Optimal Scheduling of Critically Loaded Multiclass GI/M/n+M Queues in an Alternating Renewal Environment

Abstract

In this paper, we study optimal control problems for multiclass $GI/M/n+M$ queues in an alternating renewal (up–down) random environment in the Halfin–Whitt regime. Assuming that the downtimes are asymptotically negligible and only the service processes are affected, we show that the limits of the diffusion-scaled state processes under non-anticipative, preemptive, work-conserving scheduling policies, are controlled jump diffusions driven by a compound Poisson jump process. We establish the asymptotic optimality of the infinite-horizon discounted and long-run average (ergodic) problems for the queueing dynamics. Since the process counting the number of customers in each class is not Markov, the usual martingale arguments for convergence of mean empirical measures cannot be applied. We surmount this obstacle by demonstrating the convergence of the generators of an augmented Markovian model which incorporates the age processes of the renewal interarrival times and downtimes. We also establish long-run average moment bounds of the diffusion-scaled queueing processes under some (modified) priority scheduling policies. This is accomplished via Foster–Lyapunov equations for the augmented Markovian model.

Appendices

Appendix A. Proofs of Lemma 3.1 and Proposition 3.1

Proof of Lemma 3.1

By [13, Lemma 5.1], ${\hat{S}}^n_i(t)$ and ${\hat{R}}^n_i(t)$ in (3.1) are martingales with respect to the filtration ${{\mathcal {F}}}^n_t$ in (2.9), having predictable quadratic variation processes given by

$$\begin{aligned} \langle {\hat{S}}^n_{i} \rangle (t) \,=\, \mu ^n_i\int _0^tn^{-1}{Z}^n_i(s)\Psi ^n(s)\,\mathrm {d}{s} \quad \text {and} \quad \langle {\hat{R}}^n_{i} \rangle (t) \,=\, \gamma ^n_i\int _0^tn^{-1}{Q}^n_i(s)\,\mathrm {d}{s}, \quad t\ge 0, \end{aligned}$$

respectively. By (2.7), we have the crude inequality

$$\begin{aligned} 0 \,\le \, n^{-1}{X}^n_i(t) \,\le \, n^{-1}{X}^n_i(0) + n^{-1}A^n_i(t), \quad t\ge 0. \end{aligned}$$

Using the balance equation in (2.5), we see that the same inequalities hold for $n^{-1}{Z}_i^n$ and $n^{-1}{Q}_i^n$. Since $\Psi ^n(s) \in \{0,1\}$, it follows by Lemma 5.8 in [30] that $\{{\hat{W}}^n_i:n\in \mathbb {N}\}$ is stochastically bounded in $(\mathbb {D}^d,J_1)$. Also, $\{{\hat{L}}^n_i:n\in \mathbb {N}\}$ is stochastically bounded in $(\mathbb {D}^d,M_1)$ by (2.4). On the other hand, it is evident that

$$\begin{aligned} {\hat{Y}}^n_i(t) \,\le \, C\int _0^t (1 + \Vert n^{-1}{X}^n(s)\Vert )\,\mathrm {d}{s}, \quad t\ge 0, \end{aligned}$$

where C is some positive constant. Thus, we obtain

$$\begin{aligned} \Vert {\hat{X}}^n(t)\Vert \,\le \, \Vert {\hat{X}}^n(0)\Vert + \Vert {\hat{W}}^n(t)\Vert + \Vert {\hat{L}}^n(t)\Vert + C\int _0^t (1 + \Vert {\hat{X}}^n(s)\Vert )\,\mathrm {d}{s} \quad \forall \,t\ge 0. \end{aligned}$$

(A.1)

Since ${\hat{X}}^n(0)$ is uniformly bounded, applying Lemma 5.3 in [30] and Gronwall’s inequality, we deduce that $\{{\hat{X}}^n:n\in \mathbb {N}\}$ is stochastically bounded in $(\mathbb {D}^d,M_1)$. Using Lemma 5.9 in [30], we see that

$$\begin{aligned} n^{-\nicefrac {1}{2}}{\hat{X}}^n \,=\, n^{-1}{X}^n - \rho \;\Rightarrow \; {\mathfrak {e}}_0 \quad \text {in} \quad (\mathbb {D}^d,M_1) \quad \text {as } n\rightarrow \infty , \end{aligned}$$

which implies that $n^{-1}{X}^n \Rightarrow {\mathfrak {e}}_\rho$ in $(\mathbb {D}^d,M_1)$. By (2.5), and the fact $\langle e,n^{-1}{Q}^n \rangle = (\langle e,n^{-1}{X}^n\rangle - 1)^ + \Rightarrow {\mathfrak {e}}_0$, we have $n^{-1}{Q}^n \Rightarrow {\mathfrak {e}}_0$, and thus $n^{-1}{Z}^n \Rightarrow {\mathfrak {e}}_\rho$. This completes the proof. $\square$

To prove Proposition 3.1, we first consider a modified process. Let $\check{X}^n = (\check{X}^n_1,\dotsc ,\check{X}^n_d)'$ be the d-dimensional process defined by

$$\begin{aligned} \begin{aligned} \check{X}^n_i(t)&\,:=\, {\hat{X}}^n(0) + \ell ^n_it+ {\hat{W}}^n_i(t) + {\hat{L}}^n_i(t) \\ {}&\quad - \int _0^t\mu ^n_i \bigl (\check{X}^n_i(s)- \langle e,\check{X}^n(s) \rangle ^+ {U}^n_i(s)\bigr )\,\mathrm {d}{s} \\&\quad - \int _0^t\gamma ^n_i\langle e,\check{X}^n(s) \rangle ^+ {U}^n_i(s)\,\mathrm {d}{s}, \quad \text {for } i\in {{\mathscr {I}}}. \end{aligned} \end{aligned}$$

(A.2)

Lemma A.1

As $n\rightarrow \infty$, $\check{X}^n$ and ${\hat{X}}^n$ are asymptotically equivalent, that is, if either of them converges in distribution as $n\rightarrow \infty$, then so does the other, and both of them have the same limit.

