Resolvent of Random Matrix is N^2 Lipschitz For $\tau > 0$, let S($\tau$) = {E + i$\eta$ | $\tau^{-1} \leq E \leq \tau$, $\eta \geq N^{-1 + \tau}$}, and let $H$ be an $N\times N$ Wigner random matrix (i.i.d entries up to Hermitian condition). The resolvent is $G(z) = (H-z)^{-1}$ and what I am trying to clarify is that on $S(\tau)$, $G_{ij}(z)$ is $N^2$-Lipschitz, i.e. each entry of $G$ is $N^2$-Lipschitz as a function of $z$.
Here is what I have tried so far. Let $u_k$, $k = 1, ..., N$ be the normalized eigenvectors of $H$ with eigenvalues $\lambda_k$, i.e. $Hu_k = \lambda_k u_k$ and $||u_k||_2 = 1$. Then we can write $G(z) = \sum_{k=1}^N \frac{u_ku_k^*}{\lambda_k -z}$. So $G_{ij}(z) - G_{ij}(z') = \sum_{k=1}^N \frac{(u_ku_k^*)_{ij}}{(\lambda_k - z)(\lambda_k-z')}(z - z')$ Taking magnitude on both sides, I would like to have $\left|\sum_{k=1}^N \frac{(u_ku_k^*)_{ij}}{(\lambda_k - z)(\lambda_k-z')}\right| \leq N^2$. Since we are on the domain $S(\tau)$, $\left|\frac{1}{(\lambda_k - z)(\lambda_k - z')}\right| \leq \frac{1}{\eta^2} \leq N^2$ We also have $\sum_{k}(u_ku_k^*)_{ij} = \delta_{ij}$. So then I write $\sum_{k=1}^N \left|\frac{(u_ku_k^*)_{ij}}{(\lambda_k - z)(\lambda_k-z')}\right| \leq \sum_{k=1}^N \left|\frac{(u_ku_k^*)_{ij}}{\eta^2}\right|$ but I don't see how to proceed from here because the fact that $\sum_{k}(u_ku_k^*)_{ij} = \delta_{ij}$ depends on cancellation in the sum, and with absolute value it may be as large as $N$ which would be too large to make the bound I want. What can I do to get around this?
 A: It is probably a bit late for this, however I stumbled upon the same problem and ended up here looking for an answer, and there wasn't any. However, I was able to find an answer myself, and for the next that ends up here, here it is.
Let $z_1,z_2\in\mathbb{C}_+.$ Since for invertible matrices $A,B,$ we have
$$A^{-1}-B^{-1}=A^{-1}(B-A)B^{-1},$$
we obtain using the definition of $G$
$$G(z_1)-G(z_2)=G(z_1)(z_2-z_1)G(z_2),$$
and therefore
$$|G_{ij}(z_1)-G_{ij}(z_2)|\leq|z_2-z_1|\sum\limits_{k=1}^N|G_{ik}(z_1)||G_{kj}(z_2)|.$$
Using Cauchy-Schwarz to decouple the sum,
$$\sum\limits_{k=1}^N|G_{ik}(z_1)||G_{kj}(z_2)|\leq\sqrt{\sum\limits_{k=1}^N|G_{ik}(z_1)|^2\sum\limits_{k=1}^N|G_{kj}(z_2)|^2}.$$
Then, there is this really helpful and simple identity called the Ward's Identity, that says
$$\sum\limits_{j=1}^N|G_{ij}|^2=\frac{\mathrm{Im}~G_{ii}}{\eta}.$$
The proof of this uses the fact that the LHS is $(GG^*)_{ii}$ and $2\mathrm{i}\mathrm{Im}~G_{ii}=(G-G^*)_{ii}=\ldots,$
where you should use the same matrix identity as above on $G-G^*$.
So, from the Ward's identity,
$$\sqrt{\sum\limits_{k=1}^N|G_{ik}(z_1)|^2\sum\limits_{k=1}^N|G_{kj}(z_2)|^2}=\sqrt{\frac{\mathrm{Im}~G_{ii}(z_1)\mathrm{Im}~G_{jj}(z_2)}{\eta_1\eta_2}}\leq\sqrt{\frac{|G_{ii}(z_1)||G_{jj}(z_2)|}{\eta_1\eta_2}}.$$
From the eigendecomposition of $H$ (what you used above), one has the trivial bound
$$|G_{ij}|\leq\frac{1}{\eta},$$
so that we finally have
$$|G_{ij}(z_1)-G_{ij}(z_2)|\leq\frac{1}{\eta_1\eta_2}|z_1-z_2|.$$
On your domain, it holds that
$$\frac{1}{\eta_1\eta_2}\leq N^{2-2\tau}\leq N^2,$$ so that
$G$ is $N^2$-lipschitz.
