Suppose that we have two functions of several complex variables that are holomorphic on the whole euclidean space. If these two functions are equal on an open and connected subset of the REAL euclidean space, can we say that they are equal everywhere? This is true for holomorphic functions of one complex variable, but does it hold for several variables holomorphic functions?
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+1. The open subset of $\mathbb R^n$ needn't be connected but it should be non-empty :-) – Georges Elencwajg Jul 07 '13 at 17:10
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Yes, it does hold. You can argue componentwise. I assume, without loss of generality, that the second function is $\equiv 0$, and the open subset of $\mathbb{R}^n$ on which $f$ vanishes is a neighbourhood of $0$.
For $x_2,\,\ldots,\, x_n \in \mathbb{R}$ close enough to $0$, consider
$$ h(z) = f(z,\,x_2,\,\ldots,\, x_n).$$
By the identity theorem (in one variable), $h \equiv 0$. So you know $f(z_1,\, z_2,\, \ldots,\, z_n)$ is $0$ whenever the last $n-1$ arguments are real and of small enough norm.
Then for arbitrary $z_1 \in \mathbb{C}$ and $x_3,\,\ldots,\,x_n \in \mathbb{R}$ of small enough norm, consider
$$k(w) = f(z_1,\,w,\,x_3,\,\ldots,\,x_n).$$
By the identity theorem, $k \equiv 0$.
Thus $f(z_1,\, z_2,\, \ldots,\, z_n) = 0$ whenever the last $n-2$ arguments are real and of small enough norm.
Repeat until the last argument.
Daniel Fischer
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Thank you very much! I think it should also imply the following fact: suppose we have two functions of several real variables such that they coincide on an open connected subset. If these two functions admit an olomorphic extension on the complex euclidean space, then they coincide everywhere, don't they? – Lav Jul 07 '13 at 15:00
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Yes. Maybe I should have better started with the local argument. The same argument, applied locally, shows that the two functions coincide on a nonempty open subset of the complex space. Then the identity theorem shows they coincide on the connected component of the intersection of their domains that contains that open set. If both functions are entire, that means all of $\mathbb{C}^n$. – Daniel Fischer Jul 07 '13 at 15:07
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We may assume that the second function is identically zero.
The first function is entire on $\mathbb C^n$, and thus has a series expansion of the form $$f(z)=\sum c_Iz^I=\sum a_Iz^I+i\sum b_Iz^I$$ (where $I\in \mathbb N^n$ is a multi-index, $c_I=a_I+ib_I$ and $a_I\in \mathbb R, b_I\in \mathbb R$).
Assuming that the open subset of $\mathbb R^n$ contains zero, we have $f(r)=\sum c_I r^I=\sum a_I r^I+i\sum b_I r^I=0$ for all sufficiently small $r=(r_1,\cdots,r_n)\in \mathbb R^n$.
This implies $\sum a_I r^I=0,\sum b_I r^I=0$ which forces all $a_I,b_I$ to be zero.
Thus all $c_I=a_I+ib_I=0$ and $f=0$.
Georges Elencwajg
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