Proof

Let $K = K(\epsilon _1)>0$ be the constant satisfying ${{\,\mathrm{{\mathbb {P}}}\,}}(\Vert {\hat{X}}^n\Vert _T > K) < \epsilon _1$ for $T>0$ and any $\epsilon _1 > 0$, where $\Vert {\hat{X}}^n\Vert _T :=\sup _{0\le t\le T}\Vert {\hat{X}}^n(t)\Vert$. Since ${\hat{U}}^n(s)\in {{\mathcal {S}}}$ for $s\ge 0$, on the event $\{\Vert {\hat{X}}^n\Vert _T \le K\}$, we obtain

$$\begin{aligned} \begin{aligned} \Vert \check{X}^n(t) - {\hat{X}}^n(t)\Vert&\,\le \, C_1\int _0^{t} \Vert {\hat{X}}^n(s)\Vert \bigl (1-\Psi ^n(s)\bigr )\,\mathrm {d}{s} + C_2\int _0^{t} \Vert \check{X}^n(s) - {\hat{X}}^n(s)\Vert \,\mathrm {d}{s} \\&\,\le \, C_1K C^n_{{\mathsf {d}}}(t) + C_2\int _0^{t} \Vert \check{X}^n(s) - {\hat{X}}^n(s)\Vert \,\mathrm {d}{s} \quad \forall \,t\in [0,T], \end{aligned} \end{aligned}$$

where $C_1$ and $C_2$ are some positive constants. Then, by Gronwall’s inequality, on the event $\{\Vert {\hat{X}}^n\Vert _T \le K\}$, we have

$$\begin{aligned} \Vert \check{X}^n(t) - {\hat{X}}^n(t)\Vert \,\le \, C_1K C^n_{{\mathsf {d}}}(t) \mathrm {e}^{C_2 T} \quad \forall \,t\in [0,T]. \end{aligned}$$

Thus, applying [13, Lemma 2.2], we deduce that for any $\epsilon _2 > 0$, there exist $\epsilon _3>0$ and $n_{\circ } = n_{\circ }(\epsilon _1,\epsilon _2,\epsilon _3,T)$ such that

$$\begin{aligned} \Vert \check{X}^n - {\hat{X}}^n\Vert _T \,\le \, \epsilon _2 \end{aligned}$$

on the event $\{\Vert {\hat{X}}^n\Vert _T \le K\}\cap \{\Vert C^n_{{\mathsf {d}}}\Vert _T\le \epsilon _3\}$, for all $n\ge n_{\circ }$, which implies that

$$\begin{aligned} {{\,\mathrm{{\mathbb {P}}}\,}}(\Vert \check{X}^n - {\hat{X}}^n\Vert _T > \epsilon _2) < \epsilon _1, \quad \forall \,n\ge n_{\circ }. \end{aligned}$$

As a consequence, $\Vert \check{X}^n-{\hat{X}}^n\Vert _T\Rightarrow 0$, as $n\rightarrow \infty$, and this completes the proof. $\square$

Proof of Proposition 3.1

We first prove (i). Define the processes

$$\begin{aligned} \uptau ^n_{1,i}(t) \,:=\, \frac{\mu ^n_i}{n}\int _0^tZ^n(s)\Psi ^n(s)\,\mathrm {d}{s}, \quad \uptau ^n_{2,i}(t) \,:=\, \frac{\gamma ^n_i}{n}\int _0^t Q^n(s)\,\mathrm {d}{s}, \end{aligned}$$

${\tilde{S}}_i^n(t) :=n^{-\nicefrac {1}{2}}(S^n(nt) - nt)$, and ${\tilde{R}}_i^n(t) :=n^{-\nicefrac {1}{2}}(R^n(nt) - nt)$, for $i\in {{\mathscr {I}}}$. Then, since $\Psi ^n(s)\in \{0,1\}$ for $s\ge 0$, applying Lemma 3.1 and Lemma 2.2 in [13], we have

$$\begin{aligned} \uptau ^n_{1,i}(\cdot ) \,=\, \mu ^n_i\int _0^\cdot (n^{-1}{Z}_i^n(s) - \rho _i)\Psi ^n(s)\,\mathrm {d}{s} + \mu ^n_i\int _0^\cdot \rho _i\Psi ^n(s)\,\mathrm {d}{s} \;\Rightarrow \; \lambda _i{\mathfrak {e}}(\cdot ). \end{aligned}$$

in $(\mathbb {D}, M_1)$, as $n\rightarrow \infty$, and that $\uptau ^n_{2,i}$ weakly converges to the zero process. Since $\{A_i^n,S_i^n,R^n_i,\Psi ^n:i\in {{\mathscr {I}}}, n\in \mathbb {N}\}$ are independent processes, and $\tau ^n_{1,i}$ and $\tau ^n_{2,i}$ converge to deterministic functions, we have joint weak convergence of $({\hat{A}}^n,{\hat{S}}^n,{\hat{R}}^n,{\hat{L}}^n,\uptau ^n_{1},\uptau ^n_2)$, where $\uptau ^n_1:=(\uptau ^n_{1,1},\dotsc ,\uptau ^n_{1,d})'$, and $\uptau ^n_2$ is defined analogously. On the other hand, since the second moment of $A^n$ is finite, it follows that ${\hat{A}}^n$ converges weakly to a d-dimensional Wiener process with mean 0 and covariance matrix $\mathrm{diag}\bigl (\sqrt{\lambda _1c^2_{a,1}},\dotsc ,\sqrt{\lambda _dc^2_{a,d}}\bigr )$ (see, e.g., [31]). Therefore, by the FCLT for the Poisson processes ${\tilde{S}}^n$ and ${\tilde{R}}^n$, and using the random time change lemma in [21, Page 151], we obtain (i).

Using (A.1) and Proposition 3.1 (i), the proof of (ii) is same as the proof of [1, Lemma 4 (iii)].

To prove (iii), we first show any limit of $\check{X}^n$ in (A.2) satisfies (3.3). Following an argument similar to the proof of Lemma 5.2 in [13], one can easily show that the d-dimensional integral mapping $x = \Lambda (y,u) :\mathbb {D}^d\times \mathbb {D}^d \rightarrow \mathbb {D}^d$ defined by

$$\begin{aligned} x(t) = y(t) + \int _0^{t} h\bigl (x(s),u(s)\bigr )\,\mathrm {d}{s} \end{aligned}$$

is continuous in $(\mathbb {D}^d,M_1)$, provided that the function $h:{\mathbb {R}^{d}}\times {\mathbb {R}^{d}}\rightarrow {\mathbb {R}^{d}}$ is Lipschitz continuous in each coordinate. Since

$$\begin{aligned} \check{X}^n \,=\, \Lambda ({\hat{X}}^n(0) + {\hat{W}}^n + {\hat{L}}^n, {U}^n), \end{aligned}$$

then, by the tightness of ${U}^n$ and the continuous mapping theorem, any limit of $\check{X}^n$ satisfies (A.2), and the same result holds for ${\hat{X}}^n$ by Lemma A.1.

