Théorème de Liouville (analyse complexe)

Théorème de Liouville (analyse complexe) This article is about Liouville's theorem in complex analysis. Pour d'autres usages, voir le théorème de Liouville (désambiguïsation).

En analyse complexe, Théorème de Liouville, named after Joseph Liouville (although the theorem was first proven by Cauchy in 1844[1]), states that every bounded entire function must be constant. C'est-à-dire, every holomorphic function {style d'affichage f} for which there exists a positive number {style d'affichage M} tel que {style d'affichage |F(z)|leq M} pour tous {style d'affichage avec} dans {style d'affichage mathbb {C} } est constant. De manière équivalente, non-constant holomorphic functions on {style d'affichage mathbb {C} } have unbounded images.

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

Contenu 1 Preuve 2 Corollaires 2.1 Fundamental theorem of algebra 2.2 No entire function dominates another entire function 2.3 If f is less than or equal to a scalar times its input, then it is linear 2.4 Non-constant elliptic functions cannot be defined on ℂ 2.5 Entire functions have dense images 3 On compact Riemann surfaces 4 Remarques 5 Voir également 6 Références 7 External links Proof This important theorem has several proofs.

A standard analytical proof uses the fact that holomorphic functions are analytic.

Proof If f is an entire function, it can be represented by its Taylor series about 0: {style d'affichage f(z)=somme _{k=0}^{infime }un_{k}z ^{k}} où (by Cauchy's integral formula) {style d'affichage a_{k}={frac {f ^{(k)}(0)}{k!}}={1 over 2pi i}point _{C_{r}}{frac {F(zêta )}{zeta ^{k+1}}},enfant } and Cr is the circle about 0 of radius r > 0. Suppose f is bounded: c'est à dire. there exists a constant M such that {style d'affichage |F(z)|leq M} pour tout z. We can estimate directly {style d'affichage |un_{k}|leq {frac {1}{2pi }}point _{C_{r}}{frac {|F(zêta )|}{|zêta |^{k+1}}},|enfant |leq {frac {1}{2pi }}point _{C_{r}}{frac {M}{r ^{k+1}}},|enfant |={frac {M}{2pi r ^{k+1}}}point _{C_{r}}|enfant |={frac {M}{2pi r ^{k+1}}}2pi r={frac {M}{r ^{k}}},} where in the second inequality we have used the fact that {style d'affichage |z|=r} on the circle Cr. But the choice of r in the above is an arbitrary positive number. Par conséquent, letting r tend to infinity (we let r tend to infinity since f is analytic on the entire plane) gives ak = 0 for all k ≥ 1. Thus f(z) = a0 and this proves the theorem.

Another proof uses the mean value property of harmonic functions.

Preuve[2] 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 f is bounded, the averages of it over the two balls are arbitrarily close, and so f assumes the same value at any two points.

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

Corollaries Fundamental theorem of algebra There is a short proof of the fundamental theorem of algebra based upon Liouville's theorem.[3] No entire function dominates another entire function A consequence of the theorem is that "genuinely different" entire functions cannot dominate each other, c'est à dire. if f and g are entire, et |F| ≤ |g| partout, then f = α·g for some complex number α. Consider that for g = 0 the theorem is trivial so we assume {displaystyle gneq 0.} 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−1(0). But since h is bounded and all the zeroes of g are isolated, any singularities must be removable. Thus h 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 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; on a ça {style d'affichage |F'(z)|={frac {1}{2pi }}la gauche|point _{C_{r}}{frac {F(zêta )}{(zêta -z)^{2}}}dzeta right|leq {frac {1}{2pi }}point _{C_{r}}{frac {|F(zêta )|}{la gauche|(zêta -z)^{2}droit|}}|enfant |leq {frac {1}{2pi }}point _{C_{r}}{frac {M|zêta |}{la gauche|(zêta -z)^{2}droit|}}la gauche|dzeta right|={frac {MI}{2pi }}} 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 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 ℂ The theorem can also be used to deduce that the domain of a non-constant elliptic function f cannot be {style d'affichage mathbb {C} .} Suppose it was. Alors, if a and b are two periods of f such that a / b is not real, consider the parallelogram P whose vertices are 0, un, 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, Donc, it is bounded. Alors, f is constant.

The fact that the domain of a non-constant elliptic function f can not be {style d'affichage mathbb {C} } is what Liouville actually proved, dans 1847, using the theory of elliptic functions.[4] En réalité, it was Cauchy who proved Liouville's theorem.[5][6] Entire functions have dense images If f is a non-constant entire function, then its image is dense in {style d'affichage mathbb {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. Définir {style d'affichage g(z)={frac {1}{F(z)-w}}.} Then g is a bounded entire function, since for all z, {style d'affichage |g(z)|={frac {1}{|F(z)-w|}}<{frac {1}{r}}.} So, g is constant, and therefore f is constant. On compact Riemann surfaces Any holomorphic function on a compact Riemann surface is necessarily constant.[7] 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. Remarks 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, alors {style d'affichage |un_{k}|leq lim _{rto infty }Mr^{nk}=0.} Par conséquent, ak = 0.

Liouville's theorem does not extend to the generalizations of complex numbers known as double numbers and dual numbers.[8] See also Mittag-Leffler's theorem References ^ "encyclopedia of mathematics". ^ Nelson, Edouard (1961). "A proof of Liouville's theorem". Proceedings of the AMS. 12 (6): 995. est ce que je:10.1090/S0002-9939-1961-0259149-4. ^ Benjamin Fine; Gerhard Rosenberger (1997). The Fundamental Theorem of Algebra. Springer Science & Business Media. pp. 70–71. ISBN 978-0-387-94657-3. ↑ Liouville, Joseph (1847), "Leçons sur les fonctions doublement périodiques", Journal de mathématiques pures et appliquées (publié 1879), volume. 88, pp. 277–310, ISSN 0075-4102, archivé à partir de l'original sur 2012-07-11 ^ Cauchy, Augustin-Louis (1844), "Mémoires sur les fonctions complémentaires", Œuvres complètes d'Augustin Cauchy, 1, volume. 8, Paris: Gauthiers-Villars (publié 1882) ^ Lützen, Jesper (1990), Joseph Liouville 1809–1882: Master of Pure and Applied Mathematics, Studies in the History of Mathematics and Physical Sciences, volume. 15, Springer Verlag, ISBN 3-540-97180-7 ^ a concise course in complex analysis and Riemann surfaces, Wilhelm Schlag, corollaire 4.8, p.77 http://www.math.uchicago.edu/~schlag/bookweb.pdf Archived 2017-08-30 at the Wayback Machine ^ Denhartigh, Kyle; Flim, Rachel (15 Janvier 2017). "Liouville theorems in the Dual and Double Planes". Rose-Hulman Undergraduate Mathematics Journal. 12 (2). Liens externes "Théorème de Liouville". PlanèteMath. Weisstein, Eric W. "Liouville's Boundedness Theorem". MathWorld. Catégories: Theorems in complex analysisAnalytic functions