$H^1_0$ on half space Is it true that for any $u \in H^1_0(\mathbb R^n_+)$ there holds $u = 0$ on $\partial \mathbb R^n_+$?
How can I prove it?
 A: From the comments I gather that your main issue is the continuity (or equivalently the existence) of the trace operator $T:H^1(\mathbb{R}^n_+)\to L^2(\mathbb{R}^{n-1})$. This is however easy to fix: 
We will denote by $(x,t)\in\mathbb{R}^{n-1}\times \mathbb{R}$ points in $\mathbb{R}^n$.


*

*Consider first $f\in C^\infty_c(\mathbb{R}^n)$ so that by the Fundamental Theorem of Calculus we have 
$$
|f(x,0)|^2= \left|\int_0^\infty 2 f(x,t) \partial_t f(x,t)\, dt  \right|.
$$
Integrating over $\mathbb{R}^{n-1}$ we arrive at
\begin{equation}
\| f(\cdot, 0)\|_{L^2(\mathbb{R}^{n-1})}^2 \leq 2 \, \| f\|_{L^2(\mathbb{R}^n_+)} \, \| \partial_t f\|_{L^2(\mathbb{R}^n_+)} \leq 2 \, \| f\|_{H^1(\mathbb{R}^n_+)}^2.
\end{equation} 

*Now we need to show that $C^\infty_c(\mathbb{R}^n)$ is dense in $H^1(\mathbb{R}^n_+)$. To see this we first note that the even extension of $f\in H^1(\mathbb{R}^{n}_+)$, denoted by $F$, is in $H^1(\mathbb{R}^n)$ (to prove this work first with $C^\infty(\mathbb{R}^n_+)$ functions and then extend by density**) and we have 
$$
\| F\|_{H^1(\mathbb{R}^n)} \leq 2 \| f\|_{H^1(\mathbb{R}^n_+)}.
$$
Now the result follows since $C_c^\infty(\mathbb{R}^n)$ is dense in $H^1(\mathbb{R}^n)$. 
** Here we're using the Meyers-Serrin Theorem H=W which says that $C^\infty(\Omega)$ is dense in $H^1(\Omega)$ for any open set $\Omega$ (Evans proof holds in this general case once we have partitions of unity for general open sets, such a construction can be found in many books). Alternatively we could prove this directly by decomposing the upper half-space in cubes or slices of the form $\{ (x,t): 2^{k}<t<2^{k+1}\}$ for $k\in \mathbb{Z}$ and constructing appropriate cut-off functions in the $t$-variable.
