Norm closure of $C_b^1(\mathbb{R})$

I want to determine what the closure of $$C_b^1(\mathbb{R})$$, the space of continuous differentiable functions with bounded derivative, with respect to the supremums norm is. I think that $$\overline{C_b^1(\mathbb{R})}=BUC(\mathbb{R})$$, the space of bounded uniformly continuous functions. Can someone help me? Do one has to use the fundamental theorem of calculus. I think also uniform convergence plays a big role.

• I guess that you know that $BUC(\mathbb R)$ is closed in the space of bounded continuous functions endowed with the supremum norm. It thus remains to approximate every $f\in BUC$ by elements of $C_b^1$. A standard method is mollification: Fix a positive smooth function $\phi$ with compact support and integral $1$, define $\phi_n(x)=n\phi(nx)$ and consider $f\ast \phi_n$. Using $(f\ast \phi_n)'=f\ast \phi_n'$ you see that these convolutions are in $C_b^1$, and using the uniform continuity of $f$ you will show $f\ast \phi_n \to f$ uniformly. – Jochen Wengenroth Feb 21 at 10:13