First, we note that, due to the memoryless property which both the arrival and service completion processes satisfy, the choice of
m is irrelevant and thus can be made arbitrarily. To prove (
55), we start by showing that, for every
\(\{i,j\} \in E\),
$$\begin{aligned} \mathbb {P}\left( \sup _{0 \le t_1 \le t_2 \le M} \left\{ A_{ij}^{r,m}(t_2) - A_{ij}^{r,m}(t_1) - \frac{1}{p_{ij} \lambda } (t_2-t_1) \right\} \ge \epsilon \right) \le \epsilon . \end{aligned}$$
From (
50) and (
53), we observe that
\(Y_{r,m}(t) \le t\) and is non-decreasing. Due to the properties of Poisson processes,
$$\begin{aligned} A_{ij}^{r,m}(t_2) - A_{ij}^{r,m}(t_1)&\overset{d}{=} \frac{\varLambda _{ij}\left( \sqrt{x_{r,m}} Y^{r,m}(t_2)\right) -\varLambda _{ij}\left( \sqrt{x_{r,m}} Y^{r,m}(t_1)\right) }{\sqrt{x_{r,m}}} \\&\overset{d}{\le } \frac{\varLambda _{ij}\left( \sqrt{x_{r,m}} t_2\right) -\varLambda _{ij}\left( \sqrt{x_{r,m}} t_1 \right) }{\sqrt{x_{r,m}}}. \end{aligned}$$
Therefore,
$$\begin{aligned} \mathbb {P}&\left( \sup _{0 \le t_1 \le t_2 \le M} \left\{ A_{ij}^{r,m}(t_2) - A_{ij}^{r,m}(t_1) - \frac{t_2-t_1}{p_{ij} \lambda } \right\} \ge \epsilon \right) \\&\quad \le \mathbb {P}\left( \sup _{0 \le t_1 \le t_2 \le M} \left\{ \frac{\varLambda _{ij}\left( \sqrt{x_{r,m}} t_2\right) -\varLambda _{ij}\left( \sqrt{x_{r,m}} t_1 \right) }{\sqrt{x_{r,m}}} - \frac{t_2-t_1}{p_{ij} \lambda } \right\} \ge \epsilon \right) \\&\quad \le \mathbb {P}\left( \left\Vert \frac{\varLambda _{ij}\left( \sqrt{x_{r,m}} t\right) }{\sqrt{x_{r,m}}} - \frac{t}{p_{ij} \lambda } \right\Vert _M \ge \epsilon /2 \right) \le \frac{\epsilon }{2 M^2 \sqrt{x_{r,m}}} \le \frac{\epsilon }{2 M^2 \sqrt{r}} , \end{aligned}$$
where the second-to-last inequality follows from Proposition 4.3 in [
6]. Choosing
\(N = 1/(\lambda \min _{\{i,j\} \in E} p_{ij})\) and applying the union bound twice, we obtain
$$\begin{aligned} \mathbb {P}\left( \max _{m < \sqrt{r}T} \sup _{0 \le t_1 \le t_2 \le M} \left\{ \left|A^{r,m}(t_2) - A^{r,m}(t_1) \right|- N (t_2-t_1) \right\} \ge \epsilon \right) \le \frac{\epsilon T |E|}{2 M^2}, \end{aligned}$$
which yields (
55).
