The equivalence of Euclidean norm and finite element norm There is a bijection between the functions in the finite element space $V^h$ (considering on linear finite element space $\mathbb{P}_1$) and the coefficients of the finite element functions in the space $\mathbb{R}^{n_p}$. This bijection is denoted by $P^h: \mathbb{R}^{n_p} \rightarrow V^h,\ \mathbf{v}^h \mapsto v^h$ with:
$$ v^h = \sum_{i=1}^{n_h} v_i^h\varphi_i^h(x),\ \ \mathbf{v}^h = (v_i^h)$$
If the Euclidean space $\mathbb{R}^{n_h}$ is equipped with the standard Euclidean norm, then the norm equivalence:
$$ C_0h^{1/_2}\left\| \mathbf{v}^h \right\|_2 \leq \left\| P^h\mathbf{v}^h \right\|_{L^2((0,1))} \leq C_1h^{1/_2}\left\| \mathbf{v}^h \right\|_2 $$
holds with constants that are independent of the mesh size.
Ps: I tried to use the form of $P^h\mathbb{v}^h$ directly into the inequality but it is very complicated. I think it would use some properties of Euclidean norm, but I forgot a lot of them. Thanks a lot.
 A: There are three simple steps to show this:


*

*express the norm of $v_h$ element-wise,

*compute or estimate the eigenvalues of element mass matrices,

*relate $\|\boldsymbol{v}\|_2$ to the element-wise bounds.


Each of the steps is developed below.
Assume that the partitioning of $(0,1)$ is uniform so each element is an interval of the length $h$.
Let
$$
v_h := \sum_{i=1}^n v_i \varphi_i.
$$
We can express the $L^2$-norm of $v_h$ in an element-wise form
$$\tag{1}
\|v_h\|_{L^2(0,1)}^2 = (v_h,v_h)_{L^2(0,1)}=\sum_{K_i\in\mathcal{T}_h}(v_h,v_h)_{L^2(K_i)}.
$$
If $K_i=(x_i,x_{i+1})$ and $\varphi_i(x_j)=\delta_{ij}$, we have
$$\tag{2}
(v_h,v_h)_{L^2(K_i)}
=
\boldsymbol{v}_{K_i}^T\boldsymbol{M}_{K_i}
\boldsymbol{v}_{K_i},
\quad
\boldsymbol{v}_{K_i}=[v_i,v_{i+1}]^T,
$$
where
$$
\boldsymbol{M}_{K_i}
=\begin{bmatrix}
(\phi_i,\phi_i)_{L^2(K_i)} & (\phi_i,\phi_{i+1})_{L^2(K_i)} \\
(\phi_i,\phi_{i+1})_{L^2(K_i)} & (\phi_{i+1},\phi_{i+1})_{L^2(K_i)}
\end{bmatrix}
$$
is the mass matrix associated with the element $K_i$.
We want to show that
$$\tag{3}
c_0h\boldsymbol{u}^T\boldsymbol{u}\leq\boldsymbol{u}^T\boldsymbol{M}_{K_i}\boldsymbol{u}\leq c_1h\boldsymbol{u}^T\boldsymbol{u},
$$
where $c_0$ and $c_1$ are independent of $h$, that is, the eigenvalues of $\boldsymbol{M}_{K_i}$ are contained in the interval $[c_0h, c_1h]$. Indeed, evaluating the integrals gives
$$
\boldsymbol{M}_{K_i}=h\begin{bmatrix}1/3&1/6\\1/6&1/3\end{bmatrix},
$$
so $c_0=1/6$ and $c_1=1/2$.
We need to somehow relate the element-wise sums of $\|\boldsymbol{v}_{K_i}\|_2^2$ to the 2-norm of the whole vector $\|\boldsymbol{v}\|_2^2$. 
This needs a little care because each interior node is contained in two elements and the boundary ones are only in one (unless $\mathcal{T}_h$ contains just one element). We have
$$
\|\boldsymbol{v}\|_2^2 = \sum_{i=1}^n v_i^2 
=\frac{1}{2}\left[v_1^2+v_n^2+\sum_{i=1}^{n-1}(v_i^2+v_{i+1}^2)\right]
=\frac{1}{2}\left[v_1^2+v_n^2+\sum_{K_i\in\mathcal{T}_h}\boldsymbol{v}_{K_i}^T\boldsymbol{v}_{K_i}\right]
$$
so
$$\tag{4}
\frac{1}{2}\sum_{K_i\in\mathcal{T}_h}\boldsymbol{v}_{K_i}^T\boldsymbol{v}_{K_i}\leq\|\boldsymbol{v}\|_2^2 \leq \sum_{K_i\in\mathcal{T}_h}\boldsymbol{v}_{K_i}^T\boldsymbol{v}_{K_i}.
$$
Putting (1), (2), (3), and (4) together, we have
$$
\|v_h\|^2_{L^2(0,1)}
\leq c_1h\sum_{K_i\in\mathcal{T}_h}\|\boldsymbol{v}_{K_i}\|_2^2
\leq 2c_1h\|v\|_2^2
$$
and
$$
\|v_h\|^2_{L^2(0,1)}
\geq c_0h\sum_{K_i\in\mathcal{T}_h}\|\boldsymbol{v}_{K_i}\|_2^2
\geq c_0h\|v\|_2^2
$$
so finally
$$
C_0h^{1/2}\|\boldsymbol{v}\|_2\leq\|v_h\|_{L^2(0,1)}\leq C_1h^{1/2}\|\boldsymbol{v}\|_2
$$
with
$$
C_0:=\sqrt{c_0}=\frac{1}{\sqrt{6}}, \quad
C_1:=\sqrt{2c_1} = 1.
$$
