# Convergence of the tamed-Euler–Maruyama method for SDEs with discontinuous
and polynomially growing drift

Kathrin Spendier  Michaela Szölgyenyi K. Spendier and M. Szölgyenyi are
supported by the Austrian Science Fund (FWF): DOC 78.

(Preprint, )

###### Abstract

Numerical methods for SDEs with irregular coefficients are intensively studied
in the literature, with different types of irregularities usually being
attacked separately. In this paper we combine two different types of
irregularities: polynomially growing drift coefficients and discontinuous
drift coefficients. For SDEs that suffer from both irregularities we prove
strong convergence of order $1/2$ of the tamed-Euler–Maruyama scheme from
[21].

Keywords: stochastic differential equations, discontinuous and polynomially
growing drift, tamed-Euler–Maruyama scheme, strong convergence rate
Mathematics Subject Classification (2020): 60H10, 65C30

##  1 Introduction

We consider time-homogeneous stochastic differential equations (SDEs),

$\displaystyle dX_{t}$ $\displaystyle=\mu(X_{t})dt+\sigma(X_{t})dW_{t},\quad
t\in[0,T],\quad X_{0}=\xi,$ (1)

where $\xi\in{\mathbb{R}}$, $\mu,\sigma\colon{\mathbb{R}}\to{\mathbb{R}}$ are
measurable functions, $T\in(0,\infty)$, $W=(W_{t})_{t\in[0,T]}$ is a standard
Brownian motion on the filtered probability space
$(\Omega,\mathcal{F},{\mathbb{P}},(\mathbb{F}_{t})_{t\in[0,T]})$, where the
filtration satisfies the usual conditions.

The novelty in this paper is that we allow $\mu$ to be piecewise Lipschitz and
polynomially growing at the same time. Note that the explicit Euler–Maruyama
approximation fails to satisfy moment bounds for SDEs with superlinearly
growing coefficients. Therefore, the following numerical method from [21] is
used for approximating the solution of SDE (1). Let $N\in{\mathbb{N}}$ and
define the equidistant time grid $0=t_{0}<t_{1}<\dots<t_{N}=T$ with
$t_{k+1}-t_{k}=\delta$ for all $k\in\\{0,\dots,N-1\\}$. Denote for all
$t\in[0,T]$, ${\underline{t}}:=\max\\{t_{k}\colon t\geq t_{k}\\}$. The time-
continuous tamed-Euler–Maruyama (EM) scheme is given by $X^{(\delta)}_{0}=\xi$
and

$\displaystyle
X^{(\delta)}_{t}=X^{(\delta)}_{\underline{t}}+\frac{\mu(X^{(\delta)}_{\underline{t}})(t-{\underline{t}})}{1+\delta|\mu(X^{(\delta)}_{\underline{t}})|}+\sigma(X^{(\delta)}_{\underline{t}})(W_{t}-W_{{\underline{t}}}),\quad
t\in[0,T].$ (2)

In this method the drift part $\mu(X^{(\delta)}_{\underline{t}})\delta$ is
“tamed” by the factor $1/(1+\delta|\mu(X^{(\delta)}_{\underline{t}})|)$, which
makes it bounded by $1$, preventing the drift from becoming extraordinarily
large.

In case the coefficients $\mu$ and $\sigma$ are Lipschitz, it is well known
that SDE (1) admits a unique strong solution which can be approximated with
the EM scheme at strong convergence order $1/2$. In this paper we consider two
different types of irregularities of the drift coefficient that have been
studied separately in the literature.

For existence and uniqueness results of solutions to SDEs with discontinuous
drift coefficient, see [61, 55, 56, 32, 27, 51, 28, 31, 47, 48]. Approximation
results for SDEs with discontinuous drift are studied, e.g., in [13, 14, 15,
23, 52, 46, 42, 27, 43, 28, 29, 12, 4, 30, 31, 40, 44, 5, 34, 7, 39, 60, 47,
53, 36, 45]. For lower error bounds for approximation schemes based on
finitely many evaluations of the driving Brownian motion, see [16, 35]. In
this paper we focus on SDEs that locally around the points of discontinuity
satisfy a piecewise Lipschitz condition. Convergence rates for SDEs with
piecewise Lipschitz drift have first been studied in [27, 28, 30]. There, a
transformation method and a proof technique is introduced that works by
removing the discontinuities from the drift. This idea has been taken up and
advanced in [34], who prove order $1/2$ convergence of the crude EM scheme in
the same setting. Further contributions in this spirit are [29, 30, 31, 40,
60, 47, 53, 36, 45, 37].

For SDEs with superlinearly growing coefficients, the explicit EM scheme does
not converge in the strong mean square sense, see [20]. Under local
boundedness, one-sided Lipschitz and polynomially growing drift, and non-
degeneracy of the diffusion plus technical conditions, pathwise convergence is
shown in [13]. The implicit EM scheme converges in the same setting as in
[21], that is under a continuously differentiable one-sided Lipschitz drift
with polynomially growing derivative and a globally Lipschitz continuous
diffusion. In [21] the explicit tamed-EM scheme is introduced and proven to
converge at strong order $1/2$. A similar taming technique is introduced in
[49]. For further results, see [54, 57, 25, 2, 18, 1, 50, 33, 6, 59, 3, 41, 9,
26, 22, 58, 10, 19, 11, 8, 24].

In the current paper we prove strong convergence of order $1/2$ of the tamed-
EM scheme from [21] for scalar SDEs with a drift coefficient that has finitely
many discontinuities while growing superlinearly. We reach our goal by
combining ideas from [21, 27, 30, 34].

At this point, note that parallel and independent work has been done in [38],
where existence, uniqueness, and $L_{p}$-approximation results for a similar
tamed-EM scheme is presented for SDEs for locally piecewise Lipschitz and
polynomially growing drift. The assumptions on the coefficients in [38] are
more general since not only the drift, but also the diffusion may grow
polynomially. Note that our article is cited in [38] as well. [38] was on
arXiv earlier than our paper. Both, our result and the one from [38] have been
presented at conferences already in 2021. Also in parallel and independent,
[17] has proven existence, uniqueness, and $L_{p}$-convergence of order $1/2$
of the tamed-EM scheme from [21].

##  2 Convergence result

Assumption 1
We assume the following on the coefficients of SDE (1).

  * (i-1)

The drift coefficient $\mu\colon{\mathbb{R}}\to{\mathbb{R}}$ is piecewise
Lipschitz on $[\zeta_{1},\zeta_{m}]$ with $m\in{\mathbb{N}}$ discontinuities
in the points $\zeta_{1}<\ldots<\zeta_{m}\in{\mathbb{R}}$.

  * (i-2)

The drift coefficient $\mu\colon{\mathbb{R}}\to{\mathbb{R}}$ is globally one-
sided Lipschitz on $(-\infty,\zeta_{1})$ and $(\zeta_{m},\infty)$, i.e there
exists a constant $L_{\mu}\in(0,\infty)$ such that for all
$x,y\in(-\infty,\zeta_{1})$ or $(\zeta_{m},\infty)$, $(x-y)(\mu(x)-\mu(y))\leq
L_{\mu}|x-y|^{2}$, and $\mu\in C^{1}$ on each of the intervals
$(-\infty,\zeta_{1}),(\zeta_{1},\zeta_{2}),...,(\zeta_{m},\infty)$ with,
wherever it exists, at most polynomially growing derivative.

  * (ii)

The diffusion coefficient $\sigma\colon{\mathbb{R}}\to{\mathbb{R}}$ is
Lipschitz continuous, lies in $\mathcal{C}^{1}$, $\sigma^{\prime}$ is
Lipschitz, and for all $k\in\\{1,\dots,m\\}$, $\sigma(\zeta_{k})\neq 0$.

In the following, for any one-sided or globally Lipschitz function $f$ we
denote its one-sided or global Lipschitz constant by $L_{f}$. Note that by
[47, Lemma 2.2], piecewise Lipschitz functions grow at most linearly. We
denote for any function that grows at most linearly by $c_{f}$ the smallest
constant such that $|f(x)|\leq c_{f}(1+|x|)$.

######  Theorem 2.1.

Let Assumption 1 hold. Then there exists $C^{(\text{tEM})}\in(0,\infty)$ such
that for all $\delta\in(0,1)$ sufficiently small,

$\bigg{(}{\mathbb{E}}\bigg{[}\sup_{t\in[0,T]}|X_{t}-X^{(\delta)}_{t}|^{2}\bigg{]}\bigg{)}^{\\!1/2}\leq
C^{(\text{tEM})}\delta^{1/2}.$

###  2.1 Structure of the proof

Our convergence proof is based on a transformation trick from [27, 30] in
combination with ideas from [34] and [21]. Let $Z=G(X)$, where $G$ is a
specific transformation, which we are going to introduce in Section 2.2. Due
to the fact that $G$ will be invertible and bi-Lipschitz, we have
$X=G^{-1}(Z)$ and get that

$\displaystyle\bigg{(}{\mathbb{E}}\bigg{[}\sup_{t\in[0,T]}|X_{t}-X^{(\delta)}_{t}|^{2}\bigg{]}\bigg{)}^{\\!1/2}\leq
L_{G^{-1}}\bigg{(}{\mathbb{E}}\bigg{[}\sup_{t\in[0,T]}|Z_{t}-G(X^{(\delta)}_{t})|^{2}\bigg{]}\bigg{)}^{\\!1/2}.$
(3)

By the triangle inequality,

$\displaystyle\bigg{(}{\mathbb{E}}\bigg{[}\sup_{t\in[0,T]}|Z_{t}-G(X^{(\delta)}_{t})|^{2}\bigg{]}\bigg{)}^{\\!1/2}$
$\displaystyle\leq\bigg{(}{\mathbb{E}}\bigg{[}\sup_{t\in[0,T]}|Z_{t}-Z^{(\delta)}_{t}|^{2}\bigg{]}\bigg{)}^{\\!1/2}$
(4)
$\displaystyle\quad+\bigg{(}{\mathbb{E}}\bigg{[}\sup_{t\in[0,T]}|Z^{(\delta)}_{t}-G(X^{(\delta)}_{t})|^{2}\bigg{]}\bigg{)}^{\\!1/2},$

where $Z^{(\delta)}$ is defined as in (2) with $\mu,\sigma$ replaced by
$\tilde{\mu},\tilde{\sigma}$. For estimating the first error term in (4), we
basically apply the result from [21]. We need to show that the tamed-EM
approximation $Z^{(\delta)}$ converges to $Z$ with strong rate $1/2$. For
estimating the second error term in (4), we need to extend ideas from [34] for
the estimation of discontinuity crossing probabilities of the tamed-EM scheme
and do a case by case analysis.

