Definition 2.1 (Continuity equation). A pair of measures $(\boldsymbol{\mu }, \boldsymbol{\nu }) \in \mathcal{M}_+\big ((0,1) \times{\mathbb{T}}^d\big ) \times \mathcal{M}^d\big ((0,1) \times{\mathbb{T}}^d\big )$ is said to be a solution to the continuity equation

\begin{align*} \partial _t \boldsymbol{\mu } + \nabla \cdot \boldsymbol{\nu } = 0 \end{align*}

if, for all functions $\varphi \in \mathcal{C}_c^1\big ((0,1)\times{\mathbb{T}^d}\big )$ , the identity

\begin{align*} \int _{(0,1) \times{\mathbb{T}^d}} \partial _t \varphi{\, \mathrm{d}} \boldsymbol{\mu } + \int _{(0,1) \times{\mathbb{T}^d}} \nabla \varphi \cdot{\, \mathrm{d}} \boldsymbol{\nu } = 0 \end{align*}

holds. We use the notation $(\boldsymbol{\mu }, \boldsymbol{\nu }) \in \mathsf{CE}$ .

Definition 2.3 (Action functionals). We define the action functionals by

\begin{align*} &{\mathbb{A}} : \mathcal{M}_+\big ( (0,1) \times{\mathbb{T}^d} \big ) \times \mathcal{M}^d\big ( (0,1) \times{\mathbb{T}}^d\big ) \to \mathbb{R} \cup \{ +\infty \}, \\ &{\mathbb{A}}(\boldsymbol{\mu }, \boldsymbol{\nu }) \,:\!=\, \int _{(0,1) \times{\mathbb{T}^d}} f \big ( \rho, j \big ){\, \mathrm{d}} \mathscr{L}^{\,d+1} + \int _{ (0,1) \times{\mathbb{T}^d}} f^\infty \big ( \rho ^\perp, j^\perp \big ){\, \mathrm{d}} \boldsymbol{\sigma }, \\ &{\mathbb{A}}(\boldsymbol{\mu }) \,:\!=\, \inf _{\boldsymbol{\nu }} \left \{{\mathbb{A}}(\boldsymbol{\mu }, \boldsymbol{\nu }) \, : \, (\boldsymbol{\mu }, \boldsymbol{\nu }) \in \mathsf{CE} \right \} . \end{align*}

Remark 2.4. This definition does not depend on the choice of $\boldsymbol{\sigma }$ , due to the $1$ -homogeneity of $f^\infty$ . As $f(\rho, j) \geq - C(1+\rho )$ and $f^\infty (\rho, j) \geq - C \rho$ from (2.1) and (2.3), the fact that $\boldsymbol{\mu }\bigl ((0,1) \times{\mathbb{T}^d} \bigr ) \lt \infty$ ensures that ${\mathbb{A}}(\boldsymbol{\mu }, \boldsymbol{\nu })$ is well defined in $\mathbb{R} \cup \{ + \infty \}$ .

Remark 2.5 (Lack of compatible compactness). We know from [[Reference Gladbach, Kopfer, Maas and Portinale23], Lemma 3.14] that $(\boldsymbol{\mu },\boldsymbol{\nu }) \mapsto{\mathbb{A}}(\boldsymbol{\mu },\boldsymbol{\nu })$ and $\boldsymbol{\mu } \mapsto{\mathbb{A}}(\boldsymbol{\mu })$ are lower semicontinuous w.r.t. the weak topology. Moreover, if $\boldsymbol{\mu }^n$ is a sequence with

\begin{align*} \sup _n{\mathbb{A}}(\boldsymbol{\mu }^n) \lt \infty \quad \text{and} \quad \sup _n \boldsymbol{\mu }^n\bigl ((0,1) \times{\mathbb{T}}^d\bigr ) \lt \infty \end{align*}

then $\boldsymbol{\mu }^n$ is weakly compact and any limit $\boldsymbol{\mu }$ belongs to $\mathrm{BV}_{\mathrm{KR}}\bigl ((0,1); \mathcal{M}_+({\mathbb{T}^d})\bigr )$ . This can be proved as in [[Reference Gladbach, Kopfer, Maas and Portinale23], Theorem 5.4]. Nonetheless, this property does not ensure the lower semicontinuity of $\mathbb{MA}$ because weak convergence does not preserve the boundary conditions (at time $t=0$ and $t=1$ ). For similar issues in the setting of functionals of $\mathbb{R}^d$ -valued curves with bounded variations and their minimisation, see e.g. [Reference Amar and Vitali2].

Remark 2.6 (Superlinear growth). Under the strengthened assumption of superlinear growth on $f$ (with respect to the momentum variable), it is possible to prove the lower semicontinuity property (2.5), in the same way as in the proof of the discrete-to-continuum $\Gamma$ -convergence of boundary-value problems of [[Reference Gladbach, Kopfer, Maas and Portinale23], Theorem 5.10]. More precisely, we say that $f$ is of superlinear growth if there exists a function $\theta :[0,\infty ) \to [0,\infty )$ with $\lim _{t \to \infty } \frac{\theta (t)}{t} = \infty$ and a constant $C \in \mathbb{R}$ such that

\begin{align*} f(\rho, j) \geq (\rho +1) \theta \left ( \frac{\lvert j\rvert }{\rho +1} \right ) - C (\rho +1), \qquad \forall \rho \in \mathbb{R}_+, \ \ j \in \mathbb{R}^d . \end{align*}

Arguing as in [[Reference Gladbach, Kopfer, Maas and Portinale23], Remark 5.6], one shows that any function of superlinear growth must satisfy the growth condition given by Assumption 2.2. Moreover, in this case, the recession function satisfies $f^\infty (0,j)=+\infty$ , for every $j \neq 0$ . See [[Reference Gladbach, Kopfer, Maas and Portinale23], Examples 5.7 & 5.8] for some examples belonging to this class. By arguing similarly as in the proof of [[Reference Gladbach, Kopfer, Maas and Portinale23], Theorem 5.9], assuming superlinear growth one can show that if $\boldsymbol{\mu }^n$ is a sequence with bounded action ${\mathbb{A}}(\boldsymbol{\mu }^n)$ and bounded total mass $\boldsymbol{\mu }^n \bigl ((0,1) \times{\mathbb{T}}^d \bigr )$ , then, up to a (non-relabelled) subsequence, we have $\boldsymbol{\mu }^n \to \boldsymbol{\mu }$ in $\mathcal{M}_+((0,1)\times{\mathbb{T}^d})$ and $\mu _t^n \to \mu _t$ in $\mathrm{KR}$ norm uniformly in $t \in (0,1)$ , with limit curve $\boldsymbol{\mu } \in W_{\mathrm{KR}}^{1,1}((0,1); \mathcal{M}_+({\mathbb{T}^d}))$ . Using this fact, it is clear that the problem of “jumps” in the limit explained in Remark 2.5 does not occur, and the lower semicontinuity (2.5) directly follows from the lower semicontinuity of $\mathbb{A}$ .

