Normalization of a solution by a Möbius postcomposition

Direct calculation gives $A_z=1$, $A_{\bar{z}}=\nu$, $J_A=1-\mathrm{\mid }\nu {\mathrm{\mid }}^2>0$, and $A^{-1}(w)=(w-\nu \bar{w})\mathrm{/}(1-\mathrm{\mid }\nu {\mathrm{\mid }}^2)$. Thus $A$ is a real-linear homeomorphism of $\mathbb{C}$; since $\mathrm{\mid }A(z)\mathrm{\mid }\ge (1-\mathrm{\mid }\nu \mathrm{\mid })\mathrm{\mid }z\mathrm{\mid }$, it extends by $A(\mathrm{\infty })=\mathrm{\infty }$ to a sphere homeomorphism. Its affine coordinate-line restrictions are absolutely continuous with locally square-integrable derivatives. In source and target infinity charts its expression is $a(w)=w\bar{w}\mathrm{/}(\bar{w}+\nu w)$ for $w\mathrm{\ne }0$, with $a(0)=0$. The denominator has modulus at least $(1-\mathrm{\mid }\nu \mathrm{\mid })\mathrm{\mid }w\mathrm{\mid }$, so $a$ is continuous at $0$; its degree-one homogeneity and smoothness off $0$ give bounded derivatives there. Integrating along segments, splitting at $0$ if needed, makes $a$ Lipschitz. Thus its coordinate-line restrictions are absolutely continuous with locally square-integrable derivatives, and [F5] gives local $W^{1,2}$ in both charts. The pullback law [F2] transports the equation off $0$, a null point. Hence $A$ is analytically quasiconformal with coefficient $\nu$ and $K=(1+\mathrm{\mid }\nu \mathrm{\mid })\mathrm{/}(1-\mathrm{\mid }\nu \mathrm{\mid })$; [F6] gives orientation preservation. The values $A(0)=0$ and $A(1)=1+\nu$ show it is already normalized exactly when $\nu =0$.

Since $\mathrm{\mid }\nu \mathrm{\mid }<1$, $1+\nu \mathrm{\ne }0$. By [F3], there is a unique Möbius map $M$ sending $(0,1+\nu ,\mathrm{\infty })$ to $(0,1,\mathrm{\infty })$. Write $M(w)=(aw+b)\mathrm{/}(cw+d)$ with $ad-bc\mathrm{\ne }0$. The conditions $M(0)=0$ and $M(\mathrm{\infty })=\mathrm{\infty }$ force $b=c=0$, so $M(w)=\lambda w$ with $\lambda =a\mathrm{/}d\mathrm{\ne }0$; then $M(1+\nu )=1$ gives $\lambda =(1+\nu {)}^{-1}$. Thus $M(w)=w\mathrm{/}(1+\nu )$, proving (b).

Put $f^{\nu }=M\circ A$. In the finite chart, $f_z^{\nu }=1\mathrm{/}(1+\nu )$ and $f_{\bar{z}}^{\nu }=\nu \mathrm{/}(1+\nu )$, so $f_{\bar{z}}^{\nu }=\nu f_z^{\nu }$ and ${\mu }_{f^{\nu }}=\nu$. It is a sphere homeomorphism fixing $\mathrm{\infty }$; in the infinity coordinates $w=1\mathrm{/}z$ and $\eta =1\mathrm{/}{f}^{\nu }(z)$ its expression is $\eta (w)=\frac{(1+\nu )w\bar{w}}{\bar{w}+\nu w},\eta (0)=0.$ This is $(1+\nu )a(w)$, with $a$ from step 1.1, so it has the same local $W^{1,2}$ regularity by linearity of weak derivatives. The pullback law in [F2] transports the weak equation to the infinity chart, where the coefficient still has modulus $\mathrm{\mid }\nu \mathrm{\mid }$; [F5] gives the analytic quasiconformal inequality there. Thus $f^{\nu }$ is a quasiconformal sphere homeomorphism and weak solution; its three values are $0,1,\mathrm{\infty }$. By [F7] it is the unique normalized solution, proving (c). The local orientation sign remains positive by [F6].

Let $f$ be any quasiconformal homeomorphic solution. Its homeomorphism property makes $f(0),f(1),f(\mathrm{\infty })$ distinct. By [F3], the unique Möbius map carrying this triple to $(0,1,\mathrm{\infty })$ exists. By [F7], all solutions are Möbius postcompositions of $f^{\nu }$ and the normalized solution is unique; therefore the normalizing map is the inverse of the unique Möbius map in the family, and the full solution family has exactly the stated form.