I was reading Brezis's functional analysis,Sobolev spaces,and partial differential equation book.
In page 211,there is a therem which roughly speaking says $Du_\epsilon = (Du)_\epsilon$ where $u_\epsilon = u\star \eta_\epsilon$.
Which is proved as follows:
Lemma 8.4. Let $\rho \in L^{1}(\mathbb{R})$ and $v \in W^{1, p}(\mathbb{R})$ with $1 \leq p \leq \infty$. Then $\rho \star v \in$ $W^{1, p}(\mathbb{R})$ and $(\rho \star v)^{\prime}=\rho \star v^{\prime} .$
Proof. First, suppose that $\rho$ has compact support. We already know (Theorem 4.15) that $\rho \star v \in L^{p}(\mathbb{R})$. Let $\varphi \in C_{c}^{1}(\mathbb{R}) ;$ from Propositions $4.16$ and $4.20$ we have $$ \int(\rho \star v) \varphi^{\prime}=\int v\left(\check{\rho} \star \varphi^{\prime}\right)=\int v(\check{\rho} \star \varphi)^{\prime}=-\int v^{\prime}(\check{\rho} \star \varphi)=-\int\left(\rho \star v^{\prime}\right) \varphi, $$ from which it follows that $$ \rho \star v \in W^{1, p} \quad \text { and } \quad(\rho \star v)^{\prime}=\rho \star v^{\prime} . $$
This is the first part of the proof,which assume $\rho$ to be compacted supported.The question is where do we use the assumption that $\rho$ is compact support.One possible case may be the first equality which uses the Fubini theorem.But as the post shows here it seems not?
If $\rho$ does not have compact support, then $\rho * \phi$ might not have compact support in general. Then it could not be used as test function in $$ \int v(\rho * \phi)'= -\int v' (\rho*\phi), $$ as the definition of weak derivative demands a test function with compact support.