on several complex variables on a ball

Question

I want to solve the following problem:

Let $B=B(0,R)$ be a ball in $\mathbb{C}^n, n\geq 2$. Let $f$ be holomorphic on $B$ and continuous on $\overline {B}$. if $f(a)=0$ for some $a\in B$, show that there is $p\in \mathrm bB$ (boundary of B) such that $f(p)=0$.

I assume $f(p)\neq 0$ for all $p\in \mathrm bB$ and want to derive a contradiction. But I can't.

Answer 1

If $f$ has no zero on $\mathrm{b}B$, then by continuity there is an $\varepsilon > 0$ such that $f$ has no zero in the spherical shell $S(\varepsilon) = \{z \in \mathbb{C}^n : R - \varepsilon < \lVert z\rVert < R\}$.

On $S(\varepsilon)$ we consider the holomorphic function $g \colon z \mapsto 1/f(z)$. By Hartogs' Kugelsatz (Hartogs's extension theorem), $g$ has a holomorphic extension $\tilde{g}$ to the ball $B$. By the identity theorem we have $f(z)\tilde{g}(z) = 1$ for all $z \in B$, which implies that $f$ has no zero in $B$.

While the full Kugelsatz (Theorem 2.1 in §2 of chapter IV in Holomorphic Functions and Integral Representations in Several Complex Variables by R. Michael Range)

Let $D$ be a bounded domain in $\mathbb{C}^n$ with connected boundary $\mathrm{b}D$. Assume that $n > 1$. Then every $f \in \mathscr{O}(\mathrm{b}D)$ can be extended to a holomorphic function on $\overline{D}$, i.e., there is $F \in \mathscr{O}(\overline{D})$, such that $F = f$ on $\mathrm{b}D$.

requires quite a bit of theory to prove (it would be applied with $D$ the ball of radius $R - \varepsilon/2$), here a simpler version suffices. Not quite as simple as the version for the difference between two polydisks, but simple enough, it's Theorem 1.6 in chapter II of the aforementioned work:

Let $D$ be a connected Reinhardt domain with center $0$, and suppose that $D \cap \{ z \in \mathbb{C}^n : z_l = 0\} \neq \varnothing$ for all $l = 1,2,\dotsc, n$ (this holds, in particular, if $0 \in D$). Then every $f \in \mathscr{O}(D)$ has a convergent power series expansion on $D$. Moreover, this power series defines a holomorphic extension of $f$ to the smallest complete Reinhardt domain $\bigcup_{w \in D} P(0,\tau(w))$ which contains $D$.

Since $S(\varepsilon)$ is a connected Reinhardt domain, and it contains points with $z_l = 0$ for all $l$, we can apply this theorem to $g$ and $S(\varepsilon)$. The smallest complete Reinhardt domain containing $S(\varepsilon)$ is $B$.

Answer 2

Another simple argument is the following: The set $\{ z \in \bar{B} : f(z)=0 \}$ is closed. Suppose that it does not hit the boundary, so there is some point in the interior that is furthest from the origin. You can always rotate and suppose that this is the point $(r,0,0,\ldots,0)$ for some real $r < R$. Consider the one variable functions $g_k(\xi) = f(r+\frac{1}{k},\xi,0,0,\ldots,0)$ and $g(\xi) = f(r,\xi,0,0,\ldots,0)$. These are defined in some small disk around zero, and $g_k \to g$ uniformly on a small disk. $g_k$ are zero-free, while $g$ has an isolated zero at the origin, which is a contradiction to the Hurwitz theorem.

You can generalize this sort of geometry. Basically we're touching the zero set by a sequence of analytic discs $\Delta_k$ where only the limiting disc $\Delta$ touches the zero set. Hurwitz says that's not possible. If the limiting disc touches the zero set, then for large enough $k$ all the discs have to touch it.

A more general result of this sort is this: Suppose $X \subset \Omega \subset {\mathbb{C}}^n$ ($\Omega$ an open set) is the intersection of the zero sets of $\ell$ holomorphic functions defined on $\Omega$ and $X$ is compact, then 1) $\ell \geq n$ 2) $X$ consists of finitely many points.