Local approximation of a metapopulation’s equilibrium

Abstract

We consider the approximation of the equilibrium of a metapopulation model, in which a finite number of patches are randomly distributed over a bounded subset $\Omega$ of Euclidean space. The approximation is good when a large number of patches contribute to the colonization pressure on any given unoccupied patch, and when the quality of the patches varies little over the length scale determined by the colonization radius. If this is the case, the equilibrium probability of a patch at z being occupied is shown to be close to $q_1(z)$, the equilibrium occupation probability in Levins’s model, at any point $z \in \Omega$ not too close to the boundary, if the local colonization pressure and extinction rates appropriate to z are assumed. The approximation is justified by giving explicit upper and lower bounds for the occupation probabilities, expressed in terms of the model parameters. Since the patches are distributed randomly, the occupation probabilities are also random, and we complement our bounds with explicit bounds on the probability that they are satisfied at all patches simultaneously.

Appendices

Appendix: Auxiliary results

Define the function $F(\cdot ;\tau ,\nu ):\,[0,\infty ) \rightarrow [0,1]$ by

$$\begin{aligned} F(x;\tau ,\nu ) \ :=\ \frac{f(\tau x)}{\nu + f(\tau x)}. \end{aligned}$$

This function has a fixed point at 0 and, if $f^{\prime }(0) \tau > \nu$, then it also has a non-zero fixed point. The function $q_{\alpha }(z)$ is the largest fixed point of $F(\cdot ;\rho (z),\alpha e(z))$, for fixed $\alpha > 0$.

Lemma 5.1

Suppose that Assumption E holds. Let q denote the largest fixed point of $F(\cdot ;\tau ,\nu )$. Then $q \le x$ if $(1-x) f(\tau x) \le \nu x$ and $q \ge x$ if $(1-x) f(\tau x) \ge \nu x$.

Proof

Since f is concave, increasing and not identically zero, by Assumption E, $F(\cdot ;\tau ,\nu )$ is concave, and strictly concave at 0. Hence $g(x) := F(x;\tau ,\nu ) - x$ is concave, strictly concave at zero, and has $g(0)=0$ and $g(\infty ) = -\infty$. If $g'(0) = F'(0;\tau ,\nu ) - 1 \le 0$, there is thus no other solution to $g(x)=0$. If $g'(0) > 0$, there is exactly one other solution q, and $g(x) > 0$ for $0< x < q$, and $g(x) < 0$ for $x > q$. Thus, $0< x < q$ if and only if $g(x) > 0$, and so

$$\begin{aligned} F(x;\tau ,\nu ) \ =\ \frac{f(\tau x)}{\nu + f(\tau x)} \ >\ x, \end{aligned}$$

implying that $(1-x)f(\tau x) > \nu x$; similarly, $q < x$ if and only if $g(x) < 0$ and $(1-x)f(\tau x) < \nu x$. $\square$

Lemma 5.2

Suppose that Assumptions C, D and E hold. Then, for all $\alpha \ge 1/2$, $q_{\alpha }$ is Lipschitz continuous on $\{z \in \Omega : q_{\alpha }(z) > 0\}$, with Lipschitz constant at most $L_q$, as defined in (3.4).

Proof

We write $F_\alpha (q,z) := F(q;\rho (z),\alpha e(z))$, where $F_\alpha :\,[0,1] \times \mathbb {R}^d \rightarrow \mathbb {R}_+$. For functions $g:\,[0,1] \times \mathbb {R}^d \rightarrow \mathbb {R}_+$, we denote by $D_q g$ the partial derivative of g with respect to its first argument, and by $D_j g$ the partial derivative in the direction of the jth coordinate axis in $\mathbb {R}^d$, $1\le j\le d$. By the implicit function theorem, $q_{\alpha }(z)$ is continuously differentiable in an open neighbourhood of z, with

$$\begin{aligned} (D_j q_\alpha )(z) \ =\ - \left[ (D_q F_\alpha )(q_{\alpha }(z),z) -1\right] ^{-1} (D_j F_\alpha )(q_{\alpha }(z),z),\quad 1\le j\le d, \nonumber \\ \end{aligned}$$

(5.1)

provided that

$$\begin{aligned} (D_q F_\alpha )(q_{\alpha }(z),z)\ \ne \ 1. \end{aligned}$$

(5.2)

For any $z \in \Omega$ and $q \in [0,1]$,

$$\begin{aligned} (D_q F_\alpha )(q,z)&\ =\ \frac{\alpha e(z) \rho (z) f^{\prime }(q \rho (z))}{\left( \alpha e(z) + f(q\rho (z))\right) ^{2}}. \end{aligned}$$

As $q_{\alpha }(z) = F(q_{\alpha }(z);\alpha e(z),\rho (z))$,

$$\begin{aligned} (D_q F_\alpha )(q_{\alpha }(z),z)&\ =\ \frac{(1-q_{\alpha }(z)) \rho (z) f^{\prime }(q_{\alpha }(z) \rho (z))}{\left( \alpha e(z) + f(q_{\alpha }(z)\rho (z))\right) } \\&\ =\ \frac{(1-q_{\alpha }(z)) q_{\alpha }(z)\rho (z) f^{\prime }(q_{\alpha }(z) \rho (z))}{ f(q_{\alpha }(z)\rho (z))}. \end{aligned}$$

By the mean value theorem, there exists a $\tilde{q} \in (0,q_{\alpha }(z))$ such that

$$\begin{aligned} f^{\prime }(\tilde{q} \rho (z)) \ =\ \frac{f(q_{\alpha }(z)\rho (z))}{q_{\alpha }(z) \rho (z)}. \end{aligned}$$

As f is concave, $f^{\prime }(\tilde{q} \rho (z)) \ge f^{\prime }(q_{\alpha }(z) \rho (z))$. Therefore,

$$\begin{aligned} (D_q F_\alpha )(q_{\alpha }(z),z) \ \le \ 1-q_{\alpha }(z), \end{aligned}$$

(5.3)

and (5.2) holds for any $z \in \Omega$ such that $q_{\alpha }(z) > 0$. Differentiating $F_\alpha$ in direction j yields

$$\begin{aligned} (D_j F_\alpha )(q,z)&\ =\ \frac{\alpha e(z) q (D_j \rho )(z) f^{\prime }(q\rho (z)) - \alpha (D_j e)(z) f(q\rho (z))}{\left( \alpha e(z) + f(q\rho (z)) \right) ^{2}}. \end{aligned}$$

Evaluating this derivative at $(q_{\alpha }(z),z)$ gives

$$\begin{aligned} (D_j F_\alpha )(q_{\alpha }(z),z)&\ =\ \frac{\alpha e(z) q_{\alpha }(z) (D_j \rho )(z) f^{\prime }(q_{\alpha }(z)\rho (z)) - \alpha (D_j e)(z) f(q_{\alpha }(z) \rho (z))}{\left( \alpha e(z) + f(q_{\alpha }(z)\rho (z)) \right) ^{2}} \nonumber \\&\ =\ q_{\alpha }(z)\frac{(1-q_{\alpha }(z)) (D_j \rho )(z) f^{\prime }(q_{\alpha }(z)\rho (z)) - \alpha (D_j e)(z) }{\left( \alpha e(z) + f(q_{\alpha }(z)\rho (z)) \right) }. \end{aligned}$$

(5.4)

Combining Eqs. (5.1) and (5.4) with the bound (5.3) yields

$$\begin{aligned} \left| (D_j q_\alpha )(z) \right|&\ \le \ \frac{1}{e(z)} \left( \frac{L_f}{\alpha } \left| (D_j \rho )(z) \right| + \left| (D_j e)(z) \right| \right) \ \le \ \frac{1}{e_\mathrm{min}}(\alpha ^{-1}L_fL_\rho + L_e). \end{aligned}$$

Therefore, for any $\alpha \ge 1/2$, $q_{\alpha }$ is Lipschitz on $\{z\in \Omega : q_{\alpha } > 0\}$ with the Lipschitz constant given in (3.4). $\square$

Lemma 5.3

Suppose that Assumption E holds and that $q_{1}(z) \ge \eta > 0$. Then, for any $\beta \in (1, (1-\eta )^{-1})$, $q_{\beta }(z) \ge \beta \eta + 1-\beta$, and, for any $\alpha \in (0,1)$, $q_{\alpha }(z) \ge \alpha \eta$.

Proof

For any $\beta \in (1, (1-\eta )^{-1})$, it follows that $0< \beta \eta + 1- \beta < \eta$ and that, by Assumption E,

$$\begin{aligned} f\left( \rho (z)(\beta \eta + 1 - \beta ) \right) \ \ge \ \left( \beta + \frac{1-\beta }{\eta }\right) f(\rho (z) \eta ). \end{aligned}$$

As $q_{1}(z) \ge \eta$, we can apply Lemma 5.1 to give $f(\rho (z) \eta ) \ge e(z)\eta /(1-\eta )$, and hence

$$\begin{aligned} f\left( \rho (z)(\beta \eta + 1 - \beta ) \right)\ge & {} \left( \beta + \frac{1-\beta }{\eta }\right) \frac{e(z)\eta }{1-\eta } \\= & {} (\beta \eta + 1 - \beta ) \frac{e(z)}{1-\eta }\ \ =\ \frac{\beta e(z)(\beta \eta + 1 - \beta )}{1- (\beta \eta + 1 - \beta )}. \end{aligned}$$

Applying Lemma 5.1 again, we see that $q_{\beta }(z) \ge \beta \eta +1 -\beta$.

