Quasisymmetry, Welding, and Conformal Removability: Examples and Counterexamples

For $x>0$, [F1] and [F9] show that $h_p$ is strictly increasing and continuous on $(0,\mathrm{\infty })$. As $x\downarrow 0$, $\log x\to -\mathrm{\infty }$ because $\log$ is increasing and onto, so $x^p=\exp (p\log x)\to 0$ by [F1]; hence $h_p$ is continuous at $0$. The inverse is $h_{1\mathrm{/}p}$ by [F1], and the same endpoint argument makes it continuous at $0$. Thus $h_p$ is an increasing homeomorphism of the closed half-line.

By [F3], $g_p(s)=(s+1{)}^p-s^p=p{\xi }_s^{p-1}$ for some ${\xi }_s\in (s,s+1)$, and $g_p(s+1)=p{\eta }_s^{p-1}$ for some ${\eta }_s\in (s+1,s+2)$. Thus $g_p(s+1)\mathrm{/}{g}_p(s)=({\eta }_s\mathrm{/}{\xi }_s{)}^{p-1}\to 1$ as $s\to \mathrm{\infty }$, since ${\eta }_s\mathrm{/}{\xi }_s\to 1$. This gives the displayed limit for $J_k\mathrm{/}{I}_k$. If $p=1$, $h_p(x)=x$ is affine. If $p\mathrm{\ne }1$, its second derivative $p(p-1)x^{p-2}$ is nonzero for all $x>0$, so $h_p$ cannot be affine.

Put ${\varphi }_{\delta }(\theta )=\theta +\delta \sin \theta$. By [F10], its derivative lies in $[1-\delta ,1+\delta ]$, so it is strictly increasing and satisfies ${\varphi }_{\delta }(\theta +2\pi )={\varphi }_{\delta }(\theta )+2\pi$; hence it induces an orientation-preserving circle homeomorphism. For every arc represented by $[a,b]$, the mean value theorem [F3] gives $(1-\delta )(b-a)\le {\varphi }_{\delta }(b)-{\varphi }_{\delta }(a)\le (1+\delta )(b-a)$. Adjacent equal arcs therefore have image-length ratios in either order at most $(1+\delta )\mathrm{/}(1-\delta )$, so $h_{\delta }$ is quasisymmetric by [F4].

Let $I=[x,x+t]$ and $J=[x+t,x+2t]$ with $x\ge 0,t>0$, and put $s=x\mathrm{/}t$. Their image-length ratio in this order is $R_p(s)=((s+1{)}^p-s^p)\mathrm{/}((s+2{)}^p-(s+1{)}^p)$. Set $u=1\mathrm{/}(s+1)\in (0,1]$; homogeneity gives $R_p(s)=[1-(1-u{)}^p]\mathrm{/}[(1+u{)}^p-1]$. If $p>1$, convexity makes equal-step increments nondecreasing, hence $R_p(s)\le 1$; the chord bounds $v^p\le v$ on $[0,1]$ and $(1+u{)}^p\le 1+(2^p-1)u$ on $[1,2]$ give $1\mathrm{/}{R}_p(s)\le 2^p-1$. If $0<p<1$, concavity makes the increments nonincreasing, hence $R_p(s)\ge 1$; the chord bounds $v^p\ge v$ on $[0,1]$ and $(1+u{)}^p\ge 1+(2^p-1)u$ on $[1,2]$ give $R_p(s)\le 1\mathrm{/}(2^p-1)$. For $p=1$, $R_p(s)=1$. At $s=0$ the endpoint ratio is $1\mathrm{/}(2^p-1)$ and its reverse is $2^p-1$, so these bounds give the sharp two-order constant stated in part (a).

The lift ${\psi }_p$ is a continuous increasing homeomorphism of $[0,2\pi ]$ fixing the endpoints, so identifying $0$ with $2\pi$ gives the stated circle homeomorphism fixing $1$. For $0<t<2\pi$, the arcs with parameters $[0,t]$ and $[2\pi -t,2\pi ]$ are adjacent and have equal length. Their image lengths are $2\pi (t\mathrm{/}(2\pi ){)}^p$ and $2\pi [1-(1-t\mathrm{/}(2\pi ){)}^p]$. By differentiability of $v^p$ at $v=1$, the second length divided by $t$ tends to $p$, while the first divided by $t$ is $(2\pi {)}^{1-p}{t}^{p-1}$. Their ratio tends to $0$ for $p>1$ and to $+\mathrm{\infty }$ for $0<p<1$, violating the two-order adjacent-arc bound in [F4].

