There is a uniform triangulation $\tau_h$. $v_h(x)$ is a piecewise linear function such that $$v_h(x)=\sum_{i=1}^nv_i\phi_i(x)$$ where the $v_i$'s are the nodal values, $\phi_i(x)$'s are the hat functions, and $n$ is the number of nodes. Also $\textbf{v}$ is the vector of the nodal values. Have to show that $$c_1\|v_h\|_{L^2(\Omega)}\leq h\|\textbf{v}\|_2\leq c_2\|v_h\|_{L^2(\Omega)}$$ where $\|\cdot\|_2$ is the Euclidean norm, and $c_1,c_2$ are constants.
I think I have got the right portion of the inequality using Poincare-Friedrichs inequality and the Inverse Estimate. But cannot figure out a proof for the left portion of the inequality. First I thought to go through the routine to prove that the two norms are equivalent but then I saw that of what the norms are being taken are not the same. Also the presence of $h$ means it is not saying that the norms are equivalent.
Best Answer
First prove an estimate on a fixed cell $K$ of the triangulation: $$ \|v_h\|_{L^2(K)}^2 \le c \cdot |K| \cdot (v_{K,1}^2 + \dots + v_{K,d+1}^2), $$ where $v_{K,1}... v_{K,d+1}$ are the nodes values of $v_h$ in $K\subset\mathbb R^d$. Use element mass matrix to prove this. Constant $c$ is independent of $K$.
Then use the fact that each node is element of at most $L$ cells $K$, $L$ indepedent of the actual triangulation (follows from uniformity of triangulation). $$ \|v_h\|_{L^2(\Omega)}^2 = \sum_K \|v_h\|_{L^2(K)}^2 \le \sum_K c \cdot |K| \cdot (v_{K,1}^2 + \dots + v_{K,d+1}^2) \le L \ c \ h^d \|v\|_2^2, $$ where I used $|K| \le c h^d$.