For $\alpha \in (0,1)$ we follow similar reasoning to show that

$$\begin{aligned} f(\rho (z) \alpha \eta )\ \ge \ \alpha f(\rho (z) \eta )\ \ge \ \alpha e(z) \eta /(1-\eta )\ \ge \ \alpha e(z) \alpha \eta / (1-\alpha \eta ), \end{aligned}$$

and applying Lemma 5.1 we see that $q_{\alpha }(z) \ge \alpha \eta$. $\square$

In the following we let $\sigma _{n\setminus i} := \frac{A}{(n-1)}\sum _{j\ne i} \delta _{z_j}$, which is A times the empirical measure of patches excluding patch i.

Lemma 5.4

Suppose that Assumptions A, B and D hold. Then, for any $h:\, \Omega \rightarrow [0,H]$, $0 < t\le Hc_\mathrm{max}\sigma _\mathrm{max}v_d$ and $z \in \Omega$,

$$\begin{aligned} \mathbb {P} \left( \left| \int c(z,y;r) h(y) [\sigma _{n\setminus i}(dy) - \sigma (y)\,dy] \right| \ \ge \ t\right) \le 2 \exp \left( - C_2 ((n-1) r^d/A) (t/H)^{2}\right) , \end{aligned}$$

where

$$\begin{aligned} C_2 \ :=\ \{3\{c_\mathrm{max}\}^2 \sigma _\mathrm{max}v_d\}^{-1}. \end{aligned}$$

Proof

Note first that, for patches distributed independently with density $A^{-1}\sigma (\cdot )$, we have

$$\begin{aligned} \mathbb {E}\left\{ \int c(z,y;r) h(y) \sigma _{n\setminus i}(dy) \right\} \ =\ \int c(z,y;r) h(y) \sigma (y)\,dy. \end{aligned}$$

The left hand side of this expression is a sum of i.i.d. random variables, each bounded by $Hc_\mathrm{max}A/((n-1)r^d)$, and each with variance at most $\{Hc_\mathrm{max}A/((n-1)r^d)\}^2 \sigma _\mathrm{max}v_d r^d/A$, where, as before, $v_d$ denotes the volume of the unit ball in $\mathbb {R}^d$. Hence, applying McDiarmid (1998, Theorem 2.7) it follows that, for any $t> 0$,

$$\begin{aligned}&{\mathbb {P}\left( \left| \int c(z,y;r) h(y) \sigma _{n\setminus i}(dy) - \int c(z,y;r) h(y) \sigma (y)\,dy \right| > t\right) }\\&\quad \ \le \ 2\exp \left( - \frac{t^2}{2((n-1)\{Hc_\mathrm{max}A/((n-1)r^d)\}^2 \sigma _\mathrm{max}v_d A^{-1}r^d + \{Hc_\mathrm{max}A/((n-1)r^d)\}t/3)} \right) \\&\quad \ \le \ 2\exp \left( - \frac{t^2}{3(A/(n-1)r^d)\{Hc_\mathrm{max}\}^2 \sigma _\mathrm{max}v_d } \right) \\&\quad \ =\ 2\exp \left( - C_2((n-1)r^d/A)(t/H)^2 \right) \end{aligned}$$

if $t/H \le c_\mathrm{max}\sigma _\mathrm{max}v_d$. $\square$

Lemma 5.5

Suppose Assumptions A, B and D hold. Let $N(\Omega ,r)$ be the number of balls of radius r required to cover $\Omega$. If $n > 2 N(\Omega ,r/3)$, then T is primitive with probability at least

$$\begin{aligned} 1 - N(\Omega ,r/3) \exp \left( - n \min _{z\in \Omega } A^{-1}\int _{\Omega } \mathbb {I}(\Vert y-z\Vert \le r/3) \sigma (y) dy\right) \end{aligned}$$

(5.5)

Proof

Let $\tilde{T}$ be the incidence matrix of T, that is $\tilde{T}_{ij} = 1$ if $T_{ij} > 0$ and $T_{ij} = 0$ otherwise. The matrix T is primitive if $\tilde{T}$ is both irreducible and acyclic (Seneta 1981, Theorem 1.4 of Chapter 1). By Assumptions B and D, $\tilde{T}$ is symmetric and $\tilde{T}_{ii} = 0$. Define the graph $\mathcal {G} = (V,E)$ where $V:= \{z_{1},\ldots ,z_{n}\}$ and $(z_{i},z_{j}) \in E$ if and only if $\Vert z_{i} - z_{j} \Vert \le r$. The matrix $\tilde{T}$ is the adjacency matrix of $\mathcal {G}$ and is irreducible if $\mathcal {G}$ is connected. Let $N:= N(\Omega ,r/3)$ and $y_{1},\ldots ,y_{N} \in \Omega$ such that $\Omega \subset \cup _{i}^{N} B(y_{i},r/3)$, where B(y, r) is a closed ball of radius r centered at y. Define the graph $\hat{\mathcal {G}} = (\hat{V},\hat{E})$ where $\hat{V} = \{y_{1},\ldots ,y_{N}\}$ and $(y_{i},y_{j}) \in \hat{E}$ if and only if $\Vert y_{i} - y_{j} \Vert \le r/3$. Since $\Omega$ is connected, the graph $\hat{\mathcal {G}}$ is also connected. Suppose that each ball $B(y_{i}, r/3)$ contains at least one element of V. For any $z_{i}$ and $z_{j}$, there exists a path $\{y_{k_0},y_{k_1},\ldots , y_{k_{m+1}}\}$ in $\hat{\mathcal {G}}$ such that $z_{i} \in B(y_{k_0},r/3)$ and $z_{j} \in B(y_{k_{m+1}},r/3)$. Taking any $z_{k_\ell } \in B(y_{k_\ell },r/3)$, we have constructed a path $\{z_{i},z_{k_1},\ldots ,z_{k_m},z_{j}\}$ in $\mathcal {G}$, since

$$\begin{aligned} \Vert z_{k_\ell } - z_{k_{\ell +1}} \Vert \le \Vert z_{k_\ell } - y_{k_\ell } \Vert + \Vert y_{k_\ell } - y_{k_{\ell +1}} \Vert + \Vert y_{k_{\ell +1}} - z_{k_{\ell +1}} \Vert \le r. \end{aligned}$$

Thus $\mathcal {G}$ is connected and $\tilde{T}$ is irreducible if each ball $B(y_{i}, r/3)$ contains at least one element of V. This occurs with probability at least that given in (5.5).

To show that $\tilde{T}$ is acyclic, it is sufficient to show that $\tilde{T}^{2}_{ii} > 0$ and $\tilde{T}^{3}_{ii} > 0$ for some i, since $\tilde{T}$ is irreducible (Seneta 1981, Lemma 1.2 of Chapter 1). This is true if there are three elements of V that are within distance r of each other. Since $n > 2N(\Omega ,r/3)$, there is at least one $B(y_{i},r/3)$ which contains at least three elements of V, and these are within distance 2r / 3 of each other, completing the proof. $\square$

Appendix: Proof of the upper bound

In this section, we prove Theorem 3.1. Suppose

$$\begin{aligned} f\left( \int a(y) c(z_{i},y;r) p_{{\alpha _1,\alpha _2}}^{+}(y) \sigma _{n\setminus i}(dy)\right) \ \le \ \frac{e(z_{i}) p_{{\alpha _1,\alpha _2}}^{+}(z_{i})}{1-p_{{\alpha _1,\alpha _2}}^{+}(z_{i})} \end{aligned}$$

(6.1)

for all $i = 1,\ldots ,n$. Then $E_{n} (p_{{\alpha _1,\alpha _2}}^{+}(z))_{i} \le p_{{\alpha _1,\alpha _2}}^{+}(z_{i})$ for all $i=1,\ldots ,n$. As $E_{n}$ is monotone, the sequence of iterates of $E_{n}$ starting from $p_{{\alpha _1,\alpha _2}}^{+}(z_{i}),\ i=1,\ldots ,n$ is decreasing. If T is primitive, then the cone limit set trichotomy (Hirsch and Smith 2005, Theorem 6.3) holds and each sequence of iterates starting from a non-zero inital value must converge to $p^{*}$. Hence, $p_{{\alpha _1,\alpha _2}}^{+}(z_{i}),\ i=1,\ldots ,n$ is an upper bound on $p^{*}$. The matrix T is primitive with high probability by Lemma 5.5. It remains to show that for some $1/2< \alpha _2 \le \alpha _1 < 1$ inequality (6.1) holds.

