At aspect one the reciprocal-spectrum limit diverges

With $H={\delta }_1$ and $c=1$, the population-resolvent condition holds for every unit vector, so [F1] gives convergence to $F_{1,{\delta }_1}$. The transform of the law in [F3] satisfies the same fixed-point equation; [F5] identifies $F_{1,{\delta }_1}={\mu }_1$. Substituting $a_1=0$, $b_1=4$ into the density in [F3] gives ${\rho }_1(t)=\sqrt{4-t}\mathrm{/}(2\pi \sqrt{t})$ on $(0,4)$ and zero atom.

For fixed $s>0$, $t\mapsto (t+s{)}^{-1}$ is bounded and continuous, so [F2] gives the asserted limit of the negative-axis integrals. For $0<t<1$, [F3] gives ${\rho }_1(t)\ge \sqrt{3}(2\pi {)}^{-1}{t}^{-1\mathrm{/}2}$; hence $t^{-1}{\rho }_1(t)\ge \sqrt{3}(2\pi {)}^{-1}{t}^{-3\mathrm{/}2}$, whose integral at zero diverges. Along $s_k=1\mathrm{/}k$, the nonnegative functions $(t+s_k{)}^{-1}$ increase to $t^{-1}$ for $t>0$. The law has no atom at zero by [F3], and [F6] gives $\int (t+s_k{)}^{-1}d{\mu }_1(t)\uparrow \int t^{-1}d{\mu }_1(t)=\mathrm{\infty }$. Since these integrals are nondecreasing as $s\downarrow 0$, the full limit is $+\mathrm{\infty }$.

Work on the probability-one weak-convergence event of step 1.1. For each integer $j\ge 1$, set $b_j=2^{-j}$ and $a_j=b_j\mathrm{/}2$, and define the bounded continuous real function ${\phi }_j(t)=\max \{0,1-\mathrm{\mid }2t-a_j-b_j\mathrm{\mid }\mathrm{/}(b_j-a_j)\}$ on $\mathbb{R}$. It vanishes outside $[a_j,b_j]$, is positive exactly on $(a_j,b_j)$, and is at least $1\mathrm{/}2$ on the middle half $J_j$ of that interval. On $(a_j,b_j)$, [F3] gives ${\rho }_1(t)\ge \sqrt{4-b_j}\mathrm{/}(2\pi \sqrt{b_j})>0$. Applying the order and scalar rules in [F8] first to ${\phi }_j\ge \frac{1}{2}{\mathbf{1}}_{J_j}$ under ${\mu }_1$, then to this density lower bound on $J_j$ under Lebesgue measure, gives $\int {\phi }_{j}d{\mu }_1\ge \frac{1}{2}{\mu }_1(J_j)\ge (b_j-a_j)\sqrt{4-b_j}\mathrm{/}(8\pi \sqrt{b_j})>0$, since $J_j$ has length $(b_j-a_j)\mathrm{/}2$. By [F4] and step 1.1, $\int {\phi }_{j}d{\mu }_n^{x_n}\to \int {\phi }_{j}d{\mu }_1$, so for each $j$ this sample integral is eventually positive. Since $0\le {\phi }_j\le {\mathbf{1}}_{(a_j,b_j)}$, [F8] gives $0<\int {\phi }_{j}d{\mu }_n^{x_n}\le {\mu }_n^{x_n}((a_j,b_j))$. The finite atomic set formula in [F4] makes the right side ${\sum }_{\lambda \in \mathrm{spec}(A_n)\cap (a_j,b_j)}\mathrm{\parallel }{P}_{\lambda }{x}_n{\mathrm{\parallel }}_2^2$; therefore there is an actual eigenvalue in $(a_j,b_j)$.

These conclusions hold for all $j$ on the same weak-convergence event, since its bounded-continuous-test property applies to each ${\phi }_j$. In particular $j=1$ ensures at least one positive eigenvalue for every sufficiently large $n$, so the smallest positive eigenvalue ${\lambda }_{\min }^{}(A_n)$ is eventually defined. For every fixed $j$, step 2.2 gives $0<{\lambda }_{\min }^{}(A_n)<b_j$ for all sufficiently large $n$; since $b_j\downarrow 0$, it follows that ${\lambda }_{\min }^{}(A_n)\to 0$ on that probability-one event. This excludes an eventual uniform positive lower bound and proves the stated gap failure for the full complex entry class without claiming convergence of finite-sample reciprocal traces.