Liouville's Theorem and the Maximum Modulus Principle

Theorem 11.1 (Liouville’s Theorem). Every bounded entire function is constant.

Proof. If $|f(z)| \leq M$ for all $z$, then by Cauchy’s estimates with $R$ arbitrarily large: $|f'(z_0)| \leq \frac{M}{R} \to 0$ as $R \to \infty$. So $f'(z) = 0$ for all $z$, meaning $f$ is constant. $\blacksquare$

Theorem 11.2 (Fundamental Theorem of Algebra). Every non-constant polynomial $p(z) \in \mathbb{C}[z]$ has a root in $\mathbb{C}$.

Proof. Suppose $p(z)$ has no root. Then $f(z) = 1/p(z)$ is entire. Since $|p(z)| \to \infty$ as $|z| \to \infty$, $f(z) \to 0$, so $f$ is bounded. By Liouville’s theorem, $f$ is constant, so $p$ is constant, a contradiction. $\blacksquare$

Theorem 11.3 (Maximum Modulus Principle). If $f$ is analytic and non-constant on a domain $D$, then $|f|$ has no local maximum in $D$.

Corollary 11.4. If $f$ is analytic on a bounded domain $D$ and continuous on $\bar{D} = D \cup \partial D$, then $|f|$ attains its maximum on $\partial D$.

Theorem 11.5 (Minimum Modulus Principle). If $f$ is analytic and non-zero on a bounded domain $D$ and continuous on $\bar{D}$, then $|f|$ attains its minimum on $\partial D$.

Remark. If $f$ has zeros in $D$, then $|f|$ attains its minimum of $0$ at those zeros. The minimum modulus principle requires the non-vanishing hypothesis.

Theorem 11.6 (Schwarz Lemma). If $f : \mathbb{D} \to \mathbb{D}$ is analytic with $f(0) = 0$, then

$|f(z)| \leq |z| \quad \mathrm{for\ all\ } z \in \mathbb{D}$

and $|f'(0)| \leq 1$. Equality in either case implies $f(z) = e^{i\theta} z$ for some real $\theta$.

Proof. Define $g(z) = f(z)/z$ for $z \neq 0$ and $g(0) = f'(0)$. Then $g$ is analytic on $\mathbb{D}$. For $|z| = r \lt 1$: $|g(z)| = |f(z)|/|z| \leq 1/r$. By the maximum modulus principle, $|g(z)| \leq 1/r$ for $|z| \leq r$. Letting $r \to 1$: $|g(z)| \leq 1$, so $|f(z)| \leq |z|$. Also $|f'(0)| = |g(0)| \leq 1$. If $|f'(0)| = 1$, then $|g|$ attains its maximum at an interior point, so $g$ is constant: $g(z) = e^{i\theta}$. $\blacksquare$

Theorem 11.7 (Schwarz-Pick Lemma). If $f : \mathbb{D} \to \mathbb{D}$ is analytic, then for all $z, w \in \mathbb{D}$:

$\left|\frac{f(z) - f(w)}{1 - \overline{f(w)} f(z)}\right| \leq \left|\frac{z - w}{1 - \overline{w} z}\right|$

and for all $z \in \mathbb{D}$:

$\frac{|f'(z)|}{1 - |f(z)|^2} \leq \frac{1}{1 - |z|^2}$

Equality holds iff $f$ is a conformal automorphism of $\mathbb{D}$ (a Blaschke factor).

Application 1: Polynomial growth. If $f$ is entire and $|f(z)| \leq C|z|^n$ for large $|z|$, then $f$ is a polynomial of degree at most $n$.

Proof. By Cauchy’s estimate, $|f^{(n+1)}(z_0)| \leq C(n+1)!/R \to 0$, so $f^{(n+1)} \equiv 0$. Thus $f$ is a polynomial of degree $\leq n$. $\blacksquare$

Application 2: Casorati-Weierstrass. The Casorati-Weierstrass theorem states that if $f$ has an essential singularity at $z_0$, then the image of every punctured neighborhood is dense in $\mathbb{C}$. A stronger result (Picard’s great theorem) states that the image omits at most one point. Liouville’s theorem is used in proving both.