Recall the definition of $\breve{\tau }^n$ in (2.8). It is evident that

$$\begin{aligned} \begin{aligned} {\hat{L}}^n_i(t+r) - {\hat{L}}^n_i(t)&\,=\, {\hat{L}}^n_i(\breve{\tau }^n(t) + r) - {\hat{L}}^n_i\bigl (\breve{\tau }^n(t)\bigr ) \\&\quad + {\hat{L}}^n_i(t+r) - {\hat{L}}^n_i(\breve{\tau }^n(t) + r) + {\hat{L}}^n_i\bigl (\breve{\tau }^n(t)\bigr ) - {\hat{L}}^n_i(t). \end{aligned} \end{aligned}$$

(A.3)

for all $t,r\ge 0$ and $i\in {{\mathscr {I}}}$. By Assumption 2.2, we have $\breve{\tau }^n(t) \Rightarrow t$ as $n\rightarrow \infty$, for $t\ge 0$. Then, by the random time change lemma in [21, Page 151], we deduce that the last four terms on the right-hand side of (A.3) converge to 0 in distribution. It follows by Proposition 3.1 (i) and (A.3) that

$$\begin{aligned} {\hat{L}}^n(\breve{\tau }^n(t) + r) - {\hat{L}}^n\bigl (\breve{\tau }^n(t)\bigr ) \,\Rightarrow \, \lambda L_{t+r} - \lambda L_{t} \quad \text {in }{\mathbb {R}^{d}}. \end{aligned}$$

Repeating the same argument we establish convergence of ${\hat{S}}^n$ and ${\hat{R}}^n$. Proving that U is non-anticipative follows exactly as in [1]*Lemma 6. This completes the proof of (iii). $\square$

B Proofs of Lemmas 4.1 and 5.2

In this section, we construct two functions, which are used to show the ergodicity of ${\widetilde{\Xi }}^n$. We provide two lemmas concerning the properties of these functions, respectively. The proofs of Lemmas 4.1 and 5.2 are given at the end of this section.

Definition B.1

For $z^n \in {\mathfrak {Z}}_{\mathrm {sm}}^n$, define the operator ${\mathcal {L}}^{z^n}_n:{\mathcal {C}}_b({\mathbb {R}^{d}}\times {\mathbb {R}^{d}}) \rightarrow {\mathcal {C}}_b({\mathbb {R}^{d}}\times {\mathbb {R}^{d}})$ by

$$\begin{aligned} \begin{aligned} {\mathcal {L}}^{z^n}_nf(\breve{x},h)&\,:=\, \sum _{i\in {{\mathscr {I}}}}\frac{\partial f(\breve{x},h)}{\partial h_i} + \sum _{i\in {{\mathscr {I}}}} r_{i}^n(h_i)\bigl (f(\breve{x} + e_i, h-h_i\,e_i) - f(\breve{x},h)\bigr ) \\&\quad + \sum _{i\in {{\mathscr {I}}}}\mu _i^n z^n_i\bigl (f(\breve{x} - e_i,h) - f(\breve{x},h)\bigr )\\ {}&\quad + \sum _{i\in {{\mathscr {I}}}}\gamma _i^n q^n_i\bigl (f(\breve{x} - e_i,h) - f(\breve{x},h)\bigr ) \end{aligned} \end{aligned}$$

(B.1)

for $f\in {\mathcal {C}}_b({\mathbb {R}^{d}}\times {\mathbb {R}^{d}})$ and any $(\breve{x},h)\in \mathbb {R}^d_+\times \mathbb {R}^d_+$, with $q^n :=\breve{x} - z^n$.

Note that if $d_1^n \equiv 0$ for all n, the queueing system has no interruptions. In this situation, under a Markov scheduling policy, the (infinitesimal) generator of $(X^n,H^n)$ takes the form of (B.1). Recall the scheduling policies $\check{z}^n$ in Definition 4.1, and ${\bar{x}} = \breve{x} - n\rho$ in Definition 4.2. We define the sets

$$\begin{aligned} {\tilde{{{\mathcal {K}}}}}_n(\breve{x}) \,:=\, \biggl \{i\in {{\mathscr {I}}}_0 :\breve{x}_i \,\ge \, \frac{n\rho _i}{\sum _{j\in {{\mathscr {I}}}_0}\rho _j}\biggr \} \,=\, \biggl \{i\in {{\mathscr {I}}}_0 :{\bar{x}}_i \,\ge \, \frac{n\rho _i\sum _{j\in {{\mathscr {I}}}\setminus {{\mathscr {I}}}_0}\rho _j}{\sum _{j\in {{\mathscr {I}}}_0}\rho _j}\biggr \}. \end{aligned}$$

We have the following lemma.

Lemma B.1

Grant Assumptions 2.1, 2.2, and 3.2. For any even integer $\upkappa \ge 2$, there exist a positive vector $\xi \in \mathbb {R}^d_+$, $\breve{n}\in \mathbb {N}$, and positive constants $\breve{C}_0$ and $\breve{C}_1$, such that the functions $f_n$, $n\in \mathbb {N}$, defined by

$$\begin{aligned} f_n(\breve{x},h):= & {} \sum _{i\in {{\mathscr {I}}}}\xi _i |{\bar{x}}_i|^\upkappa \nonumber \\&+ \sum _{i\in {{\mathscr {I}}}}\eta ^n_i(h_i) \xi _i \bigl (|{\bar{x}}_i+1|^\upkappa - |{\bar{x}}_i|^\upkappa \bigr ) \quad \forall \, (\breve{x},h)\in \mathbb {R}^d_+\times \mathbb {R}^d_+, \end{aligned}$$

(B.2)

with $\eta ^n_i$ as defined in (4.3), satisfy

$$\begin{aligned} \begin{aligned}&{\mathcal {L}}^{\check{z}^n}_nf_n(\breve{x},h) \,\le \, \breve{C}_0 n^{\nicefrac {\upkappa }{2}} \\&\quad - \breve{C}_1\sum _{i\in {{\mathscr {I}}}\setminus {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}\xi _i |{\bar{x}}_i|^\upkappa - \breve{C}_1\sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}\bigl (\mu ^n_i(\check{z}^n_i - n\rho _i) +\gamma ^n_i\check{q}^n_i\bigr )|{\bar{x}}_i|^{\upkappa -1} \\&\quad + \sum _{i\in {{\mathscr {I}}}} \bigl ({{\mathscr {O}}}(\sqrt{n}){{\mathscr {O}}}\bigl (|{\bar{x}}_i|^{\upkappa -1}\bigr ) + {{\mathscr {O}}}(n){{\mathscr {O}}}\bigl (|{\bar{x}}_i|^{\upkappa -2})\bigr ) \end{aligned} \end{aligned}$$

(B.3)

for all $n\ge \breve{n}$ and $(\breve{x},h)\in \mathbb {R}^d_+\times \mathbb {R}^d_+$.

