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I need to understand what is the right expression for the "discrete $H^1$-norm of a function v(which of course need to be in $H^1$).
By definition, $$||v||_{H^1}^2 = ||\nabla v||_{2}^2 + ||v||_2^2 $$
so far I'm still in the continuous case. In practice if I need to compute it I have to do the discrete $L^2$ norm of the gradient and of $v$.
For $v$ I have: $$||v||_2^2 = h \sum_{i=0}^{N} v(x_i)^2$$
But what about the $$||\nabla v||_2^2$$? How can I approximate it?
Assuming 1-D and equidistant gridpoints with spacing $h$ and some form of homogenous boundary conditions, we can use $\|\nabla v\|^2\approx -h\sum_{i=1}^nv(x_i)D_2v(x_i)$, where $D_2$ is a finite difference discretization of the Laplacian operator, which is usually some variant of a tridiagonal matrix with values $(1,-2,1)/h^2$ along the sub/main/super diagonal, respectively. I'm pretty sure this approximation is $O(h)$. This formula comes from integration by parts:
$$ -\int_\Omega v\Delta vdx = \int_\Omega\|\nabla v\|^2dx-\int_{\partial\Omega}v\nabla v\cdot dS. $$ Any boundary conditions that make the boundary terms disappear will make this approximation work. This can also be extended to higher dimensions as long as you can approximate the Laplacian and the integral well, which are usually not too hard if you are already discretizing things in some sort of scheme.
