Strong stochastic persistence of some Lévy-driven Lotka–Volterra systems

Abstract

We study a class of Lotka–Volterra stochastic differential equations with continuous and pure-jump noise components, and derive conditions that guarantee the strong stochastic persistence (SSP) of the populations engaged in the ecological dynamics. More specifically, we prove that, under certain technical assumptions on the jump sizes and rates, there is convergence of the laws of the stochastic process to a unique stationary distribution supported far away from extinction. We show how the techniques and conditions used in proving SSP for general Kolmogorov systems driven solely by Brownian motion must be adapted and tailored in order to account for the jumps of the driving noise. We provide examples of applications to the case where the underlying food-web is: (a) a 1-predator, 2-prey food-web, and (b) a multi-layer food-chain.

Appendix: Deferred proofs

Proof of Lemma 1

Since $(\mathbf{A}, \mathbf{B})$ is feasible, consider the vector $\mathbf{c}$ given by Definition 1. We will prove that there exists positive constants $K_1$, $K_2$ such that:

$$\begin{aligned} \dfrac{\sum _{i}c_i x_i (B_i +(\mathbf{A}\mathbf{x})_i)}{1+\sum _{i}c_i x_i} < K_1 - K_2 (x_1+x_2+\cdots +x_n). \end{aligned}$$

(48)

for every $\mathbf{x}\in {\mathbb {R}}^n_{+}$. Indeed:

$$\begin{aligned} \sum _{i}c_i x_i (B_i +(Ax)_i)&= \sum _{i}c_iB_i x_i - \sum _{i}c_ia_{ii}x_i^2 - \sum _{i}\sum _{j \in \text {predator}_i}x_i x_j [c_ia_{ij} -c_j a_{ji}]\\&\le -\sum _{i=1}^n c_i a_{ii} x_i^2 + \sum _{i=1}^{n-1}c_i B_i x_i \\&\le -\min _{i=1, \ldots , n} \left\{ c_i a_{ii} \right\} \sum _{i=1}^n x_i^2 + \max _{i=1, \ldots , n} \left\{ c_i b_i \right\} \sum _{i=1}^n x_i \\&\le -\dfrac{\min _{i=1, \ldots , n} \left\{ c_i a_{ii} \right\} }{n} (\sum _{i=1}^n x_i)^ 2 + \max _{i=1, \ldots , n} \left\{ c_i b_i \right\} \sum _{i=1}^n x_i, \end{aligned}$$

where we have used Jensen’s inequality to obtain the last line. Thus:

$$\begin{aligned} \dfrac{ \sum _{i}c_i x_i (B_i +(\mathbf{A}\mathbf{x})_i) }{1+\sum _{i=1}^n c_i x_i}&\le \max _{i=1, \ldots , n} \left\{ c_i b_i \right\} \dfrac{\sum _{i=1}^n x_i}{1+\sum _{i=1}^n c_i x_i}\\&\quad -\dfrac{\min _{i=1, \ldots , n} \left\{ c_i a_{ii} \right\} }{n} \dfrac{(\sum _{i=1}^n x_i)^ 2}{1+\sum _{i=1}^n c_i x_i}\\&\le \dfrac{ \max _{i=1, \ldots , n} \left\{ c_i b_i \right\} }{\min _{i=1, \ldots , n} c_i}\\&\quad -\dfrac{\min _{i=1, \ldots , n} \left\{ c_i a_{ii} \right\} }{n} \dfrac{(\sum _{i=1}^n x_i)^ 2}{1+\sum _{i=1}^n c_i x_i}\\&\le \dfrac{ \max _{i=1, \ldots , n} \left\{ c_i b_i \right\} }{\min _{i=1, \ldots , n} c_i} \\&\quad -\dfrac{\min _{i=1, \ldots , n} \left\{ c_i a_{ii} \right\} }{n} \dfrac{(\sum _{i=1}^n x_i)^ 2}{1+\max _{i=1,\ldots , n} c_i \sum _{i=1}^n x_i}\\&\le \dfrac{ \max _{i=1, \ldots , n} \left\{ c_i b_i \right\} }{\min _{i=1, \ldots , n} c_i} \\&\quad -\dfrac{\min _{i=1, \ldots , n} \left\{ c_i a_{ii} \right\} }{n} \dfrac{\sum _{i=1}^n x_i}{ (\sum _{i=1}^n x_i)^{-1}+\max _{i=1,\ldots , n} c_i) }. \end{aligned}$$

