert ^{2n} \\&\quad \le \tau ^{1-2n}2^{4n} \sum ^{K}_{\ell =1}\left( \left( \tau C_\varPhi \max _{j\in [K]}\Vert U_j \Vert ^{s}\Vert e_\ell \Vert \right) ^{2n} \right. \\&\qquad + \Vert \varPhi ^\tau (U_\ell )-\varPsi ^\tau (U_\ell ) \Vert ^{2n} + \Vert \xi _\ell (\tau ) \Vert ^{2n}\biggr ) \\&\qquad + C_6\tau \sum ^{k}_{\ell =1} \max _{j\le \ell }\Vert e_j \Vert ^{2n}. \end{aligned}$$

By Lemma 4.1,

$$\begin{aligned}&\left( \tau C_\varPhi \max _{j\in [K]}\Vert U_j \Vert ^{s}\Vert e_\ell \Vert \right) ^{2n} \\&\quad \le C_\varPhi ^{2n}(2C_1)^{sn} \\&\qquad \times \left( \tau ^{2n}\Vert e_\ell \Vert ^{2n} + \tau ^{n(2-s)}\left( \sum ^{T/\tau }_{i=1}\Vert \xi _i(\tau ) \Vert ^2\right) ^{ns}\Vert e_\ell \Vert ^{2n}\right) \end{aligned}$$

which implies that

$$\begin{aligned}&\max _{\ell \le k + 1}\Vert e_{\ell } \Vert ^{2n} \\&\quad \le \tau ^{1-ns}2^{n(4 + s)}C_\varPhi ^{2n}C_1^{ns} \sum ^{K}_{\ell =1}\left( \sum ^{T/\tau }_{i=1}\Vert \xi _i(\tau ) \Vert ^2\right) ^{ns}\Vert e_\ell \Vert ^{2n} \\&\qquad + \tau ^{1-2n}2^{4n} \sum ^{K}_{\ell =1}\left( \Vert \varPhi ^\tau (U_\ell )-\varPsi ^\tau (U_\ell ) \Vert ^{2n} + \Vert \xi _\ell (\tau ) \Vert ^{2n}\right) \\&\qquad + \left( C_6\tau + \tau ^{2n}2^{n(4 + s)}C_\varPhi ^{2n}C_1^{ns}\right) \sum ^{k}_{\ell =1} \max _{j\le \ell }\Vert e_j \Vert ^{2n}. \end{aligned}$$

Define

$$\begin{aligned} C_7=C_7(n,s,C_\varPhi ,C_1):=2^{n(4 + s)}C_\varPhi ^{2n}C_1^{ns}. \end{aligned}$$

(A.18)

Since $C_\varPhi ,C_1\ge 1$, it follows that $2^{4n}\le C_7$ ,and by Grönwall’s inequality (Theorem 2.1) we obtain

$$\begin{aligned}&\max _{\ell \in [K]}\Vert e_{\ell } \Vert ^{2n} \\&\quad \le \exp \left( T(C_6 + \tau ^{2n-1}C_7)\right) C_7 \\&\qquad \times \left( \tau ^{1-ns}\sum ^K_{\ell =1}\left( \sum ^{T/\tau }_{i=1}\Vert \xi _i(\tau ) \Vert ^2\right) ^{ns}\Vert e_\ell \Vert ^{2n}\right. \\&\qquad + \left. \tau ^{1-2n}\sum ^K_{\ell =1}\left( \Vert \varPhi ^\tau (U_\ell )-\varPsi ^\tau (U_\ell ) \Vert ^{2n} + \Vert \xi _\ell (\tau ) \Vert ^{2n}\right) \right) . \end{aligned}$$

Taking expectations completes the proof, provided that we can ensure each sum is of the right order in $\tau$. By Proposition 4.2 with the hypothesis that $R\ge 2n(2s + 1)$, and by Assumption 3.3,

$$\begin{aligned}&\tau ^{1-2n} \sum ^K_{\ell =1} {\mathbb {E}}\left[ \Vert \varPhi ^\tau (U_\ell ) - \varPsi ^\tau (U_\ell ) \Vert ^{2n} + \Vert \xi _\ell (\tau ) \Vert ^{2n}\right] \nonumber \\&\quad \le T\left( C_\varPsi \tau ^{2n} + \left( C_{\xi ,R}\tau ^{p-1/2}\right) ^{2n}\right) . \end{aligned}$$

