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Let $G$ be a region and suppose that $f : G \rightarrow \mathbb{C}$ is analytic and $a \in G$ such that $|f(a)| \leq |f(z)| \forall z \in G$

Then either $f(a) =0 $ or $f$ is constant.

I concluded that $f$ is constant by applying Liouville's theorem by taking $|f(a)| $ to be some $M$ and showng that $0 \leq |f(z)| \leq \frac{1}{M}$ , showing $f$ is bounded and hence $f$ is constant throught out $\mathbb{C}$ , but i am not able to prove the part that $f(a) = 0$, any idea guys !

Thanks!

BAYMAX
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  • Now i think we cannot apply Liouville's theorem also as the inequality given is satisfied $\forall z \in G$ not in $\forall z \in C$ , ? – BAYMAX Feb 23 '17 at 04:00
  • Yes ,I have not encountered open mapping theorem , any other method is great but this is also fine as i will encounter it near future!@Shalop – BAYMAX Feb 23 '17 at 04:02
  • How about this? Assume $f(a) \neq 0$ then choose an open ball around $a$ contained in $G$. $f$ is away form zero over this ball. Fix a homeomorphism of this ball with the whole plane. Use it to define a function over $C$ with values $\frac{1}{f}.$ This function is analytic, and bounded by $\frac{1}{|f(a}|}. Thus it must be constant. – Behnam Esmayli Feb 23 '17 at 04:09
  • You don't need to prove $f(a) = 0$. You need to prove that if $f(a) \ne 0$, then $f(z)$ is constant. In fact, you use $f(a) \ne 0$ in your proof, because otherwise $1/M$ wouldn't exist. Your proof has an error in it, however. $G$ is not assumed to be the whole plane, so Liouville's Theorem is inapplicable. – user49640 Feb 23 '17 at 04:09
  • @Behnam For that argument to work, you can't just have a homeomorphism, you have to have an analytic mapping. But there is no analytic mapping that maps a disk to the plane bijectively. – user49640 Feb 23 '17 at 04:11
  • Wow! Such a peril was awaiting me! Nice catch. – Behnam Esmayli Feb 23 '17 at 04:12
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    Baymax, assume $f(a) \ne 0$ and apply the maximum modulus principle to the function $1/f$ in some neighbourhood of $a$ on which $f(z) \ne 0$. – user49640 Feb 23 '17 at 04:14
  • OMG! You are soooo right. If this was true, then every analytic function would have to be locally constant by that same argument. I apologize for this gross mistake. – Behnam Esmayli Feb 23 '17 at 04:14
  • Corollary: http://math.stackexchange.com/q/1292067/1424 – Jonas Meyer Feb 23 '17 at 04:16
  • This is minimum modulus theorem. See here. – Error 404 Feb 23 '17 at 05:09

2 Answers2

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I'm not sure why positrón0802 deleted the answer.

If $f$ is analytic and non constant on $G$, it is an open map.

Suppose $f$ is non constant.

Since $a \in G$, if $f(a) \neq 0$, then $f(G)$ is an open set containing $f(a)$, hence it contains $t f(a)$ for some $t \in (0,1)$ and since $|tf(a)| < |f(a)|$, we have a contradiction.

Hence either $f(a) = 0$ or $f$ is constant.

copper.hat
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You cannot apply Liouville's theorem because your function is not entire (holomorphic in all $\mathbb{C}$). This excercise can be solved by applying maximum principle. Consider $g(z)=\frac{1}{f(z)}, z\in G$ .Then it is true that if $f(a)\ne 0 $,then $$|g(z)|\le|g(a)|$$ for every $z\in G.$ That is $g(z)$ is constant inside $G$. Hence, either $f(a)=0$ and you cannot have any other conclusions, or your function is constant .

Beslikas Thanos
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