= Solution
The <Loewner local growth property> gives a continuous <Loewner driving function>, and we normalize $U_0=0$. Mapping out an initial hull transforms the future driver to $U_{t+s}-U_t$. The <Conformal Markov property of SLE> therefore gives <stationary increments> and <independent increments>.
Consequently $U$ is a continuous <Lévy process>. In the <Lévy–Khintchine formula>, continuity excludes its jump measure, so its only possible components are a deterministic linear drift and <Brownian motion>:
$$
U_t=\mu t+\sigma B_t,\qquad\sigma\geq0.
$$
Conformal invariance under dilations gives $U_{r^2t}/r\overset{d}=U_t$. Comparing the <expectations> of these <Gaussian random variables> yields $\mu rt=\mu t$ for every $r>0$, hence $\mu=0$. The Brownian term already has the required law by <Brownian scaling>. Therefore the <characterization of the SLE driving function> is
$$
\boxed{U_t=\sqrt\kappa B_t\quad\text{in law},\qquad\kappa=\sigma^2\geq0.}
$$
The case $\kappa=0$ is the constant zero driver. No reflection-symmetry assumption is needed to remove the drift; scaling suffices.
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