The Stone-Weierstrass theorem has an analog for the algebras of smooth functions, called
Naсhbin's theorem: An involutive subalgebra $A$ in the algebra ${\mathcal C}^\infty(M)$ of smooth functions on a smooth manifold $M$ is dense in ${\mathcal C}^\infty(M)$ if and only if $A$ separates the points and the tangent vectors of $M$.
See details in: "L.Nachbin. Sur les algèbres denses de fonctions diffèrentiables sur une variètè, C.R. Acad. Sci. Paris 228 (1949) 1549-1551", or in J.G.Llavona's monograph, or here.
This is strange, I can't find an analog for the algebras of holomorphic functions (on complex manifolds). Did anybody think about this?
Question: let $A$ be a subalgebra in the algebra ${\mathcal O}(M)$ of holomorphic functions on a complex manifold $M$ (as a first approximation, we can think that $M$ is just an open subset in ${\mathbb C}^n$). Which conditions should $A$ satisfy for being dense in ${\mathcal O}(M)$?
Remark. By topology on ${\mathcal O}(M)$ I mean the usual topology of uniform convergence on compact sets in $M$. The algebra ${\mathcal C}^\infty(M)$ is also endowed with its usual topology, which can be described, for example, as follows.
-
For each function $f\in {\mathcal C}^\infty(M)$ let us define its support as the closure of the set of the points where $f$ does not vanish:
$$
\text{supp}f=\overline{\{x\in M:\ f(x)\ne 0\}}.
$$
An equivalent definition: $\text{supp}f$ is the set of the points in $M$ where $f$ has non-zero germs:
$$
\text{supp}f=\{x\in M:\ f\not\equiv 0\ (\text{mod}\ x)\}.
$$ -
Let us define differential operators (see e.g. S.Helgason's book) on $M$ as linear mappings $D:{\mathcal C}^\infty(M)\to {\mathcal C}^\infty(M)$ which do not extend the support of functions:
$$
\text{supp}Df\subseteq \text{supp}f,\quad f\in{\mathcal C}^\infty(M).
$$
Equivalently, $D$ is local, i.e. the value of $Df$ in a point $x\in M$ depends only on the germ of $f$ in $x$:
$$
\forall f,g\in{\mathcal C}^\infty(M)\quad \forall x\in M\qquad f\equiv g\ (\text{mod}\ x)\quad\Longrightarrow\quad Df(x)=Dg(x).
$$ -
Then we say that a sequence of functions $f_n$ converges to a function $f$ in ${\mathcal C}^\infty(M)$
$$
f_n\overset{{\mathcal C}^\infty(M)}{\underset{n\to\infty}{\longrightarrow}}f
$$
if and only if for each differential operator $D:{\mathcal C}^\infty(M)\to {\mathcal C}^\infty(M)$ the sequence of functions $Df_n$ converges to $Df$ in the space ${\mathcal C}(M)$ of continuous functions with the usual topology of uniform convergence on compact sets in $M$:
$$
Df_n\overset{{\mathcal C}(M)}{\underset{n\to\infty}{\longrightarrow}}Df
$$
Of course, this is equivalent to the convergence in ${\mathcal C}^\infty(U)$ for each smooth local chart $\varphi:U\to V$, $U\subseteq\mathbb{R}^m$, $V\subseteq M$. This is also equivalent to what Alex M. writes about vector fields:
$$
f_n\overset{{\mathcal C}^\infty(M)}{\underset{n\to\infty}{\longrightarrow}}f
\quad\Longleftrightarrow\quad \forall k\ \forall X_1,…,X_k\in{\mathcal X}(M) \quad
X_1…X_kf_n\overset{{\mathcal C}(M)}{\underset{n\to\infty}{\longrightarrow}}X_1…X_kf.
$$
Best Answer
$\def\CC{\mathbb{C}}\def\cO{\mathcal{O}}$
Here is a candidate counterexample for $M= \CC$: Is $e^{-z}$ in the closure of the algebra generated by $e^z$ and $e^{\sqrt{2}z}$? My current guess is "no", but I need to move on to actual work.
I will show that separating points and separating tangents is not enough for $M = \CC^2$.
Let $A \subset \cO(\CC^2)$ be those holomorphic functions $f$ such that $f(z,z^{-1})$ extends holomorphically to $z=0$. We observe:
$A$ is a subalgebra: This is obvious.
$A$ is closed: Proof We have $f \in A$ if and only if $\oint f(z,z^{-1}) z^n dz=0$ for all $n \geq 0$, where the integral is on a circle around $0$. This fact is preserved by uniform limits on compact sets. (Specifically, by uniform limits on that circle.)
$A$ separates points: Note that the functions $f(x,y) = x$, $g(x,y) = xy$ and $h(x,y) = y (xy-1)$ are all in $A$. The functions $f$ and $g$ alone separate $(x_1, y_1)$ and $(x_2, y_2)$ unless $x_1=x_2=0$. In that case, $h$ separates them.
$A$ separates tangent vectors: Again, $df$ and $dg$ are linearly independent at all points where $x \neq 0$, and $df$ and $dh$ are linearly independent at $x=0$.
$A \neq \cO(\CC^2)$ Clearly, $y \not \in A$.
I remembered this counterexample from an old blog post of mine.
Note that we could replace $z \mapsto (z, z^{-1})$ with any map $\phi$ from the punctured disc $D^{\ast}$ to $\CC^2$. There are many such $\phi$'s, and they all appear to impose independent conditions. This makes me pessimistic about any simple criterion for equality when $M = \CC^2$.