In order to prove (
57) and (
58), we introduce the following processes: Let
\(\{u_{ij}(l), l\ge 1\}\) be independent exponentially distributed random variables with rate
\(p_{ij} \lambda \), representing the time that a car has before it needs recharging at either station
i or
j. Let
\(\{v_j(l), l\ge 1\}\) be independent exponentially distributed random variables with rate
\(\mu \), representing the service requirement (recharging time) of a battery at station
j. Define
$$\begin{aligned} U_{ij}(n)&= \sum _{l=1}^n u_{ij}(l), \quad \{i,j\} \in E, \\ V_{j}(n)&= \sum _{l=1}^n v_{j}(l), \quad j=1,\ldots ,S, \end{aligned}$$
the aggregated interarrival time of
n cars that will choose between stations
i and
j, and the total service requirement of
n batteries at station
j, respectively. We observe the identities
$$\begin{aligned} \varLambda _{ij}(t) = \max \{n : U_{ij}(n) \le t\}, \quad S_j(t) = \max \{n : V_j(n) \le t\}, \quad t\ge 0. \end{aligned}$$
Moreover, due to (
16) and (
18), we observe
$$\begin{aligned} U_{ij}(A_{ij}^r(t))&\le Y^r(t) \le U_{ij}(A_{ij}^r(t)+1), \qquad \{i,j\} \in E , \end{aligned}$$
(59)
$$\begin{aligned} V_j(D_j^r(t))&\le T_j^r(t) \le V_j(D_j(t)+1), \,\,\,\, j=1,\ldots ,S. \end{aligned}$$
(60)
As in [
9], we define for notational convention, for
\(b=(b_1,b_2) \in \mathbb {N}\),
$$\begin{aligned} U^{r,m}_{ij} \left( A_{ij}^{r,m}(t),b\right)&= \frac{1}{\sqrt{x_{r,m}}} \left( U^r_{ij}\left( A_{ij}^{r} \left( \frac{m}{\sqrt{r}} + \frac{\sqrt{x_{r,m}} t}{r} \right) + b_1 \right) - U^r_{ij}\left( A_{ij}^{r} \left( \frac{m}{\sqrt{r}} \right) + b_2 \right) \right) , \end{aligned}$$
(61)
$$\begin{aligned} V^{r,m}_j \left( D_j^{r,m}(t),b\right)&= \frac{1}{\sqrt{x_{r,m}}} \left( V^r_j\left( D_j^{r} \left( \frac{m}{\sqrt{r}} + \frac{\sqrt{x_{r,m}} t}{r} \right) + b_1 \right) - V^r_j\left( D_j^{r} \left( \frac{m}{\sqrt{r}} \right) + b_2 \right) \right) . \end{aligned}$$
(62)
In view of (
59) and (
60), this yields the inequalities
$$\begin{aligned} U_{ij}^{r,m}(A_{ij}^r(t),(0,1))&\le Y^{r,m}(t) \le U_{ij}^{r,m}(A_{ij}^r(t),(1,0)), \quad \{i,j\} \in E, \end{aligned}$$
(63)
$$\begin{aligned} V_j^{r,m}(D_j(t),(0,1))&\le T_j^{r,m}(t) \le V_j^{r,m}(D_j(t),(1,0)), \quad j=1,\ldots ,S. \end{aligned}$$
(64)
Using these processes, we first prove the following bounds:
$$\begin{aligned}&\mathbb {P}\left( \max _{m < \sqrt{r}T} \left\Vert U_{ij}(A_{ij}^{r,m}(t),b) - \frac{1}{p_{ij}\lambda } A_{ij}^{r,m}(t) \right\Vert _M \ge \epsilon \right) \le \epsilon , \quad \forall \{i,j\} \in E, \end{aligned}$$
(65)
$$\begin{aligned}&\mathbb {P}\left( \max _{m < \sqrt{r}T} \left\Vert V_{j}(D^{r,m}(t),b) - \frac{1}{\mu } D_j^{r,m}(t) \right\Vert _M \ge \epsilon \right) \le \epsilon , \quad j=1,\ldots ,S, \end{aligned}$$
(66)
for
\(b=(1,0)\) and
\(b=(0,1)\). The proof is similar to that of (78) in [
