See this image (from this work)
The existence is done through Lax-Milgram (at least for $\sigma = \frac{1}{2}$), I think. However, why the author only includes the gradient in defining $H^1$? Is it because the only constant in $L^2(D_n)$ is zero? I am not at all sure that this defines a norm in $H^1$ in a sensible fashion.
They give no details for (2.6) to (2.7). In particular why does that trace inequality hold?
Let us take $\sigma = \frac{1}{2}$ for simplicity.
The key idea is that $U=u$ on the boundary where $u\in H^{1/2}(S^n)$ is given. Hence, the Poincare inequality works in some sense and $\|\nabla U\|_{L^2}$ will be an equivalent norm of $\|U\|_{H^1}$. To see how, firstly, we trivially have $\|\nabla U\|_{L^2}\leq \| U\|_{H^1}$.
Now for the converse. Notice that $U-u\in H_0^1$ and hence by poincare we have that $$ \|U-u\|_{L^2}\leq C\|\nabla (U-u)\|_{L^2} \leq C\|\nabla U\|_{L^2}+C\|\nabla u\|_{L^2}$$ and hence we have $$\|U\|_{L^2}\leq C\|\nabla U\|_{L^2}+C\|\nabla u\|_{L^2}+\|u\|_{L^2}$$
But here $C\|\nabla u\|_{L^2}+\|u\|_{L^2}$ is only a constant and hence you have "Poincare" and that equivalent norm works.