Let $\Omega \subset \mathbb{R}^n$ be a nonempty bounded open subset (assumed to be sufficiently smooth).
Let $\Gamma$ denote the boundary of $\Omega$, and let $\Gamma_1 \subset \Gamma$ be a (measurable) subset.
For any function $u$ defined a.e. on $\Gamma_1$, let $\tilde{u}$ denote its extension to $\Gamma$ by zero, that is, $\tilde{u} =u$ on $\Gamma_1$ and $\tilde{u} = 0$ on $\Gamma$.
Now I consider the Lions-Magenes space
$$ H^{1/2}_{00} (\Gamma_1) := \{ u \in H^{1/2}(\Gamma_1) \mid \tilde{u} \in H^{1/2}(\Gamma) \}. $$
Is it true that $H^{1/2}_{00} (\Gamma_1)$ is dense in $H^{1/2} (\Gamma_1)$?
Thank you for any assistance.
I don't think so. Take $n=2$ and assume that a portion of $\Gamma$ is a segment, say, $[-1,1]\times\{0\}$. Take $\Gamma_1=[-1,0]\times\{0\}$. Then the constant function $u=1$ belongs to $H^{1/2}(\Gamma_1)$ but if you extend it to zero outside $\Gamma_1$ you get a $BV$ function, with a jump. If you compute it directly, you should be able to prove that $\tilde u\notin H^{1/2}([-1,1]\times\{0\})$. This has to do with the embedding in fractional Sobolev spaces. In one dimension I think that $BV((0,1))$ is contained in $W^{s,p}((0,1))$ for $sp<1$ but not for $ps\ge 1$.
Update Tartar's book Sobolev has a chapter on the Lions-Magenes spaces. In particular he shows that if $u\in H^{1/2}_{00}(\mathbb{R}_+)$ then $\frac{u}{\sqrt x}\in L^2(\mathbb{R}_+)$ with a continuous bound. If you had density, then for $u\in H^{1/2}(\mathbb{R}_+)$ you could find a sequence $u_n\in H^{1/2}_{00}(\mathbb{R}_+)$ with $u_n\to u$. By pointwise convergence and Fatou's lemma you would have $\Vert \frac{u}{\sqrt x}\Vert_{L^2(\mathbb{R}_+)}\le\liminf_n \Vert \frac{u_n}{\sqrt x}\Vert_{L^2(\mathbb{R}_+)}\le C$. Take a function $u\in H^{1/2}(\mathbb{R}_+)$ such that $u=1$ in $[0,1]$. Then you have a contradiction.
Update 2 Take a look at this paper paper. The space $H^{1/2}_{00}$ is defined as the space of $u\in H^{1/2}$ such that $\frac{u}{d}\in L^2(\Omega)$ and the norm is $$\Vert u\Vert_{H^{1/2}}+\Vert \frac{u}{d}\Vert_{L^{2}}.$$ With this norm it is complete. So it all depends on which norm you take in $H^{1/2}_{00}$.