Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 36 / 3

Use the PDF's starred image hull $K_t^{\ast }$ and write $h_t={\Phi }_t=g_{K_t^{\ast }}\circ \Phi \circ g_{K_t}^{-1}$. For $t<T$, this map is analytic across a real neighborhood of the driver. SetLet $u(t)=\mathrm{hcap}(K_t^{\ast })/2$. The classical conformal change of half-plane capacity gives $\dot{u}(t)=a_t^2$, and the image driver, indexed by original time, is $h_t({\xi }_t)$. Differentiating the conjugacy relation givesThese are precisely the standard classical identities allowed in this question. The singularities cancel at $z={\xi }_t$. To calculate the remaining derivative, put $\delta =z-{\xi }_t$ and expand at fixed $t$:Thereforewhere each ${\partial }_t$ keeps the spatial variable fixed before evaluation. Applying the Itô formula to $h_t^{\prime }({\xi }_t)$ gives the boundary derivative diffusion under conformal Loewner conjugacyAt $\kappa =8/3$, the third-derivative drift cancels. A second application of the Itô formula to $a_t^p$ givesChoosing $p=1-1/\kappa =5/8$ cancels the remaining drift. Hence the SLE eight-thirds restriction martingale isLocalizing away from $T$ and from vanishing derivatives makes the displayed stochastic integral a martingale on each localized interval, so it defines a continuous local martingale up to $T$.

Put $A=\mathbb{H}\setminus U$. The neighborhood assumptions at zero and infinity mean that $A$ is a bounded hull away from the starting boundary point. For $t<T$, the conjugacy map $h_t$ is the hydrodynamically normalized mapping-out function of the remaining forbidden hull after removing $K_t$. Its closure avoids the driver, so the boundary derivative is an excursion avoidance probability yields $0<{\Sigma }_t\le 1$. Thus $0<M_t\le 1$, and the bounded local martingale criterion makes the localized martingales uniformly bounded and uniformly integrable.

Choose localization times ${\tau }_n\uparrow T$, also bounded by $n$. Optional stopping gives $\mathbb{E}{M}_{{\tau }_n}=M_0$. The supplied terminal limit and continuity of the power function giveSince these variables are bounded by one, dominated convergence yieldsAs usual for a chord starting at zero, the time-zero point is understood as the boundary prime end of $U$; the avoidance assertion concerns the subsequent curve. The boundedness step is what upgrades the local martingale calculation to an equality of probabilities.