Proof. Fix $M > 0$ and $\xi \in \mathscr{K}_M$ . Fix $J \in \mathcal{Z} \setminus \{0\}$ and define $\mathcal{Z}_J = \{1,2,\ldots,J\}$ , $t_z = z \xi(z)$ for $z \in \mathcal{Z}_J$ , and $T_z = \sum_{z^\prime \in \mathcal{Z}_J, z^\prime \geq z} t_{z^\prime}$ . Note that $T_J \leq T_{J-1} \leq \cdots \leq T_{1}$ . We shall first construct a trajectory $\varphi^J$ such that $\varphi^J(0) = \delta_0$ , $\varphi^J(T_1)(z) = \xi(z)$ for each $z \in \mathcal{Z}_J$ , and $S_{[0,T_1]}(\varphi^J | \delta_0)$ is bounded above by a constant independent of J.

Let $T_{J+1} = 0$ . For each $z \in \mathcal{Z}_J$ , starting with $z = J$ , we move the mass $\xi(z)$ from state 0 to state z using a piecewise unit velocity trajectory over the time duration $(T_{z+1}, T_{z+1}+t_z]$ . We define this trajectory $\varphi^J$ on $[0,T_1]$ as follows. Let $\varphi^J_0 = \delta_0$ . For each $z \in \mathcal{Z}_J$ and $1 \leq k \leq z$ , when $t \in (T_{z+1}+(k-1)\xi(z), T_{z+1}+k\xi(z)]$ , let

\begin{align*}\dot{\varphi}^J_t(l) = \left\{ \begin{array}{l@{\quad}l}1 & \text{ if } l = k, \\-1 & \text{ if }l = k-1, \\0 & \text { otherwise},\end{array}\right.\end{align*}

for $l \in \mathcal{Z}$ , and define $\varphi^J_t(l) = \delta_0(l) + \int_{[0,t]} \dot{\varphi}^J_u(l) du$ , $l \in \mathcal{Z}$ , $t \in [0,T]$ .

We now calculate the cost of this trajectory. Fix $z \in \mathcal{Z}$ such that $\xi(z) > 0$ , and let $1 \leq k \leq z$ . For each $t \in (T_{z+1}+(k-1)\xi(z), T_{z+1}+k\xi(z))$ and $\alpha \in C_0(\mathcal{Z})$ , note that

\begin{align*}\langle \alpha,\dot{\varphi}^J_t & - \Lambda_{\varphi^J_t}^*\varphi^J_t \rangle- \sum_{(z,z^\prime) \in \mathcal{E}} \tau(\alpha(z^\prime) - \alpha(z)) \lambda_{z,z^\prime}(\varphi^J_t) \varphi^J_t(z) \\ & = (\alpha(k) - \alpha(k-1)) - \sum_{(z,z^\prime) \in \mathcal{E}} (\!\exp\{\alpha(z^\prime) - \alpha(z)\} - 1) \lambda_{z,z^\prime} (\varphi^J_t) \varphi^J_t(z).\end{align*}

Hence,

(5.1) \begin{align}\sup_{\alpha \in C_0(\mathcal{Z})} \biggr\{\langle \alpha,\dot{\varphi}^J_t &- \Lambda_{\varphi^J_t}^*\varphi^J_t \rangle- \sum_{(z,z^\prime) \in \mathcal{E}} \tau(\alpha(z^\prime) - \alpha(z)) \lambda_{z,z^\prime}(\varphi^J_t) \varphi^J_t(z) \biggr\} \nonumber \\& \leq \sup_{x \in \mathbb{R}} (x - (\!\exp\{x\}-1) \lambda_{k-1,k}(\varphi^J_t) \varphi^J_t(k-1)) \nonumber \\& \qquad + \sup_{\alpha \in C_0(\mathcal{Z})} \left( - \sum_{(z,z^\prime) \in \mathcal{E}; (z,z^\prime) \neq (k-1,k)} (\!\exp\{\alpha(z^\prime) - \alpha(z)\} - 1) \lambda_{z,z^\prime}(\varphi^J_t) \varphi^J_t(z)\right) \nonumber \\& \leq \log\left(\frac{1}{\varphi^J_t(k-1) \lambda_{k-1,k}(\varphi^J_t)}\right) +2 \overline{\lambda} \nonumber \\& \leq \log\left(\frac{1}{\varphi^J_t(k-1)}\right) + \log k + \log\left(\frac{1}{\underline{\lambda}}\right) + 2 \overline{\lambda},\end{align}

where the last two inequalities follow from Assumption (A2). Consider the first term above. For $k > 1$ , integration of this quantity over the time duration $t \in (T_{z+1}+(k-1)\xi(z), T_{z+1}+k\xi(z))$ gives

\begin{align*}\int_{(T_{z+1}+(k-1)\xi(z), T_{z+1}+k\xi(z))} \, \log\left(\frac{1}{\varphi^J_t(k-1)} \right) dt & = -\int_{\xi(z)}^0 \log \left(\frac{1}{u}\right) \, du \\& = (u\log u - u) \biggr|_{\xi(z)}^0\\& = \xi(z) \log \left(\frac{1}{\xi(z)}\right) + \xi(z),\end{align*}

where the first equality follows from the variable change $u = \varphi^J_t(k-1)$ and the facts (i) $\dot{\varphi}^J_t(k-1) = -1$ , (ii) $\varphi^J_t(k-1) = \xi(z)$ when $t = T_{z+1}+(k-1)\xi(z)$ , (iii) $\varphi^J_t(k-1) = 0$ when $t = T_{z+1}+k\xi(z)$ , and (iv) $du = -dt$ . For $k=1$ , using the bound $\varphi^J_t(0) \geq \varphi^J_t(0) - (1 - \sum_{z^\prime = z}^J \xi(z^\prime))$ , we get

