Complex integration · Definition

Liouville's theorem

complex analysiscomplex integration

Before you start you need to know

It shows that being holomorphic on the whole plane is so restrictive that only constants fail to escape to infinity.

Definition

Liouville's theorem says that every entire function \(f\), that is, holomorphic on the whole plane, bounded by a constant \(M\) is constant.

\[|f(z)|\le M\ \ \forall z\in\mathbb{C}\ \Longrightarrow\ f\ \text{constant}\]

The proof uses Cauchy's integral formula for the derivative: if \(|f|\le M\) and we take the circle \(|z-z_0|=R\), then \(|f'(z_0)|=\left|\frac1{2\pi i}\oint\frac{f(z)}{(z-z_0)^2}dz\right|\le\frac{M}{R}\). Since \(f\) is entire, \(R\) can be as large as we like, so \(f'(z_0)=0\) at every point and \(f\) is constant.

Its contrapositive says every nonconstant entire function is unbounded: it takes arbitrarily large moduli. That is why \(\sin z\), \(\cos z\), \(e^z\) and every nonconstant polynomial grow without bound somewhere in \(\mathbb{C}\), even though \(\sin\) and \(\cos\) are bounded on the real axis. The contrast with real calculus is sharp: \(\sin x\) is differentiable on all of \(\mathbb{R}\) and bounded without being constant.

The theorem is applied to auxiliary functions. If \(f\) is entire and \(\operatorname{Re}f\le0\), then \(e^{f}\) is bounded; if \(|f|\ge1\), then \(1/f\) is entire and bounded; if \(f\) is doubly periodic, it is bounded by compactness. In every case \(f\) is constant. It is the key to the shortest proof of the fundamental theorem of algebra.

Typical mistakes: forgetting that the bound must hold on the whole plane (a disk or the real axis is not enough) and that the function must be entire, with no poles. A function like \(1/(z^2+1)\) is bounded away from \(\pm i\) and nonconstant because it is not entire.

Think about it

Why doesn't \(\sin z\) contradict Liouville's theorem even though \(|\sin x|\le1\) on the real axis?

Because Liouville demands the bound on all of \(\mathbb{C}\), and \(|\sin(iy)|=\sinh|y|\) grows without limit.

If \(f\) is entire and \(|f(z)|\le|z|\) for all \(z\), what form does \(f\) have?

It is \(f(z)=cz\) with \(|c|\le1\): \(f(0)=0\), so \(f(z)/z\) is entire (the singularity at \(0\) is removable) and bounded by \(1\), hence constant.

Exercises 10

  1. Which of these entire functions is bounded on all of \(\mathbb{C}\)?
  2. If \(f\) is entire, \(|f(z)|\le5\) for all \(z\) and \(f(0)=3\), which function is \(f\)?
  3. Since \(\sin z\) is entire and not constant, what does Liouville give?
  4. What is \(|\cos(2i)|\)? Use it to see that \(\cos z\) is unbounded.
  5. The function \(\frac{1}{z^2+1}\) is bounded on the real axis. Does it contradict Liouville's theorem?
  6. If \(f\) is entire with \(|f(z)|\le M\) and one uses \(|f'(z_0)|\le M/R\) for every \(R>0\), what follows for \(f'(z_0)\)?
  7. Which statement about entire functions is true?
  8. Let \(f\) be entire with \(\operatorname{Re} f(z)\le0\) for all \(z\). To which bounded function would you apply Liouville to show \(f\) is constant?
  9. Let \(f\) be entire with \(|f(z)|\ge1\) for all \(z\). Which bounded entire function results?
  10. Let \(f\) be entire with \(f(z+1)=f(z)\) and \(f(z+i)=f(z)\) for all \(z\). What follows?

References

  • Elias M. Stein, Rami Shakarchi, Complex Analysis, 1st edition, Princeton University Press, 2003. Cap. 2, §4.3
  • James Ward Brown, Ruel V. Churchill, Complex Variables and Applications, 9th edition, McGraw-Hill, 2014. §52
  • Lars V. Ahlfors, Complex Analysis, 3rd edition, McGraw-Hill, 1979. Cap. 4, §3.2