Quasisymmetric homeomorphisms of the line and circle

For the line, take $I=[x-t,x]$ and $J=[x,x+t]$. Since $h$ is increasing, $\mathrm{\mid }h(I)\mathrm{\mid }=h(x)-h(x-t)$ and $\mathrm{\mid }h(J)\mathrm{\mid }=h(x+t)-h(x)$; swapping the adjacent pair gives the reciprocal inequality. This proves the displayed two-sided ratio criterion with the same $L$, and conversely that criterion bounds both orders of every adjacent equal pair. If $L=1$, equality holds for every adjacent equal pair, so $h(x+t)+h(x-t)=2h(x)$. The continuous midpoint identity, first iterated for dyadic subdivisions and then extended by continuity, gives $h(x)=ax+b$; monotonicity forces $a>0$. Conversely every such affine map preserves all adjacent length ratios.

On the round circle, if an arc has angular length $s\in [0,\pi ]$, its chord has length $2\sin (s\mathrm{/}2)$; hence $2s\mathrm{/}\pi \le \chi \le s$. This proves the uniform comparison of arc and chord distances used to pass between the circle's arc metric and its chordal metric. For $L=1$, adjacent equal arcs have equal image lengths. Partitioning the circle into $n$ equal arcs shows that every such arc maps to an arc of length $2\pi \mathrm{/}n$, independently of its starting point. Additivity gives preservation of rational arc lengths; continuity of $h$ gives preservation of every arc length. Thus $h$ is a rotation, and rotations plainly have constant $1$.

If $g$ has control ${\eta }_g$ and $h$ has control ${\eta }_h$, applying the first inequality to $g$ and then to $h$ gives ${\eta }_{h\circ g}={\eta }_h\circ {\eta }_g$. For the inverse, suppose $\chi (h(x),h(y))\le t\chi (h(x),h(z))$. If $\chi (x,y)\mathrm{/}\chi (x,z)>1\mathrm{/}{\eta }_h^{-1}(1\mathrm{/}t)$, then the forward control applied to the pair $(z,y)$ at base point $x$ gives $\chi (h(x),h(z))<t^{-1}\chi (h(x),h(y))$, a contradiction. Therefore $h^{-1}$ has the stated control. These formulas prove closure and show why an inverse constant need only depend on the forward constant.

Write an automorphism as $\phi (z)=e^{i\theta }(a-z)\mathrm{/}(1-\bar{a}z)$, so $\mathrm{\mid }a\mathrm{\mid }=\mathrm{\mid }\phi (0)\mathrm{\mid }\le r$. On $\mathrm{\mid }z\mathrm{\mid }=1$, the angular derivative is $\mathrm{\mid }{\phi }^{\mathrm{\prime }}(z)\mathrm{\mid }=(1-\mathrm{\mid }a{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }1-\bar{a}z{\mathrm{\mid }}^2$, which lies between $(1-r)\mathrm{/}(1+r)$ and $(1+r)\mathrm{/}(1-r)$. Image arc length is the integral of this derivative, so the ratio for any adjacent equal arcs is at most $L(r)=((1+r)\mathrm{/}(1-r){)}^2$.

