2.1. Auxiliary lemmas

Similar to Proposition A.1 and Lemmas 3.3 and 3.4 in [Reference Pagès and Panloup34], we first give the moment estimates for $(X_t)_{t\geqslant 0}$ , $(Y_{t_n})_{n\in \mathbb{N}_0}$ and $(\tilde{Y}_{t_n})_{n\in \mathbb{N}_0}$ , which will be proved in Section 3. For simplicity, for any $i\leqslant j$ , we denote by $X_{t_{i},t_{j}}^{x}$ the value of the process at time $t_j$ while starting from x at time $t_{i}$ . If $t_i=0$ , then we denote it by $X_{t_j}^x$ . The quantities $Y_{t_{i},t_{j}}^{x}$ and $\tilde{Y}_{t_{i},t_{j}}^x$ are similarly defined.

Lemma 1. Let $(X_{t})_{t\geqslant0}$ , $(Y_{t_n})_{n\in \mathbb{N}_0}$ and $(\tilde{Y}_{t_n})_{n\in \mathbb{N}_0}$ be the processes defined by (1.1), (1.3) and (1.5), respectively. Suppose that Assumption 1 holds. Then for any $1\leqslant\kappa <\alpha$ , there exists a constant $C>0$ independent of d such that for all $x \in \mathbb{R}^d$ ,

(2.2) \begin{align}\mathbb{E} |X_{t}^x|^{\kappa} &\leqslant Cd^{\frac{5\vee(3\kappa)}{2}}(1+|x|^{\kappa})\quad \forall t\geqslant0, \\[-24pt] \nonumber \end{align}

(2.3) \begin{align}\mathbb{E} |Y_{t_n}^x|^{\kappa} &\leqslant Cd^{\frac{5\vee(3\kappa)}{2}}(1+|x|^{\kappa})\quad \forall n\in \mathbb{N}_0,\\[-24pt]\nonumber \end{align}

(2.4) \begin{align}\mathbb{E} |\tilde{Y}_{t_n}^x|^{\kappa} &\leqslant Cd^{\frac{5\vee(3\kappa)}{2}} (1+|x|^{\kappa}) \quad \forall n\in \mathbb{N}_0. \\[-24pt] \nonumber \end{align}

Lemma 2. Let $(X_{t})_{t\geqslant 0}$ be the solution to the SDE (1.1), and let $(Y_{t_{n}})_{n\in\mathbb{N}_{0}}$ be the Markov chains defined by (1.3). Suppose that Assumption 1 holds. Then for any $1\leqslant\kappa <\alpha$ , there exists a positive constant C independent of d such that for all $n\in\mathbb{N}$ and $x\in\mathbb{R}^{d}$ ,

(2.5) \begin{align}\mathbb{E} |Y_{t_{n-1},t_{n}}^{x}-x|^{\kappa} &\leqslant Cd^{\kappa} (1+|x|^{\kappa})\gamma_{n}^{\frac{\kappa}{\alpha}}, \\[-24pt] \nonumber \end{align}

(2.6) \begin{align}\mathbb{E} |X_{s,t}^{x}-x|^{\kappa}&\leqslant C d^{\frac{5\vee(3\kappa)}{2}} (1+|x|^{\kappa})(t-s)^{\frac{\kappa}{\alpha}} \quad \forall0\leqslant s\leqslant t\leqslant \gamma_{1},\\[-30pt]\nonumber \end{align}

(2.7) \begin{align}\mathbb{E} |X_{t_{n-1},t_{n}}^{x}-Y_{t_{n-1},t_{n}}^{x}|^{\kappa}&\leqslant Cd^{\frac{5\vee(3\kappa)}{2}}(1+|x|^{\kappa})\gamma_n^{\kappa+\frac{\kappa}{\alpha}}. \end{align}

Based on the above lemmas, according to [Reference Chen, Nourdin, Xu and Yang7, Lemma 18], we can derive the following bound on the difference between $P_{t_{n-1},t_{n}}f$ and $Q_{t_{n-1},t_{n}}f$ , the proof of which is postponed to Section 3.

Lemma 3. Under Assumptions 1 and 3, let $f\in \mathcal{C}^2(\mathbb{R}^d,\mathbb{R})$ satisfy $\| \nabla f \|_{\infty} < \infty$ and $\| \nabla^2 f \|_{{\rm HS}, \infty} < \infty$ . Then for $\kappa \in[1,\alpha)$ , there exists a constant $C>0$ independent of d such that for all $k\in \mathbb{N}$ and $x\in\mathbb{R}^{d}$ ,

\begin{eqnarray*}\bigl|(P_{t_{k-1},t_{k}}-Q_{t_{k-1},t_{k}})f(x)\bigr|\leqslant C(1+|x|^{\kappa}) (\| \nabla f \|_{\infty}+\| \nabla^2 f \|_{{\rm HS},\infty})d^{\frac{5}{2}\vee\frac{3\kappa}{2}\vee(\frac{3}{2}+\frac{3}{2\kappa})} \gamma_k^{2+\frac{1}{\alpha}-\frac{1}{\kappa}}.\end{eqnarray*}

In the second EM scheme, the difference between $P_{t_{n-1},t_{n}}f$ and $\tilde{Q}_{t_{n-1},t_{n}}f$ is estimated in the following lemma.

Lemma 4 Under Assumption 1, let $f\in \mathcal{C}^2(\mathbb{R}^d,\mathbb{R})$ satisfy $\| \nabla f \|_{\infty} < \infty$ and $\| \nabla^2 f \|_{{\rm HS}, \infty} $ $ < \infty$ . Then there exists a constant $C>0$ independent of d such that for all $k\in \mathbb{N}$ and $x\in\mathbb{R}^{d}$ ,

\begin{eqnarray*}\bigl| (P_{t_{k-1},t_{k}} - \tilde{Q}_{t_{k-1},t_{k}} ) f(x) \bigr|&\leqslant& C(1+|x|) (\| \nabla f \|_{\infty} + \| \nabla^2 f \|_{{\rm HS},\infty}) d^{\frac{5}{2\kappa} \vee (3-\frac{\alpha}{2}) } \gamma_k^{\frac{2}{\alpha}}.\end{eqnarray*}

The above two lemmas work similarly to [Reference Pagès and Panloup34, Proposition 3.5]. Furthermore, corresponding to [Reference Pagès and Panloup34, Proposition 3.2], we also need the following regularities for $P_{t_i,t_j}h(x)$ with $h\in {\rm Lip}(1)$ and $j\geqslant i$ , which can be proved in a similar way to Lemma 2.1 in [Reference Chen, Deng, Schilling and Xu6].

Lemma 5. Let $h\in {\rm Lip}(1)$ and suppose that Assumptions 1 and 3 hold. Then there exists a constant $C>0$ independent of d such that for all $0 < t_j-t_i \leqslant 1$ and $u_1,u_2 \in \mathbb{R}^d$ ,

\begin{align*}\| \nabla_{u_1} P_{t_i,t_j} h \|_{\infty} &\leqslant C|u_1|, \\\| \nabla_{u_2} \nabla_{u_1} P_{t_i,t_j} h \|_{{\rm HS},\infty} &\leqslant C(t_j-t_i)^{-\frac{1}{\alpha}} |u_1| |u_2|.\end{align*}

Pagès et al. [Reference Pagès and Panloup34] investigated the convergence rate in the total variation distance with the help of the Domino–Malliavin method for non-smooth functions. It seems hard to obtain a similar result in our setting due to the limitation of the Malliavin calculus for SDEs driven by $\alpha$ -stable noise.

The exponential ergodicity of the solution $(X_t)_{t\geqslant 0}$ of the SDE (1.1) under Assumption 1 can be obtained using the same argument as in the proof of Theorem 1.3 in [Reference Liang and Wang27]; the proof is postponed to Section 3.5. The following lemma, together with [Reference Pagès and Panloup34, Corollary 2.4], is crucial for establishing the exponential ergodicity for the process $(X_t)_{t\geqslant 0}$ .

Lemma 6. Under Assumption 1, there exists a constant $C>0$ independent of d such that for all $x,y\in\mathbb{R}^{d}$ and $t>0$ ,

\begin{align*}W_1 (\mathcal{L}(X_{t}^x), \mathcal{L}(X_{t}^y)) \leqslant C\exp(d^{4-2\alpha} 2^{d}) \exp(-d^{2\alpha-4}2^{-d}{\mathrm{e}}^{-2^{d}d^{4-2\alpha}} t)|x-y|.\end{align*}

In addition, the process $(X_t)_{t\geqslant 0}$ in (1.1) is exponentially ergodic.

In addition, similar to Lemma A.3 in [Reference Pagès and Panloup34], the following lemma about the decreasing steps also plays an important role in proving the main theorems. This lemma will be proved in Section 3.

Lemma 7. For some $\theta\in(0,1]$ , let $(\gamma_{n})_{n\in \mathbb{N}}$ be a decreasing positive sequence such that

\begin{eqnarray*}\omega \,:\!=\, \limsup_{k\to \infty} \frac{\gamma_k^{\theta}-\gamma_{k+1}^{\theta}}{\gamma_{k+1}^{1+\theta}} <+\infty.\end{eqnarray*}

Let $\omega<\rho= d^{2\alpha-4}2^{-d}\exp(-2^{d} d^{4-2\alpha})$ and let $(v_{n})_{n\in \mathbb{N}}$ be the sequence defined by $v_{0}=0$ and, for every $n\geqslant1$ ,

\begin{align*}v_{n}=\sum_{k=1}^{n}\gamma_{k}^{1+\theta}{\mathrm{e}}^{-\rho(t_{n}-t_{k})}.\end{align*}

Then

(2.8) \begin{align}\limsup_{n\rightarrow\infty}\frac{v_{n}}{\gamma_{n}^{\theta}}&\leqslant\frac{2}{\rho-\omega}{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_1}, \end{align}

(2.9) \begin{align}\lim_{n\rightarrow\infty}\frac{{\mathrm{e}}^{-\rho t_{n}}}{\gamma_{n}^{\theta}}&= 0. \\[-18pt]\nonumber \end{align}

Assume that n is large enough that $t_{n}>1$ and let $n^*=\max\{ i: t_n - t_i >1 \}$ . We further have

(2.10) \begin{eqnarray}\sum_{i=n^*+1}^{n-1}(t_n-t_i)^{-\frac{1}{\alpha}} \gamma_i^{1+\theta}\leqslant C {\mathrm{e}}^{\rho(1+\gamma_{1})} \gamma_{n}^{\theta}\end{eqnarray}

for some constant $C>0$ independent of d.

2.2. Proof of Theorem 1

It follows from (1.7) and (2.1) that

(2.11) \begin{eqnarray}W_1(\mathcal{L}(X_{t_n}^x), \mathcal{L}(Y_{t_n}^x))&\,=\,& \text{supp}_{h\in {\rm Lip}(1)} \{ P_{0,t_n} h(x) - Q_{0,t_n} h(x) \}\nonumber\\&\leqslant& \sum_{i=1}^{n}\text{supp}_{h\in{\rm Lip}(1)}\left|Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_{i},t_n} h(x)\right|\nonumber\\&\,=\,& \sum_{i=1}^{n-1}\text{supp}_{h\in{\rm Lip}(1)}\left| Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}}-Q_{t_{i-1},t_{i}}) \circ P_{t_{i},t_n} h(x)\right|\nonumber\\&&+\: \text{supp}_{h\in{\rm Lip}(1)}\left| Q_{0,t_{n-1}} \circ (P_{t_{n-1},t_{n}}- Q_{t_{n-1},t_{n}})h(x)\right|.\end{eqnarray}

We first bound the last term in (2.11). Since $h\in {\rm Lip}(1)$ , one can derive from the Markov property, (2.7) and (2.3) with $\kappa=1$ that

(2.12) \begin{align*}& \text{supp}_{h\in {\rm Lip}(1)} \left|Q_{0,t_{n-1}} \circ (P_{t_{n-1},t_{n}} -Q_{t_{n-1},t_{n}})h(x)\right|\nonumber\\&\quad = \text{supp}_{h\in {\rm Lip}(1)}\left|\mathbb{E}[h(X_{t_{n-1},t_{n}}^{Y_{t_{n-1}}^{x}})]-\mathbb{E}[h(Y_{t_{n-1},t_{n}}^{Y_{t_{n-1}}^{x}})]\right|\nonumber\\\end{align*}\begin{align} &\quad \leqslant\mathbb{E}\left|X_{t_{n-1},t_{n}}^{Y_{t_{n-1}}^{x}}-Y_{t_{n-1},t_{n}}^{Y_{t_{n-1}}^{x}}\right|\ \leqslant \ C d^{\frac{5}{2} } (1+\mathbb{E}|Y_{t_{n-1}}^{x}|)\gamma_{n}^{1+\frac{1}{\alpha}} \nonumber \\&\quad \leqslant Cd^{5}(1+|x|)\gamma_{n}^{1+\frac{1}{\alpha}}.\end{align}