###  2.2 Properties of the transformed SDE

All the material we use in this subsection is presented in [31, 36]. We repeat
it for the convenience of the reader. We will apply a transform
$G\colon{\mathbb{R}}\to{\mathbb{R}}$ from [36] that has the property that the
process $Z\colon\Omega\times[0,T]\rightarrow\mathbb{R}$ formally defined by
$Z_{t}=G(X_{t}),\leavevmode\nobreak\ t\in[0,T]$, satisfies the SDE

$\displaystyle
dZ_{t}=\tilde{\mu}(Z_{t})\,dt+\tilde{\sigma}(Z_{t})\,dW_{t},\qquad
Z_{0}=G(\xi),$

with $\tilde{\mu},\tilde{\sigma}$ given by

$\widetilde{\mu}=\left(G^{\prime}\cdot\mu+\tfrac{1}{2}G^{\prime\prime}\cdot\sigma^{2}\right)\circ
G^{-1}\,\text{ and }\,\widetilde{\sigma}=(G^{\prime}\cdot\sigma)\circ G^{-1}.$
(5)

The transform $G$ is chosen in a way so that $\tilde{\mu},\tilde{\sigma}$ are
sufficiently well-behaved and so that it impacts the coefficients of the SDE
(1) only locally around the points of discontinuity of the drift. The specific
form of $G$ is a non-unique choice that has the following properties.

######  Lemma 2.2.

[36, Lemma 1] Let Assumption 1 hold. Let $G$ be as in [36]. Then

  * (i)

$G$ is differentiable on ${\mathbb{R}}$ with Lipschitz continuous and bounded
derivative $G^{\prime}$ that satisfies $G^{\prime}(\zeta_{k})=1$ for all
$k\in\\{1,...,m\\}$. Furthermore, there exists $\varepsilon\in(0,\infty)$ such
that for every $x\in\mathbb{R}$ with $|x|>\varepsilon,\leavevmode\nobreak\
G^{\prime}(x)=1$.

  * (ii)

$G$ has an inverse $G^{-1}\colon\mathbb{R}\rightarrow\mathbb{R}$ that is
Lipschitz continuous.

  * (iii)

The function $G^{\prime}$ is twice differentiable on each of the intervals
$(-\infty,\zeta_{1}),(\zeta_{1},\zeta_{2}),\\\ \leavevmode\nobreak\
\leavevmode\nobreak\ \leavevmode\nobreak\ ...,(\zeta_{m},\infty)$ with
Lipschitz continuous and bounded derivatives $G^{\prime\prime}$ and
$G^{\prime\prime\prime}$.

######  Lemma 2.3.

Let Assumption 1 hold. Then the functions $\tilde{\mu}$ and $\tilde{\sigma}$
from (5) satisfy the following:

  * (i)

The drift coefficient $\tilde{\mu}\colon{\mathbb{R}}\to{\mathbb{R}}$ is
Lipschitz continuous on $[\zeta_{1},\zeta_{m}]$.

  * (ii)

The drift coefficient $\tilde{\mu}\colon{\mathbb{R}}\to{\mathbb{R}}$ is
globally one-sided Lipschitz, in $C^{1}$ on each of the intervals
$(-\infty,\zeta_{1}),(\zeta_{1},\zeta_{2}),...,(\zeta_{m},\infty)$, and has,
wherever it exists, an at most polynomially growing derivative.

  * (iii)

The diffusion coefficient $\tilde{\sigma}\colon{\mathbb{R}}\to{\mathbb{R}}$ is
Lipschitz continuous on ${\mathbb{R}}$, and for all $k\in\\{1,\dots,m\\}$,
$\tilde{\sigma}(\zeta_{k})\neq 0$.

###### Proof.

  * (i)

is proven in [31, Lemma 2.4.].

  * (ii)

The proof that $\tilde{\mu}\in C^{1}$ on
$\bigcup_{k=1}^{m-1}(\zeta_{k},\zeta_{k+1})$ can be found in [36, Lemma 2].
Due to (i) it holds that $\tilde{\mu}$ is Lipschitz continuous on
$[\zeta_{1},\zeta_{m}]$ and therefore linearly growing. It holds by Assumption
1 that there exists $\epsilon\in(0,\infty)$ such that on
$(\zeta_{1}-\epsilon,\zeta_{m}+\epsilon)^{c}$, $\tilde{\mu}=\mu$, hence
$\tilde{\mu}$ has a polynomially growing derivative on
$(\zeta_{1}-\epsilon,\zeta_{m}+\epsilon)^{c}$. On
$(\zeta_{1}-\epsilon,\zeta_{1})$ and $(\zeta_{m},\zeta_{m}+\epsilon)$ it holds
that $\tilde{\mu}$ is locally Lipschitz on a bounded domain, therefore
Lipschitz, which implies also the polynomial growth condition. Finally, the
one-sided Lipschitz property of $\tilde{\mu}$ follows from the fact that
$\tilde{\mu}$ is Lipschitz on $(\zeta_{1}-\epsilon,\zeta_{m}+\epsilon)$ and
$\tilde{\mu}=\mu$ on $(\zeta_{1}-\epsilon,\zeta_{m}+\epsilon)^{c}$.

  * (iii)

For the proof of (iii), see [31, Lemma 2.5.].

∎

######  Lemma 2.4.

Let Assumption 1 hold. Then there exists a constant $\tilde{c}\in(0,\infty)$
such that for all $x,y\in{\mathbb{R}}$,

$\displaystyle|\tilde{\mu}(x)-\tilde{\mu}(y)|\leq\tilde{c}(1+|x|^{\tilde{c}}+|y|^{\tilde{c}})|x-y|.$
(6)

Furthermore, there exists $c\in(0,\infty)$ such that for all
$x\in{\mathbb{R}}$,

$\displaystyle|\mu(x)|\leq c(1+|x|^{\tilde{c}+1}).$

###### Proof.

There exists $\epsilon\in(0,\infty)$ such that the function $\tilde{\mu}$ lies
in $C^{1}$ on $(-\infty,\zeta_{1}-\epsilon)$ and $(\zeta_{m}+\epsilon,\infty)$
and is globally Lipschitz on $[\zeta_{1}-\epsilon,\zeta_{m}+\epsilon]$. Hence,
the fundamental theorem of calculus holds for $\tilde{\mu}$ on the whole of
${\mathbb{R}}$. By the polynomial growth of $\tilde{\mu}$ there exists
$\tilde{c}\in(0,\infty)$ such that for all $x,y\in{\mathbb{R}}$, $x\geq y$
(w.l.o.g.),

$\displaystyle|\tilde{\mu}(x)-\tilde{\mu}(y)|=\left|\int_{y}^{x}\tilde{\mu}^{\prime}(t)\,dt\right|\leq\int_{y}^{x}|\tilde{\mu}^{\prime}(t)|\,dt\leq\int_{y}^{x}\tilde{c}(1+|t|^{\tilde{c}})\,dt.$

Solving the integral yields (6). For the second claim of the lemma, note that

$\displaystyle|\mu(x)|\leq|\mu(x)|\mathbb{1}_{(\zeta_{1}-\epsilon,\zeta_{m}+\epsilon)}(x)+|\mu(x)|\mathbb{1}_{(-\infty,\zeta_{1}-\epsilon]}(x)+|\mu(x)|\mathbb{1}_{[\zeta_{m}+\epsilon,\infty)}(x).$

On $(\zeta_{1}-\epsilon,\zeta_{m}+\epsilon)$ we may apply the linear growth
condition on $\mu$ and for the other two terms, choose $\epsilon$ so that
$\mu=\tilde{\mu}$ and apply (6). This gives that there exists $c\in(0,\infty)$
such that

$\displaystyle|\mu(x)|\leq c(1+|x|^{\tilde{c}+1}).$

This closes the proof. ∎

######  Theorem 2.5.

Let Assumption 1 hold. Then there exists a constant
$c\in\left(0,\infty\right)$ such that

$\displaystyle{\mathbb{E}}\left[\sup_{t\in[0,T]}|Z_{t}-Z^{(\delta)}_{t}|^{2}\right]\leq
c\delta.$ (7)

This result follows in principle from [21, Theorem 1.1.]. The only assumption
that is not satisfied in our case is that $\tilde{\mu}^{\prime}$ only exists
on each of the intervals
$(-\infty,\zeta_{1}),(\zeta_{1},\zeta_{2}),...,(\zeta_{m},\infty)$ and not on
the whole of ${\mathbb{R}}$. In the proof of [21, Theorem 1.1.] this is only
used to prove Lemma 2.4, (6) above, which we also obtain in our setting.
Hence, (7) holds.

