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?


1 Answer 1


Yes this is true and stated as Theorem 14.27 in the book Holomorphy and Calculus in Normed Spaces 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 by Seán Dineen

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    $\begingroup$ no pbm. I just added another pointer to the literature if needed. $\endgroup$ Jul 11, 2017 at 20:48

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