Can anyone explain why the Sobolev spaces based on completion are important for the finite element method?
In the lecture we discussed the Sobolev spaces $W^m(\Omega)$ defined as the set of all $L^2$ functions for which all the $\alpha$-derivative with $|\alpha| \leq m$ exists. And then we discussed the spaces $H^m$ based on completion. We showed that they are the same. I guess I do not understand what this equality means for the FE method.
Thank you in advance!
Many finite element methods are based on the weak formulation of the corresponding PDE. For example the Poisson problem $-\Delta u = f$ with homogeneous Dirichlet boundary conditions is given in weak form: we seek $u\in H^1_0$ such that
$$\int\nabla u\cdot\nabla v = \int f v\quad\forall v\in H^1_0.$$
The well posedness of this problem follows from the Lax--Milgram Theorem, which requires the completeness of the space $H^1_0$.
Therefore I would say that the completeness of Sobolev spaces is more related to the PDE than the finite element method, but generally speaking, if the PDE is not well posed then it is unlikely that a corresponding finite element method is either.