How can you extend distributions non uniquely to sobolev spaces

Question

Is it possible that there are two different elements in $(W^{k,p}(\Omega))^{'}$, $T_1$ and $T_2$ such that they agree on $C_0^{\infty}(\Omega)$. ie is it possible that $T_1{\phi}=T_2{\phi}$ for all $\phi$ in $C_0^{\infty}(\Omega)$ but $T_1{u} \neq T_2{u}$ for some $u$ in $W^{k,p}(\Omega)$.

Answer 1

If you're looking for a counterexample, work in $W^{1,p}(\Omega)$ for simplicity, clearly if suffices to show that there exists a functional $F \in (W^{1,p})'$ such that $ F \phi = 0 \ \forall \phi \in \mathcal{D}(\Omega)$ but $F$ is not the zero functional in $(W^{1,p})'$. Let $ \mathbf{g} \in (L^{p'})^n$ be such that $$ \sum_{i=1}^{n} D_{i} \mathbf{g} \equiv 0 \qquad (1) $$

and define:

$$ T u=\int_{\Omega} \mathbf g \cdot\nabla u \ \qquad \forall u \in W^{1,p}(\Omega) $$

It's easy to prove that $T$ is a bounded linear functional of $W^{1,p}(\Omega)$, on the other hand its restriction to $\mathcal{D}(\Omega)$ is identically zero, as can be shown by an integration by parts and using condition $(1)$.