If $\int_{\mathbb R^2} \frac{\vert f(x)-f(y)\vert}{\vert x-y\vert^2}dxdy<+\infty$ then $f$ is a.e. constant 
Let $f \in L^1(\mathbb R)$. If 
  $$
\int_\mathbb R \int_\mathbb R \frac{\vert f(x)-f(y)\vert}{\vert x-y\vert^2}dxdy<+\infty
$$
  then $f$ is a.e. constant.

I do not know how to begin. I thought that we are set if we show that the integrand is bounded a.e. (the function $f$ would be 2-holderian, hence constant)  but this is not true in general. I mean $\phi \in L^1(\mathbb R)$ does not imply $\phi \in L^{\infty}(\mathbb R)$. 
Would you please give me some useful hints in order to start? 
Thanks.  
 A: You can find the proof of your statement in this paper of Brezis. In fact, he proves more than you need.
I would like to quote here his Remark 1: "The conclusion of Proposition 1 is easy to state, but I do not know a direct,
elementary, proof. Our proof is not very complicated but requires an “excursion” via the
Sobolev spaces."
Note that Proposition 1 contains your claim.
A: One cheap way to do it is as follows. Let $I$ be the given integral. Notice that if $I(h)=\iint_{|x-y|< h}\frac{|f(x)-f(y)|}{|x-y|^2}\,dxdy$, then
$$
\iint_{|x-y|<h}\frac{|f(x)-f(\frac{x+y}2)|}{|x-y|^2}\,dxdy=
\frac 14\iint_{|x-y|<h}\frac{|f(x)-f(\frac{x+y}2)|}{|x-\frac{x+y}2|^2}\,dxdy
\\
=
\frac 12\iint_{|x-z|<h/2}\frac{|f(x)-f(z)|}{|x-z|^2}\,dxdz=\frac 12I(h/2)\,,
$$ 
and the same identity holds for $\iint_{|x-y|<h}\frac{|f(\frac{x+y}2)-f(y)|}{|x-y|^2}\,dxdy$. (I hope that Tonelli and linear change of variable $z=\frac{x+y}2$ in one-dimensional Lebesgue integral with respect to $y$ are among available tools).
Thus, by the triangle inequality $I(h)\le I(h/2)$. From here we can finish in 10 different ways. For instance, we can conclude that $\iint_{h/2\le|x-y|<h}\frac{|f(x)-f(y)|}{|x-y|^2}\,dxdy=0$, whence, summing over $h=2^k$, $k\in\mathbb Z$, we get $I=0$. Now it means (by Tonelli again) that for almost all $y$ we have $f(x)=f(y)$ for almost all $x$, which is pretty much the end of the story.
The point, of course, is that the original integral scales in a nice way, which is always something to watch for.
A: Let
$$ F(x) = \int_{0}^{x} f(t) \,dt $$
be an anti-derivative of $f$ and $E$ the set of Lebesgue points of $f$. Then $E^{c}$ is measure-zero and $F'(x) = f(x)$ for all $x \in E$.
The condition of the problem tells us that
$$ \int_{\Bbb{R}}\int_{\Bbb{R}} \frac{\left| f(x+y) - f(y) \right|}{x^{2}} \, dxdy < \infty. $$
So if $a < b$ are Lebesgue points, then Fubini's Theorem shows
\begin{align*}
\int_{a}^{b} \int_{\Bbb{R}} \frac{f(x+y) - f(y)}{x^{2}} \, dxdy
&= \int_{\Bbb{R}} \frac{1}{x^{2}} \int_{a}^{b} \{ f(x+y) - f(y) \} \, dydx \\
&= \int_{\Bbb{R}} \left( \frac{F(b+x)-F(b)}{x^{2}} - \frac{F(a+x)-F(a)}{x^{2}} \right) \, dx \tag{1}
\end{align*}
But since
$$ \frac{F(b+x)-F(b)}{x^{2}} - \frac{F(a+x)-F(a)}{x^{2}} \sim \frac{f(b) - f(a) + o(1)}{x}, \quad \text{as } x \to 0, $$
for the integrand of $\text{(1)}$ to be integrable near $x = 0$, we must have $f(a) = f(b)$. Since this is true for any $a, b \in E$, the proof is complete.
