Inequality on the trace of the resolvent of a matrix

Question

For a (random) hermitian matrix $M$ and a complex $z$, it is well known that

$$ \left| \int_{\mathbb{R}} \frac{1}{z-x} \text{d}\mu_M(x) \right| = \left| \frac{1}{n} \text{Tr} (z-M)^{-1} \right| \leq \frac{1}{\left|\Im(z)\right|} $$

but I am looking for a majoration involving the real part of $z$.

Could you help me, please ?

Thank you for your consideration on this matter,

Nawak

Answer 1

Let $M\in M_n(\mathbb{C})$ be with $spectrum(M)=(\lambda_j)$, with characteristic polynomial $\chi_M$ and $f(z)=\frac{1}{n}Tr((z-M)^{-1})$. Then $f(z)=\dfrac{\chi_M'(z)}{n\chi_M(z)}=1/n\sum_{j=1}^n\dfrac{1}{z-\lambda_j}$ and $|f(z)|\leq\sup_j|\dfrac{1}{z-\lambda_j}|$. If $M$ is hermitian, then the $(\lambda_j)$ are real and $|f(z)|\leq \dfrac{1}{\inf_j|Re(z)-\lambda_j|}$.