While I was messing with the triangular functions that are integrable and do not converge to $0$ and its relationship with uniform continuity, I come across the following question:
Let $f:\mathbb{R}\to \mathbb{R}$ be a bounded continuous function such that $$\lim_{r\to\infty}\frac{1}{r}\int_{-r}^r|f(x)|dx=0.$$ Does this imply that $$\lim_{r\to\infty}\frac{1}{r}\int_{-r}^r\left(\sup_{x-1\leq\theta\leq x}|f(\theta)|\right)dx=0$$?
No. Say $0<\delta_n<1/2$ for all $n\in\mathbb Z$. Say $f$ is continuous, $0\le f\le 1$, $f=0$ outside the union of the intervals $(n-\delta_n,n+\delta_n)$, but $f(n)=1$ for all $n$. If $\delta_n\to0$ you can show that $\frac1r\int_{-r}^r|f|\to0$. But $\sup_{x-1\le\theta\le x}|f(\theta)|=1$ for all $x$.