Schwarz lemma

For a holomorphic function on the unit disk fixing zero, $g(z)=f(z)/z$ has a removable singularity at zero and a bounded holomorphic extension. A bounded holomorphic function evaluated along planar Brownian motion is a bounded stopped martingale before domain exit. Stop at the first exit from a concentric radius-$r$ disk with $|z|<r<1$. Optional stopping gives $g(z)={\mathbb{E}}_{z}g(Z_{T_r})$, hence $|g(z)|\le 1/r$. Letting $r$ increase to one proves the displayed bound.

If $|g(z_0)|=1$ at a nonzero point, put $c=g(z_0)$. The same expectation identity forces $\mathrm{Re}(\overline{c}g(Z_{T_r}))=1$ almost surely. The exit density is strictly positive on every boundary arc by the Poisson kernel for the unit disk, and $g$ is continuous there, so $g=c$ on that circle. The expectation identity for every starting point inside it then gives $g=c$ throughout that smaller disk. Letting $r$ increase to one shows $f(z)=cz$ on the full disk, with $|c|=1$. This derives both the inequality and its rigidity using Brownian harmonic measure.

For a zero-free holomorphic function $f:\mathbb{D}\to \mathbb{D}$, put $c=f(0)$ and $\alpha =-\log |c|$. A holomorphic logarithm makes $-\log f$ have positive real part. The Carathéodory derivative bound for the right half-plane yields the displayed sharp bound. Equality functions are $f(z)=c\exp [-2\alpha \omega z/(1-\omega z)]$, $|\omega |=1$. The Harnack inequality on the unit disk also yields the sharp pointwise bounds $|c{|}^{(1+r)/(1-r)}\le |f(w)|\le |c{|}^{(1-r)/(1+r)}$ for $r=|w|$. Aligning $\omega w=\pm r$ attains the two endpoints.

If a holomorphic function $F$ on the unit disc has positive real part, the Cayley transform between the half-plane and disk $\varphi =(F-F(0))/(F+\overline{F(0)})$ maps the disc into itself and vanishes at zero. The Schwarz lemma gives the displayed derivative bound. Equality holds exactly when $\varphi (z)=\omega z$ with $|\omega |=1$, equivalently $F(z)=i\mathrm{Im}F(0)+\mathrm{Re}F(0)(1+\omega z)/(1-\omega z)$.

A holomorphic self-map of the unit disc contracts its pseudohyperbolic distance. Conjugating by Möbius transformations that send one chosen source point and its image to zero reduces the statement to the Schwarz lemma. In the complex upper half-plane this givesIf $f(iY)=iY+o(Y)$ at infinity, multiply the equivalent inequality between one minus the squared distances by $Y$ and let $Y\to \infty$. It yields $\mathrm{Im}f(z)\ge \mathrm{Im}z$. Applied to the inverse mapping-out function of a compact H-hull, this proves the nonnegativity of its half-plane capacity.

If a holomorphic function $f$ on the unit disc satisfies $|f|\le M$ and $f(a)=0$, the Schwarz-Pick lemma applied to $f/M$ gives $|f(0)|/M\le |a|$. Equality for a nonzero central value forces a scaled automorphism of the unit disk. If $f$ extends past the unit circle and $|f(0)|$ is smaller than its minimum boundary modulus, the maximum modulus principle applied to $1/f$ proves that a zero occurs inside the circle.

For holomorphic functions $f:\mathbb{D}\to \mathbb{D}$ with $f(z_0)=w_0$, put $r=|(z-z_0)/(1-{\overline{z}}_{0}z)|$. Their possible values at $z$ form the closed disc with centre $(1-r^2)w_0/(1-r^2|w_0{|}^2)$ and radius $r(1-|w_0{|}^2)/(1-r^2|w_0{|}^2)$. Indeed ${\psi }_{w_0}\circ f={\psi }_{z_0}h$ with holomorphic $|h|\le 1$; all constant choices of $h$ in the closed unit disc are admissible. Completing the square in the displayed pseudohyperbolic inequality gives the centre and radius.

A holomorphic function between hyperbolic Riemann surfaces lifts to a disk map. If the lift is not an automorphism, strict Schwarz-Pick lemma derivative inequality makes every nonconstant geodesic shorter. Equality of distance for one distinct pair therefore forces the map to be a covering map and a local isometry. Covering maps attain equality for sufficiently close pairs in a single lift of a small convex metric ball; they need not preserve all global distances. A punctured disk covers a thrice-punctured sphere by including a cyclic parabolic subgroup into $\Gamma (2)$, so equality is possible in that example.

// Destination: analysis.bigb

Let $L>0$ and consider holomorphic functions $f$ from the unit disc into $e^{-L}<|w|<e^L$, normalized by $f(0)=1$. Their largest possible derivative modulus is $4L/\pi$. Lift $f$ to its holomorphic logarithm $h$ with $h(0)=0$, so $|\mathrm{Re}h|<L$. The function $\tan (\pi h/(4L))$ maps the disc into itself and fixes zero. The Schwarz lemma gives $|h^{\prime }(0)|\le 4L/\pi$. Equality holds precisely forHere $\arctan$ is its holomorphic branch vanishing at zero; $(4L/\pi )\arctan z$ is a conformal map from the disc onto the vertical strip. Thus the maximal derivative modulus is attained, with a circle of distinct extremizers rather than a unique map.