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.
 A: 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 $2 \pi$.)
