Banach Space Valued Function is Analytic iff Weakly Analytic
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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 if and only if it is weakly analytic.
Proof
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.
$\Box$
Sufficient Condition
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