Beurling-Ahlfors quasiconformal extension

Theorem (Beurling-Ahlfors).

There exists a quasiconformal mapping of the upper half plane to itself if and only if the boundary correspondence mapping $\mu$ is $M$-quasisymmetric (http://planetmath.org/QuasisymmetricMapping). Furthermore there exists an extension of $\mu$ to a quasiconformal mapping of the upper half planes such that the maximal dilatation of the extension depends only on $M$ and not on $\mu$.

That is, the extension is $K$-quasiconformal (http://planetmath.org/QuasiconformalMapping) if and only if the boundary correspondence is $M$-quasisymmetric (http://planetmath.org/QuasisymmetricMapping) where $K$ depends purely on $M$.

Supposing that we have the mapping $\varphi :H\to H$ (where $H$ is the upper half plane), then the mapping $\mu :ℝ\to ℝ$,such that $\mu (x)=\varphi (x)$ where $x\in ℝ$, is the boundary correspondence of $\varphi$.

To prove the sufficiency of the above theorem Beurling and Ahlfors [2] define $\varphi$ as follows. Given a $\mu$ that is a quasisymmetric mapping of the real line onto itself and fixes $\mathrm{\infty }$, we define a map $\varphi (x,y)=u(x,y)+iu(x,y)$ where

$u(x,y)$$={\displaystyle \frac{1}{2y}}{\displaystyle {\int }_{-y}^y}\mu (x+t)𝑑t,$

$v(x,y)$$={\displaystyle \frac{1}{2y}}{\displaystyle {\int }_0^y}(\mu (x+t)-\mu (x-t))𝑑t.$

Intuitively $\varphi$ is a function which “smoothes” out any kinks in the function $\mu$ as we get further and further away from the real line. It therefore intuitively follows that $\varphi$ has the worst (highest) dilatation near the $x$ axis, which actually turns out to be true.