In this [paper](https://arxiv.org/abs/1206.2251), if we denote the $k$ largest eigenvalues by $\lambda\_N,\lambda\_{n-1},··· ,\lambda\_{N-k+1}, $ then for Gaussian ensembles the joint distribution function of rescaled eigenvalues has the limit:


$$
\lim\_{N\to\infty}P(N^{2/3}(\lambda\_N-2)\le s\_1,\dots, N^{2/3}(\lambda\_{N-k+1}-2)\le s\_k)=F\_{\beta, k}(s\_1,\dots, s\_k)
$$


Can we say the gap $\lambda\_N-\lambda\_{N-k+1}=O\_p(N^{-2/3})$ by the continuous mapping theorem?



Can we apply the continuous mapping theorem for the limiting joint distribution of the Tracy-Widom law?
$\newcommand\la\lambda$The answer is: yes, of course.


Indeed, let $X\_{N,i}:=N^{2/3}(\la\_i-2)$. By the limit theorem you cited and (say) [Example 2.3, p. 18](https://archive.org/details/convergenceofpro0000bill), the $k$-tuple $(X\_{N,N},\dots,X\_{N,N-k+1})$ converges in distribution to some $k$-tuple $(Y\_1,\dots,Y\_k)$ of random variables. So, by continuous mapping theorem, $N^{2/3}(\la\_N-\la\_{N-k+1})=X\_{N,N}-X\_{N,N-k+1}$ converges in distribution to $Y\_1-Y\_k$.


So, $N^{2/3}(\la\_N-\la\_{N-k+1})=O\_P(1)$, that is, $\la\_N-\la\_{N-k+1}=O\_P(N^{-2/3})$, as desired.


