The uniform energy decrement on the basin, bounded iteration, and the closedness of the basin inside the short polygon space

Variance estimate for the edge lengths. By [F1], ${\zeta }_i\le ({\xi }_i+{\xi }_{i+1})\mathrm{/}2$ for every $i$, whence $E(fx)={\sum }_i{\zeta }_i^2\le {\sum }_i(({\xi }_i+{\xi }_{i+1})\mathrm{/}2{)}^2$ and therefore $\mathrm{\Delta }={\sum }_i({\xi }_i^2-{\zeta }_i^2)\ge {\sum }_i({\xi }_i^2-({\xi }_i+{\xi }_{i+1}{)}^2\mathrm{/}4)=\frac{1}{4}{\sum }_i({\xi }_i-{\xi }_{i+1}{)}^2\ge 0$, where the middle identity is the algebraic expansion of the square and the cyclic sums. Consequently ${\sum }_i({\xi }_i-{\xi }_{i+1}{)}^2\le 4\mathrm{\Delta }$.

First case of the decrement. If $\mathrm{\Delta }>r^2\mathrm{/}{n}^4$, then $\mathrm{\Delta }\ge {\lambda }_n(2r)$, because ${\lambda }_n(2r)$ is half of a minimum one of whose entries is $r^2\mathrm{/}{n}^4$, so ${\lambda }_n(2r)\le r^2\mathrm{/}{n}^4<\mathrm{\Delta }$.

Openness of the basin. By [F1], $C_h^0(n)$ is the union over $k\ge 0$ of the sets $\{x\in P_h(n):L(f^{k}x)<l\mathrm{/}2\}$; each of these is open because $f^k$ and $L$ are continuous ([F7]), so the basin is open in $P_h(n)$.

The basin contains no straight equilateral tuple (iii, second part). If $x$ is nonconstant, equilateral (${\xi }_i=\xi >0$ for all $i$) and straight at every vertex, then $fx$ consists of the same closed local geodesic's points shifted by arclength $\xi \mathrm{/}2$. Every subarc of that curve of length at most $2\xi$ lies in the uniform CAT(1) ball of radius $\xi <l$ about its arclength midpoint; it minimizes there by Short local geodesics in a CAT(1) space are geodesics, and closed local geodesics have length at least $2\pi$ (ii), since $2\xi <\pi$. Thus every shifted consecutive triple is straight, and induction gives $L(f^{k}x)=L(x)=n\xi >0$ for every $k$; therefore $L(f^{k}x)$ does not tend to $0$ and $x\mathrm{\notin }{C}_h^0(n)$. Such tuples are exactly the equally spaced lists of points of a closed local geodesic by the equality analysis of F1.

Each edge is close to the mean $2r\mathrm{/}n$. Put ${\delta }_i:={\xi }_i-{\xi }_{i+1}$ and recall ${\sum }_i{\xi }_i=2r$. For each $i$, ${\xi }_i-2r\mathrm{/}n=\frac{1}{n}{\sum }_{k=0}^{n-1}({\xi }_i-{\xi }_{i+k})=\frac{1}{n}{\sum }_{k=0}^{n-1}{\sum }_{j=0}^{k-1}{\delta }_{i+j}$; by Cauchy–Schwarz the absolute value of each inner sum is at most $2\sqrt{k\mathrm{\Delta }}\le n\sqrt{\mathrm{\Delta }}$, since $k\le n-1$ and $2\sqrt{n-1}\le n$, so $\mathrm{\mid }{\xi }_i-2r\mathrm{/}n\mathrm{\mid }\le n\sqrt{\mathrm{\Delta }}$.

Second case of the decrement. If $\mathrm{\Delta }\le r^2\mathrm{/}{n}^4$, then ${\xi }_i\ge r\mathrm{/}n$ for every $i$ by step 2.1. Put $\eta =(\pi -r)\mathrm{/}2$ and $\mu =\min \{r\mathrm{/}(2n),\eta \}$, so that $0<\mu <2\eta <\pi$; by F2 and [F3] there is an index $k$ with ${\zeta }_k\le ({\xi }_k+{\xi }_{k+1})\mathrm{/}2-\delta (\eta ,\mu )$, where the constants are continuous in $(n,L)$ and independent of the perturbation. Since ${\zeta }_k\ge 0$, this forces $\delta (\eta ,\mu )\le ({\xi }_k+{\xi }_{k+1})\mathrm{/}2$, and the sharpened bound ${\zeta }_k^2\le (({\xi }_k+{\xi }_{k+1})\mathrm{/}2{)}^2-\delta ({\xi }_k+{\xi }_{k+1})+{\delta }^2$ at the index $k$, combined with the unsharpened bounds at the other indices, gives $\mathrm{\Delta }\ge \delta ({\xi }_k+{\xi }_{k+1})-{\delta }^2\ge (\delta \mathrm{/}2)({\xi }_k+{\xi }_{k+1})\ge \delta r\mathrm{/}n$, where the middle inequality uses $\delta \le ({\xi }_k+{\xi }_{k+1})\mathrm{/}2$.

