Midpoint iteration on a small equilateral spherical triangle contracts geometrically to its centre

The actual midpoint side. Represent the vertices by unit vectors $u_0,u_1,u_2$ with $u_i\cdot u_j=c:=\cos s\in (0,1)$ for $i\mathrm{\ne }j$. The midpoint of the short arc from $u_i$ to $u_j$ is $(u_i+u_j)\mathrm{/}\sqrt{2+2c}$. Two adjacent such midpoints have dot product $(u_i+u_j)\cdot (u_j+u_k)\mathrm{/}(2+2c)=(1+3c)\mathrm{/}(2+2c)$, so their spherical distance is $s_1=\arccos ((1+3c)\mathrm{/}(2+2c))$. The rotational symmetry fixes the north pole and permutes these midpoints, so the midpoint triangle is equilateral with that same centre.

Strict contraction. For $0<c<1$, the quotient $u=(1+3c)\mathrm{/}(2+2c)$ satisfies $u<1$ and $u-c=(1-c)(1+2c)\mathrm{/}(2+2c)>0$. Thus $0<s_1=\arccos u<\arccos c=s$ by [L3].

The midpoint triangle is equilateral (i). The symmetry group of the equilateral triangle acts transitively on its vertices and fixes the centre $N$; the midpoint operation is equivariant under isometries of $S^2$, so $fx$ is again an equilateral triangle with centre $N$ and side $s_1$; hence $L(fx)=3s_1$, $E(fx)=3s_1^2$ and $\mathrm{mesh}(fx)=s_1$, and step 2.1 gives the strict inequalities of (i).

The iteration contracts to zero (ii). By step 3.1 applied to each iterate, $f^{k}x$ is equilateral with side $s_k$, where $s_{k+1}=\arccos ((1+3\cos s_k)\mathrm{/}(2+2\cos s_k))$, and $(s_k)$ is strictly decreasing and bounded below by $0$; hence it converges to some $\lambda \ge 0$ by A monotone sequence converges if and only if it is bounded, and continuity of the recursion ([L1], [L3]) gives $c_{\mathrm{\infty }}=(1+3c_{\mathrm{\infty }})\mathrm{/}(2+2c_{\mathrm{\infty }})$ with $c_{\mathrm{\infty }}=\cos \lambda >0$, so $(c_{\mathrm{\infty }}-1)(2c_{\mathrm{\infty }}+1)=0$ and $c_{\mathrm{\infty }}=1$, hence $\lambda =0$. Moreover $1-\cos s_{k+1}=(1-\cos s_k)\mathrm{/}(2+2\cos s_k)\le (1-\cos s_k)\mathrm{/}2$, so $\sin (s_k\mathrm{/}2)\le 2^{-k\mathrm{/}2}\sin (s_0\mathrm{/}2)$. Since $s_k\le s_0<\pi \mathrm{/}2$, The limit of sin x divided by x at zero is one bounds $u\mathrm{/}\sin u$ near zero; away from zero its numerator is bounded and its denominator is bounded below by [L3]. Thus $s_k\le C{2}^{-k\mathrm{/}2}$ for some $C>0$, establishing the geometric rate in the title. Therefore $L(f^{k}x)=3s_k\to 0$, so $x\in C_h^0(3)$ by the description of the basin [L2]. Moreover $L(f^{k}x)=3s_k\in (0,2\pi )$ for every $k$, so the uniform decrement of [L5] applies to each iterate and gives $E(f^{k+1}x)\le E(f^{k}x)-{\lambda }_3(3s_k)$, in agreement with the strict drop computed in step 3.1.

The equality case does not occur (iii). By step 3.1, $E(fx)=3s_1^2<3s^2=E(x)$ for every $s>0$, so the tuple is neither constant nor equally spaced along a closed local geodesic; the deficit $E(x)-E(fx)$ is strictly positive and the equality characterization of L2 does not apply.

Convergence to the constant tuple. The three vertex vectors have equal dot product with the north-pole vector $N$ and sum to a positive multiple of $N$. Thus ${\cos }^2{\rho }_k=(1+2\cos s_k)\mathrm{/}3$, by squaring their sum; since $s_k\to 0$, ${\rho }_k\to 0$; hence the iterates converge uniformly to the constant tuple at the north pole.

Conclusion. Clause (i) is steps 1.1, 2.1 and 3.1, clause (ii) is steps 4.1 and 5.1, and clause (iii) is step 4.2: the midpoint iteration on a small equilateral spherical triangle stays equilateral, contracts the side strictly, converges to the centre and realises a strict energy drop at every step.