Of course, there exists $R> 0$ such that $\Vert \mathbf{x}\Vert > R$ implies $\dfrac{1}{\sum _{i=1}^{n}x_i} < 1$. Define $\tilde{K}_1= \dfrac{ \max _{i=1, \ldots , n} \left\{ c_i b_i \right\} }{\min _{i=1, \ldots , n} c_i}$ and $K_2= \dfrac{\min _{i=1, \ldots , n} \left\{ c_i a_{ii} \right\} }{n (1+\max _{i=1,\ldots , n} c_i)}$. Thus, for $\Vert \mathbf{x}\Vert > R$, we have:

$$\begin{aligned} \dfrac{\sum _{i}c_i x_i (B_i +(\mathbf{A}\mathbf{x})_i)}{1+\sum _{i}c_i x_i} <\tilde{K}_1 - K_2 (x_1+x_2+\cdots +x_n). \end{aligned}$$

Observe that the right hand side is a continuous function of $\mathbf{x}$. Let:

$$\begin{aligned} r_1&= \sup _{\Vert \mathbf{x}\Vert < R} \dfrac{\sum _{i}c_i x_i (B_i +(bA\mathbf{x})_i)}{1+\sum _{i}c_i x_i}\\ r_2&= \sup _{\Vert \mathbf{x}\Vert \le R}K_2(x_1+x_2+\cdots +x_n). \end{aligned}$$

Then, (48) holds with $K_1:=\max \{\tilde{K}_1, r_1+r_2\}$. $\square$

Proof of Lemma 2

Let $V: {\mathbb {R}}^n_{++} \mapsto {\mathbb {R}}_{+}$ be the function defined by:

$$\begin{aligned} V (\mathbf{x}):= \mathbf{c}^T \mathbf{x}- \sum _{i=1}^n \log (c_i x_i), \end{aligned}$$

where $c_i$ are the constant guaranteed by Assumption 1. Then V is log-Lyapunov and for $\mathbf{x}\in {\mathbb {R}}^n_{++}$:

$$\begin{aligned} {\mathcal {L}}V(\mathbf{x})&= \sum _{i} (c_i x_i - 1 ) (B_i + \sum _j A_{ij} x_j) \nonumber \\&\quad + \dfrac{1}{2}\sum _{i=1}^n \sigma ^2_{i} + \int _{{\mathbb {R}}^n} \mathbf {1}^{T}\mathbf{L}(\mathbf{x},\mathbf{z}) - \sum _{i=1}^n \log (1+L_{i}(\mathbf{x},\mathbf{z})) \nu (\mathrm {d}\mathbf{z}). \ \end{aligned}$$

(49)

By Assumption 2, Taylor’s theorem, and Assumption 3, the last term is bounded above. Moreover, since the intraspecies interaction are negative, we have that the above expression is not grater than:

$$\begin{aligned} K_1 \mathbf{1}^T \mathbf{x}- K_2 \mathbf{x}^T \mathbf{x}+ K_3 - \sum _{i} \sum _{j: \in \text {predator}_i} x_i x_j (c_i a_{ij}-c_j a_{ji}). \end{aligned}$$

for some positive constants $K_1, K_2, K_3$. Now, the last sum is non-negative, by definition of feasibility. In toto, thus, there exist positive constants $K_1$, $K_2$, $K_3$ such that:

$$\begin{aligned} {\mathcal {L}}V(x)\le K_1 \mathbf{1}^T \mathbf{x}- K_2 \mathbf{x}^T \mathbf{x}+ K_3. \end{aligned}$$

We conclude that there exists a constant $M> 0$ such that on ${\mathbb {R}}^n_{++}$:

$$\begin{aligned} {\mathcal {L}}V < M. \end{aligned}$$

(50)

If we set:

$$\begin{aligned} V_k := \left\{ \mathbf{x}\in {\mathbb {R}}^n_{++}: \text { exists } i \in \{1, 2, \ldots , n\} \text { such that } x_i> \dfrac{k}{c_i}e^k \text { or } x_i < \dfrac{1}{c_i}e^{-k}\right\} , \end{aligned}$$

then

$$\begin{aligned} \inf _{\mathbf{x}\in V_k} V(\mathbf{x})> k. \end{aligned}$$

(51)

