Existence of the uniform radius, continuity of the midpoint operation, the energy drop, its equality case, and convergence of zero-limit polygons

A uniform radius exists. By local CAT(1), every $z\in X$ has a closed ball $\bar{B}(z,{\rho }_z)$ that is CAT(1); replacing ${\rho }_z$ by $\min \{{\rho }_z,\pi \mathrm{/}4\}$ we may assume ${\rho }_z<\pi \mathrm{/}2$, because a closed sub-ball of radius $<\pi \mathrm{/}2$ of a CAT(1) ball is convex ([F4]) and a convex subset of a CAT(1) space with the induced metric is CAT(1) ([F3]). The open balls $B(z,{\rho }_z\mathrm{/}2)$ cover the compact space $X$, so by [F10] the cover has a Lebesgue number $\delta >0$. Put $l:=\min \{\delta \mathrm{/}4,\pi \mathrm{/}4\}<\pi \mathrm{/}2$ and let $x\in X$, $0<r\le l$: the closed ball $\bar{B}(x,r)$ has diameter at most $2r\le \delta \mathrm{/}2<\delta$, so it is contained in $B(z,{\rho }_z\mathrm{/}2)\subseteq \bar{B}(z,{\rho }_z)$ for some $z$; as a ball of radius $r\le \pi \mathrm{/}4<\pi \mathrm{/}2$ in the CAT(1) space $\bar{B}(z,{\rho }_z)$ it is convex, and a convex subset of a CAT(1) space with the induced metric is CAT(1), so the induced metric on $\bar{B}(x,r)$ is CAT(1). Hence $l$ is a uniform local CAT(1) radius and $l<\pi \mathrm{/}2$; applying this with the given $l$ replaced by the smaller number justifies the standing assumption.

Pointwise midpoint bound. Let $x\in X^n$ with $\mathrm{mesh}(x)<l$ and put ${\xi }_i:=d(x_i,x_{i+1})$. For each $i$ the three vertices $x_i,x_{i+1},x_{i+2}$ lie in the closed ball $\bar{B}(x_{i+1},l)$ of radius $l$, which is CAT(1) by [F1]; the triangle $(x_i,x_{i+1},x_{i+2})$ has perimeter at most $2({\xi }_i+{\xi }_{i+1})<4l<2\pi$ and its sides lie in the ball by convexity, so its comparison triangle in $S^2$ exists and the CAT(1) inequality gives $d(m_1,m_2)\le d_S({\bar{m}}_1,{\bar{m}}_2)$, where $m_1:=\mathrm{mid}(x_i,x_{i+1})=f(x{)}_i$, $m_2:=\mathrm{mid}(x_{i+1},x_{i+2})=f(x{)}_{i+1}$ and ${\bar{m}}_1,{\bar{m}}_2$ are the comparison points of the model triangle. In the model, the two points lie at distances ${\xi }_i\mathrm{/}2$ and ${\xi }_{i+1}\mathrm{/}2$ from the vertex ${\bar{x}}_{i+1}$ with some included angle $\bar{\theta }$, so the cosine rule of [F4] gives $\cos d_S({\bar{m}}_1,{\bar{m}}_2)=\cos ({\xi }_i\mathrm{/}2)\cos ({\xi }_{i+1}\mathrm{/}2)+\sin ({\xi }_i\mathrm{/}2)\sin ({\xi }_{i+1}\mathrm{/}2)\cos \bar{\theta }\ge \cos (({\xi }_i+{\xi }_{i+1})\mathrm{/}2)$ by $\cos \bar{\theta }\ge -1$ and the addition formula; since $({\xi }_i+{\xi }_{i+1})\mathrm{/}2<l<\pi \mathrm{/}2$ and $\cos$ is decreasing on $[0,\pi ]$, we conclude $d(f(x{)}_i,f(x{)}_{i+1})\le ({\xi }_i+{\xi }_{i+1})\mathrm{/}2$. If one of the two half-edges vanishes the midpoint coincides with the vertex and the same conclusion is immediate, so the estimate holds for all tuples of mesh $<l$.

Continuity. Endpoint distances are continuous by the reverse triangle inequality, so their finite maximum mesh and finite sums $L,E$ are continuous. To check $f$ at $x$ of mesh $<l$, fix $i$ and choose $R$ with $d(x_i,x_{i+1})<R<l$. The fixed chart $Z=\bar{B}(x_i,R)$ is CAT(1). If $x^k\to x$, then for large $k$ both endpoints $x_i^k,x_{i+1}^k$ belong to $Z$. Their short geodesic in $Z$ is also the geodesic defining $f(x^k{)}_i$: both geodesics are contained in $\bar{B}(x_i^k,l)$, since every point on either is at distance at most their endpoint distance $<l$ from $x_i^k$; uniqueness in that CAT(1) ball identifies them. Continuous endpoint dependence applied in the fixed CAT(1) space $Z$ now gives $f(x^k{)}_i\to f(x{)}_i$. There are finitely many indices, so $f(x^k)\to f(x)$.