For the first sum in (2.11), we use Lemma 6 and the Markov property to show that there exists a constant $C>0$ independent of d such that for any $x,y\in \mathbb{R}^d$ ,

(2.13) \begin{eqnarray}|P_{t_i,t_j}h(x)-P_{t_i,t_j}h(y)| \leqslant C{\mathrm{e}}^{ d^{4-2\alpha} 2^{d}}|x-y|\exp\bigl(-d^{2\alpha-4}2^{-d}{\mathrm{e}}^{-2^{d} d^{4-2\alpha}}(t_{j}-t_{i})\bigr),\end{eqnarray}

It follows from this, the Markov property, (2.7) and (2.3) that

\begin{align*}& \text{supp}_{h\in {\rm Lip}(1)} \left| Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_i,t_n} h(x) \right| \\&\quad = \text{supp}_{h\in {\rm Lip}(1)} \left|\mathbb{E}\left[P_{t_{i},t_{n}}h\left(X_{t_{i-1},t_{i}}^{Y_{t_{i-1}}^{x}}\right)\right]-\mathbb{E}\left[P_{t_{i},t_{n}}h\left(Y_{t_{i-1},t_{i}}^{Y_{t_{i-1}^{x}}}\right)\right]\right| \nonumber \\&\quad \leqslant C{\mathrm{e}}^{ d^{4-2\alpha} 2^{d}} \exp(-d^{2\alpha-4}2^{-d}{\mathrm{e}}^{-2^{d} d^{4-2\alpha}}(t_{n}-t_{i})) \mathbb{E}\left|X_{t_{i-1},t_{i}}^{Y_{t_{i-1}}^{x}}-Y_{t_{i-1},t_{i}}^{Y_{t_{i-1}^{x}}}\right|\\&\quad \leqslant C{\mathrm{e}}^{ d^{4-2\alpha} 2^{d}} \exp(-d^{2\alpha-4}2^{-d}{\mathrm{e}}^{-2^{d} d^{4-2\alpha}}(t_{n}-t_{i}))d^{\frac{5}{2}}\left(1+\mathbb{E}\left|Y_{t_{i-1}}^{x}\right|\right)\gamma_{i}^{1+\frac{1}{\alpha}}\\&\quad \leqslant C{\mathrm{e}}^{ d^{4-2\alpha} 2^{d}} \exp(-d^{2\alpha-4}2^{-d}{\mathrm{e}}^{-2^{d} d^{4-2\alpha}}(t_{n}-t_{i}))d^{5}\left(1+\left|x\right|\right)\gamma_{i}^{1+\frac{1}{\alpha}}.\end{align*}

Recall that $\rho= d^{2\alpha-4}2^{-d} {\mathrm{e}}^{-2^{d} d^{4-2\alpha}}$ . We have

\begin{align*}&\sum_{i=1}^{n-1}\text{supp}_{h\in{\rm Lip}(1)}\left|Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_{i},t_n} h(x)\right|\\& \quad \leqslant C{\mathrm{e}}^{d^{4-2\alpha} 2^{d}}d^{5}(1+|x|)\sum_{i=1}^{n-1}{\mathrm{e}}^{-\rho(t_{n}-t_{i})}\gamma_{i}^{1+\frac{1}{\alpha}}\leqslant\frac{C}{\rho-\omega}{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_{1}}{\mathrm{e}}^{d^{4-2\alpha} 2^{d}}d^{5} (1+|x|)\gamma_{n}^{\frac{1}{\alpha}},\end{align*}

where the last inequality follows from (2.8).

Combining all of the above, we deduce that there exists a constant $C>0$ independent of d and n such that

\begin{align*}W_1(\mathcal{L}(X_{t_n}^x), \mathcal{L}(Y_{t_n}^x))\leqslant&\frac{C}{\rho-\omega}{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_{1}}{\mathrm{e}}^{d^{4-2\alpha}2^{d}}d^{5}(1+|x|)\gamma_{n}^{\frac{1}{\alpha}}.\end{align*}

This proves (1.11).

Now we prove (1.12). Notice that

(2.14) \begin{eqnarray}W_{1}(\nu,\mathcal{L}(X_{t_n}^x))&\leqslant& \int_{\mathbb{R}^{d}}W_{1}(\mathcal{L}(X_{t_n}^x), \mathcal{L}(X_{t_n}^y))\nu(\mathrm{d} y)\nonumber\\&\leqslant& C{\mathrm{e}}^{d^{4-2\alpha} 2^{d}}{\mathrm{e}}^{-\rho t_{n}} \int_{\mathbb{R}^{d}}|x-y|\nu(\mathrm{d} y) \nonumber \\&\leqslant& C{\mathrm{e}}^{d^{4-2\alpha} 2^{d}} d(1+|x|){\mathrm{e}}^{-\rho t_{n}},\end{eqnarray}

where the first inequality in (2.14) is from Lemma 6. Then, by (1.11) and (2.9),

\begin{align*}W_{1}(\nu, \mathcal{L}(Y_{t_n}^x)) &\leqslant W_{1}(\nu,\mathcal{L}(X_{t_n}^x))+W_{1}(\mathcal{L}(X_{t_n}^x), \mathcal{L}(Y_{t_n}^x))\\&\leqslant\frac{C}{\rho-\omega}{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_1} {\mathrm{e}}^{d^{4-2\alpha} 2^{d}}d^{5}({\mathrm{e}}^{-\rho t_{n}}+\gamma_{n}^{\frac{1}{\alpha}}) (1+|x|) \\&\leqslant\frac{C}{\rho-\omega}{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_1} {\mathrm{e}}^{d^{4-2\alpha} 2^{d}}d^{5}\gamma_{n}^{\frac{1}{\alpha}} (1+|x|).\end{align*}

The proof is complete.

2.3. Proof of Theorem 2

According to (2.11), we have

(2.15) \begin{eqnarray}W_1(\mathcal{L}(X_{t_n}^x), \mathcal{L}(Y_{t_n}^x))&\leqslant& \sum_{i=1}^{n-1}\text{supp}_{h\in{\rm Lip}(1)}\left|Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_{i},t_n} h(x)\right| \nonumber \\&&+ \:\text{supp}_{h\in{\rm Lip}(1)}\left| Q_{0,t_{n-1}} \circ ( P_{t_{n-1},t_{n}} - Q_{t_{n-1},t_{n}} )h(x)\right|.\end{eqnarray}

For the first sum on the right-hand side of (2.15), we bound it as follows:

(2.16) \begin{align} & \sum_{i=1}^{n-1} \text{supp}_{h\in{\rm Lip}(1)} \left| Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}}- Q_{t_{i-1},t_{i}} ) \circ P_{t_i,t_n} h(x) \right| \nonumber \\&\quad\leqslant \sum_{i=1}^{n^*} \text{supp}_{h\in{\rm Lip}(1)} \left| Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_i,t_n} h(x) \right| \nonumber \\&\qquad + \sum_{i=n^*+1}^{n-1} \text{supp}_{h\in{\rm Lip}(1)} \left| Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_i,t_n} h(x) \right| \nonumber \\&\quad=: \mathcal{R}_1 + \mathcal{R}_2,\end{align}

recalling that $n^*=\max\{ i: t_n - t_i >1 \}$ .

Let us first give the estimate for $\mathcal{R}_1$ , in which $t_n-t_i>1$ . Recall that $\rho= d^{2\alpha-4}2^{-d} {\mathrm{e}}^{-2^{d} d^{4-2\alpha}}$ ; then one can derive from (2.13) that

\begin{align*}& \text{supp}_{h\in {\rm Lip}(1)} | Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_i,t_n} h(x) | \\&= \text{supp}_{h\in {\rm Lip}(1)} | Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_i,t_{i}+1} \circ P_{t_{i}+1,t_n} h(x) | \nonumber \\&= \text{supp}_{\{g=P_{t_{i}+1,t_n} h: h\in {\rm Lip}(1)\}} | Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_i,t_{i}+1} g(x) | \nonumber \\&\leqslant \text{supp}_{g\in {\rm Lip}(\ell)} | Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_i,t_{i}+1} g(x) | \nonumber \\&\leqslant C{\mathrm{e}}^{d^{4-2\alpha} 2^{d}}{\mathrm{e}}^{-\rho(t_{n}-t_{i}-1)} \text{supp}_{ g\in {\rm Lip}(1) } | Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_i,t_i+1} g(x)|,\end{align*}

where $\ell=C{\mathrm{e}}^{d^{4-2\alpha} 2^{d}}{\mathrm{e}}^{-\rho(t_{n}-t_{i}-1)}$ and the first inequality follows from (2.13). Then, it follows from Lemmas 1, 3 and 5 that

\begin{align*}& \text{supp}_{h\in {\rm Lip}(1)} | Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_i,t_n} h(x) | \\&\quad\leqslant C{\mathrm{e}}^{d^{4-2\alpha} 2^{d}}{\mathrm{e}}^{-\rho(t_{n}-t_{i}-1)} d^{5\vee(3\kappa)\vee(4+\frac{3}{2\kappa}) \vee (\frac{3}{2}+\frac{3}{2\kappa}+\frac{3\kappa}{2})}\\ &\qquad\times (1+|x|^{\kappa}) \text{supp}_{g\in {\rm Lip}(\ell)} (\| \nabla P_{t_i,t_i+1}g\|_{\infty} + \| \nabla^2 P_{t_i,t_i+1}g\|_{{\rm HS},\infty}) \gamma_i^{2+\frac{1}{\alpha}-\frac{1}{\kappa}}\nonumber \\&\quad\leqslant C{\mathrm{e}}^{d^{4-2\alpha} 2^{d}}{\mathrm{e}}^{-\rho(t_{n}-t_{i}-1)}d^{5\vee(3\kappa)\vee(4+\frac{3}{2\kappa}) \vee (\frac{3}{2}+\frac{3}{2\kappa}+\frac{3\kappa}{2})}(1+|x|^{\kappa})\gamma_i^{2+\frac{1}{\alpha}-\frac{1}{\kappa}} ,\end{align*}

which implies that there exists a positive constant C independent of d and n such that

(2.17) \begin{eqnarray}\mathcal{R}_1&\,=\,&\sum_{i=1}^{n^*}\text{supp}_{h\in{\rm Lip}(1)} \left|Q_{0,t_{i-1}} \circ (P_{t_{i-1},t_{i}}-Q_{t_{i-1},t_{i}}) \circ P_{t_i,t_n} h(x) \right| \nonumber \\&\leqslant& C{\mathrm{e}}^{d^{4-2\alpha} 2^{d}} d^{5\vee(3\kappa)\vee(4+\frac{3}{2\kappa}) \vee (\frac{3}{2}+\frac{3}{2\kappa}+\frac{3\kappa}{2})}(1+|x|^{\kappa})\sum_{i=1}^{n^*} {\mathrm{e}}^{-\rho (t_{n}-t_{i}-1) } \gamma_i^{2+\frac{1}{\alpha}-\frac{1}{\kappa}} \nonumber \\&\leqslant&\frac{ C{\mathrm{e}}^{\rho}}{\rho-\omega}{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_1} {\mathrm{e}}^{d^{4-2\alpha} 2^{d}}d^{5\vee(3\kappa)\vee(4+\frac{3}{2\kappa}) \vee (\frac{3}{2}+\frac{3}{2\kappa}+\frac{3\kappa}{2})}(1+|x|^{\kappa}) \gamma_{n}^{1+\frac{1}{\alpha}-\frac{1}{\kappa}},\end{eqnarray}

where the last inequality holds by (2.8).

