The zero-length boundary: constant tuples, collapsed edges and the degeneracy of the energy decrement at $L=0$

Constant tuples (i). If $x=(p,\dots ,p)$ then every consecutive distance is $0$, so $\mathrm{mesh}(x)=L(x)=E(x)=0$; every pair $(x_i,x_{i+1})$ is degenerate with midpoint $p$ by [L1], so $f(x)=x$, and then $L(f^{k}x)=0$ for all $k$, so $x\in C_h^0(n)$ by [L2]. The corresponding constant loop is short (length $0<2\pi$) and shrinkable through constant loops, and conversely $E(x)=0$ forces all edges to have length $0$, i.e. $x$ to be constant.

Degeneracy at zero (ii). The chosen positive minorant obeys $0<{\lambda }_n(2r)\le r^2\mathrm{/}(2n^4)$, so ${\lambda }_n(2r)\to 0$ as $r\to 0^{+}$. No boundedness assertion about $\delta$ near a parameter-domain endpoint is needed.

Degeneracy at $2\pi$ (ii). As $r\to {\pi }^{-}$, $\eta =(\pi -r)\mathrm{/}2\to 0$ and $\mu =\min \{r\mathrm{/}(2n),\eta \}\to 0$. The intrinsic separation construction gives $0<\delta (\eta ,\mu )\le 2(\mu \mathrm{/}4)=\mu \mathrm{/}2$, since its cut chord length is nonnegative. Therefore $0<{\lambda }_n(2r)\le \delta r\mathrm{/}(2n)\le \mu r\mathrm{/}(4n)\to 0$. This proves the claimed limit using an actual bound, rather than continuity at a point outside $\delta$'s domain.

Collapsed edges (iii). A zero edge contributes zero to mesh, length and energy, and its midpoint is its endpoint by [L1]. If $x$ is nonconstant, another edge is positive and $E(x)>0$; equality of energies cannot hold, since [L2] requires all positive-length equality edges to have the same positive length, contradicting the zero edge. When $x$ is in the short basin, put $r=L(x)\mathrm{/}2>0$ and $\mathrm{\Delta }=E(x)-E(fx)$. The variance estimate of [L3] gives $\mathrm{\mid }{\xi }_i-2r\mathrm{/}n\mathrm{\mid }\le n\sqrt{\mathrm{\Delta }}$; at the zero edge this forces $\mathrm{\Delta }\ge 4r^2\mathrm{/}{n}^4>r^2\mathrm{/}{n}^4$, so the first decrement case applies. The perturbation supplier is needed instead for degenerate face triples in the small-deficit case, where all edges are $\ge r\mathrm{/}n>0$. Finally, deleting consecutive repeated vertices preserves the normalized polygonal loop; if both lists have at least three entries, $X$ is geodesic and their common length is $<2\pi$, [L7] identifies both basin memberships with shrinkability of that same loop. Equal initial energies alone do not identify their different midpoint iterations.

The basin criterion (ii), (iv). The explicit ${\lambda }_n$ supplies no uniform positive decrement near either length endpoint, by steps 1.2 and 1.3. The basin's defining condition is ${\lim }_{k}L(f^{k}x)=0$, and L2 gives the equivalent eventual threshold $L(f^{m}x)<l\mathrm{/}2$. Openness follows from that strict threshold; closedness inside $\{L<2\pi \}$ uses [L3]'s bounded iteration on positive compact length bands. Neither conclusion requires a positive lower bound for ${\lambda }_n$ near zero.

Conclusion. Clause (i) is step 1.1, clause (ii) is steps 1.2, 1.3 and 2.1, clause (iii) is step 1.4, and clause (iv) is step 2.1: the zero-length boundary is a genuine limit of the basin, the decrement function degenerates at both endpoints of $(0,2\pi )$, and degenerate edges do not disturb the estimates; AC enters only through the suppliers [L3] and [L4] ([L6]).