Since V is continuous, $V_k$ is open, and thus

$$\begin{aligned} \tau _k:= \inf \{ t \ge 0: \mathbf{X}_{t} \in V_k \} \end{aligned}$$

is a ${\mathcal {F}}_t$-stopping time. Dynkin’s formula applied to V gives:

$$\begin{aligned} {\mathbb {E}}_{\mathbf{X}_0}\left( V (\mathbf{X}_{\tau _k \wedge t}) \right) \le V(\mathbf{X}_0) +M{\mathbb {E}}_{\mathbf{X}_0}(\tau _k \wedge t) \le V(\mathbf{X}_0) + Mt. \end{aligned}$$

(52)

Fix $N \in {\mathbb {N}}$, and consider the set:

$$\begin{aligned} \varOmega _N:= \bigcap _{k\ge 1} \{\omega \in \varOmega : \tau _k < N\}. \end{aligned}$$

Define:

$$\begin{aligned} \epsilon _N: = {\mathbb {P}}_{\mathbf{X}_0} (\varOmega _N) \end{aligned}$$

Observe that for every $k \ge 1$:

$$\begin{aligned} {\mathbb {E}}_{\mathbf{X}_0} \left( V (\mathbf{X}_{\tau _k \wedge N}) \right)&={\mathbb {E}}_{\mathbf{X}_0}\left( \mathbf {1}_{\varOmega _N} V (\mathbf{X}_{\tau _k \wedge N}) \right) + {\mathbb {E}}_{\mathbf{X}_0}\left( \mathbf {1}_{\varOmega \setminus \varOmega _N} V (\mathbf{X}_{\tau _k \wedge N}) \right) \\&\ge {\mathbb {E}}_{\mathbf{X}_0}\left( \mathbf {1}_{\varOmega _N} V (\mathbf{X}_{\tau _k}) \right) \\&\ge k {\mathbb {E}}_{\mathbf{X}_0} \left( \mathbf {1}_{\varOmega _N} \right) \\&\ge k \epsilon _N, \end{aligned}$$

where the second-to-last line follows from the fact that $\mathbf{X}$ has right-continuous a.s. paths and by the relation (51). This last inequality and (52) gives:

$$\begin{aligned} k \epsilon _N < V(\mathbf{X}_0) + MN \quad \text { for every } k \ge 1, \end{aligned}$$

and thus necessarily $\epsilon _N={\mathbb {P}}_{\mathbf{X}_0}(\varOmega _N)=0$ for every $N \in {\mathbb {N}}$. Consequently,

$$\begin{aligned} {\mathbb {P}}_{\mathbf{X}_0} \left( \lim _{k\rightarrow \infty } \tau _k < \infty \right) ={\mathbb {P}}_{\mathbf{X}_0} \left( \bigcup _{N \in {\mathbb {N}}} \varOmega _N \right) = 0, \end{aligned}$$

and this concludes the proof. $\square$

Recall that $(P_t: t\ge 0)$ is the semigroup associated to the process $\mathbf{X}$, i.e.,

$$\begin{aligned} P_t f(\mathbf{x}):= {\mathbb {E}}_{\mathbf{x}} (f(\mathbf{X}_t)), \end{aligned}$$

for bounded measurable real functions f.

Proof of Lemma 3

Let $F_{i}(\mathbf{x})=x_{i} (B_i+\sum _{j=1}^n A_{ij}x_j )$, $G_{i}(\mathbf{x})= x_{i}\sigma _{i}$, and $H_i(\mathbf{x}, \mathbf{z})= x_i L_{i}(\mathbf{x}, \mathbf{z})$. For two initial conditions $\mathbf{x}, \mathbf{y}$, let $\ {^\mathbf{x}}{\tilde{\mathbf{X}}}{}$ and $\ {^\mathbf{y}}{\tilde{\mathbf{X}}}{}$ be the solutions of (2) in the natural coupling, i.e., driven by the same Lévy noise $(\mathbf{W}, \tilde{N})$. Let $D_{k}:= \{\mathbf{x}\in {\mathbb {R}}^n_+: \Vert \mathbf{x}\Vert \le k\}$, and set $\eta _k:=\inf \{ t\ge 0: \ {^\mathbf{x}}{\tilde{\mathbf{X}}}{_t} \in D^C_k \text { or } \ {^\mathbf{y}}{\tilde{\mathbf{X}}}{_t} \in D^C_k \}$. For $\mathbf{u}=\mathbf{x}\text { or } \mathbf{y}$ define:

$$\begin{aligned} \ {^\mathbf{u}}{\mathbf{X}}{_t}= {\left\{ \begin{array}{ll} \ {^\mathbf{u}}{\tilde{\mathbf{X}}}{_t}, \quad t < \eta _k\\ 0, \qquad \text { otherwise.} \end{array}\right. } \end{aligned}$$

