Let $(M,g)$ be a manifold with metric $g$ parametrized by the mapping $S$ and parametric domain $\Omega$. The sobolev space of order one with respect to the $L_2(M)$-norm $H^1_2(M)$ is defined as follows: Let $C_2^1 = \{ u \in C^\infty(M) s.t. \forall j = 0,1: \int_M |\nabla^j u |^2 dv(g) < \infty \}$, where $\nabla^0 u = u$ and $\nabla u $ is the ordinary gradient in local coordinates (i.e. in $\Omega$). Note that with above metric, we have $|\nabla u|^2 = (\nabla u)^T g \nabla u$. The Sobolev space $H^1_2(M)$ is the completion of $C_2^1$ with respect to the norm \begin{align} \left \| u \right \|_{H^1_2} = \sum_{j = 0}^1 \left(\int_M | \nabla^j u |^2 dv(g) \right)^{1/2}. \end{align}
I have a subtle problem with this definition. As an example, suppose that $\Omega = [0,1]$ and \begin{align} u(x) = \left \{ \begin{array}{ll} 1-2x & \quad 0 \leq x \leq 1/2 \\ 0 & \quad 1/2 < x \leq 1 \end{array} \right. . \end{align} Suppose $M$ is just the line-segment $\Omega$ mapped onto another line in $\mathbb{R}^2$. Then, I believe, $u \in H_2^1(M)$ should be true. Now suppose that $[0,1]$ is mapped onto a closed circle (via the mapping $S(x) = (r \cos(2 \pi x),r \sin(2 \pi x)$)), then I feel like $u \notin H_2^1(M)$, this is because on the circle, the function $u$ contains a `jump' at the point where the the boundaries of $\Omega$ connect.
The problem that I am having is that $\left \| u \right \|_{H_2^1} < \infty$ even when $M$ is a circle. The integral is carried out in $\Omega$ and there the discontinuity does not exist. I believe that the reason that $u$ is not in the sobolev space on the circle must be that it is not contained in the completion of $C_2^1$ in that case. However, I have no idea how to demonstrate that.
More specifically, I'd like to demonstrate that functions that contain discontinuities on $M$, in general, are not contained in $H_2^1(M)$, also when $M$ is an n-dimensional manifold.
Can somebody help me ? Thank you in advance !