View source for Liouville's theorem (complex analysis)

{{Short description|Theorem in complex analysis}}
{{About|Liouville's theorem in complex analysis||Liouville's theorem (disambiguation)}}
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In [[complex analysis]], '''Liouville's theorem''', named after [[Joseph Liouville]] (although the theorem was first proven by [[Cauchy]] in 1844<ref>{{Eom| title = Liouville theorems
 | author-last1 =  Solomentsev| author-first1 = E.D.| author-last2 = Stepanov| author-first2 = S.A.| author-last3 = Kvasnikov| author-first3 = I.A.| oldid = 52098}}</ref>), states that every [[bounded function|bounded]] [[entire function]] must be [[Constant function|constant]]. That is, every [[holomorphic function]] <math>f</math> for which there exists a positive number <math>M</math> such that <math>|f(z)| \leq M</math> for all <math>z\in\Complex</math> is constant. Equivalently, non-constant holomorphic functions on <math>\Complex</math> have unbounded images.

The theorem is considerably improved by [[Picard theorem|Picard's little theorem]], which says that every entire function whose image omits two or more [[complex number]]s must be constant.

== Statement ==
<blockquote>'''Liouville's theorem:''' Every [[holomorphic function]] <math>f:\mathbb C \to \mathbb C</math> for which there exists a positive number <math>M</math> such that <math>|f(z)| \leq M</math> for all <math>z\in\Complex</math> is [[constant function|constant]].</blockquote>

More succinctly, Liouville's theorem states that every bounded entire function must be constant.

== Proof ==

This important theorem has several proofs.

A standard analytical proof uses the fact that [[Proof that holomorphic functions are analytic|holomorphic functions are analytic]]. 

{{Math proof|title=Proof|drop=hidden|proof=
If <math>f</math> is an entire function, it can be represented by its [[Taylor series]] about 0:

: <math>f(z) = \sum_{k=0}^\infty a_k z^k</math>

where (by [[Cauchy's integral formula]])

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

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

: <math>| a_k | 
\le \frac{1}{2 \pi} \oint_{C_r} \frac{ | f ( \zeta ) | }{ | \zeta |^{k+1} } \, |d\zeta|
\le \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},</math>

where in the second inequality we have used the fact that <math>|z|=r</math> on the circle <math>C_r</math>. (This estimate is known as [[Cauchy's estimate]].) But the choice of <math>r</math> in the above is an arbitrary positive number. Therefore, letting <math>r</math> tend to infinity (we let <math>r</math> tend to infinity since <math>f</math> is analytic on the entire plane) gives <math>a_k=0</math> for all <math>k\geq 1</math>. Thus <math>f(z)=a_0</math> and this proves the theorem.
}}
Another proof uses the mean value property of harmonic functions.
{{Math proof|title=Proof<ref>{{cite journal |last=Nelson |first=Edward |year=1961 |title=A proof of Liouville's theorem |journal=Proceedings of the American Mathematical Society |volume=12 |issue=6 |pages=995 |doi=10.1090/S0002-9939-1961-0259149-4 |doi-access=free}}</ref>|drop=hidden|proof=
Given two points, choose two balls with the given points as centers and of equal radius. If the radius is large enough, the two balls will coincide except for an arbitrarily small proportion of their volume. Since <math>f</math> is bounded, the averages of it over the two balls are arbitrarily close, and so <math>f</math> assumes the same value at any two points.  
}}

The proof can be adapted to the case where the harmonic function <math>f</math> is merely bounded above or below. See [[Harmonic function#Liouville's theorem]].

== Corollaries ==

===Fundamental theorem of algebra===
There is a short [[Fundamental theorem of algebra#Complex-analytic proofs|proof of the fundamental theorem of algebra]] using Liouville's theorem.<ref>{{cite book|author1=Benjamin Fine|author2=Gerhard Rosenberger|title=The Fundamental Theorem of Algebra|url=https://books.google.com/books?id=g0KHD7EIl4cC&pg=PA70|year=1997|publisher=Springer Science & Business Media|isbn=978-0-387-94657-3|pages=70–71}}</ref>

{{Math proof|title=Proof ([[Fundamental theorem of algebra]])|drop=hidden|proof=
[[Proof by contradiction|Suppose for the sake of contradiction]] that there is a nonconstant polynomial <math>p</math> with no complex root. Note that <math>|p(z)| \to \infty</math> as <math>z \to \infty</math>. Take a sufficiently large [[ball (mathematics)|ball]] <math>B(0, R)</math>; for some constant <math>M</math> there exists a [[sufficiently large]] <math>R</math> such that <math>1/|p(z)| < 1</math> for all <math>z \not\in B(0, R)</math>.