Remark 2.7. (Nonnegativity) Without loss of generality, we can assume that $f \geq 0$ . Indeed, thanks to the linear growth assumption 2.2, we can define a new function

(2.6) \begin{align} \widetilde f(\rho, j) \,:\!=\, f(\rho, j) + C(\rho +1) \geq c|j| \geq 0 \end{align}

which is now nonnegative and with (at least) linear growth. Furthermore, we can compute the recess $\widetilde f^\infty$ and from the definition we see that

(2.7) \begin{align} \widetilde f^\infty (\rho, j) = f^\infty (\rho, j) + C \rho . \end{align}

Denote by $\widetilde{\mathbb{A}}$ the action functional obtained by replacing $f$ with $\widetilde f$ . Thanks to (2.6), (2.7), we have that

\begin{align*} \widetilde{\mathbb{A}}(\boldsymbol{\mu }) \,:\!=\, &\inf _{\boldsymbol{\nu }} \left \{ \widetilde{\mathbb{A}}(\boldsymbol{\mu }, \boldsymbol{\nu }) \, : \, (\boldsymbol{\mu }, \boldsymbol{\nu }) \in \mathsf{CE} \right \}\\ = &\inf _{\boldsymbol{\nu }} \left \{{\mathbb{A}}(\boldsymbol{\mu }, \boldsymbol{\nu }) \, : \, (\boldsymbol{\mu }, \boldsymbol{\nu }) \in \mathsf{CE} \right \} + C(\boldsymbol{\mu }\big ( (0,1) \times{\mathbb{T}}^d\big )+1). \end{align*}

It follows that the corresponding boundary value problems are given by

\begin{align*} \widetilde{\mathbb{MA}}(\mu _0, \mu _1) ={\mathbb{MA}}(\mu _0, \mu _1) + C(\mu _0({\mathbb{T}^d})+1), \quad \text{if } \mu _0({\mathbb{T}^d}) = \mu _1({\mathbb{T}^d}) . \end{align*}

Therefore, the (weak) lower semicontinuity for $\widetilde{\mathbb{MA}}$ is equivalent to that of $\mathbb{MA}$ .

Remark 2.9. Important examples that satisfy the growth condition (2.8) are of the form

\begin{align*} F(m,J) = \frac 12 \sum _{(x,y) \in \mathcal{E}^Q} \frac{|J(x,y)|^p}{\Lambda \big (q_{xy} m(x), q_{yx} m(y)\big )^{p-1}}, \end{align*}

where $1 \leq p \lt \infty$ , the constants $q_{xy}, q_{yx} \gt 0$ are fixed parameters defined for $(x,y)\in \mathcal{E}^Q$ , and $\Lambda$ is a suitable mean. Functions of this type naturally appear in discretisations of Wasserstein gradient-flow structures [Reference Maas32], [Reference Mielke33], [Reference Chow, Huang, Li and Zhou6], see also [[Reference Gladbach, Kopfer, Maas and Portinale23], Remark2.6].

Definition 2.10 (Discrete energy functional). The rescaled energy is defined by

\begin{align*} \mathcal{F}_{\varepsilon } : \mathbb{R}_+^{\mathcal{X}_{\varepsilon }} \times \mathbb{R}_a^{\mathcal{E}_{\varepsilon }} &\to \mathbb{R} \cup \{ + \infty \}, \qquad (m, J) \stackrel{\mathcal{F}_{\varepsilon }}{\longmapsto } \sum _{z \in \mathbb{Z}_{\varepsilon }^d} \varepsilon ^d F\bigg ( \frac{\tau _{\varepsilon }^z m}{\varepsilon ^d}, \frac{\tau _{\varepsilon }^z J}{\varepsilon ^{d-1}} \bigg ) . \end{align*}

Remark 2.11. As observed in [[Reference Gladbach, Kopfer, Maas and Portinale23], Remark 2.8], the value $\mathcal{F}_{\varepsilon }(m,J)$ is well defined as an element in $\mathbb{R} \cup \{ + \infty \}$ . Indeed, the condition (2.8) yields

\begin{align*} \mathcal{F}_{\varepsilon }( m, J ) = \sum _{z \in \mathbb{Z}_{\varepsilon }^d} \varepsilon ^d F\bigg ( \frac{\tau _{\varepsilon }^z m}{\varepsilon ^d}, \frac{\tau _{\varepsilon }^z J}{\varepsilon ^{d-1}} \bigg ) & \geq - C \sum _{z \in \mathbb{Z}_{\varepsilon }^d} \varepsilon ^d \Bigg ( 1 + \sum _{\substack{ x\in \mathcal{X}\\ |x_{\mathsf{z}}|_{\infty } \leq R_{\mathrm{max}} }} \frac{\tau _{\varepsilon }^z m(x)}{\varepsilon ^d} \Bigg ) \\& \geq - C \bigg ( 1 + (2R_{\mathrm{max}} + 1)^d \sum _{x \in \mathcal{X}_{\varepsilon }} m(x) \bigg ) \gt - \infty . \end{align*}

Definition 2.12 (Discrete continuity equation). A pair $({\boldsymbol{m}}, \boldsymbol{J})$ is said to be a solution to the discrete continuity equation if ${\boldsymbol{m}} : (0,1) \to \mathbb{R}_+^{\mathcal{X}_{\varepsilon }}$ is continuous, $\boldsymbol{J} : (0,1) \to \mathbb{R}_a^{\mathcal{E}_{\varepsilon }}$ is Borel measurable, and

(2.9) \begin{align} \partial _t m_t(x) + \sum _{y\sim x}J_t(x,y) = 0 \end{align}

holds for all $x \in \mathcal{X}_{\varepsilon }$ in the sense of distributions. We use the notation

\begin{align*} ({\boldsymbol{m}}, \boldsymbol{J}) \in \mathcal{CE}_{\varepsilon } . \end{align*}

Remark 2.13. We may write (2.9) as $ \partial _t m_t + \mathsf{div} J_t = 0$ using the discrete divergence operator, given by

\begin{align*} \mathsf{div} J \in \mathbb{R}^{\mathcal{X}_{\varepsilon }}, \qquad \mathsf{div} J(x) \,:\!=\, \sum _{y \sim x} J(x,y), \qquad \forall J \in \mathbb{R}_a^{\mathcal{E}_{\varepsilon }} . \end{align*}

Definition 2.15 (Discrete action functional). For any continuous function ${\boldsymbol{m}}: (0,1) \to \mathbb{R}_+^{\mathcal{X}_{\varepsilon }}$ such that $t \mapsto \sum _{x \in \mathcal{X}_{\varepsilon }} m_t(x) \in L^1\bigl ((0,1)\bigr )$ and any Borel measurable function $\boldsymbol{J}: (0,1) \to \mathbb{R}_a^{\mathcal{E}_{\varepsilon }}$ , we define

\begin{align*} \mathcal{A}_{\varepsilon }({\boldsymbol{m}}, \boldsymbol{J}) & \,:\!=\, \int _0^1 \mathcal{F}_{\varepsilon }( m_t, J_t ){\, \mathrm{d}} t \in \mathbb{R} \cup \{ + \infty \} . \end{align*}

Furthermore, we set

\begin{align*} \mathcal{A}_{\varepsilon }({\boldsymbol{m}}) & \,:\!=\, \inf _{\boldsymbol{J}} \Big \{ \mathcal{A}_{\varepsilon }({\boldsymbol{m}}, \boldsymbol{J}) \ : \ ({\boldsymbol{m}}, \boldsymbol{J}) \in \mathcal{CE}_{\varepsilon } \Big \} . \end{align*}

Definition 2.16 (Minimal discrete action functional). For any pair of boundary data $m_0$ , $m_1 \in \mathbb{R}_+^{\mathcal{X}_{\varepsilon }}$ , we define the associated discrete boundary value problem as

\begin{align*}{\mathcal{MA}}_{\varepsilon }(m_0,m_1) & \,:\!=\, \inf \left \{ \mathcal{A}_{\varepsilon }({\boldsymbol{m}} ) \, : \,{\boldsymbol{m}}: (0,1) \to \mathbb{R}_+^{\mathcal{X}_{\varepsilon }}, \quad{\boldsymbol{m}}_{t=0} = m_0 \quad \text{ and }\quad{\boldsymbol{m}}_{t=1} = m_1 \right \} . \end{align*}

