Hahn-Banach Theorem and Elliptic PDEs Suppose $A$ is positive-definite and bounded.  For a fixed $u\in W^{1,2}(U)$, define
$$
\ell_u(v)=\int_{U}A\nabla u\nabla v
$$
for $v\in W^{1,2}(U)$.  Also suppose $\ell_u$ is bounded on some linear subspace $X$ of $W^{1,2}(U)$ (e.g., maybe $W^{1,2}_0(U)$).  Is there any conclusion I can draw regarding $\|u\|_{W^{1,2}(U)}$?  
By Hahn-Banach, I can extend $\ell_u$ uniquely to a bounded linear functional $\tilde{\ell}_u$ on $W^{1,2}(U)$ with $\|\ell_u\|=\|\tilde{\ell}_u\|$, but I don't think I'm guaranteed anymore information about $\tilde{\ell}_u$.  I can say $|\tilde{\ell}_u(u)|\leq \|\ell\|\|u\|_{W^{1,2}(U)}$, but is there anymore?  
 A: Take a constant function $h_c$ such that $h_c(x) = c$ for all $x\in U$, clearly $h \in W^{1,2}(U)$ only if $U$ is bounded, but it seems that this is what you have in mind. Then, by construction $\ell_u = \ell_{u+h_c}$. However, $\|u+h_c\|_{W^{1,2}(U)}$ depends on $c$ and explodes as $c \to \infty$. Hence, a connection of the form $\|u\|_{W^{1,2}(U)} \lesssim \|\ell_u\|$, will not hold in general. 
Everything changes if you consider $u \in W^{1,2}_0(U)$, though. Denote by $|.|_{W^{1,2}(\Omega)}$ the semi-norm 
$$
|u|_{W^{1,2}(\Omega)} : = \|\nabla u \|_{L^2(U)\times L^2(U)},
$$
where $u \in W^{1,2}(\Omega)$.
We have that 
$$
\ell_u(u) = \int_U A \nabla u \nabla u \sim |u|_{W^{1,2}(\Omega)},
$$
where the equivalence in the last step is due to the positive definiteness of $A$. The Poincare inequality yields a connection between $|u|_{W^{1,2}(\Omega)}$ and $\|u\|_{W^{1,2}(\Omega)}$ if $u\in W^{1,2}_0(\Omega)$, i.e. there exists a constant $C>0$ such that $\|u\|_{W^{1,2}(\Omega)}\leq C|u|_{W^{1,2}(\Omega)} \sim \ell_u(u)$ for all $u \in W^{1,2}_0(\Omega)$. There also exists generalizations of the Poincare inequality, which you can check out.
Finally addressing your point on Hahn-Banach, I do not think you can achieve much with that approach. First, as I have shown at the beginning, it is impossible to get the statement that you request in general. Second, you will not have any knowledge of the form of the extended functional other than its norm. 