###  2.3 Preparatory lemmas

In this section we present several lemmas, which we need for the proof of our
main result. The following lemma originates from [38, Lemma 5] and is adapted
to our situation.

######  Lemma 2.6.

There exists a constant $c\in(0,\infty)$ such that for sufficiently small
$\delta\in(0,1)$ and for all $x\in{\mathbb{R}}$ it holds that

$\displaystyle\frac{|\mu(x)|}{1+\delta|\mu(x)|}\leq
c\cdot\min(\frac{1}{\sqrt{\delta}}(1+|x|),|\mu(x)|).$

######  Lemma 2.7.

Let Assumption 1 hold and let $p\in[2,\infty)$. Then there exists a constant
$C_{p}^{(\text{M})}\in(0,\infty)$ such that for all $\delta\in(0,1)$
sufficiently small,

${\mathbb{E}}\Big{[}\sup_{t\in[0,T]}|X^{(\delta)}_{t}|^{p}\Big{]}\leq
C_{p}^{(\text{M})}$

and such that for all $\delta\in(0,1)$ sufficiently small, $t\in[0,T]$,

$\displaystyle\left({\mathbb{E}}[\sup_{s\in[t,t+\delta]}|X^{(\delta)}_{s}-X^{(\delta)}_{t}|^{p}]\right)^{1/p}$
$\displaystyle\leq C_{p}^{(\text{M})}\cdot\delta^{1/2}.$

###### Proof.

For proving the first statement, we would like to apply [21, Lemma 3.9].
However, since the assumptions in [21] are different from ours in the sense
that they assume $\mu$ to be in $C^{1}$, we need to reprove [21, Lemma 3.1 –
Lemma 3.9]. The crucial estimates are (36), (39), and (40) from [21, Lemma
3.1]. They can all be proven in our setting by choosing $\epsilon$ so that on
$(\zeta_{1}-\epsilon,\zeta_{m}+\epsilon)^{c}$, $\mu=\tilde{\mu}$ and
estimating the respective terms on $(\zeta_{1}-\epsilon,\zeta_{m}+\epsilon)$,
$(\zeta_{1}-\epsilon,\zeta_{m}+\epsilon)^{c}$ separately. By having proven the
estimates (36), (39), and (40) from [21, Proof of Lemma 3.1] we may now apply
[21, Lemma 3.9] to our setting. By [21, Lemma 3.9] there exists
$c_{1_{p}}\in(0,\infty)$ such that for all $\delta\in(0,1)$ sufficiently
small,

$\sup_{\delta\in(0,1)}\sup_{t\in[0,T]}{\mathbb{E}}[|X^{(\delta)}_{\underline{t}}|^{p}]\leq
c_{1_{p}}.$ (8)

We will use this and follow the steps from [49, Proof of Lemma 3.1] in our
setup. By Lemma 2.6 and (8) we get that there exist
$c_{2},c_{3},c_{4}\in(0,\infty)$ such that

$\displaystyle\sup_{t\in[0,T]}{\mathbb{E}}\big{[}|X^{(\delta)}_{t}-X^{(\delta)}_{\underline{t}}|^{p}\big{]}$
(9)
$\displaystyle\leq\sup_{t\in[0,T]}2^{p-1}\Bigg{(}{\mathbb{E}}\Bigg{[}\Bigg{|}\frac{\mu(X^{(\delta)}_{\underline{t}})}{1+\delta|\mu(X^{(\delta)}_{\underline{t}})|}\Bigg{|}^{p}\Bigg{]}\cdot\delta^{p}+c_{2}\cdot{\mathbb{E}}[|\sigma(X^{(\delta)}_{\underline{t}})|^{p}]\cdot\delta^{p/2}\Bigg{)}$
$\displaystyle\leq 2^{p-1}(c_{3}^{p}+c_{\sigma}^{p}\cdot
c_{2})\cdot\delta^{p/2}\cdot\left(1+\sup_{\delta\in(0,1)}\sup_{t\in[0,T]}{\mathbb{E}}[|X^{(\delta)}_{\underline{t}}|^{p}]\right)\leq
c_{4}\delta^{p/2}.$

By this, (8), Hölder’s inequality, and Lemma 2.6, there exists
$c_{5}\in(0,\infty)$ such that

$\displaystyle{\mathbb{E}}\Bigg{[}|X^{(\delta)}_{t}-X^{(\delta)}_{\underline{t}}|^{p}\cdot\Bigg{|}\frac{\mu(X^{(\delta)}_{\underline{t}})}{1+\delta|\mu(X^{(\delta)}_{\underline{t}})|}\Bigg{|}^{p}\Bigg{]}$
(10) $\displaystyle\leq c_{4}\cdot\delta^{p/2}\cdot
c_{3}^{p}\cdot\delta^{-p/2}\left(1+\sup_{\delta\in(0,1)}\sup_{t\in[0,T]}{\mathbb{E}}[|X^{(\delta)}_{\underline{t}}|^{2p}]\right)^{1/2}$
$\displaystyle\leq c_{4}\cdot\delta^{p/2}\cdot
c_{3}^{p}\cdot\delta^{-p/2}\cdot(1+c_{1_{2p}})^{1/2}\leq c_{5}.$

Now we proceed as in [49, Lemma 3.3], by carrying out an inductive argument.
For $p=2$ we have that (8), (9), and (10) are fullfilled. Itô’s formula gives

$\displaystyle|X_{t}^{(\delta)}|^{2}=|\xi|^{2}+2\int_{0}^{t}X_{\underline{s}}^{(\delta)}\cdot\left(\frac{\mu(X_{\underline{s}}^{(\delta)})}{1+\delta|\mu(X_{\underline{s}}^{(\delta)})|}\right)\,ds+\int_{0}^{t}|\sigma(X_{\underline{s}}^{(\delta)})|^{2}\,ds$
(11)
$\displaystyle+2\int_{0}^{t}\left(X_{s}^{(\delta)}-X_{\underline{s}}^{(\delta)}\right)\cdot\left(\frac{\mu(X_{\underline{s}}^{(\delta)})}{1+\delta|\mu(X_{\underline{s}}^{(\delta)})|}\right)\,ds+2\int_{0}^{t}X_{s}^{(\delta)}\cdot\sigma(X_{\underline{s}}^{(\delta)})\,dW_{s}.$

By (11), the one-sided Lipschitz condition on $\mu$, the Burkholder-Davis-
Gundy inequality, Young’s inequality, and Hölder’s inequality, for $q=2p$,
i.e. $q=4$, there exists $c_{6}\in(0,\infty)$ such that

$\displaystyle{\mathbb{E}}[\sup_{s\in[0,t]}|X_{s}^{\delta}|^{q}\leq
5^{q-1}\Bigg{(}{\mathbb{E}}[|\xi|^{q}]+t^{q-1}\cdot{\mathbb{E}}\left[\int_{0}^{t}|\sigma(X_{\underline{s}}^{(\delta)})|^{q}\,ds\right]$
$\displaystyle\quad+t^{\frac{q}{2}-1}\cdot{\mathbb{E}}\left[\int_{0}^{t}\left|X_{\underline{s}}^{(\delta)}\cdot\left(\frac{\mu(X_{\underline{s}}^{(\delta)})}{1+\delta|\mu(X_{\underline{s}}^{(\delta)})|}\right)\right|^{q/2}\,ds\right]$
$\displaystyle\quad+t^{\frac{q}{2}-1}\cdot{\mathbb{E}}\left[\int_{0}^{t}|X_{s}^{(\delta)}-X_{\underline{s}}^{(\delta)}|^{q/2}\cdot\left|\frac{\mu(X_{\underline{s}}^{(\delta)})}{1+\delta|\mu(X_{\underline{s}}^{(\delta)})|}\right|^{q/2}\,ds\right]$
$\displaystyle\quad+{\mathbb{E}}\left[\sup_{s\in[0,t]}\left|\int_{0}^{s}X_{s}^{(\delta)}\cdot\sigma(X_{\underline{s}}^{(\delta)})\,dW_{s}\right|^{q/2}\right]\Bigg{)}$
$\displaystyle\leq
c_{6}\Bigg{(}1+{\mathbb{E}}[|\xi|^{q}]+\int_{0}^{t}{\mathbb{E}}[|X_{\underline{s}}^{(\delta)}|^{q}]\,ds$
(12)
$\displaystyle\quad+{\mathbb{E}}\left[\left(\int_{0}^{t}|X_{s}^{(\delta)}|^{2}\cdot|\sigma(X_{\underline{s}}^{(\delta)})|^{2}\,ds\right)^{q/4}\right]\Bigg{)}$
$\displaystyle\leq
c_{6}\Bigg{(}1+{\mathbb{E}}[|\xi|^{q}]+\int_{0}^{t}\sup_{u\in[0,s]}{\mathbb{E}}[|X_{\underline{u}}^{(\delta)}|^{q}]\,ds$
$\displaystyle\quad\color[rgb]{0,0,0}+\frac{1}{2c_{6}}\sup_{s\in[0,t]}{\mathbb{E}}[|X_{s}^{(\delta)}|^{q}]+\frac{c_{6}}{2}{\mathbb{E}}\left[\left(\int_{0}^{t}|\sigma(X_{\underline{s}}^{(\delta)})|^{2}\,ds\right)^{q/2}\right]\Bigg{)}.$

By the linear growth condition on $\sigma$, Hölder’s inequality, and (8),
there exists $c_{7}\in(0,\infty)$ such that

