Norm Closure of a Set Let $q:\mathbb{R}\rightarrow\mathbb{C}$ be continuous and suppose we look to the set $$A:=\{f\in C_b(\mathbb{R}):\ q\cdot f\in C_b(\mathbb{R})\}$$ as subspace of $C_b(\mathbb{R})$ equipped with the supremum norm. Now I want to figure out what $\overline{A}^{||\cdot||_{\infty}}$ is. Of course it has to be a subspace of $C_b(\mathbb{R})$ again, since the uniform limit of bounded functions is again bounded. But what about the second condition. I think this set has to be something like $C_0(\mathbb{R})$. Does someone has an idea about that? Thanks a lot.
 A: Well, if $q$ is bounded then you just recover $C_b(\mathbb{R})$. If $q$ goes to infinity at $\pm \infty$ then you get $C_0(\mathbb{R})$ as you suggest. The general answer is that $C_b(\mathbb{R}) \cong C(\beta \mathbb{R})$ and your construction produces a C*-subalgebra of this, which must therefore be either $C(\beta\mathbb{R})$ itself or else the continuous functions on some quotient of $\beta\mathbb{R}$ which vanish at some point. (If $q$ is bounded we are in the first case, if unbounded, the second.) One additional thing I can say are that $A$ contains the continuous functions with compact support, so the quotient does not identify any distinct points of $\mathbb{R}$ and the point at which the functions vanish must lie outside of $\mathbb{R}$.
To see that there are other possibilities besides $C_0(\mathbb{R})$ and $C_b(\mathbb{R})$, consider a function $q$ which is $0$ on every interval $[2n,2n+1]$ for $n \in \mathbb{Z}$ but whose value at $2n + 1.5$ goes to infinity as $n \to \pm \infty$. Then $\overline{A}$ does not contain $1$ but it does contain a function whose value at $2n+.5$ is $1$ for all $n$. So you don't get either $C_b(\mathbb{R})$ or $C_0(\mathbb{R})$.