Suppose a Möbius map $M$ restricts to $h_{\delta }$. By [F5], it is a holomorphic sphere homeomorphism; because it preserves ${\mathbb{S}}^1$, it maps $\mathbb{D}$ to one of the two complementary components, and the orientation-preserving boundary map forces $M(\mathbb{D})=\mathbb{D}$. The disk-automorphism form in [F5] is $M(z)=e^{i\vartheta }(a-z)\mathrm{/}(1-\bar{a}z)$. Since $h_{\delta }$ fixes $1$ and $-1$, the equations $M(1)=1$ and $M(-1)=-1$, namely $e^{i\vartheta }(a-1)=1-\bar{a}$ and $e^{i\vartheta }(a+1)=-(1+\bar{a})$, give $e^{i\vartheta }=-1$ and $a\in (-1,1)$. Hence $M(z)=(z-a)\mathrm{/}(1-az)$. Its angular derivative at $\theta$ is $(1-a^2)\mathrm{/}(1-2a\cos \theta +a^2)$; at $0$ and $\pi$ these derivatives multiply to $1$. The corresponding derivatives of $h_{\delta }$ are $1+\delta$ and $1-\delta$, whose product is $1-{\delta }^2\mathrm{\ne }1$. This contradiction proves that $h_{\delta }$ is not Möbius.

The lift $f_{\delta }(t)=t+\delta \sin t$ is increasing, has $f_{\delta }(t+2\pi )=f_{\delta }(t)+2\pi$, and is line-quasisymmetric with the constant from step 1.3. For $z=x+iy$ with $y>0$, its Ahlfors–Beurling extension is $G_{\delta }(z)=\frac{1}{2y}{\int }_{x-y}^{x+y}{f}_{\delta }(t)dt+\frac{i}{y}({\int }_x^{x+y}{f}_{\delta }(t)dt-{\int }_{x-y}^x{f}_{\delta }(t)dt)$; below the line set $G_{\delta }(z)=\overline{G_{\delta }(\stackrel{‾}{z})}$, and on the line set $G_{\delta }(t)=f_{\delta }(t)$. Its translation covariance gives $G_{\delta }(z+2\pi )=G_{\delta }(z)+2\pi$, so ${\tilde{h}}_{\delta }(e^{iz})=e^{i{G}_{\delta }(z)}$ is well defined on ${\mathbb{C}}^{\ast }$ and has boundary values $h_{\delta }$. The descent and its quasiconformal extension across $0$ and $\mathrm{\infty }$ are exactly the construction in the circle clause [F7], which supplies a quasiconformal sphere homeomorphism preserving $\mathbb{D}$ and ${\mathbb{S}}^1$; restricting it gives the claimed quasiconformal disc extension. Its AC and Countable Choice assumptions follow by [F6].

The maps in [F2] send $T$ into itself; direct substitution of its three vertices verifies this. Their triangles meet only at consecutive retained vertices, and nonconsecutive triangles are separated by at least $1\mathrm{/}6$: their real projections lie respectively in $[-1\mathrm{/}2,-1\mathrm{/}6]$, $[-1\mathrm{/}6,0]$, $[0,1\mathrm{/}6]$, $[1\mathrm{/}6,1\mathrm{/}2]$. For adjacent triangles, their cones at the common vertex have angular separation at least $\pi \mathrm{/}3$, so their intersection is just that vertex. The same maps send $V$ into $V$, as substitution of its four vertices verifies, and their interiors lie in the corresponding disjoint open real-coordinate strips. Iteration gives level-$n$ rhombi of diameter $3^{-n}$ with disjoint interiors. Define $\rho (t)$ by the nested triangles prescribed by the base-four digits of $t\in [0,1]$, interpreting $1$ by the all-three digits. Nested diameters tend to zero, so completeness gives a unique point. At a double expansion, the two addresses end in all-three and all-zero digits; their triangles shrink to the same consecutive vertex, so $\rho$ is well-defined. Parameters within $4^{-n}$ lie in the same or adjacent level-$n$ parameter intervals, whose triangles have union diameter at most $2\cdot 3^{-n}$; hence $\rho$ is continuous. The polygonal parametrizations differ uniformly from it by at most $3^{-n}$ and have the retained vertices as interval endpoints, proving surjectivity onto the Hausdorff limit. If two parameters have different first child addresses, triangle separation forces any common image to be their shared endpoint; the endpoint's only addresses are the corresponding all-three/all-zero tails. Thus the parameters agree, proving injectivity. The three outward copies of $T$ on the initial equilateral triangle meet only at the initial vertices: their endpoint cones again have separation at least $\pi \mathrm{/}3$, and away from the vertices they lie on different exterior sides. The concatenation of the three side parametrizations is therefore continuous, with equal endpoints and injective on $[0,1)$. It induces a continuous bijection from the compact circle to $S$, a homeomorphism by A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Hence $S$ is Jordan. Every level-$n$ rhombus has inradius $3^{-n}\mathrm{/}4$, so it contains an open axis-parallel square of side $3^{-n}\mathrm{/}4$; this is valid after any rotation, since that square's circumradius is less than the inradius.