\begin{align*}\int_{(T_{z+1}, T_{z+1}+\xi(z))} & \, \log\left(\frac{1}{\varphi^J_t(0)} \right)dt \\& \leq \int_{(T_{z+1}, T_{z+1}+\xi(z))} \log\left(\frac{1}{\varphi^J_t(0) - (1 - \sum_{z^\prime = z}^J \xi(z^\prime))} \right)dt \\& = -\int_{\xi(z)}^0 \log \left(\frac{1}{u}\right) \, du,\end{align*}

where the last equality follows from the variable change $u = \varphi^J_t(0) - (1 - \sum_{z^\prime = z}^J \xi(z^\prime))$ and the facts (i) $\dot{\varphi}^J_t(0) = -1$ , (ii) $\varphi^J_t(0) = 1- \sum_{z^\prime = z+1}^J \xi(z^\prime)$ when $t = T_{z+1}$ , so that $\varphi^J_t(0) - (1 - \sum_{z^\prime = z}^J \xi(z^\prime)) = \xi(z)$ when $t = T_{z+1}$ , (iii) $\varphi^J_t(0) = 1- \sum_{z^\prime = z}^J \xi(z^\prime)$ when $t = T_{z+1}+\xi(z)$ , so that $\varphi^J_t(0) - (1 - \sum_{z^\prime = z}^J \xi(z^\prime)) = 0$ when $t = T_{z+1} + \xi(z)$ , and (iv) $du = -dt$ . Thus, proceeding as before for the case $k > 1$ , we arrive at

\begin{align*}\int_{(T_{z+1}, T_{z+1}+\xi(z))} \log\left(\frac{1}{\varphi^J_t(0)} \right) dt \leq \xi(z) \log \left(\frac{1}{\xi(z)}\right) + \xi(z).\end{align*}

Hence, integrating (5.1) over $t \in (T_{z+1}+(k-1)\xi(z), T_{z+1}+k\xi(z))$ and summing over $1 \leq k \leq z$ , we get, for each $z \in \mathcal{Z}_J$ ,

(5.2) \begin{align}\int_{(T_{z+1}, T_{z+1}+t_z)} \, \sup_{\alpha \in C_0(\mathcal{Z})} \biggr\{\langle \alpha,\dot{\varphi}^J_t &- \Lambda_{\varphi^J_t}^*\varphi^J_t \rangle- \sum_{(z,z^\prime) \in \mathcal{E}} \tau(\alpha(z^\prime) - \alpha(z)) \lambda_{z,z^\prime}(\varphi^J_t) \varphi^J_t(z) \biggr\} dt \nonumber \\& \leq z\xi(z) \log \left(\frac{1}{\xi(z)}\right) + \tilde{C}_z, \end{align}

where $\tilde{C}_z = (z \log z+z) \xi(z) + z \xi(z) \left(\log\left(\frac{1}{\underline{\lambda}}\right) + 2 \overline{\lambda}\right).$ Let $\tilde{C}^J = \sum_{z \in \mathcal{Z}_J} \tilde{C}_z$ . Then, summing the above display over $z \in \mathcal{Z}_J$ , we arrive at

\begin{align*}S_{[0,T_1]}(\varphi^J | \delta_0) \leq \sum_{z \in \mathcal{Z}_J} z \xi(z) \log \left(\frac{1}{\xi(z)}\right) + \tilde{C}^J.\end{align*}

Note that

(5.3) \begin{align}\sum_{z \in \mathcal{Z}_J} z \xi(z) \log \left(\frac{1}{\xi(z)}\right) & = \sum_{\stackrel{z \in \mathcal{Z}_J\,:\,}{\xi(z) \leq 1/z^3}}z \xi(z) \log \left(\frac{1}{\xi(z)}\right) + \sum_{\stackrel{z \in \mathcal{Z}_J\,:\,}{\xi(z) > 1/z^3}} z \xi(z) \log \left(\frac{1}{\xi(z)}\right) \nonumber \\& \leq \frac{1}{e} + \sum_{\stackrel{z \in \mathcal{Z}_J\setminus \{1\}\,:\,}{\xi(z) \leq 1/z^3}} \frac{3 \log z}{z^2} + 3 \sum_{\stackrel{z \in \mathcal{Z}_J\,:\,}{\xi(z) > 1/z^3}} (z \log z) \xi(z) \nonumber \\& \leq \frac{1}{e} + 3\sum_{z \in \mathcal{Z}_J}\left\{\frac{\log z}{z^2} + (z \log z) \xi(z)\right\},\end{align}

where the first inequality comes from the fact that the mapping $x \mapsto x \log (1/x)$ is monotonically increasing for $x \in [0, 1/e]$ . Hence,

\begin{align*}S_{[0,T_1]}(\varphi^J | \delta_0) \leq \frac{1}{e} + 3\sum_{z \in \mathcal{Z}_J}\left\{\frac{\log z}{z^2} + (z \log z) \xi(z)\right\} + \tilde{C}^J, \qquad J \geq 1.\end{align*}

