Suppose thatt $u(x,y)$ is a real valued non constant harmonic function on a bounded domain D. Then $u(x,y)$ can not attain its maximum or minimum value in $D$.

I am studing complex analysys in $S. Ponnusammy$. It used the following result

$\bullet$ (Maximum Modulus Theorem ) Suppose $f$ is analytic function in a bounded domain $\overline D$. Then $|f(z)|$ attains its maximum at some point on the boundary $\partial D$ of $D$.

$\bullet$ ( Maximum Modulus Principal) Suppose $f$ is analytic function on a domain $D$ and a is a point in $D$ such that $|f(z)| < |f(a) |$ hold for all $ z \in D$. Then $f$ is constant.

Suppose $D$ is simply connected , then there exist analytic function $f = u + \iota v$. He apply the maximum modulus theorem to $g(z) = e^{f(z)}$.

Why he apply Maximum Modulus Theorem to $g(z)$ rather than $f(z)$. Can I use directly Maximum Modulus Principal. Please Clear my doubt. Any help would be appreciated. Thank you

  • $\begingroup$ Because $|g| = |e^f| = |e^u|$. And since $u$ is real-valued, $u(x)$ takes its maximum iff $e^u$ does. $\endgroup$
    – user171326
    May 20, 2015 at 17:52
  • $\begingroup$ @ N.H. : Can I use Maximum Modulus Principal on $f(z)$. $\endgroup$
    – user120386
    May 21, 2015 at 1:15
  • $\begingroup$ Of course you can, but it will be not very interesting for you. It will tell you that $u(z)^2 + v(z)^2$ has a maximum only on the boundary, which not allow you to conclude immediately. $\endgroup$
    – user171326
    May 21, 2015 at 7:09

1 Answer 1


Using the maximum principle on $f$ is of course allowed because the hypothesis are satisfied. But this is not very concluding : the only information we know is that $u^2 + v^2$ take its maximum on $\partial D$ .

Using $g$ is more appropriate : $|g| = |e^u|$ and therefore $g$ verifie the Maximum Modulus Principle $\Leftrightarrow$ $u$ verifies the MMP too.

But in fact it's true for $v$ too : take $f' = if$ and $g' = e^{f'}$. Because $g'$ is analytic, the maximum principle applies, and by the same argument (namely that $|g'| = |v|$) you can conclude that $v$ also take its maximum on $\partial D$ (until it's constant).

  • $\begingroup$ Why is $|g'| = |v|?$ When I calculate this I get that $|g'| = e^{-v}$ $\endgroup$ Apr 7, 2021 at 2:51

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