For $\mathcal{R}_2$ , it follows from Lemmas 1, 3, and 5 that

(2.18) \begin{eqnarray}\mathcal{R}_2&\,=\,&\sum_{i=n^*+1}^{n-1} \text{supp}_{h\in{\rm Lip}(1)}\left| Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_i,t_n} h(x) \right| \nonumber \\&\leqslant& C {\mathrm{e}}^{\rho(1+\gamma_{1})} d^{5\vee(3\kappa)\vee(4+\frac{3}{2\kappa}) \vee (\frac{3}{2}+\frac{3}{2\kappa}+\frac{3\kappa}{2})} (1+|x|^{\kappa})\sum_{i=n^*+1}^{n-1}(t_n-t_i)^{-\frac{1}{\alpha}} \gamma_i^{2+\frac{1}{\alpha}-\frac{1}{\kappa}}\nonumber\\&\leqslant& C {\mathrm{e}}^{\rho(1+\gamma_{1})} d^{5\vee(3\kappa)\vee(4+\frac{3}{2\kappa}) \vee (\frac{3}{2}+\frac{3}{2\kappa}+\frac{3\kappa}{2})}(1+|x|^{\kappa})\gamma_{n}^{1+\frac{1}{\alpha}-\frac{1}{\kappa}},\end{eqnarray}

where the constant C is independent of d and the last inequality is from (2.10). Hence, (2.16), (2.17) and (2.18) imply that

(2.19) \begin{align}& \sum_{i=1}^{n-1} \text{supp}_{h\in{\rm Lip}(1)}\left| Q_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - Q_{t_{i-1},t_{i}} ) \circ P_{t_i,t_n} h(x) \right|\nonumber\\&\quad\leqslant\frac{C}{\rho-\omega}\,{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_1} {\mathrm{e}}^{d^{4-2\alpha} 2^{d}}d^{5\vee(3\kappa)\vee(4+\frac{3}{2\kappa}) \vee (\frac{3}{2}+\frac{3}{2\kappa}+\frac{3\kappa}{2})}(1+|x|^{\kappa}) \gamma_{n}^{1+\frac{1}{\alpha}-\frac{1}{\kappa}}.\end{align}

In addition, the second term in (2.15) is directly obtained from (2.12); that is,

\begin{eqnarray*}\text{supp}_{h\in{\rm Lip}(1)}\left|Q_{0,t_{n-1}}\circ ( P_{t_{n-1},t_{n}} - Q_{t_{n-1},t_{n}} )h(x)\right|\leqslant Cd^{5} (1+|x|)\gamma_{n}^{1+\frac{1}{\alpha}}.\end{eqnarray*}

Therefore, there exists a positive constant C independent of d such that

\[W_{1}(\mathcal{L}(X_{t_n}^x), \mathcal{L}(Y_{t_n}^x))\leqslant\frac{C}{\rho-\omega}{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_1}\,{\mathrm{e}}^{d^{4-2\alpha} 2^{d}}d^{5\vee(3\kappa)\vee(4+\frac{3}{2\kappa}) \vee (\frac{3}{2}+\frac{3}{2\kappa}+\frac{3\kappa}{2})} (1+|x|^{\kappa}) \gamma_{n}^{1+\frac{1}{\alpha}-\frac{1}{\kappa}}.\]

This proves (1.15).

Now we prove (1.16). By (2.14), (1.15) and (2.9), we have

\begin{eqnarray*}W_{1}(\nu, \mathcal{L}(Y_{t_n}^x))&\leqslant& W_{1}(\nu,\mathcal{L}(X_{t_n}^x))+W_{1}(\mathcal{L}(X_{t_n}^x), \mathcal{L}(Y_{t_n}^x))\\&\leqslant&\frac{C}{\rho-\omega}\,{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_1}{\mathrm{e}}^{d^{4-2\alpha} 2^{d}}d^{5\vee(3\kappa)\vee(4+\frac{3}{2\kappa}) \vee (\frac{3}{2}+\frac{3}{2\kappa}+\frac{3\kappa}{2})}({\mathrm{e}}^{-\rho t_{n}}+\gamma_{n}^{1+\frac{1}{\alpha}-\frac{1}{\kappa}}) (1+|x|^{\kappa}) \\&\leqslant&\frac{C}{\rho-\omega}\,{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_1}{\mathrm{e}}^{d^{4-2\alpha} 2^{d}} d^{5\vee(3\kappa)\vee(4+\frac{3}{2\kappa}) \vee (\frac{3}{2}+\frac{3}{2\kappa}+\frac{3\kappa}{2})} \gamma_{n}^{1+\frac{1}{\alpha}-\frac{1}{\kappa}} (1+|x|^{\kappa}).\end{eqnarray*}

The proof is complete.

2.4. Proof of Theorem 3

By (2.15), we can write

\begin{eqnarray*}W_1(\mathcal{L}(X_{t_n}^x), \mathcal{L}(\tilde{Y}_{t_n}^x))&\leqslant&\sum_{i=1}^{n-1}\text{supp}_{h\in{\rm Lip}(1)}\left|\tilde{Q}_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - \tilde{Q}_{t_{i-1},t_{i}} ) \circ P_{t_{i},t_n} h(x)\right|\\&&+\:\text{supp}_{h\in{\rm Lip}(1)}\left|\tilde{Q}_{0,t_{n-1}} \circ ( P_{t_{n-1},t_{n}} - \tilde{Q}_{t_{n-1},t_{n}} )h(x)\right|,\end{eqnarray*}

and by the same argument as in the proof of (2.19) but with Lemma 3 replaced by Lemma 4, we have

\begin{align*}&\sum_{i=1}^{n-1} \text{supp}_{h\in{\rm Lip}(1)} \left|\tilde{Q}_{0,t_{i-1}} \circ ( P_{t_{i-1},t_{i}} - \tilde{Q}_{t_{i-1},t_{i}} ) \circ P_{t_i,t_n} h(x) \right|\\&\quad\leqslant\frac{C}{\rho-\omega}\,{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_1}{\mathrm{e}}^{d^{4-2\alpha} 2^{d}}d^{\frac{5(1+\kappa)}{2\kappa}\vee(\frac{11-\alpha}{2})}(1+|x|) \gamma_{n}^{\frac{2-\alpha}{\alpha}}.\end{align*}

In addition, (2.4) and (2.6) give

\begin{align*}& \text{supp}_{h\in{\rm Lip}(1)}\left|\tilde{Q}_{0,t_{n-1}} \circ ( P_{t_{n-1},t_{n}} - \tilde{Q}_{t_{n-1},t_{n}} )h(x)\right|\\&\quad\leqslant \text{supp}_{h\in{\rm Lip}(1)}\left|\mathbb{E}\left[h\left(X_{t_{n-1},t_{n}}^{\tilde{Y}_{t_{n-1}}^{x}}\right)-h\left(\tilde{Y}_{t_{n-1}}^{x}\right)\right]\right|+ \text{supp}_{h\in{\rm Lip}(1)} \left|\mathbb{E}\left[h\left(\tilde{Y}_{t_{n-1},t_{n}}^{\tilde{Y}_{t_{n-1}}^{x}}\right)-h\left(\tilde{Y}_{t_{n-1}}^{x}\right)\right]\right|\\&\quad\leqslant \mathbb{E}\left|X_{t_{n-1},t_{n}}^{\tilde{Y}_{t_{n-1}}^{x}}-\tilde{Y}_{t_{n-1}}^{x}\right|+\mathbb{E}\left|\gamma_{n}b(\tilde{Y}_{t_{n-1}})+\frac{\gamma_{n}^{\frac{1}{\alpha}}}{\beta}A\tilde{Z}_{n}\right|\\&\quad\leqslant Cd^{\frac{5}{2}} \left(1+\mathbb{E}|\tilde{Y}_{t_{n-1}}^{x}|\right)\gamma_{n}^{\frac{1}{\alpha}}\ \leqslant \ Cd^{5} (1+|x|)\gamma_{n}^{\frac{1}{\alpha}}.\end{align*}

Combining the above, we obtain

\[W_{1}(\mathcal{L}(X_{t_n}^x), \mathcal{L}(\tilde{Y}_{t_n}^x))\leqslant\frac{C}{\rho-\omega}\,{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_1}{\mathrm{e}}^{d^{4-2\alpha} 2^{d}}d^{5}(1+|x|) \gamma_{n}^{\frac{2-\alpha}{\alpha}}.\]

This proves (1.17).

Next, we prove (1.18). By (2.14), (1.17) and (2.9), we have

\begin{eqnarray*}W_{1}(\nu, \mathcal{L}(\tilde{Y}_{t_n}^x))&\leqslant& W_{1}(\nu,\mathcal{L}(X_{t_n}^x))+W_{1}(\mathcal{L}(X_{t_n}^x), \mathcal{L}(\tilde{Y}_{t_n}^x))\\&\leqslant&\frac{C}{\rho-\omega}\,{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_1}{\mathrm{e}}^{d^{4-2\alpha} 2^{d}}d^{5}({\mathrm{e}}^{-\rho t_{n}}+\gamma_{n}^{\frac{2-\alpha}{\alpha}}) (1+|x|)\\&\leqslant&\frac{C}{\rho-\omega}\,{\mathrm{e}}^{\frac{\rho-\omega}{2}\gamma_1}{\mathrm{e}}^{d^{4-2\alpha} 2^{d}}d^{5}\gamma_{n}^{\frac{2-\alpha}{\alpha}} (1+|x|).\end{eqnarray*}

The proof is complete.

3.1. Proof of Lemma 1

For any $\kappa \in [1,\alpha)$ , define

(3.1) \begin{eqnarray}V_{\kappa}(x) = (1+|x|^2)^{\frac{\kappa}{2}};\end{eqnarray}

then $|x|^{\kappa}\leqslant V_{\kappa}(x)\leqslant1+|x|^{\kappa}$ .

We first prove (2.2). Recall that the infinitesimal generator $\mathcal{A}$ of the process $(X_{t})_{t\geqslant 0}$ is

(3.2) \begin{eqnarray}\mathcal{L} f(x)= \langle b(x), \nabla f(x) \rangle + \Delta^{\frac{\alpha}{2},A}f(x),\end{eqnarray}

where

(3.3) \begin{eqnarray}\Delta^{\frac{\alpha}{2},A}f(x)&\,=\,& \int_{ \mathbb{R}^d \setminus \{0\} } \bigl[f(x+Az)-f(x)-\langle Az, \nabla f(x) \rangle 1_{\{|z|\leqslant 1\}}\bigr] \frac{d_{\alpha}}{|z|^{d+\alpha}} \,\mathrm{d} z.\end{eqnarray}

By Itô’s formula, (1.9) and the same argument as in the proof of (A.3) in [Reference Chen, Deng, Schilling and Xu6], we have

\begin{eqnarray*}\frac{\mathrm{d} \mathbb{E}\left[V_{\kappa}(X_{t}^{x})\right]}{\mathrm{d} t}&\,=\,&\mathbb{E}\left[\left\langle b\!\left(X_{t}^{x}\right),\nabla V_{\kappa}(X_{t}^{x})\right\rangle\right]+\mathbb{E}\left[\Delta^{\frac{\alpha}{2},A}V_{\kappa}(X_{t}^{x})\right]\\&\leqslant&-\frac{\theta_{1}}{2}\mathbb{E}\left[V_{\kappa}(X_{t}^{x})\right]+C_{1},\end{eqnarray*}

where

\begin{align*}C_{1}&=\theta_{1}\kappa+\kappa K+\theta_{1}^{1-\kappa}|b(0)|^{\kappa}\\&\quad+\frac{d_{\alpha}\kappa(3-\kappa)\sqrt{d}\sigma_{d-1}}{2(2-\alpha)}\|A\|_{{\rm op}}^{2}+\frac{d_{\alpha}\kappa\sigma_{d-1}}{\alpha-\kappa}\|A\|_{{\rm op}}^{\kappa}+\left(\frac{\theta_{1}}{4}\right)^{1-\kappa}\left(\frac{d_{\alpha}\sigma_{d-1}}{\alpha-1}\|A\|_{{\rm op}}\right)^{\kappa}.\end{align*}

This inequality, together with $X_{0}=x$ , implies

(3.4) \begin{eqnarray}\mathbb{E}\!\left[V_{\kappa}(X_{t}^{x})\right]\leqslant {\mathrm{e}}^{-\frac{\theta_{1}}{2}t}V_{\kappa}(x)+C_{1}\int_{0}^{t}{\mathrm{e}}^{-\frac{\theta_{1}}{2}(t-s)}\,\mathrm{d} s\leqslant V_{\kappa}(x)+\frac{2C_{1}}{\theta_{1}}.\end{eqnarray}

Hence, the desired result follows from the fact that

\begin{align*}C_{1}\leqslant Cd^{\frac{5\vee(3\kappa)}{2}}\end{align*}

for some constant $C>0$ , which is independent of d.

For the remaining two inequalities, we only give the proof of (2.3). Inequality (2.4) can be proved in the same way.

