# A necessary condition for a multi-complex-variable holomorphic function.

Let $\Omega\subset \mathbb{C}^n$ be an open unit ball, $f:\Omega \to\mathbb{C}$ is a bounded function. For $a \in \mathbb{C}^n$, define $$\Omega_{j,a}=\{z\in\mathbb{C}:(a_1,…a_{j-1},z,a_{j+1},…,a_n)\in \Omega\}$$
If $f(a_1,…a_{j-1},z,a_{j+1},…,a_n)$is holomorphic on $\Omega_{j,a}$ for each$a\in\mathbb{C}^n$. Show that $f:\Omega \to\mathbb{C}$ is continuous. Moreover, $f$ is also holomorphic.

(Definition:$f:\Omega \to\mathbb{C}$ is said to be holomorphic if for any $a\in \Omega$ there exists a neighborhood $U$ of $a$ such that $f(z)=\sum_{\alpha\in\mathbb{N}^n} c_\alpha (z-a)^\alpha$ for $z\in U$)

#### Solutions Collecting From Web of "A necessary condition for a multi-complex-variable holomorphic function."

Originally, up to about 1905 or so, the usual definition of what it should mean that “a function of several complex variables is holomorphic” looked something like this:

A function $f : \Omega \to \mathbb{C}$ (where $\Omega \subset \mathbb{C}^n$) is holomorphic if it is

1. bounded on compact subsets of $\Omega$ and
2. holomorphic in each variable separately

Then, in 1906, Hartogs’ proved that 1. follows from 2. and nowadays we just say that a function of several complex variables is called holomorphic if it is holomorphic in each variable separately. The proof of Hartogs’ theorem (which for some reason is often called “Hartogs’ lemma”) is very technical.

The question above has to do with the original definition and is a little easier to handle.

Fix $z \in \Omega$ and choose a small polydisc $\Delta = \{ \zeta : |\zeta_j – z_j| < r \text{ for all$j$} \}$ centered at $z$ such that $\Delta \subset\subset \Omega$. Then, repeated applications of Cauchy’s integral formula, one variable at the time gives:
\begin{align}
f(z)
&= \frac{1}{2\pi i} \int_{|\zeta_n – z_n| = r} \frac{f(z_1, \ldots, z_{n-1}, \zeta_n)}{\zeta_n – z_n}\,d\zeta_n \\[12pt]
&= \Big( \frac{1}{2\pi i} \Big)^2 \int_{|\zeta_{n-1}-z_{n-1}| = r} \int_{|\zeta_n – z_n| = r} \frac{f(z_1, \ldots, z_{n-1}, \zeta_{n-1}, \zeta_n)}{(\zeta_{n-1}-z_{n-1})(\zeta_n – z_n)}\,d\zeta_{n-1}\,d\zeta_n \\[12pt]
&= \cdots \\[12pt]
&= \Big( \frac{1}{2\pi i} \Big)^n \int_{|\zeta_{1}-z_{1}| = r} \cdots \int_{|\zeta_n – z_n| = r} \frac{f(\zeta_1, \ldots, \zeta_n)}{(\zeta_{1}-z_{1})\cdots (\zeta_n – z_n)}\,d\zeta_{1} \ldots\,d\zeta_n
\end{align}
which is Cauchy’s integral formula for polydiscs. The assumption of separate holomorphicisy is enough to ensure that the integrals exist.

From the above representation, it follows that $f$ is continuous at $z$ (I’ll leave it as an exercise to estimate $|f(z)-f(\tilde z)|$. Here you will need the assumption that $f$ is bounded on $\Delta$.) and in fact $C^\infty$. (Verify that differentiation under the integral sign is permitted.)

Furtermore, it follows (just as in the one-variable proof) that $f$ can be represented as a convergent power series on $\Delta$.

Since you don’t state what exact defintion of holomorphic you are using, I can’t answer the final question, but it should follow from one or more of the facts above.