Proof. We may and will assume that ${\mathbb{MA}}_{\mathrm{hom}}(\mu _0,\mu _1) \lt \infty$ . We also claim that it suffices to prove the statement with ${\mathbb{MA}}(\mu _0,\mu _1)+1/k$ in place of the right-hand side of (3.4) for every $k \in \mathbb{N}_1$ . Indeed, assume that we know of the existence of sequences $(m_i^{\varepsilon, k})_{\varepsilon }$ such that $m_i^{\varepsilon, k} \to \mu _i$ and

\begin{align*} \limsup _{\varepsilon \to 0}{\mathcal{MA}}_{\varepsilon }(m_0^{\varepsilon, k},m_1^{\varepsilon, k}) \le{\mathbb{MA}}(\mu _0,\mu _1)+1/k, \end{align*}

for every $k \in \mathbb{N}_1$ . Since ${\mathbb{T}}^d$ is compact, the weak convergence is equivalent to convergence in the Kantorovich–Rubinstein norm. Hence, for every $k$ , we can find $\varepsilon _k$ such that when $\varepsilon \le \varepsilon _k$ ,

\begin{align*}{\mathcal{MA}}_{\varepsilon }(m_0^{\varepsilon, k},m_1^{\varepsilon, k}) \le{\mathbb{MA}}_{\mathrm{hom}}(\mu _0,\mu _1)+2/k \quad \text{ and }\quad \max _{i=0,1} \|{\iota _{\varepsilon } m_i^{\varepsilon, k}-\mu _i}\|_{\mathrm{KR}} \le 1/k . \end{align*}

We can also assume that $\varepsilon _{k+1} \le \frac{\varepsilon _k}{2}$ , for every $k$ . It now suffices to set

\begin{align*} k_{\varepsilon } \,:\!=\, \max \{ k \in \mathbb{N}_1 \, : \, \varepsilon _k \ge \varepsilon \} \quad \text{ and }\quad m_i^{\varepsilon } \,:\!=\, m_i^{\varepsilon, k_{\varepsilon }}, \end{align*}

for every $\varepsilon$ and $i=0,1$ to get

\begin{align*} \limsup _{\varepsilon \to 0}{\mathcal{MA}}_{\varepsilon }(m_0^{\varepsilon },m_1^{\varepsilon }) \le{\mathbb{MA}}_{\mathrm{hom}} (\mu _0,\mu _1)+ \limsup _{\varepsilon \to 0} \frac{2}{k_{\varepsilon }} \end{align*}

as well as

\begin{align*} \limsup _{\varepsilon \to 0} \max _{i=0,1} \|{\iota _{\varepsilon } m_i^{\varepsilon } -\mu _i}\|_{\mathrm{KR}} \le \limsup _{\varepsilon \to 0} \frac{1}{k_{\varepsilon }} . \end{align*}

The claim is proved, since $k_{\varepsilon } \to _{\varepsilon } \infty$ , as can be readily verified.

Thus, let us now choose $k$ and keep it fixed. By definition of ${\mathbb{MA}}_{\mathrm{hom}}$ , there exists $\boldsymbol{\mu } = \mu _t({\, \mathrm{d}} x){\, \mathrm{d}} t \in \mathrm{BV}_{\mathrm{KR}}\bigl ((0,1); \mathcal{M}_+({\mathbb{T}^d})\bigr )$ with $\boldsymbol{\mu }_{t=0} = \mu _0, \boldsymbol{\mu }_{t=1} = \mu _1$ and such that

\begin{align*}{\mathbb{A}}_{\mathrm{hom}}(\boldsymbol{\mu }) \le{\mathbb{MA}}_{\mathrm{hom}}(\mu _0,\mu _1)+1/k . \end{align*}

Recall from Remark 3.10 that $\mathcal{A}_{\varepsilon } \xrightarrow{\Gamma }{\mathbb{A}}_{\mathrm{hom}}$ ; in particular, there exists a recovery sequence $({\boldsymbol{m}}^{\varepsilon }, \boldsymbol{J}^{\varepsilon }) \in \mathcal{CE}_{\varepsilon }$ such that ${\boldsymbol{m}}^{\varepsilon } \to \boldsymbol{\mu }$ weakly and

\begin{align*} \limsup _{\varepsilon \to 0} \mathcal{A}_{\varepsilon }({\boldsymbol{m}}^{\varepsilon }, \boldsymbol{J}^{\varepsilon }) \le{\mathbb{A}}_{\mathrm{hom}}(\boldsymbol{\mu }) . \end{align*}

We shall prove that $\|{\iota _{\varepsilon } m_t^{\varepsilon } - \mu _t}\|_{\mathrm{KR}({\mathbb{T}}^d)} \to 0$ in ( $\mathscr{L}^1$ -)measure or, equivalently, that

(3.5) \begin{equation} \lim _{\varepsilon \to 0} \int _0^1 \min \left \{\|{\iota _{\varepsilon } m_t^{\varepsilon } - \mu _t}\|_{\mathrm{KR}({\mathbb{T}}^d)}, 1 \right \}{\, \mathrm{d}} t = 0 . \end{equation}

In order to do this, assume by contradiction that there exists a subsequence such that

\begin{align*} \int _0^1 \min \left \{\|{\iota _{\varepsilon _n} m_t^{\varepsilon _n} - \mu _t}\|_{\mathrm{KR}({\mathbb{T}}^d)}, 1 \right \}{\, \mathrm{d}} t \gt \delta, \qquad n \in \mathbb{N}, \end{align*}

for some $\delta \gt 0$ . Up to possibly extracting a further subsequence, it can be easily checked that the hypotheses of [[Reference Gladbach, Kopfer, Maas and Portinale23], Theorem 5.4] are satisfied (cfr. Remark 3.11); hence, there exists a further (not relabelled) subsequence such that, for almost every $t \in (0,1)$ , $m_t^{\varepsilon _n} \to \mu _t$ weakly and thus $\|{\iota _{\varepsilon _n}m_t^{\varepsilon _n} - \mu _t}\|_{\mathrm{KR}({\mathbb{T}}^d)} \to 0$ . The dominated convergence theorem yields an absurd.

From (3.5) we deduce that for every $T \in (0,1/2)$ there exists a sequence of times $(a_{\varepsilon }^T)_{\varepsilon } \subseteq (0,T)$ such that

\begin{align*} \lim _{\varepsilon \to 0} \|{\iota _{\varepsilon } m_{a_{\varepsilon }^T}^{\varepsilon } - \mu _{a_{\varepsilon }^T}}\|_{\mathrm{KR}({\mathbb{T}}^d)} = 0 . \end{align*}

With a diagonal argument, we find a sequence $(a_{\varepsilon })_{\varepsilon } \subseteq (0,1/2)$ such that

\begin{align*} \lim _{\varepsilon \to 0} a_{\varepsilon } = 0 \quad \text{ and }\quad \lim _{\varepsilon \to 0} \|{\iota _{\varepsilon } m_{a_{\varepsilon }}^{\varepsilon } - \mu _{a_{\varepsilon }}}\|_{\mathrm{KR}({\mathbb{T}}^d)} = \lim _{\varepsilon \to 0} \|{\iota _{\varepsilon } m_{a_{\varepsilon }}^{\varepsilon } - \mu _0}\|_{\mathrm{KR}({\mathbb{T}}^d)} = 0 . \end{align*}

Similarly, we can find another sequence $(b_{\varepsilon })_{\varepsilon } \subseteq (1/2,1)$ such that

\begin{align*} \lim _{\varepsilon \to 0} b_{\varepsilon } = 1 \quad \text{and} \quad \lim _{\varepsilon \to 0} \|{\iota _{\varepsilon } m_{b_{\varepsilon }}^{\varepsilon } - \mu _1}\|_{\mathrm{KR}({\mathbb{T}}^d)}=0 . \end{align*}

We claim the sought recovery sequences is provided by $m_0^{\varepsilon } \,:\!=\, m^{\varepsilon }_{a_{\varepsilon }}$ and $m_1^{\varepsilon } \,:\!=\, m_{b_{\varepsilon }}^{\varepsilon }$ . In order to show this, let us define $\widehat J^{\varepsilon } : \mathcal{E}_{\varepsilon } \to \mathbb{R}$ via the formulaFootnote ¹ (recall the assumption (3.3))

\begin{align*} \frac{\tau _{\varepsilon }^z \widehat J^{\varepsilon }}{\varepsilon ^{d-1}} \,:\!=\, \bar J, \qquad z \in \mathbb{Z}^d_{\varepsilon }, \end{align*}

so that $\widehat J^{\varepsilon }$ is divergence-free. Now define

\begin{align*} \widetilde m^{\varepsilon }_t \,:\!=\, \begin{cases} m^{\varepsilon }_{a_{\varepsilon }} &\text{if } t \in [0,a_{\varepsilon }) \\ m^{\varepsilon }_t &\text{if } t \in [a_{\varepsilon },b_{\varepsilon }] \\ m^{\varepsilon }_{b_{\varepsilon }} &\text{if } t \in (b_{\varepsilon },1] \end{cases} \quad \text{ and }\quad \widetilde J^{\varepsilon }_t \,:\!=\, \begin{cases} \widehat J^{\varepsilon } &\text{if } t \in [0,a_{\varepsilon }) \\ J^{\varepsilon }_t &\text{if } t \in [a_{\varepsilon },b_{\varepsilon }] \\ \widehat J^{\varepsilon } &\text{if } t \in (b_{\varepsilon },1] \end{cases} . \end{align*}

