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Let $X$ be a dense subset of $C(\mathbb{R})$ for the compact-convergence topology such that each $f \in X$ has non-compact support, for each $f \in X$ let $$ C_f :=\left\{ g \in C(\mathbb{R}):(\exists h \in C_0(\mathbb{R}))\, g=f+h \right\}, $$ and equip $C(\mathbb{R})$ with the topology generated by the (possibly infinite metric) $$ d(f,g) := \sup_{x \in \mathbb{R}^d} \|f(x)-g(x)\| . $$

Has the space $X+C_0$ in the latter topology ever been considered? If so is it and what would some references be?

ABIM
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  • Well, replacing $X$ with its closure under compact convergence just replaces the assumption that "$X$ is a subset and no $f\in X$ has compact support" to "$X$ is a subspace and some $f\in X$ has non-compact support", and then it's harmless to assume that $C_0\subset X$ (changing $X$ to $X+C_0$). The topology on $C(R)$ you're considering is just a disjoint union of cosets of the subspace $C_b(X)$ of bounded functions, and $X+C_0$ lies as some subset of this one in some way which I'm not sure there is a lot interesting to say. – YCor Apr 10 '20 at 11:41
  • Basically, I'm wondering if any function in $C(\mathbb{R})$ can be approximated uniformly by first approximating most of it on a large compact and then approximating the "extra part" separately.... somehow.. – ABIM Apr 10 '20 at 12:08
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    This seems to be essentially hopeless. Compact convergence is way too weak. For instance you have plenty of countable $X$ that satisfy your compact convergence density assumption, while $C(R)$ is not separable (since it has the uncountable discrete quotient $C(R)/C_b(R)$). – YCor Apr 10 '20 at 12:12
  • So essentially, you're saying we need the set of orbits $X/C_b(R)$ to be dense in $C(R)/C_b(R)$? Then we'd have our result... But what is $C(R)/C_b(R)$ (besides its abstract definition...) – ABIM Apr 10 '20 at 15:13
  • This quotient is discrete. I'm just saying that if $Y\subset C(R)$ is dense for the topology of uniform convergence, then the projection of $Y$ in the (huge) quotient space $C(R)/C_b(R)$ is the whose set. And also $Y\cap C_b(R)$ has to be dense in $C_b(R)$, and already $C_b(R)$ is a non-separable Banach space. – YCor Apr 10 '20 at 15:18

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