About a completion of a Sobolev space Let $\Omega$ be a bounded smooth domain and define $\mathcal{C} = \Omega \times (0,\infty)$. Below, $x$ refers to the variable in $\Omega$ and $y$ to the variable in $(0,\infty)$. The map $\operatorname{tr}_\Omega:H^1(\mathcal C) \to L^2(\Omega)$ refers to the trace operator ($\operatorname{tr}_\Omega u = u(\cdot,0)$ for smooth functions). 
How do I know that the constant functions are in that bigger space (let's just take $\epsilon =1$)? They obviously have finite $H^\epsilon(\mathcal{C})$ norm but that is not enough. 
We can approximate (see this) the constant function $1$ by $u_n$, where $u_n(x,y) = 1$for $y \in [0,n)$ and $u_n(x,y) = 0$ for $y \in [2n, \infty)$ and $u_n(x,y)$ linearly interpolates between $(n,2n)$. This is Cauchy with respect to the $H^\epsilon$ norm (edit: it's not Cauchy), but how to prove that $1$ is in $H^\epsilon$? I thought we could say $\lVert u_n - 1 \rVert_{H^\epsilon(\mathcal C)} \to 0$ but this is not sensible since $tr_\Omega$ is only defined for $H^1(\mathcal C)$ functions, and $1$ is not in $H^1(C)$.
 A: The trace only depends on values near the boundary.
That is, if $\phi\in C^\infty([0,\infty))$ is one in a neighborhood of zero, then $\operatorname{tr}_\Omega(\phi u)=\operatorname{tr}_\Omega(u)$ for every $u\in H^1(\mathcal C)$.
With this in mind, you can formally apply a cut-off to your constant function and treat the trace in that way.
Traces do indeed make sense in $H^\epsilon(\mathcal C)$ since multiplication by a compactly supported function brings your function to $H^1(\mathcal C)$.
But this is not needed if you just want to show that the inclusion is strict.
The point is that $H^\epsilon(\mathcal C)$ is a completion of $H^1(\mathcal C)$.
When showing that constant functions are in the completion but not the original space, we should of course remember that they are not in the original space – if you permit the tautology.
The norm $\|\cdot\|_{H^\epsilon(\mathcal C)}$ was only defined for functions in $H^1(\mathcal C)$ by the integral expression in the first place, so the expression $\|u_n-1\|_{H^\epsilon(\mathcal C)}$ (or $\|1\|_{H^\epsilon(\mathcal C)}$) indeed does not make sense for the norm defined on $H^1(\mathcal C)$.
(Of course the norm can be naturally extended and the integral expression is exactly the same. You just need to extend the trace map to $\operatorname{tr}_\Omega:H^\epsilon(\mathcal C)\to L^2(\Omega)$ via cut-offs or otherwise to make sense of the expression.)
Once you have confirmed (as you seem to have) that your sequence is Cauchy in the norm, then it automatically has a limit in the completion.
The sequence converges locally uniformly (in fact, it is eventually constant in any compact set), so it is easy to observe that if it had a limit in $H^1(\mathcal C)$, it would have to be the pointwise limit – the constant function.
But the constant is not in $H^1(\mathcal C)$, so you have indeed shown that the completion contains a point outside the original space.
This point is a point in a formal completion (an equivalence class of Cauchy sequences) but in this case it is natural to identify with a function (which is not in $H^1(\mathcal C)$).
A: I think that the trace is defined by the completion. Indeed for every $u \in H^1 (\Omega)$, you have $\Vert\operatorname{tr} u\Vert_{L^2 (\Omega)} \le \Vert u \Vert_{\varepsilon}$. Since $H^1 (\mathcal{C})$ is dense in $H^\varepsilon (\mathcal{C})$, the trace operator $\operatorname{tr}$ is a well-defined continuous linear operator on $H^\varepsilon (\mathcal{C})$.
By the way, it is the same argument that shows that $\nabla v$ is defined on $H^\varepsilon (\mathcal{C})$. A more refined argument (relying I think on the Hardy inequality) would show that the restriction on $\Omega \times [0, R]$ is well-defined from $H^\varepsilon (\mathcal{C})$ to $L^2 (\Omega \times [0, R])$.
Going back to the question, you just then need to observe that your approximating sequence of the constant has its traces (as traces of $H^1 (\mathcal{C})$ functions) converging in the space $L^2 (\Omega)$. That is the since the sequence is constant in a neighbourhood of the set $\Omega \times \{0\}$.
