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Liouville's theorem (complex analysis)

Statement

Every bounded entire function is constant: if f:C→Cf : \mathbb{C} \to \mathbb{C} is holomorphic on the entire complex plane C\mathbb{C} and there exists a real constant M≥0M \ge 0 such that ∣f(z)∣≤M|f(z)| \le M for all z∈Cz \in \mathbb{C}, then ff is constant.

Why is it true?

On the real line R\mathbb{R}, smooth functions like sin⁡x\sin x can oscillate forever while staying bounded between −1-1 and 11. In the complex plane C\mathbb{C}, holomorphic functions cannot do this: the mean value property forces the value at any point to be the exact average over circles of arbitrarily large radius RR centered at that point. If the function is trapped inside a disk of radius MM everywhere on C\mathbb{C}, then taking averages over larger and larger circles leaves no room for the derivative to be nonzero — to vary at all, a non-constant entire function must grow without bound in some direction (for example, ∣sin⁡(iy)∣=sinh⁡∣y∣→∞|\sin(iy)| = \sinh|y| \to \infty).

Proof sketch

Fix any z0∈Cz_0 \in \mathbb{C} and apply Cauchy's integral formula for the first derivative on the circle CR={∣z−z0∣=R}C_R = \{|z - z_0| = R\} of radius R>0R > 0: f′(z0)=12πi∮CRf(z)(z−z0)2 dzf'(z_0) = \frac{1}{2\pi i} \oint_{C_R} \frac{f(z)}{(z - z_0)^2}\,dz. Estimating the integral with ∣f(z)∣≤M|f(z)| \le M and ∣z−z0∣=R|z - z_0| = R along the circumference 2πR2\pi R gives Cauchy's estimate ∣f′(z0)∣≤12π⋅MR2⋅2πR=MR|f'(z_0)| \le \frac{1}{2\pi} \cdot \frac{M}{R^2} \cdot 2\pi R = \frac{M}{R}. Letting R→∞R \to \infty forces f′(z0)=0f'(z_0) = 0 for every z0∈Cz_0 \in \mathbb{C}; since C\mathbb{C} is connected, ff is constant. (Historically, Augustin-Louis Cauchy published this proof in 1844; Joseph Liouville presented it in his 1847 lectures, from which the theorem acquired its name.)

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Step-by-step proofs

No step-by-step proof yet for this theorem.

References

  1. Lars V. Ahlfors (1979). Complex Analysis
  2. Reinhold Remmert (1991). Theory of Complex Functions