Accelerated finite elements schemes for parabolic stochastic partial differential equations

Abstract

For a class of finite elements approximations for linear stochastic parabolic PDEs it is proved that one can accelerate the rate of convergence by Richardson extrapolation. More precisely, by taking appropriate mixtures of finite elements approximations one can accelerate the convergence to any given speed provided the coefficients, the initial and free data are sufficiently smooth.

1 Introduction

We are interested in finite elements approximations for Cauchy problems for stochastic parabolic PDEs of the form of Eq. (2.1) below. Such kind of equations arise in various fields of sciences and engineering, for example in nonlinear filtering of partially observed diffusion processes. Therefore these equations have been intensively studied in the literature, and theories for their solvability and numerical methods for approximations of their solutions have been developed. Since the computational effort to get reasonably accurate numerical solutions grow rapidly with the dimension d of the state space, it is important to investigate the possibility of accelerating the convergence of spatial discretisations by Richardson extrapolation. About a century ago Lewis Fry Richardson had the idea in [18] that the speed of convergence of numerical approximations, which depend on some parameter h converging to zero, can be increased if one takes appropriate linear combinations of approximations corresponding to different parameters. This method to accelerate the convergence, called Richardson extrapolation, works when the approximations admit a power series expansion in h at $h=0$ with a remainder term, which can be estimated by a higher power of h. In such cases, taking appropriate mixtures of approximations with different parameters, one can eliminate all other terms but the zero order term and the remainder in the expansion. In this way, the order of accuracy of the mixtures is the exponent $k+1$ of the power $h^{k+1}$, that estimates the remainder. For various numerical methods applied to solving deterministic partial differential equations (PDEs) it has been proved that such expansions exist and that Richardson extrapolations can spectacularly increase the speed of convergence of the methods, see, e.g., [16, 17] and [20]. Richardson’s idea has also been applied to numerical solutions of stochastic equations. It was shown first in [21] that by Richardson extrapolation one can accelerate the weak convergence of Euler approximations of stochastic differential equations. Further results in this direction can be found in [14, 15] and the references therein. For stochastic PDEs the first result on accelerated finite difference schemes appears in [7], where it is shown that by Richardson extrapolation one can accelerate the speed of finite difference schemes in the spatial variables for linear stochastic parabolic PDEs to any high order, provided the initial condition and free terms are sufficiently smooth. This result was extended to (possibly) degenerate stochastic PDEs in to [6, 8] and [9]. Starting with [22] finite elements approximations for stochastic PDEs have been investigated in many publications, see, for example, [3, 4, 10,11,12] and [23].

Our main result, Theorem  2.4 in this paper, states that for a class of finite elements approximations for stochastic parabolic PDEs given in the whole space an expansion in terms of powers of a parameter h, proportional to the size of the finite elements, exists up to any high order, if the coefficients, the initial data and the free terms are sufficiently smooth. Then clearly, we can apply Richardson extrapolation to such finite elements approximations in order to accelerate the convergence. The speed we can achieve depends on the degree of smoothness of the coefficients, the initial data and free terms; see Corollary 2.5. Note that due to the symmetry we require for the finite elements, in order to achieve an accuracy of order $J+1$ we only need $\lfloor \frac{J}{2} \rfloor$ terms in the mixture of finite elements approximation. As far as we know this is the first result on accelerated finite elements by Richardson extrapolation for stochastic parabolic equations. There are nice results on Richardson extrapolation for finite elements schemes in the literature for some (deterministic) elliptic problems; see, e.g., [1, 2] and the literature therein.

We note that in the present paper we consider stochastic PDEs on the whole space ${\mathbb {R}}^d$ in the spatial variable, and our finite elements approximations are the solutions of infinite dimensional systems of equations. Therefore one may think that our accelerated finite elements schemes cannot have any practical use. In fact they can be implemented if first we localise the stochastic PDEs in the spatial variable by multiplying their coefficients, initial and free data by sufficiently smooth non-negative “cut-off” functions with value 1 on a ball of radius R and vanishing outside of a bigger ball. Then our finite elements schemes for the “localised stochastic PDEs” are fully implementable and one can show that the results of the present paper can be extended to them. Moreover, by a theorem from [6] the error caused by the localization is of order $\exp (-\delta R^2)$ within a ball of radius $R'<R$. Moreover, under some further constraints about a bounded domain D and particular classes of finite elements such as those described in Sects. 6.1 and 6.2, our arguments could extend to parabolic stochastic PDEs on D with periodic boundary conditions. Note that our technique relies on finite elements defined by scaling and shifting one given mother element, and that the dyadic rescaling used to achieve a given speed of convergence is similar to that of wavelet approximation. We remark that our accelerated finite elements approximations can be applied also to implicit Euler–Maruyama time discretisations of stochastic parabolic PDEs to achieve higher order convergence with respect to the spatial mesh parameter of fully discretised schemes. However, as one can see by adapting and argument from [5], the strong rate of convergence of these fully discretised schemes with respect to the temporal mesh parameter cannot be accelerated by Richardson approximation. Dealing with weak speed of convergence of time discretisations is beyond the scope of this paper.

In conclusion we introduce some notation used in the paper. All random elements are defined on a fixed probability space $(\Omega ,{\mathcal {F}},P)$ equipped with an increasing family $({\mathcal {F}}_t)_{t\ge 0}$ of $\sigma$-algebras ${\mathcal {F}}_{t}\subset {\mathcal {F}}$. The predictable $\sigma$-algebra of subsets of $\Omega \times [0,\infty )$ is denoted by ${\mathcal {P}}$, and the $\sigma$-algebra of the Borel subsets of ${\mathbb {R}}^d$ is denoted by ${{\mathcal {B}}}({\mathbb {R}}^d)$. We use the notation

$$\begin{aligned} D_i=\frac{\partial }{\partial x_i}, \quad D_{ij}=D_iD_j=\frac{\partial ^2}{\partial x_i\partial x_j}, \quad i,j=1,2,\ldots ,d \end{aligned}$$

for first order and second order partial derivatives in $x=(x_1,\ldots ,x_d)\in {\mathbb {R}}^d$. For integers $m\ge 0$ the Sobolev space $H^m$ is defined as the closure of $C_0^{\infty }$, the space of real-valued smooth functions $\varphi$ on ${\mathbb {R}}^d$ with compact support, in the norm $|\varphi |_m$ defined by

$$\begin{aligned} |\varphi |_m^2=\sum _{|\alpha |\le m}\int _{{\mathbb {R}}^d}|D^{\alpha }\varphi (x)|^2\,dx, \end{aligned}$$

(1.1)

where $D^{\alpha }=D_1^{\alpha _1}\ldots D_d^{\alpha _d}$ and $|\alpha |=\alpha _1+\cdots +\alpha _d$ for multi-indices $\alpha =(\alpha _1,\ldots ,\alpha _d)$, $\alpha _i\in \{0,1,\ldots ,d\}$, and $D_i^0$ is the identity operator for $i=1,\ldots ,d$. Similarly, the Sobolev space $H^m(l_2)$ of $l_2$-valued functions are defined on ${\mathbb {R}}^d$ as the closure of the of $l_2$-valued smooth functions $\varphi =(\varphi _i)_{i=1}^{\infty }$ on ${\mathbb {R}}^d$ with compact support, in the norm denoted also by $|\varphi |_m$ and defined as in (1.1) with $\sum _{i=1}^{\infty }| D^{\alpha }\varphi _i(x)|^2$ in place of $|D^{\alpha }\varphi (x)|^2$. Unless stated otherwise, throughout the paper we use the summation convention with respect to repeated indices. The summation over an empty set means 0. We denote by C and N constants which may change from one line to the next, and by C(a) and N(a) constants depending on a parameter a.

For theorems and notations in the $L_2$-theory of stochastic PDEs the reader is referred to [13] or [19].

2 Framework and some notations

Let $(\Omega ,{\mathcal {F}},P,({{\mathcal {F}}}_t)_{t\ge 0})$ be a complete filtered probability space carrying a sequence of independent Wiener martingales $W=(W^{\rho })_{\rho =1}^{\infty }$ with respect to a filtration $({{\mathcal {F}}}_t)_{t\ge 0}$.

We consider the stochastic PDE problem

$$\begin{aligned} d u_t(x) =\big [ {{\mathcal {L}}}_t u_t(x) + f_t(x)\big ] dt + \big [ {{\mathcal {M}}}_t^{\rho }u_t(x) + g^{\rho }_t(x) \big ] dW_t^{\rho } ,\quad (t,x)\in [0,T]\times {\mathbb {R}}^d,\!\!\!\!\nonumber \\ \end{aligned}$$

(2.1)

with initial condition

$$\begin{aligned} u_0(x)=\phi (x),\quad x\in {\mathbb {R}}^d, \end{aligned}$$

(2.2)

for a given $\phi \in H^0=L_2({\mathbb {R}}^d)$, where

$$\begin{aligned} {{\mathcal {L}}}_t u (x)= & {} D_{i}( a^{ij}_t(x)D_{j} u(x))+b^i_t(x)D_i u(x)+c_t(x)u(x) , \\ {{\mathcal {M}}}^{\rho }_t u(x)= & {} \sigma ^{i\rho }_t(x)D_{i} u(x) +\nu ^{\rho }_t(x)u(x) \quad \text {for } u\in H^1=W^1_2({\mathbb {R}}^d), \end{aligned}$$

with ${\mathcal {P}}\otimes {{\mathcal {B}}}({\mathbb {R}}^d)$-measurable real-valued bounded functions $a^{ij}$, $b^i$, c, and $l_2$-valued bounded functions $\sigma ^{i}=(\sigma ^{i\rho })_{\rho =1}^{\infty }$ and $\nu =(\nu ^{\rho })_{\rho =1}^{\infty }$ defined on $\Omega \times [0,T]\times {\mathbb {R}}^d$ for $i,j\in \{1,\ldots , d\}$. Furthermore, $a^{ij}_t(x)=a^{j i}_t(x)$ a.s. for every $(t,x)\in [0,T]\times {\mathbb {R}}^d$. For $i=1,2,\ldots , d$ the notation $D_i = \frac{\partial }{\partial x_i}$ means the partial derivative in the i-th coordinate direction.

The free terms f and $g=(g^{\rho })_{\rho =1}^{\infty }$ are ${\mathcal {P}}\otimes {{\mathcal {B}}}({\mathbb {R}}^d)$-measurable functions on $\Omega \times [0,T]\times {\mathbb {R}}^d$, with values in ${\mathbb {R}}$ and $l_2$ respectively. Let $H^{m}(l_2)$ denote the $H^m$ space of $l_2$-valued functions on ${\mathbb {R}}^d$. We use the notation $|\varphi |_m$ for the $H^m$-norm of $\varphi \in H^m$ and of $\varphi \in H^m(l_2)$, and $|\varphi |_0$ denotes the $L_2$-norm of $\varphi \in H^0=L_2$.

Let $m\ge 0$ be an integer, $K\ge 0$ be a constant and make the following assumptions.

Assumption 2.1

The derivatives in $x\in {\mathbb {R}}^d$ up to order m of the coefficients $a^{ij}$, $b^i$, c, and of the coefficients $\sigma ^{i}$, $\nu$ are ${\mathcal {P}}\otimes {{\mathcal {B}}}({\mathbb {R}})$-measurable functions with values in ${\mathbb {R}}$ and in $l_2$-respectively. For almost every $\omega$ they are continuous in x, and they are bounded in magnitude by K.

Assumption 2.2

The function $\phi$ is an $H^m$-valued ${{\mathcal {F}}}_0$-measurable random variable, and f and $g=(g^{\rho })_{\rho =1}^{\infty }$ are predictable processes with values in $H^{m}$ and $H^m(l_2)$, respectively, such that

$$\begin{aligned} {{\mathcal {K}}}^2_m:=|\phi |^2_m+ \int _0^T \big ( |f_t|_{m}^2+|g_t|^2_{m} \big ) \,dt < \infty \,\,(a.s.). \end{aligned}$$

(2.3)

Assumption 2.3

There exists a constant $\kappa >0$, such that for $(\omega ,t,x)\in \Omega \times [0,T] \times {\mathbb {R}}^d$

$$\begin{aligned} \sum _{i,j=1}^d \big ( a^{ij}_t(x) -\tfrac{1}{2}\sum _\rho \sigma ^{i\rho }_t(x)\sigma ^{j\rho }_t(x) \big ) z^i z^j \ge \kappa |z|^2 \quad \text {for all }z=(z^1,\ldots ,z^d)\in {\mathbb {R}}.\nonumber \\ \end{aligned}$$

(2.4)

For integers $n\ge 0$ let ${\mathbb {W}}^n_2(0,T)$ denote the space of $H^n$-valued predictable processes $(u_t)_{t\in [0,T]}$ such that almost surely

$$\begin{aligned} \int _0^T|u_t|^2_{n}\,dt<\infty . \end{aligned}$$

Definition 2.1

A continuous $L_2$-valued adapted process $(u_t)_{t\in [0,T]}$ is a generalised solution to (2.1)–(2.2) if it is in ${\mathbb {W}}^1_2(0,T)$, and almost surely

$$\begin{aligned} (u_t,\varphi )=&(\phi ,\varphi ) +\int _0^t \big ( a^{ij}_sD_{j}u_s,D^{*}_{i}\varphi )+(b^i_sD_iu_s+c_su_s+f_s,\varphi \big )\,ds \\&+\,\int _0^t \big ( \sigma _s^{i\rho } D^i u_s + \nu ^\rho _s u_s + g^{\rho }_s,\varphi \big )\,dW^{\rho }_s \end{aligned}$$

for all $t\in [0,T]$ and $\varphi \in C_0^{\infty }$, where $D^{*}_{i}:=-D_{i}$ for $i\in \{1,2,\ldots ,d\}$, and (, ) denotes the inner product in $L_2$.

For $m\ge 0$ set

$$\begin{aligned} {\mathfrak {K}}_m = |\phi |_m^2 + \int _0^T \big ( |f_t|_{m-1}^2 + |g_t|_m^2\big ) dt. \end{aligned}$$

(2.5)

Then the following theorem is well-known (see, e.g., [19]).

Theorem 2.1

Let Assumptions 2.1, 2.2 and 2.3 hold. Then (2.1)–(2.2) has a unique generalised solution $u=(u_t)_{t\in [0,T]}$. Moreover, $u\in {\mathbb {W}}^{m+1}_2(0,T)$, it is an $H^{m}$-valued continuous process, and

$$\begin{aligned} E\sup _{t\in [0,T]}|u_t|^2_{m}+E\int _0^T|u_t|_{m+1}^2\,dt \le C E {\mathfrak {K}}_m, \end{aligned}$$

where C is a constant depending only on $\kappa$, d, T, m and K.

The finite elements we consider in this paper are determined by a continuous real function $\psi \in H^1$ with compact support, and a finite set $\Lambda \subset {\mathbb {Q}}^d$, containing the zero vector, such that $\psi$ and $\Lambda$ are symmetric, i.e.,

$$\begin{aligned} \psi (-x)=\psi (x)\; \text{ for } \text{ all } \; x\in {\mathbb {R}}^d, \quad \text{ and } \; \Lambda =-\Lambda . \end{aligned}$$

(2.6)

We assume that $|\psi |_{L_1}=1$, which can be achieved by scaling. For each $h\ne 0$ and $x\in {\mathbb {R}}^d$ we set $\psi ^h_x(\cdot ):=\psi ((\cdot -x)/h)$, and our set of finite elements is the collection of functions $\{\psi ^{h}_x:x\in {\mathbb {G}}_h\}$, where

$$\begin{aligned} {\mathbb {G}}_h:=\left\{ h\sum _{i=1}^nn_i\lambda _i:\lambda _i\in \Lambda , \,n_i,n\in {\mathbb {N}}\right\} . \end{aligned}$$

Let $V_h$ denote the vector space

$$\begin{aligned} V_h :=\left\{ \sum _{x\in {\mathbb {G}}_h}U(x) \psi ^{h}_{x}: (U(x))_{x\in {{\mathbb {G}}_h}}\in \ell _2({\mathbb {G}}_h)\right\} , \end{aligned}$$

where $\ell _2({\mathbb {G}}_h)$ is the space of functions U on ${\mathbb {G}}_h$ such that

$$\begin{aligned} |U|_{0,h}^2 :=|h|^d \sum _{x\in {\mathbb {G}}_h} U^2(x) <\infty . \end{aligned}$$

(2.7)

Definition 2.2

An $L_2({\mathbb {R}}^d)$-valued continuous process $u^h=(u^h_t)_{t\in [0,T]}$ is a finite elements approximation of u if it takes values in $V_h$ and almost surely

$$\begin{aligned} (u^h_t,\psi ^{h}_{x})=&(\phi ,\psi ^{h}_{x}) +\int _0^t \big [ (a^{ij}_sD_{j}u^{h}_s,D^{*}_{i}\psi ^{h}_{x}) +(b^i_sD_iu^h_s+c_su^h_s+f_s,\psi ^{h}_x)\big ] \,ds \nonumber \\&+\int _0^t (\sigma ^{i\rho }_sD_{i}u^h_s+\nu ^{\rho }_su^h_s+g^{\rho }_s,\psi ^{h}_x)\,dW^{\rho }_s, \end{aligned}$$

(2.8)

for all $t\in [0,T]$ and $\psi _x^h$ is as above for $x\in {\mathbb {G}}_h$. The process $u^h$ is also called a $V_h$-solution to (2.8) on [0, T].

Since by definition a $V_h$-valued solution $(u^h_t)_{t\in [0,T]}$ to (2.8) is of the form

$$\begin{aligned} u^h_t(x)=\sum _{y\in {\mathbb {G}}_h}U^h_t(y)\psi ^h_y(x), \quad x\in {\mathbb {R}}^d, \end{aligned}$$

we need to solve (2.8) for the random field $\{U^h_t(y):y\in {\mathbb {G}}_{h}, t\in [0,T]\}$. Remark that (2.8) is an infinite system of stochastic equations. In practice one should “truncate” this system to solve numerically a suitable finite system instead, and one should also estimate the error caused by the truncation. We will study such a procedure and the corresponding error elsewhere.

Our aim in this paper is to show that for some well-chosen functions $\psi$, the above finite elements scheme has a unique solution $u^h$ for every $h\ne 0$, and that for a given integer $k\ge 0$ there exist random fields $v^{(0)}$, $v^{(1)}$,...,$v^{(k)}$ and $r_k$, on $[0,T]\times {\mathbb {G}}_h$, such that almost surely

$$\begin{aligned} U^h_t(x)=v^{(0)}_t(x)+\sum _{1\le j\le k}v^{(j)}_t(x)\frac{h^j}{j!}+ r^{h}_t(x), \quad t\in [0,T], x\in {\mathbb {G}}_h, \end{aligned}$$

(2.9)

where $v^{(0)}$,..., $v^{(k)}$ do not depend on h, and there is a constant N, independent of h, such that

$$\begin{aligned} E\sup _{t\le T} |h|^d \sum _{x\in {\mathbb {G}}_h}|r^h_t(x)|^2 \le N |h|^{2(k+1)} E {{\mathfrak {K}}}_m^2 \end{aligned}$$

(2.10)

for all $|h|\in (0,1]$ and some $m> \frac{d}{2}$.

To write (2.8) more explicitly as an equation for $(U^h_t(y))_{y\in {\mathbb {G}}_h}$, we introduce the following notation:

$$\begin{aligned}&R^{\alpha \beta }_{\lambda }=(D_{\beta }\psi _{\lambda },D^{*}_{\alpha }\psi ), \quad \alpha ,\beta \in \{0,1,\ldots ,d\}, \nonumber \\&R^{\beta }_{\lambda }=R^{0\beta }_{\lambda }:=(D_{\beta }\psi _{\lambda },\psi ) , \quad R_{\lambda }:=R^{00}_{\lambda }:=(\psi _{\lambda },\psi ), \quad \lambda \in {\mathbb {G}}, \end{aligned}$$

(2.11)

where $\psi _{\lambda }:=\psi ^1_{\lambda }$, and ${\mathbb {G}}:={\mathbb {G}}_1$.