The closed parametrisation traverses the three side parametrisations in cyclic order. By [F2], subdividing each side parameter interval into $4^n$ equal pieces inscribes exactly the $4^n$ retained level-$n$ edges on that side. Each has length $3^{-n}$, so the concatenated partition has polygonal sum $3\cdot 4^n{3}^{-n}=3(4\mathrm{/}3{)}^n$, which tends to $+\mathrm{\infty }$ by [F10]. By [F1], the limiting boundary is not rectifiable. A chord-arc curve is rectifiable by definition, so its chord-arc condition fails here.

Put $\alpha =\log 4\mathrm{/}\log 3$. For each $n$, the $3\cdot 4^n$ level-$n$ rhombi covering the three Koch sides have diameter $3^{-n}$. Since $4^n(3^{-n}{)}^{\alpha }=1$, these covers have total $\alpha$-cost $3$ and diameters tending to zero. Thus ${\mathcal{H}}^{\alpha }(S)\le 3<\mathrm{\infty }$.

Let $K$ be one side and $(U_j)$ any countable cover of $K$ with $\mathrm{diam}{U}_j\le \delta <1$. For a nonempty $U$ with $d=\mathrm{diam}U>0$, choose $n\ge 1$ with $3^{-n}\le d<3^{-n+1}$. At most $1024$ level-$n$ rhombi meet $U$: each contains an open axis-parallel square of side $3^{-n}\mathrm{/}4$, these squares are pairwise disjoint, and all such squares lie in one axis-parallel square of side $8\cdot 3^{-n}$; finite additivity and monotonicity of planar area give $N(3^{-n}{)}^2\mathrm{/}16\le 64(3^{-n}{)}^2$. The base-four parameter intervals of the cells meeting $U$ cover ${\rho }^{-1}(U)$, so ${\lambda }_1^{\ast }({\rho }^{-1}(U))\le 1024\cdot 4^{-n}\le 1024d^{\alpha }$. If $d=0$, then $U$ is empty or a singleton, and injectivity of $\rho$ makes its preimage empty or a singleton, of outer measure zero. Countable subadditivity and ${\lambda }_1^{\ast }([0,1])=1$ now give $1\le 1024{\sum }_j(\mathrm{diam}{U}_j{)}^{\alpha }$. Taking the infimum over every such cover and then the small-scale limit yields ${\mathcal{H}}^{\alpha }(K)\ge 1\mathrm{/}1024$; monotonicity gives ${\mathcal{H}}^{\alpha }(S)\ge 1\mathrm{/}1024>0$.

First consider the side arc from a point $x$ to an endpoint $v$. If $x=v$, its diameter is zero. Otherwise let $T_w$ be the deepest nested endpoint triangle containing $x$, of diameter $a=3^{-\mathrm{\mid }w\mathrm{\mid }}$. The endpoint children are ${\psi }_0$ and ${\psi }_3$; their repeated triangles shrink to the endpoint, so this depth is finite. The other three children of $T_w$ are at distance at least $a\mathrm{/}3$ from $v$, by the real-coordinate strips in step 1.1. Thus $\mathrm{\mid }x-v\mathrm{\mid }\ge a\mathrm{/}3$, while the entire endpoint subarc lies in $T_w$ and has diameter at most $a\le 3\mathrm{\mid }x-v\mathrm{\mid }$. Now take distinct $x,y$ on one side and their deepest common triangle, of diameter $a$. If their first distinct children are nonconsecutive, their distance is at least $a\mathrm{/}6$ and the intervening subarc has diameter at most $a$, giving ratio at most $6$. If the children are consecutive with shared vertex $v$, their endpoint cones have separation at least $\pi \mathrm{/}3$ by step 1.1. Writing $r=\mathrm{\mid }x-v\mathrm{\mid }$, $s=\mathrm{\mid }y-v\mathrm{\mid }$, the cosine law gives $\mathrm{\mid }x-y{\mathrm{\mid }}^2\ge r^2+s^2-rs\ge (r+s{)}^2\mathrm{/}4$. The two endpoint tails have combined diameter at most $3(r+s)\le 6\mathrm{\mid }x-y\mathrm{\mid }$. This also includes a zero tail when one point is the vertex. Points on different initial sides admit the subarc through their shared initial vertex, with the identical cone and endpoint-tail estimate. These cases prove bounded turning with bound $6$, hence with the advertised bound $12$; no sharpness is asserted.

