Logarithmic divergence of spectra

If $A$ is a positive selfadjoint operator with compact resolvent, let $\set{\la_k(A) : k \in \bN}$ be its eigenvalues listed in increasing order, $\la_0(A) \leq \la_1(A) \leq \la_2(A) \leq\cdots$ (an eigenvalue of multiplicity $r$ occurs exactly $r$ times in the list). The counting function $N_A(\la)$, defined for $\la > 0$, is the number of eigenvalues not exceeding $\la$:

$$N_A(\la) := \char93 \set{k \in \bN : \la_k(A) \leq \la}.$$

If $A$ is invertible (i.e., if $\la_0(A) > 0$), we can define the ``zeta function''

$$\ze_A(s) := \Tr A^{-s} = \sum_{k\geq 0} \la_k(A)^{-s}, \word{for} s > 0,$$

where we understand that $\ze_A(s) = +\infty$ when $A^{-s}$ is not traceless. For real $s$, $\ze_A(s)$ is a nonnegative decreasing function.

It is actually more useful to consider finite partial sums.

We shall see later that for many spectral triples, the counting function of the positive (unbounded) operator $\vert D\vert$ has polynomial growth: for some $n$, one can verify an asymptotic relation $N_{\vert D\vert}(\la) \sim C'_n \la^n$. In that case we can take $A := \vert D\vert^{-n}$, which is compact. Then the number of eigenvalues of $A$ that are $\geq \eps$ equals $N_{\vert D\vert}(\la)$ for $\la = 1/\eps$. This suggests heuristically that for $N$ close to $N_{\vert D\vert}(1/\eps)$, the $N$-th eigenvalue is roughly $C/\eps$ for some constant $C$, so that $\sg_N(\vert D\vert^{-n}) = O(\log N)$. We now check this condition in a few examples.