Topological entropy of switched nonlinear and interconnected systems

Abstract

A general upper bound for topological entropy of switched nonlinear systems is constructed, using an asymptotic average of upper limits of the matrix measures of Jacobian matrices of strongly persistent individual modes, weighted by their active rates. A general lower bound is constructed as well, using a similar weighted average of lower limits of the traces of these Jacobian matrices. In a case of interconnected structure, the general upper bound is readily applied to derive upper bounds for entropy that depend only on “network-level” information. In a case of block-diagonal structure, less conservative upper and lower bounds for entropy are constructed. In each case, upper bounds for entropy that require less information about the switching signal are also derived. The upper bounds for entropy and their relations are illustrated by numerical examples of a switched Lotka–Volterra ecosystem model.

Appendices

Appendix A Proofs of Lemmas 4 and 5

Lemmas 4 and 5 are established based on similar properties of a linear time-varying (LTV) system

$$\begin{aligned} \dot{x} = A(t)\, x \end{aligned}$$

(A1)

with a piecewise-continuous, matrix-valued function $A: \mathbb {R}_{\ge 0} \rightarrow \mathbb {R}^{n \times n}$. The solution to (A1) at time $t \ge 0$ with initial state v is given by $\xi (t, v) = \Phi _A(t, 0)\, v$, where $\Phi _A(t, 0)$ is the state-transition matrix and satisfies Coppel’s inequalities [14, Th. 2.8.27, p. 34]

$$\begin{aligned} e^{\int _{0}^{t} -\mu (-A(s)) \textrm{d}s} |v| \le |\Phi _A(t, 0)\, v| \le e^{\int _{0}^{t} \mu (A(s)) \textrm{d}s} |v| \qquad \forall t \ge 0,\, v \in \mathbb {R}^n \end{aligned}$$

(A2)

and Liouville’s formula [9, Th. 4.1, p. 28]

$$\begin{aligned} \det (\Phi _A(t, 0)) = e^{\int _{0}^{t} {{\,\textrm{tr}\,}}(A(s)) \textrm{d}s} \qquad \forall t \ge 0. \end{aligned}$$

(A3)

Proof of Lemma 4

We prove Lemma 4 by writing the Jacobian matrix of a solution to the switched nonlinear system (2) with respect to initial state $J_x \xi _\sigma (t, x)$ as a matrix solution to the LTV system (A1) with a suitable function A(t) , using variational arguments from nonlinear systems analysis (see, e.g., [24, Sect. 4.2.4]). For all $v \in \mathbb {R}^n$, we have $J_x \xi _\sigma (0, v) = I$ and

$$\begin{aligned} \partial _t J_x \xi _\sigma (t, v) = J_x {{\dot{\xi }}}_\sigma (t, v) = J_x f_{\sigma (t)}(\xi _\sigma (t, v))\, J_x \xi _\sigma (t, v) \end{aligned}$$

for all $t \ge 0$ that are not switches. Hence, for each fixed $v \in \mathbb {R}^n$, the matrix $J_x \xi _\sigma (t, v)$ is equal to the state-transition matrix $\Phi _A(t, 0)$ of (A1) with $A(t) = J_x f_{\sigma (t)}(\xi _\sigma (t, v))$.

First, given arbitrary initial states $x, {{\bar{x}}} \in K$, let

$$\begin{aligned} \nu (\rho ):= \rho {{\bar{x}}} + (1 - \rho )\, x, \qquad \rho \in [0, 1]. \end{aligned}$$

Then $\nu (\rho ) \in {{\,\textrm{co}\,}}(K)$ and $\nu '(\rho ) = {{\bar{x}}} - x$ for all $\rho \in [0, 1]$. Hence,

$$\begin{aligned}&|\xi _\sigma (t, {{\bar{x}}}) - \xi _\sigma (t, x)|\\&\quad = |\xi _\sigma (t, \nu (1)) - \xi _\sigma (t, \nu (0))| = \bigg | \int _{0}^{1} J_x \xi _\sigma (t, \nu (\rho )) ({{\bar{x}}} - x) \textrm{d}\rho \bigg | \\&\quad \le \int _{0}^{1} |J_x \xi _\sigma (t, \nu (\rho )) ({{\bar{x}}} - x)| \textrm{d}\rho \le \int _{0}^{1} e^{\int _{0}^{t} \mu (J_x f_{\sigma (s)}(\xi _\sigma (s, \nu (\rho )))) \textrm{d}s} |{{\bar{x}}} - x| \textrm{d}\rho \\&\quad \le \Big ( \max _{v \in {{\,\textrm{co}\,}}(K)} e^{\int _{0}^{t} \mu (J_x f_{\sigma (s)}(\xi _\sigma (s, v))) \textrm{d}s} \Big ) |{{\bar{x}}} - x| = e^{{{\overline{\eta }}}_\sigma (t)} |{{\bar{x}}} - x| \end{aligned}$$

