Proof. Suppose we have a sequence \(x_n\) where every subsequence \(x_{n_k}\) itself has a subsequence \(x_{n_{k_j}}\) that converges to the same limit \(x\). For contradiction, assume that \(x_n\) does not converge to \(x\), so there is a subsequence \(x_{n_{k,0}}\) that is always at least some distance \(\epsilon\) away from \(x\). By assumption, \(x_{n_{k,0}}\) has a subsequence \(x_{n_{k,1}}\) that converges to \(x\), which contradicts that \(x_{n_{k,0}}\) is always at least some \(\epsilon\) away from \(x\). So, it must be that \(x_n\) converges to \(x\). ◻

Proof. Suppose that for some \(n\), \(\norm{\bar{x}(T_n)} > C_1 (1+\norm{\bar{y}(T_n)})\), which implies that \(\norm{\tilde{x}_n(0)} > C_1 (\frac{1}{r_n}+\norm{\tilde{y}_n(0)})\). Since \(1 \geq \norm{\tilde{x}_n(0)}\) and \(\frac{\norm{\tilde{x}_n(0)}}{C_1} > \norm{\tilde{y}_n(0)}\), we have \[\begin{align} \norm{\tilde{y}_n(0)} < \frac{1}{C_1}. \label{eq:32y32less32132over32C} \end{align}\tag{64}\]

We know from Lemma 3 that there exists some \(T_{\frac{1}{4}}\) such that for all \(t \geq T_{\frac{1}{4}}\), \[\begin{align} \norm{\eta^0_{\infty}(t,\tilde{x}_n(0))} \leq \frac{1}{4}. \label{C46832in32ball} \end{align}\tag{65}\] So if we set \(T \doteq T_{\frac{1}{4}}\), then 65 holds for all \(t \in [T, T+1)\).

From Lemma 4 we know that there exist \(\rho_{\frac{1}{4}}\) and \(c_{\frac{1}{4}}\) such that if \(y(t) \in B(0, \rho_{\frac{1}{4}})\) for all \(t \in [0, T+1)\) and \(r_n > c_\frac{1}{4}\) then \[\begin{align} \norm{\eta^{y(t)}_{r_n}(t,\tilde{x}_n(0)) - \eta^{0}_{\infty}(t,\tilde{x}_n(0))} \leq \frac{1}{4} \label{close32external32input} \end{align}\tag{66}\] for all \(t \in [0, T+1)\). Since \(\lim_{n\rightarrow \infty} r_n = \infty\), there is some finite \(n_1\) such that if \(n > n_1\), we can be sure that \(r_n > c_\frac{1}{4}\). By 63 , we know that there exists \(n_2\) such that for \(n > n_2\), the discretization error will diminish enough so that \[\begin{align} \tilde{y}_n(t) \in B \left(\tilde{y}_n(0),\min \left(\frac{\rho_{\frac{1}{4}}}{2}, \frac{1}{8} \right) \right) \label{y32tilde32in32small32ball} \end{align}\tag{67}\] for all \(t \in [0, T+1)\). So, if we choose \(C_1\) large enough so that \(C_1 > 16\) and \(\frac{1}{C_1} < \frac{\rho_{\frac{1}{4}}}{2}\), then we have \(\tilde{y}_n(t) \in B(0, \rho_{\frac{1}{4}})\) (by 64 and 67 ) for all \(t \in [0, T+1)\). So, we know that for all \(n > \max(n_1, n_2)\), 66 holds.

From 63 , we know that there is some finite \(n_3\) such that for \(n > n_3\), \[\begin{align} \norm{\tilde{x}_n(t) - \eta^{\tilde{y}_n(0)}_{r_n}(t,\tilde{x}_n(0))} \leq \frac{1}{4} \label{C46832x32discret32error} \end{align}\tag{68}\] for all \(t \in [0, T+1)\).

By 65 , 66 , and 68 , since \(\tilde{x}_n(T_{n+1}-T_n) = \frac{\bar{x}(T_{n+1})}{r_n}\), we have \(\norm{\frac{\bar{x}(T_{n+1})}{r_n}} \leq \frac{3}{4}\).

For \(n > n_1\), since \(\tilde{y}_n(T_{n+1}-T_n) = \frac{\bar{y}(T_{n+1})}{r_n}\) we also have \(\norm{\frac{\bar{y}(T_{n+1})}{r_n}} \leq \frac{1}{4}\) (by 64 , \(\frac{1}{C_1} < \frac{1}{16}\), and 67 ).

So we conclude \(\norm{\frac{\bar{z}(T_{n+1})}{r_n}} \leq 1\), telling us that \(r_{n+1} = r_n\) as desired. If we let \(n_0 = \max(n_1, n_2, n_3)\) and ensure that \(C_1 > \max_{i \leq n_0} \norm{\bar{x}(T_i)}\), then the result holds for all \(n\). ◻

Proof. From Lemma 61, we know that there are some constants \(A, B\) such that for all \(n\), \[\begin{align} \norm{\bar{z}(T_{n+1})} \leq A \norm{\bar{z}(T_n)} + B. \end{align}\]

Our argument is inductive in nature. Let \(C_1, C_2\) and ensure that \(C_1 > A + B\) and \(C_2 > 2A + B + 2\). To make sure the base case \((n=0)\), holds, we also ensure that \(C_1C_2 > \norm{z_0}\). From Lemma 15 we know that if \(\norm{\bar{x}(T_n)} \geq C_1 (1 + \norm{\bar{y}(T_n)})\), then \(r_{n+1} = r_n\). Thus, if the result holds for all \(i\) less than some \(n\), i.e., we have \(\norm{\bar{z}(T_i)} \leq C_1 C_2 (\max_{m \leq i} \norm{\bar{y}(T_m)} + 1)\) for all \(i \leq n\), then we can conclude that \(\norm{\bar{z}(T_{n+1})} \leq C_1 C_2 (\max_{m \leq n} \norm{\bar{y}(T_m)} + 1)\) also.

To address the other case, assume for some \(n\) that \(\norm{\bar{x}(T_{n})} \leq C_1(\norm{\bar{y}(T_n)}+1)\). Then we have \[\begin{align} \norm{\bar{z}(T_{n+1})} &\leq A \norm{\bar{z}(T_n)} + B \\ &\leq A \norm{\bar{x}(T_n)} + A \norm{\bar{y}(T_n)} + B \\ &\leq AC_1\norm{\bar{y}(T_n)} + AC_1 + A \norm{\bar{y}(T_n)} + B \\ &\leq C_1C_2(\norm{\bar{y}(T_n)}+1) \\ &\leq C_1C_2(\max_{m\leq n+1} \norm{\bar{y}(T_m)}+1). \end{align}\] ◻

Proof. From Lemma 61, we know that for all \(m\) such that \(m(T_n) \leq m \leq m(T_{n+1})\), there exist constants \(A, B\) such that \[\begin{align} \norm{z_m} \leq A \norm{\bar{z}(T_n)} + B. \end{align}\] Let \(C_3 > A + B\). This means that \[\begin{align} \norm{x_m} \leq \norm{z_m} &\leq A C_1C_2(\max_{l\leq n} \norm{\bar{y}(T_l)}+1) + B \\ &\leq C_1C_2C_3 (\max_{l \leq n}\norm{\bar{y}(T_l)} + 1) \\ & \leq C_1C_2C_3(\norm{y^{\text{max}}_n} + 1). \end{align}\] ◻

Proof. By Lyapunov stability, we know that there is some \(\delta\) with \(\frac{\epsilon}{2} > \delta > 0\) such that if \(\eta_{\infty}^y(t, x) \in B(\lambda_{\infty}(y), \delta)\), then for all \(t' > t\), \(\eta_{\infty}^y(t', x) \in B(\lambda_{\infty}(y), \frac{\epsilon}{2})\). By Lemma 3 we know that there exists some time \(T_{\delta}\) such that for all \(x \in \lambda_{\infty}(K_y)\) (image of a compact set under a continuous map is compact) and \(t \geq T_{\delta}\), we have \[\begin{align} \eta^y_{\infty}(t,x) \in B(\lambda_{\infty}(y), \frac{\delta}{2}). \label{eq:32all32y32ext32inp321} \end{align}\tag{69}\] By Lemma 4, we can select \(\rho_y\) small enough such that for all \(y_1 \in B(y, \rho_y)\), \[\begin{align} \norm{\eta_{\infty}^y(t,x) - \eta_{\infty}^{y_1}(t, x)} \leq \frac{\delta}{2} \label{eq:32all32y32ext32inp322} \end{align}\tag{70}\] for all \(t \in [0, T_{\delta}]\), \(x \in \lambda_{\infty}(K_y)\).

We can split up the timeline into chunks of size \(T_{\delta}\). The insight we will rely on is that if \(\eta_{\infty}^y(T_{\delta}, x) = x_1\), then \(\eta_{\infty}^y(T_{\delta}+t, x) = \eta_{\infty}^y(t, x_1)\) for all \(t \geq 0\). The way our logic proceeds is as follows: By 69 and 70 we have that \[\begin{align} \eta_{\infty}^{y_1}(T_{\delta}, x) \in B(\lambda_{\infty}(y), \delta) \end{align}\]

Let \(x_1 \doteq \eta_{\infty}^{y_1}(T_{\delta}, x)\). For all \(t \in [0, T_{\delta}]\), we know by Lyapunov stability that \[\begin{align} \eta_{\infty}^{y}(t, x_1) \in B(\lambda_{\infty}(y), \frac{\epsilon}{2}), \end{align}\] so by 70 we know (since we can reuse the \(\rho_y\) selected earlier) that \[\begin{align} \eta_{\infty}^{y_1}(t, x_1) \in B(\lambda_{\infty}(y), \frac{3\epsilon}{4}). \end{align}\] for all \(t \in [0, T_{\delta}]\), implying that \[\begin{align} \eta_{\infty}^{y_1}(t, x) \in B(\lambda_{\infty}(y), \frac{3\epsilon}{4}) \label{eq:32all32y32ext32inp323474} \end{align}\tag{71}\] for all \(t \in [T_{\delta}, 2T_{\delta}]\).

By 69 and 70 we know that \[\begin{align} \eta_{\infty}^{y_1}(T_{\delta}, x_1) \in B(\lambda_{\infty}(y), \delta). \end{align}\] We can then define \(x_2 \doteq \eta_{\infty}^{y_1}(T_{\delta}, x_1)\) and repeat the above arguments to see that 71 holds for all \(t \in [2T_{\delta}, 3T_{\delta}]\). We can continue repeating this argument for all \(x_n\) to see that 71 holds for all \(t \geq T_{\delta}\).

Let \(L_{\lambda}\) be the Lipschitz constant for \(\lambda_{\infty}\). Then for all \(y_1 \in B(y, \frac{\epsilon}{4 L_{\lambda}})\) we have \[\begin{align} \norm{\lambda_{\infty}(y) - \lambda_{\infty}(y_1)} \leq \frac{\epsilon}{4}. \label{eq:32all32y32ext32inp323} \end{align}\tag{72}\] To summarize, by 71 and 72 , we know that if \(y_1 \in B(y, \min(\rho_y, \frac{\epsilon}{4L_{\lambda}}))\) then \[\begin{align} \eta^{y_1}_{\infty}(t,x) \in B(\lambda_{\infty}(y_1), \epsilon). \label{eq:32all32y32ext32inp32last} \end{align}\tag{73}\] Each \(y\) gives us such a ball \(B(y, \min(\rho_y, \frac{\epsilon}{4L_{\lambda}}))\) along with a \(T_{\delta}\) and these balls cover \(K\). By compactness, we can obtain a finite subcover and a finite number of times \(T_1, \ldots, T_n\). Taking \(T_{\epsilon}\) to be the maximum of these times completes the proof. ◻

Proof. Fix \(y\) and \(\epsilon\). From Lemma 18 we know that the result holds for \(t \geq T_{\epsilon}\), so we just need to show that it holds for \(0 \leq t < T_{\epsilon}\). By Lyapunov stability, there exists \(\delta_y\) such that if \(x \in B(\lambda_{\infty}(y), \delta_y)\), then for all \(t \geq 0\), \(\eta^y_{\infty}(t, x) \in B(\lambda_{\infty}(y), \frac{\epsilon}{3})\).

By Lemma 4, there exists \(\rho\) small enough that for all \(y' \in B(y, \rho)\), \[\begin{align} \norm{\eta^y_{\infty}(t, x)-\eta^{y_1}_{\infty}(t, x)} < \frac{\epsilon}{3} \end{align}\] for all \(x \in B(\lambda_{\infty}(y), \delta_y)\), \(t \in [0, T_{\delta}]\). Finally, since \(\lambda_{\infty}\) is Lipschitz, if \(y' \in B(y, \frac{\epsilon}{3L_{\lambda}})\), \[\begin{align} \norm{\lambda_{\infty}(y)-\lambda_{\infty}(y_1)} < \frac{\epsilon}{3}. \end{align}\] Combining these facts, we know that if \(y_1 \in B(y, \min(\rho, \frac{\epsilon}{3L_{\lambda}}))\), then \(\eta_{\infty}^y(t,x) \in B(\lambda_{\infty}(y),\epsilon)\) for all \(t \geq 0\). Each \(y\) comes with such a neighborhood and a distance \(\delta\). By compactness, we can extract a finite subcover and a finite number of distances \(\delta_1, \ldots, \delta_n\), and taking \(\delta\) to be the smallest one gives us our result. ◻

Proof. By 19, we know that for all \(y \in [-2, 2]^{d_2}\) (chosen since \(\norm{\tilde{y}_n(0)} \leq 1\)) there is some \(\delta\) with \(\frac{\epsilon}{2} > \delta > 0\) such that if \(\eta_{\infty}^y(t, x) \in B(\lambda_{\infty}(y), \delta)\), then for all \(t' > t\), \(\eta_{\infty}^y(t', x) \in B(\lambda_{\infty}(y), \frac{\epsilon}{2})\). By Lemma 18 we know that there exists some time \(T_{\delta}\) such that for all \(x \in \lambda_{\infty}(K_y)\) (image of a compact set under a continuous map is compact) and \(t \geq T_{\delta}\), we have, for all \(y \in [-2, 2]^{d_2}\), \[\begin{align} \eta^y_{\infty}(t,x) \in B(\lambda_{\infty}(y), \frac{\delta}{4}). \label{eq:32close32lambda32asymp32stab} \end{align}\tag{74}\]

By Lemma 4, there exists \(n_1\) such that for all \(n > n_1\), \[\begin{align} \norm{\eta^y_{r_n}(t, x) - \eta^y_{\infty}(t,x)} < \frac{\delta}{4} \label{eq:32close32lambda32e40c41} \end{align}\tag{75}\] for all \(t \in [0, T_{\delta} + 1]\).

