# Ability to have function sequence converging to zero at some points

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.

• 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). Sep 11, 2016 at 18:59
• Also, an easy way to produce $g_n$'s as described in the last part is to use moving bumps. Sep 11, 2016 at 19:01
• The m.se post has been deleted. Nov 26, 2019 at 21:17

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 $$2 \pi$$.)

• $k_n$ is a sequence of integers. Mar 1, 2019 at 14:24
• 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.
– Max
Jul 19, 2022 at 18:07