Define $T = \sum_{z \in \mathcal{Z}} z \xi(z)$ . We now extend the trajectory $\varphi^J$ to $(T_1, T]$ by defining $\varphi^J_t = \varphi^J_{T_1}$ for $t \in (T_1, T]$ . Noting that $\dot{\varphi}^J_t(z) = 0$ for all $z \in \mathcal{Z}$ on $t \in (T_1, T]$ , this extension suffers an additional cost of at most $2 \overline{\lambda} T$ . Hence, we get

\begin{align*}S_{[0,T]}(\varphi^J | \delta_0) \leq \frac{1}{e} + 3\sum_{z \in \mathcal{Z}_J}\left\{\frac{\log z}{z^2} + (z \log z) \xi(z)\right\} + \tilde{C}^J + 2\overline{\lambda}T.\end{align*}

Noting that (i) the right-hand side above is bounded above by $\langle \xi, \vartheta \rangle C(\overline{\lambda}, \underline{\lambda})$ , where $C(\overline{\lambda}, \underline{\lambda})$ is a constant depending on $\overline{\lambda}$ and $\underline{\lambda}$ , and (ii) $\langle \xi, \vartheta \rangle \leq M$ , the above display yields

\begin{align*}S_{[0,T]}(\varphi^J | \delta_0) \leq C(M, \overline{\lambda}, \underline{\lambda}),\end{align*}

where $C(M, \overline{\lambda}, \underline{\lambda})$ is a constant depending on $M, \overline{\lambda}$ , and $\underline{\lambda}$ . Using the compactness of the level sets of $S_{[0,T]}$ (see Lemma 2.1), it follows that the sequence of trajectories $\{\varphi^J, J \geq 1\}$ has a convergent subsequence. Re-indexing the original sequence, let $\varphi^J \to \varphi$ in $D([0,T],\mathcal{M}_1(\mathcal{Z}))$ as $J \to \infty$ . By construction, for each $J \in \mathcal{Z} \setminus \{0\}$ , $\varphi^J_T(z) = \xi(z)$ for all $z \in \mathcal{Z}_J$ ; hence $\varphi_T(z) = \xi(z)$ for all $z \in \mathcal{Z}$ . Recall that lower semicontinuity of $S_{[0,T]}$ was proved in the course of the proof of Lemma 2.1. Therefore, it follows that

\begin{align*}S_{[0,T]}(\varphi | \delta_0) \leq \liminf_{J \to \infty} S_{[0,T]}(\varphi^J | \delta_0) \leq C(M, \overline{\lambda}, \underline{\lambda}).\end{align*}

This completes the proof of the lemma.

Proof. Let $\xi \in \mathcal{M}_1(\mathcal{Z})$ be such that $V(\xi) < \infty$ . Then there exist a $T > 0$ and a trajectory $\varphi$ on [0, T] such that $\varphi(0) = \xi^*, \varphi(T) = \xi$ , and $S_{[0,T]}(\varphi | \xi^*) \leq V(\xi) + 1$ . By Theorem 2.2, there exists a measurable function $h_\varphi$ on $[0,T]\times \mathcal{E}$ such that

(5.4) \begin{align}\langle \varphi_t, f \rangle & = \langle \varphi_0, f\rangle + \int_{[0,t]} \sum_{(z,z^\prime)\in \mathcal{E}} (f(z^\prime) - f(z)) (1+h_\varphi(u,z,z^\prime)) \lambda_{z,z^\prime}(\varphi_u) \varphi_u(z) du\end{align}

holds for all $t \in [0,T]$ and $f \in C_0(\mathcal{Z})$ , and $S_{[0,T]}(\varphi|\varphi(0))$ is given by

\begin{align*}S_{[0,T]}(\varphi|\varphi(0)) = \int_{[0,T]} \sum_{(z,z^\prime)\in \mathcal{E}} \tau^*(h_\varphi(t,z,z^\prime)) \lambda_{z,z^\prime}(\varphi_t) \varphi_t(z) dt.\end{align*}

For any $x \geq 0$ and $y \in \mathbb{R}$ , using the convex duality relation $(x-1)y \leq \tau^*(x-1)+\tau(y)$ , we get the inequality $xy \leq \tau^*(x-1) + (\!\exp\{y\}-1)$ . Hence, from the above non-variational representation for $S_{[0,T]}(\varphi | \varphi(0))$ , (5.4) implies

(5.5) \begin{align}\langle \varphi_t, f \rangle & \leq \langle \xi^*, f\rangle + \int_{[0,t]} \sum_{(z,z^\prime)\in \mathcal{E}} \tau^*(h_\varphi(u,z,z^\prime)) \lambda_{z,z^\prime}(\varphi_u) \varphi_u(z) du \nonumber \\& \qquad + \int_{[0,t]} \sum_{(z,z^\prime)\in \mathcal{E}} (\!\exp\{f(z^\prime) - f(z)\}-1) \lambda_{z,z^\prime}(\varphi_u) \varphi_u(z) du \nonumber \\& \leq \langle \xi^*, f\rangle + V(\xi) + 1 \nonumber \\& \qquad + \int_{[0,t]} \sum_{(z,z^\prime)\in \mathcal{E}} (\!\exp\{f(z^\prime) - f(z)\}-1) \lambda_{z,z^\prime}(\varphi_u) \varphi_u(z) du.\end{align}

Recall the function $\vartheta$ on $\mathcal{Z}$ . For $n \geq 1$ , define

\begin{align*}\vartheta_n(z) = \left\{\begin{array}{l@{\quad}l}\vartheta(z), & \text{ if } z \leq n,\\0, & \text{ otherwise}.\end{array}\right.\end{align*}