${\mathbb{E}}[\sup_{s\in[0,t]}|X_{s}^{\delta}|^{q}]\leq
c_{7}\Bigg{(}1+\int_{0}^{t}\sup_{u\in[0,s]}{\mathbb{E}}[|X_{\underline{u}}^{(\delta)}|^{q}]\,ds\Bigg{)}<\infty.$
(13)

This ensures

${\mathbb{E}}[\sup_{s\in[0,t]}|X_{s}^{\delta}|^{q}]\leq
c_{7}\Bigg{(}1+\int_{0}^{t}{\mathbb{E}}[\sup_{u\in[0,s]}|X_{u}^{(\delta)}|^{q}]\,ds\Bigg{)}.$
(14)

With this and (13), Gronwall’s lemma yields for all $\delta\in(0,1)$
sufficiently small,

${\mathbb{E}}[\sup_{t\in[0,T]}|X^{(\delta)}_{t}|^{p}]\leq C_{p}^{(\text{M})}.$
(15)

Therefore, (10) holds true for $p=4$ and one may use (2.3) to obtain (15) for
$q=2p$, i.e. $q=8$. Repeating the same procedure, by induction one obtains the
desired result. For the second statement by standard proof steps we obtain
that there exists $c_{8}\in(0,\infty)$ such that

$\displaystyle\left({\mathbb{E}}\Bigg{[}\sup_{s\in[t,t+\delta]}\Bigg{|}\int_{t}^{s}\frac{\mu(X^{(\delta)}_{\underline{u}})}{1+\delta|\mu(X^{(\delta)}_{\underline{u}})|}\,du\Bigg{|}^{p}\Bigg{]}\right)^{1/p}$
$\displaystyle\leq c_{8}\delta$ (16)

and hence by choosing $C_{p}^{(M)}$ appropriately,

$\displaystyle\left({\mathbb{E}}\big{[}\sup_{s\in[t,t+\delta]}|X^{(\delta)}_{s}-X^{(\delta)}_{t}|^{p}\big{]}\right)^{1/p}\leq
C_{p}^{(M)}\delta^{1/2}.\qed$ (17)

####  2.3.1 Estimation of the occupation time and the discontinuity crossing
probabilities of the tamed-Euler–Maruyama process

For all $x\in\mathbb{R}$ denote by $X^{(\delta),x}$ the solution of the time-
continuous tamed-Euler–Maruyama scheme (2) starting at $X^{(\delta),x}_{0}=x$.
In this subsection we follow the steps of [34] and adapt all the results to
our setting.

######  Lemma 2.8.

Let Assumption 1 hold. Then there exists $C^{(\text{I})}\in(0,\infty)$ such
that for all $p\in[2,\infty)$, $x\in\mathbb{R}$, $t\in[0,T]$, $\delta\in(0,1)$
sufficiently small,

$\Big{(}{\mathbb{E}}\Big{[}\sup\limits_{t\in[0,T]}|X^{(\delta),x}_{t}|^{p}\Big{]}\Big{)}^{\\!1/p}\leq
C^{(\text{I})}(1+|x|),$
$\big{(}{\mathbb{E}}\big{[}\sup_{s\in[t,t+\delta]}|X^{(\delta),x}_{s}-X^{(\delta),x}_{t}|^{p}\big{]}\big{)}^{\\!1/p}\leq
C^{(\text{I})}(1+|x|^{\tilde{c}+1})\delta^{1/2}.$

The proof is very similar to the proof of Lemma 2.7 and therefore omitted.

######  Lemma 2.9.

Let Assumption 1 hold and let $\tilde{c}\in(0,\infty)$. There exists
$C^{(O)}\in(0,\infty)$ such that for all $k\in\\{1,\dots,m\\}$,
$x\in\mathbb{R}$, $\delta\in(0,1)$, $\varepsilon\in(0,\infty)$ it holds

$\int\limits_{0}^{T}\mathbb{P}(|X^{(\delta),x}_{t}-\zeta_{k}|\leq\varepsilon)dt\leq
C^{(O)}(1+|x|^{2\tilde{c}+2})(\varepsilon+\delta^{1/2}).$

###### Proof.

The steps in this proof run along the same lines as in [34]. We obtain by
Lemma 2.4 and Lemma 2.8,

$\displaystyle{\mathbb{E}}\left[\int\limits_{0}^{t}\Bigg{|}\frac{\mu(X^{(\delta),x}_{\underline{s}})}{1+\delta|\mu(X^{(\delta),x}_{\underline{s}})|}\Bigg{|}\mathbb{1}_{[\zeta_{1},\zeta_{m}]^{c}}(X^{(\delta),x}_{\underline{s}})\,ds\right]$
(18)
$\displaystyle\leq\int\limits_{0}^{t}c\left(1+{\mathbb{E}}\left[\sup\limits_{t\in[0,T]}|X^{(\delta),x}_{t}|^{\tilde{c}+1}\right]\right)\,ds\leq
c_{1}(1+|x|^{\tilde{c}+1}).$

For the term on $[\zeta_{1},\zeta_{m}]$ our estimate works as in [34] and
gives

$\displaystyle{\mathbb{E}}\left[\int\limits_{0}^{t}\Bigg{|}\frac{\mu(X^{(\delta),x}_{\underline{s}})}{1+\delta|\mu(X^{(\delta),x}_{\underline{s}})|}\Bigg{|}\mathbb{1}_{[\zeta_{1},\zeta_{m}]}(X^{(\delta),x}_{\underline{s}})\,ds\right]\leq
c_{2}(1+|x|).$ (19)

From here on the proof runs again along the same lines as in [34], but with
$|x|^{\tilde{c}+1}$ instead of $|x|$. ∎

Now, let for all $k\in\\{1,\dots,m\\}$, $t\in[0,T]$,
$\mathcal{Z}_{k}^{t}=\\{\omega\in\Omega\colon(X^{(\delta)}_{\underline{t}}(\omega)-\zeta_{k})(X^{(\delta)}_{t}(\omega)-\zeta_{k})\leq
0\\}.$

######  Lemma 2.10.

Let Assumption 1 hold. Let $s,t\in[0,T]$ with ${\underline{t}}-s\geq\delta$.
There exists a constant $C_{1}\in(0,\infty)$ such that for all
$k\in\\{1,\dots,m\\}$, $\delta\in(0,1)$ sufficiently small,

$\displaystyle{\mathbb{P}}(\mathcal{Z}_{k}^{s}\cap\mathcal{Z}_{k}^{t})$
$\displaystyle\leq C_{1}{\mathbb{P}}(\mathcal{Z}_{k}^{s})\delta$
$\displaystyle\quad+C_{1}\cdot\int_{\mathbb{R}}{\mathbb{P}}\\!\left(\mathcal{Z}_{k}^{s}\cap\left\\{|X^{(\delta)}_{{\underline{t}}-(t-{\underline{t}})}-\zeta_{k}|\leq
C_{1}\delta^{1/2}(1+|z|)\right\\}\right)\cdot e^{-\frac{z^{2}}{2}}dz.$

###### Proof.

The proof works as in [34, Lemma 5]. The only difference lies in showing the
inclusion [34, (20)], which is a part of [34, Proof of Lemma 5], that is

$\displaystyle\mathcal{Z}_{k}^{t}\cap\bigg{\\{}\max_{i\in\\{1,2,3\\}}|\bar{W}_{i}|\leq\sqrt{2\log(T/\delta)}\bigg{\\}}$
(20)
$\displaystyle\quad\subseteq\bigg{\\{}|X^{(\delta)}_{{\underline{t}}-(t-{\underline{t}})}-\zeta_{k}|\leq\frac{12c_{2}(1+|\zeta_{k}|)\cdot(1+|\bar{W}_{1}|+|\bar{W}_{2}|)}{\sqrt{T/\delta}}\bigg{\\}},$

where

$\bar{W}_{1}=\frac{W_{t}-W_{\underline{t}}}{\sqrt{t-{\underline{t}}}},\qquad\bar{W}_{2}=\frac{W_{\underline{t}}-W_{{\underline{t}}-(t-{\underline{t}})}}{\sqrt{t-{\underline{t}}}},\qquad\bar{W}_{3}=\frac{W_{{\underline{t}}-(t-{\underline{t}})}-W_{{\underline{t}}-\delta}}{\sqrt{\delta-(t-{\underline{t}})}}.$

Choose $\omega\in\Omega$ such that

$(X^{(\delta)}_{\underline{t}}(\omega)-\zeta_{k})(X^{(\delta)}_{t}(\omega)-\zeta_{k})\leq
0\leavevmode\nobreak\ \text{\quad and
\quad}\max_{i\in\\{1,2,3\\}}|\bar{W}_{i}(\omega)|\leq\sqrt{2\log(T/\delta)}.$

Using the linear growth condition on $\sigma$, Lemma 2.6, and the fact that
for all $a,b\in{\mathbb{R}}$,

$\displaystyle 1+|a|$ $\displaystyle=1+|a-b+b|\leq 1+|a-b|+|b|\leq
1+|a-b|+|b|+|b|\cdot|a-b|$ (21) $\displaystyle=(1+|a-b|)\cdot(1+|b|),$

we obtain that there exist $c_{1},c_{2}\in(0,\infty)$ with

$\displaystyle|X^{(\delta)}_{\underline{t}}(\omega)-\zeta_{k}|\leq|(X^{(\delta)}_{\underline{t}}(\omega)-\zeta_{k})-(X^{(\delta)}_{t}(\omega)-\zeta_{k})|$
(22)
$\displaystyle\leq\frac{|\mu(X^{(\delta)}_{\underline{t}}(\omega))|\delta}{1+\delta|\mu(X^{(\delta)}_{\underline{t}}(\omega))|}+\big{|}\sigma(X^{(\delta)}_{\underline{t}}(\omega))\big{|}\cdot\big{|}W_{t}(\omega)-W_{{\underline{t}}}(\omega)\big{|}$
$\displaystyle\leq
c_{2}(1+|X^{(\delta)}_{\underline{t}}(\omega)-\zeta_{k}|)\cdot(1+|\zeta_{k}|)\cdot\sqrt{\delta}(1+|\bar{W}_{1}(\omega)|).$

