I want to ask the following question.

Given a sequence of monotone non-decreasing functions $\{f_n\}$ on $[0,1]$. For any $n$, function $f_n(x)$ is bounded, specifically $f_n(x) \in [0,2]$ for $\forall x \in [0,1]$. And $f_n(x)$ is Lipschitz continuous for any $n$. (I am not sure if this is useful for this question.)

Does there exist a pointwise convergent subsequence in $\{f_n\}$?

Here are some thoughts: I know that there is a counterexample for the bounded function sequence. See Pointwise almost everywhere convergent subsequence of $\{\sin (nx)\}$.

But, we can find a subsequence for pointwise bounded functions defined on a countable set. Does a sequence of equicontinuous functions have a pointwise convergent subsequence? Maybe we can first find a subsequence pointwise convergent on rational numbers. Then, by monotonicity and continuity, it seems these functions can not behave too weird on rest numbers. I am not sure about this.

I will be grateful for any suggestions.


A proof that is not hard to get uses Sobolev space embeddings. This amounts to an Ascoli-Arzela' argument in $L^p$ spaces (one can also use the Kolmogorov compactness theorem - see thm 5 in the cited text).

A continuous monotone function $f_n$ can be represented by:$$f_n(x) = \int_0^x p_n(y) dy,$$ for $p_n \geq 0$ integrable. Or simply $\partial_x f_n \in L^1$. In particular $$f_n \text{ is bounded in }W^{1,1}.$$

By sobolev embedding (since we are in one dimension) $W^{1,1} \subset W^{\frac 1 2, 2} \subset L^2$, where the latter embedding is compact. This means that your sequence is compact in $L^2$. In particular there exists a subsequence which converges almost surely.

Maybe almost sure convergence is not sufficient (but the phrasing of the question did not suggest this). In this case on would like to embed into $W^{1,0}.$ But this embedding is not compact, so the general theory we used so far fails. What might save the day is that the functions are also monotone.

Also, if the Lipshitz constant is uniform in $n$ (which I assume it is not) the result would be a simple consequence of Ascoli-Arzela'.

Finally, for a definition of Sobolev spaces, with fractional regularity, see here.

  • $\begingroup$ Hi, thank you for your help. I am not familiar with Sobolev embedding, so I need some time to understand your solution. I only need almost surely convergence. And you are right, we assume the Lipschitz constant is not uniform in $n$ here. $\endgroup$
    – Simon Shan
    Feb 20 '20 at 22:17
  • $\begingroup$ I'm curious if there's a direct proof without all the machinery. I can't think of one yet. $\endgroup$
    – Jake Mirra
    Feb 20 '20 at 22:24
  • $\begingroup$ I think I understand your proof now. Just one thing to ask, is it because the Sobolev embedding is continuous, this sequence after embedding is still bounded? $\endgroup$
    – Simon Shan
    Feb 21 '20 at 0:34
  • $\begingroup$ @SimonShan yes, the first embedding is continuous, have the sequence is bounded. The second one is compact, hence you get strong convergence in $L^2$. Note that $L^2$ is arbitrary. You can show compactness in $L^p$ for any $p\in [1, \infty)$. $\endgroup$
    – Kore-N
    Feb 21 '20 at 7:58

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