Lemma 2.2

For $\alpha , \beta \in \{1,\ldots , d\}$ and $\lambda \in {\mathbb {G}}$ we have:

$$\begin{aligned} R^{\alpha \beta }_{-\lambda } =R^{\alpha \beta }_{\lambda }, \quad R^\beta _{-\lambda }=-R^\beta _{\lambda }, \quad R_{-\lambda }=R_{\lambda }. \end{aligned}$$

Proof

Since $\psi (-x)=\psi (x)$ we deduce that for any $\alpha \in \{1,\ldots , d\}$ we have $D_\alpha \psi (-x)= - D_\alpha \psi (x)$. Hence for any $\alpha , \beta \in \{1,\ldots , d\}$ and $\lambda \in {\mathbb {G}}$, a change of variables yields

$$\begin{aligned} R_{-\lambda }^{\alpha \beta }&= \int _{{\mathbb {R}}^d} D_\beta \psi (z+\lambda ) D^{*}_\alpha \psi (z) dz = \int _{{\mathbb {R}}^d} D_\beta \psi (-z+\lambda )D^{*}_\alpha \psi (-z) dz \\&= \int _{{\mathbb {R}}^d} D_\beta \psi (z-\lambda ) D^{*}_\alpha \psi (z) dz =R^{\alpha \beta }_\lambda , \\ R^\beta _{-\lambda }&= \int _{{\mathbb {R}}^d} D_\beta \psi (-z+\lambda ) \psi (-z) dz = - \int _{{\mathbb {R}}^d} D_\beta \psi (z-\lambda ) \psi (z) dz = -R^\beta _\lambda ,\\ R_{-\lambda }&= \int _{{\mathbb {R}}^d} \psi (-z+\lambda ) \psi (-z) dz = \int _{{\mathbb {R}}^d} \psi (z-\lambda ) \psi (z) dz =R_\lambda ; \end{aligned}$$

this concludes the proof. $\square$

To prove the existence of a unique $V_h$-valued solution to (2.8), and a suitable estimate for it, we need the following condition.

Assumption 2.4

There is a constant $\delta >0$ such that

$$\begin{aligned} \sum _{\lambda ,\mu \in {\mathbb {G}}}R_{\lambda -\mu } z^{\lambda }z^{\mu }\ge \delta \sum _{\lambda \in {\mathbb {G}}}|z^{\lambda }|^2, \quad \text{ for } \text{ all } (z^{\lambda })_{\lambda \in {\mathbb {G}}}\in \ell _2({\mathbb {G}}). \end{aligned}$$

Remark 2.1

Note that since $\psi \in H^1$ has compact support, there exists a constant M such that

$$\begin{aligned} |R_\lambda ^{\alpha ,\beta }|\le M \quad \text{ for } \alpha , \beta \in \{0,\ldots , d\} \; \text{ and } \, \lambda \in {\mathbb {G}}. \end{aligned}$$

Remark 2.2

Due to Assumption 2.4 for $h\ne 0$, $u:=\sum _{y\in {\mathbb {G}}_h}U(y)\psi ^h_y$, $U=(U(y))_{y\in {\mathbb {G}}_h}\in \ell _2({\mathbb {G}}_h)$ we have

$$\begin{aligned} |u|_0^2=&\sum _{x,y\in {\mathbb {G}}_h}U(x)U(y)(\psi ^h_x,\psi ^h_y) \nonumber \\ =&\sum _{x,y\in {\mathbb {G}}_h}R_{(x-y)/h}U(x)U(y) |h|^d \ge \delta \sum _{x\in {\mathbb {G}}_h}U^2(x)|h|^d =\delta |U|^2_{0,h} . \end{aligned}$$

(2.12)

Clearly, since $\psi$ has compact support, only finitely many $\lambda \in {\mathbb {G}}$ are such that $(\psi _\lambda ,\psi )\ne 0$; hence

$$\begin{aligned} |u|_0^2 \le \sum _{x,y\in {\mathbb {G}}_h} |R_{(x-y)/h}|\, |U(x)U(y)| |h|^d \le N |h|^d \sum _{x\in {\mathbb {G}}_h}U^2(x)=N |U|^2_{0,h} , \end{aligned}$$

where N is a constant depending only on $\psi$.

By virtue of this remark for each $h\ne 0$ the linear mapping ${\varvec{\Phi }}_h$ from $\ell _2({\mathbb {G}}_h)$ to $V_h\subset L_2({\mathbb {R}}^d)$, defined by

$$\begin{aligned} {\varvec{\Phi }}_hU:=\sum _{x\in {\mathbb {G}}_h}U(x)\psi ^h_x \quad \text {for }U=(U(x))_{x\in {\mathbb {G}}_h}\in \ell _2({\mathbb {G}}_h), \end{aligned}$$

is a one-to-one linear operator such that the norms of U and ${\varvec{\Phi }}_hU$ are equivalent, with constants independent of h. In particular, $V_h$ is a closed subspace of $L_2({\mathbb {R}}^d)$. Moreover, since $D_i\psi$ has compact support, (2.12) implies that

$$\begin{aligned} |D_i u|_0 \le \frac{N}{|h|}|u|_0 \quad \text {for all }u\in V_h, \quad i\in \{1,2,\ldots ,d\}, \end{aligned}$$

where N is a constant depending only on $D_i\psi$ and $\delta$. Hence for any $h>0$

$$\begin{aligned} |u|_{1}\le N(1+|h|^{-1})|u|_0 \quad \text {for all }u\in V_h \end{aligned}$$

(2.13)

with a constant $N=N(\psi ,d,\delta )$ which does not depend on h.

Theorem 2.3

Let Assumptions 2.1 through 2.4 hold with $m=0$. Then for each $h\ne 0$ Eq. (2.8) has a unique $V_h$-solution $u^h$. Moreover, there is a constant $N=N(d,K,\kappa )$ independent of h such that

$$\begin{aligned}&E\sup _{t\in [0,T]}|u^h_t|_0^2+E\int _0^T|u^h_t|^2_{1}\,dt \nonumber \\&\qquad \le NE|\pi ^h \phi |_0^2+NE\int _0^T\big ( |\pi ^hf_s|_0^2 +\sum _{\rho }|\pi ^h g^{\rho }_s|_0^2\big ) \,ds\le NE{\mathcal {K}}_0^2 \end{aligned}$$

(2.14)

for all $h\ne 0$, where $\pi ^h$ denotes the orthogonal projection of $H^0=L_2$ into $V_h$.

Proof

We fix $h\ne 0$ and define the bilinear forms $A^h$ and $B^{h\rho }$ by

$$\begin{aligned} A^h_s(u,v):= & {} (a^{ij}_sD_{j}u,D^{*}_{i}v) +(b^i_sD_iu+c_su,v)\\ B^{h \rho }_s (u,v):= & {} (\sigma ^{i\rho }_sD_iu+ \nu _s^{\rho } u ,v) \end{aligned}$$

for all $u,v\in V_h$. Using Assumption 2.1 with $m=0$, by virtue of (2.13) we have a constant $C=C(|h|,K,d,\delta , \psi )$, such that

$$\begin{aligned} A^h_s(u,v)\le C|u|_0 |v|_0 \quad B^{h \rho }_s(u,v)\le C|u|_0 |v|_0\quad \text {for all }u,v\in V_h. \end{aligned}$$

Hence, identifying $V_h$ with its dual space $(V_h)^{*}$ by the help of the $L_2({\mathbb {R}}^d)$ inner product in $V_h$, we can see there exist bounded linear operators ${\mathbb {A}}^h_s$ and ${\mathbb {B}}^{h\rho }_s$ on $V_h$ such that

$$\begin{aligned} A_s^h(u,v)=({\mathbb {A}}^h_su,v),\quad B^{h\rho }_s(u,v)=({\mathbb {B}}^{h\rho }_su,v) \quad \text {for all }u,v\in V_h, \end{aligned}$$

and for all $\omega \in \Omega$ and $t\in [0,T]$. Thus (2.8) can be rewritten as

$$\begin{aligned} u^h_t=\pi ^h\phi +\int _0^t({\mathbb {A}}_s^hu^h_s+\pi ^hf_s)\,ds +\int _0^t({\mathbb {B}}_s^{h\rho }u^h_s+\pi ^hg^{\rho }_s)\,dW_s^{\rho }, \end{aligned}$$

(2.15)

which is an (affine) linear SDE in the Hilbert space $V_h$. Hence, by classical results on solvability of SDEs with Lipschitz continuous coefficients in Hilbert spaces we get a unique $V_h$-solution $u^h=(u^h_t)_{t\in [0,T]}$. To prove estimate (2.14) we may assume $E{\mathcal {K}}_0^2<\infty$. By applying Itô’s formula to $|u^h|_0^2$ we obtain

$$\begin{aligned} |u^h(t)|_0^2=|\pi ^h\phi |_0^2+ \int _0^t I^h_s\,ds+\int _0^tJ^{h,\rho }_s\,dW_s^{\rho }, \end{aligned}$$

(2.16)

with

$$\begin{aligned} I^h_s:=&\,2({\mathbb {A}}_s^hu^h_s+\pi ^hf_s,u^h_s) +\sum _{\rho }|{\mathbb {B}}_s^{h\rho }u^h_s+\pi ^hg^{\rho }_s|_0^2 , \\ J^{h\rho }_s:=&\, 2({\mathbb {B}}_s^{h\rho }u^h_s+\pi ^hg^{\rho }_s,u^{h}_s). \end{aligned}$$

Owing to Assumptions 2.1, 2.2 and 2.3, by standard estimates we get

$$\begin{aligned} I^h_s\le -\kappa |u^h(s)|^2_{1}+N\Big ( |u^h_s|_0^2+|f_s|_0^2 +\sum _{\rho }|g^{\rho }_s|_0^2 \Big ) \end{aligned}$$

(2.17)

with a constant $N=N(K,\kappa ,d)$; thus from (2.16) using Gronwall’s lemma we obtain

$$\begin{aligned} E|u^h_t|_0^2+\kappa E\int _0^T|u^h_s|^2_{1}\,ds \le NE{\mathcal {K}}_0^2\quad t\in [0,T] \end{aligned}$$

(2.18)

with a constant $N=N(T,K,\kappa ,d)$. One can estimate $E\sup _{t\le T} |u^h_t|_0^2$ also in a standard way. Namely, since

$$\begin{aligned} \sum _{\rho }|J^{h\rho }_s|^2 \le N^2 \, \big (|u^h_s|_1^2+ |g_s|_0^2 \big ) \, \sup _{s\in [0,T]}|u^h_s|_0^2 \end{aligned}$$

with a constant $N=N(K,d)$, by the Davis inequality we have

$$\begin{aligned} E\sup _{t\le T}\Big |\int _0^tJ^h_s\,dW_s^{\rho }\Big |&\le \, 3E\Big (\int _0^T\sum _{\rho }|J^{h,\rho }_s|^2\,ds\Big )^{1/2} \nonumber \\&\le \, 3NE\Big (\sup _{s\in [0,T]}|u^h_s|_0^2\int _0^T\big ( |u^h_s|^2_1+| g_s |_0^2\big ) \,ds\Big )^{1/2} \nonumber \\&\le \, \frac{1}{2}E\sup _{s\in [0,T]}|u^h_s|_0^2 +5N^2E\int _0^T \big ( |u_s^h |^2_1+| g_s |_0^2\big ) \,ds. \end{aligned}$$

(2.19)

Taking supremum in t in both sides of (2.16) and then using (2.17), (2.18) and (2.19), we obtain estimate (2.14). $\square$

Remark 2.3

An easy computation using the symmetry of $\psi$ imposed in (2.6) shows that for every $x\in {\mathbb {R}}^d$ and $h\ne 0$ we have $\psi ^{-h}_x=\psi ^h_x$. Hence the uniqueness of the solution to (2.8) proved in Theorem 2.3 implies that the processes $u^{-h}_t$ and $u^h_t$ agree for $t\in [0,T]$ a.s.

To prove rate of convergence results we introduce more conditions on $\psi$ and $\Lambda$.

Notation Let $\Gamma$ denote the set of vectors $\lambda$ in ${\mathbb {G}}$ such that the intersection of the support of $\psi _{\lambda }:=\psi ^1_{\lambda }$ and the support of $\psi$ has positive Lebesgue measure in ${\mathbb {R}}^d$. Then $\Gamma$ is a finite set.

Assumption 2.5

Let $R_\lambda$, $R^i_\lambda$ and $R^{ij}_\lambda$ be defined by (2.11); then for $i,j,k,l\in \{1,2,\ldots ,d\}$:

$$\begin{aligned}&\sum _{\lambda \in \Gamma }R_{\lambda }=1, \quad \sum _{\lambda \in \Gamma }R_{\lambda }^{ij}=0, \end{aligned}$$

(2.20)

$$\begin{aligned}&\sum _{\lambda \in \Gamma }\lambda _k R^{i}_{\lambda }=\delta _{i,k}, \end{aligned}$$

(2.21)

$$\begin{aligned}&\sum _{\lambda \in \Gamma }\lambda _k \lambda _l R^{ij}_{\lambda } =\delta _{\{i,j\},\{k,l\}} \quad \text{ for } \, i\ne j, \quad \sum _{\lambda \in \Gamma }\lambda _k\lambda _l R^{ii}_{\lambda } = 2 \delta _{\{i,i\},\{k,l\}}, \end{aligned}$$

(2.22)

$$\begin{aligned}&\sum _{\lambda \in \Gamma }Q^{ij,kl}_{\lambda }=0 \quad \text{ and } \quad \sum _{\lambda \in \Gamma }{\tilde{Q}}^{i,k}_{\lambda }=0, \end{aligned}$$

(2.23)

where

$$\begin{aligned} Q_\lambda ^{ij,kl}:=\int _{{\mathbb {R}}^d}z_k z_l D_j\psi _{\lambda }(z) D_i^{*}\psi (z)\,dz, \quad {\tilde{Q}}_\lambda ^{i,k}:= \int _{{\mathbb {R}}^d}z_k D_i\psi _{\lambda }(z) \psi (z) \,dz, \end{aligned}$$

and for sets of indices A and B the notation $\delta _{A,B}$ means 1 when $A=B$ and 0 otherwise.

Note that if Assumption 2.5 holds true, then for any family of real numbers $X_{ij,kl}, i,j,k,l\in \{1,\ldots , d\}$ such that $X_{ij,kl}=X_{ji,kl}$ we deduce from the identities (2.22) that

$$\begin{aligned} \frac{1}{2}\sum _{i,j=1}^d \sum _{k,l=1}^d X_{ij,kl} \sum _{\lambda \in \Gamma }\lambda _k \lambda _l R^{ij}_{\lambda } =\sum _{i,j=1}^d X_{ij,ij}. \end{aligned}$$

(2.24)

Our main result reads as follows.

Theorem 2.4

Let $J\ge 0$ be an integer. Let Assumptions 2.1 and 2.2 hold with $m>2J+\frac{d}{2}+2$. Assume also Assumption 2.3 and Assumptions 2.4 and 2.5 on $\psi$ and $\Lambda$. Then expansion (2.9) and estimate (2.10) hold with a constant $N=N(m, J,\kappa ,K,d,\psi ,\Lambda )$, where $v^{(0)}=u$ is the solution of (2.1) with initial condition $\phi$ in (2.2). Moreover, in the expansion (2.9) we have $v^{(j)}_t=0$ for odd values of j.

Set

$$\begin{aligned} {\bar{u}}^h_t(x)=\sum _{j=0}^{\bar{J}} c_ju^{h/2^j}_t(x) \quad t\in [0,T], \quad x\in {\mathbb {G}}_h, \end{aligned}$$

with $\bar{J}:= \lfloor \frac{J}{2} \rfloor$, $(c_0,\ldots ,c_{\bar{J}}) =(1,0\ldots ,0)V^{-1}$, where $V^{-1}$ is the inverse of the $(\bar{J}+1)\times (\bar{J}+1)$ Vandermonde matrix

$$\begin{aligned} V^{ij}=2^{-4(i-1)(j-1)},\quad i,j=1,2,\ldots ,\bar{J}+1. \end{aligned}$$

We make also the following assumption.

Assumption 2.6

$$\begin{aligned} \psi (0)=1\quad \text {and }\psi (\lambda )=0\text { for } \lambda \in {\mathbb {G}}\setminus \{0\}. \end{aligned}$$

Corollary 2.5

Let Assumption 2.6 and the assumptions of Theorem 2.4 hold. Then

$$\begin{aligned} E\sup _{t\in [0,T]} \sum _{x\in {\mathbb {G}}_h}|u_t(x)- \bar{u}^h_t(x) |^2 |h|^{d} \le |h|^{2J+2} N E {{\mathfrak {K}}}_m^2 \end{aligned}$$

for $|h|\in (0,1]$, with a constant $N=N(m,K,\kappa ,J,T,d,\psi ,\Lambda )$ independent of h, where u is the solution of (2.1)–(2.2).

3 Preliminaries

Assumptions 2.1, 2.2 and 2.4 are assumed to hold throughout this section. Recall that $|\cdot |_{0,h}$ denotes the norm, and $(\cdot ,\cdot )_{0,h}$ denotes the inner product in $\ell _2({\mathbb {G}}_h)$, i.e.,

$$\begin{aligned} |\varphi _1|^2_{0,h}:=|h|^d \sum _{x\in {\mathbb {G}}_h} \varphi _1^2(x)\, , \quad (\varphi _1,\varphi _2)_{0,h}:=|h|^d \sum _{x\in {\mathbb {G}}_h}\varphi _1(x)\varphi _2(x) \end{aligned}$$

for functions $\varphi _1, \varphi _2 \in \ell _2({\mathbb {G}}_h)$.

Dividing by $|h|^d$, it is easy to see that the Eq. (2.8) for the finite elements approximation

$$\begin{aligned} u^h_t(y)=\sum _{x\in {\mathbb {G}}_h}U^h_t(x)\psi _x(y), \quad t\in [0,T],\,y\in {\mathbb {R}}^d, \end{aligned}$$

can be rewritten for $(U^h_t(x))_{x\in {\mathbb {G}}_h}$ as

$$\begin{aligned} {\mathcal {I}}^hU^h_t(x) = \,&\phi ^{h}(x)+\int _0^t \big ( {\mathcal {L}}^h_sU^h_s(x)+f^h_s(x) \big )\,ds \nonumber \\&+\int _0^t \big ( {\mathcal {M}}^{h,\rho }_sU^h_s(x)+g^{h,\rho }_s(x) \big )\,dW^{\rho }_s, \end{aligned}$$

(3.1)

$t\in [0,T],\,x\in {\mathbb {G}}_h$, where

$$\begin{aligned} \phi ^h(x)&=\int _{{\mathbb {R}}^d}\phi (x+hz)\psi (z)\,dz, \quad f^h_t(x)=\int _{{\mathbb {R}}^d}f_t(x+hz)\psi (z)\,dz, \nonumber \\ g^{h,\rho }_t(x)&=\int _{{\mathbb {R}}^d}g^{\rho }_t(x+hz)\psi (z)\,dz, \end{aligned}$$

(3.2)

and for functions $\varphi$ on ${\mathbb {R}}^d$

$$\begin{aligned} {\mathcal {I}}^h\varphi (x)&=\sum _{\lambda \in \Gamma } R_{\lambda }\varphi (x+h\lambda ), \end{aligned}$$

(3.3)

$$\begin{aligned} {\mathcal {L}}^h\varphi (x)&= \sum _{\lambda \in \Gamma } \Big [ \frac{1}{h^2}A_t^h(\lambda ,x) + \frac{1}{h}B_t^h(\lambda ,x) + C_t^h(\lambda ,x) \Big ] \varphi (x+h\lambda ), \end{aligned}$$

(3.4)

$$\begin{aligned} {\mathcal {M}}^{h,\rho }\varphi (x)&=\sum _{\lambda \in \Gamma } \Big [ \frac{1}{h} S^{h,\rho }_t(\lambda , x) +N^{h,\rho }_t(\lambda ,x) \Big ] \varphi (x+h\lambda ) , \end{aligned}$$

(3.5)

with

$$\begin{aligned} A_t^h(\lambda ,x)&=\int _{{\mathbb {R}}^d}a^{ij}_t(x+hz)D_j\psi _{\lambda }(z)D_i^{*}\psi (z)\,dz,\\ B_t^h(\lambda ,x)&=\int _{{\mathbb {R}}^d}b^{i}_t(x+hz)D_i\psi _{\lambda }(z)\psi (z)\,dz,\\ C_t^h(\lambda ,x)&=\int _{{\mathbb {R}}^d}c_t(x+hz)\psi _{\lambda }(z)\psi (z)\,dz, \\ S_t^{h,\rho }(\lambda ,x)&=\int _{{\mathbb {R}}^d}\sigma ^{i\rho }_t(x+hz)D_i\psi _{\lambda }(z)\psi (z)\,dz, \quad \\ N_t^{h,\rho }(\lambda ,x)&=\int _{{\mathbb {R}}^d}\nu ^{\rho }_t(x+hz)\psi _{\lambda }(z)\psi (z)\,dz. \end{aligned}$$

Remark 3.1

Notice that due to the symmetry of $\psi$ and $\Lambda$ required in (2.6), Eq. (3.1) is invariant under the change of h to $-h$.

