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?