Consistency of Dirac delta representations in $H^{-1}(\mathbb{R})$ It is well-known that $\delta_{0} \in H^{-1}(\mathbb{R})$ and that through  the Fourier transform characterization of $H^{1}(\mathbb{R})$ we can obtain the following:
$$ \langle \delta_{0}, \phi \rangle_{H^{-1} \times H^1}=\int_{\mathbb{R}} u \phi    +\int_{\mathbb{R}} u' \phi ' \qquad (1) $$
where $u(x)=\frac{1}{2} e^{-|x|}$. In particular, $(1)$ holds for every $\phi \in \mathcal{D}(\mathbb{R})$. 
On the other hand, a characterization of $H^{-1}$ states that $F \in H^{-1}$ iff there exist $f_0, \mathbf{f} \in L^2$ such that:
$$ F= f_0 + \text{div}(\mathbf{f}) \qquad \text{in} \ \ \mathcal{D}' \qquad (2)$$ 
Letting $F=\delta_{0}$, in a one-dimensional scenario  $(2)$ is equivalent to:
$$ \delta_{0}=f_0+f' \qquad \text{in} \ \ \mathcal{D}'(\mathbb{R}) $$
where $f_0,f \in L^2(\mathbb{R})$. With some attempts and calculations I found out that:
$$ \langle \delta_{0}, \phi \rangle_{H^{-1} \times H^1}=\int_{\mathbb{R}} -\frac{\text{sgn(x)}e^{-|x|}}{2} \phi \, dx = \langle \left( \frac{e^{-|x|}}{2} \right)', \phi \rangle \qquad (3) $$
I cannot quite get the right connection between $(1),(2)$ and $(3)$: it seems like $(2)$ holds with $f_0=0$ and $f= \frac{e^{-|x|}}{2}$, is it then possible to get $(3)$ from $(1)$ with some manipulations? I wasn't able to do that.  Any suggestion or comment is appreciated.
 A: On the one hand, by equation (1), for every $\phi \in \mathcal{D}(\mathbb{R})$,
$$
 \langle \delta_0, \phi \rangle_{\mathcal{D}^\prime \times \mathcal{D}} = \langle \delta_0,\phi\rangle_{H^{-1}\times H^1} = \langle u, \phi \rangle_{L^2} + \langle u^\prime, \phi^\prime \rangle_{L^2} = \langle u, \phi \rangle_{\mathcal{D}^\prime \times \mathcal{D}} + \langle u^\prime, \phi^\prime \rangle_{\mathcal{D}^\prime \times \mathcal{D}}\\ = \langle u, \phi \rangle_{\mathcal{D}^\prime \times \mathcal{D}} - \langle u^{\prime\prime}, \phi\rangle_{\mathcal{D}^\prime \times \mathcal{D}} = \langle u + (-u^\prime)^\prime,\phi\rangle_{\mathcal{D}^\prime \times \mathcal{D}},
$$
so that $\delta_0 = u + (-u^\prime)^\prime$ in $\mathcal{D}^\prime(\mathbb{R})$, where $u, -u^\prime \in L^2(\mathbb{R})$. This, then, is the representation of $\delta_0$ in the form of equation (2).
On the other hand, equation (3) cannot be true: since $\mathcal{D}(\mathbb{R})$ is dense in $L^2(\mathbb{R})$ and $u \in H^1(\mathbb{R})$,
$$
 \forall \phi \in \mathcal{D}(\mathbb{R}), \quad \langle \delta_0,\phi \rangle_{H^{-1} \times H^1} = \langle u^\prime, \phi \rangle_{L^2}
$$
would mean that $\delta_0$ extends to $\langle u^\prime, \cdot \rangle_{L^2}$ in $L^2(\mathbb{R})^\prime \cong L^2(\mathbb{R})$, but this is known to be impossible.