Proof

Using the estimate

$$\begin{aligned} (a \pm 1)^m - a^{\upkappa } \,=\, \pm \upkappa a^{\upkappa -1} + {{\mathscr {O}}}(a^{\upkappa -2}) \qquad \forall \, a\in \mathbb {R}, \end{aligned}$$

(B.4)

an easy calculation shows that

$$\begin{aligned} {\mathcal {L}}^{\check{z}^n}_n f_n(\breve{x},h)&\,=\, \sum _{i\in {{\mathscr {I}}}} {{\dot{\eta }}}^n_i(h_i) \xi _i\bigl (|{\bar{x}}_i+1|^\upkappa - |{\bar{x}}_i|^\upkappa \bigr ) + \sum _{i\in {{\mathscr {I}}}}r^n_i(h_i)\eta ^n_i(0)\xi _i \bigl (({\bar{x}}_i+2)^{\upkappa } - ({\bar{x}}_i+1)^{\upkappa }\bigr )\nonumber \\&\quad - \sum _{i\in {{\mathscr {I}}}}r^n_i(h_i)\eta ^n_i(h_i)\xi _i \bigl (|{\bar{x}}_i+1|^\upkappa - |{\bar{x}}_i|^\upkappa \bigr )\nonumber \\&\quad + \sum _{i\in {{\mathscr {I}}}} \eta ^n_i(h_i)(\mu ^n_i\check{z}^n_i + \gamma ^n_i\check{q}^n_i){{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -2}) + \sum _{i\in {{\mathscr {I}}}}r^n_i(h_i)\xi _i(|{\bar{x}}_i + 1|^\upkappa - |{\bar{x}}_i|^\upkappa )\nonumber \\&\quad + \sum _{i\in {{\mathscr {I}}}}(\mu ^n_i\check{z}^n_i + \gamma ^n_i\check{q}^n_i)\xi _i (|{\bar{x}}_i - 1|^\upkappa - |{\bar{x}}_i|^\upkappa ), \end{aligned}$$

(B.5)

where for the fourth term on the right-hand side we also used the fact that

$$\begin{aligned} \bigl (|{\bar{x}}_i|^\upkappa - |{\bar{x}}_i - 1|^\upkappa \bigr ) - \bigl (|{\bar{x}}_i+1|^\upkappa - |{\bar{x}}_i|^\upkappa \bigr ) \,=\, {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -2}). \end{aligned}$$

It is clear that $\eta ^n_i(0) = 0$, since $F_i(0) = 0$ and ${{\,\mathrm{{\mathbb {E}}}\,}}[G_i] = 1$. On the other hand, $\eta ^n_i(t)$ is bounded for all $n\in \mathbb {N}$ and $t\ge 0$ by Assumption 3.2. Thus, applying (B.4), (B.5) and (4.4), it follows that

$$\begin{aligned} \begin{aligned} {\mathcal {L}}^{\check{z}^n}_n f_n(\breve{x},h) \,=\, \sum _{i\in {{\mathscr {I}}}} \bigl [\xi _i (\lambda ^n_i - \mu _i^n\check{z}^n_i - \gamma ^n_i\check{q}^n_i) \bigl (\upkappa ({\bar{x}}_i)^{\upkappa -1} + {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -2})\bigr )\\ + \eta ^n_i(h_i)(\mu ^n_i\check{z}^n_i + \gamma ^n_i\check{q}^n_i) {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -2})\bigr ]. \end{aligned} \end{aligned}$$

(B.6)

Since $\eta ^n_i(h_i)$ is uniformly bounded, and $\check{z}^n_i, \check{q}^n_i \le {\bar{x}}_i + n\rho _i$, it follows that the last term in (B.6) is equal to ${{\mathscr {O}}}(n){{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -2}) + {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -1})$. Note that for $i\in {{\mathscr {I}}}\setminus {{\mathscr {I}}}_0$, $\check{z}^n_i$ is equivalent to the static priority scheduling policy. Note also, that

$$\begin{aligned} {\bar{x}}_i \,\ge \, \check{z}^n_i - n\rho _i \,\ge \, \frac{n\rho _i\sum _{j\in {{\mathscr {I}}}\setminus {{\mathscr {I}}}_0}\rho _j}{\sum _{j\in {{\mathscr {I}}}_0}\rho _j} \,>\, 0 \qquad \forall \,i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x}), \end{aligned}$$

(B.7)

and for $i\in {{\mathscr {I}}}_0 \setminus {\tilde{{{\mathcal {K}}}}}_n(\breve{x})$, we have $\check{z}^n_i - n\rho _i = {\bar{x}}_i$ and $\check{q}^n_i = 0$. By using (B.6), and the identity in (5.20), we obtain

$$\begin{aligned} \begin{aligned} {\mathcal {L}}^{\check{z}^n}_n f_n(\breve{x},h)&\,\le \, \sum _{i\in {{\mathscr {I}}}\setminus {{\mathscr {I}}}_0}\xi _i \bigl (-\mu ^n_i {\bar{x}}_i + (\mu ^n_i - \gamma ^n_i) \check{q}^n_i\bigr )m({\bar{x}}_i)^{\upkappa -1}\\&\quad - \sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}\xi _i\bigl (\mu ^n_i (\check{z}^n_i - n\rho _i) + \gamma ^n_i\check{q}^n_i\bigr ) |{\bar{x}}_i|^{\upkappa -1} \\&\quad - \sum _{i\in {{\mathscr {I}}}_0\setminus {\tilde{{{\mathcal {K}}}}}_n(\breve{x})} \xi _i\mu ^n_i|{\bar{x}}_i|^\upkappa + \sum _{i\in {{\mathscr {I}}}}\bigl ({{\mathscr {O}}}(\sqrt{n}){{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -1})\\&\quad + {{\mathscr {O}}}(n){{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -2})\bigr ). \end{aligned} \end{aligned}$$

(B.8)

