Quasisymmetry, Welding, and Conformal Removability

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.

Apply [F1] to regard f as an analytic K-QC map. The exact Borel-set area formula in [F2] gives ${\int }_E{J}_f=\mathrm{\mid }f(E)\mathrm{\mid }$, hence the claimed lower inequality. No unrecovered Gehring–Lehto source is used; the complete differentiability and signed-degree arguments are in the earlier12 suppliers.

The Beltrami bound gives $J_f\ge (1-k^2)\mathrm{\mid }{f}_z{\mathrm{\mid }}^2$, and [F3] gives $\mathrm{\mid }Df{\mathrm{\mid }}_{\mathrm{H}\mathrm{S}}^2\le 2(1+k^2)\mathrm{\mid }{f}_z{\mathrm{\mid }}^2\le 2(1+k^2)(1-k^2{)}^{-1}{J}_f$. Integrate and use step 1.1 to obtain the stated constant.

If $K=\mathrm{\varnothing }$, the assertion is the hypothesis. Otherwise, [F10] makes the image under $h$ of each component of $U\setminus K$ open, so both it and the source component are complex domains. By [F1], the map on each component is analytically $K_0$-quasiconformal. Its weak coordinate derivatives are locally square integrable there and satisfy the same Beltrami bound.

The set $K$ has planar measure zero by [F3]. Fix an open square $R$ with $\bar{R}\subset U$ and put $O=R\setminus K$. Choose the maximal dyadic squares $Q_j$ whose closures lie in $O$. They have disjoint interiors and cover $O$ except for the countable union of dyadic grid lines; each grid-line segment inside $R$ is null because it lies in a degenerate rectangle of measure zero. Every point of $O$ off those grid lines belongs to a sufficiently small dyadic square with closure in $O$, and then to a maximal such square. Each $\overline{Q_j}$ lies in one component of $U\setminus K$, so [F2] gives ${\int }_{Q_j}\mathrm{\mid }Dh{\mathrm{\mid }}_{\mathrm{H}\mathrm{S}}^{2}dA\le C(K_0){\lambda }_2(h(Q_j))$. The images $h(Q_j)$ are pairwise disjoint open sets because $h$ is a homeomorphic embedding and [F10] makes it open; they lie in $h(\bar{R})$, which is bounded by [F11]. Hence [F9] gives ${\sum }_j{\lambda }_2(h(Q_j))={\lambda }_2\text{\hspace{0.17em}⁣}({\bigcup }_{j}h(Q_j))\le {\lambda }_2(h(R))<\mathrm{\infty }\mathrm{.}$ Countable additivity and the null grid lines therefore give ${\int }_{R\setminus K}\mathrm{\mid }Dh{\mathrm{\mid }}_{\mathrm{H}\mathrm{S}}^{2}dA\le C(K_0){\lambda }_2(h(R))<\mathrm{\infty }\mathrm{.}$

Extend each weak coordinate derivative $g_i$ from $O$ by $0$ on $K\cap R$. It is measurable, and step 1.2 gives $g_i\in L^2(R)$. By [F4], for almost every horizontal line and almost every vertical line through $R$, the restriction of $g_i$ is in $L^2$; [F8] then puts it in $L^1$ on that bounded line interval.

Also discard the null line families on a countable rational-box cover of $O$ where the ACL representative or its agreement almost everywhere with $h$ fails; [F4]–[F5] justify this common exceptional family. Fix one of the remaining good coordinate lines. Its intersection with $K$ is finite except for at most one exceptional line when $K$ lies in a straight line parallel to the chosen direction; that one line is a null member of the parallel family. For a circle there are at most two intersection points on every coordinate line. On each open interval left after removing these finitely many points, [F1] and [F5] give an absolutely continuous representative with derivative $g_i$ a.e. The representative agrees a.e. with the continuous restriction of $h$, so the two agree everywhere on each such interval. Since $g_i\in L^1$ on the whole line interval, [F6] and continuity of $h$ at the finitely many missing points show that the restriction of $h$ on the full interval is the indefinite integral of $g_i$ plus a constant. It is therefore absolutely continuous across every point of $K$.

Applying step 2.1 in both coordinate directions on a countable rational-box cover of $U$ shows that $h$ itself is ACL on $U$. On every relatively compact rational box, the derivatives $g_i$ belong to $L^2$ by step 1.2; the ACL characterization [F5] therefore identifies them as the weak derivatives of $h$, so $h\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}(U)$. Since $K$ has planar measure zero, the inequality $\mathrm{\mid }{h}_{\bar{z}}\mathrm{\mid }\le k_0\mathrm{\mid }{h}_z\mathrm{\mid }$ continues to hold almost everywhere on $U$. Hence $h$ is analytically $K_0$-quasiconformal on $U$, and [F1] gives geometric $K_0$-quasiconformality with the same bound.

