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Let $X$ be a (complex) banach space, $U$ be an open subset of $\mathbb{C}$ and $f: U \to X$ be a function that is completely arbitrary except that it satisfies the property that for any continuous linear functional $l$ on $X$, $l \circ f$ is complex analytic in the usual sense. Is it possible to deduce from this that $f$ is continuous? What about strongly analytic? (This means that the usual limit of the difference quotient exists in the norm of $X$.)

Can strong analyticity be concluded if I assume the weak analyticity condition plus continuity?

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    It seems this approach deadends at precisely where I stopped. But I found this http://www.math.ubc.ca/~feldman/m511/analytic.pdf which instead uses the uniform boundedness principle.2012-09-29

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This may be a variation on the reference you linked above. If you shift so that $f(0) = 0$, we can check continuity at $0$. Apply linear functionals $l$ to the values of $f(z)/z$, which up to removable singularities is holomorphic, apply Cauchy's integral formula in the classical case, and then use the Cauchy estimate $|l \circ f(z)/z)| \le C/r$, with $C$ dependent on $l$. This means the set of values $f(z)/z$ is weakly bounded, so by Uniform Boundedness Principle or Banach-Alaoglu, the set is bounded, so $f(z)$ must be continuous at $0$. Thus indeed $f$ is continuous.

Then, as suggested by Qiaochu, there are no worries about verifying the Cauchy integral formula on $f$.

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    Well, if I just want to prove continuity, then everything is local.2012-09-30