Since $c(z,y;r) = 0$ for all y such that $\Vert y-z\Vert > r$, and since $p_{{\alpha _1,\alpha _2}}^{+}$ is Lipschitz with constant $L_q$, as given in (3.4), we have

$$\begin{aligned}&f\left( \int a(y)c(z,y;r) p_{{\alpha _1,\alpha _2}}^{+}(y) \sigma _{n\setminus i}(dy)\right) \ \le \ f\left( \int a(y)c(z,y;r) \left[ p_{{\alpha _1,\alpha _2}}^{+}(z) + L_q r\right] \sigma _{n\setminus i}(dy)\right) \nonumber \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \le \ f\left( \rho _{n\setminus i}(z) p_{{\alpha _1,\alpha _2}}^{+}(z)\right) + L_{f} L_q \rho _{n\setminus i}(z) r, \end{aligned}$$

(6.2)

for $\alpha _{2} > 1/2$, where $\rho _{n\setminus i}(z) := \int a(y) c(z,y;r) \sigma _{n\setminus i}(dy)$.

U1 For all z such that $q_{\alpha _{1}}(z) < 1 - \alpha _{2}$,

$$\begin{aligned}&{ \frac{e(z) p_{{\alpha _1,\alpha _2}}^{+}(z)}{1-p_{{\alpha _1,\alpha _2}}^{+}(z)} - f\left( \int a(y)c(z,y;r) p_{{\alpha _1,\alpha _2}}^{+}(y) \sigma _{n\setminus i}(dy)\right) } \nonumber \\&\quad \ge \ \frac{e(z) (1-\alpha _{2})}{\alpha _{2}} - f(\rho _{n}(z) (1-\alpha _{2})) - L_{f} L_q \rho _{n\setminus i}(z) r \nonumber \\&\quad \ge \ \frac{e(z) (1-\alpha _{2})}{\alpha _{2}} - f(\rho (z) (1-\alpha _{2})) -\,L_{f} ((1-\alpha _{2}) + L_q r) \left| \rho (z)\right. \left. -\, \rho _{n\setminus i}(z) \right| - L_{f} L_q \rho (z) r.\nonumber \\ \end{aligned}$$

(6.3)

From Lemma 5.1, if $q_{\alpha _{1}}(z) \le (1-\alpha _{2})$, then $f(\rho (z)(1-\alpha _{2})) \le \alpha _{1}(1-\alpha _{2}) e(z)/\alpha _{2}$. Combining this bound with inequality (6.3) gives

$$\begin{aligned}&{ \frac{e(z) p_{{\alpha _1,\alpha _2}}^{+}(z)}{1-p_{{\alpha _1,\alpha _2}}^{+}(z)} - f\left( \int a(y)c(z,y;r) p_{{\alpha _1,\alpha _2}}^{+}(y) \sigma _{n\setminus i}(dy)\right) } \nonumber \\&\quad \ge \ \frac{e(z)(1-\alpha _{1})(1-\alpha _{2})}{\alpha _{2}} - L_{f} ((1-\alpha _{2}) + L_qr) \left| \rho (z) - \rho _{n\setminus i}(z) \right| - L_{f} L_q \rho (z) r \nonumber \\&\quad \ge \ p_{{\alpha _1,\alpha _2}}^{+}(z) \left( \frac{(1-\alpha _{1}) e(z)}{\alpha _{2}} - \frac{L_{f} L_q r \rho (z) }{ p_{{\alpha _1,\alpha _2}}^{+}(z)} - L_{f} \left( 1 + \frac{L_q r}{ p_{{\alpha _1,\alpha _2}}^{+}(z)} \right) \left| \rho (z) - \rho _{n\setminus i}(z) \right| \right) , \end{aligned}$$

(6.4)

where the last inequality follows as $q_{\alpha _{1}}(z) < 1-\alpha _{2}$ implies $p_{{\alpha _1,\alpha _2}}^{+}(z) = 1-\alpha _{2}$.

U2 We now consider the case where $q_{\alpha _{1}} (z) \ge 1- \alpha _{2}$. Using the fact that $q_{\alpha _{1}}(z)$ is a fixed point of $F(\cdot ;\rho (z),\alpha _{1} e(z))$ and inequality (6.2), it follows that

$$\begin{aligned}&{ \frac{e(z) p_{{\alpha _1,\alpha _2}}^{+}(z)}{1-p_{{\alpha _1,\alpha _2}}^{+}(z)} - f\left( \int a(y) c(z,y;r) p_{{\alpha _1,\alpha _2}}^{+}(y) \sigma _{n\setminus i}(dy)\right) } \nonumber \\&\quad \ge \ \frac{f(\rho (z) q_{\alpha _{1}}(z))}{\alpha _{1}} - f(\rho _{n\setminus i}(z) q_{\alpha }(z)) - L_{f}L_q r \rho _{n\setminus i}(z) \nonumber \\&\quad \ge \ \frac{(1-\alpha _{1})}{\alpha _{1}} f(\rho (z) q_{\alpha _{1}}(z)) - L_{f} L_q r \rho (z) - L_{f} (q_{\alpha }(z) + L_q r) \left| \rho (z) - \rho _{n\setminus i}(z)\right| . \end{aligned}$$

(6.5)

Since $f(\rho (z) q_{\alpha _{1}}(z)) = \alpha _{1} e(z) q_{\alpha _{1}}(z)/(1-q_{\alpha _{1}}(z)) \ge \alpha _{1} e(z) q_{\alpha _{1}}(z)$ and $q_{\alpha _{1}}(z) \ge 1-\alpha _{2}$ here, inequality (6.5) becomes

$$\begin{aligned}&{ \frac{e(z) p_{{\alpha _1,\alpha _2}}^{+}(z)}{1-p_{{\alpha _1,\alpha _2}}^{+}(z)} - f\left( \int a(y) c(z,y;r) p_{{\alpha _1,\alpha _2}}^{+}(y) \sigma _{n\setminus i}(dy)\right) } \nonumber \\&\quad \ge \ q_{\alpha _{1}}(z) \left( (1-\alpha _{1}) e(z) - \frac{L_{f} L_q r \rho (z) }{q_{\alpha _{1}}(z)} - L_{f} \left( 1 + \frac{L_q r}{q_{\alpha _{1}}(z)}\right) \left| \rho (z) - \rho _{n\setminus i}(z)\right| \right) \nonumber \\&\quad \ge \ p^{+}_{\alpha _{1},\alpha _{2}}(z) \left( (1-\alpha _{1}) e(z) - \frac{L_{f} L_q r \rho (z) }{p_{{\alpha _1,\alpha _2}}^{+}(z)} - L_{f} \left( 1 + \frac{L_q r}{p_{{\alpha _1,\alpha _2}}^{+}(z)}\right) \left| \rho (z) - \rho _{n\setminus i}(z)\right| \right) . \end{aligned}$$

(6.6)

U3 Combining inequalities (6.4) and (6.6), we see that inequality (6.1) holds if

$$\begin{aligned} (1-\alpha _{1}) e(z_{i}) - \frac{L_{f} L_q r \rho (z_{i})}{p_{{\alpha _1,\alpha _2}}^{+}(z_{i})} - L_{f} \left( 1 + \frac{L_q r}{p_{{\alpha _1,\alpha _2}}^{+}(z_{i})}\right) \left| \rho (z_{i}) - \rho _{n\setminus i}(z_{i})\right| \ge 0 \end{aligned}$$

which is equivalent to

$$\begin{aligned} \frac{(1-\alpha _{1}) e(z_{i}) p_{{\alpha _1,\alpha _2}}^{+}(z_{i}) - L_{f} L_q r \rho (z_{i})}{L_{f}\left( p_{{\alpha _1,\alpha _2}}^{+}(z_{i}) + L_qr\right) } \ \ge \ \left| \rho (z_{i}) - \rho _{n\setminus i}(z_{i})\right| . \end{aligned}$$

By Lemma 5.3, $p_{{\alpha _1,\alpha _2}}^{+}(z) \ge \bigl (\alpha _{1} \eta _{\Omega } \vee (1-\alpha _{2})\bigr )$, and from inequality (3.3) together with $L_f\rho _\mathrm{max}/e_\mathrm{min}> 1/2$, we see that inequality (6.1) is satisfied if

$$\begin{aligned} \frac{(1-\alpha _{1}) e_\mathrm{min}}{4L_{f}}\ \ge \ \left| \rho (z_{i}) - \rho _{n\setminus i}(z_{i})\right| . \end{aligned}$$

Applying Lemma 5.4 yields the bound

$$\begin{aligned}&{\mathbb {P} \left( \max _{i} \left| \int a(y)c(z_{i},y;r) [\sigma _{n\backslash i}(dy) - \sigma (y)\,dy] \right| \le \frac{(1-\alpha _{1})e_\mathrm{min}}{4L_{f}}\right) }\\&\quad \ge \ 1 - n \sup _{z\in \Omega } \mathbb {P} \left( \left| \int a(y)c(z,y;r) [\sigma _{n\backslash i}(dy) - \sigma (y)\,dy] \right| > \frac{(1-\alpha _{1})e_\mathrm{min}}{4L_{f}} \right) \\&\quad \ge \ 1 - 2n \exp \left( -C_2\{(n-1)r^d/A\}\frac{e_\mathrm{min}^{2} (1-\alpha _{1})^{2}}{16 a_\mathrm{max}^2 L_{f}^{2}} \right) , \end{aligned}$$

if $\frac{(1-\alpha _{1})e_\mathrm{min}}{4L_{f}} \le \rho _\mathrm{max}$, which is the case if $L_f\rho _\mathrm{max}/e_\mathrm{min}> 1/2$.

