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The Stone–Weierstrass theorem gives an easy to check criterion on a (algebra) set of functions $D\subseteq C(X)$ which ensures that $D$ is dense.

Are any similar results for density in $C_b(X)$ when it is equipped with the topology of uniform convergence? Here, I'm now assuming that $X$ is a Euclidean space.

Edit: Due to the comments this type of result seems unlikely. However, I'm tempted to ask what are such criteria on $C^1(X)$ or $C^{\infty}(X)$; assuming that $X$ is a compact sub-manifold with boundary of $\mathbb{R}^K$, for some finite $K$. Here we take the completion of the metric topology $$ d(f,g) := \sum_{n =0}^k \frac1{2^{n+A(n)}} \sum_{|\alpha|\leq n}\frac{ \sup_{x \in X} \frac{\partial}{\partial x^{\alpha}} \|f(x)-g(x)\| }{ 1+ \sup_{x \in X} \frac{\partial}{\partial x^{\alpha}} \|f(x)-g(x)\| } $$ as described by this nice post.

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    $\begingroup$ Do you mean $C_b(X)$ if $X$ is not compact? $\endgroup$
    – Dieter Kadelka
    Commented Apr 15, 2020 at 11:39
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    $\begingroup$ Many naive generalizations fail because $C_b(\mathbf{R}^n)$ for $n\ge 1$ is not separable. $\endgroup$
    – YCor
    Commented Apr 15, 2020 at 12:26
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    $\begingroup$ I'm no specialist but I doubt if there are easily practicable criteria. Let $\beta X$ be the Stone-Cech compactification of $X$. Then $C_b(X)$ and $C(\beta X)$ are almost indistinguishable. If you want to apply Weierstrass-Stone you have to work with $H \subset C(\beta X)$, which separates points of $\beta X$ (not of $X$ alone). And here is the problem. Are there "natural" candidates for $H$? We have similar problems even with $\ell_\infty$. $\endgroup$
    – Dieter Kadelka
    Commented Apr 15, 2020 at 12:58
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    $\begingroup$ There is a natural lc topology on $C^b(X)$ for which the Stone-Weierstraß theorem holds—the so-called strict topology. This was introduced by R.C. Buch in the 50’s for the special case of locally compact spaces and has several natural definitions. For example, it is the finest lc topology which coincides with that of compact convergence on the unit ball for the supremum norm. Another natural property is that the dual space is naturally identifiable with the space of bounded, radon measures. It can also be defined using weighted seminorms (as was done by Buck). $\endgroup$
    – user131781
    Commented Apr 15, 2020 at 13:30
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    $\begingroup$ @AnnieTheKatsu If you are referring to my comment, the answer is indeed “no”. The dual of your space consists of all radon measures, i.e., also unbounded ones. $\endgroup$
    – user131781
    Commented Apr 15, 2020 at 17:33

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