By convexity, note that $\vartheta_n(z+1) - \vartheta_n(z) \leq 1 + \log(z+1)$ and $\vartheta_n(0) - \vartheta_n(z) \leq 0$ for each $z \in \mathcal{Z}$ . Therefore, using the upper bound for the transition rates from Assumption (A2), we have

\begin{align*}\int_{[0,t]} \sum_{(z,z^\prime)\in \mathcal{E}} (\!\exp\{\vartheta_n(z^\prime) - \vartheta_n(z)\}-1) \lambda_{z,z^\prime}(\varphi_u) \varphi_u(z) du \leq \overline{\lambda}(e-1)t\end{align*}

for each $t \in [0,T]$ and $n \geq 1$ . It follows from (5.5) with f replaced by $\vartheta_n$ that

\begin{align*}\langle \varphi_t, \vartheta_n \rangle \leq \langle \xi^*, \vartheta_n \rangle + V(\xi) + 1 + \overline{\lambda}(e-1)T\end{align*}

for each $t \in [0,T]$ and $n \geq 1$ . Letting $n \to \infty$ and using monotone convergence, we conclude that

(5.6) \begin{align}\sup_{t \in [0,T]} \langle \varphi_t, \vartheta \rangle = \sup_{t \in [0,T]} \lim_{n\to \infty} \langle \varphi_t, \vartheta_n \rangle \leq \langle \xi^*, \vartheta \rangle + V(\xi) + 1 + \overline{\lambda}(e-1)T.\end{align}

In particular, $\langle \xi, \vartheta \rangle \leq \langle \xi^*, \vartheta \rangle + V(\xi) + 1 + \overline{\lambda}(e-1)T$ . It follows that $\xi \in \mathscr{K}$ .

Conversely, let $\xi \in \mathscr{K}$ . Let $M> 0$ be such that $\xi \in \mathscr{K}_M$ . By Lemma 5.1, there exist a $T > 0$ and a trajectory $\varphi^{(2)}$ on [0, T] such that $\varphi^{(2)}(0) = \delta_0$ , $\varphi^{(2)}(T) = \xi$ , and $S_{[0,T]}(\varphi^{(2)} | \delta_0) \leq C_M$ for some constant $C_M > 0$ depending on M. Let $t_0 = 0$ , $t_z = \sum_{z^\prime =1}^{z}\xi^*(z^\prime)$ , $z \in \mathcal{Z} \setminus\{0\}$ , and $T_1 = \sum_{z^\prime \neq 0}\xi^*(z^\prime)$ . We now construct another trajectory $\varphi^{(1)}$ on $[0,T_1]$ such that $\varphi^{(1)}(0) = \xi^*$ , $\varphi^{(1)}(T_1) = \delta_0$ , and $S_{[0,T_1]}(\varphi^{(1)}| \xi^*) < \infty$ . This trajectory is constructed using piecewise-constant-velocity paths, and its cost $S_{[0,T_1]}(\varphi^{(1)}| \xi^*)$ is computed using arguments similar to those used in the proof of Lemma 5.1; we provide the details here for completeness. When $t \in (t_{z-1}, t_z]$ for some $z \in \mathcal{Z} \setminus \{0\}$ , let

\begin{align*}\dot{\varphi}^{(1)}_t(l) = \left\{\begin{array}{l@{\quad}l}-1 & \text{ if } l = z, \\1 & \text{ if } l = 0, \\0 & \text{ otherwise},\end{array}\right.\end{align*}

for $l \in \mathcal{Z}$ , and define $\varphi^{(1)}_t(l) = \varphi^{(1)}_0(l) + \int_{[0,t]} \dot{\varphi}^{(1)}_u(l) du$ , $l \in \mathcal{Z}$ , $t \in [0,T_1]$ . Note that for each $\alpha \in C_0(\mathcal{Z})$ , when $t \in (t_{z-1}, t_z)$ , we have

\begin{align*}\biggr\{ \langle \alpha, & \dot{\varphi}^{(1)}_t - \Lambda_{\varphi^{(1)}_t}^*\varphi^{(1)}_t \rangle- \sum_{(z,z^\prime) \in \mathcal{E}} \tau(\alpha(z^\prime) - \alpha(z)) \lambda_{z,z^\prime}(\varphi^{(1)}_t) \varphi^{(1)}_t(z) \biggr\} \\& = (\alpha(0) - \alpha(z)) - (\!\exp\{\alpha(0) - \alpha(z)\}-1) \lambda_{z,0}(\varphi^{(1)}_t) \varphi^{(1)}_t(z) \\& \qquad - \sum_{(z_0,z^\prime) \in \mathcal{E}\,:\, (z_0,z^\prime) \neq (z,0)} (\!\exp\{\alpha(z^\prime) - \alpha(z_0)\}-1) \lambda_{z_0,z^\prime}(\varphi^{(1)}_t) \varphi^{(1)}_t(z_0) \biggr\},\end{align*}

so that optimising the left-hand side of the above display over $\alpha \in C_0(\mathcal{Z})$ yields

\begin{align*}\sup_{\alpha \in C_0(\mathcal{Z})}\biggr\{ \langle \alpha, & \dot{\varphi}^{(1)}_t - \Lambda_{\varphi^{(1)}_t}^*\varphi^{(1)}_t \rangle- \sum_{(z,z^\prime) \in \mathcal{E}} \tau(\alpha(z^\prime) - \alpha(z)) \lambda_{z,z^\prime}(\varphi^{(1)}_t) \varphi^{(1)}_t(z) \biggr\} \\& \leq \log\left(\frac{1}{ \varphi^{(1)}_t(z) \lambda_{z,0}(\varphi^{(1)}_t)}\right) + 2\bar{\lambda} \\& \leq \log\left(\frac{1}{ \varphi^{(1)}_t(z)}\right) + \log\left(\frac{1}{\underline{\lambda}}\right) + 2\overline{\lambda},\end{align*}

where the last inequality follows from the lower bound on the backward transition rates in Assumption (A2). Integrating the above over $(t_{z-1}, t_z)$ and summing over $z \in \mathcal{Z} \setminus \{0\}$ , we arrive at

