Let $X$ be any topological space and denote by $\tau\_X$ the topology on $C\_b(X;\mathbb{R})$ that is induced by the family of seminorms $(\|\cdot\|\_\psi\mid\psi\in B\_0(X))$ with $\|f\|\_\psi:=\sup\_{x\in X}|f(x)\psi(x)|$ and


$$\tag{1}B\_0(X):=\{\psi:X\rightarrow\mathbb{R} \text{ bounded}\mid \forall\,\varepsilon>0\, : \,\exists\, K\subset X \text{ compact} \ : \ \sup\nolimits\_{x\in X\setminus K}|\psi(x)|<\varepsilon\}.$$


A result of [[Giles](https://www.ams.org/journals/tran/1971-161-00/S0002-9947-1971-0282206-4/S0002-9947-1971-0282206-4.pdf), Thm. 3.1] asserts that if $\mathfrak{A}$ is a point-separating and pointwise non-vanishing subalgebra of $C\_b(X;\mathbb{R})$, then $\mathfrak{A}$ is dense in $C\_b(X;\mathbb{R})$ wrt. $\tau\_X$.


**Question:** Given $c : X\rightarrow \mathbb{R}$ continuous, can we infer that also
$$\tag{2}\mathfrak{A}\_c:=\{p\in\mathfrak{A} \mid p\leq c\} \quad\text{is $\tau\_X$-dense in} \quad C\_{b\,|\,c}(X):=\{f\in C\_b(X;\mathbb{R})\mid f\leq c\} \ ?$$


(The above inequalities are understood to hold pointwise on $X$.) More regularity on $c$ can be assumed if necessary.



Does global boundedness ruin Stone-Weierstrass denseness?
If $c\notin\mathfrak{A}$ in general it is not true: e.g.: consider the case where $X$ is the real line , $\mathfrak{A}$ is the algebra of the polynomials, and $c$ is the function $-e^x$. Then $\mathfrak{A}\_c$ is empty.


If $\mathfrak{A}$ is also a lattice, it is true:
Note that for any $u,v\in C\_b(X;\mathbb{R})$ one has $|u\wedge c-v\wedge c|\le |u-v|$ point-wise. Therefore $\|u\wedge c-v\wedge c\|\_\psi\le \|u-v\|\_\psi
$ for any weight $\psi$, and $F:u\mapsto u\wedge c$ is a continuous self-map on $\big(C\_b(X;\mathbb{R}), \tau\_X\big)$. So $ C\_{b|c}(X) =F(C\_b(X;\mathbb{R}))=F(\overline{\mathfrak{A}}) \subset \overline{F (\mathfrak{A})}= \overline{ \mathfrak{A}\_c}$.