Let $\breve{c}_1:=\sup _{i,n}\{\gamma ^n_i,\mu ^n_i\}$, and $\breve{c}_2$ be some constant such that $\inf \{\mu _i^n,\gamma ^n_j :i\in {{\mathscr {I}}},j\in {{\mathscr {I}}}\setminus {{\mathscr {I}}}_0, n \in \mathbb {N}\} \ge \breve{c}_2 > 0$. We select a positive vector $\xi \in \mathbb {R}^d_+$ such that $\xi _1 :=1$, $\xi _{i}:=\frac{\kappa ^m_1}{d^\upkappa } \min _{i^{\prime }\le i-1}\xi _{i^{\prime }}$, $i\ge 2$, with $\kappa _1 :=\frac{\breve{c}_1}{8\breve{c}_2}$. Compared with [4, Lemma 5.1], the important difference here is that, for $i\in {{\mathscr {I}}}\setminus {{\mathscr {I}}}_0$, we have

$$\begin{aligned} \check{q}^n_i \,=\, \Biggl (\breve{x}_i - \biggl (n - \sum _{j\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}\check{z}^n_j - \sum _{j\in {{\mathscr {I}}}_0\setminus {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}x_j - \sum _{j=|{{\mathscr {I}}}_0|+1}^{i-1}x_j\biggr )^+\Biggr )^+. \end{aligned}$$

Repeating the argument in the proof of [4, Lemma 5.1], it follows by (B.8) that

$$\begin{aligned} \begin{aligned} {\mathcal {L}}^{\check{z}^n}_n f_n(\breve{x},h)&\,\le \, c_3n^{\nicefrac {\upkappa }{2}} - c_4 \sum _{i\in {{\mathscr {I}}}\setminus {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}\xi _i|{\bar{x}}_i|^\upkappa - c_5\sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}\xi _i\bigl (\mu ^n_i(\check{z}^n_i -n\rho _i) \\&\quad + \gamma ^n_i\check{q}^n_i\bigr )|{\bar{x}}_i|^{\upkappa -1} + \frac{c_5}{2} \sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}\xi _i\mu ^n_i \bigl (\check{z}^n_i -n\rho _i\bigr )^\upkappa \\ {}&\quad + \sum _{i\in {{\mathscr {I}}}}\bigl ({{\mathscr {O}}}(\sqrt{n}){{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -1}) + {{\mathscr {O}}}(n){{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -2})\bigr ) \end{aligned} \end{aligned}$$

(B.9)

for some positive constants $c_3$, $c_4$ and $c_5$. Therefore, (B.3) follows by (B.7) and (B.9), and this completes the proof. $\square$

Let

$$\begin{aligned}&{\tilde{g}}_n(\breve{x},h,\psi ,k)\nonumber \\&\quad :=\frac{\psi + \upalpha ^n(k)}{\vartheta ^n} \sum _{i\in {{\mathscr {I}}}}\mu _i^n\xi _i\Bigl ({\tilde{g}}_{n,i}(\breve{x}_i) + \eta ^n_i(h_i)\bigl ({\tilde{g}}_{n,i}(\breve{x}_i + 1) - {\tilde{g}}_{n,i}(\breve{x}_i)\bigr )\Bigr )\qquad \quad \end{aligned}$$

(B.10)

for $(\breve{x},h,\psi ,k)\in {{\mathfrak {D}}}$, where ${\tilde{g}}_{n,i}(\breve{x}_i) :=- |{\bar{x}}_i|^{\upkappa }$ for $i\in {{\mathscr {I}}}\setminus {{\mathscr {I}}}_0$, and

$$\begin{aligned} {\tilde{g}}_{n,i}(\breve{x}_i) \,:=\, {\left\{ \begin{array}{ll} - |{\bar{x}}_i|^\upkappa , &{}\quad \text {if } {\bar{x}}_i \,<\, \frac{n\rho _i\sum _{j\in {{\mathscr {I}}}\setminus {{\mathscr {I}}}_0}\rho _j}{\sum _{j\in {{\mathscr {I}}}_0}\rho _j}, \\ - \frac{n\rho _i\sum _{j\in {{\mathscr {I}}}\setminus {{\mathscr {I}}}_0}\rho _j}{\sum _{j\in {{\mathscr {I}}}_0}\rho _j}|{\bar{x}}_i|^{\upkappa -1}, &{}\quad \text {if } {\bar{x}}_i \,\ge \, \frac{n\rho _i\sum _{j\in {{\mathscr {I}}}\setminus {{\mathscr {I}}}_0}\rho _j}{\sum _{j\in {{\mathscr {I}}}_0}\rho _j}. \end{array}\right. } \quad \forall \,i\in {{\mathscr {I}}}_0. \end{aligned}$$

Recall ${\overline{{\mathcal {L}}}}^{z^n}_{n,\psi }$ in (4.13). We also define

$$\begin{aligned} {\overline{q}}^{n,k}_i(\breve{x},z^n) \,=\, \int _{\mathbb {R}_*} q^n_i\bigl (\breve{x} - n\upmu ^n(y - k),z^n\bigr )\, {\tilde{F}}^{d^n_1}_{\breve{x},k}(\mathrm {d}{y}). \end{aligned}$$

Lemma B.2

Grant Assumptions 2.1, 2.2 and 3.2, and let $\xi \in \mathbb {R}^d_+$ be as in (B.2). Then, for any even integer $\upkappa \ge 2$ and any $\varepsilon >0$, there exist a positive constant ${\overline{C}}$, and ${\bar{n}}\in \mathbb {N}$, such that

$$\begin{aligned} \begin{aligned}&{\overline{{\mathcal {L}}}}^{{z}^n}_{n,\psi }\, {\tilde{g}}_n(\breve{x},h,\psi ,k) \,\le \, {\overline{C}} n^{\nicefrac {\upkappa }{2}} +\varepsilon \sum _{i\in {{\mathscr {I}}}\setminus {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}|{\bar{x}}_i|^{\upkappa } + \sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})} {{\mathscr {O}}}\bigl (|{\bar{x}}_i|^{\upkappa -1}\bigr ) \\&\quad + \frac{1}{\sqrt{n}}\sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})} \bigl (\psi \mu ^n_i(|{z}^n_i - n\rho _i|) + \psi \gamma ^n_i {q}^n_i + (1 - \psi )\gamma ^n_i{\overline{q}}^{n,k}_i\bigr ) {{\mathscr {O}}}\bigl (|{\bar{x}}_i|^{\upkappa -1}\bigr ) \end{aligned} \end{aligned}$$

(B.11)

for any $z^n\in {\mathfrak {Z}}_{\mathrm {sm}}^n$, and all $(\breve{x},h,\psi ,k)\in {{\mathfrak {D}}}$ and $n>{\bar{n}}$.

