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. The easiest way to describeBy 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, I thinkwhich can be described, the followingfor 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: $$ f_n\overset{{\mathcal C}^\infty(M)}{\underset{n\to\infty}{\longrightarrow}}f \quad\Longleftrightarrow\quad \forall k\ \forall X_1,...,X_k\in{\mathcal X}(M) \quad X_1...X_kf_n\overset{{\mathcal C}(M)}{\underset{n\to\infty}{\longrightarrow}}X_1...X_kf. $$
This is strangeOf course, I can't find an analog forthis is equivalent to the algebras of holomorphic functionsconvergence in (on complex manifolds)${\mathcal C}^\infty(U)$ for each smooth local chart $\varphi:U\to V$, $U\subseteq\mathbb{R}^m$, $V\subseteq M$. Did anybody thinkThis is also equivalent to what Alex M. writes about this?vector fields: $$ f_n\overset{{\mathcal C}^\infty(M)}{\underset{n\to\infty}{\longrightarrow}}f \quad\Longleftrightarrow\quad \forall k\ \forall X_1,...,X_k\in{\mathcal X}(M) \quad X_1...X_kf_n\overset{{\mathcal C}(M)}{\underset{n\to\infty}{\longrightarrow}}X_1...X_kf. $$
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 ${\mathcal O}(M)$?