By 63 , there exists \(n_2\) such that for all \(n > n_2\), \[\begin{align} \norm{\tilde{x}_n(t) - \eta^{\tilde{y}_n(0)}_{r_n}(t,\tilde{x}_n(0))} < \frac{\delta}{4} \label{eq:32close32lambda32x32discret} \end{align}\tag{76}\] and \[\begin{align} \norm{\lambda_{\infty}(\tilde{y}_n(t)) - \lambda_{\infty}(\tilde{y}_n(0))} \leq L_{\lambda} \norm{\tilde{y}_n(t) - \tilde{y}_n(0)} < \frac{\delta}{4} \label{eq:32close32lambda32y32discret} \end{align}\tag{77}\] for all \(t \in [0, T_{\delta} + 1]\). Combining 74 , 75 , 76 , 77 , and the fact that for \(n\) large enough, \(T_{n+1}-T_n < T + 1\), we have \[\begin{align} \norm{\tilde{x}_n(T_{n+1}-T_n) - \lambda_{\infty}(\tilde{y}_n(T_{n+1} - T_n))} < \delta \end{align}\] for all \(n > \max(n_1, n_2)\). This means that, by homogeneity of \(\lambda_{\infty}\), \[\begin{align} \norm{\tilde{x}_{n+1}(0)-\lambda_{\infty}(\tilde{y}_{n+1}(0))} = \frac{r_n}{r_{n+1}} \norm{\tilde{x}_n(T_{n+1} - T_n) - \lambda_{\infty}(\tilde{y}_n(T_{n+1} - T_n))} < \delta \end{align}\] when \(n > \max(n_1, n_2)\), giving us \[\begin{align} \norm{\tilde{x}_n(0)-\lambda_{\infty}(\tilde{y}_n(0))} < \delta \end{align}\] for all \(n > N_{\epsilon} = \max(n_1, n_2) + 1\). So, \[\begin{align} \norm{\eta^{\tilde{y}_n(0)}_{r_n}(t, \tilde{x}_n(0))-\lambda_{\infty}(\tilde{y}_n(0))} < \frac{\epsilon}{2} \end{align}\] for all \(t \geq 0\). Combining this with 76 and 77 gives us \[\begin{align} \norm{\tilde{x}_n(t) - \lambda_{\infty}(\tilde{y}_n(t))} \leq \epsilon \end{align}\] for all \(t \in [0, T_{\delta}]\), as desired. Taking \(T_{\epsilon} \doteq T_{\delta}\) completes the proof. ◻

Proof. For clarity, since we will work with both fast and slow timescale objects, we will use \(\alpha\) and \(\beta\) in superscripts to indicate the fast and slow timescale variants respectively. We also define some notation to help in converting between timescales: \[\begin{align} \tau^{\alpha}(t) \doteq t^{\alpha}(m^{\beta}(t)) \\ \tau^{\beta}(t) \doteq t^{\beta}(m^{\alpha}(t)) \end{align}\] By the difference in how the rescaling is defined between the timescales, we know that \(\norm{\tilde{z}^{\beta}_n(0)} \leq \norm {\tilde{z}^{\alpha}_{m}(\tau^{\alpha}(T_n^{\beta})-T_m^{\alpha})}\), where \(m\) is the largest \(v\) such that \(T_v^{\alpha} \leq \tau^{\alpha}(T_n^{\beta})\). Essentially, whenever the slow timescale is rescaled, the iterate has a smaller norm than the fast timescale’s iterate at that index. Now we need to show that for the rest of that period in the slow timescale (i.e. when \(T^{\beta}_n \leq t \leq T^{\beta}_{n+1}\)) the slow timescale iterates, despite not being rescaled during this period, never get too much larger than the fast timescale iterates which may be rescaled many times.

Thus, given \(t \in [0, T^{\beta}]\), we are interested in the ratio \(\frac{r^{\alpha}_l}{r^{\beta}_n}\), where \(l\) is the largest \(v\) where \(T_v^{\alpha} \leq \tau^{\alpha}(T_n^{\beta}+t)\). By Lemma 65 we know that \[\begin{align} r^{\alpha}_l \leq \norm{z^{max}_{m^{\beta}(T^{\beta}_n+t)}} \leq C \norm{z^{max}_{m^{\beta}(T_n^{\beta})}} + D = C \cdot r^{\beta}_n + D \end{align}\] implying that \[\begin{align} \frac{r^{\alpha}_l}{r^{\beta}_n} \leq C + \frac{D}{r^{\beta}_n}. \end{align}\] There exists \(n_1\) such that for all \(n> n_1\), \(\frac{D}{r^{\beta}_n} \leq C\), so for all \(n > n_1\), \[\begin{align} \frac{r^{\alpha}_l}{r^{\beta}_n} \leq 2C. \end{align}\] By Lemma 20, there exist \(n_2\) and \(T_{\epsilon}\) such that for all \(n> n_{\epsilon}\), we have \[\begin{align} \norm{\tilde{x}^{\alpha}_n(t) - \lambda_{\infty}(\tilde{y}^{\alpha}_n(t))} \leq \frac{\epsilon}{2C} \end{align}\] for all \(t \in [0, T_{\epsilon}]\). Now, set \(\require{physics} N = \max \qty{n_1, n_2}\), where \(n_2\) is the smallest \(v\) such that \(T_v^{\beta} \geq \tau^{\beta}(T_{n_{\epsilon}}^{\alpha})\) and \(T^{\alpha}\) is set to \(T_{\epsilon}\). We then have, for all \(t \in [0, T^{\beta}]\), by homogeneity of \(\lambda_{\infty}\), \[\begin{align} \norm{\tilde{x}^{\beta}_n(t) - \lambda_{\infty}(\tilde{y}^{\beta}_n(t))} \leq 2C \norm{\tilde{x}^{\alpha}_l(\tau^{\alpha}(T_n^{\beta}+t)) - \lambda_{\infty}(\tilde{y}^{\alpha}_l(\tau^{\alpha}(T_n^{\beta}+t)))} \leq \epsilon \end{align}\] for all \(n > N\). Since we set \(T^{\alpha}\) to \(T_{\epsilon}\) without modifying \(T^{\beta}\), the result holds for all \(\epsilon\), regardless of what value we set \(T^{\beta}\) to at the start. ◻

Proof. \[\begin{align} \lim_{k \to \infty} &\norm{ \sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \beta(i)[G_{r_{n_k}}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1}) - G_{r_{n_k}}(\lambda_{\infty}(\tilde{y}_{n_k}(t(i)-T_{n_k})),\tilde{y}_{n_k}(t(i)-T_{n_k}),W_{i+1})]} \\ \leq& \lim_{k \to \infty} \sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \beta(i) L(W_i) \norm{\tilde{x}_{n_k}(t(i)-T_{n_k})-\lambda_{\infty}(\tilde{y}_{n_k}(t(i)-T_{n_k}))} \\ \leq& \lim_{k \to \infty} \epsilon(k) \sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \beta(i) L(W_i) \tag*{(Lemma \ref{lemma:32slow4432x32close32lambda32y})} \\ \leq& C \lim_{k \to \infty} \epsilon(k) \tag*{(\ref{eq:32L32property32bounded})} \\ =& 0. \end{align}\] ◻

Proof. By Lyapunov stability, there is a \(\delta > 0\) such that any trajectory beginning within \(B(0, \delta)\) stays within \(\frac{\epsilon}{2}\) of the equilibrium \(0\).

For an initial condition \(y\), let \(T_y\) be a time at which the trajectory is within \(\frac{\delta}{2}\) of the equilibrium. Let \(y_1\) be some other initial condition. By definition, Lipschitzness, and the Gronwall inequality, we have \[\begin{align} \norm{\chi_{\infty}(t, y) - \chi_{\infty}(t, y_1) } &\leq \norm{y-y_1} + L(L_{\lambda}+1) \int_0^t \norm{\chi_{\infty}(s, y) - \chi_{\infty}(s, y_1)}ds \\ &\leq \norm{y-y_1}e^{L (L_{\lambda}+1) T_y} \end{align}\] for all \(t \leq T_y\).

So there is a neighborhood \(V_y\) such that for all \(y_1 \in V_y\), \(\chi_{\infty}(T_y, y_1)\) is within \(\delta\) of the equilibrium, which by Lyapunov stability implies that it will always be within \(\epsilon\) of the equilibrium after \(T_y\).

By compactness, we can cover the set \(K\) by a finite number of such intervals and obtain a finite number of times \(T_{y_1}, \ldots, T_{y_n}\) and then take the maximum time to be the value of \(T_{\epsilon}\). ◻

Proof. We have \[\begin{align} \chi_{c}(t,y) &= y + \int_0^t g_c(\lambda_{\infty}(\chi_c(s, y)), \chi_c(s, y)) ds, \\ \chi_{\infty}(t,y) &= y + \int_0^t g_{\infty}(\lambda_{\infty}(\chi_{\infty}(s, y)), \chi_{\infty}(s, y)) ds. \end{align}\] Let us define the error term: \[\begin{align} e(t) = \norm{\chi_{c}(t,y) - \chi_{\infty}(t,y)}. \end{align}\]

We can bound \(e(t)\) by two terms: \[\begin{align} e(t) \leq& \int_0^t \norm{g_c(\lambda_{\infty}(\chi_c(s, y)), \chi_c(s, y)) - g_c(\lambda_{\infty}(\chi_{\infty}(s, y)), \chi_{\infty}(s, y))} ds \\ &+ \int_0^t \norm{g_c(\lambda_{\infty}(\chi_{\infty}(s, y)), \chi_{\infty}(s, y)) - g_{\infty}(\lambda_{\infty}(\chi_{\infty}(s, y)), \chi_{\infty}(s, y))} ds \end{align}\] To bound the first term, we use Lipschitzness: \[\begin{align} &\int_0^t \norm{g_c(\lambda_{\infty}(\chi_c(s, y)), \chi_c(s, y)) - g_c(\lambda_{\infty}(\chi_{\infty}(s, y)), \chi_{\infty}(s, y))} ds \\ &\leq L(L_{\lambda}+1) \int_0^t \norm{\chi_{c}(t,y) - \chi_{\infty}(t,y)} ds \\ &= L \int_0^t e(s) ds. \end{align}\]

To bound the second term, we use Lipschitzness and Assumption 3:

\[\begin{align} \int_0^t &\norm{g_c(\lambda_{\infty}(\chi_{\infty}(s, y)), \chi_{\infty}(s, y)) - g_{\infty}(\lambda_{\infty}(\chi_{\infty}(s, y)), \chi_{\infty}(s, y))} ds \\ \leq& \int_0^t \epsilon(c) ds\\ \leq& T\epsilon(c). \end{align}\]

To conclude, we will use the Gronwall Inequality (Appendix 5): \[\begin{align} e(t) \leq& T \epsilon(c) + L(L_{\lambda}+1) \int_0^t e(s)ds \\ \leq& \epsilon(c) \cdot Te^{L(L_{\lambda}+1)T}. \end{align}\] ◻

Proof. By Lyapunov stability there is some \(\delta\) with \(\frac{1}{4(L_{\lambda}+1)} > \delta > 0\) such that if \(\chi_{\infty}(t, y) \in B(0, \delta)\) at some time \(t\), then for all times \(t' > t\), we have \[\begin{align} \chi_{\infty}(t', y) \in B \left(0, \frac{1}{4(L_{\lambda}+1)} \right). \label{eq:32slow32chi32lyapunov} \end{align}\tag{96}\] By Lemma 25 we know that there is a \(T_{\delta}\) such that when we set \(T=T_{\delta}\), for all \(t > T\), \[\begin{align} \chi_{\infty}(t, y) \in B(0, \frac{\delta}{3}) \label{eq:32slow32chi32fall} \end{align}\tag{97}\] for all \(y \in [-2, 2]^{d_2}\).

By Lemma 26 we know that there is an \(n_1\) such that for all \(n > n_1\), \[\begin{align} \norm{\chi_{r_n}(t,y) - \chi_{\infty}(t,y)} < \frac{\delta}{3} \label{eq:32slow32chi32close} \end{align}\tag{98}\] for all \(y \in [-2, 2]^{d_2}\) and \(t \in [0, T+1]\).

By 95 , we know that there is an \(n_2\) such that for all \(n > n_2\), we have \[\begin{align} \norm{\tilde{z}_n(t)-(\lambda_{\infty}(\chi_{r_n}(t, \tilde{y}_n(0))), \chi_{r_n}(t, \tilde{y}_n(0)))} < \frac{\delta}{3} \label{eq:32finishing32slow32disc} \end{align}\tag{99}\] for all \(t \in [0, T+1]\).

From 97 , 98 , and 99 , and the fact that when \(n\) is large enough, \(T_{n+1}-T_n < T + 1\), we have \[\begin{align} \tilde{y}_n(0) \in B(0, \delta) \end{align}\] for all \(\require{physics} n \geq \max\qty{n_1, n_2}+1\). Thus, from 96 , we have \[\begin{align} \chi_{\infty}(t, \tilde{y}_n(0)) \in B \left(0, \frac{1}{4(L_{\lambda}+1)} \right) \end{align}\] for all \(t \in [0, T + 1]\), which implies that \[\begin{align} (\lambda_{\infty}(\chi_{\infty}(t, \tilde{y}_n(0))), \chi_{\infty}(t, \tilde{y}_n(0))) \in B \left(0, \frac{1}{4}\right). \end{align}\] Combining this with 99 gives us \[\begin{align} \tilde{z}_n(t) \in B(0, \frac{1}{2}), \end{align}\] for all \(\require{physics} n \geq \max\qty{n_1,n_2}+1\) and \(t \in [0,T+1]\), which, by definition, prevents the sequence \(r_n\) from increasing, contradicting the assumption that \(r_n\) increases to infinity. ◻

Proof. To prove convergence results on \(t \in(-\infty, \infty)\) in Corollary 1, we fix an arbitrary sample path \(\require{physics} \qty{z_0,\qty{W_i}_{i=1}^\infty}\). The stability results from Theorem 1 hold. To prove properties on \(t \in(-\infty, \infty)\), we first fix an arbitrary \(\tau > 0\) and show properties on \(\forall t \in [-\tau, \tau]\).

\(\bar{z}_n(t)\) can also be written as \[\begin{align} \bar{z}_n(t) &= \bar{z}(t(n)) + \int_0^t (h(\bar{z}_n(s)),0)ds, \quad \forall t \in (-\infty, \infty). \label{cor-def:32z32n32integral} \end{align}\tag{102}\]

We need to show that the error of Euler’s discretization diminishes asymptotically. With 44 and 45 , \(\forall \tau > 0\), \(\forall t \in [-\tau, \tau]\), \[\begin{align} \label{eq:32bar32z32extend} &\bar{z}(t(n) + t) = z_{m(t(n) + t)} \\ &= \begin{cases} \bar{z}(t(n)) + \sum_{i = n}^{m(t(n)+t) - 1} (\alpha(i) H(\bar{z}(t(i)), W_{i+1}),\beta(i)G(\bar{z}(t(i)),W_{i+1})) & \text{if } t \geq 0 \\ \bar{z}(t(n)) - \sum_{i = m(t(n)+t)}^{n - 1} (\alpha(i) H(\bar{z}(t(i)), W_{i+1}),\beta(i)G(\bar{z}(t(i)),W_{i+1})) & \text{if } t < 0. \end{cases} \end{align}\tag{103}\] Notably, the property 44 that \(\forall t<0, m(t) = 0\) in 103 ensures \(\bar{z}(t(n) + t)\) is well-defined when \(t(n) + t < 0\). Precisely speaking, \(\forall \tau > 0\), \(\forall t \in [-\tau, \tau]\), the discretization error is defined as \[\begin{align} \bar{f}_n(t) \doteq \bar{z}(t(n) + t) - \bar{z}_n(t). \label{cor-def:32f} \end{align}\tag{104}\] and we need \(\bar{f}_n(t)\) to diminish to 0 as \(n\to\infty\). To this end, we study the following three sequences of functions \[\require{physics} \begin{align} \label{cor-eq32three32seq} \qty{ \bar z(t(n) + t)}_{n=0}^\infty, \qty{\bar{z}_n(t)}_{n=0}^\infty, \qty{\bar{f}_n(t)}_{n=0}^\infty. \end{align}\tag{105}\] Equicontinuity in the extended sense on the domain \((-\infty, \infty)\) is defined as following (Section 4.2.1 in [3]).

We show \(\require{physics} \qty{\bar{z}(t(n)+t)}\), \(\require{physics} \qty{\bar{z}_n(t)}\) and \(\require{physics} \qty{\bar{f}_n(t)}\) are all equicontinuous in the extended sense.

