Vtgyjfy

$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 ?

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.

We have that $h(z)= sum_{i+1}^n |f_i|$ is a subharmonic function. So it obeys the maximum modulus principle. Hence, $h$ is a constant. If we take $h_j= sum_{i neq j } |f_i|$, we see that $|h_j| leq |h|$ for each $j$. If an $f_j$ has a zero in our domain, notice that $h_j$ attains its maximum as well. This would imply that $f_j equiv 0$ in our domain. So, if we suppose each $f_j$ is non-constant and non-zero in our domain, we have that each $f_j = e^{g_j(z)}$ for some analytic function $g_j$ on our domain. This reduces $h=sum_{i=1}^n e^{Re(g_j(z)}$. Since each $g_j(z)$ is non-constant, by the open mapping theorem, we can take a $z_0$ in our domain contained in an open set $U_j$ such that each $U_j$ contain infinitely many points, $z$, such that $e^{Re(g_j(z)} < e^{Re(g_j(z_0)} $ . Since we only have finitely many $U_j$, we may take their intersection to be $U$ and select a point $w in U$ such that $e^{Re(g_j(w)} < e^{Re(g_j(z_0)} $ for each $j$. This gives us $h(w) < h(z_0)$ - a contradiction.