It is readily verified that $(\widetilde{\boldsymbol{m}}^{\varepsilon }, \widetilde{\boldsymbol{J}}^{\varepsilon })$ solves the continuity equation for every $\varepsilon$ . Therefore,

(3.6) \begin{align} {\mathcal{MA}}_{\varepsilon }(m_0^{\varepsilon },m_1^{\varepsilon }) &\leq \mathcal{A}_{\varepsilon }(\widetilde{\boldsymbol{m}}^{\varepsilon },\widetilde{\boldsymbol{J}}^{\varepsilon }) = \int _0^1 \sum _{z \in \mathbb{Z}^d_{\varepsilon }} \varepsilon ^d F\left ( \frac{\tau ^z_{\varepsilon } \widetilde m_t^{\varepsilon }}{\varepsilon ^d}, \frac{\tau ^z_{\varepsilon } \widetilde J^{\varepsilon }_t}{\varepsilon ^{d-1}} \right ) \,{\, \mathrm{d}} t\\ &= \int _0^{a_{\varepsilon }} \sum _{z \in \mathbb{Z}^d_{\varepsilon }} \varepsilon ^d F\left ( \frac{\tau ^z_{\varepsilon } m_{a_{\varepsilon }}^{\varepsilon }}{\varepsilon ^d}, \bar J \right ) \,{\, \mathrm{d}} t + \int _{a_{\varepsilon }}^{b_{\varepsilon }} \sum _{z \in \mathbb{Z}^d_{\varepsilon }} \varepsilon ^d F\left ( \frac{\tau ^z_{\varepsilon } m_t^{\varepsilon }}{\varepsilon ^d}, \frac{\tau ^z_{\varepsilon } J^{\varepsilon }_t}{\varepsilon ^{d-1}} \right ) \,{\, \mathrm{d}} t\nonumber \\ &\quad + \int _{b_{\varepsilon }}^1 \sum _{z \in \mathbb{Z}^d_{\varepsilon }} \varepsilon ^d F\left ( \frac{\tau ^z_{\varepsilon } m_{b_{\varepsilon }}^{\varepsilon }}{\varepsilon ^d}, \bar J \right ) \,{\, \mathrm{d}} t\nonumber \\ & =: I_1 + I_2 + I_3.\nonumber \end{align}

The first and last integral can be estimated using the assumption (3.3). Indeed,

\begin{align*} I_1+I_3 &\le C \sum _{z \in \mathbb{Z}^d_{\varepsilon }} \Biggl ((a_{\varepsilon }+1-b_{\varepsilon }) \varepsilon ^d + \sum _{\substack{x \in \mathcal X \\ |x_{\mathsf{z}}|_{\infty } \le R}} \left ( a_{\varepsilon } (\tau ^z_{\varepsilon } m_{a_{\varepsilon }}^{\varepsilon })(x) + (1-b_{\varepsilon }) (\tau ^z_{\varepsilon } m_{b_{\varepsilon }}^{\varepsilon })(x) \right )\Biggr ) \\ &\le C \left ( (a_{\varepsilon }+1-b_{\varepsilon }) + (2R+1)^d \sum _{x \in \mathcal{X}_{\varepsilon }} \left (a_{\varepsilon } m_{a_{\varepsilon }}^{\varepsilon }(x) + (1-b_{\varepsilon }) m_{b_{\varepsilon }}^{\varepsilon }(x)\right ) \right ) \\ &= C \left ( (a_{\varepsilon }+1-b_{\varepsilon }) + (2R+1)^d \left (a_{\varepsilon } \iota _{\varepsilon } m_{a_{\varepsilon }}^{\varepsilon }({\mathbb{T}}^d) + (1-b_{\varepsilon })\iota _{\varepsilon } m_{b_{\varepsilon }}^{\varepsilon }({\mathbb{T}}^d)\right ) \right ), \end{align*}

and in the limit we find

(3.7) \begin{align} \limsup _{\varepsilon \to 0} I_1+I_3 \le C\left (0 + (2R+1)^d(0\cdot \mu _0({\mathbb{T}}^d) + 0 \cdot \mu _1({\mathbb{T}}^d))\right ) = 0 . \end{align}

As for the second integral, thanks to Assumption 2.8(c) we have that

(3.8) \begin{align} I_2 - \mathcal A_{\varepsilon }({\boldsymbol{m}}^{\varepsilon },\boldsymbol{J}^{\varepsilon }) &= - \int _{(0,a_{\varepsilon })\cup (b_{\varepsilon },1)} \sum _{z \in \mathbb{Z}^d_{\varepsilon }} \varepsilon ^d F\left ( \frac{\tau ^z_{\varepsilon } m_t^{\varepsilon }}{\varepsilon ^d}, \frac{\tau ^z_{\varepsilon } J^{\varepsilon }_t}{\varepsilon ^{d-1}} \right ) \,{\, \mathrm{d}} t\\ &\le C' \left ((a_{\varepsilon }+1-b_{\varepsilon })+(2R_{\mathrm{max}}+1)^d \iota _{\varepsilon }{\boldsymbol{m}}^{\varepsilon }\left (\left ((0,a_{\varepsilon })\cup (b_{\varepsilon },1)\right ) \times{\mathbb{T}}^d\right ) \right ) .\nonumber \end{align}

Since $(\iota _{\varepsilon }{\boldsymbol{m}}^{\varepsilon })_{\varepsilon }$ converges weakly, for every $a, b \in (0,1)$ , we have that

\begin{align*} \limsup _{\varepsilon \to 0} \iota _{\varepsilon }{\boldsymbol{m}}^{\varepsilon } \left ( \left ( (0,a_{\varepsilon })\cup (b_{\varepsilon },1) \right ) \times{\mathbb{T}}^d \right ) &\leq \limsup _{\varepsilon \to 0} \iota _{\varepsilon }{\boldsymbol{m}}^{\varepsilon } \left ( \big ( (0,a] \cup [b,1) \big ) \times{\mathbb{T}^d} \right ) \\ &\leq \boldsymbol{\mu } \left ( \big ( (0,a] \cup [b,1) \big ) \times{\mathbb{T}^d} \right ) . \end{align*}

Using the fact that the previous estimate holds for every $a,b \in (0,1)$ , we obtain that

\begin{align*} \limsup _{\varepsilon \to 0} \iota _{\varepsilon }{\boldsymbol{m}}^{\varepsilon } \left ( \left ( (0,a_{\varepsilon })\cup (b_{\varepsilon },1) \right ) \times{\mathbb{T}}^d \right ) = 0 . \end{align*}

This, together with the estimate obtained in (3.8), gives us the inequality

(3.9) \begin{align} \limsup _{\varepsilon \to 0} I_2 \le \limsup _{\varepsilon \to 0} \mathcal{A}_{\varepsilon }({\boldsymbol{m}}^{\varepsilon }, \boldsymbol{J}^{\varepsilon }) . \end{align}

In conclusion, from (3.6), (3.7), and (3.9), we find

\begin{align*} \limsup _{\varepsilon \to 0}{\mathcal{MA}}_{\varepsilon }(m_0^{\varepsilon },m_1^{\varepsilon }) \le \limsup _{\varepsilon \to 0} \mathcal{A}_{\varepsilon }({\boldsymbol{m}}^{\varepsilon }, \boldsymbol{J}^{\varepsilon }) \le{\mathbb{A}}(\boldsymbol{\mu }) \le{\mathbb{MA}}(\mu _0,\mu _1) + 1/k, \end{align*}

which is sought upper bound.