To prove those lemmas, we need the Gronwall inequality in the reverse time in Appendix 6. Compared to lemmas in the main text which have the domain \(t \in [0,T+1)\), lemmas in this section have similar proofs because we first fix an arbitrary \(\tau\) and prove properties on the domain \(t \in [-\tau, \tau]\). We omit proofs for Lemma 28 because they are extremely similar to the proofs of equicontinuity in the fast timescale. Similar to Lemma 2, we now construct a particular subsequence of interest.

Its proof is very similar to the proof of Lemma 2 and is omitted. We use the subsequence \(\require{physics} \qty{n_k}\) intensively in the remaining proofs. Recall that \(\bar{f}_n(t)\) denotes the discretization error between \(\bar z(t(n) + t)\) and \(\bar{z}_n(t)\). We now proceed to prove that this discretization error diminishes along \(\require{physics} \qty{n_k}\). In particular, we aim to prove that \(\forall \tau\), \(\forall t \in [-\tau, \tau]\),

\[\begin{align} \lim_{k \rightarrow \infty} \norm{\bar{f}_{n_k}(t)} = \norm{\bar{f}^{\lim}(t)} = 0. \end{align}\] This means \(\bar{z}(t(n_k) + t)\) is close to \(\bar{z}_{n_k}(t)\) as \(k \rightarrow \infty\). For \(t \in (0,\tau]\), the proof for this part is the same as the proof we have done in Section 12.4. Thus, we only discuss the proof for \(t \in [-\tau, 0]\). \(\forall \tau\), \(\forall t \in [-\tau, 0]\), \[\begin{align} &\lim_{k \rightarrow \infty} \norm{ \bar{f}_{n_k}(t)} \\ =& \lim_{k \rightarrow \infty} \norm{ \bar{z}(t(n_k)) -\sum_{i = m(t(n_k)+t)}^{n_k - 1} (\alpha(i)H(\bar{z}(t(i)),W_{i+1}), \beta(i)G(\bar{z}(t(i)),W_{i+1})) - \bar{z}_{n_k}(t)} \tag*{(by \eqref{eq:32bar32z32extend} and \eqref{cor-def:32f})}\\ =& \lim_{k \rightarrow \infty} \norm{ -\sum_{i = m(t(n_k)+t)}^{n_k - 1} (\alpha(i)H(\bar{z}(t(i)),W_{i+1}), \beta(i)G(\bar{z}(t(i)),W_{i+1})) - \int_0^t (h(\bar{z}_{n_k}(s)),0)ds} \tag*{(by \eqref{cor-def:32z32n32integral})}\\ \leq&\lim_{k \to \infty} \norm{ -\sum_{i = m(t(n_k)+t)}^{n_k - 1} \alpha(i)H(\bar{z}(t(i)),W_{i+1}) - \int_0^t h(\bar{z}^{\lim}(s))ds } \\ &+ \lim_{k \to \infty} \norm{ -\sum_{i = m(t(n_k)+t)}^{n_k - 1} \beta(i)G(\bar{z}(t(i)),W_{i+1}) } \\ &+ \lim_{k \to \infty} \norm{\int_0^t h(\bar{z}^{\lim}(s))ds - \int_0^t h(\bar{z}_{n_k}(s))ds } . \label{cor-eq:32lim32f32k} \\ \end{align}\tag{111}\]

The second term in the RHS of 111 is 0.

The proof is very similar to the proof of Lemma 12 so is omitted.

The first term in the RHS of 111 is also 0.

Its proof is very similar to the proof of Lemma 44 and is omitted. This convergence is also simpler than 148 because here we have only a single \((H, h)\). But in 148 , we have a sequence \(\require{physics} \qty{(H_{n_k}, h_{n_k})}\), for which we have to split it to a double limit 149 and then invoke the Moore-Osgood theorem to reduce it to the single \((H, h)\) case.

Lemma 31 confirms that the first term in the RHS of 111 is 0. Moreover, it also enables us to rewrite \(\bar{z}^{\lim}(t)\) from a summation form to an integral form. \(\forall \tau\), \(\forall t \in [-\tau, 0]\) \[\begin{align} &\bar{z}^{\lim}(t) \\ =& \lim_{k \to \infty} \bar{z}(t(n_k)) -\sum_{i = m(t(n_k)+t)}^{n_k - 1} (\alpha(i)H(\bar{z}(t(i)),W_{i+1}), \beta(i)G(\bar{z}(t(i)),W_{i+1})) \\ =& \makebox[0.85\textwidth]{\displaystyle\lim_{k \to \infty} \bar{z}(t(n_k)) + \int_0^t (h (\bar{z}^{\lim}(s)),0)ds.\hfill(by Lemma \ref{cor-lemma:32double32limit322})} \label{cor-eq:32z32lim32definition322} \end{align}\tag{112}\] Thus, we can show the following diminishing discretization error.

Its proof is very similar to the proof of Lemma 46 and is omitted. This immediately implies that for any \(t \in (-\infty, \infty)\) \[\begin{align} \label{eq32corollary32ode32convergence} \lim_{k \to \infty} \bar{z}(t(n_k) + t) = z^{\lim}(t). \end{align}\tag{113}\] Theorem 1 then yields that \[\begin{align} \sup_{t \in (-\infty, \infty)} \norm{ z^{\lim}(t)} < \infty. \end{align}\]

Let \(Z\) be the set of limit points of \(\require{physics} \qty{z_n}\). By Theorem 1, \(\sup_n \norm{z_n} < \infty\), so \(Z\) is bounded and nonempty. We now prove \(Z\) is an invariant set of the ODE 100 . For any \(z \in Z\), there exists a subsequence \(\require{physics} \qty{z_{n_k}}\) such that \[\begin{align} \lim_{k\to\infty} z_{n_k} = z. \end{align}\] Since \(\require{physics} \qty{\bar z(t(n_k) + t)}\) is equicontinuous in the extended sense, following the way we arrive at 113 , we can construct a subsequence \(\require{physics} \qty{n'_k} \subseteq \qty{n_k}\) such that \[\begin{align} \label{eq:32z32n32k3932limit} \lim_{k\to\infty} \bar z(t(n_k') + t) = z_*^{\lim}(t), \end{align}\tag{114}\] where \(z_*^{\lim}(t)\) is a solution to the ODE 100 and \(z_*^{\lim}(0) = z\). The remaining is to show that \(z_*^{\lim}(t)\) lies entirely in \(Z\). For any \(t \in (-\infty, \infty)\), by the piecewise constant nature of \(\bar z\) in 103 , the above limit 114 implies that there exists a subsequence of \(\require{physics} \qty{z_n}\) that converges to \(z_*^{\lim}(t)\), indicating \(z_*^{\lim}(t) \in Z\) by the definition of the limit set. We now have proved \(\forall z \in Z\), there exists a solution \(z_*^{\lim}(t)\) to the ODE 100 such that \(z_*^{\lim}(0) = z\) and \(\forall t \in (-\infty, \infty), z_*^{\lim}(t) \in Z\). This means \(Z\) is an invariant set, by definition. In particular, \(Z\) is a bounded invariant set.

We now prove that \(\require{physics} \qty{z_n}\) converges to \(Z\). Let \(\require{physics} \qty{z_{n_k}}\) be any convergent subsequence of \(\require{physics} \qty{z_n}\) with its limit denoted by \(z\). We must have \(z \in Z\) by the definition of the limit set. So we have proved that all convergent subsequences of \(\require{physics} \qty{z_n}\) converge to a point in the bounded invariant set \(Z\). If \(\require{physics} \qty{z_n}\) does not converge to \(Z\), there must exist a subsequence \(\require{physics} \qty{z_{n_k'}}\) such that \(\require{physics} \qty{z_{n_k'}}\) is always away from \(Z\) by some small \(\epsilon_0 > 0\), i.e., \(\forall k\), \[\begin{align} \inf_{z\in Z} \norm{z_{n_k'} - z} \geq \epsilon_0. \label{eq:32z32n32k32epsilon320} \end{align}\tag{115}\] But \(\require{physics} \qty{z_{n_k'}}\) is bounded, so by the Bolzano-Weierstrass Theorem, it must have a convergent subsequence, which, by the definition of the limit set, converges to some point in \(Z\). This contradicts 115 . So we must have \(\require{physics} \qty{z_n}\) converge to \(Z\), which is a bounded invariant set of the ODE 100 . This completes the proof. ◻

Proof. To prove convergence results on \(t \in(-\infty, \infty)\) in Corollary 2, we fix an arbitrary sample path \(\require{physics} \qty{z_0,\qty{W_i}_{i=1}^\infty}\). The stability results from Theorem 1 hold. To prove properties on \(t \in(-\infty, \infty)\), we first fix an arbitrary \(\tau > 0\) and show properties on \(\forall t \in [-\tau, \tau]\).

\(\bar{y}_n(t)\) can also be written as \[\begin{align} \bar{y}_n(t) &= \bar{y}(t(n)) + \int_0^t g(\lambda(\bar{y}_n(s)),\bar{y}_n(s))ds, \quad \forall t \in (-\infty, \infty). \label{cor-def:32slow32y32n32integral} \end{align}\tag{119}\]

We need to show that the error of Euler’s discretization diminishes asymptotically. With 44 and 45 , \(\forall \tau > 0\), \(\forall t \in [-\tau, \tau]\), \[\begin{align} \label{eq:32slow32bar32z32extend} &\bar{z}(t(n) + t) = z_{m(t(n) + t)} \\ &= \begin{cases} \bar{z}(t(n)) + \sum_{i = n}^{m(t(n)+t) - 1} (\alpha(i) H(\bar{z}(t(i)), W_{i+1}),\beta(i)G(\bar{z}(t(i)),W_{i+1})) & \text{if } t \geq 0 \\ \bar{z}(t(n)) - \sum_{i = m(t(n)+t)}^{n - 1} (\alpha(i) H(\bar{z}(t(i)), W_{i+1}),\beta(i)G(\bar{z}(t(i)),W_{i+1})) & \text{if } t < 0. \end{cases} \end{align}\tag{120}\] Notably, the property 44 that \(\forall t<0, m(t) = 0\) in 120 ensures \(\bar{z}(t(n) + t)\) is well-defined when \(t(n) + t < 0\). Precisely speaking, \(\forall \tau > 0\), \(\forall t \in [-\tau, \tau]\), the discretization error is defined as \[\begin{align} \bar{f}_n(t) \doteq \bar{y}(t(n) + t) - \bar{y}_n(t). \label{cor-def:32slow32f} \end{align}\tag{121}\] and we need \(\bar{f}_n(t)\) to diminish to 0 as \(n\to\infty\). To this end, we study the following three sequences of functions \[\require{physics} \begin{align} \label{cor-eq32slow32three32seq} \qty{ \bar y(t(n) + t)}_{n=0}^\infty, \qty{\bar{y}_n(t)}_{n=0}^\infty, \qty{\bar{f}_n(t)}_{n=0}^\infty. \end{align}\tag{122}\]

We show \(\require{physics} \qty{\bar{y}(t(n)+t)}\), \(\require{physics} \qty{\bar{y}_n(t)}\) and \(\require{physics} \qty{\bar{f}_n(t)}\) are all equicontinuous in the extended sense on the domain \((-\infty, \infty)\).

To prove those lemmas, we need the Gronwall inequality in the reverse time in Appendix 6. Compared to lemmas in the main text which have the domain \(t \in [0,T+1)\), lemmas in this section have similar proofs because we first fix an arbitrary \(\tau\) and prove properties on the domain \(t \in [-\tau, \tau]\). We omit proofs for Lemma 33 because they are extremely similar to the proofs of equicontinuity in the slow timescale. Similar to Lemma 22, we now construct a particular subsequence of interest.

Its proof is very similar to the proof of Lemma 22 and is omitted. We use the subsequence \(\require{physics} \qty{n_k}\) intensively in the remaining proofs. Recall that \(\bar{f}_n(t)\) denotes the discretization error between \(\bar y(t(n) + t)\) and \(\bar{y}_n(t)\). We now proceed to prove that this discretization error diminishes along \(\require{physics} \qty{n_k}\). In particular, we aim to prove that \(\forall \tau\), \(\forall t \in [-\tau, \tau]\), \[\begin{align} \lim_{k \rightarrow \infty} \norm{\bar{f}_{n_k}(t)} = \norm{\bar{f}^{\lim}(t)} = 0. \end{align}\] This means \(\bar{y}(t(n_k) + t)\) is close to \(\bar{y}_{n_k}(t)\) as \(k \rightarrow \infty\). For \(t \in (0,\tau]\), the proof for this part is the same as the proof we have done in Section 13.4. Thus, we only discuss the proof for \(t \in [-\tau, 0]\). \(\forall \tau\), \(\forall t \in [-\tau, 0]\), \[\begin{align} &\lim_{k \rightarrow \infty} \norm{ \bar{f}_{n_k}(t)} \\ =& \lim_{k \rightarrow \infty} \norm{ \bar{y}(t(n_k)) -\sum_{i = m(t(n_k)+t)}^{n_k - 1} \beta(i)G(\bar{z}(t(i)),W_{i+1}) - \bar{y}_{n_k}(t)} \tag*{(by \eqref{eq:32slow32bar32z32extend} and \eqref{cor-def:32slow32f})}\\ =& \lim_{k \rightarrow \infty} \norm{ -\sum_{i = m(t(n_k)+t)}^{n_k - 1} \beta(i)G(\bar{z}(t(i)),W_{i+1}) - \int_0^t g(\lambda(\bar{y}_{n_k}),\bar{y}_{n_k})ds} \tag*{(by \eqref{cor-def:32slow32y32n32integral})}\\ \leq&\lim_{k \to \infty} \norm{ -\sum_{i = m(t(n_k)+t)}^{n_k - 1} \beta(i)G(\bar{z}(t(i)),W_{i+1}) + \sum_{i = m(t(n_k)+t)}^{n_k - 1} \beta(i)G(\lambda(\bar{y}(t(i))), \bar{y}(t(i)),W_{i+1}) } \\ &+ \lim_{k \to \infty} \norm{- \sum_{i = m(t(n_k)+t)}^{n_k - 1} \beta(i)G(\lambda(\bar{y}(t(i))), \bar{y}(t(i)),W_{i+1}) - \int_0^t g(\lambda(\bar{y}_{n_k}),\bar{y}_{n_k})ds } . \label{cor-eq:32slow32lim32f32k} \\ \end{align}\tag{127}\]

The first term in the RHS of 127 is 0.

Its proof is very similar to the proof of Lemma 24 (except that we use 116 instead of Lemma 23) and is omitted. This convergence is also simpler than 148 because here we have only a single \((G, g)\).

