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In the second answer here one considers a function $w\in H^{\frac{1}{2}}(\Gamma_g)$. What is this space and what is the finite element functions that one should use that belongs to this space? I have tetrahedral elements.

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Suppose that $\Omega$ is your domain and $\Gamma_g \subset \partial \Omega$. You can think of $H^{1/2}(\Gamma_g)$ as the space of $H^1(\Omega)$ functions restricted to $\Gamma_g$. So any function belonging to $u \in H^1(\Omega)$ has its trace, $u|_{\Gamma_g}$, in $H^{1/2}(\Gamma_g)$. Therefore you can use, for example, piecewise linear elements defined on the boundary (triangular $P_1$ element in your case).

Piecewise constant elements do not work since such a function would belong to $H^{1/2-\varepsilon}(\Gamma_g)$ , $\varepsilon > 0$.

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  • $\begingroup$ so it has to be conitunous on element vertices? $\endgroup$
    – badmf
    Nov 30, 2017 at 13:32
  • $\begingroup$ Continuous on $\Gamma_g$. In particular, continuous on element vertices. $\endgroup$
    – knl
    Nov 30, 2017 at 13:33
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    $\begingroup$ This answer is not quite correct. $H^1$ can be discontinuous, and so their traces can be discontinuous as well at individual points. Whether $H^{1/2}$ consists only of functions that are continuous or not depends on how exactly you define this space. $\endgroup$ Dec 1, 2017 at 20:23
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    $\begingroup$ As usual, I define $H^{1/2}(\partial \Omega)$ as the space for which the norm $\|v\|_{1/2} = \inf_{w|_{\partial \Omega}=v} \|\nabla w\|_{0,\Omega}$ is finite. It's quite trivial to test computationally that this is not the case for piecewise constant functions. Thus, can you point where the error is in my answer? I did not say that all discontinous functions do not belong to $H^{1/2}$. I just said that piecewise constant would not. $\endgroup$
    – knl
    Dec 2, 2017 at 13:14

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