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.
Best Answer
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...