This shows for $\delta\in(0,1)$ sufficiently small,

$\displaystyle|X^{(\delta)}_{\underline{t}}(\omega)-\zeta_{k}|\leq\frac{c_{2}\cdot(1+|\zeta_{k}|)\cdot\sqrt{\delta}(1+|\bar{W}_{1}(\omega)|)}{1-c_{2}\cdot(1+|\zeta_{k}|)\cdot\sqrt{\delta}(1+|\bar{W}_{1}(\omega)|)}.$
(23)

For $\delta\in(0,1)$ sufficiently small we have

$c_{2}\cdot(1+|\zeta_{k}|)\cdot\sqrt{\delta}(1+|\bar{W}_{1}(\omega)|)\leq
c_{2}\cdot(1+|\zeta_{k}|)\cdot\sqrt{\delta}\cdot\left(1+\sqrt{2\log(T/\delta)}\right)\leq\frac{1}{2}.$

This and (23) show

$\displaystyle|X^{(\delta)}_{\underline{t}}(\omega)-\zeta_{k}|$
$\displaystyle\leq\frac{c_{2}\cdot(1+|\zeta_{k}|)\cdot\sqrt{\delta}(1+|\bar{W}_{1}(\omega)|)}{1-\frac{1}{2}}$
(24)
$\displaystyle=2c_{2}\cdot(1+|\zeta_{k}|)\cdot\sqrt{\delta}(1+|\bar{W}_{1}(\omega)|).$

Similar to (22), again using Lemma 2.6, we obtain the estimates

$\displaystyle|X^{(\delta)}_{{\underline{t}}-(t-{\underline{t}})}(\omega)-\zeta_{k}|$
$\displaystyle\leq\frac{6c_{2}\cdot(1+|\zeta_{k}|)\cdot\sqrt{\delta}\cdot(1+|\bar{W}_{1}(\omega)|+|\bar{W}_{2}(\omega)|)}{1-6c_{2}\cdot(1+|\zeta_{k}|)\cdot\sqrt{\delta}\cdot(1+|\bar{W}_{1}(\omega)|+|\bar{W}_{2}(\omega)|)},$

and

$\displaystyle|X^{(\delta)}_{{\underline{t}}-(t-{\underline{t}})}(\omega)-\zeta_{k}|$
$\displaystyle\leq\frac{6c_{2}\cdot(1+|\zeta_{k}|)\cdot\sqrt{\delta}\cdot(1+|\bar{W}_{1}(\omega)|+|\bar{W}_{2}(\omega)|)}{1-\frac{1}{2}}$
$\displaystyle=12c_{2}\cdot(1+|\zeta_{k}|)\cdot\sqrt{\delta}\cdot(1+|\bar{W}_{1}(\omega)|+|\bar{W}_{2}(\omega)|).$

This finishes the proof of (20). ∎

######  Lemma 2.11.

Let Assumption 1 hold. Let $s\in[0,T)$, and let $\tilde{c}\in(0,\infty)$.
There exists a constant ${C_{2}}\in(0,\infty)$ such that for all
$k\in\\{1,\dots,m\\}$, $\delta\in(0,1)$ sufficiently small,

$\int_{s}^{T}{\mathbb{P}}(\mathcal{Z}_{k}^{s}\cap\mathcal{Z}_{k}^{t})dt\leq{C_{2}}\delta^{1/2}\cdot\left({\mathbb{P}}(\mathcal{Z}_{k}^{s})+{\mathbb{E}}\\!\left[\mathbb{1}_{\mathcal{Z}_{k}^{s}}\cdot(X^{(\delta)}_{{\underline{s}}+\delta}-\zeta_{k})^{2\tilde{c}+2}\right]\right).$

The proof runs along the same lines as [34, Proof of Lemma 6] or [47, Proof of
Lemma 4.4.]. The only difference is that we apply Lemma 2.9 instead of [34,
Lemma 4] or [47, Lemma 4.2.], which changes one term in the statement of the
Lemma to $(X^{(\delta)}_{{\underline{s}}+\delta}-\zeta_{k})^{2\tilde{c}+2}$.

######  Lemma 2.12.

Let Assumption 1 hold. There exists a constant ${C_{3}}\in(0,\infty)$ such
that for all $k\in\\{1,\dots,m\\}$, $\tilde{c}\in(0,\infty)$,

$\int_{0}^{T}{\mathbb{E}}\\!\left[\mathbb{1}_{\mathcal{Z}_{k}^{s}}\cdot(X^{(\delta)}_{{\underline{s}}+\delta}-\zeta_{k})^{2\tilde{c}+2}\right]\,ds\leq{C_{3}}\delta.$

###### Proof.

First, note that

$\displaystyle\mathbb{1}_{\mathcal{Z}_{k}^{s}}\cdot|X^{(\delta)}_{{\underline{s}}+\delta}-\zeta_{k}|$
$\displaystyle\leq\mathbb{1}_{\mathcal{Z}_{k}^{s}}\cdot(|X^{(\delta)}_{{\underline{s}}+\delta}-X^{(\delta)}_{s}|+|X^{(\delta)}_{s}-X^{(\delta)}_{\underline{s}}|).$

With this and by Lemma 2.7 we get that

$\displaystyle\int_{0}^{T}{\mathbb{E}}\\!\left[\mathbb{1}_{\mathcal{Z}_{k}^{s}}\cdot(X^{(\delta)}_{{\underline{s}}+\delta}-\zeta_{k})^{2\tilde{c}+2}\right]\,ds$
$\displaystyle\quad\leq
2^{2\tilde{c}+1}\cdot\sum_{k=0}^{N-1}\int_{t_{k}}^{t_{k+1}}\left({\mathbb{E}}\\!\left[|X^{(\delta)}_{{\underline{s}}+\delta}-X^{(\delta)}_{s}|^{2\tilde{c}+2}\right]+{\mathbb{E}}\\!\left[|X^{(\delta)}_{s}-X^{(\delta)}_{\underline{s}}|^{2\tilde{c}+2}\right]\right)\,ds$
$\displaystyle\quad\leq
2^{2\tilde{c}+1}\cdot(2C_{2\tilde{c}+2}^{(\text{M})}\cdot\delta)\cdot\int_{0}^{T}1\,ds=2^{2\tilde{c}+1}2TC_{2\tilde{c}+2}^{(M)}\cdot\delta.$

∎

######  Proposition 2.13.

Let Assumption 1 hold. There exists a constant
$C^{(\text{cross})}\in(0,\infty)$ such that for all $k\in\\{1,\dots,m\\}$,
$\tilde{c}\in(0,\infty)$, $\delta\in(0,1)$ sufficiently small,

$\left({\mathbb{E}}\\!\left[\left|\int_{0}^{T}\mathbb{1}_{\\{(x,y)\in{\mathbb{R}}^{2}\colon(x-\zeta_{k})(y-\zeta_{k})\leq
0\\}}(X^{(\delta)}_{\underline{s}},X^{(\delta)}_{s})ds\right|^{2}\right]\right)^{1/2}\leq
C^{(\text{cross})}\cdot\delta.$

The proof works exactly as the one of [34, Proposition 1] or [47, Proposition
4.6.].

###  2.4 Proof of Theorem 2.1

As already shown in Section 2.1, we have to estimate the two terms in (4). To
estimate the first part of (4), we may apply [21, Theorem 1.1.], see Section 1
herein, i.e. there exists a constant $c_{0}\in(0,\infty)$ such that

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,T]}|Z_{t}-Z^{(\delta)}_{t}|^{2}\bigg{]}^{1/2}\leq
c_{0}\delta^{1/2}.$ (25)

Hence our task is to estimate the second term in (4). For all $\tau\in[0,T]$,
let

$u(\tau):={\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|Z^{(\delta)}_{t}-G(X^{(\delta)}_{t})|^{2}\bigg{]}$
(26)

and let for all $x_{1},x_{2}\in{\mathbb{R}}$,

$\nu(x_{1},x_{2})=G^{\prime}(x_{1})\frac{\mu(x_{2})}{1+\delta|\mu(x_{2})|}+\frac{1}{2}G^{\prime\prime}(x_{1})\sigma(x_{2})^{2}.$