(5.7) \begin{align}S_{[0,T_1]}(\varphi^{(1)} | \xi^* ) \leq \sum_{z \in \mathcal{Z} \setminus\{0\}} \left\{\xi^*(z) \log \frac{1}{\xi^*(z)} + \xi^*(z)\left( 1 + \log\left(\frac{1}{\underline{\lambda}}\right) + 2\overline{\lambda}\right) \right\}. \end{align}

Since $\xi^* \in \mathscr{K}$ , proceeding via the steps in (5.3), we conclude that the right-hand side of the above display is finite. We combine $\varphi^{(1)}$ and $\varphi^{(2)}$ and define a new trajectory $\tilde{\varphi}$ on $[0,T_1+T]$ as follows: $\tilde{\varphi}(t) = \varphi^{(1)}(t)$ on $t \in [0, T_1]$ ; $\tilde{\varphi}(t) = \varphi^{(2)}(t-T_1)$ on $t \in (T_1, T_1+T]$ . Note that $\tilde{\varphi}(0) = \xi^*$ , $\tilde{\varphi}(T_1+T) = \xi$ , and $S_{[0,T_1+T]}(\tilde{\varphi} | \xi^*) < \infty$ . Hence $V(\xi) < \infty$ .

To prove the second statement, we note that given any $M > 0$ , for any $\xi \in \mathscr{K}_M$ , the cost of the trajectory $\tilde{\varphi}$ constructed in the previous paragraph is bounded above by a constant depending only on M (and not on $\xi$ ). This completes the proof of the lemma.

Proof. We first prove that $\limsup_{n \to \infty} V(\xi_n) \leq V(\xi)$ . Fix $\varepsilon > 0$ . We shall move from $\xi$ to $\xi_n$ in five steps. The outline of this construction is as follows:

• $\varphi^{(0)}$ : This trajectory starts with $\xi$ and moves all the mass for all states $z > z_0$ , for a suitable large enough $z_0$ , back to state 0. This backward movement results in a cost of $O(\varepsilon)$ .

• $\varphi^{(1)}$ : Next, we move any additional mass, if required, from the states $\{1,2,\ldots, z_0\}$ back to state 0 so that there is enough mass at state 0 to fill up all the states beyond $z_0$ . Again, this backward movement results in a cost of $O(\varepsilon)$ .

• $\varphi^{(2)}$ : Next, we construct a trajectory of piecewise constant velocity to move the mass $\sum_{z^\prime > z_0}\xi_n(z)$ from state 0 to state $z_0+1$ . After this movement, state $z_0+1$ contains all the mass required to fill up the states beyond it. This forward movement results in a cost of $O(\varepsilon \log (1/\varepsilon))$ , instead of $O(\varepsilon)$ , because we move the total mass for all the states beyond $z_0$ .

• $\varphi^{(3)}$ : Then, for each $z > z_0$ , we move the required mass (i.e., $\xi_n(z)$ ) from state 0 to state z using a trajectory of piecewise constant velocity. At the end of this procedure, for each $z > z_0$ , the mass at state z becomes $\xi_n(z)$ . This forward movement results in a cost of $O(\varepsilon)$ .

• $\varphi^{(4)}$ : Finally, we adjust the mass within the finite set $\{1,2,\ldots, z_0\}$ to match with $\xi_n$ . This also results in a cost of at most $O(\varepsilon \log (1/\varepsilon))$ . Again, this cost is $O(\varepsilon \log (1/\varepsilon))$ instead of $O(\varepsilon)$ because we move, for each $z \in \{1,2,\ldots, z_0\}$ , the sum of the additional mass (under $\xi_n$ as opposed to $\xi$ ) in the states $\{z, z+1, \ldots, z_0\}$ from state 0 to state z.

Therefore, the total cost of all these trajectories is at most $O(\varepsilon \log (1/\varepsilon))$ , which vanishes as $\varepsilon \to 0$ . We now define these trajectories in detail and evaluate their costs.

Let $z_0 \geq 2$ be such that

\begin{align*}\sum_{z > z_0} \vartheta(z) \xi(z) < \varepsilon/6 \quad \text{and} \quad \sum_{z > z_0}\frac{\log z}{z^2} < \varepsilon.\end{align*}

Then choose $n_1 \geq 1$ such that $\sum_{z > z_0} \vartheta(z) \xi_n(z) < \varepsilon/3$ holds for all $n \geq n_1$ ; this is possible since $\xi_n \to \xi$ in $\mathcal{M}_1(\mathcal{Z})$ and $\langle \xi_n, \vartheta \rangle \to \langle \xi, \vartheta \rangle$ as $n \to \infty$ . Let

\begin{align*}t_{z_0} = 0, \quad t_z = \sum_{z^\prime = z_0+1}^z \xi(z^\prime), \ z > z_0, \quad \text{ and} \quad T_0 = \sum_{z^\prime > z_0} \xi(z^\prime).\end{align*}

