Suppose $f$ is Lebesgue integrable on $\mathbb R$ and satisfies:
$$\lim_{t\to 0} \frac1{t} \int_{\mathbb R} |f(x+t)-f(x)|dx = 0$$
Prove that $f = 0$ almost everywhere.
It seems to me that there is no way to solve this problem using elementary methods. If $f$ is differentiable we can use Fatou's lemma as follows: let $g_n(x) = n\left( f\left(x + \frac1n \right) - f(x) \right)$.
$$\int_{\mathbb R} |f'(x)| dx = \int_{\mathbb R} \liminf |g_n(x)| dx \le \liminf \int_{\mathbb R} |g_n(x)|dx = 0$$
so $f' = 0$ a.e. and $f = c$. Since $f$ is integrable, $c = 0$.
We can always use Fatou's lemma to obtain: $\liminf g_n(x) = 0$ but this doesn't look very useful to me. How to solve this problem?
Maybe the Fourier transform? Let $h_t(x)=\frac{f(x+t)-f(x)}{t}$, with $\widehat h_t(\xi)=\frac{e^{it\xi}-1}{t}\widehat f(\xi)$. The condition $\|h_t\|_1\to 0$ implies $\|\widehat h_t\|_\infty\to 0$, which seems to imply $\widehat f(\xi)=0$, since $\widehat h_t(\xi)\to i\xi \widehat f(\xi)$. By continuity, it would seem $\widehat f(0)=0$ as well.