Let $D\subset\mathbb{R}$ be a bounded interval and $f: D\times D \rightarrow \mathbb{R}$ a real-valued analytic function of two variables such that $f\in L_2(D\times D)$. Suppose we have upper bounds on the eigenvalues of the operator,
$ \mathcal{F} : L_2(D)\rightarrow L_2(D), \hspace{.2in} \mathcal{F}(u)(\cdot) = \int_{D}f(\cdot,y)u(y)dy. $
That is $\lambda_m(\mathcal{F})\leq C_m$ for some constant $C_m$ for all $m$. Suppose now, we pick $N\in\mathbb{N}$ real numbers $x_1,\ldots, x_N$ in $D$ and form the matrix, $F\in\mathbb{R}^{N\times N}$ such that,
$ F_{jk} = f(x_j,x_k). $
This is in some sense the discretised version of $\mathcal{F}$. Do we automatically have upper bounds on the eigenvalues of $F$? Ideally, bounds of the form
$ \lambda_m(F) \leq \lambda_m(\mathcal{F}) + \delta, \text{ } 1\leq m\leq N, \text{ } \delta<<1. $
As $N\rightarrow\infty$ and $\{x_k\}_{1\leq k\leq N}$ becomes dense in $D$, do the eigenvalues of $F$ approximate $\mathcal{F}$?
Wrong First attempt
For all $\epsilon>0$ there is an $N$ and points $\{x_k\}_{1\leq k\leq N}$ sufficiently dense in $D$ with non-negative weights $w_1,\ldots, w_N$ such that,
$ \left|\int_{D}\int_{D} f(x,y)u(y)u(x)dxdy - \sum_{j,k=1}^N f(x_j,x_k)u(x_j)w_ju(x_k)w_k\right|<\epsilon. $
Letting $\alpha_j=u(x_j)w_j$ we have,
$ \sum_{j,k=1}^N f(x_j,x_k)\alpha_j\alpha_k < \int_{D}\int_{D} f(x,y)u(y)u(x)dxdy + \epsilon $
Can this be used to bound the Rayleigh quotient and hence the eigenvalues of $F$?
