Let $\phi: [0,1 ] \times \mathbb{R} \rightarrow \mathbb{R}$ be bounded and continuous function. For each $n\in \mathbb{N}$, let $F_n: [0,1] \rightarrow \mathbb{R}$ satisfy that for $t \in [0,1]$, $F_n(0)=1/n$ and $F'_n(t)= \phi(t, F_n(t))$.
Show that there exists a $F$ such that $F(0)=0$ and $F'(t)=\phi(t, F(t))$.
First using Arzela-Ascoli, I showed that there exists a subsequence $(F_{n_k})_n$ such that it converges uniformly to a limit $F$.
Clearly, we have $ \lim_{n \rightarrow \infty} F_n(0)=0$. Now we just have to show that $F'(t)= \phi(t, F(t))$.
Using triangle inequality, we have $|F(t)-F_n(t)| \leq |F(t)-F_{n_k}(t)| + |F_{n_k} -F_n(t)| < \epsilon +|F_{n_k} -F_n(t)| $, where the $\epsilon$ came from the subsequence $(F_{n_k})_n$ converging uniformly to a limit $F$.
Now we just have to show that $|F_{n_k} -F_n(t)| < \epsilon $. I thought of showing that $(F_n)_n$ is a Cauchy sequence so that since $\mathbb{R}$ is complete, we will know that $(F_n)_n$ is converge.
But I'm stuck on finding a constant to prove that it's a Cauchy sequence.
Any help will be appreciated!