Bloch's theorem

Remarks

The supremum of all such $b$ is the Bloch constant $B$, one of Landau's constants. Lars Ahlfors proved $B \ge \sqrt{3}/4$, and it is known that $B < 0.472$; the exact value is unknown. One proof first replaces $f$ by $f(rz)/r$ for some $r < 1$, so that $f$ is holomorphic on the closed disc. Then one picks a point where $(1 - \abs{z}^2)\abs{f'(z)}$ is maximal and rescales around it, which gives a function whose derivative is bounded on a disc of fixed size. A bounded derivative with $\abs{f'(0)} = 1$ forces $f$ to be injective on a smaller disc of explicit radius, and the image of that disc contains a disc of explicit radius. The theorem gives a short proof of Picard's little theorem. The normalisation $f'(0) = 1$ is needed: the functions $f(z) = \varepsilon z$ only cover discs of radius $\varepsilon$.

Going deeper

For a student

The theorem is a covering theorem with no assumption of injectivity on $f$. For injective $f$, the Koebe quarter theorem gives much more: a disc of radius $1/4$ around $f(0)$. Bloch's theorem is scale invariant in the right way, so applying it to $f(rz)$ with large $r$ shows that a nonconstant entire function covers schlicht discs of every radius. This is the key step in the proof of Picard's little theorem by Bloch's method.