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}) \mkern 1.5mu\overline{\mkern-1.5muf(re^{i\theta}) \mkern-1.5mu}\mkern 1.5mu \,d\theta\\ &= \int_{[0, 2\pi]} \sum_{k\geq 0} c_k r^k e^{ik\theta} \sum_{j\geq 0} \mkern 1.5mu\overline{\mkern-1.5muc_j\mkern-1.5mu}\mkern 1.5mu r^j e^{-ij\theta} \,d\theta\\ &= \int_{[0, 2\pi]} \sum_{k,j\geq 0} c_k\mkern 1.5mu\overline{\mkern-1.5muc_j\mkern-1.5mu}\mkern 1.5mu r^{k+j} e^{i(k-j)\theta} \,d\theta\\ &= \sum_{k,j\geq 0} c_k\mkern 1.5mu\overline{\mkern-1.5muc_j\mkern-1.5mu}\mkern 1.5mu r^{k+j} \int_{[0, 2\pi]} e^{i(k-j)\theta} \,d\theta\\ &= \sum_{k,j\geq 0} c_k\mkern 1.5mu\overline{\mkern-1.5muc_j\mkern-1.5mu}\mkern 1.5mu r^{k+j} \chi_{i=j}\cdot 2\pi \\ &= \sum_{k\geq 0} c_k\mkern 1.5mu\overline{\mkern-1.5muc_k\mkern-1.5mu}\mkern 1.5mu 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 \(z\in {\mathbb{C}}\), 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 \({\mathbb{C}}\).

  • 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 \({\mathbb{C}}\) and thus \(1/p\) is bounded above on \({\mathbb{C}}\).

• 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 \({\mathbb{C}}\setminus Z_f\).
• 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 \({\mathbb{C}}\)
  • If not, then \({\left\lvert {\tilde g(0)} \right\rvert} = 1+{\varepsilon}> 1\), but then there would be a domain \(\Omega \subseteq {\mathbb{C}}\setminus\left\{{0}\right\}\) 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 \(c\in {\mathbb{C}}\) 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.