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.

  1. 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: $$ \text{supp}f=\{x\in M:\ f\not\equiv 0\ (\text{mod}\ x)\}. $$

  2. 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 $x$: $$ \forall f,g\in{\mathcal C}^\infty(M)\quad \forall x\in M\qquad f\equiv g\ (\text{mod}\ x)\quad\Longrightarrow\quad Df(x)=Dg(x). $$

  3. 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: $$ f_n\overset{{\mathcal C}^\infty(M)}{\underset{n\to\infty}{\longrightarrow}}f \quad\Longleftrightarrow\quad \forall k\ \forall X_1,...,X_k\in{\mathcal X}(M) \quad X_1...X_kf_n\overset{{\mathcal C}(M)}{\underset{n\to\infty}{\longrightarrow}}X_1...X_kf. $$

  • $\begingroup$ Can you formulate Runge's theorem in general terms of the algebra of rational functions with enough poles (one in each component of the complement)? $\endgroup$ Dec 9, 2015 at 14:12
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    $\begingroup$ Runge's theorem is about a concrete subalgebra $A$, the algebra of rational functions, but I am asking about an arbitrary subalgebra $A\subseteq{\mathcal O}(M)$. I don't know, maybe there were generalizations... $\endgroup$ Dec 9, 2015 at 14:19
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    $\begingroup$ If you are looking for an abstract condition it should apply to the concrete situation. I believe it is thus a good idea to try to formulate the concrete situation in general abstract terms. $\endgroup$ Dec 9, 2015 at 15:09
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    $\begingroup$ Does someone know how to deal with the case of a simply connected open subset of $\mathbb{C}$ ? or basically with the unit disc in fact by Riemann's theorem. Because of Runge's theorem this sugest that this is the simplest case and that bounded connected component in the complement will create additional difficulties... $\endgroup$ Dec 10, 2015 at 15:13
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    $\begingroup$ @AlexM. I edited. There is no need to take into account the stereotype theory, when you define topology on ${\mathcal C}^\infty(M)$. This is equivalent to your condition with vector fields $X_i$. $\endgroup$ Feb 2, 2016 at 21:10

1 Answer 1



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 $M = \CC^2$.

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    $\begingroup$ David, your example is interesting, but I think, you did not understand my question. First, I have no hypotheses on what the answer should be. Second, separating points does not mean the equality of spectra. For example, from the Stone-Weierstrass theorem it follows that the algebra $A$ of almost periodic functions is dense in ${\mathcal C}({\mathbb R})$, but $\text{Spec}(A)\cong\text{Bohr compactification of} \ {\mathbb R}\ne{\mathbb R}\cong\text{Spec}({\mathcal C}({\mathbb R}))$ (if $\text{Spec}$ is defined somewhat reasonably, say, if it means continuous involutive characters). $\endgroup$ Dec 10, 2015 at 20:13
  • $\begingroup$ I deleted remarks about Spec since I decided I didn't know what I was talking about. But would you agree that this suggests that there is no simple answer for $M = \mathbb{C}^2$? $\endgroup$ Dec 10, 2015 at 20:20
  • $\begingroup$ David, actually, I would not expect that the criteria for ${\mathcal O}(M)$ and for ${\mathcal C}^\infty(M)$ will be similar. I would suppose that separating points will be their common part, but separating tangent vectors sounds strange for me. $\endgroup$ Dec 10, 2015 at 20:25
  • $\begingroup$ Separating tangent vectors (meaning that $\{ df \}_{f \in A}$ spans the space of complex linear functionals on the tangent space) is certainly necessary. Note that we can write $f'(0)$ as $\oint f(z) dz/z^2$, so this is a closed condition. $\endgroup$ Dec 10, 2015 at 20:33
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    $\begingroup$ I think @DavidESpeyer's candidate counterexample for $\mathbb C$ does not work and $e^{-z}$ is in the closure of the subalgebra generated by $e^z,e^{\sqrt 2 z}$ - otherwise there is a compactly-supported complex measure $\mu$ such that $f(u):=\int_{\mathbb C} e^{uz} d\mu(z)$ has zeroes at $\mathbb N+\sqrt 2 \mathbb N$ but not at $-1$. But $f$ has order $\leq 1$ and the size of $\left(\mathbb N+\sqrt 2 \mathbb N\right)\cap B(0,r)$ grows quadratically as $r\to\infty$, which contradicts Jensen's formula $\endgroup$
    – user49822
    Aug 16, 2021 at 8:54

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