$L^{\alpha -1}$ distance between two one-dimensional stochastic differential equations driven by a symmetric $\alpha$-stable process

Abstract

In this article, we consider a coefficient stability problem for one-dimensional stochastic differential equations driven by an $\alpha$-stable process with $\alpha \in (1,2)$. More precisely, we find an upper bound for the $L^{\alpha -1}(\varOmega ,{\mathbb {P}})$ distance between two solutions in terms of the $L^{\alpha }\left( {\mathbb {R}},\mu^{\alpha }_{x_0}\right)$ distance of the coefficients for an appropriate measure $\mu^{\alpha} _{x_0}$ which characterizes symmetric stable laws and depends on the initial value of the stochastic differential equation. We obtain this result using the method introduced by Komatsu (Proc Jpn Acad Ser A Math Sci 58(8):353–356, 1982) which is used in the proof of uniqueness of solutions together with an upper bound for the transition density function of the solution of the stochastic differential equation obtained by Kulik (The parametrix method and the weak solution to an SDE driven by an $\alpha$-stable noise. arXiv:1412.8732, 2014).

Appendix

In this section, we prove some lemmas used in the proof of Theorems 3.1, 4.1 and we introduce some results concomitant with these theorems.

1.1 Proof of the martingale property for $M^{\delta ,\varepsilon }$

In this section, we prove that $M^{\delta ,\varepsilon }$ is a martingale. For this, we need to show that the function $u_{\delta ,\varepsilon }$ introduced in Lemma 3.1 is Lipschitz continuous.

Lemma 5.1

The function $u_{\delta ,\varepsilon }$ is Lipschitz continuous.

Proof

From the mean value theorem, we know that a differentiable function is Lipschitz if and only if its derivative is bounded. We show that $u_{\delta ,\varepsilon }'$ is bounded. Since the support of $\psi _{\delta ,\varepsilon }$ is $[\varepsilon \delta^{-1},\varepsilon ]$ and using Jensen’s inequality, we have

$$\begin{aligned} \left| u_{\delta ,\varepsilon }'(x)\right|&=\left| (\alpha -1)\int _{\mathbb {R}}sgn (x-y)|x-y|^{\alpha -2}\psi _{\delta ,\varepsilon }(y)dy\right| \\&\le (\alpha -1)\int _{\varepsilon \delta^{-1}}^{\varepsilon }|x-y|^{\alpha -2}\psi _{\delta ,\varepsilon }(y)dy. \end{aligned}$$

Here, when $x>2\varepsilon$ or $0\ge x$, we observe that for any $y\in [\varepsilon \delta^{-1},\varepsilon ]$,

$$\begin{aligned} |x-y|\ge |y| \,\, \text {so that}\,\, |x-y|^{\alpha -2}\le |y|^{\alpha -2}. \end{aligned}$$

Hence we have that for any $x>2\varepsilon$ or $0>x$,

$$\begin{aligned} \left| u_{\delta ,\varepsilon }'(x)\right|&\le (\alpha -1)\int _{\varepsilon \delta^{-1}}^{\varepsilon }y^{\alpha -2}\psi _{\delta ,\varepsilon }(y)dy\le (\alpha -1)\left( \frac{\delta }{\varepsilon }\right)^{2-\alpha }\int _{\varepsilon \delta^{-1}}^{\varepsilon }\psi _{\delta ,\varepsilon }(y)dy\\&=(\alpha -1)\left( \frac{\delta }{\varepsilon }\right)^{2-\alpha }. \end{aligned}$$

