Show that either $f\left(c\right)=0$ or $f$ is constant in $D$.

Question

Let $D$ be a domain and suppose $f$ is analytic on $D$ and $c\in D$ such that $\left|f\left(c\right)\right|\leq\left|f\left(z\right)\right|$ for all $z$ in $D$. Show that either $f\left(c\right)=0$ or $f$ is constant in $D$.

Proof. If $f\left(c\right)=0$ then we are done. If $f\left(c\right)\neq0$ and $\left|f\left(c\right)\right|\leq\left|f\left(z\right)\right|$ for every $z\in D$. Then ${\displaystyle \left|\frac{1}{f\left(z\right)}\right|\leq\left|\frac{1}{f\left(c\right)}\right|}$. Thus, ${\displaystyle \frac{1}{f\left(z\right)}}$ is entire $\left(?\right)$ and bounded. By Liouvilles' Theorem, $f$ is constant.

I think something is wrong with this proof. Can we say that ${\displaystyle \frac{1}{f\left(z\right)}}$ is entire?

Answer 1

We definitely can't say that $1/f(z)$ is entire, because it's not even defined outside of $D$. But your idea to take the reciprocal is definitely on the right track:

Let $A$ be a small disk centered at $c$ and contained in $D$. Then $1/f(z)$ is analytic on this disk, and $|1/f(z)|$ takes its maximum value at $c$ which is in the interior of the disk. By the maximum modulus principle, $1/f(z)$ must be constant on this disk, and so $f(z)$ is also constant on the disk. But $f$ is analytic on $D$ and constant on a subset of $D$ with an accumulation point, and therefore $f$ is constant on $D$ itself (by uniqueness of analytic continuation).

(The small disk is only to ensure that there is a boundary available to invoke the maximum modulus principle with.)

Answer 2

There exists a disc $c\in U_r(c)\subset D$ of radius $r$ and center $c$, we have by Cauchy integral $$f(c) = \frac{1}{2\pi i} \int_{| z-c| = r} \frac{f(z)}{z-c}\,dz$$ Write $z:= c+re^{i\theta} $ then $dz=ir e^{i\theta}d\theta$ and $$f(c) = \frac{1}{2\pi i} \int_0^{2\pi} \frac{f(c+re^{i\theta})}{re^{i\theta}}\; ire^{i\theta}d\theta$$ Thus $$f(c) = \frac{1}{2\pi } \int_0^{2\pi} f(c +re^{i\theta})\;d\theta $$

Hence $$\displaystyle \left\vert{f \left({c}\right)}\right\vert \le \frac 1 {2 \pi} \int_0^{2 \pi} \left\vert{f \left({c + r e^{i \theta} }\right)}\right\vert \, \mathrm d \theta \le \left\vert{f \left({c + r e^{i \theta} }\right)}\right\vert$$

i.e, for all $w\in \partial U_r(c)$ we have $|f(c)|=|f(w)|$

Now let $r\to 0$ we conclude that $f|_{U_r(c)}$ is constant. By the Maximum Modulus Principle (not Liouvilles' Theorem) we have $f$ is constant in $D$.

Note that Liouvilles' Theorem for entire functions (analytic on $\mathbb C$) but the Maximum Modulus Principle works for any domain $D$.