Relative stability of the HHL filter on supported eigenvalues

At $u=1\mathrm{/}(2\kappa )$ the transition angle is $0$, so [F5] makes the transition values $(0,1\mathrm{/}2)$, matching the lower plateau; at $u=1\mathrm{/}\kappa$ the angle is $\pi \mathrm{/}2$, so [F4] makes them $(\mathrm{sgn}(z)\mathrm{/}2,0)$, matching the reciprocal branch. Hence $m$ is continuous on each fixed-sign half-line. On the reciprocal branch, for $u,u^{\mathrm{\prime }}\ge 1\mathrm{/}\kappa$, $\mid \frac{1}{2\kappa u}-\frac{1}{2\kappa u^{\mathrm{\prime }}}\mid =\frac{\mathrm{\mid }u-u^{\mathrm{\prime }}\mathrm{\mid }}{2\kappa u{u}^{\mathrm{\prime }}}\le \frac{\kappa }{2}\mathrm{\mid }u-u^{\mathrm{\prime }}\mathrm{\mid }\mathrm{.}$ On the transition branch, [F2] bounds each coordinate change by $(\pi \kappa \mathrm{/}2)\mathrm{\mid }u-u^{\mathrm{\prime }}\mathrm{\mid }$, so the two-coordinate norm changes by at most $(\pi \kappa \mathrm{/}\sqrt{2})\mathrm{\mid }u-u^{\mathrm{\prime }}\mathrm{\mid }$. The lower plateau is constant, and $\kappa \mathrm{/}2\le \pi \kappa \mathrm{/}\sqrt{2}$ by [F3]. Splitting any same-sign interval at the two breakpoints and adding these bounds proves the global same-sign Lipschitz estimate $\mathrm{\parallel }m(z)-m(z^{\mathrm{\prime }}){\mathrm{\parallel }}_2\le (\pi \kappa \mathrm{/}\sqrt{2})\mathrm{\mid }z-z^{\mathrm{\prime }}\mathrm{\mid }$.

Write $\mathrm{\Delta }=\mathrm{\mid }\tilde{\lambda }-\lambda \mathrm{\mid }=\mathrm{\mid }\delta \mathrm{\mid }\mathrm{/}{t}_0$. If $\tilde{\lambda }$ has the same sign as $\lambda$ and $\mathrm{\mid }\tilde{\lambda }\mathrm{\mid }\ge 1\mathrm{/}\kappa$, both values use the reciprocal branch, and $\frac{\mathrm{\parallel }m(\tilde{\lambda })-m(\lambda ){\mathrm{\parallel }}_2}{\mathrm{\mid }{f}_{\kappa }(\lambda )\mathrm{\mid }}=\frac{\mathrm{\Delta }}{\mathrm{\mid }\tilde{\lambda }\mathrm{\mid }}\le \kappa \mathrm{\Delta }\mathrm{.}$ If the signs differ (including zero), then $\mathrm{\Delta }\ge \mathrm{\mid }\lambda \mathrm{\mid }$, and the norm difference is at most $1\mathrm{/}2+\mathrm{\mid }{f}_{\kappa }(\lambda )\mathrm{\mid }$ by [F1]. Dividing by $\mathrm{\mid }{f}_{\kappa }(\lambda )\mathrm{\mid }=1\mathrm{/}(2\kappa \mathrm{\mid }\lambda \mathrm{\mid })$ gives at most $1+\kappa \mathrm{\mid }\lambda \mathrm{\mid }\le 2\kappa \mathrm{\Delta }$.

It remains that the signs agree and $\mathrm{\mid }\tilde{\lambda }\mathrm{\mid }<1\mathrm{/}\kappa$. If $\mathrm{\mid }\lambda \mathrm{\mid }\le 2\mathrm{/}\kappa$, step 1.1 gives a norm difference at most $\pi \kappa \mathrm{\Delta }\mathrm{/}\sqrt{2}$. Since $\mathrm{\mid }{f}_{\kappa }(\lambda )\mathrm{\mid }=1\mathrm{/}(2\kappa \mathrm{\mid }\lambda \mathrm{\mid })\ge 1\mathrm{/}4$, the ratio is at most $2\sqrt{2}\pi \kappa \mathrm{\Delta }<(48\mathrm{/}5)\kappa \mathrm{\Delta }<10\kappa \mathrm{\Delta }$, using $\sqrt{2}<3\mathrm{/}2$ and [F3]. If $\mathrm{\mid }\lambda \mathrm{\mid }>2\mathrm{/}\kappa$, crossing below $1\mathrm{/}\kappa$ implies $\mathrm{\Delta }>\mathrm{\mid }\lambda \mathrm{\mid }\mathrm{/}2$ and $\kappa \mathrm{\Delta }>1$. By [F1] the ratio is at most $1+\kappa \mathrm{\mid }\lambda \mathrm{\mid }<1+2\kappa \mathrm{\Delta }<3\kappa \mathrm{\Delta }<10\kappa \mathrm{\Delta }$. These cases establish the claimed inequality.