Observe that $\mathbf{F}=(F_{1}, F_2, \ldots , F_n)$ is locally Lipschitz continuous, and thus for every $k\in {\mathbb {N}}$ there exist a constant $M_k$ such that for $\mathbf{x}, \mathbf{y}\in D_k$:

$$\begin{aligned} \Vert \mathbf{F}(\mathbf{x})- \mathbf{F}(\mathbf{y}) \Vert ^2 < M_k \Vert \mathbf{x}-\mathbf{y}\Vert ^2. \end{aligned}$$

(53)

Fix $T\ge 0$. For any time $0 \le t \le T$ define $t_{k}:= t \wedge \eta _k$. For $t \le T_k$:

$$\begin{aligned} \Vert \ {^\mathbf{x}}{\mathbf{X}}{_{t}} - \ {^\mathbf{y}}{\mathbf{X}}{_{t}}\Vert ^2&\le 4 \left( \Vert \mathbf{x}-\mathbf{y}\Vert ^2 + \left| \int _{0}^{t} \mathbf {F}(\ {^\mathbf{x}}{\mathbf{X}}{_s}) -\mathbf {F}(\ {^\mathbf{y}}{\mathbf{X}}{_s}) ds \right| ^2 \right. \end{aligned}$$

(54)

$$\begin{aligned}&\quad \quad + \left| \int _{0}^{t} \mathbf {G}(\ {^\mathbf{x}}{\mathbf{X}}{_s}) - \mathbf {G}(\ {^\mathbf{y}}{\mathbf{X}}{_s}) d\mathbf{W}_s \right| ^2 \nonumber \\&\quad \quad \left. + \left| \int _{0}^{t} \int _{{\mathbb {R}}^n} \mathbf {H}(\ {^\mathbf{x}}{\mathbf{X}}{_s}, \mathbf{z}) - \mathbf {H}(\ {^\mathbf{y}}{\mathbf{X}}{_s}, \mathbf{z})\tilde{N}(ds, d\mathbf{z}) \right| ^2 \right) . \end{aligned}$$

(55)

Cauchy-Schwartz inequality yields:

$$\begin{aligned} \left\| \int _{0}^{t} F(\ {^\mathbf{x}}{\mathbf{X}}{_s}) -F(\ {^\mathbf{y}}{\mathbf{X}}{_s} ) ds \right\| ^2 \le T \int _{0}^{t} \Vert F(\ {^\mathbf{x}}{\mathbf{X}}{_s}) - F(\ {^\mathbf{y}}{\mathbf{X}}{_s}) \Vert ^ 2 ds. \end{aligned}$$

On $s \le T_k$, by (53) we have that:

$$\begin{aligned} \Vert \mathbf{F}(\ {^\mathbf{x}}{\mathbf{X}}{_s}) - \mathbf{F}(\ {^\mathbf{y}}{\mathbf{X}}{_s}) \Vert ^ 2 \le M_k \Vert \ {^\mathbf{x}}{\mathbf{X}}{_s}-\ {^\mathbf{y}}{\mathbf{X}}{_s} \Vert ^ 2 , \end{aligned}$$

and trivially, for $s \le T_k$:

$$\begin{aligned} \Vert \ {^\mathbf{x}}{\mathbf{X}}{_s}-\ {^\mathbf{y}}{\mathbf{X}}{_s}\Vert ^ 2 \le \sup _{u \le s_k} \Vert \ {^\mathbf{x}}{\mathbf{X}}{_u}-\ {^\mathbf{y}}{\mathbf{X}}{_u}\Vert ^ 2. \end{aligned}$$

