I am reading Lemma 8.4 of the book Functinal Analysis, Sobolev Spaces and PDEs by Haim Brezis. The lemma states:
Lemma 8.4 Let $p\in L^1(\mathbb R)$ and $v\in W^{1,p}(\mathbb R)$ with $1\leq p\leq\infty$. Then $\rho*v\in W^{1,p}(\mathbb R)$ and $(\rho*v)'=\rho*v'$ ($v'$ here is the weak derivative of $v$.)
Proof. First, suppose that $\rho$ has compact support. We already know (Theorem 4.15) that $\rho*v\in L^p(\mathbb R)$. Let $\varphi\in C_c^1(\mathbb R)$; from Proposition 4.16 and 4.20 we have $$\int(\rho*v)\varphi'=\int v(\breve\rho*\varphi)'=\int v(\breve\rho*\varphi)'=-\int v'(\breve\rho*\varphi)=-\int(\rho*v')\varphi$$...(the omitted part of the proof generalize this to the case when $\rho$ may not have compact support)
Notations and Used Theorems:
$f*g$: convolution
$C_c^1$: $C^1$-functions of compact support
$\breve\rho$: defined by $\breve\rho(x)=\rho(-x)$
$v',\varphi'$: derivative or weak derivative
Theorem 4.15: $p\geq1,\ f\in L^1,\ g\in L^p\implies f*g\in L^p$.
Proposition 4.16: $p\geq1,\ \frac{1}{p}+\frac{1}{q}=1,\ f\in L^1,\ g\in L^p,\ h\in L^q\implies\int(f*g)h=\int g(\breve f*h)$
Proposition 4.20: $f\in C_C^1,\ g\in L^p\implies f*g\in C^1$ and $(f*g)'=f'*g$
I don't see why $\rho$ needs to have compact support in the first place. All the equalities hold without this assumption:
First equality: Proposition 4.16.
Second equality: Proposition 4.20 + $\varphi$ has compact support.
Third equality: The fact that $v\in W^{1,p}$.
Fourth equality: Proposition 4.16 again
So am I missing something or has the proof done some unneeded work?