The Beurling–Ahlfors extension theorem for circles and lines

Put $T=2\pi$ and $q(t)=e^{it}$. The continuous circle map $t\mapsto h(q(t))$ has a continuous argument lift $f:\mathbb{R}\to \mathbb{R}$: choose one argument at $0$, subdivide each compact interval into finitely many pieces whose images lie in open semicircles, and match the local arguments at successive endpoints. Orientation preservation makes $f$ strictly increasing. Since $f(t+T)-f(t)$ is a continuous integer multiple of $T$, it is constant; strict increase and injectivity of $h$ on the circle force that integer to be $1$. Thus $q(f(t))=h(q(t))$ and $f(t+T)=f(t)+T$.

Suppose $H$ satisfies (b), and put $a=H(0)$. The Möbius map ${\mathrm{\Phi }}_a$ in [F3] makes $F={\mathrm{\Phi }}_a\circ H$ fix $0$ and map the closed disc to itself. By [F4], $F$ is $K$-quasiconformal in the disc. Define its exterior extension by $\tilde{F}(z)=1\mathrm{/}\overline{F(1\mathrm{/}\stackrel{‾}{z})}$ for $\mathrm{\mid }z\mathrm{\mid }>1$ and set $\tilde{F}=F$ on the closed disc. The formulas agree on ${\mathbb{S}}^1$; the pasted map is a sphere homeomorphism fixing $0$ and $\mathrm{\infty }$, and [F6] makes it $K$-quasiconformal. Apply [F5] to see that $\tilde{F}{\mathrm{\mid }}_{{\mathbb{S}}^1}$ is quasisymmetric with control depending only on $K$. The map ${\mathrm{\Phi }}_a^{-1}{\mathrm{\mid }}_{{\mathbb{S}}^1}$ has control depending only on $r$ by [F3]. Composing these controls gives the claimed $L(K,r)$ for $H{\mathrm{\mid }}_{{\mathbb{S}}^1}={\mathrm{\Phi }}_a^{-1}\circ \tilde{F}{\mathrm{\mid }}_{{\mathbb{S}}^1}$.

Let $I=[x,x+s]$ and $J=[x+s,x+2s]$. If $s\le T\mathrm{/}2$, their projections are adjacent equal arcs and their image-length ratio in either order is at most $L$. If $T\mathrm{/}2<s<T$, put $d=T-s$, $\alpha =f(x+T)-f(x+s)$, and $\beta =f(x+T-d)-f(x+T-2d)$. Then $\mathrm{\mid }f(I)\mathrm{\mid }=T-\alpha$ and $\mathrm{\mid }f(J)\mathrm{\mid }=T-\beta$. Each of $\alpha ,\beta$ is the image length of an arc of length $d<T\mathrm{/}2$; comparing it with an adjacent arc of the same length gives $\alpha ,\beta \le LT\mathrm{/}(L+1)$, since the two image lengths sum to at most $T$. Therefore both $\mathrm{\mid }f(I)\mathrm{\mid }$ and $\mathrm{\mid }f(J)\mathrm{\mid }$ lie in $[T\mathrm{/}(L+1),T]$. If $s\ge T$, write $s=nT+u$ with integer $n\ge 1$ and $0\le u<T$; each image increment lies in $[nT,(n+1)T]$. In every case the adjacent image-length ratio in either order is at most $L+1$, so $f$ is $(L+1)$-quasisymmetric on $\mathbb{R}$.