The last equality follows from $\int _{\varepsilon \delta^{-1}}^\varepsilon \psi _{\delta ,\varepsilon }(y)dy=1$. When $x\in (0,2\varepsilon ]$, then we have from $\psi _{\delta ,\varepsilon }(x)\le 2/(x\log \delta )$,

$$\begin{aligned} \left| u_{\delta ,\varepsilon }'(x)\right|&\le (\alpha -1)\int _{\varepsilon \delta^{-1}}^{\varepsilon }|x-y|^{\alpha -2}\frac{2}{y \log \delta }dy\\&\le \frac{2(\alpha -1)\delta }{\varepsilon \log \delta }\int _{\varepsilon \delta^{-1}}^{\varepsilon }|x-y|^{\alpha -2}dy\\&=\frac{2\delta }{\varepsilon \log \delta }\left( |x-\varepsilon |^{\alpha -1}-\left| x-\varepsilon \delta^{-1}\right|^{\alpha -1}\right) \\&\le \frac{2\delta }{\varepsilon^{2-\alpha }\log \delta }\\&<\infty . \end{aligned}$$

Thus $u_{\delta ,\varepsilon }'$ is bounded for each $\varepsilon >0$ and $\delta >1$. This concludes the proof. $\square$

Now, we will prove that $M^{\delta ,\varepsilon }$ is a martingale. We set $M_t^{\delta ,\varepsilon ,1}$ and $M_t^{\delta ,\varepsilon ,2}$ for each $t\in [0,T]$ as

$$\begin{aligned} M_t^{\delta ,\varepsilon ,1}&=\int _{0}^{t}\int _{|z|\ge 1}\{ u_{\delta ,\varepsilon }\left( Y_{s-}+\left( \sigma (X_{s-})-{\widetilde{\sigma }}({\widetilde{X}}_{s-})\right) z\right) -u_{\delta ,\varepsilon }\left( Y_{s-}\right) \}{\widetilde{N}}(dz,ds),\\ M_t^{\delta ,\varepsilon ,2}&=\int _{0}^{t}\int _{|z|<1}\{ u_{\delta ,\varepsilon }\left( Y_{s-}+\left( \sigma (X_{s-})-{\widetilde{\sigma }}({\widetilde{X}}_{s-})\right) z\right) -u_{\delta ,\varepsilon }\left( Y_{s-}\right) \}{\widetilde{N}}(dz,ds). \end{aligned}$$

Then, $M_t^{\delta ,\varepsilon }=M_t^{\delta ,\varepsilon ,1}+M_t^{\delta ,\varepsilon ,2}$. We need to prove that $\left( M_t^{\delta ,\varepsilon ,1}\right) _{0\le t \le T}$ and $\left( M_t^{\delta ,\varepsilon ,2}\right) _{0\le t \le T}$ are martingales. First, we treat the term $\left( M_t^{\delta ,\varepsilon ,2}\right) _{0\le t \le T}$. Since $u_{\delta ,\varepsilon }$ is Lipschitz continuous by Lemma 5.1, there exists a constant $U_{\delta ,\varepsilon }$ such that

$$\begin{aligned}&{\mathbb {E}}\left[ \int _{0}^{t}\int _{|z|<1} \left| u_{\delta ,\varepsilon }\left( Y_s+(\sigma (X_s)-{\widetilde{\sigma }}({\widetilde{X}}_s))z\right) -u_{\delta ,\varepsilon }\left( Y_{s-}\right) \right|^2\frac{c_\alpha }{|z|^{1+\alpha }}dzds\right] \\&\quad \le U_{\delta ,\varepsilon }^2 {\mathbb {E}}\left[ \int _{0}^{t}\int _{|z|<1}\left| (\sigma (X_s)-{\widetilde{\sigma }}({\widetilde{X}}_s))z\right|^2\frac{c_\alpha }{|z|^{1+\alpha }}dzds\right] .\\ \end{aligned}$$

