Every quasisymmetric circle homeomorphism is a conformal welding

By [F1], choose a $K_0(L)$-quasiconformal homeomorphism $E:\bar{\mathbb{D}}\to \bar{\mathbb{D}}$ extending $h$. Let ${\mu }_E$ be its Beltrami coefficient on $\mathbb{D}$ and define a sphere coefficient $\mu$ by $\mu ={\mu }_E$ on $\mathbb{D}$, $\mu =0$ on ${\mathbb{D}}^{\ast }$, and $\mu =0$ on ${\mathbb{S}}^1$. The circle has area zero, so this is a measurable coefficient with $\mathrm{\parallel }\mu {\mathrm{\parallel }}_{\mathrm{\infty }}\le (K_0(L)-1)\mathrm{/}(K_0(L)+1)<1$.

By [F3], let $H:\widehat{\mathbb{C}}\to \widehat{\mathbb{C}}$ solve the Beltrami equation for $\mu$. It is conformal on ${\mathbb{D}}^{\ast }$ by [F4]. On $\mathbb{D}$, $H$ and $E$ have the same Beltrami coefficient, so $H\circ E^{-1}:\mathbb{D}\to H(\mathbb{D})$ has zero coefficient by [F5] and is conformal by [F4]. Also $K_H\le K_0(L)$.

Let $E^{\mathrm{\prime }}$ be another orientation-preserving quasiconformal disk extension of $h$ and put $\psi =E^{-1}\circ E^{\mathrm{\prime }}$ on $\bar{\mathbb{D}}$. Then $\psi {\mathrm{\mid }}_{{\mathbb{S}}^1}=\mathrm{i}\mathrm{d}$. Paste $\psi$ on $\bar{\mathbb{D}}$ to the identity on $\overline{{\mathbb{D}}^{\ast }}$ to obtain a sphere homeomorphism $\mathrm{\Psi }$; [F6] makes it quasiconformal. The map $H^{\mathrm{\prime }}=H\circ \mathrm{\Psi }$ has coefficient ${\mu }_{E^{\mathrm{\prime }}}$ on $\mathbb{D}$ and zero on ${\mathbb{D}}^{\ast }$ by [F5], so it is a solution for the coefficient constructed from $E^{\mathrm{\prime }}$. Because $\mathrm{\Psi }$ fixes ${\mathbb{S}}^1$, $H^{\mathrm{\prime }}({\mathbb{S}}^1)=H({\mathbb{S}}^1)$, proving extension independence for compatible choices of solutions.

Put ${\mathrm{\Gamma }}_h=H({\mathbb{S}}^1)$, ${\mathrm{\Omega }}_0=H(\mathbb{D})$, and ${\mathrm{\Omega }}_1=H({\mathbb{D}}^{\ast })$, and define $f=H\circ E^{-1}:\mathbb{D}\to {\mathrm{\Omega }}_0$ and $g=H{\mathrm{\mid }}_{{\mathbb{D}}^{\ast }}:{\mathbb{D}}^{\ast }\to {\mathrm{\Omega }}_1$. The sphere homeomorphism $H$ makes ${\mathrm{\Gamma }}_h$ a Jordan curve with complementary components ${\mathrm{\Omega }}_0,{\mathrm{\Omega }}_1$; both maps are conformal by step 2.1 and extend homeomorphically to the closures by their formulas. For $t\in {\mathbb{S}}^1$, $f^{-1}(g(t))=E(H^{-1}(H(t)))=h(t)$, so $({\mathrm{\Gamma }}_h,f,g)$ is a welding in the convention of The welding homeomorphism of a Jordan curve. Since $H$ is $K_0(L)$-quasiconformal, ${\mathrm{\Gamma }}_h$ is a $K_0(L)$-quasicircle.

Let $A,B\in \mathrm{Aut}(\mathbb{D})$ and replace $h$ by $h^{\mathrm{\prime }}=A\circ h\circ B^{-1}$. Choose $E^{\mathrm{\prime }}=A\circ E\circ B^{-1}$ and $H^{\mathrm{\prime }}=H\circ B^{-1}$, using the sphere Möbius extensions from [F7]. Postcomposition by $A$ preserves the coefficient and precomposition by $B^{-1}$ pulls it back, so $H^{\mathrm{\prime }}$ solves the coefficient for $E^{\mathrm{\prime }}$ on $\mathbb{D}$ and has zero coefficient on ${\mathbb{D}}^{\ast }$. Since $B^{-1}({\mathbb{S}}^1)={\mathbb{S}}^1$, $H^{\mathrm{\prime }}({\mathbb{S}}^1)=H({\mathbb{S}}^1)$. Finally, [F3] says that a different choice of uniformizing solution changes the curve only by postcomposition with a Möbius map; no uniqueness among other weldings is used.