Let $f^N_n: [0,T]\to [0,2]$ be continuous functions such that for all $n=1,\ldots, N-1$
$$f_n^N(t) \le C_{n}\int_0^t \big(f_{n+1}^N(u)+f_n^N(u)\big)du + \frac{C_{n}}{\sqrt{N-n}}t,\quad \forall t\in [0,T],$$
where $C_n>0$ denotes some constant depending on $n$ but independent of $N$. In my context $C_n=cn^2$ for some $c>0$.
Assume that $f^N_n(0)=0$. Can we prove that for every fixed $n$ and $t$
$$\lim_{N\to\infty} f^N_n(t)=0?$$
The answer is no, in general. Suppose, e.g., $T = 1$ and $c > 2\pi^2/3$. Let $u_1 = 1/2$ and $$ u_{n + 1} = u_n - \frac{2}{C_n} = u_n - \frac2{cn^2}, \quad n \in \mathbb{N}. $$ Note that $$ u_n = \frac12 - \sum_{k = 1}^{n - 1} \frac2{ck^2} > \frac12 - \frac{\pi^2}{3c} > 0, \quad n \in \mathbb{N}. $$ Let $$ f_n(x) = \begin{cases} 0,& 0 \leqslant x \leqslant u_n, \\ \frac{x-u_n}{1-u_n},& u_n < x \leqslant 1. \end{cases} $$ Then for every $n \in \mathbb{N}$ and $t > u_n$ $$ \int_0^{t}f_{n+1}(x)\,dx \geqslant (t - u_n) f_{n+1}(u_n), $$ so $$ f_n(t) = \frac{t-u_n}{1-u_n} = (t - u_n) f_{n+1}(u_n) \cdot \frac{C_n(1 - u_{n+1})}{2(1-u_n)} \leqslant C_n \int_0^{t}f_{n+1}(x)\,dx. $$ So required inequality is true for $f_n^N(t) = f_n(t)$, but $f_n(t) \not \equiv 0$.