Its proof is very similar to the proof of the diminishing discretization error in the slow timescale. This immediately implies that for any \(t \in (-\infty, \infty)\) \[\begin{align} \label{eq32corollary32slow32ode32convergence} \lim_{k \to \infty} \bar{z}(t(n_k) + t) = (\lambda(y^{\lim}(t)), y^{\lim}(t)). \end{align}\tag{128}\] Theorem 1 then yields that \[\begin{align} \sup_{t \in (-\infty, \infty)} \norm{ (\lambda(y^{\lim}(t)), y^{\lim}(t))} < \infty. \end{align}\]

Let \(Y\) be the set of limit points of \(\require{physics} \qty{y_n}\). By Theorem 1, \(\sup_n \norm{y_n} < \infty\), so \(Y\) is bounded and nonempty. We now prove \(Y\) is an invariant set of the ODE 117 . For any \(y \in Y\), there exists a subsequence \(\require{physics} \qty{y_{n_k}}\) such that \[\begin{align} \lim_{k\to\infty} y_{n_k} = y. \end{align}\] Since \(\require{physics} \qty{\bar y(t(n_k) + t)}\) is equicontinuous in the extended sense, following the way we arrive at 128 , we can construct a subsequence \(\require{physics} \qty{n'_k} \subseteq \qty{n_k}\) such that \[\begin{align} \label{eq:32slow32y32n32k3932limit} \lim_{k\to\infty} \bar y(t(n_k') + t) = y_*^{\lim}(t), \end{align}\tag{129}\] where \(y_*^{\lim}(t)\) is a solution to the ODE 117 and \(y_*^{\lim}(0) = y\). The remaining is to show that \(y_*^{\lim}(t)\) lies entirely in \(Y\). For any \(t \in (-\infty, \infty)\), by the piecewise constant nature of \(\bar y\) in 120 , the above limit 129 implies that there exists a subsequence of \(\require{physics} \qty{y_n}\) that converges to \(y_*^{\lim}(t)\), indicating \(y_*^{\lim}(t) \in Y\) by the definition of the limit set. We now have proved \(\forall y \in Y\), there exists a solution \(y_*^{\lim}(t)\) to the ODE 117 such that \(y_*^{\lim}(0) = y\) and \(\forall t \in (-\infty, \infty), y_*^{\lim}(t) \in Y\). This means \(Y\) is an invariant set, by definition. Additionally, \(Y\) is a bounded invariant set.

We now prove that \(\require{physics} \qty{y_n}\) converges to \(Y\). Let \(\require{physics} \qty{y_{n_k}}\) be any convergent subsequence of \(\require{physics} \qty{y_n}\) with its limit denoted by \(y\). We must have \(y \in Y\) by the definition of the limit set. So we have proved that all convergent subsequences of \(\require{physics} \qty{y_n}\) converge to a point in the bounded invariant set \(Y\). If \(\require{physics} \qty{y_n}\) does not converge to \(Y\), there must exist a subsequence \(\require{physics} \qty{y_{n_k'}}\) such that \(\require{physics} \qty{y_{n_k'}}\) is always away from \(Y\) by some small \(\epsilon_0 > 0\), i.e., \(\forall k\), \[\begin{align} \inf_{y\in Y} \norm{y_{n_k'} - y} \geq \epsilon_0. \label{eq:32slow32y32n32k32epsilon320} \end{align}\tag{130}\] But \(\require{physics} \qty{y_{n_k'}}\) is bounded, so by the Bolzano-Weierstrass Theorem, it must have a convergent subsequence, which, by the definition of the limit set, converges to some point in \(Y\). This contradicts 130 . So we must have \(\require{physics} \qty{y_n}\) converge to \(Y\), which is a bounded invariant set of the ODE 117 . This completes the proof. ◻

Proof. We show that all the assumptions are satisfied such that Theorem 2 applies.

Assumption 1 follows immediately from Lemma 37.

The assumption 2 follows immediately from the choice of appropriate learning rates.

For Assumption 3, define \[\begin{align} H_\infty(x, y, w) \doteq& \mqty[-C(w) & A(w)] \mqty[x \\ y], \\ G_\infty(x, y, w) \doteq& \mqty[-D^\top(w) & A(w)] \mqty[x \\ y]. \end{align}\] Then we have \[\begin{align} H_c(x, y, w) - H_\infty(x, y, w) =\frac{b(w)}{c} , \\ G_c(x, y, w) - G_{\infty}(x, y, w) = \frac{b(w)}{c} \end{align}\] After noticing \[\begin{align} \norm{b((s, a, s', e)) - b((s, a, s', e'))} = \rho(s, a) \abs{r(s, a)} \norm{e - e'}, \quad \forall s, a, s', e, e', \end{align}\] Assumption 3 follows immediately from Lemma 37.

For Assumption 4, it can be easily verified that \(H(x, y, w), G(x, y, w), H_\infty(x, y, w),\) and \(G_\infty(x, y, w)\) are Lipschitz continuous in \((x,y)\) for each \(w\), with the Lipschitz constant being \[\require{physics} \begin{align} L(w) =& \max \qty{\norm{\mqty[-C(w) & A(w) ]},\norm{\mqty[-D^\top(w) & A(w) ]}} \end{align}\]

Since \(A(w), b(w), C(w), D(w)\) are Lipschitz continuous in \(e\) for each \((s, a, s')\), Lemma 37 implies that \[\require{physics} \begin{align} h(x, y) =& \mqty[-C & A ] \mqty[x \\ y] + b, \\ h_\infty(x, y) =& \mqty[-C & A ] \mqty[x \\ y], \\ g(x, y) =& \mqty[-D^\top & A ] \mqty[x \\ y] + b \\ g_\infty(x, y) =& \mqty[-D^\top & A ] \mqty[x \\ y], \\ L =& \max \qty{\norm{\mqty[-C & A ]},\norm{\mqty[-D^\top & A ]}} . \end{align}\] Assumption 4 then follows.

For Assumption 5, we have \[\begin{align} \norm{h_c(x, y) - h_\infty(x, y)} =& \frac{1}{c}\norm{b}, \\ \norm{g_c(x, y) - g_\infty(x, y)} =& \frac{1}{c}\norm{b}, \end{align}\] so the uniform convergence of \(h_c\) to \(h_\infty\) and \(g_c\) to \(g_\infty\) follow immediately.

Since \(\Phi\) has full column rank and \(D_{\mu}\) is positive definite (due to irreducibility of \(\require{physics} \qty{S_t}\)), \(C=\Phi^\top D_{\mu} \Phi\) is positive definite, so \(-C\) is negative definite. This implies the global asymptotic stability of these two ODEs: \[\begin{align} \dv{x(t)}{t} = h(x(t),y) = -Cx(t) +Ay+ b, \quad \dv{x(t)}{t} = h_{\infty}(x(t),y)= -C x(t) + Ay. \end{align}\] The unique globally asymptotically stable equilibrium of the first is \(C^{-1}(Ay+b)\), and of the second is \(C^{-1}Ay\). This means that we can define \(\lambda(y) = C^{-1}(Ay+b)\) and \(\lambda_{\infty}(y) = C^{-1}Ay\). Note that \(\lambda_{\infty}(cy) = c\lambda_{\infty}(y)\) as required.

Now we will analyze two ODEs: \[\begin{align} \dv{y(t)}{t} =& g(\lambda(y(t)), y(t)) \\ =& -D^\top \lambda(y(t))+Ay(t)+b\\ =& -D^\top (C^{-1}(Ay(t)+b))+Ay(t)+b \\ =& (I-D^\top C^{-1})(Ay(t)+b) \\ =& (I-(A^\top + C)C^{-1})(Ay(t)+b) \tag*{(by A+C=D, C=C^\top)} \\ =& -A^\top C^{-1} (Ay(t)+b), \end{align}\]

\[\begin{align} \dv{y(t)}{t} =& g_{\infty}(\lambda_{\infty}(y(t)), y(t)) \\ =& -D^\top \lambda_{\infty}(y(t))+Ay(t)\\ =& -D^\top C^{-1} Ay(t)+Ay(t) \\ =& (I-D^\top C^{-1})Ay(t) \\ =& (I-(A^\top + C)C^{-1})Ay(t) \tag*{(by A+C=D, C=C^\top)} \\ =& -A^\top C^{-1} Ay(t), \end{align}\]

Since \(A\) is nonsingular and \(C\) is positive definite, \(-A^\top C^{-1} A\) is negative definite. This implies the global asymptotic stability of the previous two ODEs. The unique globally asymptotically stable equilibrium of the first is \(-A^{-1}b\), and of the second is \(0\). Assumption 5 then follows.

Assumption 6 follows immediately from Lemma 37 and Assumption 2.

Theorem 2 then implies that \[\begin{align} \lim_{t\to\infty} x_t &= 0 \qq{a.s.} \\ \lim_{t\to\infty} y_t &= -A^{-1}b \qq{a.s.} \end{align}\] which completes the proof. ◻

Proof. Fix an arbitrary sample path \(\require{physics} \qty{x_0,y_0, \qty{W_i}_{i=1}^\infty}\). Use \(\mathcal{B}\) to denote an arbitrary compact subset of \(\mathbb{R}^{d_1+d_2}\). \[\begin{align} &\lim_{c \to \infty} \sup_{z \in \mathcal{B}} \sup_n \sup_{t \in [0,T]} \norm{ \sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) \left[H_c(z, W_{i+1}) - H_\infty(z, W_{i+1}) \right] } \\ =&\lim_{c \to \infty} \sup_{z \in \mathcal{B}} \sup_n \sup_{t \in [0,T]} \norm{\sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) \kappa(c) b(z, W_{i+1})} \tag*{(by \eqref{eq:32H32c32H32inf32kappa})} \\ =&\lim_{c \to \infty} \kappa(c) \sup_{z \in \mathcal{B}} \sup_n\sup_{t \in [0,T]} \norm{\sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) b(z,W_{i+1})} \\ =& 0 \sup_{z \in \mathcal{B}} \sup_n \sup_{t \in [0,T]} \norm{\sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) b(z, W_{i+1})} \label{eq:32H32c32H32inf32032finite} \end{align}\tag{143}\] We now show that the function \[\begin{align} \label{eq32a32function32of32x} z \mapsto \sup_n \sup_{t \in [0,T]} \norm{\sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) b(z, W_{i+1})} \end{align}\tag{144}\] is Lipschitz continuous. \(\forall z, z'\),

\[\begin{align} &\abs{\sup_n \sup_{t \in [0,T]} \norm{\sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) b(z, W_{i+1})} - \sup_n \sup_{t \in [0,T]} \norm{\sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) b(z', W_{i+1})}} \\ \leq&\sup_n \sup_{t \in [0,T]} \abs{\norm{\sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) b(z, W_{i+1})} - \norm{\sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) b(z', W_{i+1})}} \tag*{(by \abs{\sup_z f(z) - \sup_z g(z)} \leq \sup_z \abs{f(z) - g(z)} )} \\ \leq&\sup_n \sup_{t \in [0,T]} \norm{\sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) b(z, W_{i+1}) - \sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) b(z', W_{i+1})} \\ \leq&\sup_n \sup_{t \in [0,T]} \sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) \norm{ b(z, W_{i+1}) - b(z', W_{i+1})} \\ \leq& \sup_n \sup_{t \in [0,T]} \left(\sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) L_b(W_{i+1})\right) \norm{z - z'} \end{align}\] By Lemma 38 and 180 , \[\begin{align} \sup_n \sup_{t \in [0,T]} \left(\sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) L_b(W_{i+1})\right) < \infty \end{align}\] can be viewed as the Lipschitz constant. Thus, 144 is a continuous function. Since \(\mathcal{B}\) is compact, the extreme value theorems asserts that the supremum of 144 in \(\mathcal{B}\) is attainable at some \(z_{\mathcal{B}}\) and is finite. This means the RHS of 143 is 0, so

\[\begin{align} &\lim_{c \to \infty} \sup_{z \in \mathcal{B}} \sup_n \sup_{t \in [0,T]} \norm{ \sum_{i=m(T_n)}^{m(T_n+t) - 1} \alpha(i) \left[H_c(z, W_{i+1}) - H_\infty(z, W_{i+1}) \right] } = 0. \label{eq:32b32e32b320} \end{align}\tag{145}\] The proofs for the other statements follow similar arguments and so are omitted. ◻

Proof. By 47 , \[\begin{align} \sup_n \norm{\tilde{z}_n(0)} \leq 1 . \end{align}\]

Without loss of generality, let \(t_1 \leq t_2\). \[\begin{align} &\limsup_n \sup_{ 0\leq t_2-t_1 \leq \delta } \norm{ \tilde{z}_n(t_1) - \tilde{z}_n(t_2)} \\ =& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} (\alpha(i) H_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}), \beta(i) G_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1})) }\\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1})} \\ &+ \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1})}. \end{align}\]

We bound each term individually. We start with the first term.

\(\forall \xi > 0\), by 172 , \(\exists \delta_0,\) such that \(\forall 0< \delta \leq \delta_0\), \[\begin{align} \sup_{c\geq 1} \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{ \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_c(0, W_{i+1}) } \leq \xi. \label{eq:32lemma32hat32x32equicontinuous32xi321} \end{align}\tag{146}\] By 179 , \(\exists \delta_1,\) such that \(\forall 0< \delta \leq \delta_1\), \[\begin{align} \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) L(W_{i+1}) \leq \xi. \label{eq:32lemma32hat32x32equicontinuous32xi322} \end{align}\tag{147}\] Without loss of generality, let \(t_1 \leq t_2\). Then \(\require{physics} \forall \delta \leq \min \qty{\delta_0,\delta_1}\), we have \[\begin{align} &\limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) }\\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) } - \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_{r_n}(0, W_{i+1}) } \\ &+ \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_{r_n}(0, W_{i+1}) } \\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) } - \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_{r_n}(0, W_{i+1}) } \\ &+ \sup_{c\geq 1} \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_c(0, W_{i+1}) } \\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) } - \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_{r_n}(0, W_{i+1}) } \\ &+ \xi \tag*{(by \eqref{eq:32lemma32hat32x32equicontinuous32xi321})}\\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) - \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_{r_n}(0, W_{i+1}) } \\ &+ \xi \\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) \norm{H_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) - H_{r_n}(0, W_{i+1}) } + \xi\\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) L(W_{i+1})\norm{\tilde{z}_n(t(i)-T_n)} + \xi \\ \leq& (C_{\hat{x}}+C_{\hat{y}}) \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) L(W_{i+1}) + \xi \tag*{(by Lemma~\ref{lemma:32bound32z32hat})} \\ \leq& (C_{\hat{x}}+C_{\hat{y}}) \xi + \xi. \tag*{(by \eqref{eq:32lemma32hat32x32equicontinuous32xi322})} \end{align}\] A similar argument bounds the other term, implying that \(\require{physics} \qty{\tilde{z}_n(t)}\) is equicontinuous in the extended sense. ◻

Proof. By 47 and 50 , \[\begin{align} \sup_n \norm{z_n(0)} \leq 1. \end{align}\] Without loss of generality, let \(t_1 \leq t_2\). Then \(\forall \delta > 0\), we have \[\begin{align} & \sup_n \sup_{0\leq \abs{t_1-t_2} \leq \delta, \, 0 \leq t_1 \leq t_2 < T} \norm{z_n(t_1) - z_n(t_2)} \\ =&\sup_n \sup_{0\leq \abs{t_1-t_2} \leq \delta, \, 0 \leq t_1 \leq t_2 < T} \norm{ \int_{t_1}^{t_2} h_{r_n}(z_n(s))ds }\\ =& \sup_n \sup_{0\leq \abs{t_1-t_2} \leq \delta, \, 0 \leq t_1 \leq t_2 < T} \norm{\int_{t_1}^{t_2} \left[ h_{r_n}(z_n(s)) - h_{r_n}(0) \right]ds + \int_{t_1}^{t_2} h_{r_n}(0) ds }& \\ \leq &\sup_n \sup_{0\leq \abs{t_1-t_2} \leq \delta, \, 0 \leq t_1 \leq t_2 < T} \int_{t_1}^{t_2}\norm{h_{r_n}(z_n(s)) - h_{r_n}(0) }ds + \sup_n \sup_{0\leq \abs{t_1-t_2} \leq \delta, \, 0 \leq t_1 \leq t_2 < T} \int_{t_1}^{t_2} \norm{ h_{r_n}(0) }ds & \\ \leq & \sup_n \sup_{0\leq \abs{t_1-t_2} \leq \delta, \, 0 \leq t_1 \leq t_2 < T} \int_{t_1}^{t_2}L\norm{z_n(s)} ds + \sup_n \sup_{0\leq \abs{t_1-t_2} \leq \delta, \, 0 \leq t_1 \leq t_2 < T} \int_{t_1}^{t_2} \norm{ h_{r_n}(0) }ds& \tag*{(by Lemma \ref{lemma:32h32Lipschitz})} \\ \leq & \delta LC_{\hat{x}} + \sup_n \sup_{0\leq \abs{t_1-t_2} \leq \delta, \, 0 \leq t_1 \leq t_2 < T} \int_{t_1}^{t_2} \norm{ h_{r_n}(0) }ds& \tag*{(by Lemma~\ref{lemma:32bound32z})} \\ \leq & \delta (L C_{\hat{x}} + C_H),& \tag*{(by \eqref{eq:32h32c32c95H})} \end{align}\] which implies that \(\require{physics} \qty{z_n}\) is equicontinuous. ◻

