Real-analytic function of two complex variables, holomorphic in first and anti-holo in second, which vanishes on the diagonal is identically zero.

Question

The following theorem is stated as being a well-known result of the theory of several complex variables in a book I am reading (on a more or less unrelated subject):

Let $f:\mathbb C^2\to\mathbb C$ be a real-analytic function such that $f$ is holomorphic in the first variable and anti-holomorphic in the second variable. If $f(z,z) = 0$ for all $z\in\mathbb C$, then $f=0$ identically.

1) Can somebody point me to a reference where this is proved, or provide a proof that does not use extensive machinery of several complex variables?

2) What is the necessity of specifying real-analyticity? Is it false that holomorphicity and anti-holomorphicitiy respectively in the two variables implies real-analyticity?

Answer 1

Let's start with part 2). It is indeed redundant to explicitly demand that $f$ be real-analytic. That is however a nontrivial theorem. Hartogs proved that if a function (of several complex variables) is separately holomorphic, then the function is holomorphic. Here, applying that theorem to $g\colon (z,w) \mapsto f(z,\overline{w})$ yields the holomorphicity of $g$ and then the real-analyticity of $f$, as a composition of real-analytic functions.

For part 1), we also use the function $g$. The premise that $f$ vanish on the diagonal means that $g$ vanishes on the conjugate-diagonal $\{(z,\overline{z}) : z \in \mathbb{C}\}$. A modicum of several-variable theory tells us that if $h$ is a holomorphic function on an open subset $U$ of $\mathbb{C}^n$, and there is a $z_0 \in U$ with $h(z_0) = 0$ and $\operatorname{grad} h(z_0) \neq 0$, then the zero set of $h$ is in a neighbourhood of $z_0$ a complex submanifold of dimension $n-1$. Since the conjugate-diagonal is nowhere a complex submanifold of $\mathbb{C}^2$, it follows that $\partial_k g$ vanishes identically there. Iterating the reasoning, it follows that all partial derivatives of $g$ vanish on the conjugate-diagonal, and hence the Taylor series of $g$ with centre $0$ is the zero series, i.e. $g \equiv 0$.

Answer 2

Assuming $f$ is real-analytic leads to a very simple proof; I wasn't aware of this result and know more or less nothing about several complex variables, but I found a proof more or less automatically:

At least in some neighborhood of the origin we have $$f(z,w)=\sum_{n,m\ge0}a_{n,m}z^n\overline{w}^m.$$ So for $r>0$ small enough and $t\in\Bbb R$ we have $$0=\sum_{n,m\ge0}a_{n,m}r^{n+m}e^{i(n-m)t}.$$Uniqueness for Fourier series shows that for small enough $r>0$ and every $k\in\Bbb Z$ we have $$\sum_{n-m=k}a_{n,m}r^{n+m}=0.$$

BUT given $N\ge0$ there exists at most one pair $(n,m)$ with $n-m=k$ and $n+m=N$. Which is to say that the coefficient of $r^{n+m}$ in that last series really is $a_{n,m}$, as it would appear. So one complex variable shows $$a_{n,m}=0.$$