Consider trace map $T:H^1(\Omega) \to H^{\frac 12}(\partial\Omega)$ on a sufficiently smooth domain $\Omega$. It has a partial inverse $E$.
If we have the statement $$F(u,Eu) = 0\quad\text{for all $u \in H^{\frac 12}(\partial\Omega)$}$$ does this imply $$F(Tv,v) = 0\quad\text{for all $v \in H^{1}(\Omega)$}?$$
We do know that $T(H^1) = H^{\frac 12}$. So it seems like this result should be true. Here $F$ is just some weak formulation so some integral over $\Omega$ and $\partial\Omega$. For concretness take $$F(u,Eu) = \int_\Omega \nabla w \nabla Eu + \int_{\partial\Omega}w u$$ for $w \in H^1(\Omega)$ fixed.
This is not true, even for this particular $F$. To see it, let $Q:H^1_0(\Omega)\to\mathbb{R}$ be the linear functional, defined by
$$Q\phi=\int_\Omega \nabla w\nabla \phi.$$
If $w\neq 0$ then, the image of $Q$ is $\mathbb{R}$. Therefore, there is $v\in H_0^1(\Omega)$ such that $Qv\neq 0$. Note that for this particular $v$, we have that $Tv=0$, because $v\in H_0^1(\Omega)$. We must conclude that $$\int_\Omega\nabla w\nabla v\neq 0=\int_{\partial\Omega}w Tv.$$