Possible density of Schwartz Functions in the space of continuous functions vanishing at infinity While reading about Function Spaces here ; from the chain of inclusions I had some questions that: whether $\mathcal{S}(\mathbb R^{n})$ (the Schwartz class of functions) is included in $C_{0} (\mathbb R^{n})$ (Space of all continuous functions vanishing at infinity) AND if affirmative, is the inclusion is dense?? i.e. Can we approximate a function in $C_{0} (\mathbb R^{n})$ by a sequence of functions in $\mathcal{S}(\mathbb R^{n})$ uniformly??? 
I found an article here .
So my question is: Can we approximate a function in $C_{0} (\mathbb R^{n})$ by a sequence of functions in $\mathcal{S}(\mathbb R^{n})$ uniformly??? 
 A: Yes, $\mathcal{S}(\mathbb{R}^n)$ is dense in $C_0(\mathbb{R}^n)$.
Let $f$ be in $C_0(\mathbb{R}^n)$. Recall that this implies that $f$ is uniformly continuous.
Your claim is easy using mollifiers. Let me show you how this works.
Pick a smooth positive function $\psi$ with support contained in the open unit ball $B_1(0)$ such that $\int_{\mathbb{R}^n} \psi = 1$. Notice that $\psi$ is clearly a Schwartz function. For $\delta>0$ we define $\psi_\delta(x)=\frac1{\delta^n} \psi(\frac x{\delta})$. 
Then $\psi_{\delta}$ is supported inside $B_\delta(0)$ and $\int_{\mathbb{R}^n} \psi_\delta=1$ (this is why we multiplied by $\delta^{-n}$). So as $\delta$ tends to zero, $\psi_\delta$ becomes more concentrated around the origin. Define
$$f_\delta := f*\psi_\delta.$$
Then $f_\delta\in\mathcal{S}(\mathbb{R}^n)$, because $\psi_\delta$ is Schwartz. We claim that $f_\delta\to f$ in $C_0(\mathbb{R}^n)$ as $\delta\to 0$. 
Let $\varepsilon>0$. Since $f$ is uniformly continuous, there exists $\delta_0>0$ such that $|f(x)-f(y)|<\varepsilon$ for all $|x-y|<\delta_0$. Thus, if $\delta<\delta_0$ we have for $x\in\mathbb{R}^n$ arbitrary:
$$|f*\psi_\delta(x)-f(x)| \le \int_{\mathbb{R}^n} |\psi_\delta(y)(f(x-y)-f(x))|  dy< \varepsilon\int_{\mathbb{R}^n} \psi_\delta(y) dy=\varepsilon.$$
In the first inequality we used the triangle inequality and used that $\int_{\mathbb{R}^n} \psi_\delta = 1$. In the second inequality we used that $\psi_\delta$ is supported in $B_\delta(0)$.
This proves the claim.