Proof

It is straightforward to verify that

$$\begin{aligned} \begin{aligned}&|{g}_{n,i}(\breve{x}_i \pm 1) - {g}_{n,i}(\breve{x}_i)| \,=\, {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -1}), \\&|\bigl ({g}_{n,i}(\breve{x}_i) - {g}_{n,i}(\breve{x}_i - 1)\bigr ) - \bigl ({g}_{n,i}(\breve{x}_i + 1) - {g}_{n,i}(\breve{x}_i)\bigr )| \,=\, {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -2}), \end{aligned} \end{aligned}$$

(B.12)

for $i\in {{\mathscr {I}}}$. Repeating the calculation in (B.5) and (B.6), and applying (B.4) and (B.12), we have

$$\begin{aligned}&{\overline{{\mathcal {L}}}}^{{{z}}^n}_{n,\psi }{\tilde{g}}_{n}(\breve{x},h,\psi ,k) \,\le \, \frac{\psi + \upalpha ^n(k)}{\vartheta ^n} \nonumber \\&\quad \Biggl [\sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}\mu ^n_i\xi _i \Bigl [ \bigl (|\lambda ^n_i - n\mu ^n_i\rho _i| + \psi \mu ^n_i|{z}^n_i - n\rho _i| + \psi \gamma _i^n {q}^n_i + (1 - \psi )\gamma ^n_i{\overline{q}}^{n,k}_i\bigr ) {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -1}) \nonumber \\&\qquad + \eta ^n_i(h_i)\bigl (\psi \mu ^n_i {z}^n_i + \psi \gamma ^n_i {q}^n_i + (1 - \psi )\gamma ^n_i{\overline{q}}^{n,k}_i\bigr ) {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -2})\Bigr ] \nonumber \\&\qquad + \sum _{i\in {{\mathscr {I}}}\setminus {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}\mu ^n_i\xi _i\Bigl [ \bigl (\lambda ^n_i + (1 - \psi ) n\mu ^n_i\rho _i \nonumber \\&\qquad + \bigl (1 + \eta ^n_i(h_i)\bigr ) (\psi \mu ^n_i {z}^n_i + \psi \gamma ^n_i {q}^n_i + (1 - \psi )\gamma ^n_i{\overline{q}}^{n,k}_i \bigr ) {{\mathscr {O}}}\bigl (|{\bar{x}}_i|^{\upkappa -1}\bigr )\Bigr ]\Biggr ]. \end{aligned}$$

(B.13)

Note that ${\overline{q}}^{n,k}_i \le c(1 + \langle e,{\bar{x}}\rangle ^{+})$ for some positive constant c, by (5.21). Since ${z}^n_i, {q}^n_i \le {\bar{x}}_i + n\rho _i$, $(\vartheta ^n)^{-1}$ is of order $n^{-\nicefrac {1}{2}}$ by Assumption 2.2, and $\eta ^n_i$ and $\upalpha ^n$ are bounded, it follows by (5.20) and (B.13) that

$$\begin{aligned} \begin{aligned}&{\overline{{\mathcal {L}}}}^{ {z}^n}_{n,\psi }{\tilde{g}}_{n}(\breve{x},h,\psi ,k)\\&\le \sum _{i\in {{\mathscr {I}}}\setminus {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}\frac{1}{\sqrt{n}} \bigl ({{\mathscr {O}}}(n){{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -1})+ {{\mathscr {O}}}(|{\bar{x}}_i|^\upkappa ) \bigr ) + \sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}{{\mathscr {O}}}(\sqrt{n}) {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -2})\\&\quad + \sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})} \frac{1}{\sqrt{n}}\bigl ({{\mathscr {O}}}(\sqrt{n}) + \psi \mu ^n_i|{z}^n_i - n\rho _i| + \psi \gamma _i^n {q}^n_i + (1 - \psi )\gamma ^n_i{\overline{q}}^{n,k}_i\bigr ) {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -1}). \end{aligned} \end{aligned}$$

Thus, applying Young’s inequality, we obtain (B.11), and this completes the proof. $\square$

Proof of Lemma 4.1

We define the function ${\tilde{f}}_n \in {\mathcal {C}}({\mathbb {R}^{d}}\times \mathbb {R}^d_+\times \{0,1\}\times \mathbb {R}_+)$ by

$$\begin{aligned} {\tilde{f}}_n(\breve{x},h,\psi ,k) \,:=\, f_n(\breve{x},h) + {\tilde{g}}_n(\breve{x},h,\psi ,k), \end{aligned}$$

with $f_n$ and ${\tilde{g}}_n$ in (B.2) and (B.10), respectively. Recall ${\widetilde{{{\mathcal {V}}}}}^n_{\upkappa ,\xi }$ in (4.17). With $\xi \in \mathbb {R}^d_+$ as in (B.2), we have

$$\begin{aligned} n^{\nicefrac {\upkappa }{2}} {\widetilde{{{\mathcal {V}}}}}^n_{\upkappa ,\xi }({\tilde{x}}^n(\breve{x}),h,\psi ,k) \,=\, {\tilde{f}}_n(\breve{x},h,\psi ,k) \qquad \forall \, (\breve{x},h,\psi ,k)\in {{\mathfrak {D}}}. \end{aligned}$$

Hence, to prove (4.18), it suffices to show that

$$\begin{aligned}&\breve{{\mathcal {L}}}^{\check{z}^n}_n {\tilde{f}}_{n}(\breve{x},h,\psi ,k) \,\le \, {\widetilde{C}}_0 n^{\nicefrac {\upkappa }{2}}\nonumber \\&- {\widetilde{C}}_1 \sum _{i\in {{\mathscr {I}}}\setminus {\tilde{{{\mathcal {K}}}}}_n(x)}\xi _i|{\bar{x}}_i|^{\upkappa } - {\widetilde{C}}_1 \sqrt{n}\sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})} \xi _i|{\bar{x}}_i|^{\upkappa -1} \qquad \forall \, n > \breve{n}, \end{aligned}$$