Because <math>p</math> has no roots, the function <math>q(z) = 1/p(z)</math> is [[entire function|entire]] and [[holomorphic function|holomorphic]] inside <math>B(0, R)</math>, and thus it is also [[continuous function|continuous]] on its [[closed ball|closure]] <math>\overline B(0, R)</math>. By the [[extreme value theorem]], a continuous function on a closed and bounded set obtains its extreme values, implying that <math>1/|p(z)| \le C</math> for some constant <math>C</math> and <math>z \in \overline B(0, R)</math>.

Thus, the function <math>q(z)</math> is bounded in <math>\mathbb C</math>, and by Liouville's theorem, is [[constant function|constant]], which contradicts our assumption that <math>p</math> is nonconstant.
}}

===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 <math>f</math> and <math>g</math> are entire, and <math>|f|\leq |g|</math> everywhere, then <math>f=\alpha g</math> for some complex number <math>\alpha</math>. Consider that for <math>g=0</math> the theorem is trivial so we assume <math>g\neq 0</math>. Consider the function <math>h=f/g</math>. It is enough to prove that <math>h</math> can be extended to an entire function, in which case the result follows by Liouville's theorem. The holomorphy of <math>h</math> is clear except at points in <math>g^{-1}(0)</math>. But since <math>h</math> is bounded and all the zeroes of <math>g</math> are isolated, any singularities must be removable. Thus <math>h</math> can be extended to an entire bounded function which by Liouville's theorem implies it is constant.

===If ''f'' is less than or equal to a scalar times its input, then it is linear===
Suppose that <math>f</math> is entire and <math>|f(z)|\leq M|z|</math>, for <math>M>0</math>. We can apply Cauchy's integral formula; we have that

:<math>|f'(z)|=\frac{1}{2\pi}\left|\oint_{C_r}\frac{f(\zeta)}{(\zeta-z)^2}d\zeta\right|\leq \frac{1}{2\pi} \oint_{C_r} \frac{|f(\zeta)|}{\left|(\zeta-z)^2\right|} |d \zeta|\leq \frac{1}{2\pi} \oint_{C_r} \frac{M |\zeta|}{\left| (\zeta-z)^2\right|} \left|d\zeta\right|=\frac{MI}{2\pi}</math>

where <math>I</math> is the value of the remaining integral. This shows that <math>f'</math> is bounded and entire, so it must be constant, by Liouville's theorem. Integrating then shows that <math>f</math> is [[Affine transformation|affine]] and then, by referring back to the original inequality, we have that the constant term is zero.

===Non-constant elliptic functions cannot be defined on the complex plane===
The theorem can also be used to deduce that the domain of a non-constant [[elliptic function]] <math>f</math> cannot be <math>\Complex</math>. Suppose it was. Then, if <math>a</math> and <math>b</math> are two periods of <math>f</math> such that <math>\tfrac{a}{b}</math> is not real, consider the [[parallelogram]] <math>P</math> whose [[Vertex (geometry)|vertices]] are 0, <math>a</math>, <math>b</math>, and <math>a+b</math>. Then the image of <math>f</math> is equal to <math>f(P)</math>. Since <math>f</math> is [[continuous functions|continuous]] and <math>P</math> is [[Compact space|compact]], <math>f(P)</math> is also compact and, therefore, it is bounded. So, <math>f</math> is constant.

The fact that the domain of a non-constant [[elliptic function]] <math>f</math> cannot be <math>\Complex</math> is what Liouville actually proved, in 1847, using the theory of elliptic functions.<ref>{{Citation|last = Liouville|first = Joseph|author-link = Joseph Liouville|publication-date = 1879|year = 1847|title = Leçons sur les fonctions doublement périodiques|periodical = [[Crelle's Journal|Journal für die Reine und Angewandte Mathematik]]|volume = 88|pages = 277–310|issn = 0075-4102|url = http://gdz.sub.uni-goettingen.de/no_cache/en/dms/load/img/?IDDOC=266004|archive-url = https://archive.today/20120711004552/http://gdz.sub.uni-goettingen.de/no_cache/en/dms/load/img/?IDDOC=266004|url-status = dead|archive-date = 2012-07-11}}</ref> In fact, it was [[Augustin Louis Cauchy|Cauchy]] who proved Liouville's theorem.<ref>{{Citation|last = Cauchy|first = Augustin-Louis|authorlink = Augustin Louis Cauchy|year = 1844|publication-date = 1882|contribution = Mémoires sur les fonctions complémentaires|contribution-url = http://visualiseur.bnf.fr/StatutConsulter?N=VERESS5-1212867208163&B=1&E=PDF&O=NUMM-90188|title = Œuvres complètes d'Augustin Cauchy|series = 1|volume = 8|place = Paris|publisher = Gauthiers-Villars}}</ref><ref>{{Citation|last = Lützen|first = Jesper|year = 1990|title = Joseph Liouville 1809–1882: Master of Pure and Applied Mathematics|series = Studies in the History of Mathematics and Physical Sciences|volume = 15|publisher = Springer-Verlag|isbn = 3-540-97180-7}}</ref>