Proof. \[\begin{align} \sup_n f_n(0) = \sup_n \tilde{z}_n(0) - z_n(0) = \sup_n \tilde{z}_n(0) - \tilde{z}_n(0) = 0 < \infty. \end{align}\] By Lemma 40 and Lemma 41, \(\forall \epsilon > 0\), \(\exists \delta\) such that \[\begin{align} \limsup_n \sup_{0\leq t_2-t_1 \leq \delta} \norm{\tilde{z}_n(t_1) - \tilde{z}_n(t_2)} & \leq \frac{\epsilon}{2},\\ \sup_n \sup_{0\leq t_2-t_1 \leq \delta} \norm{ z_n(t_1) - z_n(t_2)} &\leq \frac{\epsilon}{2}. \end{align}\] Without loss of generality let \(t_1 \leq t_2\). Then \(\forall \epsilon\), \(\exists \delta\) such that \[\begin{align} &\limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{f_n(t_1) - f_n(t_2)} \\ =&\limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\tilde{z}_n(t_1) - \tilde{z}_n(t_2) - (z_n(t_1) - z_n(t_2))} \\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\tilde{z}_n(t_1) - \tilde{z}_n(t_2)} + \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{ z_n(t_1) - z_n(t_2)} \\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\tilde{z}_n(t_1) - \tilde{z}_n(t_2)} + \sup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{ z_n(t_1) - z_n(t_2)} \\ \leq& \epsilon, \end{align}\] which implies that \(\require{physics} \qty{f_n}\) is equicontinuous in the extended sense. ◻

Proof. Since \(\sup_n r_n = \infty\) and \(r_n\) is monotonic, \(\lim_{n \rightarrow \infty} r_n = \infty\), and every subsequence also converges to infinity.

Since \(\{f_{n_{k,0}}\}\) is equicontinuous, by the Arzelà-Ascoli theorem (see Appendix 8), there exists a subsequence \(n_{k,1} \subseteq n_{k,0}\) such that \(\{f_{n_{k,1}}\}\) converges uniformly to a continuous limit \(f^{\text{lim}}\). Similarly, since \(\require{physics} \qty{\tilde{z}_{n_{k,1}}(t)}\) is equicontinuous, there is a subsequence \(n_k \subseteq n_{k,1}\) such that \(\require{physics} \qty{\tilde{z}_{n_k}(t)}\) converges uniformly in \(t\) to a continuous limit \(\tilde{z}^{\text{lim}}(t)\).

The proof that \(\lim_{n \rightarrow \infty} z_{n_k}(t) = z^{\text{lim}}(t)\) uniformly is by lemma 57. ◻

Proof. \[\begin{align} &\lim_{k \to \infty} \norm{ \sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \beta(i)G_{r_{n_k}}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1})} \\ \leq& \lim_{k \to \infty} \sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \beta(i)\norm{G_{r_{n_k}}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1})} \\ \leq& \lim_{k \to \infty} \sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \beta(i)L(W_{i+1})\norm{\tilde{z}_{n_k}(t(i)-T_{n_k})} \\ \leq& C_{\hat{z}} \lim_{k \to \infty} \sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \beta(i)L(W_{i+1}) \tag*{(by Lemma \ref{lemma:32bound32z32hat})} \\ \leq& 0. \tag*{(\ref{eq:32beta32L32property32bounded})} \end{align}\] ◻

Proof. \(\forall j\), \(\forall k\), \(\forall t \in [0,T+1)\), \[\begin{align} & \delimitershortfall=-1pt \left \vert \norm{\sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \alpha(i)H_{r_{n_j}}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1}) - \int_0^t h_{r_{n_j}} (\tilde{z}^{\lim}(s))ds} \right. \\ & \delimitershortfall=-1pt \left. -\norm{\sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \alpha(i)H_{\infty}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1}) - \int_0^t h_{\infty} (\tilde{z}^{\lim}(s))ds} \right \vert \\ \leq& \Bigg\lVert\sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \alpha(i)H_{r_{n_j}}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1}) - \int_0^t h_{r_{n_j}} (\tilde{z}^{\lim}(s))ds \\ &-\sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \alpha(i)H_{\infty}(\tilde{z}_{n_k}(t(i)),W_{i+1}) + \int_0^t h_{\infty} (\tilde{z}^{\lim}(s))ds \Bigg\rVert \tag*{(by \abs{\norm{a} - \norm{b}} \leq \norm{a-b})} \\ \leq& \norm{\sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \alpha(i)(H_{r_{n_j}}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1}) - H_{\infty}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1}))} \\ &+ \norm{\int_0^t h_{r_{n_j}} (\tilde{z}^{\lim}(s)) - h_{\infty} (\tilde{z}^{\lim}(s))ds }\\ \leq& \norm{ \sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \alpha(i)(H_{r_{n_j}}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1}) - H_{\infty}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1}))} \\ &+ \int_0^t \norm{ h_{r_{n_j}} (\tilde{z}^{\lim}(s)) - h_{\infty} (\tilde{z}^{\lim}(s))} ds \label{eq:32H32h32j32H32h32inf321} \end{align}\tag{150}\] By Lemma 52, \(\tilde{z}_{n_k}(t(i)-T_{n_k})\) is in a compact set \(\mathcal{B}_{\hat{z}}\). By Lemma 39, for the compact set \(\mathcal{B}_{\hat{z}}\), \(\forall \epsilon>0\), \(\exists j_1\) such that \(\forall j \geq j_1\), \(\forall k\), \(\forall z \in \mathcal{B}\), \(\forall t \in [0,T+1)\), \[\begin{align} \norm{\sum_{i=m(T_{n_k})}^{m(T_{n_k}+t) - 1} \alpha(i) \left[H_{r_{n_j}}(z, W_{i+1}) - H_{\infty}(z, W_{i+1})\right]} \leq \epsilon \label{eq:32H32y95123k95012532delta32T}. \end{align}\tag{151}\]

Similar to the proof of Lemma 56, we have \[\begin{align} \label{eq:32h32hat32z32uniform} \lim_{j \to \infty}h_{r_{n_j}}(\tilde{z}_k(t)) = h_{\infty}(\tilde{z}_k(t)) \end{align}\tag{152}\] uniformly in \(k\) and \(t \in [0,T+1)\). By 152 , \(\forall \epsilon>0\), \(\exists j_2\) such that \(\forall j > j_2\), \(\forall k\), \(\forall t \in [0,T+1)\), \[\begin{align} \norm{h_{r_{n_j}}(\tilde{z}_k(t)) - h_{\infty}(\tilde{z}_k(t))} \leq \epsilon. \label{eq:32uniform32k32h32epsilon} \end{align}\tag{153}\] Define \(\require{physics} j_0 \doteq \max \qty{j_1, j_2}\). \(\forall j \geq j_0\), \(\forall k\), \(\forall t \in [0,T+1)\), \[\begin{align} & \delimitershortfall=-1pt \left \vert \norm{\sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \alpha(i)H_{r_{n_j}}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1}) - \int_0^t h_{r_{n_j}} (\tilde{z}^{\lim}(s))ds} \right. \\ & \delimitershortfall=-1pt \left. -\norm{\sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \alpha(i)H_{\infty}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1}) - \int_0^t h_{\infty} (\tilde{z}^{\lim}(s))ds} \right \vert \\ \leq& \norm{\sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \alpha(i)(H_{r_{n_j}}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1}) - H_{\infty}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1}))} + (T+1) \epsilon \tag*{(by \eqref{eq:32H32h32j32H32h32inf321}, \eqref{eq:32uniform32k32h32epsilon})} \\ \leq& \epsilon + (T+1) \epsilon \tag*{(by \eqref{eq:32H32h32j32H32h32inf321}, \eqref{eq:32H32y95123k95012532delta32T})} \\ \leq& (T+2) \epsilon. \end{align}\] This completes the proof of uniform convergence. ◻

Proof. It follows immediately from Lemmas 43 & 44, the Moore-Osgood theorem, and Lemma 58. ◻

Proof.

\(\forall t\in [0,T+1)\), \[\begin{align} &\lim_{k \to \infty} \norm{f_{n_k}(t)} \\ \leq&\lim_{k \to \infty} \norm{ \sum_{i = m(T_{n_k})}^{m(T_{n_k} +t) -1 } \alpha(i)H_{r_{n_k}}(\tilde{z}_{n_k}(t(i)-T_{n_k}),W_{i+1}) - \int_0^t h_{r_{n_k}}(\tilde{z}^{\lim}(s))ds } \\ &+\lim_{k \to \infty} \norm{\int_0^t h_{r_{n_k}}(\tilde{z}^{\lim}(s))ds - \int_0^t h_{r_{n_k}}(z_{n_k}(s))ds } + 0 \tag*{(by \eqref{eq:32lim32f32k} and Lemma {\ref{lemma:32third32term32disc32error}})} \\ =& \lim_{k \to \infty} \norm{ \int_0^t h_{r_{n_k}} (\tilde{z}^{\lim}(s))ds - \int_0^t h_{r_{n_k}}(z_{n_k}(s))ds } \\ =& \makebox[0.85\textwidth]{\displaystyle\norm{\int_0^t h_{\infty} (\tilde{z}^{\lim}(s))ds - \int_0^t h_{\infty}(z^{\lim}(s))ds}.\hfill(by Lemma \ref{lemma:32h32k32x32lim32h32inf32x32lim32uniform} and Lemma \ref{lemma:32h32k32z32k32h32inf32z32lim32uniform})} \label{eq:32reduce32f32limit321} \end{align}\tag{155}\] We now show the relationship between \(\tilde{z}^{\lim}(t)\) and \(z^{\lim}(t)\). \[\begin{align} &\norm{\tilde{z}^{\lim}(t) - z^{\lim}(t)} \tag{156} \\ =& \norm{\lim_{k \to \infty} \left[\tilde{z}_{n_k}(0) + \int_0^t h_{r_{n_k}} (\tilde{z}^{\lim}(s))ds \right] - \left[\tilde{z}^{\lim}(0) + \int_0^t h_{\infty}(z^{\lim}(s))ds\right] } \tag*{(by \eqref{eq:32z32lim32t} and \eqref{eq:32z32lim32definition322})} \\ =& \norm{\tilde{z}^{\lim}(0) + \int_0^t h_{\infty} (\tilde{z}^{\lim}(s))ds - \left[\tilde{z}^{\lim}(0) + \int_0^t h_{\infty}(z^{\lim}(s))ds\right] } \tag*{(by Lemma \ref{lemma:32h32k32x32lim32h32inf32x32lim32uniform})}\\ =& \norm{ \int_0^t h_{\infty} (\tilde{z}^{\lim}(s))ds - \int_0^t h_{\infty}(z^{\lim}(s))ds } \tag{157}\\ \leq & \int_0^t L \norm{ \tilde{z}^{\lim}(s) - z^{\lim}(s)}ds \tag*{(by Lemma \ref{lemma:32h32Lipschitz})} \\ \leq& 0. \tag*{(by Gronwall inequality in Theorem \ref{theorem:32Gronwall})} \end{align}\] Thus, \[\begin{align} &\norm{\lim_{k \to \infty} f_{n_k}(t) } \\ \leq& \norm{\int_0^t h_{\infty} (\tilde{z}^{\lim}(s))ds - \int_0^t h_{\infty}(z^{\lim}(s))ds} \tag*{(by \eqref{eq:32reduce32f32limit321})} \\ =& \norm{\tilde{z}^{\lim}(t) - z^{\lim}(t)} \tag*{(by \eqref{eq:32f32limit32construct32Gronwall321})} \\ \leq& 0. \tag*{(by \eqref{eq:32norm32z32lim32z32lim})} \end{align}\] ◻

Proof. By Lyapunov stability, there is a \(\delta > 0\) such that any trajectory beginning within \(B(\lambda_{\infty}(y), \delta)\) stays within \(\frac{\epsilon}{2}\) of the equilibrium \(\lambda_{\infty}(y)\).

For an initial condition \(x\), let \(T_x\) be a time at which the trajectory is within \(\frac{\delta}{2}\) of the equilibrium. Let \(x_1\) be some other initial condition. By definition, Lipschitzness, and the Gronwall inequality, we have \[\begin{align} \norm{\eta^y_{\infty}(t, x) - \eta^y_{\infty}(t, x_1) } &\leq \norm{x-x_1} + L \int_0^t \norm{\eta^y_{\infty}(s, x) - \eta^y_{\infty}(s, x_1)}ds \\ &\leq \norm{x-x_1}e^{L T_x} \end{align}\] for all \(t \leq T_x\).

So there is a neighborhood \(V_x\) such that for all \(x_1 \in V_x\), \(\eta^y_{\infty}(T_x, x_1)\) is within \(\delta\) of the equilibrium, which by Lyapunov stability implies that it will always be within \(\epsilon\) of the equilibrium after \(T_x\).