By [F4], the finite positive $\alpha$-measure gives ${\dim }_{H}S=\alpha$; since $\alpha >1$, the same theorem gives ${\mathcal{H}}^1(S)=+\mathrm{\infty }$.

The Jordan curve $S$ has bounded turning by step 2.4, so [F7] makes it a quasicircle. Applying [F8] then proves that $S$ is conformally removable.

The identity maps on $\mathbb{D}$ and ${\mathbb{D}}^{\ast }$ are conformal bijections, and their boundary extensions are both ${\mathrm{i}\mathrm{d}}_{\mathbb{T}}$. By [F1], their welding is ${\mathrm{i}\mathrm{d}}_{\mathbb{T}}^{-1}\circ {\mathrm{i}\mathrm{d}}_{\mathbb{T}}={\mathrm{i}\mathrm{d}}_{\mathbb{T}}$.

By [F2], $\tilde{A}$ and $\tilde{B}$ preserve $\mathbb{D}$ and ${\mathbb{D}}^{\ast }$, so the restrictions in the statement are conformal bijections of the two sides and extend to $\mathbb{T}$. Applying [F1] gives $h={\tilde{A}}^{-1}\circ \tilde{B}{\mathrm{\mid }}_{\mathbb{T}}$. Its factors preserve the orientation of $\mathbb{T}$, so $h$ is an orientation-preserving Möbius circle homeomorphism.

If $A=B$, their sphere extensions agree and the formula in step 1.2 gives $h={\mathrm{i}\mathrm{d}}_{\mathbb{T}}$. Conversely, if $h={\mathrm{i}\mathrm{d}}_{\mathbb{T}}$, the Möbius transformation $M={\tilde{A}}^{-1}\circ \tilde{B}$ fixes every $\zeta \in \mathbb{T}$. Write $M(z)=(az+b)\mathrm{/}(cz+d)$ with $ad-bc\mathrm{\ne }0$. Its denominator has no zero on $\mathbb{T}$, and each fixed point satisfies $c{\zeta }^2+(d-a)\zeta -b=0$. A polynomial of degree at most two that vanishes at three distinct points of $\mathbb{T}$ is the zero polynomial; hence $c=b=0$ and $d=a$, so $M$ is the identity. Thus $\tilde{A}=\tilde{B}$ and $A=B$.

Let $({\mathrm{\Gamma }}^{\mathrm{\prime }},f^{\mathrm{\prime }},g^{\mathrm{\prime }})$ be any other welding of ${\mathrm{i}\mathrm{d}}_{\mathbb{T}}$. By step 1.1, $(\mathbb{T},{\mathrm{i}\mathrm{d}}_{\mathbb{D}},{\mathrm{i}\mathrm{d}}_{{\mathbb{D}}^{\ast }})$ is a welding of the same homeomorphism, and [F3] makes its first curve removable. Apply [F4] with this round welding first: a Möbius transformation $M$ satisfies $f^{\mathrm{\prime }}=M\circ {\mathrm{i}\mathrm{d}}_{\mathbb{D}}$ and $g^{\mathrm{\prime }}=M\circ {\mathrm{i}\mathrm{d}}_{{\mathbb{D}}^{\ast }}$, so ${\mathrm{\Gamma }}^{\mathrm{\prime }}=M(\mathbb{T})$. By [F5], this is a generalized round circle. The inherited AC premise is recorded in [F6].