By Assumption 1 and the same argument as in the proof of Lemma 1.8 in [Reference Chen, Deng, Schilling and Xu6], we have

\begin{align*}\mathbb{E}\left[V_{\kappa}(Y_{t_{n+1}}^{x})\right]\leqslant\left(1-\frac{\theta_{1}}{2}\gamma_{n+1}\right)\mathbb{E}[V_{\kappa}(Y_{t_{n}}^x)]+C_{2}\gamma_{n+1},\end{align*}

where

\begin{align*}C_{2}&=\frac{\theta_{1}\kappa}{2} \left(\frac{2\gamma_{n+1}|b(0)|^{2}}{\theta_{1}}+2\gamma_{n+1}^{2}|b(0)|^{2}+1+2\gamma_{n+1} K\right)+ \frac{\kappa |b(0)|^{2}}{\theta_{1}}+2\kappa \gamma_{n+1} |b(0)|^{2}\\&\quad+\kappa K+d_{\alpha}\kappa\sigma_{d-1}\left(\frac{(3-\kappa)\sqrt{d}}{2(2-\alpha)}\|A\|_{{\rm op}}^{2}+\frac{1}{\alpha-\kappa}\|A\|_{{\rm op}}^{\kappa}+\frac{|b(0)|^{\kappa-1}}{\alpha-1}\|A\|_{{\rm op}}\right)\\&\quad+d_{\alpha}\kappa\sigma_{d-1}\frac{\mathbb{E}|Z_{1}|^{\kappa-1}}{\alpha-1}\|A\|_{{\rm op}}+\left(\frac{d_{\alpha}\sigma_{d-1}(1+ L^{\kappa-1})}{\alpha-1}\|A\|_{{\rm op}}\right)^{\kappa}\left(\frac{2}{\theta_{1}}\right)^{\kappa-1}.\end{align*}

By induction, we further have

\begin{align*}\mathbb{E}\left[V_{\kappa}(Y_{t_{n+1}}^{x})\right]&\leqslant\prod_{i=1}^{n+1}\left(1-\frac{\theta_{1}}{2}\gamma_{i}\right)V_{\kappa}(x)+C_{2}\sum_{i=1}^{n+1}\prod_{j=i+1}^{n+1}\left(1-\frac{\theta_{1}}{2}\gamma_{j}\right)\gamma_{i}\\&\leqslant V_{\kappa}(x)+C_{2}\sum_{i=1}^{n+1}\prod_{j=i+1}^{n+1}\left(1-\frac{\theta_{1}}{2}\gamma_{j}\right)\gamma_{i},\end{align*}

where we have used the convention that $\prod_{j=n+2}^{n+1}\left(1-\frac{\theta_{1}}{2}\gamma_{j}\right)=1$ . In addition,

\begin{align*}\frac{\theta_{1}}{2}\sum_{i=1}^{n+1}\prod_{j=i+1}^{n+1}\left(1-\frac{\theta_{1}}{2}\gamma_{j}\right)\gamma_{i}=1-\prod_{j=1}^{n+1}\left(1-\frac{\theta_{1}}{2}\gamma_{j}\right)\leqslant 1,\end{align*}

which implies

\begin{align*}\mathbb{E}\left[V_{\kappa}(Y_{t_{n+1}}^{x})\right]\leqslant V_{\kappa}(x)+\frac{2C_{2}}{\theta_{1}}.\end{align*}

Since $V_{\kappa}(x)\leqslant1+|x|^{\kappa}$ , the desired result follows from the fact that

\begin{align*}C_{2}\leqslant Cd^{\frac{5\vee(3\kappa)}{2}}\end{align*}

for some constant $C>0$ , which is independent of d.

3.2. Proof of Lemma 2

The first inequality follows immediately from

\begin{align*}\mathbb{E} |Y_{t_{n-1},t_{n}}^{x}-x|^{\kappa}&= \mathbb{E}\left|\gamma_{n} b(x) + A\left(Z_{t_{n}}-Z_{t_{n-1}}\right)\right|^{\kappa}\\&\leqslant 2\left[\gamma_{n}^{\kappa}|b(x)|^{\kappa}+\|A\|_{{\rm op}}^{\kappa} \mathbb{E}|Z_{\gamma_{n}}|^{\kappa}\right]\leqslant Cd^{\kappa}(1+|x|^{\kappa})\gamma_{n}^{\frac{\kappa}{\alpha}}.\end{align*}

By Hölder’s inequality and (2.2), we obtain

\begin{eqnarray*}\mathbb{E}|X_{s,t}^{x}-x|^{\kappa}&\,=\,&\mathbb{E}\left|\int_{s}^{t}b\!\left(X_{s,r}^{x}\right)\mathrm{d} r+A\left(Z_{t}-Z_{s}\right)\right|^{\kappa}\\&\leqslant& 2\mathbb{E}\left|\int_{s}^{t}b\!\left(X_{s,r}^{x}\right)\mathrm{d} r\right|^{\kappa}+2\|A\|_{{\rm op}}^{\kappa}|Z_{t-s}|^{\kappa}\\&\leqslant & 2(t-s)^{\kappa-1}\int_{s}^{t}\mathbb{E}\left|b\!\left(X_{s,r}^{x}\right)\right|^{\kappa}\,\mathrm{d} r+2d^{\frac{\kappa}{2}}\mathbb{E}|Z_{1}|^{\kappa}(t-s)^{\frac{\kappa}{\alpha}}\\&\leqslant & Cd^{\frac{5\vee(3\kappa)}{2}}(1+|x|^{\kappa})(t-s)^{\frac{\kappa}{\alpha}},\end{eqnarray*}

which implies the second inequality.

For the last inequality, notice that $\gamma_{n}=t_{n}-t_{n-1}$ , Hölder’s inequality, (1.13) and (2.6) imply

\begin{eqnarray*}\mathbb{E} |X_{t_{n-1},t_{n}}^{x}-Y_{t_{n-1},t_{n}}^{x}|^{\kappa}&\,=\,&\mathbb{E} \left|\int_{t_{n-1}}^{t_{n}}\left[b\!\left(X_{t_{n-1},s}^{x}\right)-b(x)\right]\mathrm{d} s\right|^{\kappa}\\&\leqslant& \gamma_{n}^{\kappa-1}\int_{t_{n-1}}^{t_{n}}\mathbb{E}\left|b\!\left(X_{t_{n-1},s}^{x}\right)-b(x)\right|^{\kappa}\,\mathrm{d} s\\&\leqslant& L\gamma_{n}^{\kappa-1}\int_{t_{n-1}}^{t_{n}}\mathbb{E}\left|X_{t_{n-1},s}^{x}-x\right|^{\kappa}\,\mathrm{d} s\\&\leqslant & Cd^{\frac{5\vee(3\kappa)}{2}}(1+|x|^{\kappa})\gamma_{n}^{\kappa-1}\int_{t_{n-1}}^{t_{n}}(s-t_{n-1})^{\frac{\kappa}{\alpha}}\,\mathrm{d} s\\&\,=\,& Cd^{\frac{5\vee(3\kappa)}{2}}(1+|x|^{\kappa})\gamma_{n}^{\kappa+\frac{\kappa}{\alpha}}.\end{eqnarray*}

The proof is complete.

3.3. Proof of Lemma 3

By the Taylor expansion formula, we have

\begin{align*}& \mathbb{E} f\!\left(X_{t_{k-1},t_{k}}^{x}\right)-\mathbb{E} f\!\left(Y_{t_{k-1},t_{k}}^{x}\right)\\&\quad=\mathbb{E}\left\langle\nabla f\!\left(Y_{t_{k-1},t_{k}}^{x}\right),X_{t_{k-1},t_{k}}^{x}-Y_{t_{k-1},t_{k}}^{x}\right\rangle\\&\qquad+\mathbb{E}\int_{0}^{1}\!\!\left\langle\nabla f\!\left(Y_{t_{k-1},t_{k}}^{x}+r\left(X_{t_{k-1},t_{k}}^{x}-Y_{t_{k-1},t_{k}}^{x}\right)\right)\!-\!\nabla f\!\left(Y_{t_{k-1},t_{k}}^{x}\right),X_{t_{k-1},t_{k}}^{x}-Y_{t_{k-1},t_{k}}^{x}\right\rangle\,\mathrm{d} r\\&\quad=:\mathcal{I}+\mathcal{II}+\mathcal{III},\end{align*}

where

\begin{align*}\mathcal{I}&=\mathbb{E}\left\langle\nabla f\!\left(x+\gamma_{k}b(x)+A(Z_{t_{k}}-Z_{t_{k-1}})\right)-\nabla f\!\left(x+\gamma_{k}b(x)\right),X_{t_{k-1},t_{k}}^{x}-Y_{t_{k-1},t_{k}}^{x}\right\rangle, \\\mathcal{II}&=\mathbb{E}\left\langle\nabla f\!\left(x+\gamma_{k}b(x)\right),X_{t_{k-1},t_{k}}^{x}-Y_{t_{k-1},t_{k}}^{x}\right\rangle, \\\mathcal{III}&=\mathbb{E}\int_{0}^{1}\left\langle\nabla f\!\left(Y_{t_{k-1},t_{k}}^{x}+r\left(X_{t_{k-1},t_{k}}^{x}-Y_{t_{k-1},t_{k}}^{x}\right)\right)-\nabla f\!\left(Y_{t_{k-1},t_{k}}^{x}\right),X_{t_{k-1},t_{k}}^{x}-Y_{t_{k-1},t_{k}}^{x}\right\rangle\mathrm{d} r.\end{align*}

We divide the first term, $\mathcal{I}$ , into two parts as follows:

\begin{align*} \mathcal{I} &=\mathbb{E}\left[\left\langle\nabla f\!\left(x+\gamma_{k}b(x)+A(Z_{t_{k}}-Z_{t_{k-1}})\right)-\nabla f\!\left(x+\gamma_{k}b(x)\right),X_{t_{k-1},t_{k}}^{x}-Y_{t_{k-1},t_{k}}^{x}\right\rangle\right.\\ &\left.\qquad\qquad\qquad\qquad\times\left({\bf 1}_{(0,1]}(|Z_{t_{k}}-Z_{t_{k-1}}|)+{\bf 1}_{(1,\infty)}(|Z_{t_{k}}-Z_{t_{k-1}}|)\right)\right]\\ & =:\mathcal{I}_{1}+\mathcal{I}_{2}.\end{align*}

According to [Reference Chen, Nourdin, Xu and Yang7, Lemma 18], it is straightforward to derive the following estimates:

\begin{align*}\mathbb{P}\!\left(|Z_{\gamma_{k}}| > 1\right) &\leqslant C\pi^{-\frac{d}{2}}\Gamma\!\left(\frac{d+\alpha}{2}\right)\int_{|z|\geqslant\gamma_{k}^{-\frac{1}{\alpha}}}\frac{1}{|z|^{\alpha+d}}\,\mathrm{d} z\\&\leqslant C\pi^{-\frac{d}{2}}\Gamma\!\left(\frac{d+\alpha}{2}\right)\sigma_{d-1}\int_{\gamma_{k}^{-\frac{1}{\alpha}}}^{\infty}\frac{1}{r^{\alpha+1}}\,\mathrm{d} r\leqslant C\frac{\Gamma\!\left(\frac{d+\alpha}{2}\right)}{\Gamma\!\left(\frac{d}{2}\right)}\gamma_{k}\leqslant Cd\gamma_{k}\end{align*}

and, for any $\lambda>\alpha$ ,

\begin{align*}\mathbb{E}\left[|Z_{\gamma_{k}}|^\lambda \textbf{1}_{(0,1]}(|Z_{\gamma_{k}}|)\right]&\leqslant C\pi^{-\frac{d}{2}}\Gamma\!\left(\frac{d+\alpha}{2}\right)\gamma_{k}^{\frac{\lambda}{\alpha}}\int_{|z|\leqslant\gamma_{k}^{-\frac{1}{\alpha}}}\frac{|z|^{\lambda}}{|z|^{\alpha+d}}\,\mathrm{d} z\\&\leqslant C\pi^{-\frac{d}{2}}\Gamma\!\left(\frac{d+\alpha}{2}\right)\sigma_{d-1}\gamma_{k}^{\frac{\lambda}{\alpha}}\int_{0}^{\gamma_{k}^{-\frac{1}{\alpha}}}\frac{r^{\lambda}}{r^{\alpha+1}}\,\mathrm{d} r\\&\leqslant C\frac{\Gamma\!\left(\frac{d+\alpha}{2}\right)}{\Gamma\!\left(\frac{d}{2}\right)}\gamma_{k}\leqslant Cd\gamma_{k}.\end{align*}

