Why is it sufficient to prove this inequality? Let $\phi\in C_0^\infty(\mathbb{R}^N)$ and $\int \phi\, dx = 1$, let $v\in L^2(\mathbb{R}^N)$ have compact support and let $a$ be a $C^1$ function in a neighborhood of the support of $v$. Put
$$(J_\epsilon v)(x)=\int v(x-\epsilon y)\phi(y) \, dy$$
The problem is to show that $aD_kJ_\epsilon v - J_\epsilon(aD_kv)\to 0$ in $L^2$ as $\epsilon\to 0$. (Here, $aD_kv$ is defined in the sense of distribution theory, $D_k=\partial/\partial_k$.)
It is written in my text that "since the statement is obvious if $v\in C^1$, it is therefore enough to prove the inequality $\|aD_kJ_\epsilon v - J_\epsilon(aD_kv)\|\leq C\|v\|_{L^2}$". I know that the statement is true if $v\in C^1$, and $C_c^1$ is dense in $L^2$. However, I still do not get why it is sufficient to prove the above inequality. Why?
 A: Yes, it suffices to show that there is a constant $C$ such that for any $w\in L^2(\mathbb{R}^N)$ with compact support AND for any $\epsilon>0$,
$$||aD_kJ_\epsilon w - J_\epsilon(aD_kw)||\leq C||w||_{L^2}\tag{1}.$$
Let $(v_n)$ be a sequence $C_c^1$ such that  $\|v-v_n\|_{L^2}\to 0$. Then given $\delta>0$, there is $N$ such that for any $n\geq N$, $\|v-v_n\|_{L^2}<\delta/(2C)$. Now for $v_N\in C_c^1$ there is a $r>0$ such that for all $0<\epsilon<r$, $\|aD_kJ_\epsilon v_N - J_\epsilon(aD_kv_N)\|_{L^2}<\delta/2$. 
Hence, by linearity, by taking $w=v-v_N$ in (1) (which is $L^2$ with compact support), for $0<\epsilon<r$,
\begin{align*}\|aD_kJ_\epsilon v - J_\epsilon(aD_kv)\|_{L^2}&\leq \|aD_kJ_\epsilon (v-v_N) - J_\epsilon(aD_k(v-v_N))\|_{L^2}+\|aD_kJ_\epsilon v_N - J_\epsilon(aD_kv_N)\|_{L^2}\\
&\leq C\|v-v_N\|_{L^2}+\|aD_kJ_\epsilon v_N - J_\epsilon(aD_kv_N)\|_{L^2}
<\frac{\delta}{2}+\frac{\delta}{2}=\delta.
\end{align*}
Therefore $aD_kJ_\epsilon v - J_\epsilon(aD_kv)\to 0$ as $\epsilon\to 0$.
