A two-node Gaussian Nyström rule solves a rank-two separable equation exactly

Given: Countable Choice, $\mu$ Lebesgue measure on $[0,1]$, the kernel $k(x,y)=\alpha +\beta xy$ with $\alpha =1\mathrm{/}4$, $\beta =1\mathrm{/}2$, and $f(x)=1$.

The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$) and A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral: for continuous integrands, the Riemann and Lebesgue integrals on $[0,1]$ agree under Countable Choice.

Positive-measure Gaussian quadrature moment data: an order-$2m-1$ datum is a finite positive measure of positive mass whose support has at least $m+1$ points and whose moments through degree $2m-1$ are finite; for this example, the Lebesgue measure on $[0,1]$ meets these conditions by [F6] and [F7].

The m-node Gaussian quadrature rule for a positive measure: the Gaussian rule $Q_2$ uses its two nodes and weights to approximate integration against $\mu$.

Gaussian quadrature is exact through degree 2m−1 and has positive weights: for this order-3 datum, $Q_2(p)=\int pd\mu$ for every polynomial $p$ of degree at most $3$.

Nyström quadrature operators and their reconstructed equations: the Nyström equation replaces each integral $\int k(x,y)u(y)d\mu (y)$ by its quadrature rule and reconstructs $u$ from its nodal values.

Finite-node Nyström equations are equivalent to their reconstructed equations: solutions of the grid system reconstruct function solutions, and the nullspaces of its matrix and reconstructed operator are linearly isomorphic.

The relative Lebesgue measure of $[0,1]$ is $1$, and every relatively open neighborhood of $0$, $1\mathrm{/}2$, or $1$ contains a closed subinterval of positive length. Each such interval has its length as measure (A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included), so all three points lie in the relative support of $\mu$.

The functions $y^j$ for $0\le j\le 3$ are continuous, bounded, and Riemann integrable (A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion). Under [A1] they are Lebesgue integrable with finite moments, so the order-3 moment condition holds.

For $j=0,1,2$, the Riemann moment is ${\int }_0^1{y}^{j}dy=1\mathrm{/}(j+1)$ by the Riemann–Stieltjes fundamental theorem applied to $y^{j+1}\mathrm{/}(j+1)$ (The two FTC forms for a Riemann–Stieltjes integral with a $C^1$ integrator, For a natural $n\ge 1$ the function $x\mapsto x^n$ is differentiable everywhere with derivative $\iota (n)x^{n-1}$; for $n=0$ it is the constant $1$, with derivative $0$; for a natural $n\ge 1$ the function $x\mapsto x^{-n}$ is differentiable at every $x\mathrm{\ne }0$ with derivative $-\iota (n)x^{-n-1}$; consequently every polynomial function is differentiable at every real, with the derivative computed term by term); [A1] gives the same Lebesgue moments.

The Riemann integral is linear on continuous functions on $[0,1]$ (Integrable functions on $[a,b]$ form a set closed under sums and scalar multiples, and ${\int }_a^b(\lambda f+\mu g)=\lambda {\int }_a^{b}f+\mu {\int }_a^{b}g$).