By compactness, we can cover the set \(K\) by a finite number of such intervals and obtain a finite number of times \(T_{x_1}, \ldots, T_{x_n}\) and then take the maximum time to be the value of \(T_{\epsilon}\). ◻

Proof. We have \[\begin{align} \eta^{y'(t)}_{c}(t,x) &= x + \int_0^t h_c(\eta^{y'(s)}_c(s, x), y'(s)) ds, \\ \eta^{y}_{\infty}(t,x) &= x + \int_0^t h_{\infty}(\eta^{y}_{\infty}(s, x), y) ds. \end{align}\] Let us define the error term: \[\begin{align} e(t) = \norm{\eta^{y'(t)}_{c}(t,x) - \eta^{y}_{\infty}(t,x)}. \end{align}\]

We can bound \(e(t)\) by two terms: \[\begin{align} e(t) \leq& \int_0^t \norm{h_c(\eta_c^{y'(s)}(s, x), y'(s)) - h_c(\eta_{\infty}^{y}(s, x), y'(s))} ds \\ &+ \int_0^t \norm{h_c(\eta_{\infty}^{y}(s, x), y'(s)) - h_{\infty}(\eta^{y}_{\infty}(s, x), y)} ds \end{align}\] To bound the first term, we use Lipschitzness: \[\begin{align} \int_0^t \norm{h_c(\eta_c^{y'(s)}(s, x), y'(s)) - h_c(\eta_{\infty}^{y}(s, x), y'(s))} ds &\leq L \int_0^t \norm{\eta^{y'(t)}_{c}(t,x) - \eta^{y}_{\infty}(t,x)} ds \\ &= L \int_0^t e(s) ds. \end{align}\]

To bound the second term, we use Lipschitzness and Assumption 3:

\[\begin{align} \int_0^t \norm{h_c(\eta_{\infty}^{y}(s, x), y'(s)) - h_{\infty}(\eta^{y}_{\infty}(s, x), y)} ds \leq& \int_0^t \norm{h_c(\eta_{\infty}^{y}(s, x), y'(s)) - h_{\infty}(\eta^{y}_{\infty}(s, x), y'(s))} ds \\ &+ \int_0^t \norm{h_{\infty}(\eta_{\infty}^{y}(s, x), y'(s)) - h_{\infty}(\eta^{y}_{\infty}(s, x), y)} ds \\ \leq& \int_0^t \epsilon(c) ds + L\int_0^t \norm{y'(s)-y}ds \\ \leq& T\epsilon(c) + TL\rho. \end{align}\]

To conclude, we will use the Gronwall Inequality (Appendix 5): \[\begin{align} e(t) \leq& T \epsilon(c) + TL \rho + L \int_0^t e(s)ds \\ \leq& (L \rho + \epsilon(c))T e^{LT}. \end{align}\] ◻

Proof. We know that \[\begin{align} \sup_n \norm{\tilde{y}_n(0)} \leq 1 . \end{align}\]

Without loss of generality, let \(t_1 \leq t_2\). \[\begin{align} &\limsup_n \sup_{ 0\leq t_2-t_1 \leq \delta } \norm{ \tilde{y}_n(t_1) - \tilde{y}_n(t_2)} = \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) }\\ \end{align}\] \(\forall \xi > 0\), by 172 , \(\exists \delta_0,\) such that \(\forall 0< \delta \leq \delta_0\), \[\begin{align} \sup_{c\geq 1} \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{ \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_c(0, W_{i+1}) } \leq \xi. \label{eq:32lemma32hat32y32equicontinuous32xi321} \end{align}\tag{158}\] By 179 , \(\exists \delta_1,\) such that \(\forall 0< \delta \leq \delta_1\), \[\begin{align} \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) L(W_{i+1}) \leq \xi. \label{eq:32lemma32hat32y32equicontinuous32xi322} \end{align}\tag{159}\] Without loss of generality, let \(t_1 \leq t_2\). Then \(\require{physics} \forall \delta \leq \min \qty{\delta_0,\delta_1}\), we have \[\begin{align} &\limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) }\\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) } - \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_{r_n}(0, W_{i+1}) } \\ &+ \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_{r_n}(0, W_{i+1}) } \\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) } - \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_{r_n}(0, W_{i+1}) } \\ &+ \sup_{c\geq 1} \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_c(0, W_{i+1}) } \\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) } - \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_{r_n}(0, W_{i+1}) } \\ &+ \xi \tag*{(by \eqref{eq:32lemma32hat32y32equicontinuous32xi321})}\\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{\sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) - \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) G_{r_n}(0, W_{i+1}) } \\ &+ \xi \\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) \norm{G_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) - G_{r_n}(0, W_{i+1}) } + \xi\\ \leq& \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) L(W_{i+1})\norm{\tilde{z}_n(t(i)-T_n)} + \xi \\ \leq& E \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \beta(i) L(W_{i+1}) + \xi \tag*{(by Lemma~\ref{lemma:32slow32gronwall})} \\ \leq& E \xi + \xi. \tag*{(by \eqref{eq:32lemma32hat32y32equicontinuous32xi322})} \end{align}\] ◻

Proof. Since \(\sup_n r_n = \infty\) and \(r_n\) is monotonic, \(\lim_{n \rightarrow \infty} r_n = \infty\), and every subsequence also converges to infinity.

Since \(\{f_{n_{k,0}}\}\) is equicontinuous, by the Arzelà-Ascoli theorem (see Appendix 8), there exists a subsequence \(n_{k,1} \subseteq n_{k,0}\) such that \(\{f_{n_{k,1}}\}\) converges uniformly to a continuous limit \(f^{\text{lim}}\). Similarly, since \(\require{physics} \qty{\tilde{y}_{n_{k,1}}(t)}\) is equicontinuous, there is a subsequence \(n_k \subseteq n_{k,1}\) such that \(\require{physics} \qty{\tilde{z}_{n_k}(t)}\) converges uniformly in \(t\) to a continuous limit \(\tilde{y}^{\text{lim}}(t)\).

The proof that \(\lim_{n \rightarrow \infty} y_{n_k}(t) = y^{\text{lim}}(t)\) uniformly is by lemma 64. ◻

Proof. \(\forall n\), \[\begin{align} &T_{n+1} - T_n \\ =&t(m(T_n+T) + 1) - T_n \\ \geq&T_n + T - T_n \\ \geq&T. \end{align}\] Thus, \[\begin{align} &\lim_{n\to\infty} T_{n+1} - T_n \geq T. \end{align}\] With \[\begin{align} &\lim_{n\to\infty} T_{n+1} - T_n \\ =&\lim_{n\to\infty} t(m(T_n+T) + 1) - T_n \\ =&\lim_{n\to\infty} t(m(T_n+T)) + \alpha(m(T_n+T)) - T_n \\ \leq&\lim_{n\to\infty} T_n + T + \alpha(m(T_n+T)) - T_n \\ =& T, \end{align}\] by the squeeze theorem, we have \(\lim_{n\to\infty} T_{n+1} - T_n = T\).

To prove 160 , \(\forall \tau\), \(\forall -\tau \leq t_1 \leq t_2 \leq \tau\), it suffices to only consider large \(n\) such that \(t(n) - \tau \geq 0\). We have \[\begin{align} &\lim_{n \to \infty} \sum_{i = m(t(n)+t_1)}^{m(t(n) + t_2) - 1 } \alpha(i) \\ =& \lim_{n \to \infty} t(m(t(n)+t_2)) - t(m(t(n)+t_1)) \\ \leq& \lim_{n \to \infty} t(n)+t_2 - t(m(t(n)+t_1)) \\ \leq& \lim_{n \to \infty} t(n)+t_2 - ( t(n)+t_1 - \alpha(m(t(n)+t_1)) ) \\ =& t_2 - t_1 + \lim_{n \to \infty} \alpha(m(t(n)+t_1)) \\ =& t_2 - t_1 \tag*{(by \eqref{eq32147n32alpha})} \end{align}\] and \[\begin{align} &\lim_{n \to \infty} \sum_{i = m(t(n)+t_1)}^{m(t(n) + t_2) - 1 } \alpha(i) \\ =& \lim_{n \to \infty} t(m(t(n)+t_2)) - t(m(t(n)+t_1)) \\ \geq& \lim_{n \to \infty} t(n)+t_2 - \alpha(m(t(n)+t_2)) - t(m(t(n)+t_1)) \\ \geq& \lim_{n \to \infty} t(n)+t_2 - \alpha(m(t(n)+t_2)) - (t(n)+t_1) \\ =& \lim_{n \to \infty} t_2 - t_1 - \alpha(m(t(n)+t_2)) \\ =&t_2 - t_1 \tag*{(by \eqref{eq32147n32alpha})} . \end{align}\] By the squeeze theorem, we have \[\begin{align} \lim_{n \rightarrow \infty} \sum_{i = m(t(n)+t_1)}^{m(t(n)+t_2) - 1} \alpha(i) = t_2 - t_1. \end{align}\]

To prove 161 , fix \(\epsilon > 0\). By 17 , there exists \(N_1 \in \mathbb{N}\) such that for all \(n > m(t(N_1)+t_1)\), \[\begin{align} \frac{\beta(n)}{\alpha(n)} < \epsilon. \label{ineq32ratio326032eps} \end{align}\tag{162}\] By 160 , there exists \(N_2 \in \mathbb{N}\) such that for all \(n > N_2\), \[\begin{align} \sum_{i = m(t(n)+t_1)}^{m(t(n)+t_2) - 1} \alpha(i) < t_2 - t_1 + \epsilon. \label{lim32t95132t95232eps} \end{align}\tag{163}\] Let \(\require{physics} N = \max \qty{N_1, N_2}\). Then for all \(n > N\), \[\begin{align} &\sum_{i = m(t(n)+t_1)}^{m(t(n)+t_2) - 1} \beta(i) \\ <& \epsilon \sum_{i = m(t(n)+t_1)}^{m(t(n)+t_2) - 1} \alpha(i) \tag*{(by \eqref{ineq32ratio326032eps})} \\ <& \epsilon(t_2-t_1 + \epsilon) \tag*{(by \eqref{lim32t95132t95232eps})}. \end{align}\] Since \(t_2 - t_1\) is a constant and \(\epsilon\) can be made arbitrarily small, we have \[\begin{align} \lim_{n \to \infty} \sum_{i = m(t(n)+t_1)}^{m(t(n)+t_2) - 1} \beta(i) &= 0. \end{align}\] ◻

Proof. To prove 164 , we first consider \(1 \leq c < \infty\), \[\begin{align} &\norm{H_c(x,y,w) - H_c(x',y',w)} \\ =& \norm{\frac{H(cx,cy,w)}{c} - \frac{H(cx',cy,w)}{c}} \tag*{(by \eqref{def:32H32c})}\\ \leq& \frac{\norm{H(cx,cy, w)- H(cx',cy',w)}}{c}\\ \leq& L(w) \frac{\norm{(cx,cy) - (cx',cy')}}{c} \tag*{(by \eqref{eq:32H32L})} \\ =& L(w)\norm{(x,y) - (x',y')}. \label{eq:32H32c32L32x32x39} \end{align}\tag{168}\] By 26 , \[\begin{align} \norm{H_\infty(x,y,w) - H_\infty(x',y',w)} &\leq L(w) \norm{(x,y) - (x',y')}. \end{align}\]

To prove 165 , \(\forall x,x',y,y'\), \(\forall c \geq 1\) including \(c= \infty\), \[\begin{align} &\norm{h_c(x,y) - h_c(x',y')} \\ =& \norm{ \mathbb{E}_{w\sim \omega}\left[H_c(x,y,w) - H_c(x',y',w)\right]} \\ \leq& \mathbb{E}_{w\sim \omega}\left[\norm{ H_c(x,y,w) - H_c(x',y',w) } \right] \\ \leq& \mathbb{E}_{w\sim \omega}\left[ L(w) \norm{(x,y) - (x',y')}\right] \\ \leq& L \norm{(x,y) - (x',y')}. \end{align}\]

By similar arguments, 166 and 167 also follow. ◻

Proof. Proof of 169 : \[\begin{align} \sup_{c\geq 1} \norm{h_c(0)} = \sup_{c\geq 1} \norm{ \frac{h(0)}{c} } \leq \sup_{c\geq 1} \norm{h(0)} = \norm{h(0)} < \infty \tag{173}, \\ \sup_{c\geq 1} \norm{g_c(0)} = \sup_{c\geq 1} \norm{ \frac{g(0)}{c} } \leq \sup_{c\geq 1} \norm{g(0)} = \norm{g(0)} < \infty \tag{174}. \end{align}\] Proof of 170 : \(\forall x\), \[\begin{align} &\sup_{c\geq 1} \limsup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) \left[H_c(x,y,W_{i+1}) - h_c(x,y) \right] } \\ =&\sup_{c\geq 1} \limsup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) \left[\frac{H(cx,cy,W_{i+1})}{c} - \frac{h(cx,cy)}{c} \right] }\\ =& \sup_{c\geq 1} \frac{1}{c} \limsup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) \left[H(cx, cy,W_{i+1}) - h(cx,cy) \right] }\\ \leq& \sup_{c\geq 1} \frac{1}{c} \limsup_n \sup_{0\leq t_1 \leq t_2 \leq T + \sup_j \alpha(j)} \norm{ \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) \left[H(cx, cy,W_{i+1}) - h(cx,cy) \right] } \tag*{(\forall n, T_{n+1} - T_n \leq T + \sup_j \alpha(j))} \\ =& \sup_{c\geq 1} \frac{1}{c} \cdot 0 \\ =& 0 . \label{eq:32lim32H32h32t320}\\ \end{align}\tag{175}\] Proof of 171 : \[\begin{align} &\limsup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \alpha(i) H(0, W_{i+1}) } \\ =& \limsup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \alpha(i) [H(0, W_{i+1}) - h(0) + h(0)] } \\ \leq& \limsup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \alpha(i) [H(0, W_{i+1}) - h(0)] } \\ &+ \limsup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \alpha(i) h(0) } \\ \leq& \limsup_n \sup_{0\leq t_1 \leq t_2 \leq T + \sup_j \alpha(j)} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \alpha(i) [H(0, W_{i+1}) - h(0)] } \\ &+ \limsup_n \sup_{0\leq t_1 \leq t_2 \leq T + \sup_j \alpha(j)} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \alpha(i) h(0) } \tag*{(\forall n, T_{n+1} - T_n \leq T + \sup_j \alpha(j))} \\ =& \limsup_n \sup_{0\leq t_1 \leq t_2 \leq T + \sup_j \alpha(j)} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \alpha(i) h(0) } \\ =&\norm{h(0) } \limsup_n \sup_{0\leq t_1 \leq t_2 \leq T+\sup_j \alpha(j)} \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \alpha(i) \\ =&\norm{h(0) } (T + \sup_j \alpha(j)) \tag*{(by Lemma~\ref{lemma:32T-minus})} \\ <& \infty. \label{eq:32bounded-h032inf32y32sample32path} \end{align}\tag{176}\] We now consider \(c\) in the above bounds. We first get \[\begin{align} &\sup_{c\geq 1}\sup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_c(0, W_{i+1}) } \\ =&\sup_{c\geq 1}\sup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \alpha(i) \frac{H(0, W_{i+1}) }{c}} \tag*{(by \eqref{def:32H32c})} \\ =&\sup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \alpha(i) H(0, W_{i+1})} \tag*{(by c\geq 1)} \\ <& \infty. \tag*{(by \eqref{eq:32bounded-h032inf32y32sample32path})} \end{align}\] Proof of 172 : \[\begin{align} &\lim_{\delta \to 0^+} \sup_{c\geq 1} \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{ \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_c(0, W_{i+1}) } \\ \leq &\lim_{\delta \to 0^+} \sup_{c\geq 1} \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{ \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) \left[ H_c(0, W_{i+1}) -h_c(0)\right] } \\ &+ \lim_{\delta \to 0^+} \sup_{c\geq 1} \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{ \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) h_c(0) } \\ \leq & 0 + \lim_{\delta \to 0^+} \sup_{c\geq 1} \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{ \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) h_c(0) } \tag*{(by \eqref{eq:32lim32H32h32t320})} \\ \leq & 0 + \lim_{\delta \to 0^+} \sup_{c\geq 1} \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \norm{ \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) \frac{h(0)}{c} } \\ \leq & 0 + \norm{ h(0)} \lim_{\delta \to 0^+} \sup_{c\geq 1} \frac{1}{c} \limsup_n \sup_{0\leq t_2-t_1 \leq \delta } \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) \\ \leq& \norm{ h(0)} \lim_{\delta \to 0^+} \sup_{c\geq 1} \frac{1}{c} \delta \tag*{(by \eqref{eq:32lim32lr32t95132t952})} \\ =& \norm{ h(0)} \lim_{\delta \to 0^+} \delta \\ =& 0. \end{align}\] ◻

Proof. Here we show 178 : \[\begin{align} &\lim_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \beta(i) L(W_{i+1}) } \\ =& \lim_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \beta(i) [L(W_{i+1}) - L + L] } \\ \leq& \lim_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \beta(i) [L(W_{i+1}) - L] } \\ &+ \lim_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \beta(i) L } \\ \leq& \lim_n \sup_{0\leq t_1 \leq t_2 \leq T + \sup_j \alpha(j)} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \beta(i) [L(W_{i+1}) - L] } \\ &+ \lim_n \sup_{0\leq t_1 \leq t_2 \leq T + \sup_j \alpha(j)} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \beta(i) L } \tag*{(\forall n, T_{n+1} - T_n \leq T + \sup_j \alpha(j))} \\ =& \lim_n \sup_{0\leq t_1 \leq t_2 \leq T + \sup_j \alpha(j)} \norm{ \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \beta(i) L } \\ =&\norm{L } \lim_n \sup_{0\leq t_1 \leq t_2 \leq T+\sup_j \alpha(j)} \sum_{i = m(T_n + t_1)}^{m(T_n+t_2) - 1} \beta(i) \\ =&0. \tag*{(by Lemma~\ref{lemma:32T-minus})} \\ \end{align}\] ◻