(A.19)

Thus, we need $p-1/2\ge 1$ to hold. Next, using the bound $\Vert e_{\ell } \Vert \le \max _{j\in [K]}\Vert e_j \Vert$, Young’s inequality (2.1) with $a=(\sum ^{K}_{i=1}\Vert \xi _{i}(\tau ) \Vert ^2)^{ns}$, $b=\Vert e_{\ell } \Vert ^{2n}$, and some $\delta >0$ and conjugate exponent pair $(r,r^*)\in (1,\infty )^2$ to be determined later, we obtain with (3.2) that

$$\begin{aligned}&{\mathbb {E}}\left[ \left( \sum ^{T/\tau }_{i=1} \Vert \xi _{i}(\tau ) \Vert ^{2}\right) ^{ns} \Vert e_{\ell } \Vert ^{2n}\right] \\&\quad \le \frac{\delta }{r} {\mathbb {E}}\left[ \left( \sum ^{T/\tau }_{i=1} \Vert \xi _{i}(\tau ) \Vert ^2\right) ^{nrs} \right] + \frac{1}{\delta ^{r^*/r}r^*}{\mathbb {E}}\left[ \max _{\ell \in [K]} \Vert e_{\ell } \Vert ^{2nr^*}\right] \\&\quad \le \frac{\delta }{r} \left( TC^2_{\xi ,R}\tau ^{2p}\right) ^{nrs} + \frac{1}{\delta ^{r^*/r}r^*}{\mathbb {E}}\left[ \max _{\ell \in [K]} \Vert e_{\ell } \Vert ^{2nr^*}\right] . \end{aligned}$$

Since $R\ge 2n(2s + 1)$, the maximal value of r compatible with integrability of $(\sum ^{K}_{i=1}\Vert \xi _{i}(\tau ) \Vert ^2)^{nrs}$ is $r=2 + s^{-1}$. Since we are not interested in optimal estimates, we shall set $r=r^*=2$ and $\delta =\tau ^{-n(2 + s)}$. We thus obtain

$$\begin{aligned}&\tau ^{1-ns}\sum ^K_{\ell =1}~{\mathbb {E}}\left[ \left( \sum ^{T/\tau }_{i=1} \Vert \xi _{i}(\tau ) \Vert ^{2}\right) ^{ns} \Vert e_{\ell } \Vert ^{2n}\right] \\&\quad \le \frac{T}{2}\tau ^{-ns} \left( (TC^2_{\xi ,R})^{nrs}\tau ^{-n(2 + s) + 2p(2ns)} \right. \\&\qquad \left. + \,\tau ^{n(2 + s)}{\mathbb {E}}\left[ \max _{\ell \in [K]}\Vert e_\ell \Vert ^{4n}\right] \right) . \end{aligned}$$

For the exponent of $\tau$ of the first term in the parentheses, we want to ensure that $-n(2 + s) + 2p(2ns)-ns\ge 2n$, or equivalently that $p\ge \tfrac{1}{s} + \tfrac{1}{2}$. Comparing this condition with the condition $p-\tfrac{1}{2}\ge 1$ that arose from (A.19), and recalling that $s\ge 1$, we observe that if $p\ge \tfrac{3}{2}$, then the preceding estimates yield

$$\begin{aligned}&{\mathbb {E}}\left[ \max _{\ell \in [K]}\Vert e_{\ell } \Vert ^{2n}\right] \\&\quad \le \exp \left( T(C_6 + \tau ^{2n-1}C_7)\right) \frac{C_7 T}{2} \\&\qquad \times \left( \left( TC^2_{\xi ,R}\right) ^{2ns} + {\mathbb {E}}\left[ \max _{\ell \in [K]}\Vert e_\ell \Vert ^{4n}\right] \right) \tau ^{2n}. \end{aligned}$$