Since $\frac{\kappa}{\kappa-1} > \alpha$ , we can use Hölder’s inequality and (2.7) to get

\begin{align*} |\mathcal{I}_{1}| &\leqslant \|\nabla^{2}f\|_{\mathrm{HS},\infty}\|A\|_{{\rm op}}\mathbb{E}\left[|Z_{t_{k}}-Z_{t_{k-1}}|{\bf 1}_{(0,1]}(|Z_{t_{k}}-Z_{t_{k-1}}|)\left|X_{t_{k-1},t_{k}}^{x}-Y_{t_{k-1},t_{k}}^{x}\right|\right]\\ &\leqslant\|\nabla^{2}f\|_{\mathrm{HS},\infty}\sqrt{d} \left(\mathbb{E}\left[|Z_{\gamma_{k}}|^{\frac{\kappa}{\kappa-1}}{\bf 1}_{(0,1]}(|Z_{\gamma_{k}}|)\right]\right)^{\frac{\kappa-1}{\kappa}} \left(\mathbb{E}\left|X_{t_{k-1},t_{k}}^{x}-Y_{t_{k-1},t_{k}}^{x}\right|^{\kappa}\right)^{\frac{1}{\kappa}}\\ &\leqslant C (1+|x|)\|\nabla^{2}f\|_{\mathrm{HS},\infty} d^{(\frac{3}{2}+\frac{3}{2\kappa})\vee (3-\frac{1}{\kappa})} \gamma_{k}^{2+\frac{1}{\alpha}-\frac{1}{\kappa}},\end{align*}

and Hölder’s inequality gives

\begin{align*} |\mathcal{I}_{2}| &\leqslant2\|\nabla f\|_{\infty} \mathbb{E}\left[{\bf 1}_{(1,\infty)}(|Z_{t_{k}}-Z_{t_{k-1}}|)\left|X_{t_{k-1},t_{k}}^{x}-Y_{t_{k-1},t_{k}}^{x}\right|\right]\\ &\leqslant2\|\nabla f\|_{\infty} \left(\mathbb{E}{\bf 1}_{(1,\infty)}(|Z_{\gamma_{k}}|)\right)^{\frac{\kappa-1}{\kappa}} \left(\mathbb{E}\left|X_{t_{k-1},t_{k}}^{x}-Y_{t_{k-1},t_{k}}^{x}\right|^{\kappa}\right)^{\frac{1}{\kappa}}\\ &\leqslant C (1+|x|) \|\nabla f\|_{\infty}d^{(1+\frac{3}{2\kappa})\vee (\frac{5}{2}-\frac{1}{\kappa})}\gamma_{k}^{\frac{\kappa-1}{\kappa}}\gamma_{k}^{1+\frac{1}{\alpha}}.\end{align*}

Hence, we have

\begin{align*} |\mathcal{I}| \leqslant C (1+|x|)\left(\|\nabla f\|_{\infty}+\|\nabla^{2}f\|_{\mathrm{HS},\infty}\right) d^{(\frac{3}{2}+\frac{3}{2\kappa})\vee (3-\frac{1}{\kappa})} \gamma_{k}^{2+\frac{1}{\alpha}-\frac{1}{\kappa}}.\end{align*}

For $\mathcal{II}$ , one can deduce from (1.1), (1.3), Itô’s formula and (3.2) that

\begin{align*} |\mathcal{II}| &=\left|\mathbb{E}\left\langle\nabla f\!\left(x+\gamma_{k} b(x)\right),\int_{t_{k-1}}^{t_{k}}\left[b(X_{t_{k-1},s}^{x})-b(x)\right] \mathrm{d} s\right\rangle\right|\\ &=\left|\left\langle\nabla f\!\left(x+\gamma_{k} b(x)\right),\int_{t_{k-1}}^{t_{k}}\mathbb{E}\left[b(X_{t_{k-1},s}^{x})-b(x)\right] \mathrm{d} s\right\rangle\right|\\ &=\left|\left\langle\nabla f\!\left(x+\gamma_{k} b(x)\right),\int_{t_{k-1}}^{t_{k}}\int_{t_{k-1}}^{s}\mathbb{E}\left[\mathcal{L}b(X_{t_{k-1},r}^{x})\right] \mathrm{d} r\,\mathrm{d} s\right\rangle\right|\\ &\leqslant C\|\nabla f\|_{\infty} (1+|x|)d^{\frac{5\vee(3\kappa)}{2}} \int_{t_{k-1}}^{t_{k}}(s-t_{k-1})\,\mathrm{d} s \nonumber \\ &\leqslant C\|\nabla f\|_{\infty} (1+|x|)d^{\frac{5\vee(3\kappa)}{2}}\gamma_{k}^{2},\end{align*}

where the second-to-last inequality is from (1.13), the moment estimate (2.2) and the inequality

\begin{eqnarray*}\left|\mathbb{E}\left[\mathcal{L}b(X_{t_{k-1},r}^{x})\right]\right|&\leqslant&\mathbb{E}\left[\left|\nabla b(X_{t_{k-1},r}^{x})\right|\left|b(X_{t_{k-1},r}^{x})\right|\right]+\mathbb{E}\left|\Delta^{\frac{\alpha}{2},A}b(X_{t_{k-1},r}^{x})\right|\\&\leqslant&C\left[1+\mathbb{E}\left|X_{t_{k-1},r}^{x}\right|+d_{\alpha}\sigma_{d-1}(\sqrt{d}\|A\|_{{\rm op}}^{2}+\|A\|_{{\rm op}})\right]\\&\leqslant& C(1+|x|)d^{\frac{5\vee(3\kappa)}{2}},\end{eqnarray*}

which is proved by the same argument as in the proof of (A.2) in [Reference Chen, Deng, Schilling and Xu6].

Finally, by [Reference Chen, Deng, Schilling and Xu6, Lemma 2.6] and (2.7), we have

\begin{align*} \mathcal{III} &\leqslant C\left(\|\nabla f\|_{\infty}+\|\nabla^{2}f\|_{\mathrm{HS},\infty}\right)\mathbb{E}\left|X_{t_{k-1},t_{k}}^{x}-Y_{t_{k-1},t_{k}}^{x}\right|^{\kappa}\\ &\leqslant C (1+|x|^{\kappa}) \left(\|\nabla f\|_{\infty}+\|\nabla^{2}f\|_{\mathrm{HS},\infty}\right) d^{\frac{5\vee(3\kappa)}{2}} \gamma_{k}^{\kappa+\frac{\kappa}{\alpha}}.\end{align*}

This finishes the proof.

3.4. Proof of Lemma 4

Before proving Lemma 4, we first present the following lemma.

Lemma 8 Let $\alpha\in(1,2)$ and let $\Delta^{\frac{\alpha}{2},A}$ be defined by (3.3). Then, for any $f:\mathbb{R}^{d}\to \mathbb{R}$ satisfying $\|\nabla f\|_{\infty}<\infty$ and $\|\nabla^{2}f\|_{\mathrm{HS},\infty}<\infty$ , we have

(3.5) \begin{eqnarray} \left|\Delta^{\frac{\alpha}{2},A}f(x) -\Delta^{\frac{\alpha}{2},A}f(y)\right| \leqslant\frac{d_{\alpha}\|\nabla^{2}f\|_{\mathrm{HS},\infty}\|A\|_{{\rm op}}^{2}\sigma_{d-1}}{(2-\alpha)(\alpha-1)}\,|x-y|^{2-\alpha}\quad \forall x,y\in\mathbb{R}^{d}.\end{eqnarray}

Proof. According to (3.3) and the symmetry of the Lebesgue measure, for any $R>0$ we have

\begin{align*}\Delta^{\frac{\alpha}{2},A}f(x)& =d_{\alpha}\int_{\mathbb{S}^{d-1}}\int_{0}^{\infty}\frac{f(x+Ar\theta)-f(x)-Ar\left\langle\theta,\nabla f(x)\right\rangle{\bf 1}_{(0,R)}(r)}{r^{\alpha+1}}\,\mathrm{d} r\,\mathrm{d}\theta\\&= d_{\alpha}\int_{\mathbb{S}^{d-1}} \int_{0}^{R}\int_{0}^{r}\frac{\left\langle A\theta,\nabla f(x+A\theta s)-\nabla f(x)\right\rangle}{r^{\alpha+1}}\,\mathrm{d} s\,\mathrm{d} r\,\mathrm{d} \theta\\ &\quad\mbox{}+d_{\alpha}\int_{\mathbb{S}^{d-1}} \int_{R}^{\infty}\int_{0}^{r}\frac{\left\langle A\theta,\nabla f(x+A\theta s)\right\rangle}{r^{\alpha+1}}\,\mathrm{d} s\,\mathrm{d} r\,\mathrm{d} \theta.\end{align*}

Then, for all $x,y\in\mathbb{R}^{d}$ ,

(3.6) \begin{align}& \big|\Delta^{\frac{\alpha}{2},A}f(x) -\Delta^{\frac{\alpha}{2},A}f(y)\big| \nonumber\\&\quad\leqslant d_{\alpha}\|A\|_{{\rm op}}\int_{\mathbb{S}^{d-1}} \int_{0}^{R}\int_{0}^{r}\frac{\left|\nabla f(x+A\theta s)-\nabla f(x)-\nabla f(y+A\theta s)+\nabla f(y)\right|}{r^{\alpha+1}}\,\mathrm{d} s\,\mathrm{d} r\,\mathrm{d} \theta \nonumber\\&\qquad+d_{\alpha}\|A\|_{{\rm op}}\int_{\mathbb{S}^{d-1}} \int_{R}^{\infty}\int_{0}^{r}\frac{\left|\nabla f(x+A\theta s)-\nabla f(y+A\theta s)\right|}{r^{\alpha+1}}\,\mathrm{d} s\,\mathrm{d} r\,\mathrm{d} \theta.\end{align}

For the first integral in (3.6), we have

\begin{align*}& \int_{\mathbb{S}^{d-1}} \int_{0}^{R}\int_{0}^{r}\frac{\left|\nabla f(x+A\theta s)-\nabla f(x)-\nabla f(y+A\theta s)+\nabla f(y)\right|}{r^{\alpha+1}}\,\mathrm{d} s\,\mathrm{d} r\,\mathrm{d} \theta\\&\quad\leqslant \int_{\mathbb{S}^{d-1}} \int_{0}^{R}\int_{0}^{r}\frac{\left|\nabla f(x+A\theta s)-\nabla f(x)\right|+\left|\nabla f(y+A\theta s)-\nabla f(y)\right|}{r^{\alpha+1}}\,\mathrm{d} s\,\mathrm{d} r\,\mathrm{d} \theta\\&\quad\leqslant 2\|\nabla^{2}f\|_{\mathrm{HS},\infty}\|A\|_{{\rm op}}\int_{\mathbb{S}^{d-1}} \int_0^R \int_{0}^{r}\frac{s}{r^{\alpha+1}}\,\mathrm{d} s\,\mathrm{d} r\,\mathrm{d}\theta\ = \ \frac{\|\nabla^{2}f\|_{\mathrm{HS},\infty}\|A\|_{{\rm op}}\sigma_{d-1}}{2-\alpha}\,R^{2-\alpha}.\end{align*}

For the second integral in (3.6), we have

\begin{align*}&\int_{\mathbb{S}^{d-1}} \int_{R}^{\infty}\int_{0}^{r}\frac{\left|\nabla f(x+A\theta s)-\nabla f(y+A\theta s)\right|}{r^{\alpha+1}}\,\mathrm{d} s\,\mathrm{d} r\,\mathrm{d} \theta\\&\quad\leqslant \|\nabla^{2}f\|_{\mathrm{HS},\infty}\|A\|_{{\rm op}}\int_{\mathbb{S}^{d-1}}\int_{R}^{\infty}\int_{0}^{r}\frac{|x-y|}{r^{\alpha+1}}\,\mathrm{d} s\,\mathrm{d} r\,\mathrm{d}\theta\\&\quad= \frac{\|\nabla^{2}f\|_{\mathrm{HS},\infty}\|A\|_{{\rm op}}\sigma_{d-1}}{\alpha-1}\,|x-y|R^{1-\alpha}.\end{align*}

Hence, the assertion follows upon taking $R=|x-y|$ .

Now, we proceed to prove Lemma 4.