Let $\tilde{M}_1$ be a Lipschitz constant for G . By Burkholder-Davis-Gundy inequality, Lipschitz-continuity of G and Fubini’s theorem:

$$\begin{aligned}&{\mathbb {E}}_{\mathbf{x},\mathbf{y}}\left( \sup _{0 \le t \le T_k}\left| \int _{0}^{t} \mathbf {G}(\ {^\mathbf{x}}{\mathbf{X}}{_s}) - \mathbf {G}(\ {^\mathbf{y}}{\mathbf{X}}{_s}) d\mathbf{W}_s \right| ^2 \right) \le {\mathbb {E}}_{\mathbf{x},\mathbf{y}} \left( \int _{0}^{T_k} \Vert \mathbf {G}(\ {^\mathbf{x}}{\mathbf{X}}{_s}) - \mathbf {G}(\ {^\mathbf{y}}{\mathbf{X}}{_s})\Vert ^2 ds \right) \\&\le \tilde{M}_1 {\mathbb {E}}_{\mathbf{x},\mathbf{y}} \left( \int _{0}^{T_k} \sup _{u\le s_k} \Vert \ {^\mathbf{x}}{\mathbf{X}}{_u} - \ {^\mathbf{y}}{\mathbf{X}}{_u}\Vert ^2 ds \right) \\&\le \tilde{M}_1 {\mathbb {E}}_{\mathbf{x},\mathbf{y}} \left( \int _{0}^{T} \sup _{u\le s_k} \Vert \ {^\mathbf{x}}{\mathbf{X}}{_u} - \ {^\mathbf{y}}{\mathbf{X}}{_u}\Vert ^2 ds \right) \\&\le \tilde{M}_1 \int _{0}^{T} {\mathbb {E}}_{\mathbf{x},\mathbf{y}} \left( \sup _{u\le s_k} \Vert \ {^\mathbf{x}}{\mathbf{X}}{_u} - \ {^\mathbf{y}}{\mathbf{X}}{_u}\Vert ^2 \right) ds. \end{aligned}$$

Analogously, for the jump component, we have:

$$\begin{aligned} {\mathbb {E}}_{\mathbf{x},\mathbf{y}}&\left( \sup _{0 \le t \le T_k}\left| \int _{0}^{t} \int _{{\mathbb {R}}^n} \mathbf {H}(\ {^\mathbf{x}}{\mathbf{X}}{_s}, z) - \mathbf {H}(\ {^\mathbf{y}}{\mathbf{X}}{_s}, \mathbf{z}) \tilde{N}(ds, d\mathbf{z}) \right| ^2 \right) \\&\quad \le \tilde{M}_2 \int _{0}^{T} {\mathbb {E}}_{\mathbf{x},\mathbf{y}} \left( \sup _{u\le s_k} \Vert \ {^\mathbf{x}}{\mathbf{X}}{_u} - \ {^\mathbf{y}}{\mathbf{X}}{_u}\Vert ^2 \right) ds, \end{aligned}$$

where the constant $\tilde{M}_2$ is guaranteed by Assumption3(c). Thus, if we take supremum on $0 \le t \le T_{k}$ and then take expectation at both sides of the inequality (54), we obtain:

$$\begin{aligned} {\mathbb {E}}_{\mathbf{x}, \mathbf{y}}\left( \sup _{0 \le t \le T_k} \Vert \ {^\mathbf{x}}{\mathbf{X}}{_t} - \ {^\mathbf{y}}{\mathbf{X}}{_t}\Vert ^2 \right)&\le C \left( \Vert \mathbf{x}-\mathbf{y}\Vert \right. \\&\quad \left. + C_k \int _{0}^T {\mathbb {E}}_{\mathbf{x},\mathbf{y}}\left( \sup _{0 \le u \le s_k} \Vert \ {^\mathbf{x}}{\mathbf{X}}{_u} - \ {^\mathbf{y}}{\mathbf{X}}{_u}\Vert ^2\right) ds \right) , \end{aligned}$$

where C is a universal constant and $(C_k: k\ge 0)$ is an increasing positive sequence. We conclude by Gromwall’s Lemma that:

$$\begin{aligned} {\mathbb {E}}_{\mathbf{x}, \mathbf{y}}\left( \sup _{0 \le t \le T_k} \Vert \ {^\mathbf{x}}{\mathbf{X}}{_t} - \ {^\mathbf{y}}{\mathbf{X}}{_t} \Vert ^2 \right) \le C \Vert \mathbf{x}-\mathbf{y}\Vert ^2 e^{C_k T}. \end{aligned}$$

(56)

