# Proving Liouville for entire functions using MVT for analytic functions

I am trying to prove Liouville's theorem: An entire bounded function is constant. I'm trying to use the Mean Value Theorem from my textbook.

MVT: $$\space\space$$If $$f$$ is analytic in $$D$$ and $$a \in D$$, then $$f(a)$$ is equal to the mean value of $$f$$ taken around the boundary of any disk centered at $$a$$ and contained in $$D$$. That is,

$$f(a)$$ = $$\frac{1}{2\pi} \int_{0}^{2\pi} f(a + re^{i\theta}) d\theta$$ when $$D(a; r) \subset D$$

Proof:

Assumtion: Suppose that $$f(z)$$ is analytic and NOT constant on a circle $$C$$.

Clearly $$f(z)$$ is bounded, giving us the necessary assumption (hold your horses for the entire part).

$$(*) \space \space \space \space \space \space\space \space \space|f(z)| \leq{\frac{1}{2\pi}\int_{0}^{2\pi}|f(z + re^{i\theta})| d\theta} \leq{\max{_\theta(|f(z + re^{i\theta})}}|)$$ $$(**)$$ Since $$f(z)$$ is bounded, $$\exists{}$$ $$z$$, s.t.equality is achieved in $$(*)$$

The only way equality holds, is if $$f$$ is constant throughout the circle. So f is constant throughout the circle

Let $$f$$ be entire. Since entire functions are by definition holomorphic and analytic on their domain, our statement $$(**)$$ still holds since $$f$$ is still bounded by assumption.

So $$f$$ is constant. Contradiction. $$QED$$

• Clarification: for the entire part of the theorem to work, take a circle but let $r$ => $\infty$ Dec 17 '18 at 4:47

It is not at all clear why boundedness of $$f$$ must imply equality is achieved in (*).

• Since $f$ is bounded, $f(z)$ is less or equal to the right hand side of $(*)$ for some $z$. Same holds for absolute value of each side. Do you disagree? Dec 17 '18 at 4:40
• No, you just prove that a bounded function is bounded. Surprise. -- Note that the circle moves with $z$, so that your claim that "equality must hold" because of some compactness argument fails because for a fixed circle your theorem only makes a claim on one point $z$, the center of that circle. Dec 17 '18 at 17:40
• What if you use the MVT directly without the inequality and take r goes to infinity. I'm trying to not use another point $|f(w)|$ (and that the difference between the two approaches 0 as r => $\infty$) Dec 17 '18 at 19:39