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?

  • $\begingroup$ That first paragraph needs a rewrite. $\endgroup$ – zhw. Dec 17 '15 at 0:49
  • $\begingroup$ Could you edit it or elaborate? $\endgroup$ – FTem Dec 17 '15 at 2:26
  • $\begingroup$ No it would be better if you did it. You can't see the problem? $\endgroup$ – zhw. Dec 17 '15 at 2:30
  • $\begingroup$ I fixed everything I could find... did I miss something? Also, any thoughts on the problem? I think I can show the this all true if $\{f'_n\}$ is uniformly bounded... $\endgroup$ – FTem Dec 17 '15 at 2:46
  • $\begingroup$ I edited the question to make it simpler. I am not sure why you think this comes down to showing $f_n'$ is equicontinuous. $\endgroup$ – zhw. Dec 17 '15 at 20:19

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.