By Itô’s formula we obtain

$\displaystyle G(X^{(\delta)}_{t})$
$\displaystyle=G(X^{(\delta)}_{0})+\int_{0}^{t}\nu(X^{(\delta)}_{s},X^{(\delta)}_{\underline{s}})\,ds+\int_{0}^{t}G^{\prime}(X^{(\delta)}_{s})\sigma(X^{(\delta)}_{\underline{s}})\,dW_{s},$

and

$\displaystyle|Z^{(\delta)}_{\tau}-G(X^{(\delta)}_{\tau})|^{2}=\sum_{i=1}^{8}E_{i,\tau},$
(27)

where for all $s\in[0,T]$,

$\displaystyle
E_{1,s}=-2\int_{0}^{s}\Bigg{(}\frac{\delta\tilde{\mu}(Z^{(\delta)}_{\underline{t}})|\tilde{\mu}(Z^{(\delta)}_{\underline{t}})|}{1+\delta|\tilde{\mu}(Z^{(\delta)}_{\underline{t}})|}\Bigg{)}(Z^{(\delta)}_{t}-G(X^{(\delta)}_{t}))\,dt$
$\displaystyle
E_{2,s}=2\int_{0}^{s}\Bigg{(}(G^{\prime}(X^{(\delta)}_{\underline{t}})-G^{\prime}(X^{(\delta)}_{t}))\frac{\mu(X^{(\delta)}_{\underline{t}})}{1+\delta|\mu(X^{(\delta)}_{\underline{t}})|}\Bigg{)}(Z^{(\delta)}_{t}-G(X^{(\delta)}_{t}))\,dt$
$\displaystyle
E_{3,s}=2\int_{0}^{s}\Bigg{(}G^{\prime}(X^{(\delta)}_{\underline{t}})\left(\frac{\delta\mu(X^{(\delta)}_{\underline{t}})|\mu(X^{(\delta)}_{\underline{t}})|}{1+\delta|\mu(X^{(\delta)}_{\underline{t}})|}\right)\Bigg{)}(Z^{(\delta)}_{t}-G(X^{(\delta)}_{t}))\,dt$
$\displaystyle
E_{4,s}=2\int_{0}^{s}\Bigg{(}\tilde{\sigma}(Z^{(\delta)}_{\underline{t}})-G^{\prime}(X^{(\delta)}_{t})\sigma(X^{(\delta)}_{\underline{t}})\Bigg{)}(Z^{(\delta)}_{t}-G(X^{(\delta)}_{t}))\,dW_{t}$
$\displaystyle
E_{5,s}=\int_{0}^{s}\Big{(}\tilde{\sigma}(Z^{(\delta)}_{\underline{t}})-G^{\prime}(X^{(\delta)}_{t})\sigma(X^{(\delta)}_{\underline{t}})\Big{)}^{2}\,dt$
$\displaystyle
E_{6,s}=\int_{0}^{s}\left(G^{\prime\prime}(X^{(\delta)}_{\underline{t}})\sigma^{2}(X^{(\delta)}_{\underline{t}})-G^{\prime\prime}(X^{(\delta)}_{t})\sigma^{2}(X^{(\delta)}_{\underline{t}})\right)(Z^{(\delta)}_{t}-G(X^{(\delta)}_{t}))\,dt$
$\displaystyle
E_{7,s}=2\int_{0}^{s}\Big{(}(\tilde{\mu}(Z^{(\delta)}_{\underline{t}})-\tilde{\mu}(G(X^{(\delta)}_{\underline{t}}))-(\tilde{\mu}(Z^{(\delta)}_{t})-\tilde{\mu}(G(X^{(\delta)}_{t}))\Big{)}$
$\displaystyle\qquad\cdot(Z^{(\delta)}_{t}-G(X^{(\delta)}_{t}))\,dt$
$\displaystyle
E_{8,s}=2\int_{0}^{s}(\tilde{\mu}(Z^{(\delta)}_{t})-\tilde{\mu}(G(X^{(\delta)}_{t}))(Z^{(\delta)}_{t}-G(X^{(\delta)}_{t}))\,dt.$

The fact that $ab\leq\frac{a^{2}}{2}+\frac{b^{2}}{2}$ for all
$a,b\in{\mathbb{R}}$ give

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{1,t}|\bigg{]}\leq\delta\int_{0}^{\tau}{\mathbb{E}}\left[|\tilde{\mu}(Z^{(\delta)}_{\underline{s}})|^{4}\right]\,ds+\int_{0}^{\tau}u(s)\,ds,$

respectively

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{3,t}|\bigg{]}\leq\|G^{\prime}\|_{\infty}^{2}\delta\int_{0}^{\tau}{\mathbb{E}}\left[|\tilde{\mu}(Z^{(\delta)}_{\underline{s}})|^{4}\right]\,ds+\|G^{\prime}\|_{\infty}^{2}\int_{0}^{\tau}u(s)\,ds.$

By [21, Lemma 3.10.] there exists $c_{1}\in(0,\infty)$ such that

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{1,t}|\bigg{]}+{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{3,t}|\bigg{]}\leq
c_{1}\left(\delta+\int_{0}^{\tau}u(t)\,dt\right).$ (28)

For $E_{5,\tau}$ it holds by the estimate $(a+b)^{2}\leq 2a^{2}+2b^{2}$ that

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{5,t}|\bigg{]}$
$\displaystyle\quad\leq
2\int_{0}^{\tau}\bigg{(}{\mathbb{E}}\left[|\tilde{\sigma}(Z^{(\delta)}_{\underline{s}})-G^{\prime}(X^{(\delta)}_{\underline{s}})\sigma(X^{(\delta)}_{\underline{s}})|^{2}\right]$
(29)
$\displaystyle\qquad+{\mathbb{E}}\left[|G^{\prime}(X^{(\delta)}_{\underline{s}})\sigma(X^{(\delta)}_{\underline{s}})-G^{\prime}(X^{(\delta)}_{s})\sigma(X^{(\delta)}_{\underline{s}})|^{2}\right]\bigg{)}\,ds.$

By using the Lipschitz continuity of $\tilde{\sigma}$ and the fact that
$\tilde{\sigma}(G(X^{(\delta)}_{\underline{s}}))\\\
=G^{\prime}(X^{(\delta)}_{\underline{s}})\sigma(X^{(\delta)}_{\underline{s}})$,
we get for the first term in (29) that there exists $c_{2}\in(0,\infty)$ such
that

$\displaystyle\int_{0}^{\tau}{\mathbb{E}}\left[|\tilde{\sigma}(Z^{(\delta)}_{\underline{s}})-G^{\prime}(X^{(\delta)}_{\underline{s}})\sigma(X^{(\delta)}_{\underline{s}})|^{2}\right]\,ds\leq
c_{2}\int_{0}^{\tau}u(s)\,ds.$ (30)

The linear growth of $\sigma$, the Lipschitz continuity of $G^{\prime}$, and
the fact that $X^{(\delta)}_{\underline{s}}$ is
$\mathbb{F}_{\underline{s}}$-measurable give for the second term in (29),

$\displaystyle\int_{0}^{\tau}{\mathbb{E}}\\!\left[|G^{\prime}(X^{(\delta)}_{\underline{s}})\sigma(X^{(\delta)}_{\underline{s}})-G^{\prime}(X^{(\delta)}_{s})\sigma(X^{(\delta)}_{\underline{s}})|^{2}\right]\,ds$
(31)
$\displaystyle\leq(c_{\sigma})^{2}(L_{G^{\prime}})^{2}\cdot\int_{0}^{\tau}{\mathbb{E}}\\!\left[\left(1+|X^{(\delta)}_{\underline{s}}|\right)^{2}{\mathbb{E}}\left[|X^{(\delta)}_{\underline{s}}-X^{(\delta)}_{s}|^{2}|\mathbb{F}_{\underline{s}}\right]\right]ds.$

Since $W_{s}-W_{\underline{s}}$ is independent of
$\mathbb{F}_{\underline{s}}$, $X^{(\delta)}_{\underline{s}}$ is
$\mathbb{F}_{\underline{s}}$-measurable and by the linear growth of $\sigma$,
we get for $s\in[0,T]$,

$\displaystyle{\mathbb{E}}\\!\left[|X^{(\delta)}_{\underline{s}}-X^{(\delta)}_{s}|^{2}\,\big{|}\,\mathbb{F}_{\underline{s}}\right]$
$\displaystyle=2\Big{|}\int_{{\underline{s}}}^{s}\frac{\mu(X^{(\delta)}_{\underline{s}})}{1+\delta|\mu(X^{(\delta)}_{\underline{s}})|}\,ds\Big{|}^{2}+2|\sigma(X^{(\delta)}_{\underline{s}})|^{2}{\mathbb{E}}\left[|W_{s}-W_{{\underline{s}}}|^{2}\right]$
$\displaystyle\leq
4|s-{\underline{s}}|\bigg{[}\int_{{\underline{s}}}^{s}|\mu(X^{(\delta)}_{\underline{s}})|^{2}\left(\mathbb{1}_{[\zeta_{1},\zeta_{m}]}(X^{(\delta)}_{\underline{s}})+\mathbb{1}_{[\zeta_{1},\zeta_{m}]^{c}}(X^{(\delta)}_{\underline{s}})\right)+c_{\sigma}^{2}(1+|X^{(\delta)}_{\underline{s}}|^{2})\bigg{]}.$

With the linear growth of $\mu\mathbb{1}_{[\zeta_{1},\zeta_{m}]}$ and the
estimate $|\mu(x)\mathbb{1}_{[\zeta_{1},\zeta_{m}]^{c}}(x)|^{2}\leq
2c^{2}\left(1+|x|^{2(\tilde{c}+1)}\right)$ we get

$\displaystyle{\mathbb{E}}\\!\left[|X^{(\delta)}_{\underline{s}}-X^{(\delta)}_{s}|^{2}\,\big{|}\,\mathbb{F}_{\underline{s}}\right]$
(32) $\displaystyle\leq
8c_{\mu}^{2}(1+|X^{(\delta)}_{\underline{s}}|^{2})\delta+8\delta
c^{2}\left(1+|X^{(\delta)}_{\underline{s}}|^{2(\tilde{c}+1)}\right)+4c_{\sigma}^{2}(1+|X^{(\delta)}_{\underline{s}}|^{2})\delta.$