Proof. Let $(\boldsymbol{\mu }^n,\boldsymbol{\nu }^n) \in \mathsf{CE}$ be (almost-)optimal solutions associated to ${\mathbb{MA}}(\mu _0^n,\mu _1^n)$ , i.e.

(3.26) \begin{align} \liminf _{n \to \infty }{\mathbb{MA}}(\mu _0^n,\mu _1^n) = \liminf _{n \to \infty }{\mathbb{A}}(\boldsymbol{\mu }^n,\boldsymbol{\nu }^n) . \end{align}

With no loss of generality, we can assume $\sup _n{\mathbb{A}}(\boldsymbol{\mu }^n, \boldsymbol{\nu }^n) \lt \infty$ and that the limits inferior are true limits. By the compactness of Remark 2.5, we know that, up to a non-relabelled subsequence, $\boldsymbol{\mu }^n \to \boldsymbol{\mu }$ weakly in $\mathcal{M}_+\bigl ( (0,1) \times{\mathbb{T}^d} \bigr )$ . Moreover, we also have ${\, \mathrm{d}} \boldsymbol{\mu }(t,x) = \mu _t({\, \mathrm{d}} x){\, \mathrm{d}} t \in \mathrm{BV}_{\mathrm{KR}}\bigl ((0,1); \mathcal{M}_+({\mathbb{T}^d})\bigr )$ for some measurable curve $t \mapsto \mu _t \in \mathcal{M}_+({\mathbb{T}^d})$ of constant, finite mass. Once again, due to the lack of continuity of the trace operators in $\mathrm{BV}$ , we cannot ensure that $\boldsymbol{\mu }_{t=0} = \mu _0$ and $\boldsymbol{\mu }_{t=1} = \mu _1$ . To solve this issue, we rescale our measures $\boldsymbol{\mu }^n$ in time in the same spirit as in (3.12). For a given $\delta \gt 0$ , we define $\mathcal{I}_\delta \,:\!=\, (\delta, 1-\delta )$ and $\boldsymbol{\mu }^{n,\delta } \in \mathrm{BV}_{\mathrm{KR}}\bigl ((0,1); \mathcal{M}_+({\mathbb{T}^d})\bigr )$ as

\begin{align*} \mu _t^{n,\delta } \,:\!=\, \begin{cases} \mu _0^n &\text{if } t \in (0,\delta ] \\ \mu _{\frac{t-\delta }{1-2\delta }}^n &\text{if } t \in \mathcal{I}_\delta \\ \mu _1^n &\text{if } t \in [1-\delta, 1) \end{cases}, \quad{\, \mathrm{d}} \boldsymbol{\mu }^{n,\delta }(t,x) \,:\!=\, \mu _t^{n,\delta }({\, \mathrm{d}} x){\, \mathrm{d}} t . \end{align*}

By construction, it is not hard to see that $\boldsymbol{\mu }^{n,\delta } \to \widehat{\boldsymbol{\mu }}^\delta$ weakly, where

\begin{align*} \widehat \mu _t^\delta \,:\!=\, \begin{cases} \mu _0 &\text{if } t \in (0,\delta ] \\ \mu _{\frac{t-\delta }{1-2\delta }} &\text{if } t \in \mathcal{I}_\delta \\ \mu _1 &\text{if } t \in [1-\delta, 1) \end{cases}, \quad{\, \mathrm{d}} \widehat{\boldsymbol{\mu }}^\delta (t,x) \,:\!=\, \widehat \mu _t^\delta ({\, \mathrm{d}} x){\, \mathrm{d}} t . \end{align*}

We stress that, as in (3.13), the rescaled curve $t \mapsto \widehat \mu _t^\delta$ could have discontinuities at $t=\delta$ and $t= 1-\delta$ , corresponding to the possible jumps in the limit as $n \to \infty$ for $\boldsymbol{\mu }^n$ at $\{0,1\}$ . Nevertheless, $\widehat{\boldsymbol{\mu }}^\delta$ is a competitor for ${\mathbb{MA}}(\mu _0,\mu _1)$ , which, by lower semicontinuity of $\mathbb{A}$ , ensures that

(3.27) \begin{align} \liminf _{n \to \infty }{\mathbb{A}}(\boldsymbol{\mu }^{n,\delta }) \geq{\mathbb{A}}(\widehat{\boldsymbol{\mu }}^\delta ) \geq{\mathbb{MA}}(\mu _0,\mu _1) . \end{align}

In order to estimate the left-hand side of the latter displayed equation, we seek a suitable vector meausure $\boldsymbol{\nu }^{n,\delta }$ so that $(\boldsymbol{\mu }^{n,\delta }, \boldsymbol{\nu }^{n,\delta }) \in \mathsf{CE}$ and whose action ${\mathbb{A}}(\boldsymbol{\mu }^{n,\delta },\boldsymbol{\nu }^{n,\delta })$ is comparable with ${\mathbb{A}}(\boldsymbol{\mu }^n,\boldsymbol{\nu }^n)$ for small $\delta \gt 0$ .

It is useful to introduce the following notation: for $\delta \in (0,1/2)$ ,

\begin{align*} r_\delta : (0,1) \to \mathcal{I}_\delta, \quad r_\delta (s) \,:\!=\, (1-2\delta )s + \delta, \end{align*}

\begin{align*} R_\delta :(0,1) \times{\mathbb{T}^d} \to \mathcal{I}_\delta \times{\mathbb{T}^d}, \quad R_\delta (s,x) \,:\!=\, (r_\delta (s), x) . \end{align*}

Define $\widehat \iota _\delta : \mathcal{M}^d(\mathcal{I}_\delta \times{\mathbb{T}^d}) \to \mathcal{M}^d\bigl ((0,1)\times{\mathbb{T}^d}\bigr )$ as the natural embedding obtained by extending to $0$ any measure outside $\mathcal{I}_\delta$ , and set

\begin{align*} \boldsymbol{\nu }^{n,\delta } \,:\!=\, \widehat \iota _\delta \big [ (R_\delta )_{\#} \boldsymbol{\nu }^n \big ] \in \mathcal{M}^d((0,1)\times{\mathbb{T}^d}) . \end{align*}

The proof that $( \boldsymbol{\mu }^{n,\delta }, \boldsymbol{\nu }^{n,\delta } ) \in \mathsf{CE}$ works as in (3.16). In the same spirit as in (3.15), we claim that

(3.28) \begin{align} {\mathbb{A}}(\boldsymbol{\mu }^{n,\delta }, \boldsymbol{\nu }^{n,\delta }) \leq \bigl (1+C(f)\delta \bigr ){\mathbb{A}}(\boldsymbol{\mu }^n, \boldsymbol{\nu }^n) + C(f) \delta \Big ( 1 + \boldsymbol{\mu }^n ((0,1) \times{\mathbb{T}^d}) \Big ), \end{align}

where $C(f) \in \mathbb{R}_+$ only depends on $f$ . The combination of (3.26), (3.27) and (3.28), and the arbitrariness of $\delta$ would then suffice to conclude the proof.

We are left with the proof of the claim (3.28), which is a bit more involved, compared to that of (3.15), due to the presence of the singular part at the continuous level. We need the following.

