Let $f\in L^1([-1,2])$ and suppose $\lim_{h\to 0} \frac{1}{h}\int_0^1|f(x+h)-f(x)|=0$. I want to prove that f is constant a.e.
I want to check if my approach is correct and fill out any gaps. My idea is to prove this for absolutely cont. functions and use the fact that they 're dense in $L^1$ .
First, we know $f$ is abs. cont. if-f $\text{Diff}_h(f)=\frac{f(x+h)-f(x)}{h}$ is uniformly integrable. Assume $f$ is abs. cont. By Vitali convergence th., we can pass the limit $\lim_{n\to\infty}\int_0^1 |\text{Diff}_{1/n}(f)|=\int_0^1 |f'|=0$ which implies $f'=0$ a.e.
Now in the general case, $f$ can be approximated by an abs. cont $g$. Then $\int_0^1 |\text{Diff}_{1/n}(g)|\leq \int_0^1 |\text{Diff}_{1/n}(f)|+2\int_0^1 |\text{Diff}_{1/n}(f-g)|$. The first term goes to $0$ as $n\to \infty$ while the second term goes to zero as $g$ approaches $f$. I am a bit confused however if this would imply that $f$ is constant.
I would appreciate any feedback on this, if this approach is doable, if there are any big gaps and if not, how to solve the last step, or of course, if there are simpler solutions.