for all $t \ge 0$, where the second inequality follows from the second inequality in (A2) with $A(t) = J_x f_{\sigma (t)}(\xi _\sigma (t, \nu (\rho )))$ and $\Phi _A(t, 0) = J_x \xi _\sigma (t, \nu (\rho ))$, and the last equality follows from the transformation (58). Hence, (56) holds.

Second, for each fixed $v \in \mathbb {R}^n$, Liouville’s formula (A3) with $A(t) = J_x f_{\sigma (t)}(\xi _\sigma (t, v))$ and $\Phi _A(t, 0) = J_x \xi _\sigma (t, v)$ implies that

$$\begin{aligned} \det (J_x \xi _\sigma (t, v)) = e^{\int _{0}^{t} {{\,\textrm{tr}\,}}(J_x f_{\sigma (s)}(\xi _\sigma (s, v))) \textrm{d}s} \qquad \forall t \ge 0. \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} {{\,\textrm{vol}\,}}(\xi _\sigma (t, K))&= \int _{K} |\det (J_x \xi _\sigma (t, v))| \textrm{d}v \ge \Big ( \min _{v \in K} |\det (J_x \xi _\sigma (t, v))| \Big ) {{\,\textrm{vol}\,}}(K) \\&= \Big ( \min _{v \in K} e^{\int _{0}^{t} {{\,\textrm{tr}\,}}(J_x f_{\sigma (s)}(\xi _\sigma (s, v))) \textrm{d}s} \Big ) {{\,\textrm{vol}\,}}(K) = e^{\gamma _\sigma (t)} {{\,\textrm{vol}\,}}(K) \end{aligned} \end{aligned}$$

for all $t \ge 0$, where the last equality follows from the transformation (58). Hence, (57) holds. $\square$

Proof of Lemma 5

We prove Lemma 5 by writing the difference between two solutions to the switched nonlinear system (2) as a solution to the LTV system (A1) with a suitable function A(t) , using variational arguments similar to those in the proof of Lemma 4. Given arbitrary initial states $x, {{\bar{x}}} \in K$, let

$$\begin{aligned} \nu (t, \rho ):= \rho \xi _\sigma (t, {{\bar{x}}}) + (1 - \rho ) \xi _\sigma (t, x), \qquad \rho \in [0, 1],\, t \ge 0. \end{aligned}$$

Then $\nu (t, \rho ) \in {{\,\textrm{co}\,}}(\xi _\sigma (t, K))$ and $\partial _\rho \nu (t, \rho ) = \xi _\sigma (t, {{\bar{x}}}) - \xi _\sigma (t, x)$ for all $\rho \in [0, 1]$ and $t \ge 0$. Hence,

$$\begin{aligned} \begin{aligned} \partial _t (\xi _\sigma (t, {{\bar{x}}}) - \xi _\sigma (t, x))&= f_{\sigma (t)}(\xi _\sigma (t, {{\bar{x}}})) - f_{\sigma (t)}(\xi _\sigma (t, x)) \\&= f_{\sigma (t)}(\nu (t, 1)) - f_{\sigma (t)}(\nu (t, 0)) \\&= \bigg ( \int _{0}^{1} J_x f_{\sigma (t)}(\nu (t, \rho )) \textrm{d}\rho \bigg ) (\xi _\sigma (t, {{\bar{x}}}) - \xi _\sigma (t, x)) \end{aligned} \end{aligned}$$

for all $t \ge 0$ that are not switches. Therefore, $\xi _\sigma (t, {{\bar{x}}}) - \xi _\sigma (t, x)$ is the solution to (A1) with $A(t) = \int _{0}^{1} J_x f_{\sigma (t)}(\nu (t, \rho )) \textrm{d}\rho$ at time t with initial state ${{\bar{x}}} - x$. Consequently, (A2) implies that

