# Liouville’s Theorem

## Spring 2020.3, Extras Fall 2009 #complex/qual/completed

• Assume $$f(z)=\sum_{n=0}^{\infty} c_{n} z^{n}$$ converges in $$|z|<R$$. Show that for $$r<R$$,

\begin{align*} \frac{1}{2 \pi} \int_{0}^{2 \pi}\left|f\left(r e^{i \theta}\right)\right|^{2} d \theta=\sum_{n=0}^{\infty}\left|c_{n}\right|^{2} r^{2 n} \end{align*}

• Deduce Liouville’s theorem from (a).

Computing the LHS: \begin{align*} \int_{[0, 2\pi]} {\left\lvert {f(re^{i\theta})} \right\rvert}^2 \,d\theta &= \int_{[0, 2\pi]} f(re^{i\theta}) \overline{f(re^{i\theta}) } \,d\theta\\ &= \int_{[0, 2\pi]} \sum_{k\geq 0} c_k r^k e^{ik\theta} \sum_{j\geq 0} \overline{c_j} r^j e^{-ij\theta} \,d\theta\\ &= \int_{[0, 2\pi]} \sum_{k,j\geq 0} c_k\overline{c_j} r^{k+j} e^{i(k-j)\theta} \,d\theta\\ &= \sum_{k,j\geq 0} c_k\overline{c_j} r^{k+j} \int_{[0, 2\pi]} e^{i(k-j)\theta} \,d\theta\\ &= \sum_{k,j\geq 0} c_k\overline{c_j} r^{k+j} \chi_{i=j}\cdot 2\pi \\ &= \sum_{k\geq 0} c_k\overline{c_k} r^{2k} \cdot 2\pi \\ &= 2\pi \sum_{k\geq 0}{\left\lvert {c_k} \right\rvert}^2 r^{2k} ,\end{align*} where we’ve used that the series converges uniformly in its radius of convergence to commute sums and integrals.

Now supposing $${\left\lvert {f(z)} \right\rvert}\leq M$$ for all $$z\in {\mathbf{C}}$$, if $$f$$ is entire then $$\sum_{k\geq 0} c_k z^k$$ converges for all $$r$$, so \begin{align*} \sum_{k\geq 0} {\left\lvert {c_k} \right\rvert}^2 r^{2k} = {1\over 2\pi }\int_{[0, 2\pi]} {\left\lvert {f(re^{i\theta})} \right\rvert}^2 \,d\theta\leq {1\over 2\pi}\int_{[0, 2\pi]} M^2 \,d\theta= M^2 .\end{align*} Thus for all $$r$$, \begin{align*} {\left\lvert {c_0} \right\rvert}^2 + {\left\lvert {c_1} \right\rvert}^2 r^2 + {\left\lvert {c_2} \right\rvert}^2 r^{4} + \cdots \leq M^2 ,\end{align*} and taking $$r\to\infty$$ forces $${\left\lvert {c_1} \right\rvert}^2 = {\left\lvert {c_2} \right\rvert}^2 = \cdots = 0$$. So $$f(z) = c_0$$ is constant.

## FTA via Liouville #complex/exercise/completed

Prove the Fundamental Theorem of Algebra (using complex analysis).

• Strategy: By contradiction with Liouville’s Theorem
• Suppose $$p$$ is non-constant and has no roots.
• Claim: $$1/p(z)$$ is a bounded holomorphic function on $${\mathbf{C}}$$.
• Holomorphic: clear? Since $$p$$ has no roots.

