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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

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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.

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First, when you say "dense" you need to specify with respect to what norm. I'm assuming you mean the uniform norm (the sup norm).

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 $C(\mathbb{R})\cap L^\infty(\mathbb{R})$.

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  • $\begingroup$ @kahen for every function $\varphi\in C^\infty_0$, we have $||\varphi-f||\geq c$. Can you please explain, why this statement is true? $\endgroup$ – Unknown x Nov 1 '17 at 16:43
  • $\begingroup$ @ManeeshNarayanan, $\varphi \in C^\infty_0$ means that $\varphi=0$ outside of some interval $[-a,a]$. So, for $|x|>a$ we have $f(x)-\varphi(x)=f(x)$. $\endgroup$ – bartgol Nov 1 '17 at 17:28

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