By using this fact and the boundedness of $\sigma$ and ${\widetilde{\sigma }}$, we have

$$\begin{aligned}&{\mathbb {E}}\left[ \int _{0}^{t}\int _{|z|<1} \left| u_{\delta ,\varepsilon }\left( Y_s+(\sigma (X_s)-{\widetilde{\sigma }}({\widetilde{X}}_s))z\right) -u_{\delta ,\varepsilon }\left( Y_{s-}\right) \right|^2\frac{c_\alpha }{|z|^{1+\alpha }}dzds\right] \\&\quad \le U_{\delta ,\varepsilon }^2 (\sup _{x\in {\mathbb {R}}}|\sigma (x)|^2+m_2^2)\int _{0}^{t}\int _{|z|<1}|z|^2\frac{c_\alpha }{|z|^{1+\alpha }}dzds\\&\quad \le U_{\delta ,\varepsilon }^2 (\sup _{x\in {\mathbb {R}}}|\sigma (x)|^2+m_2^2)T\int _{|z|<1}|z|^2\frac{dz}{|z|^{1+\alpha }}\\&\quad =\frac{2}{\alpha }U_{\delta ,\varepsilon }^2(\sup _{x\in {\mathbb {R}}}|\sigma (x)|^2+m_2^2)T\\&\quad <\infty . \end{aligned}$$

Hence, $\left( M_t^{\delta ,\varepsilon ,2}\right) _{0\le t \le T}$ is a $L^2(\varOmega ,{\mathbb {P}})$-martingale (see [1], Theorem 4.2.3 and P.231). Similarly, $\left( M_t^{\delta ,\varepsilon ,1}\right) _{0\le t \le T}$ is a $L^1(\varOmega ,{\mathbb {P}})$-martingale since

$$\begin{aligned}&{\mathbb {E}}\left[ \int _{0}^{t}\int _{|z|\ge 1} \left| u_{\delta ,\varepsilon }\left( Y_s+(\sigma (X_s)-{\widetilde{\sigma }}({\widetilde{X}}_s))z\right) -u_{\delta ,\varepsilon }\left( Y_{s-}\right) \right| \frac{c_\alpha }{|z|^{1+\alpha }}dzds\right] \\&\quad \le U_{\delta ,\varepsilon } (\sup _{x\in {\mathbb {R}}}|\sigma (x)|+m_2)T\int _{|z|\ge 1}|z|\frac{dz}{|z|^{1+\alpha }}\\&\quad = U_{\delta ,\varepsilon } (\sup _{x\in {\mathbb {R}}}|\sigma (x)|+m_2)T(\alpha +1)^{-1}\\&\quad <\infty . \end{aligned}$$

Therefore $M^{\delta ,\varepsilon }$ is a martingale since it is the sum of two martingales.

1.2 Hölder continuity of $\sigma^{\alpha}$

In this section, we prove that $\sigma^{\alpha}$ satisfies the Hölder continuity property stated in Lemma 2.1.

Lemma 5.2

Fix $\eta \in (0,1]$and $\rho >0$. If $\sigma$satisfies $\Vert \sigma \Vert _\infty <\infty$and $|\sigma (x)-\sigma (y)|<\rho |x-y|^{\eta }$for any $x,y\in {\mathbb {R}}$, then the function $\sigma^{\alpha}$is $\eta (\alpha -1)$-Hölder continuous.

Proof

By the triangle inequality and $\Vert \sigma \Vert _\infty <\infty$, we obtain

$$\begin{aligned} \left| \sigma^{\alpha} (x)-\sigma^{\alpha} (y)\right|&=\left| \sigma (x)\sigma^{\alpha -1}(x)-\sigma (x)\sigma^{\alpha -1}(y)+\sigma (x)\sigma^{\alpha -1}(y)-\sigma (y)\sigma^{\alpha -1}(y)\right| \\&\le \left| \sigma (x)\right| \left| \sigma^{\alpha -1}(x)-\sigma^{\alpha -1}(y)\right| +\left| \sigma^{\alpha -1}(y)\right| \left| \sigma (x)-\sigma (y)\right| \\&\le \Vert \sigma \Vert _\infty \left| \sigma^{\alpha -1}(x)-\sigma^{\alpha -1}(y)\right| +\Vert \sigma \Vert _\infty^{\alpha -1}\left| \sigma (x)-\sigma (y)\right| \\&\le \Vert \sigma \Vert _\infty \left| \sigma (x)-\sigma (y)\right|^{\alpha -1}+\Vert \sigma \Vert _\infty^{\alpha -1}\left| \sigma (x)-\sigma (y)\right| . \end{aligned}$$