The situation in which $L_f\rho _\mathrm{max}/e_\mathrm{min}\le 1/2$ is one in which the metapopulation is nowhere viable, so the conclusion is not surprising. We begin by noting that $q_\alpha (z) = 0$ for all $\alpha > 1/2$ if $L_f\rho _\mathrm{max}/e_\mathrm{min}\le 1/2$, so that $p^{+}_{\alpha _{1},\alpha _{2}}(z) = (1-\alpha _2)$ for all $z\in \Omega$. Lemma 5.4 with $t = \rho _\mathrm{max}/2$ then shows that

$$\begin{aligned} \left| \rho (z_{i}) - \rho _{n\setminus i}(z_{i})\right| \ \le \ \frac{1}{2} \rho _\mathrm{max}, \end{aligned}$$

on an event of probability at least

$$\begin{aligned} 1 - 2 \exp \left\{ -C_2 \frac{(n-1) r^d}{A} \Bigl (\frac{ \rho _\mathrm{max}}{2 a_\mathrm{max}}\Bigr )^2 \right\} . \end{aligned}$$

Hence, on this event, we have

$$\begin{aligned}&f\left( (1-\alpha _2)\int a(y)c(z_{i},y;r) \sigma _{n\setminus i}(dy)\right) \ \le \ \frac{3}{2}(1-\alpha _2)L_f \rho _\mathrm{max}\\&\quad \ \le \ (1-\alpha _2)e_\mathrm{min}\ \le \ \frac{e(z_i)(1-\alpha _2)}{\alpha _2}. \end{aligned}$$

This establishes 6.1, on an event with probability as given in Theorem 3.1, for any choice of $1/2< \alpha _2 < 1$, since $p^{+}_{\alpha _{1},\alpha _{2}}(z) = (1-\alpha _2)$ for all $z\in \Omega$. This completes the proof of Theorem 3.1.

Appendix: Proof of the lower bound

To find a good lower bound on $p^{*}_{n}$, we introduce a modification of $E_{n}$. For any $\Theta \subseteq \Omega$ and $\beta >1$ define the operator $E_{n,\Theta ,\beta }: [0,1]^{n} \rightarrow [0,1]^{n}$ by

$$\begin{aligned} E_{n,\Theta ,\beta }(p)_{i} \ :=\ \frac{f\left( (A/(n-1)) \sum _{j\ne i} a(z_{j}) c(z_{i},z_{j};r) \mathbb {I}(z_{j} \in \Theta ) p_{j}\right) }{ \beta e(z_{i}) + f\left( (A/(n-1)) \sum _{j\ne i} a(z_{j}) c(z_{i},z_{j};r) \mathbb {I}(z_{j} \in \Theta ) p_{j}\right) }. \end{aligned}$$

Denote the largest fixed point of $E_{n,\Theta ,\beta }$ by $p_{n,\Theta ,\beta }^{*}$. Since f is an increasing function, for any $\Theta \subseteq \Omega$ and any $\beta >1$, $E _{n,\Theta ,\beta } (p) \le E_{n,\Theta ,1}(p) \le E_{n}(p)$ for all $p \in [0,1]^{n}$, which implies that $p_{n,\Theta ,\beta }^{*} \le p^{*}_{n,\Theta ,1} \le p_{n}^{*}$. Thus a lower bound on $p_{n,\Theta ,\beta }^{*}$ yields a lower bound on $p_{n}^{*}$. To construct a lower bound on $p_{n,\Theta ,\beta }^{*}$, we examine the limiting form of $E_{n,\Theta ,\beta }$ as $n \rightarrow \infty$. Let $C^{+}(\Theta )$ be the set of non-negative functions on $\Theta$ and define $E_{\Theta ,\beta } :\, C^{+}(\Theta ) \rightarrow C^{+}(\Theta )$ by

$$\begin{aligned} E_{\Theta ,\beta }(p) \ :=\ \frac{f\left( \int a(y) c(z,y;r) \mathbb {I}(y \in \Theta ) p(y) \sigma (dy) \right) }{ \beta e(z) + f\left( \int a(y) c(z,y;r) \mathbb {I}(y \in \Theta ) p(y) \sigma (dy) \right) }. \end{aligned}$$

Let $p_{\Theta ,\beta }^{*}$ denote the largest fixed point of $E_{\Theta ,\beta }$. Our aim now is to find a $\beta > 1$ such that with high probability

$$\begin{aligned} p_{i,n}^{*}\ \ge \ p^{*}_{\Theta ,\beta }(z_{i}), \end{aligned}$$

(7.1)

for all $z_{i} \in \Theta$.

Lemma 7.1

Suppose that Assumptions A, B, D and E hold. Suppose also that, for a given $\Theta \subseteq \Omega$ and $\beta > 1$, there exists an $\epsilon _{\Theta ,\beta } > 0$ such that $p_{\Theta ,\beta }^{*} (z) \ge \epsilon _{\Theta ,\beta }$ for all $z \in \Theta$. Assume that

$$\begin{aligned} e_\mathrm{min}\epsilon _{\Theta ,\beta } (\beta -1)&\ \le \ L_f\rho _\mathrm{max}. \end{aligned}$$

(7.2)

Then

$$\begin{aligned} \mathbb {P}\left( p_{i,n}^{*} \ge p^{*}_{\Theta ,\beta }(z_{i}) \text{ for } \text{ all } z_{i} \in \Theta \right) \ \ge \ 1- 2n \exp \left( -C_2 \frac{(n-1)r^d}{A} \frac{ e_\mathrm{min}^2 \epsilon _{\Theta ,\beta }^{2} (\beta -1)^{2}}{4a_\mathrm{max}^2L_{f}^{2}} \right) .\nonumber \\ \end{aligned}$$

(7.3)

Proof

Suppose that

$$\begin{aligned} \frac{ e(z_{i}) p_{\Theta ,\beta }^{*}(z_{i})}{1 - p_{\Theta ,\beta }^{*}(z_{i})} \ \le \ f\left( \int a(y) c(z_{i},y;r) \mathbb {I}(y \in \Theta ) p_{\Theta ,\beta }^{*}(y) \sigma _{n\setminus i} (dy) \right) \end{aligned}$$

(7.4)

for all $z_{i} \in \Theta$. Then $E_{n,\Theta ,1}$ maps the set $\{p: p_{\Theta ,\beta }^{*}(z_{i}) \le p_{i} \le 1\}$ into itself as the map is monotone. Applying the Brouwer fixed point theorem, we see $p_{\Theta ,\beta }^{*}(z_{i}) \le p_{n,\Theta ,1,i}$ for all $z_{i} \in \Theta$. Since $p^{*}_{n,\Theta ,1} \le p^{*}_{n}$, it remains to verify that inequality (7.4) holds.

Now

$$\begin{aligned}&{ f\left( \int a(y) c(z_{i},y;r) \mathbb {I}(y \in \Theta ) p_{\Theta ,\beta }^{*}(y) \sigma _{n\setminus i} (dy) \right) } \\&\quad =\ f\biggl (\int a(y) c(z_{i},y;r) \mathbb {I}(y \in \Theta ) p_{\Theta ,\beta }^{*}(y) \sigma (y)\,dy \\&\qquad \quad \qquad + \int a(y) c(z_{i},y;r) \mathbb {I}(y \in \Theta ) p_{\Theta ,\beta }^{*} (y) [\sigma _{n\setminus i}(dy) - \sigma (y)\,dy] \biggr ) \\&\quad \ge \ f\biggl (\int a(y) c(z_{i},y;r) \mathbb {I}(y \in \Theta ) p_{\Theta ,\beta }^{*}(y) \sigma (y)\,dy \biggr ) \\&\qquad - L_{f} \left| \int a(y) c(z_{i},y;r) \mathbb {I}(y \in \Theta ) p_{\Theta ,\beta }^{*}(y) [\sigma _{n\setminus i}(dy) - \sigma (y)\,dy] \right| \\&\quad \ge \ \frac{\beta e(z_{i}) p_{\Theta ,\beta }^{*}(z_{i})}{1-p_{\Theta ,\beta }^{*}(z_{i})} - L_{f} \left| \int a(y) c(z_{i},y;r) \mathbb {I}(y \in \Theta ) p_{\Theta ,\beta }^{*}(y)[\sigma _{n\setminus i}(dy) - \sigma (y)\,dy] \right| . \end{aligned}$$

Therefore,

$$\begin{aligned}&{f\left( \int a(y) c(z_{i},y;r) \mathbb {I}(y \in \Theta ) p_{\Theta ,\beta }^{*}(y) \sigma _{n\setminus i} (dy) \right) - \frac{ e(z_{i}) p_{\Theta }^{*}(z_{i})}{1 - p_{\Theta ,\beta }^{*}(z_{i})}} \\&\quad \ge \ \frac{(\beta -1) e(z_{i}) p_{\Theta ,\beta }^{*}(z_{i})}{1-p_{\Theta ,\beta }^{*}(z_{i})} - L_{f} \left| \int a(y) c(z_{i},y;r) \mathbb {I}(y \in \Theta ) p_{\Theta ,\beta }^{*}(y)[\sigma _{n\setminus i}(dy) - \sigma (y)\,dy] \right| . \end{aligned}$$

