$f \in C_{00}(\mathbb{R^p},\mathbb{C})$. $ \mapsto f_t \in L_\infty(\mathbb{R}^p, \mathcal B_p, \lambda_p, \mathbb{C})$ uniformly continuous? Continuing from here
Let $f_t(x):=f(x+t)$
Consider $f \mapsto f_t$ which is a linear, isometric bijection from $L_\infty(\mathbb{R}^p, \mathcal B_p, \lambda_p, \mathbb{C})\to L_\infty(\mathbb{R}^p, \mathcal B_p, \lambda_p, \mathbb{C})$ for every $t \in \mathbb{R}^p$

How can I show that for $f \in C_{00}(\mathbb{R^p},\mathbb{C})$ (continuous and compact support) the mapping $\mathbb{R}^p\ni t \mapsto f_t \in L_\infty(\mathbb{R}^p, \mathcal B_p, \lambda_p, \mathbb{C})$ is uniformly continuous?


This follows from the fact that a continuous function with compact support is uniformly continuous. For a fixed $\varepsilon$, there exists $\delta$ such that if $s,t\in\mathbb R^p$ satisfy $\lVert t-s\rVert\lt\delta$, then $\left\lvert f(t)-f(s)\right\rvert\lt\varepsilon$. 
For any $x\in\mathbb R^p$ and $s,t$ satisfying $\lVert t-s\rVert\lt\delta$, the following inequality holds 
$$
\left\lvert f(x+t)-f(x+s)\right\rvert\lt\varepsilon
$$
hence 
$$\left\lVert f_t-f_s\right\rVert_\infty   ...   $$
