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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).

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    $\begingroup$ I don't know about measurable functions, but here's a relatively easy proof for when $f$ is continuously differentiable. Suppose $f$ is not constant. Then $|f'(y)| > c > 0$ on some small interval. For each such $y$, we have $|f(y) - f(x)| > c|y-x|$ on a neighborhood of $y$. Then your integrand (for that particular $y$) is $>\! c/|y-x|$ on a neighborhood of $y$, and therefore goes to $\infty$. Since this happens for an interval of $y$'s, the whole double integral is also $\infty$. $\endgroup$ – Will Brian May 13 at 18:00
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    $\begingroup$ Such a function $f$ is in $L^1_{loc}$. Assuming $\Omega=R^N$ we can take convolutions with smooth cut-off function $\phi_\epsilon$ and the integrability condition is preserved. Then $f*\phi_\epsilon$ is a constant (by the argument of @Will Brian) and letting $\epsilon \to 0$, $f$ is constant, as well. For general $\Omega$ this does not work, but in 1d $\Omega$ is an interval $[a,b]$ and we can extend $f$ outside, using the values at the endpoints. I did not check the details but maybe you see immediately if there is a silly mistake in the argument. $\endgroup$ – Giorgio Metafune May 13 at 18:22
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    $\begingroup$ @GiorgioMetafune What worries me in your argument is this: we start with a measurable function $f$ for which the double integral from the statement of the problem converges. How do you conclude, that the same holds for $f\ast\phi_\epsilon$? $\endgroup$ – Vahe May 13 at 19:52
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    $\begingroup$ @Vahe If $g=f*\phi$, then $\int |g(y)-g(x)|/|y-x|^{N+1} dy dx \le \int |f(y-z)-f(x-z)/|y-x|^{N+1}|\phi (z) dx dy dz$. Next write $|y-x|=|(y-z)-(x-z)|$ and change variables by setting $u=y-z, v=x-z$ in the $dx\, dy$ integral. Then you get the original double integral to be integrated against $\phi(z)\, dz$. $\endgroup$ – Giorgio Metafune May 13 at 20:58
  • $\begingroup$ @GiorgioMetafune You're right, thanks! $\endgroup$ – Vahe May 13 at 22:11
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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 $k\rightarrow\infty$.

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