# Tag Info

1

Split $[0,1]$ into two elements $[0,1/2]$ and $[1/2,1]$. Consider the function $f$ satisfying $f(x)=0$ for $x < 1/2$ and $f(x)=1$ otherwise. For this function the $H^1$ norm is infinity but if you calculate the $H^1$ norm over the two elements separately you get 0 and $1/2$, respectively. In particular, the derivative of the Heaviside function is a delta ...

2

The statement as given is indeed correct (i.e., left and right hand side are different) for almost any function $f(x)$. What it says, once you write out what these norms are, is that $$\sum_m \sqrt{\int_{T_m} f(x)^2 } \neq \sqrt{ \sum_m \int_{T_m} f(x)^2 }.$$

1

I think $u - \Pi u$ is zero on a part of the boundary. Then you can use Poincare inequality to bound $\| u - \Pi u \|_0 \leq C| u - \Pi u|_1$. It could also be possible to look at the case $m = 0$ and argue the $L^2$ part is of higher order and, hence, smaller in the asymptotic limit.

6

The other answer has everything you already need, but it's also worth pointing out that $u_h$ is computable whereas $I_hu$ is not: The latter requires you to know the exact solution, which in general we of course don't know (and if we did, we didn't need to compute a Galerkin approximation $u_h$).

6

Let's assume $u$ is the solution to the variational problem and $u_h$ is the Galerkin approximation of the solution on the subspace $V_h \subset V$. By Cea's lemma you have: $$||u-u_h||_V \leq C\inf\limits_{v_h \in V_h}||u-v_h||_V$$ for some positive constant $C$. Now you define the projection $I_h:V \rightarrow V_h$. As you ...

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