the vector ${\widetilde}\tau^{k}$ associated with ${\widetilde}\ell_{k}$ for this case of $\tau$ are given by ${\widetilde}\tau_{i}^{k} = \frac{\sum_{p=0}^{N-i}({\varepsilon}+1)^p}{\sum_{p=0}^{N-1}({\varepsilon}+1)^p}{\bar{\tau}}$, which by Lemma \[lem:consum1\] is equal to $\frac{({\varepsilon}+1)^{N - i +1} - 1}{({\varepsilon}+1)^{N} - 1}{\bar{\tau}}$. After the jump, $G(\tau)$ is single valued and is such that its elements are ordered as follows: $g_{2}(\tau) > g_{3}(\tau) > \ldots > g_{N}(\tau) > g_{1}(\tau) = 0.$ Specifically, the jump map is $G(\tau) = [0,(1+{\varepsilon})\tau_{2},\ldots, (1+{\varepsilon})\tau_{N}]^{\top}$. Then, the formulation in Section \[sec:AN\] and [Lemma \[eqn:tauSsolution\]]{} leads to a case of ${\widetilde}\tau^{k}$ denoted as ${\widetilde}\tau^{k'}$. By [Lemma \[lem:consum1\]]{}, the elements of the vector ${\widetilde}\tau^{k'}$ are given by ${\widetilde}\tau^{k'}_{1} = \frac{{\varepsilon}}{({\varepsilon}+1)^{N} - 1}{\bar{\tau}}$ and ${\widetilde}\tau^{k'}_{i} = \frac{({\varepsilon}+1)^{N - i +2} - 1}{({\varepsilon}+1)^{N} - 1}{\bar{\tau}}$ for $i > 1$. Due to the ordering of $\tau$ and $G(\tau)$, ${\widetilde}\tau^{k'}$ is a one-element shifted (to the right) version of ${\widetilde}\tau^{k}$.

From the definition of ${\widetilde}\tau^{k}$ above, $V$ at $\tau$ reduces to $$V(\tau) = |\tau|_{{\widetilde}\ell_{k}} = \left|({\widetilde}\tau^{k} - \tau) - \frac{1}{N}(({\widetilde}\tau^{k} - \tau)^{\top}{\mathbf 1}){\mathbf 1}\right|$$ for some $k$. Note that $$({\widetilde}\tau^{k} - \tau)^{\top}{\mathbf 1}= \sum_{i = 1}^{N} {\widetilde}\tau^{k}_{i} - \sum_{i=1}^{N} \tau_{i}$$ reduces to $\sum_{i = 2}^{N} {\widetilde}\tau^{k}_{i} - \sum_{i=2}^{N} \tau_{i}$ since $\tau_{1} = {\widetilde}\tau_{1}^{k} = {\bar{\tau}}$. Using [Lemmas \[lem:consum1\] and \[lem:consum2\]]{}, it follows 