Spring 2015 - Problem 8

Phragmén-Lindelöf

Let f ⁣:CC\func{f}{\C}{\C} be holomorphic and suppose

supxR{f(x)2+f(ix)2}andf(z)ezfor all zC.\sup_{x \in \R} \,\set{\abs{f\p{x}}^2 + \abs{f\p{ix}}^2} \quad\text{and}\quad \abs{f\p{z}} \leq e^{\abs{z}} \quad\text{for all } z \in \C.

Deduce that f(z)f\p{z} is constant.

Solution.

Let M=supxR{f(x)2+f(ix)2}M = \sup_{x \in \R} \set{\abs{f\p{x}}^2 + \abs{f\p{ix}}^2}. Recall that a Phragmén-Lindelöf function for subharmonic functions on a quarter-plane is zk\abs{z}^k for 0<k<20 < k < 2. In particular, z\abs{z} is a PL function.

Observe that by assumption, logf(z)z\log{\abs{f\p{z}}} \leq \abs{z} and so logf(z)\log{\abs{f\p{z}}} obeys the maximum principle on any quarter-plane. Thus, on each quadrant, we have logf(z)logM\log{\abs{f\p{z}}} \leq \log{M}, since the boundary of each quadrant is contained in the coordinate axes. Hence, we see that logf(z)logM\log{\abs{f\p{z}}} \leq \log{M} in the entire plane, and so ff itself is bounded on C\C. By Liouville's theorem, ff is constant, as desired.