$f_n$ forms a sequence of equicontinuous functions in $C^0$, and domain is $[a,b]$. For $p$ belongs to $[a,b]$, we have $f_n(p)$ to be a bounded sequence in $R$. Prove $f_n$ to be uniformly bounded.

What I can do is to use connectedness of $[a,b]$ (chain connected) and pointwise bounded at one point to prove pointwise boundedness at every point. Then use pointwise boundedness and compactness of $[a,b]$ to show uniformly bounded.

But is there any method that we can avoid the use of chain connected to show $f_n$ is pointwise bounded at every point (i.e. only using equicontinuous and compactness of the domain). We can have another condition for the Arzela-Ascoli theorem to be true, if we can avoid the use of connectedness. (Clearly, one condition is that pointwise bounded at every point, but I am wondering if we can extend it to pointwise bounded at one single point, given the compact domain).


Here is the question, and part b asks a weaker condition. So is boundedness at one point enough or we need pointwise bounded for all points? enter image description here

And the chain-connected argument comes from a previous question: enter image description here


Take the statement for equicontinuity instantiated with $\varepsilon=1$. This creates an open cover of $[a,b]$. Compactness allows you to extract a finite subcover. Then $|f_n(x)|=|f_n(p)-f_n(p)+f_n(x)| \leq |f_n(p)|+|f_n(x)-f_n(p)|$. But $|f_n(x)-f_n(p)| \leq N$ where $N$ is the number of elements in our subcover, and $|f_n(p)|$ is bounded by hypothesis.

This is the standard trick used to prove things like the Heine-Cantor theorem. I'm not sure if this is actually what you meant by "chain connectedness", though.

Edit: Yeah, this is pretty much the chain-connectedness argument, just expanded out in this special case.

  • $\begingroup$ I have added a picture for the chain connected part $\endgroup$
    – Toad Jiang
    Feb 20 '15 at 21:29
  • $\begingroup$ So if the domain is just a general compact set, we do not have such argument? $\endgroup$
    – Toad Jiang
    Feb 20 '15 at 21:31
  • $\begingroup$ And answer to part b question 18 is just pointwise bounded at all points? $\endgroup$
    – Toad Jiang
    Feb 20 '15 at 21:32
  • $\begingroup$ @12345 It couldn't be with just a general compact set, because in that case you could have boundedness in one connected component and unboundedness in the other. As for part b, I assume the best you can do is a decomposition into chain-connected components. $\endgroup$
    – Ian
    Feb 20 '15 at 21:35
  • 2
    $\begingroup$ @12345 It doesn't follow just from compactness. What happens is that you get a finite subcover, then chain connectedness lets you "jump" from one open set in the cover to the next, each time increasing $|f_n|$ by at most $1$. Since there are only finitely many sets in the cover and you also have chain connectedness, you can reach any other point in $[a,b]$ in this manner and obtain a bound. As I said, just compactness is not sufficient. $\endgroup$
    – Ian
    Feb 20 '15 at 23:18

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