$$\begin{aligned} \begin{aligned} e^{\int _{0}^{t} -\mu \big ( -\int _{0}^{1} J_x f_{\sigma (s)}(\nu (s, \rho )) \textrm{d}\rho \big ) \textrm{d}s} |{{\bar{x}}} - x|&\le |\xi _\sigma (t, {{\bar{x}}}) - \xi _\sigma (t, x)| \\&\le e^{\int _{0}^{t} \mu \big ( \int _{0}^{1} J_x f_{\sigma (s)}(\nu (s, \rho )) \textrm{d}\rho \big ) \textrm{d}s} |{{\bar{x}}} - x| \end{aligned} \end{aligned}$$

for all $t \ge 0$. Moreover, as the matrix measure $\mu$ is a convex function, Jensen’s inequality [3, Th. 11.24, p. 417] implies that

$$\begin{aligned} \begin{aligned} \mu \bigg ( \int _{0}^{1} J_x f_{\sigma (t)}(\nu (t, \rho )) \textrm{d}\rho \bigg )&\le \int _{0}^{1} \mu (J_x f_{\sigma (t)}(\nu (t, \rho )) \textrm{d}\rho \\&\le \max _{v \in {{\,\textrm{co}\,}}(\xi _\sigma (t, K))} \mu (J_x f_{\sigma (t)}(v)) \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} -\mu \bigg ( {-\int _{0}^{1}} J_x f_{\sigma (t)}(\nu (t, \rho )) \textrm{d}\rho \bigg )&\ge \int _{0}^{1} -\mu (-J_x f_{\sigma (t)}(\nu (t, \rho )) \textrm{d}\rho \\&\ge \min _{v \in {{\,\textrm{co}\,}}(\xi _\sigma (t, K))} -\mu (-J_x f_{\sigma (t)}(v)) \end{aligned} \end{aligned}$$

for all $t \ge 0$. Then (59) follows from the transformation (58). $\square$

Appendix B Proof of Lemma 6

• 1. If (62) holds, then

  $$\begin{aligned} K \subset \bigcup _{x \in G(\theta )} R(x) \subset \bigcup _{x \in G(\theta )} B_{f_\sigma }(x, \varepsilon , T). \end{aligned}$$

  Hence, $G(\theta )$ is $(T, \varepsilon )$-spanning following the definition (4), and thus,

  $$\begin{aligned} S(f_\sigma , \varepsilon , T, K) \le \#G(\theta ) \le \prod _{i=1}^{n} \bigg ( \bigg \lfloor \frac{2 r_2}{\theta _i} \bigg \rfloor + 1 \bigg ) \le \prod _{i=1}^{n} \bigg ( \frac{2 r_2}{\theta _i} + 1 \bigg ). \end{aligned}$$

  If (62) holds for all $T \ge 0$ and $\varepsilon > 0$, then the definition of entropy (5) implies that

  $$\begin{aligned} h(f_\sigma , K)&\le \limsup _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \sum _{i=1}^{n} \frac{\log (2 r_2/\theta _i + 1)}{T} \nonumber \\&\le \limsup _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \sum _{i=1}^{n} \frac{\log (1/\theta _i)}{T} \nonumber \\&\quad \, + \sum _{i=1}^{n} \limsup _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \frac{\log (\theta _i + 2 r_2)}{T}, \end{aligned}$$

  (B4)

  where the last inequality holds as the upper limit is a subadditive function. Moreover, for all $i \in \{1, \ldots , n\}$, the summands in the last term in (B4) satisfy

  $$\begin{aligned} \begin{aligned}&\limsup _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \frac{\log (\theta _i + 2 r_2)}{T} \le \limsup _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \frac{\max \{\log (2\theta _i),\, \log (4 r_2)\}}{T} \\&\qquad = \max \left\{ \limsup _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \frac{\log (2 \theta _i)}{T},\, \lim _{T \rightarrow \infty } \frac{\log (4 r_2)}{T} \right\} \\&\qquad = \max \left\{ \limsup _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \frac{\log \theta _i}{T},\, 0 \right\} , \end{aligned} \end{aligned}$$

  where the inequality holds as the logarithm is an increasing function and $\theta _i, r_2 > 0$, and the last equality holds in part because $r_2$ is independent of T. Then (63) implies (64).⁹

• 2. If (65) holds, then for all distinct points $x, {{\bar{x}}} \in G(\theta )$, we have ${{\bar{x}}} \notin B_{f_\sigma }(x, \varepsilon , T)$ as ${{\bar{x}}} \notin R(x)$. Hence, $G(\theta )$ is $(T, \varepsilon )$-separated following the definition (6), and thus,

  $$\begin{aligned} N(f_\sigma , \varepsilon , T, K) \ge \#G(\theta ) \ge \prod _{i=1}^{n} \bigg \lfloor \frac{2 r_1}{\theta _i} \bigg \rfloor \ge \prod _{i=1}^{n} \max \left\{ \frac{2 r_1}{\theta _i} - 1,\, 1 \right\} . \end{aligned}$$

