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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?


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Maybe I overlook something, but I think it's easy to show the integrand is bounded by contradiction, then just following your argument we have $f$ is constant. – L. Xu Sep 9 '13 at 23:16
Would you please expand a little bit your comment? I can't see how to show that the integrand is bounded... Thanks. – Romeo Sep 10 '13 at 10:35
The assumptions give the convergence of $\int_0^1\frac 1{x^2}\int_{\mathbb R}|f(x+y)-f(y)|\mathrm dy\mathrm dx$. So the answer relies on the behaviour of $\int_{\mathbb R}|f(x+y)-f(y)|\mathrm dy$ when $x$ approaches $0$. Probably Lebesgue differentiation theorem will help (a standard approximation argument says it goes to $0$, but we have to give an estimation of the rate). – Davide Giraudo Sep 10 '13 at 10:59
Hi Davide, your hint seems fine but I do not understand why the second integral is from $0$ to $1$; shouldn't it be on the whole $\mathbb R$? Furthermore, I know that the translation is continuous in $L^1$ but I've never heard of an estimation of its rate of convergence... I'll think about it. Thanks. – Romeo Sep 10 '13 at 12:01
The part $\int_1^infty$ is convergent, because the integrand is bounded by $\frac{2\lVert f\rVert_{L^1}}{x^2}$. – Davide Giraudo Sep 10 '13 at 12:36
up vote 11 down vote accepted


$$ F(x) = \int_{0}^{x} f(t) \,dt $$

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.

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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.

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+1-that's neat! – leshik Oct 30 '13 at 2:52
+1 for a nice idea! – Sangchul Lee Oct 30 '13 at 2:56

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.

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"The conclusion of Proposition 1 is easy to state, but I do not know a direct, elementary, proof." I guess I just gave one: that the dimension is $1$ and not $N$ doesn't matter. – fedja Oct 30 '13 at 2:48
I think you are right @fedja, I have verified your proof and it seems to me that you have not used the hypothesis $f\in L^1(\mathbb{R})$. That's a great answer (+1). – Tomás Oct 30 '13 at 10:54

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