solution of this equation in (\[Gdiag\]) one finds $$\begin{aligned}
\langle\vec{K}\,\vec{k}| {\cal G}(z&)& |\vec{K}\,\vec{k}'\rangle  =
\frac{1}{z-E_0(\vec{K},\vec{k})} \left\{ \vphantom{\frac{1}{2^2}
\sum_{3^4}}
\delta_{\vec{k},\vec{k}'}\right.
\nonumber \\
&& +\left.
\frac{2U/N}{1-2U{\cal R}(z,\vec{K})} \frac{1}{z-E_0(\vec{K},\vec{k}')}
\right\}.\nonumber\\
\label{gsol}\end{aligned}$$ Note that the off-diagonal elements are of order $N^{-1}$ while the diagonal elements are of order 1. With (\[gsol\]) we can calculate the level density from $$\mathop{\rm Tr}\nolimits {\cal G}(z)=\sum_{\vec{K},\vec{k}}
\langle\vec{K}\,\vec{k}| {\cal G}(z)
|\vec{K}\,\vec{k} \rangle,
\label{trace}$$ leading to the compact expression $$\mathop{\rm Tr}\nolimits {\cal G}(z) =\mathop{\rm Tr}\nolimits
{\cal G}_0(z) +\frac{\partial}{\partial z} \sum_{\vec{K}}
\ln \left[ 1-2U{\cal R}(z,\vec{K}) \right] ,
\label{trdz}$$ which in turn yields for the difference $\rho_2-\rho_2^0$ $$\begin{aligned}
\rho_2(E)&-&\rho_2^0(E)=\nonumber\\
&-&\frac{1}{\pi}\mathop{\rm Im}\nolimits\left[ \frac{\partial}
{\partial z}
\sum_{\vec{K}}\ln \left( 1
-2U{\cal R}(z,\vec{K})\right) \right]_{z=E+i\epsilon}.\nonumber\\
\label{rhodiff}\end{aligned}$$

This expression holds generally for any on-site repulsion $U$. We may let $U\rightarrow\infty$, by which it will disappear from the formula as the $U$ term under the logarithm starts to dominate the argument for any $z$ and $\vec{K}$. Omitting the 1 in the argument of the logarithm, the term $\ln( -2U)$ drops out after differentiation with respect to $z$. So for (\[rhodiff\]) we have in the limit $U\rightarrow\infty$ the equivalent expression $$\rho_2(E)-\rho_2^0(E)=-\frac{1}{\pi}\mathop{\rm Im}\nolimits\left[
\frac{\partial}{\partial z}
\sum_{\vec{K}}\ln {\cal R}(z,\vec{K}) \right]_{z=E+i\epsilon}.
\label{rhodiff2}$$ Hereby the problem is essentially reduced to the evaluation of ${\cal 