Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 203 / 2

For Schramm–Loewner evolution in $(\mathbb{H},0,\infty )$, the Scaling invariance of SLE states that, for every $a>0$,has the same law as $K_t$. The scaled Loewner driving function is ${\tilde{U}}_t=a^{-1}{U}_{a^{2}t}$. Since $U_t=\sqrt{\kappa }{B}_t$, the Brownian scaling identity $a^{-1}{B}_{a^{2}t}\stackrel{d}{=}{B}_t$ proves the claim.

The Conformal Markov property of SLE states that, conditionally on the hull through time $t$, the future hull mapped by $g_t-U_t$ is an independent ${\mathrm{SLE}}_{\kappa }$ in $(\mathbb{H},0,\infty )$. More precisely,has driving function ${\tilde{U}}_s=U_{t+s}-U_t$. The stationary increments and independent increments of Brownian motion show that $\tilde{U}$ is independent of ${\mathcal{F}}_t$ and has the same law as $\sqrt{\kappa }B$. The deterministic correspondence between continuous drivers and Loewner chains completes the proof.

Put $w=g_T^{-1}(U_T+z)$ and defineThen $f_0(z)=z$. Differentiating with the Chordal Loewner equation givesUniqueness for this ordinary differential equation shows that $f_s=h_s$. At $s=T$,which is the endpoint identity for the Reverse Loewner flow.

For $0<s<T$, the pathwise identity $h_s(z)=g_s^{-1}(U_s+z)-U_s$ is generally false. The left side is built from the reversed final driver segment $(U_{T-r}-U_T{)}_{0\le r\le s}$, whereas the right side is built from the initial segment $(U_r{)}_{0\le r\le s}$. By time reversal and symmetry of Brownian motion they have the same probability distribution, but they are not equal for the given Brownian path.

For $z=x+iy$ with $|x|\le 1$ and $0<y\le 1$, the given Reverse SLE derivative martingale starts fromIt is a nonnegative local martingale and therefore a supermartingale. Since its second factor is at least one,

Choosewhich 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,There are $O(y_n^{-1})$ grid points, so the probability that the bound fails anywhere on level $n$ is at most $C{y}_n^{1-2\alpha -8/\kappa }$. 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 $C$ satisfiesfor all $x\in [-1,1]$ and $0<y\le 1$. This proves the Reverse SLE derivative bound above the space-filling threshold.

Apply the derivative criterion for Hölder continuity up to a boundary to the estimate from part (c). For two points at distance $r$, move each vertically to height at least $r$, join them horizontally there, and move back. The two vertical integrals are bounded byand the horizontal integral is at most $Cr{r}^{-1+\alpha }=C{r}^{\alpha }$. Thus $h_T$ extends continuously to the bottom edge and satisfieson the half-rectangle. In particular, it is almost surely a Hölder continuous function there.