Now, let $f: {\mathbb {R}}^n_{+} \mapsto {\mathbb {R}}$ be a bounded continuous function, and fix $\mathbf{x}\in {\mathbb {R}}^n_{+}$, $t \ge 0$, $\varepsilon > 0$. Let $r > 0$ such that $\bar{B} ( \mathbf{x}, r ) \subset {\mathbb {R}}^n_{++}$. Plainly, for $\mathbf{y}\in \bar{B} ( \mathbf{x}, r)$:

$$\begin{aligned} \vert {\mathbb {E}}_{\mathbf{x}}(f(\mathbf{X}_t))- {\mathbb {E}}_{\mathbf{y}}(f(\mathbf{X}_{t}))\vert \le {\mathbb {E}}_{\mathbf{x}, \mathbf{y}} \left( \vert f(\ {^\mathbf{x}}{\tilde{\mathbf{X}}}{_t}) - f(\ {^\mathbf{y}}{\tilde{\mathbf{X}}}{_t})\vert \right) . \end{aligned}$$

On the other hand, observe that from inequality (52), we deduce that for $\mathbf{y}\in \bar{B}(\mathbf{x}, r)$:

$$\begin{aligned} {\mathbb {P}}_{\mathbf{y}} (\tau _k < t)k&\le {\mathbb {E}}_{\mathbf{y}} (V(\mathbf{X}_{\tau _k\wedge t}))\le V (\mathbf{y})+Mt \\&\le \sup _{\mathbf{z}\in \bar{B} (\mathbf{x}, r)} V(\mathbf{z}) + Mt, \end{aligned}$$

and thus, we can choose a $k_0$ such that uniformly on $\mathbf{y}\in \bar{B} (\mathbf{x}, r)$, we have:

$$\begin{aligned} {\mathbb {P}}_{\mathbf{x},\mathbf{y}} (\eta _k < t) \le \dfrac{\varepsilon }{6 \Vert f \Vert }. \end{aligned}$$

Since f is continuous, it is uniformly continuous on $D_{k_0}$. Let $\delta : {\mathbb {R}}_{+} \mapsto {\mathbb {R}}_{+}$ be a modulus of continuity of f on $D_{k_0}$, i.e. a function that satisfies that for every $\mathbf{y}_0 \in D_k$, $f(B(\mathbf{y}_0, \delta (\varepsilon )) \cap D_k) \subseteq B (f(\mathbf{y}_0), \varepsilon )$. Let $\varDelta = \delta (\varepsilon /3)$. Then, again for $\mathbf{y}\in \bar{B} (\mathbf{x}, r)$:

$$\begin{aligned} {\mathbb {E}}_{\mathbf{x}, \mathbf{y}} \left( \vert f(\ {^\mathbf{x}}{\tilde{\mathbf{X}}}{_t}) - f(\ {^\mathbf{y}}{\tilde{\mathbf{X}}}{_t})\vert \right)&\le {\mathbb {E}}_{\mathbf{x}, \mathbf{y}} \left( \mathbf {1}_{\eta _k<t}\vert f(\ {^\mathbf{x}}{\tilde{\mathbf{X}}}{_t}) - f(\ {^\mathbf{y}}{\tilde{\mathbf{X}}}{_t})\vert \right) \end{aligned}$$

(57)

$$\begin{aligned}&\qquad + {\mathbb {E}}_{\mathbf{x}, \mathbf{y}} \left( \mathbf {1}_{t<\eta _k}\vert f(\ {^\mathbf{x}}{\tilde{\mathbf{X}}}{_t}) - f(\ {^\mathbf{y}}{\tilde{\mathbf{X}}}{_t})\vert \right) \nonumber \\&\le \dfrac{\varepsilon }{3} + {\mathbb {E}}_{\mathbf{x}, \mathbf{y}} \left( \mathbf {1}_{t<\eta _k}\vert f(\ {^\mathbf{x}}{\mathbf{X}}{_t}) -f(\ {^\mathbf{y}}{\mathbf{X}}{_t})\vert \right) \nonumber \\&= \dfrac{\varepsilon }{3} + {\mathbb {E}}_{\mathbf{x}, \mathbf{y}} \left( \mathbf {1}_{t<\eta _k}\mathbf {1}_{\vert \ {^\mathbf{x}}{\mathbf{X}}{_t} - \ {^\mathbf{y}}{\mathbf{X}}{_t} \vert< \varDelta }\vert f(\ {^\mathbf{x}}{\mathbf{X}}{_t}) -f(\ {^\mathbf{y}}{\mathbf{X}}{_t})\vert \right) \nonumber \\&\quad + {\mathbb {E}}_{\mathbf{x}, \mathbf{y}} \left( \mathbf {1}_{t<\eta _k}\mathbf {1}_{\vert \ {^\mathbf{x}}{\mathbf{X}}{_t} - \ {^\mathbf{x}}{\mathbf{X}}{_t} \vert> \varDelta }\vert f(\ {^\mathbf{x}}{\mathbf{X}}{_t}) -f(\ {^\mathbf{y}}{\mathbf{X}}{_t})\vert \right) \nonumber \\&\le \dfrac{\varepsilon }{3} + \dfrac{\varepsilon }{3} + 2 \Vert f \Vert {\mathbb {P}}_{\mathbf{x},\mathbf{y}} \left( \sup _{0 \le s \le t_k} \vert \ {^\mathbf{x}}{\mathbf{X}}{_s} - \ {^\mathbf{y}}{\mathbf{X}}{_s}\vert > \varDelta \right) . \end{aligned}$$

