Power maps, endpoint distortion, and a non-Möbius quasisymmetric circle map

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].