The Möbius extension $H$ carries $\mathbb{T}$ onto itself, so it permutes the two complementary components $\mathbb{D}$ and ${\mathbb{D}}^{\ast }$. If $H(\mathbb{D})={\mathbb{D}}^{\ast }$, its orientation-preserving sphere map carries the counterclockwise boundary orientation of $\mathbb{D}$ to the clockwise boundary orientation induced by ${\mathbb{D}}^{\ast }$; then $h=H{\mathrm{\mid }}_{\mathbb{T}}$ would reverse the circle orientation. Since $h$ is orientation-preserving, $H(\mathbb{D})=\mathbb{D}$ and $H({\mathbb{D}}^{\ast })={\mathbb{D}}^{\ast }$.

For any Möbius map $M$, postcomposition gives $(M\circ f{)}^{-1}\circ (M\circ g)=f^{-1}\circ M^{-1}\circ M\circ g=f^{-1}\circ g$; the welding map is unchanged while the curve becomes $M(\mathrm{\Gamma })$. Closure and inversion in [F7] make this composition calculation valid in the Möbius group.

Let $(\mathbb{T},f_1,g_1)$ and $(\mathbb{T},f_2,g_2)$ weld the same $h$. The first curve is globally removable by [F4], so [F5] gives one Möbius map $M$ with $f_2=M\circ f_1$ and $g_2=M\circ g_1$. Since both curves are $\mathbb{T}$, $M(\mathbb{T})=\mathbb{T}$. This proves the stated ambiguity for weldings with the round curve fixed.

The maps $f={\mathrm{i}\mathrm{d}}_{\mathbb{D}}$ and $g=H{\mathrm{\mid }}_{{\mathbb{D}}^{\ast }}$ are conformal bijections of the two complementary components and extend continuously to $\mathbb{T}$. Therefore [F1] gives $f^{-1}\circ g{\mathrm{\mid }}_{\mathbb{T}}=H{\mathrm{\mid }}_{\mathbb{T}}=h$. In particular, for the displayed Blaschke map, its disk automorphism extension restricted to ${\mathbb{D}}^{\ast }$ is the required exterior parameter map.

Take any welding $({\mathrm{\Gamma }}^{\mathrm{\prime }},f^{\mathrm{\prime }},g^{\mathrm{\prime }})$ of $h$. Compare it by [F5] with the round welding $(\mathbb{T},{\mathrm{i}\mathrm{d}}_{\mathbb{D}},H{\mathrm{\mid }}_{{\mathbb{D}}^{\ast }})$ from step 2.1, using $\mathbb{T}$ first. Then ${\mathrm{\Gamma }}^{\mathrm{\prime }}=M(\mathbb{T})$ and both parameter maps are postcomposed by $M$. If two such weldings have the same images of three distinct boundary points under their first parameter map, their relative Möbius map fixes those three distinct points; [F6] forces it to be the identity. Hence the normalized maps and curve are unique. The inherited AC premise is recorded in [F8].

Write $M(z)=(az+b)\mathrm{/}(cz+d)$ with $ad-bc\mathrm{\ne }0$. For $w=M(z)$, the condition $\mathrm{\mid }z\mathrm{\mid }=1$ becomes $\mathrm{\mid }dw-b\mathrm{\mid }=\mathrm{\mid }a-cw\mathrm{\mid }$, which is a nondegenerate circle equation or line equation in the finite plane; the line case includes $\mathrm{\infty }$ on the sphere. Therefore every curve ${\mathrm{\Gamma }}^{\mathrm{\prime }}=M(\mathbb{T})$ is a generalized round circle, proving the Statement.

By [F1], $\bar{\mathbb{D}}$ is compact.

By [F3], its boundary ${\mathbb{S}}^1$ is globally conformally removable.

By [F2], every compact sphere set with positive area in its finite chart is globally conformally nonremovable.

For $0\le r\le 1$, put $\rho (r)=r(1+r)\mathrm{/}2$. This function is continuous and strictly increasing from $[0,1]$ onto $[0,1]$, with inverse $q(s)=(\sqrt{1+8s}-1)\mathrm{/}2$. The map $F$ in the Statement sends each radius $r\le 1$ to $\rho (r)$ with the same argument and is the identity for $r\ge 1$; the inside and outside formulas agree at $r=1$. Its inverse uses $q$ on radii in $[0,1]$ and is the identity outside. Both maps are continuous at $0$, at radius $1$, and at $\mathrm{\infty }$, so [F6] makes $F$ a sphere homeomorphism.