Analytic Function on Banach Space is Continuous

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 an analytic function.

Then $f$ is continuous.

Proof
Let $x \in U$.

Since $f : U \to X$ is analytic function, the limit:
 * $\ds \lim_{y \mathop \to x} \frac {\map f y - \map f x} {y - x} = \map {f'} x$ exists.

We have, from :
 * $\ds \norm {\map f y - \map f x} = \cmod {y - x} \norm {\frac {\map f y - \map f x} {y - x} }$

From Modulus of Limit: Normed Vector Space, we have:
 * $\ds \lim_{y \mathop \to x} \norm {\frac {\map f y - \map f x} {y - x} } = \norm {\map {f'} x}$

We also have:
 * $\ds \lim_{y \mathop \to x} \cmod {y - x} = 0$

So from Product Rule for Complex Sequences, we have:
 * $\ds \lim_{y \mathop \to x} \norm {\map f y - \map f x} = 0$

So, we have that $f$ is continuous at $x$.

Since $x \in U$ was arbitrary, we have that $f$ is continuous.