Suppose $F:\mathbb R^2 \to \mathbb R$ is continuous. Assume $f_n$ is a uniformly bounded sequence of real-valued functions on $[0,1]$ such that for each $n, f_n'(x) = F(x,f_n(x)),x\in [0,1].$
Is there a subsequence $f_{n_k}$ that converges uniformly on $[0,1]?$
This question seems to be coming down to prove $\{f'_n(x)\}$ is equicontinuous. If so, then it has a uniformly convergent subsequence and then so does $\{f_n\}$. We know each $f_n'(x)$ is uniformly continuous on $[0,1]$ by compactness. But in general, uniform continuity for each function doesn't imply equicontinuity, correct?
Best Answer
The answer is yes. Proof: We are given that there exists a constant $M$ such that $|f_n|\le M$ on $[0,1]$ for all $n.$ Now $F$ is continuous on $[0,1]\times [-M,M],$ a compact set. Therefore $|F|$ is bounded by some constant $C$ on this set. It follows that
$$|f_n'(x)| = |F(x,f_n(x))| \le C$$
for all $n$ and all $x\in [0,1].$ By the mean value theorem, we then have $|f_n(y)-f_n(x)| \le C|y-x|$ for all $n$ and all $x,y \in [0,1].$ This shows $(f_n)$ is equicontinuous, and since $(f_n)$ is uniformly bounded, Arzela-Ascoli gives the desired uniformly convergent subsequence.