A near-resonant Nyström equation amplifies midpoint quadrature error

Substituting $u=1+cx$ into $u=1+Ku$ gives $c=\lambda (1\mathrm{/}2+c\mathrm{/}3)$, so $c(1-\lambda \mathrm{/}3)=\lambda \mathrm{/}2$. Since $1-\lambda \mathrm{/}3=\delta >0$, the unique coefficient is $c=\lambda \mathrm{/}(2\delta )$, proving the exact formula.

On each midpoint panel, direct integration gives ${\int }_{jh}^{(j+1)h}ydy=h(j+\frac{1}{2})h$ and ${\int }_{jh}^{(j+1)h}{y}^{2}dy-h((j+\frac{1}{2})h{)}^2=h^3\mathrm{/}12.$ Summing the $N$ panels yields $Q_N(y)=1\mathrm{/}2$ and $Q_N(y^2)=1\mathrm{/}3-h^2\mathrm{/}12$.

Any discrete solution has the form $u_N(x)=1+c_{N}x$ because the range of $K_N$ is contained in the span of $x$. Its coefficient satisfies $c_N=\lambda (1\mathrm{/}2+c_N(1\mathrm{/}3-h^2\mathrm{/}12))$, hence $c_N=\lambda \mathrm{/}(2(\delta +\lambda h^2\mathrm{/}12))$; the denominator is positive, so the discrete solution is unique.

On the direction $x$, $Kx=(\lambda \mathrm{/}3)x$, so $I-K$ scales that direction by $\delta$. Also $K_{N}x=\lambda (1\mathrm{/}3-h^2\mathrm{/}12)x$, so $I-K_N$ scales it by $\delta +\lambda h^2\mathrm{/}12$. Directly using the exact and discrete coefficients from steps 1.1 and 2.1 gives $\mathrm{\parallel }{K}_{N}u-Ku{\mathrm{\parallel }}_{\mathrm{\infty }}=\frac{{\lambda }^2{h}^2}{24\delta },\mathrm{\parallel }{u}_N-u{\mathrm{\parallel }}_{\mathrm{\infty }}=\frac{{\lambda }^2{h}^2}{24\delta (\delta +\lambda h^2\mathrm{/}12)}\mathrm{.}$ The ratio of solution error to quadrature residual is exactly $(\delta +\lambda h^2\mathrm{/}12{)}^{-1}$, and the stated lower bound follows when $\lambda h^2\mathrm{/}12\le \delta$.