Let $f\in L^1(\mathbb{R})$. Show that $\lim_{x\to 0}\int |f(y-x) - f(y)|dy = 0$
attempted proof - Let $f\in L^1(\mathbb{R})$, then
\begin{align*} \lim_{x\to 0} \int |f(y-x) - f(y)|dy &= \lim_{x\to 0} \int |f(y) - f(x) - f(y)|dy\\ &= \lim_{x\to 0} \int |-f(x)|dy\\ &= \int |-f(0)|dy = 0 \end{align*}
I am not sure if I can simply assume $f(y-x) = f(y) - f(x)$. Any suggestions or comments are greatly appreciated.
It is not a valid proof because $f(y - x)$ need not equal $f(y) - f(x)$. To prove the result, first deal with the case when $f$ is a continuous function with compact support, since $C_c(\Bbb R)$ is dense in $L^1(\Bbb R)$.
Suppose $f\in C_c(\Bbb R)$. Let $a > 0$ such that $\operatorname{supp}(f) \subset [-a,a]$. Since $f$ is uniformly continuous, given $\epsilon > 0$ there corresponds $\delta$ with $0 < \delta < a$ such that for all $s, t\in \Bbb R$, $\lvert s - t\rvert < \delta$ implies $\lvert f(s) - f(t)\rvert < \frac{\epsilon}{3a}$. If $\lvert x \rvert < \delta$, then for all $y\in \Bbb R$, $\lvert f(y - x) - f(y)\rvert < \frac{\epsilon}{3a}$. Hence
$$\int \lvert f(y - x) - f(y)\rvert\, dy = \int_{[x-a,x+a]\cup[-a,a]} \lvert f(y - x) - f(y)\rvert\, dy < \frac{\epsilon}{3a}(2a + \delta) <\epsilon.$$
Since $\epsilon$ was arbitrary, the result is proved for continuous functions with compact support. Now prove the general case.