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In "Weierstrass-Stone, the Theorem" by Joao Prolla, there is a Stone-Weierstrass theorem for modules, stated as the following:

Let $\mathcal{A}$ be a subalegebra of $C(X, \mathbb{R})$ and $(E, \|\cdot\|)$ be a normed space over $\mathbb{R}$. Let $W\subset C(X, E)$ be a vector subspace which is an $\mathcal{A}$-module. For each $f\in C(X, E)$ and $\epsilon>0$, there exists $g\in W$ such that $\|f-g\|<\epsilon$ if and only if for each $x\in X$, there exists $g_x\in W$ such that $\|f(t) - g_x(t)\| < \epsilon$ for all $t\in [x]_{\mathcal{A}}$, where $[x]_\mathcal{A}$ is the equivalent class of $x$ under $\mathcal{A}$.

I know that the above theorem can be extended to $\mathcal{A}\subset C(X, \mathbb{C})$ with $\mathcal{A}$ being a self-adjoint subalgebra. I wonder whether there are some similar results for modules of non-self-adjoint algebras.

I'm interested in generalizing the above theorem into the following case. Let $\mathcal{S}$ be a finite subset of $C([0, 1], E)$, denoted as $S:=\{s_1, \ldots, s_m\}$, and $\mathcal{A}\subset C([0, 1], \mathbb{C})$ be a subalgebra (not necessarily self-adjoint). Then $W := \mathrm{span}\{as : a\in \mathcal{A}, s\in \mathcal{S}\}$ is a vector subspace which is an $\mathcal{A}$-module. Shall we still claim that $f\in \overline{W}$ if and only if $f\big\vert_{[x]_{\mathcal{A}}} \in \overline{W}\big\vert_{[x]_{\mathcal{A}}}$? Is there any counter-example to this statement? Or is it an open problem in general?

Note: For any $x\in X$, the equivalent class $[x]_{\mathcal{A}}$ is a subset of $X$ such that $\forall u, v\in [x]_{\mathcal{A}}$, we have $a(u) = a(v)$ for all $a\in \mathcal{A}$.

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  • $\begingroup$ Sorry to derail, but why did you delete your previous question? It might have been of interest to future readers, and someone might have answered it at a later date $\endgroup$ – Yemon Choi Oct 17 at 21:26
  • $\begingroup$ @YemonChoi I just felt there was no canonical answer and the post seemed to stop drawing attention. $\endgroup$ – potionowner Oct 17 at 21:30
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    $\begingroup$ It was only up for less than a month! Sometimes on MathOverflow people may want ttime o think about questions; sometimes people notice an old question and know how to make progress. That said, it is of course up to you $\endgroup$ – Yemon Choi Oct 17 at 21:40
  • $\begingroup$ Regarding your current question, if $S$ is not closed under addition then I don't see how you know $W$ is a vector space... $\endgroup$ – Yemon Choi Oct 17 at 21:41
  • $\begingroup$ I admit I don't quite follow all of the technical definitions in your setup, but for non-self-adjoint subalgebras of C(X) (complex scalars) one usually doesn't get Stone-Weierstrass; the natural place to look for a counterexample is the disc algebra $A({\bf D})$ ,which can be viewed as a subalgebra of $C({\bf T})$ since a function $h\in A({\bf D})$ is uniquely determined by its boundary values $\endgroup$ – Yemon Choi Oct 17 at 21:46
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If I have understood the definitions correctly, then the answer is still negative, because one can transfer the "disc algebra counterexample" over to $[0,1]$.

In what follows I shall write $C[0,1]$ rather than $C([0,1];{\mathbb C})$, just as a convenient shorthand. $\newcommand{\cA}{{\mathcal A}}$ $\newcommand{\cB}{{\mathcal B}}$ $\newcommand{\cS}{{\mathcal S}}$

Let $\cB=\{ f\in C[0,1] \colon f(0)=f(1)\}$. For $f\in \cB$ and $n\in \mathbb Z$ let $$ \widehat{f}(n)= \int_0^1 f(t) e^{-2\pi in t}\,dt $$ (This is the $n$th Fourier coefficient of $f$, if we identify functions in $\cB$ with continuous complex-valued functions on the unit circle in the natural way.) Now let $\cA=\{ f\in \cB \colon \widehat{f}(n)=0\,\forall\,n < 0 \}$. This is a closed subalgebra of $\cB$ and hence a closed subalgebra of $C[0,1]$.

Taking $\cS=\{ {\bf 1} \}$, we have $W=\overline{W}=\cA$.

The equivalence relation on $X=[0,1]$ defined by $\cA$ has the following explicit description: $0\sim_{\cA} 1$; and all other equivalence classes are singletons. This last claim follows by considering the function $t\mapsto e^{2\pi it}$.

In particular, the function $g(t)=e^{-2\pi it}$ belongs to $\cB$ and for every $t\in [0,1]$ we can find $f\in \cA$ such that $f$ agrees with $g$ on $[t]_{\cA}$. On the other hand, it does not belong to $\cA$, since $\widehat{g}(-1)=1$.

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  • $\begingroup$ Thanks for the answer here! I have two questions: (1) How shall we argue that $\mathcal{A}$ is a subalgebra? (2) How can we find $f\in \mathcal{A}$ agrees with $g$ on $[t]_{\mathcal{A}}$? $\endgroup$ – potionowner Oct 21 at 18:32
  • $\begingroup$ Also, just out of curiosity, $\mathcal{A}$ is not finitely generated, right? Does the statement of OP become true if we restrict $\mathcal{A}$ to be a finitely generated subalgebra? $\endgroup$ – potionowner Oct 21 at 18:34
  • $\begingroup$ @potionowner When you say finitely generated, do you mean "topologically finitely generated"? In any case, $\mathcal A$ is the closed linear span of $\{ f_n: n\in {\mathbb Z}, n\geq 0 \}$ where $f_n(t)=e^{2\pi in t}$ -- this follows from basic facts in Fourier analysis. So if you want you can restrict yourself to the incomplete algebra generated by $f_0$ and $f_1$, which is dense in ${\mathcal A}$ $\endgroup$ – Yemon Choi Oct 21 at 18:38
  • $\begingroup$ Regarding your first question: I already observed that the equivalence clases for $\sim_{\mathcal A}$ are precisely the singletons $\{t\}$ for each $0< t<1$, together with the equivalence class $\{0,1\}$. You should be able to work out for yourself how to find, for each equivalence class, a function in ${\mathcal A}$ which agrees with $g$ on that equivalence class; this is fairly basic. $\endgroup$ – Yemon Choi Oct 21 at 18:40
  • $\begingroup$ For the second question, I mean $\mathcal{A}$ is a subalgebra generated by finite elements. The simplest case would be $\mathcal{A} = \mathrm{span}\{a^k, k = 0, 1, \ldots\}$ for some $a\in C([0, 1], \mathbb{C})$. It seems that the example in your comment, say $\mathcal{A} = \mathrm{span} \{f_n\}$ is finitely generated with $\mathcal{A} = \{(e^{2\pi i t)}^{k}, k = 0, 1, \}$, right? For the first question, I've figured it out. Somehow I forgot that the equivalent classes are singletons. Anyway, thanks again for the discussion here! $\endgroup$ – potionowner Oct 22 at 17:06

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