Consider the continuous and non negative function $c : \mathbb R \to [0,1]$ defined by $$ c(x) = \begin{cases} \cos \frac{\pi x}{2} &\text{for } x \in [-1,1]\\ 0 &\text{otherwise} \end{cases}$$ Let also $(r_n)$ be an enumeration of the rational points of the interval $[0,1]$ and $(k_n)$ be a strictly increasing sequence of integers.

Based on those elements, one can build the sequence of functions $g_n : [0,1] \to [0,1]$ defined by $g_n(x) = c(k_n(x-r_n))$. Is it possible to have $$\lim\limits_{n \to \infty} g_n(y) = 0$$ for some $y \in [0,1]$?

The origin of the question is the construction of a sequence of continuous functions $g_n$ defined on $ [0, 1]$ such that $0 \le g_n \le 1$ and $$ \lim\limits_{n \to \infty} \int_0^1 g_n(x) \ dx = 0,$$ but such that the sequence $(g_n(x))$ converges for no $x \in [0,1]$.

This is question I raised at Mathematics.

  • 2
    $\begingroup$ Sure. Pick a $y$ ($y=0$, say), fix an enumeration $r_n$ and just choose the $k_n$'s accordingly now. In the last step, we use that the $kr\bmod\pi$, $k\ge N$, are dense in $[0,\pi]$ for any $r\in\mathbb Q$ (and $\not= 0$ I guess). $\endgroup$ Sep 11, 2016 at 18:59
  • 2
    $\begingroup$ Also, an easy way to produce $g_n$'s as described in the last part is to use moving bumps. $\endgroup$ Sep 11, 2016 at 19:01
  • $\begingroup$ The m.se post has been deleted. $\endgroup$ Nov 26, 2019 at 21:17

1 Answer 1


According to your description, the sequences $(r_n)$ and $(k_n)$ are given. Then there is not always such an $y$.

Indeed, we know that $(r_n)$ has 0 as well as 1 as accumulation point. This implies that in general, for any given $y$, there will be two subsequences of $g_n(y) = c(k_ny - k_nr_n)$, which cannot converge both to zero. Assume for example that $k_n \to 1$ (e.g., if $k_n=1-1/n$). Then there is one subsequence of $g_n(y)$ that tends to $c(y-0)$, and one that tends to $c(y-1)$, and these values cannot both equal zero. (They are different unless $y = 1/2$ in which case the value is $c(1/2)=\pi/4\ne 0$.)

Now, if your question is rather whether it is possible that for some particular $(k_n)$ and $(r_n)$ there may exist such an $y$, then the answer is yes, and even better, for any $y$ and any $(r_n)$, there is some $(k_n)$ such that $\lim_{n\to\infty} g_n(y) = 0$. For this it is sufficient to take $k_n$ such $c(k_n(y-r_n))$ becomes smaller and smaller, i.e., $k_n(y-r_n)$ closer and closer to some odd multiple of $\pi/2$. It is easy to see that for each $n$ (except possibly for one single $n$ if $y$ is equal to the rational $r_n$), one can choose such a $k_n$, since $y-r_n$ is a nonzero number. (The requirement that k be increasing is obviously no restriction, since one can always take it larger to get the same value modulo $2 \pi$.)

  • 2
    $\begingroup$ $k_n$ is a sequence of integers. $\endgroup$
    – Wojowu
    Mar 1, 2019 at 14:24
  • $\begingroup$ oh yes, I missed that, so my counter-example in the first part doesn't work. Will think it over. However, I guess OP is rather interested in the second part, which remains valid: you can take k_n to be integers so that k_n (y-r_n) gets always closer to some (increasingly large) odd multiple of pi/2. $\endgroup$
    – Max
    Jul 19, 2022 at 18:07

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.