Supposing $X_{t_i}$ is a real-valued stochastic process defined for $i=1,\ldots, N$, where $N=N_n\rightarrow\infty$, and $a_n\rightarrow 0$ as $n\rightarrow\infty$, I am trying to see if the following holds: \begin{equation} \sum_{i=1}^{N}(X^2_{t_i}-a_n^2)=o_p(a_n)\tag{Result} \end{equation} when we have \begin{equation} \max_{1\leq i\leq N}\frac{X_{t_i}^2}{a_n^2}\overset{p}\rightarrow c\tag{$*$} \end{equation} as $n\rightarrow\infty$.
Open to amend the condition ($*$) if necessary.
Some thoughts: obviously, since we have
\begin{align*} &\frac{\sum_{i=1}^N(X_{t_i}^2-a_n^2)}{a_n}\\ \qquad&\leq N\cdot \max_{1\leq i\leq N}\left(\frac{X_{t_i}^2-a_n^2}{a_n}\right)=Na_n\cdot \max_{1\leq i\leq N}\left(\frac{X_{t_i}^2-a_n^2}{a_n^2}\right)\\ &\leq Na_n\cdot \max_{1\leq i\leq N}\left(\frac{X_{t_i}^2}{a_n^2}-1\right), \end{align*} if $c=1$ and $N=O(a_n^{-1})$, we have the desired result by ($*$).
But I am not sure if we really need to restrict $c=1$. Also, I am looking for a way to relax $N=O(a_n^{-1})$ as well. Perhaps the bounds above in my solution is not tight enough somehow