Finite element method -- projection onto finite-dimensional subspace I am curious about the projection operator people use in finite element method where they project a function onto the finite-dimensional subspace $V_N$.
I read there the basis they use to generate $V_N$ is not orthogonal or orthnormal. In this case, what projection operator is used? Is it bounded? What properties does it have?
 A: It depends on how to define the "Degrees of Freedom" (the coefficient) associated with that subspace, and what kind of projection you want. 
When the basis functions $v_i$ are orthonormal, the coefficients are just $\langle\cdot,v_i\rangle$. Finite element space's basis is not meant to be orthogonal but rather be locally-supported. 
Here are two examples in a triangulation made of tetrahedra in $\mathbb{R}^3$:


*

*First degree piecewise continuous polynomial $\mathbb{P}^1 \subset H^1$. We can project using a pointwise interpolant:
$$v\mapsto \Pi v : = \sum_{z\in \{\text{Vertices}\}} v(z) \phi_{z},$$
where $\phi_z = 1$ on vertex $z$ and linearly decreases to zero on its neighbor vertices, zero elsewhere. Quasi-interpolant (when pointwise value is not well-defined) is used for projection sometimes as well:
$$
\mathcal{I}v : = \sum_{z\in \{\text{Vertices}\}} (\frac{1}{|\omega_z|}\int_{\omega_z}v )\phi_{z},
$$
where $\omega_z$ is the collection of tetrahedra sharing this vertex $z$.

*Whitney elements (lowest order Nédélec elements for electric fields) space which is a subspace of $H(\mathbf{curl})$, the projection can be this interpolant:
$$
\mathbf{v} \mapsto \Pi\mathbf{v}:= \sum_{e\in \{\text{Edges}\}} (\int_{e}\mathbf{v}\cdot \boldsymbol{\tau}\,ds)\boldsymbol{\psi}_{e},
$$ 
where $\boldsymbol{\psi}_{e}$ is the Whitney 1-form for edge $e$ consisting two vertices $z_i$, $z_j$: 
$$
\boldsymbol{\psi}_{e} = \phi_{z_i}\nabla \phi_{z_j} - \phi_{z_j}\nabla \phi_{z_i}.
$$
This lives in a bigger picture of discrete de Rham cohomology, if you are interested, Higher-order finite element methods by Šolín, Segeth, and Doležel is a good read.
The interpolant (not the quasi- one) is constructed in a way so that
$$
\Pi (\Pi v) = \Pi v,
$$
i.e., the interpolant is a projection. Also we have the commutative diagram to inherit the geometrical aspect of de Rham complex:
$$\require{AMScd}
\begin{CD}
H\Lambda^0 @>{\mathrm{d}^0}>> H\Lambda^1 @>{\mathrm{d}^1}>>H\Lambda^2 @>{\mathrm{d}^2}>> H\Lambda^3
\\
@V{\Pi^0_h}VV @V{\Pi^1_h}VV @V{\Pi^2_h}VV @V{\Pi^3_h}VV
\\
P\Lambda^0_h @>{\mathrm{d}^0}>> P\Lambda^1_h @>{\mathrm{d}^1}>>P\Lambda^2_h @>{\mathrm{d}^2}>> P\Lambda^3_h
\end{CD}$$
Top complex is for the continuous function spaces, bottom is for the finite element subspaces, it is commutative in the sense that
$$
P\Lambda^{i+1}_h = \Pi^{i+1}_h\big(\mathrm{d}^i H\Lambda^i\big) = \mathrm{d}^i\big( \Pi^{i}_h H\Lambda^i\big).
$$

EDIT: leshik pointed out $L^2$-projection. He said they usually use orthogonal basis, I doubt that. In computation community, using orthogonal basis normally is accredited with spectral methods. 
(weighted) $L^2$-projection is certainly a projection too, the coefficients of the basis again are not just $\langle\cdot,v_i\rangle$. Instead, we have to solve a linear system for the projection $\Pi u = \sum u_i v_i$:
$$
\langle \Pi u,v_j\rangle_A = \int_{\Omega} A(\sum u_i v_i) v_j = \langle  u,v_j\rangle_A =  \int_{\Omega} A u v_j \quad j= 1,\ldots,N,
$$
where $A$ is a weight, and $\Pi(\Pi u) = \Pi u$ as well.
