A small Levin collocation residual can hide a large endpoint error

Direct differentiation gives $p_{\epsilon }(0)=1$, $p_{\epsilon }(1)=0$, $p_{\epsilon }^{\mathrm{\prime }}(0)=-i\omega$, and $p_{\epsilon }^{\mathrm{\prime }}(1)=\epsilon$. Therefore $L_{\omega }{p}_{\epsilon }(0)=0$ and $L_{\omega }{p}_{\epsilon }(1)=\epsilon$. The basis functions are linearly independent because $p_{\epsilon }$ is nonconstant, and [F1] gives $A_{\epsilon }=\left[\begin{array}{cc}i\omega & i\omega \\ i\omega & i\omega +\epsilon \end{array}\right],\det A_{\epsilon }=i\omega \epsilon \mathrm{\ne }0.$

For $f=0$, the exact right-hand side and solution are zero. Multiplying $A_{\epsilon }$ by $\widehat{c}=(-1,1)$ gives $\rho =(0,\epsilon )$, while the associated polynomial is $\widehat{p}=-{\psi }_1+{\psi }_2=p_{\epsilon }$. Its endpoint values give $Q(\widehat{c})=0\cdot e^{i\omega }-1=-1$. Hence the exact integral is $0$ and the endpoint error has magnitude $1$ although $\mathrm{\parallel }\rho {\mathrm{\parallel }}_2=\epsilon$.

Solving $A_{\epsilon }z=(0,1)$ gives $z=(-1\mathrm{/}\epsilon ,1\mathrm{/}\epsilon )$, so $\mathrm{\parallel }{A}_{\epsilon }^{-1}{\mathrm{\parallel }}_2\ge \sqrt{2}\mathrm{/}\epsilon$. Also $\mathrm{\parallel }{A}_{\epsilon }{\mathrm{\parallel }}_2\ge \mathrm{\parallel }{A}_{\epsilon }(1,0{)}^T{\mathrm{\parallel }}_2=\sqrt{2}\omega$. Thus ${\kappa }_2(A_{\epsilon })\ge 2\omega \mathrm{/}\epsilon \to \mathrm{\infty }$ as $\epsilon \downarrow 0$. The factor in [F2] records this amplification; the residual norm alone cannot certify a small quadrature error.