(B.14)

and all $(\breve{x},h,\psi ,k)\in {{\mathfrak {D}}}$, where the generator $\breve{{\mathcal {L}}}^{\check{z}^n}_n$ is given in (4.12). It is clear that ${{\mathcal {Q}}}_{n,\psi }f_n(\breve{x},h) = 0$. Since $(\vartheta ^n)^{-1}$ is of order $n^{\nicefrac {-1}{2}}$, it follows by (4.10) and (4.15) that

$$\begin{aligned} {{\mathcal {Q}}}_{n,0}{\tilde{g}}_n(\breve{x},h,0,k)&\,\le \, \sum _{i\in {{\mathscr {I}}}\setminus {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}-\mu ^n_i\xi _i|{\bar{x}}_i|^\upkappa \nonumber \\&\quad + \sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}- \mu ^n_i\xi _i\frac{n\rho _i\sum _{j\in {{\mathscr {I}}}\setminus {{\mathscr {I}}}_0}\rho _j}{\sum _{j\in {{\mathscr {I}}}_0}\rho _j} |{\bar{x}}_i|^{\upkappa -1} \nonumber \\&\quad + \epsilon _n\sum _{i\in {{\mathscr {I}}}\setminus {\tilde{{{\mathcal {K}}}}}_n(\breve{x})} {{\mathscr {O}}}(|{\bar{x}}_i|^\upkappa ) + \sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}{{\mathscr {O}}}(\sqrt{n}) {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -1}), \end{aligned}$$

(B.15)

where C is some positive constant and $\epsilon _n\rightarrow 0$ as $n\rightarrow \infty$. Since all the moments of $d_1$ are finite by (3.10) and $(a+z)^\upkappa - a^\upkappa = {{\mathscr {O}}}(z){{\mathscr {O}}}(a^{\upkappa -1}) + {{\mathscr {O}}}(z^2){{\mathscr {O}}}(a^{\upkappa -2})+ \dots + {{\mathscr {O}}}(z^\upkappa )$ for any $a,z\in \mathbb {R}$, it is easy to verify that

$$\begin{aligned} {{\mathcal {I}}}_{n,1} {\hat{f}}_n(\breve{x},h,1,0) \,=\, \sum _{i\in {{\mathscr {I}}}} \sum ^\upkappa _{j=1}{{\mathscr {O}}}(n^{\nicefrac {j}{2}}) {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -j}), \end{aligned}$$

(B.16)

using also the fact that

$$\begin{aligned} \beta ^n_{{\mathsf {u}}}\int _{R_*} \biggl (\frac{n}{\vartheta ^n}\mu ^n_i\rho _iz\biggr )^j F^{d_1}(\mathrm {d}{z}) \,=\, \beta ^n_{{\mathsf {u}}}\biggl (\frac{n}{\vartheta ^n}\biggr )^j(\mu ^n_i\rho _i)^j {{\,\mathrm{{\mathbb {E}}}\,}}\bigl [(d_1)^j\bigr ] \,=\, {{\mathscr {O}}}(n^{\nicefrac {j}{2}})\quad \forall \,j>0, \end{aligned}$$

which follows by by Assumptions 2.1, 2.2, and (3.10). Then, for $\psi =1$, it follows by (B.16) and Young’s inequality that

$$\begin{aligned} \begin{aligned} \breve{{\mathcal {L}}}^{\check{z}^n}_n {\tilde{f}}_n(\breve{x},h,1,0)&\,\le \, {\mathcal {L}}^{\check{z}^n}_n f_n(\breve{x},h) + {\overline{{\mathcal {L}}}}^{\check{z}^n}_{n,1}{\tilde{g}}_n(\breve{x},h,1,0) \\&\quad + Cn^{\nicefrac {\upkappa }{2}} + \epsilon _n\sum _{i\in {{\mathscr {I}}}\setminus {\tilde{{{\mathcal {K}}}}}_n(\breve{x})} {{\mathscr {O}}}(|{\bar{x}}_i|^\upkappa )\\ {}&\quad + \sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}{{\mathscr {O}}}(\sqrt{n}) {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -1}). \end{aligned} \end{aligned}$$

(B.17)

Note that the last two terms in (B.3) and the last term in (B.11) are of smaller order than the second and third terms on the right-hand side of (B.3), respectively. Thus, applying Lemmas B.1 and B.2, and using (B.17), we obtain

$$\begin{aligned}&n^{-\nicefrac {\upkappa }{2}} \breve{{\mathcal {L}}}^{\check{z}^n}_n {\tilde{f}}_n(\breve{x},h,1,0) \,\le \, {\widetilde{C}}_0 - {\widetilde{C}}_1\sum _{i\in {{\mathscr {I}}}\setminus {\tilde{{{\mathcal {K}}}}}_n({\tilde{x}})}|{\bar{x}}_i|^\upkappa \nonumber \\&- {\widetilde{C}}_1\sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})}n^{-\nicefrac {1}{2}} \bigl (\mu ^n_i(\check{z}^n_i - n\rho _i) +\gamma ^n_i\check{q}^n_i\bigr )|{\tilde{x}}_i|^{\upkappa -1} \end{aligned}$$

(B.18)

for all large enough n, where ${\tilde{x}}$ is defined in Definition 4.2. Since $\check{q}^n_i \ge 0$ and $\check{z}^n_i - n\rho _i>0$ for $i\in {{\tilde{{{\mathcal {K}}}}}}_n(\breve{x})$, then by using (B.7) and (B.18), we see that (B.14) holds when $y = 1$.

For $\psi = 0$, using (B.15), Young’s inequality, and the fact that for $i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})$, ${\bar{x}}_i > 0$, we obtain

$$\begin{aligned}&\breve{{\mathcal {L}}}^{\check{z}^n}_n {\tilde{f}}_{n}(\breve{x},h,0,k) \\&\le \sum _{i\in {{\mathscr {I}}}}{{\mathscr {O}}}(\sqrt{n}){{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -1}) + \sum _{i\in {{\mathscr {I}}}}{{\mathscr {O}}}(n){{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -2}) + Cn^{\nicefrac {\upkappa }{2}} \nonumber \\&\quad + (\epsilon + \epsilon _n)\sum _{i\in {{\mathscr {I}}}\setminus {\tilde{{{\mathcal {K}}}}}_n(\breve{x})} \xi _i|{\bar{x}}_i|^\upkappa \nonumber \\&\quad + \sum _{i\in {{\mathscr {I}}}\setminus {\tilde{{{\mathcal {K}}}}}_n(\breve{x})} \biggl (- \mu ^n_i\xi _i|{\bar{x}}_i|^\upkappa + \gamma ^n_i\xi _i{\overline{q}}^{n,k}_i\bigl (-\upkappa ({\bar{x}}_i)^{\upkappa -1} + {{\mathscr {O}}}(|{\bar{x}}_i|^{\upkappa -2})\bigr )\biggr ) \nonumber \\&\quad + \sum _{i\in {\tilde{{{\mathcal {K}}}}}_n(\breve{x})} -\frac{n\rho _i\sum _{j\in {{\mathscr {I}}}\setminus {{\mathscr {I}}}_0}\rho _j}{\sum _{j\in {{\mathscr {I}}}_0}\rho _j}\mu ^n_i\xi _i |{\bar{x}}_i|^{\upkappa -1} + {\overline{{\mathcal {L}}}}^{\check{z}^n}_{n,0}{\tilde{g}}_n(\breve{x},h,0,k) \end{aligned}$$