Define the trajectory $\varphi^{(0)}$ on $[0,T_0]$ as follows. When $t \in (t_{z-1}, t_{z}]$ for some $z > z_0$ , let

\begin{align*}\dot{\varphi}^{(0)}_t(l) = \left\{\begin{array}{l@{\quad}l}-1 & \text{ if } l = z, \\1 & \text{ if } l = 0, \\0 & \text{ otherwise},\end{array}\right.\end{align*}

for $l \in \mathcal{Z}$ , and define

\begin{align*}\varphi^{(0)}_t(l) = \xi(l) + \int_{[0,t]} \dot{\varphi}^{(0)}_u(l) du, \quad l \in \mathcal{Z}, \, t \in [0, T_0].\end{align*}

Note that $\varphi^{(0)}_{T_0}(z) = \xi(z)$ for $1 \leq z \leq z_0$ , $\varphi^{(0)}_{T_0}(z) = 0$ for $z > z_0$ , and $\varphi^{(0)}_{T_0}(0) = \xi(0) + \sum_{z > z_0} \xi(z)$ . Let $M = (\!\sup_{n \geq n_1} \langle \xi_n, \vartheta \rangle) \vee \langle \xi, \vartheta \rangle + 1$ . Using ideas similar to those used in the proof of Lemma 5.2, it can be checked that $S_{[0,T_0]}(\varphi^{(0)} | \xi) \leq C_0(M, \overline{\lambda}, \underline{\lambda}) \varepsilon$ , for some constant $C_1(M, \overline{\lambda}, \underline{\lambda})$ depending on M, $\overline{\lambda}$ , and $\underline{\lambda}$ . Indeed, the cost is $O(\sum_{z > z_0}\xi(z) \log (1/\xi(z)))$ , which, using the argument used to arrive at the bound (5.7) and the choice of $z_0$ , is bounded by

\begin{align*}O\left(\sum_{z > z_0}\left((z \log z)\xi(z) + \frac{\log z}{z^2} \right) \right) = O(\varepsilon).\end{align*}

Let $\varepsilon_n = \sum_{z > z_0} \xi_n(z)$ . If $\varepsilon_n > \varphi_{T_0}^{(0)}(0)$ , then we move the extra mass $\varepsilon_n - \varphi_{T_0}^{(0)}(0)$ from the states $\{1,2,\ldots, z_0\}$ to state 0 as follows. Let $T_1 = T_0 + \varepsilon_n - \varphi_{T_0}^{(0)}(0)$ . When t is between $T_0 + \sum_{z^\prime = z+1}^{z_0} \varphi_{T_0}^{(0)}(z^\prime)$ and $(T_0 + \sum_{z^\prime = z}^{z_0} \varphi_{T_0}^{(0)}(z^\prime)) \wedge T_1$ for some $z \leq z_0$ , let

\begin{align*}\dot{\varphi}^{(1)}_t(l) = \left\{\begin{array}{l@{\quad}l}-1 & \text{ if } l = z, \\1 & \text{ if } l = 0,\\0 & \text{ otherwise},\end{array}\right.\end{align*}

for $l \in \mathcal{Z}$ . Define the trajectory $\varphi^{(1)}$ on $[0, T_1]$ as follows: $\varphi^{(1)}_t = \varphi^{(0)}_t$ when $t \in [0, T_0]$ ; $\varphi^{(1)}_t(l) = \varphi^{(0)}_{T_0}(l) + \int_{[0,t]} \dot{\varphi}^{(1)}_u(l) du$ , $l \in \mathcal{Z}$ , $t \in (T_0, T_1]$ . Note that $\varphi^{(1)}$ depends on n, but we suppress this in the notation for ease of reading. Again, since $\varepsilon_n$ is smaller than $\varepsilon/3$ , by using calculations similar to those used in the proof of Lemma 5.2, we see that $S_{[T_0, T_1]}(\varphi^{(1)}|\varphi_{T_0}^{(0)}) \leq C_1(M,\overline{\lambda}, \underline{\lambda}) \varepsilon$ for some constant $C_1(M,\overline{\lambda}, \underline{\lambda})$ depending on M, $\overline{\lambda}$ , and $\underline{\lambda}$ . On the other hand, if $\varepsilon_n \leq \varphi_{T_0}^{(0)}(0)$ , we set $T_1 = T_0$ and $\varphi^{(1)}_t = \varphi^{(0)}_t$ on $[0, T_1]$ . In both cases, we have $\varphi^{(1)}_{T_1}(0) \geq \varepsilon_n $ .

Let $T_2 = (z_0+1) \varepsilon_n$ . We now construct another trajectory $\varphi^{(2)}$ on $[0,T_2]$ to transfer the mass $\varepsilon_n$ from state 0 (in $\varphi^{(1)}_{T_1}$ ) to state $z_0 + 1$ . Let $\varphi^{(2)}_0 = \varphi^{(1)}_{T_1}$ . When $t \in ((z-1)\varepsilon_n, z\varepsilon_n]$ for some $z \in \{1,2,\ldots, z_0+1\}$ , let

\begin{align*}\dot{\varphi}^{(2)}_t(l) = \left\{\begin{array}{l@{\quad}l}-1 & \text{ if } l = z-1, \\1 & \text{ if } l = z,\\0 & \text{ otherwise},\end{array}\right.\end{align*}

