On the trace of Sobolev function

Question

Let $\Omega \subset \mathbb{R}^n$ be a $C^2$ domain. Consider any $f \in H^1 (\Omega)$. Does there exist some $ g \in H^2(\Omega)$ such that trace of $f$ and $g$ are same?

Answer 1

This depends on $f$. A basic example is any function in $H^1_0$, which means that the trace vanishes. For these functions the answer is trivially affirmative; $g=0$.

However, the range of the trace operator on $H^1(\Omega)$ is exactly the space $H^{1/2}(\partial \Omega)$. On the other hand, the range of the trace operator on $H^2(\Omega)$ is the space $H^{3/2}(\partial \Omega)$. These are not straigtforward theorems, and frankly I don't know the proof in detail, but if needed I could try and find some reference.

Anyway, the bottom line is that there are functions $f\in H^1(\Omega)$ such that their trace is in $H^{1/2}(\partial \Omega)$ and not in $H^{3/2}(\partial \Omega)$. For these functions, the answer is negative.