# Question about Maximum Modulus Principle applied to $|f|+|g|$

Let the functions $$f$$ and $$g$$ be holomorphic in $$U$$ and continuous in $$\overline{U}$$. Show that $$|f(z)| + |g(z)|$$ attains its maximum on $$\{|z| = 1\}$$. Hint: consider the function $$h = e^{iα}f + e ^{iβ}g$$ with suitably chosen constants $$α$$ and $$β$$.

Even with the hint I am lost. Of course $$|h(z)|$$ must attain its maximum when $$|z|=1$$, and $$|h| \le |f|+|g|$$, but that doesn't mean that $$|f|+|g|$$ couldn't attain its maximum inside the unit circle.

ny help?

• Show that for each $z$ in the unit circle, there are $\alpha, \beta$ (depending on $z$ of course) s.t. $|f(z)|+|g(z)|=e^{i\alpha}f(z)+e^{i\beta}g(z)$ and then show that implies the result Commented Nov 14, 2019 at 1:29
• What kind of set is $U$? Was that the unit disk? Commented Nov 14, 2019 at 2:29
• The hint is not clear to me either as $|f|$ and $|g|$ could achieve maximum at different $z$'s on the boundary. However since $|f|+|g|$ is subharmonic, the maximum principle remains valid. Commented Nov 14, 2019 at 5:19
• You can use the proof given here: math.stackexchange.com/a/429226. Commented Nov 14, 2019 at 8:10
• @Pythagoras: The idea is to assume that $|f(z)| + |g(z)|$ assumes its maximum at a point $z_0$ in the interior of the disk, and apply the hint at that point $z_0$. But that is essentially what is written in the answer referenced in my previous comment. Commented Nov 14, 2019 at 9:22

Proposition. Assume $$f,g$$ are holomorphic in a domain $$U \subset \mathbb C$$, and $$|f(z)|+|g(z)|$$ achieves maximum at an interior point. Then $$f$$ and $$g$$ are constant.
Proof. Suppose $$|f(z)|+|g(z)|$$ assumes maximum at an interior point $$z_0$$. Let $$\alpha =-\arg f(z_0)$$ and $$\beta=-\arg g(z_0)$$. Define $$h(z)=e^{i\alpha}f(z)+e^{i\beta}g(z),$$ which is holomorphic in $$U$$. Then $$h(z_0)=|f(z_0)|+|g(z_0)|=|h(z_0)|.$$ Since $$|h(z)|\leq |f(z)|+|g(z)|\leq |f(z_0)|+|g(z_0)|,$$ and $$h(z)$$ achieves maximum modulus at an interior point, $$h(z)$$ is a constant, and necessarily $$h(z)=h(z_0).$$ It follows that $$|h(z_0)|=|h(z)|\leq |f(z)|+|g(z)|\leq |f(z_0)|+|g(z_0)|=|h(z_0)|$$ $$\Rightarrow |f(z)|+|g(z)|~{\rm is~a~constant.}$$ By the result proven here, both $$f$$ and $$g$$ are constant. QED