This last inequality follows as $|a^{\alpha -1}-b^{\alpha -1}|\le |a-b|^{\alpha -1}$ for any $a,b\ge 0$. Here, since $|\sigma (x)-\sigma (y)|<\rho |x-y|^{\eta }$ and $|\sigma (x)|<m_2$ for any $x,y\in {\mathbb {R}}$, we have that any $x\ne y$

$$\begin{aligned} \frac{\left| \sigma (x)-\sigma (y)\right| }{|x-y|^{\eta (\alpha -1)}}&=\left( \frac{\left| \sigma (x)-\sigma (y)\right| }{|x-y|^{\eta }}\right)^{\alpha -1}\left| \sigma (x)-\sigma (y)\right|^{2-\alpha }\\&\le \rho^{\alpha -1}(|\sigma (x)|+|\sigma (y)|)^{2-\alpha }\\&\le 2\rho^{\alpha -1} \Vert \sigma \Vert _\infty^{2-\alpha }. \end{aligned}$$

Thus, $\sigma$ is $\eta$-Hölder continuous and $\eta (\alpha -1)$-Hölder continuous. Hence, we get

$$\begin{aligned} \left| \sigma^{\alpha} (x)-\sigma^{\alpha} (y)\right|&\le \Vert \sigma \Vert _\infty \rho^{\alpha -1}|x-y|^{\eta (\alpha -1)}+2\rho^{\alpha -1} \Vert \sigma \Vert _\infty^{2-\alpha }|x-y|^{\eta (\alpha -1)}\\&\le \max \left\{ \Vert \sigma \Vert _\infty \rho^{\alpha -1},\ 2\rho^{\alpha -1} \Vert \sigma \Vert _\infty^{2-\alpha }\right\} |x-y|^{\eta (\alpha -1)}. \end{aligned}$$

We conclude the proof. $\square$

1.3 The limit of subsequences of solutions $(X^{(n)},\sigma _n)_{n\in {\mathbb {N}}}$

In this section, we consider that the subsequential limit of the solution of SDE (2) is the solution of SDE (1) which the coefficient is the subsequential limit of $(\sigma _n)_{n\in {\mathbb {N}}}$. Suppose that $x_0=x_0^{(n)}$ and $\left( \sigma _n\right) _{n\in {\mathbb {N}}}$ is a Cauchy sequence in the norm $\Vert \cdot \Vert _{_{L^{\alpha} \left( {\mathbb {R}},\mu _{x_0}^{\alpha} \right) }}$ and satisfies following conditions.

$$\begin{aligned} 0<\inf _{n\in {\mathbb {N}}}\inf _{x\in {\mathbb {R}}}\sigma _{n}(x) \,\, \text {and}\,\, \sup _{n\in {\mathbb {N}}}\sup _{x\in {\mathbb {R}}}\sigma _{n}(x)<\infty . \end{aligned}$$

(33)

Furthermore, there exist constants ${\check{\rho }}>0$ and $\gamma \in [1/\alpha ,1]$ such that for any $x,y \in {\mathbb {R}}$ and $n\in {\mathbb {N}}$

$$\begin{aligned} \sup _{n\in {\mathbb {N}}}|\sigma _{n}(x)-\sigma _{n}(y)|\le {\check{\rho }}|x-y|^{\gamma }. \end{aligned}$$

(34)

We prove the following corollary by using Theorems 3.1, 4.1.

