= Solution
For $z=x+iy$ with $|x|\leq1$ and $0<y\leq1$, the given <Reverse SLE derivative martingale> starts from
$$
M_0=\left(1+\frac{x^2}{y^2}\right)^{4/\kappa}\leq C_0y^{-8/\kappa}.
$$
It is a nonnegative local martingale and therefore a <supermartingale>. Since its second factor is at least one,
$$
\mathbb E|h_T'(x+iy)|^2\leq\mathbb E M_T\leq C_0y^{-8/\kappa}.
$$
Choose
$$
0<\alpha<\frac12\left(1-\frac8\kappa\right),
$$
which is possible exactly because $\kappa>8$. At height $y_n=2^{-n}$, take a horizontal grid of spacing comparable to $y_n$ in $[-1,1]$. The <Markov inequality> gives, at each grid point,
$$
\mathbb P\bigl(|h_T'(z)|>y_n^{-1+\alpha}\bigr)\leq C y_n^{2-2\alpha-8/\kappa}.
$$
There are $O(y_n^{-1})$ grid points, so the probability that the bound fails anywhere on level $n$ is at most $Cy_n^{1-2\alpha-8/\kappa}$. These probabilities are summable. The <Borel-Cantelli lemmas> therefore give an almost surely finite random constant controlling every sufficiently fine grid, and enlarging it handles the finitely many remaining levels.
Every point of the half-rectangle lies within a fixed hyperbolic distance of one of these grid points at comparable height. The <Koebe distortion theorem> compares the two derivatives by a universal factor. Hence an almost surely finite random $C$ satisfies
$$
|h_T'(x+iy)|\leq Cy^{-1+\alpha}
$$
for all $x\in[-1,1]$ and $0<y\leq1$. This proves the <Reverse SLE derivative bound above the space-filling threshold>.
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