As $p_{\Theta }^{*} (z) \ge \epsilon _{\Theta ,\beta }$ for all $z \in \Theta$, inequality (7.1) will hold if

$$\begin{aligned} \frac{(\beta -1)e_\mathrm{min}\epsilon _{\Theta ,\beta }}{L_{f}} - \left| \int a(y) c(z_{i},y;r) \mathbb {I}(y \in \Theta ) p_{\Theta ,\beta }^{*} (y) [\sigma _{n\setminus i}(dy) - \sigma (y)\,dy] \right| \ \ge \ 0,\nonumber \\ \end{aligned}$$

(7.5)

for all $z_{i} \in \Theta$. Applying Lemma 5.4 yields the bound

$$\begin{aligned}&\mathbb {P} \left( \max _{i: z_{i} \in \Theta } \left| \int a(y) c(z_{i},y;r) \mathbb {I}(y \in \Theta ) p_{\Theta ,\beta }^{*} (y) [\sigma _{n\backslash i}(dy) - \sigma (y)\,dy] \right| \right. \left. > \frac{e_\mathrm{min}\epsilon _{\Theta ,\beta } (\beta -1)}{L_{f}} \right) \\&\quad \le 2n \exp \left( -C_2 \frac{(n-1)r^d}{A} \frac{ e_\mathrm{min}^2 \epsilon _{\Theta ,\beta }^{2} (\beta -1)^{2}}{a_\mathrm{max}^2L_{f}^{2}} \right) , \end{aligned}$$

if inequality (7.2) holds. $\square$

Lemma 7.1 shows that inequality (7.1) holds with high probability if $p^{*}_{\Theta ,\beta }$ can be bounded away from zero. We now establish a lower bound on $p^{*}_{\Theta ,\beta }$.

To state the lemma that we need, some further notation is necessary. With $\Theta := \Theta _{x,t}$ as before, suppose that $\eta _{\Theta }:= \min _{z\in \Theta }q_{1}(z) > 0$. Recall $C_1$, as introduced following Assumption E, and set

$$\begin{aligned} {c}_{\Theta }:= & {} \min _{z\in \Theta } \int ^{1}_{0} c_{z}(\lambda ) \lambda ^{d}\, d\lambda ; \end{aligned}$$

(7.6)

$$\begin{aligned} C_3:= & {} v_dc_\mathrm{max}(a_\mathrm{max}L_{\sigma } + \sigma _\mathrm{max}L_{a}). \end{aligned}$$

(7.7)

Lemma 7.2

Suppose that Assumptions B–E hold. Define

$$\begin{aligned} q_{\Theta ,\beta ^{\prime },m}(z) := \left( m(t-\Vert z-x\Vert ) \wedge q_{\beta ^{\prime }}(z) \right) . \end{aligned}$$

If there exists constants $\beta ^{\prime } \in (\beta , (1-\eta _{\Theta })^{-1}),\ \theta _{1} \in (1,\infty ) ,\ \theta _{2} \in (0,1)$ and $m \in (0,\infty )$ such that

$$\begin{aligned}&(1+ \theta _{1})mr \ \le \ (\beta '\eta _{\Theta }+ 1 - \beta '); \end{aligned}$$

(7.8)

$$\begin{aligned}&L_{f} \rho _\mathrm{max}(m\vee L_q) r \ \le \ (\beta ^{\prime } - \beta ) e_\mathrm{min}\theta _{1} mr; \end{aligned}$$

(7.9)

$$\begin{aligned}&\frac{L_{f} (C_3 + \rho _\mathrm{max}/t) r}{\theta _{2}} + \rho _\mathrm{max}(C_1 \rho _\mathrm{max}+L_f) \theta _{1} mr \ \le \ e_\mathrm{min}(\beta \eta _{\Theta }+ 1 - \beta ); \end{aligned}$$

(7.10)

$$\begin{aligned}&\frac{r}{t} \ \le \ \min \Bigl \{ \theta _{2},\frac{1}{2(2+\theta _{1})} \Bigr \}; \end{aligned}$$

(7.11)

$$\begin{aligned}&(C_3 + \rho _\mathrm{max}/t)r \ \le \ a_\mathrm{min}v_{d-1} \sigma _\mathrm{min}\left( {c}_{\Theta }-2c_\mathrm{max}\theta _{2}\right) (1-\theta ^{2}_{2})^{(d-1)/2}, \end{aligned}$$

(7.12)

then $q_{\Theta ,\beta ^{\prime },m}(z) \le p_{\Theta ,\beta }^{*}(z)$ for all $z \in \Theta$.

Proof

Suppose that

$$\begin{aligned} \frac{\beta e(z) q_{\Theta ,\beta ^{\prime },m}(z)}{1-q_{\Theta ,\beta ^{\prime },m}(z)} \ \le \ f\left( \int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \mathbb {I}(y \in \Theta ) \sigma (y)\,dy \right) \end{aligned}$$

(7.13)

for all $z \in \Theta$. Then $E_{\Theta ,\beta }$ maps $\{p \in C^{+}(\Theta ): q_{\Theta ,\beta ^{\prime },m} \le p\}$ into itself. The map $E_{\Theta ,\beta }$ is compact by Assumption C. By the Schauder fixed point theorem, $q_{\Theta ,\beta ',m} \le p^{*}_{\Theta ,\beta }$. We now verify that inequality (7.13) holds.

L1 For any z such that $\Vert z -x\Vert \le t-r$,

$$\begin{aligned} \int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \mathbb {I}(y\in \Theta ) \sigma (y)\,dy \ =\ \int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \sigma (y)\,dy. \end{aligned}$$

From Lemma 5.2, $q_{\Theta ,\beta ^{\prime },m}$ is Lipschitz continuous with constant $(m \vee L_q)$. Hence,

$$\begin{aligned}&{f\left( \int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \mathbb {I}(y \in \Theta ) \sigma (y)\,dy \right) } \nonumber \\&\quad \ =\ f\left( q_{\Theta ,\beta ^{\prime },m}(z) \int a(y) c(z,y;r) \sigma (y)\,dy\right. \nonumber \\&\qquad \quad \qquad \left. + \int a(y) c(z,y;r) [q_{\Theta ,\beta ^{\prime },m}(y) - q_{\Theta ,\beta ^{\prime },m}(z)] \sigma (y)\,dy \right) \nonumber \\&\quad \ge \ f\left( \rho (z) q_{\Theta ,\beta ^{\prime },m}(z)\right) - L_{f} \rho (z) (m\vee L_q) r. \end{aligned}$$

(7.14)

As $q_{\Theta ,\beta ^{\prime },m}(z) \le q_{\beta ^{\prime }}(z)$, we can apply Lemma 5.1 with inequality (7.14) to show

$$\begin{aligned}&{f\left( \int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \mathbb {I}(y\in \Theta ) \sigma (y)\,dy\right) - \frac{\beta e(z) q_{\Theta ,\beta ^{\prime },m}(z)}{1-q_{\Theta ,\beta ^{\prime },m}(z)}} \nonumber \\&\qquad \ge \frac{(\beta ^{\prime } - \beta ) e(z) q_{\Theta ,\beta ^{\prime },m}(z)}{1-q_{\Theta ,\beta ^{\prime },m}(z)} - L_{f}\rho (z) (m\vee L_q) r. \end{aligned}$$

(7.15)

From the definition of $\eta _{\Theta }$, $q_{\beta '}(z) \ge (\beta '\eta _{\Theta }+ 1 - \beta ')$ for all $z\in \Theta$, by Lemma 5.3. Set $\Theta _{1} := \{y: \Vert y-x\Vert \le t - \theta _{1} r \}$. Then $q_{\Theta ,\beta ^{\prime },m}(z)\ \ge \ \theta _{1} mr$ for all $z \in \Theta _{1}$ by inequality (7.8). Applying this lower bound to inequality (7.15), we see that inequality (7.13) holds for all $z \in \Theta _{1}$ if inequality (7.9) holds.