===Entire functions have dense images===
If <math>f</math> is a non-constant entire function, then its image is [[Dense set|dense]] in <math>\Complex</math>. This might seem to be a much stronger result than Liouville's theorem, but it is actually an easy corollary. If the image of <math>f</math> is not dense, then there is a complex number <math>w</math> and a real number 
<math>r > 0 </math> such that the open disk centered at <math>w</math> with radius <math>r</math> has no element of the image of <math>f</math>. Define 

:<math>g(z) = \frac{1}{f(z) - w}.</math>

Then <math>g</math> is a bounded entire function, since for all <math>z</math>,

:<math>|g(z)|=\frac{1}{|f(z)-w|} < \frac{1}{r}.</math>

So, <math>g</math> is constant, and therefore <math>f</math> is constant.

== On compact Riemann surfaces ==
Any holomorphic function on a [[compact space|compact]] [[Riemann surface]] is necessarily constant.<ref>a concise course in complex analysis and Riemann surfaces, Wilhelm Schlag, corollary 4.8, p.77 http://www.math.uchicago.edu/~schlag/bookweb.pdf {{Webarchive|url=https://web.archive.org/web/20170830063422/http://www.math.uchicago.edu/~schlag/bookweb.pdf |date=2017-08-30 }}</ref>

Let <math>f(z)</math> be holomorphic on a compact Riemann surface <math>M</math>. By compactness, there is a point <math>p_0 \in M</math> where <math>|f(p)|</math> attains its maximum. Then we can find a chart from a neighborhood of <math>p_0</math> to the unit disk <math>\mathbb{D}</math> such that <math>f(\varphi^{-1}(z))</math> is holomorphic on the unit disk and has a maximum at <math>\varphi(p_0) \in \mathbb{D}</math>, so it is constant, by the [[maximum modulus principle]].

== Remarks ==

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

Similarly, if an entire function has a [[Pole (complex analysis)|pole]] of order <math>n</math> at <math>\infty</math> &mdash;that is, it grows in magnitude comparably to <math>z^n</math> in some neighborhood of <math>\infty</math> 
&mdash;then <math>f</math> is a polynomial. This extended version of Liouville's theorem can be more precisely stated: if <math>|f(z)|\leq M|z|^n</math> for <math>|z|</math> sufficiently large, then <math>f</math> is a polynomial of degree at most <math>n</math>. This can be proved as follows. Again take the Taylor series representation of <math>f</math>,

:<math> f(z) = \sum_{k=0}^\infty a_k z^k.</math>

The argument used during the proof using [[Cauchy estimates]] shows that for all <math>k\geq 0</math>,

:<math>|a_k| \leq Mr^{n-k}.</math>

So, if <math>k > n </math>, then

:<math>|a_k| \leq \lim_{r\to\infty}Mr^{n-k} = 0.</math>

Therefore, <math>a_k = 0</math>.

Liouville's theorem does not extend to the generalizations of complex numbers known as [[split-complex number|double numbers]] and [[dual number]]s.<ref>{{Cite journal|url=https://scholar.rose-hulman.edu/rhumj/vol12/iss2/4/|title=Liouville theorems in the Dual and Double Planes|journal=Rose-Hulman Undergraduate Mathematics Journal|date=15 January 2017|volume=12|issue=2|last1=Denhartigh|first1=Kyle|last2=Flim|first2=Rachel}}</ref>

==See also==
* [[Mittag-Leffler's theorem]]

==References==
<references/>

==External links==
*{{planetmath reference|urlname=LiouvillesTheorem|title=Liouville's theorem}}
* {{MathWorld | urlname= LiouvillesBoundednessTheorem | title= Liouville’s Boundedness Theorem}}

[[Category:Theorems in complex analysis]]
[[Category:Articles containing proofs]]
[[Category:Analytic functions|holomorphic functions]]