Smooth Beltrami coefficients admit quasiconformal solutions

In each standard source chart the coefficient has a smooth representative and essential norm at most $k$ by [F1]. By [F8], its pointwise modulus is at most $k$. Apply [F2] with the bound $k$, regularity order $0$ and exponent $1\mathrm{/}2$ at every point. The family of all local solution charts so obtained covers the compact sphere; [F4] gives a finite subcover, which we denote $(V_j,{\mathrm{\Phi }}_j{)}_{j=1}^m$. Each ${\mathrm{\Phi }}_j$ is a $C^{1,1\mathrm{/}2}$ diffeomorphism onto an open subset and solves the Beltrami equation for the coefficient expressed in its source chart.

On an overlap, fix a point and restrict to a standard source chart $z$ around it. Write $a_i=({\mathrm{\Phi }}_i{)}_z$, $b_i=({\mathrm{\Phi }}_i{)}_{\bar{z}}$ and likewise $a_j,b_j$. The pullback law [F1] makes both local equations use the same chart representative ${\mu }_z$, so $b_i={\mu }_z{a}_i$ and $b_j={\mu }_z{a}_j$. The transition $\tau ={\mathrm{\Phi }}_i\circ {\mathrm{\Phi }}_j^{-1}$ is $C^1$; the real chain rule and inverse derivative formula [F7] give ${\tau }_{\bar{w}}=(b_i{a}_j-a_i{b}_j)\mathrm{/}(\mathrm{\mid }{a}_j{\mathrm{\mid }}^2-\mathrm{\mid }{b}_j{\mathrm{\mid }}^2)=0$, whose denominator is $J_{{\mathrm{\Phi }}_j}>0$ by [F2]. Thus $\tau$ is holomorphic by [F7]. Interchanging $i$ and $j$ proves its inverse is holomorphic as well. These transitions are holomorphic near every overlap point, so the finite family $(V_j,{\mathrm{\Phi }}_j)$ is a holomorphic atlas on the topological sphere. Since $J_{{\mathrm{\Phi }}_j}>0$, [F9] shows these charts induce the usual sphere orientation.

Let $X$ be the sphere with this atlas. Its topological properties are unchanged: it is a Riemann surface by [F3], compact and simply connected by [F4]. Uniformization [F5] gives a biholomorphism from $X$ to one of $\widehat{\mathbb{C}}$, $\mathbb{C}$, or $\mathbb{D}$. The last two targets are not compact: the disks $D(0,n)$ for $n\ge 1$ cover $\mathbb{C}$ with no finite subcover, and the disks $D(0,1-1\mathrm{/}n)$ for $n\ge 2$ cover $\mathbb{D}$ with no finite subcover. By [F10], neither is compact. A biholomorphism is a homeomorphism [F3], so it preserves compactness. Hence its target must be $\widehat{\mathbb{C}}$; write $H:X\to \widehat{\mathbb{C}}$ for this biholomorphism.

Regard $H$ as a homeomorphism $f$ of the underlying sphere. In a source chart $z$ and a target chart, its local expression has the form $\chi \circ {\mathrm{\Phi }}_j$, where $\chi$ is holomorphic with nonzero derivative because $H$ and its inverse are holomorphic. The chain rule [F7] and the local equation give $f_{\bar{z}}={\mu }_z{f}_z$ and $J_f=\mathrm{\mid }{\chi }^{\mathrm{\prime }}\circ {\mathrm{\Phi }}_j{\mathrm{\mid }}^2{J}_{{\mathrm{\Phi }}_j}>0$. Moreover $f_z\mathrm{\ne }0$: from $J_{{\mathrm{\Phi }}_j}=(1-\mathrm{\mid }{\mu }_z{\mathrm{\mid }}^2)\mathrm{\mid }{a}_j{\mathrm{\mid }}^2>0$ we have $a_j\mathrm{\ne }0$, and ${\chi }^{\mathrm{\prime }}\mathrm{\ne }0$. Since $f$ is locally $C^1$, [F12] makes it locally $W^{1,2}$; the derivatives are locally square-integrable because they are continuous on compact subcharts. With $K=(1+k)\mathrm{/}(1-k)$, the equation and [F8] give $\mathrm{\mid }{f}_{\bar{z}}\mathrm{\mid }\le (K-1)\mathrm{/}(K+1)\mathrm{\mid }{f}_z\mathrm{\mid }$. By [F6], $f$ is analytically quasiconformal. The positive Jacobian makes it preserve local orientation by [F9], so it is orientation-preserving; the nonvanishing $f_z$ gives ${\mu }_f=\mu$ a.e.

The three points $f(0),f(1),f(\mathrm{\infty })$ are distinct because $f$ is a homeomorphism. By [F11], choose a Möbius map $M$ carrying them to $0,1,\mathrm{\infty }$. Its derivative is nonzero by composing with its holomorphic inverse and applying the chain rule [F7]. In local target charts, $(M\circ f{)}_{\bar{z}}=(M^{\mathrm{\prime }}\circ f)f_{\bar{z}}$ and $(M\circ f{)}_z=(M^{\mathrm{\prime }}\circ f)f_z$, so postcomposition preserves the derivative ratio and Beltrami coefficient. Since $M\circ f$ is still a local $C^1$ diffeomorphism, it remains in $W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}$ and the same bound in [F6] applies; its local Jacobian is positive because $M$ is conformal. Therefore $M\circ f$ has all claimed properties and the prescribed normalization.