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Suppose that we have $f_{1},\cdots, f_{n}$ holomorphic in a connected open set $D$. For $$f(z):=|f_{1}(z)|+|f_{2}(z)|+\cdot+|f_{n}(z)|\ \ \text{for}\ \ z\in D.$$ I want to show that $f(z)$ cannot achieve a strict local maximum inside $D$.

I know that this must have something to do with the maximum modulus principle, especially we have some connected open set, this will enable us to directly conclude that a function $g(z)$ is constant all over $D$ if it is holomorphic in $D$ and $|g|$ achieves a local maximum at $z_{0}\in D$.

However, I cannot connect this theorem to $f(z)$ because even thought $f_{k}(z)$'s are holomorphic, the summation of moduli of them may not be holomorphic.

I know a way to show that

Let $D$ be a disc with radius $R$ and center $z_{0}$. Suppose $f_{k}(z)$'s are holomorphic on $D$ and continuous on $\partial D$ such that $f(z_{0})\geq z$ for all $z\in D\cup C$, then $f$ is constant in $D\cup C$.

I can prove this claim using Cauchy's Integral formula, but it is long. This lemma can certainly conclude the proof quickly, but I realized that I did not really use the "connectedness" of $D$.

Is there any way for me to drive a shorter proof using maximum modulus principle? Thank you!

Edit 1:

The post is duplicated, but I'd still like to post a relatively complete proof.

Thanks to Conrad, the following is a relatively complete proof:

Suppose $\omega$ is the local maximum of $f(z):=\sum_{k=1}^{n}|f_{k}(z)|$. Note that as $f_{k}(\omega)$ is a complex number, unless it is zero, we can represent (uniquely) as $$f_{k}(\omega)=|f_{k}(\omega)|e^{it_{k}},\ \ t_{k}\in [0,2\pi).$$ Take $\alpha_{k}:=e^{-itk}$, and if $f_{k}(\omega)=0$ for some $k$, we tae $\alpha_{k}:=1$.

Then, we have $$f_{k}(\omega)=\alpha_{k}^{-1}|f_{k}(\omega)|\ \text{with}\ |\alpha_{k}|=1.$$

Now, define $g(z):=\sum_{k=1}^{n}\alpha_{k}f_{k}(z).$ Then, we have \begin{align*} |g(z)|\leq \sum_{k=1}^{n}|\alpha_{k}||f_{k}(z)|=\sum_{k=1}^{n}|f_{k}(z)|&=f(z)\\ &\leq \sum_{k=1}^{n}|f_{k}(\omega)|=f(\omega)\\ &=\sum_{k=1}^{n}f_{k}(\omega)\alpha_{k}=g(\omega)\leq|g(\omega)|\ \ \ (*). \end{align*}

Hence, $\omega$ is also the point at which $|g|$ achieves local max. As $g$ is holomorphic and $D$ is connected open, by the global maximum modulus principle, we conclude that $g(z)=C$ for some constant $C$ and for all $z\in D.$

But then, $(*)$ implies that $$|C|\leq C\leq |C|\implies C=|C|\implies C\geq 0.$$

So we have $g(z)=g(\omega)\geq 0$. But then (*) implies that $$g(\omega)=g(z)=|g(z)|\leq f(z)\leq f(\omega)\leq g(\omega),$$ and thus we must have $$g(\omega)=g(z)=f(z)=f(\omega)=C\geq 0\ \ \text{for all}\ \ z\in D,$$ and thus the local maximum of $f$ cannot be strict.

I really appreciate the patient and inspiring help from Conrad.

