Let \(f\) and \(g\) be functions holomorphic defined on a domain \(U \subseteq \mathbb{C}\). Set \(\varphi(z) = |f(z)| + |g(z)|\) for \(z \in U\). If \(\varphi\) assumes a maximum value on \(U\), show that both \(f\) and \(g\) are constants on \(U\).
Solution
Choose \(z_0\in U\) where the maximum is attained and set \(M=|f(z_0)|+|g(z_0)|\). If \(M=0\), the inequality \(0\leq|f(z)|+|g(z)|\leq M\) makes both functions zero. Assume henceforth that \(M>0\).
1. A suitable holomorphic linear combination is constant.
Choose complex numbers \(\alpha,\beta\) of modulus one with \(\alpha f(z_0)=|f(z_0)|\) and \(\beta g(z_0)=|g(z_0)|\). For a nonzero value, take its conjugate divided by its modulus; for a zero value, any unit complex number works. The holomorphic function \(H=\alpha f+\beta g\) satisfies \[ |H(z)|\leq|f(z)|+|g(z)|\leq M, \qquad H(z_0)=M. \] The maximum modulus principle on the connected domain \(U\) therefore gives \(H\equiv M\) [@SS03].
2. Each summand must be constant separately.
For every \(z\in U\), \[ \bigl(|\alpha f(z)|-\operatorname{Re}(\alpha f(z))\bigr) +\bigl(|\beta g(z)|-\operatorname{Re}(\beta g(z))\bigr) =\varphi(z)-M\leq0. \] Each term on the left is nonnegative, so both vanish. A complex number whose real part equals its modulus is real and nonnegative. Thus \(\alpha f\) and \(\beta g\) take values in \([0,\infty)\). A nonconstant holomorphic function on a domain has open image [@SS03], which cannot be contained in this real half-line. Hence both \(\alpha f\) and \(\beta g\) are constant. Since the phases are nonzero, \(f\) and \(g\) are constant as well.