Proof of Lemma 4. From (1.1) and (1.5), we see that

\begin{align*}&\mathbb{E} [f(X_{t_{k-1},t_{k}}^{x})-f(\tilde{Y}_{t_{k-1},t_{k}}^{x})]\\&\quad=\mathbb{E} \left[f\!\left(x+\int_{t_{k-1}}^{t_{k}}b(X_{t_{k-1},r}^{x})\,\mathrm{d} r +A(Z_{t_{k}}-Z_{t_{k-1}})\right)- f\!\left(x+\gamma_{k}b(x) +\frac{\gamma_{k}^{\frac{1}{\alpha}}}{\beta}A\tilde{Z}\right)\right]\\&\quad=\mathcal{J}_{1}+\mathcal{J}_{2},\end{align*}

where

\begin{align*} \mathcal{J}_{1} \,:\!=\,\mathbb{E} \left[f\!\left(x+\int_{t_{k-1}}^{t_{k}} b(X_{t_{k-1},r}^{x})\,\mathrm{d} r+AZ_{\gamma_{k}}\right) -f\!\left(x+\gamma_{k}b(x)+AZ_{\gamma_{k}}\right)\right]\end{align*}

and

\begin{align*}\mathcal{J}_{2}&\,:\!=\, \mathbb{E} \left[f\!\left(x+\gamma_{k}b(x)+AZ_{\gamma_{k}}\right)-f\!\left(x+\gamma_{k} b(x)\right)\right]\\&\quad -\mathbb{E} \left[f\!\left(x+\gamma_{k} b(x)+\frac{\gamma_{k}^{\frac{1}{\alpha}}}{\beta} A\tilde{Z}\right)-f\!\left(x+\gamma_{k} b(x)\right)\right].\end{align*}

We can bound $\mathcal{J}_{1}$ by using (2.6) with $\kappa=1$ :

\begin{align*}|\mathcal{J}_{1}|&\leqslant\|\nabla f\|_{\infty}\mathbb{E} \left|\int_{t_{k-1}}^{t_{k}}b(X_{t_{k-1},r}^{x})\,\mathrm{d} r-\gamma_{k} b(x)\right|\\&\leqslant\|\nabla f\|_{\infty}\int_{t_{k-1}}^{t_{k}}\mathbb{E} |b(X_{t_{k-1},r}^{x})-b(x)|\,\mathrm{d} r\\&\leqslant L\|\nabla f\|_{\infty}\int_{t_{k-1}}^{t_{k}}\mathbb{E} |X_{t_{k-1},r}^{x}-x|\,\mathrm{d} r\\&\leqslant C(1+|x|)\|\nabla f\|_{\infty}d^{\frac{5}{2\kappa} \vee \frac{3}{2}} \int_{t_{k-1}}^{t_{k}}(r-t_{k-1})^{\frac{1}{\alpha}}\,\mathrm{d} r \\&\leqslant C(1+|x|)\| \, \nabla f\|_{\infty}d^{\frac{5}{2\kappa} \vee \frac{3}{2}} \gamma_{k}^{1+\frac{1}{\alpha}}.\end{align*}

For the first term of $\mathcal{J}_{2}$ , we use Dynkin’s formula (see e.g. [Reference Chen and Xu8]) to get

\begin{align*}\mathbb{E} \big[f\big(x+\gamma_{k}b(x)+AZ_{\gamma_{k}}\big)-f\big(x+\gamma_{k} b(x)\big)\big]= \int_{0}^{\gamma_{k}}\mathbb{E} \big[\Delta^{\frac{\alpha}{2},A}f\big(x+\gamma_{k} b(x)+Z_{r}\big)\big]\,\mathrm{d} r.\end{align*}

For the second part of $\mathcal{J}_{2}$ , observing that $d_{\alpha}= \alpha \sigma_{d-1}^{-1} \beta^{-\alpha}$ , Taylor’s formula implies that

\begin{align*}& \mathbb{E}\left[f\!\left(x+\gamma_{k} b(x)+\frac{\gamma_{k}^{\frac{1}{\alpha}}}{\beta}A\tilde{Z}\right)-f\big(x+\gamma_{k} b(x)\big)\right]\\&\quad= \frac{\gamma_{k}^{\frac{1}{\alpha}}}{\beta}\mathbb{E}\left[\int_{0}^{1}\left\langle\nabla f\!\left(x+\gamma_{k} b(x)+\frac{\gamma_{k}^{\frac{1}{\alpha}}}{\beta}tA\tilde{Z}\right),\: A\tilde{Z}\right\rangle \mathrm{d} t\right]\\&\quad= \frac{\gamma_{k}^{\frac{1}{\alpha}}}{\beta} \int_{|z|\geqslant1} \int_{0}^{1} \alpha \left\langle\nabla f\big(x+\gamma_{k} b(x)+\frac{\gamma_{k}^{\frac{1}{\alpha}}} {\beta}tAz\big),\: Az\right\rangle \frac{\mathrm{d} t\,\mathrm{d} z}{\sigma_{d-1} |z|^{\alpha+d}}\\&\quad= \frac{\alpha\gamma_{k}}{\sigma_{d-1} \beta^{\alpha}}\int_{|z|\geqslant\beta^{-1}\gamma_{k}^{\frac{1}{\alpha}}}\int_{0}^{1}\left\langle\nabla f\big(x+\gamma_{k} b(x)+tAz\big),\:Az\right\rangle \frac{\mathrm{d} t\,\mathrm{d} z}{|z|^{\alpha+d}}\\&\quad= \gamma_{k}\Delta^{\frac{\alpha}{2},A}f(x+\gamma_{k} b(x))-\mathcal{R},\end{align*}

where

\begin{eqnarray*}\mathcal{R} \,:\!=\, \gamma_{k}d_{\alpha}\int_{|z|<\beta^{-1}\gamma_{k}^{\frac{1}{\alpha}}}\int_{0}^{1}\left\langle\nabla f\big(x+\gamma_{k}b(x)+tAz\big),\: Az\right\rangle\frac{\mathrm{d} t\,\mathrm{d} z}{|z|^{\alpha+d}}.\end{eqnarray*}

Together, the above estimates yield

\begin{eqnarray*}|\mathcal{J}_{2}|\leqslant |\mathcal{R}| + \left|\int_{0}^{\gamma_{k}}\mathbb{E} \big[\Delta^{\frac{\alpha}{2},A} f\big(x+\gamma_{k} b(x)+Z_{r}\big)\big]\,\mathrm{d} r-\gamma_{k}\Delta^{\frac{\alpha}{2},A}f(x+\gamma_{k} b(x))\right|.\end{eqnarray*}

Further, we have

\begin{eqnarray*}|\mathcal{R}|&\,=\,& \gamma_{k}d_{\alpha} \left|\int_{|z|<\beta^{-1}\gamma_{k}^{\frac{1}{\alpha}}}\int_{0}^{1} \left\langle\nabla f\big(x+\gamma_{k} b(x)+tAz\big)-\nabla f\big(x+\gamma_{k} b(x)\big),\: Az\right\rangle \frac{\mathrm{d} t\,\mathrm{d} z}{|z|^{\alpha+d}} \right|\\&\leqslant& d_{\alpha} \|A\|_{{\rm op}}\gamma_{k}\int_{|z|<\beta^{-1}\gamma_{k}^{\frac{1}{\alpha}}}\int_{0}^{1} \left|\nabla f\big(x+\gamma_{k} b(x)+tAz\big)-\nabla f\big(x+\gamma_{k} b(x)\big)\right| \frac{\mathrm{d} t\,\mathrm{d} z}{|z|^{\alpha+d-1}}\\&\leqslant& C\|\nabla^{2}f\|_{\mathrm{HS},\infty} d_{\alpha} \|A\|_{{\rm op}}^{2}\gamma_{k}\int_{|z|<\beta^{-1}\gamma_k^{\frac{1}{\alpha}}} \frac{\mathrm{d} z}{|z|^{\alpha+d-2}}\\&\leqslant& C\|\nabla^{2}f\|_{\mathrm{HS},\infty} d_{\alpha}\sigma_{d-1} \|A\|_{{\rm op}}^{2}\gamma_{k}^{\frac{2}{\alpha}}\leqslant C\|\nabla^{2}f\|_{\mathrm{HS},\infty}d^{2}\gamma_{k}^{\frac{2}{\alpha}}.\end{eqnarray*}

By Lemma 8, we also have

\begin{align*}&\left|\int_{0}^{\gamma_{k}}\mathbb{E}\left[\Delta^{\frac{\alpha}{2},A}f\big(x+\gamma_{k} b(x)+Z_{r}\big)\right] \mathrm{d} r- \gamma_{k}\Delta^{\frac{\alpha}{2},A}f(x+\gamma_{k} b(x))\right|\\&\quad\leqslant \int_{0}^{\gamma_{k}}\mathbb{E}\left|\Delta^{\frac{\alpha}{2},A}f\big(x+\gamma_{k} b(x)+Z_{r}\big)\big] -\Delta^{\frac{\alpha}{2},A}f(x+\gamma_{k} b(x))\right| \mathrm{d} r\\&\quad\leqslant C\|\nabla^{2}f\|_{\mathrm{HS},\infty}d^{2}\int_{0}^{\gamma_{k}}\mathbb{E}\left[|Z_{r}|^{2-\alpha}\right] \mathrm{d} r\\&\quad= C\|\nabla^{2}f\|_{\mathrm{HS},\infty}d^{2}\int_{0}^{\gamma_{k}}\mathbb{E}\left[|Z_{1}|^{2-\alpha}\right] r^{\frac{2}{\alpha}-1}\,\mathrm{d} r \leqslant C \|\nabla^{2}f\|_{\mathrm{HS},\infty}d^{3-\frac{\alpha}{2}} \,\gamma_{k}^{\frac{2}{\alpha}}.\end{align*}

Combining all the estimates finishes the proof.

3.5. Proof of Lemma 6

In this section, we prove Lemma 6 under a more general condition. Consider

(3.7) \begin{eqnarray}\mathrm{d} X_t = b(X_t) \,\mathrm{d} t + A \,\mathrm{d} Z_t, \quad X_0=x,\end{eqnarray}

where the drift coefficient b is assumed to satisfy the following condition: there exist positive constants $K_1,\,K_2$ and $L_0$ , all independent of the dimension d, such that

(3.8) \begin{eqnarray}\langle x-y, b(x)-b(y) \rangle\leqslant\left\{\begin{array}{l@{\quad}l@{\quad}l}K_1 |x-y|^2 & \text{if } |x-y|\leqslant L_0, \\-K_2 |x-y|^2 & \text{if } |x-y| > L_0.\end{array}\right.\end{eqnarray}

It is easy to verify that Assumption 1 satisfies this condition. The corresponding generator $\mathcal{L}$ for process $(X_t)_{t\geqslant 0}$ is defined as

\begin{eqnarray*}\mathcal{L} f(x) = \langle b(x), \nabla f(x) \rangle + \int_{ \mathbb{R}^d \setminus \{ 0 \} } [\,f\,(x+Az)-f(x)-\langle Az, \nabla f(x) \rangle1_{\{|z|\leqslant 1\}}] \frac{d_{\alpha}}{|z|^{d+\alpha}} \,\mathrm{d} z,\end{eqnarray*}

where $\alpha\in(1,2)$ . In addition, there exists a positive constant $\Lambda \asymp d^{\frac{1}{2}}$ with $\Lambda \geqslant 1$ such that

\begin{eqnarray*}\Lambda^{-1} \leqslant \| A \|_{\rm op} \vee \| A^{-1} \|_{\rm op} \leqslant \Lambda.\end{eqnarray*}

We shall use [Reference Liang and Wang27] to prove the contraction property in the lemma under the condition (3.8). Let us first recall some notation from [Reference Liang and Wang27]. For some sufficiently small constant $\tau \leqslant \frac{1}{32 \Lambda^5} \wedge L_0$ with $L_0$ and $\Lambda$ as above, for any $x,y \in \mathbb{R}^d$ define

\begin{eqnarray*}(x-y)_{\tau} = \left(1 \wedge \frac{\tau}{|x-y|} \right) (x-y),\end{eqnarray*}

and for any $z\in \mathbb{R}^d$ let $\Psi: \mathbb{R}^d \to \mathbb{R}^d$ be a continuous and bijective mapping defined by

\begin{eqnarray*}\Psi(z) = z + A^{-1} (x-y)_{\tau};\end{eqnarray*}

then the inverse function $\Psi^{-1}$ is given by

\begin{eqnarray*}\Psi^{-1}(z) = z - A^{-1} (x-y)_{\tau}.\end{eqnarray*}

Let the measure $\pi_0(\mathrm{d} z)$ be related to the Lévy measure $\pi(\mathrm{d} z)$ and satisfy $0\leqslant \pi_0(\mathrm{d} z) \leqslant \pi(\mathrm{d} z)$ and $\int_{\mathbb{R}^d} |z|^2 \pi_0(\mathrm{d} z)<\infty.$

For any integers $n\geqslant 1$ , define

\begin{eqnarray*}\mu_{\pi_0,\Psi} = \limsup_{n\to \infty} \mu_{\pi_0,n,\Psi}= \limsup_{n\to \infty} \,[\pi_{0,n} \wedge (\pi_{0,n} \Psi)],\end{eqnarray*}

where $\pi_{0,n}(B)=\int_{B\cap \{|z|>\frac{1}{n}\} }\pi_0(\mathrm{d} z)$ and $(\pi_{0,n}\Psi)(B)=\pi_{0,n}(\Psi(B))$ for all Borel-measurable sets $B\in \mathcal{B}(\mathbb{R}^d)$ .

