Let us assume that we are given two polynomials $f,g$ with real coefficients in several variables, say $x_1, \ldots, x_n \in \mathbb{R}$. Further, assume that $f_{|X} \equiv g_{|X}$, with $X$ being the open ball around some $a \in \mathbb{R}^n$, i.e. $X = B_\delta(a)$.
Question: Does this imply $f \equiv g$?
It is clear to me that this holds in the case of polynomials in one variable, i.e. $n = 1$.
According to Wikipedia the identity theorem of complex analysis carries over to several variables only in the case of special sets $X$. Unfortunately, I was not able to find a book which discusses this in detail. And even this applies, it is not clear to me whether this would yield the statement for the real domain.
Any piece of information would be appreciated.
Best wishes, Max
Use induction. Fix $(x_1,\dots,x_{n-1})\in B_\delta(a_1,\dots,a_{n-1})$. By the uniqueness theorem for holomorphic functons of one variable, you obtain $f(x_1,\dots,x_{n-1},z_n)=g(x_1,\dots,x_{n-1},z_n)$ for all $z_n\in\mathbb{C}$ and $(x_1,\dots,x_{n-1})\in B_\delta(a_1,\dots,a_{n-1})$. Then proceed to $x_{n-1}$, etc.
Alternatively: assume without loss in generality that $a=0$. Let $f=\sum a_\alpha x^\alpha$, $\alpha=(\alpha_1,\dots,\alpha_n)$ being a multiindex. Then $$ a_\alpha=(\alpha_1!\dots \alpha_n!)^{-1}\frac{\partial^{|\alpha|}f}{\partial x^\alpha}(0), $$ that is, all coefficients of a polynomial are determined by its values in an arbitrarily small ball.