(58)

Using (56) and Markov’s Inequality, the last term is not greater than $\Vert \mathbf{x}-\mathbf{y}\Vert ^2 \dfrac{2 C \Vert f \Vert e ^{C_k t}}{\varDelta ^2}$ and this is smaller than $\varepsilon /3$ for every $\mathbf{y}$ in the open ball $B \left( \mathbf{x}, r \wedge \dfrac{\sqrt{\varepsilon }\varDelta }{\sqrt{6 C \Vert f \Vert e^{C_{k}t}}} \right)$. The result follows. $\square$

Proof of Lemma 9

For fixed $\alpha > 0$ set $\varrho (\alpha ; \mathbf{x})= \exp \{\alpha (1+\mathbf{c}^T \mathbf{x}) \}$. An easy computation shows that:

$$\begin{aligned} {\mathcal {L}}\varrho (\alpha ; \mathbf{x})&= \varrho (\alpha ; \mathbf{x})\left( \alpha \sum _{i=1}^n c_ix_i (B_i +(\mathbf{A}\mathbf{x})_i) + \dfrac{\alpha ^2 }{2} \sum _{i=1}^n c_i^2 \sigma _i^2 x_i^2 \right. \\&\quad \left. + \int _{{\mathbb {R}}^n} (\exp \{\alpha \mathbf{c}^ T (\mathbf{x}\circ \mathbf{L}(\mathbf{x},\mathbf{z})) \} -1- \alpha \mathbf{x}^ T (\mathbf{x}\circ \mathbf{L}(\mathbf{x},\mathbf{z})))\nu (d\mathbf{z}) \right) . \end{aligned}$$

Fix $\alpha _1> 0$ such that for $i=1, \ldots , n$:

$$\begin{aligned} a_{ii} > \dfrac{1}{2} \alpha _1 c_i \sigma _i^2. \end{aligned}$$

and observe that for every $0 \le \alpha \le \alpha _1$, our choice of the $c_i$, Assumptions 3 and 5 imply that for every $\delta > 0$ there exists $M> 0$ such that:

$$\begin{aligned} {\mathcal {L}}\varrho (\alpha ; \mathbf{x}) \le -\delta \varrho (\alpha ;\mathbf{x})+M, \end{aligned}$$

whenever $\alpha$ is small enough. Just as in the proof of Lemma 5, Dynkin’s formula yields:

$$\begin{aligned} {\mathbb {E}}_{\mathbf{x}} (\varrho (\alpha ; \mathbf{X}_t)) \le \dfrac{M}{\delta } +\varrho (\alpha ; \mathbf{x}) e^{-\delta t}. \end{aligned}$$

(59)

for $0 \le \alpha \le \alpha _1$. Consider the random variables $\ {^\mathbf{x}}{U}{_t} = \varrho (\alpha _1, \ {^\mathbf{x}}{\mathbf{X}}{_t})$. Since $\varrho (\alpha ; \cdot )$ is continuous, De la Vallée Poussin’s Lemma implies that for every compact set $K \subset {\mathbb {R}}^n_{+}$ and $T > 0$ the family:

$$\begin{aligned} (\ {^\mathbf{x}}{U}{_t}: \mathbf{x}\in K, 0 \le t \le T), \end{aligned}$$

is a uniformly integrable (UI) family of random variables, written:

$$\begin{aligned} \lim _{N \rightarrow \infty } \sup _{\mathbf{x}\in K, 0 \le t \le T} {\mathbb {E}}_{\mathbf{x}} (U_t \mathbf {1}_{U_t > N})=0. \end{aligned}$$

(60)

