In this paper, 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?
$\newcommand\la\lambda$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, 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.