Proof. Fix a sample path \(\require{physics} \qty{x_0,y_0,\qty{W_i}_{i=1}^\infty}\), \[\begin{align} &\makebox[0.85\textwidth]{\displaystyle LT < \infty,\hfill(L and T are constants)} \\ &\makebox[0.85\textwidth]{\displaystyle\sup_{c\geq 1} \norm{ h_c(0) }T < \infty,\hfill(by \eqref{eq:32h32c32bound})} \\ &\makebox[0.85\textwidth]{\displaystyle\sup_{c\geq 1} \sup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_c(0, W_{i+1}) } < \infty,\hfill(by \eqref{eq:32h32property32H32bounded})} \\ &\makebox[0.85\textwidth]{\displaystyle\sup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) L(W_{i+1}) < \infty.\hfill(by \eqref{eq:32L32property32bounded})} \end{align}\] Thus, there exists a constant \(C_H\) such that \[\begin{align} &LT \leq C_H \\ &\sup_{c\geq 1} \norm{ h_c(0) } \leq \frac{C_H}{T}, \\ &\sup_{c\geq 1} \sup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \norm{ \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) H_c(0, W_{i+1}) } \leq C_H,\\ &\sup_n \sup_{0\leq t_1 \leq t_2 \leq T_{n+1} - T_n} \sum_{i = m(T_n+t_1)}^{m(T_n+t_2) - 1} \alpha(i) L(W_{i+1}) \leq C_H. \end{align}\] ◻

Proof. \(\forall n \in \mathbb{N}, t \in [0, T+1)\), \[\begin{align} &\norm{\tilde{z}_n(t)} & \\ =& \norm{ \tilde{z}_n(0) + \sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i) H_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}), \beta(i) G_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}))} & \\ \leq& \norm{\tilde{z}_n(0)} + \norm{\sum_{i = m(T_n)}^{m(T_n+t) - 1} \alpha(i) \left[ H_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) - H_{r_n}(0, W_{i+1}) \right] + \sum_{i = m(T_n)}^{m(T_n+t) - 1} \alpha(i) H_{r_n}(0, W_{i+1}) } & \\ &+ \norm{\sum_{i = m(T_n)}^{m(T_n+t) - 1} \beta(i) \left[ G_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) - G_{r_n}(0, W_{i+1}) \right] + \sum_{i = m(T_n)}^{m(T_n+t) - 1} \beta(i) G_{r_n}(0, W_{i+1}) } \\ \leq & \norm{\tilde{z}_n(0)} + \sum_{i = m(T_n)}^{m(T_n+t) - 1} \alpha(i)\norm{H_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) - H_{r_n}(0, W_{i+1}) } + C_H & \\ &+ \sum_{i = m(T_n)}^{m(T_n+t) - 1} \beta(i)\norm{G_{r_n}(\tilde{z}_n(t(i)-T_n), W_{i+1}) - G_{r_n}(0, W_{i+1}) } + C_G \\ \leq & \norm{\tilde{z}_n(0)} + \sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i)+\beta(i))L(W_{i+1})\norm{\tilde{z}_n(t(i)-T_n)} + C' & \tag*{(by \eqref{eq:32H32c32c95H})}\\ \leq &1 + \sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i)+\beta(i))L(W_{i+1})\norm{\tilde{z}_n(t(i)-T_n)} + C' & \tag*{(by \eqref{eq:32hat-z-norm-1})}\\ \leq &\left[1 + C' \right] e^{\sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i)+\beta(i))L(W_{i+1})} \tag*{(by \tilde{z}_n(t) = \tilde{z}_n(t(m(T_n + t)) - T_n) and discrete Gronwall inequality in Theorem \ref{theorem:32discrete32Gronwall})} & \\ \leq &\left[ 1 + C' \right] e^{C'} \tag*{(by \eqref{eq:32alpha32L32bound32C95H})} & \\ = & C_{\hat{z}}.\tag*{(by \eqref{def:32C32hat32x})} \end{align}\] ◻

Proof. \(\forall n, t \in [0, T+1)\), \[\begin{align} &\norm{z_n(t)}& \\ =& \norm{ z_n(0) + \int_0^t (h_{r_n}(z_n(s)),0)ds}& \\ \leq & \norm{z_n(0)} + \norm{\int_0^t h_{r_n}(z_n(s))ds}& \\ \leq & \norm{z_n(0)} + \int_0^t \norm{h_{r_n}(z_n(s)) - h_{r_n}(0) }ds + \int_0^t \norm{ h_{r_n}(0) }ds & \\ \leq & \norm{z_n(0)} + \int_0^t L\norm{z_n(s)} ds + \int_0^t \norm{ h_{r_n}(0) }ds& \tag*{(by Lemma \ref{lemma:32h32Lipschitz})} \\ \leq & \norm{z_n(0)} + \int_0^t L\norm{z_n(s)} ds + (T+1) \norm{ h_{r_n}(0)} & \\ \leq & \norm{z_n(0)} + \int_0^t L\norm{z_n(s)} ds + (T+1) \frac{C_H}{T+1} \tag*{(by \eqref{eq:32h32c32c95H})} \\ \leq & 1 + \int_0^t L\norm{z_n(s)} ds + C_H \tag*{(by \eqref{eq:32hat-z-norm-1}, \eqref{def:32z32n32init})} \\ \leq &\left[1 + C_H \right] e^{L(T+1)}& \tag*{(by Gronwall inequality in Theorem \ref{theorem:32Gronwall})}\\ \leq &\left[1 + C_H \right] e^{C_H}& \tag*{(by \eqref{eq:32LT32C95H})}\\ =& C_{\hat{x}} \tag*{(by \eqref{def:32C32hat32x})} \end{align}\] ◻

Proof. \(\forall n, \forall t \in [0, T+1)\), \[\begin{align} &\norm{h_{r_n}(z_n(t))} \\ \leq& \norm{h_{r_n}(z_n(t)) - h_{r_n}(0)} + \norm{h_{r_n}(0)}\\ \leq& L\norm{z_n(t)} + \norm{h_{r_n}(0)} \tag*{(by Lemma \ref{lemma:32h32Lipschitz})} \\ \leq& L C_{\hat{x}} + \norm{h_{r_n}(0)} \tag*{(by Lemma~\ref{lemma:32bound32z})} \\ \leq& L C_{\hat{x}} + \frac{C_H}{T}. \tag*{(by \eqref{def:32r32n} and \eqref{eq:32h32c32c95H})} \end{align}\] Thus, because \(C_{\hat{x}}, C_H\) are independent of \(n,t\), \(\sup_{n, t \in [0, T+1)} \norm{h_{r_n}(z_n(t))} < \infty\). ◻

Proof. \(\forall t \in [0, T+1)\), \[\begin{align} &\norm{z^{\lim}(t)}& \\ =& \norm{ z^{\lim}(0) + \int_0^t (h_{\infty}(z^{\lim}(s)), 0)ds}& \\ \leq & \norm{z^{\lim}(0)} + \norm{\int_0^t h_{\infty}(z^{\lim}(s))ds}& \\ =& \norm{z^{\lim}(0)} + \norm{\int_0^t \left[ h_{\infty}(z^{\lim}(s)) - h_{\infty}(0) \right]ds + \int_0^t h_{\infty}(0) ds }& \\ \leq & \norm{z^{\lim}(0)} + \int_0^t \norm{h_{\infty}(z^{\lim}(s)) - h_{\infty}(0) }ds + \int_0^t \norm{ h_{\infty}(0) }ds & \\ \leq & \norm{z^{\lim}(0)} + \int_0^t L\norm{z^{\lim}(s)} ds + \int_0^t \norm{ h_{\infty}(0) }ds& \tag*{(by Lemma \ref{lemma:32h32Lipschitz})} \\ \leq &1 + \int_0^t L\norm{z^{\lim}(s)} ds + \int_0^t \norm{ h_{\infty}(0) }ds& \tag*{(by \eqref{eq:32hat-z-norm-1}, \eqref{def:32z32n32init})} \\ \leq &1 + \int_0^t L\norm{z^{\lim}(s)} ds + (T+1)\norm{ h_{\infty}(0) } \\ \leq &1 + \int_0^t L\norm{z^{\lim}(s)} ds + C_H \tag*{(by Assumption \ref{assumption:32lim32h44g32uniformly32convergent} and \eqref{eq:32h32c32c95H})} \\ \leq &\left[1 + C_H \right] e^{\int_0^t L ds }& \tag*{(by Gronwall inequality in Theorem \ref{theorem:32Gronwall})}\\ \leq &\left[1 + C_H \right] e^{L(T+1)} & \\ \leq& C_{\hat{x}}. \tag*{(by \eqref{eq:32LT32C95H}, \eqref{def:32C32hat32x})}& \end{align}\] ◻

Proof. By Assumption 5, \(\lim_{k \to \infty}h_{r_{n_k}}(v) = h_{\infty}(v)\) uniformly in a compact set \(\require{physics} \qty{v \vert v\in \mathbb{R}^d, \norm{v}\leq C_x }\). By Lemma 55, \(\require{physics} \qty{z^{\lim}(t) \vert t \in [0,T+1)} \subseteq \qty{v \vert v\in \mathbb{R}^d, \norm{v}\leq C_x }\). Therefore, \(\lim_{k \to \infty}h_{r_{n_k}}(z^{\lim}(t)) = h_{\infty}(z^{\lim}(t))\) uniformly in \(\require{physics} \qty{z^{\lim}(t) \vert t \in [0,T+1)}\) and in \(t \in [0,T+1)\). ◻

Proof. By 57 , \(\forall \delta > 0\), there exists a \(k_1\) such that \(\forall k \geq k_1\), \(\forall t \in [0,T+1)\), \[\begin{align} \norm{\tilde{z}_{n_k}(t) - \tilde{z}^{\lim}(t)} \leq \delta .\label{eq:32z32lim32epsilion321} \end{align}\tag{187}\] By Lemma 56, there exists a \(k_2\) such that \(\forall k \geq k_2\), \(\forall t \in [0,T+1)\), \[\begin{align} \norm{h_{r_{n_k}}(z^{\lim}(t)) - h_{\infty}(z^{\lim}(t))} \leq \delta. \label{eq:32z32lim32epsilion322} \end{align}\tag{188}\] \(\require{physics} \forall k \geq \max\qty{ k_1, k_2}\), \(\forall t \in [0,T+1)\) \[\begin{align} & \norm{z_{n_k}(t) - z^{\lim}(t) } \\ =& \norm{\tilde{z}_{n_k}(0) + \int_0^t (h_{r_{n_k}}(z_{n_k}(s)),0)ds - \tilde{z}^{\lim}(0) - \int_0^t (h_{\infty}(z^{\lim}(s)),0)ds} \\ \leq& \norm{\tilde{z}_{n_k}(0) - \tilde{z}^{\lim}(0)} + \norm{\int_0^t h_{r_{n_k}}(z_{n_k}(s))ds - \int_0^t h_{\infty}(z^{\lim}(s))ds} \\ \leq& \delta + \norm{\int_0^t h_{r_{n_k}}(z_{n_k}(s)) - h_{\infty}(z^{\lim}(s))ds} \tag*{(by \eqref{eq:32z32lim32epsilion321})} \\ \leq& \delta + \int_0^t \norm{h_{r_{n_k}}(z_{n_k}(s)) - h_{r_{n_k}}(z^{\lim}(s)) }ds + \int_0^t \norm{ h_{r_{n_k}}(z^{\lim}(s)) - h_{\infty}(z^{\lim}(s))} ds \\ \leq& \delta + L \int_0^t \norm{z_{n_k}(s) -z^{\lim}(s) }ds + \int_0^t \norm{ h_{r_{n_k}}(z^{\lim}(s)) - h_{\infty}(z^{\lim}(s))} ds \tag*{(by Lemma \ref{lemma:32h32Lipschitz})} \\ \leq& \delta + t\delta + L \int_0^t \norm{z_{n_k}(s) -z^{\lim}(s) }ds \tag*{(by \eqref{eq:32z32lim32epsilion322})} \\ \leq& (\delta + t\delta)e^{Lt} \tag*{(by Gronwall inequality in Theorem \ref{theorem:32Gronwall})}\\ \leq& (\delta + (T+1)\delta)e^{L(T+1)}, \label{eq:32z32z32lim32difference} \end{align}\tag{189}\] which completes the proof. ◻

Proof. By definition, \(\forall \epsilon > 0, \exists a_0, b_0\) such that \(\forall a > a_0, b > b_0\), \(\norm{f(a,b) - L } < \epsilon\). Thus, \(\require{physics} \forall \epsilon > 0, \exists c_0 = \max \qty{a_0, b_0}\) such that \(\forall c > c_0\), \(\norm{f(c,c) - L } < \epsilon\). ◻

Proof. From Lemma 52, it is easy to see that \[\begin{align} \sup_{t\in[0, T+1)} \norm{\tilde{z}^{\lim}(t)} < \infty, \end{align}\] which, similar to Lemma 54, implies that \[\begin{align} \sup_{k, t\in[0, T+1)} \norm{h_{r_{n_k}}\left(\tilde{z}^{\lim}(t)\right)} < \infty. \end{align}\]

By the dominated convergence theorem, \(\forall t\in [0,T+1)\), \[\begin{align} & \lim_{k \to \infty} \int_0^t h_{r_{n_k}} (\tilde{z}^{\lim}(s))ds = \int_0^t \lim_{k \to \infty} h_{r_{n_k}} (\tilde{z}^{\lim}(s))ds = \int_0^t h_{\infty} (\tilde{z}^{\lim}(s))ds, \end{align}\] which completes the proof. ◻

Proof. \(\forall \epsilon>0\), by Lemma 56, \(\exists k_0\) such that \(\forall k \geq k_0\), \(\forall t \in [0,T)\), \[\begin{align} \norm{ h_{r_{n_k}}(z^{\lim}(s)) - h_{\infty}(z^{\lim}(s)) } \leq \epsilon. \label{eq:32f32k32032k950} \end{align}\tag{192}\] By Lemma 57, \(\exists k_1\) such that \(\forall k \geq k_1\), \(\forall t \in [0,T+1)\), \[\begin{align} \norm{ z_{n_k}(t)- z^{\lim}(t) } \leq \epsilon. \label{eq:32f32k32032k951} \end{align}\tag{193}\] Thus, \(\require{physics} \forall k \geq \max \qty{k_0,k_1}\), \(\forall t \in [0,T+1)\), \[\begin{align} & \norm{ \int_0^t h_{r_{n_k}}(z_{n_k}(s))ds - \int_0^t h_{\infty}(z^{\lim}(s))ds } \\ \leq& \norm{ \int_0^t h_{r_{n_k}}(z_{n_k}(s))ds - \int_0^t h_{r_{n_k}}(z^{\lim}(s))ds} + \norm{ \int_0^t h_{r_{n_k}}(z^{\lim}(s))ds - \int_0^t h_{\infty}(z^{\lim}(s))ds } \\ \leq& \int_0^t \norm{h_{r_{n_k}}(z_{n_k}(s)) - h_{r_{n_k}}(z^{\lim}(s))}ds + \int_0^t \norm{h_{r_{n_k}}(z^{\lim}(s)) - h_{\infty}(z^{\lim}(s)) }ds \\ \leq& \int_0^t \norm{h_{r_{n_k}}(z_{n_k}(s)) - h_{r_{n_k}}(z^{\lim}(s))} ds + (T+1)\epsilon \tag*{(by \eqref{eq:32f32k32032k950})} \\ \leq& \int_0^t L\norm{z_{n_k}(s) - z^{\lim}(s)} ds + T\epsilon \tag*{(by Lemma \ref{lemma:32h32Lipschitz})}\\ \leq& L (T+1)\epsilon + (T+1)\epsilon. \tag*{(by \eqref{eq:32f32k32032k951})} \end{align}\]