Now, by our assumptions on $\mathbf{L}$:

$$\begin{aligned} \vert \varUpsilon (\mathbf{x})\vert \le C (1+\mathbf{c}^T \mathbf{x}), \end{aligned}$$

for some positive constant C. Consequently, there exists $\alpha _0$ such that for $\alpha < \alpha _0$:

$$\begin{aligned} \exp ^{\alpha \vert \varUpsilon (\mathbf{x})\vert } \le \varrho (\alpha _1/2, \mathbf{x}). \end{aligned}$$

Consequently, there exists $R_1$ such that whenever $\varrho (\alpha _1, \mathbf{x})> R_1$ and $\alpha < \alpha _0$:

$$\begin{aligned} \dfrac{ \exp ^{\alpha \vert \varUpsilon (\mathbf{x})\vert } }{\varrho (\alpha _1, \mathbf{x})} < \dfrac{\varepsilon }{4}. \end{aligned}$$

Next, consider a sequence $(\mathbf{x}_{k})$ converging to $\mathbf{x}\in {\mathbb {R}}^n_{+}$, and set $\varepsilon > 0$. Fix $T > 0$. Since $\mathbf{x}_{k}$ is convergent, there exists a compact set K containing $(\mathbf{x}_k)$ and $\mathbf{x}$. By the UI property (60), there exists $R_2 >0$ such that for every $\mathbf{y}\in K$ and $0 \le t \le T$:

$$\begin{aligned} {\mathbb {E}}_{\mathbf{y}} ( U_t \mathbf {1}_{U_t> R_2} ) < 1. \end{aligned}$$

Fix $R_3 = \max \{R_1, R_2\}$, and let $\eta : {\mathbb {R}}^n_{+}\mapsto [0,1]$ be a smooth function such that $\eta (\mathbf{x})=1$ for $\varrho (\alpha _1; \mathbf{x})< R_3$ and $\eta (\mathbf{x})=0$ for $\varrho (\alpha _1; \mathbf{x}) > 2 R_3$. Now, for $\alpha < \alpha _0$, $0\le t \le T$ and $k \in {\mathbb {N}}$:

$$\begin{aligned} \vert {\mathbb {E}}_{\mathbf{x}_k} (e^{\alpha \varUpsilon (\mathbf{X}_t)}) - {\mathbb {E}}_{\mathbf{x}} (e^{\alpha \varUpsilon (X_t)}) \vert&\le \vert {\mathbb {E}}_{\mathbf{x}_k} (\eta (\mathbf{X}_t)e^{\alpha \varUpsilon (\mathbf{X}_t)} ) - {\mathbb {E}}_{\mathbf{x}} (\eta (\mathbf{X}_t)e^{\alpha \varUpsilon (\mathbf{X}_t)}) \vert \\&\quad + \vert {\mathbb {E}}_{\mathbf{x}_k} ((1-\eta (\mathbf{X}_t))e^{\alpha \varUpsilon (\mathbf{X}_t)} )-{\mathbb {E}}_{\mathbf{x}} ((1-\eta (\mathbf{X}_t))e^{\alpha \varUpsilon (\mathbf{X}_t)})\vert . \end{aligned}$$

The ${\mathcal {C}}_b$-Feller property (Lemma 3) applies to the first term on the right, and thus there exist a $k_0 > 0$ such that for every $k \ge k_0$ it is smaller that $\varepsilon /2$. As for the second term, observe that:

$$\begin{aligned} {\mathbb {E}}_{\mathbf{x}} ((1-\eta (\mathbf{X}_t)) e^{\alpha \varUpsilon (\mathbf{X}_t)} )&\le {\mathbb {E}}_{\mathbf{x}} ((1-\eta (\mathbf{X}_t)) e^{\alpha \vert \varUpsilon (\mathbf{X}_t)\vert })\\&\le {\mathbb {E}}_{\mathbf{x}} \left( (1-\eta (\mathbf{X}_t)) U_t \dfrac{e^{\alpha \vert \varUpsilon (\mathbf{X}_t)\vert } }{U_t} \right) \\&\le \dfrac{\varepsilon }{4} {\mathbb {E}}_{\mathbf{x}} \left( (1-\eta (\mathbf{X}_t)) U_t \right) \\&\le \dfrac{\varepsilon }{4} {\mathbb {E}}_{\mathbf{x}} \left( U_t \mathbf {1}_{U_t > R_3} \right) \\&\le \dfrac{\varepsilon }{4}. \end{aligned}$$

Of course, an analogous inequality holds for the terms ${\mathbb {E}}_{\mathbf{x}_k}(\cdot )$. This concludes the proof.