$\mathscr{S}(\mathbb{R}) \text {is dense in } L^{2}(\mathbb{R})$ This lemma was given by my professor in my notes of mathematical methods in physics. I believe with this lemma I can say that any function in the Schwartz space is $L^2$ integrable. However my professor also wrote another lemma $\mathscr{S}(\mathbb{R}) \subset L^{p}(\mathbb{R}), \forall p \in ]1,\infty[$ which already gives me said results. 
To be more precise my question is what other implications of $\mathscr{S}(\mathbb{R}) \text {is dense in } L^{2}(\mathbb{R})$ are there?  From another user I got every element in $L^{2}$
 can be approximated arbitrarily well by elements of $\mathscr{S}(\mathbb{R})$. Which comes from the definition of a dense set.
Thank you for your help.
 A: Indeed, $\mathcal{S}$ is dense in $L_p(\mathbb{R}^n)$ for all $1\leq p<\infty$.
A relative quick way to see this is by the method of approximations to the identity.
Consider say $\phi(x)=c e^{-|x|^2/2}$.
with $c$ to get integral 1. Set $\phi_\varepsilon(x)=\varepsilon^{-n}\phi(x/\varepsilon)$.
Then, for any $f\in L_p$


*

*$\|f*\phi_\varepsilon - f\|_p\xrightarrow{\varepsilon\rightarrow0}0$.

*$f*\varepsilon_\varepsilon \in\mathcal{C}^\infty(\mathbb{R}^n)$.


Choose a mollifier, that is a function $\eta\in\mathcal{C}^\infty(\mathbb{R}^n)$ and with compact support such that $0\leq \eta(x)\leq1$ which equals one on the unit circle, and vanishes outside the circle of radius $2$ for example, and define $\eta^\delta(x):=\eta(\delta x)$. Then, the double index family $\{\eta^\delta(f*\phi_\varepsilon): \delta>0,\varepsilon\}$ is in the class $\mathcal{S}$, and $\|\eta^\delta(f*\phi_\varepsilon) -f\|_p\rightarrow0$ as $\delta,\eta\rightarrow0$.
A: A set $A$ being dense in another set $B$ means not only that $A$ is a subset of $B$, but has to do with the norm or, more generally, the topology of the spaces in question.
In the case of normed spaces, it means that for every $b\in B$, you can find an $a\in A$ arbitrarily close to $b$:
$$\forall r>0\forall b\in B\exists a\in A\colon\|b-a\|_B<r.$$
Basically, think of the rational numbers and the real numbers: no rational number will ever truly equal $\pi$, for example, but one can find a sequence of rational numbers that converge towards $\pi$, or one rational that is close enough to it that the first 10 000 000 digits are the same. Since you can do this for every other real number, the rationals are dense in the reals.
So, in our case, since the 'larger' set $B=L_2(\mathbb R)$ is defined via the $L_2$-norm
$$\|f\|_2=\left(\int_\mathbb Rf^2\text d\lambda\right)^\frac12=\left(\int_{-\infty}^\infty f(x)^2\text dx\right)^\frac12,$$
what dense means is that while not every $L_2$-integrable function $f\in L_2(\mathbb R)$ might be in the Schwartz space $\mathscr S(\mathbb R)$, one can find a sequence of functions $(f_n)_{n\in\mathbb N}\in\mathscr S(\mathbb R)^\mathbb N$ in the Schwartz space, such that, for example,
$$\|f-f_n\|_{L_2}=\left(\int_{-\infty}^\infty (f(x)-f_n(x))^2\text dx\right)^\frac12<\frac1n$$
for every $n\in \mathbb N$, or just generally
$$\lim_{n\rightarrow\infty}\|f-f_n\|_{L_2}=0.$$
One could say "every element in $B$ can be approximated arbitrarily well by elements of $A$".