It remains to bound ${\mathbb {E}}[\max _{\ell \in [K]}\Vert e_\ell \Vert ^{4n}]$ by a constant that does not depend on $\tau$. By (2.4), Proposition 4.1, and the assumption that $\tau <\tau '$ for $\tau '$ in Assumption 4.4, we obtain

$$\begin{aligned}&{\mathbb {E}}\left[ \max _{\ell \in [K]}\Vert e_\ell \Vert ^{4n}\right] \nonumber \\&\quad \le 2^{4n}\left( \max _{k\in [K]}\Vert u_k \Vert ^{4n} + {\mathbb {E}}\left[ \max _{k\in [K]}\Vert U_k \Vert ^{4n}\right] \right) \nonumber \\&\quad \le 2^{4n}\left( \Vert u \Vert _{\infty }^{4n} + (2C_2)^{2n}\left( 1 + TC_{\xi ,R}^2\tau ^{2p-1}\right) ^{2n}\right) \nonumber \\&\quad \le 2^{4n}\left( \Vert u \Vert _{\infty }^{4n} + (2C_2)^{2n}\left( 1 + TC_{\xi ,R}^2(\tau ')^{2p-1}\right) ^{2n}\right) =:C_8, \end{aligned}$$

(A.20)

where $C_8=C_8(C_2,C_{\xi ,R},n,p,\tau ',T,(u_t)_{0\le t\le T})>0$ does not depend on $\tau$. Note that in applying Proposition 4.1, we have used that $s\ge 1$ for the exponent s in Assumption 4.1, since this implies that $2n(2s + 1)\ge 4n$. $\square$

Proof of Theorem 5.2

Let $k\in [K]$ and $t_{k}<t\le t_{k + 1}$. Then

$$\begin{aligned} e(t)&= \varPhi ^{t - t_{k}}(u_{k}) - \varPsi ^{t - t_{k}}(U_{k}) - \xi _{k}(t - t_{k}) \\&= \varPhi ^{t - t_{k}}(u_{k}) - \varPhi ^{t - t_{k}}(U_{k}) + \varPhi ^{t - t_{k}}(U_{k}) \\&\quad - \varPsi ^{t - t_{k}}(U_{k}) - \xi _{k}(t - t_{k}) , \end{aligned}$$

and given that Assumption 3.1 implies that $\varPhi ^{t'}$ is Lipschitz on ${\mathbb {R}}^{d}$ for every $t'\ge 0$,

$$\begin{aligned} \Vert e(t) \Vert ^{n}&\le 3^{n-1}\left( ( 1 + C_\varPhi (t - t_{k}) )^{n} \Vert e_{k} \Vert ^{n} \right. \\&\quad \left. + \left( C_\varPsi (t - t_{k})^{q + 1}\right) ^{n} + \Vert \xi _{k}(t - t_{k}) \Vert ^{n}\right) \end{aligned}$$

by applying (2.4). Since $t-t_{k}\le \tau$, it follows from the inequality above that

$$\begin{aligned}&{\mathbb {E}}\biggl [ \sup _{0 \le t \le T} \Vert e(t) \Vert ^{n} \biggr ] \\&\quad = {\mathbb {E}}\biggl [ \max _{k\in [K]} \sup _{t_{k}<t \le t_{k + 1}} \Vert e(t) \Vert ^{n} \biggr ] \\&\quad \le 3^{n-1}\left( (1 + C_\varPhi \tau )^n{\mathbb {E}}\left[ \max _{k\in [K]}\Vert e_{k} \Vert ^{n}\right] + \left( C_\varPsi \tau ^{q + 1}\right) ^n\right. \\&\qquad \left. + \,{\mathbb {E}}\left[ \max _{0 \le k \le K-1} \sup _{t_{k} <t \le t_{k + 1}} \Vert \xi _{k}(t - t_{k}) \Vert ^{n} \right] \right) . \end{aligned}$$

By Assumption 5.1,

$$\begin{aligned}&{\mathbb {E}}\left[ \max _{0 \le k \le K-1} \sup _{t_{k}< t \le t_{k + 1}} \Vert \xi _{k}(t - t_{k}) \Vert ^{n} \right] \\&\quad \le \sum _{k = 0}^{K-1}{\mathbb {E}}\left[ \sup _{t_{k}< t \le t_{k + 1}} \Vert \xi _{k}(t - t_{k}) \Vert ^{n} \right] \\&\quad \le \frac{T}{\tau }\left( C_{\xi ,R} \tau ^{p + 1/2}\right) ^{n}=TC^n_{\xi ,R}\tau ^{n(p + 1/2)-1} \\&\quad \le T C^n_{\xi ,R}\tau ^{n(p-1/2)}. \end{aligned}$$

Note that Assumption 5.1 is stronger than Assumption 3.3. Therefore, we may apply Theorem 3.5 to obtain (5.1). $\square$

Proof of Lemma 5.1

If $r=0$, then the desired statement follows immediately. Therefore, let $p,r \ge 1$. Let $\xi _0$ be the integrated ${\mathbb {P}}$-Brownian motion process scaled by $\tau ^{p-1}$, so that

$$\begin{aligned} \Vert \xi _0(t) \Vert ^r&= \tau ^{r(p-1)}t^r \left\| \frac{1}{t}\int _0^t B_s \, \mathrm {d}s \right\| ^r \\&\le \tau ^{r(p-1)}t^r\left( \frac{1}{t}\int _0^t \Vert B_s \Vert ^r \, \mathrm {d}s\right) \\&= \tau ^{r(p-1)}t^{r-1}\int _0^t \Vert B_s \Vert ^r \, \mathrm {d}s, \end{aligned}$$

where we applied Jensen’s inequality to the uniform probability measure on [0, t]. It follows that

$$\begin{aligned}&{\mathbb {E}}\biggl [ \sup _{t \le \tau } \Vert \xi _0(t) \Vert ^r \biggr ] \\&\quad \le \tau ^{r(p-1)}{\mathbb {E}}\biggl [ \sup _{t \le \tau }t^{r-1}\int _0^t \Vert B_s \Vert ^r \, \mathrm {d}s \biggr ] \\&\quad \le \tau ^{r(p-1)}\tau ^{r-1}{\mathbb {E}}\biggl [ \sup _{t \le \tau }\int _0^t \Vert B_s \Vert ^r \, \mathrm {d}s \biggr ] \\&\quad \le \tau ^{r p-1} \int _0^\tau {\mathbb {E}}\biggl [ \sup _{0 \le t \le \tau } \Vert B_t \Vert ^r \biggr ] \, \mathrm {d}s \\&\quad = \tau ^{rp} {\mathbb {E}}\biggl [ \sup _{0 \le t \le \tau } \Vert B_t \Vert ^r \biggr ]. \end{aligned}$$

Above, we used the Fubini–Tonelli theorem to interchange expectation and integration with respect to s, and the fact that ${\mathbb {E}}\bigl [ \sup _{t \le \tau } \Vert B_t \Vert ^r \bigr ]$ is constant with respect to the variable of integration s. For $r=1$, the Burkholder–Davis–Gundy martingale inequality (Peškir 1996, Equation 2.2) yields

$$\begin{aligned} {\mathbb {E}}\left[ \sup _{0\le t\le \tau }\Vert B_t \Vert ^r\right] \le \frac{4-r}{2-r}\tau ^{r/2}, \end{aligned}$$

with $(4-r)/(2-r)=3$ for $r=1$. For $r>1$, Doob’s inequality (Peškir 1996, Equation 2.1) yields

$$\begin{aligned} {\mathbb {E}}\left[ \sup _{0\le t\le \tau }\Vert B_t \Vert ^r\right] \le \left( \frac{r}{r-1}\right) ^r\tau ^{r/2}. \end{aligned}$$

Since $r\mapsto [r/(r-1)]^r$ is continuously differentiable and monotonically decreasing on $2<r<\infty$, the desired conclusion follows. $\square$