If $L^1$ average of $f$ is smaller than $t^2$ then $f=0$ a.e. 
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. 
 A: $|\int_a^{b} f(x)dx-\int_{a+t}^{b+t} f(x)dx| =|\int_a^{a+t} f(x)dx-\int_b^{b+t} f(x)dx|\leq t^{2}$ for $t>0$ sufficiently small. [ Because $\int_a^{b} f(x)dx-\int_{a+t}^{b+t} f(x)dx=\int_a^{b} f(x)dx-\int_a^{b} f(x+t)dx$]. Divide by $t$ and let $t \to 0$. Using Lebesgue's Theorem we get $f(a)-f(b)=0$ whenever $a$ and $b$ are Lebesgue points of $f$. Can you now show that $f=0$ a.e.? 
A: 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 $f=0$.
A: Consider the Fourier transformation, $\hat f(\xi)=\int_\mathbb Rf(x)e^{-2\pi ix\xi}\,dx$ for $\xi\in\mathbb R$. Denote $f_t(x)=f(x+t)$, then $\hat f_t(\xi)=e^{2\pi it\xi}\hat f(\xi)$.
By hypothesis, $\|\hat f(\cdot)(e^{2\pi it\cdot}-1)\|_{L^\infty}=\|\hat f_t-\hat f\|_{L^\infty}\leq \|f_t-f\|_{L^1}\leq |t|^2$. So for a.e. $\xi\in\mathbb R-\{0\}$,
$$|\hat f(\xi)||e^{2\pi it\xi}-1|\leq |t|^2,\ \ \text{for all } 0<|t|\leq\delta.$$
This implies 
$$|\hat f(\xi)|\frac{|e^{2\pi it\xi}-1|}{|t|}\leq |t|,\ \ \text{for all } 0<|t|\leq\delta.$$
Letting $t\to 0$ gives that $|\xi\hat f(\xi)|=0$ for a.e. $\xi\in\mathbb R$, so $\hat f\equiv0$ since $\hat f$ is continuous, which completes the proof.
A: Applying the triangle inequality we have for, each $n\in \mathbb{N}$: $\lVert f(x+n\lceil{\frac{1}{t}}\rceil t)-f(x)\rVert_1\leq n \lceil{\frac{1}{t}}\rceil t^2\leq 2nt$ for $t>0$ sufficiently small. In particular, if $f\neq 0$ a.e. we can, for sufficiently large $n$, set $t=\frac{||f||_1}{2n}$ to conclude that $\lVert f(x+n\lceil{\frac{1}{t}}\rceil t)-f(x)\rVert_1\leq \lVert f\rVert_1$ for sufficiently large $n$ (1). Using the MCT we can find $m>0$ such that $\int_{[-m,m]}|f| d\mu>(7/8)\cdot\lVert f\rVert_1$. If $m$ was sufficiently large we then have $\int|f(x+2m\lceil{\frac{1}{t}}\rceil t)-f(x)| d\mu(x)\geq \int_{[m,3m+1]}|f(x+2m\lceil{\frac{1}{t}}\rceil t)| d\mu(x)+\int_{[-m,m]}|f(x)| d\mu(x)-\int_{[m,3m+1]^C}|f(x+2m\lceil{\frac{1}{t}}\rceil t)| d\mu(x)-\int_{[-m,m]^C}|f(x)| d\mu(x)>(12/8)\lVert f\rVert_1$. 
This contradicts (1).
