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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    $\begingroup$ 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? $\endgroup$
    – Yemon Choi
    Commented Jun 27, 2012 at 8:16
  • $\begingroup$ I intended to use these examples (either in class or as exercises) to "motivate" the course. $\endgroup$
    – js21
    Commented Jun 27, 2012 at 8:49

5 Answers 5


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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    $\begingroup$ Essentially the same proof works in any complex Banach algebra with unit. $\endgroup$ Commented Jun 27, 2012 at 17:24
  • $\begingroup$ 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. $\endgroup$ Commented Jun 27, 2012 at 19:25
  • $\begingroup$ @Qiaochu and of course, that result in Lin alg follows by FTA applied to the min polynomial, while FTA follows from Liouville's theorem... $\endgroup$
    – Yemon Choi
    Commented Jun 28, 2012 at 2:06
  • $\begingroup$ @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? $\endgroup$
    – diverietti
    Commented Jun 28, 2012 at 7:32

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

See here: http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra#Complex-analytic_proofs


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.

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.

  • $\begingroup$ I'm no expert, but shouldn't the functions $f$ a $g$ be holomorphic? $\endgroup$ Commented Jun 27, 2012 at 15:24
  • $\begingroup$ 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. $\endgroup$ Commented Jun 27, 2012 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.)


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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