Assume $f : (0,\infty)\to \mathbb{R}$ is non-negative continuous function. Let $(t_{n})_{n\in\mathbb{N}}\subset (0,\infty)$ be any sequence such that $t_{n}\to\infty$. Assume further than $\lim\limits_{n\to\infty}\int_{t_{n}-1}^{t_{n}+1}|f(t)|^{2}dt=0$. I would like to show that $\lim\limits_{n\to\infty}f(t_{n})\to0$.
This is my attempt so far :
First, we fix $n\in\mathbb{N}$ large enough so that $t_{n}>1$ define $s := t-t_{n}$ and $f_{n}(s) := f(t_{n}+s)$. Then, I have $\lim\limits_{n\to\infty}\int_{-1}^{1}|f_{n}(s)|^{2}ds=0$. Of course this means that $f_{n}(s)\to 0$ for $s\in[-1,1]$ a.e.
I don't know how to use continuity to ensure that at least $f_{n}(0)\to 0$ as $n\to\infty$ which means $\lim\limits_{n\to\infty}f(t_{n})\to 0$. Any help or hint is much appreciated! Intuitively, since $f$ is nonnegative and continuous, then the maximum value of of $f_{n}$ should go to zero because $f$ does not oscillate. Again, thank you for the help!