TITLE: $G$-invariant holomorphic vs. polynomial functions
QUESTION [6 upvotes]: Let $X\subseteq\Bbb C^n$ be a smooth affine variety and $G$ a complex reductive group acting algebraically on $X$. 
Let $x_0\in X$. If there is a non-constant $G$-invariant holomorphic function $f:X\to\Bbb C$ such that $f(x_0)\neq 0$, must there also exist a non-constant $G$-invariant polynomial $p\in\Bbb C[X]$ such that $p(x_0)\neq 0$?

REPLY [4 votes]: Luna has proved that every holomorphic invariant is of the form $h(p_1,\ldots,p_n)$ where $h$ is a holomorphic function in $n$ variables and $p_1,\ldots,p_n$ are polynomial invariants. By the way, Luna proves the same with "holomorphic" replaced by "continuous". The statement is wrong for real differentiable or real analytic functions, though: Take $G=\{\pm1\}$ and $X=\mathbb C$. Then the ring of invariants is generated by $z^2$ and $f(z):=z\overline z=\|z^2\|$ is a counterexample.
Reference: Luna, Domingo, Fonctions différentiables invariantes sous l'opération d'un groupe réductif. Ann. Inst. Fourier (Grenoble) 26 (1976), no. 1, ix, 33–49