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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. $QED$

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  • $\begingroup$ Clarification: for the entire part of the theorem to work, take a circle but let $r$ => $\infty$ $\endgroup$
    – lf2225
    Dec 17 '18 at 4:47
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It is not at all clear why boundedness of $f$ must imply equality is achieved in (*).

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  • $\begingroup$ 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? $\endgroup$
    – lf2225
    Dec 17 '18 at 4:40
  • $\begingroup$ 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. $\endgroup$ Dec 17 '18 at 17:40
  • $\begingroup$ 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 => $\infty$) $\endgroup$
    – lf2225
    Dec 17 '18 at 19:39

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