Corollary 5.3

Suppose that $(\sigma _n)_{n\in {\mathbb {N}}}$satisfies (33), (34) and $(\sigma _n)_{n\in {\mathbb {N}}}$is a Cauchy sequence in the norm $\Vert \cdot \Vert _{_{L^{\alpha} ({\mathbb {R}},\mu _{x_0}^{\alpha} )}}$. Then, there exists a subsequence $(n_k)_{k\in {\mathbb {N}}}$such that the limit $\displaystyle X^\infty :=\lim _{k\rightarrow \infty }X^{(n_k)}$exists almost surely and it is the unique solution of SDE (1) which has the coefficient $\lim _{k\rightarrow \infty }\sigma _{n_k}$

Proof

First, we confirm existence of the subsequential limit of the solution of SDE (2). Since $\mu _{x_0}^{\alpha}$ is a finite measure and $\mu _{x_0}^{\alpha} \ll \text {Leb} \ll \mu _{x_0}^{\alpha}$, there exists a subsequence $(m_k)_{k\in {\mathbb {N}}}$ such that the sequence $(\sigma _{m_k})_{k\in {\mathbb {N}}}$ converges pointwise to $\displaystyle \sigma _\infty :=\lim _{k\rightarrow \infty }\sigma _{m_k}$ Lebesgue almost everywhere (see [16] A13.2 (e)). Note that the limit $\sigma _\infty$ is also $\gamma$-Hölder continuous Lebesgue almost everywhere (see (36)). Using Theorem 4.1, there exists a subsequence $(n_k)_{k\in {\mathbb {N}}}$ of subsequence $(m_k)_{k\in {\mathbb {N}}}$ such that the following limit $\displaystyle \lim _{k\rightarrow \infty }X^{(n_k)}$ exists almost surely. Note that this subsequence $(n_k)_{k\in {\mathbb {N}}}$ does not depend on $t\in [0,T]$. The limit $X^\infty$ is càdlàg since $X^{(n_k)}$ converges in supremum norm by Theorem 4.1 (see [1], P.140).

We confirm that the limit $X^\infty$ is the unique solution of SDE (1). We define $V=\left( V_t\right) _{0 \le t \le T}$ as

$$\begin{aligned} V_t=x_0+\int _0^t\sigma _{\infty }(X_{s-}^\infty )dZ_s. \end{aligned}$$

We prove that

$$\begin{aligned} {\mathbb {P}}\left( V_t=X_t^\infty ,\text { for each } t\in [0,T]\right) =1. \end{aligned}$$

From Theorem 3.1 and the bounded convergence theorem, we have

$$\begin{aligned} {\mathbb {E}}\left[ \left| V_t-X_t^\infty \right|^{\alpha -1}\right] =\lim _{k\rightarrow \infty }{\mathbb {E}}\left[ \left| V_t-X_t^{(n_k)}\right|^{\alpha -1}\right] . \end{aligned}$$

In the same way as shown in the proof of Theorem 3.1, we have

$$\begin{aligned} {\mathbb {E}}\left[ \left| V_t-X_t^{(n_k)}\right|^{\alpha -1}\right] \le 2\varepsilon^{\alpha -1}+ {\widehat{M}}_t^{\delta ,\varepsilon ,k}+ {\widehat{J}}_t^{\delta ,\varepsilon ,k}, \end{aligned}$$

