# Vector-Valued Stone-Weierstrass Theorem?

MathOverflow Asked by mw19930312 on September 26, 2020

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., $$Xin mathbb{R}^d$$. I wonder whether this theorem can be extended to vector-valued continuous functions. Specifically, let $$mathcal{A}$$ be a subalgebra of continuous functions $$Xto mathbb{R}^n$$, with the multiplication defined componentwisely, i.e., $$forall f, gin mathcal{A}$$, $$fg = (f_1g_1, ldots, f_ng_n)$$. 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 $$xmapsto (f(x), 0, ldots, 0)$$ 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’?

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.

Answered by bathalf15320 on September 26, 2020

I think that you want something like this:

Let $$Eto X$$ 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 $$Eto X$$. Suppose that, at every point $$xin X$$, the set $${,e(x) | einmathcal{E} }$$ 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 $$xin X$$. Then $$mathcal{E}$$ contains all the sections of the form $$a_1, e_1 + cdots + a_m,e_m$$ where $$a_iinmathcal{A}$$, and every section $$ein C(X,E)$$ can be written in the form $$e = f_1, e_1 + cdots + f_m,e_m$$ for some functions $$f_iin C(X,mathbb{R})$$. By the Stone-Weierstrass Theorem, for any given $$delta>0$$, we can choose $$a_iin mathcal{A}$$ so that $$|f_i-a_i| for all $$1le ile 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$$.

Answered by Robert Bryant on September 26, 2020

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