For the solution $(X_t)_{t\geqslant 0}$ of the SDE (3.7) and its operator $\mathcal{L}$ , it follows from [Reference Liang and Wang27, Theorem 2.3 and Proposition 2.6] that there exist the coupling process for $(X_t)_{t\geqslant 0}$ and the coupling operator $\widetilde{\mathcal{L}}$ defined as follows: for any $h\in \mathcal{C}_b^2(\mathbb{R}^d \times \mathbb{R}^d)$ and $x,y\in \mathbb{R}^d$ ,

\begin{eqnarray*}\widetilde{\mathcal{L}} h(x,y)&\,=\,& \langle \nabla_x h(x,y),b(x) \rangle+\langle \nabla_y h(x,y),b(y) \rangle \\&& +\:\frac{1}{2} \int_{ \mathbb{R}^d \setminus \{ 0 \} } ( h(x+ Az,y+A \Psi(z))-h(x,y)-\langle \nabla_x h(x,y),A z \rangle 1_{\{ |z|\leqslant 1 \} } \\&& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\:\langle \nabla_y h(x,y), A \Psi(z) \rangle 1_{ \{ |\Psi(z)| \leqslant 1 \} } ) \mu_{\pi_0,\Psi}(\mathrm{d} z) \\&& +\:\frac{1}{2} \int_{ \mathbb{R}^d \setminus \{ 0 \} } ( h(x+A z,y+A \Psi^{-1}(z))-h(x,y)-\langle \nabla_x h(x,y),A z \rangle 1_{\{ |z|\leqslant 1 \} } \\&& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\:\langle \nabla_y h(x,y), A \Psi^{-1}(z) \rangle 1_{ \{ |\Psi^{-1}(z)| \leqslant 1 \} } ) \mu_{\pi_0,\Psi^{-1}}(\mathrm{d} z) \\&&+ \:\int_{ \mathbb{R}^d \setminus \{ 0 \} } ( h(x+A z,y+A z)-h(x,y)-\langle \nabla_x h(x,y),A z \rangle 1_{\{ |z|\leqslant 1 \} } \\&& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -\:\langle \nabla_y h(x,y), A z \rangle 1_{ \{ |z| \leqslant 1 \} } ) \left[ \pi- \frac{1}{2}\mu_{\pi_0,\Psi} -\frac{1}{2}\mu_{\pi_0,\Psi^{-1}} \right](\mathrm{d} z),\end{eqnarray*}

where $\nabla_x$ and $\nabla_y$ denote the gradients with respect to variables x and y, respectively.

Let $\pi_0(\mathrm{d} z) = q(z) \,\mathrm{d} z$ and $(\pi_0 \Psi^{-1})(\mathrm{d} z) = q_{\Psi}(z)\, \mathrm{d} z$ , and for $r\geqslant 0$ let

\begin{eqnarray*}J(r) = \inf_{x,y \in \mathbb{R}^d, |x-y|\leqslant r} \mu_{\pi_0,\Psi}(\mathbb{R}^d) = \inf_{x,y\in \mathbb{R}^d, |x-y|\leqslant r} \int_{\mathbb{R}^d} [q(z) \wedge q_{\Psi}(z) ] \,\mathrm{d} z.\end{eqnarray*}

For $r\in (0,2L_0]$ and constants $K_1$ and $K_2$ as in (3.8), define

\begin{eqnarray*}\psi(r) \leqslant \frac{1}{2r} J(\tau \wedge r)(\tau \wedge r)^2, \quad g_1(r) = \int_0^r \frac{1}{\psi(s)}\,\mathrm{d} s\end{eqnarray*}

and

\begin{eqnarray*}g(r) = \left(1+\frac{2K_1}{(2K_2) \wedge g_1^{-1}(2L_0)} \right)g_1(r).\end{eqnarray*}

The proof of exponential ergodicity is standard; for more details see [Reference Liang, Schilling and Wang26, Reference Liang and Wang27, Reference Wang46] and references therein. By applying Theorem 4.4 of [Reference Liang and Wang27], we prove the following exponential ergodicity for the process $(X_t)_{t\geqslant 0}$ under the Wasserstein-1 distance. Lemma 6 follows from this lemma.

Lemma 9. Let the process $(X_t)_{t\geqslant 0}$ be as in (3.7). Under condition (3.8), there exist positive constants $c_{1}$ and $c_{2}$ independent of d such that for any $x,y\in \mathbb{R}^d$ and $t>0$ ,

\begin{eqnarray*}W_1( \mathcal{L}(X_t^x), \mathcal{L}(X_t^y) )\leqslant c_{1}\exp( c_{2}d^{4-2\alpha}2^{d}) \exp(-c_{1}d^{2\alpha-4} 2^{-d} {\mathrm{e}}^{-c_{2}d^{4-2\alpha}2^d}t) |x-y|.\end{eqnarray*}

Proof. Recall that $\pi(\mathrm{d} z) = \frac{d_{\alpha}}{|z|^{d+\alpha}} \,\mathrm{d} z$ . We choose

\begin{eqnarray*}\pi_0 (\mathrm{d} z) = 1_{\{-\frac{1}{8} < z_1 < \frac{1}{8}, |z|\leqslant 1 \}} \frac{d_{\alpha}}{|z|^{d+\alpha}} \,\mathrm{d} z,\end{eqnarray*}

where $z_1$ is the first element of vector z, that is, $z=(z_1,z_2,\ldots,z_d)$ . Thus, we know that

\begin{align*}q(z) &= 1_{\{-\frac{1}{8} < z_1 < \frac{1}{8}, |z|\leqslant 1 \}} \frac{d_{\alpha}}{|z|^{d+\alpha}}, \\q_{\Psi}(z) &= 1_{\{-\frac{1}{8} < (\Psi^{-1}(z))_1 < \frac{1}{8}, |\Psi^{-1}(z)|\leqslant 1 \}} \frac{d_{\alpha}}{|\Psi^{-1}(z)|^{d+\alpha}}\end{align*}

satisfy

\begin{eqnarray*}q(z) \wedge q_{\Psi}(z) \geqslant d_{\alpha} 1_{\{-\frac{1}{8}< z_1< \frac{1}{8}, -\frac{1}{8} < (\Psi^{-1}(z))_1 < \frac{1}{8}, |z|\leqslant 1,|\Psi^{-1}(z)|\leqslant 1 \}} \left[\frac{1}{|z|^{d+\alpha}} \wedge \frac{1}{|\Psi^{-1}(z)|^{d+\alpha}} \right].\end{eqnarray*}

We give the estimate for the function J(r) first. For $x,y\in \mathbb{R}^d$ with $|\tilde{x}-\tilde{y}| \leqslant \tau$ , we consider the following two cases.

1. (i) If $(A^{-1}(\tilde{x}-\tilde{y}))_1 \leqslant 0$ , then after some calculations we get

   \begin{eqnarray*}q(z) \wedge q_{\Psi}(z) \geqslant \frac{d_{\alpha}}{2^{d+\alpha}} 1_{\{z_1>0, \Lambda|\tilde{x}-\tilde{y}|\leqslant |z| \leqslant \frac{1}{16\Lambda^2} \} } \frac{1}{|z|^{d+\alpha}},\end{eqnarray*}
   that is,
   \begin{eqnarray*}\int_{\mathbb{R}^d} [q(z) \wedge q_{\Psi}(z)] \,\mathrm{d} z&\,\geqslant& \frac{d_{\alpha} \sigma_{d-1}}{\alpha 2^{d+\alpha+1}} \left[ \frac{1}{(\Lambda|\tilde{x}-\tilde{y}|)^{\alpha}} - \frac{1}{(\frac{1}{16\Lambda^2})^{\alpha}}\right] \\&\,\geqslant& \frac{d_{\alpha} \sigma_{d-1}}{\alpha 2^{d+\alpha+1}}\left(\frac{1}{\Lambda^{\alpha}}-\frac{1}{2^{\alpha}\Lambda^{3\alpha}}\right) |\tilde{x}-\tilde{y}|^{-\alpha},\end{eqnarray*}
   where the last inequality holds provided $|\tilde{x}-\tilde{y}| \leqslant \tau \leqslant \frac{1}{32 \Lambda^5}$ .

2. (ii) If $(A^{-1}(\tilde{x}-\tilde{y}))_1>0$ , then

   \begin{eqnarray*}q(z) \wedge q_{\Psi}(z) \geqslant \frac{d_{\alpha}}{2^{d+\alpha}} 1_{\{ (\Psi^{-1}(z))_1>0, \Lambda|\tilde{x}-\tilde{y}|\leqslant |\Psi^{-1}(z)|\leqslant \frac{1}{16\Lambda^2} \} } \frac{1}{|\Psi^{-1}(z)|^{d+\alpha}},\end{eqnarray*}
   that is,
   \begin{eqnarray*}\int_{\mathbb{R}^d} [q(z) \wedge q_{\Psi}(z)] \,\mathrm{d} z&\,\geqslant& \frac{d_{\alpha} \sigma_{d-1}}{\alpha 2^{d+\alpha+1}} \left[ \frac{1}{(\Lambda|\tilde{x}-\tilde{y}|)^{\alpha}} - \frac{1}{(\frac{1}{16\Lambda^2})^{\alpha}}\right] \\&\,\geqslant& \frac{d_{\alpha} \sigma_{d-1}}{\alpha 2^{d+\alpha+1}} \left(\frac{1}{\Lambda^{\alpha}}-\frac{1}{2^{\alpha}\Lambda^{3\alpha}}\right) |\tilde{x}-\tilde{y}|^{-\alpha},\end{eqnarray*}
   where the last inequality holds provided $|\tilde{x}-\tilde{y}| \leqslant \tau \leqslant \frac{1}{32\Lambda^5}$ .

Combining the above two cases, for any $0 < r\leqslant \tau$ we know that

\begin{eqnarray*}\int_{\mathbb{R}^d} [q(z) \wedge q_{\Psi}(z)] \,\mathrm{d} z\geqslant \frac{d_{\alpha} \sigma_{d-1}}{\alpha 2^{d+\alpha+1}} \left(\frac{1}{\Lambda^{\alpha}}-\frac{1}{2^{\alpha}\Lambda^{3\alpha}}\right) |\tilde{x}-\tilde{y}|^{-\alpha} = \widehat{C}_1 |\tilde{x}-\tilde{y}|^{-\alpha}\end{eqnarray*}

with $\widehat{C}_1=\frac{d_{\alpha} \sigma_{d-1}}{\alpha 2^{d+\alpha+1}} (\frac{1}{\Lambda^{\alpha}}-\frac{1}{2^{\alpha}\Lambda^{3\alpha}})$ . Thus, for any $0 < r\leqslant \tau$ , we know that

\begin{eqnarray*}J(r) =\inf_{\tilde{x},\tilde{y}\in \mathbb{R}^d,|\tilde{x}-\tilde{y}|\leqslant r} \int_{\mathbb{R}^d} [q(z) \wedge q_{\Psi}(z)] \,\mathrm{d} z \geqslant \widehat{C}_1 r^{-\alpha}.\end{eqnarray*}

This implies that we can choose the function $\psi$ as follows:

\begin{eqnarray*}\psi(r)=\left\{\begin{array}{lll}\frac{\widehat{C}_1}{2} r^{1-\alpha} & \text{if } r\in (0,\tau], \\[4pt] \frac{\widehat{C}_1}{2}\tau^{2-\alpha} r^{-1} & \text{if } r\in (\tau,2L_0].\end{array}\right.\end{eqnarray*}

Then we know that

\begin{eqnarray*}g_1(2L_0) &\,=\,& \left(\int_0^{\tau} + \int_{\tau}^{2L_0} \right) \frac{1}{\psi(s)} \,\mathrm{d} s\ \ = \ \ 2\alpha^{-1} \widehat{C}_1^{-1} \tau^{\alpha}+ \tau^{\alpha-2} \widehat{C}_1^{-1} (4L_0^2 - \tau^2), \\g(2L_0) &\,=\,& \left(1+\frac{2K_1}{(2K_2) \wedge g_1^{-1}(2L_0)} \right)g_1(2L_0).\end{eqnarray*}

Using Theorem 4.4 from [Reference Liang and Wang27], we obtain that

\begin{eqnarray*}W_1( \mathcal{L}(X_t^x), \mathcal{L}(X_t^y) ) \leqslant \widehat{C} {\mathrm{e}}^{-\lambda t} |x-y|,\end{eqnarray*}

where

\begin{eqnarray*}\widehat{C} &\,=\,&\frac{1}{2}\left( 1+\exp\!\left\{ g(2L_0) [ (2K_2) \wedge g_1^{-1}(2L_0) ] \right\} \right) ,\\\lambda &\,=\,& \frac{(2K_2) \wedge g_1^{-1}(2L_0)}{1+\exp\!\left\{ g(2L_0)[ (2K_2) \wedge g_1^{-1}(2L_0) ] \right\} }.\end{eqnarray*}

Recall that $\widehat{C}_1=\frac{d_{\alpha} \sigma_{d-1}}{\alpha 2^{d+\alpha+1}} (\frac{1}{\Lambda^{\alpha}}-\frac{1}{2^{\alpha}\Lambda^{3\alpha}})$ , $d_{\alpha}= 2^{\alpha}\Gamma(\frac{d+\alpha}{2})\pi^{-\frac{d}{2}}|\Gamma(-\frac{\alpha}{2})|^{-1}$ and $\sigma_{d-1}=2\pi^{\frac{d}{2}} \Gamma^{-1}(\frac{d}{2})$ ; then there exists a positive constant C independent of d such that

\begin{eqnarray*}\widehat{C}_1 \leqslant Cd^{1-\frac{\alpha}{2}} 2^{-d}.\end{eqnarray*}

Moreover, there exist positive constants $c_{1}$ and $c_{2}$ both independent of d such that

\begin{eqnarray*}\widehat{C} \leqslant c_{1}\exp\!\left( c_{2}d^{4-2\alpha }2^{d} \right), \quad\lambda \geqslant c_{1}d^{2\alpha-4}2^{-d} \exp\!\left(-c_{2}d^{4-2\alpha} 2^d \right).\end{eqnarray*}

The proof is complete.

