In complex analysis one learns Hartogs' theorem:



> 
> Let $U\subseteq \mathbb{C}^n$ open and $f: U \rightarrow \mathbb{C}$ a function. Then $f$ is analytic iff for all $1\leq i \leq n$
>  $$ z \mapsto f(z\_1, \dots, z\_{i-1}, z, z\_{i+1}, \dots, z\_n) $$
>  is analytic.
> 
> 
> 


Can we generalize this theorem to the Banach space setting? I.e. does the following statement hold?



> 
> Let $X,Y, Z$ be complex Banach spaces, $U\times V \subseteq X \times Y$ open and $f: U\times V \rightarrow Z$. Then $f$ is analytic iff $u \mapsto f(u, v\_0) $ and $v\mapsto f(u\_0, v)$ are analytic.
> 
> 
> 


If the statement is false, what would be reasonable assumption such that the statement becomes true?



Hartogs' theorem in Banach spaces
Yes this is true and stated as Theorem 14.27 in the book [Holomorphy and Calculus in Normed Spaces](https://books.google.com/books?id=wtrvhCdbsd8C&printsec=frontcover#v=onepage&q&f=false) by Soo Bong Chae. If you need even more sophisticated generalizations, you can also look at the notes related to Section 2.76 on Page 415 of [Complex Analysis in Locally Convex Spaces](https://books.google.com/books?id=4yEm92kChEoC&printsec=frontcover#v=onepage&q&f=false) by Seán Dineen