for $l \in \mathcal{Z}$ , and define $\varphi^{(2)}_t(l) = \varphi^{(1)}_{T_1}(l) + \int_{[0,t]} \dot{\varphi}^{(2)}_u(l) du$ , $l \in \mathcal{Z}$ , $t \in (0, T_2]$ . Note that $|x \log (\frac{1}{x})- y \log (\frac{1}{y})| \leq \delta + \delta \log(1/\delta)$ whenever $|x - y| \leq \delta$ , and that $\varepsilon_n \leq \varepsilon/(z_0 \log z_0)$ . Hence, using calculations similar to those done in the proof of Lemma 5.1, we see that $S_{[0,T_2]}(\varphi^{(2)}|\varphi^{(1)}_{T_1})$ can be bounded above by $C_2(M, \overline{\lambda}, \underline{\lambda})\varepsilon \log (1/\varepsilon)$ , where $C_2(M, \overline{\lambda}, \underline{\lambda})$ is a constant depending on M, $\overline{\lambda}$ , and $\underline{\lambda}$ , for each $n \geq n_1$ (recall that $\varphi^{(2)}$ depends on n). Indeed, the cost is bounded by the order of

\begin{align*}z_0 \varepsilon_n \log \left(\frac{1}{\varepsilon_n}\right) + (z_0 \log z_0) \varepsilon_n & \leq z_0 \frac{\varepsilon}{z_0 \log z_0} \log \left(\frac{z_0 \log z_0}{\varepsilon} \right) + \varepsilon\\& \leq \varepsilon \log (1/\varepsilon) + 3 \varepsilon\end{align*}

(see the bound in (5.2)), where the first inequality uses the fact that $\varepsilon_n \leq \varepsilon/(z_0 \log z_0)$ , and the second inequality uses the fact that $z_0 \geq 2$ , so that $z_0 \log z_0 > 1$ .

Note that $\varphi^{(2)}_{T_2}(z_0+1) = \varepsilon_n$ . We now construct a trajectory that distributes this mass $\varepsilon_n$ from state $z_0 + 1$ to all the states $z \geq z_0 + 1$ to match with $\xi_n(z)$ . Let $t^\prime_z = z \xi_n(z)$ for $z \geq z_0+2$ and $T_3 = \sum_{z \geq z_0+2} t^\prime_z$ . Similarly to the construction in the proof of Lemma 5.1, we can now construct a trajectory $\varphi^{(3)}$ on $[0,T_3]$ such that $\varphi^{(3)}_0 = \varphi^{(2)}_{T_2}$ , $\varphi^{(3)}_{T_3}(z) = \xi_n(z)$ for each $z \geq z_0+1$ , and $S_{[0,T_3]}(\varphi^{(3)}|\varphi^{(2)}_{T_2}) \leq C_3(M, \overline{\lambda}, \underline{\lambda}) \varepsilon$ for some constant $C_3(M, \overline{\lambda}, \underline{\lambda})$ depending on M, $\overline{\lambda}$ , and $\underline{\lambda}$ , for all $n \geq n_1$ . Indeed, using the bounds in (5.2) and (5.3), the total cost is bounded by the order of

\begin{align*}\sum_{z > z_0 + 1} \left( (z \log z) \xi_n(z) + \frac{\log z}{z^2} \right) \leq \frac{\varepsilon}{3} + \varepsilon,\end{align*}

where the inequality follows from the choice of $z_0$ .

Finally, we construct a trajectory that connects $\varphi^{(3)}_{T_3}$ to $\xi_n$ by adjusting the mass within the states $\{0,1,\ldots, z_0\}$ . Note that $\varphi^{(3)}_{T_3}(z) = \xi_n(z)$ for each $z \geq z_0+1$ . Let $\mathcal{Z}_0 \subset \{1,2,\ldots, z_0\}$ denote the set of all $z \in \{1,2,\ldots, z_0\}$ such that $\varphi^{(3)}_{T_3}(z) > \xi_n(z)$ . Similarly to the construction of $\varphi^{(1)}$ , for each $z \in \mathcal{Z}_0$ , we move the mass $\varphi^{(3)}_{T_3}(z) - \xi_n(z)$ from state z to state 0 using unit velocity over a time duration $\varphi^{(3)}_{T_3}(z) - \xi_n(z)$ . Once these mass transfers are complete, starting with $z =1$ , we move the mass

\begin{align*}\sum_{z^\prime \geq z, z^\prime \notin \mathcal{Z}_0, z^\prime \leq z_0} (\xi_n(z^\prime) - \varphi_{T_3}^{(3)}(z^\prime))\end{align*}

from state $z-1$ to state z at unit rate. Let

\begin{align*}T_4 = \sum_{z \in \mathcal{Z}_0}(\varphi^{(3)}_{T_3}(z) - \xi_n(z)) + \sum_{z \notin \mathcal{Z}_0, z \leq z_0} (\xi_n(z) - \varphi^{(3)}_{T_3}(z)),\end{align*}

and let $\varphi^{(4)}$ denote this piecewise-constant-velocity trajectory. Let $\tilde{\varepsilon}_n = \sum_{z \notin \mathcal{Z}_0, z \leq z_0} (\xi_n(z) - \varphi_{T_3}^{(3)}(z))$ . At each step of $\varphi^{(4)}$ , since we move a mass of at most $\tilde{\varepsilon}_n$ from state $z-1$ to state z, the cost of $\varphi^{(4)}$ is at most of the order of

\begin{align*}z_0 \tilde{\varepsilon}_n \log \left(\frac{1}{\tilde{\varepsilon}_n}\right) + (z_0 \log z_0) \tilde{\varepsilon}_n\end{align*}

(see (5.2)). Since $\tilde{\varepsilon}_n \to 0$ as $n \to \infty$ , we may choose $n_2 \geq n_1$ so that $\tilde{\varepsilon}_n \leq \varepsilon/(z_0 \log z_0)$ for all $n \geq n_2$ . Therefore, for $n \geq n_2$ , the above display is bounded by

\begin{align*}z_0 \frac{\varepsilon}{z_0 \log z_0} \log \left(\frac{z_0 \log z_0}{\varepsilon}\right)+\varepsilon \leq \varepsilon \log (1/\varepsilon) + 3 \varepsilon,\end{align*}

which is $O(\varepsilon \log (1/\varepsilon))$ . Therefore, $S_{[0,T_4]}(\varphi^{(4)} | \varphi^{(3)}_{T_3}) \leq C_4(M, \overline{\lambda}, \underline{\lambda}) \varepsilon \log (1/\varepsilon)$ for all $n \geq n_2$ , for some constant $C_4(M, \overline{\lambda}, \underline{\lambda})$ depending on M, $\overline{\lambda}$ , and $\underline{\lambda}$ .