Combining this with (31) yields for the second term in (29) that

$\displaystyle\int_{0}^{\tau}{\mathbb{E}}\\!\left[|G^{\prime}(X^{(\delta)}_{\underline{s}})\sigma(X^{(\delta)}_{\underline{s}})-G^{\prime}(X^{(\delta)}_{s})\sigma(X^{(\delta)}_{\underline{s}})|^{2}\right]\,ds$
(33)
$\displaystyle\quad\leq(c_{\sigma})^{2}(L_{G^{\prime}})^{2}\cdot\int_{0}^{\tau}{\mathbb{E}}\bigg{[}|\left(1+|X^{(\delta)}_{\underline{s}}|\right)^{2}\bigg{\\{}\left(1+|X^{(\delta)}_{\underline{s}}|^{2}\right)(8c_{\mu}^{2}+4c_{\sigma}^{2})\delta$
$\displaystyle\qquad+8\delta
c^{2}\left(1+|X^{(\delta)}_{\underline{s}}|^{2(\tilde{c}+1)}\right)\bigg{\\}}\bigg{]}ds.$

Hence, (30), (33), and Lemma 2.7 establish that there exist
$c_{3},c_{4}\in(0,\infty)$ such that

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{5,t}|\bigg{]}$
$\displaystyle\leq c_{3}\delta+c_{4}\int_{0}^{\tau}u(s)\,ds.$ (34)

For $E_{4,\tau}$ the Burkholder-Davis-Gundy inequality and the estimate
$a\cdot b\leq\frac{a^{2}}{2}+\frac{b^{2}}{2}$ give

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{4,t}|\bigg{]}$
$\displaystyle\leq\int_{0}^{\tau}u(t)\,dt$ (35)
$\displaystyle\quad+{\mathbb{E}}\left[\int_{0}^{\tau}|\tilde{\sigma}(Z^{(\delta)}_{\underline{s}})-G^{\prime}(X^{(\delta)}_{s})\sigma(X^{(\delta)}_{\underline{s}})|^{2}\,ds\right].$

By estimating the second term in (35) analog to (29), this yields that there
exists $c_{5}\in(0,\infty)$ such that

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{4,t}|\bigg{]}\leq\int_{0}^{\tau}u(t)\,dt+c_{5}\delta.$
(36)

For $E_{7,\tau}$ we have by Hölder’s inequality and Lemma 2.4 that

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{7,t}|\bigg{]}$
$\displaystyle\quad\leq
2\int_{0}^{\tau}\Bigg{(}\left({\mathbb{E}}\left[\tilde{c}^{2}(1+|Z^{(\delta)}_{\underline{s}}|^{\tilde{c}}+|Z^{(\delta)}_{s}|^{\tilde{c}})^{2}\cdot|Z^{(\delta)}_{\underline{s}}-Z^{(\delta)}_{s}|^{2}\right]\right)^{1/2}$
$\displaystyle\qquad\cdot\left({\mathbb{E}}\left[|Z^{(\delta)}_{s}-G(X^{(\delta)}_{s})|^{2}\right]\right)^{1/2}$
$\displaystyle\qquad+\left({\mathbb{E}}\left[\tilde{c}^{2}(1+|G(X^{(\delta)}_{\underline{s}})|^{\tilde{c}}+|G(X^{(\delta)}_{s})|^{\tilde{c}})^{2}\cdot|G(X^{(\delta)}_{\underline{s}})-G(X^{(\delta)}_{s})|^{2}\right]\right)^{1/2}$
$\displaystyle\qquad\cdot\left({\mathbb{E}}\left[|Z^{(\delta)}_{s}-G(X^{(\delta)}_{s})|^{2}\right]\right)^{1/2}\Bigg{)}\,ds.$

By Lipschitz continuity of $G$, Hölder’s inequality, the estimate $a\cdot
b\leq\frac{a^{2}}{2}+\frac{b^{2}}{2}$, and Lemma 2.7, we get that there exist
$c_{6},c_{7}\in(0,\infty)$ such that

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{7,t}|\bigg{]}\leq\int_{0}^{\tau}\left(2c_{6}\sqrt{\delta}\cdot\left({\mathbb{E}}\left[|Z^{(\delta)}_{s}-G(X^{(\delta)}_{s})|^{2}\right]\right)^{1/2}\right)\,ds$
(37)
$\displaystyle\quad\leq\int_{0}^{\tau}\left(c_{6}\delta+\frac{1}{2}{\mathbb{E}}\left[|Z^{(\delta)}_{s}-G(X^{(\delta)}_{s})|^{2}\right]\right)\,ds\leq
c_{7}\delta+\frac{1}{2}\int_{0}^{\tau}u(t)\,dt.$

By the one-sided Lipschitz continuity of $\tilde{\mu}$ we get that there
exists $L_{\tilde{\mu}}\in(0,\infty)$ such that for all $t\in[0,T]$,

$\displaystyle E_{8,\tau}\leq
2L_{\tilde{\mu}}\int_{0}^{\tau}|Z^{(\delta)}_{t}-G(X^{(\delta)}_{t})|^{2}\,dt.$

Hence, there exists $c_{8}\in(0,\infty)$ such that

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{8,t}|\bigg{]}\leq
c_{8}\int_{0}^{\tau}u(t)\,dt.$ (38)

For $E_{2,\tau}$ we have by the estimate $a\cdot
b\leq\frac{a^{2}}{2}+\frac{b^{2}}{2}$,

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{2,t}|\bigg{]}$
$\displaystyle\leq\int_{0}^{\tau}{\mathbb{E}}\left[|(G^{\prime}(X^{(\delta)}_{\underline{s}})-G^{\prime}(X^{(\delta)}_{s})|^{2}\cdot|\mu(X^{(\delta)}_{\underline{s}})|^{2}\right]\,ds$
(39)
$\displaystyle\quad+\int_{0}^{\tau}{\mathbb{E}}\left[|Z^{(\delta)}_{s}-G(X^{(\delta)}_{s})|^{2}\right]\,ds.$

By using the linear growth condition of $\mu$ on $[\zeta_{1},\zeta_{m}]$, the
polynomial growth condition of $\mu$ on $[\zeta_{1},\zeta_{m}]^{c}$, and the
fact that $X^{(\delta)}_{\underline{s}}$ is
$\mathbb{F}_{\underline{s}}$-measurable, we obtain for the first term in (39),

$\displaystyle\int_{0}^{T}{\mathbb{E}}\bigg{[}|\mu(X^{(\delta)}_{\underline{s}})|^{2}|G^{\prime}(X^{(\delta)}_{\underline{s}})-G^{\prime}(X^{(\delta)}_{s})|^{2}\,ds\bigg{]}$
$\displaystyle\leq(L_{G^{\prime}})^{2}\int_{0}^{T}{\mathbb{E}}\bigg{[}|\mu(X^{(\delta)}_{\underline{s}})|^{2}\left(\mathbb{1}_{[\zeta_{1},\zeta_{m}]}(X^{(\delta)}_{\underline{s}})+\mathbb{1}_{[\zeta_{1},\zeta_{m}]^{c}}(X^{(\delta)}_{\underline{s}})\right)|X^{(\delta)}_{\underline{s}}-X^{(\delta)}_{s}|^{2}\bigg{]}\,ds$
$\displaystyle=2(L_{G^{\prime}})^{2}\int_{0}^{T}{\mathbb{E}}\bigg{[}\left(c_{\mu}^{2}\left(1+|X^{(\delta)}_{\underline{s}}|^{2}\right)+c^{2}\left(1+|X^{(\delta)}_{\underline{s}}|^{2(\tilde{c}+1)}\right)\right)$
$\displaystyle\quad{\mathbb{E}}\left[|X^{(\delta)}_{\underline{s}}-X^{(\delta)}_{s}|^{2}\Big{|}\mathbb{F}_{\underline{s}}\right]\bigg{]}\,ds.$

Combining this with (LABEL:eq:cond-exp2) and Lemma 2.7 shows that there exists
$c_{9}\in(0,\infty)$ such that

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{2,t}|\bigg{]}\leq
c_{9}\delta+\int_{0}^{\tau}u(s)\,ds.$ (40)

For the term $E_{6,\tau}$ the inequality $a\cdot
b\leq\frac{a^{2}}{2}+\frac{b^{2}}{2}$ gives

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{6,t}|\bigg{]}\leq\frac{1}{2}{\mathbb{E}}\bigg{[}\sup_{r\in[0,\tau]}|Z^{(\delta)}_{r}-G(X^{(\delta)}_{r})|^{2}\bigg{]}$
(41)
$\displaystyle\quad+\frac{1}{2}{\mathbb{E}}\bigg{[}\Big{|}\int_{0}^{\tau}|G^{\prime\prime}(X^{(\delta)}_{\underline{s}})\sigma^{2}(X^{(\delta)}_{\underline{s}})-G^{\prime\prime}(X^{(\delta)}_{s})\sigma^{2}(X^{(\delta)}_{\underline{s}})|\,ds\Big{|}^{2}\bigg{]}.$

For recovering the optimal convergence rate $1/2$, it is essential that we do
not apply the Cauchy-Schwarz inequality to the second term in (41). Instead,
we adopt an idea from [34]. That is, we define for all $k\in\\{1,\dots,m\\}$
the sets

$\mathcal{Z}_{k}=\\{(x,y)\in{\mathbb{R}}^{2}\colon(x-\zeta_{k})(y-\zeta_{k})\leq
0\\},\qquad\mathcal{Z}=\bigcup_{k=1}^{m}\mathcal{Z}_{k}.$

$\mathcal{Z}$ is the set of all $x,y\in{\mathbb{R}}$ which lie on different
sides of a point of discontinuity of the drift. So for
$(X^{(\delta)}_{\underline{s}},X^{(\delta)}_{s})\in\mathcal{Z}^{c}$ we have
that a Lipschitz condition is satisfied by $G^{\prime\prime}$. With this,