Remark 3.2

Recall the definition of $\Gamma$ introduced before Assumption 2.5. Clearly

$$\begin{aligned} R_{\lambda }&=0, \,\,A^h_t(\lambda ,x)=B^h_t(\lambda ,x) =C_t^h(\lambda ,x)=S^{h, \rho } _t(\lambda , x)\\&=N^{h , \rho }_t(\lambda ,x)=0 \quad \text {for }\lambda \in {\mathbb {G}}\setminus \Gamma , \end{aligned}$$

i.e., the definition of ${\mathcal {I}}^h$, ${\mathcal {L}}^h_t$ and ${\mathcal {M}}^{h,\rho }_t$ does not change if the summation there is taken over $\lambda \in {\mathbb {G}}$. Owing to Assumption 2.1 with $m=0$ and the bounds on $R^{\alpha \beta }_\lambda$, the operators ${\mathcal {L}}^h_t$ and ${\mathcal {M}}^{h,\rho }_t$ are bounded linear operators on $\ell _2({\mathbb {G}}_h)$ such that for each $h\ne 0$ and $t\in [0,T]$

$$\begin{aligned} |{\mathcal {L}}^h_t\varphi |_{0,h}\le N_h |\varphi |_{0,h}, \quad \sum _{\rho }|{\mathcal {M}}^{h,\rho }_t\varphi |^2_{0,h}\le N_h^2 |\varphi |^2_{0,h} \end{aligned}$$

for all $\varphi \in \ell _2({\mathbb {G}}_h)$, with a constant $N_h =N(|h|,K,d,\psi ,\Lambda )$. One can similarly show that

$$\begin{aligned} |{\mathcal {I}}^h\varphi |_{0,h}\le N|\varphi |_{0,h} \quad \text {for } \varphi \in \ell _2({\mathbb {G}}_h), \end{aligned}$$

(3.6)

with a constant $N=N(K,d,\Lambda ,\psi )$ independent of h. It is also easy to see that for every $\phi \in L_2$ and $\phi ^h$ defined as in (3.2) we have

$$\begin{aligned} |\phi ^h|_{0,h}\le N|\phi |_{L_2} \end{aligned}$$

with a constant $N=N(d, \Lambda ,\psi )$ which does not depend on h; therefore

$$\begin{aligned} |\phi ^h|^2_{0,h} +\int _0^T \Big ( |f_t^h|^2_{0,h} +\sum _{\rho }|g_t^{h,\rho }|^2_{0,h}\Big ) \,dt\le N^2{\mathcal {K}}_0^2. \end{aligned}$$

Lemma 3.1

The inequality (3.6) implies that the mapping ${\mathcal {I}}^h$ is a bounded linear operator on $\ell _2({\mathbb {G}}_h)$. Owing to Assumption 2.4 it has an inverse $({\mathcal {I}}^h)^{-1}$ on $\ell _2({\mathbb {G}}_h)$, and

$$\begin{aligned} |({\mathcal {I}}^h)^{-1}\varphi |_{0,h}\le \frac{1}{\delta } |\varphi |_{0,h}\quad \text {for }\varphi \in \ell _2({\mathbb {G}}_h). \end{aligned}$$

(3.7)

Proof

For $\varphi \in \ell _2({\mathbb {G}}_h)$ and $h \ne 0$ we have

$$\begin{aligned} (\varphi ,{\mathcal {I}}^h\varphi )_{0,h}=&\,|h|^d \sum _{x\in {\mathbb {G}}_h}\varphi (x){\mathcal {I}}^h\varphi (x) = |h|^d \sum _{x\in {\mathbb {G}}_h}\sum _{\lambda \in {\mathbb {G}}} \varphi (x)(\psi _{\lambda },\psi )\varphi (x+h\lambda )\\ =&\, |h|^d \sum _{x\in {\mathbb {G}}_h}\sum _{y-x\in {\mathbb {G}}_h} \varphi (x)(\psi _{\frac{y-x}{h}},\psi )\varphi (y) = |h|^d \sum _{\lambda ,\mu \in {\mathbb {G}}} \varphi (h\mu )R_{\lambda -\mu }\varphi (h\lambda ) \\ \ge&\, \delta |h|^d \sum _{\lambda \in {\mathbb {G}}}|\varphi (h\lambda )|^2=\delta |\varphi |^2_{0,h}. \end{aligned}$$

Together with (3.6), this estimate implies that ${\mathcal {I}}^h$ is invertible and that (3.7) holds. $\square$

Remark 3.2 and Lemma 3.1 imply that Eq. (3.1) is an (affine) linear SDE in the Hilbert space $\ell _2({\mathbb {G}}_h)$, and by well-known results on solvability of SDEs with Lipschitz continuous coefficients in Hilbert spaces, Eq. (3.1) has a unique $\ell _2({\mathbb {G}}_h)$-valued continuons solution $(U_t)_{t\in [0,T]}$, which we call an $\ell _2$-solution to (3.1).

Now we formulate the relationship between Eqs. (2.8) and (3.1).

Theorem 3.2

Let Assumption 2.4 hold. Then the following statements are valid.

(i) Let Assumptions 2.1 and 2.2 be satisfied with $m=0$, and

$$\begin{aligned} u^h_t=\sum _{x\in {\mathbb {G}}_h}U^h_t(x)\psi ^h_x,\quad t\in [0,T] \end{aligned}$$

(3.8)

be the unique $V_h$-solution of (2.8); then $(U^h_t)_{t\in [0,T]}$ is the unique $\ell _2$-solution of (3.1).

(ii) Let Assumption 2.1 hold with $m=0$. Let $\Phi$ be an $\ell _2({\mathbb {G}}_h)$-valued ${\mathcal {F}}_0$-measurable random variable, and let $F=(F_t)_{t\in [0,T]}$ and $G^{\rho }=(G^{\rho }_t)_{[0,T]}$ be $\ell _2({\mathbb {G}}_h)$-valued adapted processes such that almost surely

$$\begin{aligned} {\mathcal {K}}^2_{0,h} :=|\Phi |^2_{0,h} +\int _0^T \Big ( |F_t|^2_{0,h}+\sum _{\rho }|G_t^{\rho }|^2_{0,h} \Big ) \,dt<\infty . \end{aligned}$$

Then Eq. (3.1) with $\Phi$, F and $G^{\rho }$ in place of $\phi ^h$, $f^h$ and $g^{\rho ,h}$, respectively, has a unique $\ell _2({\mathbb {G}}_h)$-solution $U^h=(U^h_t)_{t\in [0,T]}$. Moreover, if Assumption 2.3 also holds then

$$\begin{aligned} E\sup _{t\in [0,T]}|U_t^h|^2_{0,h}\le NE{\mathcal {K}}^2_{0,h} \end{aligned}$$

(3.9)

with a constant $N=N(K,d,\kappa ,\delta ,\Lambda , \psi )$ which does not depend on h.

Proof

(i) Substituting (3.8) into Eq. (2.8), then dividing both sides of the equation by $|h|^d$ we obtain Eq. (3.1) for $U^h$ by simple calculation. Hence by Remark 3.2 we can see that $U^h$ is the unique $\ell _2({\mathbb {G}})$-valued solution to (3.1).

To prove (ii) we use Remark 3.1 on the invertibility of ${\mathcal {I}}^h$ and a standard result on solvability of SDEs in Hilbert spaces to see that Eq. (3.1) with $\Phi$, F and $G^{\rho }$ has a unique $\ell _2({\mathbb {G}})$-valued solution $U^h$. We claim that $u^h_t(\cdot )=\sum _{y\in {\mathbb {G}}} U^h_t (y)\psi ^h_{y}(\cdot )$ is the $V_h$-valued solution of (2.8) with

$$\begin{aligned} \phi (\cdot ):=\sum _{y\in {\mathbb {G}}_h}({\mathcal {I}}^{h})^{-1}\Phi (y)\psi ^h_y(\cdot ), \quad f_t(\cdot ):=\sum _{y\in {\mathbb {G}}_h}({\mathcal {I}}^h)^{-1}F_t(y)\psi ^h_y(\cdot ), \end{aligned}$$

and

$$\begin{aligned} g_t^{\rho }(\cdot ):=\sum _{y\in {\mathbb {G}}_h}({\mathcal {I}}^h)^{-1}G^{\rho }_t(y)\psi ^h_y(\cdot ), \end{aligned}$$

respectively. Indeed, (3.3) yields

$$\begin{aligned} |h|^{-d} ( \phi , \psi ^h_x)&= |h|^{-d} \sum _{y\in {\mathbb {G}}_h}(\psi ^h_y,\psi ^h_x)({\mathcal {I}}^h)^{-1}\Phi (y) =\sum _{y\in {\mathbb {G}}_h}R_{\frac{y-x}{h}}({\mathcal {I}}^h)^{-1}\Phi (y) \\&={\mathcal {I}}^h\{({\mathcal {I}}^h)^{-1}\Phi \}(x)= \Phi (x), \quad x\in {\mathbb {G}}_h. \end{aligned}$$

In the same way we have

$$\begin{aligned} |h|^{-d} (f_t,\psi _{x}^h)=F_t(x), \quad |h|^{-d} (g^{\rho }_t,\psi _{x}^h)=G^{\rho }_t(x)\quad \text {for }x\in {\mathbb {G}}_h, \end{aligned}$$

which proves the claim. Using Remarks 2.2 and 3.1 we deduce

$$\begin{aligned} \Vert \phi \Vert&\le N|({\mathcal {I}}^{h})^{-1}\Phi |_{0,h}\le \frac{N}{\delta }| \Phi |_{0,h}, \quad \Vert f_t\Vert \le N|({\mathcal {I}}^{h})^{-1}F_t|_{0,h}\le \frac{N}{\delta }|F_t|_{0,h}, \\ \sum _{\rho }\Vert g^{\rho }_t\Vert ^2&\le N^2\; \sum _{\rho } |({\mathcal {I}}^{h})^{-1}G^{\rho }_t|^2_{0,h} \le \frac{N^2}{\delta ^2}\sum _{\rho } |G^{\rho }_t|_{0,h}^2 \end{aligned}$$

with a constant $N=N(\psi ,\Lambda )$. Hence by Theorem 2.3

$$\begin{aligned} E\sup _{t\le T}\Vert u_t^h\Vert ^2\le NE|\phi |^2_{0,h} +NE\int _0^T\Big ( |F_t|^2_{0,h}+\sum _{\rho }|G^{\rho }_t|^2_{0,h}\Big ) \,dt \end{aligned}$$

with $N=N(K,T,\kappa ,d,\psi ,\Lambda ,\delta )$ independent of h, which by virtue of Remark 2.2 implies estimate (3.9). $\square$

4 Coefficients of the expansion

Notice that the lattice ${\mathbb {G}}_h$ and the space $V_h$ can be “shifted” to any $y\in {\mathbb {R}}^d$, i.e., we can consider ${\mathbb {G}}_h(y):={\mathbb {G}}_h+y$ and

$$\begin{aligned} V_h(y) :=\Big \{\sum _{x\in {\mathbb {G}}_h(y)}U(x) \psi ^{h}_{x}: (U(x))_{x\in {{\mathbb {G}}_h(y)}}\in \ell _2({\mathbb {G}}_h(y))\Big \}. \end{aligned}$$

Thus Eq. (2.8) for $u^h=\sum _{x\in {\mathbb {G}}_h(y)}U(x)\psi ^h_x$ should be satisfied for $\psi _x$, $x\in {\mathbb {G}}_h(y)$. Correspondingly, Eq. (3.1) can be considered for all $x\in {\mathbb {R}}^d$ instead of $x\in {\mathbb {G}}_h$.

To determine the coefficients $(v^{(j)})_{j=1}^{k}$ in the expansion (2.9) we differentiate formally (3.1) in the parameter h, j times, for $j=1,2,\ldots ,J$, and consider the system of SPDEs we obtain for the formal derivatives

$$\begin{aligned} u^{(j)}=D_h^jU^h\big |_{h=0}, \end{aligned}$$

(4.1)

where $D_h$ denotes differentiation in h. To this end given an integer $n\ge 1$ let us first investigate the operators ${\mathcal {I}}^{(n)}$, ${\mathcal {L}}^{(n)}_t$ and ${\mathcal {M}}^{(n)\rho }_t$ defined by

$$\begin{aligned}&{\mathcal {I}}^{(n)}\varphi (x)=D^n_h{\mathcal {I}}^h\varphi (x)\big |_{h=0}, \quad {\mathcal {L}}^{(n)}_t\varphi (x)=D^n_h{\mathcal {L}}^h_t\varphi (x)\big |_{h=0}, \nonumber \\&{\mathcal {M}}^{(n)\rho }_t\varphi (x)=D^n_h{\mathcal {M}}^{h,\rho }_t\varphi (x)\big |_{h=0} \end{aligned}$$

(4.2)

for $\varphi \in C_0^{\infty }$.

Lemma 4.1

Let Assumption 2.1 hold with $m\ge n+l+2$ for nonnegative integers l and n. Let Assumption 2.5. also hold. Then for $\varphi \in C_0^{\infty }$ and $t\in [0,T]$ we have

$$\begin{aligned} |{\mathcal {I}}^{(n)}\varphi |_l\le N|\varphi |_{l+n}, \quad |{\mathcal {L}}^{(n)}_t\varphi |_{l} \le N|\varphi |_{l+2+n}, \quad \sum _\rho |{\mathcal {M}}^{(n)\rho }_t\varphi |_{l}^2 \le N^2|\varphi |_{l+1+n}\nonumber \\ \end{aligned}$$

(4.3)

with a constant $N=N(K,d,l, n,\Lambda ,\Psi )$ which does not depend on h.

Proof

Clearly, ${\mathcal {I}}^{(n)} =\sum _{\lambda \in \Gamma } R_{\lambda }\partial ^n_{\lambda }\varphi$, where

$$\begin{aligned} \partial _{\lambda }\varphi :=\sum _{i=1}^d\lambda ^iD_i \varphi . \end{aligned}$$

(4.4)

This shows the existence of a constant $N=N(\Lambda , \psi ,d,n)$ such that the first estimate in (4.3) holds. To prove the second estimate we first claim the existence of a constant $N=N(K,d,l,\Lambda ,\psi )$ such that

$$\begin{aligned} \Big |D^n_h \Phi _t (h,\cdot )\big |_{h=0}\Big |_l\le N|\varphi |_{l+n+2} \end{aligned}$$

(4.5)

for

$$\begin{aligned} \Phi _t (h,x):=h^{-2}\sum _{\lambda \in \Gamma }\varphi (x+h\lambda ) \int _{{\mathbb {R}}^d}a^{ij}_t(x+hz)D_j\psi _{\lambda }(z)D_i^{*}\psi (z)\,dz. \end{aligned}$$

Recall the definition of $R^{ij}_\lambda$ given in (2.11). To prove (4.5) we write $\Phi _t(h,x)=\sum _{i=1}^3 \Phi ^{(i)}_t(h,x)$ for $h\ne 0$ with

$$\begin{aligned} \Phi ^{(1)}_t(h,x)&=h^{-2}\sum _{\lambda \in \Gamma }\varphi (x+h\lambda ) \int _{{\mathbb {R}}^d}a^{ij}_t(x)D_j\psi _{\lambda }(z)D_i^{*}\psi (z)\,dz\\&=h^{-2}a^{ij}_t(x)\sum _{\lambda \in \Gamma }\varphi (x+h\lambda )R_{\lambda }^{ij},\\ \Phi ^{(2)}_t(h,x)&=h^{-1}\sum _{\lambda \in \Gamma }\varphi (x+h\lambda ) \int _{{\mathbb {R}}^d} \sum _{k=1}^d D_k a^{ij}_t(x) z_k D_j\psi _{\lambda }(z)D_i^{*}\psi (z)\,dz,\\&=h^{-1}\sum _{\lambda \in \Gamma }\varphi (x+h\lambda ) Da^{ij}_t(x)S^{ij}_{\lambda }, \end{aligned}$$

for

$$\begin{aligned} S^{ij}_{\lambda }:=\int _{{\mathbb {R}}^d}zD_j\psi _{\lambda }(z)D_i^{*}\psi (z)\,dz \in {{\mathbb {R}}^d} , \end{aligned}$$

and

$$\begin{aligned}&\Phi ^{(3)}_t(h,x)\\ {}&~~=\sum _{\lambda \in \Gamma }\varphi (x+h\lambda ) \int _{{\mathbb {R}}^d} \int _0^1(1-\vartheta )D_{kl}a^{ij}_t(x+h\vartheta z)z^kz^lD_j\psi _{\lambda }(z)D_i^{*}\psi (z) \,d\vartheta \,dz, \end{aligned}$$

where $D_{kl}:=D_kD_l$. Here we used Taylor’s formula

$$\begin{aligned} f(h)=\sum _{i=0}^{n}\frac{h^{i}}{i!} f^{(i)}(0)+\frac{h^{n+1}}{n!} \int _{0}^{1}(1-\theta )^{n}f^{(n+1)}(h\theta ) \,d\theta \end{aligned}$$

(4.6)

with $n=1$ and $f(h):=a^{ij}_t(x+h\lambda )$.

Note that Lemma 2.2 and (2.20) in Assumption 2.5 imply

$$\begin{aligned} \Phi ^{(1)}_t(h,x)&= \frac{1}{2}a^{ij}_t(x)\sum _{\lambda \in \Gamma } R^{ij}_{\lambda }h^{-2}(\varphi (x+h\lambda )-2\varphi (x)+\varphi (x-h\lambda )) \nonumber \\&=\frac{1}{2}a^{ij}_t(x)\sum _{\lambda \in \Gamma } R^{ij}_{\lambda }\int _{0}^{1}\int _{0}^{1} \partial ^2_{ \lambda } \varphi (x+h\lambda (\theta _{1}-\theta _{2})) \,d\theta _{1}d\theta _{2}. \end{aligned}$$

(4.7)

To rewrite $\Phi ^{(2)}_t(h,x)$ note that $S^{ij}_{-\lambda } = -S^{ij}_{\lambda }$; indeed since $\psi (-x)=\psi (x)$ the change of variables $y=-z$ implies that

$$\begin{aligned} S^{ij}_{-\lambda }&=\int _{{\mathbb {R}}^d} z D_j\psi (z+\lambda ) D_i^*\psi (z) dz =-\int _{{\mathbb {R}}^d} y D_j\psi (-y+\lambda ) D_i^*\psi (-y) dy \nonumber \\&=-\int _{{\mathbb {R}}^d} y D_j\psi (y-\lambda ) D_i^*\psi (y) dy=-S^{ij}_{\lambda }. \end{aligned}$$

(4.8)

Furthermore, an obvious change of variables, (4.8) and Lemma 2.2 yield

$$\begin{aligned} S^{ji}_\lambda&=\int _{{\mathbb {R}}^d} z D_i\psi (z-\lambda ) D_j^*\psi (z) dz = \int _{{\mathbb {R}}^d} (z+\lambda ) D_i\psi (z) D_j^*\psi (z+\lambda ) dz\\&=\int _{{\mathbb {R}}^d} z D_i^*\psi (z) D_j \psi _{-\lambda }(z) dz + \lambda \int _{{\mathbb {R}}^d} D_i^*\psi (z) D_j\psi _{-\lambda }(z) dz \\&=S^{ij}_{-\lambda } +\lambda R^{ij}_{-\lambda } =-S^{ij}_{\lambda } +\lambda R^{ij}_{\lambda } . \end{aligned}$$

This implies

$$\begin{aligned} S^{ji}_\lambda + S^{ij}_\lambda =\lambda R^{ij}_\lambda , \quad i,j=1,\ldots , d . \end{aligned}$$

Note that since $a^{ij}_t(x)=a^{ji}_t(x)$, we deduce

$$\begin{aligned} D a^{ij}_t(x) S^{ij}_\lambda = D a^{ij}_t(x) S^{ji}_\lambda =\frac{1}{2} D a^{ij}_t(x) \lambda R^{ij}_\lambda = \frac{1}{2} R^{ij}_\lambda \partial _\lambda a^{ij}_t(x), \end{aligned}$$

(4.9)

for $\partial _\lambda F$ defined by (4.4). Thus the Eqs. (4.8) and (4.9) imply

$$\begin{aligned} \Phi ^{(2)}_t(h,x)&= \frac{1}{2} \sum _{\lambda \in \Gamma }h^{-1} (\varphi (x+h\lambda )-\varphi (x-h\lambda )) Da^{ij}_t(x)S^{ij}_{\lambda } \nonumber \\&= \frac{1}{4} \sum _{\lambda \in \Gamma } R^{ij}_{\lambda }\partial _{\lambda }a^{ij}_t(x) \; 2 \int _{0}^{1}\partial _{\lambda }\varphi (x+h\lambda (2\theta -1)) \,d\theta . \end{aligned}$$

(4.10)

From (4.7) and (4.10) we get

$$\begin{aligned} D^n_h\Phi ^{(1)}_t(h,x)\big |_{h=0}&= \frac{1}{2}a^{ij}_t(x)\sum _{\lambda \in \Gamma }R^{ij}_{\lambda } \int _{0}^{1}\int _{0}^{1} \partial ^{n+2}_{\lambda } {\varphi (x)}(\theta _1-\theta _2)^n \,d\theta _{1}d\theta _{2}, \\ D^n_h\Phi ^{(2)}_t(h,x)\big |_{h=0}&= \frac{1}{2} \sum _{\lambda \in \Gamma }R^{ij}_{\lambda }\partial _{\lambda }a^{ij} \int _{0}^{1} \partial ^{n+1}_{\lambda }{\varphi (x)} (2\theta -1)^n \,d\theta . \end{aligned}$$

Furthermore, the definition of $\Phi ^{(3)}_t(h,x)$ yields

$$\begin{aligned}&D^n_h\Phi ^{(3)}_t(h,x)\big |_{h=0}\\&\quad = \sum _{\lambda \in \Gamma } \sum _{k=0}^n{n\atopwithdelims ()k}\partial _{\lambda }^{n-k}\varphi {(x)} \int _{{\mathbb {R}}^d} \int _0^1(1-\theta )\theta ^{k}\partial _z^{k}D_{kl}a^{ij}_t(x)z^kz^lD_j\psi _{\lambda }(z)D_i^{*}\psi (z) \,d\theta \,dz. \end{aligned}$$

Using Assumption 2.1 and Remark 2.1, this completes the proof of (4.5).

