Let \(f,g\) be entire functions and suppose there is a real constant \(k\) such that \[ \operatorname{Re}f(z)\le k\operatorname{Re}g(z) \] for every \(z\in\mathbb C\). Show that there are constants \(a,b\) such that \[ f(z)=ag(z)+b. \]
Solution
Define \[ h\coloneqq f-kg. \]
1. The function \(h\) is entire and satisfies\[ \Re h(z)\leq0 \]for every \(z\in\CC\).
Since \(f\) and \(g\) are entire and \(k\) is constant, \(h\) is entire. Because \(k\) is real, \[ \Re h(z) = \Re f(z)-k\Re g(z) \leq0 \] by the hypothesis.
2. The entire function \(e^h\) is bounded by \(1\) in modulus.
For every \(z\in\CC\), \[ \abs{e^{h(z)}} = e^{\Re h(z)} \leq1 \] by step 1.
3. The function \(h\) is constant.
By step 2 and Liouville’s theorem, \(e^h\) is constant. Differentiating gives \[ e^{h(z)}h'(z)=0 \] for every \(z\in\CC\). Since the exponential function never vanishes, \[ h'(z)=0 \] for every \(z\). Hence \(h\) is constant on \(\CC\).
4. There are constants \(a,b\in\CC\) such that\[ f(z)=ag(z)+b \]for every \(z\in\CC\).
By step 3, there is \(b\in\CC\) with \[ h=f-kg=b. \] Thus \[ f=kg+b. \] Taking \[ a=k \] gives the required form.
5. Q.E.D.
Step 4 is the desired conclusion.