• Bounded: for $$z\neq 0$$, write \begin{align*} \frac{P(z)}{z^{n}}=a_{n}+\left(\frac{a_{n-1}}{z}+\cdots+\frac{a_{0}}{z^{n}}\right) .\end{align*}

• The term in parentheses goes to 0 as $${\left\lvert {z} \right\rvert}\to \infty$$

• Thus there exists an $$R>0$$ such that \begin{align*} {\left\lvert {z} \right\rvert} > R \implies {\left\lvert {P(z) \over z^n} \right\rvert} \geq c \coloneqq{{\left\lvert {a_n} \right\rvert} \over 2} .\end{align*}

• So $$p$$ is bounded below when $${\left\lvert {z} \right\rvert} > R$$

• Since $$p$$ is continuous and has no roots in $${\left\lvert {z} \right\rvert} \leq R$$, it is bounded below when $${\left\lvert {z} \right\rvert} \leq R$$.

• Thus $$p$$ is bounded below on $${\mathbf{C}}$$ and thus $$1/p$$ is bounded above on $${\mathbf{C}}$$.

• By Liouville’s theorem, $$1/p$$ is constant and thus $$p$$ is constant, a contradiction.

## Entire functions satisfying an inequality #complex/exercise/completed

Find all entire functions that satisfy \begin{align*} {\left\lvert {f(z)} \right\rvert} \geq {\left\lvert {z} \right\rvert} \quad \forall z\in {\mathbf{C}} .\end{align*} Prove this list is complete.

• If $$f$$ is bounded in a neighborhood of a singularity $$z_0$$, then $$z_0$$ is removable.

• Suppose $$f$$ is entire and define $$g(z) \coloneqq{z \over f(z)}$$.
• By the inequality, $${\left\lvert {g(z)} \right\rvert} \leq 1$$, so $$g$$ is bounded.
• $$g$$ potentially has singularities at the zeros $$Z_f \coloneqq f^{-1}(0)$$, but since $$f$$ is entire, $$g$$ is holomorphic on $${\mathbf{C}}\setminus Z_f$$.
• Claim: $$Z_f = \left\{{0}\right\}$$.
• If $$f(z) = 0$$, then $${\left\lvert {z} \right\rvert} \leq {\left\lvert {f(z)} \right\rvert} = 0$$ which forces $$z=0$$.
• We can now apply Riemann’s removable singularity theorem:
• Check $$g$$ is bounded on some open subset $$D\setminus\left\{{0}\right\}$$, clear since it’s bounded everywhere
• Check $$g$$ is holomorphic on $$D\setminus\left\{{0}\right\}$$, clear since the only singularity of $$g$$ is $$z=0$$.
• By Riemann’s removable singularity theorem, the singularity $$z = 0$$ is removable and $$g$$ has an extension to an entire function $$\tilde g$$.
• By continuity, we have $${\left\lvert {\tilde g(z)} \right\rvert} \leq 1$$ on all of $${\mathbf{C}}$$
• If not, then $${\left\lvert {\tilde g(0)} \right\rvert} = 1+{\varepsilon}> 1$$, but then there would be a domain $$\Omega \subseteq {\mathbf{C}}\setminus\left\{{0}\right\}$$ such that $$1 < {\left\lvert {\tilde g(z)} \right\rvert} \leq 1 +{\varepsilon}$$ on $$\Omega$$, a contradiction.
• By Liouville, $$\tilde g$$ is constant, so $$\tilde g(z) = c_0$$ with $${\left\lvert {c_0} \right\rvert} \leq 1$$
• Thus $$f(z) = c_0^{-1}z \coloneqq cz$$ where $${\left\lvert {c} \right\rvert}\geq 1$$

Thus all such functions are of the form $$f(z) = cz$$ for some $$c\in {\mathbf{C}}$$ with $${\left\lvert {c} \right\rvert}\geq 1$$.