Taylor’s formula (4.6) with $n=0$ and $f(h):=b^i_t(x+h z )$ implies

$$\begin{aligned} {\tilde{\Phi }}_t(h,x):&=h^{-1} \sum _{\lambda \in \Gamma } \varphi (x+h\lambda ) \int _{{\mathbb {R}}^d} b^i_t(x+hz) D_i\psi _\lambda (z) \psi (z) dz\\&={\Phi }^{(4)}_t(h,x) + {\Phi }^{(5)}_t(h,x), \end{aligned}$$

with

$$\begin{aligned} {\Phi }^{(4)}_t(h,x)&= h^{-1} b^i_t(x) \sum _{\lambda \in \Gamma } \varphi (x+h\lambda ) R^i_\lambda , \\ {\Phi }^{(5)}_t(h,x)&= \sum _{\lambda \in \Gamma } \varphi (x+h\lambda ) \int _{{\mathbb {R}}^d} \int _0^1 (1-\theta ) \sum _{k=1}^d D_k b^i_t(x+h\theta z) z_k D_i\psi _\lambda (z) \psi (z) d\theta dz. \end{aligned}$$

Using Lemma 2.2 and computations similar to those used to prove (4.5) we deduce that

$$\begin{aligned} \Phi ^{(4)}_t(h,x)&=\frac{1}{2} \sum _{\lambda \in \Gamma } h^{-1} \big [ \varphi (x+h\lambda )-\varphi (x-h\lambda )\big ] b^i_t(x) R^i_\lambda \\&= b^i_t(x) \sum _{\lambda \in \Gamma } R^i_\lambda \int _0^1 \partial _\lambda \varphi \big (x+ h\lambda (2\theta -1)\big ) d\theta , \end{aligned}$$

which yields

$$\begin{aligned} D^n_h\Phi ^{(4)}_t(h,x) \big |_{h=0} = b^i_t(x) \sum _{\lambda \in \Gamma } R^i_\lambda \partial _\lambda ^{n+1} \varphi ({x}) \int _0^1 (2\theta -1)^n d\theta . \end{aligned}$$

Furthermore, the definition of $\Phi ^{(5)}(h,x)$ implies

$$\begin{aligned}&D^n_h\Phi ^{(5)}_t(h,x){\big |_{h=0}}\\&\quad =\sum _{\lambda \in \Gamma } \sum _{\alpha =0}^n {n\atopwithdelims ()\alpha } \int _{{\mathbb {R}}^d} \int _0^1(1-\theta ) \partial _{\lambda }^{n-\alpha }\varphi ({x}) \theta ^{\alpha } \partial _z^{\alpha } D_{\alpha }b^{i}_t(x) z^\alpha D_i\psi _{\lambda }(z)\psi (z) \,d\theta \,dz. \end{aligned}$$

This implies the existence of a constant $N=N(K,d,l,\Lambda ,\psi )$ which does not depend on h such that

$$\begin{aligned} \Big |D^n_h {\tilde{\Phi }}_t (h,\cdot )\big |_{h=0}\Big |_l\le N|\varphi |_{l+n+1} . \end{aligned}$$

(4.11)

Finally, let

$$\begin{aligned} \Phi ^{(6)}_t(h,x):=\sum _{\lambda \in \Gamma } \varphi (x+h\lambda ) \int _{{\mathbb {R}}^d} c_t(x+hz) \psi _\lambda (z)\psi (z) dz. \end{aligned}$$

Then we have

$$\begin{aligned} D^n_h\Phi ^{(6)}_t(h,x){\big |_{h=0}} =\sum _{\lambda \in \Gamma } \sum _{\alpha =0}^n {n\atopwithdelims ()\alpha } \partial _{\lambda }^{n-\alpha }\varphi ({x}) \int _{{\mathbb {R}}^d} \partial _z^{\alpha } c_t(x) \psi _{\lambda }(z)\psi (z) \,dz, \end{aligned}$$

so that

$$\begin{aligned} \Big |D^n_h \Phi ^{(6)}_t (h,\cdot )\big |_{h=0}\Big |_l\le N|\varphi |_{l+n} \end{aligned}$$

(4.12)

for some constant N as above.

Since ${\mathcal {L}}_t^h \varphi (x)=\Phi _t(h,x)+{\tilde{\Phi }}_t(h,x)+\Phi ^{(6)}_t(h,x)$, the inequalities (4.5), (4.11) and (4.12) imply that ${\mathcal {L}}^{(n)}_t$ satisfies the estimate in (4.3); the upper estimates of ${\mathcal {M}}^{(n),\rho }_t$ can be proved similarly. $\square$

For an integer $k\ge 0$ define the operators ${\hat{L}}_t^{(k)h}$, ${\hat{M}}_t^{(k)h,\rho }$ and ${\hat{I}}^{(k)h}$ by

$$\begin{aligned} {\hat{L}}_t^{(k)h}\varphi&={\mathcal {L}}^h_t\varphi -\sum _{i=0}^{k}\frac{h^i}{i!} {\mathcal {L}}_t^{(i)}\varphi , \quad {\hat{M}}_t^{(k)h,\rho }\varphi ={\mathcal {M}}^{h,\rho }_t\varphi -\sum _{i=0}^{k}\frac{h^i}{i!}{\mathcal {M}}^{(i)\rho }_t\varphi , \nonumber \\ {\hat{I}}^{(k)h}\varphi&={\mathcal {I}}^h\varphi -\sum _{i=0}^{k}\frac{h^i}{i!}{\mathcal {I}}^{(i)}\varphi , \end{aligned}$$

(4.13)

where ${\mathcal {L}}^{(0)}_t:={\mathcal {L}}_t$, ${\mathcal {M}}^{(0),\rho }_t:={\mathcal {M}}^{\rho }_t$, and ${\mathcal {I}}^{(0)}$ is the identity operator.

Lemma 4.2

Let Assumption 2.1 hold with $m\ge k+l+3$ for nonnegative integers k and n. Let Assumption 2.5 also hold. Then for $\varphi \in C_0^{\infty }$ we have

$$\begin{aligned}&|{\hat{L}}_t^{(k)h}\varphi |_l\le N |h|^{k+1} |\varphi |_{l+k+3}, \quad \sum _{\rho }|{\hat{M}}_t^{(k)h,\rho }\varphi |^2_l\le N^2 |h|^{2k+2} |\varphi |^2_{l+k+2}. \\&|{\hat{I}}^{(k)h}\varphi |_l\le N |h|^{k+1} |\varphi |_{k+1}, \end{aligned}$$

for a constant N which does not depend on h.

Proof

We obtain the estimate for ${\hat{L}}_t^{(k)h}$ by applying Taylor’s formula (4.6) to $f(h):= \Phi ^{(i)}_t(h,x)$ for $i=1,\ldots , 6$ defined in the proof of Lemma 4.1, and by estimating the remainder term

$$\begin{aligned} \frac{h^{k+1}}{k!} \int _{0}^{1}(1-\theta )^{k}f^{(k+1)}(h\theta ) \,d\theta \end{aligned}$$

using the Minkowski inequality. Recall that ${\mathcal {L}}_t\varphi (x) = {\mathcal {L}}^{(0)}_t\varphi (x)$. Using Assumption 2.5 we prove that ${\mathcal {L}}_t^{(0)} \varphi (x)= \lim _{h\rightarrow 0} {\mathcal {L}}_t^h \varphi (x)$. We have ${\mathcal {L}}^{h}_t\varphi (x) = \sum _{i=1}^6 \Phi ^{(i)}_t(h,x)$ for $h\ne 0$. The proof of Lemma 4.1 shows that ${\tilde{\Phi }}^{(i)}_t(0,x) :=\lim _{h\rightarrow 0} \Phi ^{(i)}_t(h,x)$ exist and we identify these limits. Using (4.7), (4.10) and (2.24) with $X_{ij,kl}=a^{ij}_t(x) D_{kl}\varphi (x)$ (resp. $X_{ij,kl}=\partial _k a^{ij}_t(x) \partial _l \varphi (x)$) we deduce

$$\begin{aligned} {\tilde{\Phi }}^{(1)}_t(0,x) =&\sum _{i,j} \frac{1}{2} a^{ij}_t(x) \sum _{k,l} D_k D_l \varphi (x) \sum _{\lambda \in \Gamma } \lambda _k\lambda _l R^{ij}_\lambda = \sum _{i,j} a^{ij}_t(x) D_{ij}\varphi (x), \\ {\tilde{\Phi }}^{(2)}_t(0,x)=&\frac{1}{2} \sum _{i,j} \sum _{k,l} \partial _k a^{ij}_t(x) \partial _l \varphi (x) \sum _{\lambda \in \Gamma } \lambda _k \lambda _l R^{ij}_\lambda = \sum _{i,j}\partial _i a^{ij}_t(x) \partial _j\varphi (x), \end{aligned}$$

which implies that ${\tilde{\Phi }}^{(1)}_t(0,x)+{\tilde{\Phi }}^{(2)}_t(0,x) =D_i\big (a^{ij}_t D_j\varphi \big )(x)$. The first identity in (2.23) [resp. (2.21), the second identity in (2.23) and the first identity in (2.20)] imply

$$\begin{aligned} {\tilde{\Phi }}^{(3)}_t(0,x) =&\,\frac{1}{2}\varphi (x) \sum _{k,l} \sum _{i,j} D_{kl}a^{ij}_t(x) \sum _{\lambda \in \Gamma } Q^{ij,kl}_\lambda =0,\\ {\tilde{\Phi }}^{(4)}_t(0,x) =&\,\sum _i b^i_t(x) \sum _k \partial _k\varphi (x) \sum _{\lambda \in \Gamma } R^i_\lambda \lambda _k=\sum _i b^i_t(x) \partial _i\varphi (x),\\ {\tilde{\Phi }}^{(5)}_t(0,x) =&\, \frac{1}{2} \varphi (x) \sum _k \sum _i D_kb^i_t(x) \sum _{\lambda \in \Gamma } {\tilde{Q}}^{i,k}_\lambda =0,\\ {\tilde{\Phi }}^{(6)}_t(0,x) =&\,\varphi (x) c_t(x) \sum _{\lambda \in \Gamma } R_\lambda = \varphi (x) c_t(x). \end{aligned}$$

This completes the identification of ${\mathcal {L}}_t$ as the limit of ${\mathcal {L}}_t^h$. Using once more the Minkowski inequality and usual estimates, we prove the upper estimates of the $H^l$ norm of $\hat{L}^{(k)h}_t\varphi$. The other estimates can be proved similarly. $\square$

Assume that Assumption 2.2 is satisfied with $m\ge J+1$ for an integer $J\ge 0$. A simple computation made for differentiable functions in place of the formal ones introduced in (4.1) shows the following identities

$$\begin{aligned} \phi ^{(i)}(x)&=\, \int _{{\mathbb {R}}^d}\partial _z^{i}\phi (x)\psi (z)\,dz,\\ f^{(i)}_t(x)&= \int _{{\mathbb {R}}^d}\partial _z^{i}f_t(x)\psi (z)\,dz, \; g_t^{(i)\rho }(x)=\int _{{\mathbb {R}}^d}\partial _z^{i}g_t^{\rho }(x)\psi (z)\,dz, \end{aligned}$$

where $\partial _z^i \varphi$ is defined iterating (4.4), while $\phi ^h$, $f_t^h$ and $g_t^{h,\rho }$ are defined in (3.2). Set

$$\begin{aligned} {\hat{\phi }}^{(J)h}&:=\,\phi ^{h}-\sum _{i=0}^{J}\frac{h^i}{i!}\phi ^{(i)},\nonumber \\ {\hat{f}}^{(J)h}_t&:=f^{h}_t-\sum _{i=0}^{J}\frac{h^i}{i!}f_t^{(i)} \; \text{ and } \, {\hat{g}}^{(J)h\rho }_t :=g^{h,\rho }_t-\sum _{i=0}^{J}g^{(i)\rho }_t\frac{h^i}{i!}. \end{aligned}$$

(4.14)

Lemma 4.3

Let Assumption 2.1 holds with $m\ge l+J+1$ for nonnegative integers J and l. Then there is a constant $N=N(J,l,d,\psi )$ independent of h such that

$$\begin{aligned}&|{\hat{\phi }}^{(J)h}|_l \le |h|^{J+1}N|\phi |_{l+1+J}, \quad |{\hat{f}}^{(J)h}_t|_l \le N |h|^{J+1} |f_t|_{l+1+J}, \\&\quad |{\hat{g}}^{(J)h\rho }_t|_l \le N |h|^{J+1} |g^{\rho }_t|_{l+1+J}. \end{aligned}$$

Proof

Clearly, it suffices to prove the estimate for ${\hat{\phi }}^{(J)h}$, and we may assume that $\phi \in C_0^{\infty }$. Applying Taylor’s formula (4.6) to $f(h)=\phi ^h(x)$ for the remainder term we have

$$\begin{aligned} {\hat{\phi }}^{(J)h}(x) =\frac{h^{J+1}}{J!} \int _{0}^{1}\int _{{\mathbb {R}}^d}(1-\theta )^{J}\partial _z^{J+1}\phi (x+\theta hz)\psi (z)\,dz. \end{aligned}$$

Hence by Minkowski’s inequality and the shift invariance of the Lebesgue measure we get

$$\begin{aligned} |{\hat{\phi }}^{(J)h}(x)| \le \frac{h^{J+1}}{J!} \int _{0}^{1}\int _{{\mathbb {R}}^d}(1-\theta )^{J}|\partial _z^{J+1}\phi (\cdot +\theta hz)|_l|\psi (z)|\,dz \le N h^{J+1} |\phi |_{l+J+1} \end{aligned}$$

with a constant $N=N(J,m,d,\psi )$ which does not depend on h. $\square$

Differentiating formally Eq. (3.1) with respect to h at 0 and using the definition of ${\mathcal {I}}^{(i)}$ in (4.2), we obtain the following system of SPDEs:

$$\begin{aligned}&dv^{(i)}_t+\sum _{1\le j\le i} {i\atopwithdelims ()j} {\mathcal {I}}^{(j)}dv^{(i-j)}_t = \Big \{{\mathcal {L}}^{(0)}_tv^{(i)}_t +f^{(i)}_t+\sum _{1\le j\le i} {i\atopwithdelims ()j} {\mathcal {L}}^{(j)}_t v_t^{(i-j)} \Big \}\,dt \nonumber \\&\qquad \qquad +\Big \{{\mathcal {M}}^{(0)\rho }_tv^{(i)}_t+g^{(i)\rho }_t +\sum _{1\le j\le i} {i\atopwithdelims ()j} {\mathcal {M}}^{(j)\rho }_t v_t^{(i-j)}\Big \}\,dW^{\rho }_t , \end{aligned}$$

(4.15)

$$\begin{aligned}&v^{(i)}_0(x)=\phi ^{(i)}(x) , \end{aligned}$$

(4.16)

for $i=1,2,\ldots ,J$, $t\in [0,T]$ and $x\in {\mathbb {R}}^d$, where ${\mathcal {L}}^{(0)}_t={\mathcal {L}}_t$, ${\mathcal {M}}^{(0)\rho }_t={\mathcal {M}}^{\rho }_t$, and $v^{(0)}=u$ is the solution to (2.1)–(2.2).

Theorem 4.4

Let Assumptions 2.1 and 2.2 hold with $m\ge J+1$ for an integer $J\ge 1$. Let Assumptions 2.3 through 2.5 be also satisfied. Then (4.15)–(4.16) has a unique solution $(v^{(0)},\ldots ,v^{(J)})$ such that $v^{(n)}\in {\mathbb {W}}^{m+1-n}_2(0,T)$ for every $n=0,1,\ldots ,J$. Moreover, $v^{(n)}$ is a $H^{m-n}$-valued continuous adapted process, and for every $n=0,1,\ldots ,J$

$$\begin{aligned} E\sup _{t\le T}|v^{(n)}_{t}|^{2}_{m-n} +E\int _{0}^{T}|v^{(n)}_{t}|^{2}_{m+1-n}\,dt \le N E{\mathfrak {K}}_m^2 \end{aligned}$$

(4.17)

with a constant $N=N(m,J,d,T,\Lambda ,\psi ,\kappa )$ independent of h, and ${\mathfrak {K}}_m$ defined in (2.5).