21]. We observe that the proof of (
55) implies that in particular
$$\begin{aligned} \mathbb {P}&\left( A_{ij}^{r}\left( \frac{\sqrt{x_{r,m}}M}{r} \right) \ge \frac{2M}{p_{ij}\lambda } \sqrt{x_{r,m}} \right) \le \frac{\epsilon }{M^2 \sqrt{r}}, \end{aligned}$$
and hence also
$$\begin{aligned} \mathbb {P}&\left( A_{ij}^{r}\left( \frac{\sqrt{x_{r,m}}M}{r} \right) +1 \ge \frac{3M}{p_{ij}\lambda } \sqrt{x_{r,m}} \right) \le \frac{\epsilon }{M^2 \sqrt{r}} \end{aligned}$$
for
r large enough. Proposition 4.2 of [
6] states
$$\begin{aligned} \mathbb {P}\left( \left\Vert U_{ij}(l) - \frac{l}{p_{ij}\lambda } \right\Vert _n \ge \epsilon n \right) \le \frac{\epsilon }{n}. \end{aligned}$$
Therefore, it follows that
$$\begin{aligned} \mathbb {P}\left( \left\Vert U_{ij}(A_{ij}^r(t)) - \frac{A_{ij}^r(t)}{p_{ij}\lambda } \right\Vert _{\sqrt{x_{r,m}}M/r} \ge \epsilon \frac{2M\sqrt{x_{r,m}}}{p_{ij}\lambda } \right) \le \frac{\epsilon }{\sqrt{r}} \left( \frac{1}{M^2} + \frac{p_{ij}\lambda }{2 M} \right) , \end{aligned}$$
and
$$\begin{aligned} \mathbb {P}\left( \left\Vert U_{ij}(A_{ij}^r(t)+1) - \frac{A_{ij}^r(t)}{p_{ij}\lambda } \right\Vert _{\sqrt{x_{r,m}}M/r} \ge \epsilon \frac{3M\sqrt{x_{r,m}}}{p_{ij}\lambda } \right) \le \frac{\epsilon }{\sqrt{r}} \left( \frac{1}{M^2} + \frac{p_{ij}\lambda }{3 M} \right) . \end{aligned}$$
Increasing
\(\epsilon \) appropriately, we obtain
$$\begin{aligned} \mathbb {P}\left( \left\Vert U_{ij}^{r,m}(A_{ij}^{r,m}(t),b) - \frac{A_{ij}^{r,m}(t)}{p_{ij}\lambda } \right\Vert _{M} \ge \epsilon \right) \le \frac{\epsilon }{T\sqrt{r}} \end{aligned}$$
for both
\(b=(0,0)\) and
\(b=(1,0)\). Using the union bound yields
$$\begin{aligned} \mathbb {P}\left( \max _{m < \sqrt{r} T} \left\Vert U_{ij}^{r,m}(A_{ij}^{r,m}(t),b) - \frac{A_{ij}^{r,m}(t)}{p_{ij}\lambda } \right\Vert _{M} \ge \epsilon \right) \le \epsilon \end{aligned}$$
for both
\(b=(0,0)\) and
\(b=(1,0)\). To conclude the proof for
\(b=(0,1)\) as well, we observe
$$\begin{aligned} \mathbb {P}&\left( \max _{m< \sqrt{r} T} \left\Vert U_{ij}^{r,m}(A_{ij}^{r,m}(t),b) - U_{ij}^{r,m}(A_{ij}^{r,m}(t),(0,0))\right\Vert _{M} \ge \epsilon \right) \\&\le \mathbb {P}\left( \max _{m < \sqrt{r} T} \left|U_{ij}\left( A_{ij}^{r} \left( \frac{m}{\sqrt{r}} \right) +1 \right) - U_{ij}\left( A_{ij}^{r} \left( \frac{m}{\sqrt{r}}\right) \right) \right|\ge \epsilon \sqrt{x_{r,m}} \right) \\&\le \mathbb {P}\left( u_{ij}^{r,T,\max } \ge \sqrt{x_{r,m}} \epsilon \right) \le \epsilon , \end{aligned}$$
where the final inequality follows from Lemma 5.1 in [
6] with
$$\begin{aligned} u_{ij}^{r,T,\max }= \max \{u_{ij}(l) : U_i(l-1) \le r T\}. \end{aligned}$$
The proof of (
66) is analogous to (
65), replacing the arrival processes by the service processes. Equations (
57) and (
58) are then a direct consequence of (
63) and (
64).