$G$-invariant holomorphic vs. polynomial functions 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 $p(x_0)\neq 0$?
 A: Lemma (Dolgachev's Lectures on Invariant Theory). Let $X$ be an affine $G$-variety, and let $Z_1$ and $Z_2$ be two closed $G$-invariant
subsets with $Z_1\cap Z_2 = \emptyset$. Assume $G$ is geometrically reductive. Then there exists a $G$-invariant function $\varphi \in \mathcal{O}(X)^G$ such that $\varphi(Z_1) = 0$, $\varphi(Z_2) = 1$.
If $X\not=\overline{Gx_0}$, then letting $Z_2=\overline{Gx_0}$ and $Z_1$ any closed orbit in $X-\overline{Gx_0}$ shows that non-constant $G$-invariant polynomials that are not 0 at $x_0$ always exist.  Since you assumed $X$ is smooth, these will be holomorphic too.
A: 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
