I'm studying pde's and I'm on the topic of Sobolev spaces and I have a small question regarding the dual of a Sobolev space. I'm trying to understand a proof about the characterization of the dual of $H_0^{-1}$.
The Theorem states that:
(i) If $f\in H^{-1}(U)$. Then there exist functions $f^0,f^1,...,f^n$ in $L^2(U)$ such that
$\langle f,v \rangle=\int_U f^0 v + \sum_{i=1}^n f^i v_{x_i}$
for $v\in H_0^1(U)$
(ii) $\left||f| \right|_{H^{-1}(U)}=\inf \{\left( \int_U \sum_{i=0}^n |f^i|^2 \right)^{1/2} | \text{$f$ satisfies (i)} \} $
(iii) in particular $(v,u)_{L^2(U)} =\langle v,u \rangle$ for all $u\in H_0^1(U), v\in L^2(U) \subset H^{-1}(U)$.
In the proof it is states that (iii) follows immediately from assertion (i). I have a really hard time seeing this. Could anyway explain this? Furthermore it is stated that $H_0^1(U)\subset L^2(U) \subset H^{-1}(U)$, why is this?
The inclusions $H_0^1 \subset L^2$ holds per definition. To show the inclusion $L^2 \subset H^{-1}$ we take a $v \in L^2$, then for every $u \in H_0^1$ we have by the Cauchy–Schwarz inequality \begin{equation} (v,u)_{L^2} \leq \Vert v \Vert_{L^2} \Vert u \Vert_{L^2} \leq \Vert v \Vert_{L^2} \Vert u \Vert_{H_0^1}. \end{equation} Therefore the Map $I_v : u \mapsto \int_U uv \,dx$ belongs to the Dual Space of $H_0^1$. You can therefore identify every $v \in L^2$ as an element of $\left(H_0^1\right)'$ via $I_v$.
Now taking $f_1 = \dots = f_n = 0$ you get $(iii)$.