Can a $H^{1/2}(\partial\Omega)$ function be extended to $\Omega$ in practice? I have the following question: Suppose $\Omega$ is a bounded open subset of $\mathbb R^d$ with Lipschitz boundary. Say $g\in H^{1/2}(\partial\Omega)$ is known. 


*

*Can I then write down a function $\tilde g\in H^1(\Omega)$ which satisfies $\tilde g\Big|_{\partial\Omega}=g$? Obviously such a $\tilde g$ will not be unique in general if it exists, but that's ok.

*If the previous question is answered in the affirmative, can you name some methods by which to construct such a $\tilde g$ from knowing $g$ and $\Omega$? I don't need a full answer to this question; I'm just curious to know how one can do it.
Thanks!
 A: As Umberto said, the the answer to the first question is "yes".
In fact, $H^{1/2}(\partial \Omega)$ is the image of $H^1(\Omega)$
under the trace operator $T$.
For the second one, you can solve
$$-\Delta u = 0 \quad \text{in } \Omega\,,
\quad u = g \quad \text{on } \partial\Omega\,.$$
Otherwise, here is a way how to construct another approximate extension in the finite element framework.
Let assume that $\Omega$ is a polygon/polyhedron/... . Then


*

*mesh it with simpleces (triangles, tetrahedra, ... , depending on $d$)

*create a piecewise linear approximation $g^h$ of $g$ on $\partial\Omega$ via interpolation 

*extend $g^h$ to the volume by assigning its coefficients to linear Lagrangian basis functions that "live" in the volume


This way you can build a function $\tilde{g}^h\in H^1(\Omega)$ that
satisfies
$$\Vert T(\tilde{g}^h)-g \Vert_{H^{1/2}} = \mathcal{O}(h)\,.$$
If $\Omega$ has no flat facets, you can still use this procedure, but the first step (meshing the domain) introduces an additional error in the analysis.