Proof

The proof is based on an induction argument. We can solve this system consecutively for $i=1,2,\ldots ,J$, by noticing that for each $i=1,2,\ldots ,k$ the equation for $v^{(i)}$ does not contain $v^{(n)}$ for $n=i+1,\ldots ,J$. For $i=1$ we have $v^{(1)}_0 = \phi ^{(1)}$ and

$$\begin{aligned} dv^{(1)}_t+{\mathcal {I}}^{(1)}du_t =&\{{\mathcal {L}}_tv^{(1)}_t +f^{(1)}_t+{\mathcal {L}}^{(1)}_tu_t\}\,dt \\&+\,\{{\mathcal {M}}^{\rho }_t v^{(1)}_t+g^{(1)\rho }_t +{\mathcal {M}}^{(1)\rho }_t u_t\}\,dW^{\rho }_t, \end{aligned}$$

i.e.,

$$\begin{aligned} dv^{(1)}_t = ({\mathcal {L}}_t v^{(1)}_t+{\bar{f}}^{(1)}_t)\,dt +({\mathcal {M}}^{\rho }_t v^{(1)}_t+{\bar{g}}^{(1)\rho }_t)\,dW^{\rho }_t, \end{aligned}$$

with

$$\begin{aligned} {\bar{f}}_t^{(1)}:=&\,f_t^{(1)}-{\mathcal {I}}^{(1)} f_t +({\mathcal {L}}^{(1)}_t-{\mathcal {I}}^{(1)}{\mathcal {L}}_t) u_t, \\ {\bar{g}}^{(1)\rho }_t:=&\,g^{(1)\rho }_t-{\mathcal {I}}^{(1)}g^{\rho }_t+({\mathcal {M}}^{(1)\rho }_t-{\mathcal {I}}^{(1)}{\mathcal {M}}^{\rho }_t) u_t. \end{aligned}$$

By virtue of Theorem 2.1 this equation has a unique solution $v^{(1)}$ and

$$\begin{aligned}&E\sup _{t\le T} |v^{(1)}_t|_{m-1}^2 +E\int _0^T |v^{(1)}_t|_{m}^2 \,dt \\&\quad \le NE|\phi ^{(1)} |^2_{m-1}+NE\int _0^T\big ( |{\bar{f}}_t^{(1)}|^2_{m-2}+|{\bar{g}}^{(1)}_t|_{m-1}^2\big ) \,dt. \end{aligned}$$

Clearly, Lemma 4.1 implies

$$\begin{aligned}&|\phi ^{(1)}|^2_{m-1}\le N |\phi |^2_{m}, \quad |f^{(1)}_t |_{m-2}+|{\mathcal {I}}^{(1)}f_t | _{m-2}\le N |f_t |_{m-1}, \\&\quad |g^{(1)\rho }_t -{\mathcal {I}}^{(1)}g^{\rho }_t|_{m-1}\le N|g^{\rho }_t |_{m}, \\&\quad |({\mathcal {L}}^{(1)}_t -{\mathcal {I}}^{(1)}{\mathcal {L}}_t)u|_{m-2}\le N|u|_{m+1}, \quad \sum _{\rho }|({\mathcal {M}}^{(1)\rho }_t-{\mathcal {I}}^{(1)}{\mathcal {M}}^{\rho }_t)u|^2_{ m-1} \le N^2|u|^2_{m+1}, \end{aligned}$$

with a constant $N=N(d,K,\Lambda ,\psi ,m)$ which does not depend on h. Hence for $m\ge 1$

$$\begin{aligned}&E\sup _{t\le T}|v^{(1)}_t|^2_{m-1}+E\int _0^T|v^{(1)}_t|^2_{m}\,dt\\&\quad \le NE|\phi |^2_{m} +NE\int _0^T \big ( |f_t|^2_{m-1}+|g_t|_{m}^2+|u_t|^2_{m+1}\big ) \,dt \le N E {\mathfrak {K}}^2_{m}. \end{aligned}$$

Let $j\ge 2$. Assume that for every $i<j$ the equation for $v^{(i)}$ has a unique solution such that (4.15) holds and that its equation can be written as $v^{(i)}_0=\phi ^{(i)}$ and

$$\begin{aligned} dv^{(i)}_t=({\mathcal {L}}_t v^{(i)}_t+{\bar{f}}^{(i)}_t)\,dt+({\mathcal {M}}^{\rho }_t v^{(i)}_t +{\bar{g}}^{(i)\rho }_t)\,dW^{\rho }_t \end{aligned}$$

with adapted processes ${\bar{f}}^{(i)}$ and ${\bar{g}}^{(i) \rho }$ taking values in $H^{m-i-1}$ and $H^{m-i}$ respectively, such that

$$\begin{aligned} E\int _0^T \big ( |{\bar{f}}^{(i)}_t|^2_{m-i-1}+|{\bar{g}}^{(i)}_t|^2_{m- i }\big ) \,dt \le NE {\mathfrak {K}}^2_m \end{aligned}$$

(4.18)

with a constant $N=N(K,J,m,d,T,\kappa ,\Lambda ,\psi )$ independent of h. Hence

$$\begin{aligned} E\left( \sup _{t\in [0,T]} |v^{(i)}_t|_{m-i}^2 + \int _0^T |v^{(i)}_t|_{m+1-i}^2 dt\right) \le N E {\mathfrak {K}}^2_m ,\quad i=1,\ldots , j-1.\nonumber \\ \end{aligned}$$

(4.19)

Then for $v^{(j)}$ we have

$$\begin{aligned} dv^{(j)}_t=({\mathcal {L}}_t v^{(j)}_t+{\bar{f}}^{(j)}_t)\,dt+ ({\mathcal {M}}^{\rho }_t v^{(j)}_t+{\bar{g}}^{(j)\rho }_t)\,dW^{\rho }_t, \quad v^{(j)}_0=\phi ^{(j)}, \end{aligned}$$

(4.20)

with

$$\begin{aligned} {\bar{f}}^{(j)}_t:=&\,f^{(j)}_t +\sum _{1\le i\le j} {j\atopwithdelims ()i} \big ({\mathcal {L}}^{(i)}_t-{\mathcal {I}}^{(i)}{\mathcal {L}}_t\big )v_t^{(j-i)}- \sum _{1\le i\le j} {j\atopwithdelims ()i} {\mathcal {I}}^{(i)}{\bar{f}}_t^{(j-i)}, \\ {\bar{g}}^{(j)\rho }_t:=&\,g^{(j)\rho }_t +\sum _{1\le i\le j} {j\atopwithdelims ()i} \big ( {\mathcal {M}}^{(i)\rho }_t-{\mathcal {I}}^{(i)}{\mathcal {M}}_t^{\rho } \big ) v_t^{(j-i)}- \sum _{1\le i\le j} {j\atopwithdelims ()i} {\mathcal {I}}^{(i)}{\bar{g}}_t^{(j-i)\rho }. \end{aligned}$$

Note that $|f^{(j)}_t|_{m-1-j}\le N|f_t|_{m-1}$ ; by virtue of Lemma 4.1, and by the inequalities (4.18) and (4.19) we have

$$\begin{aligned} E\int _0^T|({\mathcal {L}}^{(i)}_t-{\mathcal {I}}^{(i)}{\mathcal {L}}_t) v^{(j-i)}_t|^2_{m-j-1}\,dt&\le NE\int _0^T|v^{(j-i)}_t|^2_{m-j+1+i}\, dt \le NE {\mathfrak {K}}^2_m , \\ E\int _0^T|{\mathcal {I}}^{(i)}{\bar{f}}^{(j-i)}_t|^2_{m-j-1}\,dt&\le NE\int _0^T|{\bar{f}}^{(j-i)}_t|_{m-j+i-1}\,dt \le NE {\mathfrak {K}}^2_m, \end{aligned}$$

where $N=N(K,J,d,T,\kappa ,\psi ,\Lambda )$ denotes a constant which can be different on each occurrence. Consequently,

$$\begin{aligned} E\int _0^T|{\bar{f}}_t^{(j)}|_{m-j-1}^2\,dt\le NE {\mathfrak {K}}^2_m, \end{aligned}$$

and we can get similarly

$$\begin{aligned} E\int _0^T|{\bar{g}}_t^{(j)}|_{m-j}^2\,dt\le NE {\mathfrak {K}}^2_m. \end{aligned}$$

Hence (4.20) has a unique solution $v^{(j)}$, and Theorem 2.1 implies that the estimate (4.17) holds for $v^{(j)}$ in place of $v^{(n)}$. This completes the induction and the proof of the theorem. $\square$

Recall that the norm $|\cdot |_{0,h}$ has been defined in (2.7).

Corollary 4.5

Let Assumptions 2.1 and 2.2 hold with $m>\frac{d}{2}+J+1$ for an integer $J\ge 1$. Let Assumptions 2.3 through 2.5 be also satisfied. Then almost surely $v^{(i)}$ is continuous in $(t,x)\in [0,T]\times {\mathbb {R}}^d$ for $i\le J$, and its restriction to ${\mathbb {G}}_h$ is an adapted continuous $\ell _2({\mathbb {G}}_h)$-valued process. Moreover, almost surely (4.15)–(4.16) hold for all $x\in {\mathbb {R}}^d$ and $t\in [0,T]$, and

$$\begin{aligned} E\sup _{t\in [0,T]}\sup _{x}|v^{( j)}_t(x)|^2+E\sup _{t\le T} |v^{(j)}_t|^2_{0,h} \le NE {\mathfrak {K}}^2_m,\quad j=1,2,\ldots , J. \end{aligned}$$

for some constant $N=N(m,J,d,T,\Lambda ,\psi ,\kappa )$ independent of h.

One can obtain this corollary from Theorem 4.4 by a standard application of Sobolev’s embedding of $H^m$ into $C^{2}_b$ for $m>2+d/2$ and by using the following known result; see, for example [7], Lemma 4.2.

Lemma 4.6

Let $\varphi \in H^m$ for $m>d/2$. Then there is a constant $N=N(d,\Lambda )$ such that

$$\begin{aligned} | I \varphi |^2_{0,h} \le N|\varphi |^2_m, \end{aligned}$$

where I denotes the Sobolev embedding operator from $H^m$ into $C_b({\mathbb {R}}^d)$.

5 Proof of theorem 2.4

Define a random field $r^h$ by

$$\begin{aligned} r^h_t(x):=u^h_t(x)-\sum _{0\le i\le J} v_t^{(i)}(x) \frac{h^i}{i!}, \end{aligned}$$

(5.1)

where $(v^{(1)},\ldots ,v^{(J)})$ is the solution of (4.15) and (4.16).

Theorem 5.1

Let Assumptions 2.1 and 2.2 hold with $m>\frac{d}{2}+2J+2$ for an integer $J\ge 0$. Let Assumptions 2.3 through 2.5 be also satisfied. Then $r^h$ is an $\ell _2({\mathbb {G}}_h)$-solution of the equation

$$\begin{aligned} {\mathcal {I}}^h dr^h_t(x)=&\,\big ( {\mathcal {L}}^h_t r^h_t(x)+F^h_t(x)\big )\,dt+\big ({\mathcal {M}}^{h,\rho }_t r^h_t(x) +G^{h,\rho }_t(x)\big ) \,dW^{\rho }_t , \end{aligned}$$

(5.2)

$$\begin{aligned} r^h_0(x)=&\,{\hat{\phi }}^{(J)h}(x), \end{aligned}$$

(5.3)

where $F^h$ and $G^h$ are adapted $\ell _2({{\mathbb {G}}}_h)$-valued such that for all $h\ne 0$ with $|h|\le 1$

$$\begin{aligned} E\int _0^T \big ( |F^h_t|^2_{\ell _2({\mathbb {G}}_h)} +|G^h_t|^2_{\ell _2({\mathbb {G}}_h)}\big )\,dt\le N |h|^{2(J+1)} E{{\mathfrak {K}}}^2_m, \end{aligned}$$

(5.4)

where $N=N(m, K,J,d,T,\kappa ,\Lambda ,\psi )$ is a constant which does not depend on h.

Proof

Using (5.1), the identity $u^h_t(x)= U^h_t(x)$ for $x\in {{\mathbb {G}}}_h$ which is deduced from Assumption 2.6 and Eq. (3.1), we deduce that for $x\in {{\mathbb {G}}}_h$,

$$\begin{aligned} d \big ( {{\mathcal {I}}}^h r^h_t(x)\big ) =&\, d {{\mathcal {I}}}^h U^h_t - \sum _{i=0}^J \frac{h^i}{i!}\, {{\mathcal {I}}}^h d v_t^{(i)}(x) \nonumber \\ =&\, \big [ {{\mathcal {L}}}^h_t U^h_t(x) + f^h_t(x) ] dt + \big [ {{\mathcal {M}}}^{h,\rho }_t U^h_t(x) + g^{h,\rho }_t(x) \big ] dW^\rho _t \nonumber \\&- \sum _{i=0}^J \frac{h^i}{i!}\, {{\mathcal {I}}}^h d v_t^{(i)}(x) \nonumber \\ =&\, {{\mathcal {L}}}^h_t r^h_t(x) dt + \Big [ {{\mathcal {L}}}^h_t \sum _{i=0}^J \frac{h^i}{i!}\, v^{(i)}_t(x) + f^h_t(x)\Big ] dt + {{\mathcal {M}}}^{h,\rho }_t r^h_t(x) dW^\rho _t \nonumber \\&\quad +\, \Big [ {{\mathcal {M}}}^{h,\rho }_t \sum _{i=0}^J \frac{h^i}{i!}\, v^{(i)}_t(x) + g^{h,\rho }_t(x)\Big ] dW^\rho _t - \sum _{i=0}^J \frac{h^i}{i!}\, {{\mathcal {I}}}^h d v_t^{(i)}(x). \end{aligned}$$

(5.5)

Taking into account Corollary 4.5, in the definition of $d v^{(i)}_t(x)$ in (4.15) we set

$$\begin{aligned} dv^{(i)}_t(x) = \big [ B(i)_t(x) + F(i)_t(x) \big ]dt + \big [ \sigma (i)^\rho _t (x) + G(i)^\rho _t (x)\big ] dW^{\rho }_t, \end{aligned}$$

(5.6)

where $B(i)_t$ (resp. $\sigma (i)^\rho _t$) contains the operators ${{\mathcal {L}}}^{(j)}$ (resp. ${{\mathcal {M}}}^{(j)\rho }_t$) for $0\le j\le i$ while $F(i)_t$ (resp. $G(i)^\rho _t$) contains all the free terms $f^{(j)}_t$ (resp. $g^{(j)\rho }_t$) for $1\le j\le i$. We at first focus on the deterministic integrals. Using the recursive definition of the processes $v^{(i)}$ in (4.15), it is easy to see that

$$\begin{aligned} B(i)_t + \sum _{1\le j\le i} {i\atopwithdelims ()j} {{\mathcal {I}}}^{(j)} B(i-j)_t =&\sum _{j=0}^i {i\atopwithdelims ()j} {{\mathcal {L}}}^{(j)}_t v^{(i-j)}_t, \end{aligned}$$

(5.7)

$$\begin{aligned} F(i)_t + \sum _{1\le j\le i} {i\atopwithdelims ()j} {{\mathcal {I}}}^{(j)} F(i-j)_t =&f^{(i)}_t. \end{aligned}$$

(5.8)

In the sequel, to ease notations we will not mention the space variable x. Using the expansion of ${{\mathcal {L}}}_t^h$, ${{\mathcal {I}}}^h$ and the definitions of $\hat{L}_t^{(J),h}$ and $\hat{I}^{(J),h}$ in (4.13), the expansion of $f^h_t$ and the definition of $\hat{f}^{(J)h}_t$ given in (4.14) together with the definition of $d v^{(i)}_t$ in (5.6), we deduce

$$\begin{aligned} \left[ {{\mathcal {L}}}^h_t \sum _{i=0}^J \frac{h^i}{i!}\, v^{(i)}_t + f^h_t\right] dt - \sum _{i=0}^J \frac{h^i}{i!}\, {{\mathcal {I}}}^h \big [ B(i)_t^h + F^{(i)}_t\big ] = \sum _{j=1}^6 {{\mathcal {T}}}^h_t(i) dt, \end{aligned}$$

where

$$\begin{aligned} {{\mathcal {T}}}^h_t(1) =&\sum _{i=0}^J \sum _{j=0}^i \frac{h^j}{j!} \frac{h^{i-j}}{(i-j)!} \big [ {{\mathcal {L}}}^{(j)}_t v^{(i-j)}_t - {{\mathcal {I}}}^{(j)} B(i)_t \big ], \\ {{\mathcal {T}}}^h_t (2) =&\sum _{i=0}^J \sum _{{\mathop {i+j\ge J+1}\limits ^{0\le j\le J}}} \frac{h^i}{i!} \frac{h^j}{j!} \big [ {{\mathcal {L}}}^{(i)}_t v^{(j)}_t - {{\mathcal {I}}}^{(i)} B(j)_t\big ], \\ {{\mathcal {T}}}^h_t (3) =&\, \hat{L}^{(J),h}_t \sum _{i=0}^J \frac{h^i}{i!} v^{(i)}_t - \hat{I}^{(J),h} \sum _{i=0}^J \frac{h^i}{i!} B(i)_t, \\ {{\mathcal {T}}}^h_t(4) =&\sum _{i=0}^J \frac{h^i}{i!} f^{(i)}_t - \sum _{i=0}^J \sum _{j=0}^i \frac{h^j}{j!} \frac{h^{i-j}}{(i-j)!} {{\mathcal {I}}}^{(j)} F(i-j)_t, \\ {{\mathcal {T}}}^h_t (5) =&- \sum _{i=0}^J \sum _{{\mathop {i+j\ge J+1}\limits ^{0\le j\le J}}} \frac{h^i}{i!} \frac{h^j}{j!} {{\mathcal {I}}}^{(j)} F(i)_t, \\ {{\mathcal {T}}}^h_t(6) =&\, \hat{f}^{(J)h}_t - \sum _{i=0}^J \frac{h^i}{i!} \hat{I}^{(J)h} f^{(i)}_t. \end{aligned}$$

Equation (4.15) implies

$$\begin{aligned} {{\mathcal {T}}}^h_t(1) = \sum _{i=0}^J \frac{h^i}{i!} \left[ {{\mathcal {L}}}^{(0)}_t v^{(i)}_t + \sum _{j=1}^{i} {i\atopwithdelims ()j} {{\mathcal {L}}}^{(j)}_t v^{(i-j)}_t - B(i)_t - \sum _{j=1}^{i} {i\atopwithdelims ()j} {{\mathcal {I}}}^{(j)} B(i-j)_t\right] . \end{aligned}$$

Using the recursive Eq. (5.7) we deduce that for every $h>0$ and $t\in [0,T]$,

$$\begin{aligned} {{\mathcal {T}}}^h_t(1)=0. \end{aligned}$$

(5.9)

A similar computation based on (5.8) implies

$$\begin{aligned} {{\mathcal {T}}}^h_t(4)=0. \end{aligned}$$

(5.10)

In ${{\mathcal {T}}}^h_t(2)$ all terms have a common factor $h^{J+1}$. We prove an upper estimate of

$$\begin{aligned} E\int _0^T | {{\mathcal {L}}}^{(i)}_t v^{(j)}_t |^2_{0,h}\,dt \end{aligned}$$

for $0\le i,j\le J$. Let I denote the Sobolev embedding operator from $H^k$ to $C_b({\mathbb {R}}^d)$ for $k>d/2$. Lemma 4.6, inequalities (4.3) and (4.17) imply that for $k>d/2$,

$$\begin{aligned} E\int _0^T | I {{\mathcal {L}}}^{(i)}_t v^{(j)}_t |_{0,h}^2 dt&\le N E\int _0^T | {{\mathcal {L}}}^{(i)}_t v^{(j)}_t |_k^2 dt \\&\le N E\int _0^T |v^{(j)}_t|_{i+k+2}^2 dt \le N E {\mathfrak {K}}_{i+j+k+1}^2, \end{aligned}$$

where the constant N does not depend on h and changes from one upper estimate to the next. Clearly, for $0\le i,j\le J$ with $i+j\ge J+1$, we have $i+j+k+1>2J+1+\frac{d}{2}$. Similar computations prove that for $i,j\in \{0,\ldots , J\}$ with $i+j\ge J+1$ and $k>\frac{d}{2}$,

$$\begin{aligned} E\int _0^T \big | I {{\mathcal {I}}}^{(i)} B(j)_t \big |_{0,h}^2 \, dt \le&\; N \sum _{l=0}^j E\int _0^T \big | {{\mathcal {L}}}^{(l)}_t v^{(j-l)}_t \big |_{k+i}^2 \, dt\\ \le&\; N \sum _{l=0}^j E\int _0^T \big |v^{(j-l)}_t \big |_{k+i+l+2}^2 \, dt\\ \le&\; N E{{\mathfrak {K}}}_{k+i+j+1}^2. \end{aligned}$$

These upper estimates imply the existence of some constant N independent of h such that for $|h|\in (0,1]$ and $k>\frac{d}{2}$

$$\begin{aligned} E\int _0^T |{{\mathcal {T}}}^h_t(2)|_{0,h}^2\,ds \le N |h|^{2(J+1)} E{{\mathfrak {K}}}_{k+2J+1}^2. \end{aligned}$$

(5.11)

We next find an upper estimate of the $|\cdot |_{0,h}$ norm of both terms in ${{\mathcal {T}}}_t^h(3)$. Using Lemmas 4.6, 4.2 and (4.17) we deduce that for $k>\frac{d}{2}$

$$\begin{aligned} E\int _0^T \Big | I \hat{L}^{(J),h}_t \sum _{i=0}^J \frac{h^i}{i!} v^{(i)}_t\Big |_{0,h}^2 dt \le&\, NE\int _0^T \Big | \hat{L}^{(J),h}_t \sum _{i=0}^J \frac{h^i}{i!} v^{(i)}_t\Big |_k^2 dt \\ \le&\, N |h|^{2(J+1)}\sum _{i=0}^J \int _0^T \big | v^{(i)}_t\big |_{k+J+3}^2 dt \\ \le&\, N |h|^{2(J+1)} E{{\mathfrak {K}}}_{k+2J+2}^2, \end{aligned}$$

where N is a constant independent of h with $|h|\in (0,1]$. Furthermore, similar computations yield for $k>\frac{d}{2}$ and $|h|\in (0,1]$

$$\begin{aligned} E\int _0^T\Big | I \hat{I}^{(J),h} \sum _{i=0}^J \frac{h^i}{i!} B(i)_t\Big |_{0,h}^2 dt \le&\, N E\int _0^T \Big | \sum _{i=0}^J \frac{h^i}{i!} \hat{I}^{(J),h} B(i)_t\Big |_{k}^2 dt \\ \le&\, N |h|^{2(J+1)} E\int _0^T \sum _{i=0}^J \Big | \sum _{l=0}^i {i\atopwithdelims ()l} {{\mathcal {L}}}^{(l)}_t v^{(i-l)}_t \Big |_{k+J+1}^2 dt \\ \le&\, N |h|^{2(J+1)} \sum _{i=0}^J \sum _{l=0}^i |v^{(i-l)}_t|_{k+J+l+3}^2 dt \\ \le&\, N |h|^{2(J+1)} E {{\mathfrak {K}}}_{k+2J+2}^2. \end{aligned}$$

Hence we deduce the existence of a constant N independent of h such that for $|h|\in (0,1]$,

$$\begin{aligned} E\int _0^T |{{\mathcal {T}}}^h_t(3)|_{0,h}^2 dt \le N |h|^{2(J+1)} E {{\mathfrak {K}}}_{k+2J+2}^2, \end{aligned}$$

(5.12)

where $k>\frac{d}{2}$.