Now we are ready to prove Lemma 6.

Proof of Lemma 6. In (3.8), let $K_{1}= L$ , $K_{2}=\frac{\theta_{1}}{2}$ and $L_{0}=\sqrt{\frac{2K}{\theta_{1}}}$ . It follows from (1.8), (1.9) and Lemma 9 that

(3.9) \begin{eqnarray}W_1( \mathcal{L}(X_t^x), \mathcal{L}(X_t^y) ) \leqslant \widehat{C} {\mathrm{e}}^{-\lambda t} |x-y|.\end{eqnarray}

For a sequence $\{x_n\}_{n=1}^{\infty} \subseteq \mathbb{R}^d$ and x such that $\lim_{n\to\infty} x_n=x$ , it follows from (3.9) and the dominated convergence theorem that for any continuous bounded function $\phi$ ,

\begin{eqnarray*}\lim_{n\to \infty} P_{t} \phi(x_n)= \lim_{n\to \infty} \mathbb{E} [ \phi(X_t^{x_n}) ] = \mathbb{E}[ \phi(X_t^{x}) ] = P_{t} \phi(x),\end{eqnarray*}

which implies that the semigroup $P_{t}$ is Feller.

The existence of an invariant measure will be established below by using the Krylov–Bogolioubov theorem (see e.g. [Reference Hairer16, Theorem 1.10]). Let $X_0=x$ , and for any measurable set A and $T>0$ define

\begin{eqnarray*}\mu_{T}^{x}(A) = \frac{1}{T} \int_0^T P_{t}(x,A) \,\mathrm{d} t.\end{eqnarray*}

Since $P_{t}$ is Feller, it suffices to show that $\{ \mu_{T}^{x} \}_{T>0}$ is tight from the Krylov–Bogolioubov theorem. For the function $V_{\kappa}$ in (3.1), one gets the inequality (3.4); there exists some positive constant c such that

\begin{eqnarray*}\mu_T^{x} (V_{\kappa})= \frac{1}{T}\int_0^T \mathbb{E} V_{\kappa}(X_t^{x})\,\mathrm{d} t \leqslant c.\end{eqnarray*}

Since $V_{\kappa}(x)\geqslant 1$ and $V_{\kappa}(x)\to\infty$ as $|x|\to \infty$ , for any fixed $\delta>0$ there exists a compact set $K_0$ such that $V_{\kappa}(x)> \frac{c}{\delta}$ for $x\in K_0^c$ , where $K_0^c$ is the complement set of $K_0$ . Then we have

\begin{eqnarray*}\frac{c}{\delta}\mu_{T}^{x}(K_0^c) < \mu_{T}^{x} (V) \leqslant c,\end{eqnarray*}

that is, $\mu_{T}^{x} (K_0^c)<\delta$ , which implies $\mu_{T}^{x} (K_0)>1-\delta$ . So $\{\mu_{T}^{x} \}_{T>0}$ is tight. Hence the limit of a convergent subsequence $\mu_{T}^{x}$ as $T\to \infty$ provides an invariant measure of $(P_{t})_{t\geqslant 0}$ .

The inequality (3.9) implies that the invariant measure is unique. Indeed, let $\mu_1$ and $\mu_2$ be two invariant measures; for any function $h\in \mathrm{Lip}(1)$ , we have

\begin{eqnarray*}W_1 (\mu_1,\mu_2)&\,=\,&\text{supp}_{h\in {\rm Lip}(1)} \left\{\mu_1(h)-\mu_2(h)\right\} \\ &\,=\,& \text{supp}_{h\in {\rm Lip}(1)} \left\{\int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \mathbb{E} h( X_t^{x})\mu_1({\mathrm{d}} x) \mu_2({\mathrm{d}} y)-\int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \mathbb{E} h(X_t^{y}) \mu_1({\mathrm{d}} x) \mu_2({\mathrm{d}} y) \right\}\\&\leqslant& \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} \text{supp}_{h\in {\rm Lip}(1)} \{ \mathbb{E} h(X_t^{x})-\mathbb{E} h(X_t^{y}) \} \mu_1({\mathrm{d}} x) \mu_2({\mathrm{d}} y) \\&\,=\,& \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} W_1 (\mathcal{L}(X_t^{x})\mathcal{L}(X_t^{y})) \mu_1({\mathrm{d}} x) \mu_2({\mathrm{d}} y) \ \ \rightarrow \ \ 0\end{eqnarray*}

as $t \rightarrow \infty$ by (3.9). The proof is complete.

3.6. Proof of Lemma 7

For $n\geqslant1$ , set $u_{n}=\frac{v_{n}}{\gamma_{n}^{\theta}}$ . Then

(3.10) \begin{eqnarray}u_{n+1} = u_{n}\lambda_n+\gamma_{n+1},\end{eqnarray}

where $\lambda_n=(\frac{\gamma_{n}}{\gamma_{n+1}})^{\theta}{\mathrm{e}}^{-\rho\gamma_{n+1}}$ . Recall that $\omega = \limsup\limits_{k\to \infty} \frac{\gamma^{\theta}_k-\gamma^{\theta}_{k+1}}{\gamma_{k+1}^{1+\theta}}< +\infty$ and $\rho>\omega$ . Then for any constant $c\in(\omega,\rho)$ , there exists an $n_{0}\in\mathbb{N}$ such that for all $n\geqslant n_{0}$ ,

(3.11) \begin{eqnarray}\frac{\gamma^{\theta}_n}{\gamma^{\theta}_{n+1}}\ =\ 1+\frac{\gamma^{\theta}_{n} - \gamma^{\theta}_{n+1}}{\gamma^{1+\theta}_{n+1}} \gamma_{n+1}\ \leqslant \ 1+ c \gamma_{n+1}\ \leqslant \ {\mathrm{e}}^{c \gamma_{n+1}}.\end{eqnarray}

Hence

\begin{eqnarray*}\lambda_n \leqslant {\mathrm{e}}^{-(\rho-c)\gamma_{n+1}}.\end{eqnarray*}

This and (3.10) imply

\begin{eqnarray*}u_{n+1} \leqslant {\mathrm{e}}^{-(\rho-c)\gamma_{n+1}}u_{n}+\gamma_{n+1}.\end{eqnarray*}

Consequently,

\begin{eqnarray*}{\mathrm{e}}^{(\rho-c)t_{n+1}} u_{n+1}\leqslant {\mathrm{e}}^{(\rho-c)t_{n}}u_n + M {\mathrm{e}}^{(\rho-c)t_{n}}\gamma_{n+1},\end{eqnarray*}

where the constant $M=\text{supp}_{k\geqslant1}{\mathrm{e}}^{(\rho-c)\gamma_{k}}$ . Hence, by induction, we obtain

\begin{align*}{\mathrm{e}}^{(\rho-c)t_{n}} u_{n}\leqslant {\mathrm{e}}^{(\rho-c)t_{n_{0}}} u_{n_{0}}+M\int_{t_{n_{0}}}^{t_{n}}{\mathrm{e}}^{(\rho-c)u}\,{\mathrm{d}} u\leqslant {\mathrm{e}}^{(\rho-c)t_{n_{0}}}u_{n_{0}}+\frac{M}{\rho-c}{\mathrm{e}}^{(\rho-c)t_{n}}.\end{align*}

Consequently,

(3.12) \begin{eqnarray} u_{n}\ \leqslant\ {\mathrm{e}}^{(\rho-c)(t_{n_{0}} - t_n)} u_{n_{0}} +\frac{M}{\rho-c}.\end{eqnarray}

We take $c=\frac{\rho+\omega}{2}$ ; then $M={\mathrm{e}}^{(\rho-c)\gamma_1} = {\mathrm{e}}^{\frac{\rho-\omega} 2 \gamma_1}$ . Letting $n \to \infty$ in (3.12) yields (2.8).

To prove (2.9), we set $\omega_{n}=\frac{{\mathrm{e}}^{-\rho t_{n}}}{\gamma_{n}^{\theta}}$ . Then, by (3.11), we see that for $n\geqslant n_{0}$ ,

(3.13) \begin{eqnarray} \omega_{n+1}\ =\ \omega_{n}{\mathrm{e}}^{-\rho\gamma_{n+1}}\left(\frac{\gamma_{n}}{\gamma_{n+1}}\right)^{\theta}\ \leqslant \ \omega_{n}{\mathrm{e}}^{(c-\rho)\gamma_{n+1}}\ \leqslant \ \omega_{n_{0}}{\mathrm{e}}^{(c-\rho)(t_{n+1}-t_{n_{0}})}.\end{eqnarray}

Hence (2.9) follows from (3.13) and the facts that $\sum_{k=n_{0}}^{\infty}\gamma_{k}=\infty$ and $c < \rho$ .

Finally, we prove the last inequality, (2.10). By the same argument as in the proof of (3.11), we have

$$\text{supp}_{n\geqslant1}\frac{\gamma_{n}^{\frac{1}{\alpha}}}{\gamma_{n+1}^{\frac{1}{\alpha}}}<1+c\gamma_{1}.$$

This implies that for any $i\leqslant n-1$ ,

\begin{align*}\frac{(t_{n}-t_{i-1})^{\frac{1}{\alpha}}}{(t_{n}-t_{i})^{\frac{1}{\alpha}}}=\frac{(t_{n}-t_{i}+\gamma_{i})^{\frac{1}{\alpha}}}{(t_{n}-t_{i})^{\frac{1}{\alpha}}}\leqslant 2+\frac{2\gamma_{i}^{\frac{1}{\alpha}}}{(t_{n}-t_{i})^{\frac{1}{\alpha}}}\leqslant2+\frac{2\gamma_{i}^{\frac{1}{\alpha}}}{\gamma_{i+1}^{\frac{1}{\alpha}}}\leqslant C.\end{align*}

Hence, we have

\begin{align*}\sum_{i=n^*+1}^{n-1}(t_n-t_i)^{-\frac{1}{\alpha}} \gamma_i^{1+\theta}&\leqslant C\sum_{i=n^*+1}^{n-1}(t_n-t_{i-1})^{-\frac{1}{\alpha}} \gamma_i^{1+\theta}\\&\leqslant C\gamma_{n^{*}}^{\theta}\int^{t_{n-1}}_{t_{n^{*}}}(t_{n}-t)^{-\frac{1}{\alpha}}\,\mathrm{d} t\leqslant C\gamma_{n^{*}}^{\theta}.\end{align*}

In addition, (3.11) implies

\begin{align*}\frac{\gamma_{n^{*}}^{\theta}}{\gamma_{n}^{\theta}}=\prod_{i=n^{*}}^{n-1}\left(\frac{\gamma_{i}}{\gamma_{i+1}}\right)^{\theta}\leqslant {\mathrm{e}}^{c(t_{n}-t_{n^{*}})}\leqslant {\mathrm{e}}^{\rho(1+\gamma_{1})}.\end{align*}

Therefore, we have

\begin{align*}\sum_{i=n^*+1}^{n-1}(t_n-t_i)^{-\frac{1}{\alpha}} \gamma_i^{1+\theta}\leqslant C{\mathrm{e}}^{\rho(1+\gamma_{1})}\gamma_{n}^{\theta}.\end{align*}

This proves (2.10).