# $H^1$-conforming finite elements

Let us consider $\Omega\subset\mathbb{R}^d$, $d=2,3$ polyhedral domain, i.e. $\bar{\Omega}$ is the union of a finite number of polyhedra.

Let $\bar{\Omega}=\bigcup_{K\in\mathcal{T}_h}K$, where $\mathcal{T}_h$ is a triangulation of simplexes.

I want to prove the following result.

A function $v:\Omega\to\mathbb{R}$ belongs to $H^1\left(\Omega\right)$ if and only if

• $v|_{K}\in H^1\left(K\right)$ for each $K\in\mathcal{T}_h$,
• $v|_{K}\in H^1\left(K\right)$ for each $K\in\mathcal{T}_h$, for each common face $F=K_1\cap K_2$, $K_1,K_2\in\mathcal{T}_h$, the trace of $v|_{K_1}$ and $v|_{K_2}$ on $F$ is the same.

I proved the reverse implication "$\leftarrow$", but I have troubles with the other one.

Assume $v\in H^1\left(K\right)$. We have, of course, that $v|_{K}\in H^1\left( K\right)$ for each $K\in\mathcal{T}_h$, which implies that the trace on $F$ is well defined. We need to prove that the last statement about the traces holds.

Any ideas?

I know that I should end up with $$\sum_F\int_F\left({v}|_{K_1}-{v}|_{K_2} \right)\varphi n_j=0$$ for every $\varphi\in C^{\infty}_0(\Omega)$.

• What you are trying to prove should be $\int_\Omega v \, \partial_j \varphi = -\sum_K \int_K \partial_j(v|_K) \, \varphi$. This is possible using integration by parts. – gerw Mar 24 '16 at 7:59
• Such sobolev functions are absolutely continuous along lines, and thus cannot jump in the interior of the domain. As such the restrictions to the the interior edges are the same. – Ellya Mar 24 '16 at 23:11

Pick a generic $\varphi\in C_0^\infty(\Omega)$.
1. The functional $T:H^1(\Omega) \to \mathbb{R}$ defined by $$T(v) := \sum_F \int_F (v\vert_{K_1}-v\vert_{K_2})\varphi n_j dS$$ is continuous.
2. $C^\infty (\bar{\Omega})$ is in the kernel of $T$ and is dense in $H^1(\Omega)$.
Thus, $H^1(\Omega)$ is in the kernel of $T$, too.