$\displaystyle{\mathbb{E}}\bigg{[}\Big{|}\int_{0}^{\tau}|G^{\prime\prime}(X^{(\delta)}_{\underline{s}})\sigma^{2}(X^{(\delta)}_{\underline{s}})-G^{\prime\prime}(X^{(\delta)}_{s})\sigma^{2}(X^{(\delta)}_{\underline{s}})|ds\Big{|}^{2}\bigg{]}\leq
F_{1}+F_{2}$ (42)

with

$\displaystyle F_{1}$
$\displaystyle={\mathbb{E}}\bigg{[}\Big{|}\int_{0}^{\tau}|G^{\prime\prime}(X^{(\delta)}_{\underline{s}})\sigma^{2}(X^{(\delta)}_{\underline{s}})-G^{\prime\prime}(X^{(\delta)}_{s})\sigma^{2}(X^{(\delta)}_{\underline{s}})|$
(43)
$\displaystyle\quad\cdot\mathbb{1}_{\mathcal{Z}}(X^{(\delta)}_{\underline{s}},X^{(\delta)}_{s})ds\Big{|}^{2}\bigg{]},$
$\displaystyle F_{2}$
$\displaystyle=T\cdot{\mathbb{E}}\bigg{[}\int_{0}^{T}|G^{\prime\prime}(X^{(\delta)}_{\underline{s}})\sigma^{2}(X^{(\delta)}_{\underline{s}})-G^{\prime\prime}(X^{(\delta)}_{s})\sigma^{2}(X^{(\delta)}_{\underline{s}})|^{2}$
$\displaystyle\quad\cdot\mathbb{1}_{\mathcal{Z}^{c}}(X^{(\delta)}_{\underline{s}},X^{(\delta)}_{s})ds\bigg{]}.$

For $F_{2}$ note that by (LABEL:eq:cond-exp2), Lemma 2.7, and the Lipschitz
continuity of $G^{\prime\prime}$ on $\mathcal{Z}^{c}$, there exists
$c_{10}\in(0,\infty)$ such that

$\displaystyle F_{2}\leq
8T(c_{\sigma})^{4}(L_{G^{\prime\prime}})^{2}\int_{0}^{T}{\mathbb{E}}\Big{[}(1+|X^{(\delta)}_{\underline{s}}|^{4})\big{(}8c_{\mu}^{2}(1+|X^{(\delta)}_{\underline{s}}|^{2})\delta$
(44) $\displaystyle\qquad+8\delta
c^{2}(1+|X^{(\delta)}_{\underline{s}}|^{2(\tilde{c}+1)})+4c_{\sigma}^{2}(1+|X^{(\delta)}_{\underline{s}}|^{2})\delta\Big{)}\Big{]}ds\leq
c_{10}\delta.$

For the term $F_{1}$, observe, as in [47], that for all $k\in\\{1,\dots,m\\}$,
$x,y\in{\mathbb{R}}$,

$\displaystyle|x|\mathbb{1}_{\mathcal{Z}_{k}}(x,y)=(|\zeta_{k}+x-\zeta_{k}|)\mathbb{1}_{\mathcal{Z}_{k}}(x,y)\leq(|\zeta_{k}|+|x-\zeta_{k}|)\mathbb{1}_{\mathcal{Z}_{k}}(x,y)$
$\displaystyle\leq(|\zeta_{k}|+|x-y|)\mathbb{1}_{\mathcal{Z}_{k}}(x,y).$

As in [47, Proof of Theorem 4.7.] we have that

$\displaystyle(1+x^{2})\cdot\mathbb{1}_{\mathcal{Z}}(x,y)$
$\displaystyle=(1+x^{2})\cdot\mathbb{1}_{\cup_{k=1}^{m}\mathcal{Z}_{k}}(x,y)$
$\displaystyle\leq
2m|x-y|^{2}+(1+2\max\\{|\zeta_{1}|,|\zeta_{m}|\\}^{2})\cdot\sum_{k=1}^{m}\mathbb{1}_{\mathcal{Z}_{k}}(x,y).$

This yields

$\displaystyle F_{1}\leq
16(c_{\sigma})^{4}\|G^{\prime\prime}\|_{\infty}^{2}\cdot{\mathbb{E}}\bigg{[}\Big{|}\int_{0}^{\tau}(1+|X^{(\delta)}_{\underline{s}}|^{2})\cdot\mathbb{1}_{\mathcal{Z}}(X^{(\delta)}_{\underline{s}},X^{(\delta)}_{s})ds\Big{|}^{2}\bigg{]}$
$\displaystyle\leq
16(c_{\sigma})^{4}\|G^{\prime\prime}\|_{\infty}^{2}\cdot{\mathbb{E}}\bigg{[}\Big{|}\int_{0}^{T}\Big{(}2m|X^{(\delta)}_{\underline{s}}-X^{(\delta)}_{s}|^{2}$
$\displaystyle\quad+(1+2\max\\{|\zeta_{1}|,|\zeta_{m}|\\}^{2})\cdot\sum_{k=1}^{m}\mathbb{1}_{\mathcal{Z}_{k}}(X^{(\delta)}_{\underline{s}},X^{(\delta)}_{s})\Big{)}ds\Big{|}^{2}\bigg{]}$
$\displaystyle\leq
128mT(c_{\sigma})^{4}\|G^{\prime\prime}\|_{\infty}^{2}m^{2}\cdot{\mathbb{E}}\bigg{[}\int_{0}^{T}|X^{(\delta)}_{\underline{s}}-X^{(\delta)}_{s}|^{4}ds\bigg{]}$
$\displaystyle\quad+32(c_{\sigma})^{4}\|G^{\prime\prime}\|_{\infty}^{2}(1+2\max\\{|\zeta_{1}|,|\zeta_{m}|\\}^{2})^{2}\cdot{\mathbb{E}}\bigg{[}\Big{|}\int_{0}^{T}\sum_{k=1}^{m}\mathbb{1}_{\mathcal{Z}_{k}}(X^{(\delta)}_{\underline{s}},X^{(\delta)}_{s})\,ds\Big{|}^{2}\bigg{]}.$

Together with Lemma 2.7 we therefore have

$\displaystyle F_{1}\leq
128(c_{\sigma})^{4}\|G^{\prime\prime}\|_{\infty}^{2}m^{2}T^{2}C^{(M)}\delta$
(45)
$\displaystyle\qquad+32(c_{\sigma})^{4}\|G^{\prime\prime}\|_{\infty}^{2}m^{2}(1+2\max\\{|\zeta_{1}|,|\zeta_{m}|\\}^{2})^{2}$
$\displaystyle\qquad\cdot{\mathbb{E}}\bigg{[}\Big{|}\int_{0}^{T}\mathbb{1}_{\mathcal{Z}}(X^{(\delta)}_{\underline{s}},X^{(\delta)}_{s})\,ds\Big{|}^{2}\bigg{]}.$

Proposition 2.13 yields

$\displaystyle{\mathbb{E}}\bigg{[}\Big{|}\int_{0}^{T}\mathbb{1}_{\mathcal{Z}_{k}}(X^{(\delta)}_{\underline{s}},X^{(\delta)}_{s})\,ds\Big{|}^{2}\bigg{]}\leq
C^{(\text{cross})}\cdot\delta.$ (46)

Combining (41), (42), (43), (44), (45), and (46) we get that there exists
$c_{11}\in(0,\infty)$ such that

$\displaystyle{\mathbb{E}}\bigg{[}\sup_{t\in[0,\tau]}|E_{6,t}|\bigg{]}$
$\displaystyle\leq\frac{1}{2}\int_{0}^{\tau}u(t)\,dt+c_{11}\delta.$ (47)

Combining (26) with (27) and with the estimates (28), (34), (36), (37), and
(38), (40), and (47) shows that there exist constants
$c_{12},c_{13}\in\left(0,\infty\right)$ such that

$\displaystyle 0\leq u(\tau)$ $\displaystyle\leq
c_{12}\int_{0}^{\tau}u(s)\,ds+c_{13}\delta.$

By Lemma 2.7 and the fact that $[0,T]\ni
t\mapsto{\mathbb{E}}\left[\sup_{s\in[0,t]}|Z^{(\delta)}_{s}-G(X^{(\delta)}_{s})|^{2}\right]$
is Borel measurable, Gronwall’s inequality yields for all
$\tau\in\left[0,T\right]$,

$\displaystyle u(\tau)\leq c_{13}\exp(c_{12}T)\cdot\delta.$

Combining this with (3), (4), and (25) finally yields

$\displaystyle\bigg{(}{\mathbb{E}}\bigg{[}\sup_{t\in[0,T]}|X_{t}-X^{(\delta)}_{t}|^{2}\bigg{]}\bigg{)}^{\\!1/2}\leq
L_{G^{-1}}(c_{0}\delta)^{1/2}+L_{G^{-1}}(c_{13}\exp(c_{12}T)\cdot\delta)^{1/2}.$

## Acknowledgements

K. Spendier and M. Szölgyenyi are supported by the Austrian Science Fund
(FWF): DOC 78. The authors thank Verena Schwarz for useful discussions and
proof reading and an anonymous referee for utmost carefully reading our paper
and making very helpful comments.

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Michaela Szölgyenyi
Department of Statistics, University of Klagenfurt, Universitätsstraße 65-67,
9020 Klagenfurt, Austria
<EMAIL_ADDRESS>

Kathrin Spendier 🖂
Department of Statistics, University of Klagenfurt, Universitätsstraße 65-67,
9020 Klagenfurt, Austria
<EMAIL_ADDRESS>