# Applications of Liouville's theorem

I'm looking for "nice" applications of Liouville's theorem (every bounded entire map is constant) outside the area of complex analysis.

An example of what I'm not looking for : a non-constant entire function has dense image (this is essentially a corollary).

An example of the kind of thing I'm looking for : a complex matrix whose conjugacy class is bounded must be a homothety (if $A$ is such a matrix and $B$ is an other matrix, then $z \mapsto e^{-z B} A e^{z B}$ is entire and bounded hence constant, but its derivative at $0$ is $[A,B]$ : thus $[A,B]=0$). In a similar vein : a subalgebra of $M_n (\mathbb{C})$ on which the spectral radius is submultiplicative is simultaneously triangularizable.

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I think the question should be community wiki. Also, is this just an enquiry out of curiosity, or are you after examples that could be used in class? –  Yemon Choi Jun 27 '12 at 8:16
I intended to use these examples (either in class or as exercises) to "motivate" the course. –  js21 Jun 27 '12 at 8:49

A very nice application of Liouville's theorem in functional analysis is the following, which is of great theoretical and practical importance.

Theorem (Spectrum). If $X\ne\lbrace0\rbrace$ is a complex Banch space and $T\colon X\to X$ a bounded linear operator, then its spectrum $\sigma(T)\ne\emptyset$.

First of all, let $X$ a complex Banach space, $B(X,X)$ the space of bounded linear operators from $X$ to $X$ and $\Lambda\subset\mathbb C$ a domain of the complex plane. Consider a function $$S\colon\Lambda\to B(X,X), \qquad\lambda\mapsto S_\lambda.$$

Definition. The map $S$ is said to be holomorphic on $\Lambda$ if for every $x\in X$ and $f\in X^*$ the function $h$ defined by $$h(\lambda)=f(S_\lambda(x))$$ is holomorphic at every $\lambda_0\in\Lambda$.

The following proposition is an easy exercise.

Proposition (Holomorphy of $R_\lambda$). The resolvent $R_\lambda(T)$ of a bounded linear operator $T\in B(X,X)$ is holomorphic at every point of the resolvent set $\rho(T)$ of $T$.

The proof of the Spectrum theorem is then quite elementary and goes as follows.

Proof. By assumption, $X\ne\lbrace0\rbrace$. If $T=0$, then $\sigma(T)=\lbrace0\rbrace\ne\emptyset$. So, let $T\ne 0$ and $$R_\lambda=(T-\lambda I)^{-1}=-\frac 1\lambda\sum_{j=0}^\infty(\frac 1\lambda T)^j.$$ This series is convergent for all $|\lambda|>||T||$, and thus it converges absolutely for instance for $|\lambda|>2||T||$. For these $\lambda$, by the formula for the sum of a geometric series, we have $$||R_\lambda||\le\frac 1{||T||}.$$ If $\sigma(T)=\emptyset$, then by definition the resolvent $\rho(T)$ is the whole complex plane. Hence, $R_\lambda$ is holomorphic for all $\lambda$. Consequently, for a fixed $x\in X$ and $f\in X^*$, the function $h$ defined by $$h(\lambda)=f(R_\lambda(x))$$ is holomorphic on $\mathbb C$, that is, it is an entire function. Now, $h$ is in particular continuous and thus bounded on the compact disk $|\lambda|\le 2||T||$. But $h$ is also bounded for $\lambda\ge 2||T||$ since $||R_\lambda||\le1/||T||$ and $$|h(\lambda)|=|f(R_\lambda(x))|\le||f||\cdot||R_\lambda(x)||\le||f||\cdot||R_\lambda||\cdot||x||\le\frac{||f||\cdot||x||}{||T||}.$$ Hence, $h$ is constant by Liouville's theorem. But this implies that $R_\lambda$ is independent of $\lambda$ and that so is $R_\lambda^{-1}=T-\lambda I$, which is a contradiction.$\qquad\square$

Observe that in the finite dimensional case, that is $X=\mathbb C^n$, the Spectrum Theorem says that the characteristic polynomial $\det(A-\lambda I)$ of a complex $(n\times n)$-matrix $A$ has a solution, which is just the fundamental theorem of algebra, which in turn follows again by Liouville's theorem...

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Essentially the same proof works in any complex Banach algebra with unit. –  Robert Israel Jun 27 '12 at 17:24
In the finite-dimensional case you get something slightly more basic than what you said, namely, you get that if $T : \mathbb{C}^n \to \mathbb{C}^n$ is a linear operator then there exists $\lambda$ such that $T - \lambda$ is not invertible. This is equivalent to $T$ having an eigenvector, and stating this doesn't require even knowing what determinants are. FTA is, of course, a corollary once you know how to construct companion matrices. –  Qiaochu Yuan Jun 27 '12 at 19:25
@Qiaochu and of course, that result in Lin alg follows by FTA applied to the min polynomial, while FTA follows from Liouville's theorem... –  Yemon Choi Jun 28 '12 at 2:06
@Quiachu. If $T-\lambda I$ is not invertible, then shouldn't its determinant vanish? And if it vanishes, doesn't this mean that $\lambda$ is a root of the characteristic polynomial? –  diverietti Jun 28 '12 at 7:32

The fundamental theorem of algebra: The field $\mathbb{C}$ is algebraically closed.

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There are many cool applications when combined with the uniformization theorem. Not sure if you count them as "complex analysis" or not - you could really think of them as algebraic geometry. For example:

The only meromorphic functions $f$ and $g$ satisfying $f^n+g^n = 1$ for $n>3$ are constant.

Proof sketch: If there were such functions they would define a map $F$ from the plane to the Riemann surface $x^n+y^n = 1$. We can compute the genus of this guy using Hurwitz - it is $(n-1)(n-2)/2$. So for $n>3$ the genus is bigger than two - the corresponding Riemann surface has negative curvature, and by uniformization it has the disk as its holomorphic universal cover. But then, $F$ would factor through the disk, and so by Louville F had to be constant.

Note:
for n=1 the equation has lots of solutions

for n=2 sine and cosine work (for instance)

for n=3 the Riemann surface has genus 1, and so its universal cover is the plane. You can find explicit solutions to the equation by using theta functions.

EDIT: another very important example of the same sort of reasoning is the little Picard theorem. At a high level, the proof just says that the plane with 2 points deleted has a holomorphic universal covering by the disk, so any entire function which misses 2 points factors through the disk - Louville gives the contradiction.

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I'm no expert, but shouldn't the functions $f$ a $g$ be holomorphic? –  Vít Tuček Jun 27 '12 at 15:24
Meromorphic functions can be extended to the projective completion x^n+y^n = z^n, which is compact. I left that bit out of my outline. –  Steven Gubkin Jun 27 '12 at 18:39

Fuglede's theorem (if $N$ and $P$ are commuting operators on Hilbert space, and $N$ is normal, then $P$ commutes with $N^*$) has a slick proof using the vector-valued Liouville theorem. I guess the special case of matrices may be sufficiently interesting to be included as an exercise. (For normal matrices one could just use the spectral theorem, I guess.)

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The first published proof of the Mazur-Gelfand theorem, due to Gelfand himself (though previously announced without proof by Mazur), is based on the vector-valued version of the Liouville theorem, which was further extended by Arens to cover a more general situation (see [1] and references therein).

[1] R. Arens (1947), Linear topological division algebras, Bull. AMS, Vol. 53, pp. 623-630.

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