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Liouville's theorem (complex analysis)

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In complex analysis, Liouville's theorem, named after Joseph Liouville, states that every bounded entire function must be constant. That is, every holomorphic function f for which there exists a positive number M such that |f(z)| ≤ M for all z in C is constant.

The theorem is considerably improved by Picard's little theorem, which says that every entire function whose image omits at least two complex numbers must be constant.

Contents

[edit] Proof

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

f(z) = \sum_{k=0}^\infty a_k z^k

where (by Cauchy's integral formula)


a_k = \frac{f^{(k)}(0)}{k!} = {1 \over 2 \pi i} \oint_{C_r} 
\frac{f( \zeta )}{\zeta^{k+1}}\,d\zeta

and Cr is the circle about 0 of radius r > 0. We can estimate directly


| 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 invoked the assumption that |f(z)| ≤ M for all z and the fact that |z|=r on the circle Cr. But the choice of r in the above is an arbitrary positive number. Therefore, letting r tend to infinity gives ak = 0 for all k ≥ 1. Thus f(z) = a0 and this proves the theorem.

[edit] Corollaries

[edit] Fundamental theorem of algebra

There is a short proof of the fundamental theorem of algebra based upon Liouville's theorem.

[edit] No entire function dominates another entire function

A consequence of the theorem is that "genuinely different" entire functions cannot dominate each other, i.e. if f and g are entire, and |f| ≤ |g| everywhere, then f = α·g for some complex number α. To show this, consider the function h = f/g. It is enough to prove that h is can be extended to an entire function, in which case the result follows by Liouville's theorem. The holomorphy of h is clear except at points in g−1(0). But since h is bounded any singularities must be removable. Thus h can be extended to an entire bounded function which by Liouville's theorem implies it is constant.

[edit] If f is less than or equal to a scalar times its input, then it is linear

Suppose that f is entire and |f(z)| is less than or equal to M|z|, for M a positive real number. We can apply Cauchy's integral formula; we have that

|f'(z)|=\frac{1}{2\pi}\left|\oint_{C_r }\frac{f(\zeta)}{(\zeta-z)^2}d\zeta\right|\leq \frac{1}{2\pi}\left|\oint_{C_r}\frac{M\zeta}{(\zeta-z)^2}d\zeta\right|=M\cdot

Now, an application of Liouville's theorem easily leads us to conclude that f's derivative is a constant. Integrating then shows that f is linear and then, by referring back to the original inequality, we have that the constant term is zero.

[edit] Non-constant elliptic functions can not be defined on C

The theorem can also be used to deduce that the domain of a non-constant elliptic function f can not be C. Suppose it was. Then, if a and b are two periods of f such that ab is not real, consider the parallelogram P whose vertices are 0, a, b and a + b. Then the image of f is equal to f(P). Since f is continuous and P is compact, f(P) is also compact and, therefore, it is bounded. So, f is constant.

The fact that the domain of a non-constant elliptic function f can not be C is what Liouville actually proved, in 1847, using the theory of elliptic functions.[1] In fact, it was Cauchy who proved Liouville's theorem.[2][3]

[edit] Entire functions have dense images

If f is a non-constant entire function, then its image is dense in C. This might seem to be a much stronger result than Liouville's theorem, but it is actually an easy corollary. If the image of f is not dense, then there is a complex number w and a real number r  > 0 such that the open disk centered at w with radius r has no element of the image of f. Define g(z) = 1/(f(z) − w). Then g is a bounded entire function, since

(\forall z\in\mathbb{C}):|g(z)|=\frac1{|f(z)-w|}<\frac1r\cdot

So, g is constant, and therefore f is constant.

[edit] Remarks

Let C ∪ {∞} be the one point compactification of the complex plane C. In place of holomorphic functions defined on regions in C, one can consider regions in C ∪ {∞}. Viewed this way, the only possible singularity for entire functions, defined on CC ∪ {∞}, 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 at ∞, i.e. blows up like 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,

 f(z) = \sum_{k=0}^\infty a_k z^k.

The argument used during the proof shows that

(\forall k\in\mathbb{N}):|a_k|\leqslant Mr^{n-k}.

So, if k > n,

|a_k|\leqslant\lim_{r\rightarrow+\infty}Mr^{n-k}=0.

Therefore, ak = 0.

[edit] References

  1. ^ Liouville, Joseph (1847), "Leçons sur les fonctions doublement périodiques", Journal für die Reine und Angewandte Mathematik 88: 277–310, 1879, ISSN 0075-4102, http://gdz.sub.uni-goettingen.de/no_cache/en/dms/load/img/?IDDOC=266004 
  2. ^ Cauchy, Augustin-Louis (1844), "Mémoires sur les fonctions complémentaires", Œuvres complètes d'Augustin Cauchy, 1, 8, Paris: Gauthiers-Villars (published 1882) 
  3. ^ Lützen, Jesper (1990), Joseph Liouville 1809–1882: Master of Pure and Applied Mathematics, Studies in the History of Mathematics and Physical Sciences, 15, Springer-Verlag, ISBN 3-540-97180-7 

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