JacobsonRadical
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  • Let $w$ such a maximum and $f_k(w)=\alpha_k^{-1}|f_k(w)|, |\alpha_k|=1$ (where one picks $\alpha_k=1$ if some $f_k(w)=0$; then look at $g=\sum \alpha_k f_k$ and show that $g$ has a local maximum at $w$ hence it is constant so the maximum which is the same as the one of $f$ cannot be strict; connectdness is used to deduce that $g$ is constant of course – Conrad Dec 10 '20 at 20:43
  • @Conrad Sure, for all $z\in D$, $$|g(z)|\leq \sum_{k=1}^{n}\alpha_{k}|f_{k}(z)|=\sum_{k=1}^{n}\alpha_{k}|f_{k}(\omega)|,$$ so $\omega$ is also a local maximum of $g(z)$. Then the maximum modulus principle with connectedness of $D$ implies that $g$ is constant over all $D$. But then $$g=\sum_{k=1}^{n}\alpha_{k}f_{k}=\sum_{k=1}^{n}|f_{k}|$$ is constant over all $D$, so is not strict maximum. However, I am not sure how can I write $f_{k}(\omega)=\alpha_{k}^{-1}|f_{k}(\omega)|$. Would you mind elaborating this? – JacobsonRadical Dec 10 '20 at 21:24
  • @Conrad I am sorry, the first line, it should be, for all $z\in D$, we have $$|g(z)|\leq \sum_{k=1}^{n}|\alpha_{k}||f_{k}(z)|=\sum_{k=1}^{n}|f_{k}(z)|\leq \sum_{k=1}^{n}|f_{k}(\omega)|=\sum_{k=1}^{n}\alpha_{k}f_{k}(\omega)=g(\omega),$$ so $\omega$ is also the point $g$ attains a local maximum in $D$. – JacobsonRadical Dec 10 '20 at 21:42
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    $f_k(w)$ is a complex number so unless it is zero, there is a unique representation $f_k(w)=|f_k(w)|e^{it_k}, t_k \in [0, 2\pi)$ say; take $\alpha_k=e^{-it_k}$; the zero case has of course no unique representation, but then you can just take $\alpha_k=1$; once you deduce $g$ constant, $g(z)=g(w) >0$ it follows that $g(w)=g(z)=|g(z)| \le f(z) \le f(w) =g(w)$ so all those inequalities become equalities and $f(z)=f(w)$ for all $z$ – Conrad Dec 10 '20 at 22:06
  • @Conrad nice. Thank you! – JacobsonRadical Dec 10 '20 at 22:24
  • happy to help - no problem – Conrad Dec 10 '20 at 22:30
  • I had closed this question as a duplicate of https://math.stackexchange.com/q/96257/42969 because I think that is the identical question and it has two good answers. If you think that those answers are not detailed enough then you can always add another answer to the other question. – The purpose of closing a question as a duplicate is to have all solutions of a problem at one place. – Martin R Dec 10 '20 at 23:24
  • @MartinR Okay. I am sorry that I did not understand this community rule clearly. I will avoid such things in the future. Thank you for pointing out the duplication :) – JacobsonRadical Dec 10 '20 at 23:58
  • @amWhy yes, similar as what Martin R pointed out. Let me close my own post. – JacobsonRadical Dec 11 '20 at 00:41

1 Answers1

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Thanks to Conrad's patient and inspiring help, I wrote the proof, as follows:

Suppose $\omega$ is the point at which local maximum of $f(z):=\sum_{k=1}^{n}|f_{k}(z)|$ is achieved. Note that as $f_{k}(\omega)$ is a complex number, unless it is zero, we can represent (uniquely) as $$f_{k}(\omega)=|f_{k}(\omega)|e^{it_{k}},\ \ t_{k}\in [0,2\pi).$$ Take $\alpha_{k}:=e^{-itk}$, and if $f_{k}(\omega)=0$ for some $k$, we tae $\alpha_{k}:=1$.

Then, we have $$f_{k}(\omega)=\alpha_{k}^{-1}|f_{k}(\omega)|\ \text{with}\ |\alpha_{k}|=1.$$

Now, define $g(z):=\sum_{k=1}^{n}\alpha_{k}f_{k}(z).$ Then, we have \begin{align*} |g(z)|\leq \sum_{k=1}^{n}|\alpha_{k}||f_{k}(z)|=\sum_{k=1}^{n}|f_{k}(z)|&=f(z)\\ &\leq \sum_{k=1}^{n}|f_{k}(\omega)|=f(\omega)\\ &=\sum_{k=1}^{n}f_{k}(\omega)\alpha_{k}=g(\omega)\leq|g(\omega)|\ \ \ (*). \end{align*}

Hence, $\omega$ is also the point at which $|g|$ achieves local max. As $g$ is holomorphic and $D$ is connected open, by the global maximum modulus principle, we conclude that $g(z)=C$ for some constant $C$ and for all $z\in D.$

But then, $(*)$ implies that $$|C|\leq C\leq |C|\implies C=|C|\implies C\geq 0.$$

So we have $g(z)=g(\omega)\geq 0$. But then (*) implies that $$g(\omega)=g(z)=|g(z)|\leq f(z)\leq f(\omega)\leq g(\omega),$$ and thus we must have $$g(\omega)=g(z)=f(z)=f(\omega)=C\geq 0\ \ \text{for all}\ \ z\in D,$$ and thus the local maximum of $f$ cannot be strict.

JacobsonRadical
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