The standard statement of the Stone-Weierstrass theorem is:

Let $X$ be compact Hausdorff topological space, and $\mathcal{A}$ a subalgebra of the continuous functions from $X$ to $\mathbb{R}$ which separates points. Then $\mathcal{A}$ is dense in $C(X, \mathbb{R})$ in sup-norm.

Most materials that I can find on the extension of Stone-Weierstrass theorem discuss only the multivariate case, i.e., $X\in \mathbb{R}^d$. I wonder whether this theorem can be extended to vector-valued continuous functions. Specifically, let $\mathcal{A}$ be a subalgebra of continuous functions $X\to \mathbb{R}^n$, with the multiplication defined componentwisely, i.e., $\forall f, g\in \mathcal{A}$, $fg = (f_1g_1, \ldots, f_ng_n)$. Then shall we claim $\mathcal{A}$ is dense in $C(X, \mathbb{R}^n)$ in sup-norm if $\mathcal{A}$ separates points?

Any direct answer or reference would greatly help me!

Edit: As Nik Weaver points out, the original conjecture is false since the functions of the form $x\mapsto (f(x), 0, \ldots, 0)$ create a counter-example. I wonder whether there are non-trivial Weierstrass-type theorems on vector-valued functions. For instance, what if we further assume $\mathcal{A}$ is dense on each `axis'?

  • 3
    $\begingroup$ What is the multiplication in an "algebra of vector valued functions"? Or maybe you mean a structure of $C(X,\mathbb{R})$-module? $\endgroup$ – Pietro Majer Sep 13 '20 at 19:23
  • 1
    $\begingroup$ The statement is trivially false: the functions of the form $x \mapsto (f(x),0,\ldots, 0)$ with $f \in C(X)$ are sup-norm closed and separate points. $\endgroup$ – Nik Weaver Sep 13 '20 at 19:46
  • $\begingroup$ @NikWeaver Thanks for the counterexample here! I've added a short comment in the post. $\endgroup$ – mw19930312 Sep 13 '20 at 20:12
  • $\begingroup$ @PietroMajer The componentwise multiplication induces an "algebra of vector valued functions", doesn't it? $\endgroup$ – mw19930312 Sep 13 '20 at 20:13
  • 2
    $\begingroup$ I see, thanks! Then it is like the algebra $C(X\times\{1,\dots,n\},\mathbb{R})$? $\endgroup$ – Pietro Majer Sep 13 '20 at 20:25

I think that you want something like this:

Let $E\to X$ be a (finite rank) vector bundle over a compact, Hausdorff topological space $X$, let $\mathcal{A}\subset C(X,\mathbb{R})$ be a subalgebra that separates points, and let $\mathcal{E}\subset C(X,E)$ be an $\mathcal{A}$-submodule of the $C(X,\mathbb{R})$-module of continuous section of $E\to X$. Suppose that, at every point $x\in X$, the set $\{\,e(x)\ |\ e\in\mathcal{E}\ \}$ spans $E_x$. Then $\mathcal{E}$ is dense in $C(X,E)$ with respect to the sup-norm defined by any norm on $E$.

Addendum: Here is a sketch of the argument: First, by an easy compactness argument, one can show that $\mathcal{E}$ contains a finite set $e_1,\ldots e_m$ such that $e_1(x),e_2(x),\ldots,e_m(x)$ spans $E_x$ for all $x\in X$. Then $\mathcal{E}$ contains all the sections of the form $$a_1\, e_1 + \cdots + a_m\,e_m$$ where $a_i\in\mathcal{A}$, and every section $e\in C(X,E)$ can be written in the form $$e = f_1\, e_1 + \cdots + f_m\,e_m$$ for some functions $f_i\in C(X,\mathbb{R})$. By the Stone-Weierstrass Theorem, for any given $\delta>0$, we can choose $a_i\in \mathcal{A}$ so that $\|f_i-a_i\|<\delta$ for all $1\le i\le m$. Now the equivalence of all norms in finite dimensional vector spaces can be applied (together with the compactness of $X$) to conclude that $\mathcal{E}$ is dense in $C(X,E)$ in any sup-norm derived from a norm on the (finite rank) vector bundle $E$.

  • $\begingroup$ Thanks for the addendum! While I'm digesting this, I have a quick question here. Does this result hold for matrix-valued subalgebras, i.e., $\mathcal{A} \subset C(X, \mathbb{R}^{n\times n})$ equipped with matrix multiplication? $\endgroup$ – mw19930312 Sep 15 '20 at 16:38
  • $\begingroup$ @mw19930312: I'm confused by your question. Could you state what you mean by 'this result' more precisely? $\endgroup$ – Robert Bryant Sep 15 '20 at 19:49
  • $\begingroup$ @mw19930312: I'm sorry, but I don't know what you mean by assuming that $E\subset \mathbb{R}^n$. In my answer, $E\to X$ is a vector bundle over $X$ and $\mathcal{A}$ is a subset of $C(X,\mathbb{R})$. Maybe you mean something like "If $\mathcal{A}\subset C(X,\mathrm{End}(E))$ is an algebra of bundle maps from $E$ to $E$ and $\mathcal{E}\subset C(X,E)$ is a space of sections that is an $\mathcal{A}$-module, then under what conditions can we conclude that $\mathcal{E}$ is dense in $C(X,E)$ in the sup-norm induced by any norm on $E$"? $\endgroup$ – Robert Bryant Sep 15 '20 at 21:27
  • $\begingroup$ Hi Robert, sorry for coming back to this question after it is posted for a month. I believe you are interpreting this correctly, I'm looking for the case of $\mathcal{A}\subset C(X, \mathrm{End}(E))$. If the same result holds, would you mind referring me a detailed proof of this? $\endgroup$ – mw19930312 Oct 17 '20 at 16:20
  • $\begingroup$ @mw19930312: That's OK. What I don't know is what you want to assume about $\mathcal{A}\subset C(X,\mathrm{End}(E))$. Is it a subalgebra? Are you asking for checkable conditions that would guarantee that $\mathcal{A}$ is $C^0$-dense in $C(X,\mathrm{End}(E))$? $\endgroup$ – Robert Bryant Oct 17 '20 at 18:58

This is a comment, not an answer but I am, alas, not entitled. Vector valued Stone-Weierstraß theorems were studied in great detail in the second half of the last century and there is a comprehensive monograph on the subject by João Prolla ("Weierstraß-Stone, the theorem", 1993). Not on topic, but he also considered the case of bounded, continuous vector-valued functions on non-compact spaces, using the strict topology of R.C. Buck.

  • $\begingroup$ Great thanks for the references here! By any chance, could you refer me to some other materials on this topic? The book by João Prollaa appears very difficult for me to find. I've got a note by him named `Approximation of Vector-Valued Functions' in 1977. Is there any recent update on this notes? $\endgroup$ – mw19930312 Sep 14 '20 at 15:43
  • 2
    $\begingroup$ The book title written by Yoda has been. $\endgroup$ – Johannes Hahn Sep 15 '20 at 21:27
  • $\begingroup$ @JohannesHahn Sorry, your comment somehow seems incomplete... $\endgroup$ – mw19930312 Sep 15 '20 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.