Help with proof that $H^{1}(\mathbb{R})$ is closed under multiplication. Edit: Prove that if $u,v \in H^{1}(\mathbb{R})$ then $uv \in H^{1}(\mathbb{R})$. 
My idea is to approximate with functions in $C^{\infty}(\mathbb{R})$ with compact support. 
Let $u,v \in H^{1}(\mathbb{R})$. Since $C^{\infty}_0(\mathbb{R})$ is dense in $H^1(\mathbb{R})$ there exists sequences $u_n,v_n \in C^{\infty}(\mathbb{R})$ such that 
$u_n \rightarrow u$ and $v_n \rightarrow v$ in $|| \cdot ||_{H^1} $. Now
$$ uv - u_n v_n = (u-u_n)v + u_n(v-v_n) \Rightarrow $$
$$ || uv - u_n v_n||_{H^1} = || (u-u_n)v + u_n(v-v_n)||_{H^1} \leq  $$
$$|| (u-u_n)v ||_{H^1} + ||u_n(v-v_n)||_{H^1} \rightarrow 0 \quad,\quad n \rightarrow \infty $$ 
Thus $u_n v_n \rightarrow uv$ when $n \rightarrow \infty$. Since $u_n v_n \in C^{\infty}_0(\mathbb{R}) $ the product $uv \in H^1(\mathbb{R})$. 
Feedback and corrections are appreciated!   
 A: You need to prove that $uv\in H^1(\mathbb{R})$, or equivalently, that $uv\in L^2(\mathbb{R})$ and there is $g\in L^2(\mathbb{R})$ such that $$\int_\mathbb{R}(uv)\varphi'=-\int_\mathbb{R}g\varphi,\ \forall\ \varphi\in C_0^\infty(\mathbb{R}).\tag{1}$$
The first question is: does $uv\in L^2(\mathbb{R})$? The answer is yes, because $H^1(\mathbb{R})$ is a subset of $L^\infty(\mathbb{R})$.
So, it remains to find some $g$, satisfying $(1)$ with $g\in L^2(\mathbb{R})$.  Once weak derivatives are so closely related with standard derivatives, we may try as a first candidate for $g$, the function $uv'+u'v$. As in the previous paragraph, this function does belong to $L^2(\mathbb{R})$.
Now let $u_n,v_n\in C_0^\infty(\mathbb{R})$ and $u_n,v_n\to u,v$ in $H^1(\mathbb{R})$. We have that $$\int_\mathbb{R}(u_nv_n)\varphi'=-\int_\mathbb{R} (u_nv_n'+u'_nv_n)\varphi,\ \forall\ \varphi\in C_0^\infty(\mathbb{R}). \tag{2}$$
To conclude, remember the fact that $H^1(\mathbb{R})\subset L^\infty(\mathbb{R})$ is not only an inclusion, it is an continuous embedding, so by using similar inequalities as in your question, we can prove that in the limit, $(2)$ converge to $(1)$. I leave to you this calculation.
