Liouville's theorem (complex analysis)

The theorem follows from the fact that holomorphic functions are analytic. If f is an entire function, it can be represented by its Taylor series about 0:

${\displaystyle f(z)=\sum _{k=0}^{\infty }a_{k}z^{k}}$

where (by Cauchy's integral formula)

${\displaystyle a_{k}={\frac {f^{(k)}(0)}{k!}}={1 \over 2\pi i}\oint _{C_{r}}{\frac {f(\zeta )}{\zeta ^{k+1}}}\,d\zeta }$

and C_r is the circle about 0 of radius r > 0. Suppose f is bounded: i.e. there exists a constant M such that |f(z)| ≤ M for all z. We can estimate directly

${\displaystyle |a_{k}|\leq {\frac {1}{2\pi }}\oint _{C_{r}}{\frac {|f(\zeta )|}{|\zeta |^{k+1}}}\,|d\zeta |\leq {\frac {1}{2\pi }}\oint _{C_{r}}{\frac {M}{r^{k+1}}}\,|d\zeta |={\frac {M}{2\pi r^{k+1}}}\oint _{C_{r}}|d\zeta |={\frac {M}{2\pi r^{k+1}}}2\pi r={\frac {M}{r^{k}}},}$

where in the second inequality we have used the fact that |z| = r on the circle C_r. But the choice of r in the above is an arbitrary positive number. Therefore, letting r tend to infinity (we let r tend to infinity since f is analytic on the entire plane) gives a_k = 0 for all k ≥ 1. Thus f(z) = a₀ and this proves the theorem.

Any holomorphic function on a compact Riemann surface is necessarily constant.[5]

Let ${\displaystyle f(z)}$ be holomorphic on a compact Riemann surface ${\displaystyle M}$. By compactness, there is a point ${\displaystyle p_{0}\in M}$ where ${\displaystyle |f(p)|}$ attains its maximum. Then we can find a chart from a neighborhood of ${\displaystyle p_{0}}$ to the unit disk ${\displaystyle \mathbb {D} }$ such that ${\displaystyle f(\phi ^{-1}(z))}$ is holomorphic on the unit disk and has a maximum at ${\displaystyle \phi (p_{0})\in \mathbb {D} }$, so it is constant, by the maximum modulus principle.

Let ${\displaystyle \mathbb {C} \cup \{\infty \}}$ be the one point compactification of the complex plane ${\displaystyle \mathbb {C} .}$ In place of holomorphic functions defined on regions in ${\displaystyle \mathbb {C} }$, one can consider regions in ${\displaystyle \mathbb {C} \cup \{\infty \}.}$ Viewed this way, the only possible singularity for entire functions, defined on ${\displaystyle \mathbb {C} \subset \mathbb {C} \cup \{\infty \},}$ is the point ∞. If an entire function f is bounded in a neighborhood of ∞, then ∞ is a removable singularity of f, i.e. f cannot blow up or behave erratically at ∞. In light of the power series expansion, it is not surprising that Liouville's theorem holds.

Similarly, if an entire function has a pole of order n at ∞—that is, it grows in magnitude comparably to zⁿ in some neighborhood of ∞—then f is a polynomial. This extended version of Liouville's theorem can be more precisely stated: if |f(z)| ≤ M|zⁿ| for |z| sufficiently large, then f is a polynomial of degree at most n. This can be proved as follows. Again take the Taylor series representation of f,

${\displaystyle f(z)=\sum _{k=0}^{\infty }a_{k}z^{k}.}$

The argument used during the proof using Cauchy estimates shows that for all k ≥ 0,

${\displaystyle |a_{k}|\leq Mr^{n-k}.}$

So, if k > n, then

${\displaystyle |a_{k}|\leq \lim _{r\to \infty }Mr^{n-k}=0.}$

Therefore, a_k = 0.

Liouville's theorem does not extend to the generalizations of complex numbers known as double numbers and dual numbers.[6]

• Mittag-Leffler's theorem

• "Liouville's theorem". PlanetMath.
• Weisstein, Eric W. "Liouville's Boundedness Theorem". MathWorld.