Liouville's Theorem (Complex Analysis)/Banach Space

Theorem
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\C$.

Let $f : \C \to X$ be an analytic function that is bounded.

Then $f$ is constant.

Proof
Take $M \ge 0$ such that:


 * $\norm {\map f x} \le M$

for each $x \in X$.

Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,} }$.

From Banach Space Valued Function is Analytic iff Weakly Analytic, $f$ is weakly analytic.

So for each $\phi \in X^\ast$, $\phi \circ f : \C \to \C$ is analytic.

Further, for each $z \in \C$ we have:


 * $\cmod {\map \phi {\map f x} } \le \norm \phi_{X^\ast} \norm {\map f x} \le M \norm \phi_{X^\ast}$

So $\phi \circ f$ is bounded.

From Liouville's Theorem (Complex Analysis), it follows that $\phi \circ f$ is constant for each $\phi \in X^\ast$.

That is, for each $z, w \in \C$ we have:


 * $\map \phi {\map f z} = \map \phi {\map f w}$

for each $\phi \in X^\ast$.

That is, since $\phi$ is linear:


 * $\map \phi {\map f z - \map f w} = 0$

From Normed Dual Space Separates Points, we have:


 * $\map f z = \map f w$

for each $z, w \in \C$.

So $f$ is constant.