Write $d$ for shortest arc distance and $\mu (I)$ for the length of the image of an oriented arc $I$. Put $q=L\mathrm{/}(1+L)<1$ and $\alpha =-{\log }_{2}q>0$. Either half of an arc has image length at most $q$ times its parent's; hence each depth-$n$ dyadic cell has image length at most $q^n\mu (I)$. Given $d(x,z)=b\le \pi$, let $B$ be a shortest arc from $x$ to $z$. Its complement contains the opposite initial arc of length $b$, whose image length is at least $\mu (B)\mathrm{/}L$. Consequently $\mu (B)\mathrm{/}L\le d(h(x),h(z))\le \mu (B)$. If $a=d(x,y)\le b$, the shortest arc from $x$ to $y$ lies in $B$ or in that opposite initial arc. With $n=\lfloor {\log }_2(b\mathrm{/}a)\rfloor$, it is covered by at most two depth-$n$ cells there, so $d(h(x),h(y))\mathrm{/}d(h(x),h(z))\le 2L^2{q}^n\le (2L^2\mathrm{/}q)(a\mathrm{/}b{)}^{\alpha }$. If $a>b$, divide the shortest arc to $y$ into $m=\lceil a\mathrm{/}b\rceil$ consecutive pieces, all of length $b$ except possibly the last. Extend the last piece to length $b$ for comparison. All comparisons use adjacent length-$b$ arcs, with $b<\pi$, so their image lengths are bounded successively by $L,L^2,\dots ,L^m$ times $\mu (B)$, including the opposite-side initial comparison when needed. Thus the image distance ratio is at most $m{L}^{m+1}$. With $C=2L^2\mathrm{/}q$, a continuous increasing control dominating both bounds is ${\eta }_d(t)=C(L+1{)}^2{t}^{\alpha }$ for $0\le t\le 1$ and ${\eta }_d(t)=Ct(L+1{)}^{t+1}$ for $t\ge 1$. The comparison $2d\mathrm{/}\pi \le \chi \le d$ in step 1.2 gives chordal control ${\eta }_{\chi }(t)=(\pi \mathrm{/}2){\eta }_d(\pi t\mathrm{/}2)$. The same subdivision and consecutive-interval argument on the line, without the arc/complement comparison, supplies metric control there as well. No external weak-to-full quasisymmetry theorem is needed.

Full chordal control implies the symmetric-triple test with $M=\max (1,{\eta }_{\chi }(1))$, because equal angular offsets less than $\pi$ give equal input chords. Conversely suppose the symmetric-triple test holds with constant $M$. Let $I,J$ be adjacent equal arcs of length $s\le \pi$ with common endpoint $x$, and put $u=\mu (I)$, $v=\mu (J)$. If $u,v\le \pi$, step 1.2 and the test give $u\le (\pi M\mathrm{/}2)v$; the endpoint case $s=\pi$ follows by continuity from $s<\pi$. If $u>\pi$, an interior point $w$ of $I$ maps to the antipode of $h(x)$. Its input offset is some $t<s\le \pi$; the point $w^{\mathrm{\prime }}$ at the same offset on the opposite side of $x$ lies in $J$. The test gives $\chi (h(x),h(w^{\mathrm{\prime }}))\ge 2\mathrm{/}M$, whence $v\ge 2\mathrm{/}M$ and $u\mathrm{/}v\le \pi M$. If $u\le \pi <v$, the ratio is at most one. Interchanging $I,J$ proves the reciprocal bound, so the arc definition holds with $L=\pi M$. This proves both equivalences with control depending only on the specified control data. Together with step 1.3 it proves composition and inversion for the original definitions.

For $a\in (0,1)$ real, take ${\phi }_a(z)=(z-a)\mathrm{/}(1-az)$. If $0<\theta <\pi$, write ${\phi }_a(e^{i\theta })=e^{i{A}_a(\theta )}$ with $0<A_a(\theta )<\pi$, and set $A_a(0)=0$. Substituting $e^{i\theta }=(1+i\tan (\theta \mathrm{/}2))\mathrm{/}(1-i\tan (\theta \mathrm{/}2))$ gives $\tan (A_a(\theta )\mathrm{/}2)=((1+a)\mathrm{/}(1-a))\tan (\theta \mathrm{/}2)$. Hence $A_a(\theta )\to \pi$ as $a\uparrow 1$ for every fixed $0<\theta <\pi$. For fixed $0<\delta <\pi \mathrm{/}2$, the image lengths of $[0,\delta ]$ and $[\delta ,2\delta ]$ are $A_a(\delta )$ and $A_a(2\delta )-A_a(\delta )$; these tend to $\pi$ and zero, respectively. Their ratio is unbounded, so the full disc-automorphism group has no common quasisymmetry constant.