I think you can pull an argument like this : define the sequence of compacts $[-n, n]$ ; since $\mathcal A$ is dense in each of them, for each function $f \in L^p(\mathbb R)$ and for all $n$, there exists a sequence $f_{n,m}(x)$ such that $f_{n,m} \to f$ in $[-n,n]$ with respect to the norm you've chosen as $m \to \infty$.
You could extract a subsequence out of the sequences $f_{n,m}$ by the following argument. Choose a sequence $\varepsilon_k \to 0$. We will find $g_k$ such that
$$
\|g_k - f\|_{\infty} \overset{def}{=} \inf \{ K > 0 \, | \, \mu(\{x \in \mathbb R \, | \, |g_k(x) - f(x)| > K\} ) = 0 \} < \varepsilon_k.
$$
(I believe this is the detailed definition of your norm? Correct me if I'm wrong.)
with $\mu$ being the Lebesgue measure.
Now $f \in L^p$, so that for $n$ big enough, $|f(x)| < \varepsilon_k/2$ almost everywhere outside $[-n,n]$. Fix this $n$.
This is the point where I am stuck at. Since $f_{n,m} \to f$ in $[-n,n]$, there exists $m$ such that $\|f_{n,m}-f \|_{\infty} < \varepsilon_k$ inside $[-n,n]$ and $|f_{n,m}(x)| < \varepsilon_k/2$ outside$^{*}$ $[-n,n]$ ($f_{n,m} \in L^p$ also, hence goes to $0$ at infinity.)
This guy ($^{*}$) is the problem ; I don't have any control over what happens to $f_{n,m}$ outside $[-n,n]$ and that is what gives me intuition that it might be false in the general case. Although your choice of functions exhibit more properties than the abstract one... but this is now at a state of intuition only. Let me finish just to see where I froze in my argument.
With those properties summed up, outside $[-n,n]$, $|f_{n,m}(x) - f(x)| \le |f_{n,m}(x)| + |f(x)| < \varepsilon_k$ almost everywhere outside $[-n,n]$ and $|f_{n,m}(x) - f(x)| < \varepsilon_k$ inside $[-n,n]$, so that $|f_{n,m}(x)-f| < \varepsilon_k$ almost everywhere over $\mathbb R$, that is, we have found $f_{n,m}$ such that $\|f_{n,m} - f\|_{\infty} < \varepsilon_k$. Just let $g_k = f_{n,m}$.
I hope it helped in some way to at least give you some ideas.
EDIT : I forgot the "$ = 0$" in my definition of the norm. Typo error.
I'll assume we are talking about functions on the real line. Fix $f\in C_0(\mathbb R)$. The fact that $f$ is $C_0$ allows us to write it as a uniform limit of continuous functions with compact support$^1$. So now we can assume without loss of generality that $f$ is continuous with compact support.
Start with $h_0(x)=\begin{cases}e^{-1/x^2},&\ x>0\\0,&\ x\leq0\end{cases}$.
Next you notice that $h(x)=h_0(x)h_0(1-x)\in C^\infty_K$, with support in $[0,1]$. We can normalize it so that $\int_{\mathbb R} h=1$. Then, for each $\varepsilon>0$, you take
$$
h_\varepsilon(x)=\frac1\varepsilon\,h\left(\frac x\varepsilon\right)
$$
and you note that $\int_{\mathbb R}h_\varepsilon=\int_{\mathbb R}h=1$.
Now form the convolutions
$$
f_\varepsilon(x)=\int_{\mathbb R}f(t)\,h_\varepsilon(x-t)\,dt.
$$
It is not hard to show that because $h_\varepsilon\in C^\infty$ we have $f_\varepsilon\in C^\infty$; and because $f$ and $h_\varepsilon$ have compact support, so does $f_\varepsilon$.
Finally, given $\varepsilon>0$, as $f$ is continuous with compact support, it is uniformly continuous. So given $c>0$ there exists $\delta>0$ such that $|f(x)-f(y)|<c$ if $|x-y|<\delta$. Then
\begin{align*}
|f_\varepsilon(x)-f(x)|&=\left|\int_{\mathbb R} [f(x-t)-f(x)]\,h_\varepsilon(t)\,dt\right|
\leq\int_{\mathbb R} |f(x-t)-f(x)|\,h_\varepsilon(t)\,dt\\
&\leq c\,\int_{|t|<\delta} h_\varepsilon(t)\,dt
+2\|f\|_\infty\,\int_{|t|\geq\delta}h_\varepsilon(t)\,dt\\
&=c
+2\|f\|_\infty\,\int_{|t|\geq\delta}h_\varepsilon(t)\,dt.
\end{align*}
with $\delta$ fixed, the last integral goes to zero as $\varepsilon\to0$ (because the support of $h_\varepsilon$ is contained in $[0,\varepsilon]$). Thus
$$
\limsup_{\varepsilon\to0}|f_\varepsilon(x)-f(x)|\leq c
$$
for all $x$ and all $c>0$. So $\|f_\varepsilon-f\|_\infty\to0$.
- Fix $f\in C_0(\mathbb R)$. For each $n\in\mathbb N$ there exists $N>0$ with $|f(x)|<\tfrac1n$ when $|x|>N$. Let $g_n$ be continuous, with $0≤g_n≤1$, $g_n(x)=1$ when $|x|≤N$ and $g_n=0$ when $|x|>N+1$. Then $fg_n\in C_c(\mathbb R)$ and $$|f-fg_n|=|f(1-g_n)|<\tfrac1n,$$ so $fg_n\to f$ uniformly.
Best Answer
The appeal to "density" is a little sloppy. I believe the following is what the author meant:
I don't have a suitable reference for (1), but this fact is not hard to prove directly: (a) smooth functions are in $c^\beta$; (b) $c^\beta$ is a closed subspace of $C^\beta$; (c) mollifying a $c^\beta$ function produces an approximating sequence that converges in $C^\beta$ norm; the key is that sufficiently small scales don't really contribute to the norm.
For reference, here is the relevant part of the source (page 25 in Selected Aspects of Fractional Brownian Motion by Ivan Nourdin).
The reference Young [67] is to the classical article An inequality of the Hölder type, connected with Stieltjes integration, Acta Math. (1936) 67, 251-282. Young actually works with larger spaces of functions of finite $p$-variation, and I couldn't locate a relevant density argument there. A more accessible source is a pair of blog posts by Fabrice Baudoin: