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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). enter image description here

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 $H^1(C)$.

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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 $H^1(\mathcal C)$).

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  • $\begingroup$ Thanks for the answer. I understand that $tr_\Omega(\phi u) = tr_\Omega(u)$ for $u \in H^1(\mathcal{C})$. You say that the norm in $H^\epsilon(\mathcal{C})$ will be given by the natural extension of the expression $$\int\int_{\mathcal{C}}\epsilon |\nabla_x v|^2 + v_y^2 + \int_\Omega (tr_\Omega v)^2.$$ But, strictly speaking, although I do understand that the trace depends only on a neighbourhood around $y=0$, we still have $tr_\Omega: H^1(\mathcal{C}) \to H^{\frac 12}(\Omega)$. $\endgroup$ – jamesC Jan 21 '15 at 11:20
  • $\begingroup$ To make this rigourous, should I not instead use as a norm $$\int\int_{\mathcal{C}}\epsilon |\nabla_x v|^2 + v_y^2 + \int_\Omega (tr_\Omega (\phi v))^2$$ and then show that this norm is independent of $\phi$ etc. Or is it the case that, upon completion wrt. the norm as it was, the norm in the completed space $H^\epsilon$ automatically will be taken care of? $\endgroup$ – jamesC Jan 21 '15 at 11:20
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    $\begingroup$ @jamesC, there are many possible ways to make things rigorous. One is to redefine the trace map to $H^\epsilon(\mathcal C)\to L^2(\Omega)$ (using a cut-off $\phi$ and showing independence of the choice of the cut-off, or try to imitate the original definition of the trace map). Another option is to use the norm in your second comment (with the invariance of $\phi$). My idea was that the norm in $H^\epsilon$ will be given by the natural expression once you have extended the trace operator (which is also very natural). $\endgroup$ – Joonas Ilmavirta Jan 21 '15 at 11:59
  • $\begingroup$ Thanks, let me think about it a bit. It seems a bit circular to me, extending the trace map to a space which is defined in terms of the trace map to be extended. Btw I made a mistake: the sequence I claimed is Cauchy is not Cauchy, so that one won't do. $\endgroup$ – jamesC Jan 22 '15 at 12:47
  • $\begingroup$ ..actually maybe the sequence is Cauchy (see the other answer).. $\endgroup$ – jamesC Jan 22 '15 at 14:42
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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 $\Omega \times \{0\}$.

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  • $\begingroup$ Thanks for the answer. I agree with your first paragraph. If I understand you, you propose: 1) Define $H^\epsilon$ as the completion given in the OP. 2) Extend $tr:H^1 \to L^2$ to an operator $T:H^{\epsilon} \to L^2$ (done via limits of Cauchy sequences and density). 3) Find a sequence $u_n$ which is Cauchy in $H^\epsilon$-norm with $tr(u_n) = c$ and which converges in some sense to the constant function $c$. Unfortunately the sequence I listed in the OP is not Cauchy, but presumably another one can be constructed... $\endgroup$ – jamesC Jan 22 '15 at 13:07
  • $\begingroup$ I think that your sequence is Cauchy: all the traces are equal and $\int_{\mathcal{C}} \vert \nabla u_n \vert^2 = O (n)$ as $n \to \infty$. $\endgroup$ – Jean Van Schaftingen Jan 22 '15 at 14:16
  • $\begingroup$ Hmm, I think you're right, we have $\int_{\mathcal{C}}|\nabla u_n|^2 = |\Omega|\frac{1}{n}$. $\endgroup$ – jamesC Jan 22 '15 at 14:41
  • $\begingroup$ @JeanVanSchaftingen is it really the case that the norm in the completion space is given by the natural extension of expression (2.15) in the op? $\endgroup$ – riem Jan 29 '15 at 12:27
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    $\begingroup$ @jamesC If you complete a space with respect to the norm (2.15), then the restriction of the norm on your completed to your original space is (2.15). For other points in your completed space, the norm can be computed by taking the limit of (2.15) on a Cauchy sequence. $\endgroup$ – Jean Van Schaftingen Jan 30 '15 at 12:54

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