Lidskii trace formula for trace-class operators

By [F1] and [F2], the eigenvalue list has the stated algebraic multiplicities and $L:={\sum }_j\mathrm{\mid }{\lambda }_j(T)\mathrm{\mid }\le \mathrm{\parallel }T{\mathrm{\parallel }}_1$. For a finite initial product $P_N(z)={\prod }_{j\le N}(1+z{\lambda }_j)$, expansion and the ordered-tuple bound for elementary symmetric sums give $\mid P_N(z)-1-z{\sum }_{j\le N}{\lambda }_j\mid \le {\sum }_{k=2}^N(\mathrm{\mid }z\mathrm{\mid }L{)}^k\mathrm{/}k!\le (\mathrm{\mid }z\mathrm{\mid }L{)}^2{e}^{\mathrm{\mid }z\mathrm{\mid }L}\mathrm{/}2$. Indeed, every unordered product of $k$ distinct absolute eigenvalues occurs $k!$ times among the ordered $k$-tuples contributing to $L^k$.

Let $N\to \mathrm{\infty }$. The local product convergence and absolute convergence of ${\sum }_j{\lambda }_j$ from [F2] preserve the bound in step 1.1, so ${\lim }_{z\to 0}(D_T(z)-1)\mathrm{/}z={\sum }_j{\lambda }_j(T)$. The left side is $D_T^{\mathrm{\prime }}(0)={\mathrm{tr}}_H(T)$ by [F2]. The same argument applies to finite and empty lists, with the empty sum equal to $0$ and the empty product equal to $1$.