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| {{About|Liouville's theorem in complex analysis||Liouville's theorem (disambiguation)}}
| | Alfonso will be the name people use to call him and he believes it sounds quite solid. Invoicing may be my regular job for a little while but I've always wanted my own home based business. Playing handball is what my family and I have fun with. Mississippi is where he's been living for years and he won't ever move. Go to her website to find out more: http://rodolfod6.deviantart.com/journal/El-Epoxi-Una-Resina-Excesivamente-Convertible-461313715<br><br>Also visit my website: [http://rodolfod6.deviantart.com/journal/El-Epoxi-Una-Resina-Excesivamente-Convertible-461313715 resinas] |
| In [[complex analysis]], '''Liouville's theorem''', named after [[Joseph Liouville]], states that every [[bounded function|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.
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| The theorem is considerably improved by [[Picard theorem|Picard's little theorem]], which says that every entire function whose image omits at least two complex numbers must be constant.
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| == Proof ==
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| The theorem follows from the fact that [[Proof that holomorphic functions are analytic|holomorphic functions are analytic]]. If ''f'' is an entire function, it can be represented by its [[Taylor series]] about 0:
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| : <math>f(z) = \sum_{k=0}^\infty a_k z^k</math>
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| where (by [[Cauchy's integral formula]])
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| :<math>
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| a_k = \frac{f^{(k)}(0)}{k!} = {1 \over 2 \pi i} \oint_{C_r}
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| \frac{f( \zeta )}{\zeta^{k+1}}\,d\zeta
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| </math>
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| and ''C''<sub>''r''</sub> 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
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| :<math>
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| | a_k |
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| \le \frac{1}{2 \pi} \oint_{C_r} \frac{ | f ( \zeta ) | }{ | \zeta |^{k+1} } \, |d\zeta|
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| \le \frac{1}{2 \pi} \oint_{C_r} \frac{ M }{ r^{k+1} } \, |d\zeta|
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| = \frac{M}{2 \pi r^{k+1}} \oint_{C_r} |d\zeta|
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| = \frac{M}{2 \pi r^{k+1}} 2 \pi r
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| = \frac{M}{r^k},
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| </math>
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| where in the second inequality we have used the fact that |''z''|=''r'' on the circle ''C''<sub>''r''</sub>. 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''<sub>''k''</sub> = 0 for all ''k'' ≥ 1. Thus ''f''(''z'') = ''a''<sub>0</sub> and this proves the theorem.
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| == Corollaries ==
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| ===Fundamental theorem of algebra===
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| There is a short [[Fundamental theorem of algebra#Complex-analytic proofs|proof of the fundamental theorem of algebra]] based upon Liouville's theorem.
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| ===No entire function dominates another entire function===
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| 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'' 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''<sup>−1</sup>(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.
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| ===If ''f'' is less than or equal to a scalar times its input, then it is linear===
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| 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
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| :<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{\left| f(\zeta) \right|}{\left| (\zeta-z)^2\right|} \left|d\zeta\right|\leq \frac{1}{2\pi} \oint_{C_r} \frac{M\left| \zeta \right|}{\left| (\zeta-z)^2\right|} \left|d\zeta\right|=\frac{MI}{2\pi}</math>
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| where ''I'' is the value of the remaining integral. This shows that ''f''' is bounded and entire, so it must be constant, by Liouville's theorem. Integrating then shows that ''f'' is [[Affine transformation|affine]] and then, by referring back to the original inequality, we have that the constant term is zero.
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| ===Non-constant elliptic functions cannot be defined on C===
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| The theorem can also be used to deduce that the domain of a non-constant [[elliptic function]] ''f'' cannot be '''C'''. Suppose it was. Then, if ''a'' and ''b'' are two periods of ''f'' such that <sup>''a''</sup>⁄<sub>''b''</sub> is not real, consider the [[parallelogram]] ''P'' whose [[Vertex (geometry)|vertices]] are 0, ''a'', ''b'' and ''a'' + ''b''. Then the image of ''f'' is equal to ''f''(''P''). Since ''f'' is [[continuous functions|continuous]] and ''P'' is [[Compact space|compact]], ''f''(''P'') is also compact and, therefore, it is bounded. So, ''f'' is constant.
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| 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.<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}}</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>
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| ===Entire functions have dense images===
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| If ''f'' is a non-constant entire function, then its image is [[Dense set|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
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| :<math>(\forall z\in\mathbb{C}):|g(z)|=\frac1{|f(z)-w|}<\frac1r\cdot</math>
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| So, ''g'' is constant, and therefore ''f'' is constant.
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| == Remarks ==
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| 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 '''C''' ⊂ '''C''' ∪ {∞}, 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.
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| Similarly, if an entire function has a [[Pole (complex analysis)|pole]] at ∞, i.e. blows up like ''z<sup>n</sup>'' 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''.|''z<sup>n</sup>''| 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'',
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| :<math> f(z) = \sum_{k=0}^\infty a_k z^k.</math>
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| The argument used during the proof using Cauchy estimates shows that
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| :<math>(\forall k\in\mathbb{N}):|a_k|\leqslant Mr^{n-k}.</math>
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| So, if ''k'' > ''n'',
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| :<math>|a_k|\leqslant\lim_{r\rightarrow+\infty}Mr^{n-k}=0.</math>
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| Therefore, ''a<sub>k</sub>'' = 0.
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| ==See also==
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| * [[Mittag-Leffler's theorem]]
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| ==References==
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| <references/>
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| ==External links==
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| *{{planetmath reference|title=Liouville's theorem|id=1145}}
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| * {{MathWorld | urlname= LiouvillesBoundednessTheorem | title= Liouville’s Boundedness Theorem}}
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| * [http://math.fullerton.edu/mathews/c2003/LiouvilleMoreraGaussMod.html Module for Liouville’s Theorem by John H. Mathews]
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| [[Category:Theorems in complex analysis]]
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| [[Category:Articles containing proofs]]
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| [[Category:Analytic functions|holomorphic functions]]
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Alfonso will be the name people use to call him and he believes it sounds quite solid. Invoicing may be my regular job for a little while but I've always wanted my own home based business. Playing handball is what my family and I have fun with. Mississippi is where he's been living for years and he won't ever move. Go to her website to find out more: http://rodolfod6.deviantart.com/journal/El-Epoxi-Una-Resina-Excesivamente-Convertible-461313715
Also visit my website: resinas