Liouville’s Theorem

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.

• 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.

• 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 ${\left\lvert {c} \right\rvert}\geq 1$.

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.

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.

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.