TITLE: A characterization of constant functions
QUESTION [23 upvotes]: In How to recognize constant functions. Connections with Sobolev spaces (Russian Math Surveys 57 (2002); MSN), H. Brezis recalls the following fact:

Let $\Omega\subset{\mathbb R}^N$ be connected and $f:\Omega\rightarrow{\mathbb R}$ be measurable, such that 
  $$\int\int_{\Omega\times\Omega}\frac{|f(y)-f(x)|}{|y-x|^{N+1}}\,dx\,dy<\infty.$$
  Then $f$ is constant.

He adds 

The conclusion 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.

My question is whether there is such an elementary proof in the special case of one space dimension ($N=1$, $\Omega$ an interval).

REPLY [9 votes]: Not quite an answer, but too long for a comment.
Let me make my life easier a bit and take $\Omega=\mathbb{R}^N$ while increasing the exponent slightly. Namely, I will assume that
$$
I:=\ \int_{\mathbb{R}^{2N}}\ \frac{|f(x)-f(y)|}{|x-y|^{\alpha}}\ d^Nx d^Ny
$$
is finite, where $\alpha>N+1$.
Then we have, just by the triangle inequality involving the midpoint, $I\le J$ where
$$
J:=\ \int_{\mathbb{R}^{2N}}\ 
\frac{\left|f(x)-f\left(\frac{x+y}{2}\right)\right|
+\left|f\left(\frac{x+y}{2}\right)-f(y)\right|}{|x-y|^{\alpha}}\ d^Nx d^Ny
$$
$$
=\ 2^{N+1-\alpha}\ I\ ,
$$
by a trivial change of variable. Since $I\in [0,\infty)$ and $I\le 2^{N+1-\alpha}\ I$ with $N+1-\alpha<0$, we immediately get $I=0$.
The OP's case is clearly a borderline/endpoint one where the above argument just happens to break down. Perhaps one can get a logarithmic improvement, by using smarter estimates.
The above simple idea is just a "Sobolev-ish"-flavored (as opposed to "Hölder-ish") adaptation of the classic proof of Hölder with exponent greater than one in 1D implies constant, by subdividing the interval $[x,y]$ into $k$ pieces and taking $k\rightarrow\infty$.