A spectral gap persists on a computable matrix ball with a uniform resolvent bound

Since ${\lambda }_{r+1}(H_0)={\lambda }_r(H_0)+g$, we have $R={\lambda }_r(H_0)-{\lambda }_1(H_0)+g\mathrm{/}2\ge g\mathrm{/}2>0$. Reversing the descending-index convention in [F1] gives the same perturbation bound for the ascending eigenvalues used here. Thus if $\mathrm{\parallel }H-H_0{\mathrm{\parallel }}_{op}\le g\mathrm{/}4$, then ${\lambda }_1(H)\ge c-g\mathrm{/}4$ and ${\lambda }_r(H)\le c+R-g\mathrm{/}4$. Since $R-g\mathrm{/}4\ge g\mathrm{/}4$, every first-$r$ eigenvalue therefore satisfies $\mathrm{\mid }{\lambda }_j(H)-c\mathrm{\mid }\le R-g\mathrm{/}4$, so it lies at least $g\mathrm{/}4$ inside the circle centered at $c$.

The same eigenvalue estimate gives ${\lambda }_{r+1}(H)\ge c+R+g\mathrm{/}4$, so all remaining eigenvalues satisfy $\mathrm{\mid }{\lambda }_j(H)-c\mathrm{\mid }\ge R+g\mathrm{/}4$ and lie at least $g\mathrm{/}4$ outside the circle. Also ${\lambda }_{r+1}(H)-{\lambda }_r(H)\ge g-2(g\mathrm{/}4)=g\mathrm{/}2$. Thus the contour separates exactly the occupied cluster and the stated gap persists.

By [F2], the spectrum of $H$ is real and $H$ diagonalizes unitarily. For every $z\in \mathrm{\Gamma }$, reverse triangle gives $\mathrm{\mid }z-{\lambda }_j(H)\mathrm{\mid }\ge \mathrm{\mid }\mathrm{\mid }z-c\mathrm{\mid }-\mathrm{\mid }{\lambda }_j(H)-c\mathrm{\mid }\mathrm{\mid }\ge g\mathrm{/}4$ by steps 1.1 and 2.1. In the unitary eigenbasis, $(zI-H{)}^{-1}$ is diagonal with entries of modulus at most $4\mathrm{/}g$. Applying [F3] to a unit vector in this basis gives $\mathrm{\parallel }(zI-H{)}^{-1}{\mathrm{\parallel }}_{op}\le 4\mathrm{/}g$.