Dual pairing on $H^1_0$ I am reading an article on optimal control by Jean-Pierre RAYMOND and on page 57, in the proof of Theorem 5.4.4 he uses the fact that
$\int_{\Omega} |\nabla((-\Delta)^{-1}f)|^2=\langle f,(-\Delta)^{-1}f\rangle_{H^{-1} \times H^1_0} $
The only way I can explain this to myself is by the following argument : If we take the norm $\|g\|_{H^1}=\int_\Omega|\nabla g|^2$ on $H^1$, then because of the Riesz representation theorem, for every $f\in H^{-1}$ there exists a $u_f\in H_0^1$ such that
$\langle f,g\rangle_{H^{-1} \times H^1_0}=\int_\Omega \nabla g\nabla u_f$
for every $g\in H_0^1$. Now we can identify $u_f$ by the isomorphism $(-\Delta)^{-1}:H^{-1}\rightarrow H^1_0$ and we finally get
$\langle f,g\rangle_{H^{-1} \times H^1_0}=\int_\Omega \nabla g\nabla (-\Delta)^{-1}f$.
By replacing $g$ with the term above yields the result. I'm not quite sure if this correct though, some help would be appreciated
Thank you in advance!

You have the right idea. Let's introduce $g=(-\Delta)^{-1}f$, so that the claim becomes more readable: 
$$\int_{\Omega} |\nabla g|^2=\langle -\Delta g,g \rangle_{H^{-1} \times H^1_0}\tag1$$ 
Recall the general definition of  $-\Delta g$: it is the distribution $T$ that acts on test functions $\varphi$ by 
$$ T(\varphi) =  -\int g\Delta \varphi\tag2$$
When $g\in H^1$, we can integrate by parts on the right, obtaining another formula for $T$: 
$$ T(\varphi) =   \int \nabla g \nabla \varphi\tag3$$
The right hand  side of (3) is a continuous linear functional on $ H^1_0$, and therefore $T$ extends to an element of $H^{-1}$. Computing $T(g)$ according to (3), we get (1).
