[example: Smooth Poisson Solutions Are Weak Solutions]
Let $U \subset \mathbb R^n$ be open, let $f \in C(U)$, and suppose $u \in C^2(U)$ satisfies $-\Delta u=f$ pointwise in $U$. Fix $\phi \in C_c^\infty(U)$. Since $\operatorname{supp}\phi$ is compact in $U$, choose an open box $Q$ such that $\operatorname{supp}\phi\subset Q$ and $\overline Q\subset U$. For each $i=1,\dots,n$, the functions $\phi$ and $\partial_{x_i}\phi$ vanish outside $\operatorname{supp}\phi$, so
\begin{align*}
\int_U \partial_{x_i}u\,\partial_{x_i}\phi\,d\mathcal L^n=\int_Q \partial_{x_i}u\,\partial_{x_i}\phi\,d\mathcal L^n.
\end{align*}
Because $\phi=0$ near $\partial Q$, the one-variable [integration by parts](/theorems/2098) formula in the $x_i$ direction has no boundary term, and therefore
\begin{align*}
\int_Q \partial_{x_i}u\,\partial_{x_i}\phi\,d\mathcal L^n=-\int_Q \partial_{x_i x_i}u\,\phi\,d\mathcal L^n.
\end{align*}
Again using that $\phi$ vanishes outside $\operatorname{supp}\phi\subset Q$, this becomes
\begin{align*}
\int_U \partial_{x_i}u\,\partial_{x_i}\phi\,d\mathcal L^n=-\int_U \partial_{x_i x_i}u\,\phi\,d\mathcal L^n.
\end{align*}
Summing over $i$ gives
\begin{align*}
\int_U \nabla u\cdot \nabla \phi\,d\mathcal L^n=\sum_{i=1}^n \int_U \partial_{x_i}u\,\partial_{x_i}\phi\,d\mathcal L^n.
\end{align*}
The identity above for each coordinate direction gives
\begin{align*}
\sum_{i=1}^n \int_U \partial_{x_i}u\,\partial_{x_i}\phi\,d\mathcal L^n=-\sum_{i=1}^n\int_U \partial_{x_i x_i}u\,\phi\,d\mathcal L^n.
\end{align*}
Since $\Delta u=\sum_{i=1}^n\partial_{x_i x_i}u$, we obtain
\begin{align*}
-\sum_{i=1}^n\int_U \partial_{x_i x_i}u\,\phi\,d\mathcal L^n=-\int_U (\Delta u)\phi\,d\mathcal L^n.
\end{align*}
Finally, the classical equation $-\Delta u=f$ gives
\begin{align*}
-\int_U (\Delta u)\phi\,d\mathcal L^n=\int_U f\phi\,d\mathcal L^n.
\end{align*}
Thus $u$ satisfies the weak identity for every $\phi\in C_c^\infty(U)$, and the final substitution is exactly where the sign convention for $-\Delta$ enters.
[/example]