Proof. By assumption, $\boldsymbol{\sigma }$ is singular with respect to $\mathscr{L}^{\,d+1}$ , which means there exists a set $A \subset (0,1) \times{\mathbb{T}^d}$ such that $\mathscr{L}^{\,d+1}(A) = 0 = \boldsymbol{\sigma }(A^c)$ . By the very definition of push-forward and the bijectivity of $R_\delta$ , we then have that

\begin{align*} (R_\delta )_{\#} \boldsymbol{\sigma } \big ((R_\delta (A))^c \big ) = \boldsymbol{\sigma } \Big ( R_\delta ^{-1} \big (R_\delta (A^c) \big ) \Big ) = \boldsymbol{\sigma }(A^c) = 0, \end{align*}

whereas, by the scaling properties of the Lebesgue measure, we have that $\mathscr{L}^{\,d+1}(R_\delta (A)) = (1-2\delta ) \mathscr{L}^{\,d+1}(A) = 0$ , which shows the claimed singularity. The second part of the lemma follows from the fact that $(R_\delta )_{\#} \mathscr{L}^{\,d+1} = (1-2\delta )^{-1} \mathscr{L}^{\,d+1}$ and the following statement: for every $\boldsymbol{\xi }' = f' \boldsymbol{\sigma }'$ with $\boldsymbol{\sigma }' \in \mathcal{M}_+((0,1) \times{\mathbb{T}^d})$ , we claim that

(3.29) \begin{align} \frac{{\, \mathrm{d}} (R_\delta )_{\#} \boldsymbol{\xi }'}{{\, \mathrm{d}} (R_\delta )_{\#} \boldsymbol{\sigma }'} (t,x) = f'(R_\delta ^{-1}(t,x)), \quad \forall (t,x) \in \mathcal{I}_\delta \times{\mathbb{T}^d} . \end{align}

Indeed, by definition of push-forward, we have for every test function $\varphi \in C_b$

\begin{align*} \int \varphi{\, \mathrm{d}} (R_\delta )_{\#} \boldsymbol{\xi }' = \int (\varphi \circ R_\delta ){\, \mathrm{d}} \boldsymbol{\xi }' = \int (\varphi \circ R_\delta ) f'{\, \mathrm{d}} \boldsymbol{\sigma }' = \int \varphi \cdot (f' \circ R_\delta ^{-1}){\, \mathrm{d}} (R_\delta )_{\#} \boldsymbol{\sigma }', \end{align*}

which indeed shows (3.29).

Let

\begin{align*} \boldsymbol{\mu }^n = \rho ^n{\, \mathrm{d}} \mathscr{L}^{\,d+1} + \rho ^{n,\perp }{\, \mathrm{d}} \boldsymbol{\sigma } \quad \text{ and }\quad \boldsymbol{\nu }^n = j^n{\, \mathrm{d}} \mathscr{L}^{\,d+1} + j^{n,\perp }{\, \mathrm{d}} \boldsymbol{\sigma } \end{align*}

be Lebesgue decompositions. We apply Lemma 3.16 to both $\boldsymbol{\mu }^n$ and $\boldsymbol{\nu }^n$ and find that, on $\mathcal{I}_\delta \times{\mathbb{T}^d}$ , we have

\begin{gather*} \boldsymbol{\mu }^{n,\delta } = \rho ^{n,\delta }{\, \mathrm{d}} \mathscr{L}^{\,d+1} + \rho ^{n, \delta, \perp }{\, \mathrm{d}} (R_\delta )_{\#} \boldsymbol{\sigma } \quad \text{ and }\quad \boldsymbol{\nu }^{n,\delta } = j^{n,\delta }{\, \mathrm{d}} \mathscr{L}^{\,d+1} + j^{n, \delta, \perp }{\, \mathrm{d}} (R_\delta )_{\#} \boldsymbol{\sigma }, \end{gather*}

with $(R_\delta )_{\#} \boldsymbol{\sigma }$ singular with respect to $\mathscr{L}^{\,d+1}$ and

(3.30) \begin{align} \rho ^{n,\delta }(t,x) &= \big ( \rho ^n \circ R_\delta ^{-1} \big ) (t,x), \qquad &&\rho ^{n, \delta, \perp }(t,x) = (1-2\delta ) \big ( \rho ^{n,\perp } \circ R_\delta ^{-1} \big ) (t,x), \\ j^{n,\delta }(t,x) &= \frac 1{1-2\delta } \big ( j^n \circ R_\delta ^{-1} \big ) (t,x), \qquad &&j^{n, \delta, \perp }(t,x) = \big ( j^{n,\perp } \circ R_\delta ^{-1} \big ) .\nonumber \end{align}

Further consider the Lebesgue decompositions

\begin{equation*} \mu _i^n = \rho _i^n {\, \mathrm {d}} \mathscr {L}^d + \rho _i^{n,\perp } {\, \mathrm {d}} \sigma _i, \qquad i \in \{0,1\} \end{equation*}

for some $\sigma _1,\sigma _2 \in \mathcal{M}_+({\mathbb{T}}^d)$ singular w.r.t. $\mathscr{L}^d$ . The action of $(\boldsymbol{\mu }^{n,\delta }, \boldsymbol{\nu }^{n,\delta })$ is given byFootnote ²

\begin{align*}{\mathbb{A}}(\boldsymbol{\mu }^{n,\delta }, \boldsymbol{\nu }^{n,\delta }) &={\mathbb{A}}^{\mathcal{I}_\delta } (\boldsymbol{\mu }^{n,\delta }, \boldsymbol{\nu }^{n,\delta }) + \sum _{i=0,1} \delta \Big ( \int _{{\mathbb{T}^d}} f(\rho _i^n,0){\, \mathrm{d}} \mathscr{L}^d + \int _{{\mathbb{T}^d}} f^\infty (\rho _i^{n,\perp },0){\, \mathrm{d}} \sigma _i \Big ), \end{align*}

where we used the notation

\begin{equation*} {\mathbb {A}}^{\mathcal {I}_\delta }(\boldsymbol {\mu }^{n,\delta }, \boldsymbol {\nu }^{n,\delta }) \,:\!=\, \int _{\mathcal {I}_\delta \times {\mathbb {T}^d}} f(\rho ^{n,\delta }, j^{n,\delta }) {\, \mathrm {d}} \mathscr {L}^{d+1} + \int _{ \mathcal {I}_\delta \times {\mathbb {T}^d}} f^\infty ( \rho ^{n,\delta, \perp }, j^{n,\delta, \perp } ) {\, \mathrm {d}} (R_\delta )_{\#} \boldsymbol {\sigma } . \end{equation*}

Making use of the formulas (3.30) and the homogeneity of $f^\infty$ , we find

(3.31) \begin{align} {\mathbb{A}}^{\mathcal{I}_\delta }&(\boldsymbol{\mu }^{n,\delta }, \boldsymbol{\nu }^{n,\delta }) = (1-2\delta ) \int _{(0,1) \times{\mathbb{T}^d}} f\Big ( \rho ^n, \frac{j^n}{1-2\delta } \Big ){\, \mathrm{d}} \mathscr{L}^{\,d+1} + \int _{(0,1) \times{\mathbb{T}^d}} f^\infty \big ( (1-2\delta ) \rho ^{n,\perp }, j^{n,\perp } \big ){\, \mathrm{d}} \boldsymbol{\sigma } \\ &\qquad = (1-2\delta ) \bigg ( \int _{(0,1) \times{\mathbb{T}^d}} f\Big ( \rho ^n, \frac{j^n}{1-2\delta } \Big ){\, \mathrm{d}} \mathscr{L}^{\,d+1} + \int _{(0,1) \times{\mathbb{T}^d}} f^\infty \Big ( \rho ^{n,\perp }, \frac{j^{n,\perp }}{1-2\delta } \Big ){\, \mathrm{d}} \boldsymbol{\sigma } \bigg ) .\nonumber \end{align}

Furthermore, it follows from the linear growth assumption that, for $i=0,1$ ,

\begin{align*} \int _{{\mathbb{T}^d}} f(\rho _i^n,0){\, \mathrm{d}} \mathscr{L}^d + \int _{{\mathbb{T}^d}} f^\infty (\rho _i^{n,\perp },0){\, \mathrm{d}} \sigma _i \leq C (\mu _i^n({\mathbb{T}^d}) +1) = C \bigl ( \boldsymbol{\mu }^n ((0,1) \times{\mathbb{T}^d}) + 1 \bigr ) \end{align*}

as well as, by (3.31), the nonnegativity of $f$ , and Assumption 2.2,

\begin{align*}{\mathbb{A}}^{\mathcal{I}_\delta }(\boldsymbol{\mu }^{n,\delta }, \boldsymbol{\nu }^{n,\delta }) &\leq{\mathbb{A}}(\boldsymbol{\mu }^n, \boldsymbol{\nu }^n) + 2\delta (\mathrm{Lip} f ) \bigg ( \int _{(0,1) \times{\mathbb{T}^d}} | j^n |{\, \mathrm{d}} \mathscr{L}^{\,d+1} + \int _{(0,1) \times{\mathbb{T}^d}} | j^{n,\perp } |{\, \mathrm{d}} \boldsymbol{\sigma } \bigg ) \\ &\leq{\mathbb{A}}(\boldsymbol{\mu }^n, \boldsymbol{\nu }^n) + \frac{2 \delta (\mathrm{Lip} f)}{c} \bigg ({\mathbb{A}}(\boldsymbol{\mu }^n,\boldsymbol{\nu }^n) + C(1+\|\boldsymbol{\mu }^n\|_{\mathrm{TV}}) \bigg ) . \end{align*}

We thus conclude (3.28).

