characterization of the dual space of the Sobolev space $H_0^1$ I am slightly confused about the properties of the dual space of the Sobolev space $H_0^1$ as outlined on page 299 in Evans.
In particular, following the notation in the book, item 3 says that $\forall u \in H_{0}^{1}(U), v \subset L^2(U) \subset H^{-1}(U)$, $$(v,u)_{L^2(U)}=\langle v,u \rangle.$$
I am not quite sure how to prove this and it might be due to a confusion with notation. Since $v \in H^{-1}(U)$, item 1 in the book states that $\exists \, v^0,v^1,
\dots , v^n$ in $L^2(U)$ such that $$\langle v,u\rangle=\int_Uv^0u+\sum_{i=1}^{n}v^{i}u_{x_i} \,dx.$$
In other words, we can identify $v$ with $(v^0,\dots,v^n)$. Since $v\in L^2$ and since this implies that $v$ is "associated" with the above functional $\langle v,u \rangle$ then one of the $v_i$'s have to be $v$ and the rest have to be $0$. Certainly, if $v^0=v$ the above statement follows. But why must this be the case? Why can't $v^1=v$ instead? 
Or is this what is meant by $L^2(U) \subset H^{-1}(U)$, i.e. that if $v \in L^2$ then the functional associated with $v$ takes on the form $\int vu\,dx$? I am hoping to clarify this part because certainly $\int v_{x_i}u\,dx$ seems legitimate too.
 A: It appears that my (first) edition of the book does not contain this statement, but I think I understand it. The elements of $H^{-1} $ are bounded linear functionals on $H^1_0$: 
$$H^{-1}=\{f:H_0^1\to\mathbb R \ ; \ |f(u)|\le C\|u\|_{H^1}\}$$
 Then what do we mean by saying that $L^2\subset H^{-1}$? It means that some functionals on $H^1_0$ admit a bound by the $L^2$ norm, and thus can be extended to a functional on $L^2$.
$$H^{-1}\supset L^2 = \{f:H_0^1\to\mathbb R \ ; \ |f(u)|\le C'\|u\|_{L^2}\}$$
Now invoke the structure theorem for $H^{-1}$, which identifies $f\in H^{-1}$ with a tuple $(f^0,\dots,f^n)$ of $L^2$ functions, via 
$$f(u)=\int f^0u+\sum_{i=1}^{n} \int f^{i}u_{x_i} \tag{1}$$
If $f^{i}$ is not a zero function for some $i\ne 0$, the functional (1) is not bounded by the $L^2$ norm of $u$, since the integral norm offers no control of the derivative $u_{x_i}$. Conversely, if $f^1=\dots=f^n=0$, then of course (1) is bounded on $L^2$. 
Conclusion: the copy of $L^2$ within $H^{-1}$ can be described as 
$$\left\{f:H_0^1\to \mathbb R \ ; \ f(u) =  \int f^0 u\right\}$$
where $f^0$ is an $L^2$ function. 
The statement you quoted identifies $f$ with $f^0$, which is shorter but less precise than  identifying it with $(f^0,0,\dots,0)$.
A: Sorry to bother here but I double the answer by @user127096. Let me quote the result from H. Brezis's book. 
This book states that, in Proposition $9.20$, for $v\in H^{-1}(\Omega)$, there $\exists \, v^0,v^1,
\dots , v^n$ in $L^2(U)$ such that $$\langle v,u\rangle=\int_Uv^0u+\sum_{i=1}^{n}v^{i}u_{x_i} \,dx.$$
and MOREOVER:
$(1)$: $ \, v^0,v^1,
\dots , v^n$ is NOT determined uniquely by $v\in H^{-1}$, that is, $v_0$ can be varied. 
$(2)$: If $\Omega$ is bounded, we could take $v^0\equiv 0$. 
Hence, not to mention $(1)$, by $(2)$ we see that there has to be some $i\neq 0$ such that $v^i\neq0$, otherwise you functional $v$ will be $0$ all the time. 
To explain your question, we have to go back to definition. How we write done 
$$ H_0^1\subset L^2 \approx (L^2)^*\subset H^{-1} $$
in the first place?
Actually, we use $(L^2)^*$ to discover what is $H^{-1}$ by applying a canonical mapping $T$: $(L^2)^*\to H^{-1}$ that is the restriction to $H_0^1$ of elements in $(L^2)^*$. i.e., 
$$<Tv,u>_{<H^{-1},\,H^1_0>}=<v,u>_{<(L^2)^*,L^2>} \,\,\,\,\,\,\,\,\,\,\,(*) $$
for all $u\in L^2$.
Hence, we have
$(1)$: $\|Tv\|_{H^{-1}}\leq C\|v\|_{L^2}$
$(2)$: $T$ is injective
$(3)$: $R(T)$ is dense since $H_0^1$ is dense in $L^2$ with respect to $L^2$ norm.
Therefore, we know already a big enough part of $H^{-1}$ by this way. Hence, your question is just the result of $(*)$
For more general explanation, please check out H. Brezis, page 136, Remark $3$.