(35)

where $c_\alpha =\pi^{-1}\varGamma (\alpha +1)\sin \left( \frac{\alpha \pi }{2}\right)$, $C_\alpha$ appears in Lemma 3.3,

$$\begin{aligned} {\widehat{M}}_t^{\delta ,\varepsilon ,k}&:=\int _{0}^{t}\int _{{\mathbb {R}}\setminus \{0\}}\left\{ u_{\delta ,\varepsilon }\left( V_{s-}-X_{s-}^{(n_k)}+\left( \sigma _\infty (X_s^\infty )-\sigma _{n_k}(X_s^{(n_k)}\right) z\right) \right. \\&\quad \left. -u_{\delta ,\varepsilon }\left( V_{s-}-X_{s-}^{(n_k)}\right) \right\} {\widetilde{N}}(dz,ds) \text { and }\\ {\widehat{J}}_t^{\delta ,\varepsilon ,k}&:=c_\alpha \left| C_{\alpha }\right| \int _0^t \left| \sigma _\infty (X_s^\infty )-\sigma _{n_k}(X_s^{(n_k)})\right|^{\alpha} \psi _{\delta ,\varepsilon }(V_s-X_s^{(n_k)})ds. \end{aligned}$$

By using the same arguments as in Sect. 5.1, it is shown that ${\widehat{M}}^{\delta ,\varepsilon ,k}:=\left( {\widehat{M}}_t^{\delta ,\varepsilon ,k}\right) _{0\le t \le T}$ is a martingale. By the inequality $|x+y|^{\alpha} \le 2^{\alpha -1}\left( |x|^{\alpha} +|y|^{\alpha} \right)$ for any $x,y\in {\mathbb {R}}$, we have

$$\begin{aligned}&{\widehat{J}}_t^{\delta ,\varepsilon ,k}\le 2^{\alpha -1}\left| C_{\alpha }\right| \int _0^t \left\{ \left| \sigma _\infty (X_s^\infty )-\sigma _\infty (X_s^{(n_k)})\right|^{\alpha} \right. \\&\quad \left. +\left| \sigma _\infty (X_s^{(n_k)})-\sigma _{n_k}(X_s^{(n_k)})\right|^{\alpha} \right\} \psi _{\delta ,\varepsilon }(V_s-X_s^{(n_k)})ds. \end{aligned}$$

Here, since the sequence $(\sigma _{n_k})_{k\in {\mathbb {N}}}$ converges pointwise to $\sigma _{\infty }$ Lebesgue almost everywhere, we have

$$\begin{aligned} \left| \sigma _\infty (x)-\sigma _\infty (y)\right|&\le \left| \sigma _\infty (x)-\sigma _{n_k}(x)\right| +\left| \sigma _{n_k}(x)-\sigma _{n_k}(y)\right| +\left| \sigma _{n_k}(y)-\sigma _\infty (y)\right| \nonumber \\&\le \left| \sigma _\infty (x)-\sigma _{n_k}(x)\right| +{\check{\rho }}\left| x-y\right|^{\gamma }+\left| \sigma _{n_k}(y)-\sigma _\infty (y)\right| \nonumber \\&\rightarrow {\check{\rho }}\left| x-y\right|^{\gamma } \text { as } k\rightarrow \infty \text { for almost all }x,\ y. \end{aligned}$$

(36)

Hence, $\sigma _\infty$ is $\gamma$-Hölder continuous Lebesgue almost everywhere. Here, since $\sigma _\infty$ is a bounded function and $\gamma \ge 1/\alpha >(\alpha -1)/\alpha$, the function $\sigma _\infty$ is also $((\alpha -1)/\alpha )$-Hölder continuous. Thus, we have

$$\begin{aligned} {\widehat{J}}_t^{\delta ,\varepsilon ,k}&\le 2^{\alpha -1}\left| C_{\alpha }\right| \int _0^t \left\{ {\check{\rho }}\left| X_s^\infty -X_s^{(n_k)}\right|^{\alpha -1}\right. \nonumber \\&\quad \left. +\left| \sigma _\infty (X_s^{(n_k)})-\sigma _{n_k}(X_s^{(n_k)})\right|^{\alpha} \right\} \psi _{\delta ,\varepsilon }(V_s-X_s^{(n_k)})ds\nonumber \\&\le \frac{{\widehat{C}}_{\alpha }}{\log \delta }\left( \frac{\delta }{\varepsilon }\right)^{ }\int _0^t \left\{ {\check{\rho }}\left| X_s^\infty -X_s^{(n_k)}\right|^{\alpha -1}+\left| \sigma _\infty (X_s^{(n_k)})-\sigma _{n_k}(X_s^{(n_k)})\right|^{\alpha} \right\} ds. \end{aligned}$$

