I would like to calculate the limit value of a linear functional \begin{equation} \lim_{n\rightarrow\infty}\mathcal{I}_n=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^n f(\lambda_i)=\lim_{n\rightarrow\infty}\int f(x)\mathrm{d}G_n(x), \end{equation} where $\lambda_i$ are eigenvalues of a $n\times n$ matrix and $G_n(x)$ is the empirical eigenvalue distribution function. In the large dimension limit, $G_n(x)$ converges to a nonrandom limit $G(x)$ but its explicit form is difficult to obtain. In stead, it is easy to obtain the Stieltjes transform of $G(x)$, say $m_G(x)$.

I hope I could obtain $\lim_{n\rightarrow\infty}\mathcal{I}_n$ using $m_G(x)$. Indeed, the equation (1.14) in "Z. D. Bai and J. W. Silverstein, CLT for Linear Spectral Statistics of Large-Dimensional Sample Covariance Matrices, Ann. Prob., 2004" shows \begin{equation} \int f(x)\mathrm{d}G(x)=-\frac{1}{2\pi i}\int f(z)m_G(z)\mathrm{d}z. \end{equation} But I don't understand how they obtain it and if it applies for general $G(x)$. I'd appreciate someone giving me some hints.


1 Answer 1


Let me write the limit value $I$ which you are seeking as

$$I=\int f(x)\rho(x)dx$$

where $\rho=dG/dx$ is the eigenvalue density, with Stieltjes transform $S_\rho$. The Stieltjes-Perron inversion formula reads

$$\rho(x)=\lim_{\epsilon\rightarrow 0^+}\frac{S_\rho(x-i\epsilon)+S_\rho(x+i\epsilon)}{2\pi i}$$

Substitute in the integral over $x$,

$$I=\frac{1}{2\pi i}\lim_{\epsilon\rightarrow 0^+}\left[\int_{-\infty-i\epsilon}^{\infty-i\epsilon}f(x)S_\rho(x)dx-\int_{-\infty+i\epsilon}^{\infty+i\epsilon}f(x)S_\rho(x)dx\right]$$

Now assume that $f(x)S_{\rho}(x)$ is analytic for ${\rm Im}\,x\leq 0$, and decays sufficiently rapidly for ${\rm Im}\,x\rightarrow -\infty$ that the integration contour of the first integral may be closed in the lower half of the complex plane. This integral then evaluates to zero, leaving only the second integral,

$$I=-\frac{1}{2\pi i}\lim_{\epsilon\rightarrow 0^+}\int_{-\infty+i\epsilon}^{\infty+i\epsilon}f(x)S_\rho(x)dx$$


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