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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.

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    $\begingroup$ 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
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    $\begingroup$ 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

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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 $2 \pi$.)

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    $\begingroup$ $k_n$ 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

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