[proofplan] The proof uses only the variational definition of the discrete minimizing movement. At each time step, compare the minimizer $x_{k+1}^\tau$ with the admissible competitor $x_k^\tau$ to obtain a one-step energy inequality. Summing those one-step inequalities over $k=m,\dots,\ell-1$ makes the intermediate energy terms telescope, leaving exactly the stated discrete dissipation estimate. [/proofplan]

[step:Compare each minimizer with the previous time step]Fix $k\in\mathbb N_0$. Define the penalized functional $\Phi_k:X\to(-\infty,+\infty]$ by \begin{align*} \Phi_k(y)=\mathcal E(y)+\frac{1}{2\tau}d^2(y,x_k^\tau). \end{align*} By the discrete minimizing movement hypothesis, $x_{k+1}^\tau$ minimizes $\Phi_k$ over $X$. Since $x_k^\tau\in X$ and $\mathcal E(x_k^\tau)<+\infty$, it is an admissible competitor. Hence \begin{align*} \mathcal E(x_{k+1}^\tau)+\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)\le \mathcal E(x_k^\tau)+\frac{1}{2\tau}d^2(x_k^\tau,x_k^\tau). \end{align*} Because $d(x_k^\tau,x_k^\tau)=0$, this becomes \begin{align*} \frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)+\mathcal E(x_{k+1}^\tau)\le \mathcal E(x_k^\tau). \end{align*}[/step]

[guided]Fix a time index $k\in\mathbb N_0$. The definition of a discrete minimizing movement says that $x_{k+1}^\tau$ is chosen by minimizing the previous energy plus a quadratic penalty for moving away from $x_k^\tau$. To use this definition, introduce the map $\Phi_k:X\to(-\infty,+\infty]$ by \begin{align*} \Phi_k(y)=\mathcal E(y)+\frac{1}{2\tau}d^2(y,x_k^\tau). \end{align*} The hypothesis says precisely that $\Phi_k(x_{k+1}^\tau)\le \Phi_k(y)$ for every $y\in X$. We choose the competitor $y=x_k^\tau$. This point is allowed because $x_k^\tau\in X$ and, by hypothesis, $\mathcal E(x_k^\tau)<+\infty$. Therefore \begin{align*} \Phi_k(x_{k+1}^\tau)\le \Phi_k(x_k^\tau). \end{align*} Expanding the definition of $\Phi_k$ gives \begin{align*} \mathcal E(x_{k+1}^\tau)+\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)\le \mathcal E(x_k^\tau)+\frac{1}{2\tau}d^2(x_k^\tau,x_k^\tau). \end{align*} The metric axiom $d(x_k^\tau,x_k^\tau)=0$ removes the penalty term on the right-hand side. Thus the one-step dissipation inequality is \begin{align*} \frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)+\mathcal E(x_{k+1}^\tau)\le \mathcal E(x_k^\tau). \end{align*} This is the entire source of the estimate: each minimizing step pays for its movement by decreasing the energy.[/guided]

[step:Sum the one-step inequalities over the desired time interval] Let $m,\ell\in\mathbb N_0$ satisfy $0\le m<\ell$. Summing the one-step inequality from $k=m$ to $k=\ell-1$ gives \begin{align*} \sum_{k=m}^{\ell-1}\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)+\sum_{k=m}^{\ell-1}\mathcal E(x_{k+1}^\tau)\le \sum_{k=m}^{\ell-1}\mathcal E(x_k^\tau). \end{align*} Reindexing the second sum on the left gives \begin{align*} \sum_{k=m}^{\ell-1}\mathcal E(x_{k+1}^\tau)=\sum_{j=m+1}^{\ell}\mathcal E(x_j^\tau). \end{align*} Thus \begin{align*} \sum_{k=m}^{\ell-1}\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)+\sum_{j=m+1}^{\ell}\mathcal E(x_j^\tau)\le \sum_{k=m}^{\ell-1}\mathcal E(x_k^\tau). \end{align*} Subtracting the common finite sum $\sum_{j=m+1}^{\ell-1}\mathcal E(x_j^\tau)$ from both sides yields \begin{align*} \sum_{k=m}^{\ell-1}\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)+\mathcal E(x_\ell^\tau)\le \mathcal E(x_m^\tau). \end{align*} Since \begin{align*} \sum_{k=m}^{\ell-1}\frac{1}{2\tau}d^2(x_{k+1}^\tau,x_k^\tau)=\frac{1}{2}\sum_{k=m}^{\ell-1}\frac{d^2(x_{k+1}^\tau,x_k^\tau)}{\tau}, \end{align*} the desired estimate follows. [/step]