## Liouville’s Theorem in Banach Spaces

The classical version of Liouville’s theorem states that if ${f: \mathbb{C} \rightarrow \mathbb{C}}$ is an entire bounded function, then ${f}$ is constant. One might ask if an analogous result holds for complex-differentiable functions taking values in a Banach space ${(X,\left\|\cdot\right\|_{X})}$. The answer is yes.

\bigskip Let ${(X,\left\|\cdot\right\|_{X})}$ be a Banach space, and let ${D \subset \mathbb{C}}$ be a domain (i.e., a connected open subset). Recall that a function ${f: D \rightarrow X}$ is said to be analyticon ${D}$ if, for each ${z \in D}$, there exists ${f'(z) \in X}$ such that

$\displaystyle \lim_{h \rightarrow 0} \left\|f'(z) - \frac{f(z+h) - f(z)}{h}\right\|_{X} = 0$

Lemma 1 If ${f: D \rightarrow X}$ is an analytic function and ${x^{*} \in X^{*}}$ is a continuous linear functional, then the function ${x^{*}(f): D \rightarrow \mathbb{C}}$ defined by

$\displaystyle (x^{*}(f))(z) = x^{*}(f(z)), \indent z \in D$

is analytic with derivative ${(x^{*}(f))'(z) = x^{*}(f'(z))}$.

Proof:Fix a continuous linear functional ${x^{*} \in X^{*}}$ and define ${x^{*}(f)}$ as above. Since ${x^{*}}$ is bounded and linear, we have for ${z \in D}$ and ${h \neq 0}$, that

$\displaystyle \left\|x^{*}(f'(z)) - \frac{x^{*}(f(z+h)) - x^{*}(f(z))}{h}\right\|_{X} = \left\|x^{*}\left(f'(z) - \frac{f(z+h) - f(z)}{h}\right)\right\|_{X} \leq \left\|x^{*}\right\|\left\|f'(z) - \frac{f(z+h) - f(z)}{h}\right\|_{X} \rightarrow 0, h \rightarrow 0$

$\Box$

In particular, we see that if ${f}$ is an entire function in the extended Banach-space-valued sense, then, for any ${x^{*} \in X^{*}}$, ${x^{*}(f)}$ is an entire function in the classical sense. We are now ready to state and prove the generalization of Liouville’s theorem to ${X}$-valued functions.

Theorem 2 Let ${f: \mathbb{C} \rightarrow X}$ be an entire function such that ${\left\|f(z)\right\|_{X} \leq M}$ for all ${z \in \mathbb{C}}$, for some finite ${M > 0}$. Then there exists ${x_{0} \in X}$ such that ${f(z) = x_{0}}$ for all ${z \in \mathbb{C}}$. In other words, ${f}$ is constant.

Proof:Assume the contrary. Then there exists ${z_{1}, z_{2} \in \mathbb{C}}$ such that ${f(z_{1}) \neq f(z_{2})}$. By the Hahn-Banach theorem, there exists ${x^{*} \in X^{*}}$ such that ${x^{*}(f(z_{1})) \neq x^{*}(f(z_{2}))}$. Since

$\displaystyle \left|x^{*}(f(z))\right| \leq \left\|x^{*}\right\|\left\|f(z)\right\|_{X} \leq M\left\|x^{*}\right\|, \indent \forall z \in \mathbb{C}$

we see that ${x^{*}(f)}$ is constant by the classical Liouville’s theorem. But this contradicts that ${x^{*}(f(z_{1})) \neq x^{*}(f(z_{2}))}$. $\Box$