Define $f(x) = |x|$. It can be shown that $f$ has a weak derivative
$$g(x) = \begin{cases}1 &: x > 0\\\ 0 &: x = 0 \\\ -1 &: x < 0\end{cases}$$
I'm trying to show that $f$ does not have a second weak derivative, but I'm not sure about all the details, specifically what a contradiction in the attempted proof by contradiction below should be.
By the way of contradiction, assume that some $h\in L_{\mathrm{loc}}^1(\mathbb{R})$ satisfies
$$\forall \phi\in C_0^\infty(\mathbb{R}):\int_\mathbb{R}\frac{\partial \phi}{\partial x}(x)g(x)dx = -\int_{\mathbb{R}}\phi(x)h(x)dx$$
For any $\phi\in C_0^\infty(\mathbb{R})$ with $\mathrm{supp}(\phi)\subset (0,\infty)$ (similar for $\mathrm{supp}(\phi)\subset (-\infty, 0)$), we have
$$\int_{\mathbb{R}}\frac{\partial \phi}{\partial x}(x)g(x)dx = \int_{\mathbb{R}}\frac{\partial \phi}{\partial x}(x)dx = \left[\phi(x)\right]_0^\infty = 0$$
as $\mathrm{supp}(\phi)$ is a compact subset of $(0, \infty)$.
For $\phi\in C_0^\infty(\mathbb{R})$ with $\phi(0)\neq 0$, we have
$$\int_{\mathbb{R}}\frac{\partial \phi}{\partial x}(x)g(x)dx = -\int_{-\infty}^0\frac{\partial \phi}{\partial x}(x) dx + \int_0^\infty \frac{\partial \phi}{\partial x}(x)dx = 2\int_0^\infty \frac{\partial \phi}{\partial x}(x)dx = -2\phi(0)$$
So on the one hand, on the space $\mathbb{R}\setminus\{0\}$, the supposed weak derivative of $g$ must be zero. And the weak derivative is unique up to a set of Lebesgue measure zero, so (and this is something I'm not sure about) since the weak derivative of $g$ exists in $\mathbb{R}$, it also exists in $\mathbb{R}\setminus\{0\}$, and hence by uniqueness it must be zero a.e. in $\mathbb{R}$. But now by the last integral equality, for any $\phi\in C_0^\infty(\mathbb{R})$ with $\phi(0)\neq 0$ we get a contradiction.
So I'm not that comfortable with embedding/extension type results, and therefore my mathematical awareness screams at me when I say out loud parts after the bolded text. Is the argument good enough, or is there some caveat I have not seen?