Past exam of the Brownian motion in complex analysis course of the University of Oxford / 2011 (C10.1b, DMAC 2B83, DMCO 2B83)

Source: released examination bundle, Part C 2007 11.pdf, pages 457–460. All three questions are covered. No mark allocations are printed.

Write $Z_t=z+B_t^1+i{B}_t^2$, where the coordinates are independent standard Brownian motions, and write $f=u+iv$. The Cauchy-Riemann equations imply $\Delta u=\Delta v=0$. Itô's formula therefore makescontinuous local martingales. Their quadratic variations and quadratic covariation areIndeed the two gradient norms are equal to $|f^{\prime }|$, and their dot product vanishes by the Cauchy-Riemann equations.

The clock is finite at every finite time, since a continuous Brownian path has compact range there and $f^{\prime }$ is continuous. To show its lifetime is infinite, choose a small disc on which $|f^{\prime }{|}^2\ge c>0$; such a disc exists because the entire function is nonconstant. A supporting recurrence result is that planar Brownian motion spends infinite total time in every nonempty open disc almost surely. One proof combines recurrent returns to a smaller disc with the strong Markov property: each visit has a uniformly positive chance of lasting a fixed positive time in the larger disc. Equivalently, use the divergence of a positive planar Brownian occupation integral. ThereforeFinally the common-clock time change of orthogonal local martingales states that two continuous local martingales starting at zero with common bracket $A$, zero cross bracket, and $A_{\infty }=\infty$ are the coordinates of a planar Brownian motion evaluated at $A_t$. Applying it gives a planar Brownian motion $\tilde{Z}$ starting at zero withThis proves conformal invariance of planar Brownian motion in the entire-function setting, including the divergence of the conformal Brownian clock for an entire function.

Apply part (a) to the entire function $f(z)=e^{i\theta }z$. Its derivative has modulus one, so its conformal Brownian clock is $A_t=t$. Thuswhich is a planar Brownian motion starting from the rotated initial point. In particular, if the initial point is zero, its law is unchanged. This is rotational invariance of planar Brownian motion.

Start a planar Brownian motion $B$ at zero and apply part (a) to the entire function $f(z)=e^z$. ThenThe left side never vanishes. The clock is continuous, starts at zero, and tends to infinity almost surely, so its range is all nonnegative times. Consequently the complex Brownian motion $1+{\tilde{Z}}_s$ never reaches zero at any time $s$. Every planar Brownian motion started at one has that same law, henceThis is the complex exponential construction of planar Brownian motion and proves that a fixed point is a polar point for planar Brownian motion. The divergence of the clock is essential: it makes this a conclusion about all Brownian times.

Let $\beta (\theta )=w-e^{i\theta }=|\beta (\theta )|e^{ig(\theta )}$. It is never zero, since $|w|<1$. On the unit circle,Dividing these identities givesAny other continuous lift of the argument differs by an integer multiple of $2\pi$, which leaves the exponential unchanged.

Differentiate the argument through the nonvanishing curve, using $g^{\prime }=\mathrm{Im}({\beta }^{\prime }/\beta )$. Since ${\beta }^{\prime }=-i{e}^{i\theta }$,Also $|w-e^{i\theta }{|}^2=1+|w{|}^2-2(w_1\cos \theta +w_2\sin \theta )$. Subtracting one therefore givesThus the lifted boundary angle $\varphi (\theta )=-\theta +2g(\theta )$ is strictly increasing. Its total change around the circle is $2\pi$, because the disk automorphism has boundary degree one. The same total change follows directly from the winding number one of $w-e^{i\theta }$. This is the boundary-angle derivative of a disk automorphism.

The disk automorphism satisfies $f(w)=0$ and maps the unit circle bijectively to itself. Its derivative $(1-|w{|}^2)/(1-\overline{w}z{)}^2$ is bounded above and away from zero on the closed disk. By conformal invariance of planar Brownian motion, the image path up to the exit time is a time-changed planar Brownian motion started at zero. Its exit point is $f(Z_T)$; the clock changes when it exits, but not where.

The original exit time is finite almost surely. For example, $|Z_{t\wedge T}{|}^2-2(t\wedge T)$ is a martingale, so $\mathbb{E}(t\wedge T)\le (1-|w{|}^2)/2$. Boundedness of $f^{\prime }$ makes the image exit clock finite too. By rotational invariance of planar Brownian motion, the image exit angle is uniform on $[0,2\pi )$.

Let $\Theta =\arg Z_T$. Parts (a) and (b) show that $\varphi (\Theta )$ modulo $2\pi$ is uniform, where $\varphi$ is strictly increasing and increases by exactly $2\pi$ in one circuit. The one-dimensional change-of-variables formula therefore givesIts integral is one because the total angle change is $2\pi$. This harmonic measure density is the Poisson kernel for the unit disk divided by $2\pi$. It is uniform for $w=0$ and concentrates near the boundary point closest to $w$ as $|w|$ increases.

A bounded holomorphic function of planar Brownian motion is a local martingale up to exit, because its real and imaginary parts are harmonic and Itô's formula has zero drift. Stop before the given $S<T_D$. Localization inside compact subdisks gives the local martingale $f(Z_{t\wedge S}^z)$, and $|f|<1$ makes it bounded. The bounded local martingale criterion therefore gives a true martingale.

Exit from the unit disk is finite almost surely, so $S<T_D$ is finite too. Taking $t\to \infty$ and using the dominated convergence theorem yieldsThe same proof works for any bounded holomorphic function, after rescaling if needed; it does not require the stopping time itself to be bounded.

Since $f(0)=0$, the function $g(z)=f(z)/z$ extends analytically to zero by $g(0)=f^{\prime }(0)$. It is bounded on the disk: outside $|z|<1/2$, $|g(z)|\le 2$, while on the closed half-radius disk its analytic extension is continuous and hence bounded.

Fix $z\ne 0$ and choose $|z|<r<1$. Let $T_r$ be the first exit time of Brownian motion started at $z$ from the radius-$r$ disk. The bounded analytic martingale identity in part (a) givesOn this exit circle, $|g(Z_{T_r})|=|f(Z_{T_r})|/r\le 1/r$. The triangle inequality gives $|g(z)|\le 1/r$. Letting $r\uparrow 1$ proves $|g(z)|\le 1$, and thereforeAt zero the claim holds by the assumed zero value. This is the Schwarz lemma, obtained from Brownian optional stopping rather than an analytic maximum principle.

Let $z_0\ne 0$ be an equality point and set $c=g(z_0)$, so $|c|=1$. For any $|z_0|<r<1$, the same Brownian exit identity givesPart (i) gives $|g|\le 1$, so the real random variable in the expectation is at most one. Equality of its mean to one forces $g(Z_{T_r})=c$ almost surely. By the scaled version of the exit density from Question 2, every nonempty arc of the exit circle has positive probability. Continuity of $g$ on that circle therefore implies $g=c$ at every boundary point of the radius-$r$ disk.

For any $w$ inside that circle, the Brownian exit identity now gives $g(w)={\mathbb{E}}_{w}g(Z_{T_r})=c$. Letting $r$ increase to one makes this true throughout $D$. Writing $c=e^{i\theta }$ yieldsThis completes the equality case of the Brownian proof of the Schwarz lemma.