$\lim_{t\to 0}\sup_{s\in \mathbb{R}}|f(t+s)-f(s)|$ uniformly on $f$

Question

Let $X$ be the space of all bounded uniformly continuous functions $f:\mathbb{R}\to \mathbb{R}$ equiped with the supremum norm $|f|_\infty$.

We know that for each $f\in X$ we have $$\sup_{s\in \mathbb{R}}|f(t+s)-f(s)|\to0,$$ when $t\to0$.

I just need a sequence $f_n \in X$ with $|f_n|_\infty\leq 1$ such that we don't have $$\sup_n\sup_{s\in \mathbb{R}}|f_n(t+s)-f_n(s)|\to0.$$

Answer 1

Consider $f_n(x):=\sin(n\pi x)$. Then $\lVert f_n\rVert_\infty=1$ and if $t\gt 0$ is fixed, and $n$ is such that $ 1(2(n+1))\leqslant t\leqslant 1(2n)$ $$|f_n(t)|\geqslant \sin(\pi/3).$$