# Inequality from eigenvalue relations

This arises from the Saddle Point theorem as found in Rabinowitz "Minimax Methods in Critical Point Theory", but I'll just provide the relevant information to what I'm trying to understand.

We consider $$-\Delta v = \mu a(x)v$$ in some smooth domain $\Omega$, $v = 0$ on $\partial \Omega$, $a$ is sufficiently smooth. This has a sequence of eigenvalues $\lambda_j \rightarrow \infty$ as is well known.

Let $V \equiv \text{span}\{v_1 \dots v_k\}$, where $v_j$ is an eigenfunction corresponding to $\lambda_j$ and normalized so that $\int_{\Omega}|\nabla v_j|^2 = 1 =\lambda_j\int_{\Omega}av_j^2$

Let $X$ = span$\{v_{k+1} \dots\}$ (so $X = V^{\perp}$)

Here's the claim I don't get: If $u\in X$, then $u = \sum_{j = k+1}^{\infty} a_jv_j$ and $$\int_{\Omega}{|\nabla u|^2-\lambda_k au^2} = \sum_{j = k+1}^{\infty}{a_j^2\left(1-\frac{\lambda_k}{\lambda_j}\right)} \geq\left(1-\frac{\lambda_k}{\lambda_{k+1}} \right)||u||^2$$

To begin with, observe that we are dealing with the eigenvalues of the operator $Tv = -a^{-1}\Delta v$ (because $Tv=\mu v$ is what we are after). The operator $T$ is self-adjoint on the weighted space $L^2_a$ with the inner product $\langle f,g\rangle =\int fga$. (Although you did not say so, I'm assuming $a>0$.) Indeed, $$\langle Tu,v\rangle= \langle u,Tv\rangle = \int \nabla u\cdot \nabla v$$ The eigenvectors of a self-adjoint operator are orthogonal, when they correspond to different eigenvalues. By normalization, $\langle v_i, v_j\rangle = \lambda_i^{-1}\,\delta_{ij}$. Therefore, $$\int au^2 = \langle u,u\rangle = \sum_{j>k} \lambda_j^{-1}a_j^2$$ which explains one piece of the puzzle. The other piece of the puzzling identity is $$\int |\nabla u|^2 = \langle Tu,u\rangle = \sum_{j>k} \langle Tu_j,u_j\rangle = \sum_{j>k}\int |\nabla u_j|^2 = \sum_{j>k} a_j^2$$ As for the inequality at the end, it needs some clarification because $j$ has no meaning outside of summation.