 Solution

For $\lambda >0$, define ${\tilde{g}}_t(z)={\lambda }^{-1/2}{g}_{\lambda t}(\sqrt{\lambda }z)$. Differentiating the Chordal Loewner equation givesBrownian scaling makes $\tilde{W}$ have the original driver law. The corresponding hull is ${\lambda }^{-1/2}{K}_{\lambda t}$, soThe time scaling applies to the whole coupled hull process. In particular, the joint law of swallowing times satisfies $(T(cx),T(cy))\stackrel{d}{=}{c}^2(T(x),T(y))$.

Thus $F(cx,cy)=F(x,y)$. A pair $0<x<y$ is determined up to dilation by $z=y/(y-x)>1$: dilating by $(y-x{)}^{-1}$ turns it into $(z-1,z)$. Define $f(z)=F(z-1,z)$. Then