Uniform continuity and translation invariance Consider the function $a(s)=\dfrac{1}{1+s^2}$ and the space  $X=\{f:\mathbb{R}\to \mathbb{R}$ such that $t\mapsto a(t)f(t)$ is bounded uniformly continuous$ \}$.
I want to show that $X$ is translation invariant, i.e. if $f\in X$ then $f_t\in X$ for all $t\in \mathbb{R}$, where $f_t$ is the $t$-translation of $f$ defined by $f_t(s)=f(t+s)$.
I showed that if the function $s\mapsto a(s)f(s)$ is bounded then the function $s\mapsto a(s)f_t(s)$ is also bounded. In fact
$$\sup_{s\in \mathbb{R}}\left\lvert a(s)f_t(s)\right\rvert=\sup_{s\in \mathbb{R}}\left\lvert a(s)f(t+s)\right\rvert=\sup_{s\in \mathbb{R}}\left\lvert a(s-t)f(s)\right\rvert\leq M_t\sup_{s\in \mathbb{R}} \left\lvert a(s)f(s)\right\rvert,$$
because we know that there's a constant $M_t$  such that for all $s\in \mathbb{R} $ 
$$a(s-t)=\dfrac{1}{1+(s-t)^2}\leq M_t \dfrac{1}{1+s^2}=M_t a(s).$$
Now I just need to prove that if the function $s\mapsto a(s)f(s)$ is uniformly continuous,  then the function $s\mapsto a(s)f_t(s)$ is also uniformly continuous.
 A: Let $t\in \mathbb R$
Claim 1: the function $\displaystyle x\to \frac{a(x)}{a(x+t)}$ is bounded (by $M$) and uniformly continuous
Claim 2: the function $x\to a(x+t)f(x+t)$ is bounded (by $N$)

Let $\epsilon >0$
By uniform continuity of $af$ there is some $\delta >0$ such that $\displaystyle |x-y|\leq \delta \implies |a(x)f(x)-a(y)f(y)|\leq \frac{\epsilon}{2M}$ 
By uniform continuity of $\displaystyle x\to \frac{a(x)}{a(x+t)}$, there is some $\beta>0$ such that $\displaystyle |x-y|\leq \beta \implies |\frac{a(x)}{a(x+t)}-\frac{a(y)}{a(y+t)}|\leq \frac{\epsilon}{2N}$

Let $\alpha := \min{(\beta,\delta)}$ 
Let $x,y$ such that $|x-y|\leq \alpha$
$\displaystyle \left| a(x)f(x+t)-a(y)f(y+t)\right|=\left| \frac{a(x)}{a(x+t)}a(x+t)f(x+t)-\frac{a(y)}{a(y+t)}a(y+t)f(y+t)\right|$
$\displaystyle \leq \left|\frac{a(x)}{a(x+t)}\right|\left| a(x+t)f(x+t)-a(y+t)f(y+t)\right| 
+ \left|\frac{a(x)}{a(x+t)}-\frac{a(y)}{a(y+t)}\right||a(y+t)f(y+t)|$
$\displaystyle \leq \frac{\epsilon}{2M}\cdot M + \frac{\epsilon}{2N}\cdot N$
$\leq \epsilon$
