Let $M$ be a non-compact manifold and denote by $C_b(M)$ the space of bounded continuous functions on $M$. Is it true that the space of Hölder functions is dense in $C_b(M)$ (in the $C^0$ norm: $||f||=\sup_{x\in M}|f(x)|$)?
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$\begingroup$ Presumably you don't want just an arbitrary manifold, but a Riemannian manifold in order to have a canonical metric w.r.t. which "Hölder continuous" is sensibly defined. $\endgroup$– Johannes HahnCommented Apr 10, 2018 at 18:23
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2$\begingroup$ What is Hölder continuity in the absence of a distance on $M$? Also: Hölder of every order, or of some arbitrary but fixed order? $\endgroup$– Alex M.Commented Apr 10, 2018 at 18:23
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2 Answers
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With $M = \mathbb{R}$, not even the uniformly continuous functions are dense in $C_b(\mathbb{R})$. As an example, take any bounded continuous functions with faster and faster oscillation, such as $x\mapsto \sin(x^2)$.
It's not hard to show that on any metric space---or more generally on any uniform space---the uniformly continuous functions are a $C^0$-closed subset of $C_b(M)$. So you can at most expect to be able to approximate these. I don't have access to the paper right now, but there is a version of the Stone-Weierstraß theorem due to Isbell that may give a useful sufficient criterion for when this actually happens.
answered Apr 10, 2018 at 19:01
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1$\begingroup$ About the Lipschitz approximation of uniformly continuous functions: there is a nice article on wikipedia: en.wikipedia.org/wiki/… $\endgroup$ Commented Apr 10, 2018 at 19:11
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It is not: for instance on the non-compact manifold $M:=\mathbb{R}_+$ the bounded continuous function $x\mapsto \sin(1/x)$ has obviously uniform distance equal to $1$ from any uniformly continuous function.
answered Apr 10, 2018 at 19:00