We next compute an upper estimate of the $|\cdot |_{0,h}$ norm of ${{\mathcal {T}}}^h_t(5)$. All terms have a common factor $h^{(J+1)}$. Recall that ${{\mathcal {I}}}^{(0)}=Id$. The induction Eq. (5.8) shows that $F(i)_t$ is a linear combination of terms of the form $\Phi (i)_t:=\big ( {{\mathcal {I}}}^{(a_1)} \big )^{k_1} \ldots \big ( {{\mathcal {I}}}^{(a_i)} \big )^{k_i} f_t$ for $a_p, k_p\in \{0,\ldots , i\}$ for $1\le p\le i$ with $\sum _{p=1}^i a_p k_p=i$, and of terms of the form $\Psi (i)_t:=\big ( {{\mathcal {I}}}^{(b_1)} \big )^{l_1} \ldots \big ( {{\mathcal {I}}}^{(b_{i-j})} \big )^{l_{i-j}} f^{(j)}_t$ for $1\le j\le i$, $b_p, l_p\in \{0,\ldots , i-j\}$ for $1\le p\le i-j$ with $\sum _{p=1}^{i-j} b_p l_p+j=i$. Using Lemmas 4.6 and 4.1 we obtain for $k>\frac{d}{2}$, $i,j=1, \ldots J$

$$\begin{aligned} E\int _0^T | I {{\mathcal {I}}}^{(j)} \Phi (i)_t|_{0,h}^2 dt \le&\, N E\int _0^T| {{\mathcal {I}}}^{(j)} \Phi (i)_t(x)|_k^2\, dt \\ \le&\, N E\int _0^T |\Phi (i)_t|_{k+j}^2 dt\\ \le&\, N E\int _0^T |f_t|_{k+j+a_1k_1+ \cdots a_ik_i}^2 dt\\ \le&\, N E\int _0^T |f_t|_{k+i+j}^2 dt \le N E{{\mathfrak {K}}}_{k+i+j}^2. \end{aligned}$$

A similar computation yields

$$\begin{aligned} E\int _0^T | I {{\mathcal {I}}}^{(j)} \Psi (i)_t|_{0,h}^2\, dt \le&\, N E\int _0^T |f^{(i)}_t|_{k+j+b_1l_1+ \cdots + b_{i-j} l_{i-j}}^2\, dt\\ \le&\, NE\int _0^T |f_t|_{k+j+(i-j)+j}^2\, dt\\ \le&\, N E{{\mathfrak {K}}}_{k+i+j}^2. \end{aligned}$$

These upper estimates imply that for $k>\frac{d}{2}$, there exists some constant N independent on h such that for $|h|\in (0,1)$

$$\begin{aligned} E\int _0^T |{{\mathcal {T}}}^h_t(5)|_{0,h}^2\,dt \le N |h|^{2(J+1)}E {{\mathfrak {K}}}_{k+2J}^2. \end{aligned}$$

(5.13)

We finally prove an upper estimate of the $|\cdot |_{0,h}$-norm of both terms in ${{\mathcal {T}}}_t^h(6)$. Using Lemmas 4.6 and 4.3, we obtain for $k>\frac{d}{2}$,

$$\begin{aligned} E\int _0^T \big | I \hat{f}^{(J)h}_t\big |_{0,h}^2\, dt \le&\, N E\int _0^T \big | \hat{f}^{(J)h}_t\big |_k^2 dt\\ \le&\, N |h|^{2(J+1)} E\int _0^T |f_t|_{k+J+1}^2 dt\\ \le&\, N |h|^{2(J+1)} E{{\mathfrak {K}}}_{k+J+1}^2, \end{aligned}$$

where N is a constant which does not depend on h. Furthermore, Lemmas 4.6 and 4.2 yield for $k>\frac{d}{2}$ and $|h|\in (0,1]$,

$$\begin{aligned} E\int _0^T \Big | I \sum _{i=0}^J \frac{h^i}{i!} \hat{I}^{(J)h} f^{(i)}_t \Big |_{0,h}^2\, dt \le&\, N E\int _0^T \Big | \sum _{i=0}^J \frac{h^i}{i!} \hat{I}^{(J)h} f^{(i)}_t \Big |_k^2\,dt \\ \le&\, N |h|^{2(J+1)}E\int _0^T \sum _{i=0}^J |f^{(i)}_t |_{k+J+1}^2 dt \\ \le&\, N |h|^{2(J+1)} E{{\mathfrak {K}}}_{k+2J+1}^2, \end{aligned}$$

for some constant N independent of h. Hence we deduce that for some constant N which does not depend on h and $k>\frac{d}{2}$, we have for $|h|\in (0,1]$

$$\begin{aligned} E\int _0^T |{{\mathcal {T}}}^h_t(6)|_{0,h}^2\,dt \le N |h|^{2(J+1)} E {{\mathfrak {K}}}_{k+2J+1}^2. \end{aligned}$$

(5.14)

Similar computations can be made for the coefficients of the stochastic integrals. The upper bounds of the corresponding upper estimates in (5.11) and (5.12) are still valid because the operators ${{\mathcal {M}}}^\rho _t$ are first order operators while the operator ${{\mathcal {L}}}_t$ is a second order one. This implies that all operators ${{\mathcal {M}}}^{h,\rho }_t$, ${{\mathcal {M}}}^{(i)\rho }_t$ and $\hat{M}^{(J)h}_t$ contain less derivatives than the corresponding ones deduced from ${{\mathcal {L}}}_t$.

Using the expansion (5.5), the upper estimates (5.9)–(5.14) for the coefficients of the deterministic and stochastic integrals, we conclude the proof. $\square$

We now complete the proof of our main result.

Proof of Theorem  2.4

By virtue of Theorems 3.2 and  5.1 we have for $|h|\in (0,1]$

$$\begin{aligned} E\sup _{t\in [0,T]}|r^h_t|_{0,h}^2&\le NE|{\hat{\phi }}^{(J)h}|^2_{0,h} +NE\int _0^T \big ( |F^h_t|^2_{0,h}+|G^h_t|^2_{0,h}\big ) \,dt\\&\le |h|^{2(J+1)} NE{{\mathfrak {K}}}^2_m. \end{aligned}$$

Using Remark 3.1 we have $U^{-h}_t=U^h_t$ for $t\in [0,T]$ a.s. Hence from the expansion (2.9) we obtain that $v^{(j)}=-v^{(j)}$ for odd j, which completes the proof of Theorem 2.4. $\square$

6 Some examples of finite elements

In this section we propose three examples of finite elements which satisfy Assumptions 2.4,  2.5 and 2.6.

6.1 Linear finite elements in dimension 1

Consider the following classical linear finite elements on ${\mathbb {R}}$ defined as follows:

$$\begin{aligned} \psi (x)=\big (1-|x|\big )\, 1_{\{ |x|\le 1\}}. \end{aligned}$$

(6.1)

Let $\Lambda = \{-1, 0, 1\}$; clearly $\psi$ and $\Lambda$ satisfy the symmetry condition (2.6). Recall that $\Gamma$ denotes the set of elements $\lambda \in {{\mathbb {G}}}$ such that the intersection of the support of $\psi _\lambda :=\psi ^1_\lambda$ and of the support of $\psi$ has a positive Lebesgue measure. Then $\Gamma =\{ -1,0,1\}$, the function $\psi$ is clearly non negative, $\int _{{\mathbb {R}}} \psi (x) dx =1$, $\psi (x)=0$ for $x\in {\mathbb {Z}}\setminus \{0\}$ and Assumption 2.6 clearly holds.

Simple computations show that

$$\begin{aligned} R_0=2\int _0^1 x^2 dx = \frac{2}{3}, \quad R_{-1}=R_1=\int _0^1 x(1-x) dx = \frac{1}{6}. \end{aligned}$$

Hence $\sum _{\lambda \in \Gamma }R_\lambda =1$. Furthermore, given any $z=(z_n)\in \ell _2({\mathbb {Z}})$ we have using the Cauchy-Schwarz inequality:

$$\begin{aligned} \sum _{n\in {\mathbb {Z}}} \Big ( \frac{2}{3} z_n^2 + \frac{1}{6} z_n z_{n-1} + \frac{1}{6} z_n z_{n+1} \Big ) \ge \frac{2}{3} \Vert z\Vert ^2 - \frac{1}{6} \sum _{n\in {\mathbb {Z}}} \big ( z_n^2 + z_{n+1}^2\big ) = \frac{1}{3}\Vert z\Vert ^2. \end{aligned}$$

Hence Assumption 2.4 is satisfied. Easy computations show that for $\epsilon \in \{-1,1\}$ we have

$$\begin{aligned} R^{11}_0 = -2,\quad R^{11}_{\epsilon }=1,\quad R^1_0=0\quad \text{ and } \; R^1_{\epsilon }=\frac{\epsilon }{2}. \end{aligned}$$

Hence $\sum _{\lambda \in \Gamma }R^{11}_\lambda =0$, which completes the proof of (2.20). Furthermore, $\sum _{\lambda \in \Gamma } \lambda R^1_\lambda =1$, which proves (2.21) while $\sum _{\lambda \in \Gamma } \lambda ^2 R^{11}_\lambda = 2$, which proves (2.22).

Finally, we have for $\epsilon \in \{-1,1\}$

$$\begin{aligned} Q^{11,11}_0= -\frac{2}{3},\quad Q^{11,11}_\epsilon = \frac{1}{3},\quad {\tilde{Q}}^{11}_0=0\quad \text{ and } \; {\tilde{Q}}^{11}_\epsilon = -\frac{\epsilon }{6}. \end{aligned}$$

This clearly implies $\sum _{\lambda \in \Gamma } Q^{11,11}_\lambda =0$ and $\sum _{\lambda \in \Gamma } {\tilde{Q}}^{11}_\lambda =0$, which completes the proof of (2.23); therefore, Assumption 2.5 is satisfied.

The following example is an extension of the previous one to any dimension.

6.2 A general example

Consider the following finite elements on ${\mathbb {R}}^d$ defined as follows: let $\psi$ be defined on ${\mathbb {R}}^d$ by $\psi (x)=0$ if $x\notin (-1,+1]^d$ and

$$\begin{aligned} \psi (x)=\prod _{k=1}^d \big (1-|x_k|\big ) \; \text{ for } \; x=(x_1,\ldots , x_d) \in (-1,+1]^d. \end{aligned}$$

(6.2)

The function $\psi$ is clearly non negative and $\int _{{\mathbb {R}}^d} \psi (x) dx =1$. Let $\Lambda = \{ 0, \, \epsilon _k e_k, \, \epsilon _k\in \{-1,+1\}, \, k=1,\ldots ,d\}$. Then $\psi$ and $\Lambda$ satisfy the symmetry condition (2.6). Furthermore, $\psi (x)=0$ for $x\in {\mathbb {Z}}^d\setminus \{0\}$; Assumption 2.6 clearly holds.

These finite elements also satisfy all requirements in Assumptions 2.4–2.5. Even if these finite elements are quite classical in numerical analysis, we were not able to find a proof of these statements in the literature. To make the paper self-contained the corresponding easy but tedious computations are provided in an Appendix.

6.3 Linear finite elements on triangles in the plane

We suppose that $d=2$ and want to check that the following finite elements satisfy Assumptions 2.4–2.6. For $i=1,\ldots , 6$, let $\tau _i$ be the triangles defined as follows:

$$\begin{aligned}&\tau _1= \{ x\in {\mathbb {R}}^2 : 0\le x_2\le x_1\le 1\}, \quad \tau _2=\{ x\in {\mathbb {R}}^2 : 0\le x_1\le x_2\le 1\}, \nonumber \\&\tau _3=\{ x\in {\mathbb {R}}^2 : 0\le x_2\le 1 ,\quad x_2-1\le x_1\le 0\},\quad \tau _4=\{ x\in {\mathbb {R}}^2 : -1\le x_1\le x_2\le 0\}, \nonumber \\&\tau _5=\{x\in {\mathbb {R}}^2 : -1\le x_2\le x_1\le 0\},\quad \tau _6=\{ x\in {\mathbb {R}}^2 : 0\le x_1\le 1,\quad x_1-1\le x_2\le 0\}. \end{aligned}$$

(6.3)

Let $\psi$ be the function defined by:

$$\begin{aligned}&\psi (x)=1-|x_1| \, \text{ on } \tau _1\cup \tau _4, \; \psi (x)=1-|x_2| \, \text{ on } \tau _2\cup \tau _5, \nonumber \\&\psi (x)=1-|x_1-x_2| \, \text{ on } \tau _3\cup \tau _6, \; \text{ and } \, \psi (x)=0 \; \text{ otherwise }. \end{aligned}$$

(6.4)

It is easy to see that the function $\psi$ is non negative and that $\int _{{\mathbb {R}}^2} \psi (x) dx=1$. Set $\Lambda = \{ 0, e_1, -e_1, e_2, -e_2\}$; the function $\psi$ and the set $\Lambda$ fulfill the symmetry condition (2.6). Furthermore, $\Gamma = \big \{ \epsilon _1 e_1 + \epsilon _2 e_2 : (\epsilon _1, \epsilon _2)\in \{-1,0,1\}^2 , \; \epsilon _1\epsilon _2\in \{0,1\} \big \}$. Hence $\psi$ satisfies Assumption 2.6.

For $\mathbf{i}=(i_1,i_2)\in {\mathbb {Z}}^2$, let $\psi _\mathbf{i}$ the function defined by

$$\begin{aligned} \psi _\mathbf{i}(x_1,x_2) = \psi \big ( (x_1,x_2) - \mathbf{i} \big ) . \end{aligned}$$

For $\gamma =1, 2,\ldots , 6$, we denote by $\tau _\gamma (\mathbf{i}) =\big \{ (x_1,x_2) : (x_1,x_2) - \mathbf{i} \in \tau _\gamma \big \}$. Then

$$\begin{aligned}&D_1 \psi _\mathbf{i}= -1 \,\, \text{ on } \,\, \tau _1(\mathbf{i}) \cup \tau _6(\mathbf{i})\,\, \text{ and } \,\, D_1 \psi _\mathbf{i}=1\,\, \text{ on } \,\,\tau _3(\mathbf{i})\cup \tau _4(\mathbf{i}), \\&D_2 \psi _\mathbf{i}= -1\,\, \text{ on } \,\,\tau _2(\mathbf{i}) \cup \tau _3(\mathbf{i}) \,\, \text{ and } \,\, D_2 \psi _\mathbf{i}=1\,\, \text{ on } \,\, \tau _5(\mathbf{i})\cup \tau _6(\mathbf{i}),\\&D_1 \psi _\mathbf{i}=0\,\, \text{ on } \,\,\tau _2(\mathbf{i})\cup \tau _5(\mathbf{i}) \,\, \text{ and } \,\, D_2 \psi _\mathbf{i}=0 \,\, \text{ on } \,\, \tau _1(\mathbf{i})\cup \tau _4(\mathbf{i}). \end{aligned}$$

Easy computations show that for $\mathbf{i}\in {\mathbb {Z}}^2$, and $\mathbf{k} \in \{ \mathbf{i}+\lambda : \lambda \in \Gamma \}$

$$\begin{aligned} (\psi _\mathbf{i}, \psi _\mathbf{i})=\frac{1}{2}, \quad (\psi _\mathbf{i} , \psi _\mathbf{k}) = \frac{1}{12}, \end{aligned}$$

and $(\psi _\mathbf{i} , \psi _\mathbf{j}) =0$ otherwise. Thus

$$\begin{aligned} \sum _{\lambda \in \Gamma } R_\lambda =\sum _{\lambda \in \Gamma } (\psi , \psi _\lambda ) = \frac{1}{2}+6\times \frac{1}{12}=1, \end{aligned}$$

which proves the first identity in (2.20). First we check that given any $\alpha \in (0,1)$ by Cauchy–Schwarz inequality we have some positive constants $C_1$ and $C_2$ such that

$$\begin{aligned} \Big |\sum _\mathbf{i} U_\mathbf{i} \psi _\mathbf{i} \Big |_{L^2}^2 \ge&\sum _\mathbf{i} \int _0^\alpha dx_1 \int _0^{x_1} \big | (1-x_1) U_\mathbf{i} + (x_1-x_2) U_{\mathbf{i}+e_1} + x_2 U_{\mathbf{i}+e_1+e_2} \big |^2 dx_2\\&+ \sum _\mathbf{i} \int _0^\alpha dx_2 \int _0^{x_2} \big | (1-x_2) U_\mathbf{i} + (x_2-x_1) U_{\mathbf{i}+e_2} + x_1 U_{\mathbf{i}+e_1+e_2} \Big |^2 dx_1\\ \ge&|U|^2_{{l_2}({\mathbb {Z}}^2)} \big ( \alpha ^2 - C_1 \alpha ^3 -C_2 \alpha ^4\big ) \end{aligned}$$

for all $(U_\mathbf{i})\in \ell _2({\mathbb {Z}}^2)$. Hence, by taking $\alpha \in (0,1)$ such that $1-C_1 \alpha -C_2 \alpha ^2 \ge \frac{1}{2}$, we see that Assumption 2.4 is satisfied.

We next check the compatibility conditions in Assumption 2.5. Easy computations prove that for $k=1,2$ and $l\in \{1,2\}$ with $l\ne k$, $\epsilon _k, \epsilon _l \in \{-1,1\}$ we have

$$\begin{aligned} (D_k\psi , D_k\psi )&=\,2, \quad (D_k\psi , D_k\psi _{\epsilon _k e_k})=-1, \quad (D_k\psi , D_k\psi _{\epsilon _le_l})=0, \\ (D_k\psi , D_k\psi _\lambda )&=\,0 \; \text{ for } \lambda =\epsilon _1 e_1+\epsilon _2e_2, \; \epsilon _1\epsilon _2=1, \end{aligned}$$

while

$$\begin{aligned} (D_k \psi , D_l\psi )&=-1,\quad (D_k\psi ,D_l\psi _{\epsilon _ke_k})=(D_k\psi , D_l\psi _{\epsilon _le_l})=\frac{1}{2},\\ (D_k\psi , D_l\psi _\lambda )&=-\frac{1}{2} \; \text{ for } \lambda =\epsilon _1 e_1+\epsilon _2e_2, \; \epsilon _1\epsilon _2=1. \end{aligned}$$

Hence for any $k,l=1,2$ and $l\ne k$ we have

$$\begin{aligned}&\sum _{\lambda \in \Gamma }(D_k\psi , D_k\psi _\lambda )=2 + 2 \times (-1)=0,\quad \\&\sum _{\lambda \in \Gamma } (D_k\psi , D_l\psi _\lambda )=-1+4\times \frac{1}{2} + 2\times \big ( - \frac{1}{2}\big ) =0. \end{aligned}$$

This completes the proof of equation $\sum _{\lambda \in \Gamma } R^{ij}_\lambda =0$ and hence of Eq. (2.20). Furthermore, given $k,l=1,2$ with $k\ne l$ we have for $\alpha =k$ or $\alpha =l$:

$$\begin{aligned}&\sum _{\lambda \in \Gamma } R^{kk}_\lambda \lambda _k\lambda _k = - \sum _{\lambda \in \Gamma } (D_k\psi , D_k\psi _\lambda ) \lambda _k\lambda _k=2\times 1^2 =2,\\&\sum _{\lambda \in \Gamma } R^{kk}_\lambda \lambda _l\lambda _l = - \sum _{\lambda \in \Gamma } (D_k\psi , D_k\psi _\lambda ) \lambda _l\lambda _l=0, \\&\sum _{\lambda \in \Gamma } R^{kk}_\lambda \lambda _k\lambda _l = - \sum _{\lambda \in \Gamma } (D_k\psi , D_k\psi _\lambda ) \lambda _k\lambda _l=0,\\&\sum _{\lambda \in \Gamma } R^{kl}_\lambda \lambda _k\lambda _l = - \sum _{\lambda \in \Gamma } (D_k\psi , D_l\psi _\lambda ) \lambda _k\lambda _l= \frac{1}{2} \times 1^2 +\frac{1}{2} (-1)^2 =1,\\&\sum _{\lambda \in \Gamma } R^{kl}_\lambda \lambda _\alpha \lambda _\alpha =- \sum _{\lambda \in \Gamma } (D_k\psi , D_l\psi _\lambda ) \lambda _\alpha \lambda _\alpha =0. \end{aligned}$$

The last identities come from the fact that $(D_k\psi , D_l\psi _{\epsilon e_k})$ , $(D_k\psi , D_l\psi _{\epsilon e_l})$ or $(D_k\psi , D_l\psi _{\epsilon (e_1+e_2)}$ agree for $\epsilon =-1$ and $\epsilon =1$. This completes the proof of Eq. (2.22).