L2 Define $\Theta _{2} := \{y: t - \theta _{1}r <\Vert y-x\Vert \le t - \theta _{2} r \}$. For any $z \in \Theta _{2}$ and y such that $\Vert y-z\Vert \le r$,

$$\begin{aligned} m(t-\Vert y-x\Vert ) \ \le \ m(t - \Vert z-x\Vert + \Vert z-y\Vert ) \ \le \ (1+ \theta _{1}) mr \ \le \ q_{\beta ^{\prime }}(y) \end{aligned}$$

by Lemma 5.3 and since $(1+\theta _{1}) mr \le \beta ^{\prime }\eta _{\Theta } + 1 - \beta ^{\prime }$ by inequality (7.8). Therefore, for any $z \in \Theta _{2}$ and $y \in \Theta$ such that $\Vert y-z\Vert \le r$, $q_{\Theta ,\beta ^{\prime },m}(y) = m(t-\Vert y-x\Vert )$. For any $z \in \Theta _{2}$ and $y \not \in \Theta$ such that $\Vert y-z\Vert \le r$, we have $m(t-\Vert y-x\Vert ) \le 0$. Hence

$$\begin{aligned}&{\int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \mathbb {I}(y\in \Theta ) \sigma (y)\,dy} \nonumber \\&\quad \ \ge \ \int a(y) c(z,y;r) m(t- \Vert y-x\Vert ) \sigma (y)\,dy \nonumber \\&\quad \ =\ m(t-\Vert z-x\Vert ) \rho (z) + m \int a(y) c(z,y;r) \left[ \Vert z-x\Vert - \Vert y - x\Vert \right] \sigma (y)\,dy. \end{aligned}$$

(7.16)

Let $\gamma (x,y,z)$ be the angle formed between the vectors $x-z$ and $y-z$. By the cosine rule

$$\begin{aligned}&{\Vert z-x\Vert - \Vert y-x\Vert - \Vert z-y\Vert \cos \gamma (x,y,z)} \\&\quad \ =\ \Vert z-x\Vert \left( 1 - \left( 1+ \left( \frac{\Vert z-y\Vert }{\Vert z-x\Vert }\right) ^{2} - 2 \left( \frac{\Vert z-y\Vert }{\Vert z-x\Vert }\right) \cos \gamma (x,y,z)\right) ^{1/2} \right) \\&\qquad \quad - \Vert z-y\Vert \cos \gamma (x,y,z) . \end{aligned}$$

Let $h(u) = (1 + u^{2} -2u\cos \gamma )^{1/2}$. Taking a Taylor expansion about 0 gives $h(u) = 1-u\cos \gamma + \frac{1}{2} u^{2} h^{\prime \prime }(\tilde{u})$ for some $\tilde{u} \in (0,u)$. Therefore,

$$\begin{aligned}&{\Vert z-x\Vert - \Vert y-x\Vert - \Vert z-y\Vert \cos \gamma (x,y,z)} \\&\quad \ge \ -\frac{\Vert z-y\Vert ^{2}}{2\Vert z-x\Vert } \sup _{\xi \in (0, \frac{\Vert z-y\Vert }{\Vert z-x\Vert })} \left( 1 + \xi ^{2} - 2\xi \cos \gamma (x,y,z)\right) ^{-1/2} \\&\quad \ge \ -\frac{r^{2}}{2\Vert z-x\Vert } \left( 1 - \frac{2r}{\Vert z-x\Vert }\right) ^{-1}. \end{aligned}$$

Noting that $\Vert z-x\Vert \ge t - \theta _{1} r$ and substituting this bound into (7.16) gives

$$\begin{aligned}&{\int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \mathbb {I}(y\in \Theta ) \sigma (y)\,dy} \\&\quad \ge \ q_{\Theta ,\beta ^{\prime },m}(z) \rho (z) + m \int a(y) c(z,y;r) \Vert z-y\Vert \cos \gamma (x,y,z) \sigma (y)\,dy\\&\qquad - \frac{mr^{2}\rho (z)}{2(t-(2+\theta _{1})r)}, \end{aligned}$$

if $t > (2+\theta _{1})r$; but this follows from inequality (7.11), which gives $t-(2+\theta _{1})r \ge t/2$. Now, from Assumption C,

$$\begin{aligned}&{\int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \mathbb {I}(y\in \Theta ) \sigma (y)\,dy} \nonumber \\&\quad \ge \ q_{\Theta ,\beta ^{\prime },m}(z) \rho (z) + m a(z) \sigma (z) \int c(z,y;r) \Vert z-y\Vert \cos \gamma (x,y,z) dy \nonumber \\&\qquad + m a(z) \int c(z,y;r) \Vert z-y\Vert \cos \gamma (x,y,z) [\sigma (y) - \sigma (z)] dy \nonumber \\&\qquad + m \int [a(y) -a(z)] c(z,y;r) \Vert z-y\Vert \cos \gamma (x,y,z) \sigma (y) dy -\frac{mr^{2}\rho (z)}{t} \nonumber \\&\quad \ge \ q_{\Theta ,\beta ^{\prime },m}(z) \rho (z) + m a(z) \sigma (z) \int c(z,y;r) \Vert z-y\Vert \cos \gamma (x,y,z) dy - (C_3 + \rho _\mathrm{max}/t) mr^{2}. \end{aligned}$$

(7.17)

By the radial symmetry of c(z, y; r) , $\int c(z,y;r) \Vert z-y\Vert \cos \gamma (x,y,z) dy = 0$. Thus we deduce that

$$\begin{aligned} f\left( \int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \sigma (y)\,dy \right) \ge \ f\left( q_{\Theta ,\beta ^{\prime },m}(z) \rho (z) \right) - L_{f} (C_3 + \rho _\mathrm{max}/t) mr^{2} . \end{aligned}$$

Therefore, applying Lemma 5.1 and noting that $f(x) \ge L_f x - C_1 x^{2}$ gives

(7.18)

We now need a lower bound on $L_f \rho (z) - \beta e(z)$. By Assumption E, we have

$$\begin{aligned} (1 - q_{\beta }(z)) L_f \rho (z) q_{\beta }(z)\ \ge \ \beta e(z) q_{\beta }(z) . \end{aligned}$$

Hence, because $L_f \rho (z) > e(z)$ whenever $q_1(z) > 0$, we deduce that

$$\begin{aligned} L_f \rho (z) - \beta e(z)\ \ge \ L_f \rho (z) q_{\beta }(z)\ \ge \ e(z) q_{\beta }(z) . \end{aligned}$$

This, together with the lower bound on $q_{\beta }(z)$ from Lemma 5.3, gives $L_f \rho (z) - \beta e(z) \ge e_\mathrm{min}(\beta \eta _{\Theta }+1 - \beta )$. As $\theta _{2} mr \le q_{\Theta ,\beta ^{\prime },m}(z) \le \theta _{1} mr$ for all $z \in \Theta _{2}$ we see that the right hand side of inequality (7.18) is positive if inequality (7.10) holds. Hence, inequality (7.13) holds for all $z \in \Theta _{2}$.

L3 Define $\Theta _{3} := \{y: t - \theta _{2}r <\Vert y-x\Vert \le t\}$. As in L2, for any $z \in \Theta _{3}$ and y such that $\Vert y-z\Vert \le r$, we have $q_{\Theta ,\beta ^{\prime },m}(y) = m(t-\Vert y-x\Vert )$. Following inequality (7.17) in L2,

$$\begin{aligned}&{\int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \mathbb {I}(y\in \Theta ) \sigma (y)\,dy - m(t-\Vert z-x\Vert )\rho (z)} \\&\quad \ge \ m a(z) \sigma (z) \int c(z,y;r) \Vert z-y\Vert \cos \gamma (x,y,z) \mathbb {I}(y \in \Theta ) dy - (C_3 + \rho _\mathrm{max}/t) mr^{2} . \end{aligned}$$

As $\theta _{2} < 1$, let w be a point of intersection of the ball $B_z(r)$ with $\partial \Theta$, and let $\phi := \gamma (x,w,z)$. Applying the change of variable $\lambda (y) = r^{-1} \Vert z - y\Vert$ and $\omega (y) = \gamma (x,y,z)$ yields

$$\begin{aligned} \int _{\{ y : -\phi \le \omega (y) \le \phi \}} c(z,y;r) \Vert z-y\Vert \cos \omega (y) \mathbb {I}(y \in \Theta ) dy \ge \ v_{d-1}r\left( \int ^{1}_{0} c_{z}(\lambda ) \lambda ^{d}\, d\lambda \right) (\sin \phi )^{d-1}. \end{aligned}$$

It remains to determine a lower bound for the integral

$$\begin{aligned} \int _{\{ y : \gamma (x,y,z) \in [-\pi ,-\phi ) \cup (\phi ,\pi ]\}} c(z,y;r) \Vert z-y\Vert \cos \gamma (x,y,z) \mathbb {I}(y \in \Theta )\, dy . \end{aligned}$$

(7.19)

The region of integration is included in a cylinder of height $(t-\Vert z-x\Vert ) + r\max \{\cos \phi ,0\}$ and radius $r\sin \phi$. As $\phi$ is determined by the intersection of two circles,

$$\begin{aligned} \cos \phi \ =\ r^{-1} \left( \Vert z-x\Vert - \frac{\Vert z-x\Vert ^{2} -r^{2} + t^{2}}{2\Vert z-x\Vert } \right) \ =\ \frac{\Vert z-x\Vert -t}{r} + \frac{r[1 - ((t-\Vert z-x\Vert )/r)^{2}]}{2 \Vert z-x\Vert }. \end{aligned}$$

The function $x + r(1-x^2)/(2(t+rx))$ is increasing when $t > r$, and so

$$\begin{aligned} -\theta _{2} \ \le \ \cos \phi \ \le \ \frac{r}{2t}. \end{aligned}$$

(7.20)

