1

I know there have been some other posts, but the always had the additional requirement, that $|f_1|^2+|f_2|^2$ is constant, which I don‘t have.\ $G\subseteq \mathbb{C}$ is a Domain and $f_1,f_2\in H(G)$.
My ideas so far have been to define $g:= f_1+f_2 $ and then we get: \begin{align*} |g(z_0)|=|(f_1+f_2)(z_0)|\leq (|f_1|+|f_2|)(z_0)=m \end{align*} The last equality is given in the exercise and I also know by triangle inequality, that $|g(z)|\leq (|f_1|+|f_2|)(z), \quad \forall z\in G$. So also $g$ has to reach its maximum in G and by maximum principle $g=d\in \mathbb{C}$.
From here on I got stuck, since I can‘t see how to conclude from $g=f_1+f_2=d$ that $f_1,f_2$ are constant. Can anyone help me with this last step?

Edit: It is important to note, that it is already given, that $f_1+f_2$ has a local maximum. The important step is the last one, which didn‘t get clarified by the other questions regarding this question, especially not the one that works with a finite sum of functions $f_i$. Also in the other posts the question was to proof, that the maximum apears on the border. That's given in this exercise.

MilesDefis
  • 177
  • Why "$g$ has to reach its maximum in G"? – Anne Bauval Jun 13 '23 at 12:23
  • @AnneBauval my idea was, since by the triangle inequality $|()|≤(|_1|+|_2|)()$ satisfy for all z and $|f_1|+|f_2|$ takes its maximum on G, also $g$ has to have a maximum on z. – MilesDefis Jun 13 '23 at 13:17
  • I don't see why the former implies the latter 2) I doubt the former https://math.stackexchange.com/questions/1180056
    • – Anne Bauval Jun 13 '23 at 13:21
  • @AnneBauval Well that's just the case with my exercise... My domain $G$ isn't bounded per se, so I think it's safe to assume that the max isn't necessarily on the boundary of G. – MilesDefis Jun 13 '23 at 13:29
  • Assuming that the max (of $|f_1|+|f_2|$) is not necessarily on $\partial G$ $\ne$ proving "$|f_1|+|f_2|$ does take its maximum on G". And even if it does, this (+ your triangle inequality) does not seem to imply that $g$ also does. – Anne Bauval Jun 13 '23 at 15:17