[proofplan] The proof passes the limiting discrete energy-dissipation estimate for the minimizing-movement scheme to the limit. First, the assumed prelimit inequality supplies the endpoint energy balance together with the discrete metric-speed and local-slope dissipation terms. Then the endpoint convergence identifies the limiting endpoint energies. Finally, the two assumed lower semicontinuity inequalities for the metric-speed and slope terms give the continuum energy dissipation inequality. [/proofplan]

[step:Record the limiting discrete energy-dissipation estimate]Fix $0\le s\le t\le T$. By hypothesis, the discrete interpolations satisfy the limiting endpoint-form energy-dissipation estimate \begin{align*} \limsup_{n\to\infty}\left(\mathcal F[\bar\rho_{\tau_n}(t)]+\frac12\int_s^t |\rho_{\tau_n}'|^2(r)\,d\mathcal L^1(r)+\frac12\int_s^t |\partial\mathcal F|^2(\tilde\rho_{\tau_n}(r))\,d\mathcal L^1(r)-\mathcal F[\bar\rho_{\tau_n}(s)]\right)\le0. \end{align*} Equivalently, if \begin{align*} A_n:=\int_s^t |\rho_{\tau_n}'|^2(r)\,d\mathcal L^1(r) \end{align*} and \begin{align*} B_n:=\int_s^t |\partial\mathcal F|^2(\tilde\rho_{\tau_n}(r))\,d\mathcal L^1(r), \end{align*} then \begin{align*} \limsup_{n\to\infty}\left(\mathcal F[\bar\rho_{\tau_n}(t)]+\frac12A_n+\frac12B_n-\mathcal F[\bar\rho_{\tau_n}(s)]\right)\le0. \end{align*}[/step]

[guided]Fix $0\le s\le t\le T$. The statement assumes the limiting discrete energy-dissipation estimate in endpoint form: \begin{align*} \limsup_{n\to\infty}\left(\mathcal F[\bar\rho_{\tau_n}(t)]+\frac12\int_s^t |\rho_{\tau_n}'|^2(r)\,d\mathcal L^1(r)+\frac12\int_s^t |\partial\mathcal F|^2(\tilde\rho_{\tau_n}(r))\,d\mathcal L^1(r)-\mathcal F[\bar\rho_{\tau_n}(s)]\right)\le0. \end{align*} This is the form needed for passage to the limit at arbitrary fixed times $s$ and $t$. It avoids asserting a pointwise-in-$n$ inequality with the piecewise constant endpoints $\bar\rho_{\tau_n}(s)$ and $\bar\rho_{\tau_n}(t)$, which need not be valid when $s$ and $t$ lie inside time steps. Define \begin{align*} A_n:=\int_s^t |\rho_{\tau_n}'|^2(r)\,d\mathcal L^1(r) \end{align*} and \begin{align*} B_n:=\int_s^t |\partial\mathcal F|^2(\tilde\rho_{\tau_n}(r))\,d\mathcal L^1(r). \end{align*} Both $A_n$ and $B_n$ belong to $[0,\infty]$ because they are integrals of nonnegative functions. With this notation, the assumed estimate is \begin{align*} \limsup_{n\to\infty}\left(\mathcal F[\bar\rho_{\tau_n}(t)]+\frac12A_n+\frac12B_n-\mathcal F[\bar\rho_{\tau_n}(s)]\right)\le0. \end{align*}[/guided]

[step:Pass the endpoint energies to the limit] For the fixed pair $0\le s\le t\le T$ chosen above, by the convergence assumption on the piecewise constant interpolations, applied at the fixed times $s$ and $t$, \begin{align*} \mathcal F[\bar\rho_{\tau_n}(s)]\to E(s) \end{align*} and \begin{align*} \mathcal F[\bar\rho_{\tau_n}(t)]\to E(t). \end{align*} Therefore the endpoint terms in the discrete inequality converge to the endpoint terms in the asserted continuum inequality. [/step]

[step:Use lower semicontinuity for the metric-speed and slope terms] Set \begin{align*} A_n:=\int_s^t |\rho_{\tau_n}'|^2(r)\,d\mathcal L^1(r) \end{align*} and \begin{align*} B_n:=\int_s^t |\partial\mathcal F|^2(\tilde\rho_{\tau_n}(r))\,d\mathcal L^1(r). \end{align*} The endpoint convergence and the assumed limiting discrete inequality give \begin{align*} E(t)+\frac12\liminf_{n\to\infty}(A_n+B_n)\le E(s). \end{align*} Since $A_n\ge0$ and $B_n\ge0$ for every $n\in\mathbb N$, the superadditivity of the lower limit for nonnegative extended-real sequences gives \begin{align*} \liminf_{n\to\infty}(A_n+B_n)\ge \liminf_{n\to\infty}A_n+\liminf_{n\to\infty}B_n. \end{align*} Therefore \begin{align*} E(t) +\frac12\liminf_{n\to\infty}\int_s^t |\rho_{\tau_n}'|^2(r)\,d\mathcal L^1(r) +\frac12\liminf_{n\to\infty}\int_s^t |\partial\mathcal F|^2(\tilde\rho_{\tau_n}(r))\,d\mathcal L^1(r) \le E(s). \end{align*} By the assumed lower semicontinuity of the metric derivative term, \begin{align*} \int_s^t |\rho'|^2(r)\,d\mathcal L^1(r)\le \liminf_{n\to\infty}\int_s^t |\rho_{\tau_n}'|^2(r)\,d\mathcal L^1(r). \end{align*} By the assumed lower semicontinuity of the relaxed-slope term, \begin{align*} \int_s^t G^2(\rho(r))\,d\mathcal L^1(r)\le \liminf_{n\to\infty}\int_s^t |\partial\mathcal F|^2(\tilde\rho_{\tau_n}(r))\,d\mathcal L^1(r). \end{align*} Substituting these two bounds into the lower-limit inequality yields \begin{align*} E(t)+\frac12\int_s^t |\rho'|^2(r)\,d\mathcal L^1(r)+\frac12\int_s^t G^2(\rho(r))\,d\mathcal L^1(r)\le E(s). \end{align*} [/step]

[step:Conclude the energy dissipation inequality on every subinterval] The preceding argument was carried out for an arbitrary pair $0\le s\le t\le T$. Hence the inequality \begin{align*} E(t)+\frac12\int_s^t |\rho'|^2(r)\,d\mathcal L^1(r)+\frac12\int_s^t G^2(\rho(r))\,d\mathcal L^1(r)\le E(s) \end{align*} holds for every $0\le s\le t\le T$. This is exactly the asserted energy dissipation inequality for the limit curve $\rho$ with relaxed slope $G$. [/step]