Explicit description of the closure of a given set

Question

Let $C$ be the subset of $C_b(\mathbb{R})$ given by $$C:=\{f\in C_b(\mathbb{R}):\ \exists f'\in C_b(\mathbb{R})\}$$ Now I want to take the closure of this set with respect to the supremum norm on $C_b(\mathbb{R})$. Hence the closure contains functions which can be uniformly approximate by sequences of functions $(f_n)_{n\in\mathbb{N}}$ such that $f_n\in C_b(\mathbb{R})$ and $f'_n$ exists for every $n\in\mathbb{N}$ in $C_b(\mathbb{R})$. Hence the limit function has first to be bounded since it is the uniform limit of bounded functions. Since every derivate is bounded, we know that $f_n$ is uniformly continuous. But what can I say more about the elements in this closure? Is there a explicit description of this closre by well-know spaces, for example $W^{2,2}(\mathbb{R})$ (which will not be true here)?

Thank you very much.

Answer 1

The closure is the set of uniformly continuous bounded functions. At first, each function $f$ in the closure is bounded, as you note, and it is uniformly continuous: If $\|f-f_n\|<\delta/3$ and $|f_n(x)-f_n(y)|<\delta/3$ whenever $|x-y|<\varepsilon$, then $||f(x)-f(y)|<\delta$ whenever $|x-y|<\varepsilon$. On the other hand, if $f$ is bounded and uniformly continuous, then $f$ is norm-approximated by the functions $f_n(x)=n\int_x^{x+1/n}f(t)dt$, which have bounded derivatives $f_n'=n(f(x+1/n)-f(x))$.