Definition. A function $u$ has weak derivative $v \in L^1_{\text{loc}}$ if for any $C^1$ function $\phi$ with compact support we have $$ -\int_U v(x)\,\phi(x)\, \mathrm{d}x = \int_U u(x)\,\phi'(x)\,\mathrm{d}x. $$
Definition. $u$ is called a Sobolev function (in the one-dimensional case) if its weak derivative exists.
Let $u:\mathbb{R}\to\mathbb{R}$, $u(x):=|x|$ be the absolute value on the real numbers. I'm looking for functions $\phi$ ($C^1$ and with compact support) and $v$ (locally integrable) such that on finite subsets $U \subset \mathbb{R}$ we have $$ \int_U (v(x)\,\phi(x) + u(x)\,\phi'(x))\,\mathrm{d}x = 0. $$
Any ideas on how to proceed?
You could try the "intuitive" idea that the derivative of $|x|$ is probably $\operatorname{sign}(x)$, i.e.
$$v(x) = \operatorname{sign}(x):=\begin{cases} 1 &: x\geq 0 \\ -1 &: x <0.\end{cases}$$
Then, for any $\phi\in C^\infty_0(\mathbb{R})$, one has:
\begin{align}\int\limits_\mathbb{R} \phi'(x) \cdot |x|dx&= \int\limits_0^\infty\phi'(x) \cdot xdx - \int\limits_{-\infty}^0 \phi'(x) \cdot xdx\\ & = [\phi(x)\cdot x]_0^\infty-\int\limits_0^\infty \phi(x)\cdot 1 dx - [\phi(x)\cdot x]_0^\infty -\int\limits_{-\infty}^0\phi(x)\cdot (-1)dx\\ &=-\int\limits_\mathbb{R} \phi(x) \cdot v(x) dx \end{align} In the second equation, one uses partial integration, and for the third equation, one uses that $\phi$ has compact support, i.e. $\phi(\pm\infty)=0$ and also of course that $\phi(0)-\phi(0)=0$.