Hence, for $z \in \Theta _3$, the volume of integration cannot exceed

$$\begin{aligned} v_{d-1}(r\sin \phi )^{d-1}(\theta _2 r + r^2/t) \ \le \ 2 v_{d-1} (r\sin \phi )^{d-1} \theta _{2} r , \end{aligned}$$

by inequality (7.11). The largest negative value of the integrand is bounded below by $-c_\mathrm{max}r^{-d+1}$. Hence this integral is bounded below by $-2c_\mathrm{max}v_{d-1}r (\sin \phi )^{d-1}\theta _2$. This leads to the lower bound

$$\begin{aligned}&{\int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \mathbb {I}(y\in \Theta ) \sigma (y)\,dy - m(t-\Vert z-x\Vert )\rho (z)} \nonumber \\&\quad \ge \ m r a(z) \sigma (z) v_{d-1} \left( {c}_{\Theta }- 2c_\mathrm{max}\theta _2 \right) (\sin \phi )^{d-1} - (C_3 + \rho _\mathrm{max}/t) mr^2. \end{aligned}$$

(7.21)

From inequalities (7.11) and (7.20), $(\sin \phi )^{d-1} \ge (1-\theta ^{2}_{2})^{(d-1)/2}$. Applying inequality (7.12), we see that the right-hand side of (7.21) is positive. Therefore,

$$\begin{aligned} f(\rho (z)q_{\Theta ,\beta ^{\prime },m}(z)) \ \le \ f\left( \int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \mathbb {I}(y\in \Theta ) \sigma (y)\,dy\right) . \end{aligned}$$

Lemma 5.1, with $\tau = \rho (z)$ and $\nu = \beta 'e(z)$, then implies that inequality (7.13) holds for all $z \in \Theta _{3}$. Hence, $q_{\Theta ,\beta ^{\prime },m}(z) \le p_{\Theta ,\beta }^{*}(z)$ for all $z \in \Theta$. $\square$

Lemma 7.3

Suppose that Assumptions B–E hold, that $\min _{z\in \Theta }q_{1}(z) =: \eta _{\Theta } > 0$ and that $\beta \in (1,1+ \eta _{\Theta }/2)$. Assume that

$$\begin{aligned} L_{q} r&\ \le \ \frac{\eta _{\Theta }^{2} e_\mathrm{min}^{2}}{32 L_{f} \rho _\mathrm{max}^{2}(C_1 \rho _\mathrm{max}+ L_f)}; \end{aligned}$$

(7.22)

$$\begin{aligned} \frac{r}{t}&\ \le \ \left( \frac{ {c}_{\Theta }}{4 c_\mathrm{max}} \wedge \frac{1}{\sqrt{2}} \wedge \frac{\eta _{\Theta }e_\mathrm{min}}{8L_{f} \rho _\mathrm{max}+ 4 \eta _{\Theta }e_\mathrm{min}}\right) ; \end{aligned}$$

(7.23)

$$\begin{aligned} (C_3 + \rho _\mathrm{max}/t)r&\ \le \ \left\{ a_\mathrm{min}\sigma _\mathrm{min}{c}_{\Theta }v_{d-1} 2^{-(d+3)/2}\right\} \wedge \left\{ \frac{\eta _{\Theta }e_\mathrm{min}}{4L_{f}} \left( \frac{{c}_{\Theta }}{4 c_\mathrm{max}} \wedge \frac{1}{\sqrt{2}} \right) \right\} , \end{aligned}$$

(7.24)

and define

$$\begin{aligned} C_4 \ :=\ \left( 1 \wedge \frac{f(a_\mathrm{min}\sigma _\mathrm{min})}{2^{(d+1)/2} e_\mathrm{max}}\right) \left( \frac{ {c}_{\Theta }}{4 c_\mathrm{max}} \wedge \frac{1}{\sqrt{2}} \right) \frac{e_\mathrm{min}^{2}}{32 L_{f} \rho _\mathrm{max}^{2}(C_1 \rho _\mathrm{max}+ L_f)} . \end{aligned}$$

(7.25)

Then $p_{\Theta ,\beta }^{*}(z) \ge C_4 \eta _{\Theta }^{2}$ for all $z \in \Theta$.

Proof

We begin by showing that the above inequalities are sufficient for the inequalities of Lemma 7.2 to hold, for suitable choices of $\beta ',\theta _1,\theta _2$ and m. Set $\beta ' = \frac{1}{2(1-\eta _{\Theta })} + \frac{\beta }{2}$. Then

$$\begin{aligned} \beta ^{\prime } - \beta \ =\ \frac{1}{2(1-\eta _{\Theta })} - \frac{\beta }{2} = \frac{\beta \eta _{\Theta }+ 1 - \beta }{2(1-\eta _{\Theta })} \ \ge \ \frac{\eta _{\Theta }(1+\eta _{\Theta })}{4(1-\eta _{\Theta })}\ \ge \ \frac{\eta _{\Theta }}{4}, \end{aligned}$$

(7.26)

and

$$\begin{aligned} \beta ^{\prime } \eta _{\Theta }+ 1 - \beta ^{\prime } \ =\ \frac{1}{2} (\beta \eta _{\Theta }+ 1 - \beta ) \ \ge \ \frac{\eta _{\Theta }}{4}. \end{aligned}$$

(7.27)

Set

$$\begin{aligned} \theta _{1}&\ :=\ \frac{4 L_{f} \rho _\mathrm{max}}{\eta _{\Theta }e_\mathrm{min}}; \\ mr&\ :=\ \frac{\eta _{\Theta }^{2} e_\mathrm{min}^{2}}{32 L_{f} \rho _\mathrm{max}^{2}(C_1 \rho _\mathrm{max}+ L_f)}; \\ \theta _{2}&\ :=\ \left( \frac{ {c}_{\Theta }}{4 c_\mathrm{max}} \wedge \frac{1}{\sqrt{2}} \right) . \end{aligned}$$

Since

$$\begin{aligned} \frac{L_{f} \rho _\mathrm{max}}{\eta _{\Theta }e_\mathrm{min}}\ \ge \ \frac{ L_f \rho (z)}{\eta _{\Theta }e(z)}\ \ge \ \frac{1 + q_{1}(z)}{\eta _{\Theta }} \ \ge \ \frac{1 + \eta _{\Theta }}{\eta _{\Theta }}\ \ge \ 1, \end{aligned}$$

it follows that $\theta _{1} > 1$. This, together with inequality (7.27), implies that inequality (7.8) is satisfied if $8 \theta _{1} mr \le \eta _{\Theta }$. This is indeed the case, since, from the choices of $\theta _1$ and mr, we have

$$\begin{aligned} 8\theta _1 mr \ \le \ \eta _{\Theta }\,\frac{e_\mathrm{min}}{L_f\rho _\mathrm{max}} \ \le \ \eta _{\Theta }, \end{aligned}$$

because $L_f \rho _\mathrm{max}/e_\mathrm{min}> 1$ if $\eta _{\Theta }> 0$.

Then $L_{q} r \le mr$, by inequality (7.22), so inequality (7.9) simplifies to give $L_{f} \rho _\mathrm{max}\le (\beta ^{\prime } - \beta ) e_\mathrm{min}\theta _{1}$; and this is seen to hold, by inequality (7.26) and the choice of $\theta _1$. The choices of $\theta _{1}$ and $\theta _{2}$, together with inequality (7.27), show further that inequality (7.10) is implied by inequality (7.24), and that inequality (7.11) is implied by inequality (7.23). Finally, the choice of $\theta _{2}$ shows that inequality (7.12) follows from inequality (7.24). Thus, inequalities (7.8)–(7.12) in Lemma 7.2 hold.

Take $\Theta _{1},\ \Theta _{2}$ and $\Theta _{3}$ as defined in the proof of Lemma 7.2. On $\Theta _{1} \cup \Theta _{2}$,

$$\begin{aligned} p^{*}_{\Theta ,\beta }(z)\ \ge \ q_{\Theta ,\beta ^{\prime },m}(z)\ \ge \ \theta _{2} mr. \end{aligned}$$

For $z \in \Theta _{3}$, note that $q_{\Theta ,\beta ^{\prime },m} \le E_{\Theta ,\beta }(q_{\Theta ,\beta ^{\prime },m}) \le p^{*}_{\Theta ,\beta }$ and that

$$\begin{aligned} E_{\Theta ,\beta }(q_{\Theta ,\beta ^{\prime },m})(z) \ \ge \ \frac{f\left( \int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \mathbb {I}(y\in \Theta ) \sigma (y)\,dy\right) }{\beta e(z)}. \end{aligned}$$

Now inequalities (7.21) and (7.24) imply that

$$\begin{aligned} f\left( \int a(y) c(z,y;r) q_{\Theta ,\beta ^{\prime },m}(y) \mathbb {I}(y\in \Theta ) \sigma (y)\,dy\right)\ge & {} f\left( mr\, \frac{a_\mathrm{min}\sigma _\mathrm{min}{c}_{\Theta }}{2^{(d+3)/2} c_\mathrm{max}} \right) \\\ge & {} f\left( \theta _2 mr\, \frac{a_\mathrm{min}\sigma _\mathrm{min}}{2^{(d-1)/2} } \right) . \end{aligned}$$

Then, from Assumption E, we have $f(ab) \ge bf(a)$ if $0 \le b \le 1$. Now $\theta _2 < 1$, $2^{(d-1)/2} \ge 1$ and $mr \le 1/32$, because $0 < \eta _{\Theta }\le 1$ and $L_f \rho _\mathrm{max}/e_\mathrm{min}> 1$, so we conclude that

$$\begin{aligned} p^{*}_{\Theta ,\beta }(z)\ \ge \ \frac{f\left( a_\mathrm{min}\sigma _\mathrm{min}\right) }{2^{(d-1)/2}}\, \theta _{2} mr\,\frac{1}{\beta e_\mathrm{max}} \ \ge \ \frac{f\left( a_\mathrm{min}\sigma _\mathrm{min}\right) }{2^{(d+1)/2}e_\mathrm{max}}\, \theta _{2} mr \end{aligned}$$

for all $z \in \Theta _{3}$.

