I'm currently trying to understand a step of a proof of the following estimation:
$\displaystyle \sum_{j=2}^{N} \left(\left< \varphi_1-R_N\varphi_1, \varphi_{j,N} \right>_{L_2}\right)^2 \ \leq \ \displaystyle \sum_{j=1}^{N} \left(\left< \varphi_1-R_N\varphi_1, \varphi_{j,N} \right>_{L_2}\right)^2 \ \leq \ \| R_N \varphi_1 - \varphi_1 \|_{L_2}^2$
Where $\{ \varphi_k \}_{k=1}^{\infty}$ is a $L_2$-orthonormalbasis of $L_2(\Omega)$ and $\{ \varphi_{k,N} \}_{k=1}^{N}$ is a $L_2$-orthonormalbasis of some finite dimensional subspace $S_N \subset L_2(\Omega)$ with $\dim(S_N) = N$. Further information about the functions $R_N(\varphi_1) \in S_N$ are below if they're required to solve this problem.
I suppose that this has something to do with the orthonormality of the functions $\varphi_{j,N}$ since it kinda looks like the $ \mathbf{Parseval \ identity}$:
$\displaystyle \sum_{j=1}^{N} \left(\left< u_N, \varphi_{j,N} \right>_{L_2}\right)^2 \ = \ \| u_N \|_{L_2}^2$
Which holds for all $u_N \in S_N$, since $\{ \varphi_{k,N} \}_{k=1}^{N}$ is an $L_2$-orthonormalbasis of $S_N$.
But here we got $\varphi_1 \in L_2(\Omega)$, not necessarily be in $S_N$. I looked this proof up in three different sources and they all said nothing to this step, so i assume this is probably not very hard but i thought a whole day over this step and i couldn't figure out why this should be true, so i need some help.
I don't know if it's required, but here are also additional information what this $R_N$ and $\varphi_k$ is supposed to be:
$R_N:H^1_0(\Omega) \to S_N$ is some projector ("Ritz-Projector") on $H^1_0 \subset L_2(\Omega)$ defined as orthoprojector according to the $H^1_0(\Omega)$-scalar-product $a( \dot{} , \dot{} ) := \left< \nabla \dot{}, \nabla \dot{} \right>_{L_2}$, i.e. $R_N$ is defined by:
$a(R_N u, w_N) = \left<u,w_N\right>_{L_2} \ \ \forall w_h \in S_N$ and $u \in H^1_0(\Omega)$
The $\varphi_k$ and $\varphi_{k,N}$ are supposed to be $L_2$-normalized eigenfunctions of the weak formulation of the Laplace-Operator with zero-boundary-conditions, i.e.:
$-\Delta \varphi_k = \lambda_k \varphi_k$ on $H^1_0(\Omega)$ in the weak sense, and
$-\Delta \varphi_{k,N} = \lambda_{k,N} \varphi_{k,N}$ on $S_N$ in the weak sense
With sorted eigenvalues
$0 < \lambda_1 < \lambda_2 \leq \lambda_3 \leq \dots$, and
$0 < \lambda_{1,N} \leq \lambda_{2,N} \leq \dots$
I would be grateful for some help!
Thank You!
Perhaps one way to see this is to notice that if $f \in L^2$, then you can write $f = f_{//} + f_{\perp}$ where $f_{//} \in S_N$ and $f_{\perp} \in S_N^{\perp}$. In particular, $\|f\|_{L^2}^2 = \|f_{//}\|_{L^2}^2 + \|f_{\perp}\|^2_{L^2}$, which implies that $\|f\|_{L^2}^2 \geq \|f_{//}\|_{L^2}^2$. Notice that $\langle f, \varphi_{j,N}\rangle = \langle f_{//}, \varphi_{j,N} \rangle$ and so $$\sum_{j=1}^N \langle f, \varphi_{j,N}\rangle = \sum_{j=1}^N \langle f_{//}, \varphi_{j,N} \rangle = \|f_{//}\|_{L^2}^2 \leq \| f\|_{L^2}^2$$