## Entire functions with an asymptotic bound #complex/exercise/completed

Find all entire functions satisfying \begin{align*} {\left\lvert {f(z)} \right\rvert} \leq {\left\lvert {z} \right\rvert}^{1\over 2} \quad\text{ for } {\left\lvert {z} \right\rvert} > 10 .\end{align*}

Since $$f$$ is entire, take a Laurent expansion at $$z=0$$, so $$f(z) = \sum_{k\geq 0} c_k z^k$$ where $${2\pi i\over k!} c_k = f^{(k)}(0)$$ by Cauchy’s integral formula. Take a Cauchy estimate on a disc of radius $$R>10$$: \begin{align*} {\left\lvert {c_k} \right\rvert} &\leq {k!\over 2\pi}\int_{{\left\lvert {z} \right\rvert} = R} {\left\lvert {f(\xi) \over (\xi - 0)^{k+1}} \right\rvert}\,d\xi\\ &\leq {k! \over 2\pi}\int_{{\left\lvert {z} \right\rvert} = R}{ {\left\lvert {\xi} \right\rvert}^{1\over 2} \over {\left\lvert {\xi} \right\rvert}^{k+1} }\,d\xi\\ &= {k! \over 2\pi} \cdot {1\over R^{k+{1\over 2}}}\cdot 2\pi R \\ &= { \mathsf{O}}(1/R^{k-{1\over 2}}) .\end{align*} So in particular, if $$k\geq 1$$ then $$k-{1\over 2}>0$$ and $$c_k = 0$$. This forces $$f = c_0$$ to be constant.

## Tie’s Extra Questions: Fall 2009 #complex/exercise/completed

Let $$f(z)$$ be entire and assume values of $$f(z)$$ lie outside a bounded open set $$\Omega$$. Show without using Picard’s theorems that $$f(z)$$ is a constant.

We have $${\left\lvert {f(z)} \right\rvert}\geq M$$ for some $$M$$, so $${\left\lvert {1/f(z)} \right\rvert} \leq M^{-1}$$ is bounded, and we claim it is entire as well. This follows from the fact that $$1/f$$ has singularities at the zeros of $$f$$, but these are removable since $$1/f$$ is bounded in every neighborhood of each such zero. So $$1/f$$ extends to a holomorphic function. But now $$1/f =c$$ is constant by Liouville, which forces $$f= 1/c$$ to be constant.

## Tie’s Extra Questions: Fall 2015 #complex/exercise/completed

Let $$f(z)$$ be bounded and analytic in $$\mathbb C$$. Let $$a \neq b$$ be any fixed complex numbers. Show that the following limit exists: \begin{align*} \lim_{R \rightarrow \infty} \int_{|z|=R} \frac{f(z)}{(z-a)(z-b)} dz .\end{align*}

Use this to show that $$f(z)$$ must be a constant (Liouville’s theorem).

Apply PFD and use that $$f$$ is holomorphic to apply Cauchy’s formula over a curve of radius $$R$$ enclosing $$a$$ and $$b$$: \begin{align*} \int_\gamma {f(z) \over (z-a)(z-b)}\,dz &= \int_\gamma f(z)\qty{{a-b \over z-a} + {b-a\over z-b} } \,dz\\ &= (a-b)^{-1}\int_\gamma {f(z) \over z-a} \,dz+ (b-a)^{-1}\int_\gamma {f(z) \over z-b}\,dz\\ &= (a-b)^{-1}\cdot 2\pi i f(a) + (b-a)\cdot 2\pi i f(b)\\ &= 2\pi i\qty{f(a) - f(b) \over a-b } .\end{align*} Since $$f$$ is bounded, this number is finite and independent of $$R$$, so taking $$R\to\infty$$ preserves this equality. On the other hand, if $${\left\lvert {f(z)} \right\rvert}\leq M$$, then we can estimate this integral directly as \begin{align*} I \leq \int_{{\left\lvert {z} \right\rvert} = R} {M \over {\left\lvert {R-a} \right\rvert} \cdot {\left\lvert {R-b} \right\rvert} } = {M\cdot 2\pi R \over {\left\lvert {R-a} \right\rvert} \cdot {\left\lvert {R-b} \right\rvert}} \ll {1\over R} \to 0 ,\end{align*} which forces $$f(a) =f(b)$$. Since $$a, b$$ were arbitrary, $$f$$ must be constant.