TITLE: If $L^1$ average of $f$ is smaller than $t^2$ then $f=0$ a.e.
QUESTION [9 upvotes]: Suppose $f\in L^1(\mathbb R)$ and there exits $\delta>0$ such that
  $$\|f(\cdot + t)-f\|_{L^1}\leq |t|^2$$
  for all $|t|\leq \delta$, where $f(\cdot + t)$ is the function $x\mapsto f(x+t)$. Show that $f=0$ a.e.

It is well known that $\lim_{t\to 0}\|f(\cdot+t)-f\|_{L^1}=0$ but this fact can't help here. Intuitively, the condition in the problem would deduce that $\int |f'|=0$ if the derivative is good enough so the conclusion must be right and $|t|^2$ can be improved to $|t|^{1+\epsilon}$. I also tried to use the separability of $L^1$, like when we prove the famous fact mentioned above, but it helps nothing. Any hints and thoughts are welcomed.

REPLY [2 votes]: This is just a special case of the fact that if $\alpha>1$ then $Lip_\alpha$ contains only constants:
Say $f_t(x)=f(x+t)$, and define $F:\Bbb R\to L^1$ by $F(t)=f_t$.
Then $$F(t)-F(0)=\sum_{j=1}^n(F(jt/n)-F((j-1)t/n)), $$so if $n$ is large enough that $t/n<\delta$ we have
$$||F(t)-F(0)||_1\le\sum_{j=1}^n||F(jt/n)-F((j-1)t/n)||_1\le n(t/n)^2=t^2/n. $$
So $||F(t)-F(0)||_1=0$; hence $f$ is constant, and so $f\in L^1$ implies $f=0$.