$\sup_{x\in \mathbb{R}} \int_0^\varepsilon e^{ay}|f(x-y)|dy\to 0$ when $\varepsilon$ goes to $0$? Let $a$ be a real number and $f:\mathbb{R}\to \mathbb{R}$ a function. If the function $f$ is bounded, then we can see that $$\sup_{x\in \mathbb{R}} \int_0^\varepsilon e^{ay}|f(x-y)|dy\to0,$$ as $\varepsilon$ goes to $0$. Now if we weaken the boudedness of $f$ by the following condition: $f\in L^p_{loc}$ and $\sup_{x\in \mathbb{R}}\int_x^{x+1} |f(y)|^pdy<\infty$ for some $p>1$, then we have the same conclusion, because by Hölder we have $$\sup_{x\in \mathbb{R}} \int_0^\varepsilon e^{ay}|f(x-y)|dy\leq  \left(\int_0^\varepsilon e^{qay}dy\right)^{\frac{1}{q}}\sup_{x\in \mathbb{R}}\left(\int_{x-\varepsilon}^x |f(y)|^pdy\right)^\frac{1}{p}\to 0,$$
as $\varepsilon \to 0$, where $\frac{1}{p}+\frac{1}{q}=1$.
But now if we weaken even more and assume that $f\in L^1_{loc}$ and $\sup_{x\in \mathbb{R}}\int_x^{x+1} |f(y)|dy<\infty$, do we have the same conclusion ?
 A: I can't comment on your comment, so I'll answer instead. I was thinking about this question earlier and immediately thought of this function when trying to construct a counter example. I didn't put all the pieces together, but your comment shows that you're obviously thinking along the same lines, so I'll contribute what I can.
Consider my favorite "counter-example to physicists making incorrect statements about $L^2$ because they're thinking about wavefunctions" function: 
Let $f$ be zero on $(-\infty, 0)$ and defined as follows on each interval $(n,n+1)$ with $n = 0, 1, 2, 3, \ldots$ On the interval $(n, n+1)$, $f$ equals 0 on $(n, n + 1 - \frac{1}{n})$ and $f$ equals $n$ on $(n + 1 - \frac{1}{n}, n)$. This function has the property that 
$$\int_n^{n + 1} | f(y) |\ dy = 1$$
for all $n$. 
Further, for any $\epsilon > 0$, we have for all $n > \frac{1}{\epsilon}$, 
$$\int_{n - \epsilon}^n |f(y)|\ dy = 1.$$
Whence it follows that $\sup_{x\in\mathbb{R}} \int_{x - \epsilon}^x |f(y)|\ dy$ does not go to zero as $\epsilon \rightarrow 0$.
edit: I didn't state it, but $f$ is clearly in $L^1_\text{Loc}$.