Application 3: Density of polynomials. The set of polynomials is dense in the space of entire functions with respect to uniform convergence on compact subsets (Runge’s theorem).

Application. If $f$ is analytic on a domain $D$ and $|f|$ is constant on $D$, then $f$ is constant.

Proof. If $|f(z)| = c$ for all $z \in D$, then if $c = 0$, $f \equiv 0$. If $c > 0$, then $f$ is never zero, so $1/f$ is analytic. Both $f$ and $1/f$ attain their maxima on $D$, so $f$ must be constant by the maximum modulus principle. $\blacksquare$

Theorem 11.8. If $f$ is entire and $\mathrm{Re}\, f(z)$ is bounded above (or below), then $f$ is constant.

Proof. Suppose $\mathrm{Re}\, f(z) \leq M$ for all $z$. Consider $g(z) = e^{f(z)}$. Then $|g(z)| = e^{\mathrm{Re}\, f(z)} \leq e^M$, so $g$ is bounded entire, hence constant. Therefore $f$ is constant. $\blacksquare$

Problem 1. Show that $f(z) = \sin z$ is not bounded on $\mathbb{C}$, consistent with Liouville.

Solution. On the imaginary axis, $\sin(iy) = i\sinh y$, and $|\sin(iy)| = |\sinh y| \to \infty$ as $y \to \infty$. $\blacksquare$

Problem 2. Prove that if $f$ is entire and $|f(z)| \to \infty$ as $|z| \to \infty$, then $f$ is a polynomial.

Problem 3. Find the maximum of $|e^{z^2}|$ on the unit disk $|z| \leq 1$.

Solution. $|e^{z^2}| = e^{\mathrm{Re}(z^2)}$. On $|z| = 1$, write $z = e^{i\theta}$, then $\mathrm{Re}(z^2) = \cos 2\theta$. Maximum of $\cos 2\theta$ is $1$, so the maximum of $|e^{z^2}|$ on $\overline{\mathbb{D}}$ is $e^1 = e$, attained at $z = \pm 1$. $\blacksquare$

Problem 4. Suppose $f$ is entire and $f(z) = f(z + 1) = f(z + i)$ for all $z$. Show that $f$ is constant.

The Phragmén-Lindelöf principle extends the maximum modulus principle to unbounded domains by imposing growth conditions.

Theorem 11.9 (Phragmén-Lindelöf). Let $f$ be analytic on the sector $S = \{z : 0 < \arg z < \pi/\alpha\}$ and continuous on its closure. If $|f(z)| \leq M$ on the boundary and $|f(z)| \leq Ce^{|z|^\beta}$ for some $\beta < \alpha$, then $|f(z)| \leq M$ for all $z \in S$.

1. Liouville’s Theorem: Bounded entire functions are constant.
2. Maximum Modulus Principle: Non-constant analytic functions have no local modulus maximum.
3. Minimum Modulus Principle: Non-zero analytic functions attain minimum modulus on boundary.
4. Schwarz Lemma: Holomorphic self-maps of the disk with $f(0)=0$ satisfy $|f(z)| \leq |z|$.
5. Schwarz-Pick Lemma: Holomorphic maps from the disk to itself are contractions in the hyperbolic metric.

Problem 5. Use Liouville’s theorem to prove that if $f$ is entire and $\mathrm{Im}(f(z)) \geq 0$ for all $z$, then $f$ is constant.

Solution. Consider $g(z) = e^{if(z)}$. Then $|g(z)| = e^{-\mathrm{Im}(f(z))} \leq 1$, so $g$ is bounded entire, hence constant. Thus $f$ is constant. $\blacksquare$