A function holomorphic on all of $\mathbb{C}^n$ cannot have a nonempty bounded set as it set of zeroes

Question

I took this question from the book 'Several Complex Variables with connections to algebraic geometry and lie groups' chapter 2. The statement to be proven is that for $n>1$, a function that holomorphic all on $\mathbb{C}^n$ cannot have an nonempty bounded set as it set of zeroes.

I tried to prove by contradiction by assuming that suppose for a function $f:\mathbb{C}^n\mapsto \mathbb{C}$ that is holomorphic its set of zeroes $Z(f)$ is nonempty and bounded by say $M>0$. I have managed to also prove that the zeroes of $f$ cannot be isolated. How can I use this fact (if applicable) to prove the statement?

Answer 1

$1/f$ is a holomorphic function on the complement of some large compact ball $B_r(0)$ that contains all of the zeroes of $f$. Hartogs' extension theorem guarantees that $1/f$ extends to the whole of $\Bbb C^n$ (it is only undefined on a compact set). Call the extension $g$. The identity theorem guarantees that $1 = g(x)f(x)$ for all $x$, as these functions agree on the complement of the compact ball.

This implies that the set of zeroes of $f$ is empty.

Answer 2

Preceding me, Mike Miller offered a very nice and short answer by using Hartogs's extension theorem. However, I would nevertheless share my answer since it is based on nearly the same argument but, instead of using the famed theorem, uses the solution of the Dirichlet problem for holomorphic functions of several complex variables.

Following Mike's argument, $1/f$ is holomorphic outside a compact ball $B_r(0)$ containing all the zeros of $f$. Now consider the trace of $1/f$ on the boundary of a slightly larger ball, say $B_{r+\epsilon}(0)$, for some $\epsilon>0$: since $1/f$ is holomorphic in a neighborhood of $\partial B_{r+\epsilon}(0)\triangleq S_{r+\epsilon}(0)$, it is a CR-function on it therefore it can be used to solve the Dirichlet problem for a holomorphic function of several variables by the use of the Martinelli-Bochner formula for the ball of radius $r+\epsilon$: $$ g(z)=\frac{(n-1)!}{(r+\epsilon)2\pi^n}\int\limits_{S_{r+\epsilon}(0)}\frac{(r+\epsilon)^2-\langle\zeta,\bar{z}\rangle}{|\zeta -z|^{2n}}\frac{1}{f(\zeta)}\mathrm{d}\sigma_\zeta. $$ By construction, $g$ is an analytic extension of $1/f$ inside $B_{r+\epsilon}(0)$, and from this point on, the argument is the same followed by Mike, invoking the identity principle for holomorphic functions.

Kytmanov, Alexander M. (1995) [1992], The Bochner-Martinelli integral and its applications, Birkhäuser Verlag, pp. xii+305, doi:10.1007/978-3-0348-9094-6, ISBN 978-3-7643-5240-0, MR 1409816, Zbl 0834.32001.

Answer 3

There is a much simpler argument than using Hartogs (a very deep theorem) or Bochner-Martinelli. If the set of zeros is bounded, then there exists a smallest $R$ such that they are in a closed ball of radius $R$. WLOG assume that $R=1$ and $f(1,0,\ldots,0)=0$. Take the analytic discs $\Delta_\epsilon \colon {\mathbb D} \to {\mathbb C}^n$ defined by $\xi \mapsto (1+\epsilon,\xi,0,\ldots)$. Consider the function $f \circ \Delta_\epsilon$. Note that $f \circ \Delta_0$ is not identically zero, but has a zero at the origin, and $f \circ \Delta_\epsilon$ has no zeros for all $\epsilon > 0$. This contradicts the classical Hurwitz theorem from one complex variable.