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The problem I had was: If $f$ is holomorphic on $U \subseteq \mathbb{C}$ and $\exists z_0$ in $D(0,1)$ (which is fully contained in $U$) such that $|f|$ has a local maximum at $z_0$ then $f$ is constant. I tried using Cauchy's integral Formula and letting the path to be integrated over be the boundary of the small disc $D(z_0,\epsilon)$- we know this little disc exists because thats what it means to be a local maximum at $z_0$-there exists a small disc where the modulus of everything in the disc is bounded by $M$ where in this case $f(z_0)=M$. Applying estimation theorem on integral formula I got everything in the whole of $D(0,1)$ must be less than or equal to $M$. But I cannot get equality. Can someone please help? I googled and reached "maximum modulus principle" which is more or less this problem on an arbitrary subset of $\mathbb{C}$ and not just unit disc.

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  • $\begingroup$ Do you mean "$\exists z_0\in D(0,1)$ such that $|f|$ has a local maximum"? $\endgroup$ – Arthur Dec 25 '15 at 10:16
  • $\begingroup$ @Arthur yes sorry ill edit that $\endgroup$ – Arcane1729 Dec 25 '15 at 10:35
  • $\begingroup$ math.uchicago.edu/~bloch/math270040506.pdf $\endgroup$ – Arashium Dec 25 '15 at 10:36
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The Maximum Modulus Principle states that if $f$ is analytic on a connected open set $D$ and continuous on the closure $\overline D$ then $|f|$ attains its maximum on the boundary of $D$. The proof I've seen is done using the Mean Value Property for discs.

Given this principle your problem is pretty simple. As you have exhibited you can find a small disc $D_r(z_0)$ such that $\overline {D_r(z_0)} \subset U$ and $|f|$ attains a maximum at $z_0$. Then, $|f|$ must be constant on $D_r(z_0)$ and hence analyticity forces $f$ to be constant on this disc.

Now if $U$ is a connected set $f$ is also constant on $U$ by the theorem stated as an "improvement" here (which is awesome btw).

The conclusion need not be true if $U$ is not connected. For an example define,

$f:D_1(-1) \cup D_1(1) \to \Bbb C $ by $f(z) = 0$ if $z \in D_1(-1)$ and $f(z) = 1$ if $z \in D_1(1)$

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