Let $H$ be a sphere homeomorphism that is $K_0$-quasiconformal off a generalized circle $\mathrm{\Sigma }$, in the chartwise sense of the Statement. Choose a finite point $p\mathrm{\notin }\mathrm{\Sigma }$ and Möbius charts $\varphi ,\psi$ sending $p,H(p)$ to $\mathrm{\infty }$, respectively (if $H(p)=\mathrm{\infty }$, take $\psi$ to be the finite chart). Then $g=\psi \circ H\circ {\varphi }^{-1}:\mathbb{C}\to \mathbb{C}$ is a homeomorphism. The set $K=\varphi (\mathrm{\Sigma })$ is a finite round circle: write $\mathrm{\Sigma }$ as $\alpha \mathrm{\mid }z{\mathrm{\mid }}^2+2\mathrm{Re}(\beta z)+\gamma =0$ with $\alpha ,\gamma \in \mathbb{R}$; under $w=1\mathrm{/}(z-p)$, multiplication by $\mathrm{\mid }w{\mathrm{\mid }}^2$ gives $P(p)\mathrm{\mid }w{\mathrm{\mid }}^2+2\mathrm{Re}((\alpha p+\bar{\beta })w)+\alpha =0$, where $P(p)=\alpha \mathrm{\mid }p{\mathrm{\mid }}^2+2\mathrm{Re}(\beta p)+\gamma \mathrm{\ne }0$. This is a Euclidean circle, hence compact by [F11]. By [F7] and the quasiconformal composition interface, $g$ is $K_0$-quasiconformal off $K$. The planar assertion applies to compact $K\subset \mathbb{C}$ with $U=\mathbb{C}$, so $g$ is $K_0$-quasiconformal on the whole finite chart. The omitted source point $p$ lies off $\mathrm{\Sigma }$, where $H$ was already quasiconformal. This proves the sphere assertion with the same bound.

Put $a(x,y):=y^{-1}{\int }_x^{x+y}h(t)dt$ and $b(x,y):=y^{-1}{\int }_{x-y}^{x}h(t)dt$, so $H=u+iv$ with $u=(a+b)\mathrm{/}2$ and $v=a-b$. These ordinary Riemann integrals exist because $h$ is continuous on every finite interval by [F13]. For example, the numerator in the difference quotient for $a_x$ is ${\int }_{x+y}^{x+y+s}h(t)dt-{\int }_x^{x+s}h(t)dt$; dividing by $s$ and using continuity gives $h(x+y)-h(x)$. The endpoint quotient for $a_y$, followed by differentiating the factor $1\mathrm{/}y$, gives $R=(h(x+y)-a)\mathrm{/}y$; the same endpoint computation for $b$ gives $Q=(h(x)-h(x-y))\mathrm{/}y$ and $-S=b_y=(h(x-y)-b)\mathrm{/}y$. Thus $P:=a_x=(h(x+y)-h(x))\mathrm{/}y$, $Q:=b_x=(h(x)-h(x-y))\mathrm{/}y$, $R:=a_y=(h(x+y)-a)\mathrm{/}y$, and $S:=-b_y=(b-h(x-y))\mathrm{/}y$. These partial derivatives are continuous for $y>0$, so [F3] makes $H$ $C^1$ on $\mathbb{H}$.

The imaginary part has the symmetric form $v(x,y)=y^{-1}{\int }_0^y(h(x+s)-h(x-s))ds>0$, since $h$ is strictly increasing. As $(x,y)\to (x_0,0)$ with $y\downarrow 0$, continuity of $h$ makes both interval averages $a(x,y)$ and $b(x,y)$ tend to $h(x_0)$; hence $u\to h(x_0)$ and $v\to 0$. Thus $H$ maps $\mathbb{H}$ into itself and has the asserted continuous boundary values.

From the formulas in step 1.1, $u_x=(P+Q)\mathrm{/}2$, $u_y=(R-S)\mathrm{/}2$, $v_x=P-Q$, and $v_y=R+S$. Strict monotonicity gives $P,Q>0$ and the averages satisfy $h(x)<a(x,y)<h(x+y)$ and $h(x-y)<b(x,y)<h(x)$, hence $R,S>0$. Therefore $J_H=u_x{v}_y-u_y{v}_x=PS+QR>0$. Applying [F4] at each point proves the local-diffeomorphism clause.

Adjacent equal intervals give $P\mathrm{/}L\le Q\le LP$. Also $R\le P$ and $S\le Q$. In fact, $R=y^{-2}{\int }_x^{x+y}(h(x+y)-h(t))dt\ge [h(x+y)-h(x+y\mathrm{/}2)]\mathrm{/}(2y)\ge P\mathrm{/}[2(1+L)]$, since the two adjacent half-increments of $[x,x+y]$ have sum $yP$ and ratio at most $L$. Similarly, $S=y^{-2}{\int }_{x-y}^x(h(t)-h(x-y))dt\ge [h(x-y\mathrm{/}2)-h(x-y)]\mathrm{/}(2y)\ge Q\mathrm{/}[2(1+L)]$. Thus, with $c:=1\mathrm{/}[2(1+L)]$ and $m:=c\mathrm{/}L=1\mathrm{/}[2L(1+L)]$, every one of $P,Q,R,S$ lies between $mP$ and $LP$.