  If (65) holds for all $T \ge 0$ and $\varepsilon > 0$, then the alternative definition of entropy (7) implies that

  $$\begin{aligned} h(f_\sigma , K)&\ge \liminf _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \sum _{i=1}^{n} \frac{\log (\max \{2 r_1/\theta _i - 1,\, 1\})}{T} \nonumber \\&\ge \liminf _{\varepsilon \searrow 0} \limsup _{T \rightarrow \infty } \sum _{i=1}^{n} \frac{\log (1/\theta _i)}{T} \nonumber \\&\quad +\, \sum _{i=1}^{n} \liminf _{\varepsilon \searrow 0} \liminf _{T \rightarrow \infty } \frac{\log (\max \{2 r_1 - \theta _i,\, \theta _i\})}{T}, \end{aligned}$$

  (B5)

  where the last inequality holds as the lower limit is a superadditive function, and for two functions $g, {{\bar{g}}}: \mathbb {R}_{\ge 0} \rightarrow \mathbb {R}$, we have

  $$\begin{aligned} \limsup _{T \rightarrow \infty } (g(T) + {{\bar{g}}}(T)) \ge \limsup _{T \rightarrow \infty } g(T) + \liminf _{T \rightarrow \infty } {{\bar{g}}}(T). \end{aligned}$$

  Moreover, for all $i \in \{1, \ldots , n\}$, the summands in the last term in (B5) satisfy

  $$\begin{aligned} \begin{aligned} \liminf _{\varepsilon \searrow 0} \liminf _{T \rightarrow \infty } \frac{\log (\max \{2 r_1 - \theta _i,\, \theta _i\})}{T}&\ge \liminf _{\varepsilon \searrow 0} \liminf _{T \rightarrow \infty } \frac{\log (\max \{r_1,\, \theta _i\})}{T} \\&\ge \liminf _{\varepsilon \searrow 0} \liminf _{T \rightarrow \infty } \frac{\log r_1}{T} = 0, \end{aligned} \end{aligned}$$

  where the first inequality holds as the logarithm is an increasing function and $r_1, \theta _i > 0$, and the equality holds as $r_1$ is independent of T. Hence, (66) holds. $\square$

Appendix C Proof of Lemma 7

We prove only the lower bound (69) here; the upper bound (68) and lower bound (70) can be established using similar arguments (the omitted proof can be found in [52]). For brevity, we define the following functions on $\mathbb {R}_{\ge 0}$:

$$\begin{aligned} \begin{aligned} \eta _p^i(t)&:= \int _{0}^{t} a_p^i(s) {{\,\mathrm{\mathbbm {1}}\,}}_p(\sigma (s)) \textrm{d}s, \quad p \in {\mathcal {P}}, \\ {{\bar{a}}}^i(T)&:= \frac{1}{T} \max _{t \in [0, T]} \sum _{p \in {\mathcal {P}}} \eta _p^i(t), \end{aligned} \qquad i \in \{1, \ldots , m\} \end{aligned}$$

with ${{\bar{a}}}^i(0):= \max \{a_{\sigma (0)}^i(0),\, 0\}$. Also, note that for two continuous functions $g, {{\bar{g}}}: \mathbb {R}_{\ge 0} \rightarrow \mathbb {R}$, we have

$$\begin{aligned} \max _{t \in [0, T]} (g(t) + {{\bar{g}}}(t)) \ge \max _{t \in [0, T]} g(t) + \min _{t \in [0, T]} {{\bar{g}}}(t) \qquad \forall T \ge 0. \end{aligned}$$

(C6)

