Let $\gamma\colon H^1(\Omega) \to H^{\frac 12}(\partial\Omega)$ be the linear trace map which has a right continuous inverse $\xi\colon H^{\frac 12}(\partial\Omega) \to H^1(\Omega)$.


Is the image of $\xi$ dense in $H^1(\Omega)$? i.e., for every $u \in H^1(\Omega)$, do there exist $w\_n \in H^{\frac 12}(\partial\Omega)$ such that $\xi(w\_n) \to u$ in $H^1(\Omega)$?


I wish $\xi$ was surjective but is this the best I can do?



Image of (right) inverse trace map $\xi\colon H^{\frac 12}(\partial\Omega) \to H^1(\Omega)$ dense in $H^1(\Omega)$?
First, $\xi$ cannot be surjective.
Consider two distinct functions in $H^1(\Omega)$ with the same boundary values; they cannot both be in the range of $\xi$.
This is a general property of quotient maps (remember that $H^{1/2}(\partial\Omega)=H^1(\Omega)/H^1\_0(\Omega)$).


But $\xi$ cannot even have a dense image.
Suppose $u\in H^1(\Omega)$ is such that there is a sequence $(w\_n)\subset H^{1/2}(\partial \Omega)$ so that $\xi(w\_n)\to u$.
This implies that $w\_n=\gamma(\xi(w\_n))\to\gamma(u)$ since $\gamma$ is continuous.
It then follows from the continuity of $\xi$ that $\xi(w\_n)\to \xi(\gamma(u))$ and by uniqueness of limits $\xi(\gamma(u))=u$.
Therefore $u$ is in the range of $\xi$.
Since $\xi$ is not surjective, this cannot hold for all $u\in H^1(\Omega)$.


If you give up continuity, I suspect you could make $\xi$ have dense range.
But I don't know if this could be done explicitly (without relying on Hamel bases or such).