Let $T = \sum_{i=1}^4 T_i$ . We now append the four paths $\varphi^{(i)}$ , $1 \leq i \leq 4$ , constructed in the previous paragraphs over the time duration [0, T] to get a path $\varphi$ such that $\varphi_0 = \xi$ , $\varphi_T = \xi_n$ , and $S_{[0,T]}(\varphi | \xi) \leq C(M, \overline{\lambda}, \underline{\lambda}) \varepsilon \log (1/\varepsilon)$ , where $C(M, \overline{\lambda}, \underline{\lambda})$ is a constant depending on M, $\overline{\lambda}$ and $\underline{\lambda}$ . Hence, for each $n \geq n_2$ , we have

\begin{align*}V(\xi_n) \leq V(\xi) + S_{[0,T_4]}(\varphi | \xi) \leq V(\xi) + C(M, \overline{\lambda}, \underline{\lambda}) \varepsilon \log (1/\varepsilon).\end{align*}

Therefore, $\limsup_{n \to \infty} V(\xi_n) \leq V(\xi) + C(M, \overline{\lambda}, \underline{\lambda}) \varepsilon \log (1/\varepsilon)$ . Letting $\varepsilon \to 0$ and noting that $\varepsilon \log (1/\varepsilon) \to 0$ , we arrive at $\limsup_{n \to \infty} V(\xi_n) \leq V(\xi)$ .

To prove $\liminf_{n \to \infty} V(\xi_n) \geq V(\xi)$ , we reverse the role of $\xi_n$ and $\xi$ in the above argument. That is, we construct a trajectory $\varphi$ on [0, T] such that $\varphi_0 = \xi_n$ , $\varphi_T = \xi$ , and $S_{[0,T]}(\varphi | \xi_n) \leq \varepsilon_n$ for all $n \geq 1$ , where $\varepsilon_n \to 0$ as $n \to \infty$ . Thus, we get

\begin{align*}V(\xi) \leq V(\xi_n) + \varepsilon_n.\end{align*}

Letting $n \to \infty$ , we conclude that $\liminf_{n \to \infty} V(\xi_n) \geq V(\xi)$ . This completes the proof of the lemma.

Proof. We first prove an inclusion property of the level sets of V, namely, that given $M> 0$ there exists $M^\prime > 0$ such that

(5.8) \begin{align} \{\xi \in \mathcal{M}_1(\mathcal{Z})\,:\, V(\xi) \leq M\} \subset \mathscr{K}_{M^\prime}.\end{align}

On one hand, using Proposition 1.1 on the exponential tightness of the family $\{\wp^N, N \geq 1\}$ , we can choose $M^\prime > 0$ (see (3.4)) such that

\begin{align*}\limsup_{N \to \infty} \frac{1}{N} \log \wp^N(\!\sim \hspace{-0.4em} \mathscr{K}_{M^\prime}) \leq -(M+1).\end{align*}

On the other hand, using the LDP lower bound established in Lemma 4.2 and the compactness of $\mathscr{K}_{M^\prime}$ , we have

\begin{align*}\liminf_{N \to \infty} \frac{1}{N} \log \wp^N(\!\sim \hspace{-0.4em} \mathscr{K}_{M^\prime}) \geq -\inf_{\xi \notin \mathscr{K}_{M^\prime}} V(\xi).\end{align*}

Combining the above two displays, we get

\begin{align*}-\inf_{\xi \notin \mathscr{K}_{M^\prime}} V(\xi) \leq \liminf_{N \to \infty} \frac{1}{N} \log \wp^N(\!\sim \hspace{-0.4em} \mathscr{K}_{M^\prime}) \leq \limsup_{N \to \infty} \frac{1}{N} \log \wp^N(\!\sim \hspace{-0.4em} \mathscr{K}_{M^\prime}) \leq -(M+1).\end{align*}

That is, $\xi \notin \mathscr{K}_{M^\prime}$ implies $V(\xi) \geq M+1 > M$ . This shows (5.8). By Prokhorov’s theorem, $\mathscr{K}_M$ is a compact subset of $\mathcal{M}_1(\mathcal{Z})$ ; hence (5.8) shows that $\Xi(s)$ is precompact for each $s > 0$ .

We now show that $\Xi(s)$ is closed in $\mathcal{M}_1(\mathcal{Z})$ . Let $\xi_n \in \Xi(s)$ for each $n \geq 1$ , and let $\xi_n \to \xi$ in $\mathcal{M}_1(\mathcal{Z})$ as $n \to \infty$ . By Fatou’s lemma, we have $\liminf_{n \to \infty} \langle \xi_n, \vartheta \rangle \geq \langle \xi, \vartheta \rangle$ . Hence, by Remark 5.1, we have $\liminf_{n \to \infty} V(\xi_n) \geq V(\xi)$ . Thus, $\xi \in \Xi(s)$ . This completes the proof of the lemma.