(37)

The last inequality follows from the explicit upper bound for $\psi _{\delta ,\varepsilon }$. By (35), (37) and Fubini’s theorem, we have

$$\begin{aligned}&\sup _{0\le t\le T}{\mathbb {E}}\left[ \left| V_t-X_t^{(n_k)}\right|^{\alpha -1}\right] \\&\quad \le 2\varepsilon^{\alpha -1}+\frac{ {\widehat{C}}_{\alpha }}{\log \delta }\left( \frac{\delta }{\varepsilon }\right)^{ }\int _0^T \left\{ {\check{\rho }}{\mathbb {E}}\left[ \left| X_s^\infty -X_s^{(n_k)}\right|^{\alpha -1}\right] \right. \\&\qquad \left. +{\mathbb {E}}\left[ \left| \sigma _\infty (X_s^{(n_k)})-\sigma _{(n_k)}(X_s^{(n_k)})\right|^{\alpha} \right] \right\} ds\\&\quad \le 2\varepsilon^{\alpha -1}+\frac{ {\widehat{C}}_{\alpha }}{\log \delta }\left( \frac{\delta }{\varepsilon }\right)^{ }\left\{ T{\check{\rho }}\sup _{0\le s\le T}{\mathbb {E}}\left[ \left| X_s^\infty -X_s^{(n_k)}\right|^{\alpha -1}\right] \right. \\&\qquad \left. +\int _0^T{\mathbb {E}}\left[ \left| \sigma _\infty (X_s^{(n_k)})-\sigma _{n_k}(X_s^{(n_k)})\right|^{\alpha} \right] \right\} ds. \end{aligned}$$

From Theorem 3.1 and the bounded convergence theorem, we obtain

$$\begin{aligned} \sup _{0\le s\le T}{\mathbb {E}}\left[ \left| X_s^\infty -X_s^{(n_k)}\right|^{\alpha -1}\right] \rightarrow 0 \,\, \text {as}\,\, k\rightarrow \infty . \end{aligned}$$

By the assumption on $\sigma _n$, Lemmas 5.2 and 2.1, $X_t^{(n)}$ has a transition density function for each $t\in (0,T]$ and each $n\in {\mathbb {N}}$. This density has an upper bound as stated in Lemma 2.2, so that we have by (29),

$$\begin{aligned} \int _0^T{\mathbb {E}}\left[ \left| \sigma _\infty (X_s^{n_k})-\sigma _{n_k}(X_s^{n_k})\right|^{\alpha} \right] ds\rightarrow 0 \text { as } k\rightarrow \infty . \end{aligned}$$

Hence, we have for any $\varepsilon >0$,

$$\begin{aligned} \lim _{k\rightarrow \infty }\sup _{0\le t\le T}{\mathbb {E}}\left[ \left| V_t-X_t^{(n_k)}\right|^{\alpha -1}\right] \le 2\varepsilon^{\alpha -1}. \end{aligned}$$

Therefore, we get

$$\begin{aligned} \lim _{k\rightarrow \infty }\sup _{0\le t\le T}{\mathbb {E}}\left[ \left| V_t-X_t^{(n_k)}\right|^{\alpha -1}\right] =0. \end{aligned}$$

Thus, we have

$$\begin{aligned} {\mathbb {P}}\left( V_t=X_t^\infty \right) =1 \text { for each } t\in [0,T]. \end{aligned}$$

H ere, since each sample path of V and $X^\infty$ is càdlàg, we have (see [14], Section I, Theorem 2)

$$\begin{aligned} {\mathbb {P}}\left( V_t=X_t^\infty ,\text { for each } t\in [0,T]\right) =1. \end{aligned}$$