First, we eliminate modes that are not persistent. The definition of ${\mathcal {P}}_\infty$ in (11) implies that $t_\infty := \sup \{t \ge 0: \sigma (t) \in {\mathcal {P}}\backslash {\mathcal {P}}_\infty \}$ is finite. Then

$$\begin{aligned} \begin{aligned} {{\bar{a}}}^i(T)&\ge \frac{1}{T} \bigg ( \max _{t \in [0, T]} \sum _{p \in {\mathcal {P}}_\infty } \eta _p^i(t) + \min _{t \in [0, T]} \sum _{p \in {\mathcal {P}}\backslash {\mathcal {P}}_\infty } \eta _p^i(t) \bigg ) \\&= \frac{1}{T} \bigg ( \max _{t \in [0, T]} \sum _{p \in {\mathcal {P}}_\infty } \eta _p^i(t) + \min _{t \in [0, t_\infty ]} \sum _{p \in {\mathcal {P}}\backslash {\mathcal {P}}_\infty } \eta _p^i(t) \bigg ) \end{aligned} \end{aligned}$$

for all $i \in \{1, \ldots , m\}$ and $T > t_\infty$, where the inequality follows from (C6). Hence,

$$\begin{aligned} \limsup _{T \rightarrow \infty } \sum _{i=1}^{m} {{\bar{a}}}^i(T) \ge \limsup _{T \rightarrow \infty } \sum _{i=1}^{m} \frac{1}{T} \max _{t \in [0, T]} \sum _{p \in {\mathcal {P}}_\infty } \eta _p^i(t). \end{aligned}$$

Second, we eliminate modes that are persistent but not strongly persistent. For all $i \in \{1, \ldots , m\}$ and $p \in {\mathcal {P}}_\infty$, as ${\check{a}}_p^i = \liminf _{t \rightarrow \infty :\, \sigma (t) = p} a_p^i(t)$ are finite, ${{\check{\alpha }}}_p^i:= \inf _{t \ge 0} a_p^i(t) {{\,\mathrm{\mathbbm {1}}\,}}_p(\sigma (t))$ are finite as well. Then

$$\begin{aligned} \begin{aligned} \frac{1}{T} \max _{t \in [0, T]} \sum _{p \in {\mathcal {P}}_\infty } \eta _p^i(t)&\ge \frac{1}{T} \max _{t \in [0, T]} \sum _{p \in {\mathcal {P}}_\infty ^+} \eta _p^i(t) + \frac{1}{T} \min _{t \in [0, T]} \sum _{p \in {\mathcal {P}}_\infty \backslash {\mathcal {P}}_\infty ^+} \eta _p^i(t) \\&\ge \frac{1}{T} \max _{t \in [0, T]} \sum _{p \in {\mathcal {P}}_\infty ^+} \eta _p^i(t) + \sum _{p \in {\mathcal {P}}_\infty \backslash {\mathcal {P}}_\infty ^+} \min \{{{\check{\alpha }}}_p^i,\, 0\} \rho _p(T) \end{aligned} \end{aligned}$$

for all $i \in \{1, \ldots , m\}$ and $T > 0$, where the first inequality follows from (C6). Moreover, for all $p \notin {\mathcal {P}}_\infty ^+$, as $\rho _p(t) \ge 0$ for all $t \ge 0$ and ${{\hat{\rho }}}_p = \limsup _{t \rightarrow \infty } \rho _p(t) = 0$, we have ${{\hat{\rho }}}_p = \lim _{t \rightarrow \infty } \rho _p(t) = 0$. Then

$$\begin{aligned} \lim _{T \rightarrow \infty } \sum _{i=1}^{m} \sum _{p \in {\mathcal {P}}_\infty \backslash {\mathcal {P}}_\infty ^+} \min \{{{\check{\alpha }}}_p^i,\, 0\} \rho _p(T) = \sum _{i=1}^{m} \sum _{p \in {\mathcal {P}}_\infty \backslash {\mathcal {P}}_\infty ^+} \min \{{{\check{\alpha }}}_p^i,\, 0\} {{\hat{\rho }}}_p = 0. \end{aligned}$$

Hence,

$$\begin{aligned} \limsup _{T \rightarrow \infty } \sum _{i=1}^{m} {{\bar{a}}}^i(T) \ge \limsup _{T \rightarrow \infty } \sum _{i=1}^{m} \frac{1}{T} \max _{t \in [0, T]} \sum _{p \in {\mathcal {P}}_\infty ^+} \eta _p^i(t). \end{aligned}$$

Finally, as ${\mathcal {P}}_\infty ^+$ is a finite set, the lower limit in the definition of ${\check{a}}_p^i$ implies that, for each $\delta > 0$, there exists a large enough $t_\delta \ge 0$ such that

$$\begin{aligned} a_p^i(t)> {\check{a}}_p^i - \delta \qquad \forall i \in \{1, \ldots , m\},\, p \in {\mathcal {P}}_\infty ^+,\, t > t_\delta : \sigma (t) = p. \end{aligned}$$

