This is related to my other question. Consider the function $a(s)=\dfrac{1}{1+s^2}$. Let $f:\mathbb{R}\to \mathbb{R}$ be a function such that $t\mapsto a(t)f(t)$ is bounded uniformly continuous.
How can we show that $$\sup_{s\in \mathbb{R}}|a(s)f(s+t)-a(s)f(s)|\to 0,$$ when $t\to0$.
It's similar to what I did yestersay.
Let $t\in \mathbb R$
Claim 1: The function $\displaystyle x\to \frac{a(x)}{a(x+t)}-1$ is bounded by $\displaystyle \frac{\sqrt{t^2+4}+t}{\sqrt{t^2+4}-t}-1$
Claim 2: The function $af$ is bounded by some $M>0$
Let $\epsilon >0$
By uniform continuity of $af$, there is some $\alpha >0$ such that $\displaystyle |x-y|\leq \alpha \implies |a(x)f(x)-a(y)f(y)|\leq \frac{\epsilon}2$
Since the function $t\to \frac{\sqrt{t^2+4}+t}{\sqrt{t^2+4}-t}-1$ goes to $0$ as $t\to 0$, there is some $\beta >0$ such that $\displaystyle |t|\leq \beta \implies \forall s\in \mathbb R, \left|\frac{a(s)}{a(s+t)}-1\right| \leq \frac{\epsilon}{2M}$
Let $\delta := \min(\alpha,\beta)$
Let $s\in \mathbb R$ and $t$ such that $|t|\leq \delta$
$\displaystyle |a(s)f(s+t)-a(s)f(s)| \leq \left|\frac{a(s)}{a(s+t)}-1\right||a(s+t)f(s+t)| + |a(s+t)f(s+t)-a(s)f(s)|$
$\displaystyle\leq \frac{\epsilon}{2M}\cdot M + \frac{\epsilon}{2}$
$\leq \epsilon$
This is valid for all $s$.
Taking $\sup$ yields $\displaystyle \sup_{s\in \mathbb R} |a(s)f(s+t)-a(s)f(s)| \leq \epsilon$
It has been proved that
$\forall \epsilon >0, \exists \delta >0, |t|\leq \delta \implies \sup_{s\in \mathbb R} |a(s)f(s+t)-a(s)f(s)| \leq \epsilon$