Proof. The inequality $\le$ directly follows by choosing unit fluxes along admissible paths: let $x_0=x,x_1,\dots, x_{k-1},x_k=y$ be a path, i.e. $(x_i,x_{i+1}) \in \mathcal{E}_{\varepsilon }$ for every $i = 0,1,\dots, k$ , and consider

(4.2) \begin{equation} J^P \,:\!=\, \sum _{i=0}^{k-1} \bigl (\delta _{(x_i,x_{i+1})}-\delta _{(x_{i+1},x_i)} \bigr ), \end{equation}

which has divergence equal to $\delta _x-\delta _y$ . Then,

\begin{align*} d_{\varepsilon }(x,y) &={\mathcal{MA}}_{\varepsilon }(\delta _x, \delta _y) = \inf \left \{ \mathcal{F}_{\varepsilon }(J) \, : \, \mathsf{div} J = \delta _x-\delta _y \right \} \\ &= \inf \left \{ \sum _{(x,y) \in \mathcal{E}_{\varepsilon }} \alpha ^{\varepsilon }_{xy} \lvert J(x, y)\rvert \, : \, \mathsf{div} J = \delta _x-\delta _y \right \} \\ &\le \sum _{(x,y) \in \mathcal{E}_{\varepsilon }} \alpha ^{\varepsilon }_{xy} \lvert J^P(x,y)\rvert \le \sum _{i=0}^{k-1} 2 \alpha _{x_i x_{i+1}}^{\varepsilon }, \end{align*}

where in the last inequality we used that $\alpha ^{\varepsilon }$ is symmetric.

To prove the converse, let $\bar J \in \mathbb{R}^{\mathcal{E}_{\varepsilon }}_a$ be an optimal flux for ${\mathcal{MA}}_{\varepsilon }(\delta _x,\delta _y)$ , that is,

\begin{equation*} \mathsf {div} \bar J = \delta _x-\delta _y \quad \text { and }\quad {\mathcal {MA}}_{\varepsilon }(\delta _x,\delta _y) = \sum _{ (x, y) \in \mathcal {E}_{\varepsilon } } \alpha _{xy}^{\varepsilon } \lvert \bar J(x,y)\rvert . \end{equation*}

Since the graph $\mathcal{E}_{\varepsilon }$ is finite, in order for $\bar J$ to satisfy the divergence condition, there must exist a simple path $x_0 = x, x_1,\dots, x_k = y$ such that $(x_i,x_{i+1}) \in \mathcal{E}_{\varepsilon }$ and $\bar J(x_i,x_{i+1}) \gt 0$ for every $i$ . Let $J^P$ be the associated vector field as in (4.2). Note that, for every $\lambda \in \mathbb{R}$ , we have $\mathsf{div} \bigl ((1-\lambda ) \bar J + \lambda J^P\bigr ) = \delta _x - \delta _y$ . Furthermore, the function

\begin{equation*} \lambda \mapsto \mathcal {F}_{\varepsilon }((1-\lambda ) \bar J + \lambda J^P\bigr ) = \sum _{(x,y) \in \mathcal {E}_{\varepsilon }} \alpha _{xy}^{\varepsilon } \lvert (1-\lambda )\bar J(x,y) + \lambda J^P(x,y)\rvert \end{equation*}

is differentiable at $\lambda = 0$ , since $(\bar J(x,y) = 0) \Rightarrow (J^P(x,y) = 0)$ . By optimality, the derivative at $\lambda = 0$ must equal $0$ , i.e.

\begin{equation*} 0= \sum _{ (x, y) \in \mathcal {E}_{\varepsilon } } \alpha _{xy}^{\varepsilon } \bigl ( J^P(x,y) - \bar J(x,y) \bigr ) {\textrm {sgn}} \bar J(x,y) = \sum _{ (x, y) \in \mathcal {E}_{\varepsilon } } \alpha _{xy}^{\varepsilon } J^P(x,y) {\textrm {sgn}} \bar J(x,y) - d_{\varepsilon }(x,y), \end{equation*}

and, since $(J^P(x,y) \neq 0) \Rightarrow ({\textrm{sgn}}\, J^P(x,y) ={\textrm{sgn}}\, \bar J(x,y))$ , we have

\begin{equation*} d_{\varepsilon }(x,y) = \sum _{(x,y) \in \mathcal {E}_{\varepsilon }} \alpha _{xy}^{\varepsilon } \lvert J^P(x,y)\rvert = 2\sum _{i=1}^k \alpha _{x_i x_{i+1}}^{\varepsilon }, \end{equation*}

where, in the last equality, we used that the path is simple (and the symmetry of $\alpha ^{\varepsilon }$ ). This shows the inequality $\geq$ and concludes the proof.

Proof of ≥. Fix $m_0,m_1 \in \mathscr{P}(\mathcal{X}_{\varepsilon })$ and set $m\,:\!=\, m_0-m_1$ . Let $\bar J \in \mathbb{R}_a^{\mathcal{E}_{\varepsilon }}$ be an optimal flux for ${\mathcal{MA}}_{\varepsilon }(m_0,m_1)$ , that is,

\begin{align*} \mathsf{div} \bar J = m \quad \text{ and }\quad{\mathcal{MA}}_{\varepsilon }(m_0,m_1) = \sum _{ (x, y) \in \mathcal{E}_{\varepsilon } } \alpha _{xy}^{\varepsilon } \lvert \bar J(x,y)\rvert . \end{align*}

Let $\varphi : \mathcal{X}_{\varepsilon } \to \mathbb{R}$ be such that $\mathrm{Lip}_{d_{\varepsilon }} \varphi \leq 1$ , i.e. $ \left | \varphi (y) - \varphi (x) \right | \leq d_{\varepsilon }(x,y)$ for $ x, y \in \mathcal{X}_{\varepsilon } .$ Then,

\begin{align*} \int _{\mathcal{X}_{\varepsilon }} \varphi{\, \mathrm{d}} m &= \int _{\mathcal{X}_{\varepsilon }} \varphi{\, \mathrm{d}} \mathsf{div} \bar J = \sum _{x \in \mathcal{X}_{\varepsilon }} \varphi (x) \sum _{y \sim x} \bar J(x,y) = \frac 12 \sum _{(x,y) \in \mathcal{E}_{\varepsilon }} \varphi (x) \big ( \bar J(x,y) - \bar J(y,x) \big )\\ &= \frac 12 \sum _{(x,y) \in \mathcal{E}_{\varepsilon }} \big ( \varphi (y) - \varphi (x) \big ) \bar J(x,y) \leq \frac 12 \sum _{(x,y) \in \mathcal{E}_{\varepsilon }} d_{\varepsilon }(x,y) \lvert \bar J(x,y)\rvert . \end{align*}

In order to conclude, we make the following crucial observation: as a consequence of the optimality of $\bar J$ , we claim that

