Why is the dual space of $H_0^1(\Omega)$ denoted $H^{-1}(\Omega)$? Why is the dual of the Sobolev space $H_0^1(\Omega)$ denoted $H^{-1}(\Omega)$ ? 
For a positive integer $k$, $H^k(\Omega)=W^{k,2}(\Omega)$. What is the motivation behind the $-1$ exponent?
 A: 
Definition Let $s\in\mathbb R$. $H^s(\mathbb R^n)$ is the vector space which consists of elements $u\in\mathcal S'(\mathbb R^n)$ such that  $\widehat u$ is measurable and  $(1+|\xi|^2)^{\frac s2}\widehat u \in 
L^2(\mathbb R^n)$. We endow $H^s(\mathbb R^n)$ with the inner product
  \begin{equation}
\langle u,v\rangle_{H^s(\mathbb R^n)}:=\int_{\mathbb R^n}(1+|\xi|^2)^s\widehat u(\xi)
\overline{\widehat  v (\xi)}d\xi
\end{equation}
  and we denote
  \begin{equation}
\lVert u\rVert_{H^s(\mathbb R^n)} :=\left(\int_{\mathbb R^n}(1+|\xi|^2)^s|\widehat u(\xi)|^2d\xi
\right)^{\frac 12}
\end{equation}
  the corresponding norm.

Let $s\in \mathbb R$ and $v\in H^{-s}(\mathbb R^n)$. If $u\in H^s(\mathbb R^n)$, the map $\xi\mapsto 
\widehat u(\xi)\widehat v(-\xi)$ is in  $L^1(\mathbb R^n)$. 
Indeed, for all $\xi\in \mathbb R^n$, $\widehat u(\xi)\widehat v(-\xi)=\widehat u(\xi)
(1+|\xi|^2)^{\frac s2}\widehat v(-\xi)(1+|\xi|^2)^{-\frac s2}$, and the RHS is the product of two map of $L^2(\mathbb R^n)$. By Cauchy-Schwarz inequality, we have
$$\left|\int_{\mathbb R^n}\widehat  u(\xi)\widehat v(-\xi)d\xi\right|\leq \lVert u\rVert_{H^s(\mathbb R^n)}
\lVert v\rVert_{H^{-s}(\mathbb R^n)}.$$
Therefore, if $v\in H^{-s}(\mathbb R^n)$, the map $L_v$ defined by 
\begin{equation}
u\in H^s(\mathbb  R^n)\mapsto L_v(u)=(2\pi)^{-n}\int_{\mathbb R^n}\widehat u(\xi)\widehat v(-\xi)d\xi
=\int_{\mathbb R^n}\widehat u(\xi)\overline{\mathcal F}v(\xi)d\xi
\end{equation}
is a continuous linear functional on $H^s(\mathbb R^n)$ (hence an element of $(H^s(\mathbb R^n))'$
and $\lVert L_v\rVert _{(H^s(\mathbb R^n))'}\leq (2\pi)^{-n}\lVert v\rVert_{H^{-s}(\mathbb R^n)}$. We can define $L\colon H^{-s}(\mathbb R^n)\to (H^s(\mathbb R^n))'$ by $L(v)=L_v$.

Theorem The map $L$ defined above is linear, bijective, and bicontinuous. It allow us to indentify the dual space of $H^s(\mathbb R^n)$ by $H^{-s}(\mathbb R^n)$.

Sketch of proof:


*

*The fact that $L$ is linear comes from linearity of the inverse Fourier transform and the integral. Continuity follows from $\lVert L_v\rVert _{(H^s(\mathbb R^n))'}\leq (2\pi)^{-n}\lVert v\rVert_{H^{-s}(\mathbb R^n)}$.

*$L$ is surjective. Indeed, let $T\in (H^s(\mathbb R^n))'$. We can show that 
there exists a constant $C$ such that for all $\varphi\in \mathcal S(\mathbb R^n)$
$$|\langle T,\varphi\rangle|\leq C\lVert (1+|\xi|^2)^{s/2}\mathcal F^{-1}\varphi\rVert_{L^2(\mathbb R^n)}.$$
We can see that $(1+|\xi|^2)^{-s/2}\widehat T$ is a linear functional over $L^2(\mathbb R^n)$ and we apply Riesz representation theorem.

*Injectivity needs only the property $\mathcal F^{-1}\mathcal F(\varphi)=\varphi$ for all $\varphi\in\mathcal S(\mathbb R^n)$.

*The inverse map is continuous using Banach isomorphism theorem.

When $\Omega$ is a arbitrary open subset of $\mathbb R^n$ we have to use charts (and $\Omega$ needs to be regular enough, since the Sobolev of integer order, defined in a classical way may be not equal to the corresponding one, even if $\Omega$ is bounded), but at least it shows the idea.  
