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