(4.4) \begin{equation} \bar J(x,y) \neq 0 \quad \Longrightarrow \quad d_{\varepsilon }(x,y) = 2 \alpha _{xy}^{\varepsilon } . \end{equation}

To this end, assume that $\bar J(x,y) \neq 0$ and consider an optimal $J^{(x,y)}$ for $d_{\varepsilon }(x,y) ={\mathcal{MA}}_{\varepsilon }(\delta _x, \delta _y)$ . Note that, by construction,

\begin{align*} \mathsf{div} \big ( J^{(x,y)} \big ) = \delta _x - \delta _y = \mathsf{div} \widetilde J, \qquad \text{where} \ \widetilde J \,:\!=\, \delta _{(x,y)} - \delta _{(y,x)}, \end{align*}

which in turns also implies that

\begin{align*} \mathsf{div} \big ( \bar J + \bar J(x,y) \big ( J^{(x,y)} - \widetilde J \big ) \big ) = \mathsf{div} \bar J . \end{align*}

By optimality of $J^{(x,y)}$ , we have

(4.5) \begin{equation} \mathcal{F}_{\varepsilon }(\widetilde J) = 2 \alpha _{xy}^{\varepsilon } \geq \mathcal{F}_{\varepsilon } \big ( J^{(x,y)} \big ) = \sum _{(\widetilde x, \widetilde y) \in \mathcal{E}_{\varepsilon }} \alpha _{\widetilde x, \widetilde y}^{\varepsilon } \lvert J^{(x,y)}(\widetilde x, \widetilde y)\rvert, \end{equation}

whereas the optimality of $\bar J$ yields

\begin{align*} \mathcal{F}_{\varepsilon } \bigl (\bar J + \bar J(x,y) \big ( J^{(x,y)} - \widetilde J \big )\bigr ) &= \sum _{(\widetilde x, \widetilde y) \in \mathcal{E}_{\varepsilon } \setminus \{(x,y), (y,x)\} } \alpha _{\widetilde x \widetilde y}^{\varepsilon } \lvert \bar J(\widetilde x, \widetilde y) + \bar J(x,y) J^{(x,y)}(\widetilde x, \widetilde y)\rvert \\ &\quad + \alpha _{xy}^{\varepsilon } \lvert \bar J(x,y) J^{(x,y)}(x,y)\rvert + \alpha _{yx}^{\varepsilon } \lvert \bar J(x,y) J^{(x,y)}(y,x)\rvert \\ &\ge \mathcal{F}_{\varepsilon }(\bar J) = \sum _{(\widetilde x, \widetilde y) \in \mathcal{E}_{\varepsilon }} \alpha _{\widetilde x\widetilde y}^{\varepsilon } \lvert \bar J(\widetilde x, \widetilde y)\rvert . \end{align*}

By applying the triangle inequality and simplifying the latter formula, we find

(4.6) \begin{equation} \sum _{(\widetilde x,\widetilde y) \in \mathcal{E}_{\varepsilon }} \alpha _{\widetilde x\widetilde y}^{\varepsilon } \lvert \bar J(x,y) J^{(x,y)}(\widetilde x,\widetilde y)\rvert \ge 2\alpha _{xy}^{\varepsilon } \lvert \bar J(x,y)\rvert . \end{equation}

The combination of (4.5) and (4.6) implies $d_{\varepsilon }(x,y) = 2\alpha _{xy}^{\varepsilon }$ . With (4.4) at hand, we can write

\begin{equation*} \int _{\mathcal {X}_{\varepsilon }} \varphi {\, \mathrm {d}} m \le \sum _{(x,y) \in \mathcal {E}_{\varepsilon }} \alpha _{xy}^{\varepsilon } \lvert \bar J(x,y)\rvert = {\mathcal {MA}}_{\varepsilon }(m_0,m_1), \end{equation*}

and we conclude by arbitrariness of $\varphi$ .

Proof of ≤. Let $\pi$ be such that $(e_i)_{\#} \pi = m_i$ for $i=0,1$ . Further, for every $x,y \in \mathcal{X}_{\varepsilon }$ , let $J^{(x,y)} \in \mathbb{R}^{\mathcal{E}_{\varepsilon }}_a$ be optimal for ${\mathcal{MA}}_{\varepsilon }(\delta _x, \delta _y)$ . It follows from a direct computation that the divergence of the asymmetric flux

\begin{equation*} J \,:\!=\, \sum _{x,y \in \mathcal {X}_{\varepsilon }} \pi (x,y) J^{(x,y)} \end{equation*}

is equal to $m_0 - m_1$ . Thus,

\begin{equation*} {\mathcal {MA}}_{\varepsilon }(m_0,m_1) \le \sum _{(\widetilde x, \widetilde y) \in \mathcal {E}_{\varepsilon }} \alpha _{\widetilde x \widetilde y}^{\varepsilon } \lvert J(\widetilde x, \widetilde y)\rvert \le \sum _{x,y \in \mathcal {X}_{\varepsilon }} \pi (x,y) \sum _{(\widetilde x,\widetilde y) \in \mathcal {E}_{\varepsilon }} \alpha _{\widetilde x \widetilde y}^{\varepsilon } \lvert J^{(x,y)}(\widetilde x, \widetilde y)\rvert = \int _{\mathcal {X}_{\varepsilon } \times \mathcal {X}_{\varepsilon }} d_{\varepsilon } {\, \mathrm {d}} \pi, \end{equation*}

and we conclude by arbitrariness of $\pi$ .

Proof. Finiteness follows from the nonemptiness of the set of representatives proved in [[Reference Gladbach, Kopfer, Maas and Portinale23], Lemma 4.5]. To prove positiveness, take any $j \in \mathbb{R}^d$ and $J \in{\mathsf{Rep}}(j)$ . For every norm $\|{\cdot }\|$ , we have

(4.8) \begin{equation} \|{j}\| = \|{\mathsf{Eff}(J)}\| \le \frac{1}{2}\sum _{(x,y) \in \mathcal{E}^Q} \lvert J(x,y)\rvert \|{y_{\mathsf{z}} - x_{\mathsf{z}}}\| \le \frac{F(J)}{2} \max _{(x,y) \in \mathcal{E}^Q} \frac{\|{y_{\mathsf{z}} - x_{\mathsf{z}}}\|}{ \alpha _{xy}} . \end{equation}

The constant that multiplies $F(J)$ at the right-hand side is finite because every $\alpha _{xy}$ is strictly positive and the graph $(\mathcal{X},\mathcal{E})$ is locally finite. Absolute homogeneity and the triangle inequality follow from the absolute homogeneity and subadditivity of $F$ and the affinity of the constraints.

Proof. Let $X$ be the vector space of all $\mathbb{Z}^d$ -periodic functions $J \in \mathbb{R}^{\mathcal{E}}_a$ such that $\mathsf{div} J \equiv 0$ . The sublevel set

\begin{equation*} X_1 \,:\!=\, \left \{ J \in X \, : \, F(J) \le 1 \right \} \end{equation*}

is clearly compact (due to the strict positivity of $(\alpha _{xy})_{x,y}$ ) and can be written as finite intersection of half-spaces, namely

\begin{equation*} X_1 = \bigcap _{r \in \{-1,1\}^{\mathcal {E}^Q} } \left \{ J \in X \, : \, \sum _{(x,y) \in \mathcal {E}^Q} \alpha _{xy} r_{xy} J(x,y) \le 1 \right \}. \end{equation*}

Thus, $X_1$ is the convex hull of some finite set of points $A$ , that is, $ X_1 ={\mathrm{conv}}(A)$ . Since $f_{\mathrm{hom}}$ is defined as a minimum, we have

\begin{equation*} B = \left \{ j \in \mathbb {R}^d \, : \, \exists J \in {\mathsf {Rep}}(j), \, F(J) \le 1 \right \} = \mathsf {Eff}(X_1) = \mathsf {Eff}({\mathrm {conv}}(A)) = {\mathrm {conv}}(\mathsf {Eff}(A)), \end{equation*}

where the last equality is due to the linearity of $\mathsf{Eff}$ .