For all $i \in \{1, \ldots , m\}$ and $t \ge 0$, consider

$$\begin{aligned} \begin{aligned} \sum _{p \in {\mathcal {P}}_\infty ^+} \eta _p^i(t)&= \sum _{p \in {\mathcal {P}}_\infty ^+} \int _{0}^{t} (a_p^i(s) - {\check{a}}_p^i + \delta + {\check{a}}_p^i - \delta ) {{\,\mathrm{\mathbbm {1}}\,}}_p(\sigma (s)) \textrm{d}s \\&= \sum _{p \in {\mathcal {P}}_\infty ^+} {\check{a}}_p^i \tau _p(t) - \delta t + \sum _{p \in {\mathcal {P}}_\infty ^+} \int _{0}^{t} (a_p^i(s) - {\check{a}}_p^i + \delta ) {{\,\mathrm{\mathbbm {1}}\,}}_p(\sigma (s)) \textrm{d}s. \end{aligned} \end{aligned}$$

If $t > t_\delta$, then

$$\begin{aligned} \sum _{p \in {\mathcal {P}}_\infty ^+} \int _{0}^{t} (a_p^i(s) - {\check{a}}_p^i + \delta ) {{\,\mathrm{\mathbbm {1}}\,}}_p(\sigma (s)) \textrm{d}s \ge \sum _{p \in {\mathcal {P}}_\infty ^+} \int _{0}^{t_\delta } (a_p^i(s) - {\check{a}}_p^i + \delta ) {{\,\mathrm{\mathbbm {1}}\,}}_p(\sigma (s)) \textrm{d}s. \end{aligned}$$

Otherwise $t \in [0, t_\delta ]$, and hence,

$$\begin{aligned} \begin{aligned}&\sum _{p \in {\mathcal {P}}_\infty ^+} \int _{0}^{t} (a_p^i(s) - {\check{a}}_p^i + \delta ) {{\,\mathrm{\mathbbm {1}}\,}}_p(\sigma (s)) \textrm{d}s \\&\quad \ge \min _{{{\bar{t}}} \in [0, t_\delta ]} \sum _{p \in {\mathcal {P}}_\infty ^+} \int _{0}^{{{\bar{t}}}} (a_p^i(s) - {\check{a}}_p^i + \delta ) {{\,\mathrm{\mathbbm {1}}\,}}_p(\sigma (s)) \textrm{d}s. \end{aligned} \end{aligned}$$

Combining the two cases, we obtain

$$\begin{aligned} \sum _{p \in {\mathcal {P}}_\infty ^+} \eta _p^i(t) \ge \sum _{p \in {\mathcal {P}}_\infty ^+} {\check{a}}_p^i \tau _p(t) - \delta t + \min _{{{\bar{t}}} \in [0, t_\delta ]} \sum _{p \in {\mathcal {P}}_\infty ^+} \int _{0}^{{{\bar{t}}}} (a_p^i(s) - {\check{a}}_p^i + \delta ) {{\,\mathrm{\mathbbm {1}}\,}}_p(\sigma (s)) \textrm{d}s. \end{aligned}$$

Then

$$\begin{aligned} \begin{aligned} \frac{1}{T} \max _{t \in [0, T]} \sum _{p \in {\mathcal {P}}_\infty ^+} \eta _p^i(t)&\ge \frac{1}{T} \max _{t \in [0, T]} \sum _{p \in {\mathcal {P}}_\infty ^+} {\check{a}}_p^i \tau _p(t) - \delta \\&\quad +\, \frac{1}{T} \min _{{{\bar{t}}} \in [0, t_\delta ]} \sum _{p \in {\mathcal {P}}_\infty ^+} \int _{0}^{{{\bar{t}}}} (a_p^i(s) - {\check{a}}_p^i + \delta ) {{\,\mathrm{\mathbbm {1}}\,}}_p(\sigma (s)) \textrm{d}s \end{aligned} \end{aligned}$$

for all $T > 0$, where the inequality follows in part from (C6). Hence,

$$\begin{aligned} \limsup _{T \rightarrow \infty } \sum _{i=1}^{m} {{\bar{a}}}_i(T) \ge \limsup _{T \rightarrow \infty } \sum _{i=1}^{m} \frac{1}{T} \max _{t \in [0, T]} \sum _{p \in {\mathcal {P}}_\infty ^+} {\check{a}}_p^i \tau _p(t) - m \delta . \end{aligned}$$

Then (69) holds as $\delta > 0$ is arbitrary. $\square$