Thus, \(\forall t \in [0,T+1)\), \[\begin{align} \lim_{k \to \infty} \int_0^t h_{r_{n_k}}(z_{n_k}(s))ds = \int_0^t h_{\infty}(z^{\lim}(s))ds. \end{align}\] ◻

Proof. We first show the difference between \(\bar{z}(T_{n+1})\) and \(\bar{z}(T_n)\) by the following derivations. \(\forall n\), \(\forall t \in [0, T_{n+1} - T_n]\), \[\begin{align} &\norm{\bar{z}(T_n + t) - \bar{z}(T_n)} \\ =& \norm{\bar{z}(t(m(T_n + t))) - \bar{z}(T_n)} \\ =& \norm{ \bar{z}(T_n) + \sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i) H(\bar{z}(t(i)), W_{i+1}), \beta(i) G(\bar{z}(t(i)), W_{i+1})) - \bar{z}(T_n)} \\ = &\norm{\sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i) H(\bar{z}(t(i)), W_{i+1}), \beta(i) G(\bar{z}(t(i)), W_{i+1}))} \\ \leq &\sum_{i = m(T_n)}^{m(T_n+t) - 1} \alpha(i)\norm{H(\bar{z}(t(i)), W_{i+1}) - H( \bar{z}(T_n), W_{i+1}) } + \beta(i)\norm{G(\bar{z}(t(i)), W_{i+1}) - G( \bar{z}(T_n), W_{i+1}) } \\ &+ \norm{\sum_{i = m(T_n)}^{m(T_n+t) - 1} \alpha(i) H( \bar{z}(T_n), W_{i+1}) } + \norm{\sum_{i = m(T_n)}^{m(T_n+t) - 1} \beta(i) G( \bar{z}(T_n), W_{i+1}) } \\ \leq &\sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i) + \beta(i))L(W_{i+1})\norm{\bar{z}(t(i)) - \bar{z}(T_n)} \\ &+ \norm{\sum_{i = m(T_n)}^{m(T_n+t) - 1} \alpha(i) H( \bar{z}(T_n), W_{i+1}) } + \norm{\sum_{i = m(T_n)}^{m(T_n+t) - 1} \beta(i) G( \bar{z}(T_n), W_{i+1}) } \\ \leq &\sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i) + \beta(i) )L(W_{i+1})\norm{\bar{z}(t(i)) - \bar{z}(T_n)} + \sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i) + \beta(i)) L(W_{i+1}) \norm{ \bar{z}(T_n)} \\ &+ \norm{ \sum_{i = m(T_n)}^{m(T_n+t) - 1} \alpha(i) H(0, W_{i+1}) } + \norm{ \sum_{i = m(T_n)}^{m(T_n+t) - 1} \beta(i) G(0, W_{i+1}) } \tag*{(by Assumption \ref{assumption:32H32Lipschitz})}\\ = &\sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i) + \beta(i))L(W_{i+1})\norm{\bar{z}(t(i)) - \bar{z}(T_n)} + \norm{ \bar{z}(T_n)} \sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i) + \beta(i)) L(W_{i+1}) \\ &+ \norm{ \sum_{i = m(T_n)}^{m(T_n+t) - 1} \alpha(i) H(0, W_{i+1}) } + \norm{ \sum_{i = m(T_n)}^{m(T_n+t) - 1} \beta(i) G(0, W_{i+1}) } \\ \leq &\sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i) + \beta(i))L(W_{i+1})\norm{\bar{z}(t(i)) - \bar{z}(T_n)} + \norm{ \bar{z}(T_n)} C' + \norm{ \sum_{i = m(T_n)}^{m(T_n+t) - 1} \alpha(i) H(0, W_{i+1}) } \\ &+ \norm{ \sum_{i = m(T_n)}^{m(T_n+t) - 1} \beta(i) G(0, W_{i+1}) } \tag*{(by \eqref{eq:32alpha32L32bound32C95H})} \\ \leq & \sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i) + \beta(i))L(W_{i+1})\norm{\bar{z}(t(i)) - \bar{z}(T_n)} + \left[\norm{ \bar{z}(T_n)} C' + C' \right] \tag*{(by \eqref{eq:32H32c32c95H})} \\ \leq &\left[\norm{ \bar{z}(T_n)} C' + C' \right] e^{\sum_{i = m(T_n)}^{m(T_n+t) - 1} (\alpha(i) + \beta(i))L(W_{i+1})} \tag*{(by discrete Gronwall inequality in Theorem \ref{theorem:32discrete32Gronwall})}\\ \leq & \makebox[0.85\textwidth]{\displaystyle\left[\norm{ \bar{z}(T_n)} C' + C' \right] e^{C'}\hfill(by \eqref{eq:32alpha32L32bound32C95H})} \label{eq:32bar32x32bound} \end{align}\tag{194}\]  ◻

Proof. \(\forall t \in [0, T+1)\), \[\begin{align} &\norm{y^{\lim}(t)}& \\ =& \norm{ y^{\lim}(0) + \int_0^t g_{\infty}(\lambda_{\infty}(y^{\lim}(s)), y^{\lim}(s))ds}& \\ \leq & \norm{y^{\lim}(0)} + \norm{\int_0^t g_{\infty}(\lambda_{\infty}(y^{\lim}(s)), y^{\lim}(s))ds}& \\ =& \norm{y^{\lim}(0)} + \norm{\int_0^t \left[ g_{\infty}(\lambda_{\infty}(y^{\lim}(s)), y^{\lim}(s)) - g_{\infty}(0) \right]ds + \int_0^t g_{\infty}(0) ds }& \\ \leq & \norm{y^{\lim}(0)} + \int_0^t \norm{g_{\infty}(\lambda_{\infty}(y^{\lim}(s)), y^{\lim}(s)) - g_{\infty}(0) }ds + \int_0^t \norm{ g_{\infty}(0) }ds & \\ \leq & \norm{y^{\lim}(0)} + \int_0^t L\norm{\lambda_{\infty}(y^{\lim}(s))} + L\norm{y^{\lim}(s)} ds + \int_0^t \norm{ g_{\infty}(0) }ds& \tag*{(by Lemma \ref{lemma:32h32Lipschitz})} \\ \leq &1 + \int_0^t L \cdot L_{\lambda} \norm{y^{\lim}(s)} + L\norm{y^{\lim}(s)} ds + \int_0^t \norm{ g_{\infty}(0) }ds& \\ \leq &1 + \int_0^t L(L_{\lambda}+1)\norm{y^{\lim}(s)} ds + (T+1)\norm{ g_{\infty}(0) } \\ \leq &1 + \int_0^t L(L_{\lambda}+1)\norm{y^{\lim}(s)} ds + C_0 \\ \leq &\left[1 + C_0 \right] e^{\int_0^t L(L_{\lambda}+1) ds }& \tag*{(by Gronwall inequality in Theorem \ref{theorem:32Gronwall})}\\ \leq &\left[1 + C_0 \right] e^{L(L_{\lambda}+1)(T+1)} & \\ \leq& C. & \end{align}\] ◻

Proof. By Assumption 5, \(\lim_{k \to \infty}g_{r_{n_k}}(v) = g_{\infty}(v)\) uniformly in a compact set \(\require{physics} \qty{v \vert v\in \mathbb{R}^{d_1+d_2}, \norm{v}\leq C }\). By Lemma 62, \(\require{physics} \qty{y^{\lim}(t) \vert t \in [0,T+1)} \subseteq \qty{v \vert v\in \mathbb{R}^d, \norm{v}\leq C }\). Additionally, \(\lambda_{\infty}\) is Lipschitz, so \(\require{physics} \qty{(\lambda_{\infty}(y^{\lim}(t)), y^{\lim}(t)) \vert t \in [0,T+1)} \subseteq \qty{v \vert v\in \mathbb{R}^{d_1+d_2}, \norm{v}\leq C }\).

Therefore, \(\lim_{k \to \infty}g_{r_{n_k}}(\lambda_{\infty}(y^{\lim}(t)), y^{\lim}(t)) = g_{\infty}(\lambda_{\infty}(y^{\lim}(t)), y^{\lim}(t))\) uniformly in \(\require{physics} \qty{y^{\lim}(t) \vert t \in [0,T+1)}\) and in \(t \in [0,T+1)\). ◻

Proof. By 90 , \(\forall \delta > 0\), there exists a \(k_1\) such that \(\forall k \geq k_1\), \(\forall t \in [0,T+1)\), \[\begin{align} \norm{\tilde{y}_{n_k}(t) - \tilde{y}^{\lim}(t)} \leq \delta .\label{eq:32y32lim32epsilion321} \end{align}\tag{195}\] By Lemma 63, there exists a \(k_2\) such that \(\forall k \geq k_2\), \(\forall t \in [0,T+1)\), \[\begin{align} \norm{g_{r_{n_k}}(\lambda_{\infty}(y^{\lim}(t)), y^{\lim}(t)) - g_{\infty}(\lambda_{\infty}(y^{\lim}(t)), y^{\lim}(t))} \leq \delta. \label{eq:32y32lim32epsilion322} \end{align}\tag{196}\] \(\require{physics} \forall k \geq \max\qty{ k_1, k_2}\), \(\forall t \in [0,T+1)\) \[\begin{align} & \norm{y_{n_k}(t) - y^{\lim}(t) } \\ =& \norm{\tilde{y}_{n_k}(0) + \int_0^t g_{r_{n_k}}(\lambda_{\infty}(y_{n_k}(s)),y_{n_k}(s))ds - \tilde{y}^{\lim}(0) - \int_0^t g_{\infty}(\lambda_{\infty}(y^{\lim}(s)),y^{\lim}(s))ds} \\ \leq& \norm{\tilde{y}_{n_k}(0) - \tilde{y}^{\lim}(0)} + \norm{\int_0^t g_{r_{n_k}}(\lambda_{\infty}(y_{n_k}(s)),y_{n_k}(s))ds - \int_0^t g_{\infty}(\lambda_{\infty}(y^{\lim}(s)), y^{\lim}(s))ds} \\ \leq& \delta + \norm{\int_0^t g_{r_{n_k}}(\lambda_{\infty}(y_{n_k}(s)),y_{n_k}(s)) - g_{\infty}(\lambda_{\infty}(y^{\lim}(s)), y^{\lim}(s))ds} \tag*{(by \eqref{eq:32y32lim32epsilion321})} \\ \leq& \delta + \int_0^t \norm{g_{r_{n_k}}(\lambda_{\infty}(y_{n_k}(s)),y_{n_k}(s)) - g_{r_{n_k}}(\lambda_{\infty}(y^{\lim}(s)), y^{\lim}(s)) }ds \\ &+ \int_0^t \norm{ g_{r_{n_k}}(\lambda_{\infty}(y^{\lim}(s)), y^{\lim}(s)) - g_{\infty}(\lambda_{\infty}(y^{\lim}(s)), y^{\lim}(s))} ds \\ \leq& \delta + L\int_0^t \norm{(\lambda_{\infty}(y_{n_k}(s)),y_{n_k}(s)) - (\lambda_{\infty}(y^{\lim}(s)), y^{\lim}(s)))}ds \\ &+ \int_0^t \norm{ g_{r_{n_k}}(\lambda_{\infty}(y^{\lim}(s)), y^{\lim}(s)) - g_{\infty}(\lambda_{\infty}(y^{\lim}(s)), y^{\lim}(s))} \tag*{(by Lemma \ref{lemma:32h32Lipschitz})} \\ \leq& \delta + t\delta + L (L_{\lambda}+1) \int_0^t \norm{y_{n_k}(s) - y^{\lim}(s) }ds \tag*{(by \eqref{eq:32y32lim32epsilion322})} \\ \leq& (\delta + t\delta)e^{ L (L_{\lambda}+1)t} \tag*{(by Gronwall inequality in Theorem \ref{theorem:32Gronwall})}\\ \leq& (\delta + T\delta)e^{ L (L_{\lambda}+1)T}, \label{eq:32y32y32lim32difference} \end{align}\tag{197}\] which completes the proof. ◻

Proof. We know that \[\begin{align} \norm{y_l} &= \norm{y_{m(T_n)} + \sum_{i=m(T_n)}^l \beta(i)G(x_i, y_i, W_{i+1})} \\ &\leq \norm{y^{\text{max}}_n} + \sum_{i = m(T_n)}^l \beta(i)\norm{G(x_i, y_i, W_{i+1})} \label{eq:32ymax32broken32up} \end{align}\tag{199}\] Since the second term of 199 is monotonically increasing in \(l\), we know that the following holds: \[\begin{align} \norm{y^{\text{max}}_l} &\leq \norm{y^{\text{max}}_{m(T_{n})}}+ \sum_{i = m(T_n)}^l \beta(i)\norm{G(x_i, y_i, W_{i+1})} \\ &\leq \norm{y^{\text{max}}_{m(T_{n})}}+\sum_{i = m(T_n)}^l \beta(i)L(W_{i+1})(\norm{x_i}+\norm{y_i}) + \sum_{i = m(T_n)}^l \beta(i) G(0, W_{i+1}) \\ &\leq \norm{y^{\text{max}}_{m(T_{n})}}+\sum_{i = m(T_n)}^l \beta(i)L(W_{i+1})(\norm{x_i}+\norm{y_i}) + E \tag*{(\ref{eq:32h32property32H32bounded})} \\ &\leq \norm{y^{\text{max}}_{m(T_{n})}}+ \sum_{i = m(T_n)}^l \beta(i)L(W_{i+1})( K(1+\norm{y^{\text{max}}_i})+\norm{y^{\text{max}}_i}) + E \tag*{(Lemma \ref{thm:32x32stability})} \\ &\leq \norm{y^{\text{max}}_{m(T_{n})}}+\sum_{i = m(T_n)}^l \beta(i)L(W_{i+1})( K+(K+1)\norm{y^{\text{max}}_i}) + E \\ &\leq \norm{y^{\text{max}}_{m(T_{n})}}+K \sum_{i = m(T_n)}^l \beta(i)L(W_{i+1}) + (K+1) \sum_{i = m(T_n)}^l \beta(i)L(W_{i+1})\norm{y^{\text{max}}_i} + E \\ &\leq \norm{y^{\text{max}}_{m(T_{n})}} + K C' + (K+1) \sum_{i = m(T_n)}^l \beta(i)L(W_{i+1})\norm{y^{\text{max}}_i} \tag*{(\ref{eq:32alpha32L32bound32C95H})} + E \\ &\leq \left(\norm{y^{\text{max}}_{m(T_{n})}}+KC' \right)e^{(K+1) \sum_{i = m(T_n)}^l \beta(i)L(W_{i+1})} \tag*{(by discrete Grownwall inequality in Appendix \ref{theorem:32discrete32Gronwall})} + E\\ &\leq \left(\norm{y^{\text{max}}_{m(T_{n})}} + KC' \right)e^{(K+1)C'} \tag*{(\ref{eq:32alpha32L32bound32C95H})} + E \end{align}\]

Now we show 198 : \[\begin{align} \norm{z^{max}_l} &\leq \norm{y^{\text{max}}_l} + K (1+\norm{y^{\text{max}}_l}) \tag*{(Lemma \ref{thm:32x32stability})} \\ &\leq K + (K+1)(A \norm{y^{\text{max}}_{m(T_n)}} + B) \\ &\leq K + (K+1)(A \norm{z^{max}_{m(T_n)}} + B) \end{align}\] ◻