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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.$\square$ 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...

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.$\square$