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Dec 10, 2022 at 20:46 comment added Hermi Sure, $l$ is a fixed number.
Dec 10, 2022 at 20:42 comment added Carlo Beenakker this holds if you keep $l$ fixed as you send $N\rightarrow\infty$; in that way you ensure that both $\lambda_1$ and $\lambda_l$ are near the edge of the spectrum; if you let $l$ grow with $N$, say by taking $l$ some fraction of $N$, then it will not hold, because $\lambda_l$ will enter the bulk; and yes, the coefficient $c$ will depend both on $\epsilon$ and on $l$.
Dec 10, 2022 at 20:40 comment added Hermi Sure, I mean like for the eigenvalues $\lambda_1\le \lambda_2\le \dots \le \lambda_l$, then we have this result for $\lambda_l-\lambda_1$, right? That is for any $\epsilon>0$, there exists $c>0$ such that $\lim_{N\to\infty} P(N^{2/3}(\lambda_l-\lambda_1)\le c)\ge 1-\epsilon$.
Dec 10, 2022 at 20:37 comment added Carlo Beenakker no, this does not hold; the inequality does hold for $\lambda_N-\lambda_{N-l}$ with fixed $l$, but not for $\lambda_N-\lambda_l$ with fixed $l$.
Dec 10, 2022 at 20:01 comment added Hermi So for a fixed number $l$, let $\delta_{Nl}:=\lambda_N-\lambda_l$, we can similar get the same answer? $P(N^{2/3}\delta_{NL}\le c)\ge 1-\epsilon$? Here $c$ will dependent on $\epsilon$ and $l$?
Dec 10, 2022 at 1:59 vote accept Hermi
Dec 9, 2022 at 20:28 comment added Carlo Beenakker ChatGPT is a bot, an automated machine that produces answers that sound reasonable but are nonsensical; the answer was deleted by the moderators, like several other similar fake answers; see meta.mathoverflow.net/questions/5531/…
Dec 9, 2022 at 18:35 comment added Hermi So why ChatGPT delete his answer? Is that one not true?
Dec 9, 2022 at 9:00 history edited Carlo Beenakker CC BY-SA 4.0
deleted 4 characters in body
Dec 9, 2022 at 7:09 history answered Carlo Beenakker CC BY-SA 4.0