Liouville's theorem (complex analysis)

Liouville's theorem: Every holomorphic function $f:\backslash mathbb\; C\; \backslash to\; \backslash mathbb\; C$ for which there exists a positive number $M$ such that $|f(z)|\; \backslash leq\; M$ for all $z\backslash in\backslash Complex$ is constant.

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

This important theorem has several proofs.

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

{{Math proof|title=Proof|drop=hidden|proof=

If $f$ is an entire function, it can be represented by its Taylor series about 0:

: $f(z)\; =\; \backslash sum\_\{k=0\}^\backslash infty\; a\_k\; z^k$

where (by Cauchy's integral formula)

: $a\_k\; =\; \backslash frac\{f^\{(k)\}(0)\}\{k!\}\; =\; \{1\; \backslash over\; 2\; \backslash pi\; i\}\; \backslash oint\_\{C\_r\}\; \backslash frac\{f(\; \backslash zeta\; )\}\{\backslash zeta^\{k+1\}\}\backslash ,d\backslash zeta$

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 $|f(z)|\backslash leq\; M$ for all $z$. We can estimate directly

: $|\; a\_k\; |$

\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},

where in the second inequality we have used the fact that $|z|=r$ on the circle $C\_r$. (This estimate is known as Cauchy's estimate.) 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 $k\backslash geq\; 1$. Thus $f(z)=a\_0$ and this proves the theorem.

}}

Another proof uses the mean value property of harmonic functions.

{{Math proof|title=Proof{{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}}|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 $f$ 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.

Another approach to prove the theorem is

{{Math proof|title=Proof{{Cite book |last1=Gamelin|first1=Theodore W.|title= Complex Analysis |publisher=Springer |year=2004|isbn=9788181281142}}|drop=hidden|proof=

Suppose $|f(z)|\; \backslash leq\; M$ for all $z$ in the complex plane, we can apply the Cauchy estimate to a disk center at any $z\_0$ of any radius $\backslash rho$ to obtain:

$|f\text{'}(z)|\; \backslash leq\; \backslash frac\{M\}\{\backslash rho\}$.

Let $\backslash rho$ tend to $+\backslash infty$, we obtain $f\text{'}(z)\; =\; 0$. Since This is true for all $z\_0$, $f(z)\; =\; 0$ is a constant.

}}

There is a short proof of the fundamental theorem of algebra using Liouville's theorem.{{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}}

{{Math proof|title=Proof (Fundamental theorem of algebra)|drop=hidden|proof=

Suppose for the sake of contradiction that there is a nonconstant polynomial $p$ with no complex root. Note that $|p(z)|\; \backslash to\; \backslash infty$ as $z\; \backslash to\; \backslash infty$. Take a sufficiently large ball $B(0,\; R)$; for some constant $M$ there exists a sufficiently large $R$ such that for all $z\; \backslash not\backslash in\; B(0,\; R)$.

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

Thus, the function $q(z)$ is bounded in $\backslash mathbb\; C$, and by Liouville's theorem, is constant, which contradicts our assumption that $p$ is nonconstant.

}}

A consequence of the theorem is that "genuinely different" entire functions cannot dominate each other, i.e. if $f$ and $g$ are entire, and $|f|\backslash leq\; |g|$ everywhere, then $f=\backslash alpha\; g$ for some complex number $\backslash alpha$. Consider that for $g=0$ the theorem is trivial so we assume $g\backslash neq\; 0$. 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.

Suppose that $f$ is entire and $|f(z)|\backslash leq\; M|z|$, for $M>0$. We can apply Cauchy's integral formula; we have that

:$|f\text{'}(z)|=\backslash frac\{1\}\{2\backslash pi\}\backslash left|\backslash oint\_\{C\_r\}\backslash frac\{f(\backslash zeta)\}\{(\backslash zeta-z)^2\}d\backslash zeta\backslash right|\backslash leq\; \backslash frac\{1\}\{2\backslash pi\}\; \backslash oint\_\{C\_r\}\; \backslash frac$

where $I$ is the value of the remaining integral. This shows that $f\text{'}$ 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.

The theorem can also be used to deduce that the domain of a non-constant elliptic function $f$ cannot be $\backslash Complex$. Suppose it was. Then, if $a$ and $b$ are two periods of $f$ such that $\backslash tfrac\{a\}\{b\}$ 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 $f$ cannot be $\backslash Complex$ is what Liouville actually proved, in 1847, using the theory of elliptic functions.{{Citation|last = Liouville|first = Joseph|author-link = Joseph Liouville|publication-date = 1879|year = 1847|title = Leçons sur les fonctions doublement périodiques|periodical = 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}} In fact, it was Cauchy who proved Liouville's theorem.{{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}}{{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}}

If $f$ is a non-constant entire function, then its image is dense in $\backslash Complex$. 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

:$g(z)\; =\; \backslash frac\{1\}\{f(z)\; -\; w\}.$

Then $g$ is a bounded entire function, since for all $z$,

:$|g(z)|=\backslash frac\{1\}$

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

Any holomorphic function on a compact Riemann surface is necessarily constant.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 }}

Let $f(z)$ be holomorphic on a compact Riemann surface $M$. By compactness, there is a point $p\_0\; \backslash in\; M$ where $|f(p)|$ attains its maximum. Then we can find a chart from a neighborhood of $p\_0$ to the unit disk $\backslash mathbb\{D\}$ such that $f(\backslash varphi^\{-1\}(z))$ is holomorphic on the unit disk and has a maximum at $\backslash varphi(p\_0)\; \backslash in\; \backslash mathbb\{D\}$, so it is constant, by the maximum modulus principle.

Let $\backslash Complex\; \backslash cup\; \backslash \{\backslash infty\backslash \}$ be the one-point compactification of the complex plane $\backslash Complex$. In place of holomorphic functions defined on regions in $\backslash Complex$, one can consider regions in $\backslash Complex\; \backslash cup\; \backslash \{\backslash infty\backslash \}$. Viewed this way, the only possible singularity for entire functions, defined on $\backslash Complex\; \backslash subset\; \backslash Complex\; \backslash cup\; \backslash \{\backslash infty\backslash \}$, is the point $\backslash infty$. If an entire function $f$ is bounded in a neighborhood of $\backslash infty$, then $\backslash infty$ is a removable singularity of $f$, i.e. $f$ cannot blow up or behave erratically at $\backslash infty$. 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 $\backslash infty$ —that is, it grows in magnitude comparably to $z^n$ in some neighborhood of $\backslash infty$

—then $f$ is a polynomial. This extended version of Liouville's theorem can be more precisely stated: if $|f(z)|\backslash leq\; M|z|^n$ 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 $f$,

:$f(z)\; =\; \backslash sum\_\{k=0\}^\backslash infty\; a\_k\; z^k.$

The argument used during the proof using Cauchy estimates shows that for all $k\backslash geq\; 0$,

:$|a\_k|\; \backslash leq\; Mr^\{n-k\}.$

So, if $k\; >\; n$, then

:$|a\_k|\; \backslash leq\; \backslash lim\_\{r\backslash to\backslash infty\}Mr^\{n-k\}\; =\; 0.$

Therefore, $a\_k\; =\; 0$.

Liouville's theorem does not extend to the generalizations of complex numbers known as double numbers and dual numbers.{{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}}

• Mittag-Leffler's theorem

• {{Cite book |last1=Gamelin|first1=Theodore W.|title= Complex Analysis |publisher=Springer |year=2004|isbn=9788181281142}}

• {{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

holomorphic functions