Schwarz Lemma

Complex Analysis · Axiom Academy

An analytic map of the disk to itself that fixes the center can only pull points inward — and equality makes it a rotation. 1. Self-Maps of the Disk Pull Inward Let be analytic with f(0)=0 . The lemma's first claim is a clean bound on how far any point can travel: its image can be no farther from the center than the point itself. every point's modulus can only shrink (or hold) and the stretch at the center is capped at 1 2. The Proof in One Move: g=f/z Why must the bound hold? Because f(0)=0 , the quotient g(z)=f(z)/z has a removable singularity at the origin (set g(0)=f'(0) ) and is analytic on all of . On the circle |z|=r , since , The Maximum-Modulus Principle says |g| cannot peak inside, so throughout the disk of radius r . Now push the circle outward, : the bound 1/r slides down to 1 . That squeeze leaves everywhere, which is exactly and . Why the growing circle is the whole argument At any fixed radius the bound 1/r is loose (bigger than 1). Only in the limit, as the boundary circle fills the disk, does the bound tighten to the sharp constant 1. What if a point refuses to shrink — |f(z_0)|=|z_0| for some , or |f'(0)|=1 ? Then |g| hits its maximum value 1 at an interior point. Maximum-Modulus now bites the other way: g must be a constant of modulus one, . So the only borderline maps are pure rotations. a rotation preserves every modulus — it spins, it never shrinks

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