- Let f be an entire function such that the image of f does not intersect \(\{ z \in \mathbb { R } : z \geq 5 \}\) . Prove that \(f\) is a constant.
Solution
1. The entire function \(5-f\) takes values in \(\CC\setminus(-\infty,0]\).
By hypothesis \(f(z)\notin[5,\infty)\) for every \(z\), so \(5-f(z)\notin(-\infty,0]\). Thus the entire function \(5-f\) takes values in \(\mathbb C\setminus(-\infty,0]\).
2. The principal square root \(g=\sqrt{5-f}\) is entire and satisfies \(\operatorname{Re}g>0\).
The plane minus the nonpositive real axis is simply connected and does not contain \(0\). On this domain the principal branch of the square root is holomorphic. Since \(5-f\) is entire and maps into that domain, the composition \(g(z)=\sqrt{5-f(z)}\) is entire, and its square is \(5-f\). Moreover \(\operatorname{Re}g(z)>0\) for every \(z\), because a square root of a point off \((-\infty,0]\) lies in the open right half-plane.
3. A Cayley transform of \(g\) is a bounded entire function.
The map \(\phi(w)=(w-1)/(w+1)\) sends the open right half-plane biholomorphically onto the unit disk, since for \(\operatorname{Re}w>0\) one has \(|w+1|^2-|w-1|^2=4\operatorname{Re}w>0\). The denominator \(g(z)+1\) never vanishes because \(\operatorname{Re}g(z)>0\). Hence \(\phi\circ g\) is an entire function with \(|\phi(g(z))|<1\) for all \(z\).
4. Liouville’s theorem forces \(f\) to be constant.
The entire function \(\phi\circ g\) is bounded, so by Liouville’s theorem it is constant [@SS03]. Because \(\phi\) is injective, \(g\) is constant, and therefore \(5-f=g^2\) is constant. Hence \(f\) itself is constant.