an $L^2$-estimate for finite element solutions Let $(\tau_h)_h$ be a quasi-uniform family of triangulations. Show that for $u_h\in V^h$ the following holds:
$$c||u_h||_{L^2(\Omega)}^2 \leq h^2 \sum_{i=1}^N |u_h(z_i)|^2 \leq C ||u_h||_{L^2(\Omega)}$$
where $z_i$ denote the nodes of the basis and the finite element space $V^h$ is chosen such that the functionals in $\Sigma$ consist of point evaluations.
I'm quite clueless with this one. I don't even know how to approach this.
 A: 
Since our triangulation is quasi-uniform, $V^h$ is a finite dimensional space.

This is correct, but not 100% accurate, whether $\dim V^h< \infty$ has more to do with the space we choose, rather than the quasi-uniformity of the triangulation. For e.g., we can choose all the Legendre polynomials on just one simplex and the dimension is infinity. "Quasi-uniform" is just a qualitative assessment of the triangulation itself, opposed to uniform (see the following pictures, first is uniform vs second quasi-uniform). 


The definition is that (1) the ratio of inscribed sphere over circumsphere is greater than some positive constant for all element, so that the elements have a rather "uniform" shape. This is called "regular". (2) The areas of the triangles have to be roughly the same, i.e., the biggest area divided by the smallest area in this triangulation is bounded by a fixed constant even when the mesh size goes to zero.
For the example of not quasi-uniformity, the triangles taking rather a very regular shape near the center of the domain, but become flatter, and flatter when approaching the boundary, eventually vanish to a line. You can refer to the link in the Prof O'Rourke's answer here to see the example.
However, if a triangulation is quasi-uniform, then the following situation can NOT happen:
$$
\text{There exists some vertex, so that there are infinitely many simplices around it.}\tag{1}
$$


I don't really know what $V^h$ is, the exercise doesn't say it.

I am fairly certain that Ciarlet means Lagrange-type finite element with fixed degree, a.k.a, piecewise locally-supported polynomial space with degrees $\leq n$ which is a fixed integer. Then for this kind of finite element spaces, for any $u_h \in V^h$, restricting on one element $T$
$$
u_h|_T(x) = \sum^{N(T)}_{i=1} u_h(z_i) \lambda_{z_i}(x) ,
$$
where $\lambda_{z_i}$ is called the nodal basis function (see Brenner-Scott chapter 3, they are called the set of nodal variables there), and $\lambda_{z_i}$ are fixed degree polynomials, the $N(T)$ is the number of the nodal basis within one element $T$. The proof is a classical result, and can be found on Ern and Guermond's book, page 386, Lemma 9.7, which uses (1).
A: I have an answer myself now:
We can define $|||u|||:=h\sqrt{\sum_{i=1}^N |u_h(z_i)|^2}$ and show that this is a norm on $V^h$, which is rather easy. Since our triangulation is quasi-uniform, $V^h$ is a finite dimensional space, i.e. $\dim V^h <\infty$, so we're dealing with a finite dimensional vector space. 
Hence, all norms on this space are equivalent, so by having shown that $|||\cdot|||$ is a norm, we can deduce that there exist constants $c,C>0$ such that the proposition holds.
