Given a $n\times n$ matrix $A_n$, the average eigenvalue can easily be computed by $$ave_i(\lambda_i(A_n)) = \frac 1n\operatorname{tr}(A_n).$$

Suppose that we have a sequence of positive-definite matrices $A_1,\dots,A_n,\dots$ such that the spectral distribution of a matrix $A_n$, $$s.d.(A_n) = \frac 1n\sum_{i=1}^n \delta_{\lambda_i(A_n)},$$ where $\delta$ is the dirac-delta, converges (as $n\to\infty$) to a limiting spectral distribution compactly supported in $(0,\infty)$. In this case the notion of "average eigenvalue" is well-defined as the expected value of the underlying random variable.

It would be useful to me to be able to think about this limiting distribution also as an operator itself, in order to prove inequalities regarding the average eigenvalue of various functions of $A_n$ and show that they hold in the limit.

General question: does it make sense to represent this limiting distribution with an infinite-dimensional operator endowed with a probability distribution on its eigenvalues?

I think it will help to clarify my question with a very simple example:

Let's say that in addition to the positive-definite $A_n$ above, we have another sequence of positive-definite matrices $B_n$ converging to a limiting spectral distribution which is also compactly support in $(0,\infty)$.

For each pair of matrices $(A_n,B_n)$ it is not hard to show that if $\alpha>1$, $$ave_i( \lambda_i\left( (A_n+\alpha B_n)^{-1} \right) )<ave_i\left( \lambda_i\left( (A_n+B_n)^{-1} \right) \right).$$

It should certainly also be the case that this *strict* inequality between the average-eigenvalues should hold for the limiting spectral distributions. If this limiting distribution could be represented as an operator with sufficiently nice properties, the same finite-dimensional proof of the above inequality should extend to the infinite-dimensional positive-definite operator case.

2more comments