If $L^1$ average of $f$ is smaller than $t^2$ then $f=0$ a.e.

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.

$|\int_a^{b} f(x)dx-\int_{a+t}^{b+t} f(x)dx| =|\int_a^{a+t} f(x)dx-\int_b^{b+t} f(x)dx|\leq t^{2}$ 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.?

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 $f=0$.