We check the third compatibility condition. Using Lemma 2.2 we deduce for $k,l=1,2$ with $k\ne l$ and $\epsilon \in \{-1,+1\}$

$$\begin{aligned}&(D_k \psi , \psi )= 0,\quad (D_k \psi _{\epsilon e_k}, \psi )=\frac{\epsilon }{3}, \\&(D_k \psi _{\epsilon e_l}, \psi )=- \frac{\epsilon }{6},\quad (D_k \psi _{\epsilon (e_1+e_2)}, \psi )= \frac{\epsilon }{6}. \end{aligned}$$

Therefore, using Lemma 2.2 we deduce that

$$\begin{aligned}&\sum _{\lambda \in \Gamma } (D_k \psi _\lambda , \psi ) \lambda _k = \frac{1}{3} + (-1)\times \big ( -\frac{1}{3}\big ) + \frac{1}{6} + (-1)\times \big ( - \frac{1}{6}\big ) =1, \\&\sum _{\lambda \in \Gamma } (D_k \psi _\lambda , \psi ) \lambda _l =- \frac{1}{6} + \frac{1}{6}\times (-1) + \frac{1}{6} - \frac{1}{6} \times (-1)=0. \end{aligned}$$

This completes the proof of Eq. (2.21).

Let us check the first identity in (2.23). Recall that

$$\begin{aligned} Q^{ij,kl}_\lambda =- \int _{{\mathbb {R}}^2} z_k z_l D_i\psi (z) D_j\psi _\lambda (z) dz, \end{aligned}$$

and suppose at first that $i=j$. Then we have for $k\ne i$, $\alpha \ne i$, $k\ne l$ and $\epsilon \in \{-1,+1\}$

$$\begin{aligned} Q^{ii,ii}_0&=\,-\,\frac{2}{3}, \quad Q^{ii,ii}_{\epsilon e_i}=\frac{1}{3},\quad Q^{ii,ii}_{\epsilon e_\alpha }=Q^{ii,ii}_{\epsilon (e_i+e_\alpha )}=0,\\ Q^{ii,kk}_0&=\,-\,\frac{1}{3}, \quad Q^{ii,kk}_{\epsilon e_i}=\frac{1}{6}, \quad Q^{ii,kk}_{\epsilon e_k}=Q^{ii,kk}_{\epsilon (e_i+e_k)}=0,\\ Q^{ii,kl}_0&=\,-\,\frac{1}{6}, \quad Q^{ii,kl}_{\epsilon e_i}=\frac{1}{12}, \quad Q^{ii,kl}_{\epsilon e_\alpha }=Q^{ii,kl}_{\epsilon (e_i+e_\alpha )}=0. \end{aligned}$$

Suppose that $i\ne j$; then for $k\ne l$ and $\epsilon \in \{-1,+1\}$ we have

$$\begin{aligned} Q^{ij,ii}_0=&\frac{1}{6},\quad Q^{ij,ii}_{\epsilon e_j}=-\frac{1}{12},\quad Q_{\epsilon e_i}^{ij,ii}=-\frac{1}{4},\quad Q^{ij,ii}_{\epsilon (e_i+e_j)}=\frac{1}{4},\\ Q^{ij,jj}_0=&\frac{1}{6},\quad Q^{ij,jj}_{\epsilon e_i}=-\frac{1}{12},\quad Q_{\epsilon e_j}^{ij,jj}=-\frac{1}{12},\quad Q^{ij,jj}_{\epsilon (e_i+e_j)}=\frac{1}{12},\\ Q^{ij,kl}_0=&-\frac{1}{12},\quad Q^{ij,kl}_{\epsilon e_j}=\frac{1}{24},\quad Q_{\epsilon e_i}^{ij,kl}=-\frac{1}{8},\quad Q^{ij, kl} _{\epsilon (e_i+e_j)}=\frac{1}{8}.\\ \end{aligned}$$

The above equalities prove $\sum _{\lambda \in \Gamma } Q^{ij,kl}_\lambda =0$ for any choice of $i,j,k,l=1,2$. Hence the first identity in (2.23) is satisfied.

We finally check the second identity in (2.23). Recall that ${\tilde{Q}}_\lambda ^{i,k}=\int _{{\mathbb {R}}^2} z_k D_i\psi _\lambda (z) \psi (z) dz$. For $i=k\in \{1,2\}$, $j\in \{1,2\}$ with $i\ne j$ and $\epsilon \in \{-1,+1\}$ we have

$$\begin{aligned} {\tilde{Q}}_0^{i,i}= - \frac{3}{12}, \quad {\tilde{Q}}_{\epsilon e_i}^{i,i}= \frac{3}{24}, \quad {\tilde{Q}}_{\epsilon e_j}^{i,i}= - \frac{1}{24},\quad {\tilde{Q}}_{\epsilon (e_i+e_j)}^{i,i}= \frac{1}{24}. \end{aligned}$$

Hence $\sum _{\lambda \in \Gamma } {\tilde{Q}}_\lambda ^{i,i}=0$. Let $i\ne k$; then for $\epsilon \in \{-1,+1\}$ we have

$$\begin{aligned} {\tilde{Q}}_0^{i,k} = {\tilde{Q}}_{\epsilon e_i}^{i,k}=0, \quad {\tilde{Q}}^{i,k}_{\epsilon e_k}=- \frac{1}{12},\quad {\tilde{Q}}^{i,k}_{\epsilon (e_i+e_k)}= \frac{1}{12}. \end{aligned}$$

Hence $\sum _{\lambda \in \Gamma } {\tilde{Q}}_\lambda ^{i,k}=0$ for any choice of $i,k=1,2$, which concludes the proof of (2.23) Therefore, the function $\psi$ defined by (6.4) satisfies all Assumptions 2.4–2.6.

Appendix

The aim of this section is to prove that the example described in Sect. 6.2 satisfies Assumptions 2.4 and 2.5.

For $k=1,\ldots , d$, let $e_k\in {\mathbb {Z}}^d$ denote the k-th unit vector of ${\mathbb {R}}^d$; then ${\mathbb {G}}={\mathbb {Z}}^d$ and

$$\begin{aligned} \Gamma =\left\{ \sum _{k=1}^d \epsilon _k e_k \; : \; \epsilon _k\in \{-1,0,1\} \, \text{ for } k=1, \ldots , d \right\} . \end{aligned}$$

For fixed $k=1, \ldots , d$ (resp. $k\ne l \in \{1, \ldots , d\}$) let

$$\begin{aligned} {{\mathcal {I}}}(k)=\{1, \ldots , d\} \setminus \{k\}, \quad \text{ resp. } {{\mathcal {I}}}(k,l)=\{1, \ldots , d\} \setminus \{k,l\}. \end{aligned}$$

(7.1)

Note that in the particular case $d=1$, the functions $\psi$ gives rise to the classical linear finite elements. Then for $\mathbf{i} \in {{\mathbb {Z}}}^d$, we have for $k=0, 1, \ldots , d$:

$$\begin{aligned} R_\mathbf{i}:= (\psi _\mathbf{i},\psi )= {d\atopwithdelims ()k} \Big (\frac{1}{6}\Big )^k \Big ( \frac{2}{3}\Big )^{d-k} \quad \text{ if } \; \sum _{l=1}^d |i_l|=k. \end{aligned}$$

(7.2)

Furthermore, given $k=0, 1, \ldots , d$, there are $2^k$ elements $\mathbf{i}\in {\mathbb {Z}}^d$ such that $\sum _{l=1}^d |i_l|=k$. Therefore, we deduce

$$\begin{aligned} \sum _{\mathbf{i}\in {\mathbb {Z}}^d} (\psi _\mathbf{i},\psi )= \sum _{k=0}^d 2^k {d\atopwithdelims ()k} \Big (\frac{1}{6}\Big )^k \Big ( \frac{2}{3}\Big )^{d-k} = \sum _{k=0}^d {d\atopwithdelims ()k} \Big (\frac{2}{6}\Big )^k \Big ( \frac{2}{3}\Big )^{d-k}=1, \end{aligned}$$

which yields the first compatibility conditon in (2.20).

We at first check that Assumption 2.4 holds true, that is

$$\begin{aligned} \delta \sum _{\mathbf{i}\in {\mathbb {Z}}^d} U_\mathbf{i}^2 = \delta |U|^2_{\ell _2({\mathbb {Z}}^d)} \le \Big |\sum _{\mathbf{i}\in {\mathbb {Z}}^d} U_\mathbf{i} \psi _\mathbf{i}\Big |_{L^2}^2 =\sum _{\mathbf{i}, \mathbf{j}\in {\mathbb {Z}}^d} R_{\mathbf{i}-\mathbf{j}} U_\mathbf{i} U_\mathbf{j}, \quad U\in \ell _2({\mathbb {Z}}^d). \end{aligned}$$

for some $\delta >0$. For $U\in \ell _2({{\mathbb {Z}}}^d)$ and $k=1, \ldots , d,$ let $T_k U=U_{e_k}$, where $e_k$ denotes the k-th vector of the canonical basis.

For $U\in \ell _2({\mathbb {Z}}^d)$ we have

$$\begin{aligned} \Big | \sum _\mathbf{i} U_\mathbf{i} \psi _\mathbf{i}\Big |_{L^2}^2&= \sum _\mathbf{i\in {\mathbb {Z}}^d} \int _{[0,1]^d} \left[ U_\mathbf{i} \prod _{j=1}^d ( 1-{x_j}) + \sum _{k=1}^d (T_k U)_\mathbf{i}\; x_k \prod _{j\in {{\mathcal {I}}}(k)}(1-x_j) \right. \\&\left. \quad + \sum _{1\le k_1<k_2\le d} (T_{k_1} \circ T_{k_2} U)_\mathbf{i}\; x_{k_1} x_{k_2}\right. \\&\left. \quad \times \prod _{j\in {{\mathcal {I}}}(k_1,k_2)}(1-x_j) +\cdots + (T_1 \circ T_2 \cdots \circ T_d U)_\mathbf{i} \prod _{k=1}^d x_k\right] ^{2} dx, \end{aligned}$$

Given $\alpha \in (0,1)$ if we let

$$\begin{aligned} I(\alpha )= & {} \int _0^\alpha (1-x)^2 dx = \alpha - \alpha ^2 + \frac{\alpha ^3}{3}, \\ J(\alpha )= & {} \int _0^\alpha x(1-x) dx = \frac{\alpha ^2}{2}-\frac{\alpha ^3}{3}, \; K(\alpha )= \int _0^\alpha x^2 dx = \frac{\alpha ^3}{3} , \end{aligned}$$

restricting the above integral on the set $[0,\alpha ]^d$, expanding the square and using the Cauchy Schwarz inequality we deduce the existence of some constants $C(\gamma _1,\gamma _2, \gamma _3)$ defined for $\gamma _i\in \{0,1, \ldots , d\}$ such that

$$\begin{aligned} \Big | \sum _\mathbf{i} U_\mathbf{i} \psi _\mathbf{i}\Big |_{L^2}^2 \ge&\sum _\mathbf{i} |U_\mathbf{i}|^2 \Big [ I(\alpha )^d + {d\atopwithdelims ()1} K(\alpha )\, I(\alpha )^{d-1} \\&+ {d\atopwithdelims ()2} K(\alpha )^2\, I(\alpha )^{d-2} + \cdots + K(\alpha )^d\Big ] \\&-\, 2 \Big ( \sum _\mathbf{i} |U_\mathbf{i}|^2 \Big ) \sum _{\gamma _1+\gamma _2+\gamma _3=d, \gamma _2+\gamma _3\ge 1} C(\gamma _1,\gamma _2, \gamma _3)\; I(\alpha )^{\gamma _1}\; J(\alpha )^{\gamma _2} \; K(\alpha )^{\gamma _3}\\ \ge&\, |U|_{\ell _2({{\mathbb {Z}}}^d)}^2 \Big ( \alpha ^d - \sum _{l=d+1}^{3d} C_l \alpha ^l \Big ), \end{aligned}$$

where $C_l$ are some positive constants. Choosing $\alpha$ small enough, we have $\big | \sum _\mathbf{i} U_\mathbf{i} \psi _\mathbf{i}\big |_{L^2}^2 \ge \frac{\alpha ^d}{2} |U|_{\ell _2({{\mathbb {Z}}}^d)}^2$, which implies the invertibility Assumption 2.4.

We now prove that the compatibility Assumption 2.5 holds true. For $l=1, \ldots , d$, $n=0, \ldots , d-1$:

$$\begin{aligned} (D_l \psi _\mathbf{i} , D_l\psi )&= - \, 2^{d-1-n} \, \Big ( \frac{1}{6}\Big )^n \, \Big ( \frac{1}{3}\Big )^{d-1-n} \; \text{ for } |i_l|=1, \; \sum _{r \ne l, 1\le n\le d} |i_r|=n, \end{aligned}$$

(7.3)

$$\begin{aligned} (D_l \psi _\mathbf{i} , D_l \psi )&= + \, 2^{d-n} \, \Big (\frac{1}{6}\Big )^n \, \Big ( \frac{1}{3}\Big )^{d-1-n} \; \text{ for } |i_l|=0, \; \sum _{r\ne l, 1\le r\le d} |i_r|=n. \end{aligned}$$

(7.4)

For $n=1, \ldots , d-1$ and $k_1<k_2<\cdots < k_n$ with $k_r\in {{\mathcal {I}}}(l)$, where ${{\mathcal {I}}}(l)$ is defined in (7.1), let

$$\begin{aligned} \Gamma _l(k_1, \ldots , k_n)&=\Big \{ \sum _{r=1}^n \epsilon _{k_r} e_{k_r} \; : \; \epsilon _{k_r}\in \{ -1, 1\}, r=1, \ldots , n \Big \} ,\\ \Gamma _l(l;k_1, \ldots , k_n)&=\Big \{ \epsilon _l e_l + \sum _{r=1}^n \epsilon _{k_r} e_{k_r} \; : \; \epsilon _l \in \{-1,1\} \; \text{ and } \epsilon _{k_r}\in \{ -1, 1\}, r=1, \ldots , n\Big \} . \end{aligned}$$

Then $| \Gamma _l (k_1, \ldots , k_n)|=2^n$ while $| \Gamma _l (l;k_1, \ldots , k_n)|=2^{n+1}$. For $l=1, \ldots , d$, the identities (7.3) and (7.4) imply

$$\begin{aligned} \sum _{\lambda \in \Gamma }&(D_l \psi , D_l \psi _\lambda ) = \left[ (D_l\psi , D_l\psi ) + \sum _{\epsilon _l\in \{-1,+1\}} (D_l \psi , D_l \psi _{\epsilon _l e_l})\right] \\&\quad + \sum _{n=1}^{d-1} \sum _{k_1<k_2< \cdots <k_n, k_j\in {{\mathcal {I}}}(l)} \left[ \sum _{\lambda \in \Gamma _l(k_1,\ldots , k_n)} (D_l\psi , D_l\psi _\lambda ) + \sum _{\lambda \in \Gamma _l(l; k_1, \ldots , k_n)} (D_l\psi , D_l\psi _\lambda ) \right] \\&= \Big [ 2^d \Big ( \frac{1}{3}\Big )^{d-1} -2\times 2^{d-1} \Big ( \frac{1}{3}\Big )^{d-1}\Big ] \\&\quad +\sum _{n=1}^{d-1} 2^n {d-1 \atopwithdelims ()n} \Big [ 2^{d-n} \Big ( \frac{1}{6} \Big )^n \Big ( \frac{1}{3}\Big )^{d-1-n} -2\times 2^{d-1-n} \Big (\frac{1}{6}\Big )^n \Big ( \frac{1}{3}\Big )^{d-1-n}\Big ] =0 . \end{aligned}$$

This proves the second identity in (2.20) when $i=j$. Furthermore, (7.3) implies

$$\begin{aligned} \sum _{\lambda \in \Gamma } R^{ll}_\lambda \lambda _l \lambda _l&= - \sum _{\lambda \in \Gamma } (D_l \psi , D_l \psi _\lambda ) \lambda _l \lambda _l = - \sum _{\epsilon _l\in \{-1,1\}} \big (D_l \psi , D_l \psi _{\epsilon _l e_l} \big ) \\&\quad - \sum _{n=1}^{d-1} \sum _{k_1<k_2< \cdots <k_n, k_j\in {{\mathcal {I}}}(l)} \sum _{\lambda \in \Gamma _l(l;k_1, \ldots , k_n)} (D_l \psi , D_l \psi _\lambda )\\&= 2\times 2^{d-1} \Big (\frac{1}{3}\Big )^{d-1} + \sum _{d=1}^n {d-1 \atopwithdelims ()n} 2^{n+1} \times 2^{d-1-n} \Big ( \frac{1}{6}\Big )^n \Big ( \frac{1}{3}\Big )^{d-1-n} \\&= 2 \sum _{n=0}^{d-1} {d-1 \atopwithdelims ()n} \Big (\frac{2}{6}\Big )^n \Big (\frac{2}{3}\Big )^{d-1-n} =2,\quad l=1,\ldots ,d. \end{aligned}$$

Furthermore, given $k\ne l \in \{1, \ldots , d\}$,

$$\begin{aligned} \sum _{\lambda \in \Gamma } R^{ll}_\lambda \lambda _k\lambda _k&= - \sum _{\lambda \in \Gamma } (D_l \psi , D_l \psi _\lambda ) \lambda _k \lambda _l=0. \end{aligned}$$

Indeed, for $n=1, \ldots , d-1$, $k_1<\cdots < k_n$ where $k_r\in {{\mathcal {I}}}(l)$ and at least one of the indices $k_r$ is equal to k for $r=1, \ldots ,n$, given $\lambda \in \Gamma _l(k_1, \ldots , k_n)$ we have using (7.3) and (7.4)

$$\begin{aligned} \sum _{\epsilon _l\in \{-1,1\}} (D_l\psi ,D_l\psi _{\epsilon _le_l+\lambda }) \lambda _k \lambda _l = - 2^{d-1-n} \Big (\frac{1}{6}\Big )^n \Big (\frac{1}{3}\Big )^{d-1-n} \times (-1+1)=0. \end{aligned}$$

This proves the second identity in (2.22) when both derivatives agree.

Also note that for $k\ne l \in \{1, \ldots , d\}$ we have $\sum _{\lambda \in \Gamma } R^{kl}_\lambda =0$. Indeed, for $\lambda$ as above

$$\begin{aligned} (D_k \psi , D_l \psi _\lambda )&+ \sum _{\epsilon _l\in \{-1,1\}} (D_k\psi , D_l\psi _{\epsilon _le_l+\lambda }) \\&= 2^{d-n} \Big (\frac{1}{6}\Big )^n \Big (\frac{1}{3}\Big )^{d-1-n} -2\times 2^{d-1-n} \Big (\frac{1}{6}\Big )^n \Big (\frac{1}{3}\Big )^{d-1-n} =0, \end{aligned}$$

while $R^{kl}_\lambda =0$ for other choices of $\lambda \in \Gamma$.

