Find all entire functions that satisfy \[ \abs{f(z)} \geq \abs{z} \quad \forall z\in \CC .\] Prove this list is complete.
Solution
Suppose \(f\) is entire and define \(g(z) \definedas {z \over f(z)}\).
By the inequality, \(\abs{g(z)} \leq 1\), so \(g\) is bounded.
\(g\) potentially has singularities at the zeros \(Z_f \definedas \inverseof{f}(0)\), but since \(f\) is entire, \(g\) is holomorphic on \(\CC\setminus Z_f\).
Claim: \(Z_f = \theset{0}\).
- If \(f(z) = 0\), then \(\abs{z} \leq \abs{f(z)} = 0\) which forces \(z=0\).
We can now apply Riemann’s removable singularity theorem:
Check \(g\) is bounded on some open subset \(D\smz\), clear since it’s bounded everywhere
Check \(g\) is holomorphic on \(D\smz\), clear since the only singularity of \(g\) is \(z=0\).
By Riemann’s removable singularity theorem, the singularity \(z = 0\) is removable and \(g\) has an extension to an entire function \(\tilde g\).
By continuity, we have \(\abs{\tilde g(z)} \leq 1\) on all of \(\CC\)
- If not, then \(\abs{\tilde g(0)} = 1+\eps > 1\), but then there would be a domain \(\Omega \subseteq \CC\smz\) such that \(1 < \abs{\tilde g(z)} \leq 1 +\eps\) on \(\Omega\), a contradiction.
By Liouville, \(\tilde g\) is constant, so \(\tilde g(z) = c_0\) with \(\abs {c_0} \leq 1\)
Thus \(f(z) = \inverseof{c_0} z \definedas cz\) where \(\abs{c}\geq 1\)
Thus all such functions are of the form \(f(z) = cz\) for some \(c\in \CC\) with \(\abs{c}\geq 1\).