# Stone-Weierstrass theorem for modules of non-self-adjoint subalgebras

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 $$a\in \mathcal{A}$$.

• 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 – Yemon Choi Oct 17 at 21:26
• @YemonChoi I just felt there was no canonical answer and the post seemed to stop drawing attention. – potionowner Oct 17 at 21:30
• 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 – Yemon Choi Oct 17 at 21:40
• Regarding your current question, if $S$ is not closed under addition then I don't see how you know $W$ is a vector space... – Yemon Choi Oct 17 at 21:41
• 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 – Yemon Choi Oct 17 at 21:46

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 $$\widehat{g}(-1)=1$$.

• 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 $[t]_{\mathcal{A}}$? – potionowner Oct 21 at 18:32
• 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? – potionowner Oct 21 at 18:34
• @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 ${\mathcal A}$ – Yemon Choi Oct 21 at 18:38
• 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. – Yemon Choi Oct 21 at 18:40
• 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! – potionowner Oct 22 at 17:06