Is it so easy to show a locally bounded function with holomorphic cross sections is holomorphic?

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Let $f: U \rightarrow \mathbb{C}$ where $U$ is open in $\mathbb{C}^n$ and suppose every coordinate cross section of $f$ is holomorphic (I hope that's not too colloquial).

I've heard it's a somewhat deep theorem that this is enough to imply f is continuous, and therefore holomorphic as a function of n variables. However when one uses Cauchy's formula (in one variable) to get a power series expansion of $f$ in a neighborhood of a point, one seems only to use that $f$ is locally bounded (so one can bound the values of f on the distinguished boundary of a polydisc). Is that correct? I.e. does {locally bounded} + {holomorphic cross sections} so easily imply holomorphic or have I made a mistake?

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Yes, if $f$ is locally bounded and separately holomorphic, it's not hard to see that $f$ is in fact continuous and jointly analytic. See this question for more details.

As zwh pointed out, $f$ being separately holomorphic is enough, but this is a lot more difficult to prove.

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You only need to assume $f$ is holomorphic in each variable separately. This is a nice result of Hartogs: http://en.wikipedia.org/wiki/Hartogs'_theorem