This concludes the proof. $\square$

1.4 A more precise estimate for Theorems 3.1 and 4.1

In this section, we give a more precise estimate for the result on Theorems 3.1 and 4.1 for $\gamma \in \left( 1/\alpha ,1\right]$. We recall (31), which states that for $p>0$,

$$\begin{aligned} \sup _{0\le t \le T}{\mathbb {E}}\left[ \left| X_t-{\widetilde{X}}_t\right|^{\alpha -1}\right] \le |x_0-{\widetilde{x}}_0|^{\alpha -1}+\frac{1}{3}C \left( 2\lambda^{p(\alpha \gamma -1)}+\lambda^{-p+\alpha }\right) . \end{aligned}$$

We consider the function $g(x)=2\lambda^{x(\alpha \gamma -1)}+\lambda^{-x+\alpha }$. First, we find the critical points of g,

$$\begin{aligned}&g'(x)=2\lambda^{x(\alpha \gamma -1)}(\alpha \gamma -1)\log \lambda -\lambda^{-x+\alpha }\log \lambda =0\\&2\lambda^{x(\alpha \gamma -1)}(\alpha \gamma -1) =\lambda^{-x+\alpha }\\&\log (2(\alpha \gamma -1))+x(\alpha \gamma -1) \log \lambda =(-x+\alpha )\log \lambda \\&x =\frac{1}{\gamma }-\frac{\log \left( 2(\alpha \gamma -1)\right) }{\log \lambda }. \end{aligned}$$

Second, since $\lim _{x\rightarrow \pm \infty }g(x)=+\infty$ and there is only one critical point, the function g takes its minimum value at $x=\frac{1}{\gamma }-\frac{\log \left( 2(\alpha \gamma -1)\right) }{\log \lambda }$.

Therefore, we get the inequality

$$\begin{aligned}&\sup _{0\le t \le T}{\mathbb {E}}\left[ \left| X_t-{\widetilde{X}}_t\right|^{\alpha -1}\right] \\&\quad \le |x_0-{\widetilde{x}}_0|^{\alpha -1}\\&\qquad +\frac{1}{3}C \left( 2\lambda^{\frac{\alpha \gamma -1}{\gamma }-\frac{(\alpha \gamma -1)\log \left( 2(\alpha \gamma -1)\right) }{\log \lambda }}+\lambda^{\frac{\alpha \gamma -1}{\gamma }+\frac{\log \left( 2(\alpha \gamma -1)\right) }{\log \lambda }}\right) \\&\quad < |x_0-{\widetilde{x}}_0|^{\alpha -1}+Cg\left( \frac{1}{\gamma }\right) \\&\quad = |x_0-{\widetilde{x}}_0|^{\alpha -1}+C\lambda^{\frac{\alpha \gamma -1}{\gamma }}. \end{aligned}$$

In the same way, we have for any $h>0$,

$$\begin{aligned}&h{\mathbb {P}}\left( \sup _{0\le t \le T}\left| X_t-{\widetilde{X}}_t\right|^{\alpha -1}>h\right) \\&\quad \le |x_0-{\widetilde{x}}_0|^{\alpha -1}\\&\qquad +\frac{1}{3}C \left( 2\lambda^{\frac{\alpha \gamma -1}{\gamma }-\frac{(\alpha \gamma -1)\log \left( 2(\alpha \gamma -1)\right) }{\log \lambda }}+\lambda^{\frac{\alpha \gamma -1}{\gamma }+\frac{\log \left( 2(\alpha \gamma -1)\right) }{\log \lambda }}\right) \\&\quad < |x_0-{\widetilde{x}}_0|^{\alpha -1}+ C\lambda^{\frac{\alpha \beta }{2(1+\beta )-\alpha }}. \end{aligned}$$