Combining this with the lower bound on $q_{\Theta ,\beta ^{\prime },m}$ for $z \in \Theta _{1} \cup \Theta _{2}$ gives the uniform lower bound. $\square$

Lemma 7.4

Suppose that Assumptions B–E hold, that $\min _{z\in \Theta } q_{1}(z) =: \eta _{\Theta } > 0$ and that $1< \beta< \beta ^{\prime } < 1+ \eta _{\Theta }/2$. Assume that, in addition to inequalities (7.23)–(7.24),

$$\begin{aligned} L_{q} r&\ \le \ \frac{e_\mathrm{min}^{2} \eta _{\Theta }(\beta ' - \beta )}{4 \rho _\mathrm{max}^{2} L_{f} (C_1\rho _\mathrm{max}+ L_f)}; \end{aligned}$$

(7.28)

$$\begin{aligned} \frac{r}{t}&\ \le \ \frac{e_\mathrm{min}(\beta ' - \beta )}{6L_{f} \rho _\mathrm{max}}. \end{aligned}$$

(7.29)

Then, choosing m as in Theorem 3.2 so that

$$\begin{aligned} mr \ =\ \frac{e_\mathrm{min}^{2} \eta _{\Theta }(\beta ' - \beta )}{4 \rho _\mathrm{max}^{2} L_{f} (C_1 \rho _\mathrm{max}+ L_f)}, \end{aligned}$$

it follows that $p_{\Theta ,\beta }^{*}(z) \ge q_{\Theta ,\beta ^{\prime },m}(z)$ for all $z \in \Theta$.

Proof

We show that the above inequalities are sufficient for the inequalities of Lemma 7.2 to hold, with suitable choices of $\theta _1$ and $\theta _2$. Set

$$\begin{aligned} \theta _{1}&\ :=\ \frac{L_{f} \rho _\mathrm{max}}{e_\mathrm{min}(\beta ' - \beta )}, \end{aligned}$$

and choose $\theta _{2}$ as in Lemma 7.3. Note that $\theta _{1} > 1$, since $\beta '-\beta \le 1\le L_{f} \rho _\mathrm{max}/e_\mathrm{min}$. Since $\beta ' \eta _{\Theta }+1 - \beta ' \ge \eta _{\Theta }/2$, inequality (7.8) holds if $4mr \le \eta _{\Theta }$; but this is true with the above choice of mr, because $(\beta ' - \beta ) < 1$ and $L_f \rho _\mathrm{max}/e_\mathrm{min}> 1$.

From inequality (7.28), $L_{q} r \le mr$, and so inequality (7.9) simplifies to give $L_{f} \rho _\mathrm{max}\le (\beta ^{\prime } - \beta ) e_{\min } \theta _{1}$, which holds with equality for $\theta _{1}$ as chosen. To show that inequality (7.10) holds, we first note that $4 \rho _\mathrm{max}(C_1 \rho _\mathrm{max}+ L_f) \theta _{1} mr \le e_\mathrm{min}\eta _{\Theta }$. Therefore, inequality (7.10) holds if $4 L_{f} (C_3 + \rho _\mathrm{max}/t) r \le \theta _{2} e_\mathrm{min}\eta _{\Theta }$, which follows from inequality (7.24).

The second part of inequality (7.11) holds by (7.29) and because

$$\begin{aligned} 2(2+\theta _{1}) \ =\ \frac{4 e_\mathrm{min}(\beta ' - \beta ) + 2 L_{f} \rho _\mathrm{max}}{e_\mathrm{min}(\beta '-\beta )} \ \le \ \frac{6L_{f}\rho _\mathrm{max}}{e_\mathrm{min}(\beta '-\beta )}, \end{aligned}$$

again since $\beta '-\beta \le 1\le L_{f} \rho _\mathrm{max}/e_\mathrm{min}$.

Finally, with $\theta _{2}$ chosen as in Lemma 7.3, inequality (7.12) follows from (7.24) as in Lemma 7.3, and the first part of inequality (7.11) follows from (7.23). $\square$

Proof of Theorem 3.2

First note that inequality (7.2) of Lemma 7.1 holds with $\epsilon _{\Theta ,\beta } = C_4 \eta _{\Theta }^{2}$, since

$$\begin{aligned} C_{4} \ \le \ \frac{e_\mathrm{min}^{2}}{32\sqrt{2} L_{f}^{2} \rho _\mathrm{max}^{2}} \ \le \ \frac{1}{8\sqrt{2}} \end{aligned}$$

when $L_{f} \rho _\mathrm{max}/ e_\mathrm{min}> 1/2$, and hence

$$\begin{aligned} C_4 \eta _{\Theta }^2 (\beta -1) \ \le \ C_4 \frac{\eta _{\Theta }^3}{2} \ \le \ \frac{1}{16\sqrt{2}} \ <\ \frac{1}{2} \ \le \ \frac{L_f\rho _\mathrm{max}}{e_\mathrm{min}}. \end{aligned}$$

Now combine Lemmas 7.1, 7.3 and 7.4. $\square$

Proof of Corollary 3.3

By Theorem 3.1, $p_{i,n}^{*} \le p^{+}_{\alpha _{1},\alpha _{2}}(z_{i})$ for all $i=1,\ldots ,n$ with high probability. Taking $\alpha _{2} = 1- \eta _{\Theta }$, we note that $q_{\alpha _{1}}(z) \ge q_{1}(z) \ge \eta _{\Theta }$, and so $p_{\alpha _{1},\alpha _{2}}^{+} (z) = q_{\alpha _{1}}(z)$ for all $z \in \Theta$. Therefore,

$$\begin{aligned} p_{n,i}^{*} - q_{1}(z_{i}) \ \le \ q_{\alpha _{1}}(z_{i}) -q_{1}(z_{i}), \end{aligned}$$

(7.30)

for all $z_{i} \in \Theta$, with high probability.

Note that $q_{\Theta ,\beta ',m}(z) = q_{\beta '}(z)$ for all $z \in \Theta _{m}$. By Theorem 3.2 for all $z_{i} \in \Theta _{m}$

$$\begin{aligned} p_{n,i}^{*} - q_{1}(z_{i})\ \ge \ q_{\beta '}(z_{i}) - q_{1}(z_{i}) \end{aligned}$$

(7.31)

with high probability. Inequalities (7.30) and (7.31) imply that, for all $z_{i} \in \Theta _{m}$,

$$\begin{aligned} \left| p_{n,i}^{*} - q_{1}(z_{i}) \right| \ \le \ q_{\alpha }(z_{i}) - q_{\beta '}(z_{i}) \end{aligned}$$

with high probability. As in the proof of Lemma 5.2

$$\begin{aligned} \frac{\partial q_{\alpha }(z)}{\partial \alpha } \ \le \ \frac{-e(z)}{\alpha e(z) + f(\rho (z) q_{\alpha }(z)}. \end{aligned}$$

Hence,

$$\begin{aligned} q_{\alpha _1}(z) - q_{\beta '}(z) \ \le \ \int _{\alpha _1}^{\beta '} \frac{e(z)\, du}{ue(z) + f(\rho (z)q_{u}(z))} \ \le \ \alpha _1^{-1}(\beta '-\alpha _1). \end{aligned}$$

$\square$

Proof of Corollary 3.4

The corollary follows from Corollary 3.3, with appropriate choices of $\alpha _1, \alpha _2, \beta$ and $\beta '$. First note that $r_n^{1-\gamma _1}\phi _n \le c_1r_n^{\gamma _1}/4$ for all n sufficiently large, if $r_n^{1-2\gamma _1}\phi _n \rightarrow 0$. Thus we can take $1-\alpha _2 = \eta _{\Omega _n}$, $1- \alpha _1 = r_n^{1-\gamma _1} \phi _n$ and $\beta '-\beta = \beta -1 = r_n^{1-\gamma _1}\phi _n$, and satisfy $\alpha _1 \ge \alpha _2$ and $\beta '-1 \le \eta _{\Omega _n}/2$ for all n sufficiently large. With these choices of $\alpha _1$ and $\alpha _2$, inequality (3.3) of Theorem 3.1 holds for all n sufficiently large; the choices of $\beta$ and $\beta '$ show that inequality (3.6) holds, fulfilling the conditions of Theorem 3.2. Then the probabilities in Corollary 3.3 converge to 1, as required, in view of (3.7) and (3.5), and $\alpha _1^{-1}(\beta '-\alpha _1) = O(r_n\phi _n)$ and $m \asymp \phi _{n}$. $\square$