The decrement (i). In view of steps 1.2 and 3.1, $\mathrm{\Delta }\ge \min \{r^2\mathrm{/}{n}^4,\delta (\eta ,\mu )r\mathrm{/}n\}\ge {\lambda }_n(2r)$ for ${\lambda }_n$ the half of the displayed minorant. Since $\delta$ is continuous and positive on its domain, $\eta (\pi -r)=(\pi -r)\mathrm{/}2>0$ and $\mu =\min \{r\mathrm{/}(2n),\eta \}>0$ depend continuously on $r\in (0,\pi )$ with $\mu <2\eta$, the function ${\lambda }_n$ is continuous and positive on $(0,2\pi )$; it depends only on $n$ and on the length, not on $h$ or on $X$. The explicit separation constant obeys $0<\delta (\eta ,\mu )\le \mu \mathrm{/}2$, since it is $2(\mu \mathrm{/}4)$ minus a nonnegative chord length. Thus $0<{\lambda }_n(2r)\le r^2\mathrm{/}(2n^4)$ and ${\lambda }_n(2r)\le \mu r\mathrm{/}(4n)$: the first bound tends to zero as $r\to 0$, and the second as $r\to \pi$ because $\mu \le (\pi -r)\mathrm{/}2$. Hence $E(fx)\le E(x)-{\lambda }_n(L(x))$ for every $x\in C_h^0(n)$ with $L(x)\in (0,2\pi )$.

Bounded iteration (ii). Let $0<a<b<2\pi$; the restriction of the continuous positive function ${\lambda }_n$ to the compact interval $[a,b]$ attains a positive minimum ${\lambda }_0$ ([F5]). Fix $x\in C_h^0(n)$ with $L(x)\le b$ and put $M=\lceil b^2\mathrm{/}{\lambda }_0\rceil +1$. Since $L$ does not increase along the iterates and the basin is invariant ([F1]), every $f^{j}x$ with $j\le M$ is in the basin with $L(f^{j}x)\le b$; as long as $L(f^{j}x)\ge a$, step 4.1 gives $E(f^{j+1}x)\le E(f^{j}x)-{\lambda }_0$. If $L(f^{j}x)\ge a$ held for all $j<M$, then $E(f^{M}x)\le E(x)-M{\lambda }_0<0$ because $E(x)\le L(x{)}^2\le b^2$ and $M{\lambda }_0>b^2$, a contradiction; hence $L(f^{j}x)<a$ for some $j<M$, and then $L(f^{M}x)\le L(f^{j}x)<a$ by monotonicity of $L$.

Closedness of the basin inside the short piece. Let $x_j\in C_h^0(n)$ converge to $x\in P_h(n)$ with $L(x)<2\pi$; choose $b$ with $\max \{L(x),l\mathrm{/}4\}<b<2\pi$ and then $j_0$ with $L(x_j)\le b$ for all $j\ge j_0$, and let $M$ be the integer of step 5.1 for the parameters $a=l\mathrm{/}4$ and $b$. Then $L(f^M{x}_j)<l\mathrm{/}4$ for all $j\ge j_0$, and continuity of $f^M$ and $L$ ([F7]) gives $L(f^{M}x)\le l\mathrm{/}4<l\mathrm{/}2$, so $x\in C_h^0(n)$ by the description [F1]. Hence the basin is closed in $\{x\in P_h(n):L(x)<2\pi \}$.

No homotopy crosses the basin boundary. Let $s\mapsto y_s$ be a continuous family in the piece $\{L<2\pi \}$, $s\in [0,1]$. The basin is open by [F1] and closed in that piece by step 6.1. Its inverse image under the family is therefore both open and closed in $[0,1]$. Connectedness of the real interval [F5] excludes a nonempty proper subset that is both open and closed. Thus membership is constant along the family. Nonconstant equally spaced closed local geodesic tuples have fixed positive length under $f$ by [F1] and lie outside the basin.

Conclusion. The finite comparison disk is built from midpoint iterates of a tuple already in the basin; no loop shrinkability assumption is used. Steps 1.1–4.1 prove the decrement, step 5.1 proves bounded iteration, and steps 6.1 and 7.1 prove closedness and separation, with openness supplied by [F1]. The constants depend only on $n$ and length; AC enters through the finite-disk supplier and its perturbation extension.