Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 4

The Cauchy integral formula says that if $f$ is holomorphic on the simply connected open set $U$, and $\gamma$ is a positively oriented simple closed curve in $U$, then, for every $z$ in its interior,Here the contour and its enclosed region lie in $U$; the contour can be taken piecewise continuously differentiable. More generally, the same formula needs only holomorphic function on an open neighbourhood of that closed region. The orientation is counterclockwise, so the winding number about an interior point is one.

Write $M=|f(z_0)|$. The local maximum property gives a disk around $z_0$ on which $|f|\le M$, contained in the unit disk. If $M=0$, then $f=0$ on that neighbourhood, and the identity theorem makes $f$ zero on the connected unit disk.

Otherwise multiply $f$ by the unit complex number $\overline{f(z_0)}/M$, obtaining a holomorphic function $g$ with $g(z_0)=M$ and $|g|\le M$ nearby. Applying the Cauchy integral formula on any sufficiently small circle of radius $r$ gives the mean value propertyTake real parts. The continuous function $M-\mathrm{Re}g$ is nonnegative on the circle and has integral zero, so it vanishes everywhere there. Since $|g|\le M$, having real part $M$ forces imaginary part zero. Thus $g=M$ throughout the circle. The zeros of $g-M$ have an accumulation point inside its domain, so the identity theorem gives $f$ constant on the entire unit disk. This proves the local maximum modulus principle directly from the Cauchy integral formula.

The closed unit disk is a compact set, and $|f|$ is continuous on it because $f$ is holomorphic on an open neighbourhood. The extreme value theorem therefore supplies a point where its maximum is attained. If such a point is on the boundary, the conclusion follows. If it is in the interior, part (i) makes $f$ constant on the open disk; continuity then gives the same value on the boundary. ConsequentlyThe assumption that the closed disk lies in $U$ is what makes the boundary values and the compactness argument available.

Because $f(0)=0$, the Taylor series of $f$ at zero shows thatis holomorphic throughout the unit disk. Fix $0<r<1$. On $|z|=r$ we have $|h(z)|<1/r$, and the maximum modulus principle on the closed disk of radius $r$ gives $|h(z)|\le 1/r$ for $|z|\le r$. For any fixed $z$ inside the unit disk we may let $r$ increase to one, obtaining $|h(z)|\le 1$. Hence the Schwarz lemma inequalities areThe derivative bound is the same estimate at $z=0$, using the holomorphic extension of the quotient.

Either equality condition says that the holomorphic function $h$ from part (i) has modulus one at an interior point: at zero in the derivative case, or at the given nonzero point in the other case. Since $|h|\le 1$ everywhere, the maximum modulus principle makes $h$ a constant $w$ with $|w|=1$. ThereforeThese maps are rotations of the unit disk and do achieve both equalities, so this also identifies all equality cases of the Schwarz lemma.