Let $T_a(f):=f(x-a)$ where $a\in\mathbb{R}$. Let also $f\in L^1(\mathbb R)$ and uniformly continuous. I would like to prove that: $$ \lim _{a \rightarrow 0}\left\|T_{a} f-f\right\|_{L^{1}}=0 . $$ This is my attempt:
Let us first assume that $f$ is continuous and has compact support in the bounded interval $[c, d]$. For $y \in]-\frac{1}{2}, \frac{1}{2}[$ the function $$ \phi(x)=f(x-y)-f(x) $$ has support in the interval $\left[-\frac{1}{2}+c, d+\frac{1}{2}\right]$. Since $f$ is uniformly continuous, we can for any given $\epsilon>0$ find $\delta>0$ such that $$ |f(x-y)-f(x)| \leq \epsilon \text { for all } x \in \mathbb{R} \text { whenever }|y| \leq \delta $$ with the choice of $\delta=d-c+1$ we thus obtain that $$ \begin{aligned} \left\|\phi(x)\right\|_1 &=\int_{-\frac{1}{2}+c}^{\frac{1}{2}+d}|f(x-y)-f(x)| d x \\ & \leq \epsilon \sqrt{d-c+1} \end{aligned} $$ I stop here. How I can deal with the case of an arbitrary function?