Apply [F2] to $f$, obtaining its reflected Ahlfors–Beurling extension $G$. The formula gives $G(z+T)=G(z)+T$: in the real average the boundary shift adds $T$, and in the imaginary difference it cancels. Thus $\mathcal{H}(e^{iz}):=e^{iG(z)}$ is well-defined on ${\mathbb{C}}^{\ast }$. If $\mathcal{H}(e^{iz})=\mathcal{H}(e^{i{z}^{\mathrm{\prime }}})$, then $G(z^{\mathrm{\prime }})=G(z)+nT=G(z+nT)$ for some integer $n$, so injectivity of $G$ gives $z^{\mathrm{\prime }}=z+nT$; surjectivity follows from that of $G$. Hence $\mathcal{H}$ is a homeomorphism of ${\mathbb{C}}^{\ast }$ and is $K(L+1)$-quasiconformal locally because the exponential covering is conformal. It maps the unit circle by $h$ and maps the punctured disc onto itself.

The function $p(t)=f(t)-t$ is continuous and $T$-periodic, so let $M={\sup }_{t\in [0,T]}\mathrm{\mid }p(t)\mathrm{\mid }<\mathrm{\infty }$. In the upper half-plane, the real part of the Ahlfors–Beurling formula differs from $x$ by the average of $p$ on $[x-y,x+y]$, hence by at most $M$; its imaginary part differs from $y$ by $y^{-1}{\int }_0^y(p(x+s)-p(x-s))ds$, hence by at most $2M$. Reflection gives the same bound below the real axis. Therefore $\mathrm{\mid }G(z)-z\mathrm{\mid }\le \sqrt{5}M$ throughout the plane. It follows that $\mathcal{H}(w)\to 0$ as $w\to 0$ and $\mathcal{H}(w)\to \mathrm{\infty }$ as $w\to \mathrm{\infty }$. The inverse of $G$ has the same bounded-displacement property, so the descended map and its inverse both extend continuously at $0$ and $\mathrm{\infty }$; hence $\mathcal{H}$ is a sphere homeomorphism with $\mathcal{H}(0)=0$.

For $y\ge 1$, the derivative formulas for the line extension give $P,Q=1+O(M\mathrm{/}y)$ and $R,S=1\mathrm{/}2+O(M\mathrm{/}y)$, where $P,Q,R,S$ are the adjacent increments and average deficits in its proof. Thus $DG$ is bounded on the upper region $y\ge 1$, and by reflection on $y\le -1$. In the coordinate $w=e^{iz}$, $\mathrm{\mid }D\mathcal{H}(w)\mathrm{\mid }\le e^{\sqrt{5}M}\mathrm{\mid }DG(z)\mathrm{\mid }$ near $0$, because $\mathrm{\mid }\mathcal{H}(w)\mathrm{\mid }\mathrm{/}\mathrm{\mid }w\mathrm{\mid }\le e^{\sqrt{5}M}$; the analogous estimate in coordinate $1\mathrm{/}w$ holds near $\mathrm{\infty }$. The continuous map therefore has bounded classical derivatives off each added point. Integration by parts on a punctured disc and passage to the limit makes these bounded derivatives its weak derivatives across the point: the boundary term is bounded by a constant times the circle radius and tends to zero. The Beltrami inequality holds away from the point and hence almost everywhere across it. So the descended sphere homeomorphism is globally $K(L+1)$-quasiconformal.

The map $\mathcal{H}$ from steps 3.1–5.1 proves (a), with $K(L)=K_{\mathrm{A}\mathrm{B}}(L+1)$. Since it preserves the two complementary components of ${\mathbb{S}}^1$, it maps $\mathbb{D}$ onto itself; continuity and bijectivity on the sphere give the asserted homeomorphism of the closed disc.

The line-extension clause in (c) is [F2]. For the converse, $H{\mathrm{\mid }}_{\mathbb{C}}$ is a plane quasiconformal homeomorphism fixing infinity, so [F5] gives a global quasisymmetry control on the plane. The restriction to $\mathbb{R}$ is either increasing or decreasing; in the latter case negate its values. Negation preserves every distance ratio, so restricting the plane metric control to triples on $\mathbb{R}$ gives the adjacent-interval ratio bound in [F1] for the resulting increasing homeomorphism, with a constant depending only on $K$. Countable Choice used by these analytic interfaces follows from AC by [F7].