To prove growth at infinity, note that $u$ is the average of $h$ on $[x-y,x+y]$ and $v\ge \frac{1}{2}[h(x+y\mathrm{/}2)-h(x-y\mathrm{/}2)]$. If $y\le \mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$, that interval lies in one tail, so $u\ge h(x\mathrm{/}2)$ for $x>0$ and $u\le h(-\mathrm{\mid }x\mathrm{\mid }\mathrm{/}2)$ for $x<0$. If $y\ge x\mathrm{/}2$ and $x>0$, the symmetric interval contains $[3x\mathrm{/}4,5x\mathrm{/}4]$; comparison across four adjacent intervals of length $x\mathrm{/}4$ gives $v\ge \frac{1}{2}[h(5x\mathrm{/}4)-h(x)]\ge (2L^4{)}^{-1}[h(x\mathrm{/}4)-h(0)]$. If $y\ge \mathrm{\mid }x\mathrm{\mid }\mathrm{/}2$ and $x<0$, it contains $[5x\mathrm{/}4,3x\mathrm{/}4]$; comparison across three adjacent intervals of length $\mathrm{\mid }x\mathrm{\mid }\mathrm{/}4$ gives $v\ge \frac{1}{2}[h(3x\mathrm{/}4)-h(x)]\ge (2L^3{)}^{-1}[h(0)-h(x\mathrm{/}4)]$. For every $A>0$, choose $X$ so these tail bounds force $\mathrm{\mid }u\mathrm{\mid }>A$ in the first case and $v>A$ in the second whenever $\mathrm{\mid }x\mathrm{\mid }>X$. When $\mathrm{\mid }x\mathrm{\mid }\le X$, choose $Y$ so large that $v\ge \frac{1}{2}[h(y\mathrm{/}2-X)-h(X-y\mathrm{/}2)]>A$ for $y>Y$. If $\mathrm{\mid }x+iy\mathrm{\mid }>\sqrt{X^2+Y^2}$, either $\mathrm{\mid }x\mathrm{\mid }>X$ or $\mathrm{\mid }x\mathrm{\mid }\le X$ and $y>Y$; the preceding estimates then give $\mathrm{\mid }H(x+iy)\mathrm{\mid }>A$. Hence $\mathrm{\mid }H(x+iy)\mathrm{\mid }\to \mathrm{\infty }$ as $\mathrm{\mid }x+iy\mathrm{\mid }\to \mathrm{\infty }$.

The derivative formulas now imply $\mathrm{\mid }{u}_x\mathrm{\mid },\mathrm{\mid }{u}_y\mathrm{\mid },\mathrm{\mid }{v}_x\mathrm{\mid }\le LP$ and $\mathrm{\mid }{v}_y\mathrm{\mid }\le 2LP$, so $\mathrm{\parallel }DH{\mathrm{\parallel }}_{HS}^2\le 7L^2{P}^2$. Moreover $PS\ge m{P}^2$ and $QR\ge (P\mathrm{/}L)cP=m{P}^2$, whence $J_H\ge 2m{P}^2\ge m{P}^2$ and $\mathrm{\parallel }DH{\mathrm{\parallel }}_{HS}^2\le C(L)J_H$ for $C(L)=7L^2\mathrm{/}m=14L^3(1+L)$. By [F5], $2(\mathrm{\mid }{H}_z{\mathrm{\mid }}^2+\mathrm{\mid }{H}_{\bar{z}}{\mathrm{\mid }}^2)=\mathrm{\parallel }DH{\mathrm{\parallel }}_{HS}^2$ and $J_H=\mathrm{\mid }{H}_z{\mathrm{\mid }}^2-\mathrm{\mid }{H}_{\bar{z}}{\mathrm{\mid }}^2$; rearranging yields $\mathrm{\mid }{H}_{\bar{z}}{\mathrm{\mid }}^2\le [(C(L)-2)\mathrm{/}(C(L)+2)]\mathrm{\mid }{H}_z{\mathrm{\mid }}^2$, which is clause (c).

Let $E\subset \mathbb{H}$ be compact and nonempty. By [F9], $\mathrm{\mid }w\mathrm{\mid }$ has a finite maximum $M$ on $E$ and $\mathrm{Im}w$ has a positive minimum $\delta$ there. Step 2.3 bounds $\mathrm{\mid }z\mathrm{\mid }$ on $H^{-1}(E)$; choose $R$ larger than that bound. The continuous extension from step 1.2 is uniformly continuous on the compact rectangle $[-R,R]\times [0,R]$ by [F9] and [F10]; since its imaginary part is zero on the bottom edge, there is $0<ϵ<R$ such that $H^{-1}(E)$ contains no point with $0<y<ϵ$. Thus $H^{-1}(E)$ lies in the compact rectangle $[-R,R]\times [ϵ,R]$ and is closed there, because $E$ is closed in $\mathbb{H}$ and $H$ is continuous. By [F9] it is compact. The empty $E$ has empty preimage, so $H$ is proper as defined in [F7].