Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 28 / 5 / b

Let $u=\Re f$, and suppose $u$ has a local maximum at $z_0\in U$. Choose $R>0$ such that the closed disk $\overline{D(z_0,R)}$ lies in $U$ and $u(z)\le u(z_0)$ throughout it. Start planar Brownian motion $B$ at $z_0$, and for $0<r<R$ let ${\tau }_r$ be its first exit from $D(z_0,r)$.

By part (a) and Itô formula, $u(B_{t\wedge {\tau }_r})$ is a local martingale; boundedness of $u$ on the compact disk makes it a true bounded martingale. The exit time is finite almost surely. One direct check applies the same formula to $|B_{t\wedge {\tau }_r}-z_0{|}^2$, giving $2\mathbb{E}(t\wedge {\tau }_r)=\mathbb{E}|B_{t\wedge {\tau }_r}-z_0{|}^2\le r^2$. Thus $\mathbb{E}{\tau }_r\le r^2/2$.

The optional stopping theorem, first at $t\wedge {\tau }_r$ and then by bounded convergence, givesThe exit point is uniformly distributed on the circle: rotational invariance of planar Brownian motion about its starting point makes its exit law rotation invariant, and normalized arc length is the unique rotation-invariant probability measure on a circle. Since every boundary value is at most $u(z_0)$, equality of the mean forces every boundary value to equal $u(z_0)$; continuity rules out a strict deficit at any boundary point. This holds for every $0<r<R$, so $u$ is constant on the entire disk.

The Cauchy-Riemann equations then give both partial derivatives of $v=\Im f$ equal to zero there. Thus $f$ is constant on that disk. The identity theorem for holomorphic functions propagates this constant to the connected component of $z_0$ in $U$. Applying the same Brownian argument to $v$ proves the assertion for the imaginary part as well:If $U$ is connected, this is constancy on all of $U$. For a disconnected open set the printed global wording needs this qualification: a function taking different constants on two disjoint disks is holomorphic and has local maxima on each disk while not being globally constant.

Suppose $|g|$ has a local maximum at $z_0$. If $g(z_0)=0$, the nonnegativity of the modulus forces $g=0$ throughout a neighbourhood, and the identity theorem for holomorphic functions makes it zero on that connected component.

If $g(z_0)\ne 0$, take $c=2i/g(z_0)$. Then $cg(z_0)=2i$, and continuity gives a disk $D$ around $z_0$ on which $\Im (cg)>1$. This lies within the given half-plane for a branch of the holomorphic logarithm. On $D$ defineComposition preserves holomorphy, and the real logarithm is strictly increasing, so $\Re f$ has a local maximum at $z_0$. Part (i) forces $f$ to be constant on the disk. Exponentiating gives $cg=e^f$, hence $g$ is constant there and, by the identity theorem for holomorphic functions, on the whole connected component. ThereforeThis proves the maximum modulus principle by the requested logarithm argument. Multiplication by $c$ is what puts the image in the supplied logarithm branch; no global logarithm of $g$ is assumed.