$P_h(n)$ is compact. The function mesh is continuous on $X^n$ by step 1.3, so $P_h(n)={\mathrm{mesh}}^{-1}([0,h])$ is closed in $X^n$; the sup metric has the finite product topology (a radius-$r$ ball is a product of coordinate radius-$r$ balls, and every finite product neighborhood contains such a ball), so $X^n$ is compact by [F9], hence $P_h(n)$, a closed subspace of a compact space, is compact by [F9].

Monotonicity and the self-map property. For $x\in P_h(n)$ and each $i$, step 1.2 gives $d(f(x{)}_i,f(x{)}_{i+1})\le ({\xi }_i+{\xi }_{i+1})\mathrm{/}2$ with ${\xi }_i=d(x_i,x_{i+1})$; summing gives $L(fx)\le \frac{1}{2}{\sum }_i({\xi }_i+{\xi }_{i+1})=L(x)$, and summing squares gives $E(fx)\le \frac{1}{4}{\sum }_i({\xi }_i+{\xi }_{i+1}{)}^2$, while ${\sum }_i({\xi }_i+{\xi }_{i+1}{)}^2=2{\sum }_i{\xi }_i^2+2{\sum }_i{\xi }_i{\xi }_{i+1}\le 2{\sum }_i{\xi }_i^2+{\sum }_i({\xi }_i^2+{\xi }_{i+1}^2)=4{\sum }_i{\xi }_i^2=4E(x)$ by $2ab\le a^2+b^2$ ([F12]), so $E(fx)\le E(x)$. Also $d(f(x{)}_i,f(x{)}_{i+1})\le ({\xi }_i+{\xi }_{i+1})\mathrm{/}2\le \mathrm{mesh}(x)$, so $\mathrm{mesh}(fx)\le \mathrm{mesh}(x)\le h$: the midpoint operation maps $P_h(n)$ into itself, and it is defined on all of $P_h(n)$ because $h<l$. This proves (i)'s midpoint assertions and clause (ii).

Equality analysis. Suppose $E(fx)=E(x)$. Combining the two bounds of step 2.2, $E(fx)\le \frac{1}{4}{\sum }_i({\xi }_i+{\xi }_{i+1}{)}^2\le E(x)$, so equality forces both $\frac{1}{4}{\sum }_i({\xi }_i+{\xi }_{i+1}{)}^2=E(x)$, i.e. ${\sum }_i({\xi }_i-{\xi }_{i+1}{)}^2=0$ and hence all ${\xi }_i$ are equal to a common $\xi$, and all the pointwise inequalities of step 1.2 to be equalities: $d(f(x{)}_i,f(x{)}_{i+1})=({\xi }_i+{\xi }_{i+1})\mathrm{/}2=\xi$ for every $i$. If $\xi =0$ then all consecutive vertices coincide and $x$ is constant. If $\xi >0$, fix $i$; in the notation of step 1.2 the equality $d(m_1,m_2)=\xi$ combined with $d(m_1,m_2)\le d_S({\bar{m}}_1,{\bar{m}}_2)\le \xi$ forces equality throughout, so $\cos \bar{\theta }=-1$ and $\bar{\theta }=\pi$; the comparison triangle is degenerate with $d(x_i,x_{i+2})=\xi +\xi =2\xi$, so equality holds in the triangle inequality $d(x_i,x_{i+2})\le d(x_i,x_{i+1})+d(x_{i+1},x_{i+2})$ and the concatenation of the two geodesics $[x_i,x_{i+1}]$, $[x_{i+1},x_{i+2}]$ is a geodesic segment with midpoint $x_{i+1}$. Conversely, if $x$ is constant then $fx=x$ and $E(fx)=E(x)=0$; and if all edge lengths equal $\xi >0$ and every triple is straight, then each $d(f(x{)}_i,f(x{)}_{i+1})$ equals $\xi$ as the distance between the two points at distance $\xi \mathrm{/}2$ from $x_{i+1}$ on a common geodesic through it, so $E(fx)=n{\xi }^2=E(x)$. In the nonconstant case the concatenation of the $n$ segments $[x_i,x_{i+1}]$, parametrized on $S_{n\xi }^1$ by arclength, is by construction locally isometric at every point (at the vertices by straightness of the corresponding triple, inside the edges by the geodesic property) and has length $n\xi$, so it is a closed local geodesic on which $x_0,\dots ,x_{n-1}$ are equally spaced. For $n=2$ the two edges are $d(x_0,x_1)=d(x_1,x_0)=\xi$ and $f(x{)}_0=f(x{)}_1$ is their midpoint, so $d(f(x{)}_0,f(x{)}_1)=0$ and equality $E(fx)=E(x)=2{\xi }^2$ forces $\xi =0$: only the constant tuple.

