The subset of $C^{\infty}$ functions with compact support in $\mathbb{R}$ in the space of bounded real valued continuos function on $\mathbb{R}$

could any one tell me the following statement is true or false? and any reference for proof or counter examples?

The subset of $C^{\infty}$ functions with compact support in $\mathbb{R}$ in the space of bounded real valued continuous function on $\mathbb{R}$ is dense

Obviously false. There is no sequence in $C_c(\mathbb R)$ (of smooth or "rough" functions) that converges uniformly to the constant function $1$.
Indeed if $(f_n)$ is any sequence in $C_c(\mathbb R)$, then $\displaystyle \lim_{n\to\infty} \lVert 1-f_n\rVert_\infty \geq 1$ since each $f_n$ is compactly supported.
That said, the statement is false. To see that, take $f(x)=c$, where $c\neq0$. Clearly $f\in C(\mathbb{R})\cap L^\infty(\mathbb{R})$ (continuous and bounded). Now, for every function $\varphi\in C^\infty_0$, we have $||\varphi-f||\geq c$. Hence, there is no function in $C^\infty_0(\mathbb{R})$ that is arbitrarily close to $f$, which means that $C^\infty_0(\mathbb{R})$ is NOT dense in $C(\mathbb{R})\cap L^\infty(\mathbb{R})$.
• @kahen for every function $\varphi\in C^\infty_0$, we have $||\varphi-f||\geq c$. Can you please explain, why this statement is true? – Unknown x Nov 1 '17 at 16:43
• @ManeeshNarayanan, $\varphi \in C^\infty_0$ means that $\varphi=0$ outside of some interval $[-a,a]$. So, for $|x|>a$ we have $f(x)-\varphi(x)=f(x)$. – bartgol Nov 1 '17 at 17:28