We now study the case of mixed derivatives. Given $k\ne l \in \{1, \ldots , d\}$ recall that ${{\mathcal {I}}}(k,l)=\{ 1, \ldots , d\} \setminus \{k,l\}$. Then for $k\ne l \in \{1, \ldots , d\}$ and $\mathbf{i} \in {\mathbb {Z}}^d$ we have for $n=0, \ldots , d-2$

$$\begin{aligned} (D_k \psi _\mathbf{i}, D_l \psi )&=\,0 \; \text{ if } |i_k \, i_l|\ne 1, \end{aligned}$$

(7.5)

$$\begin{aligned} (D_k \psi _\mathbf{i}, D_l \psi )&= - \,\Big (\frac{1}{2}\Big )^2 \; \Big ( \frac{1}{6}\Big )^n \; \Big (\frac{2}{3}\Big )^{d-n-2} \; \text{ if } \; i_k\, i_l=1, \quad \sum _{r\in {{\mathcal {I}}}(k,l)} |i_r|=n, \end{aligned}$$

(7.6)

$$\begin{aligned} (D_k \psi _\mathbf{i}, D_l \psi )&= +\, \Big (\frac{1}{2}\Big )^2 \; \Big ( \frac{1}{6}\Big )^n \; \Big (\frac{2}{3}\Big )^{d-n-2} \; \text{ if } \; i_k \, i_l=- 1, \quad \sum _{r\in {{\mathcal {I}}}(k,l)} |i_r|=n. \end{aligned}$$

(7.7)

For $n=1, \ldots , d-2$ and $k_1<\cdots <k_n$ with $k_r\in {{\mathcal {I}}}(k,l)$ for $r=1,\ldots , n$, set

$$\begin{aligned} \Gamma _{k,l}(k_1, \ldots , k_n)=\Big \{ \sum _{r=1}^n \epsilon _{k_r} e_{k_r} \; : \; \epsilon _r\in \{-1,1\} \Big \}. \end{aligned}$$

For $n=0$ there is no such family of indices $k_1<\cdots <k_n$ and we let $\Gamma _{k,l}(\emptyset )=\{0\}$. Thus for $n=0, \ldots , d-2$, $| \Lambda _{k,l}(k_1,\ldots , k_n)|=2^n$. Using the identities (7.5)–(7.7) we deduce

$$\begin{aligned} \sum _{\lambda \in \Gamma }&(D_k \psi , D_l \psi _\lambda ) = \sum _{n=0}^{d-2} \; \sum _{k_1<k_2<\cdots <k_n, k_r\in {{\mathcal {I}}}(k,l)} \sum _{\lambda \in \Gamma _{k,l}(k_1,\ldots , k_n)} \big [ (D_k\psi , D_l\psi _{e_k+e_l+\lambda }) \nonumber \\&\quad + (D_k\psi , D_l\psi _{e_k-e_l+\lambda })+ (D_k\psi , D_l\psi _{-e_k+e_l+ \lambda }) + (D_k\psi , D_l\psi _{-e_k-e_l+\lambda }) \big ] \nonumber \\ =&\sum _{n=0}^{d-2} {d-2 \atopwithdelims ()n} \, 2^n \, \Big [ - \Big ( \frac{1}{2}\Big )^2 \, \Big (\frac{1}{6}\Big )^n \, \Big (\frac{2}{3}\Big )^{d-2-n} + \Big ( \frac{1}{2}\Big )^2 \, \Big (\frac{1}{6}\Big )^n \, \Big (\frac{2}{3}\Big )^{d-2-n} \nonumber \\&\quad + \Big ( \frac{1}{2}\Big )^2 \, \Big (\frac{1}{6}\Big )^n \, \Big (\frac{2}{3}\Big )^{d-2-n} - \Big ( \frac{1}{2}\Big )^2 \, \Big (\frac{1}{6}\Big )^n \, \Big (\frac{2}{3}\Big )^{d-2-n}\Big ] = 0,\quad k\ne l. \end{aligned}$$

(7.8)

This completes the proof of the second identity in (2.20) when $i\ne j$, and hence (2.20) holds true. Furthermore, the identities (7.6) and (7.7) imply for $i\ne j \in \{1, \ldots , d\}$ and $\{i,j\}=\{k,l\}$

$$\begin{aligned} \sum _{\lambda \in \Gamma }&(D_k \psi , D_l \psi _\lambda )\lambda _k \lambda _l = \sum _{n=0}^{d-2} \; \sum _{k_1<k_2<\cdots <k_n, k_r\in {{\mathcal {I}}}(k,l)} \sum _{\lambda \in \Gamma _{k,l}(k_1,\ldots , k_n)} \big [ (D_k\psi , D_l\psi _{e_k+e_l+\lambda }) \\&\qquad -\, (D_k\psi , D_l\psi _{e_k-e_l+\lambda }) - (D_k\psi , D_l\psi _{-e_k+e_l+ \lambda }) + (D_k\psi , D_l\psi _{-e_k-e_l+\lambda }) \big ] \\&\quad = -\, 4 \, \Big ( \frac{1}{2}\Big )^2 \sum _{n=0}^{d-2} {d-2 \atopwithdelims ()n} \, 2^n \, \Big (\frac{1}{6}\Big )^n \, \Big (\frac{2}{3}\Big )^{d-2-n}\\&\quad =- \sum _{n=0}^{d-2} {d-2 \atopwithdelims ()n} \, \Big (\frac{2}{6}\Big )^n \Big ( \frac{2}{3}\Big )^{d-2-n}= -1. \end{aligned}$$

Equation (7.5) proves that $(D_k \psi , D_l \psi _\lambda )=0$ if $|\lambda _k \lambda _l|\ne 1$. Hence using (7.8) we deduce that for any $r=1, \ldots , d$,

$$\begin{aligned} \sum _{\lambda \in \Gamma } (D_k \psi , D_l \psi _\lambda )\lambda _r \lambda _r =0. \end{aligned}$$

Let $r\in {{\mathcal {I}}}(k,l)$ and for $n=1, \ldots , d-3$, let $k_1<\cdots < k_n$ be such that $k_j\in \{1, \ldots , d\} \setminus \{k,l,r\}$ and $\lambda = \sum _{j=1}^n \epsilon _{k_j} e_{k_j}$ for $\epsilon _{k_j}\in \{-1,1\}$, $j=1, \ldots , n$. Then for any choice of $\epsilon _k$ and $\epsilon _l$ in $\{-1,1\}$ the equalities (7.6) and (7.7) imply that

$$\begin{aligned} (D_k\psi , D_l \psi _{\lambda + \epsilon _k e_k + \epsilon _l e_l + e_r} ) = (D_k\psi , D_l \psi _{\lambda + \epsilon _k e_k + \epsilon _l e_l - e_r} ) . \end{aligned}$$

This clearly yields that for $r\in {{\mathcal {I}}}(k,l)$ we have

$$\begin{aligned} \sum _{\lambda \in \Gamma } (D_k \psi , D_l \psi _\lambda )\lambda _k \lambda _r = \sum _{\lambda \in \Gamma } (D_k \psi , D_l \psi _\lambda )\lambda _l \lambda _r =0. \end{aligned}$$

Finally, given $n=2, \ldots , d$ and $k_1< \cdots <k_n$ where the terms $k_j\in {{\mathcal {I}}}(k,l)$, then given any choice of $\epsilon _k$ and $\epsilon _l$ in $\{-1,1\}$, the value of $(D_k \psi , D_l \psi _{\epsilon _k e_k + \epsilon _l e_l + \lambda })$ does not depend on the value of $\lambda \in \Gamma _{k,l}(k_1,\ldots , k_n)$. Therefore, if we fix $r_1\ne r_2$ in the set ${{\mathcal {I}}}(k,l)$, for fixed n there are as many choices of indices $k_1<\cdots <k_n$ such that $\epsilon _{r_1} \epsilon _{r_2}=1$ that of such indices such that $\epsilon _{r_1} \epsilon _{r_2}=-1$. This proves

$$\begin{aligned} \sum _{\lambda \in \Gamma } (D_k\psi , D_l\psi _\lambda ) \lambda _{r_1} \lambda _{r_2}=0, \end{aligned}$$

which completes the proof of the first identity in (2.22) for mixed derivatives; hence (2.22) holds true.

We now check the compatibility condition (2.21). Fix $j\in \{1, \ldots , d\}$; then

$$\begin{aligned} (D_j\psi , \psi )= & {} 2^{d-1} \Big ( \prod _{k\ne j} \int _0^1 (1-x_k)^2 dx_k\Big ) \nonumber \\&\quad \times \Big [ \int _0^1 (-1) (1-x_j) dx_j + \int _{-1}^0 (1+x_j) dx_j\Big ] =0, \end{aligned}$$

(7.9)

while

$$\begin{aligned} (D_j \psi , \psi _{e_j})&=2^{d-1} \Big ( \prod _{k\ne j} \int _0^1 (1-x_k)^2 dx_k\Big ) \int _0^1 (-1) \big (1+(x_j-1)\big ) dx_j = -\, \frac{1}{2} \Big (\frac{2}{3}\Big )^{d-1}, \nonumber \\ (D_j \psi , \psi _{-e_j})&=2^{d-1} \Big ( \prod _{k\ne j} \int _0^1 (1-x_k)^2 dx_k\Big ) \int _{-1}^0 \big (1- (x_j+1)\big ) dx_j = \frac{1}{2} \Big (\frac{2}{3}\Big )^{d-1}. \end{aligned}$$

(7.10)

For $n=1, \ldots , d-1$ and $k_1<\cdots <k_n$ where the indexes $k_r$, $r=1, \ldots , n$ are different from j we have for any $\lambda \in \Gamma _j(k_1,\ldots , k_n)$

$$\begin{aligned} (D_j\psi , \psi _{\lambda }) =&\, 2^{d-(n+1)} \Big (\prod _{k\in \Gamma \setminus \{j,k_1, \ldots , k_n\}} \int _0^1 (1-x_k)^2 dx_k\Big ) \times \Big ( \prod _{r=1}^n \int _0^1 x_{k_r} (1-x_{k_r}) dx_{k_r} \Big ) \nonumber \\&\times \, \Big [ \int _0^1 (-1) (1-x_j) dx_j + \int _{-1}^0 (1+x_j) dx_j\Big ] = 0, \end{aligned}$$

(7.11)

while

$$\begin{aligned} (D_j\psi , \psi _{e_j+\lambda }) =&\, 2^{d-(n+1)} \Big (\prod _{k\in \Gamma \setminus \{j,k_1, \ldots , k_n\}} \int _0^1 (1-x_k)^2 dx_k\Big )\nonumber \\&\times \Big ( \prod _{r=1}^n \int _0^1 x_{k_r} (1-x_{k_r}) dx_{k_r} \Big ) \nonumber \\&\times \int _0^1 (-1) \big ( 1+ (x_j-1)\big ) dx_j = -\frac{1}{2} \; \Big ( \frac{2}{3}\Big )^{d-(n+1)} \; \Big ( \frac{1}{6}\Big )^n , \end{aligned}$$

(7.12)

and

$$\begin{aligned} (D_j\psi , \psi _{-e_l+\lambda }) =&\, 2^{d-(n+1)} \Big (\prod _{k\in \Gamma \setminus \{j,k_1,\ldots , k_n\}} \int _0^1 (1-x_k)^2 dx_k\Big )\nonumber \\&\times \Big ( \prod _{r=1}^n \int _0^1 x_{k_r} (1-x_{k_r}) dx_{k_r} \Big ) \nonumber \\&\times \int _{-1}^0 \big ( 1- (x_j+1)\big ) dx_j = \frac{1}{2} \; \Big ( \frac{2}{3}\Big )^{d-(n+1)} \; \Big ( \frac{1}{6}\Big )^n . \end{aligned}$$

(7.13)

Note that the number of terms $(D_j\psi , \psi _{\epsilon _l e_l+\lambda })$ with $\epsilon _l=-1$ or $\epsilon _l =+1$ is equal to ${d-1 \atopwithdelims ()n} 2^n$. Therefore, the identities (7.9)–(7.13) imply that for any $j=1, \ldots , d$ we have

$$\begin{aligned}&\sum _{\lambda \in \Gamma } \lambda _j R^j_\lambda =- \sum _{\lambda \in \Gamma } (D_j \psi , \psi _\lambda )\; \lambda _j = \frac{1}{2} \Big (\frac{2}{3}\Big )^{d-1} - \frac{1}{2} \Big (\frac{2}{3}\Big )^{d-1} (-1) \end{aligned}$$

(7.14)

$$\begin{aligned}&\qquad + \frac{1}{2} \sum _{n=1}^{d-1} {d-1 \atopwithdelims ()n} 2^n \Big ( \frac{2}{3}\Big )^{d-1-n} \Big ( \frac{1}{6}\Big )^n - \frac{1}{2}\sum _{n=0}^{d-1} {d-1 \atopwithdelims ()n} 2^n \Big ( \frac{2}{3}\Big )^{d-1-n} \Big ( \frac{1}{6}\Big )^n \times (-1) \nonumber \\&\quad = \sum _{n=0}^{d-1} {d-1 \atopwithdelims ()n} \Big ( \frac{2}{6}\Big )^n \Big (\frac{2}{3}\Big )^{d-1-n} = 1. \end{aligned}$$

(7.15)

This proves (2.21) when $i=k$.

Let $k\ne j \in \{1,\ldots , d\}$ and given $n=1,\ldots , d-1$ let $k_1<\cdots < k_n$ be indices that belong to ${{\mathcal {I}}}(j)$ such that one of the indices $k_r, r=1,\ldots , n$ is equal to k. Given any $\lambda \in \Gamma _j(k_1,\ldots , k_n)$ we deduce that

$$\begin{aligned} (D_j \psi , \psi _{e_l+\lambda }) \lambda _k + (D_j \psi , \psi _{-e_l+\lambda }) \lambda _k =0. \end{aligned}$$

This completes the proof of the identity (2.21).

In order to complete the proof of the validity of Assumption 2.5, it remains to check that the identities in (2.23) hold true. Recall that for $\lambda \in \Gamma$ and $i,j,k,l\in \{1,\ldots ,d\}$ we have

$$\begin{aligned} Q_\lambda ^{ij,kl}=\int _{{\mathbb {R}}^d} z_k z_l D_j\psi _\lambda (z) D_i^*\psi (z) dz =-\int _{{\mathbb {R}}^d} z_k z_l D_j\psi _\lambda (z) D_i \psi (z) dz. \end{aligned}$$

For $p=1,\ldots , 4$, $n=1,\ldots , d-p$ and $i_1,\ldots , i_p \in \{1, \ldots ,d\}$ with $i_1,\ldots , i_p$ pairwise different let

$$\begin{aligned}&{\mathcal {I}}_n(i_1,\ldots , i_p):=\Big \{ \sum _{\alpha =1}^n \epsilon _\alpha e_{k_\alpha } ; \epsilon _\alpha \in \{-1,+1\}, \; 1\le k_1< \cdots <k_n\le d, \\&\quad k_\alpha \not \in \{i_1,\ldots , i_p\} \, \text{ for } \alpha =1,\ldots , n\Big \}, \end{aligned}$$

and ${\mathcal {I}}_0(i_1,\ldots , i_p)=\{0\}$.

First suppose that $i=j$.

First let $k=l=i$; then for $n=0,\ldots , d-1$ and $\mu \in {\mathcal {I}}_n(i)$ we have

$$\begin{aligned} Q_{\mu }^{ii,ii} + Q_{\mu + e_i}^{ii,ii} + Q_{\mu - e_i}^{ii,ii}=0. \end{aligned}$$

Let $k=l$ with $k\ne i$; then then for $n=0,\ldots , d-1$ and $\mu \in {\mathcal {I}}_n(i)$ we have

$$\begin{aligned} Q_{\mu }^{ii,kk} + Q_{\mu + e_i}^{ii,kk} + Q_{\mu -e_i}^{ii,kk}=0. \end{aligned}$$

Let $l=i$ and $k\ne i$; then for $n=0,\ldots , d-2$, $\epsilon \in \{-1,+1\}$ and $\mu \in {\mathcal {I}}_n(i,k)$ we have

$$\begin{aligned} Q_{\mu + \epsilon e_i + e_k}^{ii,ki} + Q_{\mu + \epsilon e_i - e_k}^{ii,ki}=0. \end{aligned}$$

A similar result holds for $k=i$ and $l\ne i$. Furthermore, $Q_\lambda ^{ii,ki}=0$ if $\lambda$ is not equal to $\mu + \epsilon e_i + e_k$ or $\mu + \epsilon e_i - e_k$ with $\mu \in {\mathcal {I}}_n(i,k)$ for some n.

Let $k\ne l$ with $k\ne i$ and $l\ne i$; then for $n=0,\ldots , d-2$, $\epsilon \in \{-1,+1\}$ and $\mu \in {\mathcal {I}}_n(k,l)$ we have

$$\begin{aligned} Q_{\mu + \epsilon e_k + e_l}^{ii,kl} + Q_{\mu + \epsilon e_k - e_l}^{ii,kl}=0, \end{aligned}$$

while $Q_\lambda ^{ii,kl}=0$ if $\lambda$ is not equal to $\mu + \epsilon e_i + e_k$ or $\mu + \epsilon e_i - e_k$ with $\mu \in {\mathcal {I}}_n(i,k)$ for some n.

We now suppose that $i\ne j$.

First suppose that $k=i$ and $l=j$; then for $n=0,\ldots , d-1$ and $\mu \in {\mathcal {I}}_n(i)$ we have

$$\begin{aligned} Q_{\mu }^{ij,ij} + Q_{\mu + e_j}^{ij,ij} + Q_{\mu -e_j}^{ij,ij}=0. \end{aligned}$$

Let $k=l=i$; then for $n=0,\ldots , d-2$, $\epsilon \in \{-1+1\}$ and $\mu \in {\mathcal {I}}_n(i,j)$ we have

$$\begin{aligned} Q_{\mu + \epsilon e_i +e_j}^{ij,ii} + Q_{\mu + \epsilon e_i -e_j}^{ij,ii}=0, \end{aligned}$$

while $Q_\lambda ^{ij,ii}=0$ is $\lambda$ is not equal to $\mu +\epsilon e_i +e_j$ or $\mu + \epsilon e_i-e_j$ where $\mu \in {\mathcal {I}}_n(i,j)$ for some n. A similar result holds exchanging i and j for $k=l=j$.

Let $k=l$ with $k\not \in \{i,j\}$ and $l\not \in \{i,j\}$; then for $n=0,\ldots , d-2$, $\epsilon \in \{-1,+1\}$ and $\mu \in {\mathcal {I}}_n(i,j)$ we have

$$\begin{aligned} Q_{\mu + \epsilon e_i +e_j}^{ij,kk} + Q_{\mu + \epsilon e_i -e_j}^{ij,kk}=0, \end{aligned}$$

while $Q_\lambda ^{ij,kk}=0$ if $\lambda$ is not equal to $\mu +\epsilon e_i +e_j$ or $\mu + \epsilon e_i-e_j$ with $\mu \in {\mathcal {I}}_n(i,j)$ for some n.

Let $l=i$ and $k\not \in \{i,j\}$; then for $n=0,\ldots , d-2$, $\epsilon \in \{-1+1\}$ and $\mu \in {\mathcal {I}}_n(i,k)$ we have

$$\begin{aligned} Q_{\mu + \epsilon e_i +e_k}^{ij,ki} + Q_{\mu + \epsilon e_i -e_k}^{ij,ki}=0, \end{aligned}$$

while $Q_\lambda ^{ij,ki}=0$ is $\lambda$ is not equal to $\mu +\epsilon e_i +e_k$ or $\mu + \epsilon e_i-e_k$ where $\mu \in {\mathcal {I}}_n(i,k)$ for some n. A similar result holds exchanging i and j for $k=l=j$.

Finally, let $k\ne l$ with $k\not \in \{i,j\}$ and $l\not \in \{i,j\}$; then for $n=0,\ldots , d-4$, $\epsilon _i, \epsilon _j, \epsilon _k \in \{-1+1\}$ and $\mu \in {\mathcal {I}}_n(i,j,k,l)$ we have

$$\begin{aligned} Q_{\mu + \epsilon _i e_i +\epsilon _j e_j + \epsilon _k e_k +e_l}^{ij,kl} + Q_{\mu + \epsilon _i e_i +\epsilon _j e_j + \epsilon _k e_k -e_l}^{ij,kl}=0, \end{aligned}$$

while $Q_\lambda ^{ij,kl}=0$ is $\lambda$ is not equal to $\mu + \epsilon _i e_i +\epsilon _j e_j + \epsilon _k e_k +e_l$ or $\mu + \epsilon _i e_i +\epsilon _j e_j + \epsilon _k e_k -e_l$ where $\mu \in {\mathcal {I}}_n(i,j,k,l)$ for some n. These computations complete the proof of the first identity in (2.23). Recall that for $i,k\in \{1,\ldots , d\}$ and $\lambda \in \Gamma$ we let

$$\begin{aligned} {\tilde{Q}}_\lambda ^{i,k}:=\int _{{\mathbb {R}}^d} z_k D_i\psi _\lambda (z) \psi (z) dz. \end{aligned}$$

Let $k=i$; for $n=0,\ldots , d-1$ and $\mu \in {\mathcal {I}}_n(i)$ we have

$$\begin{aligned} {\tilde{Q}}^{i,i}_\mu + {\tilde{Q}}^{i,i}_{\mu + e_i} + {\tilde{Q}}^{i,i}_{\mu -e_i}=0. \end{aligned}$$

Let $k\ne i$; for $n=0,\ldots , d-2$, $\epsilon \in \{-1,0, +1\}$ and $\mu \in {\mathcal {I}}_n(i,k)$ we have

$$\begin{aligned} {\tilde{Q}}_{\mu + \epsilon e_i + e_k}^{i,k} + {\tilde{Q}}_{\mu + \epsilon e_i - e_k}^{i,k}=0 \end{aligned}$$

while ${\tilde{Q}}_\lambda ^{i,k}=0$ if $\lambda$ is not equal to $\mu +\epsilon e_i +e_k$ or $\mu + \epsilon e_i-e_k$ where $\mu \in {\mathcal {I}}_n(i,k)$ for some n. This completes the proof of the second identity in (2.23); therefore Assumption 2.5 is satisfied for these finite elements. This completes the verification of the validity of Assumptions 2.4–2.5 for the function $\psi$ defined by (6.2).