Banach Space Valued Function is Analytic iff Weakly Analytic

Theorem
Let $U$ be an open subset of $\C$.

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

Let $f : U \to X$ be a function.

Then $f$ is analytic it is weakly analytic.

Necessary Condition
Suppose that $f$ is analytic.

Define $f' : U \to X$ by:


 * $\ds \map {f'} z = \lim_{w \mathop \to z} \frac {\map f w - \map f z} {w - z}$

for each $z \in U$.

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

Let $\phi \in X^\ast$.

Then for each $z, w \in U$ with $z \ne w$ we have:


 * $\ds \map \phi {\frac {\map f w - \map f z} {w - z} } = \frac {\map {\paren {\phi \circ f} } w - \map {\paren {\phi \circ f} } z} {w - z}$

since $\phi$ is linear.

From Continuity of Linear Functionals, $\phi$ is continuous.

So, we have:


 * $\ds \lim_{w \mathop \to z} \map \phi {\frac {\map f w - \map f z} {w - z} } = \map \phi {\map {f'} z}$

So:


 * $\ds \frac {\map {\paren {\phi \circ f} } w - \map {\paren {\phi \circ f} } z} {w - z}$ exists for each $z \in U$.

So $\phi \circ f$ is analytic for each $\phi \in X^\ast$.

So $f$ is weakly analytic.