for some positive constant C and sufficiently small $\epsilon >0$. We proceed by invoking the argument in the proof of [4, Lemma 5.1]. The important difference here is that

$$\begin{aligned}&\check{q}^n_i\bigl (\breve{x} - n\upmu ^n(z - k)\bigr ) \,=\, {\tilde{\epsilon }}_i\bigl (\breve{x} - n\upmu ^n(z - k)\bigr ) \bigl ({\bar{x}}_i - n\mu _i\rho _i(z - k)\bigr )\\&\quad + {\bar{\epsilon }}_i\bigl (\breve{x} - n\upmu ^n(z - k)\bigr ) \sum _{j=1}^{i-1}\bigl ({\bar{x}}_j - n\mu _j\rho _j(z - k)\bigr ), \end{aligned}$$

where the functions ${\tilde{\epsilon }}_i, {\bar{\epsilon }}_i :{\mathbb {R}^{d}}\rightarrow [0,1]$, for $i\in {{\mathscr {I}}}$. Since ${\tilde{\epsilon }}_i$ and ${\bar{\epsilon }}_i$ are bounded, we have some additional terms which are bounded by $C \int _{\mathbb {R}_*}n\mu _i\rho _i(y - k)\,{\tilde{F}}^{d^n_1}_{\breve{x},k}(\mathrm {d}{y})$ for some positive constant C. Therefore, these are of order $\sqrt{n}$ by (5.21). Thus, repeating the argument in the proof of Lemma B.1, and applying Lemma B.2, we deduce that (B.14) holds with $\psi =0$. This completes the proof. $\square$

Proof of Lemma 5.2

The proof mimics that of Proposition 4.2. We sketch the proof when ${{\mathscr {I}}}_0$ is empty. Using the estimate

$$\begin{aligned} {{\mathscr {O}}}(q^n_i){{\mathscr {O}}}(|{\bar{x}}_i|^{m-1}) \,\le \, \epsilon ^{1-m}\bigl ({{\mathscr {O}}}(q^n_i)\bigr )^m + \epsilon \bigl ({{\mathscr {O}}}(|{\bar{x}}_i|^{m-1})\bigr )^{\nicefrac {m}{m-1}} \end{aligned}$$

(B.19)

for any $\epsilon >0$, which follows by Young’s inequality, we deduce that, for some positive constants $\{c_k :k=1,2,3\}$, we have

$$\begin{aligned} {\mathcal {L}}^{z^n}_n f_n(\breve{x},h) \,\le \, c_1 n^{\nicefrac {m}{2}} + c_2(\langle e,q^n\rangle )^m - c_3\sum _{i\in {{\mathscr {I}}}}\xi _i |{\bar{x}}_i|^m \quad \forall \, (\breve{x},h)\in \mathbb {R}^d_+\times \mathbb {R}^d_+, \end{aligned}$$

(B.20)

and all large enough n. Note that Lemma B.2 holds for all $z^n\in {\mathfrak {Z}}_{\mathrm {sm}}^n$. Then, we may repeat the steps in the proof of Lemma 4.1, except that here we use

$$\begin{aligned} \begin{aligned}&({\tilde{x}}_i)^{m-1}\int _{\mathbb {R}_*}{\hat{q}}^n_i\bigl (\breve{x} - n\upmu ^n(y - k),z^n\bigr )\, {\tilde{F}}^{d^n_1}_{\breve{x},k}(\mathrm {d}{y}) \\&\quad \le \epsilon |{\bar{x}}_i|^{m} + \epsilon ^{1-m}\Bigl ({{\,\mathrm{{\mathbb {E}}}\,}}\bigl [{\hat{q}}^n_i\bigl (\breve{x} - n\upmu ^n(d^n_1 - k),z^n\bigr ) \,|\, d^n_1 > k\bigr ]\Bigr )^m, \end{aligned} \end{aligned}$$

(B.21)

where ${\hat{q}}^n = n^{\nicefrac {-1}{2}}q^n$, with $\epsilon >0$ chosen sufficiently small. Since ${\hat{q}}^n_i(\breve{x},z^n) \le \langle e,{\tilde{x}} \rangle ^+$, it follows by (5.21) that

$$\begin{aligned} {{\,\mathrm{{\mathbb {E}}}\,}}\bigl [{\hat{q}}^n_i\bigl (\breve{x} - n\upmu ^n(d^n_1 - k),z^n\bigr ) \bigm | d^n_1 > k\bigr ] \,\le \, c_4(1 + \langle e,{\tilde{x}} \rangle ^+). \end{aligned}$$

(B.22)

Thus, by the same calculation in Proposition 4.2, and using (B.19)–(B.22), we obtain

$$\begin{aligned}&{{\,\mathrm{{\mathbb {E}}}\,}}^{z^n}\biggl [\int _0^{T}|{\widetilde{X}}^n(s)|^m\biggr ] \,\le \, C_1(T + |{\hat{X}}^n(0)|^m)\nonumber \\ {}&\quad + C_2{{\,\mathrm{{\mathbb {E}}}\,}}^{z^n} \biggl [\int _0^{T}\bigl (1 + \langle e,{\widetilde{X}}^n(s)\rangle ^{+}\bigr )^m\,\mathrm {d}{s}\biggr ] \end{aligned}$$

(B.23)

for all large enough n, and $\{z^n\in {\mathfrak {Z}}_{\mathrm {sm}}^n:n\in \mathbb {N}\}$. Since $\sup _n {\hat{J}}({\hat{X}}^n(0),{z}^n)<\infty$, it follows by (4.26) that

$$\begin{aligned} \sup _{n}\,\limsup _{T\rightarrow \infty }\,\frac{1}{T}{{\,\mathrm{{\mathbb {E}}}\,}}\biggl [ \int _0^{T}\bigl (\langle e,{\widetilde{X}}^n(s)\rangle ^{+}\bigr )^m\,\mathrm {d}{s}\biggr ] \,<\, \infty . \end{aligned}$$

Therefore, dividing both sides of (B.23) by T, taking $T\rightarrow \infty$ and using (4.26) again, we obtain (5.11). We may show that the result also holds when ${{\mathscr {I}}}_0$ is nonempty by repeating the above argument and applying Lemma B.2. This completes the proof. $\square$