-2
$\begingroup$

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka Exer 8.24

(Exer 8.24) If $f: \mathbb C \to \mathbb C$ is entire and $\Im(f)$ is constant on closed unit disc $\{|z| \le 1\}$, then $f$ is constant on $\mathbb C$.

In OSU course here, there's a proof of Exer 8.24 above where the proof makes use of the Identity Principle. I think I can prove alternatively by modifying the proof of Liouville's Thm instead of using the Identity Principle. I attempt my alternative proof as follows.

Question: What mistakes, if any, have I made, and why?

Pf:

By Cauchy-Riemann, $f \equiv: K$ is constant on closed unit disc. It remains to show $f \equiv: K$ on the rest of $\mathbb C$.

Consider any path $\gamma := \{ |z-w| = R \} \subset \mathbb C, w \in \mathbb C, R > 0$. We have by Cauchy's Integral Formula for $f'$ (Formula 5.1) that

$$|f'(w)| = |\frac{1}{2 \pi i} \int_{\gamma} \frac{f}{(z-w)^2} dz| = \frac{1}{2 \pi} |\int_{\gamma} \frac{f}{(z-w)^2} dz|$$

$$ \le \frac{1}{2 \pi} \max_{z \in \gamma}|\frac{f}{(z-w)^2}| \text{length}(\gamma) = \frac{1}{\not{2}\not{\pi}} \max_{z \in \gamma}|\frac{f}{(z-w)^2}| \not{2}\not{\pi} R$$

$$ = \max_{z \in \gamma}|\frac{f}{(z-w)^2}| R = \max_{z \in \gamma}\frac{|f|}{|z-w|^2} R = \max_{z \in \gamma}\frac{|f|}{R^2} R = \max_{z \in \gamma}\frac{|f|}{R} = \max_{z \in \gamma}\frac{K}{R} = \frac{K}{R}$$

$$\therefore, |f'(w)| = \lim_{R \to \infty} |f'(w)| \le \lim_{R \to \infty} \frac{K}{R} = 0 \ \forall w \in \mathbb C \implies f'(w) = 0 \ \forall w \in \mathbb C$$

$\therefore,$ by a theorem in the textbook (Thm 2.17), $f \equiv: K$, not only in closed unit disc, but also on the whole $\mathbb C$. QED

$\endgroup$
  • 1
    $\begingroup$ If you are happy with Taylor's theorem in the complex plane, the the power series $\sum_n f^{(n)}(0) z^n/n!$ converges to $f$ everywhere, but that's a constant. $\endgroup$ – Lord Shark the Unknown Aug 12 '18 at 9:28
  • $\begingroup$ @LordSharktheUnknown Thanks. What do you mean please? It seems $$\sum_n \frac{f^{(n)}(0)z^n}{n!} = f(z) \ \forall z \in \mathbb C$$ sooo if $f$ is constant on closed unit disc $\{|z| \le 1\}$, then $\sum_n \frac{f^{(n)}(0)z^n}{n!}$ is constant on the closed unit disc $\{|z| \le 1\}$ with the same value. Sooo then what? What is the 'that' in 'but that's a constant' ? $\endgroup$ – BCLC Aug 12 '18 at 9:44
1
$\begingroup$

Here is another approach that uses the open mapping theorem.

Since $B(0,1)$ is open and $f$ is analytic, if $f$ is non constant, then $f(B(0,1))$ is open.

However, the set $\{x+iy| y = \operatorname{im} f(0) \}$ is not open, hence $f$ must be a constant.

$\endgroup$
  • 1
    $\begingroup$ Thanks copper.hat! I made an edit to see if I understand right. I don't think this text explicitly has open mapping theorem. $\endgroup$ – BCLC Aug 12 '18 at 8:21
  • 2
    $\begingroup$ That is a pity, I think it is one of the more useful characteristics of analytic functions :-(. $\endgroup$ – copper.hat Aug 12 '18 at 8:25
  • 1
    $\begingroup$ Thanks for the info copper.hat. I expect I will encounter this in higher complex analysis. $\endgroup$ – BCLC Aug 12 '18 at 8:26
  • 2
    $\begingroup$ en.wikipedia.org/wiki/Open_mapping_theorem_(complex_analysis) $\endgroup$ – copper.hat Aug 12 '18 at 8:27
0
$\begingroup$

The mistake is

$$\max_{z \in \gamma}\frac{|f|}{R} = \max_{z \in \gamma}\frac{K}{R}$$

We know that $f=|f|\equiv:K$ on closed unit disc, so including the unit circle. How do we know that $f=|f|\equiv:K$ on some $\gamma$ seemingly pulled out of a hat?

It seems that modifying Liouville's Thm will not work as it did (and should) for some exercises in Ch5. Instead, we prove by Identity Principle (Principle 8.15) as in the OSU homework:

Pf using Identity Principle (Principle 8.15):

After proving that $f$ is constant on closed unit disc, consider $g: \mathbb C \to \mathbb C$ s.t. $g(z) := f(z) - K$. Observe that $g$ is entire and zero on closed unit disc. Now consider a sequence of distinct complex numbers $\{a_n\}_{n=1}^{\infty}$ in the closed unit disc that converges to a complex number $a$ in the closed unit disc. Observe that $g(a_n)=0 \ \forall n \in \mathbb N$. $\because \mathbb C$ is a region, by Identity Principle (Principle 8.15), $g \equiv 0$ on $\mathbb C$. $\therefore, f(z) \equiv K$ on $\mathbb C$.

QED

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.