Convergence in the basin. Let $x\in C_h^0(n)$, so $L_k:=L(f^{k}x)\to 0$ by [F1]. If $L_m=0$ for some $m$, that iterate is constant and fixed by $f$, proving convergence. Otherwise all $L_k>0$; fix $m$ with $L_m<l\mathrm{/}2$; then each vertex $(f^{m}x{)}_j$ is at distance at most $L_m\mathrm{/}2$ from $(f^{m}x{)}_0$, because the two arcs of the polygon from $(f^{m}x{)}_0$ to $(f^{m}x{)}_j$ have lengths summing to $L_m$, so the shorter one is at most $L_m\mathrm{/}2$. The closed ball $\bar{B}((f^{m}x{)}_0,L_m\mathrm{/}2)$ has radius $<l\mathrm{/}4<\pi \mathrm{/}2$ and is contained in the CAT(1) ball $\bar{B}((f^{m}x{)}_0,l)$, so it is convex by [F4]; hence it contains all vertices of $f^{m}x$ and, by induction, all vertices of $f^{k}x$ for $k\ge m$, since each next iterate has as vertices midpoints of pairs of vertices of the previous one. Therefore $d((f^{k}x{)}_i,(f^{l}x{)}_i)\le L_m$ for all $k,l\ge m$ and all $i$, so $(f^{k}x)$ is Cauchy in the compact metric space $X^n$ and converges to some $z\in X^n$ ([F8], [F11]); taking limits in $d(z_i,z_j)\le L_m$ for arbitrary $m$ and using $L_m\to 0$ shows that $z$ is constant and the diameters of the iterates tend to $0$. Since $L$ is continuous by step 1.3, $L(f^{k}x)\to L(z)=0$. Conversely, if $f^{k}x\to z$ with $z$ constant, then $L(f^{k}x)\to L(z)=0$ by continuity of $L$, so $x\in C_h^0(n)$.

Degenerate and zero-length cases. If $E(x)=0$ then every ${\xi }_i=0$, so all consecutive vertices of $x$ coincide and $x$ is constant; then $fx=x$ by the convention of [F1], and (ii) and (iii) hold for it. If some consecutive vertices of $x$ coincide while $x$ is not constant, the corresponding edge contributes $0$ to mesh, $L$ and $E$, the midpoint of a degenerate pair is the point itself by [F1], and the proofs of steps 1.2, 2.2 and 3.1 go through verbatim: if a half-edge vanishes the model midline estimate reduces to the immediate bound, and a collapsed comparison triangle is the limiting case of degenerate comparison triangles, in which the cosine-rule identity and the CAT(1) inequality remain valid by continuity; the equality case then includes the possibility that a midpoint coincides with a vertex, and the straightness conclusion is unchanged.

The eventually-short set is the basin and is open. Write $U=\{x\in P_h(n):L(f^{m}x)<l\mathrm{/}2\text{ for some }m\}$. For $x\in U$, put $y=f^{m}x$, $c=y_0$ with $L(y)<l\mathrm{/}2$. Its vertices lie in the convex CAT(1) ball $B_0=\bar{B}(c,l\mathrm{/}4)$, so all later vertices do too. For $\epsilon >0$ the set $K_{\epsilon }=\{z\in P_h(n):z_i\in B_0\text{ for every }i,E(z)\ge \epsilon \}$ is compact. If nonempty, the continuous deficit $D(z)=E(z)-E(fz)$ is strictly positive there: equality would give a nonconstant closed local geodesic by step 3.1; every edge remains in $B_0$ by convexity, so the entire curve would be a closed local geodesic in the CAT(1) space $B_0$, contradicting [F2] since $\mathrm{diam}{B}_0\le l\mathrm{/}2<\pi$. Thus $D$ has a positive minimum on nonempty $K_{\epsilon }$, and the nonnegative energy of $f^{k}y$ must eventually fall below $\epsilon$. If $K_{\epsilon }$ is empty, it is already below $\epsilon$. Hence $E(f^{k}y)\to 0$ and $L(f^{k}y)\to 0$ by $L^2\le nE$. This proves $U\subseteq C_h^0(n)$; the reverse inclusion follows from $L(f^{k}x)\to 0$. Finally $U$ is the union of the open sets $\{L\circ f^m<l\mathrm{/}2\}$, since $f$ and $L$ are continuous.

Conclusion. Clause (i) is step 1.1 for the radius, step 2.2 for the self-map, step 2.1 for compactness and step 1.3 for continuity; clause (ii) is step 2.2; clause (iii) is step 3.1; clause (iv) is step 4.2; clause (v) is step 3.2 together with step 4.2 for the identification of $U$ with the basin; clause (vi) is step 4.1. This proves all assertions.