Fix $z$ and write
$$
L_t(z)=\log(g_t(z)-U_t).
$$
By assumption, $L(z)$ is a continuous local martingale, so $Z_t(z)=e^{L_t(z)}$ is a <semimartingale>. The <Chordal Loewner equation> gives
$$
g_t(z)=z+\int_0^t\frac2{Z_s(z)}ds,
$$
which has <finite variation>. Therefore
$$
U_t=g_t(z)-Z_t(z)
$$
is a semimartingale. Thus \b[the Loewner driver $U$ is a continuous semimartingale].

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 203 / 3 / d / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-203/3/d/i/solution