 Solution

After the fixed time $s$, the mapping-out function of the uncentered mapped future isIts expansion is ${\bar{g}}_t(z)=z+2t/z+O(z^{-2})$, so the elapsed time $t$ is still the half-plane-capacity parameterization. Differentiating in $t$ yieldsTherefore the Loewner transform of ${\bar{\gamma }}_t=g_s({\gamma }_{s+t})$ is ${\xi }_{s+t}$, with starting point ${\xi }_s$.

Centering gives ${\tilde{g}}_t(w)={\bar{g}}_t(w+{\xi }_s)-{\xi }_s$, whose driver is ${\tilde{\xi }}_t={\xi }_{s+t}-{\xi }_s$. By independent stationary Brownian increments, $(B_{s+t}-B_s{)}_{t\ge 0}$ is a standard Brownian motion independent of the past through time $s$. HenceThis proves the domain Markov property of a chordal Loewner chain from the composition rule for chordal Loewner driving functions. For possible contacts with the old hull, the mapped trace is understood through the continuous boundary extension or prime ends; on the open half-plane the displayed maps are ordinary conformal maps. The PDF asks first for the driver of the uncentered curve $\bar{\gamma }$; the centered driver is included as well.