First of all let me rephrase the statement:
Proposition Let $U \subset \mathbb{C}^n$ be open and connected, $f \in \mathcal{O}(U)$ non constantly zero, $V(f)$ its zero locus in $U$, $g : U-V(f) \longrightarrow \mathbb{C}$ holomorphic and bounded. Then $g$ can be extended to the whole $U$.
Proof
By Identity Principle we need only to show that $g$ can be extended to an open neighborhood of every point $x \in V(f)$. Without loss of generality we can suppose $x = 0 \in \mathbb{C}^n$ and $f(0,z_n)$ non constantly zero. By restricting to a polydisc $\Delta(0, (\tau_1, \dots, \tau_n))$ we know that $f$ can be written as $f = e \cdot h$, where $e$ is invertible and $h$ is a Weierstrass polynomial of degree $d$. Surely if $\rvert z_n \lvert = \tau_n$ then $h \neq 0$.
Being
$$ \pi : \{ \rvert z_n \lvert = \tau_n \} \longrightarrow \mathbb{C}^{n-1}$$ a proper map then exists $\delta > 0$ such that if $\rvert z_i \lvert \leq \delta, \rvert z_n \lvert = \tau_n$ then $h(z) \neq 0$
Then it ends by saying that in $\Delta(0, (\delta, \dots, \delta, \tau_n))$ we can define $g$ by Cauchy's formula for one variable.
I have two problems:
- Where it is needed the bound condition on $g$ in previous the proof?
- I don't understand the quoted passage in the proof. I guess I can reach the same result by saying: for each $z_n$ such that $\rvert z_n \lvert = \tau_n$ it exists $\delta_{z_n} >0 $ such that $h \neq 0 $ on $\Delta(0, (\delta_{z_n}, \dots, \delta_{z_n}))$ by continuity. By compactness I need only finitely many discs in $\mathbb{C}$ to cover $\{ \rvert z_n \lvert = \tau_n\} \subset \mathbb{C}$, then I take $\delta = \min{\{\delta_{z_{n_k}}\}}$, right?
Thank you in advance for the help!
To answer your two questions:
1) The boundedness is used in the application of the one-dimensional result, that is the use of Cauchy's formula. You are really applying the Riemann extension theorem from one variable. If $g$ is not bounded, you cannot apply that.
2) Your proof is correct. What the author uses seems a little more complicated (and uses a bit heavier machinery). Here is where properness is used: The projection by $\pi$ of a closed set (the zero set of $h$) is closed (because $\pi$ is proper) in the $z_1,\ldots,z_{n-1}$ variables, and since the origin (in ${\mathbb C}^{n-1}$) is not in this projection, there exists a $\delta > 0$ polydisc around the origin above which $h$ is not zero, which is the conclusion wanted.
By the way, Weierstrass is not needed for this proof. All you need is the extension result from one variable and the fact that in one variable zeros of holomorphic functions are isolated to apply the reasoning you did in 2). See page 33 of http://www.jirka.org/scv/scv.pdf