$f_1 , f_2 , … , f_n $ is a sequence of holomorphic function in an open set $\Omega$ , and also $$|f_1|+…+|f_n|$$ attains its maximum in $\Omega$ . Can we prove that each of $f_k$ is constant ?
My attempt :
If $n=1$ , we can find a $\theta$ such that $f_1' =f_1 e^{i \theta}$ attains its maximum in $\Omega$ , then $f_1 ' $ is a constant so $f_1$ is a constant .
If $n \gt 1$ , then we can find a sequence $\theta_1 ,…, \theta_n$ such that $f_1 e^{i \theta_1} +…+ f_n e^{i \theta_n}$ attains its maximum in $\Omega$ . Let $f_k'=e^{i \theta_k} f_k$ , we find that $$|f_1'|+…+|f_k'| =|f_1|+…+|f_n|$$ So , to prove both $f_k$ are constant , it suffice to prove that following statement :
$g_1 , … , g_n$ is a sequence of holomorphic function in an open set $\Omega$ and $g_1+…+g_n$ equal to a constant $C$ , $|g_1|+…+|g_n|$ attains its maximum in $\Omega$ , then each of $g_k$ is constant.
Can we show this ?
Best Answer
Proof for $n=2$. (The argument actually works for any $n$). Suppose $a \in \Omega$ and $|f(a)|+|g(a)|≥|f(z)|+|g(z)| ∀z∈Ω$. Replace $f$ by $e^{is}f$ and $g$ by $e^{it}g$ where real numbers $s$ and $t$ are chosen such that $e^{is}f(a)$ and $e^{it}g(a)$ both belong to [0,∞). This reduces the proof to the case when $f(a)$ and $g(a)$ both belong to [0,∞). We now have $f(a)+g(a)≥|f(z)|+|g(z)|≥Ref(z)+Reg(z)=Re(f(z)+g(z))$. Maximum Modulus principle applied to $e^{f+g}$ shows that $e^{f+g}$, and hence $f+g$ is a constant (because $|e^{f+g}|$ attains its maximum in the domain). Now $f(a)+g(a)≥|f(z)|+|g(z)|≥Ref(z)+Reg(z)=Re(f(z)+g(z))=Re(f(a)+g(a))$ which implies that equality holds throughout. In particular $|f(z)|=Re(f(z))$ and $|g(z)|=Re(g(z)) ∀z$ which implies that $f$ and $g$ are constants.