Consequence from Banach Theorem on $H^1(\Omega)$ I need to prove the following proposition:
Any continuous linear functional on $H^1(\Omega)$ is of the form
$v\mapsto\displaystyle\int_\Omega\left\{\sum_{i=1}^nq_i\,\dfrac{\partial v}{\partial x_i}+q_0v\right\}$
with $q_i\in L^2(\Omega)$, $i=0,\dots n$.
Here $\Omega$ is an open bounded subset of $\mathbb{R}^n$ with a Lipschitz continuous boundary, and $H^1(\Omega)$ is the usual Sobolev space.
The book says that is a consequence from Banach Theorem, but I can't see the proof idea.
 A: Ok, here it goes. We assume $F$ is an linear bounded operator over $H^1(\Omega)$
Let $E$ denote the space of $N+1$ fold $L^2(\Omega)$, i.e., $E(\Omega):=(L^2(\Omega))^{N+1}$. Then the operator $T$, from $H^1(\Omega)\to E(\Omega)$ is defined by $T[u]=(u,\partial_1 u,\partial_2u,\ldots,\partial_Nu)$ and we have $T[u]\in E(\Omega)$. Take $G:=T(H^1(\Omega))$ and $S:=T^{-1}$. Then, the linear operator $L$ over $G$ defined as $L(h):=\left<F,S(h)\right>$ is continuous because $F$ is continuous. 
Now is where we use Hanh-Banach extension theorem. We have $G$ is a subspace over $E$ and hence we could extend $L$ from $G$ to whole $E$. Hence, $L$ is now a linear continuous operator over $E$. To conclude, we recall Riesz representation for $L^p$ and we obtain $v_0$, $v_1$, $v_2$...$v_n\in L^2(\Omega)$ such that 
$$ L(h)=\int_\Omega h_1v_0+\int_\Omega h_2v_1+\cdots+\int_\Omega h_{n+1}v_N\,dx $$
In particular, if $h\in G$ we have 
$$ L(h)=\left<F,S(h)\right>=\sum_{i=1}^N\int_\Omega \partial_i uv_i\,dx+\int_\Omega uv_0\,dx $$
as you expected.
Remark
I don't think the boundary of $\Omega$ will matter, i.e., any open set would be good. Also, if you have $H_0^1$ you could set $v_0=0$.
Remark2
You can just extend $L$ by means of continuity but not H-B-E. And also notice that the sequence $(v_i)$ may not unique.
