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?

  • $\begingroup$ 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 $\endgroup$
    – Conrad
    Commented Nov 14, 2019 at 1:29
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    $\begingroup$ What kind of set is $U$? Was that the unit disk? $\endgroup$
    – Lubin
    Commented Nov 14, 2019 at 2:29
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    $\begingroup$ 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. $\endgroup$
    – Pythagoras
    Commented Nov 14, 2019 at 5:19
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    $\begingroup$ You can use the proof given here: math.stackexchange.com/a/429226. $\endgroup$
    – Martin R
    Commented Nov 14, 2019 at 8:10
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    $\begingroup$ @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. $\endgroup$
    – Martin R
    Commented Nov 14, 2019 at 9:22

1 Answer 1


As Martin R. pointed out, the result follows from a similar problem. It boils down to the following statement.

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


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