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

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